https://pressbooks.bccampus.ca/chem1114langaracollege Open Textbook Fri, 21 Jun 2019 01:12:40 +0000 en-US 1.2 http://pressbooks.bccampus.ca/ https://pressbooks.bccampus.ca/chem1114langaracollege 330 1 https://wordpress.org/?v=5.0.2 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-4/ Tue, 10 Apr 2018 21:55:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-4.png 25 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_cuagno3/ Tue, 10 Apr 2018 21:55:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_CuAgNO3.jpg 26 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/88x31/ Tue, 10 Apr 2018 21:55:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/88x31.png 27 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_pressure1/ Tue, 10 Apr 2018 21:55:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Pressure1.jpg 32 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-15/ Tue, 10 Apr 2018 21:55:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-15.png 33 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_pressure2/ Tue, 10 Apr 2018 21:55:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Pressure2.jpg 34 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_barometer/ Tue, 10 Apr 2018 21:55:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Barometer.jpg 35 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer/ Tue, 10 Apr 2018 21:55:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer.jpg 36 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer1_img/ Tue, 10 Apr 2018 21:55:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer1_img.jpg 37 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer2_img/ Tue, 10 Apr 2018 21:55:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer2_img.jpg 38 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer3_img/ Tue, 10 Apr 2018 21:55:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer3_img.jpg 39 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer4_img/ Tue, 10 Apr 2018 21:55:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_manometer4_img.jpg 40 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_spygmo/ Tue, 10 Apr 2018 21:55:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Spygmo.jpg 41 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_weathermap/ Tue, 10 Apr 2018 21:55:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_WeatherMap.jpg 42 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_atmosphere/ Tue, 10 Apr 2018 21:55:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Atmosphere.jpg 43 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer5_img/ Tue, 10 Apr 2018 21:55:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer5_img.jpg 44 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer6_img/ Tue, 10 Apr 2018 21:55:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer6_img.jpg 45 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer7_img/ Tue, 10 Apr 2018 21:55:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer7_img.jpg 46 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_manometer8_img/ Tue, 10 Apr 2018 21:55:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Manometer8_img.jpg 47 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_ballooning/ Tue, 10 Apr 2018 21:55:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_Ballooning.jpg 49 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_01_amontons1/ Tue, 10 Apr 2018 21:55:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_01_Amontons1.jpg 50 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_amontons2/ Tue, 10 Apr 2018 21:55:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_Amontons2.jpg 51 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-16/ Tue, 10 Apr 2018 21:55:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-16.png 52 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_charles2/ Tue, 10 Apr 2018 21:55:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_Charles2.jpg 53 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_boyleslaw1/ Tue, 10 Apr 2018 21:55:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_BoylesLaw1.jpg 54 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_boyleslaw2/ Tue, 10 Apr 2018 21:55:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_Boyleslaw2.jpg 55 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_boyleslaw4/ Tue, 10 Apr 2018 21:55:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_BoylesLaw4.jpg 56 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_scuba/ Tue, 10 Apr 2018 21:55:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_Scuba.jpg 57 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_greatbarri/ Tue, 10 Apr 2018 21:55:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_GreatBarri.jpg 58 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_henh3o2/ Tue, 10 Apr 2018 21:55:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_HENH3O2.jpg 59 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_weatherball_img/ Tue, 10 Apr 2018 21:55:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_WeatherBall_img.jpg 60 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_02_exercise25_img/ Tue, 10 Apr 2018 21:55:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_02_Exercise25_img.jpg 61 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_liquidgas/ Tue, 10 Apr 2018 21:55:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_liquidgas.jpg 63 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_daltonlaw1/ Tue, 10 Apr 2018 21:55:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_DaltonLaw1.jpg 64 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_watervapor/ Tue, 10 Apr 2018 21:55:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_WaterVapor.jpg 65 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_watervapor2/ Tue, 10 Apr 2018 21:55:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_WaterVapor2.jpg 66 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_ammonia/ Tue, 10 Apr 2018 21:55:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_Ammonia.jpg 67 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_example3_img/ Tue, 10 Apr 2018 21:55:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_Example3_img.jpg 68 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_globalwarming/ Tue, 10 Apr 2018 21:55:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_GlobalWarming.jpg 69 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_globalwarming2/ Tue, 10 Apr 2018 21:55:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_GlobalWarming2.jpg 70 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-17/ Tue, 10 Apr 2018 21:55:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-17.png 71 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_03_susansolom/ Tue, 10 Apr 2018 21:55:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_03_SusanSolom.jpg 72 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_04_kmt2/ Tue, 10 Apr 2018 21:55:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_04_KMT2.jpg 74 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_05_molspeed1/ Tue, 10 Apr 2018 21:55:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_05_MolSpeed1.jpg 75 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_05_molspeed2/ Tue, 10 Apr 2018 21:55:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_05_MolSpeed2.jpg 76 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_05_molspeed3/ Tue, 10 Apr 2018 21:55:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_05_MolSpeed3.jpg 77 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-18/ Tue, 10 Apr 2018 21:55:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-18.png 78 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_04_diffusion/ Tue, 10 Apr 2018 21:55:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_04_Diffusion.jpg 80 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_04_diffeff/ Tue, 10 Apr 2018 21:55:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_04_DiffEff.jpg 81 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_04_effusion2/ Tue, 10 Apr 2018 21:55:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_04_Effusion2.jpg 82 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_04_gasdiff/ Tue, 10 Apr 2018 21:55:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_04_GasDiff.jpg 83 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_06_zvspgraph/ Tue, 10 Apr 2018 21:55:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_06_ZvsPgraph.jpg 85 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_06_realgas3/ Tue, 10 Apr 2018 21:55:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_06_RealGas3.jpg 86 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_06_realgas2/ Tue, 10 Apr 2018 21:55:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_06_RealGas2.jpg 87 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_06_vanderwaals_img/ Tue, 10 Apr 2018 21:55:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_06_vanderWaals_img.jpg 88 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_06_exercise1_img/ Tue, 10 Apr 2018 21:55:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_06_Exercise1_img.jpg 89 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_09_06_realgases/ Tue, 10 Apr 2018 21:55:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_09_06_RealGases.jpg 90 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellulose-1/ Wed, 11 Apr 2018 16:22:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Cellulose-1.jpg 119 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_rutherford-1/ Wed, 11 Apr 2018 16:22:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Rutherford-1.jpg 120 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_hclsoln-1/ Wed, 11 Apr 2018 16:22:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_HClsoln-1.jpg 121 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_3dorbitals/ Wed, 11 Apr 2018 16:22:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_3dOrbitals.jpg 122 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkanes-1/ Wed, 11 Apr 2018 16:22:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkanes-1.jpg 123 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_10_06_fccntrdcb/ Wed, 11 Apr 2018 16:22:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_10_06_FcCntrdCb.jpg 124 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_reaction1-1/ Wed, 11 Apr 2018 16:22:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_Reaction1-1.jpg 125 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_alchemist/ Wed, 11 Apr 2018 16:22:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_Alchemist.jpg 128 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_chemweb/ Wed, 11 Apr 2018 16:22:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_ChemWeb.jpg 129 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_scimethod/ Wed, 11 Apr 2018 16:22:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_SciMethod.jpg 130 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_waterdom/ Wed, 11 Apr 2018 16:22:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_WaterDom.jpg 131 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_statesmatt/ Wed, 11 Apr 2018 16:22:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_StatesMatt.jpg 133 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_plasma/ Wed, 11 Apr 2018 16:22:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Plasma.jpg 134 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi/ Wed, 11 Apr 2018 16:22:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI.png 135 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_consmatter/ Wed, 11 Apr 2018 16:22:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_ConsMatter.jpg 136 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_goldatoms/ Wed, 11 Apr 2018 16:22:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_GoldAtoms.jpg 137 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellulose/ Wed, 11 Apr 2018 16:22:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Cellulose.jpg 138 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_molecules/ Wed, 11 Apr 2018 16:22:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Molecules.jpg 139 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_decomp/ Wed, 11 Apr 2018 16:22:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_decomp.jpg 140 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_mixtures/ Wed, 11 Apr 2018 16:22:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Mixtures.jpg 141 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_matttype/ Wed, 11 Apr 2018 16:22:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_MattType.jpg 142 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_electrolys/ Wed, 11 Apr 2018 16:22:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_Electrolys.jpg 143 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_fuelcell/ Wed, 11 Apr 2018 16:22:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_FuelCell.jpg 144 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellphone/ Wed, 11 Apr 2018 16:22:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_CellPhone.jpg 145 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_00_dailychem/ Wed, 11 Apr 2018 16:22:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_00_DailyChem.jpg 147 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_measure/ Wed, 11 Apr 2018 16:22:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Measure.jpg 149 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits5_img/ Wed, 11 Apr 2018 16:22:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits5_img.jpg 150 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits1_img/ Wed, 11 Apr 2018 16:22:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits1_img.jpg 151 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits2_img/ Wed, 11 Apr 2018 16:22:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits2_img.jpg 152 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits3_img/ Wed, 11 Apr 2018 16:22:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits3_img.jpg 153 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits4_img/ Wed, 11 Apr 2018 16:22:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits4_img.jpg 154 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_cylrebar/ Wed, 11 Apr 2018 16:22:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_CylRebar.jpg 155 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_cylgold/ Wed, 11 Apr 2018 16:22:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_CylGold.jpg 156 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_archery/ Wed, 11 Apr 2018 16:22:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Archery.jpg 157 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_archer2_img/ Wed, 11 Apr 2018 16:22:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Archer2_img.jpg 158 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-1/ Wed, 11 Apr 2018 16:22:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-1.png 160 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_mydcmin/ Wed, 11 Apr 2018 16:22:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_MYdCmIn.jpg 161 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_kilogram/ Wed, 11 Apr 2018 16:22:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_Kilogram.jpg 162 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_volume/ Wed, 11 Apr 2018 16:22:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_Volume.jpg 163 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_physchange/ Wed, 11 Apr 2018 16:22:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_PhysChange.jpg 165 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_rust/ Wed, 11 Apr 2018 16:22:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_Rust.jpg 166 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_chemchange/ Wed, 11 Apr 2018 16:22:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_ChemChange.jpg 167 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_hazdiamond/ Wed, 11 Apr 2018 16:22:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_HazDiamond.jpg 168 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_periodicpu/ Wed, 11 Apr 2018 16:22:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_PeriodicPU.jpg 169 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_06_tempscales/ Wed, 11 Apr 2018 16:22:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_06_TempScales.jpg 171 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_00_biomarkers/ Wed, 11 Apr 2018 16:22:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_00_Biomarkers.jpg 174 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton1/ Wed, 11 Apr 2018 16:22:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton1.jpg 176 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton2/ Wed, 11 Apr 2018 16:22:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton2.jpg 177 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton3/ Wed, 11 Apr 2018 16:22:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton3.jpg 178 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton6_img/ Wed, 11 Apr 2018 16:22:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton6_img.jpg 179 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton8_img/ Wed, 11 Apr 2018 16:22:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton8_img.jpg 180 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_multprop/ Wed, 11 Apr 2018 16:22:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_MultProp.jpg 181 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton10_img/ Wed, 11 Apr 2018 16:22:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton10_img.jpg 182 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_cathoderay/ Wed, 11 Apr 2018 16:22:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_CathodeRay.jpg 184 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_millikan/ Wed, 11 Apr 2018 16:22:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Millikan.jpg 185 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_atommodels/ Wed, 11 Apr 2018 16:22:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_AtomModels.jpg 186 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_rutherford/ Wed, 11 Apr 2018 16:23:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Rutherford.jpg 187 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_goldfoil3/ Wed, 11 Apr 2018 16:23:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_GoldFoil3.jpg 188 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_atomsize/ Wed, 11 Apr 2018 16:23:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_AtomSize.jpg 190 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_iodine/ Wed, 11 Apr 2018 16:23:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_Iodine.jpg 191 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_sisymbol/ Wed, 11 Apr 2018 16:23:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_SiSymbol.jpg 192 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-1/ Wed, 11 Apr 2018 16:23:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-1.png 193 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_atomsym/ Wed, 11 Apr 2018 16:23:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_AtomSym.jpg 194 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_massspec/ Wed, 11 Apr 2018 16:23:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_MassSpec.jpg 195 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_methanerep/ Wed, 11 Apr 2018 16:23:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_MethaneRep.jpg 197 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_sulfur/ Wed, 11 Apr 2018 16:23:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Sulfur.jpg 198 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-6-molecular-and-ionic-compounds/cnx_chem_02_04_hydrogen/ Wed, 11 Apr 2018 16:23:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Hydrogen.jpg 199 1350 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_tio2/ Wed, 11 Apr 2018 16:23:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_TiO2.jpg 200 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_benzene/ Wed, 11 Apr 2018 16:23:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Benzene.jpg 201 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_aceticacid/ Wed, 11 Apr 2018 16:23:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_AceticAcid.jpg 202 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-2/ Wed, 11 Apr 2018 16:23:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-2.png 203 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_leecronin/ Wed, 11 Apr 2018 16:23:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_LeeCronin.jpg 204 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_isomers/ Wed, 11 Apr 2018 16:23:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Isomers.jpg 205 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_isomers2/ Wed, 11 Apr 2018 16:23:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Isomers2.jpg 206 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3a_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3a_img.jpg 207 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3b_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3b_img.jpg 208 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3c_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3c_img.jpg 209 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3d_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3d_img.jpg 210 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4a_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4a_img.jpg 211 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4b_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4b_img.jpg 212 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4c_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4c_img.jpg 213 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4d_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4d_img.jpg 214 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question7a_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question7a_img.jpg 215 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question7b_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question7b_img.jpg 216 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question9a_img/ Wed, 11 Apr 2018 16:23:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question9a_img.jpg 217 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question9b_img/ Wed, 11 Apr 2018 16:23:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question9b_img.jpg 218 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_mendeleev/ Wed, 11 Apr 2018 16:23:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_Mendeleev.jpg 220 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/6-4-electronic-structure-of-atoms-electron-configurations/cnx_chem_02_05_pertable1/ Wed, 11 Apr 2018 16:23:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_PerTable1.jpg 221 1528 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-5-the-periodic-table/cnx_chem_02_05_pertable2/ Wed, 11 Apr 2018 16:23:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_PerTable2.jpg 222 1344 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-3/ Wed, 11 Apr 2018 16:23:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-3.png 223 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_nacation/ Wed, 11 Apr 2018 16:23:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_NaCation.jpg 225 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_ioncharges/ Wed, 11 Apr 2018 16:23:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_IonCharges.jpg 226 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_naclmolten/ Wed, 11 Apr 2018 16:23:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_NaClMolten.jpg 227 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-2/ Wed, 11 Apr 2018 16:23:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-2.png 228 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_sapphire/ Wed, 11 Apr 2018 16:23:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_Sapphire.jpg 229 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_07_erinbrocko/ Wed, 11 Apr 2018 16:23:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_07_ErinBrocko.jpg 231 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-4/ Wed, 11 Apr 2018 16:23:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-4.png 232 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_00_pool/ Wed, 11 Apr 2018 16:23:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_00_Pool.jpg 235 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_chloroform/ Wed, 11 Apr 2018 16:23:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_chloroform.jpg 237 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_aspirin/ Wed, 11 Apr 2018 16:23:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_aspirin.jpg 238 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ibuprofenmass_img/ Wed, 11 Apr 2018 16:23:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_ibuprofenmass_img.jpg 239 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_saltmass/ Wed, 11 Apr 2018 16:23:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_saltMass.jpg 240 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_alsulfatemass_img/ Wed, 11 Apr 2018 16:23:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_alsulfatemass_img.jpg 241 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_moles/ Wed, 11 Apr 2018 16:23:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_moles.jpg 242 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_compound/ Wed, 11 Apr 2018 16:23:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_compound.jpg 243 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_water/ Wed, 11 Apr 2018 16:23:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_water.jpg 244 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-5/ Wed, 11 Apr 2018 16:23:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-5.png 245 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_potassium_img/ Wed, 11 Apr 2018 16:23:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_potassium_img.jpg 246 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_argon_img/ Wed, 11 Apr 2018 16:23:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_argon_img.jpg 247 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_copper/ Wed, 11 Apr 2018 16:23:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_copper.jpg 248 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_coppermoles_img/ Wed, 11 Apr 2018 16:23:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_copperMoles_img.jpg 249 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_glycine_img/ Wed, 11 Apr 2018 16:23:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_glycine_img.jpg 250 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_glycinemass_img/ Wed, 11 Apr 2018 16:23:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_glycinemass_img.jpg 251 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_vitc_img/ Wed, 11 Apr 2018 16:23:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_vitC_img.jpg 252 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_saccharin_img/ Wed, 11 Apr 2018 16:23:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_saccharin_img.jpg 253 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_sacch_img/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_sacch_img.jpg 254 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_brain/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_brain.jpg 255 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_exocytosis/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_exocytosis.jpg 256 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05a_img/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05a_img.jpg 257 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05b_img/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05b_img.jpg 258 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05c_img/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05c_img.jpg 259 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05d_img/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05d_img.jpg 260 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06a_img/ Wed, 11 Apr 2018 16:23:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06a_img.jpg 261 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06b_img/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06b_img.jpg 262 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06c_img/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06c_img.jpg 263 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06d_img/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06d_img.jpg 264 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07a_img/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07a_img.jpg 265 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07b_img/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07b_img.jpg 266 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07c_img/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07c_img.jpg 267 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_empform/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_empform.jpg 269 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_hematite/ Wed, 11 Apr 2018 16:23:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_hematite.jpg 270 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-6/ Wed, 11 Apr 2018 16:23:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-6.png 271 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_brewtank/ Wed, 11 Apr 2018 16:23:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_BrewTank.jpg 272 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_espresso/ Wed, 11 Apr 2018 16:23:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_espresso.jpg 274 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_vinegar/ Wed, 11 Apr 2018 16:23:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_vinegar.jpg 275 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_dilution/ Wed, 11 Apr 2018 16:23:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_dilution.jpg 276 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_solution/ Wed, 11 Apr 2018 16:23:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_solution.jpg 277 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-7/ Wed, 11 Apr 2018 16:23:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-7.png 278 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_bleach/ Wed, 11 Apr 2018 16:23:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_bleach.jpg 280 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_example2_img/ Wed, 11 Apr 2018 16:23:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_Example2_img.jpg 281 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_saline/ Wed, 11 Apr 2018 16:23:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_saline.jpg 282 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_faucet/ Wed, 11 Apr 2018 16:23:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_faucet.jpg 283 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_00_rocket/ Wed, 11 Apr 2018 16:23:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_00_Rocket.jpg 286 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_rxn2/ Wed, 11 Apr 2018 16:23:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_rxn2.jpg 288 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_rxn3/ Wed, 11 Apr 2018 16:23:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_rxn3.jpg 289 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-3/ Wed, 11 Apr 2018 16:23:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-3.png 290 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_basehyd_img/ Wed, 11 Apr 2018 16:23:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_basehyd_img.jpg 291 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/olympus-digital-camera/ Wed, 11 Apr 2018 16:23:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_LeadIodide.jpg 293 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_hclsoln/ Wed, 11 Apr 2018 16:23:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_HClsoln.jpg 294 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_citrus/ Wed, 11 Apr 2018 16:23:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_Citrus.jpg 295 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_ammonia/ Wed, 11 Apr 2018 16:23:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_ammonia.jpg 296 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-4-1/ Wed, 11 Apr 2018 16:23:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-4-1.png 297 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_cuagno3-1/ Wed, 11 Apr 2018 16:23:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_CuAgNO3-1.jpg 298 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_iodine/ Wed, 11 Apr 2018 16:23:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_iodine.jpg 300 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_moleratio1_img/ Wed, 11 Apr 2018 16:23:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_moleratio1_img.jpg 301 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_moleratio2_img/ Wed, 11 Apr 2018 16:23:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_moleratio2_img.jpg 302 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_map2_img/ Wed, 11 Apr 2018 16:23:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_map2_img.jpg 303 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_map3_img/ Wed, 11 Apr 2018 16:23:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_map3_img.jpg 304 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_flowchart/ Wed, 11 Apr 2018 16:23:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_flowchart.jpg 305 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_airbag/ Wed, 11 Apr 2018 16:23:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_airbag.jpg 306 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_ethenebr_img/ Wed, 11 Apr 2018 16:23:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_etheneBr_img.jpg 307 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_ethene_img/ Wed, 11 Apr 2018 16:23:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_ethene_img.jpg 308 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_sandwich/ Wed, 11 Apr 2018 16:23:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_sandwich.jpg 310 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_limiting/ Wed, 11 Apr 2018 16:23:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_limiting.jpg 311 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-5/ Wed, 11 Apr 2018 16:23:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-5.png 312 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_greenchem/ Wed, 11 Apr 2018 16:23:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_GreenChem.jpg 313 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_saccharin_img/ Wed, 11 Apr 2018 16:23:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_saccharin_img.jpg 314 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_titration/ Wed, 11 Apr 2018 16:23:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_titration.jpg 316 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_map7_img/ Wed, 11 Apr 2018 16:23:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_map7_img.jpg 317 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_filter/ Wed, 11 Apr 2018 16:23:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_filter.jpg 318 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_map8_img/ Wed, 11 Apr 2018 16:23:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_map8_img.jpg 319 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_combustion/ Wed, 11 Apr 2018 16:23:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_combustion.jpg 320 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_combmap_img/ Wed, 11 Apr 2018 16:23:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_combmap_img.jpg 321 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_propionate_img/ Wed, 11 Apr 2018 16:23:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_propionate_img.jpg 322 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_00_match/ Wed, 11 Apr 2018 16:23:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_00_Match.jpg 325 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_thermochem/ Wed, 11 Apr 2018 16:23:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Thermochem.jpg 327 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_waterfall/ Wed, 11 Apr 2018 16:23:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Waterfall.jpg 328 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_hotcold/ Wed, 11 Apr 2018 16:23:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HotCold.jpg 329 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-6/ Wed, 11 Apr 2018 16:23:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-6.png 330 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_thermom/ Wed, 11 Apr 2018 16:23:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Thermom.jpg 331 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_heattrans1/ Wed, 11 Apr 2018 16:23:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HeatTrans1.jpg 332 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_oxyactorch/ Wed, 11 Apr 2018 16:23:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_OxyacTorch.jpg 333 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_heatcapacity/ Wed, 11 Apr 2018 16:23:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HeatCapacity.jpg 334 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_soltherm1/ Wed, 11 Apr 2018 16:23:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_SolTherm1.jpg 335 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_soltherm2/ Wed, 11 Apr 2018 16:23:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_SolTherm2.jpg 336 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_heatmeas/ Wed, 11 Apr 2018 16:23:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HeatMeas.jpg 338 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_calorim/ Wed, 11 Apr 2018 16:23:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_Calorim.jpg 339 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_calorim2/ Wed, 11 Apr 2018 16:23:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_Calorim2.jpg 340 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_heattrans2/ Wed, 11 Apr 2018 16:23:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HeatTrans2.jpg 341 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_handwarmer/ Wed, 11 Apr 2018 16:23:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HandWarmer.jpg 342 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-7/ Wed, 11 Apr 2018 16:23:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-7.png 343 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_icepack/ Wed, 11 Apr 2018 16:23:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_IcePack.jpg 344 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_bombcalor/ Wed, 11 Apr 2018 16:23:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_BombCalor.jpg 345 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_foodlabel/ Wed, 11 Apr 2018 16:23:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_FoodLabel.jpg 346 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_systemqw/ Wed, 11 Apr 2018 16:23:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_Systemqw.jpg 348 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-8/ Wed, 11 Apr 2018 16:23:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-8.png 349 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_summit/ Wed, 11 Apr 2018 16:23:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_Summit.jpg 350 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_gasburning/ Wed, 11 Apr 2018 16:23:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_GasBurning.jpg 351 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_algalfuel1/ Wed, 11 Apr 2018 16:23:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_AlgalFuel1.jpg 352 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_algalfuel2/ Wed, 11 Apr 2018 16:23:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_AlgalFuel2.jpg 353 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_hessco2/ Wed, 11 Apr 2018 16:23:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_HessCO2.jpg 354 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_00_crabneb/ Wed, 11 Apr 2018 16:23:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_00_CrabNeb.jpg 357 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_frequency/ Wed, 11 Apr 2018 16:23:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Frequency.jpg 359 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_emspectrum/ Wed, 11 Apr 2018 16:23:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_emspectrum.jpg 360 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_radiocell/ Wed, 11 Apr 2018 16:23:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_RadioCell.jpg 361 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_amfm/ Wed, 11 Apr 2018 16:23:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_AMFM.jpg 362 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_liteinterf/ Wed, 11 Apr 2018 16:23:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_LiteInterf.jpg 363 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_vibrstring/ Wed, 11 Apr 2018 16:23:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Vibrstring.jpg 364 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_vibratdrum/ Wed, 11 Apr 2018 16:23:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Vibratdrum.jpg 365 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-8/ Wed, 11 Apr 2018 16:23:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-8.png 366 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_solardist/ Wed, 11 Apr 2018 16:23:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Solardist.jpg 367 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_blackbody/ Wed, 11 Apr 2018 16:23:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Blackbody.jpg 368 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_ephoton/ Wed, 11 Apr 2018 16:23:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Ephoton.jpg 369 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_neon/ Wed, 11 Apr 2018 16:23:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_neon.jpg 370 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_2spectra/ Wed, 11 Apr 2018 16:23:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_2spectra.jpg 371 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_02_hlevels/ Wed, 11 Apr 2018 16:23:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_02_Hlevels.jpg 373 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_02_bohrarrows/ Wed, 11 Apr 2018 16:23:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_02_BohrArrows.jpg 374 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_waterw/ Wed, 11 Apr 2018 16:23:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_waterw.jpg 376 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_elecw/ Wed, 11 Apr 2018 16:23:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_elecw.jpg 377 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_angm/ Wed, 11 Apr 2018 16:23:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_angm.jpg 378 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_electrnin/ Wed, 11 Apr 2018 16:23:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Electrnin.jpg 379 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-9/ Wed, 11 Apr 2018 16:23:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-9.png 380 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_qnumbers/ Wed, 11 Apr 2018 16:23:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Qnumbers.jpg 381 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_sorbit/ Wed, 11 Apr 2018 16:23:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_sOrbit.jpg 382 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_oshapes/ Wed, 11 Apr 2018 16:23:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Oshapes.jpg 383 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_subshells/ Wed, 11 Apr 2018 16:23:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_subshells.jpg 384 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_spin/ Wed, 11 Apr 2018 16:23:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_spin.jpg 385 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_orboutline_img/ Wed, 11 Apr 2018 16:23:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_OrbOutline_img.jpg 386 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_ex9soln_img/ Wed, 11 Apr 2018 16:23:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Ex9soln_img.jpg 387 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_eleveldiag/ Wed, 11 Apr 2018 16:23:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_eLeveldiag.jpg 389 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_econfig/ Wed, 11 Apr 2018 16:24:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Econfig.jpg 390 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_efillorder/ Wed, 11 Apr 2018 16:24:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Efillorder.jpg 391 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_econtable/ Wed, 11 Apr 2018 16:24:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Econtable.jpg 392 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_hydrog1_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Hydrog1_img.jpg 393 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_helium1_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Helium1_img.jpg 394 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_lithium12_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Lithium12_img.jpg 395 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_beryll12_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Beryll12_img.jpg 396 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_boron122_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Boron122_img.jpg 397 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_carbon122_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Carbon122_img.jpg 398 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_nofne_img/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_NOFNe_img.jpg 399 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_valence/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Valence.jpg 400 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_ptableconf/ Wed, 11 Apr 2018 16:24:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Ptableconf.jpg 401 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_phosphorb_img/ Wed, 11 Apr 2018 16:24:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_PhosphOrb_img.jpg 402 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2a_img/ Wed, 11 Apr 2018 16:24:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2a_img.jpg 403 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2b_img/ Wed, 11 Apr 2018 16:24:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2b_img.jpg 404 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2c_img/ Wed, 11 Apr 2018 16:24:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2c_img.jpg 405 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2d_img/ Wed, 11 Apr 2018 16:24:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2d_img.jpg 406 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2e_img/ Wed, 11 Apr 2018 16:24:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2e_img.jpg 407 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-10/ Wed, 11 Apr 2018 16:24:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-10.png 409 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_covalradit/ Wed, 11 Apr 2018 16:24:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_CovalradiT.jpg 410 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_radiigraph/ Wed, 11 Apr 2018 16:24:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Radiigraph.jpg 411 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_ionradii/ Wed, 11 Apr 2018 16:24:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Ionradii.jpg 412 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_firstiongr/ Wed, 11 Apr 2018 16:24:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Firstiongr.jpg 413 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_firstionen/ Wed, 11 Apr 2018 16:24:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Firstionen.jpg 414 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_oxygen122_img/ Wed, 11 Apr 2018 16:24:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Oxygen122_img.jpg 415 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_elaffin/ Wed, 11 Apr 2018 16:24:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Elaffin.jpg 416 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_00_bucky/ Wed, 11 Apr 2018 16:24:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_00_Bucky.jpg 419 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_01_naclphotos/ Wed, 11 Apr 2018 16:24:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_01_NaClPhotos.jpg 421 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/microsoft-word-cnx_chem_ch07_mod01-docx/ Wed, 11 Apr 2018 16:24:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_01_NaClStruc.jpg 422 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_morse/ Wed, 11 Apr 2018 16:24:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_Morse.jpg 424 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_hclbond/ Wed, 11 Apr 2018 16:24:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_HClBond.jpg 425 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_entable/ Wed, 11 Apr 2018 16:24:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_ENTable.jpg 426 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_pauling/ Wed, 11 Apr 2018 16:24:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_Pauling.jpg 427 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_deltaen/ Wed, 11 Apr 2018 16:24:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_DeltaEN.jpg 428 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_lewisstruct1_img/ Wed, 11 Apr 2018 16:24:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Lewisstruct1_img.jpg 430 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_3rowlewis/ Wed, 11 Apr 2018 16:24:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_3rowLewis.jpg 431 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_naca_img/ Wed, 11 Apr 2018 16:24:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NaCa_img.jpg 432 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cls_img/ Wed, 11 Apr 2018 16:24:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ClS_img.jpg 433 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ionlewis/ Wed, 11 Apr 2018 16:24:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_IonLewis.jpg 434 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_c12dot_img/ Wed, 11 Apr 2018 16:24:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_C12dot_img.jpg 435 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cl2dash_img/ Wed, 11 Apr 2018 16:24:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Cl2dash_img.jpg 436 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xy4struc_img/ Wed, 11 Apr 2018 16:24:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XY4struc_img.jpg 437 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_numbonds_img/ Wed, 11 Apr 2018 16:24:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_numbonds_img.jpg 438 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_doublebond_img/ Wed, 11 Apr 2018 16:24:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_DoubleBond_img.jpg 439 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cocn_img/ Wed, 11 Apr 2018 16:24:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_COCN_img.jpg 440 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_unprelec_img/ Wed, 11 Apr 2018 16:24:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Unprelec_img.jpg 441 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_singlebond_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Singlebond_img.jpg 442 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_sih4_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_SiH4_img.jpg 443 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_oxydiflor2_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_oxydiflor2_img.jpg 444 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_formate2_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_formate2_img.jpg 445 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nitrosoni2_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_nitrosoni2_img.jpg 446 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nitrosoni3_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_nitrosoni3_img.jpg 447 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_1_img/ Wed, 11 Apr 2018 16:24:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_1_img.jpg 448 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_2_img/ Wed, 11 Apr 2018 16:24:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_2_img.jpg 449 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_3_img/ Wed, 11 Apr 2018 16:24:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_3_img.jpg 450 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_4_img/ Wed, 11 Apr 2018 16:24:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_4_img.jpg 451 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_coco2_img/ Wed, 11 Apr 2018 16:24:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_COCO2_img.jpg 452 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_smalley/ Wed, 11 Apr 2018 16:24:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Smalley.jpg 453 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nosingle_img/ Wed, 11 Apr 2018 16:24:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NOsingle_img.jpg 454 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nodouble_img/ Wed, 11 Apr 2018 16:24:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NODouble_img.jpg 455 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_edeficient_img/ Wed, 11 Apr 2018 16:24:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_edeficient_img.jpg 456 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_dative_img/ Wed, 11 Apr 2018 16:24:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_dative_img.jpg 457 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_pf5sf6_img/ Wed, 11 Apr 2018 16:24:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_PF5SF6_img.jpg 458 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_if5xef4_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_IF5XeF4_img.jpg 459 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xeskeleton_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Xeskeleton_img.jpg 460 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xef2_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XeF2_img.jpg 461 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xef6_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XeF6_img.jpg 462 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_interhal_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Interhal_img.jpg 463 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisa_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisa_img.jpg 464 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisb_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisb_img.jpg 465 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisc_img/ Wed, 11 Apr 2018 16:24:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisc_img.jpg 466 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisd_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisd_img.jpg 467 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioa_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioa_img.jpg 468 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbiob_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbiob_img.jpg 469 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioc_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioc_img.jpg 470 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbiod_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbiod_img.jpg 471 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioe_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioe_img.jpg 472 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1a_img/ Wed, 11 Apr 2018 16:24:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1a_img.jpg 473 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1b_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1b_img.jpg 474 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1d_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1d_img.jpg 475 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1g_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1g_img.jpg 476 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3a_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3a_img.jpg 477 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3b_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3b_img.jpg 478 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3c_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3c_img.jpg 479 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3d_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3d_img.jpg 480 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3e_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3e_img.jpg 481 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3f_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3f_img.jpg 482 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question5_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question5_img.jpg 483 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7a_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7a_img.jpg 484 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7b_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7b_img.jpg 485 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7c_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7c_img.jpg 486 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7d_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7d_img.jpg 487 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7e_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7e_img.jpg 488 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7f_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7f_img.jpg 489 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7g_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7g_img.jpg 490 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7h_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7h_img.jpg 491 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7i_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7i_img.jpg 492 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7j_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7j_img.jpg 493 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7k_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7k_img.jpg 494 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9a_img/ Wed, 11 Apr 2018 16:24:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9a_img.jpg 495 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9b_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9b_img.jpg 496 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9c_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9c_img.jpg 497 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9d_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9d_img.jpg 498 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question18_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question18_img.jpg 499 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_carbtet_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Carbtet_img.jpg 500 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansa_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansa_img.jpg 501 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansb_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansb_img.jpg 502 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansc_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansc_img.jpg 503 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansd_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansd_img.jpg 504 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioanse_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioanse_img.jpg 505 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise25_img/ Wed, 11 Apr 2018 16:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise25_img.jpg 506 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_icl4_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_ICL4_img.jpg 508 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO_img.jpg 509 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_brcl3_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_BRCL3_img.jpg 510 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ex070402_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ex070402_img.jpg 511 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co2pos_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO2pos_img.jpg 512 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_thiocyan_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Thiocyan_img.jpg 513 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_n2o_img/ Wed, 11 Apr 2018 16:24:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_N2O_img.jpg 514 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_n2ofc_img/ Wed, 11 Apr 2018 16:24:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_N2Ofc_img.jpg 515 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_nno_img/ Wed, 11 Apr 2018 16:24:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NNO_img.jpg 516 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2ion_img/ Wed, 11 Apr 2018 16:24:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2ion_img.jpg 517 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2res_img/ Wed, 11 Apr 2018 16:24:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2res_img.jpg 518 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2resarr_img/ Wed, 11 Apr 2018 16:24:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2resarr_img.jpg 519 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co3res_img/ Wed, 11 Apr 2018 16:24:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO3res_img.jpg 520 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-11/ Wed, 11 Apr 2018 16:24:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-11.png 521 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_chood_img/ Wed, 11 Apr 2018 16:24:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CHOOd_img.jpg 522 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_chooe_img/ Wed, 11 Apr 2018 16:24:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CHOOe_img.jpg 523 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_so2resd_img/ Wed, 11 Apr 2018 16:24:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_SO2resd_img.jpg 524 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_so2rese_img/ Wed, 11 Apr 2018 16:24:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_SO2rese_img.jpg 525 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_acet_img/ Wed, 11 Apr 2018 16:24:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Acet_img.jpg 526 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_hno2_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_HNO2_img.jpg 527 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansa_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansa_img.jpg 528 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansb_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansb_img.jpg 529 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansc_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansc_img.jpg 530 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansd_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansd_img.jpg 531 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2anse_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2anse_img.jpg 532 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques11ans_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques11ans_img.jpg 533 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques13ansb_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques13ansb_img.jpg 534 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques13ansc_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques13ansc_img.jpg 535 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12a_img/ Wed, 11 Apr 2018 16:24:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12a_img.jpg 536 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12b_img/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12b_img.jpg 537 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12c_img/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12c_img.jpg 538 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12d_img/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12d_img.jpg 539 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ohamine_img/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_OHamine_img.jpg 540 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_nf3_img/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NF3_img.jpg 541 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_h2so4/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_H2SO4.jpg 542 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ch4bond_img/ Wed, 11 Apr 2018 16:24:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_CH4bond_img.jpg 544 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ch3ohlew_img/ Wed, 11 Apr 2018 16:24:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_CH3OHLew_img.jpg 545 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ethanol_img/ Wed, 11 Apr 2018 16:24:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_Ethanol_img.jpg 546 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_bornhaber/ Wed, 11 Apr 2018 16:24:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_BornHaber.jpg 547 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_hydroxya_img/ Wed, 11 Apr 2018 16:24:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_Hydroxya_img.jpg 548 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_c6h8lew_img/ Wed, 11 Apr 2018 16:24:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_C6H8Lew_img.jpg 549 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_fno2_img/ Wed, 11 Apr 2018 16:24:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_FNO2_img.jpg 550 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_c6h8ans_img/ Wed, 11 Apr 2018 16:24:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_C6H8ans_img.jpg 551 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ch2o/ Wed, 11 Apr 2018 16:24:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CH2O.jpg 553 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bef2/ Wed, 11 Apr 2018 16:24:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BeF2.jpg 554 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_egeom/ Wed, 11 Apr 2018 16:24:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Egeom.jpg 555 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ch4/ Wed, 11 Apr 2018 16:24:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CH4.jpg 556 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh3/ Wed, 11 Apr 2018 16:24:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH3.jpg 557 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_molgeom/ Wed, 11 Apr 2018 16:24:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_molgeom.jpg 558 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_axeq/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Axeq.jpg 559 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_co2_img/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CO2_img.jpg 560 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bcl3_img/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BCl3_img.jpg 561 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bcl3mol/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BCl3mol.jpg 562 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh4_img/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH4_img.jpg 563 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh4mol/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH4mol.jpg 564 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_h2o_img/ Wed, 11 Apr 2018 16:24:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_H2O_img.jpg 565 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_h2omol/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_H2Omol.jpg 566 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sf4_img/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SF4_img.jpg 567 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sf4mol/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SF4mol.jpg 568 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_xef4_img/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_XeF4_img.jpg 569 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_xef4mol/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_XEF4mol.jpg 570 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_gly_img/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_gly_img.jpg 571 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_gly3d_img/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_gly3D_img.jpg 572 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ala_img/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_ala_img.jpg 573 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-12/ Wed, 11 Apr 2018 16:24:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-12.png 574 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bondvector/ Wed, 11 Apr 2018 16:24:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BondVector.jpg 575 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_co2h2odip/ Wed, 11 Apr 2018 16:24:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CO2H2Odip.jpg 576 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_osc_img/ Wed, 11 Apr 2018 16:24:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_OSC_img.jpg 577 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_chcl3_img/ Wed, 11 Apr 2018 16:24:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CHCl3_img.jpg 578 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sh2nh3_img/ Wed, 11 Apr 2018 16:24:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SH2NH3_img.jpg 579 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_dipolfield/ Wed, 11 Apr 2018 16:24:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Dipolfield.jpg 580 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ques23ans_img/ Wed, 11 Apr 2018 16:24:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Ques23ans_img.jpg 581 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3a_img/ Wed, 11 Apr 2018 16:24:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3a_img.jpg 582 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3b_img/ Wed, 11 Apr 2018 16:24:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3b_img.jpg 583 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3c_img/ Wed, 11 Apr 2018 16:24:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3c_img.jpg 584 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ques28ans_img/ Wed, 11 Apr 2018 16:24:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Ques28ans_img.jpg 585 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mexico/ Wed, 11 Apr 2018 16:24:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_00_Sinkhole.jpg 588 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_conjugate_img/ Wed, 11 Apr 2018 16:24:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_conjugate_img.jpg 590 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_hf_img/ Wed, 11 Apr 2018 16:24:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_HF_img.jpg 591 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_nh3_img/ Wed, 11 Apr 2018 16:24:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_NH3_img.jpg 592 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_water_img/ Wed, 11 Apr 2018 16:24:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_Water_img.jpg 593 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_phscale/ Wed, 11 Apr 2018 16:24:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_phscale.jpg 595 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_acidrain/ Wed, 11 Apr 2018 16:24:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_AcidRain.jpg 596 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_phmeter/ Wed, 11 Apr 2018 16:24:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_pHMeter.jpg 597 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_indicator/ Wed, 11 Apr 2018 16:24:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_indicator.jpg 598 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_strong/ Wed, 11 Apr 2018 16:24:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_strong.jpg 600 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-27/ Wed, 11 Apr 2018 16:24:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-27.png 601 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_strengths/ Wed, 11 Apr 2018 16:24:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_strengths.jpg 602 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_corresp/ Wed, 11 Apr 2018 16:24:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Corresp.jpg 603 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_acetate_img/ Wed, 11 Apr 2018 16:24:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_acetate_img.jpg 604 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_ammonia/ Wed, 11 Apr 2018 16:24:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ammonia.jpg 605 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_aceticacid-2/ Wed, 11 Apr 2018 16:24:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Vinegar.jpg 606 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps_img/ Wed, 11 Apr 2018 16:24:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps_img.jpg 607 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable1_img/ Wed, 11 Apr 2018 16:24:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable1_img.jpg 608 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable2_img/ Wed, 11 Apr 2018 16:24:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable2_img.jpg 609 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_antsting/ Wed, 11 Apr 2018 16:24:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_AntSting.jpg 610 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable3_img/ Wed, 11 Apr 2018 16:24:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable3_img.jpg 611 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps4_img/ Wed, 11 Apr 2018 16:24:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps4_img.jpg 612 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable4_img/ Wed, 11 Apr 2018 16:24:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable4_img.jpg 613 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps5_img/ Wed, 11 Apr 2018 16:24:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps5_img.jpg 614 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable5_img/ Wed, 11 Apr 2018 16:24:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable5_img.jpg 615 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_acidph/ Wed, 11 Apr 2018 16:24:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_AcidpH.jpg 616 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_ohbonds_img/ Wed, 11 Apr 2018 16:24:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_OHbonds_img.jpg 617 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_oxyacid/ Wed, 11 Apr 2018 16:24:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Oxyacid.jpg 618 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_fishlemon/ Wed, 11 Apr 2018 16:24:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_FishLemon.jpg 620 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_steps1_img/ Wed, 11 Apr 2018 16:24:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_steps1_img.jpg 621 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_hydronium/ Wed, 11 Apr 2018 16:24:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_hydronium.jpg 622 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_steps2_img/ Wed, 11 Apr 2018 16:24:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_steps2_img.jpg 623 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_icetable13_img/ Wed, 11 Apr 2018 16:24:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_ICETable13_img.jpg 624 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_acetic_img/ Wed, 11 Apr 2018 16:24:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_acetic_img.jpg 626 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_steps1_img/ Wed, 11 Apr 2018 16:24:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_steps1_img.jpg 627 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_steps2_img/ Wed, 11 Apr 2018 16:24:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_steps2_img.jpg 628 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_icetable1_img/ Wed, 11 Apr 2018 16:24:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_ICETable1_img.jpg 629 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_compare/ Wed, 11 Apr 2018 16:24:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_compare.jpg 631 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_bufferchrt/ Wed, 11 Apr 2018 16:24:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_bufferchrt.jpg 632 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellulose-1-1/ Wed, 11 Apr 2018 16:25:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Cellulose-1-1.jpg 633 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_rutherford-1-1/ Wed, 11 Apr 2018 16:25:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Rutherford-1-1.jpg 634 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_hclsoln-1-1/ Wed, 11 Apr 2018 16:25:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_HClsoln-1-1.jpg 635 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_3dorbitals-1/ Wed, 11 Apr 2018 16:25:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_3dOrbitals-1.jpg 636 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkanes-1-1/ Wed, 11 Apr 2018 16:25:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkanes-1-1.jpg 637 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_10_06_fccntrdcb-1/ Wed, 11 Apr 2018 16:25:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_10_06_FcCntrdCb-1.jpg 638 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_reaction1-1-1/ Wed, 11 Apr 2018 16:25:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_Reaction1-1-1.jpg 639 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_alchemist-1/ Wed, 11 Apr 2018 16:25:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_Alchemist-1.jpg 642 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_chemweb-1/ Wed, 11 Apr 2018 16:25:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_ChemWeb-1.jpg 643 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_scimethod-1/ Wed, 11 Apr 2018 16:25:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_SciMethod-1.jpg 644 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_waterdom-1/ Wed, 11 Apr 2018 16:25:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_WaterDom-1.jpg 645 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_statesmatt-1/ Wed, 11 Apr 2018 16:25:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_StatesMatt-1.jpg 647 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_plasma-1/ Wed, 11 Apr 2018 16:25:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Plasma-1.jpg 648 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-9/ Wed, 11 Apr 2018 16:25:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-9.png 649 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_consmatter-1/ Wed, 11 Apr 2018 16:25:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_ConsMatter-1.jpg 650 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_goldatoms-1/ Wed, 11 Apr 2018 16:25:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_GoldAtoms-1.jpg 651 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellulose-2/ Wed, 11 Apr 2018 16:25:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Cellulose-2.jpg 652 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_molecules-1/ Wed, 11 Apr 2018 16:25:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Molecules-1.jpg 653 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_decomp-1/ Wed, 11 Apr 2018 16:25:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_decomp-1.jpg 654 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_mixtures-1/ Wed, 11 Apr 2018 16:25:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Mixtures-1.jpg 655 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_matttype-1/ Wed, 11 Apr 2018 16:25:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_MattType-1.jpg 656 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_electrolys-1/ Wed, 11 Apr 2018 16:25:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_Electrolys-1.jpg 657 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_fuelcell-1/ Wed, 11 Apr 2018 16:25:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_FuelCell-1.jpg 658 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellphone-1/ Wed, 11 Apr 2018 16:25:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_CellPhone-1.jpg 659 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_00_dailychem-1/ Wed, 11 Apr 2018 16:25:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_00_DailyChem-1.jpg 661 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_measure-1/ Wed, 11 Apr 2018 16:25:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Measure-1.jpg 663 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits5_img-1/ Wed, 11 Apr 2018 16:25:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits5_img-1.jpg 664 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits1_img-1/ Wed, 11 Apr 2018 16:25:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits1_img-1.jpg 665 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits2_img-1/ Wed, 11 Apr 2018 16:25:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits2_img-1.jpg 666 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits3_img-1/ Wed, 11 Apr 2018 16:25:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits3_img-1.jpg 667 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits4_img-1/ Wed, 11 Apr 2018 16:25:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits4_img-1.jpg 668 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_cylrebar-1/ Wed, 11 Apr 2018 16:25:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_CylRebar-1.jpg 669 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_cylgold-1/ Wed, 11 Apr 2018 16:25:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_CylGold-1.jpg 670 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_archery-1/ Wed, 11 Apr 2018 16:25:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Archery-1.jpg 671 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_archer2_img-1/ Wed, 11 Apr 2018 16:25:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Archer2_img-1.jpg 672 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-1-1/ Wed, 11 Apr 2018 16:25:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-1-1.png 674 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_mydcmin-1/ Wed, 11 Apr 2018 16:25:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_MYdCmIn-1.jpg 675 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_kilogram-1/ Wed, 11 Apr 2018 16:26:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_Kilogram-1.jpg 676 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_volume-1/ Wed, 11 Apr 2018 16:26:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_Volume-1.jpg 677 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_physchange-1/ Wed, 11 Apr 2018 16:26:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_PhysChange-1.jpg 679 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_rust-1/ Wed, 11 Apr 2018 16:26:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_Rust-1.jpg 680 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_chemchange-1/ Wed, 11 Apr 2018 16:26:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_ChemChange-1.jpg 681 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_hazdiamond-1/ Wed, 11 Apr 2018 16:26:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_HazDiamond-1.jpg 682 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_periodicpu-1/ Wed, 11 Apr 2018 16:26:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_PeriodicPU-1.jpg 683 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_06_tempscales-1/ Wed, 11 Apr 2018 16:26:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_06_TempScales-1.jpg 685 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_00_biomarkers-1/ Wed, 11 Apr 2018 16:26:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_00_Biomarkers-1.jpg 688 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton1-1/ Wed, 11 Apr 2018 16:26:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton1-1.jpg 690 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton2-1/ Wed, 11 Apr 2018 16:26:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton2-1.jpg 691 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton3-1/ Wed, 11 Apr 2018 16:26:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton3-1.jpg 692 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton6_img-1/ Wed, 11 Apr 2018 16:26:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton6_img-1.jpg 693 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton8_img-1/ Wed, 11 Apr 2018 16:26:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton8_img-1.jpg 694 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_multprop-1/ Wed, 11 Apr 2018 16:26:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_MultProp-1.jpg 695 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton10_img-1/ Wed, 11 Apr 2018 16:26:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton10_img-1.jpg 696 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_cathoderay-1/ Wed, 11 Apr 2018 16:26:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_CathodeRay-1.jpg 698 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_millikan-1/ Wed, 11 Apr 2018 16:26:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Millikan-1.jpg 699 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_atommodels-1/ Wed, 11 Apr 2018 16:26:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_AtomModels-1.jpg 700 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_rutherford-2/ Wed, 11 Apr 2018 16:26:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Rutherford-2.jpg 701 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_goldfoil3-1/ Wed, 11 Apr 2018 16:26:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_GoldFoil3-1.jpg 702 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_atomsize-1/ Wed, 11 Apr 2018 16:26:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_AtomSize-1.jpg 704 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_iodine-1/ Wed, 11 Apr 2018 16:26:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_Iodine-1.jpg 705 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_sisymbol-1/ Wed, 11 Apr 2018 16:26:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_SiSymbol-1.jpg 706 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-1-1/ Wed, 11 Apr 2018 16:26:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-1-1.png 707 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_atomsym-1/ Wed, 11 Apr 2018 16:26:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_AtomSym-1.jpg 708 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_massspec-1/ Wed, 11 Apr 2018 16:26:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_MassSpec-1.jpg 709 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_methanerep-1/ Wed, 11 Apr 2018 16:26:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_MethaneRep-1.jpg 711 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_sulfur-1/ Wed, 11 Apr 2018 16:26:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Sulfur-1.jpg 712 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_hydrogen-1/ Wed, 11 Apr 2018 16:26:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Hydrogen-1.jpg 713 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_tio2-1/ Wed, 11 Apr 2018 16:26:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_TiO2-1.jpg 714 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_benzene-1/ Wed, 11 Apr 2018 16:26:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Benzene-1.jpg 715 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_aceticacid-1/ Wed, 11 Apr 2018 16:26:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_AceticAcid-1.jpg 716 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-2-1/ Wed, 11 Apr 2018 16:26:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-2-1.png 717 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_leecronin-1/ Wed, 11 Apr 2018 16:26:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_LeeCronin-1.jpg 718 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_isomers-1/ Wed, 11 Apr 2018 16:26:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Isomers-1.jpg 719 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_isomers2-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Isomers2-1.jpg 720 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3a_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3a_img-1.jpg 721 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3b_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3b_img-1.jpg 722 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3c_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3c_img-1.jpg 723 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3d_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3d_img-1.jpg 724 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4a_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4a_img-1.jpg 725 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4b_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4b_img-1.jpg 726 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4c_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4c_img-1.jpg 727 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4d_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4d_img-1.jpg 728 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question7a_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question7a_img-1.jpg 729 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question7b_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question7b_img-1.jpg 730 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question9a_img-1/ Wed, 11 Apr 2018 16:26:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question9a_img-1.jpg 731 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question9b_img-1/ Wed, 11 Apr 2018 16:26:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question9b_img-1.jpg 732 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_mendeleev-1/ Wed, 11 Apr 2018 16:26:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_Mendeleev-1.jpg 734 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_pertable1-1/ Wed, 11 Apr 2018 16:26:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_PerTable1-1.jpg 735 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_pertable2-1/ Wed, 11 Apr 2018 16:26:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_PerTable2-1.jpg 736 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-3-1/ Wed, 11 Apr 2018 16:26:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-3-1.png 737 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_nacation-1/ Wed, 11 Apr 2018 16:26:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_NaCation-1.jpg 739 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_ioncharges-1/ Wed, 11 Apr 2018 16:26:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_IonCharges-1.jpg 740 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_naclmolten-1/ Wed, 11 Apr 2018 16:26:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_NaClMolten-1.jpg 741 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-2-1/ Wed, 11 Apr 2018 16:26:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-2-1.png 742 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_sapphire-1/ Wed, 11 Apr 2018 16:26:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_Sapphire-1.jpg 743 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_07_erinbrocko-1/ Wed, 11 Apr 2018 16:26:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_07_ErinBrocko-1.jpg 745 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-4-1/ Wed, 11 Apr 2018 16:26:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-4-1.png 746 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_00_pool-1/ Wed, 11 Apr 2018 16:26:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_00_Pool-1.jpg 749 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_chloroform-1/ Wed, 11 Apr 2018 16:26:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_chloroform-1.jpg 751 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_aspirin-1/ Wed, 11 Apr 2018 16:26:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_aspirin-1.jpg 752 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ibuprofenmass_img-1/ Wed, 11 Apr 2018 16:26:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_ibuprofenmass_img-1.jpg 753 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_saltmass-1/ Wed, 11 Apr 2018 16:26:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_saltMass-1.jpg 754 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_alsulfatemass_img-1/ Wed, 11 Apr 2018 16:26:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_alsulfatemass_img-1.jpg 755 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_moles-1/ Wed, 11 Apr 2018 16:26:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_moles-1.jpg 756 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_compound-1/ Wed, 11 Apr 2018 16:26:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_compound-1.jpg 757 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_water-1/ Wed, 11 Apr 2018 16:26:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_water-1.jpg 758 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-5-1/ Wed, 11 Apr 2018 16:26:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-5-1.png 759 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_potassium_img-1/ Wed, 11 Apr 2018 16:26:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_potassium_img-1.jpg 760 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_argon_img-1/ Wed, 11 Apr 2018 16:26:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_argon_img-1.jpg 761 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_copper-1/ Wed, 11 Apr 2018 16:26:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_copper-1.jpg 762 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_coppermoles_img-1/ Wed, 11 Apr 2018 16:26:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_copperMoles_img-1.jpg 763 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_glycine_img-1/ Wed, 11 Apr 2018 16:26:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_glycine_img-1.jpg 764 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_glycinemass_img-1/ Wed, 11 Apr 2018 16:26:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_glycinemass_img-1.jpg 765 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_vitc_img-1/ Wed, 11 Apr 2018 16:26:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_vitC_img-1.jpg 766 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_saccharin_img-1/ Wed, 11 Apr 2018 16:26:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_saccharin_img-1.jpg 767 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_sacch_img-1/ Wed, 11 Apr 2018 16:26:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_sacch_img-1.jpg 768 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_brain-1/ Wed, 11 Apr 2018 16:26:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_brain-1.jpg 769 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_exocytosis-1/ Wed, 11 Apr 2018 16:26:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_exocytosis-1.jpg 770 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05a_img-1/ Wed, 11 Apr 2018 16:26:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05a_img-1.jpg 771 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05b_img-1/ Wed, 11 Apr 2018 16:26:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05b_img-1.jpg 772 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05c_img-1/ Wed, 11 Apr 2018 16:26:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05c_img-1.jpg 773 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05d_img-1/ Wed, 11 Apr 2018 16:26:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05d_img-1.jpg 774 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06a_img-1/ Wed, 11 Apr 2018 16:26:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06a_img-1.jpg 775 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06b_img-1/ Wed, 11 Apr 2018 16:26:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06b_img-1.jpg 776 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06c_img-1/ Wed, 11 Apr 2018 16:26:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06c_img-1.jpg 777 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06d_img-1/ Wed, 11 Apr 2018 16:26:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06d_img-1.jpg 778 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07a_img-1/ Wed, 11 Apr 2018 16:26:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07a_img-1.jpg 779 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07b_img-1/ Wed, 11 Apr 2018 16:26:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07b_img-1.jpg 780 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07c_img-1/ Wed, 11 Apr 2018 16:26:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07c_img-1.jpg 781 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_empform-1/ Wed, 11 Apr 2018 16:26:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_empform-1.jpg 783 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_hematite-1/ Wed, 11 Apr 2018 16:26:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_hematite-1.jpg 784 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-6-1/ Wed, 11 Apr 2018 16:26:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-6-1.png 785 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_brewtank-1/ Wed, 11 Apr 2018 16:26:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_BrewTank-1.jpg 786 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_espresso-1/ Wed, 11 Apr 2018 16:26:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_espresso-1.jpg 788 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_vinegar-1/ Wed, 11 Apr 2018 16:26:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_vinegar-1.jpg 789 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_dilution-1/ Wed, 11 Apr 2018 16:26:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_dilution-1.jpg 790 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_solution-1/ Wed, 11 Apr 2018 16:26:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_solution-1.jpg 791 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-7-1/ Wed, 11 Apr 2018 16:26:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-7-1.png 792 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_bleach-1/ Wed, 11 Apr 2018 16:26:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_bleach-1.jpg 794 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_example2_img-1/ Wed, 11 Apr 2018 16:26:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_Example2_img-1.jpg 795 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_saline-1/ Wed, 11 Apr 2018 16:26:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_saline-1.jpg 796 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_faucet-1/ Wed, 11 Apr 2018 16:26:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_faucet-1.jpg 797 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_00_rocket-1/ Wed, 11 Apr 2018 16:26:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_00_Rocket-1.jpg 800 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_rxn2-1/ Wed, 11 Apr 2018 16:26:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_rxn2-1.jpg 802 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_rxn3-1/ Wed, 11 Apr 2018 16:26:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_rxn3-1.jpg 803 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-3-1/ Wed, 11 Apr 2018 16:26:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-3-1.png 804 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_basehyd_img-1/ Wed, 11 Apr 2018 16:26:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_basehyd_img-1.jpg 805 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/olympus-digital-camera-2/ Wed, 11 Apr 2018 16:26:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_LeadIodide-1.jpg 807 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_hclsoln-2/ Wed, 11 Apr 2018 16:26:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_HClsoln-2.jpg 808 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_citrus-1/ Wed, 11 Apr 2018 16:26:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_Citrus-1.jpg 809 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_ammonia-1/ Wed, 11 Apr 2018 16:26:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_ammonia-1.jpg 810 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-4-2/ Wed, 11 Apr 2018 16:26:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-4-2.png 811 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_cuagno3-2/ Wed, 11 Apr 2018 16:26:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_CuAgNO3-2.jpg 812 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_iodine-1/ Wed, 11 Apr 2018 16:26:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_iodine-1.jpg 814 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_moleratio1_img-1/ Wed, 11 Apr 2018 16:26:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_moleratio1_img-1.jpg 815 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_moleratio2_img-1/ Wed, 11 Apr 2018 16:26:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_moleratio2_img-1.jpg 816 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_map2_img-1/ Wed, 11 Apr 2018 16:26:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_map2_img-1.jpg 817 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_map3_img-1/ Wed, 11 Apr 2018 16:26:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_map3_img-1.jpg 818 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_flowchart-1/ Wed, 11 Apr 2018 16:26:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_flowchart-1.jpg 819 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_airbag-1/ Wed, 11 Apr 2018 16:26:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_airbag-1.jpg 820 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_ethenebr_img-1/ Wed, 11 Apr 2018 16:26:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_etheneBr_img-1.jpg 821 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_ethene_img-1/ Wed, 11 Apr 2018 16:26:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_ethene_img-1.jpg 822 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_sandwich-1/ Wed, 11 Apr 2018 16:26:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_sandwich-1.jpg 824 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_limiting-1/ Wed, 11 Apr 2018 16:26:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_limiting-1.jpg 825 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-5-1/ Wed, 11 Apr 2018 16:26:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-5-1.png 826 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_greenchem-1/ Wed, 11 Apr 2018 16:26:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_GreenChem-1.jpg 827 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_saccharin_img-1/ Wed, 11 Apr 2018 16:26:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_saccharin_img-1.jpg 828 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_titration-1/ Wed, 11 Apr 2018 16:26:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_titration-1.jpg 830 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_map7_img-1/ Wed, 11 Apr 2018 16:26:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_map7_img-1.jpg 831 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_filter-1/ Wed, 11 Apr 2018 16:26:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_filter-1.jpg 832 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_map8_img-1/ Wed, 11 Apr 2018 16:26:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_map8_img-1.jpg 833 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_combustion-1/ Wed, 11 Apr 2018 16:26:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_combustion-1.jpg 834 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_combmap_img-1/ Wed, 11 Apr 2018 16:26:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_combmap_img-1.jpg 835 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_propionate_img-1/ Wed, 11 Apr 2018 16:26:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_propionate_img-1.jpg 836 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_00_match-1/ Wed, 11 Apr 2018 16:26:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_00_Match-1.jpg 839 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_thermochem-1/ Wed, 11 Apr 2018 16:26:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Thermochem-1.jpg 841 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_waterfall-1/ Wed, 11 Apr 2018 16:26:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Waterfall-1.jpg 842 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_hotcold-1/ Wed, 11 Apr 2018 16:26:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HotCold-1.jpg 843 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-6-1/ Wed, 11 Apr 2018 16:26:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-6-1.png 844 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_thermom-1/ Wed, 11 Apr 2018 16:26:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Thermom-1.jpg 845 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_heattrans1-1/ Wed, 11 Apr 2018 16:26:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HeatTrans1-1.jpg 846 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_oxyactorch-1/ Wed, 11 Apr 2018 16:26:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_OxyacTorch-1.jpg 847 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_heatcapacity-1/ Wed, 11 Apr 2018 16:26:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HeatCapacity-1.jpg 848 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_soltherm1-1/ Wed, 11 Apr 2018 16:26:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_SolTherm1-1.jpg 849 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_soltherm2-1/ Wed, 11 Apr 2018 16:26:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_SolTherm2-1.jpg 850 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_heatmeas-1/ Wed, 11 Apr 2018 16:26:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HeatMeas-1.jpg 852 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_calorim-1/ Wed, 11 Apr 2018 16:26:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_Calorim-1.jpg 853 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_calorim2-1/ Wed, 11 Apr 2018 16:26:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_Calorim2-1.jpg 854 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_heattrans2-1/ Wed, 11 Apr 2018 16:26:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HeatTrans2-1.jpg 855 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_handwarmer-1/ Wed, 11 Apr 2018 16:26:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HandWarmer-1.jpg 856 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-7-1/ Wed, 11 Apr 2018 16:26:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-7-1.png 857 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_icepack-1/ Wed, 11 Apr 2018 16:26:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_IcePack-1.jpg 858 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_bombcalor-1/ Wed, 11 Apr 2018 16:26:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_BombCalor-1.jpg 859 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_foodlabel-1/ Wed, 11 Apr 2018 16:26:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_FoodLabel-1.jpg 860 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_systemqw-1/ Wed, 11 Apr 2018 16:26:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_Systemqw-1.jpg 862 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-8-1/ Wed, 11 Apr 2018 16:26:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-8-1.png 863 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_summit-1/ Wed, 11 Apr 2018 16:26:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_Summit-1.jpg 864 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_gasburning-1/ Wed, 11 Apr 2018 16:26:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_GasBurning-1.jpg 865 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_algalfuel1-1/ Wed, 11 Apr 2018 16:26:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_AlgalFuel1-1.jpg 866 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_algalfuel2-2/ Wed, 11 Apr 2018 16:26:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_AlgalFuel2-1.jpg 867 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_hessco2-1/ Wed, 11 Apr 2018 16:26:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_HessCO2-1.jpg 868 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_00_crabneb-1/ Wed, 11 Apr 2018 16:26:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_00_CrabNeb-1.jpg 871 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_frequency-1/ Wed, 11 Apr 2018 16:26:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Frequency-1.jpg 873 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_emspectrum-1/ Wed, 11 Apr 2018 16:27:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_emspectrum-1.jpg 874 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_radiocell-1/ Wed, 11 Apr 2018 16:27:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_RadioCell-1.jpg 875 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_amfm-1/ Wed, 11 Apr 2018 16:27:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_AMFM-1.jpg 876 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_liteinterf-1/ Wed, 11 Apr 2018 16:27:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_LiteInterf-1.jpg 877 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_vibrstring-1/ Wed, 11 Apr 2018 16:27:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Vibrstring-1.jpg 878 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_vibratdrum-1/ Wed, 11 Apr 2018 16:27:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Vibratdrum-1.jpg 879 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-8-1/ Wed, 11 Apr 2018 16:27:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-8-1.png 880 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_solardist-1/ Wed, 11 Apr 2018 16:27:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Solardist-1.jpg 881 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_blackbody-1/ Wed, 11 Apr 2018 16:27:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Blackbody-1.jpg 882 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_ephoton-1/ Wed, 11 Apr 2018 16:27:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Ephoton-1.jpg 883 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_neon-1/ Wed, 11 Apr 2018 16:27:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_neon-1.jpg 884 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_2spectra-1/ Wed, 11 Apr 2018 16:27:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_2spectra-1.jpg 885 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_02_hlevels-1/ Wed, 11 Apr 2018 16:27:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_02_Hlevels-1.jpg 887 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_02_bohrarrows-1/ Wed, 11 Apr 2018 16:27:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_02_BohrArrows-1.jpg 888 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_waterw-1/ Wed, 11 Apr 2018 16:27:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_waterw-1.jpg 890 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_elecw-1/ Wed, 11 Apr 2018 16:27:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_elecw-1.jpg 891 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_angm-1/ Wed, 11 Apr 2018 16:27:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_angm-1.jpg 892 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_electrnin-1/ Wed, 11 Apr 2018 16:27:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Electrnin-1.jpg 893 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-9-1/ Wed, 11 Apr 2018 16:27:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-9-1.png 894 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_qnumbers-1/ Wed, 11 Apr 2018 16:27:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Qnumbers-1.jpg 895 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_sorbit-1/ Wed, 11 Apr 2018 16:27:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_sOrbit-1.jpg 896 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_oshapes-1/ Wed, 11 Apr 2018 16:27:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Oshapes-1.jpg 897 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_subshells-1/ Wed, 11 Apr 2018 16:27:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_subshells-1.jpg 898 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_spin-1/ Wed, 11 Apr 2018 16:27:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_spin-1.jpg 899 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_orboutline_img-1/ Wed, 11 Apr 2018 16:27:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_OrbOutline_img-1.jpg 900 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_ex9soln_img-1/ Wed, 11 Apr 2018 16:27:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Ex9soln_img-1.jpg 901 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_eleveldiag-1/ Wed, 11 Apr 2018 16:27:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_eLeveldiag-1.jpg 903 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_econfig-1/ Wed, 11 Apr 2018 16:27:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Econfig-1.jpg 904 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_efillorder-1/ Wed, 11 Apr 2018 16:27:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Efillorder-1.jpg 905 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_econtable-1/ Wed, 11 Apr 2018 16:27:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Econtable-1.jpg 906 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_hydrog1_img-1/ Wed, 11 Apr 2018 16:27:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Hydrog1_img-1.jpg 907 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_helium1_img-1/ Wed, 11 Apr 2018 16:27:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Helium1_img-1.jpg 908 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_lithium12_img-1/ Wed, 11 Apr 2018 16:27:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Lithium12_img-1.jpg 909 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_beryll12_img-1/ Wed, 11 Apr 2018 16:27:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Beryll12_img-1.jpg 910 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_boron122_img-1/ Wed, 11 Apr 2018 16:27:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Boron122_img-1.jpg 911 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_carbon122_img-1/ Wed, 11 Apr 2018 16:27:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Carbon122_img-1.jpg 912 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_nofne_img-1/ Wed, 11 Apr 2018 16:27:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_NOFNe_img-1.jpg 913 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_valence-1/ Wed, 11 Apr 2018 16:27:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Valence-1.jpg 914 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_ptableconf-1/ Wed, 11 Apr 2018 16:27:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Ptableconf-1.jpg 915 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_phosphorb_img-1/ Wed, 11 Apr 2018 16:27:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_PhosphOrb_img-1.jpg 916 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2a_img-1/ Wed, 11 Apr 2018 16:27:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2a_img-1.jpg 917 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2b_img-1/ Wed, 11 Apr 2018 16:27:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2b_img-1.jpg 918 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2c_img-1/ Wed, 11 Apr 2018 16:27:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2c_img-1.jpg 919 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2d_img-1/ Wed, 11 Apr 2018 16:27:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2d_img-1.jpg 920 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2e_img-1/ Wed, 11 Apr 2018 16:27:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2e_img-1.jpg 921 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-10-1/ Wed, 11 Apr 2018 16:27:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-10-1.png 923 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_covalradit-1/ Wed, 11 Apr 2018 16:27:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_CovalradiT-1.jpg 924 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_radiigraph-1/ Wed, 11 Apr 2018 16:27:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Radiigraph-1.jpg 925 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_ionradii-1/ Wed, 11 Apr 2018 16:27:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Ionradii-1.jpg 926 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_firstiongr-1/ Wed, 11 Apr 2018 16:27:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Firstiongr-1.jpg 927 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_firstionen-1/ Wed, 11 Apr 2018 16:27:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Firstionen-1.jpg 928 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_oxygen122_img-1/ Wed, 11 Apr 2018 16:27:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Oxygen122_img-1.jpg 929 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_elaffin-1/ Wed, 11 Apr 2018 16:27:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Elaffin-1.jpg 930 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_00_bucky-1/ Wed, 11 Apr 2018 16:27:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_00_Bucky-1.jpg 933 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_01_naclphotos-1/ Wed, 11 Apr 2018 16:27:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_01_NaClPhotos-1.jpg 935 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/microsoft-word-cnx_chem_ch07_mod01-docx-2/ Wed, 11 Apr 2018 16:27:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_01_NaClStruc-1.jpg 936 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_morse-1/ Wed, 11 Apr 2018 16:27:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_Morse-1.jpg 938 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_hclbond-1/ Wed, 11 Apr 2018 16:27:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_HClBond-1.jpg 939 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_entable-1/ Wed, 11 Apr 2018 16:27:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_ENTable-1.jpg 940 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_pauling-1/ Wed, 11 Apr 2018 16:27:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_Pauling-1.jpg 941 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_deltaen-1/ Wed, 11 Apr 2018 16:27:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_DeltaEN-1.jpg 942 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_lewisstruct1_img-1/ Wed, 11 Apr 2018 16:27:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Lewisstruct1_img-1.jpg 944 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_3rowlewis-1/ Wed, 11 Apr 2018 16:27:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_3rowLewis-1.jpg 945 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_naca_img-1/ Wed, 11 Apr 2018 16:27:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NaCa_img-1.jpg 946 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cls_img-1/ Wed, 11 Apr 2018 16:27:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ClS_img-1.jpg 947 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ionlewis-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_IonLewis-1.jpg 948 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_c12dot_img-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_C12dot_img-1.jpg 949 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cl2dash_img-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Cl2dash_img-1.jpg 950 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xy4struc_img-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XY4struc_img-1.jpg 951 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_numbonds_img-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_numbonds_img-1.jpg 952 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_doublebond_img-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_DoubleBond_img-1.jpg 953 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cocn_img-1/ Wed, 11 Apr 2018 16:27:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_COCN_img-1.jpg 954 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_unprelec_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Unprelec_img-1.jpg 955 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_singlebond_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Singlebond_img-1.jpg 956 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_sih4_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_SiH4_img-1.jpg 957 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_oxydiflor2_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_oxydiflor2_img-1.jpg 958 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_formate2_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_formate2_img-1.jpg 959 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nitrosoni2_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_nitrosoni2_img-1.jpg 960 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nitrosoni3_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_nitrosoni3_img-1.jpg 961 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_1_img-1/ Wed, 11 Apr 2018 16:27:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_1_img-1.jpg 962 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_2_img-1/ Wed, 11 Apr 2018 16:27:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_2_img-1.jpg 963 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_3_img-1/ Wed, 11 Apr 2018 16:27:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_3_img-1.jpg 964 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_4_img-1/ Wed, 11 Apr 2018 16:27:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_4_img-1.jpg 965 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_coco2_img-1/ Wed, 11 Apr 2018 16:27:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_COCO2_img-1.jpg 966 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_smalley-1/ Wed, 11 Apr 2018 16:27:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Smalley-1.jpg 967 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nosingle_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NOsingle_img-1.jpg 968 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nodouble_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NODouble_img-1.jpg 969 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_edeficient_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_edeficient_img-1.jpg 970 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_dative_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_dative_img-1.jpg 971 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_pf5sf6_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_PF5SF6_img-1.jpg 972 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_if5xef4_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_IF5XeF4_img-1.jpg 973 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xeskeleton_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Xeskeleton_img-1.jpg 974 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xef2_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XeF2_img-1.jpg 975 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xef6_img-1/ Wed, 11 Apr 2018 16:27:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XeF6_img-1.jpg 976 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_interhal_img-1/ Wed, 11 Apr 2018 16:27:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Interhal_img-1.jpg 977 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisa_img-1/ Wed, 11 Apr 2018 16:27:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisa_img-1.jpg 978 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisb_img-1/ Wed, 11 Apr 2018 16:27:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisb_img-1.jpg 979 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisc_img-1/ Wed, 11 Apr 2018 16:27:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisc_img-1.jpg 980 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisd_img-1/ Wed, 11 Apr 2018 16:27:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisd_img-1.jpg 981 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioa_img-1/ Wed, 11 Apr 2018 16:27:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioa_img-1.jpg 982 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbiob_img-1/ Wed, 11 Apr 2018 16:27:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbiob_img-1.jpg 983 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioc_img-1/ Wed, 11 Apr 2018 16:27:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioc_img-1.jpg 984 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbiod_img-1/ Wed, 11 Apr 2018 16:27:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbiod_img-1.jpg 985 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioe_img-1/ Wed, 11 Apr 2018 16:27:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioe_img-1.jpg 986 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1a_img-1/ Wed, 11 Apr 2018 16:27:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1a_img-1.jpg 987 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1b_img-1/ Wed, 11 Apr 2018 16:27:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1b_img-1.jpg 988 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1d_img-1/ Wed, 11 Apr 2018 16:27:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1d_img-1.jpg 989 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1g_img-1/ Wed, 11 Apr 2018 16:27:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1g_img-1.jpg 990 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3a_img-1/ Wed, 11 Apr 2018 16:27:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3a_img-1.jpg 991 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3b_img-1/ Wed, 11 Apr 2018 16:27:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3b_img-1.jpg 992 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3c_img-1/ Wed, 11 Apr 2018 16:27:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3c_img-1.jpg 993 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3d_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3d_img-1.jpg 994 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3e_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3e_img-1.jpg 995 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3f_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3f_img-1.jpg 996 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question5_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question5_img-1.jpg 997 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7a_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7a_img-1.jpg 998 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7b_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7b_img-1.jpg 999 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7c_img-1/ Wed, 11 Apr 2018 16:27:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7c_img-1.jpg 1000 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7d_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7d_img-1.jpg 1001 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7e_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7e_img-1.jpg 1002 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7f_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7f_img-1.jpg 1003 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7g_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7g_img-1.jpg 1004 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7h_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7h_img-1.jpg 1005 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7i_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7i_img-1.jpg 1006 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7j_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7j_img-1.jpg 1007 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7k_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7k_img-1.jpg 1008 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9a_img-1/ Wed, 11 Apr 2018 16:27:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9a_img-1.jpg 1009 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9b_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9b_img-1.jpg 1010 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9c_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9c_img-1.jpg 1011 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9d_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9d_img-1.jpg 1012 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question18_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question18_img-1.jpg 1013 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_carbtet_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Carbtet_img-1.jpg 1014 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansa_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansa_img-1.jpg 1015 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansb_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansb_img-1.jpg 1016 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansc_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansc_img-1.jpg 1017 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansd_img-1/ Wed, 11 Apr 2018 16:27:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansd_img-1.jpg 1018 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioanse_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioanse_img-1.jpg 1019 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise25_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise25_img-1.jpg 1020 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_icl4_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_ICL4_img-1.jpg 1022 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO_img-1.jpg 1023 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_brcl3_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_BRCL3_img-1.jpg 1024 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ex070402_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ex070402_img-1.jpg 1025 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co2pos_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO2pos_img-1.jpg 1026 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_thiocyan_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Thiocyan_img-1.jpg 1027 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_n2o_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_N2O_img-1.jpg 1028 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_n2ofc_img-1/ Wed, 11 Apr 2018 16:27:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_N2Ofc_img-1.jpg 1029 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_nno_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NNO_img-1.jpg 1030 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2ion_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2ion_img-1.jpg 1031 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2res_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2res_img-1.jpg 1032 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2resarr_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2resarr_img-1.jpg 1033 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co3res_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO3res_img-1.jpg 1034 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-11-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-11-1.png 1035 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_chood_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CHOOd_img-1.jpg 1036 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_chooe_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CHOOe_img-1.jpg 1037 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_so2resd_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_SO2resd_img-1.jpg 1038 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_so2rese_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_SO2rese_img-1.jpg 1039 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_acet_img-1/ Wed, 11 Apr 2018 16:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Acet_img-1.jpg 1040 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_hno2_img-1/ Wed, 11 Apr 2018 16:27:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_HNO2_img-1.jpg 1041 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansa_img-1/ Wed, 11 Apr 2018 16:27:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansa_img-1.jpg 1042 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansb_img-1/ Wed, 11 Apr 2018 16:27:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansb_img-1.jpg 1043 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansc_img-1/ Wed, 11 Apr 2018 16:27:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansc_img-1.jpg 1044 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansd_img-1/ Wed, 11 Apr 2018 16:27:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansd_img-1.jpg 1045 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2anse_img-1/ Wed, 11 Apr 2018 16:27:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2anse_img-1.jpg 1046 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques11ans_img-1/ Wed, 11 Apr 2018 16:27:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques11ans_img-1.jpg 1047 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques13ansb_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques13ansb_img-1.jpg 1048 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques13ansc_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques13ansc_img-1.jpg 1049 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12a_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12a_img-1.jpg 1050 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12b_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12b_img-1.jpg 1051 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12c_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12c_img-1.jpg 1052 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12d_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12d_img-1.jpg 1053 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ohamine_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_OHamine_img-1.jpg 1054 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_nf3_img-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NF3_img-1.jpg 1055 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_h2so4-1/ Wed, 11 Apr 2018 16:27:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_H2SO4-1.jpg 1056 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ch4bond_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_CH4bond_img-1.jpg 1058 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ch3ohlew_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_CH3OHLew_img-1.jpg 1059 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ethanol_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_Ethanol_img-1.jpg 1060 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_bornhaber-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_BornHaber-1.jpg 1061 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_hydroxya_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_Hydroxya_img-1.jpg 1062 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_c6h8lew_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_C6H8Lew_img-1.jpg 1063 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_fno2_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_FNO2_img-1.jpg 1064 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_c6h8ans_img-1/ Wed, 11 Apr 2018 16:27:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_C6H8ans_img-1.jpg 1065 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ch2o-1/ Wed, 11 Apr 2018 16:27:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CH2O-1.jpg 1067 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bef2-1/ Wed, 11 Apr 2018 16:27:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BeF2-1.jpg 1068 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_egeom-1/ Wed, 11 Apr 2018 16:27:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Egeom-1.jpg 1069 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ch4-1/ Wed, 11 Apr 2018 16:27:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CH4-1.jpg 1070 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh3-1/ Wed, 11 Apr 2018 16:27:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH3-1.jpg 1071 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_molgeom-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_molgeom-1.jpg 1072 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_axeq-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Axeq-1.jpg 1073 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_co2_img-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CO2_img-1.jpg 1074 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bcl3_img-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BCl3_img-1.jpg 1075 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bcl3mol-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BCl3mol-1.jpg 1076 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh4_img-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH4_img-1.jpg 1077 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh4mol-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH4mol-1.jpg 1078 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_h2o_img-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_H2O_img-1.jpg 1079 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_h2omol-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_H2Omol-1.jpg 1080 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sf4_img-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SF4_img-1.jpg 1081 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sf4mol-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SF4mol-1.jpg 1082 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_xef4_img-1/ Wed, 11 Apr 2018 16:27:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_XeF4_img-1.jpg 1083 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_xef4mol-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_XEF4mol-1.jpg 1084 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_gly_img-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_gly_img-1.jpg 1085 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_gly3d_img-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_gly3D_img-1.jpg 1086 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ala_img-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_ala_img-1.jpg 1087 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-12-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-12-1.png 1088 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bondvector-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BondVector-1.jpg 1089 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_co2h2odip-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CO2H2Odip-1.jpg 1090 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_osc_img-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_OSC_img-1.jpg 1091 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_chcl3_img-1/ Wed, 11 Apr 2018 16:27:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CHCl3_img-1.jpg 1092 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sh2nh3_img-1/ Wed, 11 Apr 2018 16:27:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SH2NH3_img-1.jpg 1093 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_dipolfield-1/ Wed, 11 Apr 2018 16:27:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Dipolfield-1.jpg 1094 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ques23ans_img-1/ Wed, 11 Apr 2018 16:27:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Ques23ans_img-1.jpg 1095 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3a_img-1/ Wed, 11 Apr 2018 16:27:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3a_img-1.jpg 1096 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3b_img-1/ Wed, 11 Apr 2018 16:27:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3b_img-1.jpg 1097 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3c_img-1/ Wed, 11 Apr 2018 16:27:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3c_img-1.jpg 1098 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ques28ans_img-1/ Wed, 11 Apr 2018 16:27:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Ques28ans_img-1.jpg 1099 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mexico-2/ Wed, 11 Apr 2018 16:27:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_00_Sinkhole-1.jpg 1102 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_conjugate_img-1/ Wed, 11 Apr 2018 16:27:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_conjugate_img-1.jpg 1104 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_hf_img-1/ Wed, 11 Apr 2018 16:27:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_HF_img-1.jpg 1105 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_nh3_img-1/ Wed, 11 Apr 2018 16:27:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_NH3_img-1.jpg 1106 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_water_img-1/ Wed, 11 Apr 2018 16:27:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_Water_img-1.jpg 1107 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_phscale-1/ Wed, 11 Apr 2018 16:27:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_phscale-1.jpg 1109 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_acidrain-1/ Wed, 11 Apr 2018 16:27:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_AcidRain-1.jpg 1110 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_phmeter-1/ Wed, 11 Apr 2018 16:27:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_pHMeter-1.jpg 1111 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_indicator-1/ Wed, 11 Apr 2018 16:27:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_indicator-1.jpg 1112 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_strong-1/ Wed, 11 Apr 2018 16:27:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_strong-1.jpg 1114 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-27-1/ Wed, 11 Apr 2018 16:27:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-27-1.png 1115 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_strengths-1/ Wed, 11 Apr 2018 16:27:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_strengths-1.jpg 1116 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_corresp-1/ Wed, 11 Apr 2018 16:27:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Corresp-1.jpg 1117 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_acetate_img-1/ Wed, 11 Apr 2018 16:27:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_acetate_img-1.jpg 1118 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_ammonia-1/ Wed, 11 Apr 2018 16:27:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ammonia-1.jpg 1119 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_aceticacid-3/ Wed, 11 Apr 2018 16:27:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Vinegar-1.jpg 1120 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps_img-1/ Wed, 11 Apr 2018 16:27:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps_img-1.jpg 1121 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable1_img-1/ Wed, 11 Apr 2018 16:27:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable1_img-1.jpg 1122 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable2_img-1/ Wed, 11 Apr 2018 16:27:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable2_img-1.jpg 1123 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_antsting-1/ Wed, 11 Apr 2018 16:27:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_AntSting-1.jpg 1124 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellulose-1-2/ Thu, 12 Apr 2018 02:50:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Cellulose-1-2.jpg 1239 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_rutherford-1-2/ Thu, 12 Apr 2018 02:50:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Rutherford-1-2.jpg 1240 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_hclsoln-1-2/ Thu, 12 Apr 2018 02:50:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_HClsoln-1-2.jpg 1241 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_3dorbitals-2/ Thu, 12 Apr 2018 02:50:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_3dOrbitals-2.jpg 1242 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkanes-1-2/ Thu, 12 Apr 2018 02:50:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkanes-1-2.jpg 1243 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_10_06_fccntrdcb-2/ Thu, 12 Apr 2018 02:50:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_10_06_FcCntrdCb-2.jpg 1244 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_reaction1-1-2/ Thu, 12 Apr 2018 02:50:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_Reaction1-1-2.jpg 1245 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-1-chemistry-in-context/cnx_chem_01_01_alchemist-2/ Thu, 12 Apr 2018 02:50:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_Alchemist-2.jpg 1248 1252 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_chemweb-2/ Thu, 12 Apr 2018 02:50:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_ChemWeb-2.jpg 1249 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_scimethod-2/ Thu, 12 Apr 2018 02:50:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_SciMethod-2.jpg 1250 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_waterdom-2/ Thu, 12 Apr 2018 02:50:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_WaterDom-2.jpg 1251 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_statesmatt-2/ Thu, 12 Apr 2018 02:50:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_StatesMatt-2.jpg 1253 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/cnx_chem_01_02_plasma-2/ Thu, 12 Apr 2018 02:50:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Plasma-2.jpg 1254 1266 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-10/ Thu, 12 Apr 2018 02:50:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-10.png 1255 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_consmatter-2/ Thu, 12 Apr 2018 02:50:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_ConsMatter-2.jpg 1256 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_goldatoms-2/ Thu, 12 Apr 2018 02:50:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_GoldAtoms-2.jpg 1257 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellulose-3/ Thu, 12 Apr 2018 02:50:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Cellulose-3.jpg 1258 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_molecules-2/ Thu, 12 Apr 2018 02:50:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Molecules-2.jpg 1259 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_decomp-2/ Thu, 12 Apr 2018 02:50:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_decomp-2.jpg 1260 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_mixtures-2/ Thu, 12 Apr 2018 02:50:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_Mixtures-2.jpg 1261 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_matttype-2/ Thu, 12 Apr 2018 02:50:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_MattType-2.jpg 1262 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_electrolys-2/ Thu, 12 Apr 2018 02:50:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_Electrolys-2.jpg 1263 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_01_fuelcell-2/ Thu, 12 Apr 2018 02:50:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_01_FuelCell-2.jpg 1264 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_02_cellphone-2/ Thu, 12 Apr 2018 02:50:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_02_CellPhone-2.jpg 1265 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_00_dailychem-2/ Thu, 12 Apr 2018 02:50:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_00_DailyChem-2.jpg 1267 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_measure-2/ Thu, 12 Apr 2018 02:51:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Measure-2.jpg 1269 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits5_img-2/ Thu, 12 Apr 2018 02:51:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits5_img-2.jpg 1270 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits1_img-2/ Thu, 12 Apr 2018 02:51:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits1_img-2.jpg 1271 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits2_img-2/ Thu, 12 Apr 2018 02:51:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits2_img-2.jpg 1272 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits3_img-2/ Thu, 12 Apr 2018 02:51:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits3_img-2.jpg 1273 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_sigdigits4_img-2/ Thu, 12 Apr 2018 02:51:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_SigDigits4_img-2.jpg 1274 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_cylrebar-2/ Thu, 12 Apr 2018 02:51:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_CylRebar-2.jpg 1275 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_cylgold-2/ Thu, 12 Apr 2018 02:51:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_CylGold-2.jpg 1276 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_archery-2/ Thu, 12 Apr 2018 02:51:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Archery-2.jpg 1277 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_05_archer2_img-2/ Thu, 12 Apr 2018 02:51:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_05_Archer2_img-2.jpg 1278 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-1-2/ Thu, 12 Apr 2018 02:51:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-1-2.png 1280 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_mydcmin-2/ Thu, 12 Apr 2018 02:51:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_MYdCmIn-2.jpg 1281 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-4-measurements/cnx_chem_01_04_kilogram-2/ Thu, 12 Apr 2018 02:51:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_Kilogram-2.jpg 1282 1284 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_04_volume-2/ Thu, 12 Apr 2018 02:51:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_04_Volume-2.jpg 1283 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_physchange-2/ Thu, 12 Apr 2018 02:51:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_PhysChange-2.jpg 1285 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_rust-2/ Thu, 12 Apr 2018 02:51:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_Rust-2.jpg 1286 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_chemchange-2/ Thu, 12 Apr 2018 02:51:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_ChemChange-2.jpg 1287 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_hazdiamond-2/ Thu, 12 Apr 2018 02:51:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_HazDiamond-2.jpg 1288 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_03_periodicpu-2/ Thu, 12 Apr 2018 02:51:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_03_PeriodicPU-2.jpg 1289 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_01_06_tempscales-2/ Thu, 12 Apr 2018 02:51:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_01_06_TempScales-2.jpg 1291 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_00_biomarkers-2/ Thu, 12 Apr 2018 02:51:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_00_Biomarkers-2.jpg 1294 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton1-2/ Thu, 12 Apr 2018 02:51:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton1-2.jpg 1296 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton2-2/ Thu, 12 Apr 2018 02:51:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton2-2.jpg 1297 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton3-2/ Thu, 12 Apr 2018 02:51:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton3-2.jpg 1298 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton6_img-2/ Thu, 12 Apr 2018 02:51:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton6_img-2.jpg 1299 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton8_img-2/ Thu, 12 Apr 2018 02:51:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton8_img-2.jpg 1300 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_multprop-2/ Thu, 12 Apr 2018 02:51:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_MultProp-2.jpg 1301 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_01_dalton10_img-2/ Thu, 12 Apr 2018 02:51:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_01_Dalton10_img-2.jpg 1302 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_cathoderay-2/ Thu, 12 Apr 2018 02:51:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_CathodeRay-2.jpg 1304 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_millikan-2/ Thu, 12 Apr 2018 02:51:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Millikan-2.jpg 1305 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_atommodels-2/ Thu, 12 Apr 2018 02:51:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_AtomModels-2.jpg 1306 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_rutherford-3/ Thu, 12 Apr 2018 02:51:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_Rutherford-3.jpg 1307 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_02_goldfoil3-2/ Thu, 12 Apr 2018 02:51:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_02_GoldFoil3-2.jpg 1308 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_atomsize-2/ Thu, 12 Apr 2018 02:51:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_AtomSize-2.jpg 1310 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_iodine-2/ Thu, 12 Apr 2018 02:51:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_Iodine-2.jpg 1311 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-3-atomic-structure-and-symbolism/cnx_chem_02_03_sisymbol-2/ Thu, 12 Apr 2018 02:51:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_SiSymbol-2.jpg 1312 1316 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-1-2/ Thu, 12 Apr 2018 02:51:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-1-2.png 1313 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_atomsym-2/ Thu, 12 Apr 2018 02:51:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_AtomSym-2.jpg 1314 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_03_massspec-2/ Thu, 12 Apr 2018 02:51:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_03_MassSpec-2.jpg 1315 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_methanerep-2/ Thu, 12 Apr 2018 02:51:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_MethaneRep-2.jpg 1317 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_sulfur-2/ Thu, 12 Apr 2018 02:51:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Sulfur-2.jpg 1318 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_hydrogen-2/ Thu, 12 Apr 2018 02:51:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Hydrogen-2.jpg 1319 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_tio2-2/ Thu, 12 Apr 2018 02:51:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_TiO2-2.jpg 1320 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_benzene-2/ Thu, 12 Apr 2018 02:51:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Benzene-2.jpg 1321 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_aceticacid-2-2/ Thu, 12 Apr 2018 02:51:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_AceticAcid-2.jpg 1322 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-2-2/ Thu, 12 Apr 2018 02:51:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-2-2.png 1323 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_leecronin-2/ Thu, 12 Apr 2018 02:51:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_LeeCronin-2.jpg 1324 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_isomers-2/ Thu, 12 Apr 2018 02:51:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Isomers-2.jpg 1325 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_isomers2-2/ Thu, 12 Apr 2018 02:51:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Isomers2-2.jpg 1326 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3a_img-2/ Thu, 12 Apr 2018 02:51:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3a_img-2.jpg 1327 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3b_img-2/ Thu, 12 Apr 2018 02:51:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3b_img-2.jpg 1328 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3c_img-2/ Thu, 12 Apr 2018 02:51:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3c_img-2.jpg 1329 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question3d_img-2/ Thu, 12 Apr 2018 02:51:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question3d_img-2.jpg 1330 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4a_img-2/ Thu, 12 Apr 2018 02:51:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4a_img-2.jpg 1331 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4b_img-2/ Thu, 12 Apr 2018 02:51:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4b_img-2.jpg 1332 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4c_img-2/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4c_img-2.jpg 1333 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question4d_img-2/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question4d_img-2.jpg 1334 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question7a_img-2/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question7a_img-2.jpg 1335 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question7b_img-2/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question7b_img-2.jpg 1336 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question9a_img-2/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question9a_img-2.jpg 1337 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_04_question9b_img-2/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_04_Question9b_img-2.jpg 1338 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_mendeleev-2/ Thu, 12 Apr 2018 02:51:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_Mendeleev-2.jpg 1340 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_pertable1-2/ Thu, 12 Apr 2018 02:51:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_PerTable1-2.jpg 1341 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_05_pertable2-2/ Thu, 12 Apr 2018 02:51:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_05_PerTable2-2.jpg 1342 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-3-2/ Thu, 12 Apr 2018 02:51:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-3-2.png 1343 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_nacation-2/ Thu, 12 Apr 2018 02:51:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_NaCation-2.jpg 1345 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_ioncharges-2/ Thu, 12 Apr 2018 02:51:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_IonCharges-2.jpg 1346 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_naclmolten-2/ Thu, 12 Apr 2018 02:51:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_NaClMolten-2.jpg 1347 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-2-2/ Thu, 12 Apr 2018 02:51:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-2-2.png 1348 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_06_sapphire-2/ Thu, 12 Apr 2018 02:51:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_06_Sapphire-2.jpg 1349 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_02_07_erinbrocko-2/ Thu, 12 Apr 2018 02:51:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_02_07_ErinBrocko-2.jpg 1351 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-4-2/ Thu, 12 Apr 2018 02:51:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-4-2.png 1352 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_00_pool-2/ Thu, 12 Apr 2018 02:51:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_00_Pool-2.jpg 1355 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/cnx_chem_03_01_chloroform-2/ Thu, 12 Apr 2018 02:51:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_chloroform-2.jpg 1357 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/cnx_chem_03_01_aspirin-2/ Thu, 12 Apr 2018 02:51:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_aspirin-2.jpg 1358 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/cnx_chem_03_01_ibuprofenmass_img-2/ Thu, 12 Apr 2018 02:51:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_ibuprofenmass_img-2.jpg 1359 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_saltmass-2/ Thu, 12 Apr 2018 02:51:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_saltMass-2.jpg 1360 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_alsulfatemass_img-2/ Thu, 12 Apr 2018 02:51:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_alsulfatemass_img-2.jpg 1361 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_moles-2/ Thu, 12 Apr 2018 02:51:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_moles-2.jpg 1362 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_compound-2/ Thu, 12 Apr 2018 02:51:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_compound-2.jpg 1363 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_water-2/ Thu, 12 Apr 2018 02:51:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_water-2.jpg 1364 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-5-2/ Thu, 12 Apr 2018 02:51:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-5-2.png 1365 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_potassium_img-2/ Thu, 12 Apr 2018 02:51:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_potassium_img-2.jpg 1366 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_argon_img-2/ Thu, 12 Apr 2018 02:51:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_argon_img-2.jpg 1367 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_copper-2/ Thu, 12 Apr 2018 02:51:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_copper-2.jpg 1368 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_coppermoles_img-2/ Thu, 12 Apr 2018 02:51:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_copperMoles_img-2.jpg 1369 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_glycine_img-2/ Thu, 12 Apr 2018 02:51:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_glycine_img-2.jpg 1370 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_glycinemass_img-2/ Thu, 12 Apr 2018 02:51:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_glycinemass_img-2.jpg 1371 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_vitc_img-2/ Thu, 12 Apr 2018 02:51:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_vitC_img-2.jpg 1372 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_02_saccharin_img-2/ Thu, 12 Apr 2018 02:51:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_saccharin_img-2.jpg 1373 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_02_sacch_img-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_02_sacch_img-2.jpg 1374 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_01_brain-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_brain-2.jpg 1375 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/cnx_chem_03_01_exocytosis-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_exocytosis-2.jpg 1376 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05a_img-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05a_img-2.jpg 1377 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05b_img-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05b_img-2.jpg 1378 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05c_img-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05c_img-2.jpg 1379 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_05d_img-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_05d_img-2.jpg 1380 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06a_img-2/ Thu, 12 Apr 2018 02:51:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06a_img-2.jpg 1381 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06b_img-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06b_img-2.jpg 1382 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06c_img-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06c_img-2.jpg 1383 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_06d_img-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_06d_img-2.jpg 1384 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07a_img-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07a_img-2.jpg 1385 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07b_img-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07b_img-2.jpg 1386 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_01_ex01_07c_img-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_01_Ex01_07c_img-2.jpg 1387 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-2-determining-empirical-and-molecular-formulas/cnx_chem_03_03_empform-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_empform-2.jpg 1389 1393 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-2-determining-empirical-and-molecular-formulas/cnx_chem_03_03_hematite-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_hematite-2.jpg 1390 1393 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-6-2/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-6-2.png 1391 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-2-determining-empirical-and-molecular-formulas/cnx_chem_03_03_brewtank-2/ Thu, 12 Apr 2018 02:51:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_BrewTank-2.jpg 1392 1393 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_03_espresso-2/ Thu, 12 Apr 2018 02:51:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_03_espresso-2.jpg 1394 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_vinegar-2/ Thu, 12 Apr 2018 02:51:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_vinegar-2.jpg 1395 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_dilution-2/ Thu, 12 Apr 2018 02:51:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_dilution-2.jpg 1396 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_04_solution-2/ Thu, 12 Apr 2018 02:51:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_04_solution-2.jpg 1397 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-7-2/ Thu, 12 Apr 2018 02:51:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-7-2.png 1398 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_bleach-2/ Thu, 12 Apr 2018 02:51:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_bleach-2.jpg 1400 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_example2_img-2/ Thu, 12 Apr 2018 02:51:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_Example2_img-2.jpg 1401 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_saline-2/ Thu, 12 Apr 2018 02:51:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_saline-2.jpg 1402 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_03_05_faucet-2/ Thu, 12 Apr 2018 02:51:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_03_05_faucet-2.jpg 1403 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_00_rocket-2/ Thu, 12 Apr 2018 02:51:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_00_Rocket-2.jpg 1406 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_rxn2-2/ Thu, 12 Apr 2018 02:51:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_rxn2-2.jpg 1408 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_rxn3-2/ Thu, 12 Apr 2018 02:51:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_rxn3-2.jpg 1409 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-3-2/ Thu, 12 Apr 2018 02:51:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-3-2.png 1410 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_01_basehyd_img-2/ Thu, 12 Apr 2018 02:51:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_01_basehyd_img-2.jpg 1411 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-2-classifying-chemical-reactions/olympus-digital-camera-3/ Thu, 12 Apr 2018 02:51:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_LeadIodide-2.jpg 1413 1419 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/neutralization-reactions/cnx_chem_04_02_hclsoln-3/ Thu, 12 Apr 2018 02:51:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_HClsoln-3.jpg 1414 2178 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/neutralization-reactions/cnx_chem_04_02_citrus-2/ Thu, 12 Apr 2018 02:51:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_Citrus-2.jpg 1415 2178 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_02_ammonia-2/ Thu, 12 Apr 2018 02:51:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_02_ammonia-2.jpg 1416 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-4-3/ Thu, 12 Apr 2018 02:51:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-4-3.png 1417 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_cuagno3-3/ Thu, 12 Apr 2018 02:51:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_CuAgNO3-3.jpg 1418 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_iodine-2/ Thu, 12 Apr 2018 02:51:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_iodine-2.jpg 1420 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_moleratio1_img-2/ Thu, 12 Apr 2018 02:51:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_moleratio1_img-2.jpg 1421 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_moleratio2_img-2/ Thu, 12 Apr 2018 02:51:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_moleratio2_img-2.jpg 1422 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_map2_img-2/ Thu, 12 Apr 2018 02:51:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_map2_img-2.jpg 1423 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_map3_img-2/ Thu, 12 Apr 2018 02:51:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_map3_img-2.jpg 1424 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_flowchart-2/ Thu, 12 Apr 2018 02:51:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_flowchart-2.jpg 1425 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_airbag-2/ Thu, 12 Apr 2018 02:51:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_airbag-2.jpg 1426 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_ethenebr_img-2/ Thu, 12 Apr 2018 02:51:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_etheneBr_img-2.jpg 1427 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_03_ethene_img-2/ Thu, 12 Apr 2018 02:51:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_03_ethene_img-2.jpg 1428 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_sandwich-2/ Thu, 12 Apr 2018 02:51:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_sandwich-2.jpg 1430 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_limiting-2/ Thu, 12 Apr 2018 02:51:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_limiting-2.jpg 1431 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-5-2/ Thu, 12 Apr 2018 02:51:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-5-2.png 1432 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_greenchem-2/ Thu, 12 Apr 2018 02:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_GreenChem-2.jpg 1433 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_04_saccharin_img-2/ Thu, 12 Apr 2018 02:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_04_saccharin_img-2.jpg 1434 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_titration-2/ Thu, 12 Apr 2018 02:51:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_titration-2.jpg 1436 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_map7_img-2/ Thu, 12 Apr 2018 02:51:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_map7_img-2.jpg 1437 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_filter-2/ Thu, 12 Apr 2018 02:51:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_filter-2.jpg 1438 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_map8_img-2/ Thu, 12 Apr 2018 02:51:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_map8_img-2.jpg 1439 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_combustion-2/ Thu, 12 Apr 2018 02:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_combustion-2.jpg 1440 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_combmap_img-2/ Thu, 12 Apr 2018 02:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_combmap_img-2.jpg 1441 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_04_05_propionate_img-2/ Thu, 12 Apr 2018 02:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_04_05_propionate_img-2.jpg 1442 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_00_match-2/ Thu, 12 Apr 2018 02:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_00_Match-2.jpg 1445 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_thermochem-2/ Thu, 12 Apr 2018 02:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Thermochem-2.jpg 1447 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_waterfall-2/ Thu, 12 Apr 2018 02:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Waterfall-2.jpg 1448 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_hotcold-2/ Thu, 12 Apr 2018 02:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HotCold-2.jpg 1449 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-6-2/ Thu, 12 Apr 2018 02:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-6-2.png 1450 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_thermom-2/ Thu, 12 Apr 2018 02:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_Thermom-2.jpg 1451 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_heattrans1-2/ Thu, 12 Apr 2018 02:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HeatTrans1-2.jpg 1452 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_oxyactorch-2/ Thu, 12 Apr 2018 02:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_OxyacTorch-2.jpg 1453 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_heatcapacity-2/ Thu, 12 Apr 2018 02:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_HeatCapacity-2.jpg 1454 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_soltherm1-2/ Thu, 12 Apr 2018 02:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_SolTherm1-2.jpg 1455 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_01_soltherm2-2/ Thu, 12 Apr 2018 02:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_01_SolTherm2-2.jpg 1456 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_heatmeas-2/ Thu, 12 Apr 2018 02:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HeatMeas-2.jpg 1458 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_calorim-2/ Thu, 12 Apr 2018 02:52:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_Calorim-2.jpg 1459 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_calorim2-2/ Thu, 12 Apr 2018 02:52:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_Calorim2-2.jpg 1460 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_heattrans2-2/ Thu, 12 Apr 2018 02:52:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HeatTrans2-2.jpg 1461 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_handwarmer-2/ Thu, 12 Apr 2018 02:52:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_HandWarmer-2.jpg 1462 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-7-2/ Thu, 12 Apr 2018 02:52:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-7-2.png 1463 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_icepack-2/ Thu, 12 Apr 2018 02:52:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_IcePack-2.jpg 1464 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_bombcalor-2/ Thu, 12 Apr 2018 02:52:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_BombCalor-2.jpg 1465 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_02_foodlabel-2/ Thu, 12 Apr 2018 02:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_02_FoodLabel-2.jpg 1466 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_systemqw-2/ Thu, 12 Apr 2018 02:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_Systemqw-2.jpg 1468 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_interactive_200dpi-8-2/ Thu, 12 Apr 2018 02:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Interactive_200DPI-8-2.png 1469 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_summit-2/ Thu, 12 Apr 2018 02:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_Summit-2.jpg 1470 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_gasburning-2/ Thu, 12 Apr 2018 02:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_GasBurning-2.jpg 1471 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_algalfuel1-2/ Thu, 12 Apr 2018 02:52:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_AlgalFuel1-2.jpg 1472 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_algalfuel2-3/ Thu, 12 Apr 2018 02:52:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_AlgalFuel2-2.jpg 1473 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_05_03_hessco2-2/ Thu, 12 Apr 2018 02:52:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_05_03_HessCO2-2.jpg 1474 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_00_crabneb-2/ Thu, 12 Apr 2018 02:52:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_00_CrabNeb-2.jpg 1477 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_frequency-2/ Thu, 12 Apr 2018 02:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Frequency-2.jpg 1479 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_emspectrum-2/ Thu, 12 Apr 2018 02:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_emspectrum-2.jpg 1480 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_radiocell-2/ Thu, 12 Apr 2018 02:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_RadioCell-2.jpg 1481 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_amfm-2/ Thu, 12 Apr 2018 02:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_AMFM-2.jpg 1482 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_liteinterf-2/ Thu, 12 Apr 2018 02:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_LiteInterf-2.jpg 1483 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_vibrstring-2/ Thu, 12 Apr 2018 02:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Vibrstring-2.jpg 1484 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_vibratdrum-2/ Thu, 12 Apr 2018 02:52:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Vibratdrum-2.jpg 1485 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-8-2/ Thu, 12 Apr 2018 02:52:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-8-2.png 1486 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_solardist-2/ Thu, 12 Apr 2018 02:52:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Solardist-2.jpg 1487 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_blackbody-2/ Thu, 12 Apr 2018 02:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Blackbody-2.jpg 1488 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_ephoton-2/ Thu, 12 Apr 2018 02:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_Ephoton-2.jpg 1489 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_neon-2/ Thu, 12 Apr 2018 02:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_neon-2.jpg 1490 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_01_2spectra-2/ Thu, 12 Apr 2018 02:52:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_01_2spectra-2.jpg 1491 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_02_hlevels-2/ Thu, 12 Apr 2018 02:52:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_02_Hlevels-2.jpg 1493 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/cnx_chem_06_02_bohrarrows-2/ Thu, 12 Apr 2018 02:52:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_02_BohrArrows-2.jpg 1494 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_waterw-2/ Thu, 12 Apr 2018 02:52:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_waterw-2.jpg 1496 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_elecw-2/ Thu, 12 Apr 2018 02:52:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_elecw-2.jpg 1497 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_angm-2/ Thu, 12 Apr 2018 02:52:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_angm-2.jpg 1498 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_electrnin-2/ Thu, 12 Apr 2018 02:52:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Electrnin-2.jpg 1499 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-9-2/ Thu, 12 Apr 2018 02:52:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-9-2.png 1500 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_qnumbers-2/ Thu, 12 Apr 2018 02:52:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Qnumbers-2.jpg 1501 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_sorbit-2/ Thu, 12 Apr 2018 02:52:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_sOrbit-2.jpg 1502 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_oshapes-2/ Thu, 12 Apr 2018 02:52:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Oshapes-2.jpg 1503 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_subshells-2/ Thu, 12 Apr 2018 02:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_subshells-2.jpg 1504 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_spin-2/ Thu, 12 Apr 2018 02:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_spin-2.jpg 1505 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_orboutline_img-2/ Thu, 12 Apr 2018 02:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_OrbOutline_img-2.jpg 1506 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_03_ex9soln_img-2/ Thu, 12 Apr 2018 02:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_03_Ex9soln_img-2.jpg 1507 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_eleveldiag-2/ Thu, 12 Apr 2018 02:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_eLeveldiag-2.jpg 1509 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_econfig-2/ Thu, 12 Apr 2018 02:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Econfig-2.jpg 1510 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_efillorder-2/ Thu, 12 Apr 2018 02:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Efillorder-2.jpg 1511 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_econtable-2/ Thu, 12 Apr 2018 02:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Econtable-2.jpg 1512 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_hydrog1_img-2/ Thu, 12 Apr 2018 02:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Hydrog1_img-2.jpg 1513 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_helium1_img-2/ Thu, 12 Apr 2018 02:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Helium1_img-2.jpg 1514 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_lithium12_img-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Lithium12_img-2.jpg 1515 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_beryll12_img-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Beryll12_img-2.jpg 1516 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_boron122_img-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Boron122_img-2.jpg 1517 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_carbon122_img-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Carbon122_img-2.jpg 1518 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_nofne_img-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_NOFNe_img-2.jpg 1519 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_valence-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Valence-2.jpg 1520 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/6-4-electronic-structure-of-atoms-electron-configurations/cnx_chem_06_04_ptableconf-2/ Thu, 12 Apr 2018 02:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_Ptableconf-2.jpg 1521 1528 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_phosphorb_img-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_PhosphOrb_img-2.jpg 1522 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2a_img-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2a_img-2.jpg 1523 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2b_img-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2b_img-2.jpg 1524 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2c_img-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2c_img-2.jpg 1525 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2d_img-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2d_img-2.jpg 1526 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_04_orbdiash2e_img-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_04_OrbDiaSh2e_img-2.jpg 1527 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-10-2/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-10-2.png 1529 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_covalradit-2/ Thu, 12 Apr 2018 02:52:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_CovalradiT-2.jpg 1530 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_radiigraph-2/ Thu, 12 Apr 2018 02:52:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Radiigraph-2.jpg 1531 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_ionradii-2/ Thu, 12 Apr 2018 02:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Ionradii-2.jpg 1532 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_firstiongr-2/ Thu, 12 Apr 2018 02:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Firstiongr-2.jpg 1533 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_firstionen-2/ Thu, 12 Apr 2018 02:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Firstionen-2.jpg 1534 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_oxygen122_img-2/ Thu, 12 Apr 2018 02:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Oxygen122_img-2.jpg 1535 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_06_05_elaffin-2/ Thu, 12 Apr 2018 02:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_06_05_Elaffin-2.jpg 1536 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_00_bucky-2/ Thu, 12 Apr 2018 02:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_00_Bucky-2.jpg 1539 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_01_naclphotos-2/ Thu, 12 Apr 2018 02:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_01_NaClPhotos-2.jpg 1541 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/microsoft-word-cnx_chem_ch07_mod01-docx-3/ Thu, 12 Apr 2018 02:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_01_NaClStruc-2.jpg 1542 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_morse-2/ Thu, 12 Apr 2018 02:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_Morse-2.jpg 1544 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_hclbond-2/ Thu, 12 Apr 2018 02:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_HClBond-2.jpg 1545 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_entable-2/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_ENTable-2.jpg 1546 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_pauling-2/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_Pauling-2.jpg 1547 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_02_deltaen-2/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_02_DeltaEN-2.jpg 1548 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_lewisstruct1_img-2/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Lewisstruct1_img-2.jpg 1550 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_3rowlewis-2/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_3rowLewis-2.jpg 1551 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_naca_img-2/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NaCa_img-2.jpg 1552 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cls_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ClS_img-2.jpg 1553 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ionlewis-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_IonLewis-2.jpg 1554 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_c12dot_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_C12dot_img-2.jpg 1555 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cl2dash_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Cl2dash_img-2.jpg 1556 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xy4struc_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XY4struc_img-2.jpg 1557 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_numbonds_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_numbonds_img-2.jpg 1558 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_doublebond_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_DoubleBond_img-2.jpg 1559 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_cocn_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_COCN_img-2.jpg 1560 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_unprelec_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Unprelec_img-2.jpg 1561 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_singlebond_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Singlebond_img-2.jpg 1562 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_sih4_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_SiH4_img-2.jpg 1563 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_oxydiflor2_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_oxydiflor2_img-2.jpg 1564 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_formate2_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_formate2_img-2.jpg 1565 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nitrosoni2_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_nitrosoni2_img-2.jpg 1566 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nitrosoni3_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_nitrosoni3_img-2.jpg 1567 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_1_img-2/ Thu, 12 Apr 2018 02:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_1_img-2.jpg 1568 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_2_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_2_img-2.jpg 1569 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_3_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_3_img-2.jpg 1570 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_ex070301_4_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_ex070301_4_img-2.jpg 1571 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_coco2_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_COCO2_img-2.jpg 1572 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_smalley-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Smalley-2.jpg 1573 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nosingle_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NOsingle_img-2.jpg 1574 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_nodouble_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_NODouble_img-2.jpg 1575 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_edeficient_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_edeficient_img-2.jpg 1576 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_dative_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_dative_img-2.jpg 1577 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_pf5sf6_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_PF5SF6_img-2.jpg 1578 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_if5xef4_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_IF5XeF4_img-2.jpg 1579 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xeskeleton_img-2/ Thu, 12 Apr 2018 02:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Xeskeleton_img-2.jpg 1580 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xef2_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XeF2_img-2.jpg 1581 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_xef6_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_XeF6_img-2.jpg 1582 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_interhal_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Interhal_img-2.jpg 1583 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisa_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisa_img-2.jpg 1584 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisb_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisb_img-2.jpg 1585 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisc_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisc_img-2.jpg 1586 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_mxlewisd_img-2/ Thu, 12 Apr 2018 02:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_MXLewisd_img-2.jpg 1587 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioa_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioa_img-2.jpg 1588 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbiob_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbiob_img-2.jpg 1589 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioc_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioc_img-2.jpg 1590 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbiod_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbiod_img-2.jpg 1591 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioe_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioe_img-2.jpg 1592 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1a_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1a_img-2.jpg 1593 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1b_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1b_img-2.jpg 1594 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1d_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1d_img-2.jpg 1595 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question1g_img-2/ Thu, 12 Apr 2018 02:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question1g_img-2.jpg 1596 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3a_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3a_img-2.jpg 1597 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3b_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3b_img-2.jpg 1598 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3c_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3c_img-2.jpg 1599 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3d_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3d_img-2.jpg 1600 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3e_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3e_img-2.jpg 1601 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise3f_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise3f_img-2.jpg 1602 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question5_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question5_img-2.jpg 1603 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7a_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7a_img-2.jpg 1604 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7b_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7b_img-2.jpg 1605 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7c_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7c_img-2.jpg 1606 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7d_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7d_img-2.jpg 1607 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7e_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7e_img-2.jpg 1608 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7f_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7f_img-2.jpg 1609 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7g_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7g_img-2.jpg 1610 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7h_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7h_img-2.jpg 1611 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7i_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7i_img-2.jpg 1612 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7j_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7j_img-2.jpg 1613 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question7k_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question7k_img-2.jpg 1614 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9a_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9a_img-2.jpg 1615 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9b_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9b_img-2.jpg 1616 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9c_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9c_img-2.jpg 1617 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question9d_img-2/ Thu, 12 Apr 2018 02:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question9d_img-2.jpg 1618 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_question18_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Question18_img-2.jpg 1619 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_carbtet_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Carbtet_img-2.jpg 1620 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansa_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansa_img-2.jpg 1621 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansb_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansb_img-2.jpg 1622 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansc_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansc_img-2.jpg 1623 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioansd_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioansd_img-2.jpg 1624 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_impbioanse_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Impbioanse_img-2.jpg 1625 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_03_exercise25_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_03_Exercise25_img-2.jpg 1626 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_icl4_img-2/ Thu, 12 Apr 2018 02:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_ICL4_img-2.jpg 1628 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO_img-2.jpg 1629 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_brcl3_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_BRCL3_img-2.jpg 1630 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ex070402_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ex070402_img-2.jpg 1631 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co2pos_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO2pos_img-2.jpg 1632 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_thiocyan_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Thiocyan_img-2.jpg 1633 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_n2o_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_N2O_img-2.jpg 1634 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_n2ofc_img-2/ Thu, 12 Apr 2018 02:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_N2Ofc_img-2.jpg 1635 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_nno_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NNO_img-2.jpg 1636 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2ion_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2ion_img-2.jpg 1637 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2res_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2res_img-2.jpg 1638 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_no2resarr_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NO2resarr_img-2.jpg 1639 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_co3res_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CO3res_img-2.jpg 1640 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-11-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-11-2.png 1641 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_chood_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CHOOd_img-2.jpg 1642 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_chooe_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_CHOOe_img-2.jpg 1643 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_so2resd_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_SO2resd_img-2.jpg 1644 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_so2rese_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_SO2rese_img-2.jpg 1645 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_acet_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Acet_img-2.jpg 1646 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_hno2_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_HNO2_img-2.jpg 1647 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansa_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansa_img-2.jpg 1648 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansb_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansb_img-2.jpg 1649 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansc_img-2/ Thu, 12 Apr 2018 02:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansc_img-2.jpg 1650 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2ansd_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2ansd_img-2.jpg 1651 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques2anse_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques2anse_img-2.jpg 1652 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques11ans_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques11ans_img-2.jpg 1653 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques13ansb_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques13ansb_img-2.jpg 1654 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ques13ansc_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Ques13ansc_img-2.jpg 1655 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12a_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12a_img-2.jpg 1656 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12b_img-2/ Thu, 12 Apr 2018 02:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12b_img-2.jpg 1657 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12c_img-2/ Thu, 12 Apr 2018 02:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12c_img-2.jpg 1658 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_exercis12d_img-2/ Thu, 12 Apr 2018 02:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_Exercis12d_img-2.jpg 1659 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_ohamine_img-2/ Thu, 12 Apr 2018 02:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_OHamine_img-2.jpg 1660 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_nf3_img-2/ Thu, 12 Apr 2018 02:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_NF3_img-2.jpg 1661 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_04_h2so4-2/ Thu, 12 Apr 2018 02:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_04_H2SO4-2.jpg 1662 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ch4bond_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_CH4bond_img-2.jpg 1664 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ch3ohlew_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_CH3OHLew_img-2.jpg 1665 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_ethanol_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_Ethanol_img-2.jpg 1666 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_bornhaber-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_BornHaber-2.jpg 1667 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_hydroxya_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_Hydroxya_img-2.jpg 1668 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_c6h8lew_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_C6H8Lew_img-2.jpg 1669 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_fno2_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_FNO2_img-2.jpg 1670 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_05_c6h8ans_img-2/ Thu, 12 Apr 2018 02:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_05_C6H8ans_img-2.jpg 1671 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ch2o-2/ Thu, 12 Apr 2018 02:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CH2O-2.jpg 1673 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bef2-2/ Thu, 12 Apr 2018 02:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BeF2-2.jpg 1674 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_egeom-2/ Thu, 12 Apr 2018 02:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Egeom-2.jpg 1675 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ch4-2/ Thu, 12 Apr 2018 02:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CH4-2.jpg 1676 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh3-2/ Thu, 12 Apr 2018 02:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH3-2.jpg 1677 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_molgeom-2/ Thu, 12 Apr 2018 02:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_molgeom-2.jpg 1678 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_axeq-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Axeq-2.jpg 1679 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_co2_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CO2_img-2.jpg 1680 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bcl3_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BCl3_img-2.jpg 1681 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bcl3mol-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BCl3mol-2.jpg 1682 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh4_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH4_img-2.jpg 1683 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_nh4mol-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_NH4mol-2.jpg 1684 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_h2o_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_H2O_img-2.jpg 1685 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_h2omol-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_H2Omol-2.jpg 1686 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sf4_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SF4_img-2.jpg 1687 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sf4mol-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SF4mol-2.jpg 1688 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_xef4_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_XeF4_img-2.jpg 1689 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_xef4mol-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_XEF4mol-2.jpg 1690 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_gly_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_gly_img-2.jpg 1691 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_gly3d_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_gly3D_img-2.jpg 1692 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ala_img-2/ Thu, 12 Apr 2018 02:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_ala_img-2.jpg 1693 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-12-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-12-2.png 1694 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_bondvector-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_BondVector-2.jpg 1695 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_co2h2odip-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CO2H2Odip-2.jpg 1696 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_osc_img-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_OSC_img-2.jpg 1697 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_chcl3_img-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CHCl3_img-2.jpg 1698 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_sh2nh3_img-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_SH2NH3_img-2.jpg 1699 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_dipolfield-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Dipolfield-2.jpg 1700 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ques23ans_img-2/ Thu, 12 Apr 2018 02:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Ques23ans_img-2.jpg 1701 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3a_img-2/ Thu, 12 Apr 2018 02:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3a_img-2.jpg 1702 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3b_img-2/ Thu, 12 Apr 2018 02:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3b_img-2.jpg 1703 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_cs3c_img-2/ Thu, 12 Apr 2018 02:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_CS3c_img-2.jpg 1704 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_07_06_ques28ans_img-2/ Thu, 12 Apr 2018 02:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_07_06_Ques28ans_img-2.jpg 1705 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mexico-3/ Thu, 12 Apr 2018 02:52:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_00_Sinkhole-2.jpg 1708 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_conjugate_img-2/ Thu, 12 Apr 2018 02:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_conjugate_img-2.jpg 1710 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_hf_img-2/ Thu, 12 Apr 2018 02:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_HF_img-2.jpg 1711 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_nh3_img-2/ Thu, 12 Apr 2018 02:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_NH3_img-2.jpg 1712 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_01_water_img-2/ Thu, 12 Apr 2018 02:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_01_Water_img-2.jpg 1713 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_phscale-2/ Thu, 12 Apr 2018 02:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_phscale-2.jpg 1715 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_acidrain-2/ Thu, 12 Apr 2018 02:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_AcidRain-2.jpg 1716 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_phmeter-2/ Thu, 12 Apr 2018 02:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_pHMeter-2.jpg 1717 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_02_indicator-2/ Thu, 12 Apr 2018 02:52:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_02_indicator-2.jpg 1718 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_strong-2/ Thu, 12 Apr 2018 02:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_strong-2.jpg 1720 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-27-2/ Thu, 12 Apr 2018 02:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-27-2.png 1721 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_strengths-2/ Thu, 12 Apr 2018 02:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_strengths-2.jpg 1722 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_corresp-2/ Thu, 12 Apr 2018 02:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Corresp-2.jpg 1723 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_acetate_img-2/ Thu, 12 Apr 2018 02:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_acetate_img-2.jpg 1724 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_ammonia-2/ Thu, 12 Apr 2018 02:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ammonia-2.jpg 1725 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-3-molarity/cnx_chem_02_04_aceticacid-4/ Thu, 12 Apr 2018 02:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Vinegar-2.jpg 1726 1399 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps_img-2/ Thu, 12 Apr 2018 02:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps_img-2.jpg 1727 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable1_img-2/ Thu, 12 Apr 2018 02:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable1_img-2.jpg 1728 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable2_img-2/ Thu, 12 Apr 2018 02:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable2_img-2.jpg 1729 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_antsting-2/ Thu, 12 Apr 2018 02:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_AntSting-2.jpg 1730 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable3_img-1/ Thu, 12 Apr 2018 02:52:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable3_img-1.jpg 1731 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps4_img-1/ Thu, 12 Apr 2018 02:52:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps4_img-1.jpg 1732 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable4_img-1/ Thu, 12 Apr 2018 02:52:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable4_img-1.jpg 1733 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_steps5_img-1/ Thu, 12 Apr 2018 02:52:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_steps5_img-1.jpg 1734 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_icetable5_img-1/ Thu, 12 Apr 2018 02:52:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_ICETable5_img-1.jpg 1735 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_acidph-1/ Thu, 12 Apr 2018 02:52:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_AcidpH-1.jpg 1736 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_ohbonds_img-1/ Thu, 12 Apr 2018 02:52:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_OHbonds_img-1.jpg 1737 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_oxyacid-1/ Thu, 12 Apr 2018 02:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_Oxyacid-1.jpg 1738 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_03_fishlemon-1/ Thu, 12 Apr 2018 02:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_03_FishLemon-1.jpg 1740 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_steps1_img-1/ Thu, 12 Apr 2018 02:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_steps1_img-1.jpg 1741 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_hydronium-1/ Thu, 12 Apr 2018 02:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_hydronium-1.jpg 1742 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_steps2_img-1/ Thu, 12 Apr 2018 02:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_steps2_img-1.jpg 1743 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_04_icetable13_img-1/ Thu, 12 Apr 2018 02:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_04_ICETable13_img-1.jpg 1744 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_acetic_img-1/ Thu, 12 Apr 2018 02:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_acetic_img-1.jpg 1746 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_steps1_img-1/ Thu, 12 Apr 2018 02:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_steps1_img-1.jpg 1747 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_steps2_img-1/ Thu, 12 Apr 2018 02:52:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_steps2_img-1.jpg 1748 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_05_icetable1_img-1/ Thu, 12 Apr 2018 02:52:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_05_ICETable1_img-1.jpg 1749 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_compare-1/ Thu, 12 Apr 2018 02:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_compare-1.jpg 1751 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_bufferchrt-1/ Thu, 12 Apr 2018 02:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_bufferchrt-1.jpg 1752 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_steps1_img/ Thu, 12 Apr 2018 02:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_steps1_img.jpg 1753 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_icetable16_img/ Thu, 12 Apr 2018 02:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_ICETable16_img.jpg 1754 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_steps2_img/ Thu, 12 Apr 2018 02:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_steps2_img.jpg 1755 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_steps3_img/ Thu, 12 Apr 2018 02:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_steps3_img.jpg 1756 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_exhaust/ Thu, 12 Apr 2018 02:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_exhaust.jpg 1757 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_06_buffer/ Thu, 12 Apr 2018 02:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_06_buffer.jpg 1758 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-28/ Thu, 12 Apr 2018 02:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-28.png 1759 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_07_titration/ Thu, 12 Apr 2018 02:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_07_titration.jpg 1761 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_07_indicators/ Thu, 12 Apr 2018 02:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_07_indicators.jpg 1762 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_07_titration2/ Thu, 12 Apr 2018 02:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_07_titration2.jpg 1763 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_14_07_exercise5_img/ Thu, 12 Apr 2018 02:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_14_07_Exercise5_img.jpg 1764 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_00_fluorite/ Thu, 12 Apr 2018 02:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_00_Fluorite.jpg 1767 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_agcl/ Thu, 12 Apr 2018 02:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_AgCl.jpg 1769 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_icetable1_img/ Thu, 12 Apr 2018 02:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_ICETable1_img.jpg 1770 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_icetable7_img/ Thu, 12 Apr 2018 02:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_ICETable7_img.jpg 1771 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_oilpaints/ Thu, 12 Apr 2018 02:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_OilPaints.jpg 1772 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_pbcro4_img/ Thu, 12 Apr 2018 02:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_PbCrO4_img.jpg 1773 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_format_img/ Thu, 12 Apr 2018 02:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_Format_img.jpg 1774 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_icetable2_img/ Thu, 12 Apr 2018 02:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_ICETable2_img.jpg 1775 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_bariumxray/ Thu, 12 Apr 2018 02:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_BariumXray.jpg 1776 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_blood/ Thu, 12 Apr 2018 02:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_Blood.jpg 1777 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_wastewater/ Thu, 12 Apr 2018 02:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_Wastewater.jpg 1778 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-29/ Thu, 12 Apr 2018 02:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-29.png 1779 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_01_icetable3_img/ Thu, 12 Apr 2018 02:53:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_01_ICETable3_img.jpg 1780 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_03_hydronium_img/ Thu, 12 Apr 2018 02:53:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_03_hydronium_img.jpg 1782 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_bf3-la_img/ Thu, 12 Apr 2018 02:53:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_BF3-LA_img.jpg 1783 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_nh3-lbase_img/ Thu, 12 Apr 2018 02:53:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_NH3-LBase_img.jpg 1784 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_nonmetalox_img/ Thu, 12 Apr 2018 02:53:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_NonmetalOx_img.jpg 1785 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_displace_img/ Thu, 12 Apr 2018 02:53:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Displace_img.jpg 1786 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_agnh32_img/ Thu, 12 Apr 2018 02:53:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_AgNH32_img.jpg 1787 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_aloh4_img/ Thu, 12 Apr 2018 02:53:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_AlOH4_img.jpg 1788 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_cucn2-_img/ Thu, 12 Apr 2018 02:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_CuCN2-_img.jpg 1789 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_format_img/ Thu, 12 Apr 2018 02:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Format_img.jpg 1790 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_icetable1_img/ Thu, 12 Apr 2018 02:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_ICETable1_img.jpg 1791 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_icetable3_img/ Thu, 12 Apr 2018 02:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_ICETable3_img.jpg 1792 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18a_img/ Thu, 12 Apr 2018 02:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18a_img.jpg 1793 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18b_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18b_img.jpg 1794 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18c_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18c_img.jpg 1795 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18d_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18d_img.jpg 1796 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18e_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18e_img.jpg 1797 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24a_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24a_img.jpg 1798 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24b_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24b_img.jpg 1799 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24c_img/ Thu, 12 Apr 2018 02:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24c_img.jpg 1800 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24d_img/ Thu, 12 Apr 2018 02:53:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24d_img.jpg 1801 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_03_hydronium_img-1/ Thu, 12 Apr 2018 03:47:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_03_hydronium_img-1.jpg 1802 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_bf3-la_img-1/ Thu, 12 Apr 2018 03:47:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_BF3-LA_img-1.jpg 1803 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_nh3-lbase_img-1/ Thu, 12 Apr 2018 03:47:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_NH3-LBase_img-1.jpg 1804 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_nonmetalox_img-1/ Thu, 12 Apr 2018 03:47:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_NonmetalOx_img-1.jpg 1805 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_displace_img-1/ Thu, 12 Apr 2018 03:47:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Displace_img-1.jpg 1806 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_agnh32_img-1/ Thu, 12 Apr 2018 03:47:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_AgNH32_img-1.jpg 1807 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_aloh4_img-1/ Thu, 12 Apr 2018 03:47:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_AlOH4_img-1.jpg 1808 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_cucn2-_img-1/ Thu, 12 Apr 2018 03:47:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_CuCN2-_img-1.jpg 1809 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_format_img-1/ Thu, 12 Apr 2018 03:47:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Format_img-1.jpg 1810 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_icetable1_img-1/ Thu, 12 Apr 2018 03:47:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_ICETable1_img-1.jpg 1811 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_icetable3_img-1/ Thu, 12 Apr 2018 03:47:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_ICETable3_img-1.jpg 1812 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18a_img-1/ Thu, 12 Apr 2018 03:47:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18a_img-1.jpg 1813 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18b_img-1/ Thu, 12 Apr 2018 03:47:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18b_img-1.jpg 1814 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18c_img-1/ Thu, 12 Apr 2018 03:47:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18c_img-1.jpg 1815 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18d_img-1/ Thu, 12 Apr 2018 03:47:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18d_img-1.jpg 1816 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer18e_img-1/ Thu, 12 Apr 2018 03:47:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer18e_img-1.jpg 1817 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24a_img-1/ Thu, 12 Apr 2018 03:47:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24a_img-1.jpg 1818 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24b_img-1/ Thu, 12 Apr 2018 03:47:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24b_img-1.jpg 1819 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24c_img-1/ Thu, 12 Apr 2018 03:47:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24c_img-1.jpg 1820 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_02_answer24d_img-1/ Thu, 12 Apr 2018 03:47:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_02_Answer24d_img-1.jpg 1821 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_03_coralreef/ Thu, 12 Apr 2018 03:47:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_03_CoralReef.jpg 1823 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-30/ Thu, 12 Apr 2018 03:47:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-30.png 1824 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_03_apatite/ Thu, 12 Apr 2018 03:47:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_03_Apatite.jpg 1825 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_03_toothpaste/ Thu, 12 Apr 2018 03:47:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_03_Toothpaste.jpg 1826 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_15_03_agbr_img/ Thu, 12 Apr 2018 03:47:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_15_03_AgBr_img.jpg 1827 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_00_orgcompnd/ Thu, 12 Apr 2018 03:47:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_00_OrgCompnd.jpg 1830 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkanes/ Thu, 12 Apr 2018 03:47:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkanes.jpg 1832 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct1/ Thu, 12 Apr 2018 03:47:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct1.jpg 1833 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct2_img/ Thu, 12 Apr 2018 03:47:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct2_img.jpg 1834 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct3_img/ Thu, 12 Apr 2018 03:47:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct3_img.jpg 1835 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct4_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct4_img.jpg 1836 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct5_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct5_img.jpg 1837 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct6_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct6_img.jpg 1838 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct7_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct7_img.jpg 1839 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_linestruct8_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_LineStruct8_img.jpg 1840 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_butaneisom_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_butaneIsom_img.jpg 1841 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_butane_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_butane_img.jpg 1842 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_substitu_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_substitu_img.jpg 1843 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_halalkane_img/ Thu, 12 Apr 2018 03:47:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_HalAlkane_img.jpg 1844 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_halalkane2_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_HalAlkane2_img.jpg 1845 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_halalkane3_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_HalAlkane3_img.jpg 1846 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkyl_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkyl_img.jpg 1847 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_octane1_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_octane1_img.jpg 1848 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_octane2_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_octane2_img.jpg 1849 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_propane_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_propane_img.jpg 1850 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkyls/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkyls.jpg 1851 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alklnm_img/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alklnm_img.jpg 1852 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_fracdistil/ Thu, 12 Apr 2018 03:47:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_FracDistil.jpg 1853 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ethylchlor_img/ Thu, 12 Apr 2018 03:47:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ethylChlor_img.jpg 1854 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-39/ Thu, 12 Apr 2018 03:47:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-39.png 1855 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_propene1/ Thu, 12 Apr 2018 03:47:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_propene1.jpg 1856 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_monomer/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_monomer.jpg 1857 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_recycle/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_recycle.jpg 1858 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkenes_img/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkenes_img.jpg 1859 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_geoisomers_img/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_geoIsomers_img.jpg 1860 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_halogen_img/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_halogen_img.jpg 1861 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_haloreact_img/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_HaloReact_img.jpg 1862 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_haloreact2_img/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_HaloReact2_img.jpg 1863 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_reactprod1_img/ Thu, 12 Apr 2018 03:47:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ReactProd1_img.jpg 1864 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_alkyne_img/ Thu, 12 Apr 2018 03:47:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_alkyne_img.jpg 1865 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_geomhybr1_img/ Thu, 12 Apr 2018 03:47:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_GeomHybr1_img.jpg 1866 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hybrbangl_img/ Thu, 12 Apr 2018 03:47:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_HybrBAngl_img.jpg 1867 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_acetylene_img/ Thu, 12 Apr 2018 03:47:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_acetylene_img.jpg 1868 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_aromatic_img/ Thu, 12 Apr 2018 03:47:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_aromatic_img.jpg 1869 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_benzene/ Thu, 12 Apr 2018 03:47:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_benzene.jpg 1870 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_subbenzene_img/ Thu, 12 Apr 2018 03:47:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_subbenzene_img.jpg 1871 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_aromatiso1_img/ Thu, 12 Apr 2018 03:47:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_AromatIso1_img.jpg 1872 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_aromatiso2_img/ Thu, 12 Apr 2018 03:47:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_AromatIso2_img.jpg 1873 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_aromatiso3_img/ Thu, 12 Apr 2018 03:47:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_AromatIso3_img.jpg 1874 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_notiso1_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_notiso1_img.jpg 1875 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_notiso2_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_notiso2_img.jpg 1876 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_11_c_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_11_c_img.jpg 1877 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_11_e_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_11_e_img.jpg 1878 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_11_f_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_11_f_img.jpg 1879 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_12_c_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_12_c_img.jpg 1880 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_12_e_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_12_e_img.jpg 1881 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_18a_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_18a_img.jpg 1882 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_1a_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_1a_img.jpg 1883 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_1b_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_1b_img.jpg 1884 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_1c_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_1c_img.jpg 1885 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hexane_a_img/ Thu, 12 Apr 2018 03:47:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_hexane_a_img.jpg 1886 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hexane_b_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_hexane_b_img.jpg 1887 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hexane_c_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_hexane_c_img.jpg 1888 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hexane_d_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_hexane_d_img.jpg 1889 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hexane_e_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_hexane_e_img.jpg 1890 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_hexane_f_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_hexane_f_img.jpg 1891 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_14_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_14_img.jpg 1892 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_16_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_16_img.jpg 1893 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_18b_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_18b_img.jpg 1894 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_20_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_20_img.jpg 1895 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_21_img/ Thu, 12 Apr 2018 03:47:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_21_img.jpg 1896 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_01_ex1_23_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_01_ex1_23_img.jpg 1897 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_reaction1b_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_reaction1b_img.jpg 1898 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_reaction1c_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_reaction1c_img.jpg 1899 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_ferment_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_ferment_img.jpg 1901 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_ethanol_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_ethanol_img.jpg 1902 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_polyols_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_polyols_img.jpg 1903 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_alcohol1_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_alcohol1_img.jpg 1904 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_alcohol2_img/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_alcohol2_img.jpg 1905 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_nameethers_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_NameEthers_img.jpg 1906 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_ethers1_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_ethers1_img.jpg 1907 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_ethers2_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_ethers2_img.jpg 1908 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_ether_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_ether_img.jpg 1909 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-40/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-40.png 1910 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_sugars/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_sugars.jpg 1911 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_diabetes/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_diabetes.jpg 1912 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_exercise3a_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_Exercise3a_img.jpg 1913 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_exercise3b_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_Exercise3b_img.jpg 1914 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_exercise3c_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_Exercise3c_img.jpg 1915 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_exercise4a_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_Exercise4a_img.jpg 1916 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_exercise4b_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_Exercise4b_img.jpg 1917 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_exercise4c_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_Exercise4c_img.jpg 1918 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_mtbe_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_MTBE_img.jpg 1919 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_react2a_img/ Thu, 12 Apr 2018 03:47:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_react2a_img.jpg 1920 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_react2e_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_react2e_img.jpg 1921 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_carbonyl_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_carbonyl_img.jpg 1923 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_aldehyde_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_aldehyde_img.jpg 1924 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_aldket_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_aldket_img.jpg 1925 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_cogeom_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_COgeom_img.jpg 1926 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxidation_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_oxidation_img.jpg 1927 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiredu2_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiRedu2_img.jpg 1928 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiredu_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiRedu_img.jpg 1929 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxaldehyde_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_oxaldehyde_img.jpg 1930 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxketone_img/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_oxketone_img.jpg 1931 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_carboxest1_img/ Thu, 12 Apr 2018 03:47:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_CarboxEst1_img.jpg 1932 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_carboxest2_img/ Thu, 12 Apr 2018 03:47:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_CarboxEst2_img.jpg 1933 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_carboxylic_img/ Thu, 12 Apr 2018 03:47:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_carboxylic_img.jpg 1934 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxtoacid_img/ Thu, 12 Apr 2018 03:47:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_oxtoacid_img.jpg 1935 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_esterform_img/ Thu, 12 Apr 2018 03:47:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_esterform_img.jpg 1936 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_strawberry/ Thu, 12 Apr 2018 03:47:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_strawberry.jpg 1937 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiorder_img/ Thu, 12 Apr 2018 03:47:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiOrder_img.jpg 1938 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiprod1a_a_img/ Thu, 12 Apr 2018 03:47:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiProd1a_a_img.jpg 1939 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiprod1a_b_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiProd1a_b_img.jpg 1940 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiprod1a_c_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiProd1a_c_img.jpg 1941 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_reduprod1a_a_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_ReduProd1a_a_img.jpg 1942 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_reduprod1a_b_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_ReduProd1a_b_img.jpg 1943 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_reduprod1a_c_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_ReduProd1a_c_img.jpg 1944 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiprod1b_a_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiProd1b_a_img.jpg 1945 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiprod1b_b_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiProd1b_b_img.jpg 1946 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_oxiprod1b_c_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_OxiProd1b_c_img.jpg 1947 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_acetate_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_acetate_img.jpg 1948 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_react2b_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_react2b_img.jpg 1949 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_react2c_img/ Thu, 12 Apr 2018 03:47:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_react2c_img.jpg 1950 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_03_alcoholabc_img/ Thu, 12 Apr 2018 03:47:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_03_alcoholABC_img.jpg 1951 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_amines_img/ Thu, 12 Apr 2018 03:47:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_amines_img.jpg 1953 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_pyridine_img/ Thu, 12 Apr 2018 03:47:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_pyridine_img.jpg 1954 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_dna/ Thu, 12 Apr 2018 03:47:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_DNA.jpg 1955 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-41/ Thu, 12 Apr 2018 03:47:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-41.png 1956 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_ammonia_img/ Thu, 12 Apr 2018 03:47:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_ammonia_img.jpg 1957 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_alkaloids_img/ Thu, 12 Apr 2018 03:47:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_alkaloids_img.jpg 1958 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_poppies/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_poppies.jpg 1959 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_amide1_img/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_amide1_img.jpg 1960 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_amide2_img/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_amide2_img.jpg 1961 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_peptide/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_peptide.jpg 1962 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_phenylh_img/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_PhenylH_img.jpg 1963 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_kevlar1/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_kevlar1.jpg 1964 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_kevpoly/ Thu, 12 Apr 2018 03:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_kevpoly.jpg 1965 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_kevlar2/ Thu, 12 Apr 2018 03:47:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_kevlar2.jpg 1966 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_02_functgroup_img/ Thu, 12 Apr 2018 03:47:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_02_FunctGroup_img.jpg 1967 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_pyridinium_img/ Thu, 12 Apr 2018 03:47:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_pyridinium_img.jpg 1968 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_20_04_reaction1f_img/ Thu, 12 Apr 2018 03:47:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_20_04_reaction1f_img.jpg 1969 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_00_nuclearmed/ Thu, 12 Apr 2018 03:47:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_00_NuclearMed.jpg 1972 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-42/ Thu, 12 Apr 2018 03:47:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-42.png 1974 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_01_bandstable/ Thu, 12 Apr 2018 03:47:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_01_BandStable.jpg 1975 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_01_bindenergy/ Thu, 12 Apr 2018 03:47:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_01_BindEnergy.jpg 1976 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_02_nuclearrxs/ Thu, 12 Apr 2018 03:47:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_02_Nuclearrxs.jpg 1978 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_reaction1/ Thu, 12 Apr 2018 03:47:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_Reaction1.jpg 1980 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-43/ Thu, 12 Apr 2018 03:47:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-43.png 1981 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_radiation/ Thu, 12 Apr 2018 03:47:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_Radiation.jpg 1982 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_radiodecay/ Thu, 12 Apr 2018 03:47:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_RadioDecay.jpg 1983 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_petscan/ Thu, 12 Apr 2018 03:47:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_PETScan.jpg 1984 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_decays/ Thu, 12 Apr 2018 03:47:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_DecayS.jpg 1985 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_halflife/ Thu, 12 Apr 2018 03:47:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_HalfLife.jpg 1986 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_carbondate/ Thu, 12 Apr 2018 03:47:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_CarbonDate.jpg 1987 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_03_dsscrolls/ Thu, 12 Apr 2018 03:47:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_03_DSScrolls.jpg 1988 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_cern/ Thu, 12 Apr 2018 03:47:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_CERN.jpg 1990 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osc_interactive_200-44/ Thu, 12 Apr 2018 03:47:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OSC_Interactive_200-44.png 1991 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_fission1/ Thu, 12 Apr 2018 03:47:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_Fission1.jpg 1992 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_fission2/ Thu, 12 Apr 2018 03:47:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_Fission2.jpg 1993 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_chnreact1/ Thu, 12 Apr 2018 03:47:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_ChnReact1.jpg 1994 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_critmass/ Thu, 12 Apr 2018 03:47:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_CritMass.jpg 1995 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_fissnbomb/ Thu, 12 Apr 2018 03:47:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_FissnBomb.jpg 1996 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_nuclearpwr/ Thu, 12 Apr 2018 03:47:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_NuclearPwr.jpg 1997 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_controlrod/ Thu, 12 Apr 2018 03:47:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_ControlRod.jpg 1998 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_3mileislnd/ Thu, 12 Apr 2018 03:47:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_3MileIslnd.jpg 1999 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_fukushima/ Thu, 12 Apr 2018 03:47:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_Fukushima.jpg 2000 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_04_fusion/ Thu, 12 Apr 2018 03:47:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_04_Fusion.jpg 2001 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_05_thallium/ Thu, 12 Apr 2018 03:47:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_05_Thallium.jpg 2003 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_05_tc-99/ Thu, 12 Apr 2018 03:47:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_05_Tc-99.jpg 2004 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_05_radtherapy/ Thu, 12 Apr 2018 03:47:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_05_RadTherapy.jpg 2005 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_05_co60decay/ Thu, 12 Apr 2018 03:47:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_05_Co60Decay.jpg 2006 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_05_usesofrad/ Thu, 12 Apr 2018 03:47:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_05_UsesOfRad.jpg 2007 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_05_smokealarm/ Thu, 12 Apr 2018 03:47:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_05_SmokeAlarm.jpg 2008 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_damage1/ Thu, 12 Apr 2018 03:47:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Damage1.jpg 2010 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_ionradspec/ Thu, 12 Apr 2018 03:47:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_IonRadSpec.jpg 2011 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_hydroxyrad_img/ Thu, 12 Apr 2018 03:47:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Hydroxyrad_img.jpg 2012 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_damage2/ Thu, 12 Apr 2018 03:47:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Damage2.jpg 2013 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_penetrate/ Thu, 12 Apr 2018 03:47:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Penetrate.jpg 2014 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_radonexpos/ Thu, 12 Apr 2018 03:47:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_RadonExpos.jpg 2015 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_monitors/ Thu, 12 Apr 2018 03:47:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Monitors.jpg 2016 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_exposure1/ Thu, 12 Apr 2018 03:47:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Exposure1.jpg 2017 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_21_06_exposure2/ Thu, 12 Apr 2018 03:47:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_21_06_Exposure2.jpg 2018 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_aa_periodicpu_img/ Thu, 12 Apr 2018 03:47:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_AA_PeriodicPU_img.jpg 2020 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_bb_dependence_img/ Thu, 12 Apr 2018 03:47:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_BB_Dependence_img.jpg 2022 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_bb_function_img/ Thu, 12 Apr 2018 03:47:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_BB_Function_img.jpg 2023 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ee_density_img/ Thu, 12 Apr 2018 03:47:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_EE_Density_img.jpg 2027 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ee_vapor_img/ Thu, 12 Apr 2018 03:47:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_EE_Vapor_img.jpg 2028 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ee_waterpkw_img/ Thu, 12 Apr 2018 03:47:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_EE_WaterpKw_img.jpg 2029 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ee_liqwatabso/ Thu, 12 Apr 2018 03:47:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_EE_LiqWatAbso.jpg 2030 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sacetic_img/ Thu, 12 Apr 2018 03:47:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sacetic_img.jpg 2034 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sarsenic_img/ Thu, 12 Apr 2018 03:47:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sarsenic_img.jpg 2035 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform1_img/ Thu, 12 Apr 2018 03:47:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform1_img.jpg 2036 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform2_img/ Thu, 12 Apr 2018 03:47:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform2_img.jpg 2037 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sarsenous_img/ Thu, 12 Apr 2018 03:47:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sarsenous_img.jpg 2038 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sboric_img/ Thu, 12 Apr 2018 03:47:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sboric_img.jpg 2039 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1scarbonic_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1scarbonic_img.jpg 2040 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform3_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform3_img.jpg 2041 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1scyanic_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1scyanic_img.jpg 2042 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sformic_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sformic_img.jpg 2043 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1shydrazoi_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1shydrazoi_img.jpg 2044 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1shydroper_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1shydroper_img.jpg 2045 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform4_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform4_img.jpg 2046 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1shydrosul_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1shydrosul_img.jpg 2047 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1snitrous_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1snitrous_img.jpg 2048 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1soxalic_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1soxalic_img.jpg 2049 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform5_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform5_img.jpg 2050 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sphosphor_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sphosphor_img.jpg 2051 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform6_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform6_img.jpg 2052 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform7_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform7_img.jpg 2053 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1sphospho2_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1sphospho2_img.jpg 2054 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform8_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform8_img.jpg 2055 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_1ssulfurou_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_1ssulfurou_img.jpg 2056 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_hh_chemform9_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_HH_chemform9_img.jpg 2057 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ii_lsammonia_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_II_lsammonia_img.jpg 2059 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ii_lsdimethyl_img/ Thu, 12 Apr 2018 03:47:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_II_lsdimethyl_img.jpg 2060 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ii_lsmethylam_img/ Thu, 12 Apr 2018 03:47:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_II_lsmethylam_img.jpg 2061 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ii_lsphenylam_img/ Thu, 12 Apr 2018 03:47:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_II_lsphenylam_img.jpg 2062 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cnx_chem_00_ii_lstrimethy_img/ Thu, 12 Apr 2018 03:47:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CNX_Chem_00_II_lstrimethy_img.jpg 2063 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chemistry-1024x696-1/ Thu, 12 Apr 2018 03:51:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/chemistry-1024x696-1.jpg 2070 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/water-1024x296-1/ Thu, 12 Apr 2018 03:51:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/water-1024x296-1.jpg 2072 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/800px-match_stick_lit_a_match_match_box_fire-1/ Thu, 12 Apr 2018 03:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/800px-Match_stick_lit_a_match_match_box_fire-1.jpg 2073 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/5012344135_7979fec9fb_o-1024x682-1/ Thu, 12 Apr 2018 03:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/5012344135_7979fec9fb_o-1024x682-1.jpg 2074 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mix-1024x450-1/ Thu, 12 Apr 2018 03:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/mix-1024x450-1.jpg 2075 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/element-1024x512-1/ Thu, 12 Apr 2018 03:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/element-1024x512-1.jpg 2076 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/describing-matter-1/ Thu, 12 Apr 2018 03:51:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Describing-Matter-1.png 2077 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/everyday-1024x276-1/ Thu, 12 Apr 2018 03:51:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/everyday-1024x276-1.jpg 2078 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/qrcode-23437421-150x150-1/ Thu, 12 Apr 2018 03:51:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/qrcode.23437421-150x150-1.png 2079 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/blue-square-1/ Thu, 12 Apr 2018 03:51:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Blue-Square-1.png 2081 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/drinks-1024x276-1/ Thu, 12 Apr 2018 03:51:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/drinks-1024x276-1.jpg 2082 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/stature-percentile-1/ Thu, 12 Apr 2018 03:51:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Stature-Percentile-1.png 2085 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/expressing-numbers/calc1-635x1024-1/ Thu, 12 Apr 2018 03:51:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/calc1-635x1024-1.jpg 2087 2088 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/print/ Thu, 12 Apr 2018 03:51:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/eb7d58a9777dbb124396d7c8bcb75793-1.jpg 2089 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/151d73b2e1318e4386f1be7579009b82-1/ Thu, 12 Apr 2018 03:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/151d73b2e1318e4386f1be7579009b82-1.jpg 2090 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/ruler-1/ Thu, 12 Apr 2018 03:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ruler-1.png 2091 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/rectangle-1/ Thu, 12 Apr 2018 03:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Rectangle-1.png 2092 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/sig-figs-1-1/ Thu, 12 Apr 2018 03:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Sig-Figs-1-1.png 2093 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/sig-figs-2-1/ Thu, 12 Apr 2018 03:51:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Sig-Figs-2-1.png 2094 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/qrcode-23437479-150x150-1/ Thu, 12 Apr 2018 03:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/qrcode.23437479-150x150-1.png 2095 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f85b7d0b1d3f3563a5b973ef04349df3-1/ Thu, 12 Apr 2018 03:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/f85b7d0b1d3f3563a5b973ef04349df3-1.jpg 2096 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ruler-2-1/ Thu, 12 Apr 2018 03:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ruler-2-1.png 2097 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/06979677f6f13c11ea559e495ebcbf85-1/ Thu, 12 Apr 2018 03:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/06979677f6f13c11ea559e495ebcbf85-1.jpg 2098 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ruler-3-1/ Thu, 12 Apr 2018 03:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ruler-3-1.png 2099 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_1/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Converting_Units_1.png 2101 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_2/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_2.png 2102 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_3/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_3.png 2103 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_4/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_4.png 2104 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_5/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_5.png 2105 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_6/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_6.png 2106 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_7/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_7.png 2107 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_8/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_8.png 2108 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_23/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_23.png 2109 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_10/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_10.png 2110 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_11/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_11.png 2111 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_12/ Thu, 12 Apr 2018 03:51:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_12.png 2112 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_13/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_13.png 2113 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_14/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_14.png 2114 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_15/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_15.png 2115 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-6-mathematical-treatment-of-measurement-results/800px-grapevinesnail_01-1/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/800px-Grapevinesnail_01-1.jpg 2116 1292 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_16/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_16.png 2117 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_17/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_17.png 2118 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_18/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_18.png 2119 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_21/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_21.png 2120 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_19/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_19.png 2121 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_untis_24/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_untis_24.png 2122 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/converting_units_22/ Thu, 12 Apr 2018 03:51:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/converting_units_22.png 2123 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-6-mathematical-treatment-of-measurement-results/800px-aircanada-b767-300er-c-ggmx-arp_-1/ Thu, 12 Apr 2018 03:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/800px-Aircanada.b767-300er.c-ggmx.arp_-1.jpg 2124 1292 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/qrcode-23437561-150x150-1/ Thu, 12 Apr 2018 03:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/qrcode.23437561-150x150-1.png 2125 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_1/ Thu, 12 Apr 2018 03:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_1.png 2127 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_2/ Thu, 12 Apr 2018 03:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_2.png 2128 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/temperatures-1/ Thu, 12 Apr 2018 03:52:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Temperatures-1.png 2129 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_3/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_3.png 2130 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_4/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_4.png 2131 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_5/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_5.png 2132 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_6/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_6.png 2133 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/other_units_7/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/other_units_7.png 2134 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/fwk-ball-fig02_004/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/67d75b028a8e56c36f0622ce6b20547e-1.jpg 2136 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/6866a70a4b52e0fd95f9dd3e1f3426a2-1/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/6866a70a4b52e0fd95f9dd3e1f3426a2-1.jpg 2137 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/size-of-a-liter-1/ Thu, 12 Apr 2018 03:52:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Size-of-a-Liter-1.png 2138 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/the-sun-1/ Thu, 12 Apr 2018 03:52:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/The-Sun-1.png 2142 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/4377164898_fcc9a2407f_o-1/ Thu, 12 Apr 2018 03:52:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/4377164898_fcc9a2407f_o-1.jpg 2145 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nutrition-facts1-1/ Thu, 12 Apr 2018 03:52:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nutrition-Facts1-1.png 2146 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/molecular-art-1/ Thu, 12 Apr 2018 03:52:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Molecular-Art-1.png 2149 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/the-atom-1/ Thu, 12 Apr 2018 03:52:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/The-Atom-1.png 2151 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-a-the-periodic-table/periodic-table-1/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Periodic-Table-1.png 2152 2021 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/generic_representation_of_atomic_symbol/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/generic_representation_of_atomic_symbol.png 2153 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carbon_isotope_symbol/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Carbon_isotope_symbol.png 2154 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/13_a_question/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/13_a_question.png 2155 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nucleus-1/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nucleus-1.png 2156 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/answer_11/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/answer_11.png 2157 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/15a_answer_oxygen_symbol/ Thu, 12 Apr 2018 03:52:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/15a_answer_oxygen_symbol.png 2158 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/15c_lithium_symbol/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/15c_lithium_symbol.png 2159 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-to-chemical-reactions-and-equations/400px-delta_iv_medium_rocket_dscs-1/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/400px-Delta_IV_Medium_Rocket_DSCS-1.jpg 2163 2164 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/halogens-1/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Halogens-1.png 2166 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/double-replacement-reaction-1/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Double-Replacement-Reaction-1.png 2167 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-1-writing-and-balancing-chemical-equations/ionic-compounds-1/ Thu, 12 Apr 2018 03:52:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ionic-Compounds-1.png 2169 1412 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/800px-grand_canyon_yavapal_point_2010-1/ Thu, 12 Apr 2018 03:52:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/800px-Grand_canyon_yavapal_point_2010-1.jpg 2170 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/oxidation-reduction-reactions/5345065044_0d15179564_b-1/ Thu, 12 Apr 2018 03:52:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/5345065044_0d15179564_b-1.jpg 2172 2177 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nacl-11-1/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaCl-11-1.png 2174 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nacl-21-1/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaCl-21-1.png 2175 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/redox-reaction-1/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Redox-Reaction-1.png 2176 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-44-33-am-1/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.44.33-AM-1.png 2182 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-45-13-am-1/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.45.13-AM-1.png 2183 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-45-40-am-1/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.45.40-AM-1.png 2184 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-46-06-am-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.46.06-AM-1.png 2185 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-46-31-am-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.46.31-AM-1.png 2186 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss2-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss2-1.png 2187 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss1-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss1-1.png 2188 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-49-14-am-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.49.14-AM-1.png 2189 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-49-40-am1-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.49.40-AM1-1.png 2190 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/qrcode-23437574-150x150-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/qrcode.23437574-150x150-1.png 2191 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-50-06-am-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.50.06-AM-1.png 2192 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-51-35-am-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.51.35-AM-1.png 2193 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-17-at-10-52-09-am-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-17-at-10.52.09-AM-1.png 2194 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss3-1/ Thu, 12 Apr 2018 03:52:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss3-1.png 2196 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss4-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss4-1.png 2197 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-20-at-3-09-54-pm-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-20-at-3.09.54-PM-1.png 2198 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss5-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss5-1.png 2199 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss6-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss6-1.png 2200 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss7-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss7-1.png 2201 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-16-11-pm-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.16.11-PM-1.png 2203 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-16-18-pm-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.16.18-PM-1.png 2204 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-16-21-pm-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.16.21-PM-1.png 2205 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-16-27-pm-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.16.27-PM-1.png 2206 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/187mol-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/187mol-1.png 2207 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/6molhcl-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/6molhcl-1.png 2208 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/187mol1-1/ Thu, 12 Apr 2018 03:52:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/187mol1-1.png 2209 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/249galcl31-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/249galcl31-1.png 2210 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/339molh2-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/339molh2-1.png 2211 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/226molnh3-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/226molnh3-1.png 2212 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-17-00-pm-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.17.00-PM-1.png 2213 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-17-03-pm-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.17.03-PM-1.png 2214 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-17-07-pm-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.17.07-PM-1.png 2215 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-21-at-5-17-13-pm-1024x163-1/ Thu, 12 Apr 2018 03:52:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-21-at-5.17.13-PM-1024x163-1.png 2216 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/864gk-1/ Thu, 12 Apr 2018 03:52:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/864gk-1.png 2217 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/making-water-1/ Thu, 12 Apr 2018 03:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Making-Water-1.png 2219 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-1-09-10-pm-1/ Thu, 12 Apr 2018 03:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-1.09.10-PM-1.png 2220 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-1-09-16-pm-1/ Thu, 12 Apr 2018 03:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-1.09.16-PM-1.png 2221 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-1-14-42-pm-1/ Thu, 12 Apr 2018 03:52:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-1.14.42-PM-1.png 2222 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-1-09-26-pm-1024x139-1/ Thu, 12 Apr 2018 03:52:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-1.09.26-PM-1024x139-1.png 2223 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/5oogrb-1024x269-1/ Thu, 12 Apr 2018 03:52:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/5OOGRB-1024x269-1.png 2224 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/500grb-1024x107-1/ Thu, 12 Apr 2018 03:52:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/500GRB-1024x107-1.png 2225 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mgos-1/ Thu, 12 Apr 2018 03:52:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/MGOS-1.png 2226 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/limiting-reagent-1/ Thu, 12 Apr 2018 03:52:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Limiting-Reagent-1.png 2227 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/making-water-2-1/ Thu, 12 Apr 2018 03:52:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Making-Water-2-1.png 2228 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-1-49-34-pm-1/ Thu, 12 Apr 2018 03:52:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-1.49.34-PM-1.png 2229 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-1-49-39-pm-1/ Thu, 12 Apr 2018 03:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-1.49.39-PM-1.png 2230 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-20-at-3-16-01-pm-1/ Thu, 12 Apr 2018 03:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-20-at-3.16.01-PM-1.png 2232 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ss8-1/ Thu, 12 Apr 2018 03:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ss8-1.png 2233 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-12-54-26-pm-1/ Thu, 12 Apr 2018 03:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-12.54.26-PM-1.png 2235 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/305gzn-1024x100-1/ Thu, 12 Apr 2018 03:52:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/305GZN-1024x100-1.png 2236 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/642gzn-1/ Thu, 12 Apr 2018 03:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/642GZN-1.png 2237 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-18-07-pm-1/ Thu, 12 Apr 2018 03:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.18.07-PM-1.png 2239 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-18-11-pm-1/ Thu, 12 Apr 2018 03:52:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.18.11-PM-1.png 2240 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-18-16-pm-1/ Thu, 12 Apr 2018 03:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.18.16-PM-1.png 2241 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-18-22-pm-1/ Thu, 12 Apr 2018 03:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.18.22-PM-1.png 2242 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/air-1024x281-1/ Thu, 12 Apr 2018 03:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/air-1024x281-1.png 2245 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-25-36-pm-1/ Thu, 12 Apr 2018 03:52:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.25.36-PM-1.png 2247 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/595torr-1/ Thu, 12 Apr 2018 03:52:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/595TORR-1.png 2248 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/601mmhg-1/ Thu, 12 Apr 2018 03:52:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/601MMHG-1.png 2249 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/figure-6-1-1/ Thu, 12 Apr 2018 03:52:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Figure-6-1-1.png 2251 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/244atm-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/244ATM-1.png 2252 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/244atm2-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/244ATM2-1.png 2253 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/244401-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/244401-1.png 2254 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/0633l-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/0633L-1.png 2255 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/722torr-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/722TORR-1.png 2256 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-35-38-pm-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.35.38-PM-1.png 2257 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-35-42-pm-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.35.42-PM-1.png 2258 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/figure-6-2-1/ Thu, 12 Apr 2018 03:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Figure-6-2-1.png 2259 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/348ml-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/348ml-1.png 2260 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/559k-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/559k-1.png 2261 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/559k21-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/559k21-1.png 2262 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-36-09-pm-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.36.09-PM-1.png 2263 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-2-36-13-pm-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-2.36.13-PM-1.png 2264 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/206k-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/206k-1.png 2265 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/250l-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/250l-1.png 2266 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-14-07-pm-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.14.07-PM-1.png 2268 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-14-11-pm-1/ Thu, 12 Apr 2018 03:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.14.11-PM-1.png 2269 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/245l-1/ Thu, 12 Apr 2018 03:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/245L-1.png 2270 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n2-1/ Thu, 12 Apr 2018 03:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/n2-1.png 2271 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-14-27-pm-1/ Thu, 12 Apr 2018 03:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.14.27-PM-1.png 2272 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/182-1/ Thu, 12 Apr 2018 03:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/182-1.png 2273 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/182atm2-1/ Thu, 12 Apr 2018 03:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/182ATM2-1.png 2274 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-23-00-pm-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.23.00-PM-1.png 2276 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-23-04-pm-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.23.04-PM-1.png 2277 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-01-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-01-1.png 2278 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-05-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-05-1.png 2279 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-07-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-07-1.png 2280 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example10eq2-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example10eq2-1.png 2281 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example10eq3-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example10eq3-1.png 2282 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-06-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-06-1.png 2283 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/107atm-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/107atm-1.png 2284 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-23-15-pm-1/ Thu, 12 Apr 2018 03:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.23.15-PM-1.png 2285 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/figure-6-3-300x204-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Figure-6-3-300x204-1.png 2286 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-11-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-11-1.png 2287 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-04-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-04-1.png 2288 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-08-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-08-1.png 2289 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-23-22-pm-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.23.22-PM-1.png 2290 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-23-24-pm-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.23.24-PM-1.png 2291 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-02-1/ Thu, 12 Apr 2018 03:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-02-1.png 2292 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equations-09-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equations-09-1.png 2293 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/figure-6-4-300x236-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Figure-6-4-300x236-1.png 2294 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-23-34-pm-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.23.34-PM-1.png 2295 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-51-15-pm-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.51.15-PM-1.png 2297 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/713torr-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/713torr-1.png 2298 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-51-24-pm-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.51.24-PM-1.png 2299 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-51-28-pm-1024x120-1/ Thu, 12 Apr 2018 03:52:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.51.28-PM-1024x120-1.png 2300 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-51-35-pm-1/ Thu, 12 Apr 2018 03:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.51.35-PM-1.png 2301 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-51-40-pm-1/ Thu, 12 Apr 2018 03:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.51.40-PM-1.png 2302 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/xhe-1/ Thu, 12 Apr 2018 03:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/xhe-1.png 2303 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/398px-champagne_uncorking_photographed_with_a_high_speed_air-gap_flash-1/ Thu, 12 Apr 2018 03:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/398px-Champagne_uncorking_photographed_with_a_high_speed_air-gap_flash-1.jpg 2304 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/figure-6-6-300x173-1/ Thu, 12 Apr 2018 03:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Figure-6-6-300x173-1.png 2306 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/stylized-molecular-energy-diagram-with-labels-1/ Thu, 12 Apr 2018 03:52:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Stylized-molecular-energy-diagram-with-labels-1.png 2307 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/distribution-of-molecular-speeds-of-o2-gas-at-three-temperatures-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Distribution-of-molecular-speeds-of-O2-gas-at-three-temperatures-1.jpg 2308 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/molecular-speed-distribution-for-noble-gases-jak-edit-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Molecular-Speed-Distribution-for-Noble-Gases-JAK-edit-1.jpg 2309 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/effusion-diagram-jak-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Effusion-Diagram-JAK-1.jpg 2311 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/318px-thomas_graham_litho-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/318px-Thomas_Graham_Litho-1.jpg 2312 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/diffusion-diagram-jak-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Diffusion-Diagram-JAK-1.jpg 2313 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/particlemeanfreepath-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ParticleMeanFreePath-1.png 2314 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/qrcode-23437599-150x150-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/qrcode.23437599-150x150-1.png 2315 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/compressibility-factor-three-gas-approximation-1/ Thu, 12 Apr 2018 03:52:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Compressibility-factor-three-gas-approximation-1.jpg 2317 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/compressibility-factor-nitrogen-approximation-at-three-temperatures-1/ Thu, 12 Apr 2018 03:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Compressibility-factor-nitrogen-approximation-at-three-temperatures-1.jpg 2318 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/johannes_diderik_van_der_waals-1/ Thu, 12 Apr 2018 03:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Johannes_Diderik_van_der_Waals-1.jpg 2319 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-5-59-06-pm-1/ Thu, 12 Apr 2018 03:52:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-5.59.06-PM-1.png 2321 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/385px-soyuz_tma-5_launch-1/ Thu, 12 Apr 2018 03:52:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/385px-Soyuz_TMA-5_launch-1.jpg 2324 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ion-engine-1/ Thu, 12 Apr 2018 03:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ion-Engine-1.png 2325 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-7-57-34-pm-1/ Thu, 12 Apr 2018 03:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-7.57.34-PM-1.png 2327 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/formation-reaction-1/ Thu, 12 Apr 2018 03:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Formation-Reaction-1.png 2328 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/fe2o3-1/ Thu, 12 Apr 2018 03:52:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/fe2o3-1.png 2329 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2so2-1/ Thu, 12 Apr 2018 03:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2so2-1.png 2330 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ahrxb-1/ Thu, 12 Apr 2018 03:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ahrxb-1.png 2331 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2hbr-1/ Thu, 12 Apr 2018 03:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2hbr-1.png 2332 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cog-1/ Thu, 12 Apr 2018 03:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/cog-1.png 2333 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/fda_nutrition_facts_label_2014-1/ Thu, 12 Apr 2018 03:52:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/FDA_Nutrition_Facts_Label_2014-1.jpg 2335 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/38kcal-1/ Thu, 12 Apr 2018 03:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/38kcal-1.png 2336 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-7-37-43-pm-1/ Thu, 12 Apr 2018 03:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-7.37.43-PM-1.png 2338 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2224g-1/ Thu, 12 Apr 2018 03:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2224g-1.png 2339 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/558kj-1/ Thu, 12 Apr 2018 03:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/558kj-1.png 2340 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c3e366ad3cf23dc1de728bb0e30e1e04-1/ Thu, 12 Apr 2018 03:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/c3e366ad3cf23dc1de728bb0e30e1e04-1.jpg 2341 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/reaction-energy-1/ Thu, 12 Apr 2018 03:52:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Reaction-Energy-1.png 2343 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/calorimeter-1/ Thu, 12 Apr 2018 03:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Calorimeter-1.png 2344 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-7-32-26-pm-1/ Thu, 12 Apr 2018 03:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-7.32.26-PM-1.png 2345 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-7-32-31-pm-1024x120-1/ Thu, 12 Apr 2018 03:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-7.32.31-PM-1024x120-1.png 2346 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/q100g-1/ Thu, 12 Apr 2018 03:52:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/q100g-1.png 2347 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/energy-graph-1/ Thu, 12 Apr 2018 03:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Energy-Graph-1.png 2348 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/volume-vs-pressure-1/ Thu, 12 Apr 2018 03:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Volume-vs-Pressure-1.png 2350 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/347-1/ Thu, 12 Apr 2018 03:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/347-1.png 2351 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/q250-1/ Thu, 12 Apr 2018 03:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/q250-1.png 2352 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c5408-1/ Thu, 12 Apr 2018 03:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/c5408-1.png 2353 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c0139-1/ Thu, 12 Apr 2018 03:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/c0139-1.png 2354 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-7-42-02-pm-1/ Thu, 12 Apr 2018 03:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-7.42.02-PM-1.png 2356 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-7-42-06-pm-1/ Thu, 12 Apr 2018 03:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-7.42.06-PM-1.png 2357 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/enthalpy-changes-1/ Thu, 12 Apr 2018 03:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Enthalpy-Changes-1.png 2358 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/enthalpy-changes-2-1/ Thu, 12 Apr 2018 03:52:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Enthalpy-Changes-2-1.png 2359 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electron-shells-1/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electron-Shells-1.png 2364 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/generic-energy-diagram-of-orbitals-in-multi-electron-atom-1/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/generic-energy-diagram-of-orbitals-in-multi-electron-atom-1.jpg 2365 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/boron_electron_configuration_energy_diagram-1/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Boron_electron_configuration_energy_diagram-1.png 2366 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electron-configuration-energy-diagrams-for-carbon-nitrogen-and-oxygen-e1411755889659-1/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electron-configuration-energy-diagrams-for-carbon-nitrogen-and-oxygen-e1411755889659-1.jpg 2367 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electron_configuration_potassium-svg_-1/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electron_configuration_potassium.svg_-1.png 2368 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/1000px-electron_configuration_bromine-svg_-1/ Thu, 12 Apr 2018 03:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/1000px-Electron_configuration_bromine.svg_-1.png 2369 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-a-the-periodic-table/periodic-table1-1/ Thu, 12 Apr 2018 03:52:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Periodic-Table1-1.png 2371 2021 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/1s-subshell-1/ Thu, 12 Apr 2018 03:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/1s-Subshell-1.png 2372 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2s-subshell-1/ Thu, 12 Apr 2018 03:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2s-Subshell-1.png 2373 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2p-subshell-1/ Thu, 12 Apr 2018 03:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2p-Subshell-1.png 2374 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3s-subshell-1/ Thu, 12 Apr 2018 03:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3s-Subshell-1.png 2375 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3p-subshell-1/ Thu, 12 Apr 2018 03:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3p-Subshell-1.png 2376 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/4s-subshell-1/ Thu, 12 Apr 2018 03:52:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/4s-Subshell-1.png 2377 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3d-subshell-1/ Thu, 12 Apr 2018 03:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3d-Subshell-1.png 2378 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/6-4-electronic-structure-of-atoms-electron-configurations/periodic-table-blocks-1/ Thu, 12 Apr 2018 03:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Periodic-Table-Blocks-1.png 2379 1528 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/periodic-table-selenium-1/ Thu, 12 Apr 2018 03:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Periodic-Table-Selenium-1.png 2380 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/6-4-electronic-structure-of-atoms-electron-configurations/periodic-table-various-elements-1/ Thu, 12 Apr 2018 03:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Periodic-Table-Various-Elements-1.png 2381 1528 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/food-colouring-1/ Thu, 12 Apr 2018 03:52:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Food-Colouring-1.png 2382 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/prisms-and-light-1/ Thu, 12 Apr 2018 03:52:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Prisms-and-Light-1.png 2384 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-04-37-pm-1/ Thu, 12 Apr 2018 03:52:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.04.37-PM-1.png 2385 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrogen-spectrum-1/ Thu, 12 Apr 2018 03:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hydrogen-Spectrum-1.png 2386 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/bohrs-hydrogen-atom-1/ Thu, 12 Apr 2018 03:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Bohrs-Hydrogen-Atom-1.png 2387 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electron-orbitals-1/ Thu, 12 Apr 2018 03:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electron-Orbitals-1.png 2388 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/450px-neon_internet_cafe_open_24_hours-1/ Thu, 12 Apr 2018 03:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/450px-Neon_Internet_Cafe_open_24_hours-1.jpg 2389 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/light/light-waves-1/ Thu, 12 Apr 2018 03:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Light-Waves-1.png 2391 2393 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electromagnetic-spectrum-1/ Thu, 12 Apr 2018 03:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electromagnetic-Spectrum-1.png 2392 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/zeff-trend-e1411757072427-1/ Thu, 12 Apr 2018 03:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Zeff-Trend-e1411757072427-1.jpg 2394 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/atomic-radii-trends-1/ Thu, 12 Apr 2018 03:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Atomic-Radii-Trends-1.png 2395 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ionization-energy-1/ Thu, 12 Apr 2018 03:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ionization-Energy-1.png 2396 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electron-affinity-1/ Thu, 12 Apr 2018 03:52:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electron-Affinity-1.png 2397 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/rough_diamond-1/ Thu, 12 Apr 2018 03:52:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Rough_diamond-1.jpg 2401 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrogen-1/ Thu, 12 Apr 2018 03:52:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hydrogen-1.png 2403 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrogen-sides-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hydrogen-Sides-1.png 2404 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/helium-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Helium-1.png 2405 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/lithium-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Lithium-1.png 2406 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/beryllium-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Beryllium-1.png 2407 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/boron-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Boron-1.png 2408 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carbon-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Carbon-1.png 2409 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nitrogen-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nitrogen-1.png 2410 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/oxygen-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Oxygen-1.png 2411 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/fluoride-neon-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Fluoride-Neon-1.png 2412 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/aluminium-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Aluminium-1.png 2413 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/selenium-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Selenium-1.png 2414 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/phosphorus-argon-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Phosphorus-Argon-1.png 2415 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/iron-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Iron-1.png 2416 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/first-column-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/First-Column-1.png 2417 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/lewis-dot-sodium-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Lewis-Dot-Sodium-1.png 2418 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/lewis-dot-iron-1/ Thu, 12 Apr 2018 03:52:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Lewis-Dot-Iron-1.png 2419 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/lewis-dot-chlorine-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Lewis-Dot-Chlorine-1.png 2420 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/oxygen-ion-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Oxygen-Ion-1.png 2421 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/thallium-ion-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Thallium-Ion-1.png 2422 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/strontium-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Strontium-1.png 2423 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/silicone-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Silicone-1.png 2424 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/titanium-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Titanium-1.png 2425 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/phosphorus-1/ Thu, 12 Apr 2018 03:52:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Phosphorus-1.png 2426 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/sulfur-ion-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Sulfur-Ion-1.png 2427 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nitrogen-ion-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nitrogen-Ion-1.png 2428 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nacl-1-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaCl-1-1.png 2430 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nacl-2-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaCl-2-1.png 2431 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nacl-3-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaCl-3-1.png 2432 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nacl-4-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaCl-4-1.png 2433 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mgo-1-1/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/MgO-1-1.png 2434 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mgo-2-1/ Thu, 12 Apr 2018 03:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/MgO-2-1.png 2435 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nao-1-1/ Thu, 12 Apr 2018 03:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaO-1-1.png 2436 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nao-2-1/ Thu, 12 Apr 2018 03:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaO-2-1.png 2437 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cacl-1-1/ Thu, 12 Apr 2018 03:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CaCl-1-1.png 2438 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ks-1-1/ Thu, 12 Apr 2018 03:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/KS-1-1.png 2439 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/element-1024x477-1/ Thu, 12 Apr 2018 03:52:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/element-1024x477-1.png 2440 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/kcl-1-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/KCl-1-1.png 2441 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/scf-1-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ScF-1-1.png 2442 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-h-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-H-1.png 2444 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-h-2-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-H-2-1.png 2445 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-h-3-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-H-3-1.png 2446 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-f-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-F-1.png 2447 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-f-2-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-F-2-1.png 2448 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-f-3-1/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-F-3-1.png 2449 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-f-4-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-F-4-1.png 2450 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-f-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-F-1.png 2451 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-f-2-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-F-2-1.png 2452 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-br-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-Br-1.png 2453 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-br-2-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-Br-2-1.png 2454 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cl-cl-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cl-Cl-1.png 2455 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-o-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-O-1.png 2456 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-o-2-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-O-2-1.png 2457 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-o-3-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-O-3-1.png 2458 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-1.png 2459 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-h-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-H-1.png 2460 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cl-p-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cl-P-1.png 2461 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-b-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-B-1.png 2462 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-b-2-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-B-2-1.png 2463 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/f-b-3-1/ Thu, 12 Apr 2018 03:52:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/F-B-3-1.png 2464 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-o-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-O-1.png 2465 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-o-2-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-O-2-1.png 2466 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-o-3-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-O-3-1.png 2467 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-o-4-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-O-4-1.png 2468 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-o-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-O-1.png 2469 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-o-2-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-O-2-1.png 2470 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-s-o-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-S-O-1.png 2471 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/triple-bond-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Triple-Bond-1.png 2472 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nutrition-facts-1/ Thu, 12 Apr 2018 03:52:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nutrition-Facts-1.png 2473 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/i-i-1/ Thu, 12 Apr 2018 03:52:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/I-I-1.png 2474 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-cl-1/ Thu, 12 Apr 2018 03:52:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-Cl-1.png 2475 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/s-f-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/S-F-1.png 2476 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-b-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-B-1.png 2477 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-ge-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-Ge-1.png 2478 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cl-f-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cl-F-1.png 2479 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/si-o-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Si-O-1.png 2480 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-1.png 2481 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/s-c-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/S-C-1.png 2482 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-n-h-o-1/ Thu, 12 Apr 2018 03:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-N-H-O-1.png 2483 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-h1-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-H1-1.png 2485 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-f1-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-F1-1.png 2486 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-f-charge-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-F-Charge-1.png 2487 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-f_dipole_arrow-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-F_dipole_arrow-1.png 2488 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h-2-o-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H-2-O-1.png 2489 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-2-h-6-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-2-H-6-1.png 2490 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-2-h-6-2-1/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-2-H-6-2-1.png 2491 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-o-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-O-1.png 2493 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/be-cl-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Be-Cl-1.png 2494 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/b-f-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/B-F-1.png 2495 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/p-f-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/P-F-1.png 2496 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cl-o-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cl-O-1.png 2497 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/s-f1-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/S-F1-1.png 2498 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/xe-f-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Xe-F-1.png 2499 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-o-2-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-O-2-1.png 2500 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/xe-f-4-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Xe-F-4-1.png 2501 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/p-f-o-1/ Thu, 12 Apr 2018 03:53:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/P-F-O-1.png 2502 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cl-f1-1/ Thu, 12 Apr 2018 03:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cl-F1-1.png 2503 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/180-1/ Thu, 12 Apr 2018 03:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/180-1.png 2505 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/120-1/ Thu, 12 Apr 2018 03:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/120-1.png 2506 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ge-f-1/ Thu, 12 Apr 2018 03:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ge-F-1.png 2507 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/tetrahedral-1/ Thu, 12 Apr 2018 03:53:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Tetrahedral-1.png 2508 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h1-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H1-1.png 2509 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-h1-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-H1-1.png 2510 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/o-h-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/O-H-1.png 2511 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-o1-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-O1-1.png 2512 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-o-21-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-O-21-1.png 2513 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/p-cl-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/P-Cl-1.png 2514 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/n-f-o-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/N-F-O-1.png 2515 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/water_lewis_structure-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/water_lewis_structure-1.png 2516 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/water_dipoles-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/water_dipoles-1.png 2517 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/tailtohead1-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/tailtohead1-1.png 2518 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/net_dipole_tailtohead-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/net_dipole_tailtohead-1.png 2519 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/net_molecular_dipole-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/net_molecular_dipole-1.png 2520 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/vector_components-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/vector_components-1.png 2521 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/vector_components_2-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/vector_components_2-1.png 2522 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/vector_components_3-1/ Thu, 12 Apr 2018 03:53:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/vector_components_3-1.png 2523 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/co2_no_net_dipole-1/ Thu, 12 Apr 2018 03:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/CO2_no_net_dipole-1.png 2524 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/bf3_dipoles_cancel-1/ Thu, 12 Apr 2018 03:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/BF3_dipoles_cancel-1.png 2525 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nh3_dipole-1/ Thu, 12 Apr 2018 03:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NH3_dipole-1.png 2526 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/1000px-s-p-orbitals-svg_-1/ Thu, 12 Apr 2018 03:53:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/1000px-S-p-Orbitals.svg_-1.png 2528 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h2_orbital_overlap_-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/H2_orbital_overlap_-1.png 2529 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/adapted-pi-bond-diagram-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/adapted-pi-bond-diagram-1.jpg 2530 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/methane-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/methane-1.png 2531 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/l_pauling-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/L_Pauling-1.jpg 2532 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/sp3_hybridization_of_carbon_2-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/sp3_hybridization_of_carbon_2-1.png 2533 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/sp3_orbital-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/sp3_orbital-1.png 2534 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/1000px-ae4h-jak-edit-1/ Thu, 12 Apr 2018 03:53:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/1000px-AE4h-jak-edit-1.png 2535 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethene-1/ Thu, 12 Apr 2018 03:53:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ethene-1.png 2536 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carbon_sp2_hybridization-1/ Thu, 12 Apr 2018 03:53:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/carbon_sp2_hybridization-1.png 2537 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/1000px-ae3h-jak-edit-1/ Thu, 12 Apr 2018 03:53:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/1000px-AE3h-jak-edit-1.png 2538 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethyne-1/ Thu, 12 Apr 2018 03:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ethyne-1.png 2539 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carbon_sp_hybridization-1/ Thu, 12 Apr 2018 03:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/carbon_sp_hybridization-1.png 2540 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ae2h-jak-edit-1/ Thu, 12 Apr 2018 03:53:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/AE2h-jak-edit-1.png 2541 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hybrid_orbitals_and_geometry-1/ Thu, 12 Apr 2018 03:53:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hybrid_Orbitals_and_Geometry-1.jpg 2542 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h2_mo_combination_diagram-1/ Thu, 12 Apr 2018 03:53:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/h2_mo_combination_diagram-1.png 2544 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/h2_mo_energy_diagram-1/ Thu, 12 Apr 2018 03:53:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/h2_MO_energy_diagram-1.png 2545 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/li2_mo_electron_config_diagram-1/ Thu, 12 Apr 2018 03:53:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Li2_MO_electron_config_diagram-1.png 2546 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2p_head_to_head_overlap-1/ Thu, 12 Apr 2018 03:53:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2p_head_to_head_overlap-1.png 2547 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2p_sideways_overlap-1/ Thu, 12 Apr 2018 03:53:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2p_sideways_overlap-1.png 2548 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2p_molecular_orbitals_electron_config-1/ Thu, 12 Apr 2018 03:53:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2p_molecular_orbitals_electron_config-1.png 2549 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2p_molecular_orbital_energy_diagram_swapped-1/ Thu, 12 Apr 2018 03:53:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2p_molecular_orbital_energy_diagram_swapped-1.png 2550 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/frontier_molecular_orbitals_1-1/ Thu, 12 Apr 2018 03:53:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Frontier_Molecular_Orbitals_1-1.png 2551 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/homo_and_lumo_reaction-1/ Thu, 12 Apr 2018 03:53:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/HOMO_and_LUMO_reaction-1.png 2552 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/dipole_moment_questions-1/ Thu, 12 Apr 2018 03:53:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/dipole_moment_questions-1.png 2554 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/p-p-1/ Thu, 12 Apr 2018 03:53:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/P-P-1.png 2555 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/s-o-1/ Thu, 12 Apr 2018 03:53:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/S-O-1.png 2556 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/dipole_moment_solutions-1/ Thu, 12 Apr 2018 03:53:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/dipole_moment_solutions-1.png 2557 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/he2_mo_config_001-1/ Thu, 12 Apr 2018 03:53:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/He2_MO_config_001-1.png 2558 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2253052296_3728a5a115_o-1024x680-1/ Thu, 12 Apr 2018 03:53:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2253052296_3728a5a115_o-1024x680-1.jpg 2561 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/evaporation-1/ Thu, 12 Apr 2018 03:53:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Evaporation-1.png 2563 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equilibrium-1/ Thu, 12 Apr 2018 03:53:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Equilibrium-1.png 2564 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pressure-vs-temperature-1/ Thu, 12 Apr 2018 03:53:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Pressure-vs-Temperature-1.png 2565 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pressure-vs-temperature-2-1/ Thu, 12 Apr 2018 03:53:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Pressure-vs-Temperature-2-1.png 2566 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/phase-diagram-e1412017172724-1/ Thu, 12 Apr 2018 03:53:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/phase-diagram-e1412017172724-1.png 2567 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example-question-phase-diagram-1/ Thu, 12 Apr 2018 03:53:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example-question-phase-diagram-1.png 2568 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/surface-tension-1/ Thu, 12 Apr 2018 03:53:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Surface-Tension-1.png 2569 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/10310821203_65c9e7b179_b-e1412017214713-1/ Thu, 12 Apr 2018 03:53:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/10310821203_65c9e7b179_b-e1412017214713-1.jpg 2570 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/5286250264_0a7d2640a6_b-e1412017226144-1/ Thu, 12 Apr 2018 03:53:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/5286250264_0a7d2640a6_b-e1412017226144-1.jpg 2571 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/capillary-action-1/ Thu, 12 Apr 2018 03:53:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Capillary-Action-1.png 2572 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/305966315_01ffe2af00_o-e1412017262545-1/ Thu, 12 Apr 2018 03:53:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/305966315_01ffe2af00_o-e1412017262545-1.jpg 2573 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/sodium-e1412017290552-1/ Thu, 12 Apr 2018 03:53:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/sodium-e1412017290552-1.png 2575 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/ionic-solid-1/ Thu, 12 Apr 2018 03:53:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ionic-Solid-1.png 2576 1266 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/molecular-solid-1/ Thu, 12 Apr 2018 03:53:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Molecular-Solid-1.png 2577 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/covalent-network-solid-1/ Thu, 12 Apr 2018 03:53:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Covalent-Network-Solid-1.png 2578 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/water-1024x539-1/ Thu, 12 Apr 2018 03:53:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/water-1024x539-1.png 2580 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/liquids-and-gases-1/ Thu, 12 Apr 2018 03:53:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Liquids-and-Gases-1.png 2581 1266 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/heating-curve-diagram-1/ Thu, 12 Apr 2018 03:53:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/heating-curve-diagram-1.jpg 2582 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/attraction-1/ Thu, 12 Apr 2018 03:53:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Attraction-1.png 2584 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrogen-bonding-1/ Thu, 12 Apr 2018 03:53:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hydrogen-Bonding-1.png 2585 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/glucose1-1/ Thu, 12 Apr 2018 03:53:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Glucose1-1.png 2586 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/60929326_732ef95ec2_o-1024x768-1/ Thu, 12 Apr 2018 03:53:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/60929326_732ef95ec2_o-1024x768-1.jpg 2590 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-9-35-04-am-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-9.35.04-AM-1.png 2592 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/120gc10h8-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/120gc10h8-1.png 2593 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/450gc6g6-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/450gc6g6-1.png 2594 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/xc10h8-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/xc10h8-1.png 2595 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-9-35-25-am-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-9.35.25-AM-1.png 2596 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/xc6h6-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/xc6h6-1.png 2597 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-09-58-am-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.09.58-AM-1.png 2598 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/392424994_b96ce6a787_b-e1412018672372-1/ Thu, 12 Apr 2018 03:53:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/392424994_b96ce6a787_b-e1412018672372-1.jpg 2599 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/semipermeable-membrane-1/ Thu, 12 Apr 2018 03:53:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Semipermeable-Membrane-1.png 2600 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/03333m-1/ Thu, 12 Apr 2018 03:53:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/03333m-1.png 2601 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/0333moll-1/ Thu, 12 Apr 2018 03:53:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/0333moll-1.png 2602 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/osmotic-pressure-1/ Thu, 12 Apr 2018 03:53:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Osmotic-Pressure-1.png 2603 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/4269191327_7f6f150338_b-e1412018548367-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/4269191327_7f6f150338_b-e1412018548367-1.jpg 2605 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-26-pm-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.26-PM-1.png 2607 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-31-pm-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.31-PM-1.png 2608 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-35-pm-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.35-PM-1.png 2609 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-40-pm-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.40-PM-1.png 2610 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-44-pm-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.44-PM-1.png 2611 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-08-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-08-1.png 2612 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-07-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-07-1.png 2613 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-01-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-01-1.png 2614 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-04-1/ Thu, 12 Apr 2018 03:53:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-04-1.png 2615 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-06-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-06-1.png 2616 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-05-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-05-1.png 2617 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-54-pm-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.54-PM-1.png 2618 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-37-58-pm-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.37.58-PM-1.png 2619 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-02-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-02-1.png 2620 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-38-07-pm-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.38.07-PM-1.png 2621 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-38-11-pm1-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.38.11-PM1-1.png 2622 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-22-at-8-38-15-pm-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-22-at-8.38.15-PM-1.png 2623 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-09-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-09-1.png 2624 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-03-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-03-1.png 2625 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-11-1/ Thu, 12 Apr 2018 03:53:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-11-1.png 2626 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-10-1/ Thu, 12 Apr 2018 03:53:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-10-1.png 2627 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/450px-infuuszakjes-1/ Thu, 12 Apr 2018 03:53:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/450px-Infuuszakjes-1.jpg 2631 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/wine-1024x512-1/ Thu, 12 Apr 2018 03:53:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/wine-1024x512-1.jpg 2635 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/titration-equipment-1/ Thu, 12 Apr 2018 03:53:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Titration-Equipment-1.png 2637 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-49-25-am-1/ Thu, 12 Apr 2018 03:53:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.49.25-AM-1.png 2638 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-49-29-am-1/ Thu, 12 Apr 2018 03:53:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.49.29-AM-1.png 2639 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/000390-1/ Thu, 12 Apr 2018 03:53:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/000390-1.png 2640 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/000195-1/ Thu, 12 Apr 2018 03:53:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/000195-1.png 2641 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/qrcode-23437635-150x150-1/ Thu, 12 Apr 2018 03:53:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/qrcode.23437635-150x150-1.png 2642 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/updated-bronsted-lowry-example-e1412018930649-1/ Thu, 12 Apr 2018 03:53:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/updated-bronsted-lowry-example-e1412018930649-1.png 2645 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nh3-with-water-1/ Thu, 12 Apr 2018 03:53:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NH3-with-water-1.png 2646 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nh3-with-water-dynamic-e1412019102725-1/ Thu, 12 Apr 2018 03:53:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NH3-with-water-dynamic-e1412019102725-1.png 2647 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/bl-acid-base-reaction-1/ Thu, 12 Apr 2018 03:53:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/BL-Acid-Base-Reaction-1.png 2648 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/bl-acid-base-reaction-2-1/ Thu, 12 Apr 2018 03:53:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/BL-Acid-Base-Reaction-2-1.png 2649 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/buffer-action-1/ Thu, 12 Apr 2018 03:53:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Buffer-Action-1.png 2653 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/aspirin-1/ Thu, 12 Apr 2018 03:53:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Aspirin-1.png 2654 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/x-and-y-1/ Thu, 12 Apr 2018 03:53:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/X-and-Y-1.png 2656 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-53-59-am-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.53.59-AM-1.png 2662 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-54-21-am-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.54.21-AM-1.png 2663 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-54-25-am-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.54.25-AM-1.png 2664 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-54-31-am-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.54.31-AM-1.png 2665 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-051-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-051-1.png 2666 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-071-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-071-1.png 2667 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-021-1/ Thu, 12 Apr 2018 03:53:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-021-1.png 2668 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-13-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-13-1.png 2669 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-041-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-041-1.png 2670 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-031-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-031-1.png 2671 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-55-15-am-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.55.15-AM-1.png 2672 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-011-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-011-1.png 2673 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/equation-14-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/equation-14-1.png 2674 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-11-47-am-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.11.47-AM-1.png 2675 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-57-05-am-e1406089292977-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.57.05-AM-e1406089292977-1.png 2676 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-57-12-am-e1406089317592-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.57.12-AM-e1406089317592-1.png 2677 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-57-17-am-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.57.17-AM-1.png 2678 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-57-31-am-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.57.31-AM-1.png 2679 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-57-39-am-1/ Thu, 12 Apr 2018 03:53:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.57.39-AM-1.png 2680 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-10-57-44-am-1/ Thu, 12 Apr 2018 03:53:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-10.57.44-AM-1.png 2681 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/5996174498_eb660e46c2_b-e1412019701873-1/ Thu, 12 Apr 2018 03:53:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/5996174498_eb660e46c2_b-e1412019701873-1.jpg 2683 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-04-am-1/ Thu, 12 Apr 2018 03:53:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.04-AM-1.png 2685 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-08-am-1/ Thu, 12 Apr 2018 03:53:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.08-AM-1.png 2686 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-12-am-1/ Thu, 12 Apr 2018 03:53:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.12-AM-1.png 2687 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-15-am-1/ Thu, 12 Apr 2018 03:53:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.15-AM-1.png 2688 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-18-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.18-AM-1.png 2689 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-22-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.22-AM-1.png 2690 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-27-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.27-AM-1.png 2691 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-31-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.31-AM-1.png 2692 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-23-35-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.23.35-AM-1.png 2693 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/untitled-1-14-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Untitled-1-14-1.png 2694 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/untitled-1-09-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Untitled-1-09-1.png 2695 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/untitled-1-08-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Untitled-1-08-1.png 2696 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/untitled-1-12-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Untitled-1-12-1.png 2697 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-08-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.08-AM-1.png 2698 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-12-am-1/ Thu, 12 Apr 2018 03:53:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.12-AM-1.png 2699 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-16-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.16-AM-1.png 2700 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-20-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.20-AM-1.png 2701 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-29-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.29-AM-1.png 2702 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-33-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.33-AM-1.png 2703 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-38-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.38-AM-1.png 2704 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-49-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.49-AM-1.png 2705 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-54-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.54-AM-1.png 2706 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-24-57-am-1/ Thu, 12 Apr 2018 03:53:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.24.57-AM-1.png 2707 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cork-1/ Thu, 12 Apr 2018 03:53:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cork-1.png 2709 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2593160569_19c028d796_b-e1412019982879-1/ Thu, 12 Apr 2018 03:53:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2593160569_19c028d796_b-e1412019982879-1.jpg 2713 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/electrolysis-1/ Thu, 12 Apr 2018 03:53:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Electrolysis-1.png 2717 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/test-cell-1/ Thu, 12 Apr 2018 03:53:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Test-Cell-1.png 2718 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/redox-reaction-with-voltmeter-1/ Thu, 12 Apr 2018 03:53:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Redox-Reaction-With-Voltmeter-1.png 2720 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/dry-cell-1/ Thu, 12 Apr 2018 03:53:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Dry-Cell-1.png 2721 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/coin-cells-1/ Thu, 12 Apr 2018 03:53:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Coin-cells-1.jpg 2722 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/fuel-cell-1/ Thu, 12 Apr 2018 03:53:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Fuel-Cell-1.png 2723 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/honda-fcx-1/ Thu, 12 Apr 2018 03:53:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Honda-FCX-1.png 2724 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/zinc-silver-reaction-1/ Thu, 12 Apr 2018 03:53:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Zinc-Silver-Reaction-1.png 2725 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nabr-redox-1/ Thu, 12 Apr 2018 03:53:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/NaBr-Redox-1.png 2727 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/redox1-1/ Thu, 12 Apr 2018 03:53:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/redox1-1.png 2728 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ingredients-1/ Thu, 12 Apr 2018 03:53:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ingredients-1.png 2729 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/585px-residential_smoke_detector-1/ Thu, 12 Apr 2018 03:53:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/585px-Residential_smoke_detector-1.jpg 2732 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/800px-geiger_counter-e1412020533930-1/ Thu, 12 Apr 2018 03:53:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/800px-Geiger_counter-e1412020533930-1.jpg 2734 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/shroud-of-turin-1/ Thu, 12 Apr 2018 03:53:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Shroud-of-Turin-1.png 2736 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/medical-diagnostics-1/ Thu, 12 Apr 2018 03:53:43 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Medical-Diagnostics-1.png 2737 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/wine-radioactivity-1/ Thu, 12 Apr 2018 03:53:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Wine-Radioactivity-1.png 2738 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/radioactive-decay-1/ Thu, 12 Apr 2018 03:53:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Radioactive-Decay-1.png 2740 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-50-20-am-1/ Thu, 12 Apr 2018 03:53:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.50.20-AM-1.png 2741 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/amountremaining-1/ Thu, 12 Apr 2018 03:53:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/amountremaining-1.png 2742 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/screen-shot-2014-07-23-at-11-50-31-am-1/ Thu, 12 Apr 2018 03:53:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2014-07-23-at-11.50.31-AM-1.png 2743 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuclear_equation_1-1/ Thu, 12 Apr 2018 03:53:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuclear_equation_1-1.png 2746 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_2-1/ Thu, 12 Apr 2018 03:53:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_2-1.png 2747 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_3-1/ Thu, 12 Apr 2018 03:53:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_3-1.png 2748 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_4-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_4-1.png 2749 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_5-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_5-1.png 2750 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_6-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_6-1.png 2751 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuclear-power-plant-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nuclear-Power-Plant-1.png 2752 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_7-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_7-1.png 2753 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/nuc_energ_eq_8-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/nuc_energ_eq_8-1.png 2754 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/opening_essay_photosynth_eq-1/ Thu, 12 Apr 2018 03:53:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/opening_essay_photosynth_eq-1.png 2758 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/4798126912_381f653438_b-1/ Thu, 12 Apr 2018 03:53:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/4798126912_381f653438_b-1.jpg 2759 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h2-1/ Thu, 12 Apr 2018 03:53:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H2-1.png 2761 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/513px-methane-crc-mw-3d-balls-1/ Thu, 12 Apr 2018 03:53:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/513px-Methane-CRC-MW-3D-balls-1.png 2762 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethane-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Ethane-1.png 2763 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/propane-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Propane-1.png 2764 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethene_structure-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ethene_structure-1.png 2765 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/propene_lewis_struct-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/propene_lewis_struct-1.png 2766 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/but-1-ene_and_but-2-ene_lewis_struct-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/but-1-ene_and_but-2-ene_lewis_struct-1.png 2767 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pent-2-ene-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Pent-2-Ene-1.png 2768 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hex-3-ene-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hex-3-Ene-1.png 2769 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/propyne-1/ Thu, 12 Apr 2018 03:53:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Propyne-1.png 2770 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/butynes_lewis_struct-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/butynes_lewis_struct-1.png 2771 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/benzene-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Benzene-1.png 2772 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/methane_chlorination_reaction_equation-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/methane_chlorination_reaction_equation-1.png 2773 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-cl-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-Cl-1.png 2774 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethene_hydrogenation_reaction_equation-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ethene_hydrogenation_reaction_equation-1.png 2775 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/441px-first_gas_from_the_oselvar_module_on_the_ula_platform_on_april_14th_2012-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/441px-First_gas_from_the_Oselvar_module_on_the_Ula_platform_on_April_14th_2012-1.jpg 2776 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-3-1-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-3-1-1.png 2777 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-3-2-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-3-2-1.png 2778 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrocarbons_ex_3-3-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hydrocarbons_ex_3.3-1.png 2779 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-4-11-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-4-11-1.png 2780 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-4-2-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-4-2-1.png 2781 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrocarbons_ex_4-3-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hydrocarbons_ex_4.3-1.png 2782 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-5-1-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-5-1-1.png 2783 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-5-2-1/ Thu, 12 Apr 2018 03:53:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-5-2-1.png 2784 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-5-3-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-5-3-1.png 2785 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-6-1-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-6-1-1.png 2786 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-6-2-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-6-2-1.png 2787 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-6-3-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-6-3-1.png 2788 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/answer-7-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Answer-7-1.png 2789 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hydrocarbons_ex_sol_13-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hydrocarbons_ex_sol_13-1.png 2790 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/c-h-br-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/C-H-Br-1.png 2791 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/answer-191-1/ Thu, 12 Apr 2018 03:53:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Answer-191-1.png 2792 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hexane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Hexane-1.png 2794 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/heptane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Heptane-1.png 2795 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pentane-hydrocarbon-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Pentane-Hydrocarbon-1.png 2796 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/iupac-nomenclature-guide-diagram-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/IUPAC-Nomenclature-guide-diagram-1.jpg 2797 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3-methylpentane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3-methylpentane-1.png 2798 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branched_hydroc_example_2-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branched_hydroc_example_2-1.png 2799 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branched_hydroc_example_2b-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branched_hydroc_example_2b-1.png 2800 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3-ethyl-2-methylpentane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3-ethyl-2-methylpentane-1.png 2801 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/23-dimethylbutane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/23-dimethylbutane-1.png 2802 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3-ethyl-22-dimethylheptane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3-ethyl-22-dimethylheptane-1.png 2803 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/445-tripropyloctane-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/445-tripropyloctane-1.png 2804 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/24-dimethylhept-3-ene-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/24-dimethylhept-3-ene-1.png 2805 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/22-dimethylhex-3-yne-1/ Thu, 12 Apr 2018 03:53:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/22-dimethylhex-3-yne-1.png 2806 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/234-trimethylpent-2-ene-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/234-trimethylpent-2-ene-1.png 2807 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hept-2-ene-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hept-2-ene-1.png 2808 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/23-dimethylhept-2-ene-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/23-dimethylhept-2-ene-1.png 2809 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/23-dimethyl-4-propylhept-2-ene-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/23-dimethyl-4-propylhept-2-ene-1.png 2810 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pentane-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/pentane-1.png 2811 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/234-trimethylpentane-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/234-trimethylpentane-1.png 2812 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3-ethyl-67-dimethyloct-2-ene-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3-ethyl-67-dimethyloct-2-ene-1.png 2813 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chloro_and_ethyl_benzene-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/chloro_and_ethyl_benzene-1.png 2814 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/dichlorobenzene_and_bromoethylbenzene-e1411666173419-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/dichlorobenzene_and_bromoethylbenzene-e1411666173419-1.png 2815 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/3-phenylheptane-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/3-phenylheptane-1.png 2816 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_3-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_Exc_3-1.png 2817 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_4-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_4-1.png 2818 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_5-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_5-1.png 2819 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_6-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_6-1.png 2820 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_7-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_7-1.png 2821 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_8-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_8-1.png 2822 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_9-1/ Thu, 12 Apr 2018 03:53:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_9-1.png 2823 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_10-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_10-1.png 2824 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-11-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-11-1.png 2825 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-12-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-12-1.png 2826 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_sol_13a-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_sol_13a-1.png 2827 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_sol_13b-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_sol_13b-1.png 2828 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_sol_15a-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_sol_15a-1.png 2829 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_sol_15b-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_sol_15b-1.png 2830 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/branch_hydro_exc_sol_17-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/branch_hydro_exc_sol_17-1.png 2831 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2-bromobutane-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2-bromobutane-1.png 2833 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/23-dichloropentane-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/23-dichloropentane-1.png 2834 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/23-dichloro-3-methylpentane-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/23-dichloro-3-methylpentane-1.png 2835 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/112-tribromopropane-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/112-tribromopropane-1.png 2836 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/isopropyl-1/ Thu, 12 Apr 2018 03:53:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Isopropyl-1.png 2837 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2-chloro-2-methylbutan-1-ol-1/ Thu, 12 Apr 2018 03:53:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2-chloro-2-methylbutan-1-ol-1.png 2838 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/222-trichloroethanol-1/ Thu, 12 Apr 2018 03:53:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/222-trichloroethanol-1.png 2839 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/elimination_rxn_diagram-e1411666472503-1/ Thu, 12 Apr 2018 03:53:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/elimination_rxn_diagram-e1411666472503-1.png 2840 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example-8-1/ Thu, 12 Apr 2018 03:53:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Example-8-1.png 2841 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/alk_hal_and_alc_example_8_soln-1/ Thu, 12 Apr 2018 03:53:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/alk_hal_and_alc_example_8_soln-1.png 2842 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example-8-2-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Example-8-2-1.png 2843 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_3-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_3-1.png 2844 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_4-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_4-1.png 2845 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_5-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_5-1.png 2846 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_6-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_6-1.png 2847 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_7-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_7-1.png 2848 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_8-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_8-1.png 2849 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_9-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_9-1.png 2850 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_10-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_10-1.png 2851 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_11-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_11-1.png 2852 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_12-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_12-1.png 2853 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_13-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_13-1.png 2854 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/hal_and_alc_exc_14-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/hal_and_alc_exc_14-1.png 2855 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carbonyl-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/carbonyl-1.png 2857 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/aldehydes-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/aldehydes-1.png 2858 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/propanone-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/propanone-1.png 2859 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pentan-3-one-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/pentan-3-one-1.png 2860 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pentan-2-one-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/pentan-2-one-1.png 2861 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/methyl_butyl_ketone-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/methyl_butyl_ketone-1.png 2862 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carboxyl_group-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/carboxyl_group-1.png 2863 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/carboxylic_acids-1/ Thu, 12 Apr 2018 03:53:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/carboxylic_acids-1.png 2864 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/generic_carboxylate_formation-e1411667826451-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/generic_carboxylate_formation-e1411667826451-1.png 2865 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_10a-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_10a-1.png 2866 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_10_solution-e1411667835919-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_10_solution-e1411667835919-1.png 2867 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_10_test_yourself_a-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_10_test_yourself_a-1.png 2868 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_10_test_yourself_solution-e1411667844839-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_10_test_yourself_solution-e1411667844839-1.png 2869 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/generic_ester_formation-e1411667851288-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/generic_ester_formation-e1411667851288-1.png 2870 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/methyl_propanoate-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/methyl_propanoate-1.png 2871 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_3a-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_3a-1.png 2872 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_3b-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_3b-1.png 2873 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_4a-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_4a-1.png 2874 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_4b-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_4b-1.png 2875 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_5a-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_5a-1.png 2876 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_5b-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_5b-1.png 2877 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_6a-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_6a-1.png 2878 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_6b-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_6b-1.png 2879 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_7-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_7-1.png 2880 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_8-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_8-1.png 2881 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_11-1/ Thu, 12 Apr 2018 03:54:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_11-1.png 2882 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_12-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_12-1.png 2883 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_13-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_13-1.png 2884 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_14-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_14-1.png 2885 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_15-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_15-1.png 2886 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_16-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_16-1.png 2887 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/methylamine-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Methylamine-1.png 2889 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/dimethylamine-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Dimethylamine-1.png 2890 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/trimethylamine-1/ Thu, 12 Apr 2018 03:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/trimethylamine-1.png 2891 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/diethylamine-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/diethylamine-1.png 2892 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethyldipropylamine-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ethyldipropylamine-1.png 2893 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_11-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_11-1.png 2894 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_11_test_yourself-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_11_test_yourself-1.png 2895 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/proton-acceptor-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Proton-Acceptor-1.png 2896 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/natural_products_with_amine_functional_group-e1411668131295-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Natural_products_with_amine_functional_group-e1411668131295-1.png 2897 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/amide-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Amide-1.png 2898 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/amide-formation-1/ Thu, 12 Apr 2018 03:54:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Amide-Formation-1.png 2899 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethylamine_and_butanoic_acid-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ethylamine_and_butanoic_acid-1.png 2900 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_12_solution-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_12_solution-1.png 2901 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example_12_test_yourself_solution-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/example_12_test_yourself_solution-1.png 2902 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/thiols-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Thiols-1.png 2903 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/mercaptans-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Mercaptans-1.png 2904 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/grapefruit-mercaptan-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Grapefruit-Mercaptan-1.png 2905 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cysteine-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/cysteine-1.png 2906 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cysteine1-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/cysteine1-1.png 2907 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-3-21-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-3-21-1.png 2908 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_3c-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_3c-1.png 2909 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_4a1-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_4a1-1.png 2910 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-4-21-1/ Thu, 12 Apr 2018 03:54:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-4-21-1.png 2911 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_8a-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_8a-1.png 2912 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_8b-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_8b-1.png 2913 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_9-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_9-1.png 2914 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_10-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_10-1.png 2915 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-13-1-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-13-1-1.png 2916 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_11_sol-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_11_sol-1.png 2917 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_15_sol-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_15_sol-1.png 2918 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ethylene_polym-e1411669429724-1/ Thu, 12 Apr 2018 03:54:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ethylene_polym-e1411669429724-1.png 2920 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/common_monomers_and_polymers-e1411669437391-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/common_monomers_and_polymers-e1411669437391-1.png 2921 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/tetrafluoroethylene-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/tetrafluoroethylene-1.png 2922 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/tetrafluoroethylene-hydrocarbon-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Tetrafluoroethylene-Hydrocarbon-1.png 2923 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/vinyl_chloride-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/vinyl_chloride-1.png 2924 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/vinyl-chloride-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Vinyl-Chloride-1.png 2925 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/amide_bond_condensation-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/amide_bond_condensation-1.png 2926 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/synthetic-rubber-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Synthetic-Rubber-1.png 2927 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/silicone1-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Silicone1-1.png 2928 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/glucose-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Glucose-1.png 2929 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/starch-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Starch-1.png 2930 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/cellulose-1/ Thu, 12 Apr 2018 03:54:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Cellulose-1.png 2931 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/1393832634_e1d80ab169_o-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/1393832634_e1d80ab169_o-1.jpg 2932 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_3-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_3-1.png 2933 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_4-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_4-1.png 2934 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_91-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_91-1.png 2935 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_101-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_101-1.png 2936 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_3_sol-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_3_sol-1.png 2937 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/answer-9-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Answer-9-1.png 2938 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_19_tnt-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_19_tnt-1.png 2940 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_28-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_28-1.png 2941 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/question-37-1/ Thu, 12 Apr 2018 03:54:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Question-37-1.png 2942 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_3_sol1-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_3_sol1-1.png 2943 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_5_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_5_sol-1.png 2944 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_7_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_7_sol-1.png 2945 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_9_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_9_sol-1.png 2946 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_13_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_13_sol-1.png 2947 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_17_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_17_sol-1.png 2948 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_19_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_19_sol-1.png 2949 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_21_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_21_sol-1.png 2950 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_23_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_23_sol-1.png 2951 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_27_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_27_sol-1.png 2952 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_29_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_29_sol-1.png 2953 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_31_sol-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_31_sol-1.png 2954 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ex_35_sol-e1411670852714-1/ Thu, 12 Apr 2018 03:54:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ex_35_sol-e1411670852714-1.png 2955 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/answer-37-1/ Thu, 12 Apr 2018 03:54:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Answer-37-1.png 2956 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/fireworks_at_night_over_river-1/ Thu, 12 Apr 2018 03:54:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Fireworks_at_night_over_river-1.jpg 2960 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/olympus-digital-camera-4/ Thu, 12 Apr 2018 03:54:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/barbed-wire-1.jpg 2961 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/effective-collision-1/ Thu, 12 Apr 2018 03:54:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Effective-collision-1.jpg 2962 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/activation-energy-1-1/ Thu, 12 Apr 2018 03:54:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/activation-energy-1-1.jpg 2963 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/temperature-and-kinetic-energy-distribution-1/ Thu, 12 Apr 2018 03:54:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Temperature-and-Kinetic-Energy-Distribution-1.jpg 2964 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/instantaneous-and-initial-rates-figure-1/ Thu, 12 Apr 2018 03:54:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/instantaneous-and-initial-rates-figure-1.jpg 2966 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/first-order-reaction-plot-of-ln-conc-vs-time-ver-1-1/ Thu, 12 Apr 2018 03:54:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/First-Order-reaction-plot-of-ln-conc-vs-time-ver-1-1.jpg 2969 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/second-order-reaction-plot-1/ Thu, 12 Apr 2018 03:54:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Second-order-reaction-plot-1.jpg 2970 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/zero-order-reaction-plot-concentration-vs-time-1/ Thu, 12 Apr 2018 03:54:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Zero-order-reaction-plot-concentration-vs-time-1.jpg 2971 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/table-17-4-1-integrated-rate-law-summary-1/ Thu, 12 Apr 2018 03:54:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/table-17.4.1-integrated-rate-law-summary-1.jpg 2972 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/example-plots-1/ Thu, 12 Apr 2018 03:54:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Example-plots-1.jpg 2973 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/first-order-half-life-1/ Thu, 12 Apr 2018 03:54:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/First-Order-Half-life-1.jpg 2974 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/arrhenius2-239x300-1/ Thu, 12 Apr 2018 03:54:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Arrhenius2-239x300-1.jpg 2976 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/17-5-graphical-determination-of-ea-example-plot-1/ Thu, 12 Apr 2018 03:54:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/17.5-graphical-determination-of-Ea-example-plot-1.jpg 2977 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/multistep-potential-energy-diagram-large-1/ Thu, 12 Apr 2018 03:54:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Multistep-potential-energy-diagram-large-1.jpg 2979 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/rate-determining-hourglass-1/ Thu, 12 Apr 2018 03:54:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/rate-determining-hourglass-1.jpg 2980 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/but-2-ene-isomerization-large-1/ Thu, 12 Apr 2018 03:54:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/but-2-ene-isomerization-large-1.png 2982 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/catalyzed-vs-uncatalyzed-reaction-mechanism-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/catalyzed-vs-uncatalyzed-reaction-mechanism-1.jpg 2983 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/butene-catalysis-reaction-steps-ver-2-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/butene-catalysis-reaction-steps-ver-2-1.jpg 2984 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/heterogeneous-catalysis-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/heterogeneous-catalysis-1.jpg 2985 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/enzyme_mechanism_1-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Enzyme_mechanism_1-1.jpg 2986 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/end-of-ch-rxn-mech-question-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/end-of-ch-rxn-mech-question-1.jpg 2988 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/end-of-chapter-question-instantaneous-rate-graph-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/end-of-chapter-question-instantaneous-rate-graph-1.jpg 2989 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/end-of-ch-prob-15-1/ Thu, 12 Apr 2018 03:54:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/end-of-ch-prob-15-1.png 2990 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/double_flask_gas_atom_diagram-1/ Thu, 12 Apr 2018 03:54:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/double_flask_gas_atom_diagram-1.png 2995 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/boltzmann_age31-221x300-1/ Thu, 12 Apr 2018 03:54:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Boltzmann_age31-221x300-1.jpg 2996 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/standard_molar_entropy_table_-1/ Thu, 12 Apr 2018 03:54:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Standard_Molar_Entropy_Table_-1.png 2998 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1/ Thu, 12 Apr 2018 03:54:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1.jpg 2999 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/gibbs_josiah_willard-1/ Thu, 12 Apr 2018 03:54:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Gibbs_Josiah_Willard-1.jpg 3001 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/signs_of_enthalpy_and_entropy_terms_and_spontaneity-1/ Thu, 12 Apr 2018 03:54:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/signs_of_enthalpy_and_entropy_terms_and_spontaneity-1.png 3003 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/jessie-in-the-lab-1/ Thu, 12 Apr 2018 03:54:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Jessie-in-the-lab-1.jpg 3007 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/dr-ball-1/ Thu, 12 Apr 2018 03:54:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Dr-Ball-1.png 3008 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/999992a1a87b3d301b2407777ee5f950-1/ Thu, 12 Apr 2018 03:54:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/999992a1a87b3d301b2407777ee5f950-1.jpg 3010 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ae1d9db8c6cef79a4db550933c8c02cd-1/ Thu, 12 Apr 2018 03:54:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ae1d9db8c6cef79a4db550933c8c02cd-1.jpg 3011 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/pancake-1/ Thu, 12 Apr 2018 03:54:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/pancake-1.jpg 3018 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/2-mol-1/ Thu, 12 Apr 2018 03:54:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/2-mol-1.jpg 3019 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/q-1/ Thu, 12 Apr 2018 03:54:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/q-1.jpg 3020 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/backan/ Thu, 12 Apr 2018 03:54:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/backan.jpg 3021 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/baack/ Thu, 12 Apr 2018 03:54:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/baack.jpg 3022 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/so/ Thu, 12 Apr 2018 03:54:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/so.jpg 3023 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/go/ Thu, 12 Apr 2018 03:54:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/go.jpg 3024 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image1/ Thu, 12 Apr 2018 18:08:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image1.jpg 3033 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image78/ Thu, 12 Apr 2018 18:08:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image78.png 3034 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image86/ Thu, 12 Apr 2018 18:08:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image86.jpg 3035 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image87/ Thu, 12 Apr 2018 18:08:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image87.jpeg 3036 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image223/ Thu, 12 Apr 2018 18:08:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image223.jpg 3040 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image360/ Thu, 12 Apr 2018 18:08:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image360.jpg 3041 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image397/ Thu, 12 Apr 2018 18:08:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image397.jpeg 3042 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image461/ Thu, 12 Apr 2018 18:09:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image461.jpeg 3043 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image462/ Thu, 12 Apr 2018 18:09:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image462.png 3044 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image476/ Thu, 12 Apr 2018 18:09:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image476.jpeg 3045 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image478/ Thu, 12 Apr 2018 18:09:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image478.jpg 3046 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image521/ Thu, 12 Apr 2018 18:09:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image521.jpg 3053 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/image616/ Thu, 12 Apr 2018 18:09:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/image616.jpg 3064 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-1-chemistry-in-context/scientificmethod/ Fri, 13 Apr 2018 18:43:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ScientificMethod.jpg 3117 1252 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-1-chemistry-in-context/scienticficmethod/ Fri, 13 Apr 2018 18:45:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/ScienticficMethod.png 3118 1252 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/sigfigs/ Tue, 24 Apr 2018 21:42:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/SigFigs.jpg 3163 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/beaker1/ Tue, 24 Apr 2018 22:00:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Beaker1.jpg 3170 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/thermo1/ Tue, 24 Apr 2018 22:01:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Thermo1.jpg 3171 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/distillationapparatus/ Mon, 30 Apr 2018 17:34:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/DistillationApparatus.jpg 3242 3184 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/evaporation-method/ Mon, 30 Apr 2018 17:58:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Evaporation-Method.png 3243 3184 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/filtration/ Mon, 30 Apr 2018 18:07:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Filtration.png 3244 3184 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-physical-and-chemical-properties/phasechanges/ Mon, 30 Apr 2018 22:09:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/PhaseChanges.png 3250 1290 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/tlcplate/ Mon, 30 Apr 2018 22:42:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/TLCplate.png 3253 3184 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/tlc-in-developing-chamber/ Mon, 30 Apr 2018 22:43:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/TLC-in-Developing-Chamber.png 3254 3184 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/solid-liquid-gas/ Tue, 01 May 2018 02:35:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Solid-Liquid-Gas.jpg 3261 1266 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/langara-end-of-chapter-problems/scimet-blank/ Tue, 01 May 2018 03:00:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/SciMet-blank.png 3267 3263 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-3-atomic-structure-and-symbolism/screen-shot-2018-05-03-at-12-53-17-pm/ Thu, 03 May 2018 19:54:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-03-at-12.53.17-PM.png 3378 1316 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-3-atomic-structure-and-symbolism/screen-shot-2018-05-03-at-12-54-51-pm/ Thu, 03 May 2018 19:55:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-03-at-12.54.51-PM.png 3379 1316 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-5-the-periodic-table/names-of-families-on-the-periodic-table/ Mon, 07 May 2018 19:02:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Names-of-Families-on-the-Periodic-Table.png 3426 1344 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-6-molecular-and-ionic-compounds/classification-of-pure-substances/ Mon, 07 May 2018 19:48:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Classification-of-Pure-Substances.png 3432 1350 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-7-chemical-nomenclature/flowchart/ Fri, 11 May 2018 18:17:11 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Flowchart.jpg 3484 1353 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/moles1/ Tue, 15 May 2018 18:47:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Moles1.png 3561 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/mole2/ Tue, 15 May 2018 19:02:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Mole2.png 3563 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/mole3/ Tue, 15 May 2018 19:20:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Mole3.png 3565 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/mole4/ Tue, 15 May 2018 19:21:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Mole4.png 3566 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-17-06-am/ Fri, 18 May 2018 18:19:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.17.06-AM.png 3653 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-22-16-am/ Fri, 18 May 2018 18:23:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.22.16-AM.png 3654 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-17-42-am/ Fri, 18 May 2018 18:25:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.17.42-AM.png 3655 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-28-14-am/ Fri, 18 May 2018 18:29:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.28.14-AM.png 3656 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-18-36-am/ Fri, 18 May 2018 18:31:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.18.36-AM.png 3657 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-44-39-am/ Fri, 18 May 2018 18:45:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.44.39-AM.png 3660 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/screen-shot-2018-05-18-at-11-18-15-am/ Fri, 18 May 2018 19:06:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-11.18.15-AM.png 3664 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/development-of-quantum-theory/screen-shot-2018-05-18-at-1-22-32-pm/ Fri, 18 May 2018 20:24:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-18-at-1.22.32-PM.png 3693 3668 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/development-of-quantum-theory/screen-shot-2018-05-18-at-1-22-42-pm/ Fri, 18 May 2018 20:38:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-18-at-1.22.42-PM.png 3694 3668 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/development-of-quantum-theory/screen-shot-2018-05-18-at-1-22-54-pm/ Fri, 18 May 2018 20:40:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-18-at-1.22.54-PM.png 3695 3668 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/development-of-quantum-theory/3s-and-3p/ Fri, 18 May 2018 20:54:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/3s-and-3p.png 3697 3668 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/screen-shot-2018-05-18-at-8-01-09-pm/ Sat, 19 May 2018 03:03:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-8.01.09-PM.png 3732 2399 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/screen-shot-2018-05-18-at-8-09-06-pm/ Sat, 19 May 2018 03:10:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-8.09.06-PM.png 3734 2399 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/screen-shot-2018-05-18-at-8-09-12-pm/ Sat, 19 May 2018 03:11:07 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-8.09.12-PM.png 3735 2399 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/screen-shot-2018-05-18-at-8-09-19-pm/ Sat, 19 May 2018 03:11:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-8.09.19-PM.png 3736 2399 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/screen-shot-2018-05-18-at-8-09-26-pm/ Sat, 19 May 2018 03:12:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-18-at-8.09.26-PM.png 3737 2399 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/screen-shot-2018-05-22-at-12-24-18-pm/ Tue, 22 May 2018 19:25:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-22-at-12.24.18-PM.png 3795 3792 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/screen-shot-2018-05-22-at-12-24-47-pm/ Tue, 22 May 2018 19:26:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-22-at-12.24.47-PM.png 3796 3792 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/screen-shot-2018-05-22-at-12-26-52-pm/ Tue, 22 May 2018 19:27:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-22-at-12.26.52-PM.png 3797 3792 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/screen-shot-2018-05-22-at-12-28-07-pm/ Tue, 22 May 2018 19:28:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-22-at-12.28.07-PM.png 3798 3792 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/screen-shot-2018-05-22-at-12-29-12-pm/ Tue, 22 May 2018 19:29:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-22-at-12.29.12-PM.png 3799 3792 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/screen-shot-2018-05-22-at-12-30-01-pm/ Tue, 22 May 2018 19:30:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-22-at-12.30.01-PM.png 3800 3792 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-7/graphite-233436/ Fri, 25 May 2018 17:42:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Graphite-233436.jpg 3831 1540 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-7/627px-diamond_and_graphite/ Fri, 25 May 2018 18:17:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/627px-Diamond_and_graphite.jpg 3833 1540 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-7/800px-eight_allotropes_of_carbon/ Fri, 25 May 2018 18:27:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/800px-Eight_Allotropes_of_Carbon.png 3834 1540 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-1-00-48-pm/ Fri, 25 May 2018 20:02:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-1.00.48-PM.png 3838 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/screen-shot-2018-05-25-at-1-39-00-pm/ Fri, 25 May 2018 20:39:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-1.39.00-PM.png 3851 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/screen-shot-2018-05-25-at-1-44-51-pm/ Fri, 25 May 2018 20:45:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-1.44.51-PM.png 3853 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/funcgroups/ Fri, 25 May 2018 20:59:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/FuncGroups.jpg 3858 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/screen-shot-2018-05-25-at-3-01-53-pm/ Fri, 25 May 2018 22:02:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-3.01.53-PM.png 3861 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/screen-shot-2018-05-25-at-3-04-14-pm/ Fri, 25 May 2018 22:10:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-3.04.14-PM.png 3862 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-21-21-pm/ Sat, 26 May 2018 01:27:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.21.21-PM.png 3872 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-21-49-pm/ Sat, 26 May 2018 01:30:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.21.49-PM.png 3873 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-41-31-pm/ Sat, 26 May 2018 01:43:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.41.31-PM.png 3876 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-41-51-pm/ Sat, 26 May 2018 01:44:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.41.51-PM.png 3877 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-46-45-pm/ Sat, 26 May 2018 01:49:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.46.45-PM.png 3878 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-47-00-pm/ Sat, 26 May 2018 01:50:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.47.00-PM.png 3879 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-54-07-pm/ Sat, 26 May 2018 01:56:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.54.07-PM.png 3881 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-54-26-pm/ Sat, 26 May 2018 01:57:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.54.26-PM.png 3882 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-6-54-45-pm/ Sat, 26 May 2018 01:59:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-6.54.45-PM.png 3883 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-7-02-32-pm/ Sat, 26 May 2018 02:04:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-7.02.32-PM.png 3885 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-7-03-06-pm/ Sat, 26 May 2018 02:04:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-7.03.06-PM.png 3886 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-7-06-33-pm/ Sat, 26 May 2018 02:13:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-7.06.33-PM.png 3887 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-7-07-02-pm/ Sat, 26 May 2018 02:14:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-7.07.02-PM.png 3888 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-25-at-7-07-10-pm/ Sat, 26 May 2018 03:20:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-25-at-7.07.10-PM.png 3890 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-29-at-11-22-49-am/ Tue, 29 May 2018 18:25:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-29-at-11.22.49-AM.png 3893 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-29-at-11-22-59-am/ Tue, 29 May 2018 18:28:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-29-at-11.22.59-AM.png 3894 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-29-at-11-23-10-am/ Tue, 29 May 2018 18:31:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-29-at-11.23.10-AM.png 3895 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-29-at-11-23-18-am/ Tue, 29 May 2018 18:33:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-29-at-11.23.18-AM.png 3896 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/screen-shot-2018-05-29-at-11-23-46-am/ Tue, 29 May 2018 18:35:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-29-at-11.23.46-AM.png 3897 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-1-hydrocarbons/screen-shot-2018-05-29-at-1-51-20-pm/ Tue, 29 May 2018 20:52:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-29-at-1.51.20-PM.png 3921 1900 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/fattyacid/ Wed, 30 May 2018 18:01:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/FattyAcid.jpg 3952 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/acetone-acetic-acid-isopropanol/ Wed, 30 May 2018 18:05:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Acetone-Acetic-Acid-Isopropanol.jpg 3953 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/l-ascorbic-acid/ Wed, 30 May 2018 18:10:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/L-Ascorbic-Acid.jpg 3955 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/ex1/ Wed, 30 May 2018 18:15:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Ex1.jpg 3956 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/ex2/ Wed, 30 May 2018 18:16:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Ex2.jpg 3957 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/acetaminophen/ Wed, 30 May 2018 18:19:58 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Acetaminophen.jpg 3958 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/acetaminophen-2/ Wed, 30 May 2018 18:23:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Acetaminophen-2.jpg 3959 3849 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-4-amines-and-amides/screen-shot-2018-05-30-at-1-06-47-pm/ Wed, 30 May 2018 20:07:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-1.06.47-PM.png 3976 1970 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-4-amines-and-amides/screen-shot-2018-05-30-at-1-06-56-pm/ Wed, 30 May 2018 20:07:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-1.06.56-PM.png 3977 1970 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-4-amines-and-amides/screen-shot-2018-05-30-at-1-07-01-pm/ Wed, 30 May 2018 20:08:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-1.07.01-PM.png 3978 1970 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-3-aldehydes-ketones-carboxylic-acids-and-esters/screen-shot-2018-05-30-at-1-49-08-pm/ Wed, 30 May 2018 20:50:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-1.49.08-PM.png 3993 1952 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-3-aldehydes-ketones-carboxylic-acids-and-esters/screen-shot-2018-05-30-at-2-36-09-pm/ Wed, 30 May 2018 21:38:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-2.36.09-PM.png 4001 1952 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-3-aldehydes-ketones-carboxylic-acids-and-esters/screen-shot-2018-05-30-at-2-36-32-pm/ Wed, 30 May 2018 21:38:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-2.36.32-PM.png 4002 1952 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-3-aldehydes-ketones-carboxylic-acids-and-esters/screen-shot-2018-05-30-at-2-37-09-pm/ Wed, 30 May 2018 21:39:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-30-at-2.37.09-PM.png 4003 1952 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-35-23-pm/ Thu, 31 May 2018 04:39:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.35.23-PM.png 4042 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-38-00-pm/ Thu, 31 May 2018 04:40:18 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.38.00-PM.png 4043 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-38-08-pm/ Thu, 31 May 2018 04:40:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.38.08-PM.png 4044 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-38-30-pm/ Thu, 31 May 2018 04:41:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.38.30-PM.png 4045 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-38-53-pm/ Thu, 31 May 2018 04:41:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.38.53-PM.png 4046 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-43-23-pm/ Thu, 31 May 2018 04:46:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.43.23-PM.png 4048 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-05-30-at-9-49-15-pm/ Thu, 31 May 2018 04:50:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-30-at-9.49.15-PM.png 4050 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-4-chemical-reactions-and-equations/screen-shot-2018-05-22-at-12-30-01-pm-2/ Mon, 04 Jun 2018 23:41:44 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-05-22-at-12.30.01-PM.png 4164 2162 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-4-chemical-reactions-and-equations/equation-6-1/ Mon, 04 Jun 2018 23:42:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Equation-6.1.png 4165 2162 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-06-05-at-10-43-45-am/ Tue, 05 Jun 2018 17:45:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-06-05-at-10.43.45-AM.png 4181 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/screen-shot-2018-06-05-at-10-43-59-am/ Tue, 05 Jun 2018 17:46:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-06-05-at-10.43.59-AM.png 4182 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-2-classifying-chemical-reactions/screen-shot-2018-06-05-at-10-19-29-am/ Tue, 05 Jun 2018 19:39:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-05-at-10.19.29-AM.png 4190 1419 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-2-classifying-chemical-reactions/screen-shot-2018-06-05-at-12-40-15-pm/ Tue, 05 Jun 2018 19:40:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-05-at-12.40.15-PM.png 4191 1419 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-a-the-periodic-table/periodictablesm/ Fri, 08 Jun 2018 19:19:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/PeriodicTableSM.jpg 4446 2021 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/screen-shot-2018-05-07-at-12-31-40-pm/ Sat, 09 Jun 2018 01:05:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-05-07-at-12.31.40-PM.png 4499 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/screen-shot-2018-06-08-at-6-04-26-pm/ Sat, 09 Jun 2018 01:06:12 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-06-08-at-6.04.26-PM.png 4500 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/screen-shot-2018-06-08-at-6-08-26-pm/ Sat, 09 Jun 2018 01:09:53 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-06-08-at-6.08.26-PM.png 4502 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/screen-shot-2018-06-08-at-6-08-37-pm/ Sat, 09 Jun 2018 01:12:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-06-08-at-6.08.37-PM.png 4503 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/screen-shot-2018-06-08-at-6-08-45-pm/ Sat, 09 Jun 2018 01:13:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Screen-Shot-2018-06-08-at-6.08.45-PM.png 4504 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/screen-shot-2018-06-12-at-11-37-57-am/ Tue, 12 Jun 2018 18:38:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-12-at-11.37.57-AM.png 4517 1388 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/screen-shot-2018-06-14-at-1-17-30-pm/ Thu, 14 Jun 2018 20:18:16 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-14-at-1.17.30-PM.png 4632 3184 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/screen-shot-2018-06-14-at-1-50-03-pm/ Thu, 14 Jun 2018 20:50:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-14-at-1.50.03-PM.png 4640 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/screen-shot-2018-06-14-at-1-52-07-pm/ Thu, 14 Jun 2018 20:52:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-14-at-1.52.07-PM.png 4641 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/screen-shot-2018-06-14-at-1-53-58-pm/ Thu, 14 Jun 2018 20:54:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-14-at-1.53.58-PM.png 4642 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/screen-shot-2018-06-14-at-1-54-47-pm/ Thu, 14 Jun 2018 20:55:15 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Screen-Shot-2018-06-14-at-1.54.47-PM.png 4643 1279 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/openstax/ Fri, 22 Jun 2018 15:48:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/OPENSTAX.jpg 4807 1266 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-7-chemical-nomenclature/nomenclature-flowchart/ Tue, 26 Jun 2018 01:39:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Nomenclature-Flowchart.pdf 4829 1353 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-7-chemical-nomenclature/flowchart2/ Tue, 26 Jun 2018 01:49:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Flowchart2.jpg 4830 1353 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/?attachment_id=4838 Fri, 06 Jul 2018 17:37:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?attachment_id=4838 4838 16 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/light/wavelength/ Mon, 13 May 2019 18:09:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Wavelength.png 4846 2393 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/emission/ Mon, 13 May 2019 19:10:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Emission.png 4848 2390 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/sarin/ Mon, 13 May 2019 22:00:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Sarin.png 4854 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/sf6/ Mon, 13 May 2019 22:06:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/SF6.png 4855 3495 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-1-writing-and-balancing-chemical-equations/equation/ Tue, 14 May 2019 17:06:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Equation.png 4866 1412 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-1-writing-and-balancing-chemical-equations/equation-2/ Tue, 14 May 2019 17:12:19 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/04/Equation-1.png 4870 1412 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/dna-part1/ Tue, 14 May 2019 21:14:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/DNA-part1.png 4903 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/dna-part2/ Tue, 14 May 2019 21:14:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/DNA-part2.png 4904 3836 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=5 Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/2018/04/10/chapter-1/ 5 3 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-1-chemistry-in-context/ Thu, 12 Apr 2018 02:50:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-1-chemistry-in-context/

Learning Objectives

By the end of this module, you will be able to:

• Outline the historical development of chemistry
• Provide examples of the importance of chemistry in everyday life
• Describe the scientific method
• Differentiate among hypotheses, theories, and laws
• Provide examples illustrating macroscopic, microscopic, and symbolic domains

Throughout human history, people have tried to convert matter into more useful forms. Our Stone Age ancestors chipped pieces of flint into useful tools and carved wood into statues and toys. These endeavors involved changing the shape of a substance without changing the substance itself. But as our knowledge increased, humans began to change the composition of the substances as well—clay was converted into pottery, hides were cured to make garments, copper ores were transformed into copper tools and weapons, and grain was made into bread.

Humans began to practice chemistry when they learned to control fire and use it to cook, make pottery, and smelt metals. Subsequently, they began to separate and use specific components of matter. A variety of drugs such as aloe, myrrh, and opium were isolated from plants. Dyes, such as indigo and Tyrian purple, were extracted from plant and animal matter. Metals were combined to form alloys—for example, copper and tin were mixed together to make bronze—and more elaborate smelting techniques produced iron. Alkalis were extracted from ashes, and soaps were prepared by combining these alkalis with fats. Alcohol was produced by fermentation and purified by distillation.

Attempts to understand the behavior of matter extend back for more than 2500 years. As early as the sixth century BC, Greek philosophers discussed a system in which water was the basis of all things. You may have heard of the Greek postulate that matter consists of four elements: earth, air, fire, and water. Subsequently, an amalgamation of chemical technologies and philosophical speculations were spread from Egypt, China, and the eastern Mediterranean by alchemists, who endeavored to transform “base metals” such as lead into “noble metals” like gold, and to create elixirs to cure disease and extend life (Figure 1).

[caption id="attachment_1248" align="aligncenter" width="500"] Figure 1. This portrayal shows an alchemist’s workshop circa 1580. Although alchemy made some useful contributions to how to manipulate matter, it was not scientific by modern standards. (credit: Chemical Heritage Foundation)[/caption]

From alchemy came the historical progressions that led to modern chemistry: the isolation of drugs from natural sources, metallurgy, and the dye industry. Today, chemistry continues to deepen our understanding and improve our ability to harness and control the behavior of matter. This effort has been so successful that many people do not realize either the central position of chemistry among the sciences or the importance and universality of chemistry in daily life.

Chemistry: The Central Science

Chemistry is sometimes referred to as “the central science” due to its interconnectedness with a vast array of other STEM disciplines (STEM stands for areas of study in the science, technology, engineering, and math fields). Chemistry and the language of chemists play vital roles in biology, medicine, materials science, forensics, environmental science, and many other fields (Figure 2). The basic principles of physics are essential for understanding many aspects of chemistry, and there is extensive overlap between many subdisciplines within the two fields, such as chemical physics and nuclear chemistry. Mathematics, computer science, and information theory provide important tools that help us calculate, interpret, describe, and generally make sense of the chemical world. Biology and chemistry converge in biochemistry, which is crucial to understanding the many complex factors and processes that keep living organisms (such as us) alive. Chemical engineering, materials science, and nanotechnology combine chemical principles and empirical findings to produce useful substances, ranging from gasoline to fabrics to electronics. Agriculture, food science, veterinary science, and brewing and wine making help provide sustenance in the form of food and drink to the world’s population. Medicine, pharmacology, biotechnology, and botany identify and produce substances that help keep us healthy. Environmental science, geology, oceanography, and atmospheric science incorporate many chemical ideas to help us better understand and protect our physical world. Chemical ideas are used to help understand the universe in astronomy and cosmology.

[caption id="" align="aligncenter" width="1300"] Figure 2. Knowledge of chemistry is central to understanding a wide range of scientific disciplines. This diagram shows just some of the interrelationships between chemistry and other fields.[/caption]

What are some changes in matter that are essential to daily life? Digesting and assimilating food, synthesizing polymers that are used to make clothing, containers, cookware, and credit cards, and refining crude oil into gasoline and other products are just a few examples. As you proceed through this course, you will discover many different examples of changes in the composition and structure of matter, how to classify these changes and how they occurred, their causes, the changes in energy that accompany them, and the principles and laws involved. As you learn about these things, you will be learning chemistry, the study of the composition, properties, and interactions of matter. The practice of chemistry is not limited to chemistry books or laboratories: It happens whenever someone is involved in changes in matter or in conditions that may lead to such changes.

The Scientific Method

It is important to understand that chemistry as a science is a procedure as well as sets of facts. The ancient Greeks developed some powerful methods of acquiring knowledge, particularly in mathematics, by using the process of deduction. Deduction starts with certain basic assumptions or premises and then certain conclusions must logically follow. But deduction alone is NOT enough for obtaining scientific knowledge!  The scientific method originated in the seventeenth century with such people as Galileo, Francis Bacon, Robert Boyle and Isaac Newton. The key to the method was to make NO initial assumptions.

Chemistry is a science based on observation and experimentation. Doing chemistry involves attempting to answer questions and explain observations in terms of the laws and theories of chemistry, using procedures that are accepted by the scientific community. There is no single route to answering a question or explaining an observation, but there is an aspect common to every approach: Each uses knowledge based on experiments that can be reproduced to verify the results. Some routes involve a hypothesis, a tentative explanation of observations that acts as a guide for gathering and checking information. We test a hypothesis by experimentation, calculation, and/or comparison with the experiments of others and then refine it as needed.

Some hypotheses are attempts to explain the behavior that is summarized in laws. The laws of science summarize a vast number of experimental observations, and describe or predict some facet of the natural world. If such a hypothesis turns out to be capable of explaining a large body of experimental data, it can reach the status of a theory. Scientific theories are well-substantiated, comprehensive, testable explanations of particular aspects of nature. Theories are accepted because they provide satisfactory explanations, but they can be modified if new data become available. The path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory, is called the scientific method (Figure 3).

[caption id="attachment_3118" align="aligncenter" width="511"] Figure 3. The scientific method follows a process similar to the one shown in this diagram. All the key components are shown, in roughly the right order. Scientific progress is seldom neat and clean: It requires open inquiry and the reworking of questions and ideas in response to findings.[/caption]

Finally, understand that science can be either qualitative or quantitative. Qualitative observations are descriptive, they involve a description of the quality of an object. For example, physical properties are generally qualitative observations: sulfur is yellow, your math book is heavy, or that statue is pretty. Quantitative observations represent the specific amount of something; it means knowing how much of something is present, usually by counting or measuring it. Therefore, quantitative observations involve a quantity or number and units.As such, some quantitative observation would include 25 students in a class, 650 pages in a book, or a velocity of 66 miles per hour. Quantitative expressions are very important in science; they are also very important in chemistry.

Example 1

Identify each statement as either a qualitative description or a quantitative description.

a) Gold metal is yellow.

b) A ream of paper has 500 sheets in it.

c) The weather outside is snowy.

d) The temperature outside is 24 degrees Fahrenheit.

 

Solution

a) Because we are describing a physical property of gold, this statement is qualitative.

b) This statement mentions a specific amount, so it is quantitative.

c) The word snowy is a description of how the day is; therefore, it is a qualitative statement.

d) In this case, the weather is described with a specific quantity—the temperature. Therefore, it is quantitative.

 

Test Yourself

Are these qualitative or quantitative statements?

a) Roses are red, and violets are blue.

b) Four score and seven years ago….

 

Answers

a) qualitative          b) quantitative

Steps of the scientific method:

1. Clearly state the problem as an observation with as much information as possible.  Observations should never explain why.
2. Propose a hypothesis.
3. Set up and perform a controlled experiment: “controlled” means you only alter one thing at a time.
4. State your conclusions.
5. Repeat as necessary.

Example 2

MULTIPLE CHOICE QUESTION: Which one of the following is NOT a part of the process of the scientific method? A) Theory          B) Hypothesis   C) Opinion    D) Basic assumptions  

Solution

Opinion and basic assumptions are not part of the scientic method

 

Test Yourself

MULTIPLE CHOICE QUESTION:  A student notices that the evening sky is usually red.  She wants to find out why.  Which of the following would be the next step based on the scientific method? A) She publishes an article on sky conditions. B) She starts to experiment to see how dust affects light rays. C) She thinks that dust in the sky makes the sky red. D) Other scientists confirm her experiments.  

Answer

C

Once MANY experiments confirm various hypotheses about something, it may be developed into a theory. A theory can be used to make further predictions about natural phenomena.  Only negative evidence is conclusive, therefore theories evolve and are changed over time. We keep testing theories by using them to make predictions.

The Domains of Chemistry

Chemists study and describe the behavior of matter and energy in three different domains: macroscopic, microscopic, and symbolic. These domains provide different ways of considering and describing chemical behavior.

Macro is a Greek word that means “large.” The macroscopic domain is familiar to us: It is the realm of everyday things that are large enough to be sensed directly by human sight or touch. In daily life, this includes the food you eat and the breeze you feel on your face. The macroscopic domain includes everyday and laboratory chemistry, where we observe and measure physical and chemical properties, or changes such as density, solubility, and flammability.

The microscopic domain of chemistry is almost always visited in the imagination. Micro also comes from Greek and means “small.” Some aspects of the microscopic domains are visible through a microscope, such as a magnified image of graphite or bacteria. Viruses, for instance, are too small to be seen with the naked eye, but when we’re suffering from a cold, we’re reminded of how real they are.

However, most of the subjects in the microscopic domain of chemistry—such as atoms and molecules—are too small to be seen even with standard microscopes and often must be pictured in the mind. Other components of the microscopic domain include ions and electrons, protons and neutrons, and chemical bonds, each of which is far too small to see. This domain includes the individual metal atoms in a wire, the ions that compose a salt crystal, the changes in individual molecules that result in a color change, the conversion of nutrient molecules into tissue and energy, and the evolution of heat as bonds that hold atoms together are created.

The symbolic domain contains the specialized language used to represent components of the macroscopic and microscopic domains. Chemical symbols (such as those used in the periodic table), chemical formulas, and chemical equations are part of the symbolic domain, as are graphs and drawings. We can also consider calculations as part of the symbolic domain. These symbols play an important role in chemistry because they help interpret the behavior of the macroscopic domain in terms of the components of the microscopic domain. One of the challenges for students learning chemistry is recognizing that the same symbols can represent different things in the macroscopic and microscopic domains, and one of the features that makes chemistry fascinating is the use of a domain that must be imagined to explain behavior in a domain that can be observed.

A helpful way to understand the three domains is via the essential and ubiquitous substance of water. That water is a liquid at moderate temperatures, will freeze to form a solid at lower temperatures, and boil to form a gas at higher temperatures (Figure 4) are macroscopic observations. But some properties of water fall into the microscopic domain—what we cannot observe with the naked eye. The description of water as comprised of two hydrogen atoms and one oxygen atom, and the explanation of freezing and boiling in terms of attractions between these molecules, is within the microscopic arena. The formula H₂O, which can describe water at either the macroscopic or microscopic levels, is an example of the symbolic domain. The abbreviations (g) for gas, (s) for solid, and (l) for liquid are also symbolic.

[caption id="" align="aligncenter" width="1300"] Figure 4. (a) Moisture in the air, icebergs, and the ocean represent water in the macroscopic domain. (b) At the molecular level (microscopic domain), gas molecules are far apart and disorganized, solid water molecules are close together and organized, and liquid molecules are close together and disorganized. (c) The formula H₂O symbolizes water, and (g), (s), and (l) symbolize its phases. Note that clouds are actually comprised of either very small liquid water droplets or solid water crystals; gaseous water in our atmosphere is not visible to the naked eye, although it may be sensed as humidity. (credit a: modification of work by “Gorkaazk”/Wikimedia Commons)[/caption]

Key Concepts and Summary

Chemistry deals with the composition, structure, and properties of matter, and the ways by which various forms of matter may be interconverted. Thus, it occupies a central place in the study and practice of science and technology. Chemists use the scientific method to perform experiments, pose hypotheses, and formulate laws and develop theories, so that they can better understand the behavior of the natural world. To do so, they operate in the macroscopic, microscopic, and symbolic domains. Chemists measure, analyze, purify, and synthesize a wide variety of substances that are important to our lives.

Exercises

1. Describe the scientific method.

2. Why do scientists need to perform experiments?

3. What is the scientific definition of a law? How does it differ from the everyday definition of a law?

4. Explain how you could experimentally determine whether the outside temperature is higher or lower than 0 °C (32 °F) without using a thermometer.

5. Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.

a) The pressure of a sample of gas is directly proportional to the temperature of the gas.

b) Matter consists of tiny particles that can combine in specific ratios to form substances with specific properties.

c) At a higher temperature, solids (such as salt or sugar) will dissolve better in water.

6. Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For those in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.

a) A certain molecule contains one H atom and one Cl atom.

b) Copper wire has a density of about 8 g/cm³.

c) The bottle contains 15 grams of Ni powder.

d) A sulfur molecule is composed of eight sulfur atoms.

7. The amount of heat required to melt 2 lbs of ice is twice the amount of heat required to melt 1 lb of ice. Is this observation a macroscopic or microscopic description of chemical behavior? Explain your answer. 8. Which of these statements are qualitative observations?

a)  The Titanic was the largest passenger ship build at that time. b)  The population of the United States is about 306,000,000 people. c)  The peak of Mount Everest is 29,035 feet above sea level.

  Answers 1. Simply stated, the scientific method includes three steps: (1) stating a hypothesis, (2) testing the hypothesis, and (3) refining the hypothesis. 2. Scientists perform experiments to test their hypotheses because sometimes the nature of natural universe is not obvious. 3. A scientific law is a specific statement that is thought to be never violated by the entire natural universe. Everyday laws are arbitrary limits that society puts on its members.

4. Place a glass of water outside. It will freeze if the temperature is below 0 °C.

5. a) law (states a consistently observed phenomenon, can be used for prediction);   b) theory (a widely accepted explanation of the behavior of matter);    c) hypothesis (a tentative explanation, can be investigated by experimentation)

6. a) symbolic, microscopic;   b) macroscopic;   c) symbolic, macroscopic;   d) microscopic

7. Macroscopic. The heat required is determined from macroscopic properties.

8. a) yes    b) no    c) no

Glossary

chemistry: study of is the study of matter and its properties, the changes that matter undergoes, and the energy associated with these changes. hypothesis: tentative explanation of observations that acts as a guide for gathering and checking information law: statement that summarizes a vast number of experimental observations, and describes or predicts some aspect of the natural world macroscopic domain: realm of everyday things that are large enough to sense directly by human sight and touch microscopic domain: realm of things that are much too small to be sensed directly qualitative observations: they are descriptions of the quality of an object. quantitative observations: they represent the specific amount of something; they involve a quantity or number and units. scientific method: path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory symbolic domain: specialized language used to represent components of the macroscopic and microscopic domains, such as chemical symbols, chemical formulas, chemical equations, graphs, drawings, and calculations theory: well-substantiated, comprehensive, testable explanation of a particular aspect of nature

]]> 1252 1247 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/ Thu, 12 Apr 2018 02:50:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-2-phases-and-classification-of-matter/

Learning Objectives

By the end of this section, you will be able to:

• Describe the basic properties of each physical state of matter: solid, liquid, and gas
• Define and give examples of atoms and molecules
• Classify matter as an element, compound, homogeneous mixture, or heterogeneous mixture with regard to its physical state and composition
• Distinguish between mass and weight
• Apply the law of conservation of matter

Matter is defined as anything that occupies space and has mass, and it is all around us. Solids and liquids are more obviously matter: We can see that they take up space, and their weight tells us that they have mass. Gases are also matter; if gases did not take up space, a balloon would stay collapsed rather than inflate when filled with gas.

Solids, liquids, and gases are the three states of matter commonly found on earth (Figure 1). A solid is rigid and possesses a definite shape. A liquid flows and takes the shape of a container, except that it forms a flat or slightly curved upper surface when acted upon by gravity. (In zero gravity, liquids assume a spherical shape.) Both liquid and solid samples have volumes that are very nearly independent of pressure. A gas takes both the shape and volume of its container.

[caption id="" align="aligncenter" width="1300"] Figure 1. The three most common states or phases of matter are solid, liquid, and gas.[/caption]

On the molecular level: Solid: Atoms or molecules are in close contact, often in a highly organized arrangement. Solids have the strongest forces holding atoms or molecules together. Liquid: Atoms or molecules are in close proximity (although generally not as close as solids). Gas: Atoms or molecules are far apart and move freely (very few forces between them). [caption id="attachment_3261" align="aligncenter" width="501"] Figure 2. A molecular view of a solid, a liquid, and a gas.[/caption] A fourth state of matter, plasma, occurs naturally in the interiors of stars. A plasma is a gaseous state of matter that contains appreciable numbers of electrically charged particles (Figure 3). The presence of these charged particles imparts unique properties to plasmas that justify their classification as a state of matter distinct from gases. In addition to stars, plasmas are found in some other high-temperature environments (both natural and man-made), such as lightning strikes, certain television screens, and specialized analytical instruments used to detect trace amounts of metals.

[caption id="attachment_1254" align="aligncenter" width="300"] Figure 3. A plasma torch can be used to cut metal. (credit: “Hypertherm”/Wikimedia Commons)[/caption]

In a tiny cell in a plasma television, the plasma emits ultraviolet light, which in turn causes the display at that location to appear a specific color. The composite of these tiny dots of color makes up the image that you see. Watch this video to learn more about plasma and the places you encounter it.

Some samples of matter appear to have properties of solids, liquids, and/or gases at the same time. This can occur when the sample is composed of many small pieces. For example, we can pour sand as if it were a liquid because it is composed of many small grains of solid sand. Matter can also have properties of more than one state when it is a mixture, such as with clouds. Clouds appear to behave somewhat like gases, but they are actually mixtures of air (gas) and tiny particles of water (liquid or solid).

The mass of an object is a measure of the amount of matter in it. One way to measure an object’s mass is to measure the force it takes to accelerate the object. It takes much more force to accelerate a car than a bicycle because the car has much more mass. A more common way to determine the mass of an object is to use a balance to compare its mass with a standard mass.

Although weight is related to mass, it is not the same thing. Weight refers to the force that gravity exerts on an object. This force is directly proportional to the mass of the object. The weight of an object changes as the force of gravity changes, but its mass does not. An astronaut’s mass does not change just because she goes to the moon. But her weight on the moon is only one-sixth her earth-bound weight because the moon’s gravity is only one-sixth that of the earth’s. She may feel “weightless” during her trip when she experiences negligible external forces (gravitational or any other), although she is, of course, never “massless.”

The law of conservation of matter summarizes many scientific observations about matter: It states that there is no detectable change in the total quantity of matter present when matter converts from one type to another (a chemical change) or changes among solid, liquid, or gaseous states (a physical change). Brewing beer and the operation of batteries provide examples of the conservation of matter (Figure 4). During the brewing of beer, the ingredients (water, yeast, grains, malt, hops, and sugar) are converted into beer (water, alcohol, carbonation, and flavoring substances) with no actual loss of substance. This is most clearly seen during the bottling process, when glucose turns into ethanol and carbon dioxide, and the total mass of the substances does not change. This can also be seen in a lead-acid car battery: The original substances (lead, lead oxide, and sulfuric acid), which are capable of producing electricity, are changed into other substances (lead sulfate and water) that do not produce electricity, with no change in the actual amount of matter.

[caption id="" align="aligncenter" width="1300"] Figure 4. (a) The mass of beer precursor materials is the same as the mass of beer produced: Sugar has become alcohol and carbonation. (b) The mass of the lead, lead oxide plates, and sulfuric acid that goes into the production of electricity is exactly equal to the mass of lead sulfate and water that is formed.[/caption]

Although this conservation law holds true for all conversions of matter, convincing examples are few and far between because, outside of the controlled conditions in a laboratory, we seldom collect all of the material that is produced during a particular conversion. For example, when you eat, digest, and assimilate food, all of the matter in the original food is preserved. But because some of the matter is incorporated into your body, and much is excreted as various types of waste, it is challenging to verify by measurement.

Atoms and Molecules

An atom is the smallest particle of an element that has the properties of that element and can enter into a chemical combination. Consider the element gold, for example. Imagine cutting a gold nugget in half, then cutting one of the halves in half, and repeating this process until a piece of gold remained that was so small that it could not be cut in half (regardless of how tiny your knife may be). This minimally sized piece of gold is an atom (from the Greek atomos, meaning “indivisible”) (Figure 5). This atom would no longer be gold if it were divided any further.

[caption id="" align="aligncenter" width="1300"] Figure 5. (a) This photograph shows a gold nugget. (b) A scanning-tunneling microscope (STM) can generate views of the surfaces of solids, such as this image of a gold crystal. Each sphere represents one gold atom. (credit a: modification of work by United States Geological Survey; credit b: modification of work by “Erwinrossen”/Wikimedia Commons)[/caption]

The first suggestion that matter is composed of atoms is attributed to the Greek philosophers Leucippus and Democritus, who developed their ideas in the 5th century BCE. However, it was not until the early nineteenth century that John Dalton (1766–1844), a British schoolteacher with a keen interest in science, supported this hypothesis with quantitative measurements. Since that time, repeated experiments have confirmed many aspects of this hypothesis, and it has become one of the central theories of chemistry. Other aspects of Dalton’s atomic theory are still used but with minor revisions (details of Dalton’s theory are provided in the chapter on atoms and molecules).

An atom is so small that its size is difficult to imagine. One of the smallest things we can see with our unaided eye is a single thread of a spider web: These strands are about 1/10,000 of a centimeter (0.0001 cm) in diameter. Although the cross-section of one strand is almost impossible to see without a microscope, it is huge on an atomic scale. A single carbon atom in the web has a diameter of about 0.000000015 centimeter, and it would take about 7000 carbon atoms to span the diameter of the strand. To put this in perspective, if a carbon atom were the size of a dime, the cross-section of one strand would be larger than a football field, which would require about 150 million carbon atom “dimes” to cover it. (Figure 6) shows increasingly close microscopic and atomic-level views of ordinary cotton.

[caption id="" align="aligncenter" width="1301"] Figure 6. These images provide an increasingly closer view: (a) a cotton boll, (b) a single cotton fiber viewed under an optical microscope (magnified 40 times), (c) an image of a cotton fiber obtained with an electron microscope (much higher magnification than with the optical microscope); and (d and e) atomic-level models of the fiber (spheres of different colors represent atoms of different elements). (credit c: modification of work by “Featheredtar”/Wikimedia Commons)[/caption]

An atom is so light that its mass is also difficult to imagine. A billion lead atoms (1,000,000,000 atoms) weigh about 3 × 10⁻¹³ grams, a mass that is far too light to be weighed on even the world’s most sensitive balances. It would require over 300,000,000,000,000 lead atoms (300 trillion, or 3 × 10¹⁴) to be weighed, and they would weigh only 0.0000001 gram.

It is rare to find collections of individual atoms. Only a few elements, such as the gases helium, neon, and argon, consist of a collection of individual atoms that move about independently of one another. Other elements, such as the gases hydrogen, nitrogen, oxygen, and chlorine, are composed of units that consist of pairs of atoms (Figure 7). One form of the element phosphorus consists of units composed of four phosphorus atoms. The element sulfur exists in various forms, one of which consists of units composed of eight sulfur atoms. These units are called molecules. A molecule consists of two or more atoms joined by strong forces called chemical bonds. The atoms in a molecule move around as a unit, much like the cans of soda in a six-pack or a bunch of keys joined together on a single key ring. A molecule may consist of two or more identical atoms, as in the molecules found in the elements hydrogen, oxygen, and sulfur, or it may consist of two or more different atoms, as in the molecules found in water. Each water molecule is a unit that contains two hydrogen atoms and one oxygen atom. Each glucose molecule is a unit that contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Like atoms, molecules are incredibly small and light. If an ordinary glass of water were enlarged to the size of the earth, the water molecules inside it would be about the size of golf balls.

[caption id="" align="aligncenter" width="1300"] Figure 7. The elements hydrogen, oxygen, phosphorus, and sulfur form molecules consisting of two or more atoms of the same element. The compounds water, carbon dioxide, and glucose consist of combinations of atoms of different elements.[/caption]

Classifying Matter

We can classify matter into several categories. Two broad categories are mixtures and pure substances. A pure substance has a constant composition. All specimens of a pure substance have exactly the same makeup and properties. Any sample of sucrose (table sugar) consists of 42.1% carbon, 6.5% hydrogen, and 51.4% oxygen by mass. Any sample of sucrose also has the same physical properties, such as melting point, color, and sweetness, regardless of the source from which it is isolated.

We can divide pure substances into two classes: elements and compounds. Pure substances that cannot be broken down into simpler substances by chemical changes are called elements. Iron, silver, gold, aluminum, sulfur, oxygen, and copper are familiar examples of the more than 100 known elements, of which about 90 occur naturally on the earth, and two dozen or so have been created in laboratories.

Pure substances that can be broken down by chemical changes are called compounds. This breakdown may produce either elements or other compounds, or both. Mercury(II) oxide, an orange, crystalline solid, can be broken down by heat into the elements mercury and oxygen (Figure 8). When heated in the absence of air, the compound sucrose is broken down into the element carbon and the compound water. (The initial stage of this process, when the sugar is turning brown, is known as caramelization—this is what imparts the characteristic sweet and nutty flavor to caramel apples, caramelized onions, and caramel). Silver(I) chloride is a white solid that can be broken down into its elements, silver and chlorine, by absorption of light. This property is the basis for the use of this compound in photographic films and photochromic eyeglasses (those with lenses that darken when exposed to light).

[caption id="" align="aligncenter" width="975"] Figure 8. (a) The compound mercury(II) oxide, (b) when heated, (c) decomposes into silvery droplets of liquid mercury and invisible oxygen gas. (credit: modification of work by Paul Flowers)[/caption]

Many compounds break down when heated. This site shows the breakdown of mercury oxide, HgO. You can also view an example of the photochemical decomposition of silver chloride (AgCl), the basis of early photography.

The properties of combined elements are different from those in the free, or uncombined, state. For example, white crystalline sugar (sucrose) is a compound resulting from the chemical combination of the element carbon, which is a black solid in one of its uncombined forms, and the two elements hydrogen and oxygen, which are colorless gases when uncombined. Free sodium, an element that is a soft, shiny, metallic solid, and free chlorine, an element that is a yellow-green gas, combine to form sodium chloride (table salt), a compound that is a white, crystalline solid.

A mixture is composed of two or more types of matter that can be present in varying amounts and can be physically separated by using methods that use physical properties to separate the components of the mixture, such as evaporation, distillation, and filtration (you will learn more about this later). A mixture with a composition that varies from point to point is called a heterogeneous mixture. Italian dressing is an example of a heterogeneous mixture (Figure 9). Its composition can vary because we can make it from varying amounts of oil, vinegar, and herbs. It is not the same from point to point throughout the mixture—one drop may be mostly vinegar, whereas a different drop may be mostly oil or herbs because the oil and vinegar separate and the herbs settle. Other examples of heterogeneous mixtures are chocolate chip cookies (we can see the separate bits of chocolate, nuts, and cookie dough) and granite (we can see the quartz, mica, feldspar, and more).

A homogeneous mixture, also called a solution, exhibits a uniform composition and appears visually the same throughout. An example of a solution is a sports drink, consisting of water, sugar, coloring, flavoring, and electrolytes mixed together uniformly (Figure 9). Each drop of a sports drink tastes the same because each drop contains the same amounts of water, sugar, and other components. Note that the composition of a sports drink can vary—it could be made with somewhat more or less sugar, flavoring, or other components, and still be a sports drink. Other examples of homogeneous mixtures include air, maple syrup, gasoline, and a solution of salt in water.

[caption id="" align="aligncenter" width="975"] Figure 9. (a) Oil and vinegar salad dressing is a heterogeneous mixture because its composition is not uniform throughout. (b) A commercial sports drink is a homogeneous mixture because its composition is uniform throughout. (credit a “left”: modification of work by John Mayer; credit a “right”: modification of work by Umberto Salvagnin; credit b “left: modification of work by Jeff Bedford)[/caption]

Although there are just over 100 elements, tens of millions of chemical compounds result from different combinations of these elements. Each compound has a specific composition and possesses definite chemical and physical properties by which we can distinguish it from all other compounds. And, of course, there are innumerable ways to combine elements and compounds to form different mixtures. A summary of how to distinguish between the various major classifications of matter is shown in (Figure 10).

[caption id="" align="aligncenter" width="1300"] Figure 10. Depending on its properties, a given substance can be classified as a homogeneous mixture, a heterogeneous mixture, a compound, or an element.[/caption]

Eleven elements make up about 99% of the earth’s crust and atmosphere (Table 1). Oxygen constitutes nearly one-half and silicon about one-quarter of the total quantity of these elements. A majority of elements on earth are found in chemical combinations with other elements; about one-quarter of the elements are also found in the free state.

Element	Symbol	Percent Mass	Element	Symbol	Percent Mass
oxygen	O	49.20	chlorine	Cl	0.19
silicon	Si	25.67	phosphorus	P	0.11
aluminum	Al	7.50	manganese	Mn	0.09
iron	Fe	4.71	carbon	C	0.08
calcium	Ca	3.39	sulfur	S	0.06
sodium	Na	2.63	barium	Ba	0.04
potassium	K	2.40	nitrogen	N	0.03
magnesium	Mg	1.93	fluorine	F	0.03
hydrogen	H	0.87	strontium	Sr	0.02
titanium	Ti	0.58	all others	-	0.47
Table 1. Elemental Composition of Earth

Example 1

For the following substances, decide if it is a pure element, a pure compound, a homogeneous mixture or a heterogeneous mixture. Briefly explain your choice.

a)  muddy water      b)  copper     c)  distilled water      d)  sea water

 

Solution  

a)   Heterogeneous mixture.Muddy water is a mixture because the there are at least two components (mud and water) and the amount of mud in the water can vary (not fixed proportions). It is heterogeneous because the mud is likely not evenly distributed throughout.

b)   Pure element.Copper is found on the periodic table, therefore it is an element. No other atom types are mentioned, so we presume it is pure.

c)   Pure compound.The process of distillation is meant to purify the water, so we presume no other substances are in the sample. Water is made of two atom types so this is a compound.

d)   Homogeneous mixture.This is a mixture, as we know there is salt and water, and they can be present in different proportions (more or less salty water). Presuming there are no floating bits or living things in the sample, the salt is uniform in its distribution in the water (being dissolved or mixed at the atomic level), therefore this is homogeneous.

 

Test Yourself

For the following substances, decide if it is a pure element, a pure compound, a homogeneous mixture or a heterogeneous mixture. Briefly explain your choice.

a)  gold              b)  ice              c)  beer                   d)  vitamin C

 

Answers

a) Pure element.Gold is found on the periodic table, therefore it is an element. No other atom types are mentioned, so we presume it is pure. b) Pure compound. If we presume the water was purified. If we imagine there was dissolved material in the water (typical for tap water), when it froze it would contain these other substances in a non-uniform distribution, in which case we would call it a heterogeneous mixture. c) Homogeneous mixture.Beer is a mixture, and the components are uniformly distributed. If we presume there are bubbles, certainly they would not be uniformly distributed, and therefore we would classify it as a heterogeneous mixture. d) Pure compound. Vitamin C is comprised of several different atom types present in a fixed proportion (distinct molecules of a particular construct).

Example 2

Identify the following combinations as heterogeneous mixtures or homogenous mixtures.

a) soda water (Carbon dioxide is dissolved in water.)

b) a mixture of iron metal filings and sulfur powder (Both iron and sulfur are elements.)

 

Solution

a) Because carbon dioxide is dissolved in water, we can infer from the behaviour of salt crystals dissolved in water that carbon dioxide dissolved in water is (also) a homogeneous mixture.

b) Assuming that the iron and sulfur are simply mixed together, it should be easy to see what is iron and what is sulfur, so this is a heterogeneous mixture.

 

Test Yourself

Are the following combinations homogeneous mixtures or heterogeneous mixtures?

a) the human body

b) an amalgam, a combination of some other metals dissolved in a small amount of mercury

 

Answers

a) heterogeneous mixture          b) homogeneous mixture

Chemistry Is Everywhere: In the Morning

Most people have a morning ritual, a process that they go through every morning to get ready for the day. Chemistry appears in many of these activities.

• If you take a shower or bath in the morning, you probably use soap, shampoo, or both. These items contain chemicals that interact with the oil and dirt on your body and hair to remove them and wash them away. Many of these products also contain chemicals that make you smell good; they are called fragrances.
• When you brush your teeth in the morning, you usually use toothpaste, a form of soap, to clean your teeth. Toothpastes typically contain tiny, hard particles called abrasives that physically scrub your teeth. Many toothpastes also contain fluoride, a substance that chemically interacts with the surface of the teeth to help prevent cavities.
• Perhaps you take vitamins, supplements, or medicines every morning. Vitamins and other supplements contain chemicals your body needs in small amounts to function properly. Medicines are chemicals that help combat diseases and promote health.
• Perhaps you make some fried eggs for breakfast. Frying eggs involves heating them enough so that a chemical reaction occurs to cook the eggs.
• After you eat, the food in your stomach is chemically reacted so that the body (mostly the intestines) can absorb food, water, and other nutrients.
• If you drive or take the bus to school or work, you are using a vehicle that probably burns gasoline, a material that burns fairly easily and provides energy to power the vehicle. Recall that burning is a chemical change.

[caption id="attachment_3196" align="aligncenter" width="634"] Figure 11. Chemistry in Real Life - Examples of chemistry can be found everywhere—such as in personal hygiene products, food, and motor vehicles.  “Soaps and Shampoos” by Takashi Ota is licensed under Creative Commons Attribution 2.0 Generic; “English Breakfast” is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported; “Langley, Trans-Canada Highway” by James is licensed under the Creative Commons Attribution- Share Alike 3.0 Unported.[/caption]   These are just a few examples of how chemistry impacts your everyday life. And we haven’t even made it to lunch yet!

Video source: The chemical world by keyj   (https://viuvideos.viu.ca/media/The+Chemical+World/0_ixlxmwe8)  

Chemistry of Cell Phones

Imagine how different your life would be without cell phones (Figure 12) and other smart devices. Cell phones are made from numerous chemical substances, which are extracted, refined, purified, and assembled using an extensive and in-depth understanding of chemical principles. About 30% of the elements that are found in nature are found within a typical smart phone. The case/body/frame consists of a combination of sturdy, durable polymers comprised primarily of carbon, hydrogen, oxygen, and nitrogen [acrylonitrile butadiene styrene (ABS) and polycarbonate thermoplastics], and light, strong, structural metals, such as aluminum, magnesium, and iron. The display screen is made from a specially toughened glass (silica glass strengthened by the addition of aluminum, sodium, and potassium) and coated with a material to make it conductive (such as indium tin oxide). The circuit board uses a semiconductor material (usually silicon); commonly used metals like copper, tin, silver, and gold; and more unfamiliar elements such as yttrium, praseodymium, and gadolinium. The battery relies upon lithium ions and a variety of other materials, including iron, cobalt, copper, polyethylene oxide, and polyacrylonitrile.

[caption id="" align="aligncenter" width="1178"] Figure 12. Almost one-third of naturally occurring elements are used to make a cell phone. (credit: modification of work by John Taylor)[/caption]

Key Concepts and Summary

Matter is anything that occupies space and has mass. The basic building block of matter is the atom, the smallest unit of an element that can enter into combinations with atoms of the same or other elements. In many substances, atoms are combined into molecules. On earth, matter commonly exists in three states: solids, of fixed shape and volume; liquids, of variable shape but fixed volume; and gases, of variable shape and volume. Under high-temperature conditions, matter also can exist as a plasma. Most matter is a mixture: It is composed of two or more types of matter that can be present in varying amounts and can be separated by physical means. Heterogeneous mixtures vary in composition from point to point; homogeneous mixtures have the same composition from point to point. Pure substances consist of only one type of matter. A pure substance can be an element, which consists of only one type of atom and cannot be broken down by a chemical change, or a compound, which consists of two or more types of atoms.

Exercises

1. What properties distinguish solids from liquids? Liquids from gases? Solids from gases? 2. How does a homogeneous mixture differ from a pure substance? How are they similar? 3. How do molecules of elements and molecules of compounds differ? In what ways are they similar? 4. Many of the items you purchase are mixtures of pure compounds. Select three of these commercial products and prepare a list of the ingredients that are pure compounds. 5. Classify each of the following as an element, a compound, or a mixture:

a) iron

b) oxygen

c) mercury oxide

d) pancake syrup

e) carbon dioxide

f) a substance composed of molecules each of which contains one hydrogen atom and one chlorine atom

g) baking soda

h) baking powder

6. How are the molecules in oxygen gas, the molecules in hydrogen gas, and water molecules similar? How do they differ? 7. As we drive an automobile, we don't think about the chemicals consumed and produced. Prepare a list of the principal chemicals consumed and produced during the operation of an automobile. 8. When elemental iron corrodes it combines with oxygen in the air to ultimately form red brown iron(III) oxide which we call rust.   a) If a shiny iron nail with an initial mass of 23.2 g is weighed after being coated in a layer of rust, would you expect the mass to have increased, decreased, or remained the same? Explain.   b) If the mass of the iron nail increases to 24.1 g, what mass of oxygen combined with the iron? 9. Yeast converts glucose to ethanol and carbon dioxide during anaerobic fermentation as depicted in the simple chemical equation here:

$latex \text{glucose} \rightarrow \text{ethanol + carbon dioxide} $

a) If 200.0 g of glucose is fully converted, what will be the total mass of ethanol and carbon dioxide produced?

b) If the fermentation is carried out in an open container, would you expect the mass of the container and contents after fermentation to be less than, greater than, or the same as the mass of the container and contents before fermentation? Explain.

c) If 97.7 g of carbon dioxide is produced, what mass of ethanol is produced?

10. Identify each as either matter or not matter. a)  a book   b)  hate   c)  light   d)  a car   e)  a fried egg 11. Distinguish between an element and a compound. About how many of each are known? 12. What is the difference between a homogeneous mixture and a heterogeneous mixture? 13. Identify each as a heterogeneous mixture or a homogeneous mixture.a)  air   b)  dirt   c)  a television set 14. Identify each as a heterogeneous mixture or a homogeneous mixture. a)  Salt is mixed with pepper.   b)  Sugar is dissolved in water.   c)  Pasta is cooked in boiling water.   Answers

1. Liquids can change their shape (flow); solids can’t. Gases can undergo large volume changes as pressure changes; liquids do not. Gases flow and change volume; solids do not.

2. The mixture can have a variety of compositions; a pure substance has a definite composition. Both have the same composition from point to point.

3. Molecules of elements contain only one type of atom; molecules of compounds contain two or more types of atoms. They are similar in that both are comprised of two or more atoms chemically bonded together.

4. Answers will vary. Sample answer: Gatorade contains water, sugar, dextrose, citric acid, salt, sodium chloride, monopotassium phosphate, and sucrose acetate isobutyrate.

5. a) element;   b) element;   c) compound;   d) mixture,   e) compound;   f) compound;   g) compound;

h) mixture

6. In each case, a molecule consists of two or more combined atoms. They differ in that the types of atoms change from one substance to the next.

7. Gasoline (a mixture of compounds), oxygen, and to a lesser extent, nitrogen are consumed. Carbon dioxide and water are the principal products. Carbon monoxide and nitrogen oxides are produced in lesser amounts.

8. a) Increased as it would have combined with oxygen in the air thus increasing the amount of matter and therefore the mass.   b) 0.9 g

9. a) 200.0 g;   b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container.   c) 102.3 g

10. a)  matter   b)  not matter   c)  not matter   d)  matter

11. An element is a fundamental chemical part of a substance; there are about 115 known elements. A compound is a combination of elements that acts as a different substance; there are over 50 million known substances.

12. A homogeneous mixture, also called a solution, exhibits a uniform composition and appears visually the same throughout.  A mixture with a composition that varies from point to point is called a heterogeneous mixture. 13. a)  homogeneous   b)  heterogeneous   c)  heterogeneous 14. a)  heterogeneous   b)  homogeneous   c)  heterogeneous

Glossary

atom: smallest particle of an element that can enter into a chemical combination compound: pure substance that can be decomposed into two or more elements element: substance that is composed of a single type of atom; a substance that cannot be decomposed by a chemical change gas: state in which matter has neither definite volume nor shape heterogeneous mixture: combination of substances with a composition that varies from point to point homogeneous mixture: (also, solution) combination of substances with a composition that is uniform throughout liquid: state of matter that has a definite volume but indefinite shape law of conservation of matter: when matter converts from one type to another or changes form, there is no detectable change in the total amount of matter present mass: fundamental property indicating amount of matter matter: anything that occupies space and has mass mixture: matter that can be separated into its components by physical means molecule: bonded collection of two or more atoms of the same or different elements plasma: gaseous state of matter containing a large number of electrically charged atoms and/or molecules pure substance: homogeneous substance that has a constant composition solid: state of matter that is rigid, has a definite shape, and has a fairly constant volume weight: force that gravity exerts on an object

]]> 1266 1247 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction/ Thu, 12 Apr 2018 02:51:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction/ Your alarm goes off and, after hitting “snooze” once or twice, you pry yourself out of bed. You make a cup of coffee to help you get going, and then you shower, get dressed, eat breakfast, and check your phone for messages. On your way to school, you stop to fill your car’s gas tank, almost making you late for the first day of chemistry class. As you find a seat in the classroom, you read the question projected on the screen: “Welcome to class! Why should we study chemistry?”

[caption id="" align="aligncenter" width="1300"] Figure 1. Chemical substances and processes are essential for our existence, providing sustenance, keeping us clean and healthy, fabricating electronic devices, enabling transportation, and much more. (credit “left”: modification of work by “vxla”/Flickr; credit “left middle”: modification of work by “the Italian voice”/Flickr; credit “right middle”: modification of work by Jason Trim; credit “right”: modification of work by “gosheshe”/Flickr)[/caption]

Do you have an answer? You may be studying chemistry because it fulfills an academic requirement, but if you consider your daily activities, you might find chemistry interesting for other reasons. Most everything you do and encounter during your day involves chemistry. Making coffee, cooking eggs, and toasting bread involve chemistry. The products you use—like soap and shampoo, the fabrics you wear, the electronics that keep you connected to your world, the gasoline that propels your car—all of these and more involve chemical substances and processes. Whether you are aware or not, chemistry is part of your everyday world. In this course, you will learn many of the essential principles underlying the chemistry of modern-day life.

]]> 1268 1247 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/ Thu, 12 Apr 2018 02:51:03 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-5-measurement-uncertainty-accuracy-and-precision/

Learning Objectives

By the end of this section, you will be able to:

• Define accuracy and precision
• Distinguish exact and uncertain numbers
• Correctly represent uncertainty in quantities using significant figures
• Apply proper rounding rules to computed quantities

When recording a measurement, how you record it is just as important as what you record. You must consider the precision of the instrument you are using.  Accuracy is the extent to which a measured value coincides with the true or accepted value.  Precision refers to the “fineness” (i.e. the number of digits) of the measurement as well as the reproducibility.  Significant figures are those digits in an experimentally measured quantity that establish the precision with which the value is known.  A precise measurement may not be accurate!!

Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number. If we count eggs in a carton, we know exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used.

Significant Figures in Measurement

The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.

[caption id="" align="aligncenter" width="1300"] Figure 1. To measure the volume of liquid in this graduated cylinder, you must mentally subdivide the distance between the 21 and 22 mL marks into tenths of a milliliter, and then make a reading (estimate) at the bottom of the meniscus.[/caption]

Refer to the illustration in Figure 1. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL.

This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If we weigh the quarter on a more sensitive balance, we may find that its mass is 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty, which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits. Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.

Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them.

Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point.

Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located.

The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407 × 10⁻³; then the number 8.32407 contains all of the significant figures, and 10⁻³ locates the decimal point.

The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 × 10³ (two significant figures), 1.30 × 10³ (three significant figures, if the tens place was measured), or 1.300 × 10³ (four significant figures, if the ones place was also measured). In cases where only the normal notation is used (1300 g), all trailing zeros are not significant, therefore the measurement would have two significant figures.  Though, if the measurement was express with an explicit decimal place (1300. g), then the trailing zeros would be significant, and therefore the measurement would have four significant figures.

  [caption id="attachment_3163" align="aligncenter" width="535"] Figure 2: Summary: How to determine the number of significant figures in measurements.[/caption]

Example 1

Determine the correct measurement reading for the following volume:   Solution Because the beaker has gradations of 100 mL, we know that it is at least 200 mL. It is the tens position that is uncertain. Thus we can only record to that position. The reading should be 220 mL. Better still—clarify the significant figures by using scientific notation, which would be 2.2 x 10² mL.  

Test Yourself

Determine the correct reading for the following temperature:  

Answer

14.8^oC

Example 2

Use each diagram to report a measurement to the proper number of significant figures.

 

Solution

a) The arrow is between 4.0 and 5.0, so the measurement is at least 4.0. The arrow is between the third and fourth small tick marks, so it’s at least 0.3. We will have to estimate the last place. It looks like about one-third of the way across the space, so let us estimate the hundredths place as 3. Combining the digits, we have a measurement of 4.33 psi (psi stands for “pounds per square inch” and is a unit of pressure, like air in a tire). We say that the measurement is reported to three significant figures.

b)The rectangle is at least 1.0 cm wide but certainly not 2.0 cm wide, so the first significant digit is 1. The rectangle’s width is past the second tick mark but not the third; if each tick mark represents 0.1, then the rectangle is at least 0.2 in the next significant digit. We have to estimate the next place because there are no markings to guide us. It appears to be about halfway between 0.2 and 0.3, so we will estimate the next place to be a 5. Thus, the measured width of the rectangle is 1.25 cm. Again, the measurement is reported to three significant figures.

 

Test Yourself

What would be the reported width of this rectangle?

Answer

0.63 cm

Example 3

Give the number of significant figures in each measurement.

a) 36.7 m          b) 0.006606 s          c) 2,002 kg          d) 306,490,000 people

 

Solution

a) By rule 1, all nonzero digits are significant, so this measurement has three significant figures.

b) By rule 4, the first three zeros are not significant, but by rule 2 the zero between the sixes is; therefore, this number has four significant figures.

c) By rule 2, the two zeros between the twos are significant, so this measurement has four significant figures.

d) The four trailing zeros in the number are not significant, but the other five numbers are, so this number has five significant figures.

 

Test Yourself

Give the number of significant figures in each measurement.

a) 0.000601 m          b) 65.080 kg

 

Answers

a) three significant figures          b) five significant figures

Example 4

Determine the number of significant figures in the following measurements a) 0.002040 g    b) 300 mL    c) 3.021 x 10⁵ g    d) 31 pencils (counted exactly, not an estimate)   Solution a) 4 significant figures. The leading zeros do not count, but the trailing zero does (as there’s a decimal point). b) By the above rules, 1 significant figure, because there’s no decimal point showing; 300. mL would be 3 sig figs. (To emphasize 1 sig fig it would be better to write the measurement in scientific notation, as 3 x 10² mL, rather than just “300 mL”.) c) 4 significant figures. (No leading or trailing zeros.) Note that we only consider the digits in the numerical portion of the scientific notation, not the power of 10. d) Unlimited significant figures. This is an exact number, as it has been counted.   Test Yourself

Determine the number of significant figures in the following measurements. a)  0.00100 m              b)  2.0900 x 10³ mL                 c) 100.0 ^oC

 

Answers

a) 3 significant figures          b) 5 significant figures          c) 4 significant figures

When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 3.17 × 10⁸ people.

Significant Figures in Calculations

A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:

1. When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
2. When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)

The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:

• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)

Let’s work through these rules with a few examples.

Example 5

Round the following to the indicated number of significant figures:

a) 31.57 (to two significant figures)

b) 8.1649 (to three significant figures)

c) 0.051065 (to four significant figures)

d) 0.90275 (to four significant figures)

 

Solution a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)

b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)

c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)

d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)

 

Test Yourself Round the following to the indicated number of significant figures:

a) 0.424 (to two significant figures)

b) 0.0038661 (to three significant figures)

c) 421.25 (to four significant figures)

d) 28,683.5 (to five significant figures)

  Answers a) 0.42           b) 0.00387          c) 421.2          d) 28,684

Example 6

Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).

Perform the following calculations taking significant figures into account.

a) Add 1.0023 g and 4.383 g.

b) Subtract 421.23 g from 486 g.

 

Solution

a) $latex \displaystyle \begin{array}{r}1.0023 \text{g} \\ +4.383 \;\;\text{g} \\ \hline 5.3853 \text{g} \end{array} $

Answer is 5.385 g (round to the thousandths place; three decimal places)

b) $latex \displaystyle \begin{array}{r}486 \;\;\;\;\; \text{g} \\ -421.23 \text{g} \\ \hline 64.77 \text{g} \end{array} $

Answer is 65 g (round to the ones place; no decimal places)

 

Test Yourself

a) Add 2.334 mL and 0.31 mL.

b) Subtract 55.8752 m from 56.533 m.

  Answers a) 2.64 mL          b) 0.658 m

Example 7

Express the final answer to the proper number of significant figures.

a) 101.2 + 18.702 = ?

b) 202.88 − 1.013 = ?

 

Solution

a) If we use a calculator to add these two numbers, we would get 119.902. However, most calculators do not understand significant figures, and we need to limit the final answer to the tenths place. Thus, we drop the 02 and report a final answer of 119.9 (rounding down).

b) A calculator would answer 201.867. However, we have to limit our final answer to the hundredths place. Because the first number being dropped is 7, which is greater than 7, we round up and report a final answer of 201.87.

 

Test Yourself

Express the answer for 3.445 + 90.83 − 72.4 to the proper number of significant figures.

 

Answer

21.9

Example 8

Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).

Perform the following calculations taking significant figures into account.

a) Multiply 0.6238 cm by 6.6 cm.

b) Divide 421.23 g by 486 mL.

 

Solution

a) $latex 0.6238 \text{ cm} \times 6.6 \text{ cm} = 4.11708 \text{ cm}^2 \rightarrow \text{result is 4.1} \text{cm}^2 \text{(round to two significant figures)} \\[0.75em] \text{four significant figures} \times \text{two significant figures} \rightarrow \text{two significant figures answer} $

b) $latex \frac{421.23 \text{g}}{486 \text{mL}}=0.86728 \dots \text{g/mL} \rightarrow \text{result is 0.867 g/mL (round to three significant figures)} \\[0.75em] \frac{\text{five significant figures}}{\text{three significant figures}} \rightarrow \text{three significant figures answer} $

 

Test Yourself

a) Multiply 2.334 cm and 0.320 cm.

b) Divide 55.8752 m by 56.53 s.

  Answers a) 0.747 cm²           b) 0.9884 m/s

Example 9

Express the final answer to the proper number of significant figures.

a) 76.4 × 180.4 = ?

b) 934.9 ÷ 0.00455 = ?

 

Solution

a) The first number has three significant figures, while the second number has four significant figures. Therefore, we limit our final answer to three significant figures: 76.4 × 180.4 = 13,782.56 = 13,800.

b) The first number has four significant figures, while the second number has three significant figures. Therefore we limit our final answer to three significant figures: 934.9 ÷ 0.00455 = 205,472.5275… = 205,000.

 

Test Yourself

Express the final answer to the proper number of significant figures.

a) 22.4 × 8.314 = ?

b) 1.381 ÷ 6.02 = ?

 

Answers

a) 186          b) 0.229

In the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.

Need a refresher or more practice with significant figures? Visit this site (https://viuvideos.viu.ca/media/Significant+Figures/0_t8xwe4s9) to go over the basics of significant figures.

Video source: Significant figures by keyj

Example 10

One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.

 

Solution

$latex \begin{array}{r @{{}={}} l}V & l \times w\times d \\ & 13.44 \text{ dm} \times 5.920 \text{ dm} \times 2.54 \text{ dm} \\ & 202.09459\dots \text{ dm}^3 \text{(value from calculator)} \\ & 202 \text{ dm}^3,\;\text{or}\;202\text{ L (answer rounded to three significant figures)} \end{array}$

 

Test Yourself

What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm³?   Answer 1.034 g/mL

Example 11

Experimental Determination of Density Using Water Displacement A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.

a) Use these values to determine the density of this piece of rebar.

b) Rebar is mostly iron. Does your result in (a) support this statement? How?

 

Solution The volume of the piece of rebar is equal to the volume of the water displaced:

$latex \text{volume} = 22.4 \text{mL} - 13.5 \text{mL} = 8.9 \text{mL} = 8.9 \text{cm}^3 $

(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)

The density is the mass-to-volume ratio:

$latex \text{density} = \frac{\text{mass}}{\text{volume}} = \frac{69.658 \text{g}}{8.9 \text{cm}^3} = 7.8 \text{g/cm}^3 $

(rounded to two significant figures, per the rule for multiplication and division)

From Table 3 in Chapter 2.2 Measurements, the density of iron is 7.9 g/cm³, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.

 

Test Yourself An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.

a) Use these values to determine the density of this material.

b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.

  Answers a) 19 g/cm³; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in Table 3 in Chapter 2.2 Measurements.

Example 12

Determine the answer and report to the appropriate number of significant figures. a) What is the area of a card that measures 12.74 cm by 7.60 cm ? b) What is the total mass when three samples weighing 120.0 g, 24.318 g and 15 g are combined? c) (3.02 x 10³) + (4 x 10²) = ?   Solution a) The area is width x length. 12.74 cm x 7.60 cm = 96.824 according to a calculator. The 12.74 has 4 significant figures and 7.60 has three significant figures, thus the solution should be reported to 3 significant figures. Don’t forget the units! cm x cm = cm² The answer is 96.8 cm² b) When adding, be sure that all units are the same (all g in this case). 0 g + 24.318 g + 15 g = 159.318 g according to a calculator. We must look at the decimal place (or place value). 120.0 is reported to the tenths position, 24.318 to the thousandths position, whereas 15 is reported to the ones position. The ones position is furthest to the left, thus the final answer must be rounded to that position. The answer is 159 g c) Be careful! When adding and subtracting, the power of 10 must be the same. If we put them both to 103, we have (3.02 x 10³) + (0.4 x 10³). The answer can then be reported to the first decimal place (tenths) when using 10³ (= 3.4 x 10³). We could also look at the non-scientific notation: 3.02 x 10³ = 3020, and 4 x 10² = 400 (sig. figs. in bold, for emphasis). Thus the first number is recorded to the tens position, whereas the second is to the hundreds position. final answer must be reported to the hundreds position. The answer is 3.4 x 10³   Test Yourself Determine the answer and report to the appropriate number of significant figures. a)  17.1 + 0.24 – 241               b)  (1.32 x 10⁴) x (2 x 10²)   Answers a) -224          b) 3 x 10⁶

Example 13

If a bagel has a mass of 28.3162 g when fresh and then a mass of 28.3094 g once dried out, what was the percent of moisture in the fresh bagel? Where, % moisture =   (mass of moisture/original mass of object)  x 100% , determine the solution and report to the appropriate number of significant figures.   Solution To determine the mass of moisture, we must subtract the dried mass from the original mass: 28.3162 g – 28.3094 g =  0.0068 g. Note that if the question stopped here, the answer would be reported to the 4th decimal place, and has two significant figures. The next step involves division:    0.0068 g    x  100%  = 0.0240145…% 28.3162 g The value 0.0068 g has 2 significant figures, 28.3162 g has 6 significant figures, and 100% as used here is an exact number, so the final answer can only be reported to 2 significant figures. Thus the solution is rounded to 0.024% (or 2.4 x 10^-2%, in scientific notation)   Test Yourself If a jogger runs for 2.0 hrs at 12.21 km/hr, then again for another 2.0 hrs at 12.16 km/hr, how far did she run in total? (Determine the solution and report to the appropriate number of significant figures.)   Answer 49 km

Accuracy and Precision

Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure 3).

[caption id="" align="aligncenter" width="1300"] Figure 3. (a) These arrows are close to both the bull’s eye and one another, so they are both accurate and precise. (b) These arrows are close to one another but not on target, so they are precise but not accurate. (c) These arrows are neither on target nor close to one another, so they are neither accurate nor precise.[/caption]

Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table 1.

Dispenser #1	Dispenser #2	Dispenser #3
283.3	298.3	296.1
284.1	294.2	295.9
283.9	296.0	296.1
284.0	297.8	296.0
284.1	293.9	296.1
Table 1. Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers

Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).

Key Concepts and Summary

Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the measurement. The uncertainty of a calculated value depends on the uncertainties in the values used in the calculation and is reflected in how the value is rounded. Measured values can be accurate (close to the true value) and/or precise (showing little variation when measured repeatedly).

Exercises

1. Express each measurement to the correct number of significant figures.

2. Express each measurement to the correct number of significant figures.

3. How many significant figures do these numbers have?

a)  23     b)  23.0     c)  0.00023     d)  0.0002302 4.  How many significant figures do these numbers have? a)  5.44 × 10⁸     b)  1.008 × 10⁻⁵     c)  43.09     d)  0.0000001381 5.  How many significant figures do these numbers have? a)  765,890     b)  765,890.0     c)  1.2000 × 10⁵     d)  0.0005060

6)  How many significant figures do these numbers have?

a)  0.009     b)  0.0000009     c)  65,444     d)  65,040

7.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  56.0 + 3.44 = ?                 b)  0.00665 + 1.004 = ? c)  45.99 − 32.8 = ?               d)  45.99 − 32.8 + 75.02 = ?

8.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  1.005 + 17.88 = ?              b)  56,700 − 324 = ? c)  405,007 − 123.3 = ?          d)  55.5 + 66.66 − 77.777 = ?

9.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  56.7 × 66.99 = ?                 b)  1.000 ÷ 77 = ? c)  1.000 ÷ 77.0 = ?                 d)  6.022 × 1.89 = ?

10.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  0.000440 × 17.22 = ?         b)  203,000 ÷ 0.044 = ? c)  67 × 85.0 × 0.0028 = ?       d)  999,999 ÷ 3,310 = ?

11.  Write the number 87,449 in scientific notation with four significant figures. 12.  Write the number 0.000066600 in scientific notation with five significant figures.

13.  Write the number 306,000,000 in scientific notation to the proper number of significant figures. 14.  Write the number 0.0000558 in scientific notation with two significant figures.

15.  Perform each calculation and limit each answer to three significant figures.

a)  67,883 × 0.004321 = ? b)  (9.67 × 10³) × 0.0055087 = ?

16.  Perform each calculation and limit each answer to four significant figures.

a)  18,900 × 76.33 ÷ 0.00336 = ? b)  0.77604 ÷ 76,003 × 8.888 = ? 17.  Express each of the following numbers in exponential notation with correct significant figures:

a) 704     b) 0.03344     c) 547.9     d) 22086     e) 1000.00     f) 0.0000000651     g) 0.007157

18.  Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:

a) the number of seconds in an hour

b) the number of pages in this book

c) the number of grams in your weight

d) the number of grams in 3 kilograms

e) the volume of water you drink in one day

f) the distance from San Francisco to Kansas City

19.  How many significant figures are contained in each of the following measurements?

a) 53 cm     b) 2.05 × 10⁸ m     c) 86,002 J     d) 9.740 × 10⁴ m/s     e) 10.0613 m³

f) 0.17 g/mL     g) 0.88400 s

20.  Round off each of the following numbers to two significant figures:

a) 0.436     b) 9.000     c) 27.2     d) 135     e) 1.497 × 10⁻³     f) 0.445

21.  Perform the following calculations and report each answer with the correct number of significant figures.

a) 628 × 342                                           b) (5.63 × 10²) × (7.4 × 10³)

c) $latex \frac{28.0}{13.483} $            d) 8119 × 0.000023

e) 14.98 + 27,340 + 84.7593                  f) 42.7 + 0.259

22.  Consider the results of the archery contest shown in this figure.

a) Which archer is most precise?

b) Which archer is most accurate?

c) Who is both least precise and least accurate?

  Answers 1.  a)  375 psi     b)  1.30 cm 2.  a) 32.4 psi     b) 0.90 cm 3.  a)  two     b)  three     c)  two     d)  four 4.  a)  three   b)  four      c)  four     d)  four 5.  a)  five     b)  seven     c)  five     d)  four 6.  a)  one     b)  one        c)  five      d) four 7.  a)  59.4     b)  1.011     c)  13.2     d)  88.2 8.  a)  18.88   b)  56,400    c)  404,884   d)  44.4 9.  a)  3.80 × 10³     b)  0.013     c)  0.0130     d)  11.4 10.  a)  0.00758      b)  4,600,000     c)  16     d)  302 11.  8.745 × 10⁴     12.  6.6600 x 10⁻⁵ 13.  3.06 x 10⁸ 14.  5.6 x 10^-5 15.  a)  293     b)  53.3

16.  a)  4.294 x 10⁸      b)  9.060x10^-5

17.  a) 7.04 × 10²      b) 3.344 × 10⁻²      c) 5.479 × 10²      d) 2.2086 × 10⁴      e) 1.00000 × 10³      f) 6.51 × 10⁻⁸ g) 7.157 × 10⁻³

18.  a) exact      b) exact      c) uncertain      d) exact      e) uncertain      f) uncertain

19.  a) two      b) three      c) five      d) four      e) six      f) two      g) five

20.  a) 0.44      b) 9.0      c) 27      d) 140      e) 1.5 × 10⁻³      f) 0.44

21.  a) 2.15 × 10⁵      b) 4.2 × 10⁶      c) 2.08      d) 0.19      e) 27,440      f) 43.0

22.  a) Archer X      b) Archer W      c) Archer Y

Glossary

accuracy: how closely a measurement aligns with a correct value exact number: number derived by counting or by definition precision: how closely a measurement matches the same measurement when repeated rounding: procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation significant figures: (also, significant digits) all of the measured digits in a determination, including the uncertain last digit uncertainty: estimate of amount by which measurement differs from true value

]]> 1279 2084 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-4-measurements/ Thu, 12 Apr 2018 02:51:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-4-measurements/

Learning Objectives

By the end of this section, you will be able to:

• Explain the process of measurement
• Identify the three basic parts of a quantity
• Describe the properties and units of length, mass, volume, density, temperature, and time
• Perform basic unit calculations and conversions in the metric and other unit systems

The development of modern chemistry is often attributed to 18th century Frenchman Antoine-Laurent de Lavoisier, who was able though meticulous and careful scientific measurements that during a chemical reaction mass is neither consumed or created, the principle that led to the law of conservation of mass, one of the important fundamental principles in your study of chemistry. It is fundamentally important to realize that a science is for the most part a quantitative endeavor. Our ability to make observations through numerical measures is one of the cornerstones of the scientific method.

Measurements provide the macroscopic information that is the basis of most of the hypotheses, theories, and laws that describe the behavior of matter and energy in both the macroscopic and microscopic domains of chemistry. Every measurement provides three kinds of information: a number (quantitative observation), a unit (describes how it was measured), and the degree of reliability (uncertainty of the measurement). While the number and unit are explicitly represented when a quantity is written, the uncertainty is an aspect of the measurement result that is more implicitly represented and will be discussed later.

The number in the measurement can be represented in different ways, including decimal form and scientific notation. For example, the maximum takeoff weight of a Boeing 777-200ER airliner is 298,000 kilograms, which can also be written as 2.98 × 10⁵ kg. The mass of the average mosquito is about 0.0000025 kilograms, which can be written as 2.5 × 10⁻⁶ kg.

Units, such as liters, pounds, and centimeters, are standards of comparison for measurements. When we buy a 2-liter bottle of a soft drink, we expect that the volume of the drink was measured, so it is two times larger than the volume that everyone agrees to be 1 liter. The meat used to prepare a 0.25-pound hamburger is measured so it weighs one-fourth as much as 1 pound. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.

We usually report the results of scientific measurements in SI units, an updated version of the metric system, using the units listed in Table 1. Other units can be derived from these base units. The standards for these units are fixed by international agreement, and they are called the International System of Units or SI Units (from the French, Le Système International d’Unités). SI units have been used by the United States National Institute of Standards and Technology (NIST) since 1964.

Property Measured	Name of Unit	Symbol of Unit
length	meter	m
mass	kilogram	kg
time	second	s
temperature	kelvin	K
electric current	ampere	A
amount of substance	mole	mol
luminous intensity	candela	cd
Table 1. Base Units of the SI System

Sometimes we use units that are fractions or multiples of a base unit. Ice cream is sold in quarts (a familiar, non-SI base unit), pints (0.5 quart), or gallons (4 quarts). We also use fractions or multiples of units in the SI system, but these fractions or multiples are always powers of 10. Fractional or multiple SI units are named using a prefix and the name of the base unit. For example, a length of 1000 meters is also called a kilometer because the prefix kilo means “one thousand,” which in scientific notation is 10³ (1 kilometer = 1000 m = 10³ m). The prefixes used and the powers to which 10 are raised are listed in Table 2.

Need a refresher or more practice with scientific notation? Visit this site to go over the basics of scientific notation.

 

Prefix	Symbol	Factor	Example
femto	f	10−15	1 femtosecond (fs) = 1 × 10−15 s (0.000000000000001 s)
pico	p	10−12	1 picometer (pm) = 1 × 10−12 m (0.000000000001 m)
nano	n	10−9	4 nanograms (ng) = 4 × 10−9 g (0.000000004 g)
micro	µ	10−6	1 microliter (μL) = 1 × 10−6 L (0.000001 L)
milli	m	10−3	2 millimoles (mmol) = 2 × 10−3 mol (0.002 mol)
centi	c	10−2	7 centimeters (cm) = 7 × 10−2 m (0.07 m)
deci	d	10−1	1 deciliter (dL) = 1 × 10−1 L (0.1 L )
kilo	k	103	1 kilometer (km) = 1 × 103 m (1000 m)
mega	M	106	3 megahertz (MHz) = 3 × 106 Hz (3,000,000 Hz)
giga	G	109	8 gigayears (Gyr) = 8 × 109 yr (8,000,000,000 Gyr)
tera	T	1012	5 terawatts (TW) = 5 × 1012 W (5,000,000,000,000 W)
Table 2. Common Unit Prefixes

SI Base Units

The initial units of the metric system, which eventually evolved into the SI system, were established in France during the French Revolution. The original standards for the meter and the kilogram were adopted there in 1799 and eventually by other countries. This section introduces four of the SI base units commonly used in chemistry. Other SI units will be introduced in subsequent chapters.

Length

The standard unit of length in both the SI and original metric systems is the meter (m). A meter was originally specified as 1/10,000,000 of the distance from the North Pole to the equator. It is now defined as the distance light in a vacuum travels in 1/299,792,458 of a second. A meter is about 3 inches longer than a yard (Figure 1); one meter is about 39.37 inches or 1.094 yards. Longer distances are often reported in kilometers (1 km = 1000 m = 10³ m), whereas shorter distances can be reported in centimeters (1 cm = 0.01 m = 10⁻² m) or millimeters (1 mm = 0.001 m = 10⁻³ m).

[caption id="" align="aligncenter" width="1300"] Figure 1. The relative lengths of 1 m, 1 yd, 1 cm, and 1 in. are shown (not actual size), as well as comparisons of 2.54 cm and 1 in., and of 1 m and 1.094 yd.[/caption]

Mass

The standard unit of mass in the SI system is the kilogram (kg). A kilogram was originally defined as the mass of a liter of water (a cube of water with an edge length of exactly 0.1 meter). It is now defined by a certain cylinder of platinum-iridium alloy, which is kept in France (Figure 2). Any object with the same mass as this cylinder is said to have a mass of 1 kilogram. One kilogram is about 2.2 pounds. The gram (g) is exactly equal to 1/1000 of the mass of the kilogram (10⁻³ kg).

[caption id="attachment_1282" align="aligncenter" width="200"] Figure 2. This replica prototype kilogram is housed at the National Institute of Standards and Technology (NIST) in Maryland. (credit: National Institutes of Standards and Technology)[/caption]

Temperature

Temperature is an intensive property. The SI unit of temperature is the kelvin (K). The IUPAC convention is to use kelvin (all lowercase) for the word, K (uppercase) for the unit symbol, and neither the word “degree” nor the degree symbol (°). The degree Celsius (°C) is also allowed in the SI system, with both the word “degree” and the degree symbol used for Celsius measurements. Celsius degrees are the same magnitude as those of kelvin, but the two scales place their zeros in different places. Water freezes at 273.15 K (0 °C) and boils at 373.15 K (100 °C) by definition, and normal human body temperature is approximately 310 K (37 °C). The conversion between these two units and the Fahrenheit scale will be discussed later in this chapter.

Time

The SI base unit of time is the second (s). Small and large time intervals can be expressed with the appropriate prefixes; for example, 3 microseconds = 0.000003 s = 3 × 10⁻⁶ and 5 megaseconds = 5,000,000 s = 5 × 10⁶ s. Alternatively, hours, days, and years can be used.

Derived SI Units

We can derive many units from the seven SI base units. For example, we can use the base unit of length to define a unit of volume, and the base units of mass and length to define a unit of density.

Volume

Volume is the measure of the amount of space occupied by an object. The standard SI unit of volume is defined by the base unit of length (Figure 3). The standard volume is a cubic meter (m³), a cube with an edge length of exactly one meter. To dispense a cubic meter of water, we could build a cubic box with edge lengths of exactly one meter. This box would hold a cubic meter of water or any other substance.

A more commonly used unit of volume is derived from the decimeter (0.1 m, or 10 cm). A cube with edge lengths of exactly one decimeter contains a volume of one cubic decimeter (dm³). A liter (L) is the more common name for the cubic decimeter. One liter is about 1.06 quarts.

A cubic centimeter (cm³) is the volume of a cube with an edge length of exactly one centimeter. The abbreviation cc (for cubic centimeter) is often used by health professionals. A cubic centimeter is also called a milliliter (mL) and is 1/1000 of a liter.

[caption id="" align="aligncenter" width="1200"] Figure 3 (a) The relative volumes are shown for cubes of 1 m³, 1 dm³ (1 L), and 1 cm³ (1 mL) (not to scale). (b) The diameter of a dime is compared relative to the edge length of a 1-cm³ (1-mL) cube.[/caption]

Density

We use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.

The density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter (kg/m³). For many situations, however, this as an inconvenient unit, and we often use grams per cubic centimeter (g/cm³) for the densities of solids and liquids, and grams per liter (g/L) for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 g/cm³ (the density of gasoline) to 19 g/cm³ (the density of gold). The density of air is about 1.2 g/L. Table 3 shows the densities of some common substances.

Solids	Liquids	Gases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3	water 1.0 g/cm3	dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3	ethanol 0.79 g/cm3	oxygen 1.31 g/L
iron 7.9 g/cm3	acetone 0.79 g/cm3	nitrogen 1.14 g/L
copper 9.0 g/cm3	glycerin 1.26 g/cm3	carbon dioxide 1.80 g/L
lead 11.3 g/cm3	olive oil 0.92 g/cm3	helium 0.16 g/L
silver 10.5 g/cm3	gasoline 0.70–0.77 g/cm3	neon 0.83 g/L
gold 19.3 g/cm3	mercury 13.6 g/cm3	radon 9.1 g/L
Table 3. Densities of Common Substances

While there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume. In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.

$latex \text{density} = \frac{\text{mass}}{\text{volume}} $

Example 1

Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm³. What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?  

Solution The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.

$latex \text{volume of lead cube}=2.00\text{cm}\times2.00\text{cm}\times2.00\text{cm}=9.00\text{cm}^3 $

$latex \text{density}=\frac{\text{mass}}{\text{volume}}=\frac{90.7\text{g}}{8.00\text{cm}^3}=\frac{11.3\text{g}}{1.00\text{cm}^3}=11.3\;\text{g}/\text{cm}^3 $

(We will discuss the reason for rounding to the first decimal place in the next section.)

 

Test Yourself a) To three decimal places, what is the volume of a cube (cm³) with an edge length of 0.843 cm?

b) If the cube in part a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?

  Answers a) 0.599 cm³          b) 8.91 g/cm³

 

To learn more about the relationship between mass, volume, and density, use this interactive simulator to explore the density of different materials, like wood, ice, brick, and aluminum.

Example 2

This PhET simulation illustrates another way to determine density, using displacement of water. Determine the density of the red and yellow blocks.

 

Solution When you open the density simulation and select Same Mass, you can choose from several 5.00-kg colored blocks that you can drop into a tank containing 100.00 L water. The yellow block floats (it is less dense than water), and the water level rises to 105.00 L. While floating, the yellow block displaces 5.00 L water, an amount equal to the weight of the block. The red block sinks (it is more dense than water, which has density = 1.00 kg/L), and the water level rises to 101.25 L.

The red block therefore displaces 1.25 L water, an amount equal to the volume of the block. The density of the red block is:

$latex \text{density}=\frac{\text{mass}}{\text{volume}}=\frac{5.00\;\text{kg}}{1.25\;\text{L}}=4.00 \text{kg/L} $

Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:

$latex \text{density}=\frac{\text{mass}}{\text{volume}}=\frac{\text{5.00 kg}}{\text{10.00 L}}=0.500 \text{kg/L} $

 

Test Yourself Remove all of the blocks from the water and add the green block to the tank of water, placing it approximately in the middle of the tank. Determine the density of the green block.

 

Answer

2.00 kg/L

The Angstrom Unit

Although not an SI unit, the angstrom (Å) is a useful unit of length. It is one ten-billionth of a meter, or 10⁻¹⁰ m. Why is it a useful unit? The ultimate particles that compose all matter are about 10⁻¹⁰ m in size, or about 1 Å. This makes the angstrom a natural—though not approved—unit for describing these particles.

The angstrom unit is named after Anders Jonas Ångström, a nineteenth-century Swedish physicist. Ångström’s research dealt with light being emitted by glowing objects, including the sun. Ångström studied the brightness of the different colors of light that the sun emitted and was able to deduce that the sun is composed of the same kinds of matter that are present on the earth. By extension, we now know that all matter throughout the universe is similar to the matter that exists on our own planet.

Anders Jonas Ångstrom, a Swedish physicist, studied the light coming from the sun. His contributions to science were sufficient to have a tiny unit of length named after him, the angstrom, which is one ten-billionth of a meter.

Source: Photo of the sun courtesy of NASA’s Solar Dynamics Observatory, http://commons.wikimedia.org/wiki/File:The_Sun_by_the_Atmospheric_Imaging_Assembly_of_NASA%27s_Solar_Dynamics_Observatory_-_20100801.jpg.

Key Concepts and Summary

Measurements provide quantitative information that is critical in studying and practicing chemistry. Each measurement has an amount, a unit for comparison, and an uncertainty. Measurements can be represented in either decimal or scientific notation. Scientists primarily use the SI (International System) or metric systems. We use base SI units such as meters, seconds, and kilograms, as well as derived units, such as liters (for volume) and g/cm³ (for density). In many cases, we find it convenient to use unit prefixes that yield fractional and multiple units, such as microseconds (10⁻⁶ seconds) and megahertz (10⁶ hertz), respectively.

Key Equations

• $latex \text{density}=\frac{\text{mass}}{\text{volume}} $

Activity

Make yourself a stack of small sized Qcards to help you learn your common unit prefixes, which is important because you will use later as conversion factors for unit conversions.  On one side have the common unit prefix associated with a base unit (e.g. 1 kg) and on the other side have its equivalence in terms of the base unit (e.g. 10³ g).  Make a complete set of using all the common unit prefixes from Table 2 and pick and choose different base units from Table 1.  Then use these Qcards to quiz yourself.

Exercises

1.  Identify the unit in each quantity.

a)  2 boxes of crayons     b)  3.5 grams of gold

2.  Identify the unit in each quantity.

a)  32 oz of cheddar cheese     b)  0.045 cm³ of water 

3.  Identify the unit in each quantity.

a)  9.58 s (the current world record in the 100 m dash) b)  6.14 m (the current world record in the pole vault)

4.  Identify the unit in each quantity.

a)  2 dozen eggs b)  2.4 km/s (the escape velocity of the moon, which is the velocity you need at the surface to escape the moon’s gravity)

5.  Indicate what multiplier each prefix represents.

a)  k     b)  m     c)  M

6.  Indicate what multiplier each prefix represents.

a)  c     b)  G     c)  μ

7.  Give the prefix that represents each multiplier.

a)  1/1,000th ×     b)  1,000 ×      c)  1,000,000,000 ×

8.  Give the prefix that represents each multiplier.

a)  1/1,000,000,000th ×     b)  1/100th ×     c)  1,000,000 ×   9. Complete the following table with the missing information.

Unit	Abbreviation
kilosecond
	mL
	Mg
centimeter

 

10.Complete the following table with the missing information.

Unit	Abbreviation
kilometer per second
second
	cm3
	μL
nanosecond

  11.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  3.44 × 10⁻⁶ s     b)  3,500 L     c)  0.045 m 12.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  0.000066 m/s (Hint: you need consider only the unit in the numerator.) b)  4.66 × 10⁶ s c)  7,654 L

13.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  43,600 mL     b)  0.0000044 m     c)  1,438 ms

14.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  0.000000345 m³     b)  47,000,000 mm³     c)  0.00665 L

15.  Multiplicative prefixes are used for other units as well, such as computer memory. The basic unit of computer memory is the byte (b). What is the unit for one million bytes? 16.  You may have heard the terms microscale or nanoscale to represent the sizes of small objects. What units of length do you think are useful at these scales? What fractions of the fundamental unit of length are these units? 17.  Acceleration is defined as a change in velocity per time. Propose a unit for acceleration in terms of the fundamental SI units. 18.  Density is defined as the mass of an object divided by its volume. Propose a unit of density in terms of the fundamental SI units.

19.  Is a meter about an inch, a foot, a yard, or a mile?

20.  Indicate the SI base units or derived units that are appropriate for the following measurements:

a) the mass of the moon

b) the distance from Dallas to Oklahoma City

c) the speed of sound

d) the density of air

e) the temperature at which alcohol boils

f) the area of the state of Delaware

g) the volume of a flu shot or a measles vaccination

21.  Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.

a) c     b) d     c) G     d) k     e) m     f) n      g) p     h) T

22.  Visit this PhET density simulation and select the Same Volume Blocks.

a) What are the mass, volume, and density of the yellow block?

b) What are the mass, volume and density of the red block?

c) List the block colors in order from smallest to largest mass.

d) List the block colors in order from lowest to highest density.

e) How are mass and density related for blocks of the same volume?

23.  Visit this PhET density simulation and select Mystery Blocks.

a) Pick one of the Mystery Blocks and determine its mass, volume, density, and its likely identity.

b) Pick a different Mystery Block and determine its mass, volume, density, and its likely identity.

c) Order the Mystery Blocks from least dense to most dense. Explain.

  Answers 1. a)  boxes of crayons     b)  grams of gold

2.  a)  oz of cheddar cheese     b) cm³ of water

3.  a)  seconds     b)  meters

4.  a)  dozen of eggs     b)  km/s

5.  a)  1,000 ×     b)  1/1,000 ×     c)  1,000,000 ×

6.  a)  1/100 x     b) 1,000,000,000 x     c) 1/1,000,000 x

7.  a)  milli-     b)  kilo-     c)  giga-

8.  a)  nano-     b)  centi-     c)  mega-

9.

Unit	Abbreviation
kilosecond	ks
milliliter	mL
megagram	Mg
centimeter	cm

10.

Unit	Abbreviation
kilometer per second	km/s
second	s
cubic centimeter	cm3
microliter	μL
nanosecond	ns

11.  a)  3.44 μs     b)  3.5 kL     c)  4.5 cm 12.  a) 66 µm/s     b) 4.66 Ms     c) 7.654 kL 13.  a)  43.6 L     b)  4.4 µm     c)  1.438 s 14.  a) 345 mm³     b) 47 dm³     c) 6.65 mL 15. megabytes (Mb) 16.  microscale = µm, 1/1,000,000       nanoscale = nm, 1/1,000,000,000 17. meters/second² 18. kg/m³

19. about a yard

20. a) kilograms      b) meters      c) kilometers/second      d) kilograms/cubic meter      e) kelvin      f) square meters      g) cubic meters

21. a) centi-, × 10⁻²      b) deci-, × 10⁻¹      c) Giga-, × 10⁹      d) kilo-, × 10³      e) milli-, × 10⁻³                          f) nano-, × 10⁻⁹      g) pico-, × 10⁻¹²      h) tera-, × 10¹²

22. a) 8.00 kg, 5.00 L, 1.60 kg/L      b) 2.00 kg, 5.00 L, 0.400 kg/L      c) red < green < blue < yellow              d) If the volumes are the same, then the density is directly proportional to the mass.

23. a) and b) answer is one of the following. A/yellow: mass = 65.14 kg, volume = 3.38 L, density = 19.3 kg/L, likely identity = gold. B/blue: mass = 0.64 kg, volume = 1.00 L, density = 0.64 kg/L, likely identity = apple. C/green: mass = 4.08 kg, volume = 5.83 L, density = 0.700 kg/L, likely identity = gasoline. D/red: mass = 3.10 kg, volume = 3.38 L, density = 0.920 kg/L, likely identity = ice; and E/purple: mass = 3.53 kg, volume = 1.00 L, density = 3.53 kg/L, likely identity = diamond. (c) B/blue/apple (0.64 kg/L) < C/green/gasoline (0.700 kg/L) < C/green/ice (0.920 kg/L) < D/red/diamond (3.53 kg/L) < A/yellow/gold (19.3 kg/L)

Glossary

Celsius (°C): unit of temperature; water freezes at 0 °C and boils at 100 °C on this scale cubic centimeter (cm³ or cc): volume of a cube with an edge length of exactly 1 cm cubic meter (m³): SI unit of volume density: ratio of mass to volume for a substance or object kelvin (K): SI unit of temperature; 273.15 K = 0 ºC kilogram (kg): standard SI unit of mass; 1 kg = approximately 2.2 pounds length: measure of one dimension of an object liter (L): (also, cubic decimeter) unit of volume; 1 L = 1,000 cm³ meter (m): standard metric and SI unit of length; 1 m = approximately 1.094 yards milliliter (mL): 1/1,000 of a liter; equal to 1 cm³ second (s): SI unit of time SI units (International System of Units): standards fixed by international agreement in the International System of Units (Le Système International d’Unités) unit: standard of comparison for measurements volume: amount of space occupied by an object

]]> 1284 2084 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-physical-and-chemical-properties/ Thu, 12 Apr 2018 02:51:08 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-physical-and-chemical-properties/

Learning Objectives

By the end of this section, you will be able to:

• Identify properties of and changes in matter as physical or chemical
• Identify properties of matter as extensive or intensive

Recall that chemistry is the study of matter, its properties, the changes that matter undergoes and the energy associated with these changes. In this chapter, we’ll take a closer look at matter and energy and how they are related. When matter undergoes change, the process is often accompanied by a change in energy — heat, light, sound, kinetic energy of moving matter, etc...  If heat is evolved during a change (is released) the change is exothermic. If heat is needs to be supplied, the change is endothermic. An important distinction, is that heat is energy that flows due to a temperature difference, while temperature is a measure of the average kinetic energy of the molecules in a substance. The faster they move, the “hotter” it is.

The characteristics that enable us to distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change. A physical change is a change in the state (Figure 1) or properties of matter without any accompanying change in its chemical composition (the identities of the substances contained in the matter), such as dissolution and dilution.

[caption id="attachment_3250" align="aligncenter" width="543"] Figure 1. The different phase changes that matter can undergo.[/caption]

We observe a physical change when wax melts, when sugar dissolves in coffee, and when steam condenses into liquid water (Figure 2). Other examples of physical changes include magnetizing and demagnetizing metals (as is done with common antitheft security tags) and grinding solids into powders (which can sometimes yield noticeable changes in color). In each of these examples, there is a change in the physical state, form, or properties of the substance, but no change in its chemical composition.

[caption id="" align="aligncenter" width="624"] Figure 2. (a) Wax undergoes a physical change when solid wax is heated and forms liquid wax. (b) Steam condensing inside a cooking pot is a physical change, as water vapor is changed into liquid water. (credit a: modification of work by “95jb14”/Wikimedia Commons; credit b: modification of work by “mjneuby”/Flickr)[/caption]

The change of one type of matter into another type (or the inability to change) is a chemical property. Examples of chemical properties include flammability, toxicity, acidity, reactivity (many types), and heat of combustion. Iron, for example, combines with oxygen in the presence of water to form rust; chromium does not oxidize (Figure 3). Nitroglycerin is very dangerous because it explodes easily; neon poses almost no hazard because it is very unreactive.

[caption id="" align="aligncenter" width="580"] Figure 3. (a) One of the chemical properties of iron is that it rusts; (b) one of the chemical properties of chromium is that it does not. (credit a: modification of work by Tony Hisgett; credit b: modification of work by “Atoma”/Wikimedia Commons)[/caption]

To identify a chemical property, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure 4).

[caption id="" align="aligncenter" width="581"] Figure 4. (a) Copper and nitric acid undergo a chemical change to form copper nitrate and brown, gaseous nitrogen dioxide. (b) During the combustion of a match, cellulose in the match and oxygen from the air undergo a chemical change to form carbon dioxide and water vapor. (c) Cooking red meat causes a number of chemical changes, including the oxidation of iron in myoglobin that results in the familiar red-to-brown color change. (d) A banana turning brown is a chemical change as new, darker (and less tasty) substances form. (credit b: modification of work by Jeff Turner; credit c: modification of work by Gloria Cabada-Leman; credit d: modification of work by Roberto Verzo)[/caption]

Example 1

Classify each of the following as either a physical property, or a chemical property: a)  The boiling point of water is 100^oC b)  Oxygen is a gas c)  Sugar ferments to form alcohol

 

Solution  

a)   Although this property describes a change, this change does not involve a change in substance. H₂O remains H₂O despite what state it is in. Thus, this is a physical property.

b)   This is an inherent property, and is therefore a physical property.

c)   This property involves a change in substance, from sugar to alcohol. This is a chemical property.

 

Test Yourself

Classify each of the following as either a physical property, or a chemical property: a) This page is white     b) Wood burns      c) Milk curdles if left out

 

Answers

a) physical property   b) chemical property    c) chemical property

Properties of matter fall into one of two categories. If the property depends on the amount of matter present, it is an extensive property. The mass and volume of a substance are examples of extensive properties; for instance, a gallon of milk has a larger mass and volume than a cup of milk. The value of an extensive property is directly proportional to the amount of matter in question. If the property of a sample of matter does not depend on the amount of matter present, it is an intensive property. Temperature is an example of an intensive property. If the gallon and cup of milk are each at 20 °C (room temperature), when they are combined, the temperature remains at 20 °C. As another example, consider the distinct but related properties of heat and temperature. A drop of hot cooking oil spattered on your arm causes brief, minor discomfort, whereas a pot of hot oil yields severe burns. Both the drop and the pot of oil are at the same temperature (an intensive property), but the pot clearly contains much more heat (extensive property).

Example 2

Classify each of the following as either a physical change, or a chemical change: a)  Steam condensing on a shower mirror b)  Iron forming rust c)  An antacid tablet fizzes when it comes in contact with stomach acid d)  Salt dissolves in water

 

Solution  

a)   The steam is water vapor, and when it condenses, it forms liquid water on the mirror. This is a physical change.

b)   Iron reacts with the oxygen in air, forming an iron oxide, which is rust. This is a chemical change.

c)   The fizzing in the water is the release of carbon dioxide gas when it comes in contact with acid. This is a chemical change.

d)   Dissolving is considered a physical change. Even though the bonds of salt are pulled apart when dissolved, they do not form new bonds, or a new substance. If you evaporate the water, salt will remain.

 

Test Yourself

Classify each of the following as either a physical change, or a chemical change: a)  A rubber band stretches when you pull it b)  Acetone removes nail polish c)  Copper is melted at high temperatures d)  Silver metal tarnishes over time

  Answers

a) physical change        b) physical change (dissolving) c) physical change        d) chemical change

Example 3

Describe each process as a physical change or a chemical change.

a) Water in the air turns into snow.

b) A person’s hair is cut.

c) Bread dough becomes fresh bread in an oven.

 

Solution

a) Because the water is going from a gas phase to a solid phase, this is a physical change.

b) Your long hair is being shortened. This is a physical change.

c) Because of the oven’s temperature, chemical changes are occurring in the bread dough to make fresh bread. These are chemical changes. (In fact, a lot of cooking involves chemical changes.)

 

Test Yourself

Identify each process as a physical change or a chemical change.

a) A fire is raging in a fireplace.

b) Water is warmed to make a cup of coffee.

 

Answers

a) chemical change          b) physical change

Hazard Diamond

You may have seen the symbol shown in Figure 5 on containers of chemicals in a laboratory or workplace. Sometimes called a “fire diamond” or “hazard diamond,” this chemical hazard diamond provides valuable information that briefly summarizes the various dangers of which to be aware when working with a particular substance.

[caption id="" align="aligncenter" width="1200"] Figure 5. The National Fire Protection Agency (NFPA) hazard diamond summarizes the major hazards of a chemical substance.[/caption]

The National Fire Protection Agency (NFPA) 704 Hazard Identification System was developed by NFPA to provide safety information about certain substances. The system details flammability, reactivity, health, and other hazards. Within the overall diamond symbol, the top (red) diamond specifies the level of fire hazard (temperature range for flash point). The blue (left) diamond indicates the level of health hazard. The yellow (right) diamond describes reactivity hazards, such as how readily the substance will undergo detonation or a violent chemical change. The white (bottom) diamond points out special hazards, such as if it is an oxidizer (which allows the substance to burn in the absence of air/oxygen), undergoes an unusual or dangerous reaction with water, is corrosive, acidic, alkaline, a biological hazard, radioactive, and so on. Each hazard is rated on a scale from 0 to 4, with 0 being no hazard and 4 being extremely hazardous.

Decomposition of Water / Production of Hydrogen

Water consists of the elements hydrogen and oxygen combined in a 2 to 1 ratio. Water can undergo a chemical change involving the water molecules being broken down into hydrogen and oxygen gases by the addition of energy. One way to do this is with a battery or power supply, as shown in (Figure 6).

[caption id="" align="aligncenter" width="1200"] Figure 6. The decomposition of water is shown at the macroscopic, microscopic, and symbolic levels. The battery provides an electric current (microscopic) that decomposes water. At the macroscopic level, the liquid separates into the gases hydrogen (on the left) and oxygen (on the right). Symbolically, this change is presented by showing how liquid H₂O separates into H₂ and O₂ gases.[/caption]

The breakdown of water involves a rearrangement of the atoms in water molecules into different molecules, each composed of two hydrogen atoms and two oxygen atoms, respectively. Two water molecules form one oxygen molecule and two hydrogen molecules. The representation for what occurs, $latex 2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g) $, will be explored in more depth in later chapters.

The two gases produced have distinctly different properties. Oxygen is not flammable but is required for combustion of a fuel, and hydrogen is highly flammable and a potent energy source. How might this knowledge be applied in our world? One application involves research into more fuel-efficient transportation. Fuel-cell vehicles (FCV) run on hydrogen instead of gasoline (Figure 7). They are more efficient than vehicles with internal combustion engines, are nonpolluting, and reduce greenhouse gas emissions, making us less dependent on fossil fuels. FCVs are not yet economically viable, however, and current hydrogen production depends on natural gas. If we can develop a process to economically decompose water, or produce hydrogen in another environmentally sound way, FCVs may be the way of the future.

[caption id="" align="aligncenter" width="975"] Figure 7. A fuel cell generates electrical energy from hydrogen and oxygen via an electrochemical process and produces only water as the waste product.[/caption]

While many elements differ dramatically in their chemical and physical properties, some elements have similar properties. We can identify sets of elements that exhibit common behaviors. For example, many elements conduct heat and electricity well, whereas others are poor conductors. These properties can be used to sort the elements into three classes: metals (elements that conduct well), nonmetals (elements that conduct poorly), and metalloids (elements that have properties of both metals and nonmetals).

The periodic table is a table of elements that places elements with similar properties close together (Figure 6). You will learn more about the periodic table as you continue your study of chemistry.

[caption id="" align="aligncenter" width="1300"] Figure 6. The periodic table shows how elements may be grouped according to certain similar properties. Note the background color denotes whether an element is a metal, metalloid, or nonmetal, whereas the element symbol color indicates whether it is a solid, liquid, or gas.[/caption]

Key Concepts and Summary

All substances have distinct physical and chemical properties, and may undergo physical or chemical changes. Physical properties, such as hardness and boiling point, and physical changes, such as melting or freezing, do not involve a change in the composition of matter. Chemical properties, such flammability and acidity, and chemical changes, such as rusting, involve production of matter that differs from that present beforehand.

Measurable properties fall into one of two categories. Extensive properties depend on the amount of matter present, for example, the mass of gold. Intensive properties do not depend on the amount of matter present, for example, the density of gold. Heat is an example of an extensive property, and temperature is an example of an intensive property.

Exercises

1. Classify each of the following changes as physical or chemical:

a) condensation of steam

b) burning of gasoline

c) souring of milk

d) dissolving of sugar in water

e) melting of gold

2. The volume of a sample of oxygen gas changed from 10 mL to 11 mL as the temperature changed. Is this a chemical or physical change? 3. Explain the difference between extensive properties and intensive properties. 4. The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (V).

$latex \text{density}= \frac{\text{mass}}{\text{volume}} $ $latex \text{d} = \frac{\text{m}}{\text{V}} $

Considering that mass and volume are both extensive properties, explain why their ratio, density, is intensive.

5. Does each statement represent a physical property or a chemical property? a)  Sulfur is yellow. b)  Steel wool burns when ignited by a flame. c)  A gallon of milk weighs over eight pounds. 6. Does each statement represent a physical property or a chemical property? a)  A pile of leaves slowly rots in the backyard. b)  In the presence of oxygen, hydrogen can interact to make water. c)  Gold can be stretched into very thin wires. 7. Does each statement represent a physical change or a chemical change? a)  Water boils and becomes steam. b)  Food is converted into usable form by the digestive system. c)  The alcohol in many thermometers freezes at about −40 degrees Fahrenheit. 8. Does each statement represent a physical change or a chemical change? a) Graphite, a form of elemental carbon, can be turned into diamond, another form of carbon, at very high temperatures and pressures. b)  The elements sodium and chlorine come together to make a new substance called sodium chloride.   Answers

1. a) physical;   b) chemical;   c) chemical;   d) physical;   e) physical

2. physical

3. The value of an extensive property depends upon the amount of matter being considered, whereas the value of an intensive property is the same regardless of the amount of matter being considered.

4. Being extensive properties, both mass and volume are directly proportional to the amount of substance under study. Dividing one extensive property by another will in effect “cancel” this dependence on amount, yielding a ratio that is independent of amount (an intensive property).

5. a)  physical property   b)  chemical property    c)  physical property 6. a)  chemical property   b)  chemical property    c)  physical property

7. a)  physical change   b)  chemical change   c) physical change 8. a)  physical change   b)  chemical change

Glossary

chemical change: change producing a different kind of matter from the original kind of matter chemical property: behavior that is related to the change of one kind of matter into another kind of matter endothermic: if heat is needs to be supplied, for a change to occur

energy: the ability to do “work”— that is, for a force to act on something and push some distance exothermic: if heat is released during a change extensive property: property of a substance that depends on the amount of the substance intensive property: property of a substance that is independent of the amount of the substance physical change: change in the state or properties of matter that does not involve a change in its chemical composition physical property: characteristic of matter that is not associated with any change in its chemical composition]]> 1290 1247 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-6-mathematical-treatment-of-measurement-results/ Thu, 12 Apr 2018 02:51:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-6-mathematical-treatment-of-measurement-results/

Learning Objectives

By the end of this section, you will be able to:

• Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities
• Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties
• Learn about the various temperature scales that are commonly used in chemistry and how to convert from one scale to another.

It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:

$latex \text{speed}= \frac{\text{distance}}{\text{time}} $

An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of

$latex \frac{\text{100 m}}{\text{10 s}} = \text{10 m/s} $

Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:

$latex \text{time} = \frac{\text{distance}}{\text{speed}} $

The time can then be computed as:

$latex \frac{\text{25 m}}{\text{10 m/s}} = \text{2.5 s} $

Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”

These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.

Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

$latex \frac{\text{2.54 cm}}{\text{1 in.}}\;\text{(2.54 cm = 1 in.) or 2.54}\frac{\text{cm}}{\text{in}} $

Several other commonly used conversion factors are given in Table 1.

Length	Volume	Mass
1 m = 1.0936 yd	1 L = 1.0567 qt	1 kg = 2.2046 lb
1 in. = 2.54 cm (exact)	1 qt = 0.94635 L	1 lb = 453.59 g
1 km = 0.62137 mi	1 ft3 = 28.317 L	1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m	1 tbsp = 14.787 mL	1 (troy) oz = 31.103 g
Table 1. Common Conversion Factors

When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

$latex \text{34 in.} \times \frac{\text{2.54 cm}}{\text{1}\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{in}} = \text{86 cm} $

Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield $latex \frac{\text{in.} \times \text{cm}}{\text{in.}}$. Just as for numbers, a ratio of identical units is also numerically equal to one, $latex \frac{\text{in.}}{\text{in.}}=\text{1} $, and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.

Example 1

The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table 1).

 

Solution If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.

$latex x \ \text{oz} = \text{125 \text{g}} \times \text{unit conversion factor}$

We write the unit conversion factor in its two forms:

$latex \frac{1 \text{oz}}{28.349 \text{g}} \;\text{and}\;\frac{28.349 \text{g}}{1 \text{oz}}$

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

$latex \begin{array}{r @{{}={}} l} x\;\text{oz} & 125\; \rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{g}\;\times\;\frac{\text{1 oz}}{\text{28.349 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}g}} \\[1em] & (\frac{125}{\text{28.349}})\text{oz} \\[1em] & \text{4.41 oz (three significant figures)} \end{array}$

 

Test Yourself Convert a volume of 9.345 qt to liters.

  Answer 8.844 L

Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

Example 2

What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.

 

Solution Since density = $latex \frac{\text{mass}}{\text{volume}} $, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A × unit conversion factor. The necessary conversion factors are given in Table 1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:

$latex 9.26\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{lb} \times \frac{453.59\;\text{g}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{lb}} = 4.20 \times 10^3\;\text{g} $

We need to use two steps to convert volume from quarts to milliliters.

1. Convert quarts to liters.
   $latex 4.00\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{qt} \times \frac{1 \text{L}}{1.0567 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{qt}} = 3.78 \text{L} $
2. Convert liters to milliliters.
   $latex 3.78 \;\rule[0.75ex]{0.75em}{0.1ex}\hspace{-0.75em}\text{L} \times \frac{1000 \;\text{mL}}{\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} = 3.78 \;\text{L} \times 10^3 \;\text{mL} $

Then,

$latex \text{density} = \frac{4.20 \times 10^3 \;\text{g}}{3.78 \times 10^3 \;\text{mL}} = 1.11 \;\text{g/mL} $

Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:

$latex \frac{9.26 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{lb}}{4.00 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{qt}} \times \frac{453.59 \text{g}}{1 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1.0567 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{qt}}{1 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}}{1000 \text{mL}} = 1.11 \text{g/mL} $

 

Test Yourself What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?

  Answer $latex 2.956 \times 10^{-2} \text{L} $

Example 3

While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.

a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?

b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?

 

Solution a) We first convert distance from kilometers to miles:

$latex 1250\;\text{km} \times \frac{0.62137\;\text{mi}}{1\;\text{km}} = 777\;\text{mi} $

and then convert volume from liters to gallons:

$latex 213\;\rule[0.75ex]{0.65em}{0.1ex}\hspace{-0.65em}\text{L} \times \frac{1.0567\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1\;\text{gal}}{4\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}} = 56.3\;\text{gal} $

Then,

$latex \text{(average) mileage} = \frac{777 \;\text{mi}}{56.3\;\text{gal}} = 13.8\;\text{miles/gallon} = 13.8\;\text{mpg} $

Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:

$latex \frac{1250\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{km}}{213\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{0.62137\;\text{mi}}{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{km}} \times \frac{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}}{1.0567\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}} \times \frac{4\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}}{1\;\text{gal}} = 13.8\;\text{mpg} $

b) Using the previously calculated volume in gallons, we find:

$latex \text{56.3 gal} \times \frac{\$3.80}{1\;\text{gal}} = \$214 $

 

Test Yourself A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).

a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?

b) If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?

  Answers a) 51 mpg          b) $62

Example 4

a) Convert 35.9 kL to liters. b)Convert 555 nm to meters.  

Solution

a) We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1,000 L/1 kL. Applying this conversion factor, we get

b) We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 10⁹ nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: 1 m/10⁹ nm. Applying this conversion factor, we get

In the final step, we expressed the answer in scientific notation.

 

Test Yourself

a) Convert 67.08 μL to liters.          b) Convert 56.8 m to kilometers.

 

Answers

a) 6.708 × 10⁻⁵ L          b) 5.68 × 10⁻² km

Example 5

Complete the following conversions       (use table 1.1 for any metric conversions;  Note: 1 L = 1.0567 qt )

a) 125 m = ? mm          b) 2.3 x 10^-6 Mg = ? g              c) 2.5 L = ? qt

 

Solution  

a)   $latex 125\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{m} \times \frac{1\;\text{mm}}{0.001\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}} = 1.25 \times 10^5\;\text{mm} $

Note that this equivalence statement comes from the literal “meaning” of milli in Table 1.1. Many students prefer to use the equivalence statement 1000 mm = 1 m. That is fine too, as long as you always remember to put the appropriate unit and number.

b)  $latex 2.3\times 10^-6\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{Mg} \times \frac{1\times 10^6\;\text{g}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{Mg}} = 2.3\;\text{g} $

c)  $latex 2.5\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{L} \times \frac{1.0567\;\text{qt}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} = 2.6\;\text{qt} $

Test Yourself

Complete the following conversions             (note: 1 in = 2.54 cm exactly)

a) 124 mL = ? L           b) 256 days = ? hrs                   c) 63.2 cm = ? in

Answers

a) 0.124 L          b) 6.14 x 10³ hrs          c) 24.9 in

Example 6

How many nm are in 26.5 feet? (12 in = 1 ft ,  and 2.54 cm = 1 in exactly)

 

Solution  

Likely, you do not have a direct conversion between feet and nm. So, instead, ask yourself “where can I go from feet”. The only possibility is inches. Then, where can you go from inches? …and so on. The overall path becomes:

 feet $latex \longrightarrow$ inches $latex \longrightarrow$ cm $latex \longrightarrow$ m $latex \longrightarrow$ nm

        

$latex 26.5\;\rule[0.75ex]{0.65em}{0.1ex}\hspace{-0.65em}\text{feet} \times \frac{12\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{in}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{foot}} \times \frac{2.54\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{cm}}{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{in}} \times \frac{0.01\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{cm}} \times \frac{1\;\text{nm}}{10^-9\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}} = 8.08 \times 10^9\;\text{nm} $  

Test Yourself

A marathon is 26.4 miles. If 1 mile = 1760 yards, and 1 m = 1.094 yards, how many km are in a marathon?

  Answer 42.5 km

Example 7

a)   A car is moving at 35.2 km/h. How many cm/s is this speed? b)   How many cm³are in 5 x 10²m³?

 

Solution  

a)  We must convert km $latex \longrightarrow$ m $latex \longrightarrow$ cm AND convert h $latex \longrightarrow$ min $latex \longrightarrow$ s. It does not matter which order we do this in. Note that if a unit is on the bottom of a fraction, we cancel it by putting that undesired unit on the top of a conversion factor.

           

$latex \frac{35.2\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{km}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{h}} \times \frac{1000\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{km}} \times \frac{1\;\text{cm}}{0.01\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}} \times \frac{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{h}}{60\;\rule[0.5ex]{0.5em}{0.2ex}\hspace{-0.5em}\text{min}} \times \frac{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{min}}{60\;\text{sec}} = 978\;\text{cm/s} $  

Remember… 1 m³= 1 m x 1 m x 1 m, so  1 m³= 100 cm x 100 cm x 100 cm, NOT 100 cm³

          

$latex 5\times 10^2\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{m}^{3} \times \frac{100\;\text{cm}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}}\times \frac{100\;\text{cm}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}}\times \frac{100\;\text{cm}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}} = 5 \times 10^8\;\text{cm}^{3} $

 

Test Yourself

Complete the following conversions            

a)  75 mi/h = ? m/s  (1760 yd = 1 mi and 1 m = 1.094 yd) b)  4.1 g/cm³= ? kg/m³

  Answers a) 34 m/s          b) 4.1 x 10³ kg/m³

Example 8

How many cubic centimeters are in 0.883 m³?

 

Solution

With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have

You should demonstrate to yourself that the three meter units do indeed cancel.

 

Test Yourself

How many cubic millimeters are present in 0.0923 m³?

 

Answer

9.23 × 10⁷ mm³

Example 9

Convert 88.4 m/min to meters/second.

 

Solution

We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1 min/60 s. Apply and perform the math:

Notice how the 88.4 automatically goes in the numerator. That’s because any number can be thought of as being in the numerator of a fraction divided by 1.

 

Test Yourself

Convert 0.203 m/min to meters/second.

 

Answer

0.00338 m/s or 3.38 × 10⁻³ m/s

[caption id="attachment_2116" align="alignnone" width="300"] Figure 1. How fast is fast? A common garden snail moves at a rate of about 0.2 m/min, which is about 0.003 m/s, which is 3 mm/s!  Source: “Grapevine snail”by Jürgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.[/caption]

Example 10

How many nanoseconds are in 368.09 μs?

 

Solution

You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-,

 

Test Yourself

How many milliliters are in 607.8 kL?

 

Answer

6.078 × 10⁸ mL

Example 11

A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.

 

Solution

Area is defined as the product of the two dimensions, which we then have to convert to square meters and express our final answer to the correct number of significant figures, which in this case will be three.

The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.

 

Test Yourself

What is the volume of a block in cubic meters whose dimensions are 2.1 cm × 34.0 cm × 118 cm?

 

Answer

0.0084 m³

Chemistry Is Everywhere: The Gimli Glider

On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed “the Gimli Glider.”

[caption id="attachment_2124" align="aligncenter" width="300"] Figure 2. The Gimli Glider is the Boeing 767 that ran out of fuel and glided to safety at Gimli Airport. The aircraft ran out of fuel because of confusion over the units used to express the amount of fuel. “Aircanada.b767′′ is in the the public domain.[/caption]

The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.

What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.

Conversion of Temperature Units

We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes. Because the volume of the liquid changes more than the volume of the glass, we can see the liquid expand when it gets warmer and contract when it gets cooler.

To mark a scale on a thermometer, we need a set of reference values: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).

Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:

$latex \text{length in feet} = \left( \frac {1\;\text{ft}}{12\;\text{in.}}\right) \times \text{length in inches} $

where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (y = mx + b). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (b).

The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as x and the Fahrenheit temperature as y, the slope, m, is computed to be:

$latex m = \frac{\Delta y}{\Delta x} = \frac{212 \;^{\circ}\text{F} - 32 \;^{\circ}\text{F}}{100 \;^{\circ}\text{C} - 0 \;^{\circ}\text{C}} = \frac{180 \;^{\circ}\text{F}}{100 \;^{\circ}\text{C}} = \frac{9 \;^{\circ}\text{F}}{5 \;^{\circ}\text{C}} $

The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:

$latex b = y - mx = 32\;^{\circ}\text{F} - \frac{9\;^{\circ}\text{F}}{5\;^{\circ}\text{C}} \times 0\;^{\circ}\text{C} = 32\;^{\circ}\text{F} $

The equation relating the temperature scales is then:

$latex T_{^\circ\text{F}} = (\frac{9 \;^\circ\text{F}}{5 \;^\circ\text{C}} \times T_{^\circ\text{C}}) + 32\;^\circ\text{C} $

An abbreviated form of this equation that omits the measurement units is:

$latex T_{^\circ\text{F}} = \frac{9}{5} \times T_{^\circ\text{C}} + 32 $

Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:

$latex T_{^\circ\text{C}} = \frac{5}{9} (T_{^\circ\text{F}} - 32) $

As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).

The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of $latex 1\;\frac{\text{K}}{^{\circ}\text{C}}$. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:

$latex T_{\text{K}} = T_{^\circ\text{C}} + 273.15 $

$latex T_{^\circ\text{C}} = T_{\text{K}} - 273.15 $

The 273.15 in these equations has been determined experimentally, so it is not exact. Figure 3 shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.

Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.

Example 12

Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?

 

Solution

$latex \text{K} = ^\circ\text{C} + \text{273.15} = \text{37.0} + \text{273.2} = \text{310.2 K} $

$latex ^\circ\text{F} = \frac{9}{5} ^\circ\text{C} + \text{32.0} = (\frac{9}{5} \times \text{37.0}) + \text{32.0} = \text{66.6} + \text{32.0} = \text{98.6}^\circ\text{F} $

 

Test Yourself Convert 80.92 °C to K and °F.

  Answers 354.07 K, 177.7 °F

Example 13

Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?

 

Solution

$latex ^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - \text{32}) = \frac{5}{9} \text{(450 - 32)} = \frac{5}{9} \times \text{418} = \text{232} \;^\circ\text{C} \longrightarrow \text{set oven to 230} \;^\circ\text{C} \text{(two significant figures)} $

$latex \text{K} = ^\circ\text{C} + \text{273.15} = \text{230} + \text{273} = \text{503 K} \longrightarrow \text{5.0} \times \text{10}^2\text{K} \text{(two significant figures)}$

 

Test Yourself Convert 50 °F to °C and K.

  Answers 10 °C, 280 K

Example 14

a) What is 98.6 °F in degrees Celsius? b) What is 25.0 °C in degrees Fahrenheit?  

Solution

a) Using the first formula from above, we have

$latex ^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - \text{32}) = \frac{5}{9} \text{(98.6 - 32)} = \frac{5}{9} \times \text{66.6} = \text{37.0} \;^\circ\text{C} $

  b) Using the second formula from above, we have

$latex ^\circ\text{F} = \frac{9}{5} ^\circ\text{C} + \text{32.0} = (\frac{9}{5} \times \text{25.0}) + \text{32.0} = \text{45.0} + \text{32.0} = \text{77.0}^\circ\text{F} $

 

Test Yourself

a) Convert 0 °F to degrees Celsius.

b) Convert 212 °C to degrees Fahrenheit.

 

Answers

a) −17.8 °C          b) 414 °F

Example 15

If normal room temperature is 72.0 °F, what is room temperature in degrees Celsius and kelvins?

 

Solution

First, we use the formula to determine the temperature in degrees Celsius:

$latex ^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - \text{32}) = \frac{5}{9} \text{(72.0 - 32)} = \frac{5}{9} \times \text{40.0} = \text{22.2} \;^\circ\text{C} $

Then we use the appropriate formula above to determine the temperature in the Kelvin scale:

K = 22.2 °C + 273.15 = 295.4 K

So, room temperature is about 295 K.

 

Test Yourself

What is 98.6 °F on the Kelvin scale?

 

Answer

310.2 K

Food and Drink App: Cooking Temperatures

Because degrees Fahrenheit is the common temperature scale in the United States, kitchen appliances, such as ovens, are calibrated in that scale. A cool oven may be only 150°F, while a cake may be baked at 350°F and a chicken roasted at 400°F. The broil setting on many ovens is 500°F, which is typically the highest temperature setting on a household oven.

People who live at high altitudes, typically 2,000 ft above sea level or higher, are sometimes urged to use slightly different cooking instructions on some products, such as cakes and bread, because water boils at a lower temperature the higher in altitude you go, meaning that foods cook slower. For example, in Cleveland water typically boils at 212°F (100°C), but in Denver, the Mile-High City, water boils at about 200°F (93.3°C), which can significantly lengthen cooking times. Good cooks need to be aware of this.

At the other end is pressure cooking. A pressure cooker is a closed vessel that allows steam to build up additional pressure, which increases the temperature at which water boils. A good pressure cooker can get to temperatures as high as 252°F (122°C); at these temperatures, food cooks much faster than it normally would. Great care must be used with pressure cookers because of the high pressure and high temperature. (When a pressure cooker is used to sterilize medical instruments, it is called an autoclave.)

Other countries use the Celsius scale for everyday purposes. Therefore, oven dials in their kitchens are marked in degrees Celsius. It can be confusing for US cooks to use ovens abroad—a 425°F oven in the United States is equivalent to a 220°C oven in other countries. These days, many oven thermometers are marked with both temperature scales.

Need a refresher or more practice with unit conversion? Visit this site (https://viuvideos.viu.ca/media/Unit+Conversion/0_o671v9j6) to go over the basics of unit conversions.

Video source: Unit conversion by keyj

Key Concepts and Summary

Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.

Key Equations

• $latex T_{^\circ\text{C}} = \frac{5}{9} \times T_{^\circ\text{F}} - 32 $
• $latex T_{^\circ\text{F}} = \frac{9}{5} \times T_{^\circ\text{C}} + 32 $
• $latex T_\text{K} = {^\circ\text{C}} + 273.15 $
• $latex T_{^\circ\text{C}} = \text{K} - 273.15 $

Exercises

1.  Perform the following conversions.

a)  255°F to degrees Celsius                 b)  −255°F to degrees Celsius c)  50.0°C to degrees Fahrenheit          d)  −50.0°C to degrees Fahrenheit

2.  Perform the following conversions.

a)  100.0°C to kelvins                           b)  −100.0°C to kelvins c)  100 K to degrees Celsius                 d)  300 K to degrees Celsius

3.  Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius?

4.  The hottest temperature ever recorded on the surface of the earth was 136°F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins?

5.  Write the two conversion factors that exist between the two given units.

a)  milliliters and laters      b)  microseconds and seconds       c)  kilometers and meters

6.  Perform the following conversions.

a)  5.4 km to meters      b)  0.665 m to millimeters      c)  0.665 m to kilometers

7.  Perform the following conversions.

a)  17.8 μg to grams      b)  7.22 × 10² kg to grams      c)  0.00118 g to nanograms

8.  Perform the following conversions.

a)  9.44 m² to square centimetres      b)  3.44 × 10⁸ mm³ to cubic meters

9.  Why would it be inappropriate to convert square centimeters to cubic meters?

10.  Perform the following conversions.

a)  45.0 m/min to meters/second       b)  0.000444 m/s to micrometers/second c)  60.0 km/h to kilometers/second

11.  Perform the following conversions.

a)  0.674 kL to milliliters      b)  2.81 × 10¹² mm to kilometers      c)  94.5 kg to milligrams

12.  Perform the following conversions.

a)  6.77 × 10¹⁴ ms to kilo seconds       b)  34,550,000 cm to kilometers

13.  Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.

a)  88 ft/s to miles/hour (Hint: use 5,280 ft = 1 mi.)      b)  0.00667 km/h to meters/second

14.  What is the area in square millimeters of a rectangle whose sides are 2.44 cm × 6.077 cm? Express the answer to the proper number of significant figures.

15.  The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square centimeters if its base is 1.007 m and its height is 0.665 m? Express the answer to the proper number of significant figures.

16.  Write conversion factors (as ratios) for the number of:

a) yards in 1 meter      b) liters in 1 liquid quart      c) pounds in 1 kilogram

17.  The label on a soft drink bottle gives the volume in two units: 2.0 L and 67.6 fl oz. Use this information to derive a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor? 18.  Soccer is played with a round ball having a circumference between 27 and 28 in. and a weight between 14 and 16 oz. What are these specifications in units of centimeters and grams? 19.  How many milliliters of a soft drink are contained in a 12.0-oz can? 20.  The diameter of a red blood cell is about 3 × 10⁻⁴ in. What is its diameter in centimeters? 21.  Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less? 22.  Many medical laboratory tests are run using 5.0 μL blood serum. What is this volume in milliliters? 23.  Use scientific notation to express the following quantities in terms of the SI base units:

a) 0.13 g      b) 232 Gg      c) 5.23 pm      d) 86.3 mg      e) 37.6 cm      f) 54 μm      g) 1 Ts      h) 27 ps               i) 0.15 mK

24.  Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank? 25.  A long ton is defined as exactly 2240 lb. What is this mass in kilograms? 26.  Make the conversion indicated in each of the following:

a) the length of a soccer field, 120 m (three significant figures), to feet

b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers

c) the area of an 8.5 t 11-inch sheet of paper in cm²

d) the displacement volume of an automobile engine, 161 in.³, to liters

e) the estimated mass of the atmosphere, 5.6 t 10¹⁵ tons, to kilograms

f) the mass of a bushel of rye, 32.0 lb, to kilograms

g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)

27.  A chemist’s 50-Trillion Angstrom Run would be an archeologist’s 10,900 cubit run. How long is one cubit in meters and in feet? (1 Å = 1 × 10⁻⁸ cm) 28.  As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, whose density is 1.83 g/mL? 29.  A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds? 30.  Solve these problems about lumber dimensions.

a) To describe to a European how houses are constructed in the US, the dimensions of “two-by-four” lumber must be converted into metric units. The thickness × width × length dimensions are 1.50 in. × 3.50 in. × 8.00 ft in the US. What are the dimensions in cm × cm × m?

b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?

31.  Calculate the density of aluminum if 27.6 cm³ has a mass of 74.6 g. 32.  Calculate these masses.

a) What is the mass of 6.00 cm³ of mercury, density = 13.5939 g/cm³?

b) What is the mass of 25.0 mL octane, density = 0.702 g/cm³?

33.  Calculate these volumes.

a) What is the volume of 25 g iodine, density = 4.93 g/cm³?

b) What is the volume of 3.28 g gaseous hydrogen, density = 0.089 g/L?

34.  Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin. 35.  Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin. 36.  Convert the boiling temperature of liquid ammonia, −28.1 °F, into degrees Celsius and kelvin. 37.  The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 °C. What was the temperature on the Fahrenheit scale?

  Answers 1.  a)  124°C      b)  −159°C      c)  122°F      d)  −58°F 2.  a)  373 K      b)  173 K      c)  −173°C      d)  27°C 3.   −273°C. This is the lowest possible temperature in degrees Celsius. 4.  57.8°C;  331 K 5.  a)  1,000 mL/1 L and 1 L/1,000 mL      b)  1,000,000 μs/1 s and 1 s/1,000,000 μs c)  1,000 m/1 km and 1 km1,000 m 6.  a)  5,400 m      b)  665 mm      c)  6.65 × 10⁻⁴ km 7.  a)  1.78 × 10⁻⁵ g      b)  7.22 × 10⁵ g      c) 1.18 × 10⁶ ng 8.  a)  94,400 cm²      b)  0.344 m³ 9.  One is a unit of area, and the other is a unit of volume. 10.  a)  0.75 m/s       b)  444 µm/s      c)  1.666 × 10⁻² km/s 11.  a)  674,000 mL      b)  2.81 × 10⁶ km      c)  9.45 × 10⁷ mg 12.  a)  6.77 × 10⁸ ks      b)  345.5 km 13.  a)  6.0 × 10¹ mi/h      b)  0.00185 m/s 14.  1.48 × 10³ mm² 15.  3.35 × 10³ cm²

16.  a) $latex \frac{\text{1.0936 yd}}{\text{1 m}} $   b) $latex \frac{\text{0.94635 L}}{\text{1 qt}} $          c) $latex \frac{\text{2.2046 lb}}{\text{1 kg}} $

17. $latex \frac{\text{2.0 L}}{\text{67.6 fl oz}} = \frac{\text{0.030 L}}{\text{1 fl oz}}$ Only two significant figures are justified.

18. 68–71 cm; 400–450 g

19. 355 mL

20. 8 × 10⁻⁴ cm

21. yes; weight = 89.4 kg

22. 5.0 × 10⁻³ mL

23. a) 1.3 × 10⁻⁴ kg      b) 2.32 × 10⁸ kg      c) 5.23 × 10⁻¹² m      d) 8.63 × 10⁻⁵ kg      e) 3.76 × 10⁻¹ m          f) 5.4 × 10⁻⁵ m      g) 1 × 10¹² s      h) 2.7 × 10⁻¹¹ s      i) 1.5 × 10⁻⁴ K

24. 45.4 L

25. 1.0160 × 10³ kg

26.  a) 394 ft      b) 5.9634 km      c) 6.0 × 10²      d) 2.64 L      e) 5.1 × 10¹⁸ kg      f) 14.5 kg      g) 324 mg

27. 0.46 m; 1.5 ft/cubit

28. Yes, the acid's volume is 123 mL.

29. 62.6 in (about 5 ft 3 in.) and 101 lb

30. (a) 3.81 cm × 8.89 cm × 2.44 m; (b) 40.6 cm

31. 2.70 g/cm³

32. (a) 81.6 g; (b) 17.6 g

33. (a) 5.1 mL; (b) 37 L

34. 5371 °F, 3239 K

35. −23 °C, 250 K

36. −33.4 °C, 239.8 K

37. 113 °F

Glossary

dimensional analysis: (also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities Fahrenheit: unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale unit conversion factor: ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit]]> 1292 2084 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-2/ Thu, 12 Apr 2018 02:51:10 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-2/ Your overall health and susceptibility to disease depends upon the complex interaction between your genetic makeup and environmental exposure, with the outcome difficult to predict. Early detection of biomarkers, substances that indicate an organism’s disease or physiological state, could allow diagnosis and treatment before a condition becomes serious or irreversible. Recent studies have shown that your exhaled breath can contain molecules that may be biomarkers for recent exposure to environmental contaminants or for pathological conditions ranging from asthma to lung cancer. Scientists are working to develop biomarker “fingerprints” that could be used to diagnose a specific disease based on the amounts and identities of certain molecules in a patient’s exhaled breath. An essential concept underlying this goal is that of a molecule’s identity, which is determined by the numbers and types of atoms it contains, and how they are bonded together. This chapter will describe some of the fundamental chemical principles related to the composition of matter, including those central to the concept of molecular identity.

]]> 1295 1293 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-1-early-ideas-in-atomic-theory/ Thu, 12 Apr 2018 02:51:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-1-early-ideas-in-atomic-theory/

Learning Objectives

By the end of this section, you will be able to:

• State the postulates of Dalton’s atomic theory
• Use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions

The language used in chemistry is seen and heard in many disciplines, ranging from medicine to engineering to forensics to art. The language of chemistry includes its own vocabulary as well as its own form of shorthand. Chemical symbols are used to represent atoms and elements. Chemical formulas depict molecules as well as the composition of compounds. Chemical equations provide information about the quality and quantity of the changes associated with chemical reactions.

This chapter will lay the foundation for our study of the language of chemistry. The concepts of this foundation include the atomic theory, the composition and mass of an atom, the variability of the composition of isotopes, ion formation, chemical bonds in ionic and covalent compounds, the types of chemical reactions, and the naming of compounds. We will also introduce one of the most powerful tools for organizing chemical knowledge: the periodic table.

Atomic Theory through the Nineteenth Century

The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos, a term derived from the Greek word for “indivisible.” They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four “elements”—fire, earth, air, and water—and could be infinitely divided. Interestingly, these philosophers thought about atoms and “elements” as philosophical concepts, but apparently never considered performing experiments to test their ideas.

The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton’s hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton’s atomic theory.

1. Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change.
2. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure 1). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties.

   [caption id="" align="aligncenter" width="496"] Figure 1. A pre-1982 copper penny (left) contains approximately 3 × 10²² copper atoms (several dozen are represented as brown spheres at the right), each of which has the same chemical properties. (credit: modification of work by “slgckgc”/Flickr)[/caption]

3. Atoms of one element differ in properties from atoms of all other elements.
4. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure 2).

   [caption id="" align="aligncenter" width="492"] Figure 2. Copper(II) oxide, a powdery, black compound, results from the combination of two types of atoms—copper (brown spheres) and oxygen (red spheres)—in a 1:1 ratio. (credit: modification of work by “Chemicalinterest”/Wikimedia Commons)[/caption]

5. Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure 3).

   [caption id="" align="aligncenter" width="1200"] Figure 3. When the elements copper (a shiny, red-brown solid, shown here as brown spheres) and oxygen (a clear and colorless gas, shown here as red spheres) react, their atoms rearrange to form a compound containing copper and oxygen (a powdery, black solid). (credit copper: modification of work by http://images-of-elements.com/copper.php)[/caption]

Dalton’s atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you’ve learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter).

Overall this gives us a “billiard ball” view of atoms. The shortcomings of his model, however are that it does not account for WHY elements combine as they do and does not explain the observed electrical charge of particles. More discoveries were needed to go beyond this model.

Example 1

In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?

 

Solution The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)

 

Test Yourself

In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?   Answer The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.

Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass. This statement is known as the law of definite proportions or the law of constant composition. The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table 1.

Sample	Carbon	Hydrogen	Mass Ratio
A	14.82 g	2.78 g	$latex \frac{14.82 \text{g carbon}}{2.78 \text{g hydrogen}} = \frac{5.33 \text{g carbon}}{1.00 \text{g hydrogen}} $
B	22.33 g	4.19 g	$latex \frac{22.33 \text{g carbon}}{4.19 \text{g hydrogen}} = \frac{5.33 \text{g carbon}}{1.00 \text{g hydrogen}} $
C	19.40 g	3.64 g	$latex \frac{19.40 \text{g carbon}}{3.64 \text{g hydrogen}} = \frac{5.33 \text{g carbon}}{1.00 \text{g hydrogen}} $
Table 1. Constant Composition of Isooctane

It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.

Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers. For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1.00 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1.00 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio.

$latex \frac{\frac{1.116 \text{g Cl}}{1.00 \text{g Cu}}}{\frac{0.558 \text{g Cl}}{1.00 \text{g Cu}}} = \frac{2}{1} $

This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.  This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure 4).

Overall Dalton's atomic theory gives us a “billiard ball” view of atoms. The shortcomings of his model, however are that it does not account for WHY elements combine as they do and does not explain the observed electrical charge of particles. More discoveries were needed to go beyond this model.

[caption id="" align="aligncenter" width="1300"] Figure 4. Compared to the copper chlorine compounds, where copper is represented by brown spheres and chlorine by green spheres, in (a) where the copper chlorine compound is CuCl and (b) where the copper chlorine compound CuCl₂ which has twice as many chlorine atoms per copper atom. (credit a: modification of work by “Benjah-bmm27”/Wikimedia Commons; credit b: modification of work by “Walkerma”/Wikimedia Commons)[/caption]

Example 2

A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?

 

Solution In compound A, the mass ratio of carbon to oxygen is:

$latex \frac{1.33 \text{g O}}{1.00 \text{g C}} $

In compound B, the mass ratio of carbon to oxygen is:

$latex \frac{2.67 \text{g O}}{1.00 \text{g C}} $

The ratio of these ratios is:

$latex \frac{\frac{1.33 \text{g O}}{1.00 \text{g C}}}{\frac{2.67 \text{g O}}{1.00 \text{g C}}} = \frac{1}{2} $

This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much oxygen per amount of carbon as B.  A possible pair of compounds that would fit this relationship would be A = CO and B = CO₂.

 

Test Yourself A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y?

  Answers In compound X, the mass ratio of carbon to hydrogen is $latex \frac{14.13 \text{g C}}{2.96 \text{g H}} $. In compound Y, the mass ratio of carbon to oxygen is $latex \frac{19.91 \text{g C}}{3.34 \text{g H}}$. The ratio of these ratios is $latex \frac{\frac{14.13 \text{g C}}{2.96 \text{g H}}}{\frac{19.91 \text{g C}}{3.34 \text{g H}}} = \frac{4.77 \text{g C/g H}}{5.96 \text{g C/g H}} = 0.800 = \frac{4}{5} $. This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds.

Key Concepts and Summary

The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed.

Exercises

1. In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following chemical change represented by these spheres may violate one of the ideas of Dalton’s atomic theory. Which one? 2. Which postulate of Dalton’s theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced. 3. Identify the postulate of Dalton’s theory that is violated by the following observations: 59.95% of one sample of titanium dioxide is titanium; 60.10% of a different sample of titanium dioxide is titanium. 4. Samples  X, Y, and Z are analyzed, with results shown here.

Samples	Description	Mass of Carbon	Mass of Hydrogen
X	clear, colorless, liquid with strong odor	1.776 g	0.148 g
Y	clear, colorless, liquid with strong odor	1.974 g	0.329 g
Z	clear, colorless, liquid with strong odor	7.812 g	0.651 g
Table 2.

Do these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and Z?

  Answers 1. The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed. 2. The law of conservation of matter - the total mass of matter present when matter changes from one type to another remains constant.

3. This statement violates Dalton’s fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio.

4.  Samples X and Z provide an example of the law of definite proportions (law of constant composition) for both samples are found to have a carbon-to-hydrogen mass ratio of 12:1. Samples X and Y provide an example of the law of multiple proportions,

since in sample X, the mass ratio of carbon to hydrogen is:

$latex \frac{0.0833 \text{g H}}{1.00 \text{g C}} $

and in sample Y, the mass ratio of carbon to hydrogen is:

$latex \frac{0.167 \text{g H}}{1.00 \text{g C}} $

The ratio of these ratios is:

$latex \frac{\frac{0.0833 \text{g H}}{1.00 \text{g C}}}{\frac{0.167 \text{g H}}{1.00 \text{g C}}} = \frac{1}{2} $

This supports the law of multiple proportions. This means that sample X and Y are different compounds, with X having one-half as much hydrogen per amount of carbon as Y. Samples Z and Y also provide an example of the law of multiple proportions,

since in sample Z, the mass ratio of carbon to hydrogen is:

$latex \frac{0.0833 \text{g H}}{1.00 \text{g C}} $

and in sample Y, the mass ratio of carbon to hydrogen is:

$latex \frac{0.167 \text{g H}}{1.00 \text{g C}} $

The ratio of these ratios is:

$latex \frac{\frac{0.0833 \text{g H}}{1.00 \text{g C}}}{\frac{0.167 \text{g H}}{1.00 \text{g C}}} = \frac{1}{2} $

This supports the law of multiple proportions. This means that sample Z and Y are different compounds, with Z having one-half as much hydrogen per amount of carbon as Y. Note: Samples X and Z may be the same compound or they may be different (isomers - a topic to be discussed later).

Glossary

Dalton’s atomic theory: set of postulates that established the fundamental properties of atoms law of conservation of matter: the total mass of matter present when matter changes from one type to another remains constant. law of constant composition: (also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass law of definite proportions: (also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass law of multiple proportions: when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers

]]> 1303 1293 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-2-evolution-of-atomic-theory/ Thu, 12 Apr 2018 02:51:17 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-2-evolution-of-atomic-theory/

Learning Objectives

By the end of this section, you will be able to:

• Outline milestones in the development of modern atomic theory
• Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford
• Describe the three subatomic particles that compose atoms
• Define isotopes and give examples for several elements

In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be quite interesting, it is most important to understand the concepts resulting from their work.

Atomic Theory after the Nineteenth Century

If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure 1).

[caption id="" align="aligncenter" width="1300"] Figure 1. (a) J. J. Thomson produced a visible beam in a cathode ray tube. (b) This is an early cathode ray tube, invented in 1897 by Ferdinand Braun. (c) In the cathode ray, the beam (shown in yellow) comes from the cathode and is accelerated past the anode toward a fluorescent scale at the end of the tube. Simultaneous deflections by applied electric and magnetic fields permitted Thomson to calculate the mass-to-charge ratio of the particles composing the cathode ray. (credit a: modification of work by Nobel Foundation; credit b: modification of work by Eugen Nesper; credit c: modification of work by “Kurzon”/Wikimedia Commons)[/caption]

Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron, a negatively charged, subatomic particle with a mass approximately two thousands that of a hydrogen atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “electric ion.”

 

Click here to hear Thomson describe his discovery in his own voice.

 

In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure 2).

Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 × 10⁻¹⁹ C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 × 10⁻¹⁹ C), two electrons (2 times 1.6 × 10⁻¹⁹ C), three electrons (3 times 1.6 × 10⁻¹⁹ C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 × 10¹¹ C/kg), it only required a simple calculation to determine the mass of the electron as well.

$latex \text{Mass\;of\;electron} = 1.602 \times 10^{-19} \text{C} \times \frac{1 \text{kg}}{1.759 \times 10^{11} \text{C}} = 9.107 \times 10^{-31} \text{kg} $

[caption id="" align="aligncenter" width="1300"] Figure 2. Millikan’s experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible values.[/caption]

Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure 3).

The next major development in understanding the atom came from Ernest Rutherford, a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle.

What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure 4). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you”[footnote]Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science, eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/items/backgroundtomode032734mbp/backgroundtomode032734mbp.pdf.[/footnote] (p. 68).

[caption id="" align="aligncenter" width="1300"] Figure 4. Geiger and Rutherford fired α particles at a piece of gold foil and detected where those particles went, as shown in this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and a very small number were significantly deflected.[/caption]

Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:

1. The volume occupied by an atom must consist of a large amount of empty space.
2. A small, relatively heavy, positively charged body, the nucleus, must be at the center of each atom.

View this simulation of the Rutherford gold foil experiment. Adjust the slit width to produce a narrower or broader beam of α particles to see how that affects the scattering pattern.

This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure 5). After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton, the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today.

[caption id="" align="aligncenter" width="1300"] Figure 5. The α particles are deflected only when they collide with or pass close to the much heavier, positively charged gold nucleus. Because the nucleus is very small compared to the size of an atom, very few α particles are deflected. Most pass through the relatively large region occupied by electrons, which are too light to deflect the rapidly moving particles.[/caption]

The Rutherford Scattering simulation allows you to investigate the differences between a “plum pudding” atom and a Rutherford atom by firing α particles at each type of atom.

 

Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes—atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery.

One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this chapter.

Key Concepts and Summary

Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes.

Exercises

1. The existence of isotopes violates one of the original ideas of Dalton’s atomic theory. Which one? 2. How are electrons and protons similar? How are they different? 3. How are protons and neutrons similar? How are they different? 4. Predict and test the behavior of α particles fired at a Rutherford atom model.

a) Predict the paths taken by α particles that are fired at atoms with a Rutherford atom model structure. Explain why you expect the α particles to take these paths.

b) If α particles of higher energy than those in (a) are fired at Rutherford atoms, predict how their paths will differ from the lower-energy α particle paths. Explain your reasoning.

c) Predict how the paths taken by the α particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why?

d) Now test your predictions from (a), (b), and (c). Open the Rutherford Scattering simulation and select the “Rutherford Atom” tab. Due to the scale of the simulation, it is best to start with a small nucleus, so select “20” for both protons and neutrons, “min” for energy, show traces, and then start firing α particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Pause or reset, set energy to “max,” and start firing α particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual path as shown in the simulation. Pause or reset, select “40” for both protons and neutrons, “min” for energy, show traces, and fire away. Does this match your prediction from (c)? If not, explain why the actual path would be that shown in the simulation. Repeat this with larger numbers of protons and neutrons. What generalization can you make regarding the type of atom and effect on the path of α particles? Be clear and specific.

  Answers

1. Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties.

2. Both are subatomic particles but protons reside in an atom’s nucleus and electrons reside in the electron cloud that surrounds the nucleus. The mass of protons are approximately 2000 times greater than the mass of electrons. Protons are positively charged, whereas electrons are negatively charged.

3. Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.

4. a) The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. b) Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection. c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. d) The paths followed by the α particles match the predictions from a), b), and c).

Glossary

alpha particle (α particle): positively charged particle consisting of two protons and two neutrons electron: negatively charged, subatomic particle of relatively low mass located outside the nucleus isotopes: atoms that contain the same number of protons but different numbers of neutrons neutron: uncharged, subatomic particle located in the nucleus nucleus: massive, positively charged center of an atom made up of protons and neutrons proton: positively charged, subatomic particle located in the nucleus

]]> 1309 1293 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-3-atomic-structure-and-symbolism/ Thu, 12 Apr 2018 02:51:20 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-3-atomic-structure-and-symbolism/

Learning Objectives

By the end of this section, you will be able to:

• Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
• Define the atomic mass unit and average atomic mass
• Calculate average atomic mass and isotopic abundance

The development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space containing negatively charged electrons. The nucleus contains the majority of an atom’s mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom’s volume. The diameter of an atom is on the order of 10⁻¹⁰ m, whereas the diameter of the nucleus is roughly 10⁻¹⁵ m—about 100,000 times smaller. For a perspective about their relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium (Figure 1).

Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10⁻²³ g, and an electron has a charge of less than 2 × 10⁻¹⁹ C (coulomb). When describing the properties of tiny objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu. (This isotope is known as “carbon-12” as will be discussed later in this module.) Thus, one amu is exactly $latex \frac{1}{12} $ of the mass of one carbon-12 atom: 1 amu = 1.6605 × 10⁻²⁴ g. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 × 10⁻¹⁹ C.

A proton has a mass of 1.0073 amu and a charge of 1+. A neutron is a slightly heavier particle with a mass 1.0087 amu and a charge of zero; as its name suggests, it is neutral. The electron has a charge of 1− and is a much lighter particle with a mass of about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton. The properties of these fundamental particles are summarized in Table 1. (An observant student might notice that the sum of an atom’s subatomic particles does not equal the atom’s actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu, slightly larger than 12.00 amu. This “missing” mass is known as the mass defect, and you will learn about it in the chapter on nuclear chemistry.)

The number of protons in the nucleus of an atom is its atomic number (Z). This is the defining trait of an element: Its value determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number (A). The number of neutrons is therefore the difference between the mass number and the atomic number: A – Z = number of neutrons.

$latex \begin{array}{r @ {{}={}} l} \text{atomic number (Z)} & \text{number of protons} \\[1em] \text{mass number (A)} & \text{number of protons + number of neutrons} \\[1em] \text{A - Z} & \text{number of neutrons} \end{array}$

Name	Location	Charge (C)	Unit Charge	Mass (amu)	Mass (g)
electron	outside nucleus	−1.602 × 10−19	1−	0.00055	0.00091 × 10−24
proton	nucleus	1.602 × 10−19	1+	1.00727	1.67262 × 10−24
neutron	nucleus	0	0	1.00866	1.67493 × 10−24
Table 1. Properties of Subatomic Particles

Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:

Atomic charge = number of protons − number of electrons

As will be discussed in more detail later in this chapter, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom (Z = 11) has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge (11 − 10 = 1+). A neutral oxygen atom (Z = 8) has eight electrons, and if it gains two electrons it will become an anion with a 2− charge (8 − 10 = 2−).

Example 1

Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland (Figure 2).

The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as 40% of the world’s population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1− charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions.

 

Solution

The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is 74 (127 − 53 = 74). Since the iodine is added as a 1− anion, the number of electrons is 54 [53 – (1–) = 54].   Test Yourself An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?   Answers 78 protons; 117 neutrons; charge is 4+

Example 2

1. The most common carbon atoms have six protons and six neutrons in their nuclei. What are the atomic number and the mass number of these carbon atoms?
2. An isotope of uranium has an atomic number of 92 and a mass number of 235. What are the number of protons and neutrons in the nucleus of this atom?

 

Solution

1. If a carbon atom has six protons in its nucleus, its atomic number is 6. If it also has six neutrons in the nucleus, then the mass number is 6 + 6, or 12.
2. If the atomic number of uranium is 92, then that is the number of protons in the nucleus. Because the mass number is 235, then the number of neutrons in the nucleus is 235 − 92, or 143.

 

Test Yourself

The number of protons in the nucleus of a tin atom is 50, while the number of neutrons in the nucleus is 68. What are the atomic number and the mass number of this isotope?

 

Answer

Atomic number = 50, mass number = 118

Chemical Symbols

A chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for mercury is Hg (Figure 3). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a container of many atoms of the element mercury (macroscopic domain).

[caption id="attachment_1312" align="aligncenter" width="200"] Figure 3. The symbol Hg represents the element mercury regardless of the amount; it could represent one atom of mercury or a large amount of mercury.[/caption]

The symbols for several common elements and their atoms are listed in Table 2. Some symbols are derived from the common name of the element; others are abbreviations of the name in another language. Most symbols have one or two letters, but three-letter symbols have been used to describe some elements that have atomic numbers greater than 112. To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the elements carbon (C) and oxygen (O). All known elements and their symbols are in the periodic table in Figure 2 in Chapter 3.5 The Periodic Table (also found in Appendix A).

Element	Symbol	Element	Symbol
aluminum	Al	iron	Fe (from ferrum)
bromine	Br	lead	Pb (from plumbum)
calcium	Ca	magnesium	Mg
carbon	C	mercury	Hg (from hydrargyrum)
chlorine	Cl	nitrogen	N
chromium	Cr	oxygen	O
cobalt	Co	potassium	K (from kalium)
copper	Cu (from cuprum)	silicon	Si
fluorine	F	silver	Ag (from argentum)
gold	Au (from aurum)	sodium	Na (from natrium)
helium	He	sulfur	S
hydrogen	H	tin	Sn (from stannum)
iodine	I	zinc	Zn
Table 2. Some Common Elements and Their Symbols

Traditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium (Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists (or occasionally locations); for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in the discovery of several heavy elements.

Visit this site to learn more about IUPAC, the International Union of Pure and Applied Chemistry, and explore its periodic table.

 

Isotopes

The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure 4). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as ²⁴Mg, ²⁵Mg, and ²⁶Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading. For instance, ²⁴Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” ²⁵Mg is read as “magnesium 25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only because a ²⁴Mg atom has 12 neutrons in its nucleus, a ²⁵Mg atom has 13 neutrons, and a ²⁶Mg has 14 neutrons.  Therefore the masses of isotopes of an element, isotopic mass, differ - see Table 3.

 

[caption id="" align="aligncenter" width="492"] Figure 4. The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.[/caption]

Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table 3. Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen-2, symbolized ²H, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized ³H, is also called tritium and sometimes symbolized T.

Element	Symbol	Atomic Number	# of Protons	# of Neutrons	Isotopic Mass (amu)	% Natural Abundance
hydrogen	$latex _1^1\text{H}$ (protium)	1	1	0	1.0078	99.989
	$latex _1^2\text{H}$ (deuterium)	1	1	1	2.0141	0.0115
	$latex _1^3\text{H}$ (tritium)	1	1	2	3.01605	— (trace)
helium	$latex _2^3\text{He}$	2	2	1	3.01603	0.00013
	$latex _2^4\text{He}$	2	2	2	4.0026	100
lithium	$latex _3^6\text{Li}$	3	3	3	6.0151	7.59
	$latex _3^7\text{Li}$	3	3	4	7.0160	92.41
beryllium	$latex _4^9\text{Be}$	4	4	5	9.0122	100
boron	$latex _5^{10}\text{B}$	5	5	5	10.0129	19.9
	$latex _5^{11}\text{B}$	5	5	6	11.0093	80.1
carbon	$latex _6^{12}\text{C}$	6	6	6	12.0000	98.89
	$latex _6^{13}\text{C}$	6	6	7	13.0034	1.11
	$latex _6^{14}\text{C}$	6	6	8	14.0032	— (trace)
nitrogen	$latex _7^{14}\text{N}$	7	7	7	14.0031	99.63
	$latex _7^{15}\text{N}$	7	7	8	15.0001	0.37
oxygen	$latex _8^{16}\text{O}$	8	8	8	15.9949	99.757
	$latex _8^{17}\text{O}$	8	8	9	16.9991	0.038
	$latex _8^{18}\text{O}$	8	8	10	17.9992	0.205
fluorine	$latex _9^{19}\text{F}$	9	9	10	18.9984	100
neon	$latex _{10}^{20}\text{Ne}$	10	10	10	19.9924	90.48
	$latex _{10}^{21}\text{Ne}$	10	10	11	20.9938	0.27
	$latex _{10}^{22}\text{Ne}$	10	10	12	21.9914	9.25
Table 3. Nuclear Compositions of Atoms of the Very Light Elements

Example 3

1. What is the symbol for an isotope of uranium that has an atomic number of 92 and a mass number of 235?
2. How many protons and neutrons are in $latex _{26}^{56}\text{Fe}$?

 

Solution

1. The symbol for this isotope is $latex _{92}^{235}\text{U}$.
2. This iron atom has 26 protons and 56 − 26 = 30 neutrons.

 

Test Yourself

How many protons are in $latex _{23}^{11}\text{Na}$?

 

Answer

11 protons

Example 4

Determine the number of protons, neutrons and electrons for the ion:

 

Solution  

The atomic number is 17, thus the ion contains 17 protons. The mass number is 35, therefore it contains 35 – 17 = 18 neutrons. Because it is negatively charged (-1), it must have one more electron as compared to protons, thus 17 + 1 = 18 electrons.

 

Test Yourself

Determine the number of electrons in each of the following ions. Hint: Use the periodic table to first determine the number of protons based on its elemental identity.   a)  Mg²⁺           b) Fe³⁺             c) O^2-

 

Answers

a) 10    b) 23   c) 10

Example 5

Determine the number of protons, neutrons and electrons for the following atom, as well as its identity (chemical symbol) for:

 

Solution  

The atomic number is 92 and mass number is 238. From the atomic number 92 we know that this must be Uranium (chemical symbol = U). The atomic number is equal to the number of protons, thus the number of protons is 92. Because the mass number is equal to the sum of the protons and neutrons, we know that n + 92 = 238. Thus, the number of neutrons is 238 – 92 = 146. Finally, the given symbol must represent an atom, not an ion (no electric charge is shown) and any atom is neutral, thus the number of electrons must be the same as the number of protons, or 92 .

 

Test Yourself

a)  Write the complete atomic symbol for krypton, which contains 48 neutrons/

b)  How many protons, neutrons and electrons are in ¹³²Cs?

 

Answers

a) ⁸⁴Kr  b) protons = 55, neutrons = 77, electrons = 55

Use this Build an Atom simulator to build atoms of the first 10 elements, see which isotopes exist, check nuclear stability, and gain experience with isotope symbols.

 

Atomic Mass

Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.

The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass multiplied by its fractional abundance.

$latex \displaystyle{} \text{average mass} = \sum_{i} (\text{fractional abundance} \times \text{isotopic mass})_{i} $

For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are ¹⁰B with a mass of 10.0129 amu, and the remaining 80.1% are ¹¹B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:

$latex \begin{array}{r @{{}={}} l} \text{boron average mass} & (0.199 \times 10.0129 \;\text{amu}) + (0.801 \times 11.0093 \;\text{amu}) \\[1em] & 1.99 \;\text{amu} + 8.82 \;\text{amu} \\[1em] & 10.81 \;\text{amu} \end{array}$

It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.

Example 6

A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% ²⁰Ne (mass 19.9924 amu), 0.47% ²¹Ne (mass 20.9940 amu), and 7.69% ²²Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?

 

Solution

$latex \begin{array}{r @{{}={}} l} \text{average mass} & (0.9184 \times 19.9924 \;\text{amu}) + (0.0047 \times 20.9940 \;\text{amu})+(0.0769 \times 21.9914 \;\text{amu}) \\[1em] & (18.36+0.099+1.69) \;\text{amu} \\[1em] & 20.15 \;\text{amu} \end{array}$

The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)

 

Test Yourself A sample of magnesium is found to contain 78.70% of ²⁴Mg atoms (mass 23.98 amu), 10.13% of ²⁵Mg atoms (mass 24.99 amu), and 11.17% of ²⁶Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.

  Answer 24.31 amu

We can also do variations of this type of calculation, as shown in the next example.

Example 7

Naturally occurring chlorine consists of ³⁵Cl (mass 34.96885 amu) and ³⁷Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?

 

Solution The average mass of chlorine is the fraction that is ³⁵Cl times the mass of ³⁵Cl plus the fraction that is ³⁷Cl times the mass of ³⁷Cl.

$latex \text{average mass} = (\text{fraction of} \ ^{35}\text{Cl} \ \times \ \text{mass of} \ ^{35}\text{Cl}) + (\text{fraction of} \ ^{37}\text{Cl} \ \times \ \text{mass of} \ ^{37}\text{Cl}) $

If we let x represent the fraction that is ³⁵Cl, then the fraction that is ³⁷Cl is represented by 1.00 − x.

(The fraction that is ³⁵Cl + the fraction that is ³⁷Cl must add up to 1, so the fraction of ³⁷Cl must equal 1.00 − the fraction of ³⁵Cl.)

Substituting this into the average mass equation, we have:

$latex \begin{array}{r @{{}={}} l}35.453 \;\text{amu} & (x \times 34.96885 \;\text{amu}) + [(1.00 - x) \times 36.96590\;\text{amu}] \\[1em] 35.453 & 34.96885x + 36.96590 - 36.96590x \\[1em] 1.99705x & 1.513 \\[1em] x & \frac{1.513}{1.99705} = 0.7576 \end{array}$

So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% ³⁵Cl and 24.24% ³⁷Cl.

 

Test Yourself

Naturally occurring copper consists of ⁶³Cu (mass 62.9296 amu) and ⁶⁵Cu (mass 64.9278 amu), with an average mass of 63.546 amu. What is the percent composition of Cu in terms of these two isotopes? Answers 69.15% Cu-63 and 30.85% Cu-65

Visit this site to make mixtures of the main isotopes of the first 18 elements, gain experience with average atomic mass, and check naturally occurring isotope ratios using the Isotopes and Atomic Mass simulation.

The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure 5), the sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.

[caption id="" align="aligncenter" width="1300"] Figure 5. Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different isotopes of Zr.[/caption]

See an animation that explains mass spectrometry. Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.

Key Concepts and Summary

An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $latex \frac{1}{12}$ of the mass of a carbon-12 atom and is equal to 1.6605 × 10⁻²⁴ g.

Protons are relatively heavy particles with a charge of 1+ and a mass of 1.0073 amu. Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu. Electrons are light particles with a charge of 1− and a mass of 0.00055 amu. The number of protons in the nucleus is called the atomic number (Z) and is the property that defines an atom’s elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.

Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.

Key Equations

• $latex \displaystyle{} \text{average mass} = \sum_{i} (\text{fractional abundance} \times \text{isotopic mass})_i$

Exercises

1. Write the symbol for each of the following ions:

a) the ion with a 1+ charge, atomic number 55, and mass number 133

b) the ion with 54 electrons, 53 protons, and 74 neutrons

c) the ion with atomic number 15, mass number 31, and a 3− charge

d) the ion with 24 electrons, 30 neutrons, and a 3+ charge

2. Open the Build an Atom simulation and click on the Atom icon.

a) Pick any one of the first 10 elements that you would like to build and state its symbol.

b) Drag protons, neutrons, and electrons onto the atom template to make an atom of your element.

State the numbers of protons, neutrons, and electrons in your atom, as well as the net charge and mass number.

c) Click on “Net Charge” and “Mass Number,” check your answers to (b), and correct, if needed.

d) Predict whether your atom will be stable or unstable. State your reasoning.

e) Check the “Stable/Unstable” box. Was your answer to (d) correct? If not, first predict what you can do to make a stable atom of your element, and then do it and see if it works. Explain your reasoning.

3. Open the Build an Atom simulation

a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Lithium-6 and give the isotope symbol for this atom.

b) Now remove one electron to make an ion and give the symbol for the ion you have created.

4. The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name them.

a) atomic number 26, mass number 58, charge of 2+

b) atomic number 53, mass number 127, charge of 1−

5. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:

a) $latex _3^7\text{Li}$

b) $latex _{52}^{125}\text{Te}$

c) $latex _{47}^{109}\text{Ag}$

d) $latex _{7}^{15}\text{N}$

e) $latex _{15}^{31}\text{P}$

6. Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes ⁷⁹Br and ⁸¹Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.

7. The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consists of two isotopes with accurately known masses (¹⁰B, 10.0129 amu and ¹¹B, 11.0931 amu). The average atomic mass of boron can vary from 10.807 to 10.819, depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries.

8. Explain Dalton's atomic theory.

9. Which is larger, a proton or an electron?

10. Which is larger, a neutron or an electron?

11. What are the charges for each of the three subatomic particles?

12. Where is most of the mass of an atom located?

13. Sketch a diagram of a boron atom, which has five protons and six neutrons in its nucleus.

14. Define atomic number. What is the atomic number for a boron atom?

15. Define isotope and give an example.

16. What is the difference between deuterium and tritium?

17. Which pair represents isotopes?

a)        b)  ₂₆F and ₂₅M        c)  ₁₄S and ₁₅P

18.  Which pair represents isotopes?a)  ₂₀C and ₁₉K      b)  ₂₆F and ₂₇F        c)  ₉₂U and ₉₂U     d) $latex _7^{14}\text{N}$ and $latex _8^{14}\text{N}$ 

19.  Give complete symbols of each atom, including the atomic number and the mass number.

a)  an oxygen atom with 8 protons and 8 neutrons

b)  a potassium atom with 19 protons and 20 neutrons

c)  a lithium atom with 3 protons and 4 neutrons

20.  Give complete symbols of each atom, including the atomic number and the mass number.

a)  a magnesium atom with 12 protons and 12 neutrons b)  a magnesium atom with 12 protons and 13 neutrons c)  a xenon atom with 54 protons and 77 neutrons 21.    Americium-241 is an isotope used in smoke detectors. What is the complete symbol for this isotope? 22.  Carbon-14 is an isotope used to perform radioactive dating tests on previously living material. What is the complete symbol for this isotope? 23.  Give atomic symbols for each element. a)  sodium     b)  argon     c)  nitrogen     d)  radon 24.  Give atomic symbols for each element. a)  silver     b)  gold      c)  mercury      d)  iodine 25.  Give the name of the element. a)  Si      b)  Mn      c)  Fe      d)  Cr 26.  Give the name of the element. a)  F      b)  Cl      c)  Br      d)  I 27. Determine the atomic mass of each element, given the isotopic composition. a)  lithium, which is 92.4% lithium-7 (mass 7.016 u) and 7.60% lithium-6 (mass 6.015 u) b)  oxygen, which is 99.76% oxygen-16 (mass 15.995 u), 0.038% oxygen-17 (mass 16.999 u), and 0.205% oxygen-18 (mass 17.999 u)   Answers

1. (a) ¹³³Cs⁺; (b) ¹²⁷I⁻; (c) ³¹P³⁻; (d) ⁵⁷Co³⁺

2. (a) Carbon-12, ¹²C; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral ¹²C atom. The net charge of such a neutral atom is zero, and the mass number is 12. (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen.

3. (a) Lithium-6 contains three protons, three neutrons, and three electrons. The isotope symbol is ⁶Li or $latex _3^6\text{Li}$. (b) ⁶Li⁺ or $latex _3^6 \text{Li}^+$

4. (a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons

5. (a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons

6. 79.904 amu

7. Turkey source: 0.2649 (of 10.0129 amu isotope); US source: 0.2537 (of 10.0129 amu isotope)

8. All matter is composed of atoms; atoms of the same element are the same, and atoms of different elements are different; atoms combine in whole-number ratios to form compounds. 9. A proton is larger than an electron. 10. A neutron is larger than an electron. 11. proton: 1+; electron: 1−; neutron: 0 12.  Most of the mass of an atom is located in the nucleus. 13. 14. The atomic number is the number of protons in a nucleus. Boron has an atomic number of five. 15. Isotopes are atoms of the same element but with different numbers of neutrons. 16. They are isotopes, therefore the difference between deuterium and tritium is the number of neutrons.  Deuterium has one, and tritium has two. 17. (a) 18.  (b) - note: there is an error with option (d) for the atomic number of nitrogen can only be 7. 19.  a) $latex _{8}^{16}\text{O}$      b)  $latex _{19}^{39}\text{K}$      c)  $latex _{3}^{7}\text{Li}$ 20.  Give complete symbols of each atom, including the atomic number and the mass number.

a)  $latex _{12}^{24}\text{Mg}$      b)  $latex _{12}^{25}\text{Mg}$      c)  $latex _{54}^{131}\text{Xe}$

21.   $latex _{95}^{241}\text{Am}$

22.  $latex _{6}^{14}\text{C}$ 23.  a)  Na     b)  Ar     c)  N     d)  Rn 24.  a)  Ag     b)  Au      c)  Hg      d)  I 25.  a)  silicon      b)  manganese      c)  iron      d)  chromium 26.  a)  fluorine      b)  chlorine      c)  bromine      d)  iodine 27.  a)  6.940 u     b)  16.000 u

Glossary

electron: negatively charged, subatomic particle of relatively low mass located outside the nucleus anion: negatively charged atom or molecule (contains more electrons than protons) atomic mass: average mass of atoms of an element, expressed in amu atomic mass unit (amu): (also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to $latex \frac{1}{12}$ of the mass of a ¹²C atom atomic number (Z): number of protons in the nucleus of an atom cation: positively charged atom or molecule (contains fewer electrons than protons) chemical symbol: one-, two-, or three-letter abbreviation used to represent an element or its atoms Dalton (Da): alternative unit equivalent to the atomic mass unit fundamental unit of charge: (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 × 10⁻¹⁹ C ion: electrically charged atom or molecule (contains unequal numbers of protons and electrons) isotopic mass: mass of an isotope of an element, expressed in amu mass number (A): sum of the numbers of neutrons and protons in the nucleus of an atom unified atomic mass unit (u): alternative unit equivalent to the atomic mass unit

]]> 1316 1293 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-4-chemical-formulas/ Thu, 12 Apr 2018 02:51:24 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-4-chemical-formulas/

Learning Objectives

By the end of this section, you will be able to:

• Symbolize the composition of molecules using molecular formulas and empirical formulas
• Represent the bonding arrangement of atoms within molecules using structural formulas

A molecular formula is a representation of a molecule that uses chemical symbols to indicate the types of atoms followed by subscripts to show the number of atoms of each type in the molecule. (A subscript is used only when more than one atom of a given type is present.) Molecular formulas are also used as abbreviations for the names of compounds.

The structural formula for a compound gives the same information as its molecular formula (the types and numbers of atoms in the molecule) but also shows how the atoms are connected in the molecule. The structural formula for methane contains symbols for one C atom and four H atoms, indicating the number of atoms in the molecule (Figure 1). The lines represent bonds that hold the atoms together. (A chemical bond is an attraction between atoms or ions that holds them together in a molecule or a crystal.) We will discuss chemical bonds and see how to predict the arrangement of atoms in a molecule later. For now, simply know that the lines are an indication of how the atoms are connected in a molecule. A ball-and-stick model shows the geometric arrangement of the atoms with atomic sizes not to scale, and a space-filling model shows the relative sizes of the atoms.

[caption id="" align="aligncenter" width="975"] Figure 1. A methane molecule can be represented as (a) a molecular formula, (b) a structural formula, (c) a ball-and-stick model, and (d) a space-filling model. Carbon and hydrogen atoms are represented by black and white spheres, respectively.[/caption]

Although many elements consist of discrete, individual atoms, some exist as molecules made up of two or more atoms of the element chemically bonded together. For example, most samples of the elements hydrogen, oxygen, and nitrogen are composed of molecules that contain two atoms each (called diatomic molecules) and thus have the molecular formulas H₂, O₂, and N₂, respectively. Other elements commonly found as diatomic molecules are fluorine (F₂), chlorine (Cl₂), bromine (Br₂), and iodine (I₂). The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is S₈ (Figure 2).

[caption id="" align="aligncenter" width="975"] Figure 2. A molecule of sulfur is composed of eight sulfur atoms and is therefore written as S₈. It can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. Sulfur atoms are represented by yellow spheres.[/caption]

It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, H₂ and 2H represent distinctly different species. H₂ is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2H, on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression 2H₂ represents two molecules of diatomic hydrogen (Figure 3).

Compounds are formed when two or more elements chemically combine, resulting in the formation of bonds. For example, hydrogen and oxygen can react to form water, and sodium and chlorine can react to form table salt. We sometimes describe the composition of these compounds with an empirical formula, which indicates the types of atoms present and the simplest whole-number ratio of the number of atoms (or ions) in the compound. For example, titanium dioxide (used as pigment in white paint and in the thick, white, blocking type of sunscreen) has an empirical formula of TiO₂. This identifies the elements titanium (Ti) and oxygen (O) as the constituents of titanium dioxide, and indicates the presence of twice as many atoms of the element oxygen as atoms of the element titanium (Figure 4).

[caption id="" align="aligncenter" width="975"] Figure 4. (a) The white compound titanium dioxide provides effective protection from the sun. (b) A crystal of titanium dioxide, TiO₂, contains titanium and oxygen in a ratio of 1 to 2. The titanium atoms are gray and the oxygen atoms are red. (credit a: modification of work by “osseous”/Flickr)[/caption]

As discussed previously, we can describe a compound with a molecular formula, in which the subscripts indicate the actual numbers of atoms of each element in a molecule of the compound. In many cases, the molecular formula of a substance is derived from experimental determination of both its empirical formula and its molecular mass (the sum of atomic masses for all atoms composing the molecule). For example, it can be determined experimentally that benzene contains two elements, carbon (C) and hydrogen (H), and that for every carbon atom in benzene, there is one hydrogen atom. Thus, the empirical formula is CH. An experimental determination of the molecular mass reveals that a molecule of benzene contains six carbon atoms and six hydrogen atoms, so the molecular formula for benzene is C₆H₆ (Figure 5).

[caption id="" align="aligncenter" width="1300"] Figure 5. Benzene, C₆H₆, is produced during oil refining and has many industrial uses. A benzene molecule can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. (d) Benzene is a clear liquid. (credit d: modification of work by Sahar Atwa)[/caption]

If we know a compound’s formula, we can easily determine the empirical formula. (This is somewhat of an academic exercise; the reverse chronology is generally followed in actual practice.) For example, the molecular formula for acetic acid, the component that gives vinegar its sharp taste, is C₂H₄O₂. This formula indicates that a molecule of acetic acid (Figure 6) contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. The ratio of atoms is 2:4:2. Dividing by the lowest common denominator (2) gives the simplest, whole-number ratio of atoms, 1:2:1, so the empirical formula is CH₂O. Note that a molecular formula is always a whole-number multiple of an empirical formula.

[caption id="" align="aligncenter" width="975"] Figure 6. (a) Vinegar contains acetic acid, C₂H₄O₂, which has an empirical formula of CH₂O. It can be represented as (b) a structural formula and (c) as a ball-and-stick model. (credit a: modification of work by “HomeSpot HQ”/Flickr)[/caption]

Example 1

Molecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose?

 

Solution The molecular formula is C₆H₁₂O₆ because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH₂O.

 

Test Yourself A molecule of metaldehyde (a pesticide used for snails and slugs) contains 8 carbon atoms, 16 hydrogen atoms, and 4 oxygen atoms. What are the molecular and empirical formulas of metaldehyde?

  Answers  Molecular formula, C₈H₁₆O₄; empirical formula, C₂H₄O

 

You can explore molecule building using an online simulation.

What is it that chemists do? According to Lee Cronin, chemists make very complicated molecules by “chopping up” small molecules and “reverse engineering” them. He wonders if we could “make a really cool universal chemistry set” by what he calls “app-ing” chemistry. Could we “app” chemistry?

In a 2012 TED talk, Lee describes one fascinating possibility: combining a collection of chemical “inks” with a 3D printer capable of fabricating a reaction apparatus (tiny test tubes, beakers, and the like) to fashion a “universal toolkit of chemistry.” This toolkit could be used to create custom-tailored drugs to fight a new superbug or to “print” medicine personally configured to your genetic makeup, environment, and health situation. Says Cronin, “What Apple did for music, I’d like to do for the discovery and distribution of prescription drugs.”[footnote]Lee Cronin, “Print Your Own Medicine,” Talk presented at TED Global 2012, Edinburgh, Scotland, June 2012.[/footnote] View his full talk at the TED website.

It is important to be aware that it may be possible for the same atoms to be arranged in different ways: Compounds with the same molecular formula may have different atom-to-atom bonding and therefore different structures. For example, could there be another compound with the same formula as acetic acid, C₂H₄O₂? And if so, what would be the structure of its molecules?

If you predict that another compound with the formula C₂H₄O₂ could exist, then you demonstrated good chemical insight and are correct. Two C atoms, four H atoms, and two O atoms can also be arranged to form a methyl formate, which is used in manufacturing, as an insecticide, and for quick-drying finishes. Methyl formate molecules have one of the oxygen atoms between the two carbon atoms, differing from the arrangement in acetic acid molecules. Acetic acid and methyl formate are examples of isomers—compounds with the same chemical formula but different molecular structures (Figure 7). Note that this small difference in the arrangement of the atoms has a major effect on their respective chemical properties. You would certainly not want to use a solution of methyl formate as a substitute for a solution of acetic acid (vinegar) when you make salad dressing.

[caption id="" align="aligncenter" width="505"] Figure 7. Molecules of (a) acetic acid and methyl formate (b) are structural isomers; they have the same formula (C₂H₄O₂) but different structures (and therefore different chemical properties).[/caption]

Key Concepts and Summary

A molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms.

Exercises

1. Explain why the symbol for an atom of the element oxygen and the formula for a molecule of oxygen differ. 2. Explain why the symbol for the element sulfur and the formula for a molecule of sulfur differ. 3. Write the molecular and empirical formulas of the following compounds:

a)

b) c) d) 4. Write the molecular and empirical formulas of the following compounds:

a)

b) c) d) 5. Determine the empirical formulas for the following compounds:

a) caffeine, C₈H₁₀N₄O₂

b) fructose, C₁₂H₂₂O₁₁

c) hydrogen peroxide, H₂O₂

d) glucose, C₆H₁₂O₆

e) ascorbic acid (vitamin C), C₆H₈O₆

6. Determine the empirical formulas for the following compounds:

a) acetic acid, C₂H₄O₂

b) citric acid, C₆H₈O₇

c) hydrazine, N₂H₄

d) nicotine, C₁₀H₁₄N₂

e) butane, C₄H₁₀

7. Write the empirical formulas for the following compounds:

a)

b) 8. Use the Build a Molecule simulation to build a molecule with two carbons, six hydrogens, and one oxygen.

a) Draw the structural formula of this molecule and state its name.

b) Can you arrange these atoms to make a different molecule? If so, draw its structural formula and state its name.

c) How are the molecules drawn in (a) and (b) the same? How do they differ? What are they called (the type of relationship between these molecules, not their names).

  Answers

1. The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O₂, contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.

2.  The symbol for the element sulfur is S.  Elemental sulfur is a polyatomic element S₈, a molecule of sulfur contains eight atoms of sulfur.

3. a) molecular CO₂, empirical CO₂      b) molecular C₂H₂, empirical CH

c) molecular C₂H₄, empirical CH₂       d) molecular H₂SO₄, empirical H₂SO₄

4. a) molecular C₄H₈, empirical CH₂      b) molecular C₄H₆, empirical C₂H₃

c) molecular Si₂H₂Cl₄, empirical SiHCl₂      d) molecular H₃PO₄, empirical H₃PO₄ 5. a) C₄H₅N₂O      b) C₁₂H₂₂O₁₁      c) HO     d) CH₂O      e) C₃H₄O₃ 6.  a) CH₂O      b) C₆H₈O₇      c) NH₂      d) C₅H₇N     e) C₂H₅ 7. a) CH₂O     b) C₂H₄O 8. a) ethanol

b) methoxymethane, more commonly known as dimethyl ether

c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers.

Glossary

empirical formula: formula showing the composition of a compound given as the simplest whole-number ratio of atoms isomers: compounds with the same chemical formula but different structures molecular formula: formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in a molecule of the compound. spatial isomers: compounds in which the relative orientations of the atoms in space differ structural formula: shows the atoms in a molecule and how they are connected structural isomer: one of two substances that have the same molecular formula but different physical and chemical properties because their atoms are bonded differently

]]> 1339 1293 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-5-the-periodic-table/ Thu, 12 Apr 2018 02:51:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-5-the-periodic-table/

Learning Objectives

By the end of this section, you will be able to:

• State the periodic law and explain the organization of elements in the periodic table
• Predict the general properties of elements based on their location within the periodic table
• Identify metals, nonmetals, and metalloids by their properties and/or location on the periodic table

There are many known elements, both naturally occurring and manmade. In ancient times the known elements were carbon, iron, sulfur, gold, silver, copper, tin, lead, mercury and zinc.  It was not until the late 1700s that new elements began to be discovered by Martin Klaproth (Ti, Zr, U, Te, Sr, Ce, Cr) and Jons Berzelius (Si, Se, Ce, Li, V, Th).  In the 1800s Sir Humphrey Davy discovered several alkali and alkaline earth metals and halogens through the use of electricity. Also in the 1800s new elements (Cesium and Rubidium) were discovered through the development of spectroscopy by Robert Bunsen (who also invented the Bunsen burner) and Gustav Kirchhoff.  Through the use of spectroscopy Helium was discovered by analyzing light from the sun in 1868 before it was discovered here on Earth in 1882 through the spectral analysis of lava from Mount Vesuvius. It is noteworthy to mention that the spectroscopy revolutionized our ability to identify elements and is the cornerstone of modern methods in chemical analysis.  As early chemists worked to purify ores and discovered more elements, they realized that various elements could be grouped together by their similar chemical behaviours.  In 1789, Antoine Lavoisier arranged the 33 known chemical elements into four groups: gases, metals, nonmetals and earths.  In 1829, Johann Wolfgang Döbereiner observed that many of the elements could be arranged in groups of three based on their chemical properties.  These groups were known as "Döbereiner's triads". One such grouping includes lithium (Li), sodium (Na), and potassium (K): These elements all are shiny, conduct heat and electricity well, and have similar chemical properties. A second grouping includes calcium (Ca), strontium (Sr), and barium (Ba), which also are shiny, good conductors of heat and electricity, and have chemical properties in common. However, the specific properties of these two groupings are notably different from each other. For example: Li, Na, and K are much more reactive than are Ca, Sr, and Ba; Li, Na, and K form compounds with oxygen in a ratio of two of their atoms to one oxygen atom, whereas Ca, Sr, and Ba form compounds with one of their atoms to one oxygen atom. Fluorine (F), chlorine (Cl), bromine (Br), and iodine (I) also exhibit similar properties to each other, but these properties are drastically different from those of any of the elements above.

Dimitri Mendeleev in Russia (1869) and Lothar Meyer in Germany (1870) independently recognized that there was a periodic relationship among the properties of the elements known at that time. Both published tables with the elements arranged according to increasing atomic mass. But Mendeleev went one step further than Meyer: He used his table to predict the existence of elements that would have the properties similar to aluminum and silicon, but were yet unknown. The discoveries of gallium (1875) and germanium (1886) provided great support for Mendeleev’s work. Although Mendeleev and Meyer had a long dispute over priority, Mendeleev’s contributions to the development of the periodic table are now more widely recognized (Figure 1).

[caption id="" align="aligncenter" width="1300"] Figure 1. (a) Dimitri Mendeleev is widely credited with creating (b) the first periodic table of the elements. (credit a: modification of work by Serge Lachinov; credit b: modification of work by “Den fjättrade ankan”/Wikimedia Commons)[/caption]

By the twentieth century, it became apparent that the periodic relationship involved atomic numbers rather than atomic masses. The modern statement of this relationship, the periodic law, is as follows: the properties of the elements are periodic functions of their atomic numbers. A modern periodic table arranges the elements in increasing order of their atomic numbers and groups atoms with similar properties in the same vertical column (Figure 2). Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. Groups are labeled at the top of each column. In the United States, the labels traditionally were numerals with capital letters. However, IUPAC recommends that the numbers 1 through 18 be used, and these labels are more common. For the table to fit on a single page, parts of two of the rows, a total of 14 columns, are usually written below the main body of the table.

Many elements differ dramatically in their chemical and physical properties, but some elements are similar in their behaviors. For example, many elements appear shiny, are malleable (able to be deformed without breaking) and ductile (can be drawn into wires), and conduct heat and electricity well. Other elements are not shiny, malleable, or ductile, and are poor conductors of heat and electricity. We can sort the elements into large classes with common properties: metals (elements that are shiny, malleable, good conductors of heat and electricity—shaded yellow); nonmetals (elements that appear dull, poor conductors of heat and electricity—shaded green); and metalloids (elements that conduct heat and electricity moderately well, and possess some properties of metals and some properties of nonmetals—shaded purple).

[caption id="" align="aligncenter" width="1300"] Figure 2. Elements in the periodic table are organized according to their properties.[/caption]

The elements can also be classified into the main-group elements (or representative elements) in the columns labeled 1, 2, and 13–18; the transition metals in the columns labeled 3–12; and inner transition metals in the two rows at the bottom of the table (the top-row elements are called lanthanides and the bottom-row elements are actinides; Figure 3). The elements can be subdivided further by more specific properties, such as the composition of the compounds they form. For example, the elements in group 1 (the first column) form compounds that consist of one atom of the element and one atom of hydrogen. These elements (except hydrogen) are known as alkali metals, and they all have similar chemical properties. The elements in group 2 (the second column) form compounds consisting of one atom of the element and two atoms of hydrogen: These are called alkaline earth metals, with similar properties among members of that group. Other groups with specific names are the pnictogens (group 15), chalcogens (group 16), halogens (group 17), and the noble gases (group 18, also known as inert gases). The groups can also be referred to by the first element of the group: For example, the chalcogens can be called the oxygen group or oxygen family. Hydrogen is a unique, nonmetallic element with properties similar to both group 1A and group 7A elements. For that reason, hydrogen may be shown at the top of both groups, or by itself.

[caption id="" align="aligncenter" width="1300"] Figure 3. The periodic table organizes elements with similar properties into groups.[/caption] You should also be familiar with the natural states of elements. Most metals occur as solids. An exception to this is mercury (Hg) which occurs as liquid. Noble gases, in the far right column, occur naturally as gas.  Many non-metals occur as multi-atomic molecules:  (i.e. more than one atom together is the natural state):  H₂, O₂, N₂, F₂, Cl₂ which are all gases, S₈, P₄, Se₈, I₂ which are all solids and Br₂ which is a liquid.

Click on this link for an interactive periodic table, which you can use to explore the properties of the elements (includes podcasts and videos of each element). You may also want to try this one that shows photos of all the elements.

Example 1

Atoms of each of the following elements are essential for life. Give the group name for the following elements:

a) chlorine      b) calcium      c) sodium      d) sulfur

 

Solution The family names are as follows:

a) halogen      b) alkaline earth metal      c) alkali metal      d) chalcogen

 

Test Yourself Give the group name for each of the following elements:

a) krypton      b) selenium      c) barium      d) lithium

  Answers a) noble gas      b) chalcogen      c) alkaline earth metal      d) alkali metal

Example 2

For the following elements, list their symbol, their natural state, classify them as metal, nonmetal or metalloid, and specify their group name (when applicable):

      a)  magnesium        b)  silver          c)  uranium        d)  chlorine

 

Solution  

a)   Magnesium = Mg, occurs as a solid, is a metal (main group metal) in the alkaline earth metals group.

b)   Silver = Ag, occurs as a solid, is a metal (transition metal)

c)   Uranium = U, occurs as a solid, is a metal (inner transition metal) in the actinide group

d)   Chlorine = Cl, occurs as Cl₂in the gas state, is a nonmetal and is in the halogen group

 

Test Yourself

For the following elements, list their symbol, their natural state, classify them as metal, nonmetal or metalloid, and specify their group name (when applicable):

      a) germanium           b)  lead            c) nitrogen          d)  potassium

 

Answers

a) Germanium = Ge, solid, metalloid            b) lead = Pb, solid, metal (main group) c) nitrogen = N, N­₂gas, nonmetal                 d) potassium = K, solid, metal (main group), alkali metal

In studying the periodic table, you might have noticed something about the atomic masses of some of the elements. Element 43 (technetium), element 61 (promethium), and most of the elements with atomic number 84 (polonium) and higher have their atomic mass given in square brackets. This is done for elements that consist entirely of unstable, radioactive isotopes (you will learn more about radioactivity in the nuclear chemistry chapter). An average atomic weight cannot be determined for these elements because their radioisotopes may vary significantly in relative abundance, depending on the source, or may not even exist in nature. The number in square brackets is the atomic mass number (and approximate atomic mass) of the most stable isotope of that element.

Key Concepts and Summary

The discovery of the periodic recurrence of similar properties among the elements led to the formulation of the periodic table, in which the elements are arranged in order of increasing atomic number in rows known as periods and columns known as groups. Elements in the same group of the periodic table have similar chemical properties. Elements can be classified as metals, metalloids, and nonmetals, or as a main-group elements, transition metals, and inner transition metals. Groups are numbered 1–18 from left to right. The elements in group 1 are known as the alkali metals; those in group 2 are the alkaline earth metals; those in 15 are the pnictogens; those in 16 are the chalcogens; those in 17 are the halogens; and those in 18 are the noble gases.

Activity

Make yourself this Qcard to help you learn the name of the groups in the periodic table and add it to your collection.  Then use the Qcards to quiz yourself.

Side 1: 

 

Side 2: 

Exercises

1. Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:

a) uranium      b) bromine      c) strontium      d) neon

e) gold            f) americium   g) rhodium        h) sulfur

i) carbon         j) potassium

2. Using the periodic table, identify the lightest member of each of the following groups:

a) noble gases      b) alkaline earth metals

c) alkali metals    d) chalcogens

3. Use the periodic table to give the name and symbol for each of the following elements:

a) the noble gas in the same period as germanium

b) the alkaline earth metal in the same period as selenium

c) the halogen in the same period as lithium

d) the chalcogen in the same period as cadmium

4. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.

a) the alkali metal with 11 protons and a mass number of 23

b) the noble gas element with 75 neutrons in its nucleus and 54 electrons in the neutral atom

c) the isotope with 33 protons and 40 neutrons in its nucleus

d) the alkaline earth metal with 88 electrons and 138 neutrons

  Answers 1. a) metal, inner transition metal;  b) nonmetal, representative element;  c) metal, representative element;  d) nonmetal, representative element;  e) metal, transition metal;  f) metal, inner transition metal;  g) metal, transition metal;  h) nonmetal, representative element;  i) nonmetal, representative element;  j) metal, representative element

2. a) He       b) Be      c) Li     d) O

3. a) krypton, Kr      b) calcium, Ca      c) fluorine, F      d) tellurium, Te

4. a) $latex _{11}^{23}\text{Na}$      b) $latex _{54}^{129}\text{Xe}$ c) $latex _{33}^{73}\text{As}$          d) $latex _{88}^{226}\text{Ra}$

Glossary

actinide: inner transition metal in the bottom of the bottom two rows of the periodic table alkali metal: element in group 1 alkaline earth metal: element in group 2 chalcogen: element in group 16 group: vertical column of the periodic table halogen: element in group 17 inert gas: (also, noble gas) element in group 18 inner transition metal: (also, lanthanide or actinide) element in the bottom two rows; if in the first row, also called lanthanide, or if in the second row, also called actinide lanthanide: inner transition metal in the top of the bottom two rows of the periodic table main-group element: (also, representative element) element in columns 1, 2, and 12–18 metal: element that is shiny, malleable, good conductor of heat and electricity metalloid: element that conducts heat and electricity moderately well, and possesses some properties of metals and some properties of nonmetals noble gas: (also, inert gas) element in group 18 nonmetal: element that appears dull, poor conductor of heat and electricity period: (also, series) horizontal row of the periodic table periodic law: properties of the elements are periodic function of their atomic numbers. periodic table: table of the elements that places elements with similar chemical properties close together pnictogen: element in group 15 representative element: (also, main-group element) element in columns 1, 2, and 12–18 series: (also, period) horizontal row of the period table transition metal: element in columns 3–11

]]> 1344 1293 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-a-the-periodic-table/ Thu, 12 Apr 2018 03:47:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-a-the-periodic-table/ ]]> 2021 0 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-b-essential-mathematics/ Thu, 12 Apr 2018 03:47:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-b-essential-mathematics/

Exponential Arithmetic

Exponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:

$latex \begin{array}{r @{{}={}} l} 1000 & 1\;\times\;10^3 \\[0.75em] 100 & 1\;\times\;10^2 \\[0.75em] 10 & 1\;\times\;10^1 \\[0.75em] 1 & 1\;\times\;10^0 \\[0.75em] 0.1 & 1\;\times\;10^{-1} \\[0.75em] 0.001 & 1\;\times\;10^{-3} \\[0.75em] 2386 & 2.386\;\times\;1000 = 2.386\;\times\;10^3 \\[0.75em] 0.123 & 1.23\;\times\;0.1 = 1.23\;\times\;10^{-1} \end{array}$

The power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for every large and very small numbers. For example, 1,230,000,000 = 1.23 × 10⁹, and 0.00000000036 = 3.6 × 10⁻¹⁰.

Addition of Exponentials

Convert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.

Example 1

Adding Exponentials Add 5.00 × 10⁻⁵ and 3.00 × 10⁻³.

 

Solution

$latex 3.00\;\times\;10^{-3} = 300\;\times\;10^{-5}$ $latex (5.00\;\times\;10^{-5})\;+\;(300\;\times\;10^{-5}) = 305\;\times\;10^{-5} = 3.05\;\times\;10^{-3} $

Subtraction of Exponentials

Convert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.

Example 2

Subtracting Exponentials Subtract 4.0 × 10⁻⁷ from 5.0 × 10⁻⁶.

 

Solution

$latex 4.0\;\times\;10^{-7} = 0.40\;\times\;10^{-6}$ $latex (5.0\;\times\;10^{-6})\;-\;(0.40\;\times\;10^{-6}) = 4.6\;\times\;10^{-6}$

Multiplication of Exponentials

Multiply the digit terms in the usual way and add the exponents of the exponential terms.

Example 3

Multiplying Exponentials Multiply 4.2 × 10⁻⁸ by 2.0 × 10³.

 

Solution

$latex (4.2\;\times\;10^{-8})\;\times\;(2.0\;\times\;10^3) = (4.2\;\times\;2.0)\;\times\;10^{(-8)+(+3)} = 8.4\;\times\;10^{-5}$

Division of Exponentials

Divide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.

Example 4

Dividing Exponentials Divide 3.6 × 10⁵ by 6.0 × 10⁻⁴.

 

Solution

$latex \frac{3.6\;\times\;10^{-5}}{6.0\;\times\;10^{-4}} = (\frac{3.6}{6.0})\;\times\;10^{(-5)-(-4)} = 0.60\;\times\;10^{-1} = 6.0\;\times\;10^{-2}$

Squaring of Exponentials

Square the digit term in the usual way and multiply the exponent of the exponential term by 2.

Example 5

Squaring Exponentials Square the number 4.0 × 10⁻⁶.

 

Solution

$latex (4.0\;\times\;10^{-6})^2 = 4\;\times\;4\;\times\;10^{2\;\times\;(-6)} = 16\;\times\;10^{-12} = 1.6\;\times\;10^{-11}$

Cubing of Exponentials

Cube the digit term in the usual way and multiply the exponent of the exponential term by 3.

Example 6

Cubing Exponentials Cube the number 2 × 10⁴.

 

Solution

$latex (2\;\times\;10^4)^3 = 2\;\times\;2\;\times\;2\;\times\;10^{3\;\times\;4} = 8\;\times\; 10^{12}$

Taking Square Roots of Exponentials

If necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2. Extract the square root of the digit term and divide the exponential term by 2.

Example 7

Finding the Square Root of Exponentials Find the square root of 1.6 × 10⁻⁷.

 

Solution

$latex 1.6\;\times\;10^{-7} = 16\;\times\;10^{-8}$ $latex \sqrt{16\;\times\;10^{-8}} = \sqrt{16}\;\times\;\sqrt{10^{-8}} = \sqrt{16}\;\times\;\sqrt{10}^{-\frac{8}{2}} = 4.0\;\times\;10^{-4}$

Significant Figures

A beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more accurate if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no meaning useful in this situation. In reporting any information as numbers, use only as many significant figures as the accuracy of the measurement warrants.

The importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example).

Example 8

Addition and Subtraction with Significant Figures Add 4.383 g and 0.0023 g.

 

Solution

$latex \begin{array}{l} 4.38\underline{3}\;\text{g} \\ 0.002\underline{3}\;\text{g} \\ \hline 4.38\underline{5}\;\text{g} \end{array}$

In multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures.

Example 9

Multiplication and Division with Significant Figures Multiply 0.6238 by 6.6.

 

Solution

$latex 0.623\underline{8}\;\times\;6.\underline{6} = 4.\underline{1}$

When rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (“round up”). Do not change the retained digit if the digits that follow are less than 5 (“round down”). If the retained digit is followed by 5, round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).

]]> 2024 0 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-c-units-and-conversion-factors/ Thu, 12 Apr 2018 03:47:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-c-units-and-conversion-factors/ Units of Length meter (m) = 39.37 inches (in.)

= 1.094 yards (yd) angstrom (Å) = 10^–8 cm (exact, definition)

= 10^–10 m (exact, definition) centimeter (cm) = 0.01 m (exact, definition) yard (yd) = 0.9144 m millimeter (mm) = 0.001 m (exact, definition) inch (in.) = 2.54 cm (exact, definition) kilometer (km) = 1000 m (exact, definition) mile (US) = 1.60934 km

Units of Volume
liter (L)	= 0.001 m3 (exact, definition) = 1000 cm3 (exact, definition) = 1.057 (US) quarts	liquid quart (US)	= 32 (US) liquid ounces (exact, definition) = 0.25 (US) gallon (exact, definition) = 0.9463 L
milliliter (mL)	= 0.001 L (exact, definition) = 1 cm3 (exact, definition)	dry quart	= 1.1012 L
microliter (μL)(μL)	= 10–6 L (exact, definition) = 10–3 cm3 (exact, definition)	cubic foot (US)	= 28.316 L

Units of Mass
gram (g)	= 0.001 kg (exact, definition)	ounce (oz) (avoirdupois)	= 28.35 g
milligram (mg)	= 0.001 g (exact, definition)	pound (lb) (avoirdupois)	= 0.4535924 kg
kilogram (kg)	= 1000 g (exact, definition) = 2.205 lb	ton (short)	=2000 lb (exact, definition) = 907.185 kg
ton (metric)	=1000 kg (exact, definition) = 2204.62 lb	ton (long)	= 2240 lb (exact, definition) = 1.016 metric ton

Units of Energy
4.184 joule (J)	= 1 thermochemical calorie (cal)
1 thermochemical calorie (cal)	= 4.184 × 107  erg
erg	= 10–7 J (exact, definition)
electron-volt (eV)	= 1.60218 × 10−19 J = 23.061 kcal mol−1
liter∙atmosphere	= 24.217 cal = 101.325 J (exact, definition)
nutritional calorie (Cal)	= 1000 cal (exact, definition) = 4184 J
British thermal unit (BTU)	= 1054.804 J[footnote]BTU is the amount of energy needed to heat one pound of water by one degree Fahrenheit. Therefore, the exact relationship of BTU to joules and other energy units depends on the temperature at which BTU is measured. 59 °F (15 °C) is the most widely used reference temperature for BTU definition in the United States. At this temperature, the conversion factor is the one provided in this table.[/footnote]

Units of Pressure
torr	= 1 mm Hg (exact, definition)
pascal (Pa)	= N m–2 (exact, definition) = kg m–1 s–2 (exact, definition)
atmosphere (atm)	= 760 mm Hg (exact, definition) = 760 torr (exact, definition) = 101,325 N m–2 (exact, definition) = 101,325 Pa (exact, definition)
bar	= 105 Pa (exact, definition) = 105 kg m–1 s–2 (exact, definition)

 ]]> 2025 0 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-d-fundamental-physical-constants/ Thu, 12 Apr 2018 03:47:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/back-matter/appendix-d-fundamental-physical-constants/ Fundamental Physical Constants Name and Symbol Value atomic mass unit (amu) 1.6605402 × 10⁻²⁷ kg Avogadro’s number 6.0221367 × 10²³ mol⁻¹ electron charge (e) 1.60217733 × 10⁻¹⁹ C electron rest mass (m_e) 9.1093897 × 10⁻³¹ kg neutron rest mass (m_n) 1.6749274 × 10⁻²⁷ kg Planck’s constant (h) 6.6260755 × 10⁻³⁴ J s proton rest mass (m_p) 1.6726231 × 10⁻²⁷ kg Rydberg constant (R) 1.0973731534 × 10⁷ m⁻¹ = 2.1798736 × 10⁻¹⁸ J speed of light (in vacuum) (c) 2.99792458 × 10⁸ m s⁻¹  ]]> 2026 0 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/isooctane/ Tue, 14 May 2019 22:02:59 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Isooctane.jpg 4909 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/isooctane-2/ Tue, 14 May 2019 22:05:34 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/wp-content/uploads/sites/387/2018/05/Isooctane-1.jpg 4910 3870 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-6-molecular-and-ionic-compounds/ Thu, 12 Apr 2018 02:51:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-6-molecular-and-ionic-compounds/

Learning Objectives

By the end of this section, you will be able to:

• Define ionic and molecular (covalent) compounds
• Predict the type of compound formed from elements based on their location within the periodic table
• Learn the characteristic charges that ions have.
• Determine formulas for simple ionic compounds

In ordinary chemical reactions, the nucleus of each atom (and thus the identity of the element) remains unchanged. Electrons, however, can be added to atoms by transfer from other atoms, lost by transfer to other atoms, or shared with other atoms. The transfer and sharing of electrons among atoms govern the chemistry of the elements. During the formation of some compounds, atoms gain or lose electrons, and form electrically charged particles called ions (Figure 1).

You can use the periodic table to predict whether an atom will form an anion or a cation, and you can often predict the charge of the resulting ion. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. To illustrate, an atom of an alkali metal (group 1) loses one electron and forms a cation with a 1+ charge; an alkaline earth metal (group 2) loses two electrons and forms a cation with a 2+ charge, and so on. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. This results in a cation with 20 protons, 18 electrons, and a 2+ charge. It has the same number of electrons as atoms of the preceding noble gas, argon, and is symbolized Ca²⁺. The name of a metal ion is the same as the name of the metal atom from which it forms, so Ca²⁺ is called a calcium ion.

When atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as an atom of the next noble gas in the periodic table. Atoms of group 17 gain one electron and form anions with a 1− charge; atoms of group 16 gain two electrons and form ions with a 2− charge, and so on. For example, the neutral bromine atom, with 35 protons and 35 electrons, can gain one electron to provide it with 36 electrons. This results in an anion with 35 protons, 36 electrons, and a 1− charge. It has the same number of electrons as atoms of the next noble gas, krypton, and is symbolized Br⁻. (A discussion of the theory supporting the favored status of noble gas electron numbers reflected in these predictive rules for ion formation is provided in a later chapter of this text.)

Note the usefulness of the periodic table in predicting likely ion formation and charge (Figure 2). Moving from the far left to the right on the periodic table, main-group elements tend to form cations with a charge equal to the group number. That is, group 1 elements form 1+ ions; group 2 elements form 2+ ions, and so on. Moving from the far right to the left on the periodic table, elements often form anions with a negative charge equal to the number of groups moved left from the noble gases. For example, group 17 elements (one group left of the noble gases) form 1− ions; group 16 elements (two groups left) form 2− ions, and so on. This trend can be used as a guide in many cases, but its predictive value decreases when moving toward the center of the periodic table. In fact, transition metals and some other metals often exhibit variable charges that are not predictable by their location in the table. For example, copper can form ions with a 1+ or 2+ charge, and iron can form ions with a 2+ or 3+ charge.

[caption id="" align="aligncenter" width="1300"] Figure 2. Some elements exhibit a regular pattern of ionic charge when they form ions.[/caption]

Example 1

An ion found in some compounds used as antiperspirants contains 13 protons and 10 electrons. What is its symbol?

 

Solution Because the number of protons remains unchanged when an atom forms an ion, the atomic number of the element must be 13. Knowing this lets us use the periodic table to identify the element as Al (aluminum). The Al atom has lost three electrons and thus has three more positive charges (13) than it has electrons (10). This is the aluminum cation, Al³⁺.

 

Test Yourself Give the symbol and name for the ion with 34 protons and 36 electrons.

  Answer Se²⁻, the selenide ion

Example 2

Magnesium and nitrogen react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.

 

Solution Magnesium’s position in the periodic table (group 2) tells us that it is a metal. Metals form positive ions (cations). A magnesium atom must lose two electrons to have the same number electrons as an atom of the previous noble gas, neon. Thus, a magnesium atom will form a cation with two fewer electrons than protons and a charge of 2+. The symbol for the ion is Mg²⁺, and it is called a magnesium ion.

Nitrogen’s position in the periodic table (group 15) reveals that it is a nonmetal. Nonmetals form negative ions (anions). A nitrogen atom must gain three electrons to have the same number of electrons as an atom of the following noble gas, neon. Thus, a nitrogen atom will form an anion with three more electrons than protons and a charge of 3−. The symbol for the ion is N³⁻, and it is called a nitride ion.

 

Test Yourself Aluminum and carbon react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.

  Answers Al will form a cation with a charge of 3+: Al³⁺, an aluminum ion. Carbon will form an anion with a charge of 4−: C⁴⁻, a carbide ion.

The ions that we have discussed so far are called monatomic ions, that is, they are ions formed from only one atom. We also find many polyatomic ions. These ions, which act as discrete units, are electrically charged molecules (a group of bonded atoms with an overall charge). Some of the more important polyatomic ions are listed in Table 1. Oxyanions are polyatomic ions that contain one or more oxygen atoms. At this point in your study of chemistry, you should memorize the names, formulas, and charges of the most common polyatomic ions. Because you will use them repeatedly, they will soon become familiar.

Note that there is a system for naming some polyatomic ions; -ate and -ite are suffixes designating polyatomic ions containing more or fewer oxygen atoms. Per- (short for “hyper”) and hypo- (meaning “under”) are prefixes meaning more oxygen atoms than -ate and fewer oxygen atoms than -ite, respectively. For example, perchlorate is ClO₄⁻, chlorate is ClO₃⁻, chlorite is ClO₂⁻ and hypochlorite is ClO⁻. Unfortunately, the number of oxygen atoms corresponding to a given suffix or prefix is not consistent; for example, nitrate is NO₃⁻ while sulfate is SO₄²⁻. This will be covered in more detail in the next module on nomenclature.

Name	Formula	Related Acid	Formula
ammonium	$latex \text{NH}_4^{\;\;+}$
hydronium	$latex \text{H}_3\text{O}^{+}$
oxide	$latex \text{O}_2^{\;\;-}$
peroxide	$latex \text{O}_2^{\;\;2-}$
hydroxide	$latex \text{OH}^{-}$
acetate	$latex \text{CH}_3\text{COO}^{-}$	acetic acid	$latex \text{CH}_3\text{COOH}$
cyanide	$latex \text{CN}^{-}$	hydrocyanic acid	$latex \text{HCN}$
azide	$latex \text{N}_3^{\;\;-}$	hydrazoic acid	$latex \text{HN}_3$
carbonate	$latex \text{CO}_3^{\;\;2-}$	carbonic acid	$latex \text{H}_2\text{CO}_3$
bicarbonate	$latex \text{HCO}_3^{\;\;-}$
nitrate	$latex \text{NO}_3^{\;\;-}$	nitric acid	$latex \text{HNO}_3$
nitrite	$latex \text{NO}_2^{\;\;-}$	nitrous acid	$latex \text{HNO}_2$
sulfate	$latex \text{SO}_4^{\;\;2-}$	sulfiric acid	$latex \text{H}_2\text{SO}_4$
hydrogen sulfate	$latex \text{HSO}_4^{\;\;-}$
sulfite	$latex \text{SO}_3^{\;\;2-}$	sulfurous acid	$latex \text{H}_2\text{SO}_3$
hydrogen sulfite	$latex \text{HSO}_3^{\;\;-}$
phosphate	$latex \text{PO}_4^{\;\;3-}$	phosphoric acid	$latex \text{H}_3\text{PO}_4$
hydrogen phosphate	$latex \text{HPO}_4^{\;\;2-}$
dihydrogen phosphate	$latex \text{H}_2\text{PO}_4^{\;\;-}$
phosphite	$latex \text{PO}_3^{\;\;3-}$	phosphorous acid	$latex \text{H}_3\text{PO}_3$
hydrogen phosphite	$latex \text{HPO}_3^{\;\;2-}$
dihydrogen phosphite	$latex \text{H}_2\text{PO}_3^{\;\;-}$
perchlorate	$latex \text{ClO}_4^{\;\;-}$	perchloric acid	$latex \text{HClO}_4$
chlorate	$latex \text{ClO}_3^{\;\;-}$	chloric acid	$latex \text{HClO}_3$
chlorite	$latex \text{ClO}_2^{\;\;-}$	chlorous acid	$latex \text{HClO}_2$
hypochlorite	$latex \text{ClO}^{-}$	hypochlorous acid	$latex \text{HClO}$
chromate	$latex \text{CrO}_4^{\;\;2-}$	chromic acid	$latex \text{H}_2\text{Cr}_2\text{O}_4$
dichromate	$latex \text{Cr}_2\text{O}_7^{\;\;2-}$	dichromic acid	$latex \text{H}_2\text{Cr}_2\text{O}_7$
permanganate	$latex \text{MnO}_4^{\;\;-}$	permanganic acid	$latex \text{HMnO}_4$
Table 1. Common Polyatomic Ions

The nature of the attractive forces that hold atoms or ions together within a compound is the basis for classifying chemical bonding. When electrons are transferred and ions form, ionic bonds result. Ionic bonds are electrostatic forces of attraction, that is, the attractive forces experienced between objects of opposite electrical charge (in this case, cations and anions). When electrons are “shared” and molecules form, covalent bonds result. Covalent bonds are the attractive forces between the positively charged nuclei of the bonded atoms and one or more pairs of electrons that are located between the atoms. Compounds are classified as ionic or molecular (covalent) on the basis of the bonds present in them.

Ionic Compounds

When an element composed of atoms that readily lose electrons (a metal) reacts with an element composed of atoms that readily gain electrons (a nonmetal), a transfer of electrons usually occurs, producing ions. The compound formed by this transfer is stabilized by the electrostatic attractions (ionic bonds) between the ions of opposite charge present in the compound. For example, when each sodium atom in a sample of sodium metal (group 1) gives up one electron to form a sodium cation, Na⁺, and each chlorine atom in a sample of chlorine gas (group 17) accepts one electron to form a chloride anion, Cl⁻, the resulting compound, NaCl, is composed of sodium ions and chloride ions in the ratio of one Na⁺ ion for each Cl⁻ ion. Similarly, each calcium atom (group 2) can give up two electrons and transfer one to each of two chlorine atoms to form CaCl₂, which is composed of Ca²⁺ and Cl⁻ ions in the ratio of one Ca²⁺ ion to two Cl⁻ ions.

A compound that contains ions and is held together by ionic bonds is called an ionic compound. The periodic table can help us recognize many of the compounds that are ionic: When a metal is combined with one or more nonmetals, the compound is usually ionic. This guideline works well for predicting ionic compound formation for most of the compounds typically encountered in an introductory chemistry course. However, it is not always true (for example, aluminum chloride, AlCl₃, is not ionic).

You can often recognize ionic compounds because of their properties. Ionic compounds are solids that typically melt at high temperatures and boil at even higher temperatures. For example, sodium chloride melts at 801 °C and boils at 1413 °C. (As a comparison, the molecular compound water melts at 0 °C and boils at 100 °C.) In solid form, an ionic compound is not electrically conductive because its ions are unable to flow (“electricity” is the flow of charged particles). When molten, however, it can conduct electricity because its ions are able to move freely through the liquid (Figure 3).

[caption id="" align="aligncenter" width="975"] Figure 3. Sodium chloride melts at 801 °C and conducts electricity when molten. (credit: modification of work by Mark Blaser and Matt Evans)[/caption]

 

Watch this video to see a mixture of salts melt and conduct electricity.

In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.

Example 3

The gemstone sapphire (Figure 4) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al³⁺, and oxygen anions, O²⁻. What is the formula of this compound?

[caption id="" align="alignleft" width="187"] Figure 4. Although pure aluminum oxide is colorless, trace amounts of iron and titanium give blue sapphire its characteristic color. (credit: modification of work by Stanislav Doronenko)[/caption]

 

Solution Because the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2−, would give us six negative charges. The formula would be Al₂O₃.

 

Test yourself Predict the formula of the ionic compound formed between the sodium cation, Na⁺, and the sulfide anion, S²⁻.

  Answer     Na₂S

Example 4

Write the proper ionic formula for each of the two given ions.

a) Ca²⁺ and Cl⁻            b) Al³⁺ and F⁻          c) Al³⁺ and O²⁻

 

Solution

a) We need two Cl⁻ ions to balance the charge on one Ca²⁺ ion, so the proper ionic formula is CaCl₂.

b) We need three F⁻ ions to balance the charge on the Al³⁺ ion, so the proper ionic formula is AlF₃.

c) With Al³⁺ and O²⁻, note that neither charge is a perfect multiple of the other. This means we have to go to a least common multiple, which in this case will be six. To get a total of 6+, we need two Al³⁺ ions; to get 6−, we need three O²⁻ ions. Hence the proper ionic formula is Al₂O₃.

 

Test Yourself

Write the proper ionic formulas for each of the two given ions.

a)  Fe²⁺ and S²⁻          b) Fe³⁺ and S²⁻

 

Answers

a) FeS          b) Fe₂S₃

Many ionic compounds contain polyatomic ions (Table 1) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca₃(PO₄)₂. This formula indicates that there are three calcium ions (Ca²⁺) for every two phosphate (PO₄³⁻) groups. The PO₄³⁻ groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3−. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.

Example 5

Baking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca²⁺ and H₂PO₄⁻. What is the formula of this compound?

 

Solution The positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca²⁺ ion to two H₂PO₄⁻ ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H₂PO₄)₂.

 

Test Yourself Predict the formula of the ionic compound formed between the lithium ion and the peroxide ion, O₂²⁻ (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)

  Answer Li₂O₂

Because an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a molecular formula. Instead, ionic compounds must be symbolized by a formula unit, a formula indicating the relative numbers of its constituent ions. For compounds containing only monatomic ions (such as NaCl) and for many compounds containing polyatomic ions (such as CaSO₄), these formulas are just the empirical formulas introduced earlier in this chapter. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of Na⁺ and C₂O₄²⁻ ions combined in a 2:1 ratio, and its formula is written as Na₂C₂O₄. The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, NaCO₂. This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound’s polyatomic anion, C₂O₄²⁻.

Molecular Compounds

Many compounds do not contain ions but instead consist solely of discrete, neutral molecules. These molecular compounds (covalent compounds) result when atoms share, rather than transfer (gain or lose), electrons. Covalent bonding is an important and extensive concept in chemistry, and it will be treated in considerable detail in a later chapter of this text. We can often identify molecular compounds on the basis of their physical properties. Under normal conditions, molecular compounds often exist as gases, low-boiling liquids, and low-melting solids, although many important exceptions exist.

Whereas ionic compounds are usually formed when a metal and a nonmetal combine, covalent compounds are usually formed by a combination of nonmetals. Thus, the periodic table can help us recognize many of the compounds that are covalent. While we can use the positions of a compound’s elements in the periodic table to predict whether it is ionic or covalent at this point in our study of chemistry, you should be aware that this is a very simplistic approach that does not account for a number of interesting exceptions. Shades of gray exist between ionic and molecular compounds, and you’ll learn more about those later.

Figure 5. Classification of Pure Substances.  Examples include atomic hydrogen (H), molecular oxygen (O₂), water (H₂O) and sodium chloride (NaCl).

Example 6

Predict whether the following compounds are ionic or molecular:

a) KI, the compound used as a source of iodine in table salt

b) H₂O₂, the bleach and disinfectant hydrogen peroxide

c) CHCl₃, the anesthetic chloroform

d) Li₂CO₃, a source of lithium in antidepressants

 

Solution a) Potassium (group 1) is a metal, and iodine (group 17) is a nonmetal; KI is predicted to be ionic.

b) Hydrogen (group 1) is a nonmetal, and oxygen (group 16) is a nonmetal; H₂O₂ is predicted to be molecular.

c) Carbon (group 14) is a nonmetal, hydrogen (group 1) is a nonmetal, and chlorine (group 17) is a nonmetal; CHCl₃ is predicted to be molecular.

d) Lithium (group 1) is a metal, and carbonate is a polyatomic ion; Li₂CO₃ is predicted to be ionic.

 

Test Yourself Using the periodic table, predict whether the following compounds are ionic or covalent:

a) SO₂

b) CaF₂

c) N₂H₄

d) Al₂(SO₄)₃

  Answers a) molecular          b) ionic          c) molecular          d) ionic

Key Concepts and Summary

Metals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the preceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table. Thus, nonmetals tend to form negative ions. Positively charged ions are called cations, and negatively charged ions are called anions. Ions can be either monatomic (containing only one atom) or polyatomic (containing more than one atom).

Compounds that contain ions are called ionic compounds. Ionic compounds generally form from metals and nonmetals. Compounds that do not contain ions, but instead consist of atoms bonded tightly together in molecules (uncharged groups of atoms that behave as a single unit), are called covalent compounds. Covalent compounds usually form from two nonmetals.

Exercises

1.  Explain how cations form. 2.  Explain how anions form. 3.  Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.

1. a)  K         b)  O         c)  Cu

4.  Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.

a)  Ag         b)  Au         c)  Br

5.  Name the ions from Exercise 3. 6.  Name the ions from Exercise 4. 7.  Give the formula for each ionic compound formed between the two listed ions.

a)  Mg²⁺ and Cl⁻         b)  Fe²⁺ and O²⁻         c)  Fe³⁺ and O²⁻

8.  Give the formula for each ionic compound formed between the two listed ions.

a)  Cu²⁺ and F⁻         b)  Ca²⁺ and O²⁻         c)  K⁺ and P³⁻

9.  Give the formula for each ionic compound formed between the two listed ions.

a)  K⁺ and SO₄²⁻         b)  NH₄⁺ and S²⁻         c)  NH₄⁺ and PO₄³⁻

10.  Give the formula for each ionic compound formed between the two listed ions.

a)  Pb⁴⁺ and SO₄²⁻         b)  Na⁺ and I₃⁻         c)  Li⁺ and Cr₂O₇²⁻

11.  Give the formula for each ionic compound formed between the two listed ions.

a)  Ag⁺ and SO₃²⁻         b)  Na⁺ and HCO₃⁻         c)  Fe³⁺ and ClO₃⁻

12.  What is the difference between SO₃ and SO₃²⁻?

13.  Using the periodic table, predict whether the following chlorides are ionic or covalent: KCl, NCl₃, ICl, MgCl₂, PCl₅, and CCl₄. 14.  Using the periodic table, predict whether the following chlorides are ionic or covalent: SiCl₄, PCl₃, CaCl₂, CsCl, CuCl₂, and CrCl₃. 15.  For each of the following compounds, state whether it is ionic or covalent. If it is ionic, write the symbols for the ions involved:

a) NF₃          b) BaO          c) (NH₄)₂CO₃          d) Sr(H₂PO₄)₂          e) IBr          f) Na₂O

16.  For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions involved:

a) KClO₄          b) Mg(C₂H₃O₂)₂          c) H₂S          d) Ag₂S          e) N₂Cl₄          f) Co(NO₃)₂

17.  For each of the following pairs of ions, write the symbol for the formula of the compound they will form:

a) Ca²⁺, S²⁻          b) NH₄⁺, SO₄²⁻          c) Al³⁺, Br⁻          d) Na⁺, HPO₄²⁻          e) Mg²⁺, PO₄³⁻

18.  For each of the following pairs of ions, write the symbol for the formula of the compound they will form: a) K⁺, O²⁻          b) NH₄+, PO₄³⁻          c) Al³⁺, O²⁻          d) Na⁺, CO₃²⁻          e) Ba²⁺, PO₄³⁻   Answers

1.  Cations form by losing electrons.

2.  Anions form by gaining electrons.

3.  a)  1+         b)  2−         c)  1+, 2+

4.  a)  1+         b)  1+, 3+         c)  1−

5.  a)  the potassium ion         b)  the oxide ion         c)  the cobalt(II) and cobalt(III) ions, respectively

6.  a)  the silver ion         b)  the gold(I) and gold(III) ions, respectively         c)  the bromide ion

7.  a)  MgCl₂         b)  FeO         c)  Fe₂O₃

8.  a)  CuF₂         b)  CaO         c)  K₃P9.  a)  K₂SO₄         b)  (NH₄)₂S          c)  (NH₄)₃PO₄ 10.  a)  Pb(SO₄)₂         b)  NaI₃         c)  Li₂Cr₂O₇ 11.  a)  Ag₂SO₃         b)  NaHCO₃         c)  Fe(ClO₃)₃ 12.  SO₃ is sulfur trioxide, while SO₃²⁻ is the sulfite ion.

13. Ionic: KCl, MgCl₂; Covalent: NCl₃, ICl, PCl₅, CCl₄

14. Ionic: CaCl₂, CsCl, CuCl₂, CrCl₃.; Covalent: SiCl₄, PCl₃

15. a) covalent                       b) ionic, Ba²⁺, O²⁻            c) ionic, NH₄⁺,CO₃²⁻

d) ionic, Sr²⁺, H₂PO₄⁻          e) covalent                       f) ionic, Na⁺, O²⁻ 16. a) ionic, K⁺, ClO₄^-          b) ionic, Mg⁺², C₂H₃O₂^-          c) covalent          d) ionic, Ag⁺, S^-2 e) covalent          f) ionic, Co⁺², NO₃^-

17. a) CaS          b) (NH₄)₂SO₄          c) AlBr₃          d) Na₂HPO₄          e) Mg₃ (PO₄)₂

18. a) K₂O          b) (NH₄)₃PO₄          c) Al₂O₃          d) Na₂CO₃          e) Ba₃(PO₄)₂

Glossary

isomers: compounds with the same chemical formula but different structures covalent bond: attractive force between the nuclei of a molecule’s atoms and pairs of electrons between the atoms covalent compound: (also, molecular compound) composed of molecules formed by atoms of two or more different elements formula unit: a formula indicating the relative numbers of its constituent ions ionic bond: electrostatic forces of attraction between the oppositely charged ions of an ionic compound ionic compound: compound composed of cations and anions combined in ratios, yielding an electrically neutral substance molecular compound: (also, covalent compound) composed of molecules formed by atoms of two or more different elements monatomic ion: ion composed of a single atom oxyanion: polyatomic anion composed of a central atom bonded to oxygen atoms polyatomic ion: ion composed of more than one atom

]]> 1350 2141 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-7-chemical-nomenclature/ Thu, 12 Apr 2018 02:51:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-7-chemical-nomenclature/

Learning Objectives

By the end of this module, you will be able to:

• Derive names for common types of inorganic compounds and simple molecular compounds using a systematic approach

Nomenclature, a collection of rules for naming things, is important in science and in many other situations. This module describes an approach that is used to name simple ionic and molecular compounds, such as NaCl, CaCO₃, and N₂O₄. The simplest of these are binary compounds, those containing only two elements, but we will also consider how to name ionic compounds containing polyatomic ions, and one specific, very important class of compounds known as acids - subsequent chapters in this text will focus on these compounds in great detail. We will limit our attention here to inorganic compounds, compounds that are composed principally of elements other than carbon, and will follow the nomenclature guidelines proposed by IUPAC. The rules for organic compounds, in which carbon is the principle element, will be treated in a later chapter on organic chemistry.

Nomenclature of Ionic Compounds

To name an inorganic compound, we need to consider the answers to several questions. First, is the compound ionic or molecular? If the compound is ionic, does the metal form ions of only one type (fixed charge) or more than one type (variable charge)? Are the ions monatomic or polyatomic? If the compound is molecular, does it contain hydrogen? If so, does it also contain oxygen? From the answers we derive, we place the compound in an appropriate category and then name it accordingly.

Compounds Containing Only Monatomic Ions

The name of a binary compound containing monatomic ions consists of the name of the cation (the name of the metal) followed by the name of the anion (the name of the nonmetallic element with its ending replaced by the suffix –ide). Some examples are given in Table 1.

NaCl, sodium chloride	Na2O, sodium oxide
KBr, potassium bromide	CdS, cadmium sulfide
CaI2, calcium iodide	Mg3N2, magnesium nitride
CsF, cesium fluoride	Ca3P2, calcium phosphide
LiCl, lithium chloride	Al4C3, aluminum carbide
Table 1. Names of Some Ionic Compounds

Compounds Containing Polyatomic Ions

Compounds containing polyatomic ions are named similarly to those containing only monatomic ions, except there is no need to change to an –ide ending, since the suffix is already present in the name of the anion. Examples are shown in Table 2.  See Table 1 in Chapter 4.2 Ionic and Molecular Compounds for the list of common polyatomic ions.

KC2H3O2, potassium acetate	(NH4)Cl, ammonium chloride
NaHCO3, sodium bicarbonate	CaSO4, calcium sulfate
Al2(CO3)3, aluminum carbonate	Mg3(PO4)2, magnesium phosphate
Table 2. Names of Some Polyatomic Ionic Compounds

Example 1

Name the following ionic compounds:

a)  NaCl           b) AlBr₃          c)  BaH₂

Solution  

a)  Identify the cation and anion. Na is a Group 1 metal, and thus it forms the cation Na⁺, called “sodium” ion. Cl is a nonmetal, and forms the anion Cl^-, chloride.  Thus, NaCl = sodium chloride.

b)  AlBr₃ consists of aluminum and bromine; we call it aluminum bromide.

c)  BaH₂is called barium hydride.

 

Test Yourself

Name the following ionic compounds:

      a)  Al₂S₃          b) ZnS            c)  MgI₂

 

Answers

a) aluminum sulfide         b) zinc sulfide         c) magnesium iodide

Compounds Containing a Metal Ion with a Variable Charge

Most of the transition metals can form two or more cations with different charges. Compounds of these metals with nonmetals are named with the same method as compounds in the first category, except the charge of the metal ion is specified by a Roman numeral in parentheses after the name of the metal. The charge of the metal ion is determined from the formula of the compound and the charge of the anion. For example, consider binary ionic compounds of iron and chlorine. Iron typically exhibits a charge of either 2+ or 3+ (see Figure 2 in Chapter 4.2 Ionic and Molecular Compounds), and the two corresponding compound formulas are FeCl₂ and FeCl₃. The simplest name, “iron chloride,” will, in this case, be ambiguous, as it does not distinguish between these two compounds. In cases like this, the charge of the metal ion is included as a Roman numeral in parentheses immediately following the metal name. These two compounds are then unambiguously named iron(II) chloride and iron(III) chloride, respectively. Other examples are provided in Table 4.

Transition Metal Ionic Compound	Name
FeCl3	iron(III) chloride
Hg2O	mercury(I) oxide
HgO	mercury(II) oxide
Cu3(PO4)2	copper(II) phosphate
Table 4. Names of Some Transition Metal Ionic Compounds

An old naming convention used the suffixes –ic and –ous to designate metals with higher and lower charges, respectively: Iron(III) chloride, FeCl₃, can be called called ferric chloride, and iron(II) chloride, FeCl₂, is also known as ferrous chloride.  This older naming convention remains in use by some segments of industry. For example, you may see the words stannous fluoride on a tube of toothpaste. This represents the formula SnF₂, which is also named tin(II) fluoride following the more current convention. The other fluoride of tin is SnF₄, is now named tin(IV) fluoride but is still often referred to as stannic fluoride.  Knowing both convention remains important.

Element	Common Ions	Common Names for Ions
Cu	Cu+/Cu2+	cuprous/cupric
Fe	Fe2+/Fe3+	ferrous/ferric
Co	Co2+/Co3+	cobaltous/cobaltic
Cr	Cr2+/Cr3+	chromous/chromic
Sn	Sn2+/Sn4+	stannous/stannic
Pb	Pb2+/Pb4+	plumbous/plumbic
Hg	Hg22+/Hg2+	mercurous/mercuric
Table 5. Common Names for Some Metal Ions with Variable Charges

Example 2

Name each species.

a) O²⁻         b) Co         c) Co²⁺

 

Solution

a) This species has a 2− charge on it, so it is an anion. Anions are named using the stem of the element name with the suffix -ide added. This is the oxide anion.

b) Because this species has no charge, it is an atom in its elemental form. This is cobalt.

c) In this case, there is a 2+ charge on the atom, so it is a cation. We note from Figure 2 in Chapter 4.2 Ionic and Molecular Compounds), that cobalt cations can have two possible charges, so the name of the ion must specify which charge the ion has. This is the cobalt(II) cation.

 

Test Yourself

Name each species: P³⁻  and Sr²⁺

 

Answers

the phosphide anion and the strontium cation

Example 3

Name the following ionic compounds:

      a)  SnBr₄         b) CoCl₃          c)  Fe₂O₃

 

Solution  

a)   First, identify the charge on the cation (Sn).

Because Br has a charge of –1, we know that Sn must have a charge of +4.

                  0 = 1(x) + 4(-1)                       x = +4         Therefore SnBr₄=  tin(IV) bromide

b)  Cl adopts a charge of –1

                  0 = 1(x) + 3(-1)                       x = +3         Therefore CoCl₃= cobalt(III) chloride

c)  O adopts a charge of –2

                  0 = 2(x) + 3(-2)                       x = +3         Therefore Fe₂O₃= iron(III) oxide

 

Test Yourself

Name the following ionic compounds:

      a)  HgO            b)  PbCl₄          c) PbS           d) Sc₂O₃         e) AgCl

 

Answers

a) mercury(II) oxide         b) lead(IV) chloride         c) lead(II) sulphide

d) scandium oxide         e) silver chloride

Example 4

Name the following ionic compounds:

a) Fe₂S₃         b) CuSe         c) GaN         d) CrCl₃         e) Ti₂(SO₄)₃         f) Co₂O₃         g) CaCl₂         h) AlF₃

 

Solution The anions in these compounds have a fixed negative charge (S²⁻, Se²⁻ , N³⁻, Cl⁻, SO₄²⁻, O⁻², and F⁻), and the compounds must be neutral. Because the metal ions in questions a) to f) have a variable charge, we must figure out the charge of the metal ion by ensuring that the total number of positive charges in each compound must equal the total number of negative charges.  Therefore the positive ions must be Fe³⁺, Cu²⁺, Ga³⁺, Cr³⁺, Ti³⁺ and Co³⁺. These charges are used in the names of the metal ions:

a) iron(III) sulfide                     b) copper(II) selenide           c) gallium(III) nitride

d) chromium(III) chloride         e) titanium(III) sulfate          f) cobalt(III) oxide In questions g) and h) the metal ions do not have a variable charge, therefore

g) Using the names of the ions, this ionic compound is named calcium chloride. It is not calcium(II) chloride because calcium forms only one cation when it forms an ion, and it has a characteristic charge of 2+.

h)The name of this ionic compound is aluminum fluoride.

 

Test Yourself

Write the formulas of the following ionic compounds:

a) chromium(III) phosphide        b) mercury(II) sulfide         c) manganese(II) phosphate

d) copper(I) oxide         e) chromium(VI) fluoride

  Answers a) CrP         b) HgS        c) Mn₃(PO₄)₂         d) Cu₂O         e) CrF₆

Erin Brockovich and Chromium Contamination

In the early 1990s, legal file clerk Erin Brockovich (Figure 1) discovered a high rate of serious illnesses in the small town of Hinckley, California. Her investigation eventually linked the illnesses to groundwater contaminated by Cr(VI) used by Pacific Gas & Electric (PG&E) to fight corrosion in a nearby natural gas pipeline. As dramatized in the film Erin Brokovich (for which Julia Roberts won an Oscar), Erin and lawyer Edward Masry sued PG&E for contaminating the water near Hinckley in 1993. The settlement they won in 1996—$333 million—was the largest amount ever awarded for a direct-action lawsuit in the US at that time.

[caption id="" align="aligncenter" width="1200"] Figure 1. (a) Erin Brockovich found that Cr(VI), used by PG&E, had contaminated the Hinckley, California, water supply. (b) The Cr(VI) ion is often present in water as the polyatomic ions chromate, CrO₄²⁻ (left), and dichromate, Cr₂O₇²⁻ (right).[/caption]

Chromium compounds are widely used in industry, such as for chrome plating, in dye-making, as preservatives, and to prevent corrosion in cooling tower water, as occurred near Hinckley. In the environment, chromium exists primarily in either the Cr(III) or Cr(VI) forms. Cr(III), an ingredient of many vitamin and nutritional supplements, forms compounds that are not very soluble in water, and it has low toxicity. But Cr(VI) is much more toxic and forms compounds that are reasonably soluble in water. Exposure to small amounts of Cr(VI) can lead to damage of the respiratory, gastrointestinal, and immune systems, as well as the kidneys, liver, blood, and skin.

Despite cleanup efforts, Cr(VI) groundwater contamination remains a problem in Hinckley and other locations across the globe. A 2010 study by the Environmental Working Group found that of 35 US cities tested, 31 had higher levels of Cr(VI) in their tap water than the public health goal of 0.02 parts per billion set by the California Environmental Protection Agency.

Example 5

Write the proper formula and give the proper name for each ionic compound formed between the two listed ions.

a) NH₄⁺ and S²⁻          b) Al³⁺ and PO₄³⁻          c) Fe²⁺ and PO₄³⁻

 

Solution

a) Because the ammonium ion has a 1+ charge and the sulfide ion has a 2− charge, we need two ammonium ions to balance the charge on a single sulfide ion. Enclosing the formula for the ammonium ion in parentheses, we have (NH₄)₂S. The compound’s name is ammonium sulfide.

b) Because the ions have the same magnitude of charge, we need only one of each to balance the charges. The formula is AlPO₄, and the name of the compound is aluminum phosphate.

c) Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6. To get 6+, we need three iron(II) ions, and to get 6−, we need two phosphate ions. The proper formula is Fe₃(PO₄)₂, and the compound’s name is iron(II) phosphate.

 

Test Yourself

Write the proper formula and give the proper name for each ionic compound formed between the two listed ions.

a) NH₄⁺ and PO₄³⁻          b) Co³⁺ and NO₂⁻

 

Answers

a) (NH₄)₃PO₄, ammonium phosphate          b) Co(NO₂)₃, cobalt(III) nitrite

Ionic Compounds in Your Cabinets

Every day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table 3. Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula.

Ionic Compound	Use
NaCl, sodium chloride	ordinary table salt
KI, potassium iodide	added to “iodized” salt for thyroid health
NaF, sodium fluoride	ingredient in toothpaste
NaHCO3, sodium bicarbonate	baking soda; used in cooking (and as antacid)
Na2CO3, sodium carbonate	washing soda; used in cleaning agents
NaOCl, sodium hypochlorite	active ingredient in household bleach
CaCO3 calcium carbonate	ingredient in antacids
Mg(OH)2, magnesium hydroxide	ingredient in antacids
Al(OH)3, aluminum hydroxide	ingredient in antacids
NaOH, sodium hydroxide	lye; used as drain cleaner
K3PO4, potassium phosphate	food additive (many purposes)
MgSO4, magnesium sulfate	added to purified water
Na2HPO4, sodium hydrogen phosphate	anti-caking agent; used in powdered products
Na2SO3, sodium sulfite	preservative
Table 3. Everyday Ionic Compounds

 

Nomenclature of Molecular (Covalent) Compounds

The bonding characteristics of inorganic molecular compounds are different from ionic compounds, and they are named using a different system as well. The charges of cations and anions dictate their ratios in ionic compounds, so specifying the names of the ions provides sufficient information to determine chemical formulas. However, because covalent bonding allows for significant variation in the combination ratios of the atoms in a molecule, the names for molecular compounds must explicitly identify these ratios.

Compounds Composed of Two Elements

When two nonmetallic elements form a molecular compound, several combination ratios are often possible. For example, carbon and oxygen can form the compounds CO and CO₂. Since these are different substances with different properties, they cannot both have the same name (they cannot both be called carbon oxide). To deal with this situation, we use a naming method that is somewhat similar to that used for ionic compounds, but with added prefixes to specify the numbers of atoms of each element. The name of the more metallic element (the one farther to the left and/or bottom of the periodic table) is first, followed by the name of the more nonmetallic element (the one farther to the right and/or top) with its ending changed to the suffix –ide. The numbers of atoms of each element are designated by the Greek prefixes shown in Table 6.

Number	Prefix	Number	Prefix
1 (sometimes omitted)	mono-	6	hexa-
2	di-	7	hepta-
3	tri-	8	octa-
4	tetra-	9	nona-
5	penta-	10	deca-
Table 6. Nomenclature Prefixes

When only one atom of the first element is present, the prefix mono- is usually deleted from that part. Thus, CO is named carbon monoxide, and CO₂ is called carbon dioxide. When two vowels are adjacent, the a in the Greek prefix is usually dropped. Some other examples are shown in Table 7.

Compound	Name	Compound	Name
SO2	sulfur dioxide	BCl3	boron trichloride
SO3	sulfur trioxide	SF6	sulfur hexafluoride
NO2	nitrogen dioxide	PF5	phosphorus pentafluoride
N2O4	dinitrogen tetroxide	P4O10	tetraphosphorus decaoxide
N2O5	dinitrogen pentoxide	IF7	iodine heptafluoride
Table 7. Names of Some Molecular Compounds Composed of Two Elements

There are a few common names that you will encounter as you continue your study of chemistry. For example, although NO is often called nitric oxide, its proper name is nitrogen monoxide. Similarly, N₂O is known as nitrous oxide even though our rules would specify the name dinitrogen monoxide. (And H₂O is usually called water, not dihydrogen monoxide.) You should commit to memory the common names of compounds as you encounter them.

Example 6

Name the following covalent compounds:

a) SF₆         b) N₂O₃         c) Cl₂O₇         d) P₄O₆         e) PF₃         f) CO         g) Se₂Br₂

 

Solution Because these compounds consist solely of nonmetals, they are molecular compounds, therefore according to the rules, we use prefixes to designate the number of atoms of each element:

a) sulfur hexafluoride                    b) dinitrogen trioxide              c) dichloride heptoxide

d) tetraphosphorus hexoxide         e) phosphorus trifluoride         f) carbon monoxide (not carbon monooxide) g) diselenium dibromide  

Test Yourself Write the formulas for the following compounds:

a) phosphorus pentachloride          b) dinitrogen monoxide         c) iodine heptafluoride

d) carbon tetrachloride                   e) disulfur difluoride              f) iodine pentabromide   Answers a) PCl₅         b) N₂O         c) IF₇         d) CCl₄      e) S₂F₂         f) IBr₅

The following website provides practice with naming chemical compounds and writing chemical formulas. You can choose binary, polyatomic, and variable charge ionic compounds, as well as molecular compounds.

Binary Acids

Some compounds containing hydrogen are members of an important class of substances known as acids, and these compounds have interesting chemical properties. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, H⁺, when dissolved in water.  To indicate that something is dissolved in water, we will use the phase label (aq) next to a chemical formula (where aq stands for “aqueous,” a word that describes something dissolved in water).  To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound’s name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):

1. The word “hydrogen” is changed to the prefix hydro-
2. The other nonmetallic element name is modified by adding the suffix -ic
3. The word “acid” is added as a second word

For example, when the gas HCl (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table 8.

Name of Gas	Name of Acid
HF(g), hydrogen fluoride	HF(aq), hydrofluoric acid
HCl(g), hydrogen chloride	HCl(aq), hydrochloric acid
HBr(g), hydrogen bromide	HBr(aq), hydrobromic acid
HI(g), hydrogen iodide	HI(aq), hydroiodic acid
H2S(g), hydrogen sulfide	H2S(aq), hydrosulfuric acid
HCN(g), hydrogen cyanide	HCN(aq), hydrocyanic acid
Table 8. Names of Some Simple Acids

Oxyacids

Many compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly discuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:

1. Omit “hydrogen”
2. Start with the root name of the anion
3. Replace –ate with –ic, or –ite with –ous
4. Add “acid”

For example, consider H₂CO₃ (which you might be tempted to call “hydrogen carbonate”). To name this correctly, “hydrogen” is omitted; the –ate of carbonate is replace with –ic; and acid is added—so its name is carbonic acid. Other examples are given in Table 9. There are some exceptions to the general naming method (e.g., H₂SO₄ is called sulfuric acid, not sulfic acid, and H₂SO₃ is sulfurous, not sulfous, acid).

Formula	Anion Name	Acid Name
HC2H3O2	acetate	acetic acid
HNO3	nitrate	nitric acid
HNO2	nitrite	nitrous acid
HClO4	perchlorate	perchloric acid
HClO3	chlorate	chloric acid
HClO2	chlorite	chlorous acid
HClO	hypochlorite	hypochlorous acid
H2CO3	carbonate	carbonic acid
H2SO4	sulfate	sulfuric acid
H2SO3	sulfite	sulfurous acid
H3PO4	phosphate	phosphoric acid
H3PO3	phosphite	phosphorous acid
H2CrO4	chromate	chromic acid
Table 9. Names of Common Oxyacids

Example 7

Name each acid without consulting the tables.

a) HBr(aq)         b) H₂SO₄         c)  HF(g)         d)  HCN(aq)         e) H₂S(aq)

 

Solution

a) As an aqueous binary acid, the acid’s name is hydro- + stem name + -ic acid. Because this acid contains a bromine atom, the name is hydrobromic acid.

b) Because this acid is derived from the sulfate ion, the name of the acid is the stem of the anion name + -ic acid. The name of this acid is sulfuric acid.

c) Because HF(g) is in gaseous form, we name it hydrogen fluoride.

d)  HCN(aq) contains the polyatomic ion cyanide. The root is “cyan”, thus HCN(aq) = hydrocyanic acid.

e)  H₂S(aq) contains the ion sulfide. In this case, however, the root takes a slightly different form of “sulfur” (the same as the element name). Thus H₂S(aq) = hydrosulfuric acid.

 

Test Yourself

Name each acid.

a) HF(aq)         b) HNO₂         c) HClO₄       d) H₂SO₄         e) H₂CrO₄(aq)          f)  H₃PO₄(aq)      g) HClO(aq)

 

Answers

a) hydrofluoric acid         b) nitrous acid         c) perchloric acid         d) sulphuric acid         e) chromic acid

f) phosphoric acid         g) hypochlorous acid

All acids have some similar properties. For example, acids have a sour taste; in fact, the sour taste of some of our foods, such as citrus fruits and vinegar, is caused by the presence of acids in food. Many acids react with some metallic elements to form metal ions and elemental hydrogen. Acids make certain plant pigments change colors; indeed, the ripening of some fruits and vegetables is caused by the formation or destruction of excess acid in the plant. In a later chapter, we will explore the chemical behaviour of acids.

Acids are very prevalent in the world around us. We have already mentioned that citrus fruits contain acid; among other compounds, they contain citric acid, H₃C₆H₅O₇(aq). Oxalic acid, H₂C₂O₄(aq), is found in spinach and other green leafy vegetables. Hydrochloric acid not only is found in the stomach (stomach acid) but also can be bought in hardware stores as a cleaner for concrete and masonry. Phosphoric acid is an ingredient in some soft drinks.

Sodium in Your Food

The element sodium, at least in its ionic form as Na⁺, is a necessary nutrient for humans to live. In fact, the human body is approximately 0.15% sodium, with the average person having one-twentieth to one-tenth of a kilogram in their body at any given time, mostly in fluids outside cells and in other bodily fluids.

Sodium is also present in our diet. The common table salt we use on our foods is an ionic sodium compound. Many processed foods also contain significant amounts of sodium added to them as a variety of ionic compounds. Why are sodium compounds used so much? Usually sodium compounds are inexpensive, but, more importantly, most ionic sodium compounds dissolve easily. This allows processed food manufacturers to add sodium-containing substances to food mixtures and know that the compound will dissolve and distribute evenly throughout the food. Simple ionic compounds such as sodium nitrite (NaNO₂) are added to cured meats, such as bacon and deli-style meats, while a compound called sodium benzoate is added to many packaged foods as a preservative. Table 10 "Some Sodium Compounds Added to Food" is a partial list of some sodium additives used in food. Some of them you may recognize after reading this chapter. Others you may not recognize, but they are all ionic sodium compounds with some negatively charged ion also present.

Sodium Compound	Use in Food
Sodium acetate	preservative, acidity regulator
Sodium adipate	food acid
Sodium alginate	thickener, vegetable gum, stabilizer, gelling agent, emulsifier
Sodium aluminum phosphate	acidity regulator, emulsifier
Sodium aluminosilicate	anticaking agent
Sodium ascorbate	antioxidant
Sodium benzoate	preservative
Sodium bicarbonate	mineral salt
Sodium bisulfite	preservative, antioxidant
Sodium carbonate	mineral salt
Sodium carboxymethylcellulose	emulsifier
Sodium citrates	food acid
Sodium dehydroacetate	preservative
Sodium erythorbate	antioxidant
Sodium erythorbin	antioxidant
Sodium ethyl para-hydroxybenzoate	preservative
Sodium ferrocyanide	anticaking agent
Sodium formate	preservative
Sodium fumarate	food acid
Sodium gluconate	stabilizer
Sodium hydrogen acetate	preservative, acidity regulator
Sodium hydroxide	mineral salt
Sodium lactate	food acid
Sodium malate	food acid
Sodium metabisulfite	preservative, antioxidant, bleaching agent
Sodium methyl para-hydroxybenzoate	preservative
Sodium nitrate	preservative, color fixative
Sodium nitrite	preservative, color fixative
Sodium orthophenyl phenol	preservative
Sodium propionate	preservative
Sodium propyl para-hydroxybenzoate	preservative
Sodium sorbate	preservative
Sodium stearoyl lactylate	emulsifier
Sodium succinates	acidity regulator, flavour enhancer
Sodium salts of fatty acids	emulsifier, stabilizer, anticaking agent
Sodium sulfite	mineral salt, preservative, antioxidant
Sodium sulfite	preservative, antioxidant
Sodium tartrate	food acid
Sodium tetraborate	preservative

Table 10. Some Sodium Compounds Added to Food The use of so many sodium compounds in prepared and processed foods has alarmed some physicians and nutritionists. They argue that the average person consumes too much sodium from his or her diet. The average person needs only about 500 mg of sodium every day; most people consume more than this—up to 10 times as much. Some studies have implicated increased sodium intake with high blood pressure; newer studies suggest that the link is questionable. However, there has been a push to reduce the amount of sodium most people ingest every day: avoid processed and manufactured foods, read labels on packaged foods (which include an indication of the sodium content), don’t oversalt foods, and use other herbs and spices besides salt in cooking.

Key Concepts and Summary

Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to –ide. For example, K₂O is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge. Thus, FeCl₂ is iron(II) chloride and FeCl₃ is iron(III) chloride. Some compounds contain polyatomic ions; the names of common polyatomic ions should be memorized. Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include SF₆, sulfur hexafluoride, and N₂O₄, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the –ide suffix to –ic, and adding “acid;” HCl is hydrochloric acid. Oxyacids are named by changing the ending of the anion to –ic, and adding “acid;” H₂CO₃ is carbonic acid.

 

Figure 3. Flowchart illustrating the thought process involved in naming simple ionic and covalent compounds and the rules needed to follow.

Activity

Make yourself a stack of small sized Qcards.  On one side have the name of an ionic compound (e.g. sodium hydroxide) and on the other side have its chemical formula (e.g. NaOH).  Use every example found in this chapter - including those in the exercises.  Then use these Qcards to quiz yourself.

Exercises

1.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Mg²⁺ and Cl⁻         b)  Fe²⁺ and O²⁻         c)  Fe³⁺ and O²⁻

2.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Cu²⁺ and F⁻         b)  Ca²⁺ and O²⁻         c)  K⁺ and P³⁻

3.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  K⁺ and SO₄²⁻         b)  NH₄⁺ and S²⁻         c)  NH₄⁺ and PO₄³⁻

4.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Pb⁴⁺ and SO₄²⁻         b)  Na⁺ and I₃⁻         c)  Li⁺ and Cr₂O₇²⁻

5.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Ag⁺ and SO₃²⁻         b)  Na⁺ and HCO₃⁻         c)  Fe³⁺ and ClO₃⁻

6. .Which of these formulas represent molecules? State how many atoms are in each molecule. a)  Fe         b)  PCl₃         c)  P₄         d)  Ar 7.  What is the difference between CO and Co? 8.  Give the proper formula for each diatomic element. 9.  What is the stem of fluorine used in molecule names? CF₄ is one example. 10.  Give the proper name for each molecule.

a)  PF₃         b)  TeCl₂         c)  N₂O₃

11.  Give the proper name for each molecule.

a)  XeF₂         b)  O₂F₂         c)  SF₆

12.  Give the proper name for each molecule.

a)  N₂O         b)  N₂O₄         c)  N₂O₅

13.  Give the proper formula for each name.

a)  dinitrogen pentoxide         b)  tetraboron tricarbide         c)  phosphorus pentachloride

14.  Give the proper formula for each name.

a)  dioxygen dichloride         b)  dinitrogen trisulfide         c)  xenon tetrafluoride

15.  Give the proper formula for each name.

a)  iodine trifluoride         b)  xenon trioxide         c)  disulfur decafluoride 16. Give the formula for each acid.

a)  perchloric acid         b)  hydroiodic acid

17. Name each acid.

a)  HF(aq)         b)  HNO₃(aq)         c)  H₂C₂O₄(aq) 18. Name the following compounds:

a) CsCl         b) BaO         c) K₂S         d) BeCl₂         e) HBr         f) AlF₃

19. Write the formulas of the following compounds:

a) rubidium bromide         b) magnesium selenide         c) sodium oxide         d) calcium chloride

e) hydrogen fluoride         f) gallium phosphide         g) aluminum bromide         h) ammonium sulfate

20. Write the formulas of the following compounds:

a) chlorine dioxide         b) dinitrogen tetraoxide         c) potassium phosphide

d) silver(I) sulfide         e) aluminum nitride         f) silicon dioxide

21. Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:

a) Cr₂O₃         b) FeCl₂         c) CrO₃         d) TiCl₄         e) CoO         f) MoS₂

22. The following ionic compounds are found in common household products. Write the formulas for each compound:

a) potassium phosphate         b) copper(II) sulfate         c) calcium chloride

d) titanium dioxide         e) ammonium nitrate

f) sodium bisulfate (the common name for sodium hydrogen sulfate) 23. What are the IUPAC names of the following compounds?

a) manganese dioxide         b) mercurous chloride (Hg₂Cl₂)         c) ferric nitrate [Fe(NO₃)₃]

d) titanium tetrachloride         e) cupric bromide (CuBr₂)

 

Answers 1. a)  magnesium chloride, MgCl₂         b)  iron(II) oxide, FeO         c)  iron(III) oxide, Fe₂O₃ 2. a)  copper(II) fluoride, CuF₂         b)  calcium oxide, CaO         c)  potassium phosphide, K₃P 3. a)  potassium sulfate, K₂SO₄         b)  ammonium sulfide, (NH₄)₂S         c)  ammonium phosphate, (NH₄)₃PO₄ 4. a)  lead(IV) sulfate, Pb(SO₄)₂         b)  sodium triiodide, NaI₃         c)  lithium dichromate, Li₂Cr₂O₇ 5. a)  silver sulfite, Ag₂SO₃         b)  sodium hydrogen carbonate, NaHCO₃         c)  iron(III) chlorate, Fe(ClO₃)₃ 6. a)  not a molecule         b)  a molecule; four atoms total         c)  a molecule; four atoms total d)  not a molecule 7. CO is a compound of carbon and oxygen; Co is the element cobalt. 8. H₂, O₂, N₂, F₂, Cl₂, Br₂, I₂ 9.  fluor- 10. a)  phosphorus trifluoride         b)  tellurium dichloride         c)  dinitrogen trioxide 11. a)  xenon difluoride         b)  dioxygen difluoride         c)  sulfur hexafluoride 12. a)  dinitrogen monoxide         b)  dinitrogen tetroxide         c)  dinitrogen pentoxide 13. a)  N₂O₅         b)  B₄C₃         c)  PCl₅ 14. a)  O₂Cl₂         b)  N₂S₃         c)  XeF₄ 15. a)  IF₃         b)  XeO₃         c)  S₂F₁₀

16. a)  HClO₄(aq)         b)  HI(aq)

17. a)  hydrofluoric acid         b)  nitric acid         c)  oxalic acid

18. a) cesium chloride         b) barium oxide         c) potassium sulfide

d) beryllium chloride         e) hydrogen bromide         f) aluminum fluoride

19. a) RbBr         b) MgSe         c) Na₂O         d) CaCl₂         e) HF         f) GaP         g) AlBr₃         h) (NH₄)₂SO₄

20. a) ClO₂         b) N₂O₄         c) K₃P         d) Ag₂S         e) AlN         f) SiO₂

21. a) chromium(III) oxide         b) iron(II) chloride         c) chromium(VI) oxide

d) titanium(IV) chloride         e) cobalt(II) oxide         f) molybdenum(IV) sulfide

22. a) K₃PO₄         b) CuSO₄         c) CaCl₂         d) TiO₂         e) NH₄NO₃         f) NaHSO₄

23. a) manganese(IV) oxide         b) mercury(I) chloride         c) iron(III) nitrate

d) titanium(IV) chloride         e) copper(II) bromide

Glossary

binary acid: compound that contains hydrogen and one other element, bonded in a way that imparts acidic properties to the compound (ability to release H⁺ ions when dissolved in water) binary compound: compound containing two different elements. nomenclature: system of rules for naming objects of interest oxyacid: compound that contains hydrogen, oxygen, and one other element, bonded in a way that imparts acidic properties to the compound (ability to release H⁺ ions when dissolved in water)

]]> 1353 2141 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-3/ Thu, 12 Apr 2018 02:51:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-3/ Swimming pools have long been a popular means of recreation, exercise, and physical therapy. Since it is impractical to refill large pools with fresh water on a frequent basis, pool water is regularly treated with chemicals to prevent the growth of harmful bacteria and algae. [caption id="" align="aligncenter" width="1300"] Figure 1. The water in a swimming pool is a complex mixture of substances whose relative amounts must be carefully maintained to ensure the health and comfort of people using the pool. (credit: modification of work by Vic Brincat)[/caption]

Proper pool maintenance requires regular additions of various chemical compounds in carefully measured amounts. For example, the relative amount of calcium ion, Ca²⁺, in the water should be maintained within certain limits to prevent eye irritation and avoid damage to the pool bed and plumbing. To maintain proper calcium levels, calcium cations are added to the water in the form of an ionic compound that also contains anions; thus, it is necessary to know both the relative amount of Ca²⁺ in the compound and the volume of water in the pool in order to achieve the proper calcium level. Quantitative aspects of the composition of substances (such as the calcium-containing compound) and mixtures (such as the pool water) are the subject of this chapter.

]]> 1356 1354 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/ Thu, 12 Apr 2018 02:51:36 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-1-formula-mass-and-the-mole-concept/

Learning Objectives

By the end of this section, you will be able to:

• Define the amount unit mole and Avogadro’s number
• Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations to derive these quantities from one another  

Counting by Weighing

Atomic and molecular mass provides a way of understanding/predicting the weights of substances in a reaction. For example, consider the reaction:

          C_(s)  + O_2(g)  →  CO_2(g)

This equation tells us that carbon reacts with oxygen gas to produce carbon dioxide in a 1:1:1 ratio.  In other words, the equation tells us that 1 atom of carbon reacts with 1 molecule of O₂ to form 1 molecule of CO₂, or 500 C atoms will react with 500 O₂ molecules to form 500 molecules for CO₂, and so on. That is, we need an equal number of C atoms and O₂ molecules. Now, it’s impractical to actually count out a number of atoms or molecules, they are just too small, but we can do this in effect by weighing.  Looking at an everyday item as an example of counting by weight, see example 1.

Example 1

How many grams should you weigh to get 5000 nails if each nail has an average mass of 0.25g?   Solution $latex 5000\;\rule[0.75ex]{2.0em}{0.1ex}\hspace{-2.0em}\text{nails} \times \frac{0.25\;\text{g}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{nail}} = 1.2\times 10^3\;\text{g are needed} $

   

Test Yourself If nails have an average mass of 0.25g, how many nails are present in 62.5g of nails?   Answer 2.5x10² nails

Furthermore, if two different types of things are measured by weighing, the same number of each will be present if the ratio of the masses weighed out is equal to the ratio of masses of the individual units. For example:  If 1 nail weighs 0.25 g and 1 screw weighs 0.50 g, then 300 g of nails will contain the same number of items as 600 g of screws. Why?  1 screw weighs twice as much as 1 nail, therefore you must have the same number of each item when you weigh out a total mass of screws that’s twice the total mass of nails. Proof: $latex 300\;\rule[0.75ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g nails} \times \frac{1\;\text{nail}}{0.25\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g nails}} = 1200\;\text{nails} $   $latex 600\;\rule[0.75ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g screws} \times \frac{1\;\text{screw}}{0.50\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g screws}} = 1200\;\text{screws} $   It is the same idea, with:

2 kg screws & 1 kg nails, or 2 tons screws & 1 ton nails, or 2 dozen tons of screws & 1 dozen tons of nails…

We know that in every case the number of screws is the same as the number of nails. Note: in each example, we don’t necessarily know what the number of items is, but we can be sure that it’s the same number of each item. So, for the above reaction, C_(s)  + O_2(g)  →  CO_2(g), as long as we weigh out amounts of C and O₂ in the same ratio as the weights of 1 C atom (12.011 amu) and 1 O₂ molecule (31.9988 amu), we can be sure we’ll have an equal number of C atoms and O₂ molecules. Now, the easiest way to choose total weights in the desired ratio—without doing any math—is to just mimic the numerical values of the known atomic and molecular weights. For example, to make sure we have a 12.011 to 31.9988 weight ratio of C and O₂, we could just weigh out: 12.011 grams of C and 31.9988 grams of O₂ or   12.011 kg of C and 31.9988 kg of O₂ or   12.011 lbs of C and 31.9988 lbs of O₂ or   12.011 tons of C and 31.9988 tons of O₂ or etc..  They all will have equal numbers of C atoms and O₂ molecules.

The Mole

The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H₂O, and hydrogen peroxide, H₂O₂, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

Estimating the numerical value of Avogadro’s number

The mole, the SI unit for “substance”, is a unit for counting, and it’s often said that we can think of the mole as like a dozen. A subtle difference, however is that a dozen is defined as a specific number (12) of items, but a mole (Avogadro’s number of items) is defined as whatever number of H atoms you’d have if you weighed out 1.00794 g of H, (or S atoms in 32.066 g of S, or ¹²C atoms in 12 g of ¹²C, and so on).   This usually isn’t a problem, as these terms are normally encountered. If we bought a dozen donuts, we’d want to know how many we got—is there one for each person, etc.. But if we weigh out a mole of Na (22.9898 g) and a mole of Cl (35.4527 g), we probably don’t care how many atoms we have—we’re just happy to know we have the same number of atoms of each, in order to make NaCl without any atoms of Na or Cl leftover.   However, in some relatively rare circumstances, we may need to know how many atoms or molecules do we have when we weigh out a certain amount. For this, we’d actually like to know a numerical value for Avogadro’s number. To get this value, someone has to measure how many H atoms are there in 1.00794 g of H, (or S atoms in 32.066 g of S, or ¹²C atoms in 12 g of ¹²C, or…). To do that, someone has to measure the mass, in grams, of a single atom.   Many different ways have been dreamed up for estimating, in effect, the mass of a single atom. One very good one is to weigh, in grams, a crystal of a measured volume, and use the diffraction of x-rays to measure the distance between adjacent atoms, and thus estimate the number of atoms in the crystal. Calculating the number of things in a mole (Avogadro’s number) can be illustrated using ¹²C as an example. ¹²C provides a good basis because it’s the only substance where we can start with the mass of an atom in amu exactly (12 g).   First, from the definition of a mole:  one ¹²C  atom weighs 12 amu’s (exactly), so a mole of ¹²C means 12 g of ¹²C (exactly).   The current best measurements give the mass of a ¹²C atom to 8 significant figures as 1.9926465 x 10^-23g, so to determine Avogadro’s number we ask how many times 1.99264654 x 10^-23g goes into 12 g (exactly).   This works out as:   $latex \frac{12\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g (exactly) /mole}}{1.99264654\times 10^{-23}\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g/atom}} = 6.0221418\times 10^{23}\;\text{atoms/mole} $   So, based on current best measurements, we can write: 1 Mole of things  = Avogadro’s number of them Avogadro’s number  =  6.0221418 x 10²³  =  6.022 x 10²³ approximately   Remember: this value, 6.022… x 10²³, refers to entities/particles/atoms/ molecules/ions —whatever you have a mole of—per mole, just like dozen can be for muffins, donuts, cookies, etc....

Therefore, a mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure ¹²C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.

The number of entities composing a mole has been experimentally determined to be 6.0221418 × 10²³, a fundamental constant named Avogadro’s number (N_A) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022 × 10²³/mol.

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 1).

[caption id="attachment_1362" align="aligncenter" width="500"] Figure 1. Each sample contains 6.022 × 10²³ atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott)[/caption]

Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, ¹²C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single ¹²C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of ¹²C contains 1 mole of ¹²C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, ¹²C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2).

[caption id="attachment_757" align="aligncenter" width="650"] Figure 2. Each sample contains 6.022 × 10²³ molecules or formula units—1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of C₈H₁₇OH (1-octanol, formula mass 130.2 amu), 454.4 g of HgI₂ (mercury(II) iodide, formula mass 454.4 amu), 32.0 g of CH₃OH (methanol, formula mass 32.0 amu) and 256.5 g of S₈ (sulfur, formula mass 256.5 amu). (credit: Sahar Atwa)[/caption]

Element	Average Atomic Mass (amu)	Molar Mass (g/mol)	Atoms/Mole
C	12.011	12.011	6.022 × 1023
H	1.00794	1.00794	6.022 × 1023
O	15.9994	15.9994	6.022 × 1023
Na	22.9898	22.9898	6.022 × 1023
Cl	35.4527	35.4527	6.022 × 1023
Table 1.

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 3). Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.

[caption id="attachment_758" align="aligncenter" width="250"] Figure 3. The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. (credit: “tanakawho”/Wikimedia commons)[/caption]

The mole is used in chemistry to represent 6.022 × 10²³ of something, but it can be difficult to conceptualize such a large number. Watch this video and then complete the “Think” questions that follow. Explore more about the mole by reviewing the information under “Dig Deeper.”

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

Example 2

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

 

Solution The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.0983 amu, and so its molar mass is 39.0983 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.0983 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

$latex 4.7 \rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g} \;\text{K} \times \frac{\text{mol K}}{39.0983 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}} = 0.12 \;\text{mol of potassium}$

 

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

 

Test Yourself Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?

  Answer 0.360mol

Example 3

A liter of air contains 9.2 × 10⁻⁴ mol argon. What is the mass of Ar in a liter of air?

 

Solution The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10⁻³) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

$latex 9.2 \times 10^{-4} \;\rule[0.5ex]{1.75em}{0.1ex}\hspace{-1.75em}\text{mol} \;\text{Ar} \times \frac{39.948 \;\text{g}}{\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol} \;\text{Ar}} = 0.037 \;\text{g of argon}$

The result is in agreement with our expectations, around 0.04 g of argon.

 

Test Yourself What is the mass of 2.561 mol of gold?

  Answer 504.4 g

Example 4

Copper is commonly used to fabricate electrical wire (Figure 4). How many copper atoms are in 5.00 g of copper wire?

[caption id="attachment_1368" align="aligncenter" width="150"] Figure 4. Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)[/caption]

Solution The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (N_A) to convert this molar amount to number of Cu atoms:

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth N_A, or approximately 10²² Cu atoms. Carrying out the two-step computation yields:

$latex 5.00 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g} \;\text{Cu} \times \frac{\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol} \;\text{Cu}}{63.546 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}} \times \frac{6.022 \times 10^{23} \;\text{atoms}}{\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol}} = 4.74 \times 10^{22} \;\text{atoms of copper}$

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10²² as expected.

 

Test Yourself A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?

  Answer 4.586 × 10²² Au atoms

We can calculate the total molar mass of a compound the same way we calculated total molecular mass—just replace amu (for molecular mass) with grams (for molar mass)—but keep in mind the difference between what these two terms mean.

Example 5

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C₂H₅O₂N. How many moles of glycine molecules are contained in 28.35 g of glycine?

 

Solution We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 2:

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C₂H₅O₂N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

2 C mass = 2 x 12.011 g/mol	= 24.022 g/mol
5 H masses = 5 x 1.00794 g/mol	= 5.0397 g/mol
2 O masses = 2 x 15.9994 g/mol	= 31.9988 g/mol
1 N masses = 1 x 14.0067 g/mol	= 14.0067 g/mol
Total	= 75.067 g/mol = the molar mass of C2H5O2N

Figure 5. The average mass of a mole of glycine, C₂H₅O₂N, is 75.067 g/mol, which is the sum of the average molar masses of each of its constituent atoms. The molecular structure of glycine.

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

$latex 28.35 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g} \;\text{glycine} \; \times \frac{\text{mol glycine}}{75.067 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}} = 0.3777 \;\text{mol of glycine} $

This result is consistent with our rough estimate.

 

Test Yourself How many moles of sucrose, C₁₂H₂₂O₁₁, are in a 25-g sample of sucrose?

  Answer 0.073 mol

Example 6

Vitamin C is a covalent compound with the molecular formula C₆H₈O₆. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10⁻⁴ mol. What is the mass of this allowance in grams?

 

Solution As for elements, the mass of a compound can be derived from its molar amount as shown:

The molar mass for this compound is computed to be 176.126 g/mol. The given number of moles is a very small fraction of a mole (~10⁻⁴ or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:

$latex 1.42 \times 10^{-4} \;\rule[0.5ex]{1.75em}{0.1ex}\hspace{-1.75em}\text{mol} \;\text{vitamin C} \times \frac{176.126 \text{g}}{\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol} \;\text{vitamin C}} = 0.0250 \;\text{g of vitamin C}$

This is consistent with the anticipated result.

 

Test Yourself What is the mass of 0.443 mol of hydrazine, N₂H₄?

  Answer 14.2 g

Just as we can use the chemical formula to go between molecules of a compound and atoms of an element in the compound (and vice versa): we can also use chemical formulas as conversion factors to convert from moles of a compound to moles of an element in the compound (and vice versa). Be particularly careful when looking at the mass of an element in a compound. Remember, the only way you can change from molecule to atoms, or vice-versa, is by using the chemical formula, and the chemical formula relates ONLY to particles or moles, not to mass.

Example 7

A packet of an artificial sweetener contains 40.0 mg of saccharin (C₇H₅NO₃S), which has the structural formula:

Given that saccharin has a molar mass of 183.188 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?

 

Solution The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example 5, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

$latex 0.0400 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g} \; \text{C}_7\text{H}_5\text{NO}_3\text{S} \times \frac{\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol} \;\text{C}_7\text{H}_5\text{NO}_3\text{S}}{183.188 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g} \;\text{C}_7\text{H}_5\text{NO}_3\text{S}} \times \frac{6.022 \times 10^{23} \;\text{C}_7\text{H}_5\text{NO}_3\text{S} \;\text{molecules}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol} \;\text{C}_7\text{H}_5\text{NO}_3\text{S}}$ $latex = \underline{1.31}49 \times 10^{20} \;\text{C}_7\text{H}_5\text{NO}_3\text{S} \;\text{molecules with 3 sig figs}$

 

The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:

$latex \underline{1.31}49 \times 10^{20}\;\rule[0.5ex]{4.0em}{0.1ex}\hspace{-4.0em}\text{molecules}\; \text{C}_7\text{H}_5\text{NO}_3\text{S} \times \frac{7\;\text{C atoms}} {1 \;\rule[0.25ex]{3.0em}{0.1ex}\hspace{-3.0em}\text{molecules}\; \text{C}_7\text{H}_5\text{NO}_3\text{S}}$ $latex = \underline{9.20}43 \times 10^{21} \;\text{C molecules with 3 sig figs}$ $latex = 9.20 \times 10^{21} \;\text{C molecules}$

 

Test Yourself How many C₄H₁₀ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?

  Answers 9.545 × 10²² molecules C₄ H₁₀; 9.545 × 10²³ atoms H

Counting Neurotransmitter Molecules in the Brain

The brain is the control center of the central nervous system (Figure 6). It sends and receives signals to and from muscles and other internal organs to monitor and control their functions; it processes stimuli detected by sensory organs to guide interactions with the external world; and it houses the complex physiological processes that give rise to our intellect and emotions. The broad field of neuroscience spans all aspects of the structure and function of the central nervous system, including research on the anatomy and physiology of the brain. Great progress has been made in brain research over the past few decades, and the BRAIN Initiative, a federal initiative announced in 2013, aims to accelerate and capitalize on these advances through the concerted efforts of various industrial, academic, and government agencies (more details available at www.whitehouse.gov/share/brain-initiative).

[caption id="attachment_1375" align="aligncenter" width="500"] Figure 6. (a) A typical human brain weighs about 1.5 kg and occupies a volume of roughly 1.1 L. (b) Information is transmitted in brain tissue and throughout the central nervous system by specialized cells called neurons (micrograph shows cells at 1600× magnification).[/caption]

Specialized cells called neurons transmit information between different parts of the central nervous system by way of electrical and chemical signals. Chemical signaling occurs at the interface between different neurons when one of the cells releases molecules (called neurotransmitters) that diffuse across the small gap between the cells (called the synapse) and bind to the surface of the other cell. These neurotransmitter molecules are stored in small intracellular structures called vesicles that fuse to the cell wall and then break open to release their contents when the neuron is appropriately stimulated. This process is called exocytosis (see Figure 7). One neurotransmitter that has been very extensively studied is dopamine, C₈H₁₁NO₂. Dopamine is involved in various neurological processes that impact a wide variety of human behaviors. Dysfunctions in the dopamine systems of the brain underlie serious neurological diseases such as Parkinson’s and schizophrenia.

[caption id="attachment_1376" align="aligncenter" width="500"] Figure 7. (a) Chemical signals are transmitted from neurons to other cells by the release of neurotransmitter molecules into the small gaps (synapses) between the cells. (b) Dopamine, C₈H₁₁NO₂, is a neurotransmitter involved in a number of neurological processes.[/caption]

One important aspect of the complex processes related to dopamine signaling is the number of neurotransmitter molecules released during exocytosis. Since this number is a central factor in determining neurological response (and subsequent human thought and action), it is important to know how this number changes with certain controlled stimulations, such as the administration of drugs. It is also important to understand the mechanism responsible for any changes in the number of neurotransmitter molecules released—for example, some dysfunction in exocytosis, a change in the number of vesicles in the neuron, or a change in the number of neurotransmitter molecules in each vesicle.

Significant progress has been made recently in directly measuring the number of dopamine molecules stored in individual vesicles and the amount actually released when the vesicle undergoes exocytosis. Using miniaturized probes that can selectively detect dopamine molecules in very small amounts, scientists have determined that the vesicles of a certain type of mouse brain neuron contain an average of 30,000 dopamine molecules per vesicle (about 5 × 10⁻²⁰ mol or 50 zmol). Analysis of these neurons from mice subjected to various drug therapies shows significant changes in the average number of dopamine molecules contained in individual vesicles, increasing or decreasing by up to three-fold, depending on the specific drug used. These studies also indicate that not all of the dopamine in a given vesicle is released during exocytosis, suggesting that it may be possible to regulate the fraction released using pharmaceutical therapies.[footnote]Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.[/footnote]

Summary of using the mole idea:

Note:

• there is no direct conversion between number of atoms (or other entities) and grams of a substance. You need to first change to moles!
• which entities you’re expressing a number of will depend on what you are discussing. For example: Fe would be discussed in number of atoms, whereas Fe³⁺ would be expressed in ions, and H₂O would be expressed in molecules.

Example 8

a) How many atoms are present in a 25.0 g sample of Na? b) What is the mass (in grams) of 5.00 x 10²⁵atoms of Cr?   Solution a) Convert grams to mol using molar mass, then to atoms with Avogadro’s no. (g$latex \longrightarrow$ mol$latex \longrightarrow$ atoms):

$latex 25.0 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Na}\; \times \frac{1\; \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Na}} {22.9898 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Na}} \times \frac{6.022 \times 10^{23}\; \text{atoms Na}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol of Na}}$ $latex = 6.55 \times 10^{23}\; \text{atoms of sodium}$

  b) Now we are going from atoms $latex \longrightarrow$mol $latex \longrightarrow$ g

$latex 5.00\times 10^{25}\; \rule[0.5ex]{2.0em}{0.1ex}\hspace{-2.0em}\text{atoms of Cr}\; \times \frac{1\; \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Cr}} {6.022 \times 10^{23} \;\rule[0.25ex]{3.0em}{0.1ex}\hspace{-3.0em}\text{atoms Cr}} \times \frac{51.996\; \text{g Cr}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol of Cr}}$ $latex = 4.32 \times 10^{3}\; \text{g of chromium}$

  Test Yourself How many atoms are present in 21.2 mg of Ag?   Answer 1.18x10²⁰ atoms of Ag

Notice that the Test Yourself  question above included a metric conversion. We can add metric and other conversions from Chapter 2 to our “road maps”.  Remember that there is often more than one approach for a question.

Example 9

How many atoms are in a piece of gold measuring 1 cm x 2 cm x 0.5 cm? (Density of gold is 19.32 g/mL)   Solution Our “road map” for this question:

Measurements $latex \longrightarrow$ volume $latex \longrightarrow$ g $latex \longrightarrow$ mol $latex \longrightarrow$ atoms

Keep in mind that 1 cm³ = 1 mL, we use the density to get the mass based on the measurements, and then follow the standard procedure to atoms using the molar mass and Avogadro’s number. 1 cm x 2 cm x 0.5 cm = 1 cm³ of Au   $latex 1\; \rule[0.75ex]{1.25em}{0.1ex} \hspace{-1.25em} \text{cm}^3\; \times \frac{1\; \rule[0.5ex]{1.0em}{0.1ex} \hspace{-1.0em} \text{mL Au}} {1\; \rule[0.5ex]{1.25em}{0.1ex} \hspace{-1.25em} \text{cm}^3 \text{ Au}} \times \frac{19.32\; \rule[0.5ex]{0.5em}{0.1ex} \hspace{-0.5em}\text{g Au}} {1\; \rule[0.5ex]{0.8em}{0.1ex} \hspace{-0.8em} \text{mL Au}} \times \frac{1\; \rule[0.5ex]{1.25em}{0.1ex} \hspace{-1.25em} \text{mol Au}} {196.967\; \rule[0.5ex]{0.5em}{0.1ex} \hspace{-0.5em} \text{g Au}} \times \frac{6.022 \times 10^{23}\; \text{atoms of Au}}{1\; \rule[0.5ex]{1.25em}{0.1ex} \hspace{-1.25em} \text{mol Au}} = 6 \times 10^{22}\; \text{atoms of Au} $   Test Yourself The density of mercury (Hg) is 13.53 g/mL. How many litres will 4.2 x 10²¹ atoms of Hg occupy?   Answer 1.0x10^-4 L

More Worked Out Problems

Example 10

How many molecules are present in 2.76 mol of H₂O? How many atoms is this?

 

Solution

The definition of a mole is an equality that can be used to construct a conversion factor. Also, because we know that there are three atoms in each molecule of H₂O, we can also determine the number of atoms in the sample.

 

To determine the total number of atoms, we have

Test Yourself

How many molecules are present in 4.61 × 10⁻² mol of O₂?

 

Answer

2.78 × 10²² molecules

Example 11

What is the molar mass of C₆H₁₂O₆?

 

Solution

To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table:

6 C = 6 × 12.011	= 72.066
12 H = 12 × 1.00794	= 12.09528
6 O = 6 × 15.9994	= 95.9964
TOTAL	= 180.158 g/mol

Per convention, the unit grams per mole is written as a fraction.

 

Test Yourself

What is the molar mass of AgNO₃?

 

Answer

169.873 g/mol

Example 12

What is the mass of 3.56 mol of HgCl₂? The molar mass of HgCl₂ is 271.50 g/mol.

 

Solution

Use the molar mass as a conversion factor between moles and grams. Because we want to cancel the mole unit and introduce the gram unit, we can use the molar mass as given:

$latex 3.56\; \rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol}\; \text{HgCl}_2 \times \frac{271.50\; \text{g HgCl}_2} {1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol}\; \text{HgCl}_2}$ $latex = 967\; \text{g}\; \text{HgCl}_2$

 

Test Yourself

What is the mass of 33.7 mol of H₂O?

 

Answer

607 g

Example 13

How many moles of H₂O are present in 240.0 g of water (about the mass of a cup of water)?

 

Solution

Use the molar mass of H₂O as a conversion factor from mass to moles. The molar mass of water is (1.00794 + 1.00794 + 15.9994) = 18.0153 g/mol. However, because we want to cancel the gram unit and introduce moles, we need to take the reciprocal of this quantity, or 1 mol/18.0153 g:

$latex 240.0\; \rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}\; \text{H}_2\text{O} \times \frac{1\; \text{mol H}_2\text{O}} {18.0153\; \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}\; \text{H}_2\text{O}}$ $latex = 13.32\; \text{mol}\; \text{H}_2\text{O}$

 

Test Yourself

How many moles are present in 35.6 g of H₂SO₄ (molar mass = 98.079 g/mol)?

 

Answer

0.363 mol

Example 14

The density of ethanol is 0.789 g/mL. How many moles are in 100.0 mL of ethanol? The molar mass of ethanol is 46.069 g/mol.

 

Solution

Here, we use density to convert from volume to mass and then use the molar mass to determine the number of moles.

$latex 100.0\; \rule[0.5ex]{0.8em}{0.1ex}\hspace{-0.8em}\text{mL ethanol}\; \times \frac{0.789\; \text{g}} {1\; \rule[0.25ex]{0.8em}{0.1ex}\hspace{-0.8em}\text{mL}} \times \frac{1\; \text{mol}} {46.069\; \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}}$ $latex = 1.71\; \text{mol ethanol}$

 

Test Yourself

If the density of benzene, C₆H₆, is 0.879 g/mL, how many moles are present in 17.9 mL of benzene?

 

Answer

0.201 mol

Key Concepts and Summary

A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be 6.022 × 10²³, a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H₂O molecule weighs approximately 18 amu and 1 mole of H₂O molecules weighs approximately 18 g).

Exercises

1. Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula. 2. Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C₂H₅OH), 0.60 mol of formic acid (HCO₂H), or 1.0 mol of water (H₂O)? Explain why. 3. How are the molecular mass and the molar mass of a compound similar and how are they different? 4. Calculate the molar mass of each of the following:

a) S₈      b) C₅H₁₂      c) Sc₂(SO₄)₃      

d) CH₃COCH₃ (acetone)      e) C₆H₁₂O₆ (glucose) 5. Calculate the molar mass of each of the following:

a) the anesthetic halothane, C₂HBrClF₃

b) the herbicide paraquat, C₁₂H₁₄N₂Cl₂

c) caffeine, C₈H₁₀N₄O₂

d) urea, CO(NH₂)₂

e) a typical soap, C₁₇H₃₅CO₂Na

6. Determine the mass of each of the following:

a) 0.0146 mol KOH

b) 10.2 mol ethane, C₂H₆

c) 1.6 × 10⁻³ mol Na₂ SO₄

d) 6.854 × 10³ mol glucose, C₆ H₁₂ O₆

e) 2.86 mol Co(NH₃)₆Cl₃

7. Determine the mass of each of the following:

a) 2.345 mol LiCl

b) 0.0872 mol acetylene, C₂H₂

c) 3.3 × 10⁻² mol Na₂ CO₃

d) 1.23 × 10³ mol fructose, C₆ H₁₂ O₆

e) 0.5758 mol FeSO₄•(H₂O)₇

8. Determine the mass in grams of each of the following:

a) 0.600 mol of oxygen atoms

b) 0.600 mol of oxygen molecules, O₂

c) 0.600 mol of ozone molecules, O₃

9. Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO₄, a semiprecious stone. 10. Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO₄, 266 g of Al₂C1₆, or 225 g of Al₂S₃. 11. The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone? 12. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C₁₂H₂₂O₁₁) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar? 13. Which of the following represents the least number of molecules?

a) 20.0 g of H₂O (18.02 g/mol)

b) 77.0 g of CH₄ (16.06 g/mol)

c) 68.0 g of CaH₂ (42.09 g/mol)

d) 100.0 g of N₂O (44.02 g/mol)

e) 84.0 g of HF (20.01 g/mol)

14. How many atoms are present in 4.55 mol of Fe? 15. How many molecules are present in 2.509 mol of H₂S? 16. How many moles are present in 3.55 × 10²⁴ Pb atoms? 17. How many moles are present in 1.00 × 10²³ PF₃ molecules? 18. Determine the molar mass of each substance.

a)  Si      b)  SiH₄      c)  K₂O

19.  Determine the molar mass of each substance.

a)  Al      b)  Al₂O₃      c)  CoCl₃ 20.  What is the mass of 4.44 mol of Rb? 21.  What is the mass of 12.34 mol of Al₂(SO₄)₃? 22.  How many moles are present in 45.6 g of CO? 23.  How many moles are present in 1.223 g of SF₆? 24.  How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?

Answers

1. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.

2. Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.

3. The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 10²³ molecules.

4. a) 256.528 g/mol         b) 72.150 g mol⁻¹         c) 378.103 g mol⁻¹

d) 58.080 g mol⁻¹          e) 180.158 g mol⁻¹

5. a) 197.382 g mol⁻¹         b) 257.162 g mol⁻¹         c) 194.193 g mol⁻¹

d) 60.056 g mol⁻¹         e) 306.464 g mol⁻¹

6. a) 0.819 g         b) 307 g         c) 0.23 g         d) 1.235 × 10⁶ g (1235 kg)

e) 765 g 7. a) 99.41         b) 2.27 g         c) 3.5 g         d) 222 kg         e) 160.1 g

8. a) 9.60 g         b) 19.2 g         c) 28.8 g

9. zirconium: 2.038 × 10²³ atoms; 30.87 g; silicon: 2.038 × 10²³ atoms; 9.504 g; oxygen: 8.151 × 10²³ atoms; 21.66 g

10. 122 g of AlPO₄ = 27.0 g of Al        266 g of Al₂Cl₆ = 53.8 g of Al         225 g of Al₂S₃ = 80.8 g Al; therefore 225 g of Al₂S₃ has the greatest mass of Al.

11. 3.113 × 10²⁵ C atoms

12. 0.865 servings, or about 1 serving.

13. 20.0 g H₂O represents the least number of molecules since it has the least number of moles.

14. 2.74 × 10²⁴ atoms 15. 1.511 × 10²⁴ molecules 16. 5.90 mol 17. 0.166 mol 18. a)  28.0855 g          b)  32.1172 g            c)  94.1960 g 19. a)  26.9815 g          b)  101.9612 g          c)  165.2913 g 20. 379 g 21. 4,222 g 22. 1.63 mol 23. 0.008373 mol 24. 3.72 mol

Glossary

Avogadro’s number (N_A): experimentally determined value of the number of entities comprising 1 mole of substance, equal to 6.022 × 10²³ mol⁻¹ formula mass: sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass molar mass: mass in grams of 1 mole of a substance mole: amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ¹²C

]]> 1388 1354 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-2-determining-empirical-and-molecular-formulas/ Thu, 12 Apr 2018 02:51:37 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-2-determining-empirical-and-molecular-formulas/

Learning Objectives

By the end of this section, you will be able to:

• Determine the empirical formula of a compound
• Determine the molecular formula of a compound

In the previous section, we discussed how to calculate percent compositon from experimental mass measurements.  Now we will see how to apply this to the determination of of the chemical formula of a compound.

Empirical Formulas

The empirical formula is the simplest formula of a compound. It is the smallest whole number ratio of atoms, but does not necessarily represent the arrangement of atoms in the actual molecule. For example: a molecule of hydrogen peroxide is made up of two atoms of O and two atoms of H bonded together—the molecular formula is thus H₂O₂. Since the simplest ratio of H and O atoms is 1:1, the empirical formula (though not the actual arrangement of atoms within the molecule) is HO.

Example 1

Determine the empirical formula for dioxin (C₁₂H₄Cl₄O₂), a very powerful poison.

 

Solution  

The subscripts are 12, 4, 4, and 2. These are all divisible by 2. Thus, the empirical formula = C_12/2H_4/2Cl_4/2O_2/2= C₆H₂Cl₂O

 

Test Yourself

Determine the empirical formula for the following compounds:

      a)  C₆H₁₆N₂                  b)  CCl₄                        c) C₄H₁₀

  Answers

      a)  C₃H₈N                  b)  CCl₄                        c) C₂H₅

Determination of Empirical Formulas

As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

$latex 1.17 \;\text{g C} \times \frac{1 \;\text{mol C}}{12.011 \;\text{g C}} = 0.0\underline{974}11 \;\text{mol C with 3 sig figs} $

$latex 0.287 \;\text{g H} \times \frac{1 \;\text{mol H}}{1.00794 \;\text{g H}} = 0.\underline{284}74 \;\text{mol H with 3 sig figs} $

Thus, we can accurately represent this compound with the formula C_0.974H_0.284.  Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

$latex \text{C}_{\frac{0.0\underline{974}11}{0.0\underline{974}11}} \; \text{H}_{\frac{0.\underline{284}74}{0.0\underline{974}11}} \;\text{or CH}_3$

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

The empirical formula for this compound is thus CH₃. This may or may not be the compound’s molecular formula; however, we would need additional information to make that determination (as discussed later in this section).

Consider another example, a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

$latex \text{Cl}_{0.150}\text{O}_{0.525} \; = \; \text{Cl}_{\frac{0.150}{0.150}} \; \text{O}_{\frac{0.525}{0.150}} = \text{ClO}_{3.5}$

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl₂O₇ as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

1. Deriving the number of moles of each element from its mass
2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Note that it is important to follow the usual rule not to round-off values in the middle of a calculation. (Keep track of the significant figures after each step, but then use digits in your calculator beyond the last significant figure when you carry a value into a subsequent calculation.) In the end, you will round to whole numbers.

Figure 1 outlines this procedure in flow chart fashion for a substance containing elements A and X.

[caption id="attachment_1389" align="aligncenter" width="1300"] Figure 1. The empirical formula of a compound can be derived from the masses of all elements in the sample.[/caption]

Example 2

A sample of the black mineral hematite (Figure 2), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

[caption id="attachment_1390" align="aligncenter" width="300"] Figure 2. Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)[/caption]

 

Solution For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

$latex 34.97\; \text{g Fe} \times \frac{\text{mol Fe}} {55.847\; \text{g}} = 0.\underline{6261}8\; \text{mol Fe with 4 sig figs}$

$latex 15.03\; \text{g O} \times \frac{\text{mol O}} {15.9994 \; \text{g}} = 0.\underline{9394}1\; \text{mol O with 4 sig figs}$

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

$latex \begin{array}{r @{{}={}} l} \frac{0.\underline{6261}8}{0.\underline{6261}8} & \underline{1.000}0 \;\text{mol Fe with 4 sig figs} \\[1em] \frac{0.\underline{9394}1}{0.\underline{6261}8} & \underline{1.500}2 \;\text{mol O with 4 sig figs} \end{array}$

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe₁O_1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

$latex 2(\text{Fe}_1\text{O}_{1.5}) = \text{Fe}_2\text{O}_3$

The empirical formula is Fe₂O₃.

 

Test Yourself What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

  Answer N₂O₅

It is helpful to remember some common fractions in decimal form so you can identify when you should multiply through: 1/2 = 0.500, therefore multiply by 2 1/3 = 0.333…, (x3)                  1/4 = 0.25, (x4) 2/3 = 0.666…, (x3)                  3/4 = 0.75, (x4) Note: if you have a formula that has more than one of these decimals, you can do the multiplication stepwise: For example: C_1.50H_1.33O₁x 2 = C₃H_2.66O₂x 3 = C₉H₈O₆

Example 3

A 2.500 g sample of a compound containing only carbon and hydrogen is found to contain 2.002 g of carbon. Determine the empirical formula.

 

Solution  

Step 1: We are told the compound contains 2.002 g of C. To find H, subtract the C from the total amount         2.500 g – 2.002 g = 0.498 g of H

Step 2: Convert each to mol:

            $latex 2.002\; \text{g C} \times \frac{1\; \text{mol C}} {12.011\; \text{g C}} = 0.\underline{1666}8\; \text{mol C with 4 sig figs}$

            $latex 0.498\; \text{g H} \times \frac{1\; \text{mol H}} {1.00794\; \text{g H}} = 0.\underline{494}1\; \text{mol H with 3 sig figs}$

Step 3: C_{0.16668/0.16668}H_{0.4941/0.16668}= C_1.000H_2.96  Because 2.96 is within a few hundredths of the whole number 3, we can round to C₁H₃.  [Note: As long as the ratio value is  < ± 0.2 of a whole number, it can be rounded.]

We have the empirical formula:  CH₃

 

Test Yourself

A sample of para-dichlorobenzene contains 11.314 g of carbon, 0.6330 g of hydrogen and 11.132 g of chlorine. What is the empirical formula of this compound?

  Answer C₃H₂Cl

For additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip.

 

Deriving Empirical Formulas from Percent Composition

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.

Example 4

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure 3). What is the empirical formula for this gas? [caption id="attachment_1392" align="aligncenter" width="300"] Figure 3. An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: “Dual Freq”/Wikimedia Commons)[/caption]  

Solution Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

$latex \begin{array}{r @{{}={}} l} 27.29\% \;\text{C} & \frac{27.29 \;\text{g C}}{100 \;\text{g compound}} \\[1em] 72.71\% \;\text{O} & \frac{72.71 \;\text{g O}}{100 \;\text{g compound}} \end{array}$

The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

$latex \begin{array}{r @{{}={}} l} 27.29 \;\text{g of C} (\frac{\text{mol C}}{12.011 \;\text{g}}) & \underline{2.272}08 \;\text{mol C with 4 sig figs} \\[1em] 72.71\;\text{g of O} (\frac{\text{mol O}}{15.9994 \;\text{g}}) & \underline{4.544}54 \;\text{mol O with 4 sig figs} \end{array}$

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

$latex \begin{array}{r @{{}={}} l} \frac{\underline{2.272}08 \;\text{mol C}}{\underline{2.272}08} & 1.000 \\[1em] \frac{\underline{4.544}54\; \text{mol O}}{\underline{2.272}08} & 2.000 \end{array}$

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO₂.

 

Test Yourself What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?

  Answer CH₂O

Example 5

An iron oxide contains 69.9% Fe by mass. Find its empirical formula. Note: “oxide” means it contains oxygen, as well as iron.

 

Solution  

Step 1: If we assume 100 g sample, it must contain 69.9 g of Fe and thus 100-69.9 = 30.1 g of O.

Step 2: Convert each to mol:

      $latex 69.9\; \text{g Fe} \times \frac{1\; \text{mol Fe}} {55.847\; \text{g Fe}} = \underline{1.25}16\; \text{mol Fe with 3 sig figs}$

$latex 30.1\; \text{g O} \times \frac{1\; \text{mol O}} {15.9994\; \text{g O}} = \underline{1.88}13\; \text{mol O with 3 sig figs}$

Step 3: 1.2516… is the smallest number, so we will divide each subscript by that value:

      Fe_1.2516O_1.8813= Fe_{1.2516/1.2516} O_{1.8813/1.2516 =} Fe_1.0000O_1.503

 

Step 4: Since the ratio for oxygen is 1.503 and does not fall into the acceptable limit where we can round off [Note: As long as the ratio value is  < ± 0.2 of a whole number, it can be rounded.], we cannot interpret 1.503 as a whole number (i.e. NOT 1 or 2!).  We recognize that 1.503 is approximately 1.50, so we can multiply each subscript by 2 to get a whole numbers: Fe_1x2O_1.5x2= Fe₂O₃.

      The empirical formula is Fe₂O₃

 

Test Yourself

Nylon-6 contains 63.68% C, 12.38% N and 9.80% H and 14.14% O by mass. What is the empirical formula for Nylon-6?

  Answer C₆H₁₁NO

Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n:

$latex \frac{\text{molecular or molar mass (amu or} \;\frac{\text{g}}{\text{mol}})}{\text{empirical formula mass (amu or} \;\frac{\text{g}}{\text{mol}})} = n \;\text{formula units/molecule} $

The molecular formula is then obtained by multiplying each subscript in the empirical formula by n, as shown by the generic empirical formula A_xB_y:

$latex (\text{A}_{\text{x}} \text{B}_{\text{y}})_{\text{n}} = \text{A}_{\text{nx}} \text{B}_{\text{nx}} $

For example, consider a covalent compound whose empirical formula is determined to be CH₂O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

$latex \frac{180 \;\text{amu/molecule}}{30\;\frac{\text{amu}}{\text{formula unit}}} = 6 \;\text{formula units/molecule}$

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

$latex \text{(CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6$

Note that this same approach may be used with the molar mass (g/mol) instead of the molecular mass (amu).

Example 6

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

 

Solution Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

$latex \begin{array}{r @{{}={}} l} (74.02 \;\text{g C}) (\frac{1 \;\text{mol C}}{12.011 \;\text{g C}}) & \underline{6.162}68 \;\text{mol C with 4 sig figs} \\[1em] (8.710 \;\text{g H}) (\frac{1 \;\text{mol H}}{1.00794 \;\text{g H}}) & \underline{8.641}39 \;\text{mol H with 4 sig figs} \\[1em] (17.27 \;\text{g N}) (\frac{1 \;\text{mol N}}{14.0067 \;\text{g N}}) & \underline{1.232}98 \;\text{mol N with 4 sig figs} \end{array}$

Next, we calculate the molar ratios of these elements relative to the least abundant element, N.

$latex \frac{\underline{1.232}98}{\underline{1.232}98} = 1.000 \;\text{mol N} = 1 \;\text{mol N}$

$latex \frac{\underline{6.162}68}{\underline{1.232}98} = 4.999 \;\text{mol C} = 5 \;\text{mol C}$

$latex \frac{\underline{8.641}39}{\underline{1.232}98} = 7.008 \;\text{mol H} = 7 \;\text{mol H}$

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C₅H₇N. The empirical formula mass for this compound is therefore 81.117 amu/formula unit, or 81.117 g/mol formula unit.

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

$latex \frac{40.57 \;\text{g nicotine}}{0.2500 \;\text{mol nicotine}} = \frac{162.3 \;\text{g}}{\text{mol}}$

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

$latex \frac{162.3 \;\text{g/mol}}{81.117 \;\frac{\text{g}}{\text{formula unit}}} = 2 \;\text{formula units/molecule} $

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

$latex (\text{C}_5\text{H}_7\text{N})_2 = \text{C}_{10}\text{H}_{14}\text{N}_2$

 

Test Yourself What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

  Answer C₈H₁₀N₄O₂

Example 7

A 13.8 g sample of a compound containing only nitrogen and oxygen produced 4.2 g of nitrogen upon decomposition. What is its empirical formula and molecular formula if the molar mass is 90.3 g/mol?

 

Solution  

First, we need to determine the empirical formula:

Step 1: We are told it contains 4.2 g of N. Thus it must contain 13.8 g – 4.2 g = 9.6 g of O.

Step 2: Convert each to mol:

$latex 4.3\; \text{g N} \times \frac{1\; \text{mol N}} {14.0067\; \text{g N}} = 0.\underline{30}7\; \text{mol N with 2 sig figs}$ $latex 9.5\; \text{g O} \times \frac{1\; \text{mol O}} {15.9994\; \text{g O}} = 0.\underline{59}4\; \text{mol O with 2 sig figs}$

Step 3: N_0.307/0.307O_0.594/0.307= N_1.00O_1.93

Step 4: The value 1.93 (calc. to 2 sig fig’s) is close enough to round to the whole number, or  NO₂

Then, we need to use the molar mass values to determine the molecular formula:

Step 1:

Empirical formula molar mass = 14.0067 g/mol + 2(15.9994 g/mol) = 46.0055 g/mol

Step 2:

Molar mass of molecular formula / Molar mass empirical formula = 90.3 g/mol / 46.0055 g/mol = 1.963 (to 3 SF) = 2

Step 3: N_1x2O_2x2= N₂O₄

 

Test Yourself

Caffeine contains hydrogen, carbon, nitrogen and oxygen. The mass % composition is as follows: C = 49.47%; H = 5.191%; N = 28.86%; O = 16.48 %. The molar mass is approximately 194 g/mol. Determine the molecular formula.

  Answer C₈H₁₀N₄O₂

Key Concepts and Summary

The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.

Determining an Empirical Formula

STEP 1: Determine the mass of each element in a particular sample. If you are given the data in percent composition, it’s easiest to assume a total sample size of 100g. (If the percent of one element is 64%, 100 g of the compound would contain 64 g of that particular element.)

STEP 2: For each element, change  from g $latex \longrightarrow$ mol. These mole values become first-try subscripts for the formula—for example, C_0.1H_0.4O_0.1

STEP 3: Now divide each element’s subscript by the lowest subscript value. E.g.: C_0.1H_0.4O_0.1, divide each subscript by 0.1 and get CH₄O

STEP 4: If any subscript is NOT a whole number (or easily rounded to one), then multiply ALL subscripts by the smallest integer that will turn all subscripts to whole integers. This is now the empirical formula. E.g.: C_1.5N_0.5H₄ multiply each by 2 and get C₃NH₈

   

Determining the Molecular Formula from the Empirical Formula

STEP 1: Calculate the molar mass of the empirical formula. STEP 2: Divide the given molecular molar mass by the molar mass calculated for the empirical formula. STEP 3: Multiply each subscript by the whole number that resulted from step 2. This is now the molecular formula.

Key Equations

• $latex \%\text{X} = \frac{\text{mass X}}{\text{mass commpound}} \times 100\% $
• $latex \frac{\text{molecular or molar mass ( amu or} \;\frac{\text{g}}{\text{mol}})}{\text{empirical formula mass ( amu or} \;\frac{\text{g}}{\text{mol}})} = n \;\text{formula units/molecule}$
• (A_xB_y)_n = A_nxB_ny

Exercises

1. Determine the empirical formulas for compounds with the following percent compositions:

a) 15.8% carbon and 84.2% sulfur

b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen

2. A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula? 3. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol. 4. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.   Answers

1. a) CS₂         b) CH₂O

2. C₆H₆

3. Mg₃Si₂H₃O₈ (empirical formula), Mg₆Si₄H₆O₁₆ (molecular formula)

4. C₁₅H₁₅N₃

Glossary

empirical formula: is the simplest formula of a compound. It is the smallest whole number ratio of atoms, but does not necessarily represent the arrangement of atoms in the actual molecule. empirical formula mass: sum of average atomic masses for all atoms represented in an empirical formula percent composition: percentage by mass of the various elements in a compound

]]> 1393 1354 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-3-molarity/ Thu, 12 Apr 2018 02:51:38 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-3-molarity/

Learning Objectives

By the end of this section, you will be able to:

• Describe the fundamental properties of solutions
• Calculate solution concentrations using molarity
• Perform dilution calculations using the dilution equation

In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances.

Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

[caption id="" align="aligncenter" width="281"] Figure 1. Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage’s sweetness. (credit: Jane Whitney)[/caption]

 Solutions

We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions.

The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution.

A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).

Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:

$latex M = \frac{\text{mol solute}}{\text{L solution}}$

Example 1

A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?

 

Solution Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:

$latex M = \frac{\text{mol solute}}{\text{L solution}} = \frac{0.133 \;\text{mol}}{355 \;\text{mL} \times \frac{1 \;\text{L}}{1000 \;\text{mL}}} = 0.375 \; M$

 

Test Yourself A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?

  Answer 0.05 M

Example 2

How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 1?

 

Solution In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 1, 0.375 M:

$latex M = \frac{\text{mol solute}}{\text{L solution}}$ $latex \text{mol solute} = M \times \text{L solution}$

$latex \text{mol solute} = 0.375 \;\frac{\text{mol sugar}}{\text{L}} \times (10 \;\text{mL} \times \frac{1 \text{L}}{1000 \;\text{mL}}) = 0.004 \;\text{mol sugar}$

 

Test Yourself What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?

  Answer 80 mL

Example 3

Distilled white vinegar (Figure 2) is a solution of acetic acid, CH₃CO₂H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

[caption id="attachment_1726" align="aligncenter" width="300"] Figure 2. Distilled white vinegar is a solution of acetic acid in water.[/caption]

 

Solution As in previous textbox shaded, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:

$latex M = \frac{\text{mol solute}}{\text{L solution}} = \frac{25.2 \;\text{g CH}_3\text{CO}_2\text{H} \times \frac{1 \;\text{mol CH}_3\text{CO}_2\text{H}}{60.052 \;\text{g CH}_3\text{CO}_2\text{H}}}{0.500 \;\text{L solution}} = 0.839 \; M$

 

Test Yourself Calculate the molarity of 6.52 g of CoCl₂ (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

  Answer 0.674 M

Example 4

How many grams of NaCl are contained in 0.250 L of a 5.30-M solution?

 

Solution The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 2:

$latex M = \;\frac{\text{mol solute}}{\text{L solution}}$ $latex \text{mol solute} = M \times \text{L solution}$ $latex \text{mol solute} = 5.30 \;\frac{\text{mol NaCl}}{\text{L}} \times 0.250 \;\text{L} = \underline{1.32}50 \;\text{mol NaCl with 3 sig figs}$

Finally, this molar amount is used to derive the mass of NaCl:

$latex \underline{1.32}50 \;\text{mol NaCl} \times \frac{58.4425 \;\text{g NaCl}}{\text{mol NaCl}} = 77.4 \;\text{g NaCl}$

 

Test Yourself How many grams of CaCl₂ (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?

 

Answer

5.55 g CaCl₂

When performing calculations stepwise, as in Example 4, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see Example 5). This eliminates intermediate steps so that only the final result is rounded.

Example 5

In Example 3, we found the typical concentration of vinegar to be 0.839 M. What volume of vinegar contains 75.6 g of acetic acid?

 

Solution First, use the molar mass to calculate moles of acetic acid from the given mass:

$latex \text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} = \text{mol solute}$

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

$latex \text{mol solute} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}$

Combining these two steps into one yields:

$latex \text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}$$latex 75.6 \;\text{g CH}_3\text{CO}_2\text{H} (\frac{\text{mol CH}_3\text{CO}_2\text{H}}{60.053 \;\text{g}}) (\frac{\text{L solution}}{0.839 \;\text{mol CH}_3\text{CO}_2\text{H}}) = 1.50 \;\text{L solution}$

 

Test Yourself What volume of a 1.50-M KBr solution contains 66.0 g KBr?

  Answer 0.370 L

Dilution of Solutions

Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 3).

[caption id="" align="aligncenter" width="422"] Figure 3. Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)[/caption]

Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure 4).

 

A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters:

$latex n = ML $

Expressions like these may be written for a solution before and after it is diluted:

$latex n_1 = M_1L_1$

$latex n_2 = M_2L_2$

where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution,n₁ = n₂. Thus, these two equations may be set equal to one another:

$latex M_1L_1 = M_2L_2$

This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:

$latex C_1V_1 = C_2V_2$

where C and V are concentration and volume, respectively.

 

Use the simulation to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.

 

Example 6

If 0.850 L of a 5.00-M solution of copper nitrate, Cu(NO₃)₂, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?

 

Solution We are given the volume and concentration of a stock solution, V₁ and C₁, and the volume of the resultant diluted solution, V₂. We need to find the concentration of the diluted solution, C₂. We thus rearrange the dilution equation in order to isolate C₂:

$latex C_1V_1 = C_2V_2$ $latex C_2 = \frac{C_1V_1}{V_2}$

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

$latex C_2 = \frac{0.850 \;\text{L} \times 5.00 \frac{\text{mol}}{\text{L}}}{1.80 \;\text{L}} = 2.36 \;M$

This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M).

 

Test Yourself What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-M solution of CH₃OH to 500.0 mL?

  Answer 0.102 M CH₃OH

Example 7

What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr?

 

Solution We are given the volume and concentration of a stock solution, V₁ and C₁, and the concentration of the resultant diluted solution, C₂. We need to find the volume of the diluted solution, V₂. We thus rearrange the dilution equation in order to isolate V₂:

$latex C_1V_1 = C_2V_2$ $latex V_2 = \frac{C_1V_1}{C_2}$

Since the diluted concentration (0.12 M) is slightly more than one-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:

$latex V_2 = \frac{(0.45\;M)(0.011 \;\text{L})}{0.12 \; M}$ $latex V_2 = 0.041 \;\text{L}$

The volume of the 0.12-M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.

 

Test Yourself A laboratory experiment calls for 0.125 M HNO₃. What volume of 0.125 M HNO₃ can be prepared from 0.250 L of 1.88 M HNO₃?

  Answer 3.76 L

Example 8

What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?

 

Solution We are given the concentration of a stock solution, C₁, and the volume and concentration of the resultant diluted solution, V₂ and C₂. We need to find the volume of the stock solution, V₁. We thus rearrange the dilution equation in order to isolate V₁:

$latex C_1V_1 = C_2V_2$ $latex V_2 = \frac{C_2V_2}{C_2}$

Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:

$latex V_1 = \frac{(0.100\;M)(5.00 \;\text{L})}{1.59 \; M}$ $latex V_1 = 0.314 \;\text{L}$

Thus, we would need 0.314 L of the 1.59-M solution to prepare the desired solution. This result is consistent with our rough estimate.

 

Test Yourself What volume of a 0.575-M solution of glucose, C₆H₁₂O₆, can be prepared from 50.00 mL of a 3.00-M glucose solution?

  Answer 0.261 L

Key Concepts and Summary

Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.

Key Equations

• $latex M = \frac{\text{mol solute}}{\text{L solution}}$
• C₁V₁ = C₂V₂

Exercises

1. What information do we need to calculate the molarity of a sulfuric acid solution? 2. Determine the molarity for each of the following solutions:

a) 0.444 mol of CoCl₂ in 0.654 L of solution

b) 98.0 g of phosphoric acid, H₃PO₄, in 1.00 L of solution

c) 0.2074 g of calcium hydroxide, Ca(OH)₂, in 40.00 mL of solution

d) 10.5 kg of Na₂SO₄·10H₂O in 18.60 L of solution

e) 7.0 × 10⁻³ mol of I₂ in 100.0 mL of solution

f) 1.8 × 10⁴ mg of HCl in 0.075 L of solution

3. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C₆H₁₂O₆, used for intravenous injection?

a) Outline the steps necessary to answer the question.

b) Answer the question.

4. Calculate the number of moles and the mass of the solute in each of the following solutions:

a) 2.00 L of 18.5 M H₂SO₄, concentrated sulfuric acid

b) 100.0 mL of 3.8 × 10⁻⁵M NaCN, the minimum lethal concentration of sodium cyanide in blood serum

c) 5.50 L of 13.3 M H₂CO, the formaldehyde used to “fix” tissue samples

d) 325 mL of 1.8 × 10⁻⁶M FeSO₄, the minimum concentration of iron sulfate detectable by taste in drinking water

5. Consider this question: What is the molarity of KMnO₄ in a solution of 0.0908 g of KMnO₄ in 0.500 L of solution?

a) Outline the steps necessary to answer the question.

b) Answer the question.

6. Calculate the molarity of each of the following solutions:

a) 0.195 g of cholesterol, C₂₇H₄₆O, in 0.100 L of serum, the average concentration of cholesterol in human serum

b) 4.25 g of NH₃ in 0.500 L of solution, the concentration of NH₃ in household ammonia

c) 1.49 kg of isopropyl alcohol, C₃H₇OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol

d) 0.029 g of I₂ in 0.100 L of solution, the solubility of I₂ in water at 20 °C

7. There is about 1.0 g of calcium, as Ca²⁺, in 1.0 L of milk. What is the molarity of Ca²⁺ in milk? 8. If 0.1718 L of a 0.3556-M C₃H₇OH solution is diluted to a concentration of 0.1222 M, what is the volume of the resulting solution? 9. What volume of a 0.33-M C₁₂H₂₂O₁₁ solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M? 10. What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?

a) 1.00 L of a 0.250-M solution of Fe(NO₃)₃ is diluted to a final volume of 2.00 L

b) 0.5000 L of a 0.1222-M solution of C₃H₇OH is diluted to a final volume of 1.250 L

c) 2.35 L of a 0.350-M solution of H₃PO₄ is diluted to a final volume of 4.00 L

d) 22.50 mL of a 0.025-M solution of C₁₂H₂₂O₁₁ is diluted to 100.0 mL

11. A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution? 12. What volume of a 0.20-M K₂SO₄ solution contains 57 g of K₂SO₄?   Answers 

1. We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

2. a) 0.679 M;      b) 1.00 M;        c) 0.06998 M; d) 1.75 M;            e) 0.070 M;      f) 6.6 M

3. a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass;

b) 27 g

4. a) 37.0 mol H₂SO₄;   3.63 × 10³ g H₂SO₄; b) 3.8 × 10⁻⁶ mol NaCN;    1.9 × 10⁻⁴ g NaCN; c) 73.2 mol H₂CO;   2.20 kg H₂CO; d) 5.9 × 10⁻⁷ mol FeSO₄;    8.9 × 10⁻⁵ g FeSO₄

5. a) Determine the molar mass of KMnO₄; determine the number of moles of KMnO₄ in the solution; from the number of moles and the volume of solution, determine the molarity;

b) 1.15 × 10⁻³M

6. a) 5.04 × 10⁻³M;      b) 0.499 M; c) 9.92 M;                     d) 1.1 × 10⁻³M

7. 0.025 M

8. 0.5000 L

9. 1.9 mL

10. a) 0.125 M;      b) 0.04888 M; c) 0.206 M;            d) 0.0056 M

11. 11.9 M

12. 1.6 L

Glossary

aqueous solution: solution for which water is the solvent concentrated: qualitative term for a solution containing solute at a relatively high concentration concentration: quantitative measure of the relative amounts of solute and solvent present in a solution dilute: qualitative term for a solution containing solute at a relatively low concentration dilution: process of adding solvent to a solution in order to lower the concentration of solutes dissolved: describes the process by which solute components are dispersed in a solvent molarity (M): unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution solute: solution component present in a concentration less than that of the solvent solvent: solution component present in a concentration that is higher relative to other components

]]> 1399 1405 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-4-other-units-for-solution-concentrations/ Thu, 12 Apr 2018 02:51:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/3-4-other-units-for-solution-concentrations/

Learning Objectives

By the end of this section, you will be able to:

• Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-per-million (ppm), and parts-per-billion (ppb)
• Perform computations relating a solution’s concentration and its components’ volumes and/or masses using these units

In the previous section, we introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. In this section, we will introduce some other units of concentration that are commonly used in various applications, either for convenience or by convention.

Mass Percentage

Earlier in this chapter, we introduced percent composition as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component’s mass to the solution’s mass, expressed as a percentage:

$latex \text{mass percentage} = \frac{\text{mass of component}}{\text{mass of solution}} \times 100\% $

We are generally most interested in the mass percentages of solutes, but it is also possible to compute the mass percentage of solvent.

Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).

Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure 1) cites the concentration of its active ingredient, sodium hypochlorite (NaOCl), as being 7.4%. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl.

Example 1

A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of glucose. What is the percent by mass of glucose in spinal fluid?

 

Solution The spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about 0.1%. Substituting the given masses into the equation defining mass percentage yields:

$latex \% \;\text{glucose} = \frac{3.75 \;\text{mg glucose} \times \frac{1 \;\text{g}}{1000 \;\text{mg}}}{5.0 \;\text{spinal fluid}} = 0.075\% $

The computed mass percentage agrees with our rough estimate (it’s a bit less than 0.1%).

Note that while any mass unit may be used to compute a mass percentage (mg, g, kg, oz, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, we converted the units of solute in the numerator from mg to g to match the units in the denominator. We could just as easily have converted the denominator from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.

 

Test Yourself A bottle of a tile cleanser contains 135 g of HCl and 775 g of water. What is the percent by mass of HCl in this cleanser?

  Answer 14.8%

Example 2

“Concentrated” hydrochloric acid is an aqueous solution of 37.2% HCl that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of HCl is contained in 0.500 L of this solution?

 

Solution The HCl concentration is near 40%, so a 100-g portion of this solution would contain about 40 g of HCl. Since the solution density isn’t greatly different from that of water (1 g/mL), a reasonable estimate of the HCl mass in 500 g (0.5 L) of the solution is about five times greater than that in a 100 g portion, or 5 ×× 40 = 200 g. In order to derive the mass of solute in a solution from its mass percentage, we need to know the corresponding mass of the solution. Using the solution density given, we can convert the solution’s volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart:

For proper unit cancellation, the 0.500-L volume is converted into 500. mL, and the mass percentage is expressed as a ratio, 37.2 g HCl/g solution:

$latex \underline{500} \;\text{mL solution} \; (\frac{1.19 \;\text{g solution}}{\text{mL solution}})(\frac{37.2 \;\text{g HCl}}{100 \;\text{g solution}}) = 221 \;\text{g HCl}$

This mass of HCl is consistent with our rough estimate of approximately 200 g.

 

Test Yourself What volume of concentrated HCl solution contains 125 g of HCl?

  Answer 282 mL

Volume Percentage

Liquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, %vol or (v/v)%:

$latex \text{volume percentage} = \frac{\text{volume solute}}{\text{volume solution}} \times 100\% $

Example 3

Rubbing alcohol (isopropanol) is usually sold as a 70.0%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol?

 

Solution Per the definition of volume percentage, the isopropanol volume is 70.0% of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:

$latex (335 \;\text{mL solution})(\frac{70.0\;\text{isopropryl alcohol}}{100\;\text{mL solution}})(\frac{0.785\;\text{g isopropryl alcohol}}{1\;\text{mL isopropyl alcohol}}) = 195\;\text{g isopropyl alcohol}$

 

Test Yourself Wine is approximately 12% ethanol (CH₃CH₂OH) by volume. Ethanol has a molar mass of 46.06 g/mol and a density 0.789 g/mL. How many moles of ethanol are present in a 750-mL bottle of wine?

  Answer 1.5 mol ethanol

Mass-Volume Percentage

“Mixed” percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute’s mass to the solution’s volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of 0.9% mass/volume (m/v), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as “blood sugar”) is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter (100 mL) of blood (Figure 2).

Parts per Million and Parts per Billion

Very low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage (“part per hundred”) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.

The mass-based definitions of ppm and ppb are given here:

$latex \text{ppm} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^6 \;\text{ppm}$ $latex \text{ppb} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^9 \;\text{ppm}$

Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water (Figure 3).

Example 4

According to the EPA, when the concentration of lead in tap water reaches 15 ppb, certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead (μg) would be contained in a typical glass of water (300 mL)?

 

Solution The definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm. Comparing these two unit definitions shows that ppm is 1000 times greater than ppb (1 ppm = 10³ ppb). Thus:

$latex 15 \;\text{ppb} \times \frac{1 \;\text{ppm}}{10^3 \;\text{ppb}} = 0.015 \;\text{ppm}$

The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:

$latex \text{ppb} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^9 \;\text{ppb}$ $latex \text{mass solute} = \frac{\text{ppb} \;\times \;\text{mass solution}}{10^9 \;\text{ppb}}$ $latex \text{mass solute} = \frac{15 \;\text{ppb} \;\times \;300 \;\text{mL} \times \frac{1.00 \;\text{g}}{\text{mL}}}{10^9 \;\text{ppb}} = \underline{4.5}0 \times 10^{-6} \;\text{g with 2 sig figs}$

Finally, convert this mass to the requested unit of micrograms:

$latex \underline{4.5}0 \times 10^{-6} \;\text{g} \times \frac{1 \mu\text{g}}{10^{-6}\;\text{g}} = 4.5 \mu\text{g}$

 

Test Yourself A 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units.

  Answer 9.6 ppm, 9600 ppb

Key Concepts and Summary

In addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components’ masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.

Key Equations

• $latex \text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 \%$
• $latex \text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6 \;\text{ppm} $
• $latex \text{ppb} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^9 \;\text{ppb} $

Exercises

1. Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO₃ by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO₃ by mass?

a) Outline the steps necessary to answer the question.

b) Answer the question.

2. What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL. 3. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of CaCO₃, which is equivalent to milligrams of CaCO₃ per liter of water. What is the molar concentration of Ca²⁺ ions in a water sample with a hardness count of 175 mg CaCO₃/L? 4. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose (C₆H₁₂O₆) in mg/dL? 5. Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass? 6. D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose (C₆H₁₂O₆) in water. If the density of D5W is 1.029 g/mL, calculate the molarity of dextrose in the solution.   Answers

1. a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: $latex \%\text{mass}_1 \times \;\text{mass}_1 = \%\text{mass}_2 \times \;\text{mass}_2$

This equation can be rearranged to isolate mass₁ and the given quantities substituted into this equation. b) 58.8 g

2. 114 g

3. 1.75 × 10⁻³M

4. 95 mg/dL

5. 2.38 × 10⁻⁴ mol

6. 0.29 mol

Glossary

mass percentage: ratio of solute-to-solution mass expressed as a percentage mass-volume percent: ratio of solute mass to solution volume, expressed as a percentage parts per billion (ppb): ratio of solute-to-solution mass multiplied by 10⁹ parts per million (ppm): ratio of solute-to-solution mass multiplied by 10⁶ volume percentage: ratio of solute-to-solution volume expressed as a percentage

]]> 1404 1405 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-4/ Thu, 12 Apr 2018 02:51:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-4/

Solid-fuel rockets are a central feature in the world’s space exploration programs, including the new Space Launch System being developed by the National Aeronautics and Space Administration (NASA) to replace the retired Space Shuttle fleet (Figure 1). The engines of these rockets rely on carefully prepared solid mixtures of chemicals combined in precisely measured amounts. Igniting the mixture initiates a vigorous chemical reaction that rapidly generates large amounts of gaseous products. These gases are ejected from the rocket engine through its nozzle, providing the thrust needed to propel heavy payloads into space. Both the nature of this chemical reaction and the relationships between the amounts of the substances being consumed and produced by the reaction are critically important considerations that determine the success of the technology. This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions—that is, the reaction stoichiometry.

]]> 1407 1405 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-1-writing-and-balancing-chemical-equations/ Thu, 12 Apr 2018 02:51:42 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-1-writing-and-balancing-chemical-equations/

Learning Objectives

By the end of this section, you will be able to:

• Derive chemical equations from narrative descriptions of chemical reactions.
• Write and balance chemical equations in molecular, total ionic, and net ionic formats.

The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH₄) and two diatomic oxygen molecules (O₂) to produce one carbon dioxide molecule (CO₂) and two water molecules (H₂O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.

This example illustrates the fundamental aspects of any chemical equation:

1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
2. The substances generated by the reaction are called products, and their formulas are placed on the right sight of the equation.
3. Plus signs (+) separate individual reactant and product formulas, and an arrow ($latex \longrightarrow$) separates the reactant and product (left and right) sides of the equation.
4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 2). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:

• One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
• One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
• One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules.

Balancing Equations

The chemical equation described in Figure 1 is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO₂ and H₂O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

$latex (1 \;\text{CO}_2 \;\text{molecule} \times \frac{2 \;\text{O atoms}}{\text{CO}_2 \;\text{molecule}}) + (2\;\text{H}_2\text{O molecule} \times \frac{1 \;\text{O atom}}{\text{H}_2\text{O molecule}}) = 4 \;\text{O atoms}$

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

$latex \text{CH}_4 + 2\text{O}_2 \longrightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Element	Reactants	Products	Balanced?
C	1 × 1 = 1	1 × 1 = 1	1 = 1, yes
H	4 × 1 = 4	2 × 2 = 4	4 = 4, yes
O	2 × 2 = 4	(1 × 2) + (2 × 1) = 4	4 = 4, yes
Table 1.

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

$latex \text{H}_2\text{O} \longrightarrow \text{H}_2 + \text{O}_2 \;(\text{unbalanced})$

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element	Reactants	Products	Balanced?
H	1 × 2 = 2	1 × 2 = 2	2 = 2, yes
O	1 × 1 = 1	1 × 2 = 2	1 ≠ 2, no
Table 2.

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H₂O to H₂O₂ would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H₂O to 2.

$latex 2\text{H}_2\text{O} \longrightarrow \text{H}_2 + \text{O}_2 \;(\text{unbalanced})$

Element	Reactants	Products	Balanced?
H	2 × 2 = 4	1 × 2 = 2	4 ≠ 2, no
O	2 × 1 = 2	1 × 2 = 2	2 = 2, yes
Table 3.

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H₂ product to 2.

$latex 2\text{H}_2\text{O} \longrightarrow 2\text{H}_2 + \text{O}_2 \;(\text{balanced})$

Element	Reactants	Products	Balanced?
H	2 × 2 = 4	2 × 2 = 4	4 = 4, yes
O	2 × 1 = 2	1 × 2 = 2	2 = 2, yes
Table 4.

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

$latex 2\text{H}_2\text{O} \longrightarrow 2\text{H}_2 + \text{O}_2 $

Example 1

Write a balanced equation for the reaction of molecular nitrogen (N₂) and oxygen (O₂) to form dinitrogen pentoxide.

 

Solution First, write the unbalanced equation.

$latex \text{N}_2 + \text{O}_2 \longrightarrow \text{N}_2 \text{O}_5 \;(\text{unbalanced})$

Next, count the number of each type of atom present in the unbalanced equation.

Element	Reactants	Products	Balanced?
N	1 × 2 = 2	1 × 2 = 2	2 = 2, yes
O	1 × 2 = 2	1 × 5 = 5	2 ≠ 5, no
Table 5.

Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O₂ and N₂O₅ to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

$latex \text{N}_2 + 5\text{O}_2 \longrightarrow 2\text{N}_2\text{O}_5 \;(\text{unbalanced})$

Element	Reactants	Products	Balanced?
N	1 ×× 2 = 2	2 × 2 = 4	2 ≠ 4, no
O	5 × 2 = 10	2 × 5 = 10	10 = 10, yes
Table 6.

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N₂ to 2.

$latex 2\text{N}_2 + 5\text{O}_2 \longrightarrow 2\text{N}_2 \text{O}_5 $

Element	Reactants	Products	Balanced?
N	2 × 2 = 4	2 × 2 = 4	4 = 4, yes
O	5 × 2 = 10	2 × 5 = 10	10 = 10, yes
Table 7.

The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.

 

Test Yourself Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)

  Answer $latex 2\text{NH}_4 \text{NO}_3 \longrightarrow 2\text{N}_2 + \text{O}_2 + 4\text{H}_2\text{O} $

It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C₂H₆) with oxygen to yield H₂O and CO₂, represented by the unbalanced equation:

$latex \text{C}_2 \text{H}_6 + \text{O}_2 \longrightarrow \text{H}_2 \text{O} + \text{C} \text{O}_2 \;(\text{unbalanced}) $

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:

$latex \text{C}_2 \text{H}_6 + \text{O}_2 \longrightarrow 3\text{H}_2 \text{O} + 2\text{C} \text{O}_2 \;(\text{unbalanced}) $

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O₂ reactant to yield an odd number, so a fractional coefficient, $latex \frac{7}{2}$, is used instead to yield a provisional balanced equation:

$latex \text{C}_2 \text{H}_6 + \frac{7}{2}\text{O}_2 \longrightarrow 3\text{H}_2 \text{O} + 2\text{C} \text{O}_2 \; $

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

$latex 2\text{C}_2 \text{H}_6 + 7\text{O}_2 \longrightarrow 6\text{H}_2 \text{O} + 4\text{C} \text{O}_2 \; $

Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,

$latex 3\text{N}_2 + 9\text{H}_2 \longrightarrow 6\text{N} \text{H}_3 $

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:

$latex \text{N}_2 + 3\text{H}_2 \longrightarrow 2\text{N} \text{H}_3 $

 

Use this interactive tutorial for additional practice balancing equations.

 

Additional Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here:

$latex 2\text{Na}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow 2\text{NaOH}(aq) + \text{H}_2(g) $

This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).

Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow.

$latex \text{CaCO}_3(s) \;\xrightarrow{\Delta} \; \text{CaO}(s) + \text{CO}_2(g) $

Other examples of these special conditions will be encountered in more depth in later chapters.

Ionic Compounds in Solution

One important aspect about ionic compounds that differs from molecular compounds has to do with dissolving in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, the ions physically separate from each other. We can use a chemical equation to represent this process—for example, with NaCl:

 

Figure 3. The dissolution of sodium chloride.

When NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (Figure 3 "Ionic Solutions"). This process is called dissociation; we say that the ions dissociate.

When an ionic compound dissociates in water, water molecules surround each ion and separate it from the rest of the solid. Each ion goes its own way in solution.

All ionic compounds that dissolve behave this way. (This behaviour was first suggested by the Swedish chemist Svante August Arrhenius [1859–1927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry.)

Keep in mind that when the ions separate, all of the ions separate. Thus, when CaCl₂ dissolves, the one Ca²⁺ ion and the two Cl⁻ ions separate from each other:

CaCl₂(s) $latex \longrightarrow$ Ca²⁺(aq) + Cl^-(aq) + Cl^-(aq)

or

CaCl₂(s) $latex \longrightarrow$ Ca²⁺(aq) + 2Cl^-(aq)

That is, the two chloride ions go off on their own. They do not remain as Cl₂ (that would be elemental chlorine; these are chloride ions); they do not stick together to make Cl₂⁻ or Cl₂²⁻. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved.

Example 2

Write the chemical equation that represents the dissociation of each ionic compound.

a) KBr      b) Na₂SO₄

 

Solution

a) KBr(s) $latex \longrightarrow$ K⁺(aq) + Br⁻(aq)

b) Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is

Na₂SO₄(s) $latex \longrightarrow$ 2Na⁺(aq) + SO₄²⁻(aq)  

Test Yourself

Write the chemical equation that represents the dissociation of (NH₄)₂S.

 

Answer

(NH₄)₂S(s) $latex \longrightarrow$ 2NH₄⁺(aq) + S²⁻(aq)

Equations for Ionic Reactions

Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl₂ and AgNO₃ are mixed, a reaction takes place producing aqueous Ca(NO₃)₂ and solid AgCl:

$latex \text{CaCl}_2(aq) + 2\text{AgNO}_3(aq) \longrightarrow \text{Ca(NO}_3)_2(aq) + 2\text{AgCl}(s)$

This balanced equation, derived in the usual fashion, is called a molecular equation because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:

$latex \begin{array}{r @{{}\longrightarrow{}} l} \text{CaCl}_2(aq) & \text{Ca}^{2+}(aq) + 2 \text{Cl}^{-}(aq) \\[0.5em] 2 \text{AgNO}_3(aq) & 2\text{Ag}^{+}(aq) + 2 {\text{NO}_3}^{-}(aq) \\[0.5em] \text{Ca(NO}_3)_2(aq) & \text{Ca}^{2+}(aq) + 2 {\text{NO}_3}^{-}(aq) \end{array}$

Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, s.

Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:

$latex \text{Ca}^{2+}(aq) + 2\text{Cl}^{-}(aq) + 2\text{Ag}^{+}(aq) + 2{\text{NO}_3}^{-}(aq) \longrightarrow \text{Ca}^{2+}(aq) + 2{\text{NO}_3}^{-}(aq) + 2\text{AgCl}(s) $

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca²⁺(aq) and NO₃⁻(aq).NO₃⁻(aq). These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:

$latex \rule[0.5ex]{4em}{0.1ex}\hspace{-4em} \text{Ca}^{2+}(aq) + 2\text{Cl}^{-}(aq) + 2\text{Ag}^{+}(aq) + \rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em} 2\text{NO}_3^{-}(aq) \longrightarrow \rule[0.5ex]{4em}{0.1ex}\hspace{-4em} \text{Ca}^{2+}(aq) + \rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em} 2{\text{NO}_3}^{-}(aq) + 2\text{AgCl}(s) $

$latex 2\text{Cl}^{-}(aq) + 2\text{Ag}^{+}(aq) \longrightarrow 2\text{AgCl}(s) $

Following the convention of using the smallest possible integers as coefficients, this equation is then written:

$latex \text{Cl}^{-}(aq) + \text{Ag}^{+}(aq) \longrightarrow \text{AgCl}(s) $

This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl⁻ and Ag⁺.

Example 3

Write the complete ionic equation for each chemical reaction.

a) KBr(aq) + AgC₂H₃O₂(aq) $latex \longrightarrow$ KC₂H₃O₂(aq) + AgBr(s)

b) MgSO₄(aq) + Ba(NO₃)₂(aq) $latex \longrightarrow$ Mg(NO₃)₂(aq) + BaSO₄(s)

 

Solution

For any ionic compound that is aqueous, we will write the compound as separated ions.

a) The complete ionic equation is

K⁺(aq) + Br⁻(aq) + Ag⁺(aq) + C₂H₃O₂⁻(aq) $latex \longrightarrow$ K⁺(aq) + C₂H₃O₂⁻(aq) + AgBr(s) b) The complete ionic equation is Mg²⁺(aq) + SO₄²⁻(aq) + Ba²⁺(aq) + 2NO₃⁻(aq) $latex \longrightarrow$ Mg²⁺(aq) + 2NO₃⁻(aq) + BaSO₄(s)  

Test Yourself

Write the complete ionic equation for

CaCl₂(aq) + Pb(NO₃)₂(aq) $latex \longrightarrow$ Ca(NO₃)₂(aq) + PbCl₂(s)  

Answer

Ca²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) $latex \longrightarrow$ Ca²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

Example 4

Write the net ionic equation for each chemical reaction.

a) K⁺(aq) + Br⁻(aq) + Ag⁺(aq) + C₂H₃O₂⁻(aq) $latex \longrightarrow$ K⁺(aq) + C₂H₃O₂⁻(aq) + AgBr(s)

b) Mg²⁺(aq) + SO₄²⁻(aq) + Ba²⁺(aq) + 2 NO₃⁻(aq) $latex \longrightarrow$ Mg²⁺(aq) + 2 NO₃⁻(aq) + BaSO₄(s)

 

Solution

a) In the first equation, the K⁺(aq) and C₂H₃O₂⁻(aq) ions are spectator ions, so they are canceled:

 K⁺(aq) + Br⁻(aq) + Ag⁺(aq) + C₂H₃O₂⁻(aq) $latex \longrightarrow$ K⁺(aq) + C₂H₃O₂⁻(aq) + AgBr(s)

The net ionic equation is

Br⁻(aq) + Ag⁺(aq) $latex \longrightarrow$ AgBr(s) b) In the second equation, the Mg²⁺(aq) and NO₃⁻(aq) ions are spectator ions, so they are canceled: Mg²⁺(aq) + SO₄²⁻(aq) + Ba²⁺(aq) + 2 NO₃⁻(aq) $latex \longrightarrow$ Mg²⁺(aq) + 2 NO₃⁻(aq) + BaSO₄(s)

The net ionic equation is

SO₄²⁻(aq) + Ba²⁺(aq) $latex \longrightarrow$ BaSO₄(s)  

Test Yourself

Write the net ionic equation for

CaCl₂(aq) + Pb(NO₃)₂(aq) $latex \longrightarrow$ Ca(NO₃)₂(aq) + PbCl₂(s)  

Answer

Pb²⁺(aq) + 2Cl⁻(aq) $latex \longrightarrow$ PbCl₂(s)

Example 5

When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.

 

Solution Begin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:

$latex \text{CO}_2(aq) + \text{NaOH}(aq) \longrightarrow \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l) \;(\text{unbalanced}) $

Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:

$latex \text{CO}_2(aq) + 2\text{NaOH}(aq) \longrightarrow \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l) $

The two dissolved ionic compounds, NaOH and Na₂CO₃, can be represented as dissociated ions to yield the complete ionic equation:

$latex \text{CO}_2(aq) + 2\text{Na}^{+}(aq) + 2\text{OH}^{-}(aq) \longrightarrow 2\text{Na}^{+}(aq) + {\text{CO}_3}^{2-}(aq) + \text{H}_2 \text{O}(l) $

Finally, identify the spectator ion(s), in this case Na⁺(aq), and remove it from each side of the equation to generate the net ionic equation:

$latex \text{CO}_2(aq) + \rule[0.5ex]{4.25em}{0.1ex}\hspace{-4.25em} 2\text{Na}^{+}(aq) + 2\text{OH}^{-}(aq) \longrightarrow \rule[0.5ex]{4.25em}{0.1ex}\hspace{-4.25em} 2\text{Na}^{+}(aq) + {\text{CO}_3}^{2-}(aq) + \text{H}_2 \text{O}(l) $$latex \text{CO}_2(aq) + 2\text{OH}^{-}(aq) \longrightarrow {\text{CO}_3}^{2-}(aq) + \text{H}_2 \text{O}(l) $

 

Test Yourself Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:

$latex \text{NaCl}(aq) + \text{H}_2 \text{O}(l) \;\;\xrightarrow{\text{electricity}}\;\; \text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(g) $

Write balanced molecular, complete ionic, and net ionic equations for this process.

  Answers $latex 2\text{NaCl}(aq) + 2\text{H}_2 \text{O} \longrightarrow 2 \text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(g) (\text{molecular}) $ $latex 2\text{Na}^{+}(aq) + 2\text{Cl}^{-}(aq) + 2\text{H}_2 \text{O} \longrightarrow 2\text{Na}^{+}(aq) + 2\text{OH}^{-}(aq) + \text{H}_2(g) + \text{Cl}_2(g) (\text{complete ionic}) $ $latex 2\text{Cl}^{-}(aq) + 2\text{H}_2 \text{O} \longrightarrow 2\text{OH}^{-}(aq) + 2\text{H}_2(g) + \text{Cl}_2(g) (\text{net ionic}) $

Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds

The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider NaCl soluble but AgCl insoluble.

[caption id="attachment_3210" align="aligncenter" width="385"] Figure 4. The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. “Grand canyon yavapal point 2010′′ by chensiyuan is licensed under Creative Commons[/caption]

One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO₃). However, CaCO₃ has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO₃ can precipitate if there is enough of it in the water. This precipitate, called limescale, can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction.

Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility!

Key Concepts and Summary

Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations.

Exercises

1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced? 2. Balance the following equations:

a) $latex \text{PCl}_5(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{POCl}_3(l) + \text{HCl}(aq) $

b) $latex \text{Cu}(s) + \text{HNO}_3(aq) \longrightarrow \text{Cu(NO}_3)_2(aq) + \text{H}_2 \text{O}(l) + \text{NO}(g) $

c) $latex \text{H}_2(g) + \text{I}_2(s) \longrightarrow \text{HI}(s) $

d) $latex \text{Fe}(s) + \text{O}_2(g) \longrightarrow \text{Fe}_2 \text{O}_3(s) $

e) $latex \text{Na}(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{NaOH}(aq) + \text{H}_2(g) $

f) $latex \text{(NH}_4)_2 \text{Cr}_2\text{O}_7(s) \longrightarrow \text{Cr}_2\text{O}_3(s) + \text{N}_2(g) + \text{H}_2 \text{O}(g) $

g) $latex \text{P}_4(s) + \text{Cl}_2(g) \longrightarrow \text{PCl}_3(l) $

h) $latex \text{PtCl}_4(s) \longrightarrow \text{Pt}(s) + \text{Cl}_2(g) $

3. Balance the following equations:

a) $latex \text{Ag}(s) + \text{H}_2 \text{S}(g) + \text{O}_2(g) \longrightarrow \text{Ag}_2 \text{S}(s) + \text{H}_2 \text{O}(l)$

b) $latex \text{P}_4(s) + \text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(s)$

c) $latex \text{Pb}(s) + \text{H}_2 \text{O}(l) + \text{O}_2(g) \longrightarrow \text{Pb(OH)}_2(s) $

d) $latex \text{Fe}(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{Fe}_3 \text{O}_4(s) + \text{H}_2(g) $

e) $latex \text{Sc}_2 \text{O}_3(s) + \text{SO}_3(l) \longrightarrow \text{Sc}_2 \text{(SO}_4)_3(s) $

f) $latex \text{Ca}_3 \text{(PO}_4)_2(aq) + \text{H}_3 \text{PO}_4(aq) \longrightarrow \text{Ca(H}_2 \text{PO}_4)_2(aq) $

g) $latex \text{Al}(s) + \text{H}_2 \text{SO}_4(aq) \longrightarrow \text{Al}_2 \text{(SO}_4)_3(s) + \text{H}_2(g) $

h) $latex \text{TiCl}_4(s) + \text{H}_2 \text{O}(g) \longrightarrow \text{TiO}_2(s) + \text{HCl}(g) $

4. Write a balanced molecular equation describing each of the following chemical reactions.

a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.

b) Gaseous butane, C₄H₁₀, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.

c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.

d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.

5. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.

a) Write the formulas of barium nitrate and potassium chlorate.

b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.

c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.

d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe³⁺ ions.)

6. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation: 7. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).

a) Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.

b) The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.

8. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:

a) $latex \text{K}_2 \text{C}_2 \text{O}_4(aq) + \text{Ba(OH)}_2(aq) \longrightarrow 2\text{KOH}(aq) + \text{BaC}_2 \text{O}_2(s) $

b) $latex {\text{Pb(NO}_3)}_2(aq) + \text{H}_2 \text{SO}_4(aq) \longrightarrow \text{PbSO}_4(s) + 2\text{HNO}_3(aq) $

c) $latex \text{CaCO}_3(s) + \text{H}_2 \text{SO}_4(aq) \longrightarrow \text{CaSO}_4(s) + \text{CO}_2(g) + \text{H}_2\text{O}(l) $

9. From the statement “nitrogen gas and hydrogen gas react to produce ammonia gas,” identify the reactants and the products.

10. From the statement “a solution of magnesium hydroxide reacts with a solution of nitric acid to produce a solution of magnesium nitrate and water,” identify the reactants and the products.

11. Write and balance the chemical equation described by Exercise 1.

12. Write and balance the chemical equation described by Exercise 2.

13. Balance: ___NaClO₃ $latex \longrightarrow$ ___NaCl + ___O₂

14. Balance: ___N₂ + ___H₂ $latex \longrightarrow$ ___N₂H₄

15. Balance: ___Al + ___O₂ $latex \longrightarrow$ ___Al₂O₃

16. Balance: ___C₂H₄ + ___O₂ $latex \longrightarrow$ ___CO₂ + ___H₂O

17. Balance: ___N_2(g) + ___H_2(g) $latex \longrightarrow$ ___NH_3(g)

18. Write a chemical equation that represents NaBr(s) dissociating in water.

19. Write a chemical equation that represents (NH₄)₃PO₄(s) dissociating in water.

20. Write the complete ionic equation for the reaction of FeCl₂(aq) and AgNO₃(aq). You may have to consult the solubility rules.

21. Write the complete ionic equation for the reaction of KCl(aq) and NaC₂H₃O₂(aq). You may have to consult the solubility rules.

22. Write the net ionic equation for the reaction of FeCl₂(aq) and AgNO₃(aq). You may have to consult the solubility rules.

23. Write the net ionic equation for the reaction of KCl(aq) and NaC₂H₃O₂(aq). You may have to consult the solubility rules.

24. Identify the spectator ions in Exercises 20 and 21.

  Answers

1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.

2. a) $latex \text{PCl}_5(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{POCl}_3(l) + 2\text{HCl}(aq) $

b) $latex 3\text{Cu}(s) + 8\text{HNO}_3(aq) \longrightarrow 3\text{Cu(NO}_3)_2(aq) + 4\text{H}_2 \text{O}(l) + 2\text{NO}(g) $ c) $latex \text{H}_2(g) + \text{I}_2(s) \longrightarrow 2\text{HI}(s) $ d) $latex 4\text{Fe}(s) + 3\text{O}_2(g) \longrightarrow 2\text{Fe}_2 \text{O}_3(s) $ e) $latex 2\text{Na}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow 2\text{NaOH}(aq) + \text{H}_2(g) $ f) $latex \text{(NH}_4)_2 \text{Cr}_2\text{2O}_7(s) \longrightarrow \text{Cr}_2\text{O}_3(s) + \text{N}_2(g) + 4\text{H}_2 \text{O}(l) $ g) $latex \text{P}_4(s) + 6\text{Cl}_2(g) \longrightarrow 4\text{PCl}_3(l) $ h) $latex \text{PtCl}_4(s) \longrightarrow \text{Pt}(s) + 2\text{Cl}_2(g) $ 3. a) $latex 4\text{Ag}(s) + 2\text{H}_2 \text{S}(g) + \text{O}_2(g) \longrightarrow 2\text{Ag}_2 \text{S}(s) + 2\text{H}_2 \text{O}(l)$

b) $latex \text{P}_4(s) + 5\text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(s)$

c) $latex 2\text{Pb}(s) + 2\text{H}_2 \text{O}(l) + \text{O}_2(g) \longrightarrow 2\text{Pb(OH)}_2(s) $

d) $latex 3\text{Fe}(s) + 4\text{H}_2 \text{O}(l) \longrightarrow \text{Fe}_3 \text{O}_4(s) + 4\text{H}_2(g) $

e) $latex \text{Sc}_2 \text{O}_3(s) + 3\text{SO}_3(l) \longrightarrow \text{Sc}_2 \text{(SO}_4)_3(s) $

f) $latex \text{Ca}_3 \text{(PO}_4)_2(aq) + 4\text{H}_3 \text{PO}_4(aq) \longrightarrow 3\text{Ca(H}_2 \text{PO}_4)_2(aq) $

g) $latex 2\text{Al}(s) + 3\text{H}_2 \text{SO}_4(aq) \longrightarrow \text{Al}_2 \text{(SO}_4)_3(s) + 3\text{H}_2(g) $

h) $latex \text{TiCl}_4(s) + 2\text{H}_2 \text{O}(g) \longrightarrow \text{TiO}_2(s) + 4\text{HCl}(g) $

4.a) $latex \text{CaCO}_3(s) \longrightarrow \text{CaO}(s) + \text{CO}_2(g) $ b) $latex 2\text{C}_4 \text{H}_{10}(g) + 13 \text{O}_2(g) \longrightarrow 8\text{CO}_2(g) + 10\text{H}_2\text{O}(g) $ c) $latex \text{MgCl}_{2}(aq) + 2 \text{NaOH}(aq) \longrightarrow \text{Mg(OH)}_2(s) + 2 \text{NaCl}(aq) $ d) $latex 2\text{H}_2 \text{O}(g) + 2 \text{Na}(s) \longrightarrow 2\text{NaOH}(s) + \text{H}_2(g) $

5. a) $latex \text{Ba(NO}_3)_2$ , $latex \text{KClO}_3 $ b) $latex 2 \text{KClO}_3(s) \longrightarrow 2 \text{KCl}(s) + 3\text{O}_2(g)$ c) $latex 2 \text{Ba(NO}_3)_2(s) \longrightarrow 2\text{BaO}(s) + 2\text{N}_2(g) + 5\text{O}_2(g)$ d) $latex 2 \text{Mg}(s) + \text{O}_2(g) \longrightarrow 2 \text{MgO}(s) $; $latex 4\text{Al}(s) + 3\text{O}_2(g) \longrightarrow 2\text{Al}_2 \text{O}_3(g) $; $latex 4\text{Fe}(s) + 3\text{O}_2(g) \longrightarrow 2\text{Fe}_2 \text{O}_3(s)$

6. H₂O 7. a) $latex 4\text{HF}(aq) + \text{SiO}_2(s) \longrightarrow \text{SiF}_4(g) + 2\text{H}_2 \text{O}(l) $ b) complete: $latex 2\text{Na}^{+}(aq) + 2\text{F}^{-}(aq) + \text{Ca}^{2+}(aq) + 2\text{Cl}^{-}(aq) \longrightarrow \text{CaF}_2(s) + 2\text{Na}^{+}(aq) + 2\text{Cl}^{-}(aq) $ net: $latex 2\text{F}^{-}(aq) + \text{Ca}^{2+}(aq) \longrightarrow \text{CaF}_2(s)$

8. a) complete: $latex 2\text{K}^{+}(aq) + {\text{C}_2 \text{O}_4}^{2-}(aq) + \text{Ba}^{2+}(aq) + 2\text{OH}^{-}(aq) \longrightarrow 2\text{K}^{+}(aq) + 2\text{OH}^{-}(aq) + \text{BaC}_2 \text{O}_4(s) $ net: $latex \text{Ba}^{2+}(aq) + {\text{C}_2 {\text{O}_4}}^{2-}(aq) \longrightarrow \text{BaC}_2 \text{O}_4(s) $

b) complete: $latex \text{Pb}^{2+}(aq) + 2{\text{NO}_3}^{-}(aq) + 2\text{H}^{+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{PbSO}_4(s) + 2\text{H}^{+}(aq) + 2{\text{NO}_3}^{-}(aq) $ net: $latex \text{Pb}^{2+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{PbSO}_4(s) $ c) complete: $latex \text{CaCO}_3(s) + 2\text{H}^{+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{CaSO}_4(s) + \text{CO}_2(g) + \text{H}_2 \text{O}(l) $ net: $latex \text{CaCO}_3(s) + 2\text{H}^{+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{CaSO}_4(s) + \text{CO}_2(g) + \text{H}_2 \text{O}(l) $

9. reactants: nitrogen and hydrogen; product: ammonia

10. reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water 11. N_2(g) + 3 H_2(g) $latex \longrightarrow$ 2 NH_3(g) 12. Mg(OH)_2(aq) + 2 HNO_3(aq) $latex \longrightarrow$ Mg(NO₃)_2(aq) + 2 H₂O_(ℓ) 13. 2 NaClO₃ $latex \longrightarrow$ 2 NaCl + 3 O₂ 14. N₂ + 2 H₂ $latex \longrightarrow$ N₂H₄ 15. 4Al + 3O₂ $latex \longrightarrow$ 2Al₂O₃ 16. C₂H₄ + 3 O₂ $latex \longrightarrow$ 2 CO₂ + 2 H₂O 17. N_2(g) + 3 H_2(g) $latex \longrightarrow$ 2 NH_3(g) 18. NaBr(s) $latex \longrightarrow$ Na⁺(aq) + Br⁻(aq) 19. (NH₄)₃PO₄(s) $latex \longrightarrow$ 3 NH₄⁺(aq) + PO₄³⁻(aq) 20. Fe²⁺(aq) + 2 Cl⁻(aq) + 2 Ag⁺(aq) + 2 NO₃⁻(aq) $latex \longrightarrow$ Fe²⁺(aq) + 2 NO₃⁻(aq) + 2 AgCl(s) 21. K⁺(aq) + Cl⁻(aq) + Na⁺(aq) + C₂H₃O₂⁻(aq) $latex \longrightarrow$ Na⁺(aq) + Cl⁻(aq) + K⁺(aq) + C₂H₃O₂⁻(aq) 22. 2 Cl⁻(aq) + 2 Ag⁺(aq) $latex \longrightarrow$ 2 AgCl(s) 23. There is no overall reaction. 24. In Exercise 20, Fe²⁺(aq) and NO₃⁻(aq) are spectator ions; in Exercise 21, Na⁺(aq) and Cl⁻(aq) are spectator ions.

Glossary

balanced equation: chemical equation with equal numbers of atoms for each element in the reactant and product chemical equation: symbolic representation of a chemical reaction coefficient: number placed in front of symbols or formulas in a chemical equation to indicate their relative amount complete ionic equation: chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions molecular equation: chemical equation in which all reactants and products are represented as neutral substances net ionic equation: chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions) product: substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation reactant: substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation spectator ion: ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality

]]> 1412 2162 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-2-classifying-chemical-reactions/ Thu, 12 Apr 2018 02:51:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-2-classifying-chemical-reactions/

Learning Objectives

By the end of this section, you will be able to:

• Define precipitation reactions
• Recognize and identify examples of precipitation reactions
• Predict the solubility of common inorganic compounds by using solubility rules

Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. The following sections of this chapter (section 6.2-6.4) will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction.

Precipitation Reactions and Solubility Rules

A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter).

The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table 1).

[caption id="attachment_4191" align="aligncenter" width="1008"] Table 1. Solubilities of Common Ionic Compounds in Water[/caption]

A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:

$latex 2\text{KI}(aq) + \text{Pb(NO}_3)_2(aq) \longrightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)$

This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.

The net ionic equation representing this reaction is:

$latex \text{Pb}^{2+}(aq) + 2\text{I}^{-}(aq) \longrightarrow \text{PbI}_2(s)$

Lead iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow (Figure 1). The properties of pure PbI₂ crystals make them useful for fabrication of X-ray and gamma ray detectors.

[caption id="attachment_1413" align="aligncenter" width="250"] Figure 1. A precipitate of PbI2 forms when solutions containing Pb2+ and I− are mixed. (credit: Der Kreole/Wikimedia Commons)[/caption]

The solubility table in Table 1 may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag⁺, NO₃⁻, Na⁺, and F⁻ ions. Aside from the two ionic compounds originally present in the solutions, AgNO₃ and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO₃ and AgF. The solubility table indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:

$latex \text{NaF}(aq) + \text{AgNO}_3(aq) \longrightarrow \text{AgF}(s) + \text{NaNO}_3(aq) \;\text{(molecular)}$$latex \text{Ag}^{+}(aq) + \text{F}^{-}(aq) \longrightarrow \text{AgF}(s) \;\text{(net ionic)}$

Example 1

Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.

a) potassium sulfate and barium nitrate

b) lithium chloride and silver acetate

c) lead nitrate and ammonium carbonate

 

Solution a) The two possible products for this combination are KNO₃ and BaSO₄. The solubility guidelines indicate BaSO₄ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

$latex \text{Ba}^{2+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{BaSO}_4(s)$

b) The two possible products for this combination are LiC₂H₃O₂ and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

$latex \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) \longrightarrow \text{AgCl}(s)$

c) The two possible products for this combination are PbCO₃ and NH₄NO₃. The solubility guidelines indicate PbCO₃ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

$latex \text{Pb}^{2+}(aq) + {\text{CO}_3}^{2-}(aq) \longrightarrow \text{PbCO}_3(s) $

 

Test Yourself Which solution could be used to precipitate the barium ion, Ba²⁺, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?

  Answers sodium sulfate, BaSO₄

Key Concepts and Summary

Chemical reactions are classified according to similar patterns of behavior.  Precipitation is one type of chemical reaction which involves the formation of one or more insoluble products.  Precipitation reactions, also called double displacement reactions can be summarized with the following reaction equation:

$latex \text{AB}(aq) + \text{CD}(aq) \longrightarrow \text{AD}(s) + \text{CB}(aq) or (s)$

The formation of the solid is the DRIVING FORCE of the reaction (the factor that makes the reaction go).

A precipitation reaction can be predicted to occur with the help of a solubility table (Table 1). There are three ways of representing a precipitation reaction, using a molecular equation, complete ionic equation or net ionic equation, as described in section 6.1.

Exercises

1. What are the general characteristics that help you recognize double replacement reactions?

2.  Assuming that each double replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Zn(NO₃)₂ + NaOH $latex \longrightarrow$ ?                  b)  HCl + Na₂S $latex \longrightarrow$ ?

3.  Assuming that each double replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Ca(C₂H₃O₂)₂ + HNO₃ $latex \longrightarrow$ ?            b)  Na₂CO₃ + Sr(NO₂)₂ $latex \longrightarrow$ ?

4.  Assuming that each double replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Pb(NO₃)₂ + KBr $latex \longrightarrow$ ?                     b)  K₂O + MgCO₃ $latex \longrightarrow$ ?

5.  Assuming that each double replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Sn(OH)₂ + FeBr₃ $latex \longrightarrow$ ?                    b)  CsNO₃ + KCl $latex \longrightarrow$ ?

6.  Use the solubility table (Table 1) to predict if each double replacement reaction will occur and, if so, write a balanced chemical equation.

a)  Pb(NO₃)₂ + KBr $latex \longrightarrow$ ?                     b)  K₂O + Na₂CO₃ $latex \longrightarrow$ ?

7.  Use the solubility table (Table 1) to predict if each double replacement reaction will occur and, if so, write a balanced chemical equation.

a)  Na₂CO₃ + Sr(NO₃)₂ $latex \longrightarrow$ ?               b)  (NH₄)₂SO₄ + Ba(NO₃)₂ $latex \longrightarrow$ ?

8.  Use the solubility rules to predict if each double replacement reaction will occur and, if so, write a balanced chemical equation.

a)  K₃PO₄ + SrCl₂ $latex \longrightarrow$ ?                       b)  NaOH + MgCl₂ $latex \longrightarrow$ ?

9.  Use the solubility rules to predict if each double replacement reaction will occur and, if so, write a balanced chemical equation.

a)  KC₂H₃O₂ + Li₂CO₃ $latex \longrightarrow$ ?             b)  KOH + AgNO₃ $latex \longrightarrow$ ?   Answers

1. A double replacement reaction occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. 2. a) Zn(NO₃)₂ + 2 NaOH $latex \longrightarrow$ Zn(OH)₂ + 2 NaNO₃ b) 2 HCl + Na₂S $latex \longrightarrow$ 2 NaCl + H₂S 3. a)  Ca(C₂H₃O₂)₂ + 2 HNO₃ $latex \longrightarrow$ Ca(NO₃)₂ + 2 HC₂H₃O₂ b)  Na₂CO₃ + Sr(NO₂)₂ $latex \longrightarrow$ 2 NaNO₂ + SrCO₃ 4.a)  Pb(NO₃)₂ + 2 KBr $latex \longrightarrow$ PbBr₂ + 2 KNO₃ b)  K₂O + MgCO₃ $latex \longrightarrow$ K₂CO₃ + MgO 5. a)  3 Sn(OH)₂ + 2 FeBr₃ $latex \longrightarrow$ 3 Sn(Br)₂  +  2 Fe(OH)₃ b)  CsNO₃ + KCl $latex \longrightarrow$ KNO₃ + CsCl 6.a)  Pb(NO₃)₂(aq) + 2 KBr(aq) $latex \longrightarrow$ PbBr₂(s) + 2 KNO₃(aq) b)  No reaction occurs. 7. a) Na₂CO₃(aq) + Sr(NO₃)₂(aq) $latex \longrightarrow$ 2 NaNO₃(aq) + SrCO₃(s) b)  (NH₄)₂SO₄(aq) + Ba(NO₃)₂(aq) $latex \longrightarrow$ BaSO₄(s) + 2 NH₄NO₃(aq) 8.a)  2 K₃PO₄(aq) + 3 SrCl₂(aq) $latex \longrightarrow$ Sr₃(PO₄)₂(s) + 6 KCl(aq) b)  2 NaOH(aq) + MgCl₂(aq) $latex \longrightarrow$ 2 NaCl(aq) + Mg(OH)₂(s)

9. a) No reaction occurs. b)  KOH(aq) + AgNO₃(aq) $latex \longrightarrow$ AgOH(s) + KNO₃(aq)

Glossary

insoluble: of relatively low solubility; dissolving only to a slight extent precipitate: insoluble product that forms from reaction of soluble reactants precipitation reaction: reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double displacement or metathesis salt: ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide soluble: of relatively high solubility; dissolving to a relatively large extent solubility: the extent to which a substance may be dissolved in water, or any solvent

]]> 1419 2162 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-3-reaction-stoichiometry/ Thu, 12 Apr 2018 02:51:50 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-3-reaction-stoichiometry/

Learning Objectives

By the end of this section, you will be able to:

• Explain the concept of stoichiometry as it pertains to chemical reactions
• Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
• Perform stoichiometric calculations involving mass, moles, and solution molarity

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $latex \frac{3}{4}$ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

$latex 1 \;\text{cup mix} + \frac{3}{4} \;\text{cup milk} + 1 \;\text{egg} \longrightarrow 8 \;\text{pancakes}$

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

$latex 24 \;\rule[0.5ex]{4em}{0.1ex}\hspace{-4em}\text{pancakes} \times \frac{1 \;\text{egg}}{8 \;\rule[0.25ex]{3em}{0.1ex}\hspace{-3em}\text{pancakes}} = 3 \;\text{eggs}$

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

$latex \text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g)$

This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:

$latex \frac{2 \;\text{NH}_3 \;\text{molecules}}{3 \;\text{H}_2 \;\text{molecules}} \;\text{or} \;\frac{2 \;\text{doz NH}_3 \;\text{molecules}}{3 \;\text{doz H}_2 \;\text{molecules}} \;\text{or} \;\frac{2 \;\text{mol NH}_3 \;\text{molecules}}{3 \;\text{mol H}_2 \;\text{molecules}}$

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.

Example 1

How many moles of I₂ are required to react with 0.429 mol of Al according to the following equation (see Figure 1)?

$latex 2\text{Al} + 3\text{I}_2 \longrightarrow 2\text{AlI}_3$

 

Solution Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is $latex \frac{3 \;\text{mol I}_2}{2 \;\text{mol Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:

$latex \begin{array}{r @{{}={}} l} \text{mol I}_2 & 0.429 \;\rule[0.5ex]{3.25em}{0.1ex}\hspace{-3.25em}\text{mol Al} \times \frac{3 \;\text{mol I}_2}{2 \;\rule[0.25ex]{2em}{0.1ex}\hspace{-2em}\text{mol Al}} \\[1em] & 0.644 \;\text{mol I}_2 \end{array}$

 

Test Yourself How many moles of Ca(OH)₂ are required to react with 1.36 mol of H₃PO₄ to produce Ca₃(PO₄)₂ according to the equation $latex 3\text{Ca(OH)}_2 + 2\text{H}_3 \text{PO}_4 \longrightarrow \text{Ca}_3 \text{(PO}_4)_2 + 6\text{H}_2 \text{O}$?

  Answer 2.04 mol

Example 2

How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?

$latex \text{C}_3 \text{H}_8 + 5\text{O}_2 \longrightarrow 3\text{CO}_2 + 4\text{H}_2 \text{O}$

 

Solution The approach here is the same as for Example 1, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number.

The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:

$latex \frac{3 \;\text{mol CO}_2}{1 \;\text{mol C}_3 \text{H}_8}$

Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,

$latex 0.75 \;\rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em}\text{mol C}_3 \text{H}_8 \times \frac{3 \;\rule[0.25ex]{3em}{0.1ex}\hspace{-3em} \text{mol CO}_2}{1 \;\rule[0.5ex]{3em}{0.1ex}\hspace{-3em}\text{mol C}_3 \text{H}_8} \times \frac{6.022 \times 10^{23} \;\text{CO}_2 \;\text{molecules}}{\rule[0.25ex]{3em}{0.1ex}\hspace{-3em} \text{mol CO}_2} = 1.4 \times 10^{24} \text{CO}_2 \;\text{molecules}$

 

Test Yourself How many NH₃ molecules are produced by the reaction of 4.0 mol of Ca(OH)₂ according to the following equation:

$latex (\text{NH}_4)_2 \text{SO}_4 + \text{Ca(OH)}_2 \longrightarrow 2\text{NH}_3 + \text{CaSO}_4 + 2\text{H}_2 \text{O}$

Answer

4.8 × 10²⁴ NH₃ molecules

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.

Example 3

What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)₂] by the following reaction?

$latex \text{MgCl}_2(aq) + 2\text{NaOH}(aq) \longrightarrow \text{Mg(OH)}_2(s) + \text{NaCl}(aq)$

 

Solution The approach used previously in Example 1 and Example 2 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:

$latex 16 \;\rule[0.5ex]{5em}{0.1ex}\hspace{-5em}\text{g Mg(OH)}_2 \times \frac{1 \;\rule[0.25ex]{4.75em}{0.1ex}\hspace{-4.75em}\text{mol Mg(OH)}_2}{58.3197 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{g Mg(OH)}_2} \times \frac{2 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol NaOH}}{1 \;\rule[0.25ex]{4.5em}{0.1ex}\hspace{-4.5em}\text{mol Mg(OH)}_2} \times \frac{39.9971 \;\text{g NaOH}}{1 \;\rule[0.25ex]{3.25em}{0.1ex}\hspace{-3.25em}\text{mol NaOH}} = 22 \;\text{g NaOH}$

 

Test Yourself What mass of gallium oxide, Ga₂O₃, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $latex 4 \text{Ga} + 3\text{O}_2 \longrightarrow 2\text{Ga}_2 \text{O}_3.$

  Answer 39.0 g

Example 4

What mass of oxygen gas, O₂, from the air is consumed in the combustion of 702 g of octane, C₈H₁₈, one of the principal components of gasoline?

$latex 2\text{C}_8 \text{H}_{18} + 25\text{O}_2 \longrightarrow 16\text{CO}_2 + 18\text{H}_2 \text{O}$

 

Solution The approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.

$latex 702 \;\rule[0.5ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{g C}_8 \text{H}_{18} \times \frac{1 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol C}_8 \text{H}_{18}}{114.231 \;\rule[0.25ex]{2.75em}{0.1ex}\hspace{-2.75em}\text{g C}_8 \text{H}_{18}} \times \frac{25 \;\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol O}_2}{2 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol C}_8 \text{H}_{18}} \times \frac{31.9988 \;\text{g O}_2}{\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol O}_2} = 2.46 \times 10^3 \;\text{g O}_2$

 

Test Yourself What mass of CO is required to react with 25.13 g of Fe₂O₃ according to the equation

$latex \text{Fe}_2 \text{O}_3 + 3\text{CO} \longrightarrow 2\text{Fe} + 3\text{CO}_2$

  Answer 13.22 g

These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 2 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.

Airbags

Airbags (Figure 3) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN₃. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN₃ to initiate its decomposition:

$latex 2 \text{NaN}_3(s) \longrightarrow 3\text{N}_2(g) + 2\text{Na}(s)$

[caption id="" align="aligncenter" width="388"] Figure 3. Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)[/caption]  

This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s).

Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN₃ will generate approximately 50 L of N₂.

More Worked Out Problems

Example 5

How many molecules of SO₃ are needed to react with 144 molecules of Fe₂O₃ given this balanced chemical equation?

Fe₂O₃(s) + 3SO₃(g) $latex \longrightarrow$ Fe₂(SO₄)₃

 

Solution

We use the balanced chemical equation to construct a conversion factor between Fe₂O₃ and SO₃. The number of molecules of Fe₂O₃ goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO₃ go on the top. Thus, the appropriate conversion factor is

Starting with our given amount and applying the conversion factor, the result is

We need 432 molecules of SO₃ to react with 144 molecules of Fe₂O₃.

 

Test Yourself

How many molecules of H₂ are needed to react with 29 molecules of N₂ to make ammonia if the balanced chemical equation is N₂ + 3H₂ $latex \longrightarrow$ 2NH₃?

 

Answer

87 molecules

Example 6

How many molecules of NH₃ can you make if you have 228 atoms of H₂?

 

Solution

From the formula, we know that one molecule of NH₃ has three H atoms. Use that fact as a conversion factor:

Test Yourself

How many molecules of Fe₂(SO₄)₃ can you make from 777 atoms of S?

 

Answer

259 molecules

Example 7

How many moles of HCl will be produced when 249 g of AlCl₃ are reacted according to this chemical equation?  The molar mass of AlCl₃ is 133.34 g/mol.

2 AlCl₃ + 3 H₂O(ℓ) $latex \longrightarrow$ Al₂O₃ + 6 HCl(g)

 

Solution

We will do this in two steps: convert the mass of AlCl₃ to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl₃ is 133.34 g/mol, which we have to invert to get the appropriate conversion factor:

$latex 249 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g AlCl}_3 \times \frac{1 \;\text{mol AlCl}_3}{133.34\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g AlCl}_3} = \underline{1.86}74 \;\text{mol of aluminium chloride with 3 sig figs} $

 

Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl₃ and the number of moles of HCl:

$latex \frac{6 \; \text{mol HCl}}{2 \; \text{mol AlCl}_3}$

Applying this conversion factor to the quantity of AlCl₃, we get

$latex \underline{1.86}74 \;\rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol AlCl}_3 \times \frac{6 \;\text{mol HCl}}{2\;\rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol AlCl}_3} =5.60 \;\text{mol HCl} $

Alternatively, we could have done this in one line:

$latex 249 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g AlCl}_3 \times \frac{1 \;\rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol AlCl}_3}{133.34\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g AlCl}_3} \times \frac{6 \;\text{mol HCl}}{2\;\rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol AlCl}_3}=5.60 \;\text{mol HCl} $

The last digit in our final answer is slightly different because of rounding differences, but the answer is essentially the same.

 

Test Yourself

How many moles of Al₂O₃ will be produced when 23.9 g of H₂O are reacted according to this chemical equation?

2 AlCl₃ + 3 H₂O(ℓ) $latex \longrightarrow$ Al₂O₃ + 6 HCl(g)

 

Answer

0.442 mol

Example 8

How many grams of NH₃ will be produced when 33.9 mol of H₂ are reacted according to this chemical equation?  Use 17.03 g/mol as the molar mass of NH₃.

N₂(g) + 3 H₂(g) $latex \longrightarrow$ 2 NH₃(g)

 

Solution

The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have

Now, using the molar mass of NH₃, which is 17.03 g/mol, we get  

Test Yourself

How many grams of N₂ are needed to produce 2.17 mol of NH₃ when reacted according to this chemical equation?

N₂(g) + 3 H₂(g) $latex \longrightarrow$ 2 NH₃(g)

 

Answer

30.4 g (Note: here we go from a product to a reactant, showing that mole-mass problems can begin and end with any substance in the chemical equation.)

Example 9

What mass of Mg will be produced when 86.4 g of K are reacted?  Use 39.09 g/mol as the molar mass of potassium and 24.31 g/mol as the molar mass of magnesium.

MgCl₂(s) + 2 K(s) $latex \longrightarrow$ Mg(s) + 2 KCl(s)

 

Solution

We will simply follow the steps

mass K $latex \longrightarrow$ mol K $latex \longrightarrow$ mol Mg $latex \longrightarrow$ mass Mg

In addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line,

Test Yourself

What mass of H₂ will be produced when 122 g of Zn are reacted?

Zn(s) + 2 HCl(aq) $latex \longrightarrow$ ZnCl₂(aq) + H₂(g)

 

Answer

3.77 g

Example 10

Interpret this balanced chemical equation in terms of moles.

P₄ + 5 O₂ $latex \longrightarrow$ P₄O₁₀

 

Solution

The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as “one mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.”

 

Test Yourself

Interpret this balanced chemical equation in terms of moles.

N₂ + 3 H₂ $latex \longrightarrow$ 2 NH₃

 

Answer

One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.

Example 11

For the balanced chemical equation

2 C₄H₁₀(g) + 13 O₂ $latex \longrightarrow$ 8 CO₂(g) + 10 H₂O(ℓ)

if 154 mol of O₂ are reacted, how many moles of CO₂ are produced?

 

Solution

We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is 13 mol O₂ to 8 mol CO₂

We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:

The mol O₂ unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don’t contribute to the number of significant figures in the final answer.

 

Test Yourself

Using the above equation, how many moles of H₂O are produced when 154 mol of O₂ react?

 

Answer

118 mol

Key Concepts and Summary

A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.

Exercises

1. Determine the number of moles and the mass requested for each of the following reactions: (Hint: Write the balanced equation for each before attempting calculations.)

a) The number of moles and the mass of chlorine, Cl₂, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.

b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.

c) The number of moles and the mass of sodium nitrate, NaNO₃, required to produce 128 g of oxygen. (NaNO₂ is the other product.)

d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.

e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. (CO₂ is the other product.)

f)

2. Determine the number of moles and the mass requested for each of the following reactions: (Hint: Write the balanced equation for each before attempting calculations.)

a) The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl₂ and H₂.

b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.

c) The number of moles and the mass of magnesium carbonate, MgCO₃, required to produce 283 g of carbon dioxide. (MgO is the other product.)

d) The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, C₂H₂, in an excess of oxygen.

e) The number of moles and the mass of barium peroxide, BaO₂, needed to produce 2.500 kg of barium oxide, BaO (O₂ is the other product.)

f)

3. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: $latex 2\text{Ga} + 6\text{HCl} \longrightarrow 2\text{GaCl}_3 + 3\text{H}_2$.

a) Outline the steps necessary to determine the number of moles and mass of gallium chloride.

b) Perform the calculations outlined.

4. Silver is often extracted from ores such as K[Ag(CN)₂] and then recovered by the reaction $latex 2 \text{K} [\text{Ag(CN)}_2](aq) + \text{Zn}(s) \longrightarrow 2\text{Ag}(s) + \text{Zn(CN)}_2(aq) + 2\text{KCN}(aq)$

a) How many molecules of Zn(CN)₂ are produced by the reaction of 35.27 g of K[Ag(CN)₂]?

b) What mass of Zn(CN)₂ is produced?

5. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO₂, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO₂ is required to produce 3.00 kg of SiC. 6. Urea, CO(NH₂)₂, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO₂ produced by combustion of 1.00×103kg1.00×103kg of carbon followed by the reaction? $latex \text{CO}_2(g) + 2\text{NH}_3(g) \longrightarrow {\text{CO(NH}_2})_2(s) + \text{H}_2 \text{O}(l)$ 7. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon). 8. What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO₃)₂ in 43.88 mL of a 0.3842 M solution of Cu(NO₃)₂?$latex 2 \text{Cu(NO}_3)_2 + 4\text{KI} \longrightarrow 2\text{CuI} + \text{I}_2 + 4{\text{KNO}_3}$ 9. The toxic pigment called white lead, Pb₃(OH)₂(CO₃)₂, has been replaced in white paints by rutile, TiO₂. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO₃) by mass?$latex 2\text{FeTiO}_3 + 4\text{HCl} + \text{Cl}_2 \longrightarrow 2\text{FeCl}_3 + 2\text{TiO}_2 + 2\text{H}_2 \text{O}$ 10. Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe? 11. What are all the conversion factors that can be constructed from the balanced chemical reaction 2H₂(g) + O₂(g) $latex \longrightarrow$ 2H₂O(ℓ)? 12. Given the chemical equation

Na(s) + H₂O(ℓ) $latex \longrightarrow$ NaOH(aq) + H₂(g)

a)  Balance the equation. b)  How many molecules of H₂ are produced when 332 atoms of Na react?

13. For the balanced chemical equation

6 H⁺(aq) + 2 MnO₄⁻(aq) + 5 H₂O₂(ℓ) $latex \longrightarrow$ 2 Mn²⁺(aq) + 5 O₂(g) + 8 H₂O(ℓ)

how many molecules of H₂O are produced when 75 molecules of H₂O₂ react?

14.  Given the balanced chemical equation

Fe₂O₃(s) + 3SO₃(g) $latex \longrightarrow$ Fe₂(SO₄)₃

how many molecules of Fe₂(SO₄)₃ are produced if 321 atoms of S are reacted?

15.  For the balanced chemical equation

Fe₂O₃(s) + 3 SO₃(g) $latex \longrightarrow$ Fe₂(SO₄)₃

suppose we need to make 145,000 molecules of Fe₂(SO₄)₃. How many molecules of SO₃ do we need?

16.  Construct the three independent conversion factors possible for these two reactions:

2 H₂ + O₂ $latex \longrightarrow$ 2 H₂O

H₂ + O₂ $latex \longrightarrow$ H₂O₂

Why are the ratios between H₂ and O₂ different?

The conversion factors are different because the stoichiometries of the balanced chemical reactions are different.

17. What mass of CO₂ is produced by the combustion of 1.00 mol of CH₄?

CH₄(g) + 2 O₂(g) $latex \longrightarrow$ CO₂(g) + 2 H₂O(ℓ)

18. What mass of HgO is required to produce 0.692 mol of O₂?

2 HgO(s) $latex \longrightarrow$ 2 Hg(ℓ) + O₂(g)

19. How many moles of Al can be produced from 10.87 g of Ag?

Al(NO₃) ₃(s) + 3 Ag $latex \longrightarrow$ Al + 3 AgNO₃

20. How many moles of O₂ are needed to prepare 1.00 g of Ca(NO₃)₂?

Ca(s) + N₂(g) + 3 O₂(g) $latex \longrightarrow$ Ca(NO₃) ₂(s)

21. What mass of O₂ can be generated by the decomposition of 100.0 g of NaClO₃?

2 NaClO₃ $latex \longrightarrow$ 2 NaCl(s) + 3 O₂(g)

22. What mass of Fe₂O₃ must be reacted to generate 324 g of Al₂O₃?

Fe₂O₃(s) + 2 Al(s) $latex \longrightarrow$ 2 Fe(s) + Al₂O₃(s)

23. What mass of MnO₂ is produced when 445 g of H₂O are reacted?

H₂O(ℓ) + 2 MnO₄⁻(aq) + Br⁻(aq) $latex \longrightarrow$ BrO₃⁻(aq) + 2 MnO₂(s) + 2 OH⁻(aq)

24. If 83.9 g of ZnO are formed, what mass of Mn₂O₃ is formed with it?

Zn(s) + 2 MnO₂(s) $latex \longrightarrow$ ZnO(s) + Mn₂O₃(s)

25. If 88.4 g of CH₂S are reacted, what mass of HF is produced?

CH₂S + 6 F₂ $latex \longrightarrow$ CF₄ + 2 HF + SF₆

26. Express in mole terms what this chemical equation means.

CH₄ + 2O₂ $latex \longrightarrow$ CO₂ + 2H₂O 27. How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles? 28. For the chemical equation

2 C₂H₆ + 7 O₂ $latex \longrightarrow$ 4 CO₂ + 6 H₂O

what equivalences can you write in terms of moles? Use the ⇔ sign.

29. Write the balanced chemical reaction for the combustion of C₅H₁₂ (the products are CO₂ and H₂O) and determine how many moles of H₂O are formed when 5.8 mol of O₂ are reacted.

30. For the balanced chemical equation

3 Cu(s) + 2 NO₃⁻(aq) + 8 H⁺(aq) $latex \longrightarrow$ 3 Cu²⁺(aq) + 4 H₂O(ℓ) + 2 NO(g)

how many moles of Cu²⁺ are formed when 55.7 mol of H⁺ are reacted?

31. For the balanced chemical reaction

4 NH₃(g) + 5 O₂(g) $latex \longrightarrow$ 4 NO(g) + 6 H₂O(ℓ)

how many moles of H₂O are produced when 0.669 mol of NH₃ react?

32. For the balanced chemical reaction

4 KO₂(s) + 2 CO₂(g) $latex \longrightarrow$ 2 K₂CO₃(s) + 3 O₂(g)

determine the number of moles of both products formed when 6.88 mol of KO₂ react.

 

Answers

1. a) 0.435 mol Na, 0.217 mol Cl₂, 15.4 g Cl₂

b) 0.005780 mol HgO, 2.890 × 10⁻³ mol O₂, 9.248 × 10⁻² g O₂ c) 8.00 mol NaNO₃, 6.8 × 10² g NaNO₃ d) 1665 mol CO₂, 73.3 kg CO₂ e) 18.86 mol CuO, 2.330 kg CuCO₃ f) 0.4580 mol C₂H₄Br₂, 86.05 g C₂H₄Br₂

2. a) 0.0686 mol Mg, 1.67 g Mg

b) 2.701 × 10⁻³ mol O₂, 0.08644 g O₂ c) 6.43 mol MgCO₃, 542 g MgCO₃ d) 713 mol H₂O, 12.8 kg H₂O e) 16.31 mol BaO₂, 2762 g BaO₂ f) 0.207 mol C₂H₄, 5.81 g C₂H₄

3. a) $latex \text{volume HCl solution} \longrightarrow \text{mol HCl} \longrightarrow \text{mol GaCl}_3$; b) 1.25 mol GaCl₃, 2.2 × 10² g GaCl₃

4. a) 5.337 × 10²² molecules      b) 10.41 g Zn(CN)₂

5. $latex \text{SiO}_2 + 3\text{C} \longrightarrow \text{SiC} + 2\text{CO}$, 4.50 kg SiO₂

6. 5.00 × 10³ kg

7. 1.28 × 10⁵ g CO₂

8. 161.40 mL KI solution

9. 176 g TiO₂

10.  are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible.

11.

 and their reciprocals are the conversion factors that can be constructed.

12.  2Na(s) + 2H₂O(ℓ) $latex \longrightarrow$ 2NaOH(aq) + H₂(g) and 166 molecules

13. 120 molecules

14. 107 molecules

15. 435,000 molecules

16.

17. 44.0 g 18. 3.00 × 10² g 19. 0.0336 mol 20. 0.0183 mol 21. 45.1 g 22. 507 g 23. 4.30 × 10³ g 24. 163 g 25. 76.7 g 26. One mole of CH₄ reacts with 2 mol of O₂ to make 1 mol of CO₂ and 2 mol of H₂O. 27. 6.022 × 10²³ molecules of CH₄, 1.2044 × 10²⁴ molecules of O₂, 6.022 × 10²³ molecules of CO₂, and 1.2044 × 10²⁴ molecules of H₂O 28. 2 mol of C₂H₆ to 7 mol of O₂ to 4 mol of CO₂ to 6 mol of H₂O 29. C₅H₁₂ + 8 O₂  $latex \longrightarrow$ 5CO₂ + 6H₂O; 4.4 mol 30. 20.9 mol 31. 1.00 mol 32. 3.44 mol of K₂CO₃; 5.16 mol of O₂

Glossary

stoichiometric factor: ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products stoichiometry: relationships between the amounts of reactants and products of a chemical reaction

]]> 1429 1405 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-4-reaction-yields/ Thu, 12 Apr 2018 02:51:52 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-4-reaction-yields/

Learning Objectives

By the end of this section, you will be able to:

• Explain the concepts of theoretical yield and limiting reactants/reagents.
• Derive the theoretical yield for a reaction under specified conditions.
• Calculate the percent yield for a reaction.

The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.

Limiting Reactant

Consider another food analogy, making grilled cheese sandwiches (Figure 1):

$latex 1 \;\text{slice of cheese} + 2 \;\text{slices of bread} \longrightarrow 1\;\text{sandwich}$

Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

$latex \text{H}_2(s) + \text{Cl}_2(g) \longrightarrow 2\text{HCl}(g)$

The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H₂ and 2 moles of Cl₂. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.

An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield

$latex \text{mol HCl produced} = 3 \;\text{mol H}_2 \times \frac{2 \;\text{mol HCl}}{1 \;\text{mol H}_2} = 6 \;\text{mol HCl}$

Complete reaction of the provided chlorine would produce

$latex \text{mol HCl produced} = 2 \;\text{mol Cl}_2 \times \frac{2 \;\text{mol HCl}}{1 \;\text{mol Cl}_2} = 4 \;\text{mol HCl}$

The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 2).

 

View this interactive simulation illustrating the concepts of limiting and excess reactants.

 

Example 1

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

$latex 3\text{Si}(s) + 2\text{N}_2(g) \longrightarrow \text{Si}_3 \text{N}_4(s)$

Which is the limiting reactant when 2.00 g of Si and 1.50 g of N₂ react?

 

Solution Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

$latex \text{mol Si} = 2.00 \;\rule[0.5ex]{1.75em}{0.1ex}\hspace{-1.75em}\text{g Si} \times \frac{1 \;\text{mol Si}}{28.0855 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{g Si}} = 0.0\underline{712}11 \;\text{mol Si with 3 sig figs}$

$latex \text{mol N}_2 = 1.50 \;\rule[0.5ex]{2em}{0.1ex}\hspace{-2em}\text{g N}_2 \times \frac{1 \;\text{mol N}_2}{28.0134 \rule[0.25ex]{1.5em}{0.1ex}\hspace{-1.5em}\;\text{g N}_2} = 0.0\underline{535}46 \;\text{mol N with 3 sig figs}_2$

The provided Si:N₂ molar ratio is:

$latex \frac{0.0\underline{712}11 \;\text{mol Si}}{0.0\underline{535}46 \;\text{mol N}_2} = \frac{1.33 \;\text{mol Si}}{1 \text{mol N}_2}$

The stoichiometric Si:N₂ ratio is:

$latex \frac{3 \;\text{mol Si}}{2 \;\text{mol N}_2} = \frac{1.5 \;\text{mol Si}}{1 \;\text{mol N}_2}$

Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.

Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

$latex \text{mol Si}_3 \text{N}_4 \;\text{produced} = 0.0\underline{712}11 \;\text{mol Si} \times \frac{1 \;\text{mol Si}_3 \text{N}_4}{3 \;\text{mol Si}} = 0.0237 \;\text{mol Si}_3 \text{N}_4$

while the 0.0535 moles of nitrogen would produce

$latex \text{mol Si}_3 \text{N}_4 \;\text{produced} = 0.0\underline{535}46 \;\text{mol N}_2 \times \frac{1 \;\text{mol Si}_3 \text{N}_4}{2 \;\text{mol N}_2} = 0.0268 \;\text{mol Si}_3 \text{N}_4$

Since silicon yields the lesser amount of product, it is the limiting reactant.

 

Test Yourself Which is the limiting reactant when 5.00 g of H₂ and 10.0 g of O₂ react and form water?

  Answer O₂

Percent Yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

$latex \text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% $

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

Example 2

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

$latex \text{CuSO}_4(aq) + \text{Zn}(s) \longrightarrow \text{Cu}(s) + \text{ZnSO}_4(aq)$

What is the percent yield?

 

Solution The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:

$latex 1.274 \;\rule[0.5ex]{3.75em}{0.1ex}\hspace{-3.75em}\text{g CuSO}_4 \times \frac{1 \;\rule[0.25ex]{4em}{0.1ex}\hspace{-4em}\text{mol CuSO}_4}{159.610 \;\rule[0.25ex]{3em}{0.1ex}\hspace{-3em}\text{g CuSO}_4} \times \frac{1 \;\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol Cu}}{1 \rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol CuSO}_4} \times \frac{63.546 \;\text{g Cu}}{1 \;\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol Cu}} = 0.\underline{5072}21 \;\text{g Cu with 4 sig figs}$

Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be

$latex \text{percent yield} = (\frac{\text{actual yield}}{\text{theoretical yield}}) \times 100\%$

$latex \text{percent yield} = (\frac{0.392 \;\text{g Cu}}{0.\underline{5072}21 \;\text{g Cu}}) \times 100\%$

$latex = 77.3\%$

Test Yourself What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF₂Cl₂ from 32.9 g of CCl₄ and excess HF?

$latex \text{CCl}_4 + 2\text{HF} \longrightarrow \text{CF}_2 \text{Cl}_2 + 2\text{HCl}$

Answer

48.3%

Green Chemistry and Atom Economy

The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry” (see details at this website). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:

$latex \text{atom economy} = \frac{\text{mass of product}}{\text{mass of reactants}} \times 100\% $

Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 3). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%.

More Worked Out Problems

Example 3

A 5.00 g quantity of Rb are combined with 3.44 g of MgCl₂ according to this chemical reaction:

2 Rb(s) + MgCl₂(s) $latex \longrightarrow$ Mg(s) + 2 RbCl(s)

What mass of Mg is formed, and what mass of what reactant is left over?

 

Solution

Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less.

$latex 5.00 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Rb} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Rb}}{85.4678 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Rb}} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Mg}}{2 \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Rb}} \times \frac{24.3050 \;\text{g Mg}}{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Mg}} = 0.\underline{711}53 \;\text{g Mg with 3 sig figs}$

 

$latex 3.44 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g MgCl}_2 \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol MgCl}_2}{95.2104 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g MgCl}_2} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Mg}}{1 \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol MgCl}_2} \times \frac{24.3050 \;\text{g Mg}}{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Mg}} = 0.\underline{878}15 \;\text{g Mg with 3 sig figs}$

 

The 0.712 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl₂ reacted with the 5.00 g of Rb and then subtract the amount reacted from the original amount.

$latex 5.00 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Rb} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Rb}}{85.4678 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Rb}} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol MgCl}_2}{2 \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Rb}} \times \frac{95.2104 \;\text{g MgCl}_2}{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol MgCl}_2} = \underline{2.78}498 \;\text{g magnesium chloride with 3 sig figs}$  

Because we started with 3.44 g of MgCl₂, we have

3.44 g MgCl₂ −  2.78498 g MgCl₂ reacted = 0.66 g MgCl₂ left  

Test Yourself

Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent?

Answer

H₂S is the limiting reagent; 1.5 g of MgO are left over.

Example 4

A worker reacts 30.5 g of Zn with nitric acid and evapourates the remaining water to obtain 65.2 g of Zn(NO₃)₂. What are the theoretical yield, the actual yield, and the percent yield?

Zn(s) + 2 HNO₃(aq) $latex \longrightarrow$ Zn(NO₃)₂(aq) + H₂(g)

 

Solution

A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.37 g/mol) and Zn(NO₃)₂ (189.38 g/mol). In three steps, the mass-mass calculation is

$latex 30.5 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Zn} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Zn}}{65.37 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g Zn}} \times \frac{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Zn(NO}_3 \text{)}_2}{1 \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Zn}} \times \frac{189.38 \;\text{g Zn(NO}_3 \text{)}_2}{1 \;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol Zn(NO}_3 \text{)}_2} = \underline{88.3}6 \;\text{g zing nitrate with 3 sig figs}$

Thus, the theoretical yield is 88.4 g of Zn(NO₃)₂. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO₃)₂. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:

$latex \text{percent yield} = (\frac{65.2 \;\text{g Zn(NO}_3 \text{)}_2}{\underline{88.3}6 \;\text{g Zn(NO}_3 \text{)}_2}) \times 100\%$

$latex = 73.8\%$

The worker achieved almost three-fourths of the possible yield.

 

Test Yourself

A synthesis produced 2.05 g of NH₃ from 16.5 g of N₂. What is the theoretical yield and the percent yield?

N₂(g) + 3 H₂(g) $latex \longrightarrow$ 2 NH₃(g)

 

Answer

theoretical yield = 20.1 g; percent yield = 10.2%

Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification

Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.

Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C₁₃H₂₁NO₂; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:

impure albuterol → intermediate A	percent yield = 70%
intermediate A → intermediate B	percent yield = 100%
intermediate B → intermediate C	percent yield = 40%
intermediate C → intermediate D	percent yield = 72%
intermediate D → purified albuterol	percent yield = 35%
overall percent yield = 70% × 100% × 40% × 72% × 35% = 7.5%

That is, only about one-fourteenth of the original material was turned into the purified drug. This gives you one reason why some drugs are so expensive; a lot of material is lost in making a high-purity pharmaceutical.

Key Concepts and Summary

When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.

Key Equations

• $latex \text{percent yield} = (\frac{\text{actual yield}}{\text{theoretical yield}}) \times 100\%$

Exercises

1. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine? 2. A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was his percent yield? 3. Freon-12, CCl₂F₂, is prepared from CCl₄ by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl₂F₂ from 32.9 g of CCl₄. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield. 4. Toluene, C₆H₅CH₃, is oxidized by air under carefully controlled conditions to benzoic acid, C₆H₅CO₂H, which is used to prepare the food preservative sodium benzoate, C₆H₅CO₂Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?

$latex 2\text{C}_6\text{H}_5\text{CH}_3\;+\;3\text{O}_2\;{\longrightarrow}\;2\text{C}_6\text{H}_5\text{CO}_2\text{H}\;+\;2\text{H}_2\text{O}$

5. Outline the steps needed to solve the following problem, then do the calculations. Ether, (C₂H₅)₂O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.

$latex 2\text{C}_2\text{H}_5\text{OH}\;+\;\text{H}_2\text{SO}_4\;{\longrightarrow}\;(\text{C}_2\text{H}_5)_2\;+\;\text{H}_2\text{SO}_4{\cdot}\text{H}_2\text{O}$

What is the percent yield of ether if 1.17 L (d = 0.7134 g/mL) is isolated from the reaction of 1.500 L of C₂H₅OH (d = 0.7894 g/mL)?

6. Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H₃PO₄ react according to the following chemical equation.

$latex 2\text{Cr}\;+\;2\text{H}_3\text{PO}_4\;{\longrightarrow}\;2\text{CrPO}_4\;+\;3\text{H}_2$

Determine the limiting reactant.

7. Uranium can be isolated from its ores by dissolving it as UO₂(NO₃)₂, then separating it as solid UO₂(C₂O₄)·3H₂O. Addition of 0.4031 g of sodium oxalate, Na₂C₂O₄, to a solution containing 1.481 g of uranyl nitrate, UO₂(NO₃)₂, yields 1.073 g of solid UO₂(C₂O₄)·3H₂O.

$latex \text{Na}_2\text{C}_2\text{O}_4\;+\;\text{UO}_2(\text{NO}_3)_2\;+\;3\text{H}_2\text{O}\;{\longrightarrow}\;\text{UO}_2(\text{C}_2\text{O}_4){\cdot}3\text{H}_2\text{O}\;+\;2\text{NaNO}_3$

Determine the limiting reactant and the percent yield of this reaction.

8. How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms? 9. Would you agree to buy 1 trillion (1,000,000,000,000) gold atoms for $5? Explain why or why not. Find the current price of gold at http://money.cnn.com/data/commodities/ (1 troy ounce = 31.1 g)

10. The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH₃. What is the limiting reagent?

11. Given the statement “20.0 g of methane is burned in excess oxygen,” is it obvious which reactant is the limiting reagent?

12. Acetylene (C₂H₂) is formed by reacting 7.08 g of C and 4.92 g of H₂.

2 C(s) + H₂(g) $latex \longrightarrow$ C₂H₂(g)

What is the limiting reagent? How much of the other reactant is in excess?

13. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?

14. To form the precipitate PbCl₂, 2.88 g of NaCl and 7.21 g of Pb(NO₃)₂ are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?

15. What is the difference between the theoretical yield and the actual yield?

16. A worker isolates 2.675 g of SiF₄ after reacting 2.339 g of SiO₂ with HF. What are the theoretical yield and the actual yield?

SiO₂(s) + 4 HF(g) $latex \longrightarrow$ SiF₄(g) + 2 H₂O(ℓ)

17. A chemist decomposes 1.006 g of NaHCO₃ and obtains 0.0334 g of Na₂CO₃. What are the theoretical yield and the actual yield?

2 NaHCO₃(s) $latex \longrightarrow$ Na₂CO₃(s) + H₂O(ℓ) + CO₂(g)

18. What is the percent yield in Exercise 16? 19. What is the percent yield in Exercise 17?

 

  Answers

1. The limiting reactant is Cl₂.

2. Percent yield = 31%

3. g CCl₄ $latex \longrightarrow$ mol CCl₄ $latex \longrightarrow$ mol CCl₂F₂ $latex \longrightarrow$ g CCl₂F₂, percent yield = 48.3%

4. percent yield = 91.3%

5. Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6%

6. The conversion needed is $latex \text{mol\;Cr}\;{\longrightarrow}\;\text{mol\;H}_3\text{PO}_4$. Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.

7. Na₂C₂O₄ is the limiting reactant. percent yield = 86.6%

8. Only four molecules can be made.

9. This amount cannot be weighted by ordinary balances and is worthless.

10. Nitrogen is the limiting reagent. 11. Yes; methane is the limiting reagent. 12. C is the limiting reagent; 4.33 g of H₂ are left over. 13. H₂O is the limiting reagent; 25.9 g of P₄O₆ are left over. 14. 6.06 g of PbCl₂ are formed; 0.33 g of NaCl is left over. 15. Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction. 16. theoretical yield = 4.052 g; actual yield = 2.675 g 17. theoretical yield = 0.635 g; actual yield = 0.0334 g 18. 66.02% 19. 5.26%

Glossary

actual yield: amount of product formed in a reaction excess reactant: reactant present in an amount greater than required by the reaction stoichiometry limiting reactant: reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated percent yield: measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield theoretical yield: amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry

]]> 1435 1405 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-5-quantitative-chemical-analysis/ Thu, 12 Apr 2018 02:51:57 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-5-quantitative-chemical-analysis/

Learning Objectives

By the end of this section, you will be able to:

• Describe the fundamental aspects of titrations and gravimetric analysis.
• Perform stoichiometric calculations using typical titration and gravimetric data.

In the 18th century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, K₂CO₃, which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar.

We now know that the effervescence that occurred during this process was due to reaction with acetic acid, CH₃CO₂H, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:

$latex 2\text{CH}_3 \text{CO}_2 \text{H}(aq) + \text{K}_2\text{CO}_3(s) \longrightarrow \text{KCH}_3 \text{CO}_3(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)$

The bubbling was due to the production of CO₂.

The test of vinegar with potassium carbonate is one type of quantitative analysis—the determination of the amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.

Titration

The described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret (Figure 1) to make incremental additions of a solution containing a known concentration of some substance (the titrant) to a sample solution containing the substance whose concentration is to be measured (the analyte). The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration. The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction. The halt of bubble formation in the classic vinegar analysis is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration. Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration’s equivalence point, the volume of titrant actually measured is called the end point. Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.

Example 1

The end point in a titration of a 50.00-mL sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is:

$latex \text{HCl}(aq) + \text{NaOH}(aq) \longrightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$

What is the molarity of the HCl?

 

Solution

As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants provided and requested are expressed as solution concentrations.

For this exercise, the calculation will follow the following outlined steps:

The molar amount of HCl is calculated to be:

$latex 35.23 \;\rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em}\text{mL NaOH} \times \frac{1 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}}{1000 \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\;\text{mL}} \times \frac{0.250 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol NaOH}}{1 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1 \;\text{mol HCl}}{1 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol NaOH}} = \underline{8.80}75 \times 10^{-3} \;\text{mol HCl with 3 sig figs}$

Using the provided volume of HCl solution and the definition of molarity, the HCl concentration is:

$latex \begin{array}{r @{{}={}} l} M & \frac{\text{mol HCl}}{\text{L solution}} \\[1em] M & \frac{\underline{8.80}75 \times 10^{-3} \;\text{mol HCl}}{50.00 \;\text{mL} \times \frac{1 \;\text{L}}{1000 \;\text{mL}}} \\[1em] M & 0.176 \;M \end{array}$

Note: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:

$latex M = \frac{\text{mol solute}}{\text{L solution}} \times \frac{\frac{10^3 \;\text{mmol}}{\text{mol}}}{\frac{10^3 \;\text{mL}}{\text{L}}} = \frac{\text{mmol solute}}{\text{mL solution}}$

Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors:

$latex \frac{35.23 \;\text{mL NaOH} \times \;\frac{0.250 \;\text{mmol NaOH}}{\text{mL NaOH}} \times \frac{1 \;\text{mmol HCl}}{1 \;\text{mmol NaOH}}}{50.00 \;\text{mL solution}} = 0.176 \;M \;\text{HCl}$

 

Test Yourself A 20.00-mL sample of aqueous oxalic acid, H₂C₂O₄, was titrated with a 0.09113-M solution of potassium permanganate.

$latex {2\text{MnO}_4}^{-}(aq) + 5\text{H}_2 \text{C}_2 \text{O}_4(aq) + 6\text{H}^{+}(aq) \longrightarrow 10\text{CO}_2(g) + 2\text{Mn}^{2+}(aq) + 8\text{H}_2 \text{O}(l)$

A volume of 23.24 mL was required to reach the end point. What is the oxalic acid molarity?

  Answer 0.2648 M

Gravimetric Analysis

A gravimetric analysis is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.

The required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water. Also common are gravimetric techniques in which the analyte is subjected to a precipitation reaction of the sort described earlier in this chapter. The precipitate is typically isolated from the reaction mixture by filtration, carefully dried, and then weighed (Figure 2). The mass of the precipitate may then be used, along with relevant stoichiometric relationships, to calculate analyte concentration.

[caption id="" align="aligncenter" width="217"] Figure 2. Precipitate may be removed from a reaction mixture by filtration.[/caption]

Example 2

A 0.4550-g solid mixture containing MgSO₄ is dissolved in water and treated with an excess of Ba(NO₃)₂, resulting in the precipitation of 0.6168 g of BaSO₄.

$latex \text{MgSO}_4(aq) + \text{Ba(NO}_3)_2(aq) \longrightarrow \text{BaSO}_4(s) + \text{Mg(NO}_3)_2(aq)$

What is the concentration (percent) of MgSO₄ in the mixture?

 

Solution

The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO₄ and MgSO₄ through their stoichiometric factor. Once the mass of MgSO₄ is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.

The mass of MgSO₄ that would yield the provided precipitate mass is

$latex 0.6168 \;\rule[0.5ex]{3.75em}{0.1ex}\hspace{-3.75em}\text{g BaSO}_4 \times \frac{1 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol BaSO}_4}{233.391 \;\rule[0.25ex]{2.75em}{0.1ex}\hspace{-2.75em}\text{g BaSO}_4} \times \frac{1 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol MgSO}_4}{1\;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol BaSO}_4} \times \frac{120.369 \;\text{g MgSO}_4}{1 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol MgSO}_4} = 0.\underline{3181}08 \;\text{g MgSO}_4 \text{with 4 sig figs}$

The concentration of MgSO₄ in the sample mixture is then calculated to be

$latex \text{percent MgSO}_4 = \frac{\text{mass MgSO}_4}{\text{mass sample}} \times 100\% $ $latex \frac{\underline{0.3181}08 \;\text{g}}{0.4550 \;\text{g}} \times 100\% = 69.91\% $

 

Test Yourself What is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess Ag⁺?

$latex \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) \longrightarrow \text{AgCl}(s)$

Answer

23.76%

The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure 3). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.

Example 3

Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO₂ and 0.00161 g of H₂O. What is the empirical formula of polyethylene?

 

Solution The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

$latex \text{C}_\text{x} \text{H}_\text{y}(s) + \text{excess O}_2(g) \longrightarrow x\text{CO}_2(g) + \frac{y}{2}\text{H}_2\text{O}(g)$

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:

$latex \begin{array} {r @{{}={}} l} \text{mol C} & 0.00394 \;\text{g CO}_2 \times \frac{1 \;\text{mol CO}_2}{44.010 \;\text{g/mol}} \times \frac{1 \;\text{mol C}}{1 \;\text{mol CO}_2} = \underline{8.95}3 \times 10^{-5} \;\text{mol C with 3 sig figs} \\[1em] \text{mol H} & 0.00161 \;\text{g H}_2 \text{O} \times \frac{1 \;\text{mol H}_2 \text{O}}{18.0153 \;\text{g/mol}} \times \frac{2 \;\text{mol H}}{1 \;\text{mol H}_2 \text{O}} = \underline{1.78}74 \times 10^{-4} \;\text{mol H with 3 sig figs} \end{array}$

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is

$latex \frac{\text{mol H}}{\text{mol C}} = \frac{\underline{1.78}74 \times 10^{-4}\;\text{mol H}}{\underline{8.95}3 \times 10^{-5}\;\text{mol C}} = \frac{2 \;\text{mol H}}{1 \;\text{mol C}}$

and the empirical formula for polyethylene is CH₂.

 

Test Yourself A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO₂ and 0.00148 g of H₂O in a combustion analysis. What is the empirical formula for polystyrene?

  Answer CH

Key Concepts and Summary

The stoichiometry of chemical reactions may serve as the basis for quantitative chemical analysis methods. Titrations involve measuring the volume of a titrant solution required to completely react with a sample solution. This volume is then used to calculate the concentration of analyte in the sample using the stoichiometry of the titration reaction. Gravimetric analysis involves separating the analyte from the sample by a physical or chemical process, determining its mass, and then calculating its concentration in the sample based on the stoichiometry of the relevant process. Combustion analysis is a gravimetric method used to determine the elemental composition of a compound by collecting and weighing the gaseous products of its combustion.

Exercises

1. Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain? 2. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO₃)₂ solution. $latex 2\text{Cl}^{-}(aq) + \text{Hg(NO}_3)_2(aq) \longrightarrow {2\text{NO}_3}^{-}(aq) + \text{HgCl}_2(s)$

What is the Cl⁻ concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 8.25 × 10⁻⁴M Hg(NO₃)₂(aq) to reach the end point?

3. A sample of gallium bromide, GaBr₂, weighing 0.165 g was dissolved in water and treated with silver nitrate, AgNO₃, resulting in the precipitation of 0.299 g AgBr. Use these data to compute the %Ga (by mass) GaBr₂. 4. A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B₂O₃. What are the empirical and molecular formulas of the compound? 5. What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate? $latex \text{NaHCO}_3(aq) + \text{HCl}(aq) \longrightarrow \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)$ 6. What volume of a 0.3300-M solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 M oxalic acid? $latex \text{C}_2 \text{O}_4 \text{H}_2(aq) + 2\text{NaOH}(aq) \longrightarrow \text{Na}_2 \text{C}_2 \text{O}_4(aq) + 2\text{H}_2 \text{O}(l)$ 7. A sample of solid calcium hydroxide, Ca(OH)₂, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 × 10⁻²M HCl requires 36.6 mL of the acid to reach the end point. $latex \text{Ca(OH)}_2(aq) + 2\text{HCl}(aq) \longrightarrow \text{CaCl}_2(aq) + 2\text{H}_2 \text{O}(l) $

What is the molarity?

8. How many milliliters of a 0.1500-M solution of KOH will be required to titrate 40.00 mL of a 0.0656-M solution of H₃PO₄? $latex \text{H}_3\text{PO}_4(aq) + 2\text{KOH}(aq) \longrightarrow \text{K}_2 \text{HPO}_4(aq) + 2\text{H}_2 \text{O}(l)$ 9. The reaction of WCl₆ with Al at ~400 °C gives black crystals of a compound containing only tungsten and chlorine. A sample of this compound, when reduced with hydrogen, gives 0.2232 g of tungsten metal and hydrogen chloride, which is absorbed in water. Titration of the hydrochloric acid thus produced requires 46.2 mL of 0.1051 M NaOH to reach the end point. What is the empirical formula of the black tungsten chloride?   Answers

1. 3.4 × 10⁻³M H₂SO₄

2. 9.6 × 10⁻³M Cl⁻

3. 22.4%

4. The empirical formula is BH₃. The molecular formula is B₂H₆.

5. 49.6 mL

6. 13.64 mL

7. 1.22 M

8. 34.99 mL KOH

9. The empirical formula is WCl₄.

Glossary

analyze: chemical species of interest buret: device used for the precise delivery of variable liquid volumes, such as in a titration analysis combustion analysis: gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products end point: measured volume of titrant solution that yields the change in sample solution appearance or other property expected for stoichiometric equivalence (see equivalence point) equivalence point: volume of titrant solution required to react completely with the analyte in a titration analysis; provides a stoichiometric amount of titrant for the sample’s analyte according to the titration reaction gravimetric analysis: quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and/or sample indicator: substance added to the sample in a titration analysis to permit visual detection of the end point quantitative analysis: the determination of the amount or concentration of a substance in a sample titrant: solution containing a known concentration of substance that will react with the analyte in a titration analysis titration analysis: quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample

]]> 1443 1405 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/6-4-electronic-structure-of-atoms-electron-configurations/ Thu, 12 Apr 2018 02:52:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/6-4-electronic-structure-of-atoms-electron-configurations/

Learning Objectives

By the end of this section, you will be able to:

• Derive the predicted ground-state electron configurations of atoms
• Identify and explain exceptions to predicted electron configurations for atoms and ions
• Relate electron configurations to element classifications in the periodic table

Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom.

Electronic Structure of Atoms

The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information (Figure 1):

1. The number of the principal quantum shell, n,
2. The letter that designates the orbital type also called the subshell, and
3. A superscript number that designates the number of electrons in that particular subshell.

For example, the notation 2p⁴ (read "two–p–four") indicates four electrons in a p subshell with a principal quantum number (n) of 2. The notation 3d⁸ (read "three–d–eight") indicates eight electrons in the d subshell of the principal shell for which n = 3.

The Aufbau Principle

To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure 4 in section 7.3), subject to the limitations imposed by the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure 2 illustrates the traditional way to remember the filling order for atomic orbitals. Since the arrangement of the periodic table is based on the electron configurations, Figure 3 and Figure 4 provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3d orbitals.

 

We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to Figure 2 or Figure 3, we would expect to find the electron in the 1s orbital. By convention, spin up = + \frac{1}{2}$ value is usually filled first. The electron configuration and the orbital box diagram are:

Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The two electrons will occupy the same orbital but they will have different spin states, one will be spin-up (˦) and the other spin-down (˨). This is in accord with the Pauli exclusion principle.  For orbital diagrams, this means two half-arrows go in each box (representing two electrons in each orbital) and the half-arrows must point in opposite directions (representing paired spins). The electron configuration and orbital box diagram of helium are:

The n = 1 shell is completely filled in a helium atom.

The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1s orbital.  The remaining electron must occupy the orbital of next lowest energy, the 2s orbital (Figure 2 or Figure 3). Thus, the electron configuration and orbital box diagram of lithium are:

An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2s orbital.

An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2p orbital. There are three degenerate 2p orbitals, meaning they are equal in energy, and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling.

In the orbital box diagrams, notice the space between the box for the 1s and 2s orbitals - space between boxes are used to indicate a difference in energy.  Therefore, a " lack of space" between boxes, indicate the orbitals are degenerate, meaning equal in energy.

Carbon (atomic number 6) has six electrons. Four of them fill the 1s and 2s orbitals. The remaining two electrons occupy the 2p subshell. We now have a choice of filling one of the 2p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2p orbitals occupy different p-orbitals - this minimizes electron-electron repulsion within the atom.  The electron configuration and orbital box diagram for carbon are:

Nitrogen (atomic number 7) fills the 1s and 2s subshells and has one electron in each of the three 2p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital box diagrams of these four elements are:

The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3s orbital, giving a 1s²2s²2p⁶3s¹ configuration. The electrons occupying the outermost shell orbital(s) (highest value of n) are called outer electrons, and those occupying the inner shell orbitals are called core electrons or inner electrons (Figure 4).  Valence electrons are outer electrons plus any electrons found in partially filled d or f orbitals.  Often valence electron are the same as the outer electrons.  But there are examples where there is a difference.  Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1s²2s²2p⁶) and our abbreviated or condensed electron configuration is [Ne]3s¹.

[caption id="" align="aligncenter" width="650"] Figure 4. A condensed electron configuration (right) replaces the core electrons with the noble gas symbol whose configuration matches the core electron configuration of the other element.[/caption]

Similarly, the condensed electron configuration of lithium can be represented as [He]2s¹, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells.

$latex \begin{array}{l} \text{Li}: [\text{He}] \;2s^1 \\ \text{Na}: [\text{Ne}] \;3s^1 \end{array}$

The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s² configuration, is analogous to its family member beryllium, [He]2s². Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the condensed electron configuration [Ne]3s²3p¹, is analogous to its family member boron, [He]2s²2p¹.

The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure 5 shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements.

When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3d level but is, instead, added to the 4s level (Figure 5). Thus, potassium has an electron configuration of [Ar]4s¹. Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4s subshell and calcium has an electron configuration of [Ar]4s². This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium.

Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3d subshell. This subshell is filled to its capacity with 10 electrons. The 4p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the (n – 1) shell next to the n shell to bring that (n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons are successively added to the (n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.

Example 1

What is the electron configuration and orbital box diagram for a phosphorus atom?

 

Solution The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . . The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons:

The last electron added is a 3p electron.

 

Test Yourself

Identify the atoms from the condensed electron configurations given:

a) [Ar]4s²3d⁵

b) [Kr]5s²4d¹⁰5p⁶

  Answers a) Mn          b) Xe

Example 2

a) What is the electron configuration for Na, which has 11 electrons?

b) What is the predicted electron configuration for Sn, which has 50 electrons?  

Solution

a) The first two electrons occupy the 1s subshell. The next two occupy the 2s subshell, while the next six electrons occupy the 2p subshell. This gives us 10 electrons so far, with 1 electron left. This last electron goes into the n = 3 shell, s subshell. Thus, the electron configuration of Na is 1s²2s²2p⁶3s¹.

b) We will follow the chart in Figure 2 until we can accommodate 50 electrons in the subshells in the proper order: Sn: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p²

Verify by adding the superscripts, which indicate the number of electrons: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 2 = 50, so we have placed all 50 electrons in subshells in the proper order.

 

Test Yourself

a) What is the electron configuration for Mg, which has 12 electrons?

b) What is the electron configuration for Ba, which has 56 electrons?  

Answer

a) 1s²2s²2p⁶3s²          b) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² 

Example 3

What is the abbreviated electron configuration for P, which has 15 electrons?

 

Solution

With 15 electrons, the electron configuration of P is

P: 1s²2s²2p⁶3s²3p³

The first immediate noble gas is Ne, which has an electron configuration of 1s²2s²2p⁶. Using the electron configuration of Ne to represent the first 10 electrons, the abbreviated electron configuration of P is

P: [Ne]3s²3p³  

Test Yourself

What is the abbreviated electron configuration for Rb, which has 37 electrons?

 

Answer

[Kr]5s¹

The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure 2 or Figure 3. For instance, the electron configurations (shown in Figure 5) of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.

In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4s into the 3d orbital to gain the extra stability of a half-filled 3d subshell (in Cr) or a filled 3d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5s²4d³. Experimentally, we observe that its ground-state electron configuration is actually [Kr]5s¹4d⁴. We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5s orbital are larger than the gap in energy between the 5s and 4d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells.

Electron Configurations and the Periodic Table

As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure 5), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements.

Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react.

It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure 5, which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure 5 show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell, or highest energy level orbitals of an atom.

1. Main group elements (sometimes called representative elements) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure 5. This category includes all the nonmetallic elements, as well as many metals and the intermediate semimetallic elements. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar]4s²3d¹⁰4p¹, which contains three valence electrons (in bold). The completely filled d orbitals count as core, not valence, electrons. 2. Transition elements or transition metals. These are metallic elements in which the last electron added enters a d orbital. The valence electrons are the electrons in the outermost shell and also include any electrons in partially filled d or f orbitals, as these electrons are also very reactive and have a higher energy despite their lower shell value.  The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure 5) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure 5), and we will adopt this usage in this textbook. 3. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure 5. The valence shells of the inner transition elements consist of the (n – 2)f, the (n – 1)d, and the ns subshells. There are two inner transition series: a) The lanthanide series: lanthanide (La) through lutetium (Lu) b) The actinide series: actinide (Ac) through lawrencium (Lr)

Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons.

Electron Configurations of Ions

We have seen that ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s  orbital are easier to remove than the d  or f  electrons, and so the  highest  ns  electrons are lost, and then the (n – 1)d  or  (n – 2)f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle.

Example 4

What is the electron configuration and orbital diagram of:

a) Na⁺         b) P^3–         c) Al²⁺         d) Fe²⁺         e) Sm³⁺

 

Solution First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable.

Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals.

a) Na: 1s²2s²2p⁶3s¹.

Sodium cation loses one electron, so Na⁺: 1s²2s²2p⁶3s¹ = Na⁺: 1s²2s²2p⁶.

b) P: 1s²2s²2p⁶3s²3p³.

Phosphorus trianion gains three electrons, so P³⁻: 1s²2s²2p⁶3s²3p⁶.

c) Al: 1s²2s²2p⁶3s²3p¹.

Aluminum dication loses two electrons Al²⁺: 1s²2s²2p⁶3s²3p¹ =

Al²⁺: 1s²2s²2p⁶3s¹.

d) Fe: 1s²2s²2p⁶3s²3p⁶4s²3d⁶.

Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe²⁺: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ = 1s²2s²2p⁶3s²3p⁶3d⁶.

e). Sm: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f⁶.

Samarium trication loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital. Sm³⁺: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f⁶ = 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶4f⁵.  

Test Yourself Which ion with a +2 charge has the electron configuration 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d⁵? Which ion with a +3 charge has this configuration?

Answers Tc²⁺, Ru³⁺

Food and Drink App: Artificial Colors

The color of objects comes from a different mechanism than the colors of neon and other discharge lights. Although colored lights produce their colors, objects are colored because they preferentially reflect a certain color from the white light that shines on them. A red tomato, for example, is bright red because it reflects red light while absorbing all the other colors of the rainbow.

Many foods, such as tomatoes, are highly colored; in fact, the common statement “you eat with your eyes first” is an implicit recognition that the visual appeal of food is just as important as its taste. But what about processed foods?

Many processed foods have food coloring added to them. There are two types of food coloring: natural and artificial. Natural food coloring include caramelized sugar for brown; annatto, turmeric, and saffron for various shades of orange or yellow; betanin from beets for purple; and even carmine, a deep red dye that is extracted from the cochineal, a small insect that is a parasite on cacti in Central and South America. (That’s right: you may be eating bug juice!)

Some coloring agents are artificial. In the United States, the Food and Drug Administration currently approves only seven compounds as artificial coloring in food, beverages, and cosmetics:

1. FD&C Blue #1: Brilliant Blue FCF
2. FD&C Blue #2: Indigotine
3. FD&C Green #3: Fast Green FCF
4. RD&C Red #3: Erythrosine
5. FD&C Red #40: Allura Red AC
6. FD&C Yellow #5: Tartrazine
7. FD&C Yellow #6: Sunset Yellow FCF

Lower-numbered colors are no longer on the market or have been removed for various reasons. Typically, these artificial coloring agents are large molecules that absorb certain colors of light very strongly, making them useful even at very low concentrations in foods and cosmetics. Even at such low amounts, some critics claim that a small portion of the population (especially children) is sensitive to artificial coloring and urge that their use be curtailed or halted. However, formal studies of artificial coloring and their effects on behaviour have been inconclusive or contradictory. Despite this, most people continue to enjoy processed foods with artificial coloring like those shown in Figure 6.

[caption id="attachment_4707" align="aligncenter" width="600"] Figure 6. Artificial food coloring are found in a variety of food products, such as processed foods, candies, and egg dyes. Even pet foods have artificial food coloring in them, although it’s likely that the animal doesn’t care! Source: Photo courtesy of Matthew Bland, http://www.flickr.com/photos/matthewbland/3111904731.[/caption]

Key Concepts and Summary

The relative energy of the subshells determine the order in which atomic orbitals are filled (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals).

Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements (s and p orbitals), transition elements (d orbitals), and inner transition elements (f orbitals).

Exercises

1. Read the labels of several commercial products and identify monatomic ions of at least six main group elements contained in the products. Write the complete electron configurations of these cations and anions. 2. Using complete subshell notation (1s²2s²2p⁶, and so forth), predict the electron configuration of each of the following atoms:

a) N      b) Si      c) Fe      d) Te      e) Tb

3. What additional information do we need to answer the question “Which ion has the electron configuration 1s²2s²2p⁶3s²3p⁶”? 4. Use an orbital diagram to describe the electron configuration of the valence shell of each of the following atoms:

a) N      b) Si      c) Fe      d) Te      e) Mo

5. Which atom has the electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d²? 6. Which ion with a +1 charge has the electron configuration 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶? Which ion with a –2 charge has this configuration? 7. Which of the following has two unpaired electrons?

a) Mg      b) Si      c) S      d) Both Mg and S      e) Both Si and S.

8. Which atom would be expected to have a half-filled 4s subshell? 9. Thallium was used as a poison in the Agatha Christie mystery story “The Pale Horse.” Thallium has two possible cationic forms, +1 and +3. The +1 compounds are the more stable. Write the electron structure of the +1 cation of thallium. 10. Cobalt–60 and iodine–131 are radioactive isotopes commonly used in nuclear medicine. How many protons, neutrons, and electrons are in atoms of these isotopes? Write the complete electron configuration for each isotope. 11. How many subshells are completely filled with electrons for Na? How many subshells are unfilled? 12. What is the maximum number of electrons in the entire n = 2 shell? 13. Write the complete electron configuration for each atom. a)  Si, 14 electrons         b)  Sc, 21 electrons 14.  Write the complete electron configuration for each atom.

a)  Cd, 48 electrons         b)  Mg, 12 electrons 15. Write the abbreviated electron configuration for each atom in Exercise 13.

16.  Write the abbreviated electron configuration for each atom in Exercise 17. Where on the periodic table are s subshells being occupied by electrons? 18. In what block is Ra found? 19. What are the valence shell electron configurations of the elements in the second column of the periodic table? 20. What are the valence shell electron configurations of the elements in the first column of the p block? 21. From the element’s position on the periodic table, predict the electron configuration of each atom. a)  Sr      b)  S 22. From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  V      b)  Ar

23. From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  Ge      b)  C

  Answers

1. For example, Na⁺: 1s²2s²2p⁶;     Ca²⁺: 1s²2s²2p⁶;

Sn²⁺: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²;      F^–: 1s²2s²2p⁶; O^2–: 1s²2s²2p⁶;      Cl^–: 1s²2s²2p⁶3s²3p⁶.

2. a) 1s²2s²2p³       b) 1s²2s²2p⁶3s²3p²       c) 1s²2s²2p⁶3s²3p⁶4s²3d⁶

d) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁴ e) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f⁹

3. The charge on the ion.

4. a)

b) c) d) e)

5. Zr

6. Rb⁺, Se²⁻

7. Although both b) and c) are correct, e) encompasses both and is the best answer.

8. K

9. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶6s²4f¹⁴5d¹⁰

10. Co has 27 protons, 27 electrons, and 33 neutrons: 1s²2s²2p⁶3s²3p⁶4s²3d⁷.

I has 53 protons, 53 electrons, and 78 neutrons: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁵. 11. Three subshells (1s, 2s, 2p) are completely filled, and one shell (3s) is partially filled. 12. 8 electrons 13. a)  1s²2s²2p⁶3s²3p²       b)  1s²2s²2p⁶3s²3p⁶4s²3d¹ 14. a)  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰      b)  1s²2s²2p⁶3s² 15. a)  [Ne]3s²3p²      b)  [Ar]4s²3d¹ 16. a)  [Kr]5s²4d¹⁰      b)  [Ne]3s² 17. the first two columns 18. the s block 19. ns² 20. ns²np¹ 21. a)  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²      b)  1s²2s²2p⁶3s²3p⁴ 22. a)  1s²2s²2p⁶3s²3p⁶4s²3d³      b)  1s²2s²2p⁶3s²3p⁶ 23. a)  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p²      b)  1s²2s²2p²

Glossary

Aufbau principle: procedure in which the electron configuration of the elements is determined by “building” them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time core electron: electron in an atom that occupies the orbitals of the inner shells electron configuration: electronic structure of an atom in its ground state given as a listing of the orbitals occupied by the electrons Hund’s rule: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin orbital diagram: pictorial representation of the electron configuration showing each orbital as a box and each electron as a half-arrow valence electrons: are the electrons in the outermost shell and also include any electrons in partially filled d or f orbitals, as these electrons are also very reactive and have a higher energy despite their lower shell value of a ground-state atom; they determine how an element reacts valence shell: outermost shell of electrons in a ground-state atom; for main group elements, the orbitals with the highest n level (s and p subshells) are in the valence shell, while for transition metals, the highest energy s and d subshells make up the valence shell and for inner transition elements, the highest s, d, and f subshells are included

]]> 1528 2362 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-7/ Thu, 12 Apr 2018 02:52:27 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-7/ It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds.

Graphite is brittle, whereas diamond is the hardest natural material known on Earth. Yet both are just pure carbon. What is special about this element that makes these two forms of carbon so different?

Bonds. Chemical bonds!

In graphite, each carbon is bonded to three other carbons to form a flat sheets of carbon lattices which are form layers.  These layers, called graphene, are attracted to each other through Van der Waals forces, a type of intermolecular force.  Graphite is brittle because these intermolecular forces are relatively weak.

In a perfect diamond crystal, each C atom makes four connections—bonds—to four other C atoms in a three-dimensional matrix. Four is the greatest number of bonds that is commonly made by atoms, so C atoms maximize their interactions with other atoms. This three-dimensional array of connections extends throughout the diamond crystal, making it essentially one large molecule. Breaking a diamond means breaking every bond at once.

[caption id="attachment_3833" align="aligncenter" width="458"] Figure 1. Diamond and graphite samples with their respective structures. The bottom right formation of carbon is what is known as "graphene," characterized by infinite, single atom sheets of carbon. By User:Itub (Self-made derivative work (see below)) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/)], via Wikimedia Commons[/caption]

It was not until 1985 that a new form of carbon was recognized: buckminsterfullerene, commonly known as a “buckyball.” This molecule was named after the architect and inventor R. Buckminster Fuller (1895–1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, C₆₀, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory—the topic of this chapter—which explains how individual atoms connect to form more complex structures. [caption id="attachment_3834" align="aligncenter" width="506"] Figure 2. Eight allotropes of carbon: a) diamond, b) graphite, c) Ionsdaleite, d) C₆₀ buckminsterfullerene, e) C₅₄₀, Fullerite, f) C₇₀, g) amorphous carbon, and h) single-walled carbon nanotube.By Created by Michael Ströck (mstroeck) (Created by Michael Ströck (mstroeck)) [GFDL (http://www.gnu.org/copyleft/fdl.html), CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/) or CC BY-SA 2.5 (https://creativecommons.org/licenses/by-sa/2.5)], via Wikimedia Commons[/caption]

How do atoms make compounds?

Bonds. Chemical bonds!

Typically they join together in such a way that they lose their identities as elements and adopt a new identity as a compound. These joins are called chemical bonds. But how do atoms join together? Ultimately, it all comes down to electrons. Before we discuss how electrons interact, we need to introduce a tool to simply illustrate electrons in an atom.

]]> 1540 1538 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/7-1-ionic-bonding/ Thu, 12 Apr 2018 02:52:28 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/7-1-ionic-bonding/

Learning Objectives

By the end of this section, you will be able to:

• Explain the formation of cations, anions, and ionic compounds
• Predict the charge of common metallic and nonmetallic elements, and write their electron configurations

As you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell.

Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason—the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.

Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl₂, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure 1). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.

The Formation of Ionic Compounds

Binary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table.

As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, Al₂O₃, indicates that this ionic compound contains two aluminum cations, Al³⁺, for every three oxide anions, O²⁻ [thus, (2 × +3) + (3 × –2) = 0].

It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride (NaCl) “molecule” because there is not a single ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic—the same in all directions—meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, three-dimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of Na⁺ cations and Cl^– anions (Figure 2).

[caption id="" align="aligncenter" width="519"] Figure 2. The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is “bonded” to all of the surrounding ions—six in this case.[/caption]

The strong electrostatic attraction between Na⁺ and Cl^– ions holds them tightly together in solid NaCl. It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous Na⁺ and Cl^– ions:

$latex \text{NaCl}(s) \longrightarrow \text{Na}^{+}(g) + \text{Cl}^{-}(g) \;\;\;\;\; \Delta H = 769 \;\text{kJ}$

Electronic Structures of Cations

When forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed. For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of 1s²2s²2p⁶3s²3p⁶4s². When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a 2+ charge, and an electron configuration of 1s²2s²2p⁶3s²3p⁶. The Ca²⁺ ion is therefore isoelectronic with the noble gas Ar.

For groups 12–17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full d subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms 3+ ions (Al³⁺).

Exceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions Tl³⁺, Sn⁴⁺, Pb⁴⁺, and Bi⁵⁺, a partial loss of these atoms’ valence shell electrons can also lead to the formation of Tl⁺, Sn²⁺, Pb²⁺, and Bi³⁺ ions. The formation of these 1+, 2+, and 3+ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence s-electron pair for atoms of the heavy elements of groups 13, 14, and 15. Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, $latex \text{Hg}_2^{\;\;2+}$ (an ion formed from two mercury atoms, with an Hg-Hg bond), in addition to the expected monatomic ion Hg²⁺ (formed from only one mercury atom).

Transition and inner transition metal elements behave differently than main group elements. Most transition metal cations have 2+ or 3+ charges that result from the loss of their outermost s electron(s) first, sometimes followed by the loss of one or two d electrons from the next-to-outermost shell. For example, iron (1s²2s²2p⁶3s²3p⁶3d⁶4s²) forms the ion Fe²⁺ (1s²2s²2p⁶3s²3p⁶3d⁶) by the loss of the 4s electron and the ion Fe³⁺ (1s²2s²2p⁶3s²3p⁶3d⁵) by the loss of the 4s electron and one of the 3d electrons. Although the d orbitals of the transition elements are—according to the Aufbau principle—the last to fill when building up electron configurations, the outermost s electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a 3+ charge, resulting from the loss of their outermost s electrons and a d or f electron.

Example 1

There are at least 14 elements categorized as “essential trace elements” for the human body. They are called “essential” because they are required for healthy bodily functions, “trace” because they are required only in small amounts, and “elements” in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as Cr³⁺ and Zn²⁺. Write the electron configurations of these cations.

 

Solution First, write the electron configuration for the neutral atoms:

Zn: [Ar]3d¹⁰4s²

Cr: [Ar]3d⁵4s¹

Next, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the s orbital first and then from the d orbital. For the p-block elements, electrons are removed from the p orbitals and then from the s orbital. Zinc is a member of group 12, so it should have a charge of 2+, and thus loses only the two electrons in its s orbital. Chromium is a transition element and should lose its s electrons and then its d electrons when forming a cation. Thus, we find the following electron configurations of the ions:

Zn²⁺: [Ar]3d¹⁰

Cr³⁺: [Ar]3d³

 

Test Yourself Potassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements.

  Answers K⁺: [Ar], Mg²⁺: [Ne]

Electronic Structures of Anions

Most monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer s and p orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the s and p orbitals of the parent atom. Oxygen, for example, has the electron configuration 1s²2s²2p⁴, whereas the oxygen anion has the electron configuration of the noble gas neon (Ne), 1s²2s²2p⁶. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of 2– (O^2–).

Example 2

Selenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions.

 

Solution Se^2–: [Ar]3d¹⁰4s²4p⁶

I^–: [Kr]4d¹⁰5s²5p⁶

 

Test Yourself Write the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion.

  Answers P: [Ne]3s²3p³; P^3–: [Ne]3s²3p⁶

Key Concepts and Summary

Atoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells.

Exercises

1. Does a cation gain protons to form a positive charge or does it lose electrons? 2. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co? 3. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:

a) P         b) Mg         c) Al         d) O         e) Cl         f) Cs

4. Write the electron configuration for each of the following ions:

a) As^3–         b) I^–         c) Be²⁺         d) Cd²⁺         e) O^2–         f) Ga³⁺         

g) Li⁺         h) N^3–         i) Sn²⁺         j) Co²⁺         k) Fe²⁺         l) As³⁺

5. Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element:

a) Al         b) Br         c) Sr         d) Li         e) As         f) S

  Answers

1. The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.

2. P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals.

3. a) P^3–      b) Mg²⁺      c) Al³⁺      d) O^2–      e) Cl^–      f) Cs⁺

4. a) [Ar]4s²3d¹⁰4p⁶      b) [Kr]4d¹⁰5s²5p⁶        c) 1s²         d) [Kr]4d¹⁰       e) [He]2s²2p⁶       f) [Ar]3d¹⁰

g) 1s²        h) [He]2s²2p⁶        i) [Kr]4d¹⁰5s²        j) [Ar]3d⁷        k) [Ar]3d⁶       l) [Ar]3d¹⁰4s²

5. a) 1s²2s²2p⁶3s²3p¹;     Al³⁺: 1s²2s²2p⁶

b) 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁵;     1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶ c) 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²;    Sr²⁺: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶ d) 1s²2s¹;    Li⁺: 1s² e) 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p³;    1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶ f) 1s²2s²2p⁶3s²3p⁴;    1s²2s²2p⁶3s²3p⁶

Glossary

inert pair effect: tendency of heavy atoms to form ions in which their valence s electrons are not lost ionic bond: strong electrostatic force of attraction between cations and anions in an ionic compound

]]> 1543 1538 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/7-2-covalent-bonding/ Thu, 12 Apr 2018 02:52:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/7-2-covalent-bonding/

Learning Objectives

By the end of this section, you will be able to:

• Describe the formation of covalent bonds
• Define electronegativity and assess the polarity of covalent bonds

In ionic compounds, electrons are transferred between atoms of different elements to form ions. But this is not the only way that compounds can be formed. Atoms can also make chemical bonds by sharing electrons equally between each other. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an H₂ molecule; each hydrogen atom in the H₂ molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He.

Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state.

Formation of Covalent Bonds

Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, H₂, contains a covalent bond between its two hydrogen atoms. Figure 1 illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the x-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the x-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.

It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of H₂, the covalent bond is very strong; a large amount of energy, 436 kJ, must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:

$latex \text{H}_2(g) \longrightarrow 2\text{H}(g) \;\;\;\;\; \Delta H = 436\;\text{kJ}$

Conversely, the same amount of energy is released when one mole of H₂ molecules forms from two moles of H atoms:

$latex 2\text{H}(g) \longrightarrow \text{H}_2(g) \;\;\;\;\; \Delta H = -436 \;\text{kJ}$

Pure vs. Polar Covalent Bonds

If the atoms that form a covalent bond are identical, as in H₂, Cl₂, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in pure covalent bonds have an equal probability of being near each nucleus.

In the case of Cl₂, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond:

$latex \text{Cl} + \text{Cl} \longrightarrow \text{Cl}_2$

The total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, Cl₂ also features a pure covalent bond.

When the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally. Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge. Figure 2 shows the distribution of electrons in the H–Cl bond. Note that the shaded area around Cl is much larger than it is around H. Compare this to Figure 1, which shows the even distribution of electrons in the H₂ nonpolar bond.

We sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter “delta,” δ, with a plus sign or minus sign to indicate whether the atom has a partial positive charge (δ+) or a partial negative charge (δ–). This symbolism is shown for the H–Cl molecule in Figure 2.

[caption id="" align="aligncenter" width="328"] Figure 2. (a) The distribution of electron density in the HCl molecule is uneven. The electron density is greater around the chlorine nucleus. The small, black dots indicate the location of the hydrogen and chlorine nuclei in the molecule. (b) Symbols δ+ and δ– indicate the polarity of the H–Cl bond.[/caption]

Electronegativity

Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.

Figure 3 shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling (Figure 4). In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as XeO₂ do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.)

Electronegativity versus Electron Affinity

We must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in kJ/mol. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4.

Linus Pauling

Linus Pauling, shown in Figure 4, is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.

[caption id="" align="aligncenter" width="325"] Figure 4. Linus Pauling (1901–1994) made many important contributions to the field of chemistry. He was also a prominent activist, publicizing issues related to health and nuclear weapons.[/caption]

Pauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease—the presence of a genetically inherited abnormal protein in the blood—and paved the way for the field of molecular genetics. His work was also pivotal in curbing the testing of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk.

Electronegativity and Bond Type

The absolute value of the difference in electronegativity (ΔEN) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds H–H, H–Cl, and Na–Cl are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 5 shows the relationship between electronegativity difference and bond type.

[caption id="" align="aligncenter" width="1300"] Figure 5. As the electronegativity difference increases between two atoms, the bond becomes more ionic.[/caption]

A rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure 5. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in NH₃ a difference of 0.9, yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in MnI₂ have a difference of 1.0, yet both of these substances form ionic compounds.

The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.

Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as OH^–, NO₃⁻, and NH₄⁺, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, KNO₃, contains the K⁺ cation and the polyatomic NO₃⁻ anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions K⁺ and NO₃⁻, as well as covalent between the nitrogen and oxygen atoms in NO₃⁻.

Example 1

Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Figure 3, arrange the following covalent bonds—all commonly found in amino acids—in order of increasing polarity. Then designate the positive and negative atoms using the symbols δ+ and δ–:

C–H, C–N, C–O, N–H, O–H, S–H

 

Solution The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the δ– designation is the more electronegative of the two. Table 1 shows these bonds in order of increasing polarity.

Bond	ΔEN	Polarity
C–H	0.4	$latex \overset{\delta -}{\text{C}} - \overset{\delta +}{\text{H}}$
S–H	0.4	$latex \overset{\delta -}{\text{S}} - \overset{\delta +}{\text{H}}$
C–N	0.5	$latex \overset{\delta +}{\text{C}} - \overset{\delta -}{\text{N}}$
N–H	0.9	$latex \overset{\delta -}{\text{N}} - \overset{\delta +}{\text{H}}$
C–O	1.0	$latex \overset{\delta +}{\text{C}} - \overset{\delta -}{\text{O}}$
O–H	1.4	$latex \overset{\delta -}{\text{O}} - \overset{\delta +}{\text{H}}$
Table 1. Bond Polarity and Electronegativity Difference

 

Test Yourself Silicones are polymeric compounds containing, among others, the following types of covalent bonds: Si–O, Si–C, C–H, and C–C. Using the electronegativity values in Figure 3, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols δ+ and δ–.

  Answer

Bond	Electronegativity Difference	Polarity
C–C	0.0	nonpolar
C–H	0.4	$latex \overset{\delta -}{\text{C}} - \overset{\delta +}{\text{H}}$
Si–C	0.7	$latex \overset{\delta +}{\text{Si}} - \overset{\delta -}{\text{C}}$
Si–O	1.7	$latex \overset{\delta +}{\text{Si}} - \overset{\delta -}{\text{O}}$
Table 2.

Key Concepts and Summary

Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.

Exercises

1. Why is it incorrect to speak of a molecule of solid NaCl? 2. Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table:

a) Cl₂CO         b) MnO         c) NCl₃         d) CoBr₂         e) K₂S         f) CO

g) CaF₂              h) HI              i) CaO         j) IBr            k) CO₂

3. From its position in the periodic table, determine which atom in each pair is more electronegative:

a) Br or Cl         b) N or O         c) S or O         d) P or S

e) Si or N           f) Ba or P         g) N or K

4. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:

a) C, F, H, N, O            b) Br, Cl, F, H, I          c) F, H, O, P, S

d) Al, H, Na, O, P         e) Ba, H, N, O, As

5. Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom? 6. Which is the most polar bond?

a) C–C         b) C–H         c) N–H         d) O–H         e) Se–H

7. Identify the more polar bond in each of the following pairs of bonds:

a) HF or HCl         b) NO or CO         c) SH or OH         d) PCl or SCl

e) CH or NH         f) SO or PO           g) CN or NN

 

Answers

1. NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.

2. ionic: b), d), e), g), and i);      covalent: a), c), f), h), j), and  k)

3. a) Cl       b) O        c) O       d) S       e) N         f) P       g) N

4. a) H, C, N, O, F       b) H, I, Br, Cl, F       c) H, P, S, O, F       d) Na, Al, H, P, O       e) Ba, H, As, N, O

5. N, O, F, and Cl

6.  O-H

7. a) HF       b) CO       c) OH       d) PCl       e) NH       f) PO        g) CN

Glossary

bond length: distance between the nuclei of two bonded atoms at which the lowest potential energy is achieved covalent bond: bond formed when electrons are shared between atoms electronegativity: tendency of an atom to attract electrons in a bond to itself polar covalent bond: covalent bond between atoms of different electronegativities; a covalent bond with a positive end and a negative end pure covalent bond: (also, nonpolar covalent bond) covalent bond between atoms of identical electronegativities

]]> 1549 1538 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/7-4-formal-charges-and-resonance/ Thu, 12 Apr 2018 02:52:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/7-4-formal-charges-and-resonance/

Learning Objectives

By the end of this section, you will be able to:

• Compute formal charges for atoms in any Lewis structure
• Use formal charges to identify the most reasonable Lewis structure for a given molecule
• Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule

In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.

Calculating Formal Charge

The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.

Thus, we calculate formal charge as follows:

$latex \text{formal charge} = \# \;\text{valence shell electrons (free atom)} \; - \;\# \;\text{lone pair electrons}\; - \frac{1}{2} \# \;\text{bonding electrons}$

We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.

We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.

Example 1

Assign formal charges to each atom in the interhalogen ion ICl₄⁻.

 

Solution

1. We divide the bonding electron pairs equally for all I–Cl bonds:
2. We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
3. Subtract this number from the number of valence electrons for the neutral atom:I: 7 – 8 = –1Cl: 7 – 7 = 0The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).

 

Test Yourself Calculate the formal charge for each atom in the carbon monoxide molecule:

  Answer C −1, O +1

Example 2

Assign formal charges to each atom in the interhalogen molecule BrCl₃.

 

Solution

1. Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond:
2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
3. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:Br: 7 – 7 = 0Cl: 7 – 7 = 0All atoms in BrCl₃ have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.

 

Test yourself Determine the formal charge for each atom in NCl₃.

  Answer

N: 0; all three Cl atoms: 0

Using Formal Charge to Predict Molecular Structure

The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:

1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.

To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO₂. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:

Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).

As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS^–, NCS^–, or CSN^–. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).

Example 3

Nitrous oxide, N₂O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?

 

Solution Determining formal charge yields the following:

The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:

The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.

 

Test Yourself Which is the most likely molecular structure for the nitrite (NO₂⁻) ion?

  Answer ONO^–

Resonance

You may have noticed that the nitrite anion in Example 3 can have two possible structures with the atoms in the same positions. The electrons involved in the N–O double bond, however, are in different positions:

If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in NO₂⁻ have the same strength and length, and are identical in all other properties.

It is not possible to write a single Lewis structure for NO₂⁻ in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in NO₂⁻ is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms. Thus, the electronic structure of the NO₂⁻ ion is shown as:

We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).

The carbonate anion, CO₃²⁻, provides a second example of resonance:

One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.

 

The online Lewis Structure Make includes many examples to practice drawing resonance structures.

Key Concepts and Summary

In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).

Key Equations

• $latex \text{formal charge} = \# \;\text{valence shell electrons (free atom)} \; - \;\# \;\text{lone pair electrons}\; - \frac{1}{2} \# \;\text{bonding electrons}$

Exercises

1. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.

a) sulfur dioxide, SO₂

b) carbonate ion, CO₃²⁻

c) hydrogen carbonate ion, HCO₃⁻ (C is bonded to an OH group and two O atoms)

d) pyridine:

e) the allyl ion:

2. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, NO₂^–. 3. Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.

a) CO₂         b) CO

4. Determine the formal charge of each element in the following:

a) HCl         b) CF₄         c) PCl₃         d) PF₅

5. Calculate the formal charge of chlorine in the molecules Cl₂, BeCl₂, and ClF₅. 6. Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:

a) O₃         b) SO₂         c) NO₂⁻         d) NO₃⁻

7. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH? 8. Draw the structure of hydroxylamine, H₃NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges? 9. Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound. 10. Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H₂SO₄, which has two oxygen atoms and two OH groups bonded to the sulfur.   Answers

1. a)

b) c) d) e) 2.

3. a)

b) CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO₂ has double bonds.

4. a) H: 0, Cl: 0            b) C: 0, F: 0           c) P: 0, Cl 0          d) P: 0, F: 0

5. Cl in Cl₂: 0;     Cl in BeCl₂: 0;     Cl in ClF₅: 0

6. a)

b) c) d)

7. HOCl

8. The structure that gives zero formal charges is consistent with the actual structure:

9. NF₃

10.

Glossary

formal charge: charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons) molecular structure: arrangement of atoms in a molecule or ion resonance: situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed resonance forms: two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons resonance hybrid: average of the resonance forms shown by the individual Lewis structures

]]> 1663 1538 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-10/ Thu, 12 Apr 2018 03:47:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-10/

All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and CO₂. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.

Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we will begin with the recognition of each functional group, as depicted by Lewis and line structures, and the nomenclature of simple organic compounds.

]]> 1831 1829 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-1-hydrocarbons/ Thu, 12 Apr 2018 03:47:23 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-1-hydrocarbons/

Learning Objectives

By the end of this section, you will be able to:

• Name saturated and unsaturated hydrocarbons and alkyl halides following the IUPAC rules
• From the name a saturated or unsaturated hydrocarbon or alkyl halides, draw its structure

The largest database[footnote]This is the Beilstein database, now available through the Reaxys site (www.elsevier.com/online-tools/reaxys).[/footnote] of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated[footnote]Peplow, Mark. “Organic Synthesis: The Robo-Chemist,” Nature 512 (2014): 20–2.[/footnote] at 10⁶⁰—an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities.

The simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms.

The Basics of Organic Nomenclature:

Organic chemistry nomenclature is very specific following the general format shown in Figure 1.  The International Union of Pure and Applied Chemistry (IUPAC) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures.

[caption id="attachment_2773" align="aligncenter" width="431"] Figure 1. IUPAC Nomenclature Guide[/caption]

Naming Alkanes and Alkyl Halides

The IUPAC nomenclature for alkanes and alkyl halides is based on two rules: Rule 1. Identify the longest chain of carbon atoms (PREFIX+ANE). The longest chain of carbons in the structure is referred to as the parent chain.  A two-carbon parent chain is called ethane; a three-carbon parent chain, propane; and a four-carbon parent chain, butane. Longer parent chains are named as follows: pentane (five-carbon chain), hexane (6), heptane (7), octane (8), nonane (9), and decane (10). These prefixes can be seen in the names of the alkanes described in Table 1.

Alkane	Molecular Formula	Condensed Structural Formula
methane	CH4	CH4
ethane	C2H6	CH3CH3
propane	C3H8	CH3CH2CH3
butane	C4H10	CH3CH2CH2CH3
pentane	C5H12	CH3CH2CH2CH2CH3
hexane	C6H14	CH3(CH2)4CH3
heptane	C7H16	CH3(CH2)5CH3
octane	C8H18	CH3(CH2)6CH3
nonane	C9H20	CH3(CH2)7CH3
decane	C10H22	CH3(CH2)8CH3
tetradecane	C14H30	CH3(CH2)12CH3
octadecane	C18H38	CH3(CH2)16CH3
Table 1. Some n-alkanes, meaning normal alkanes, indicating that they are straight chains of carbon units, no branching.  The prefixes indicating the number of carbons in the longest chain is underlined.

Rule 2. Names and position of the substituents:  Substituents are branches or functional groups that replace hydrogen atoms on a chain.  If there are substituents on the parent chain, their names and position on the chain must be included at the front of the name. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. We number the carbon atoms in the chain by counting from the end of the chain nearest the substituents. Multiple substituents are named individually and placed in alphabetical order at the front of the name.

When more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending -o replaces -ine at the end of the name of a halide substituent.  For example, an iodine substituent would be called iodo. The number of substituents of the same type is indicated by the prefixes di- (two), tri- (three), tetra- (four), penta- (five) and so on (for example, difluoro- indicates two fluoride substituents).

Example 1

Name the molecule whose structure is shown here:

 

Solution

The four-carbon chain is numbered from the end with the chlorine atom. This puts the substituents on positions 1 and 2 (numbering from the other end would put the substituents on positions 3 and 4). Four carbon atoms means that the base name of this compound will be butane. The bromine at position 2 will be described by adding 2-bromo-; this will come at the beginning of the name, since bromo- comes before chloro- alphabetically. The chlorine at position 1 will be described by adding 1-chloro-, resulting in the name of the molecule being 2-bromo-1-chlorobutane.

 

Test Yourself Name the following molecule:

  Answer 3,3-dibromo-2-iodopentane

We call a substituent that contains one less hydrogen than the corresponding alkane an alkyl group. The name of an alkyl group is obtained by dropping the suffix -ane of the alkane name and adding -yl:

The open bonds in the methyl and ethyl groups indicate that these alkyl groups are bonded to another atom.

Branched hydrocarbons may have more than one substituent. If the substituents are different, give each substituent a number (using the smallest possible numbers) and list the substituents in alphabetical order, with the numbers separated by hyphens and no spaces in the name. So the molecule shown here is 3-ethyl-2-methylpentane.

If the substituents are the same, use the name of the substituent only once, but use more than one number, separated by a comma and put a numerical prefix before the substituent name that indicates the number of substituents of that type. Consider this molecule:

The longest chain has four C atoms, so it is a butane. There are two substituents, each of which consists of a single C atom; they are methyl groups. The methyl groups are on the second and third C atoms in the chain (no matter which end the numbering starts from), so we would name this molecule 2,3-dimethylbutane. Note the comma between the numbers, the hyphen between the numbers and the substituent name, and the presence of the prefix di- before the methyl. Other molecules—even with larger numbers of substituents—can be named similarly.

Example 2

Name the molecule whose structure is shown here:

 

Solution The longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right—this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)—so we take our name for two carbons eth- and attach -yl at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane.

 

Test Yourself Name the following molecule:

  Answer 4-propyloctane

Example 3

Name this molecule.

Solution

The longest continuous carbon chain has seven C atoms, so this molecule is named as a heptane. There is a two-carbon substituent on the main chain, which is an ethyl group. To give the substituent the lowest numbering, we number the chain from the right side and see that the substituent is on the third C atom. So this hydrocarbon is 3-ethylheptane.

 

Test Yourself

Name this molecule.

 

Answer

2-methylpentane

Example 4

Name this molecule.

 

Solution

The longest chain has seven C atoms, so we name this molecule as a heptane. We find two one-carbon substituents on the second C atom and a two-carbon substituent on the third C atom. So this molecule is named 3-ethyl-2,2-dimethylheptane.

 

Test Yourself

Name this molecule.

 

Answer

4,4,5-tripropyloctane

Want more practice naming alkanes? Watch this brief video tutorial to review the nomenclature process.

Naming Alkenes

Ethene, C₂H₄, is the simplest alkene and is commonly called ethylene.  The second member of the series is propene (propylene).  The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:

Note: The IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an “infix” rather than a prefix. For example, the new name for 1-butene and 2-butene would be but-1-ene and but-2-ene. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols.[/footnote]

Therefore when naming alkenes following IUPAC, you follow the same two rules for alkanes with modification to "rule 1" mentioned above.

Rule 1. Identify the longest chain of carbons which contains the double bond and its position (PREFIX-#-ENE).  And when numbering the main chain, the double gets the lowest possible number. Rule 2. Names and position of the substituents.

For example, this molecule is 2,4-dimethylhept-3-ene.  Note the number and the hyphens that indicate the position of the double bond.

Recycling Plastics

Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries.

Polymers (from Greek words poly meaning “many” and mer meaning “parts”) are large molecules made up of repeating units, referred to as monomers. Polymers can be natural (starch is a polymer of sugar residues and proteins are polymers of amino acids) or synthetic [like polyethylene, polyvinyl chloride (PVC), and polystyrene]. The variety of structures of polymers translates into a broad range of properties and uses that make them integral parts of our everyday lives. Adding functional groups to the structure of a polymer can result in significantly different properties (see the discussion about Kevlar later in this chapter).

An example of a polymerization reaction is shown in Figure 2. The monomer ethylene (C₂H₄) is a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of –CH₂– units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films).

[caption id="" align="aligncenter" width="1200"] Figure 2. The reaction for the polymerization of ethylene to polyethylene is shown.[/caption]

Polyethylene is a member of one subset of synthetic polymers classified as plastics. Plastics are synthetic organic solids that can be molded; they are typically organic polymers with high molecular masses. Most of the monomers that go into common plastics (ethylene, propylene, vinyl chloride, styrene, and ethylene terephthalate) are derived from petrochemicals and are not very biodegradable, making them candidate materials for recycling. Recycling plastics helps minimize the need for using more of the petrochemical supplies and also minimizes the environmental damage caused by throwing away these nonbiodegradable materials.

Plastic recycling is the process of recovering waste, scrap, or used plastics, and reprocessing the material into useful products. For example, polyethylene terephthalate (soft drink bottles) can be melted down and used for plastic furniture, in carpets, or for other applications. Other plastics, like polyethylene (bags) and polypropylene (cups, plastic food containers), can be recycled or reprocessed to be used again. Many areas of the country have recycling programs that focus on one or more of the commodity plastics that have been assigned a recycling code (see Figure 3). These operations have been in effect since the 1970s and have made the production of some plastics among the most efficient industrial operations today.

[caption id="" align="aligncenter" width="1200"] Figure 3. Each type of recyclable plastic is imprinted with a code for easy identification.[/caption]

Once you master naming hydrocarbons from their given structures, it is rather easy to draw a structure from a given name. Just draw the parent chain with the correct number of C atoms (putting the double or triple bond in the right position, as necessary) and add the substituents in the proper positions. If you start by drawing the C atom backbone, you can go back and complete the structure by adding H atoms to give each C atom four covalent bonds.

From the name 2,3-dimethyl-4-propylhept-2-ene, we start by drawing the seven-carbon parent chain with a double bond starting at the third carbon:

We add to this structure two one-carbon substituents on the second and third C atoms:

We finish the carbon backbone by adding a three-carbon propyl group to the fourth C atom in the parent chain:

If we so choose, we can add H atoms to each C atom to give each carbon four covalent bonds, being careful to note that the C atoms in the double bond already have an additional covalent bond. Question: How many H atoms do you think are required? There will need to be 24 H atoms to complete the molecule.

Example 5

Draw the carbon backbone for 2,3,4-trimethylpentane.

 

Solution

First, we draw the five-carbon backbone that represents the pentane chain:

According to the name, there are three one-carbon methyl groups attached to the second, third, and fourth C atoms in the chain. We finish the carbon backbone by putting the three methyl groups on the pentane main chain:

 

Test Yourself

Draw the carbon backbone for 3-ethyl-6,7-dimethyloct-2-ene.

 

Answer

Naming Alkynes

The simplest member of the alkyne series is ethyne, C₂H₂, commonly called acetylene.

The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, $latex \text{CH}_3\text{CH}_2\text{C}\;{\equiv}\;\text{CH}$ is called but-1-yne.

Therefore when naming alkynes following IUPAC, you follow the same two rules for alkanes with modification to "rule 1" mentioned above.

Rule 1. Identify the longest chain of carbons which contains the triple bond and its position (PREFIX-#-YNE).  And when numbering the main chain, the triple bond gets the lowest possible number. Rule 2. Names and position of the substituents

Example 6

Name the following molecule:

 

Solution but-2-yne

 

Test Yourself Name the following molecule:

Answer pent-3-en-1-yne

Example 7

Name this molecule.

 

Solution

The longest chain that contains the C–C triple bond has six C atoms, so this is a hexyne molecule. The triple bond starts at the third C atom, so this is a hex-3-yne. Finally, there are two methyl groups on the chain; to give them the lowest possible number, we number the chain from the left side, giving the methyl groups the second position. So the name of this molecule is 2,2-dimethylhex-3-yne.

 

Test Yourself

Name this molecule.

 

Answer

2,3,4-trimethylpent-2-ene

Naming Arenes

The most commonly known arene is benzene.

There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. The following are typical examples of substituted benzene derivatives:

Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene.  Toluene, xylene and styrene are common names for these compounds.  The systematic way of naming these benzene derivatives is by following the the two rules:

Rule 1. Identify the arene ring (BENZENE). Rule 2. Names and position (if more than one) of the substituents: If there are two or more substituents on a benzene molecule, the relative positions must be numbered. The substituent that is first alphabetically is assigned position 1, and the ring is numbered in a circle to give the other substituents the lowest possible number(s).

Therefore the systematic name for toluene is methylbenzene and for xylene is 1,2-dimethylbenzene.

Key Concepts and Summary

Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes and alkynes are unsaturated hydrocarbons.  Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Arenes, also known as aromatic hydrocarbons, contain ring structures with alternating single and double bonds.

The systematic methods of naming the various hydrocarbons follow a similar procedure and the names have three main parts: 1) specifying the information about the substituents, 2) specifying the information about the parent chain (or ring), and 3) the ending which specifies what functional group is present in the structure being named. Alkanes:    #-substituents-PREFIX+ANE Alkenes:   #-substituents-PREFIX-#-ENE Alkynes:   #-substituents-PREFIX-#-YNE Arenes (specifically benzene derivatives): #-substituents-BENZENE

Exercises

1. Write the chemical formula and Lewis structure of the following, each of which contains five carbon atoms:

a) an alkane         b) an alkene         c) an alkyne

2. Name the following compounds: 3. Write the Lewis structure and molecular formula for each of the following hydrocarbons:

a) hexane                              b) 3-methylpentane         c) hex-3-ene

d) 4-methylpent-1-ene         e) hex-3-yne                      f) 4-methylpent-2-yne

4. Give the complete IUPAC name for each of the following compounds:

a) $latex \text{CH}_3\text{CH}_2\text{CBr}_2\text{CH}_3$

b) $latex (\text{CH}_3)_3\text{CCl}$

c)

d) $latex \text{CH}_3\text{CH}_2\text{C}\;{\equiv}\;\text{CH\;CH}_3\text{CH}_2\text{C}\;{\equiv}\;\text{CH}$

e)

f)

g) $latex (\text{CH}_3)_2\text{CHCH}_2\text{CH} = \text{CH}_2$

5. Butane is used as a fuel in disposable lighters. Write the Lewis structure for each isomer of butane. 6. Define hydrocarbon. What are the two general types of hydrocarbons? 7. Indicate whether each molecule is an aliphatic (open chain) or an arene. If it is aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.  

a)  

 

b)  

 

c)  

 

8. Indicate whether each molecule is an aliphatic or an arene. If it is aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.

 

a)  

 

b)  

 

c)  

 

d)  

 

e)  

 

f)  

 

g)  

 

h)  

 

i)  

 

9. Name and draw the structural formulas for the four smallest alkanes.

10.  Explain why you may see prop-1-ene written just as propene. 11.  Name and draw the structural formula of each isomer of pentene. 12.  Draw the structure of the product of the reaction of bromine with propene. 13.  Draw the structure of the product of the reaction of hydrogen with but-1-ene. 14. How does a branched hydrocarbon differ from a normal hydrocarbon? 15. Name this molecule.

16. Name this molecule.

17. Name this molecule.

18. Name this molecule.

19. Name this molecule.

20. Name this molecule.

21. Name this molecule.

22. Name this molecule.

23. Name this molecule.

24. Name this molecule.

25. Draw the carbon backbone for each molecule.

a)  3,4-diethyloctane b)  2,2-dimethyl-4-propylnonane 26. Draw the carbon backbone for each molecule.

a)  4-ethyl-4-propyloct-2-yne b)  5-butyl-2,2-dimethyldecane

27. The name 2-ethylhexane is incorrect. Draw the carbon backbone and write the correct name for this molecule.   Answers

1. There are several sets of answers; one is:

(a) $latex \text{C}_5\text{H}_{12}$ ; (b) $latex \text{C}_5\text{H}_{10}$ ; (c) $latex \text{C}_5\text{H}_8$ 2. 2-hexene and 2-methylpentane

3. (a) $latex \text{C}_6\text{H}_{14}$ ;

(b) $latex \text{C}_6\text{H}_{14}$ ; (c) $latex \text{C}_6\text{H}_{12}$ ; (d) $latex \text{C}_6\text{H}_{12}$ ; (e) $latex \text{C}_6\text{H}_{10}$ ; (f) $latex \text{C}_6\text{H}_{10}$

4. (a) 2,2-dibromobutane; (b) 2-chloro-2-methylpropane; (c) 2-methylbutane; (d) but-1-yne; (e) 4-fluoro-4-methyloct-1-yne; (f) 1-chloropropene; (g) 5-methylpent-1-ene

5. 6. an organic compound composed of only carbon and hydrogen; aliphatic hydrocarbons and aromatic hydrocarbons 7. a)   aliphatic; alkane   b)   arene   c)   aliphatic; alkene 8. a)   aliphatic; alkane   b)   aliphatic; alkene   c)   arene  d) aliphatic; alkyne   e) arene f) aliphatic; alkene   g) aliphatic; alkene   h) arene   i) aliphatic; alkyne 9.   10. The 1 is not necessary since the double bond is on the first carbon. 11. 12. 13.   14. A branched hydrocarbon does not have all of its C atoms in a single row. 15. 3-methyl-hex-2-ene 16. 2,2,3-trimethylpentane 17. 4,4-dimethylpent-1-ene 18. 4,4-dimethylheptane 19. 2,4-dimethylpent-2-ene 20. hex-3-yne 21. 3,4-diethyloctane 22. 4,5-dimethylhept-3-ene 23. 1-bromo-4-chlorobenzene 24. 1-ethyl-2,3-dimethylbenzene 25.a)      b)    26.a)    b)    27.

Glossary

alkane: molecule consisting of only carbon and hydrogen atoms connected by single (σ) bonds alkene: molecule consisting of carbon and hydrogen containing at least one carbon-carbon double bond alkyl group: substituent, consisting of an alkane missing one hydrogen atom, attached to a larger structure alkyne: molecule consisting of carbon and hydrogen containing at least one carbon-carbon triple bond aromatic hydrocarbon: cyclic molecule consisting of carbon and hydrogen with delocalized alternating carbon-carbon single and double bonds, resulting in enhanced stability functional group: part of an organic molecule that imparts a specific chemical reactivity to the molecule organic compound: natural or synthetic compound that contains carbon saturated hydrocarbon: molecule containing carbon and hydrogen that has only single bonds between carbon atoms skeletal structure or line structure: shorthand method of drawing organic molecules in which carbon atoms are represented by the ends of lines and bends in between lines, and hydrogen atoms attached to the carbon atoms are not shown (but are understood to be present by the context of the structure) substituent: branch or functional group that replaces hydrogen atoms in a larger hydrocarbon chain

]]> 1900 1829 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-2-alcohols-and-ethers/ Thu, 12 Apr 2018 03:47:25 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-2-alcohols-and-ethers/

Learning Objectives

By the end of this section, you will be able to:

• Describe the structure and properties of alcohols
• Describe the structure and properties of ethers
• Name and draw structures for alcohols and ethers

Naming Alcohols

The name of an alcohol comes from the hydrocarbon from which it was derived. The final -e in the name of the hydrocarbon is replaced by -ol, and the carbon atom to which the hydroxyl group (OH group) is bonded is indicated by a number placed before the name.[footnote] The IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an “infix” rather than a prefix. For example, the new name for 2-propanol would be propan-2-ol. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols.[/footnote]

Therefore when naming alcohols following IUPAC, you follow the similar two rules for alkanes with modification to "rule 1" mentioned above.

Rule 1. Identify the longest chain of carbons which contains the OH group and its position (PREFIX-#-ANE+OL).  And when numbering the parent chain, the hydroxyl group gets the lowest possible number. Rule 2. Names and position of the substituents.

Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):

Example 1

Consider the following example. How should it be named?

Solution The carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol.

 

Test Yourself Name the following molecule:

Answer 2-methylpentan-2-ol

Naming Ethers

Ethers are compounds that contain the functional group –O–. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named “methoxy.” The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by “ether.”  But if the two alkyl groups are the same, then the di prefix is used. The common name for methoxyethane shown below is ethylmethyl ether:

Therefore when naming ethers, the two rules are modified.  And furthermore, there are two ways that are commonly used: IUPAC method and the common method.

IUPAC method of naming ethers: Rule 1. Identify the longest carbon branch (PREFIX-ANE). Rule 2. Names of the substituent, the other carbon branch (PREFIX+OXY)   Common method of naming ethers: ALKYLALKYL ether    or   diALKYL ether

  For example, the following ether would be commonly named ethylpropyl ether.  its name following the IUPAC rules would be ethoxypropane.

Example 2

Provide the IUPAC and common name for the ether shown here:

 

Solution IUPAC: The molecule is made up of an ethoxy group attached to an ethane chain, so the IUPAC name would be ethoxyethane.

Common: The groups attached to the oxygen atom are both ethyl groups, so the common name would be diethyl ether.

 

Test Yourself Provide the IUPAC and common name for the ether shown:

Answers IUPAC: 2-methoxypropane; common: isopropylmethyl ether

Want more practice naming ethers? This brief video review summarizes the nomenclature for ethers.

Carbohydrates and Diabetes

Carbohydrates are large biomolecules made up of carbon, hydrogen, and oxygen. The dietary forms of carbohydrates are foods rich in these types of molecules, like pastas, bread, and candy. The name “carbohydrate” comes from the formula of the molecules, which can be described by the general formula C_m(H₂O)_n, which shows that they are in a sense “carbon and water” or “hydrates of carbon.” In many cases, m and n have the same value, but they can be different. The smaller carbohydrates are generally referred to as “sugars,” the biochemical term for this group of molecules is “saccharide” from the Greek word for sugar (Figure 1). Depending on the number of sugar units joined together, they may be classified as monosaccharides (one sugar unit), disaccharides (two sugar units), oligosaccharides (a few sugars), or polysaccharides (the polymeric version of sugars—polymers were described in the feature box earlier in this chapter on recycling plastics). The scientific names of sugars can be recognized by the suffix -ose at the end of the name (for instance, fruit sugar is a monosaccharide called “fructose” and milk sugar is a disaccharide called lactose composed of two monosaccharides, glucose and galactose, connected together). Sugars contain some of the functional groups we have discussed: Note the alcohol groups present in the structures and how monosaccharide units are linked to form a disaccharide by formation of an ether.

[caption id="" align="aligncenter" width="1200"] Figure 1. The illustrations show the molecular structures of fructose, a five-carbon monosaccharide, and of lactose, a disaccharide composed of two isomeric, six-carbon sugars.[/caption]

Organisms use carbohydrates for a variety of functions. Carbohydrates can store energy, such as the polysaccharides glycogen in animals or starch in plants. They also provide structural support, such as the polysaccharide cellulose in plants and the modified polysaccharide chitin in fungi and animals. The sugars ribose and deoxyribose are components of the backbones of RNA and DNA, respectively. Other sugars play key roles in the function of the immune system, in cell-cell recognition, and in many other biological roles.

Diabetes is a group of metabolic diseases in which a person has a high sugar concentration in their blood (Figure 2). Diabetes may be caused by insufficient insulin production by the pancreas or by the body’s cells not responding properly to the insulin that is produced. In a healthy person, insulin is produced when it is needed and functions to transport glucose from the blood into the cells where it can be used for energy. The long-term complications of diabetes can include loss of eyesight, heart disease, and kidney failure.

In 2013, it was estimated that approximately 3.3% of the world’s population (~380 million people) suffered from diabetes, resulting in over a million deaths annually. Prevention involves eating a healthy diet, getting plenty of exercise, and maintaining a normal body weight. Treatment involves all of these lifestyle practices and may require injections of insulin.

[caption id="" align="aligncenter" width="800"] Figure 2. Diabetes is a disease characterized by high concentrations of glucose in the blood. Treating diabetes involves making lifestyle changes, monitoring blood-sugar levels, and sometimes insulin injections. (credit: “Blausen Medical Communications”/Wikimedia Commons)[/caption]

Key Concepts and Summary

Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.

The systematic methods of naming alcohols follow a similar procedure and the names have three main parts: 1) specifying the information about the substituents, 2) specifying the information about the parent chain, and 3) the ending which specifies what functional group is present in the structure being named. Alcohols:    #-substituents-PREFIX-#-ANE+OL Ethers:    PREFIX+OXY-PREFIX-ANE Remember: there are two ways that are commonly used to name ethers.  The common method is also important to know. Common method of naming ethers: ALKYLALKYL ether    or   diALKYL ether

Exercises

1. Write condensed formulas and provide IUPAC names for the following compounds:

a) ethyl alcohol (in beverages)

b) methyl alcohol (used as a solvent, for example, in shellac)

c) ethylene glycol (antifreeze)

d) isopropyl alcohol (used in rubbing alcohol)

e) glycerine

2. Give the complete IUPAC name for each of the following compounds:

a)

b)

c)

3. Give the complete IUPAC name and the common name for each of the following compounds:

a)

b)

c)

4. Name this molecule.

5. Name this molecule.

6. Name this molecule.

  Answers

1. a) ethyl alcohol, ethanol: CH₃CH₂OH; b) methyl alcohol, methanol: CH₃OH; c) ethylene glycol, ethanediol: HOCH₂CH₂OH; d) isopropyl alcohol, 2-propanol: CH₃CH(OH)CH₃; e) glycerine, l,2,3-trihydroxypropane: HOCH₂CH(OH)CH₂OH

2. a) butan-2-ol; b) 2-iodo-3,3-dimethylpentan-2-ol; c) 3-chloro-3-ethylhexan-2-ol

3. a) 1-ethoxybutane, butyl ethyl ether; b) 1-ethoxypropane, ethyl propyl ether; c) 1-methoxypropane, methyl propyl ether 4. 3-chloropentan-3-ol 5.  octan-4-ol 6. 4-ethyl-2,5-dimethylhexan-3-ol

Glossary

alcohol: organic compound with a hydroxyl group (–OH) bonded to a carbon atom ether: organic compound with an oxygen atom that is bonded to two carbon atoms

]]> 1922 1829 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-3-aldehydes-ketones-carboxylic-acids-and-esters/ Thu, 12 Apr 2018 03:47:29 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-3-aldehydes-ketones-carboxylic-acids-and-esters/

Learning Objectives

By the end of this section, you will be able to:

• Describe the structure and properties of aldehydes, ketones, carboxylic acids, esters and amides
• Name and draw structures for aldehydes, ketones, carboxylic acids, esters and amides

Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The carbonyl group can attach to two other substituents leading to several subfamilies, some of which are: aldehydes, ketones, carboxylic acids, esters and amides.

Naming of Aldehydes and Ketones

Both aldehydes and ketones contain a carbonyl group. In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms.  The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:

Therefore when naming aldehydes following IUPAC, you follow these rules:

Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-ANE+AL).  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name. Rule 2. Names and position of the substituents.

Methanal has a common name with which you may be familiar: formaldehyde.

When naming ketones following IUPAC, you follow these rules: Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-#-ANE+ONE).  And when numbering the parent chain, the carbonyl group gets the lowest possible number.  In the smaller ketones (propanone and butanone), the locant number is not used because there is no alternative placement in these smaller ketones. Rule 2. Names and position of the substituents.

The common name for propanone is acetone. There is a non-IUPAC way to name ketones that is commonly used as well: name the alkyl groups that are attached to the carbonyl group and add the word ketone to the name. So propanone can also be called dimethyl ketone, while butan-2-one is called methyl ethyl ketone.

In condensed structure, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.

Example 1

Draw the structures of:  a) pentan-2-one;   b) hexan-2-one;    c) butane

 

Solution

a) This molecule has five C atoms in a chain, with the carbonyl group on the second C atom. Its structure is:

b) This molecule has six C atoms in a chain, with the carbonyl group on the second C atom. Its structure is:

c) This molecule has four C atoms in a chain, with the carbonyl group on the first C atom since it is an aldehyde (ends with -al). Its structure is:  

Test Yourself Give the condensed structure of the following amines:

a) propanone   b) propanal   c) heptan-3-one   d) octanal

  Answer a) CH₃COCH₃    b) CH₃CH₂CHO    c) CH₃CH₂CH₂CH₂COCH₂CH₃   or  CH₃(CH₂)₃COCH₂CH₃ d) CH₃CH₂CH₂CH₂CH₂CH₂CH₂CHO   or  CH₃(CH₂)₆CHO

Naming Carboxylic Acids and Esters

Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules.

The names for carboxylic acid and ester compounds are derived using similar nomenclature rules as seen previously with aldehydes, and include the class-identifying suffixes -oic acid and -oate, respectively:

Therefore when naming carboxylic acids following IUPAC, you follow these rules:

Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-ANE+OIC ACID).  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name. Rule 2. Names and position of the substituents.

When naming esters following IUPAC, you follow these rules:

Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-ANE+OATE).  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name.  AND then name the other carbon chain (PREFIX+YL). Rule 2. Names and position of the substituents.

The functional groups for an acid and for an ester are shown in red in these formulas.  In brackets you have the common names for ethanoic acid and methyl ethanoate.

The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt:

Carboxylic acids are weak acids (see the chapter on acids and bases), meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution.

In condensed structure, the carboxylic acid group is represented as –COOH; an ester is represented as –COO– .

Example 2

Draw the structures of:  a) 3-methylpentanoic acid;   b) ethyl ethanoate;    c) propyl 2-chlorobutanoate

 

Solution

a) Its structure is:

b)Its structure is:

c) Its structure is:  

Test Yourself Name the following compounds:

a) CH₃CH₂COOH  b) CH₃CH₂CH₂CH₂COOCH₂CH₃  c) BrCH₂(CH₂)₂COCH₃  d) (CH₃)₂CH(CH₂)₆COOH

  Answer a) propanoic acid   b) ethyl pentanoate   c) methyl 4-bromobutanoate    d) 9-methylnonanoic acid   

Example 3

Complete the chemical reaction.

 

Solution

The OH^– ion removes the H atom that is part of the carboxyl group:

The carboxylate ion, which has the condensed structural formula CH₃CO₂⁻, is the ethanoate ion, but it is commonly called the acetate ion.

 

Test Yourself

Complete the chemical reaction.

 

Answer

The ion is the methanoate ion, which is commonly called the formate ion.

Naming Simple Amides

Amides are molecules that contain nitrogen atoms connected to the carbon atom of a carbonyl group.  The names for amide compounds are derived using similar nomenclature rules as seen previously with aldehydes and carboxylic acids, and include the class-identifying suffixes -amide:

Therefore when naming amides following IUPAC, you follow these rules:

Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-ANE+AMIDE).  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name. Rule 2. Names and position of the substituents. In condensed structure, the amide group is represented as –CONH₂ or –CONHR or –CONR₂.

Example 4

Give the condensed structures of:  a) decanamide;   b) hexanamide;    c) 2-chloroethanamide

 

Solution

a) Its condensed structure is: CH₃(CH₂)₈CONH₂

b) Its condensed structure is: CH₃(CH₂)₄CONH₂ c) Its condensed structure is: ClCH₂CONH₂

 

Test Yourself Name the following compounds:

a) CH₃(CH₂)₂CONH₂  b) BrCH₂(CH₂)₃CONH₂  d) (CH₃CH₂)₂CH(CH₂)₅CONH₂

  Answer a) butanamide  b) 5-bromopentanamide   c) 7-ethylnonanamide

Key Concepts and Summary

Functional groups related to the carbonyl group include the –CHO group of an aldehyde, the –CO– group of a ketone, the –CO₂H group of a carboxylic acid, the –CO₂R group of an ester and the –CONH₂ group of an amide.

The systematic methods of naming these carbonyl containing functional groups follow a similar procedure and the names have three main parts: 1) specifying the information about the substituents, 2) specifying the information about the parent chain, and 3) the ending which specifies what functional group is present in the structure being named. Aldehydes:    #-substituents-PREFIX-ANE+AL Ketones:   #-substituents-PREFIX-ANE+ONE Carboxylic Acids:   #-substituents-PREFIX-ANE+OIC ACID Esters:   ALKYL #-substituents-PREFIX-ANE+OATE Amides:   #-substituents-PREFIX-ANE+AMIDE

Exercises

1. Write a condensed structural formula of the following compounds.

a) 2-propanol

b) acetone

c) dimethyl ether

d) acetic acid

e) 3-methyl-1-hexene

2. A peptide is a short chain of amino acids connected by amide bonds. How many amide bonds are present in this peptide?

3. How many amide bonds are present in this peptide? (See Exercise 2 for the definition of a peptide.)

4. Name a similarity between the functional groups found in aldehydes and ketones. Can you name a difference between them?

5. Name each molecule.

a)       b)  

6. Name each molecule.

a)      b)  

7. Name each molecule.

a)      b)  

8. Name each molecule.

a)      b)  

9. Name this molecule.

10. The drug known as aspirin is shown here.  Identify the functional group(s) in this molecule.

11. The drug known as naproxen sodium is the sodium salt of the molecule shown here. Identify the functional group(s) in this molecule.

Answers

1. a) CH₃CH(OH)CH₃

b) $latex \text{CH}_3\text{COCH}_3$:

c) CH₃OCH₃

d) CH₃COOH

e) CH₃CH₂CH₂CH(CH₃)CHCH₂

2. two amide bonds

3. one amide bond

4. They both have a carbonyl group, but an aldehyde has the carbonyl group at the end of a carbon chain, and a ketone's carbonyl carbon is surrounded by two other carbons. 5. a)  proposal b)  butanone 6. a) 3-chloro-3-methylbutanal  b) heptan-4-one 7. a)  3-methylbutanoic acid      b)  ethyl propionate 8. a) 2,2,2-trichlroethanoic acid    b) butyl ethanoate 9. ethyl propyl ether 10. carboxylic acid, arene and ester 11. carboxylic acid, arene and ether

Glossary

aldehyde: organic compound containing a carbonyl group bonded to two hydrogen atoms or a hydrogen atom and a carbon substituent carbonyl group: carbon atom double bonded to an oxygen atom carboxylic acid: organic compound containing a carbonyl group with an attached hydroxyl group ester: organic compound containing a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent ketone: organic compound containing a carbonyl group with two carbon substituents attached to it

]]> 1952 1829 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-4-amines-and-amides/ Thu, 12 Apr 2018 03:47:32 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/20-4-amines-and-amides/

Learning Objectives

By the end of this section, you will be able to:

• Describe the structure and properties of an amine.
• Distinguish between a primary, secondary and tertiary amine.
• Name and draw structures for primary, secondary and tertiary amines.

An amine is an organic derivative of ammonia (NH₃). In amines, one or more of the H atoms in NH₃ is substituted with an organic group.

A primary amine has one H atom substituted with an R group:	A secondary amine has two H atoms substituted with an R group:	A tertiary amine has all three H atoms substituted with R groups:

 Like ammonia, amines are weak bases due to the lone pair of electrons on their nitrogen atoms:

The basicity of an amine’s nitrogen atom plays an important role in much of the compound’s chemistry.

Naming Amines

The alkyl groups connected to the nitrogen atom are named separately and followed by “amine.”  If some alkyl groups are the same, then a prefix is used (di or tri), as illustrated here for a few simple examples:

Example 1

Name the following organic compounds:

a) (CH₃)₂NCH₂CH₃    b) CH₃CH₂CH₂NHCH₃   c) CH₃(CH₃CH₂)NCH₂CH₂CH₃  

Solution a) ethyldimethylamine    b) methylpropylamine   c) ethylmethylpropylamine

 

Test Yourself Give the condensed structure of the following amines:

a) butylamine   b) trihexylamine   c) methylpentylamine

  Answer a) CH₃CH₂CH₂CH₂NH₂    b) (CH₃CH₂CH₂CH₂CH₂CH₂)₃N   c) CH₃(CH₃CH₂CH₂CH₂CH₂)NH

Key Concepts and Summary

Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Amines are a basic functional group. An acid-base reaction occurs when an amine is mixed with and an acid.

The systematic methods of naming amines follow a simple procedure:

primary amines: ALKYLamine  

secondary amines: ALKYLALKYLamine    or   diALKYLamine

tertiary amines: ALKYLALKYLALKYLamine    or   triALKYLamine

Exercises

1. What are the structure and name of the smallest amine?

2. Identify each compound as a primary, secondary, or tertiary amine.

a)  

b)  

c)  

3. Identify each compound as a primary, secondary, or tertiary amine.

a)  

b)  

4. Write the chemical reaction between each amine in Exercise 2 and HCl.

5. Name each amine.

a)      b)  

c)    d) 

  Answers 1. CH₃NH₂; methylamine 2. a)  primary  b)  tertiary  c)  secondary 3. a) primary  b) secondary 4. a)  C₃H₃CO₂HSHNH₂ + HCl $latex \longrightarrow$ C₃H₃CO₂HSHNH₃Cl b)  (C₆H₁₁)(C₂H₅)(CH₃)N + HCl $latex \longrightarrow$ (C₆H₁₁)(C₂H₅)(CH₃)NHCl c)  (C₂H₅)(CH₃)NH + HCl $latex \longrightarrow$ (C₂H₅)(CH₃)NH₂Cl 5. a)  ethylmethylamine   b)  ethyldipropylamine  c) diethylmethylamine

Glossary

amine: organic molecule in which a nitrogen atom is bonded to one or more alkyl group

]]> 1970 1829 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-12/ Thu, 12 Apr 2018 03:51:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-12/

Introduction

Data suggest that a male child will weigh 50% of his adult weight at about 11 years of age. However, he will reach 50% of his adult height at only 2 years of age. It is obvious, then, that people eventually stop growing up but continue to grow out. Data also suggest that the average human height has been increasing over time. In industrialized countries, the average height of people increased 5.5 inches from 1810 to 1984. Most scientists attribute this simple, basic measurement of the human body to better health and nutrition.

[caption id="attachment_4607" align="alignleft" width="383"] Source: Chart courtesy of Centers for Disease Control and Prevention, http://www.cdc.gov/nchs/nhanes.htm#Set%201.[/caption]

In 1983, an Air Canada airplane had to make an emergency landing because it unexpectedly ran out of fuel; ground personnel had filled the fuel tanks with a certain number of pounds of fuel, not kilograms of fuel. In 1999, the Mars Climate Orbiter spacecraft was lost attempting to orbit Mars because the thrusters were programmed in terms of English units, even though the engineers built the spacecraft using metric units. In 1993, a nurse mistakenly administered 23 units of morphine to a patient rather than the “2–3” units prescribed. (The patient ultimately survived.) These incidents occurred because people weren’t paying attention to quantities.

Chemistry, like all sciences, is quantitative. It deals with quantities, things that have amounts and units. Making measurements is very important in chemistry, as is dealing with quantities and relating quantities to each other.

]]> 2071 2084 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/expressing-numbers/ Thu, 12 Apr 2018 03:51:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/expressing-numbers/

Learning Objective

By the end of this section, you will be able to:

• Learn to express numbers properly.

Quantities have two parts: the number and the unit. The number tells “how many.” It is important to be able to express numbers properly so that the quantities can be communicated properly.

Standard notation is the straightforward expression of a number. Numbers such as 17, 101.5, and 0.00446 are expressed in standard notation. For relatively small numbers, standard notation is fine. However, for very large numbers, such as 306,000,000, or for very small numbers, such as 0.000000419, standard notation can be cumbersome because of the number of zeros needed to place nonzero numbers in the proper position.

Scientific notation is an expression of a number using powers of 10. Powers of 10 are used to express numbers that have many zeros:

100	= 1
101	= 10
102	= 100 = 10 × 10
103	= 1,000 = 10 × 10 × 10
104	= 10,000 = 10 × 10 × 10 × 10

and so forth. The raised number to the right of the 10 indicating the number of factors of 10 in the original number is the exponent. (Scientific notation is sometimes called exponential notation.) The exponent’s value is equal to the number of zeros in the number expressed in standard notation.

Small numbers can also be expressed in scientific notation but with negative exponents:

10−1	= 0.1 = 1/10
10−2	= 0.01 = 1/100
10−3	= 0.001 = 1/1,000
10−4	= 0.0001 = 1/10,000

and so forth. Again, the value of the exponent is equal to the number of zeros in the denominator of the associated fraction. A negative exponent implies a decimal number less than one.

A number is expressed in scientific notation by writing the first nonzero digit, then a decimal point, and then the rest of the digits. The part of a number in scientific notation that is multiplied by a power of 10 is called the coefficient. Then determine the power of 10 needed to make that number into the original number and multiply the written number by the proper power of 10. For example, to write 79,345 in scientific notation,

79,345 = 7.9345 × 10,000 = 7.9345 × 10⁴

Thus, the number in scientific notation is 7.9345 × 10⁴. For small numbers, the same process is used, but the exponent for the power of 10 is negative:

0.000411 = 4.11 × 1/10,000 = 4.11 × 10⁻⁴

Typically, the extra zero digits at the end or the beginning of a number are not included.

Example 1

Express these numbers in scientific notation.

a) 306,000          b) 0.00884          c) 2,760,000          d) 0.000000559

 

Solution

a) The number 306,000 is 3.06 times 100,000, or 3.06 times 10⁵. In scientific notation, the number is 3.06 × 10⁵.

b) The number 0.00884 is 8.84 times 1/1,000, which is 8.84 times 10⁻³. In scientific notation, the number is 8.84 × 10⁻³.

c) The number 2,760,000 is 2.76 times 1,000,000, which is the same as 2.76 times 10⁶. In scientific notation, the number is written as 2.76 × 10⁶. Note that we omit the zeros at the end of the original number.

d) The number 0.000000559 is 5.59 times 1/10,000,000, which is 5.59 times 10⁻⁷. In scientific notation, the number is written as 5.59 × 10⁻⁷.

 

Test Yourself

Express these numbers in scientific notation.

a) 23,070          b) 0.0009706

 

Answers

a) 2.307 × 10⁴          b) 9.706 × 10⁻⁴

Another way to determine the power of 10 in scientific notation is to count the number of places you need to move the decimal point to get a numerical value between 1 and 10. The number of places equals the power of 10. This number is positive if you move the decimal point to the right and negative if you move the decimal point to the left.

Many quantities in chemistry are expressed in scientific notation. When performing calculations, you may have to enter a number in scientific notation into a calculator. Be sure you know how to correctly enter a number in scientific notation into your calculator. Different models of calculators require different actions for properly entering scientific notation. If in doubt, consult your instructor immediately.

Example 2

Represent the following numbers in scientific notation. a)  12,500               b)  1470                c)  0.0024   Solution a) Move the decimal until you have a number between 1 and 10: Because the decimal was moved 4 places to the left, the power of ten is 4. The answer is 25 x 10⁴ b) The decimal was moved 3 places to the left, the answer is  47 x 10³ c) The decimal was moved 3 places to the right, the answer is  4 x 10^-3   Test Yourself Represent the following numbers in scientific notation a) 247         b) 100         c) 0.2089         d) 0.000000003

 

Answers

a) 2.47 × 10²            b) 1 × 10²          b) 3 × 10^-9

Example 3

Represent the following scientific notation numbers in normal notation a)  3.06 x 10^-2              b) 2.47 x 10⁶   Solution a) You now have to move the decimal in the opposite direction, thus a negative power of 10 requires a movement to the right (2 spaces):  3.06  →  0.0306 b) The positive exponent requires a movement 6 spaces to the left, adding zeros as required. 2.47 →  2,470,000   Test Yourself Represent the following scientific notation numbers in normal notation a)  1.087 x 10^-6            b) 4 x 10⁴  

Answers

a) 0.000001087          b) 40000

Key Concepts and Summary

Standard notation expresses a number normally.  Scientific notation expresses a number as a coefficient times a power of 10.  The power of 10 is positive for numbers greater than 1 and negative for numbers between 0 and 1. This calculator shows only the coefficient and the power of 10 to represent the number in scientific notation. Thus, the number being displayed is 3.84951 × 10¹⁸, or 3,849,510,000,000,000,000.  Source: “Casio”Asim Bijarani is licensed under Creative Commons Attribution 2.0 Generic.

Exercises

1.  Express these numbers in scientific notation.

a)  56.9          b)  563,100          c)  0.0804          d)  0.00000667  

2.  Express these numbers in scientific notation.

a)  −890,000          b)  602,000,000,000          c)  0.0000004099          d)  0.000000000000011

 

3.  Express these numbers in scientific notation.

a)  0.00656          b)  65,600          c)  4,567,000          d)  0.000005507

 

4.  Express these numbers in scientific notation.

a)  65          b)  −321.09          c)  0.000077099          d)  0.000000000218

 

5.  Express these numbers in standard notation.

a)  1.381 × 10⁵          b)  5.22 × 10⁻⁷          c)  9.998 × 10⁴

 

6.  Express these numbers in standard notation.

a)  7.11 × 10⁻²          b)  9.18 × 10²          c)  3.09 × 10⁻¹⁰

 

7.  Express these numbers in standard notation.

a)  8.09 × 10⁰          b)  3.088 × 10⁻⁵           c)  −4.239 × 10²

 

8.  Express these numbers in standard notation.

a)  2.87 × 10⁻⁸           b)  1.78 × 10¹¹           c)  1.381 × 10⁻²³

 

9. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  72.44 × 10³           b)  9,943 × 10⁻⁵           c)  588,399 × 10²

 

10.  These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  0.000077 × 10⁻⁷          b)  0.000111 × 10⁸          c)  602,000 × 10¹⁸

 

11.  These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  345.1 × 10²          b)  0.234 × 10⁻³          c)  1,800 × 10⁻²

 

12.  These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  8,099 × 10⁻⁸         b)  34.5 × 10⁰          c)  0.000332 × 10⁴

 

13.  Write these numbers in scientific notation by counting the number of places the decimal point is moved.

a)  123,456.78          b)  98,490          c)  0.000000445

 

14.  Write these numbers in scientific notation by counting the number of places the decimal point is moved.

a)  0.000552          b)  1,987          c)  0.00000000887

 

15.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  456 × (7.4 × 10⁸) = ?               b)  (3.02 × 10⁵) ÷ (9.04 × 10¹⁵) = ?                c)  0.0044 × 0.000833 = ?

 

16.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  98,000 × 23,000 = ?              b)  98,000 ÷ 23,000 = ?              c)  (4.6 × 10⁻⁵) × (2.09 × 10³) = ?

 

17.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  45 × 132 ÷ 882 = ?          b) [(6.37 × 10⁴) × (8.44 × 10⁻⁴)] ÷ (3.2209 × 10¹⁵) = ?

 

18.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  (9.09 × 10⁸) ÷ [(6.33 × 10⁹) × (4.066 × 10⁻⁷)] = ?          b)  9,345 × 34.866 ÷ 0.00665 = ?   Answers 1. a)  5.69 × 10¹          b)  5.631 × 10⁵          c)  8.04 × 10⁻²          d)  6.67 × 10⁻⁶ 2. a)  −8.9 × 10⁵          b)  6.02 × 10¹¹          c)  4.099 × 10^-7          d)  1.1 × 10^-14 3. a)  6.56 × 10⁻³          b)  6.56 × 10⁴          c)  4.567 × 10⁶          d)  5.507 × 10⁻⁶ 4. a)  6.5 × 10¹          b)  −3.2109 × 10²          c)  7.7099 × 10^-5          d)  2.18 × 10^-10 5. a) 138,100          b)  0.000000522          c)  99,980 6. a)  0.0711          b)  918          c)  0.000000000309 7. a)  8.09          b) 0.00003088          c)  −423.9 8. a)  0.0000000287           b)  178,000,000,000           c)  0.00000000000000000000001381 9. a)  7.244 × 10⁴          b)   9.943 × 10⁻²          c)  5.88399 × 10⁷ 10. a)  7.7 × 10⁻¹²          b)  1.11 × 10⁴          c)  6.02000 × 10²³ 11. a)  3.451 × 10⁴          b)  2.34 × 10⁻⁴          c)  1.8 × 10¹

12. a)  8.099 × 10⁻⁵         b)  3.45 × 10¹          c)  3.32 × 10⁰ 13. a)  1.2345678 × 10⁵          b)  9.849 × 10⁴          c)  4.45 × 10⁻⁷ 14. a)  5.52 × 10^-4          b)  1.987 × 10³          c)  8.87 × 10⁹ 15. a) 3.3744 × 10¹¹          b) 3.3407 × 10⁻¹¹          c) 3.665 × 10⁻⁶ 16. a) 2.254 × 10⁹               b)  4.2609 × 10⁰             c)  9.614 × 10^-2 17. a)  6.7346 × 10⁰          b) 1.6691 × 10^-14 18. a) 3.53177 × 10⁵          b)  4.89959 × 10⁷

]]> 2088 2084 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/other-units-temperature-and-density/ Thu, 12 Apr 2018 03:52:01 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/other-units-temperature-and-density/

Learning Objectives

By the end of this section, you will be able to:

• Define density and use it as a conversion factor.

Density is a physical property that is defined as a substance’s mass divided by its volume:

density = mass/volume  or d = m/V

Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm³, g/L, kg/L, and even kg/m³. Densities for some common substances are listed in Table 1 "Densities of Some Common Substances".

Table 1. Densities of Some Common Substances

Substance	Density (g/mL or g/cm3)
water	1.0
gold	19.3
mercury	13.6
air	0.0012
cork	0.22–0.26
aluminum	2.7
iron	7.87

Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm³. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table 1 "Densities of Some Common Substances"), the volume units will cancel and leave you with mass units, telling you the mass of the sample:

7.88 cm³ × 2.7 g/cm³ = 21 g of aluminum

where we have limited our answer to two significant figures.

Example 1

What is the mass of 44.6 mL of mercury?

 

Solution

Use the density from Table 1 "Densities of Some Common Substances" as a conversion factor to go from volume to mass:

44.6 mL × 13.6 g/mL = 607 g

The mass of the mercury is 607 g.

 

Test Yourself

What is the mass of 25.0 cm³ of iron?

 

Answer

197 g

Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:

Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator: That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units but also the numbers:

This reciprocal density is still a useful conversion factor, but now the mass unit will cancel and the volume unit will be introduced. Thus, if we want to know the volume of 45.9 g of gold, we would set up the conversion as follows:

Note how the mass units cancel, leaving the volume unit, which is what we’re looking for.

Example 2

A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/cm³, what is the volume of the cork?

 

Solution

To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find

We can use this expression as the conversion factor. So

Then, taking significant figures into consideration, since the density only has two significant figures, the final answer is 17 cm³.  

Test Yourself

What is the volume of 3.78 g of gold?

 

Answer

0.196 cm³

Example 3

If a 5.00 g sample has a density of 2.50 g/mL, what volume does it occupy?

 

Solution  

First, start with what you know: 5.00 g

Look at the density value as a “conversion factor” (2.50 g/mL) and arrange it so it CANCELS what you already know. Thus, in this case we must invert it.

Multiply and cancel units…

$latex 5.00\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{g} \times \frac{1\;\text{mL}}{2.50\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g}} = 2.00\;\text{mL} $  

Test Yourself

a)   Isopropyl alcohol has a density of 0.785 g/mL. What volume should be measured to obtain 10.0 g of the liquid?

b)   A cube of metal has a mass of 1.45 kg. It is placed in 200.0 mL of water, and the water level rises to 742.1 mL. What is the density of the metal?

  Answers a) 12.7 mL          b) 2.67 g/mL

Care must be used with density as a conversion factor. Make sure the mass units are the same, or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.

Key Concepts and Summary

Density relates a substance’s mass and volume. Density can be used to calculate volume from a given mass or mass from a given volume.

Exercises

1.  Give at least three possible units for density.

2.  A sample of iron has a volume of 48.2 cm³. What is its mass?

3.  The volume of hydrogen used by the Hindenburg, the German airship that exploded in New Jersey in 1937, was 2.000 × 10⁸ L. If hydrogen gas has a density of 0.0899 g/L, what mass of hydrogen was used by the airship?

4.  A typical engagement ring has 0.77 cm³ of gold. What mass of gold is present? 5.  What is the volume of 100.0 g of lead if lead has a density of 11.34 g/cm³? 6.  What is the volume in liters of 222 g of neon if neon has a density of 0.900 g/L? 7.  Which has the greater volume, 100.0 g of iron (d = 7.87 g/cm³) or 75.0 g of gold (d = 19.3 g/cm³)?   Answers 1.  g/mL, g/L, and kg/L (answers will vary) 2.  379 g 3.  1.80 × 10⁷ g 4.  15 g 5.  8.818 cm³ 6.  247 L 7.  The 100.0 g of iron has the greater volume.

]]> 2135 2084 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-2/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-2/

1. How many electrons does it take to make the mass of one proton? 2. Dalton’s initial version of the modern atomic theory says that all atoms of the same element are the same. Is this actually correct? Why or why not? 3. Give complete atomic symbols for the three known isotopes of hydrogen. 4. Use its place on the periodic table to determine if indium, In, atomic number 49, is a metal or a nonmetal. 5. Americium-241 is a crucial part of many smoke detectors. How many neutrons are present in its nucleus 6. Determine the atomic mass of ruthenium from the given abundance and mass data.

Ruthenium-96	5.54%	95.907 u
Ruthenium-98	1.87%	97.905 u
Ruthenium-99	12.76%	98.906 u
Ruthenium-100	12.60%	99.904 u
Ruthenium-101	17.06%	100.906 u
Ruthenium-102	31.55%	101.904 u
Ruthenium-104	18.62%	103.905 u

7. One atomic mass unit has a mass of 1.6605 × 10⁻²⁴ g. What is the mass of one atom of sodium? 8. One atomic mass unit has a mass of 1.6605 × 10⁻²⁴ g. What is the mass of one molecule of H₂O? 9. From their positions on the periodic table, will Cu and I form a molecular compound or an ionic compound? 10. Mercury is an unusual element in that when it takes a 1+ charge as a cation, it always exists as the diatomic ion. a)  Propose a formula for the mercury(I) ion.  b)What is the formula of mercury(I) chloride? 11. The uranyl cation has the formula UO₂²⁺. Propose formulas and names for the ionic compounds between the uranyl cation and F⁻, SO₄²⁻, and PO₄³⁻. 12. Using a periodic table, identify the element symbol and group name for the following elements and identify it as either a metal, non-metal or metalloid: a) Rubidium      b)  Strontium      c)  Californium      d)  Aluminum (no group name required) e)  Iodine           f)   Krypton        g)  Tin (no group name required) 13. List 5 transition metals with their name, element symbol and atomic number. 14. What is a diatomic element? Give several examples. 15. Explain the experiment and findings of each of the following scientists: a) Rutherford       b) J. J. Thompson      c)  Millikan 16. Use Dalton’s theory to explain why potassium nitrate from India or Italy has the same mass percents (or ratios) of K, N and O. 17. Street drugs are often mixed with an inactive substance, such as ascorbic acid (vitamin C). By separating a drug mixture into component substances and calculating the mass of vitamin C per gram of sample, government chemists can track the drug’s distribution. For example, if different cocaine samples from New York, L. A. and Paris all contain 0.6384 g of vitamin C per gram of sample, they likely come from a common source. Is this street sample a compound, element or mixture? In this case, does the constant mass ratio of the components exemplify the law of constant composition? Explain. 18. For each of the following, based on the info given, fill in the blanks: a) Zn, Z = 30, neutrons = 34, atomic number = ?, protons = ? b) Protons = 53, neutrons = 74, symbol = ?, mass number = ?, A = ?, Z = ? c)  Eu, Z = ?, A = 153, protons = ?, neutrons = ? 19.  Complete the following table:

Symbol	Element	Protons	Neutrons	Electrons	Mass Number (A)	Atomic number (Z)
38Ar
	Magnesium		13
				18	37	17
$latex _{27}^{60}\text{Co}$
	Nickel			28	60

20.  Write the complete atomic symbol for each of the following isotopes and state the number of protons, electrons and neutrons for each: a) Fluorine with a mass number of 18       b)  Atomic number of 7 and 8 neutrons c)  Z = 18, neutrons = 22                            d)  He, with A = 3 e)  Z = 82, A = 207                                     f)  Beryllium with 5 neutrons 21.  How many electrons are present in the following ions: a)Fe³⁺   b) Cu⁺   c) Ni²⁺   d) Br^-   e) Eu³⁺   f) P^3- 22.  List the number of protons, electrons and neutrons for the following: a)  $latex _{82}^{208}\text{Pb}^{2+}$       b)  $latex _{7}^{14}\text{N}^{3-}$ 23.  List the number of atoms present in a molecule (or formula unit) of: a) NaNO₃ b) C₂H₅OH    c) Fe(ClO₄)₃

24. Gallium has 2 naturally occurring isotopes, ⁶⁹Ga (isotopic mass = 68.9256 amu, percent abundance = 60.11%) and ⁷¹Ga (isotopic mass = 70.9247 amu, percent abundance = 39.89%). Calculate the average atomic mass of Gallium.

25. Chlorine has 2 naturally occurring isotopes, ³⁵Cl (isotopic mass = 34.9689 amu) and ³⁷Cl (isotopic mass = 36.9659). If the average atomic mass of Cl is 35.4527 amu, what is the percent abundance of each isotope?

26. The two naturally occurring isotopes of nitrogen have masses of 14.0031 amu and 15.0001 amu, respectively. Determine the percentage of ¹⁵N atoms in naturally occurring nitrogen with average atomic mass of 14.0067 amu.  (Hint, the TOTAL abundance of the two isotopes must = 100%.)

 

Answers

1.  About 1,800 electrons 2.  It is not strictly correct because of the existence of isotopes. 3.  $latex _{1}^{1}\text{H}$, $latex _{1}^{2}\text{H}$, and $latex _{1}^{3}\text{H}$ 4.  It is a metal. 5.  146 neutrons 6.  101.065 u 7.  3.817 × 10⁻²³ g 8.  2.991 × 10⁻²³ g 9.  ionic 10.  a) Hg₂²⁺        b) Hg₂Cl₂ 11.  Uranyl fluoride, UO₂F₂; uranyl sulfate, UO₂SO₄; uranyl phosphate, (UO₂)₃(PO₄)₂ 12.  a) Rb, alkali metal, metal      b) Sr, alkaline earth metal, metal      c) Cf, actinide, metal d) Al (no group name required), metal      e) I, halogen, non-metal      f) Kr, nobel gas, non-metal g) Sn (no group name required), metal

13.  (Variety of answers possible)

14.  An element that exists in the natural state as 2 atoms bonded together in a molecule. Examples include H₂, O₂, N₂, halogens

15.  a) Rutherford: performed an experiment where he shot alpha particles at gold foil. He observed that many alphas were deflected and some bounced back, suggesting that Thompson’s view of the atom was incorrect. Instead, he proposed that atom had a nucleus of positive charge, surrounded by a “sea” of negative charge (electrons)

b) J. J. Thompson: used a cathode ray tube (consisting of a negatively charged beam), and calculations regarding the deflection of the beam in a magnetic or electric field, to calculate the mass/charge ratio of the electron.

c) Millikan: performed an experiment where, in a mist of oil in air, droplets were covered with electrons (resulting from an X-ray hitting gas molecules in the air). He measured the rate of fall of these charged oil droplets through an electric field, and from this he determined the actual charge of an electron, and thus (in conjunction with Thompson’s work) the mass of an electron.

16.  The law of constant composition tells us that a given compound will always have the same mass percents of its components. Dalton used this idea, and his concept of the atom, to form his forth postulate, which states that atoms combine in fixed ratios of whole numbers to form compounds. If ratios remain the same, and the masses of each constituent atom are the same, the mass percents will remain the same regardless of the size of the sample.

17.  It is a mixture. It does not exemplify the law of constant composition. The “constant” composition results from the fact that they are all part of the same mixture (same source), but the composition COULD have varied and still produced a mixture of cocaine and vitamin C (with a different amount of vitamin C per sample) depending on the manufacturer.

18.  a) Zn: atomic number = 30, protons = 30 b) symbol = I (or a more complete symbol = $latex _{53}^{127}\text{I}$),  mass number = 127, A = 127, Z = 53 c) Eu: Z = 63, protons = 63, neutrons = 90

19. 

Symbol	Element	Protons	Neutrons	Electrons	Mass Number (A)	Atomic number (Z)
38Ar	Argon	18	20	18	38	18
$latex _{12}^{25}\text{Mg}$	Magnesium	12	13	12	25	12
$latex _{17}^{37}\text{Cl}^{-}$	Chlorine	17	20	18	37	17
$latex _{27}^{60}\text{Co}$	Cobalt	27	33	27	60	27
$latex _{28}^{60}\text{Ni}$	Nickel	28	32	28	60	28

20.  a) ¹⁸₉F, p = 9, e = 9, n = 9             b) ¹⁵₇N, p = 7, e = 7, n = 8 c) ⁴⁰₁₈Ar, p = 18, n = 22, e = 18          d) ³₂He, p = 2, e = 2, n = 1 e)  ²⁰⁷₈₂Pb, p = 82, e = 82, n = 125      f) ⁹₄Be, p = 4, n = 5, e = 4

21.  a) 23      b) 28       c) 26       d) 36      e) 60       f) 18

22.  a) p = 82, n = 126, e = 80         b) p = 7, n = 7, e = 10

23.  a) 1 atom of Na, 1 atom of N and 3 atoms of O  (5 atoms total) b) 2 atoms of C, 6 atoms of H and 1 atom of O  (9 atoms total) c) 1 atom of Fe, 3 atoms of Cl and 12 atoms of O  (16 atoms total)

24.  69.72 amu

25.  75.774% ³⁵Cl and 24.226% ³⁷Cl

26.  0.36% ¹⁵N

]]> 2161 1293 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-to-chemical-reactions-and-equations/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-to-chemical-reactions-and-equations/

Introduction

The space shuttle—and any other rocket-based system—uses chemical reactions to propel itself into space and maneuver itself when it gets into orbit. The rockets that lift the orbiter are of two different types. The three main engines are powered by reacting liquid hydrogen with liquid oxygen to generate water. Then there are the two solid rocket boosters, which use a solid fuel mixture that contains mainly ammonium perchlorate and powdered aluminum.

[caption id="attachment_2163" align="alignright" width="200"] Source: “Delta IV Medium Rocket DSCS” by U.S. Air Force is is in the public domain[/caption]

The chemical reaction between these substances produces aluminum oxide, water, nitrogen gas, and hydrogen chloride. Although the solid rocket boosters each have a significantly lower mass than the liquid oxygen and liquid hydrogen tanks, they provide over 80% of the lift needed to put the shuttle into orbit—all because of chemical reactions.

Chemistry is largely about chemical changes. Indeed, if there were no chemical changes, chemistry as such would not exist! Chemical changes are a fundamental part of chemistry. Because chemical changes are so central, it may be no surprise that chemistry has developed some special ways of presenting them.

]]> 2164 2162 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/oxidation-reduction-reactions/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/oxidation-reduction-reactions/

Learning Objectives

By the end of this section, you will be able to:

• Define oxidation and reduction.
• Assign oxidation numbers to atoms in simple compounds.
• Recognize a reaction as an oxidation-reduction reaction.
• Recognize composition, decomposition, combustion and single replacement reactions.
• Predict the products of a combustion reaction.

Redox Reactions

Earth’s atmosphere contains about 20% molecular oxygen, O₂, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O₂, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.

Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

$latex 2\text{Na}(s) + \text{Cl}_2(g) \longrightarrow 2\text{NaCl}(s)$

It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:

$latex 2\text{Na}(s) \longrightarrow 2\text{Na}^{+}(s) + 2\text{e}^{-}$ $latex \text{Cl}_2(g) + 2\text{e}^{-} \longrightarrow 2\text{Cl}^{-}(s)$

These equations show that Na atoms lose electrons while Cl atoms (in the Cl₂ molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:

$latex \begin{array}{r @ {{}={}} l} \pmb{\text{oxidation}} & \text{loss of electrons} \\[1em] \pmb{\text{reduction}} & \text{gain of electrons} \end{array} $

In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.

$latex \begin{array}{r @ {{}={}} l} \pmb{\text{reducing agent}} & \text{species that is oxidized} \\[1em] \pmb{\text{oxidizing agent}} & \text{species that is reduced} \end{array} $

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:

$latex \text{H}_2(g) + \text{Cl}_2(g) \longrightarrow 2 \text{HCl}(g) $

The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic.

The following guidelines are used to assign oxidation numbers to each element in a molecule or ion:

1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion’s charge.
3. Oxidation numbers for common non-metals are usually assigned as follows:
   • Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
   • Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, O₂²⁻), very rarely $latex -\frac{1}{2}$ (so-called superoxides, O₂⁻), positive values when combined with F (values vary)
   • Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.

Example 1

Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:

a) H₂S

b) SO₃²⁻

c) Na₂SO₄

 

Solution a) According to guideline 1, the oxidation number for H is +1.

Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

$latex \text{charge on H}_2 \text{S} = 0 = (2 \times +1) + (1 \times x)$

$latex x = 0 = - (2 \times +1) = -2$

b) Guideline 3 suggests the oxidation number for oxygen is −2.

Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

$latex {\text{charge on SO}_3}^{2-} = -2 = (3 \times -2) + (1 \times x)$

$latex x = -2 - (3 \times -2) = +4$

c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately.

According to guideline 2, the oxidation number for sodium is +1.

Assuming the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:

$latex {\text{charge on SO}_4}^{2-} = -2 = (4 \times -2) + (1 \times x)$

$latex x = -2 -(4 \times -2) = +6$

 

Test Yourself Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:

a) KNO₃     b) AlH₃    c) NH₄⁺    d) H₂PO₄⁻

  Answers a) N, +5     b) Al, +3     c) N, −3    d) P, +5

Example 2

Assign oxidation numbers to the atoms in each substance.

a) Br₂      b) SiO₂      c) Ba(NO₃)₂

 

Solution

a) Br₂ is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.

b) By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.

c) The compound barium nitrate can be separated into two parts: the Ba²⁺ ion and the nitrate ion. Considering these separately, the Ba²⁺ ion has an oxidation number of +2 by rule 2. Now consider the NO₃⁻ ion. Oxygen is assigned an oxidation number of −2, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation

x + 3(−2) = −1

where x is the oxidation number of the nitrogen atom and −1 represents the charge on the species. Evaluating,

x + (−6) = −1 x = +5

Thus, the oxidation number on the N atom in the nitrate ion is +5.

 

Test Yourself

Assign oxidation numbers to the atoms in H₃PO₄.

 

Answer

H = +1, O = −2, P = +5

Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 5c). Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

$latex \pmb{\text{oxidation}} = \text{increase in oxidation number}$

$latex \pmb{\text{reduction}} = \text{decrease in oxidation number}$

Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H₂ to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in HCl).

Classification of Redox Reactions

Four classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.

1 - A composition reaction (sometimes also called a combination reaction or a synthesis reaction) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction

2 H₂(g) + O₂(g) $latex \longrightarrow$ 2 H₂O(ℓ)

water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance—water—as a product. So this is a composition reaction.

2 - A decomposition reaction starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),

2 NaHCO₃(s) $latex \longrightarrow$ Na₂CO₃(s) + CO₂(g) + H₂O(ℓ)

sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.

Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.

Example 3

Identify each equation as a composition reaction, a decomposition reaction, or neither.

a) Fe₂O₃ + 3 SO₃ $latex \longrightarrow$ Fe₂(SO₄)₃

b) NaCl + AgNO₃ $latex \longrightarrow$ AgCl + NaNO₃

c) (NH₄)₂Cr₂O₇ $latex \longrightarrow$ Cr₂O₃ + 4 H₂O + N₂

 

Solution

a) In this equation, two substances combine to make a single substance. This is a composition reaction.

b) Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.

c) A single substance reacts to make multiple substances. This is a decomposition reaction.

   

Test Yourself

Identify the equation as a composition reaction, a decomposition reaction, or neither.

C₃H₈ $latex \longrightarrow$ C₃H₄ + 2 H₂  

Answer

decomposition

3 - Combustion reactions in which the reductant, also called a fuel, and oxidant, molecular oxygen, react vigorously and produce significant amounts of heat, and often light, in the form of a flame.  Combustion reactions produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N₂. Many reactants, called fuels, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO₂ and H₂O. For example, the balanced chemical equation for the combustion of methane, CH₄, is as follows:

CH₄ + 2 O₂ $latex \longrightarrow$ CO₂ + 2 H₂O

Kerosene can be approximated with the formula C₁₂H₂₆, and its combustion equation is

2 C₁₂H₂₆ + 37 O₂ $latex \longrightarrow$ 24 CO₂ + 26 H₂O

Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C₂H₅OH, whose combustion equation is

C₂H₅OH + 3 O₂ $latex \longrightarrow$ 2 CO₂ + 3 H₂O

If nitrogen is present in the original fuel, it is converted to N₂, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C₂H₂N₂O₄, we have

2 C₂H₂N₂O₄ + O₂ $latex \longrightarrow$ 4 CO₂ + 2 H₂O + 2 N₂

Example 4

Complete and balance each combustion equation.

a) the combustion of propane, C₃H₈

b) the combustion of ammonia, NH₃

[caption id="attachment_2172" align="aligncenter" width="269"] Figure 1. Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here.  Source: “flowers and propane” by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic[/caption]    

Solution

a) The products of the reaction are CO₂ and H₂O, so our unbalanced equation is

C₃H₈ + O₂ $latex \longrightarrow$ CO₂ + H₂O

Balancing (and you may have to go back and forth a few times to balance this), we get

C₃H₈ + 5 O₂ $latex \longrightarrow$ 3 CO₂ + 4 H₂O b) The nitrogen atoms in ammonia will react to make N₂, while the hydrogen atoms will react with O₂ to make H₂O: NH₃ + O₂ $latex \longrightarrow$ N₂ + H₂O

To balance this equation without fractions (which is the convention), we get

4 NH₃ + 3 O₂ $latex \longrightarrow$ 2 N₂ + 6 H₂O  

Test Yourself

Complete and balance the combustion equation for cyclopropanol, C₃H₆O.

 

Answer

C₃H₆O + 4 O₂ $latex \longrightarrow$ 3 CO₂ + 3 H₂O

Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at 3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.

4 - Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:

$latex \text{Zn}(s) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$

Metallic elements may also be oxidized by solutions of other metal salts; for example:

$latex \text{Cu}(s) + 2 \text{AgNO}_3(aq) \longrightarrow \text{Cu(NO}_3)_2(aq) + 2 \text{Ag}(s)$

This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu²⁺ ions dissolve in the solution to yield a characteristic blue color (Figure 2).

Example 5

Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.

a) $latex \text{ZnCO}_3(s) \longrightarrow \text{ZnO}(s) + \text{CO}_2(g)$

b) $latex 2\text{Ga}(l) + 3\text{Br}_2(l) \longrightarrow 2\text{GaBr}_3(s)$

c) $latex 2\text{H}_2 \text{O}_2(aq) \longrightarrow 2\text{H}_2 \text{O}(l) + \text{O}_2(g)$ d) $latex \text{BaCl}_2(aq) + \text{K}_2 \text{SO}_4(aq) \longrightarrow \text{BaSO}_4(s) + 2\text{KCl}(aq)$

e) $latex \text{C}_2 \text{H}_4(g) + 3\text{O}_2(g) \longrightarrow 2\text{CO}_2(g) + 2\text{H}_2 \text{O}(l)$

  Solution Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.

a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(l) to +3 in GaBr₃(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br₂(l) to −1 in GaBr₃(s). The oxidizing agent is Br₂(l).

c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H₂O₂(aq) to 0 in O₂(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H₂O₂(aq) to −2 in H₂O(l). For disproportionation reactions, the same substance functions as an oxidant and a reductant.

d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C₂H₄(g) to +4 in CO₂(g). The reducing agent (fuel) is C₂H₄(g). Oxygen is reduced, its oxidation number decreasing from 0 in O₂(g) to −2 in H₂O(l). The oxidizing agent is O₂(g).

 

Test Yourself This equation describes the production of tin(II) chloride:

$latex \text{Sn}(s) + 2\text{HCl}(g) \longrightarrow \text{SnCl}_2(s) + \text{H}_2(g)$

Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.

  Answer Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant.

Key Concepts and Summary

Chemical reactions are classified according to similar patterns of behavior. Redox reactions involve a change in oxidation number for one or more reactant elements. There are four classifications of chemical reactions: composition, decomposition, combustion and single displacement.

Exercises

1. Is the reaction

2 K(s) + Br₂(ℓ) $latex \longrightarrow$ 2 KBr(s)

an oxidation-reduction reaction? Explain your answer.

2. In the reaction

2 Ca(s) + O₂(g) $latex \longrightarrow$ 2 CaO

indicate what has lost electrons and what has gained electrons.

3. In the reaction

2 Li(s) + O₂(g) $latex \longrightarrow$ Li₂O₂(s)

indicate what has been oxidized and what has been reduced.

4. Assign oxidation numbers to each atom in each substance.

a)  P₄               b)  SO₂ c)  SO₂²⁻       d)  Ca(NO₃)₂ 5. .  Assign oxidation numbers to each atom in each substance.

a)  CO          b)  CO₂ c)  NiCl₂        d)  NiCl₃

6.  Assign oxidation numbers to each atom in each substance.

a)  CH₂O      b)  NH₃ c)  Rb₂SO₄    d)  Zn(C₂H₃O₂)₂

7.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 NO + Cl₂ $latex \longrightarrow$ 2 NOCl

8.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 KrF₂ + 2 H₂O $latex \longrightarrow$ 2 Kr + 4 HF + O₂

9.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 K + MgCl₂ $latex \longrightarrow$ 2 KCl + Mg

10. Indicate what type, or types, of reaction each of the following represents:

a) $latex \text{Ca}(s) + \text{Br}_2(l) \longrightarrow \text{CaBr}_2(s)$

b) $latex \text{Ca(OH)}_2 (aq) + 2\text{HBr}(aq) \longrightarrow \text{CaBr}_2(aq) + 2\text{H}_2 \text{O}(l)$

c) $latex \text{C}_6 \text{H}_{12}(l) + 9\text{O}_2(g) \longrightarrow 6\text{CO}_2(g) + 6\text{H}_2 \text{O}(g)$

11. Indicate what type, or types, of reaction each of the following represents:

a) $latex \text{H}_2 \text{O}(g) + \text{C}(s) \longrightarrow \text{CO}(g) + \text{H}_2(g)$

b) $latex 2\text{KClO}_3(s) \longrightarrow 2\text{KCl}(s) + 3\text{O}_2(g)$

c) $latex \text{Al(OH)}_3(aq) + 3\text{HCl}(aq) \longrightarrow \text{AlCl}_3(aq) + 3\text{H}_2 \text{O}(l)$

d) $latex \text{Pb(NO}_3)_2(aq) + \text{H}_2 \text{SO}_4(sq) \longrightarrow \text{PbSO}_4(s) + 2\text{HNO}_3(aq)$

12. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer. 13. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

a) H₃PO₄      b) Al(OH)₃      c) SeO₂

d) KNO₂       e) In₂S₃            f) P₄O₆

14. Classify the following as acid-base reactions or oxidation-reduction reactions:

a) $latex \text{Na}_2 \text{S}(aq) + 2 \text{HCl}(aq) \longrightarrow 2 \text{NaCl}(aq) + \text{H}_2 \text{S}(g)$

b) $latex 2\text{Na}(s) + 2\text{HCl}(aq) \longrightarrow 2\text{NaCl}(aq) + \text{H}_2(g)$

c) $latex \text{Mg}(s) + \text{Cl}_2(g) \longrightarrow \text{MgCl}_2(aq) $

d) $latex \text{MgO}(s) + 2\text{HCl}(aq) \longrightarrow \text{MgCl}_2(s) + \text{H}_2 \text{O}(l)$

e) $latex \text{K}_3 \text{P}(s) + 2\text{O}_2(g) \longrightarrow \text{K}_3 \text{PO}_4(s)$

f) $latex 3\text{KOH}(aq) + \text{H}_3 \text{PO}_4(aq) \longrightarrow \text{K}_3\text{PO}_4(aq) + 3 \text{H}_2 \text{O}(l)$

15. Complete and balance the following acid-base equations:

a) HCl gas reacts with solid Ca(OH)₂(s).

b) A solution of Sr(OH)₂ is added to a solution of HNO₃.

16. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.

a) $latex \text{Al}(s) + \text{F}_2(g) \longrightarrow $

b) $latex \text{Al}(s) + \text{CuBr}_2(aq) \longrightarrow \;\text{(single displacement)}$

c) $latex \text{P}_4(s) + \text{O}_2(g) \longrightarrow $

d) $latex \text{Ca}(s) + \text{H}_2 \text{O}(l) \longrightarrow \;\text{(products are a strong base and a diatomic gas)}$

17. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction? 18. Great Lakes Chemical Company produces bromine, Br₂, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl₂. 19. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO₂. (Hint: Water is one of the products.) 20. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:

a) $latex \text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow $

b) $latex \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow $

21. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.

a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)

b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction

c) gaseous H₂S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)

22. Which is a composition reaction and which is not? a)  NaCl + AgNO₃ $latex \longrightarrow$ AgCl + NaNO₃ b)  CaO + CO₂ $latex \longrightarrow$ CaCO₃ 23.  Which is a composition reaction and which is not?

a)  2 SO₂ + O₂ $latex \longrightarrow$ 2 SO₃ b)  6 C + 3 H₂ $latex \longrightarrow$ C₆H₆

24.  Which is a decomposition reaction and which is not?

a)  HCl + NaOH $latex \longrightarrow$ NaCl + H₂O b)  CaCO₃ $latex \longrightarrow$ CaO + CO₂

25.  Which is a decomposition reaction and which is not?

a)  Na₂O + CO₂ $latex \longrightarrow$ Na₂CO₃ b)  H₂SO₃ $latex \longrightarrow$ H₂O + SO₂

26.  Which is a combustion reaction and which is not?

a)  C₆H₁₂O₆ + 6 O₂ $latex \longrightarrow$ 6 CO₂ + 6 H₂O b)  2 Fe₂S₃ + 9 O₂ $latex \longrightarrow$ 2 Fe₂O₃ + 6 SO₂

27.  Which is a combustion reaction and which is not?

a)  P₄ + 5 O₂ $latex \longrightarrow$ 2 P₂O₅ b)  2 Al₂S₃ + 9 O₂ $latex \longrightarrow$ 2 Al₂O₃ + 6 SO₂

28.  Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case. 29.  Complete and balance each combustion equation.

a)  C₄H₉OH + O₂ $latex \longrightarrow$ ? b)  CH₃NO₂ + O₂ $latex \longrightarrow$ ?

  Answers 1. Yes; both K and Br are changing oxidation numbers. 2. Ca has lost electrons, and O has gained electrons. 3. Li has been oxidized, and O has been reduced. 4. a)  P: 0 b)  S: +4; O: −2 c)  S: +2; O: −2 d)  Ca: 2+; N: +5; O: −2 5. a)  C: +2; O: −2 b)  C: +4; O: −2 c)  Ni: +2; Cl: −1 d)  Ni: +3; Cl: −1 6. a)  C: 0; H: +1; O: −2 b)  N: −3; H: +1 c)  Rb: +1; S: +6; O: −2 d)  Zn: +2; C: 0; H: +1; O: −2 7. N is being oxidized, and Cl is being reduced. 8. O is being oxidized, and Kr is being reduced. 9. K is being oxidized, and Mg is being reduced.

10. a) oxidation-reduction (addition); b) acid-base (neutralization); c) oxidation-reduction (combustion)

11. a) single replacement;   b) decomposition;   c) acid-base;   d) precipitation

12. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.

13. a) H +1, P +5, O −2;    b) Al +3, H +1, O −2;    c) Se +4, O −2;

d) K +1, N +3, O −2;    e) In +3, S −2;    f) P +3, O −2

14. a) acid-base;    b) oxidation-reduction: Na is oxidized, H⁺ is reduced;

c) oxidation-reduction: Mg is oxidized, Cl₂ is reduced;     d) acid-base; e) oxidation-reduction: P³⁻ is oxidized, O₂ is reduced;     f) acid-base

15. a) $latex 2\text{HCl}(g) + \text{Ca(OH)}_2(s) \longrightarrow \text{CaCl}_2(s) + 2\text{H}_2 \text{O}(l)$; b) $latex \text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \longrightarrow \text{Sr(NO}_3)_2(aq) + 2\text{H}_2 \text{O}(l)$;

16. a) $latex 2\text{Al}(s) + 3\text{F}_2 \longrightarrow 2\text{AlF}_3(s)$; b) $latex 2\text{Al}(s) + 3\text{CuBr}_2(aq) \longrightarrow 3\text{Cu}(s) + 2\text{AlBr}_3(aq)$; c) $latex \text{P}_4(s) + 5\text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(s)$; d) $latex \text{Ca}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)$;

17. $latex \text{H}_2(g) + \text{F}_2(g) \longrightarrow 2\text{HF}(g) $

18. $latex 2\text{NaBr}(aq) + \text{Cl}_2(g) \longrightarrow 2\text{NaCl}(aq) + \text{Br}_2(l) $

19. $latex 2\text{LiOH}(aq) + \text{CO}_2(g) \longrightarrow \text{Li}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l) $

20. a) $latex \text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow \text{CaS}(s) + 2\text{H}_2\text{O}(l);$ b) $latex \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow \text{Na}_2 \text{S}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)$

21. a) step 1: $latex \text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g)$,

step 2: $latex \text{NH}_3(g) + \text{HNO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3\;\text{(s) (after drying)}$ b) $latex \text{H}_2(g) + \text{Br}_2(l) \longrightarrow 2\text{HBr}(g) $ c) $latex \text{Zn}(s) + \text{S}(s) \longrightarrow \text{ZnS}(s) \;\text{and} \;\text{ZnS}(s) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2 \text{S}(g) $ 22. a)  not composition    b)  composition 23. a)  composition    b)  composition

24. a)  not decomposition    b)  decomposition 25. a)  not decomposition    b)  decomposition

26. a)  combustion   b)  combustion

27. a)  combustion    b)  combustion

28. Yes; 2 H₂ + O₂ $latex \longrightarrow$ 2 H₂O (answers will vary) 29. a)  C₄H₉OH + 6 O₂ $latex \longrightarrow$ 4 CO₂ + 5 H₂O

b)  4 CH₃NO₂ + 3 O₂ $latex \longrightarrow$ 4 CO₂ + 6 H₂O + 2 N₂

 Glossary

combustion reaction: vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light half-reaction: an equation that shows whether each reactant loses or gains electrons in a reaction. oxidation: process in which an element’s oxidation number is increased by loss of electrons oxidation-reduction reaction: (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements oxidation number: (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic oxidizing agent: (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced reduction: process in which an element’s oxidation number is decreased by gain of electrons reducing agent: (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized single-displacement reaction: (also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species

]]> 2177 2162 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/neutralization-reactions/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/neutralization-reactions/

Learning Objectives

By the end of this section, you will be able to:

• Identify common acids and bases
• Define acid-base reactions
• Recognize and identify examples of acid-base reactions
• Predict the products of acid-base reactions.

Acids and Bases

The definition of an acid is often cited as: any compound that increases the amount of hydrogen ion (H⁺) in an aqueous solution. The chemical opposite of an acid is a base. The equivalent definition of a base is that a base is a compound that increases the amount of hydroxide ion (OH⁻) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the Arrhenius definition of an acid and a base, respectively.

You may recognize that, based on the description of a hydrogen atom, an H⁺ ion is a hydrogen atom that has lost its lone electron; that is, H⁺ is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H⁺ ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the hydronium ion H₃O⁺(aq), a water molecule with an extra hydrogen ion attached to it. as H₃O⁺, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H⁺ and H₃O⁺ are treated equivalently.

For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, H₃O⁺. As an example, consider the equation shown here:

$latex \text{HCl}(aq) + \text{H}_2 \text{O}(aq) \longrightarrow \text{Cl}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)$

The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H₃O⁺ ions are produced by a chemical reaction in which H⁺ ions are transferred from HCl molecules to H₂O molecules (Figure 1).

[caption id="attachment_1414" align="aligncenter" width="400"] Figure 1. When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).[/caption]

The nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table 1).

Compound Formula	Name in Aqueous Solution
HBr	hydrobromic acid
HCl	hydrochloric acid
HI	hydroiodic acid
HNO3	nitric acid
HClO4	perchloric acid
HClO3	chloric acid
H2SO4	sulfuric acid
Table 1. Common Strong Acids

 A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions.

Compound Formula	Name in Aqueous Solution
HF	hydrofluoric acid
HCN	hydrocyanic acid
HC2H3O2	acetic acid
HNO2	nitrous acid
HClO	hypochlorous acid
HClO2	chlorous acid
H2SO3	sulfurous acid
H2CO3	carbonic acid
H3PO4	phosphoric acid
Table 2. Common Weak Acids

Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:

$latex \text{CH}_3 \text{CO}_2 \text{H}(aq) + \text{H}_2 \text{O}(l) \leftrightharpoons \text{CH}_3 {\text{CO}_2}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)$

When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, $latex \text{CH}_3 {\text{CO}_2}^{-} $(Figure 2). The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)

[caption id="attachment_1415" align="aligncenter" width="300"] Figure 2. (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Brücke-Osteuropa/Wikimedia Commons)[/caption]

A base is a substance that will dissolve in water to yield hydroxide ions, OH⁻. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion—for example, NaOH and Ca(OH)₂. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)₂ dissolve in water and dissociate completely to produce cations (K⁺ and Ba²⁺, respectively) and hydroxide ions, OH⁻. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases.

Consider as an example the dissolution of lye (sodium hydroxide) in water:

$latex \text{NaOH}(s) \longrightarrow \text{Na}^{+}(aq) + \text{OH}^{-}(aq)$

This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na⁺ and OH⁻ ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.

Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 3). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:

$latex \text{NH}_3(aq) + \text{H}_2 \text{O}(l) \rightleftharpoons {\text{NH}_4}^{+}(aq) + \text{OH}^{-}(aq)$

Under typical conditions, only about 1% of the dissolved ammonia is present as NH₄⁺ ions.

Acid-Base Reactions

An acid-base reaction is one in which a hydrogen ion, H⁺, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion.

The reaction between an acid and a base is called an acid-base reaction or a neutralization reaction. Although acids and bases have their own unique chemistries, the acid and base cancel each other’s chemistry to produce a rather innocuous substance—water. In fact, the general acid-base reaction is

acid + base $latex \longrightarrow$ water + salt

where the term salt is used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. In chemistry, the word salt refers to more than just table salt. For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is

HCl(aq) + KOH(aq) $latex \longrightarrow$ H₂O(ℓ) + KCl(aq)

where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)₂(aq), additional molecules of HCl and H₂O are required to balance the chemical equation:

2 HCl(aq) + Mg(OH)₂(aq) $latex \longrightarrow$ 2 H₂O(ℓ) + MgCl₂(aq)

Here, the salt is MgCl₂. This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid.

There are acid-base reactions that do not follow the "general acid-base" equation given above.  For example, , the balanced chemical equation for the reaction between HCl(aq) and NH₃(aq) is

HCl(aq) + NH₃(aq) $latex \longrightarrow$ NH₄Cl(aq)

Example 1

Write the neutralization reactions between each acid and base.

a) HNO₃(aq) and Ba(OH)₂(aq)             b)H₃PO₄(aq) and Ca(OH)_2(aq)

 

Solution

First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.

a) The expected products are water and barium nitrate, so the initial chemical reaction is

HNO₃(aq) + Ba(OH)₂(aq) $latex \longrightarrow$ H₂O(ℓ) + Ba(NO₃)₂(aq)

To balance the equation, we need to realize that there will be two H₂O molecules, so two HNO₃ molecules are required:

2HNO₃(aq) + Ba(OH)₂(aq) $latex \longrightarrow$ 2H₂O(ℓ) + Ba(NO₃)₂(aq)

This chemical equation is now balanced.

b) The expected products are water and calcium phosphate, so the initial chemical equation is

H₃PO₄(aq) + Ca(OH)₂(aq) $latex \longrightarrow$ H₂O(ℓ) + Ca₃(PO₄)₂(s)

According to the solubility rules, Ca₃(PO₄)₂ is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions; we end up with six water molecules to balance the equation:

2 H₃PO₄(aq) + 3 Ca(OH)₂(aq) $latex \longrightarrow$ 6 H₂O(ℓ) + Ca₃(PO₄)₂(s)

This chemical equation is now balanced.

 

Test Yourself

Write the neutralization reaction between H₂SO₄(aq) and Sr(OH)₂(aq).

 

Answer

H₂SO₄(aq) + Sr(OH)₂(aq) $latex \longrightarrow$ 2 H₂O(ℓ) + SrSO₄(aq)

Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)₃(s) still proceeds according to the equation

3 HCl(aq) + Fe(OH)₃(s) $latex \longrightarrow$ 3 H₂O(ℓ) + FeCl₃(aq)

even though Fe(OH)₃ is not soluble. When one realizes that Fe(OH)₃(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids—the neutralization reaction produces products that are soluble and wash away. Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!

Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq),

HCl(aq) + NaOH(aq) $latex \longrightarrow$ H₂O(ℓ) + NaCl(aq)

the complete ionic reaction is

H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) $latex \longrightarrow$ H₂O(ℓ) + Na⁺(aq) + Cl⁻(aq)

The Na⁺(aq) and Cl⁻(aq) ions are spectator ions, so we can remove them to have

H⁺(aq) + OH⁻(aq) $latex \longrightarrow$ H₂O(ℓ)

as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H₃O⁺(aq), we would write it as

H₃O⁺(aq) + OH⁻(aq) $latex \longrightarrow$ 2H₂O(ℓ)

With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.

However, for the reaction between HCl(aq) and Cr(OH)₂(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:

2 H⁺(aq) + 2 Cl⁻(aq) + Cr(OH)₂(s) $latex \longrightarrow$ 2 H₂O(ℓ) + Cr²⁺(aq) + 2 Cl⁻(aq)

The chloride ions are the only spectator ions here, so the net ionic equation is

2 H⁺(aq) + Cr(OH)₂(s) $latex \longrightarrow$ 2 H₂O(ℓ) + Cr²⁺(aq)

Example 2

Write balanced chemical equations for the acid-base reactions described here:

a) the weak acid hydrogen hypochlorite reacts with water

b) a solution of barium hydroxide is neutralized with a solution of nitric acid

 

Solution a) The two reactants are provided, HOCl and H₂O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H⁺ from HOCl to H₂O to generate hydronium ions, H₃O⁺ and hypochlorite ions, OCl⁻.

$latex \text{HOCl}(aq) + \text{H}_2 \text{O}(l) \rightleftharpoons \text{OCl}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)$

A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.

b) The two reactants are provided, Ba(OH)₂ and HNO₃. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba²⁺) and the anion generated when the acid transfers its hydrogen ion (NO³⁻).

$latex \text{Ba(OH)}_2(aq) + 2\text{HNO}_3(aq) \longrightarrow \text{Ba(NO}_3)_2(aq) + 2\text{H}_2 \text{O}(l)$

 

Test Yourself Write the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. Hint: Consider the ions produced when a strong acid is dissolved in water.

  Answer $latex \text{H}_3 \text{O}^{+}(aq) + \text{OH}^{-}(aq) \longrightarrow 2\text{H}_2 \text{O}(l)$

Example 3

Oxalic acid, H₂C₂O₄(s), and Ca(OH)₂(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? The anion in oxalic acid is the oxalate ion, C₂O₄²⁻.

 

Solution

The products of the neutralization reaction will be water and calcium oxalate:

H₂C₂O₄(s) + Ca(OH)₂(s) $latex \longrightarrow$ 2 H₂O(ℓ) + CaC₂O₄(s)

Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.

 

Test Yourself

What is the net ionic equation between HNO₃(aq) and Ti(OH)₄(s)?

 

Answer

4 H⁺(aq) + Ti(OH)₄(s) $latex \longrightarrow$ 4 H₂O(ℓ) + Ti⁴⁺(aq)

 

Explore the microscopic view of strong and weak acids and bases.

 

Gas-forming Acid-Base reactions

A driving force for certain acid-base reactions is the formation of a gas. Common gases formed are  H₂, O₂, and CO₂. For example:

2HCl(aq) + Na₂CO₃(aq) $latex \longrightarrow$ H₂CO₃(aq) + 2NaCl(aq) $latex \longrightarrow$ CO₂(g) + H₂O(l) + 2NaCl(aq)

The above example can be viewed as an acid-base reaction followed by a decomposition. The driving force in this case is the gas formation.  The decomposition of H₂CO₃ into CO₂ and H₂O is a very common reaction. Both Na₂CO₃ and NaHCO₃ mixed with acid result in a gas-forming acid-base reaction.

HCl(aq) + NaHCO₃(aq) $latex \longrightarrow$ H₂CO₃(aq) + NaCl(aq) $latex \longrightarrow$ CO₂(g) + H₂O(l) + NaCl(aq)

Food and Drink App: Acids in Foods

Many foods and beverages contain acids. Acids impart a sour note to the taste of foods, which may add some pleasantness to the food. For example, orange juice contains citric acid, H₃C₆H₅O₇. Note how this formula shows hydrogen atoms in two places; the first hydrogen atoms written are the hydrogen atoms that can form H⁺ ions, while the second hydrogen atoms written are part of the citrate ion, C₆H₅O₇³⁻. Lemons and limes contain much more citric acid—about 60 times as much—which accounts for these citrus fruits being more sour than most oranges. Vinegar is essentially a ~5% solution of acetic acid (HC₂H₃O₂) in water. Apples contain malic acid (H₂C₄H₄O₅; the name malic acid comes from the apple’s botanical genus name, malus), while lactic acid (HC₃H₅O₃) is found in wine and sour milk products, such as yogurt and some cottage cheeses.

Table 3 "Various Acids Found in Food and Beverages" lists some acids found in foods, either naturally or as an additive. Frequently, the salts of acid anions are used as additives, such as monosodium glutamate (MSG), which is the sodium salt derived from glutamic acid. As you read the list, you should come to the inescapable conclusion that it is impossible to avoid acids in food and beverages.

Acid Name	Acid Formula	Use and Appearance
acetic acid	HC2H3O2	flavouring; found in vinegar
adipic acid	H2C6H8O4	flavouring; found in processed foods and some antacids
alginic acid	various	thickener; found in drinks, ice cream, and weight loss products
ascorbic acid	HC6H7O6	antioxidant, also known as vitamin C; found in fruits and vegetables
benzoic acid	HC6H5CO2	preservative; found in processed foods
citric acid	H3C6H5O7	flavouring; found in citrus fruits
dehydroacetic acid	HC8H7O4	preservative, especially for strawberries and squash
erythrobic acid	HC6H7O6	antioxidant; found in processed foods
fatty acids	various	thickener and emulsifier; found in processed foods
fumaric acid	H2C4H2O4	flavouring; acid reactant in some baking powders
glutamic acid	H2C5H7NO4	flavouring; found in processed foods and in tomatoes, some cheeses, and soy products
lactic acid	HC3H5O3	flavouring; found in wine, yogurt, cottage cheese, and other sour milk products
malic acid	H2C4H4O5	flavouring; found in apples and unripe fruit
phosphoric acid	H3PO4	flavouring; found in some colas
propionic acid	HC3H5O2	preservative; found in baked goods
sorbic acid	HC6H7O2	preservative; found in processed foods
stearic acid	HC18H35O2	anticaking agent; found in hard candies
succinic acid	H2C4H4O4	flavouring; found in wine and beer
tartaric acid	H2C4H4O6	flavouring; found in grapes, bananas, and tamarinds

Table 3. Various Acids Found in Food and Beverages

Key Concepts and Summary

Chemical reactions are classified according to similar patterns of behaviour. Acid-base reactions involve the transfer of hydrogen ions between reactants. General acid-base reactions, also called neutralization reactions can be summarized with the following reaction equation:

ACID(aq) + BASE(aq) $latex \longrightarrow$ H₂O(l) + SALT(aq) or (s)

The DRIVING FORCE for a general acid-base reaction is the formation of water. Gas-forming acid-base reactions can be summarized with the following reaction equation:

ACID(aq) + NaHCO₃ or Na₂CO₃(aq) $latex \longrightarrow$ H₂O(l) + CO₂(g) + SALT(aq) or (s)

The DRIVING FORCE for a gas-forming acid-base reaction is the formation of gas. There are three ways of There are three ways of representing a neutralization reaction, using a molecular equation, complete ionic equation or net ionic equation, as described in section 6.1.

Exercises

1. What is the Arrhenius definition of an acid?

2. What is the Arrhenius definition of a base?

3. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.

a)  HCl and KOH b)  H₂SO₄ and KOH c)  H₃PO₄ and Ni(OH)₂

4.  Write a balanced chemical equation for each neutralization reaction in Exercise 3.

5.  Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.

a)  HI(aq) + KOH(aq) $latex \longrightarrow$ ? b)  H₂SO₄(aq) + Ba(OH)₂(aq) $latex \longrightarrow$ ?

6.  Write the net ionic equation for each neutralization reaction in Exercise 7.

7.  Write the complete and net ionic equations for the neutralization reaction between HClO₃(aq) and Zn(OH)₂(s). Assume the salt is soluble. 8.  Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is the same as the net ionic equation for the neutralization reaction between HNO₃(aq) and RbOH. 9.  Write the complete and net ionic equations for the neutralization reaction between HCl(aq) and KOH(aq) using the hydronium ion in place of H⁺. What difference does it make when using the hydronium ion? 10. Complete and balance the following acid-base equations:

a) HCl gas reacts with solid Ca(OH)₂(s).

b) A solution of Sr(OH)₂ is added to a solution of HNO₃.

11. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.

a) $latex \text{Mg(OH)}_2(s) + \text{HClO}_4(aq) \longrightarrow $

b) $latex \text{SrO}(s) + \text{H}_2 \text{SO}_4(l) \longrightarrow $

12. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:

a) $latex \text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow $

b) $latex \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow $

  Answers 1. An Arrhenius acid increases the amount of H⁺ ions in an aqueous solution.

2. An Arrhenius base increases the amount of OH^- ions in an aqueous solution.

3. a)  KCl and H₂O

b)  K₂SO₄ and H₂O c)  Ni₃(PO₄)₂ and H₂O

4. a)    HCl + KOH $latex \longrightarrow$ KCl + H₂O b)  H₂SO₄ + 2 KOH $latex \longrightarrow$ K₂SO₄ + 2 H₂O c)  2 H₃PO₄ + 3 Ni(OH)₂ $latex \longrightarrow$ Ni₃(PO₄)₂ + 6 H₂O 5. a)  HI(aq) + KOH(aq) $latex \longrightarrow$ KCl(aq) + H₂O(ℓ)

b)  H₂SO₄(aq) + Ba(OH)₂(aq) $latex \longrightarrow$ BaSO₄(s) + 2 H₂O(ℓ)

6. a)  H⁺(aq) + OH⁻(aq) $latex \longrightarrow$ H₂O(ℓ)

b)  2 H⁺(aq) + SO₄²⁻(aq) + Ba²⁺(aq) + 2 OH⁻(aq) $latex \longrightarrow$ BaSO₄(s) + 2 H₂O(ℓ) 7.  Complete ionic equation:

2 H⁺(aq) + 2 ClO₃⁻(aq) + Zn²⁺(aq) + 2 OH⁻(aq) $latex \longrightarrow$ Zn²⁺(aq) + 2 ClO₃⁻(aq) + 2 H₂O(ℓ)

Net ionic equation:

2 H⁺(aq) + 2 OH⁻(aq) $latex \longrightarrow$ 2 H₂O(ℓ)

8.  Because the salts are soluble in both cases, the net ionic reaction is just H⁺(aq) + OH⁻(aq) $latex \longrightarrow$ H₂O(ℓ). 9.  Complete ionic equation:

H₃O⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) $latex \longrightarrow$ 2 H₂O(ℓ) + K⁺(aq) + Cl⁻(aq)

Net ionic equation:

H₃O⁺(aq) + OH⁻(aq) $latex \longrightarrow$ 2 H₂O(ℓ)

The difference is simply the presence of an extra water molecule as a product.

10. a) $latex 2\text{HCl}(g) + \text{Ca(OH)}_2(s) \longrightarrow \text{CaCl}_2(s) + 2\text{H}_2 \text{O}(l)$;

b) $latex \text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \longrightarrow \text{Sr(NO}_3)_2(aq) + 2\text{H}_2 \text{O}(l)$;

11. a) $latex \text{Mg(OH)}_2(s) + 2\text{HClO}_4(aq) \longrightarrow \text{Mg}^{2+}(aq) + 2{\text{ClO}_4}^{-}(aq) + 2\text{H}_2 \text{O}(l); $ b) $latex \text{SrO}(s) + \text{H}_2 \text{SO}_4(l) \longrightarrow \text{SrSO}_4(s) + \text{H}_2 \text{O}$

12. a) $latex \text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow \text{CaS}(s) + 2\text{H}_2\text{O}(l);$ b) $latex \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow \text{Na}_2 \text{S}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)$

Glossary

acid: substance that produces H₃O⁺ when dissolved in water acid-base reaction: reaction involving the transfer of a hydrogen ion between reactant species base: substance that produces OH⁻ when dissolved in water neutralization reaction: reaction between an acid and a base to produce salt and water salt: ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide strong acid: acid that reacts completely when dissolved in water to yield hydronium ions strong base: base that reacts completely when dissolved in water to yield hydroxide ions weak acid: acid that reacts only to a slight extent when dissolved in water to yield hydronium ions weak base: base that reacts only to a slight extent when dissolved in water to yield hydroxide ions

]]> 2178 2162 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-3/ Thu, 12 Apr 2018 03:52:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-3/

4 Na(s) + 2 Cl₂(g) $latex \longrightarrow$ 4 NaCl(s)

4. Does the following chemical reaction proceed as written? Why or why not?

3 Zn(s) + 2 Al(NO₃)₃(aq) $latex \longrightarrow$ 3 Zn(NO₃)₂(aq) + 2 Al(s)

5. Explain what is wrong with this double-replacement reaction.

NaCl(aq) + KBr(aq) $latex \longrightarrow$ NaK(aq) + ClBr(aq)

6. Predict the products of and balance this double-replacement reaction.

Ag₂SO₄(aq) + SrCl₂(aq) $latex \longrightarrow$ ?

7. Write the complete and net ionic equations for this double-replacement reaction.

BaCl₂(aq) + Ag₂SO₄(aq) $latex \longrightarrow$ ?

8. Write the complete and net ionic equations for this double-replacement reaction.

Ag₂SO₄(aq) + SrCl₂(aq) $latex \longrightarrow$ ?

9. Identify the spectator ions in this reaction. What is the net ionic equation?

NaCl(aq) + KBr(aq) $latex \longrightarrow$ NaBr(aq) + KCl(aq)

10. Can a reaction be a composition reaction and a redox reaction at the same time? Give an example to support your answer. 11. Can a reaction be a decomposition reaction and a redox reaction at the same time? Give an example to support your answer. 12. Why is CH₄ not normally considered an acid? 13. What are the oxidation numbers of the nitrogen atoms in these substances?

a)  N₂      b)  NH₃      c)  NO      d)  N₂O e)  NO₂      f)  N₂O₄      g)  N₂O₅      h)  NaNO₃

14.  Disproportion is a type of redox reaction in which the same substance is both oxidized and reduced. Identify the element that is disproportionating and indicate the initial and final oxidation numbers of that element.

2 CuCl(aq) $latex \longrightarrow$ CuCl₂(aq) + Cu(s)

15. Write the unbalanced chemical equation for each of the following reactions.

a) solid mercury(II) oxide decomposes to produce liquid mercury metal and gaseous oxygen.

b) Solid zinc metal reacts with hydrochloric acid to produce zinc chloride dissolved in water and hydrogen gas.

c) Propane (C₃H₈) gas reacts with oxygen in air to produce gaseous carbon dioxide and water vapor.

d) Solid ammonium nitrate can be produced by bubbling ammonia gas through nitric acid solution.

e) Elemental boron can be produced by heating solid diboron trioxide with magnesium metal, also producing solid magnesium oxide as a by-product.

16. Beneath each word equation, write the formula equation and balance it:

a) zinc + sulfuric acid  $latex \longrightarrow$  zinc sulfate + hydrogen

b) carbon + oxygen  $latex \longrightarrow$  carbon dioxide

c) hydrogen + oxygen  $latex \longrightarrow$  water

d)  aluminum + hydrochloric acid  $latex \longrightarrow$ aluminum chloride + hydrogen

e) chromium + oxygen  $latex \longrightarrow$ chromium(III) oxide

f) potassium + water  $latex \longrightarrow$  potassium hydroxide + hydrogen

g) copper(II) oxide + hydrochloric acid $latex \longrightarrow$  copper(II) chloride + water 

h) sodium hydrogen carbonate + nitric acid $latex \longrightarrow$  sodium nitrate + water + carbon dioxide

17. Balance the following equations:

a) H₃PO₄  +  CaO $latex \longrightarrow$ Ca₃(PO₄)₂  +  H₂O                    

b) NH₃  +  O₂ $latex \longrightarrow$ NO₂  +  H₂O

c) Cl₂O₇  +  H₂O $latex \longrightarrow$ HClO₄                                       

d) Mg₃N₂  +  H₂O $latex \longrightarrow$ Mg(OH)₂  +  NH₃

e) FeSO₄ $latex \longrightarrow$ Fe₂O₃  +  SO₂  +  O₂                             

f)P₄  +  Cl₂ $latex \longrightarrow$ PCl₃

g) MnO₂  +  C $latex \longrightarrow$ Mn  +  CO₂                                     

h) Na₂O₂  +  H₂O $latex \longrightarrow$ NaOH   +  O₂

i) CaH₂  +  H₂O $latex \longrightarrow$ Ca(OH)₂  +  H₂                             

j)NaHCO₃ $latex \longrightarrow$ Na₂CO₃  +  CO₂  +  H₂O

18. Balance the following equations:

a)  Mg  +  O₂  $latex \longrightarrow$ MgO                                          

b)  KClO₃  $latex \longrightarrow$ KCl  +  O₂

c)  Fe  +  O₂  $latex \longrightarrow$ Fe₃O₄                                         

d)  Mg +  HCl  $latex \longrightarrow$ MgCl₂  +  H₂

e)  Na  +  H₂O  $latex \longrightarrow$ NaOH  +  H₂                            

f)  N₂+  H₂ $latex \longrightarrow$ NH₃

g)  Na₂CO₃•10H₂O   $latex \longrightarrow$  Na₂CO₃+  H₂O               

h)  Fe +  H₂O  $latex \longrightarrow$ Fe₃O₄+  H₂

i)  F₂  +  H₂O  $latex \longrightarrow$ HF +  O₂

19. Balance the following chemical equations.

a)   Ca₃(PO₄)₂   +   SiO₂   +   C   $latex \longrightarrow$    P₄  +    CaSiO₃    +   CO₂

b)   C₂H₆O₂   +   O₃  $latex \longrightarrow$    CO₂   +   H₂O

c)   Fe₃O₄    +    S₈   $latex \longrightarrow$  Fe₂S₃   +    SO₃

d)   P₂S₅   +    O₂   $latex \longrightarrow$ S₈   +    P₃O₈

e)   C₃H₈   +    SO₂ $latex \longrightarrow$ C₂H₄O₄   +   H₂O   +   S₈

20. Predict whether or not a reaction will occur when each of the following pairs of solutions are mixed. If no reaction occurs, write NR on the right hand side of the equation. If a reaction does occur, complete and balance the equations as molecular equations, and give the balanced complete ionic and net ionic equations as well. Be sure to indicate the states of all reagents.

a)  NaCl(aq)  +  AgNO₃(aq) $latex \longrightarrow$                        

b)  BaCl₂(aq)  +  H₂SO₄(aq) $latex \longrightarrow$

c)  FeCl₃(aq)  +  NH₄OH(aq) $latex \longrightarrow$                       

d)  K₂CrO₄(aq)  + Pb(NO₃)₂(aq) $latex \longrightarrow$

e)  KClO₃(aq)  +  MgCl₂(aq) $latex \longrightarrow$                        

f)  (NH₄)₂CO₃(aq)  +  CaCl₂(aq) $latex \longrightarrow$

g)  NaC₂H₃O₂(aq)  +  BaCl₂(aq) $latex \longrightarrow$

21. Identify each of the following unbalanced reaction equations as belonging to one or more of the following categories: precipitation, acid-base, or redox.

 

a)  Fe_(s)+ H₂SO_4(aq) $latex \longrightarrow$ Fe₃(SO₄)_2(aq)+ H_2(g)

b)  HClO_4(aq)+ RbOH_(aq) $latex \longrightarrow$ RbClO_4(aq)+ H₂O_(l)

c)  Ca_(s)+ O_2(g) $latex \longrightarrow$ CaO_(s)

d)  H₂SO_4(aq)+ NaOH_(aq) $latex \longrightarrow$ Na₂SO_4(aq)+ H₂O_(l)

e)  Pb(NO₃)_2(aq)+ Na₂CO_3(aq) $latex \longrightarrow$ PbCO_3(s)+ NaNO_3(aq)

f)  K₂SO_4(aq)+ CaCl_2(aq) $latex \longrightarrow$ KCl_(aq)+ CaSO_4(s)

22. Classify the following unbalanced chemical reactions by as many methods as possible.

a)  I₄O_9(s) $latex \longrightarrow$ I₂O_6(s)+ I_2(s)+ O_2(g)

b)  Mg_(s)+ AgNO_3(aq) $latex \longrightarrow$ Mg(NO₃)_2(aq)+ Ag_(s)

c)  SiCl_4(l)+ Mg_(s) $latex \longrightarrow$ MgCl_2(aq)+ Si_(s)

d)  CuCl_2(aq)+ AgNO_3(aq) $latex \longrightarrow$ Cu(NO₃)_2(aq)+ AgCl_(s)

e)  Al_(s)+ Br_2(l) $latex \longrightarrow$ AlBr_3(s)

f)  HBr_(aq)+ NaOH_(aq) $latex \longrightarrow$ NaBr_(aq)+ H₂O_(l)

Answers

1. H₂O(ℓ) $latex \longrightarrow$ H₂O(g) 2. H₂O(ℓ) $latex \longrightarrow$ H₂O(s) 3. The coefficients are not in their lowest whole-number ratio. 4. No; zinc is lower in the activity series than aluminum. 5. In the products, the cation is pairing with the cation, and the anion is pairing with the anion.

6. Ag₂SO₄(aq) + SrCl₂(aq) $latex \longrightarrow$ SrSO₄(s) + 2 AgCl(s) 

7. Complete ionic equation: Ba²⁺(aq) + 2 Cl⁻(aq) + 2 Ag⁺(aq) + SO₄²⁻(aq) $latex \longrightarrow$ BaSO₄(s) + 2 AgCl(s)

Net ionic equation: The net ionic equation is the same as the complete ionic equation. 8. Complete ionic equation: Sr²⁺(aq) + 2 Cl⁻(aq) + 2 Ag⁺(aq) + SO₄²⁻(aq) $latex \longrightarrow$ SrSO₄(s) + 2 AgCl(s) Net ionic equation: The net ionic equation is the same as the complete ionic equation. 9. Each ion is a spectator ion; there is no overall net ionic equation. 10. Yes; H₂ + Cl₂ $latex \longrightarrow$ 2 HCl (answers will vary) 11. Yes; 2 HCl $latex \longrightarrow$ H₂ + Cl₂ (answers will vary) 12. It does not increase the H⁺ ion concentration; it is not a compound of H⁺. 13. a)  0      b)  −3      c)  +2      d)  +1 e)  +4          f) +4        g)  +5      h)  +5 14. Copper is disproportionating. Initially, its oxidation number is +1; in the products, its oxidation numbers are +2 and 0, respectively.

15.   a) HgO(s) $latex \longrightarrow$ Hg(l) + O₂(g) b)  Zn(s) + HCl(aq) $latex \longrightarrow$ ZnCl₂(aq) + H₂(g) c)  C₃H₈(g) + O₂(g) $latex \longrightarrow$ CO₂(g) + H₂O(g) d)  NH₃(g) + HNO₃(aq) $latex \longrightarrow$ NH₄NO₃(s) e)  B₂O₃(s) + Mg(s) $latex \longrightarrow$ B(s) + MgO(s)

16.   a) Zn(s) + H₂SO₄(aq) $latex \longrightarrow$  ZnSO₄(aq)  +  H₂(g) b)  C(s) + O₂(g) $latex \longrightarrow$ CO₂(g) c)  2H₂(g) + O₂(g)  $latex \longrightarrow$  2H₂O(l) d)  2Al(s) + 6HCl(aq)  $latex \longrightarrow$  2AlCl₃(aq)  +  3H₂(g) e)  4Cr(s) + 3O₂(g)  $latex \longrightarrow$  2Cr₂O₃(s) f)  2K(s) + 2H₂O(aq)  $latex \longrightarrow$ 2KOH(aq)  +  H₂(g) g)  CuO(s) + 2HCl(aq)  $latex \longrightarrow$  CuCl₂(aq) + H₂O(l) h)  NaHCO₃(aq) + HNO₃(aq)  $latex \longrightarrow$  NaNO₃(aq)  +  H₂O(l) + CO₂(g)

17.   a)  2H₃PO₄  +  3CaO $latex \longrightarrow$ Ca₃(PO₄)₂  +  3H₂O b)  4NH₃  +  7O₂ $latex \longrightarrow$ 4NO₂  +  6H₂O c)  Cl₂O₇  +  H₂O $latex \longrightarrow$ 2HClO₄ d)  Mg₃N₂  +  6H₂O $latex \longrightarrow$ 3Mg(OH)₂  +  2NH₃ e)  4FeSO₄ $latex \longrightarrow$ 2Fe₂O₃  +  4SO₂  +  O₂ f)  P₄+ 6Cl₂ $latex \longrightarrow$ 4PCl₃ g)  MnO₂  +  C $latex \longrightarrow$ Mn  +  CO₂ h)  2Na₂O₂  +  2H₂O $latex \longrightarrow$ 4NaOH  +  O₂ i)  CaH₂  +  2H₂O $latex \longrightarrow$ Ca(OH)₂  +  2H₂ j)  2NaHCO₃ $latex \longrightarrow$ Na₂CO₃  +  CO₂  +  H₂O

18.   a) 2Mg  +  O₂ $latex \longrightarrow$ 2MgO b)  2KClO₃  $latex \longrightarrow$ 2KCl  +  3O₂ c)  3Fe +  2O₂  $latex \longrightarrow$ Fe₃O₄ d)  Mg +  2HCl  $latex \longrightarrow$ MgCl₂  +  H₂ e)  2Na +  2H₂O  $latex \longrightarrow$ 2NaOH  +  H₂ f)  N₂+  3H₂  $latex \longrightarrow$ 2NH₃ g)  Na₂CO₃• 10H₂O  $latex \longrightarrow$ Na₂CO₃+ 10H₂O h)  3Fe  +  4H₂O  $latex \longrightarrow$ Fe₃O₄+  4H₂ i)  2F₂  +  2H₂O  $latex \longrightarrow$ 4HF +  O₂

19. Balance the following chemical equations.

a)  2 Ca₃(PO₄)₂+  6 SiO₂+  5 C  $latex \longrightarrow$ P₄+ 6 CaSiO₃+ 5 CO₂

b)  3 C₂H₆O₂   +   5 O₃   $latex \longrightarrow$   6 CO₂   +   9 H₂O

c)  48 Fe₃O₄   +   35 S₈  $latex \longrightarrow$   72 Fe₂S₃   +   64 SO₃

d)  24 P₂S₅   +  64O₂ $latex \longrightarrow$ 15 S₈   +   16 P₃O₈

e)  16 C₃H₈  +   56 SO₂ $latex \longrightarrow$  24 C₂H₄O₄  +  16 H₂O  +  7 S₈

20.  a) NaCl_(aq)+ AgNO_3(aq) $latex \longrightarrow$ AgCl_(s)+ NaNO_3(aq) Na⁺_(aq)+ Cl^-_(aq)+ Ag⁺_(aq)+ NO₃^-_(aq)                            $latex \longrightarrow$ AgCl_(s)+ Na⁺_(aq)+ NO₃^-_(aq) Cl^-_(aq)+ Ag⁺_(aq) $latex \longrightarrow$ AgCl_(s)

b) BaCl_2(aq)+ H₂SO_4(aq) $latex \longrightarrow$ BaSO_4(s)+ 2HCl_(aq) Ba²⁺_(aq)+ 2Cl^-_(aq)+ 2H⁺_(aq)+ SO₄^2-_(aq)                            $latex \longrightarrow$ BaSO_4(s)+ 2H⁺_(aq)+ 2Cl^-_(aq) Ba²⁺_(aq)+ SO₄^2-_(aq) $latex \longrightarrow$ BaSO_4(s)

c) FeCl_3(aq)+ 3NH₄OH_(aq) $latex \longrightarrow$ Fe(OH)_3(s)+ 3NH₄Cl_(aq) Fe³⁺_(aq)+ 3Cl^-_(aq)+ 3 NH₄⁺_(aq)+ 3OH^-_(aq)                            $latex \longrightarrow$ Fe(OH)_3(s)+ 3 NH₄⁺_(aq)+ 3Cl^-_(aq) Fe³⁺_(aq)+ 3OH^-_(aq) $latex \longrightarrow$ Fe(OH)_3(s)

d) K₂CrO_4(aq)+ Pb(NO₃)_2(aq) $latex \longrightarrow$ PbCrO_4(s)+ 2KNO_3(aq) 2K⁺_(aq)+ CrO₄^2-_(aq)+ Pb²⁺_(aq)+ 2NO₃^-_(aq)                            $latex \longrightarrow$ PbCrO_4(s)+ 2K⁺_(aq)+ 2NO₃^-_(aq) CrO₄^2-_(aq)+ Pb²⁺_(aq) $latex \longrightarrow$ PbCrO_4(s)

e) KClO_3(aq)+ MgCl_2(aq) $latex \longrightarrow$ N.R.

f) (NH₄)₂CO_3(aq)+ CaCl_2(aq) $latex \longrightarrow$ CaCO_3(s)+ 2NH₄Cl_(aq) 2NH₄⁺_(aq)+ CO₃^2-_(aq)+ Ca²⁺_(aq)+ 2Cl^-_(aq)                            $latex \longrightarrow$ CaCO_3(s)+ 2NH₄⁺_(aq)+ 2Cl^-_(aq) CO₃^2-_(aq)+ Ca²⁺_(aq) $latex \longrightarrow$ CaCO_3(s)

g) NaC₂H₃O_2(aq)+ BaCl_2(aq) $latex \longrightarrow$ N.R.

21.  a) redox      b) acid-base           c) redox d) acid-base       e) precipitation      f) precipitation

22.  a) decomposition, which is a type of redox

b) single-replacement, which is a type of redox

c) single-replacement, which is a type of redox

d) precipitation, which is a type of double replacement (but not redox!)

e) composition (sometimes also called a combination reaction or a synthesis reaction), which is a type of redox

f) acid-base, which is a type of double replacement (but not redox!)]]> 2179 2162 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-4/ Thu, 12 Apr 2018 03:52:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-4/

1. How many molecules of O₂ will react with 6.022 × 10²³ molecules of H₂ to make water?

The reaction is 2 H₂(g) + O₂(g) $latex \longrightarrow$ 2 H₂O(ℓ).

2. How many moles are present in 6.411 kg of CO₂? How many molecules is this?

3. What is the mass in milligrams of 7.22 × 10²⁰ molecules of CO₂?

4. What is the mass in grams of 1 molecule of H₂O?

5. What is the volume of 3.44 mol of Ga if the density of Ga is 6.08 g/mL?

6. For the chemical reaction

2 C₄H₁₀(g) + 13 O₂(g) $latex \longrightarrow$ 8 CO₂(g) + 10 H₂O(ℓ)

assume that 13.4 g of C₄H₁₀ reacts completely to products. The density of CO₂ is 1.96 g/L. What volume in liters of CO₂ is produced?

7. Calculate the mass of each product when 100.0 g of CuCl react according to the reaction

2 CuCl(aq) $latex \longrightarrow$ CuCl₂(aq) + Cu(s)

What do you notice about the sum of the masses of the products? What concept is being illustrated here?

8. What mass of CO₂ is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C₈H₁₈. Assume that there are 2,801 g of gasoline per gallon.

9. A chemical reaction has a theoretical yield of 19.98 g and a percent yield of 88.40%. What is the actual yield?

10. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?

2 P₄ +  6 NaOH +  6 H₂O  $latex \longrightarrow$  3 Na₂HPO₄  +  5 PH₃

Initial amounts used: P₄ = 35.0 g; NaOH = 12.7 g; H₂O = 9.33 g 11. Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.

12. Chlorine can be produced by the reaction of hydrochloric acid with excess manganese(IV) oxide according to the following reaction: 4HCl(aq) + MnO₂(s) $latex \longrightarrow$ Cl₂(g) + 2H₂O(l) + MnCl₂(aq). How many moles of HCl are needed to form 12.5 mol Cl₂?

13. How many moles of aluminum oxide will be produced by reacting 9.5 mol of Al with O₂? How many moles of O₂will react?

14. Nitrogen monoxide is oxidized in air to give brown nitrogen dioxide:

2NO(g) + O₂(g) $latex \longrightarrow$ 2NO₂(g) Starting with 2.2 mol NO, how many moles and how many grams of O₂ are required for complete reaction? What mass of NO₂, in grams, is produced?

15. How many grams of Mg will react with 7.5 grams of H₂SO₄ in the reaction:       Mg +  H₂SO₄  $latex \longrightarrow$  MgSO₄  +  H₂

16. Zinc will react with hydrochloric acid producing hydrogen gas and zinc chloride:       Zn +  2HCl  $latex \longrightarrow$  ZnCl₂  +  H₂ How many grams of H₂ will be produced if 5.0 grams of zinc are used?

17. The final step in the manufacture of platinum metal (for use in automotive catalytic converters and other products) is the reaction: 3(NH₄)₂PtCl₆(s) $latex \longrightarrow$ 3Pt(s) + 2NH₄Cl(s) + 2N₂(g) + 16HCl(g) How many grams of Pt can be produced by decomposing 12.35 g (NH₄)₂PtCl₆?

18. How many kilograms of NH₃ will be produced when 25.0 kg of H₂ reacts with excess N₂?

19. A 3.00 cm³ piece of aluminum reacts with a solution of HCl and produces H₂ gas and AlCl₃. Determine the mass of H₂ formed. (Al has a density of 2.70 g/cm³)

20. One of the most important commercial reactions is the "Haber" production of ammonia:       3H₂(g)  +  N₂(g)  $latex \longrightarrow$  2NH₃(g) Chiefly by this reaction, the industrial "fixation" of nitrogen now accounts for about one-third of all the nitrogen fixed on our planet. If 3.0 kg of H₂ and 1.0 kg of N₂ are mixed and allowed to react until one or both of the reactants is used up, a)   how many moles of the NH₃ molecules are produced? b)   how many moles of the H₂ are left? c)   how many moles of the N₂ are left?

21. Iron reacts with chlorine to produce FeCl₃ as:  2Fe +  3Cl₂  $latex \longrightarrow$  2FeCl₃ If 10.6 grams of iron are mixed with 18.9 grams of chlorine and allowed to react, a)   how many grams of FeCl₃ will be produced? b)   how many grams of excess reactant will be left after the reaction is complete?

22. Silver tarnishes in the presence of hydrogen sulfide in the following reaction:       4Ag  +  2H₂S  +  O₂  $latex \longrightarrow$  2Ag₂S  +  2H₂O How many grams of Ag₂S can be obtained by this reaction from a mixture of 0.950 g Ag, 0.140 g H₂S, and 0.0800 g O₂?

23. 11.92 g of Pb(NO₃)₂ and 20.31 g of KI react as:  Pb(NO₃)₂+ 2 KI  $latex \longrightarrow$  PbI₂  +  2 KNO₃ How many grams of PbI₂ are produced if the yield of the reaction is 81%?

24. Reaction of H₂ and N₂ produces ammonia (NH₃) with 65.5%yield. If 30.0 g of NH₃ are required, how many grams of N₂ and how many of H₂ must be used?

25. Solid calcium carbonate dissolves in a solution of hydrochloric acid and reacts to form a solution of calcium chloride and water, and bubbles of carbon dioxide.

How many grams of calcium chloride will form if 40.0 g of CaCO₃ is mixed with 0.500 mol of HCl? How many grams of calcium carbonate, if any, will remain unreacted?

26. Freon-12 (CCl₂F₂) is a gas that has been used as a refrigerant. It is prepared by the reaction between carbon tetrachloride and antimony trifluoride. The other product that is produced is SbCl₃. If the percent yield is 72.0%, how many grams of antimony trifluoride must be treated with excess carbon tetrachloride to obtain an actual yield of 25.0 grams of Freon-12?

27. Aluminum chloride (Al₂Cl₆) can be made by the following reaction: 2Al(s) + 3Cl₂(g) $latex \longrightarrow$ Al₂Cl₆(s) a)  which reactant is limiting if 2.70 g Al and 4.05 g Cl₂ are mixed? b)  what mass of Al₂Cl₆ can be produced? c)  What mass of the excess reactant will remain when the reaction is complete?

28. Iron oxide can be reduced to the metal as follows: Fe₂O₃(s) + 3CO(g) $latex \longrightarrow$ 2Fe(s) + 3CO₂(g) How many grams of iron can be obtained from 1.00 kg of the iron oxide? If 654 g Fe was obtained from the reaction, what was the percent yield?

29. Disulfur dichloride can be prepared by the following reaction:       3SCl₂(l) + 4NaF(s) $latex \longrightarrow$ SF₄(g) + S₂Cl₂(l) + 4NaCl(s) What mass of SCl₂is needed to react with excess NaF to prepare 1.19 g S₂Cl₂, if the yield is 51%?

30. You have a 0.12 M solution of BaCl₂. What ions exist in the solution, and what are their concentrations?

31. Assume that 6.73 g Na₂CO₃is dissolved in enough water to make 250. mL of solution, a) what is the molarity of the sodium carbonate? b) What are the concentrations of the Na⁺and CO₃^2-ions?

32. What is the mass, in grams, of solute in 250. mL of a 0.0125 M solution of KMnO₄?

33. What volume of 0.123 M NaOH, in mL, contains 25.0 g NaOH?

34. What is the maximum mass, in grams, of AgCl that can be precipitated by mixing 50.0 mL 0.025 M AgNO₃solution with 100.0 mL of 0.025 M NaCl solution? Which reactant is in excess? What is the concentration of the excess reactant remaining in solution after the AgCl has precipitated?

35. How many moles of NaOH react with 7.80 mol of the acid H₂SO₄?

36. How many mL of 0.512 M NaOH is required to react completely with 25.0 mL of 0.234 M H₂SO₄?

37. What mass, in grams, of Na₂CO₃is required for complete reaction with 25.0 mL of 0.155 M HNO₃?      Na₂CO_3(aq)+ 2HNO_3(aq) $latex \longrightarrow$ 2NaNO_3(aq)+ CO_2(g)+ H₂O_(l)

Answers

1. 3.011 × 10²³ molecules of O₂ 2. 145.7 mol; 8.77 × 10²⁵ molecules 3. 52.8 mg 4. 2.99 × 10⁻²³ g 5. 39.4 mL 6. 20.7 L 7. 67.91 g of CuCl₂; 32.09 g of Cu. The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter. 8. 8633 g 9. 17.66 g 10. The limiting reagent is NaOH; 21.9 g of P₄ and 3.61 g of H₂O are left over. 11. Both products predict that O₂ is the limiting reagent; 20.3 g of C₃H₈ are left over.

12.   50.0 mol HCl

13.   4.8 mol Al₂O₃    7.1 mol O₂

14.   1.1 mol O₂    35 g O₂    1.0 x 10²g NO₂

15.   1.9 g Mg

16.  0.15 g H₂

17.  5.428 g Pt

18.  141 kg NH₃

19.  0.908 g H₂

20.  71 mol NH₃    1.4 x 10³mol H₂    0 mol N₂

21.  28.8 g FeCl₃    0.7 g Fe

22.  1.02 g Ag₂S

23.  13 g PbI₂

24.  37.7 g N₂    8.13 g H₂

25.  Must use a balanced equation to solve this:

1 CaCO₃(s)  +  2 HCl(aq) $latex \longrightarrow$ 1 CaCl₂(aq) + 1 H₂O(l) + 1 CO₂(g)

27.7g CaCl₂    15.0g CaCO₃ unreacted

26.  34.2 g SbF₃

27.  a) Cl₂ is limiting  b) 5.08 g Al₂Cl₆   c) 1.67 g Al unreacted

28.  699 g 93.5%

29.  5.3 g SCl₂

30.  0.12 M Ba²⁺, 0.24 M Cl^-

31.  a) 0.254 M Na₂CO₃   b) 0.508 M Na⁺, 0.254 M CO₃^2-

32.  0.494 g KMnO₄

33.  5.08 x 10³mL

34.  0.18 g AgCl, NaCl, 0.0083 M NaCl

35.  15.6 mol NaOH

36.  22.9 mL NaOH solution

37.  0.205 g Na₂CO₃

]]> 2243 1405 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-to-electronic-structure/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-to-electronic-structure/

Introduction

Normal light microscopes can magnify objects up to about 1,500 times. Electron microscopes can magnify objects up to 1,000,000 times. Why can electron microscopes magnify images so much?

A microscope’s resolution depends on the wavelength of light used. The smaller the wavelength, the more a microscope can magnify. Light is a wave, and, as such, it has a wavelength associated with it. The wavelength of visible light, which is detected by the eyes, varies from about 700 nm to 400 nm.

One of the startling conclusions about modern science is that electrons also act as waves. However, the wavelength of electrons is much, much shorter—about 0.5 to 1 nm. This allows electron microscopes to magnify 600–700 times more than light microscopes. This allows us to see even smaller features in a world that are invisible to the naked eye.

Atoms act the way they do because of their structure. We already know that atoms are composed of protons and neutrons which are located in the nucleus, and of electrons which orbit around the nucleus. But we need to know the structural details to understand why atoms react the way they do.

Virtually everything we know about atoms ultimately comes from light. Before we can understand the composition of atoms (especially electrons), we need to understand the properties of light.

]]> 2363 2362 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/ Thu, 12 Apr 2018 03:52:46 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/quantum-numbers-for-electrons/

Learning Objectives

By the end of this module, you will be able to:

• Explain what spectra are.
• Describe Bohr's Model of the hydrogen atom.
• Describe the Electron Shell Model.

There are two fundamental ways of generating light: either heat an object up so hot it glows or pass an electrical current through a sample of matter (usually a gas). Incandescent lights and fluorescent lights generate light via these two methods, respectively.

A hot object gives off a continuum of light. We notice this when the visible portion of the electromagnetic spectrum is passed through a prism: the prism separates light into its constituent colors, and all colors are present in a continuous rainbow (part (a) in Figure 1 "Prisms and Light"). This image is known as a continuous spectrum. However, when electricity is passed through a gas and light is emitted and this light is passed though a prism, we see only certain lines of light in the image (part (b) in Figure 1 "Prisms and Light"). This image is called a line spectrum. It turns out that every element has its own unique, characteristic line spectrum.

[caption id="attachment_4685" align="aligncenter" width="600"] Figure 1. Prisms and Light  (a) A glowing object gives off a full rainbow of colors, which are noticed only when light is passed through a prism to make a continuous spectrum. (b) However, when electricity is passed through a gas, only certain colors of light are emitted. Here are the colors of light in the line spectrum of Hg.[/caption] Why does the light emitted from an electrically excited gas have only certain colors, while light given off by hot objects has a continuous spectrum? For a long time, it was not well explained. Particularly simple was the spectrum of hydrogen gas, which could be described easily by an equation; no other element has a spectrum that is so predictable (Figure 2 "Hydrogen Spectrum").

 

[caption id="attachment_4687" align="aligncenter" width="600"] Figure 2. Hydrogen Spectrum[/caption]  

Late-nineteenth-century scientists found that the positions of the lines obeyed a pattern given by the equation

where n = 3, 4, 5, 6,…, but they could not explain why this was so.  The spectrum of hydrogen was particularly simple and could be predicted by a simple mathematical expression.

In 1913, the Danish scientist Niels Bohr suggested a reason why the hydrogen atom spectrum looked this way. He suggested that the electron in a hydrogen atom could not have any random energy, having only certain fixed values of energy that were indexed by the number n (the same n in the equation above and now called a quantum number) (Figure 3). Quantities that have certain specific values are called quantized. Bohr suggested that the energy of the electron in hydrogen was quantized because it was in a specific orbit. Because the energies of the electron can have only certain values, the changes in energies can have only certain values (somewhat similar to a staircase: not only are the stair steps set at specific heights but the height between steps is fixed).

  [caption id="attachment_4848" align="aligncenter" width="532"] Figure 3. Some emission possibilities from the energy levels of an atom.[/caption]

Finally, Bohr suggested that the energy of light emitted from electrified hydrogen gas was equal to the energy difference of the electron’s energy states:

E_light = hν = ΔE_electron

This means that only certain frequencies (and thus, certain wavelengths) of light are emitted. Figure 4 "Bohr’s Model of the Hydrogen Atom" shows a model of the hydrogen atom based on Bohr’s ideas.

[caption id="attachment_4688" align="aligncenter" width="372"] Figure 4. Bohr’s Model of the Hydrogen Atom[/caption]

Bohr’s description of the hydrogen atom had specific orbits for the electron, which had quantized energies.

Postulates of the Bohr Model:

1)  Electrons move in specific circular orbits only. 2)  As an atom absorbs energy, the electron jumps to a larger orbit, of higher energy (an excited state). 3)  As an atom emits energy, it “falls” to a smaller, lower energy orbit.

This model represented a great intellectual achievement by Bohr, as it was the first atom model that invoked quantization of the electron energy in some way. Also his mathematical formula  which calculated the energy of the electron in any orbit, matched the real energies observed in experiments with hydrogen. However, the theory had significant limitations.

Some Key Problems with the Bohr Model: 

• It only works for hydrogen (though can be adapted to other one electron ions). If there are 2 or more electrons, the mathematical formula does not match real data.
• It is fundamentally incorrect in that electrons do not move in fixed orbits!

The Electron Shell Model of the Atom

We can overcome one of the key objections to the Bohr Model by abandoning the concept of electrons moving in fixed diameter orbits. Instead we envision a series of spherical shells of increasing size surrounding the nucleus in which the electrons reside (Figure 5). The Electron Shell Model does not attempt to describe the movement of the electrons, only that each shell has a different size and energy and the electron moves within that space. The quantum jumps of the electron are thus the electron moving from one shell to another. [caption id="attachment_3653" align="aligncenter" width="344"] Figure 5. Electron Shell Model of the Atom (showing only the first three shells)[/caption] We also account for other experimental evidence and specify that the shells can hold a certain maximum number of electrons. Table 1 shows this maximum filling, as well as some other aspects of these shells. [caption id="attachment_3655" align="aligncenter" width="464"] Table 1. Properties of Electron Shells in Atoms[/caption]

The Electron Configuration of Atoms using the Shell Model

So, for a given atom or ion, in which shell(s) do the electrons reside? It turns out the electrons follow a simple principle, namely, they go into the lowest energy shell that is available. If a lower energy shell is full, they go into the next lowest energy shell. A crude analogy is putting water into a pail; the water always fills from the bottom! So to establish this electron configuration, first determine the number of electrons the atom has, then “put” them into the shells as the above rule dictates. Look at Figure 5 again, which represents an atom with 13 electrons. Notice how the lower energy shells are full, and the last three electrons go into shell 3, which is not full. Additional electrons would continue to go into shell 3 until it is full with 8 electrons, for a total of 18. A 19th electron would be forced to go into shell 4.

Example 1

Draw an electron shell model of an aluminum atom.   Solution   Step 1: Determine the number of electrons. Since it is not specified that the atom is charged, we presume it is neutral. Aluminum has 13 protons, so neutral aluminum would have 13 electrons. Step 2: Determine the electron configuration. Put 2 electrons in shell 1 which fills it, next put 8 electrons in shell 2 which fills it, and the last three electrons go into shell 3. Step 3: Draw the image. Test Yourself Draw an electron shell model of a calcium atom.   Answer

Electron Configurations and the Periodic Table

Look at the number of elements in each row of the periodic table. Rows 1 through 4 contain 2, 8, 8, and 18 elements respectively. Now look at Table 1. Is this a coincidence? No! In fact this shows that the patterns of elemental properties that the periodic table reflects have their basis in electron configurations. Consider Figure 6 which shows the electron shell models of hydrogen, lithium, sodium, and potassium. [caption id="attachment_3657" align="aligncenter" width="528"] Figure 6. Electron shell models of hydrogen, lithium, sodium, and potassium[/caption] See how each has one electron in its highest energy shell. Now find these elements on the periodic table. They are all in the first column of the periodic table. Consider the elements of the last column of the periodic table (draw them out for yourself). They all have full outer shells. A general relationship begins to emerge: elements in the same column on the periodic table have similar electron configurations. Originally, the periodic table was constructed based on observable chemical and physical properties. Elements that behaved similarly were placed in the same column; however the chemists had no explanation of why they were similar. Now with the electron shell model we have a theory that helps us understand the reasons for these similarities.

Chemistry Is Everywhere: Neon Lights

A neon light is basically an electrified tube with a small amount of gas in it. Electricity excites electrons in the gas atoms, which then give off light as the electrons go back into a lower energy state. However, many so-called “neon” lights don’t contain neon!

Although we know now that a gas discharge gives off only certain colors of light, without a prism or other component to separate the individual light colors, we see a composite of all the colors emitted. It is not unusual for a certain color to predominate. True neon lights, with neon gas in them, have a reddish-orange light due to the large amount of red-, orange-, and yellow-colored light emitted. However, if you use krypton instead of neon, you get a whitish light, while using argon yields a blue-purple light. A light filled with nitrogen gas glows purple, as does a helium lamp. Other gases—and mixtures of gases—emit other colors of light. Ironically, despite its importance in the development of modern electronic theory, hydrogen lamps emit little visible light and are rarely used for illumination purposes.

[caption id="attachment_3226" align="alignnone" width="450"] The different colors of these “neon” lights are caused by gases other than neon in the discharge tubes. Source: “Neon Internet Cafe open 24 hours” by JustinC is licensed under the Creative Commons Attribution- Share Alike 2.0 Generic license.[/caption]

Key Concepts and Summary

Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system.

Exercises

1. What does it mean to say that the energy of the electrons in an atom is quantized? 2. How are the Bohr model and the Rutherford model of the atom similar? How are they different? 3. Differentiate between a continuous spectrum and a line spectrum.   Answers 1. Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted. 2. Both involve a relatively heavy nucleus with electrons moving around it, although strictly speaking, the Bohr model works only for one-electron atoms or ions. According to classical mechanics, the Rutherford model predicts a miniature “solar system” with electrons moving about the nucleus in circular or elliptical orbits that are confined to planes. If the requirements of classical electromagnetic theory that electrons in such orbits would emit electromagnetic radiation are ignored, such atoms would be stable, having constant energy and angular momentum, but would not emit any visible light (contrary to observation). If classical electromagnetic theory is applied, then the Rutherford atom would emit electromagnetic radiation of continually increasing frequency (contrary to the observed discrete spectra), thereby losing energy until the atom collapsed in an absurdly short time (contrary to the observed long-term stability of atoms). The Bohr model retains the classical mechanics view of circular orbits confined to planes having constant energy and angular momentum, but restricts these to quantized values dependent on a single quantum number, n. The orbiting electron in Bohr’s model is assumed not to emit any electromagnetic radiation while moving about the nucleus in its stationary orbits, but the atom can emit or absorb electromagnetic radiation when the electron changes from one orbit to another. Because of the quantized orbits, such “quantum jumps” will produce discrete spectra, in agreement with observations. 3. A continuous spectrum is a range of light frequencies or wavelengths; a line spectrum shows only certain frequencies or wavelengths.

Glossary

Bohr’s model of the hydrogen atom: structural model in which an electron moves around the nucleus only in circular orbits, each with a specific allowed radius; the orbiting electron does not normally emit electromagnetic radiation, but does so when changing from one orbit to another. excited state: state having an energy greater than the ground-state energy ground state: state in which the electrons in an atom, ion, or molecule have the lowest energy possible

]]> 2390 2362 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/light/ Thu, 12 Apr 2018 03:52:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/light/

Learning Objectives

By the end of this module, you will be able to:

• Describe light with its frequency and wavelength.
• Describe light as a particle of energy. 

Our last “picture” of the atom consisted of a nucleus containing protons and neutrons (constituting most of the atom’s mass), surrounded by a sea of electrons. The limitations of this model are that it doesn’t show how the electrons are arranged or how they move. It turns out that this electronic structure is the primary factor controlling how an atom behaves. To get a more complete view of the atom, we need more experimental evidence and interpretation. One of the main ways to investigate electrons in atoms is to use light with two techniques; 1) by shining a light on the atoms and seeing what happens to the light (absorption spectroscopy), and 2) by heating the atoms and seeing what kind of light is given off (emission spectroscopy). Clearly, we will need to start with an understanding of the nature of light. We will then move on to describe and interpret experiments with light that give us our understanding of what electrons in atoms are doing. Finally, we see how the properties of electrons are related to the way the atoms behave.

What we know as light is more properly called electromagnetic radiation. We know from experiments that light acts as a wave. As such, it can be described as having a frequency and a wavelength.

The wavelength of light is the distance between corresponding points in two adjacent light cycles.  Wavelength is typically represented by λ, the lowercase Greek letter lambda, and has units of length (meters, centimeters, etc.).  Figure 1 shows how wavelength is defined.

[caption id="attachment_4846" align="aligncenter" width="373"] Figure 1. The wavelength of light is the distance between corresponding points in two adjacent light cycles.[/caption]

The frequency of light is the number of cycles of light that pass a given point in one second.  Frequency is represented by ν, the lowercase Greek letter nu, and has units of per second, written as s⁻¹ and sometimes called a hertz (Hz).   The amplitude (a) corresponds to the magnitude of the wave's displacement and so, in Figure 2, this corresponds to one-half the height between the peaks and troughs. The amplitude is related to the intensity of the wave, which for light is the brightness, and for sound is the loudness.

Light acts as a wave and can be described by a wavelength λ and a frequency ν.

One property of waves is that their speed is equal to their wavelength times their frequency. That means we have

speed = λν

m/s = m x s^-1

For light, however, speed is actually a universal constant when light is traveling through a vacuum (or, to a very good approximation, air). The measured speed of light (c) in a vacuum is 2.9979 × 10⁸ m/s, or about 3.00 × 10⁸ m/s. Thus, we have

c = λν

m/s = m x s^-1

Because the speed of light is a constant, the wavelength and the frequency of light are related to each other: as one increases, the other decreases and vice versa. We can use this equation to calculate what one property of light has to be when given the other property.

Example 1

What is the frequency of light if its wavelength is 5.55 × 10⁻⁷ m?

 

Solution

We use the equation that relates the wavelength and frequency of light with its speed. We have

3.00×10⁸m/s = (5.55×10^-7m)ν

We divide both sides of the equation by 5.55 × 10⁻⁷ m and get

ν = 5.41×10¹⁴ s^-1

Note how the m units cancel, leaving s in the denominator. A unit in a denominator is indicated by a −1 power—s⁻¹—and read as “per second.”

 

Test Yourself

What is the wavelength of light if its frequency is 1.55 × 10¹⁰ s⁻¹?

 

Answer

0.0194 m, or 19.4 mm

Example 2

A sodium streetlight gives off yellow light that has a wavelength of 589 nm (1 nm = 1 × 10⁻⁹ m). What is the frequency of this light?

 

Solution We can rearrange the equation c = λν to solve for the frequency:

$latex \nu = \frac{c}{\lambda}$

Since c is expressed in meters per second, we must also convert 589 nm to meters.

$latex \nu = (\frac{2.998 \times 10^8 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{ms}^{-1}}{589 \;\rule[0.25ex]{1em}{0.1ex}\hspace{-1em}\text{nm}})(\frac{1 \times 10^9 \;\rule[0.25ex]{1em}{0.1ex}\hspace{-1em}\text{nm}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}}) = 5.09 \times 10^{14}\text{s}^{-1}$

 

Test Yourself One of the frequencies used to transmit and receive cellular telephone signals in the United States is 850 MHz. What is the wavelength in meters of these radio waves?

Answer 0.353 m = 35.3 cm

Light also behaves like a package of energy. It turns out that for light, the energy of the “package” of energy is proportional to its frequency. (For most waves, energy is proportional to wave amplitude, or the height of the wave.) The mathematical equation that relates the energy (E) of light to its frequency is

E = hν

where ν is the frequency of the light, and h is a constant called Planck’s constant. Its value is 6.626 × 10⁻³⁴ J·s — a very small number that is another fundamental constant of our universe, like the speed of light. The units on Planck’s constant may look unusual, but these units are required so that the algebra works out.

Example 3

What is the energy of light if its frequency is 1.55 × 10¹⁰ s⁻¹?

 

Solution

Using the formula for the energy of light, we have

E = (6.626 × 10⁻³⁴ J·s)(1.55 × 10¹⁰ s⁻¹)

Seconds are in the numerator and the denominator, so they cancel, leaving us with joules, the unit of energy. So

E = 1.03 × 10⁻²³ J

This is an extremely small amount of energy—but this is for only one light wave.

 

Test Yourself

What is the frequency of a light wave if its energy is 4.156 × 10⁻²⁰ J?

 

Answer

6.27 × 10¹³ s⁻¹

Because a light wave behaves like a little particle of energy, light waves have a particle-type name: the photon. It is not uncommon to hear light described as photons.

Wavelengths, frequencies, and energies of light span a wide range; the entire range of possible values for light is called the electromagnetic spectrum. We are mostly familiar with visible light, which is light having a wavelength range between about 400 nm and 700 nm. Light can have much longer and much shorter wavelengths than this, with corresponding variations in frequency and energy. Figure 3 "The Electromagnetic Spectrum" shows the entire electromagnetic spectrum and how certain regions of the spectrum are labelled. You may already be familiar with some of these regions; they are all light—with different frequencies, wavelengths, and energies.

[caption id="attachment_4683" align="aligncenter" width="600"] Figure 3. The Electromagnetic Spectrum  The electromagnetic spectrum, with its various regions labelled. The borders of each region are approximate.[/caption]  

Example 4

Using Figure 3, determine which category of EM radiation has more energetic photons, UV or IR.

 

Solution  

Looking at Figure 3 we see that IR radiation has LONGER wavelengths. Applying the property that the energy of a photon is inversely proportional to the wavelength of the light, we can conclude that the IR light has LESS energetic photons.

 

Test Yourself

Which light has carries less energy in its photons, light with a frequency of 4.0 x 10¹³s^-1or light with a frequency of 1.0 x 10¹⁴s^-1?

  Answer

The lower frequency light of 4.0 x 10¹³s^-1would have the lower energy photons.

Technology and the Electromagnetic Spectrum

Figure 4 shows the electromagnetic spectrum, the range of all types of electromagnetic radiation. Each of the various colors of visible light has specific frequencies and wavelengths associated with them, and you can see that visible light makes up only a small portion of the electromagnetic spectrum.

Because the technologies developed to work in various parts of the electromagnetic spectrum are different, for reasons of convenience and historical legacies, different units are typically used for different parts of the spectrum. For example, radio waves are usually specified as frequencies (typically in units of MHz), while the visible region is usually specified in wavelengths (typically in units of nm or angstroms).

Wireless Communication

Many valuable technologies operate in the radio (3 kHz-300 GHz) frequency region of the electromagnetic spectrum (Figure 5).

At the low frequency (low energy, long wavelength) end of this region are AM (amplitude modulation) radio signals (540-2830 kHz) that can travel long distances. FM (frequency modulation) radio signals are used at higher frequencies (87.5-108.0 MHz). In AM radio, the information is transmitted by varying the amplitude of the wave (Figure 6). In FM radio, by contrast, the amplitude is constant and the instantaneous frequency varies.

Other technologies also operate in the radio-wave portion of the electromagnetic spectrum. For example, 4G cellular telephone signals are approximately 880 MHz, while Global Positioning System (GPS) signals operate at 1.228 and 1.575 GHz, local area wireless technology (Wi-Fi) networks operate at 2.4 to 5 GHz, and highway toll sensors operate at 5.8 GHz. The frequencies associated with these applications are convenient because such waves tend not to be absorbed much by common building materials.

 Key Concepts and Summary

Light and other forms of electromagnetic radiation move through a vacuum with a constant speed, c, of 2.998 × 10⁸ m s⁻¹. This radiation shows wavelike behavior, which can be characterized by a frequency, ν, and a wavelength, λ.  The frequency and wavelength of light are related by the speed of light, a constant, such that c = λν.   Light acts like a particle of energy, whose value is related to the frequency of light.

Key Equations

• c = λν
• $latex E = h\nu = \frac{hc}{\lambda}$, where h = 6.626 × 10⁻³⁴ J·s

Exercises

1. Describe the characteristics of a light wave.

2. What is the frequency of light if its wavelength is 7.33 × 10⁻⁵ m?

3. What is the frequency of light if its wavelength is 733 nm?

4. What is the wavelength of light if its frequency is 8.19 × 10¹⁴ s⁻¹?

5. What is the wavelength of light if its frequency is 1.009 × 10⁶ Hz?

6. What is the energy of a photon if its frequency is 5.55 × 10¹³ s⁻¹?

7. What is the energy of a photon if its wavelength is 5.88 × 10⁻⁴ m?

8. FM-95, an FM radio station, broadcasts at a frequency of 9.51 × 10⁷ s⁻¹ (95.1 MHz). What is the wavelength of these radio waves in meters?

9. One of the radiographic devices used in a dentist's office emits an X-ray of wavelength 2.090 × 10⁻¹¹ m. What is the energy, in joules, and frequency of this X-ray?

10. RGB color television and computer displays use cathode ray tubes that produce colors by mixing red, green, and blue light. If we look at the screen with a magnifying glass, we can see individual dots turn on and off as the colors change. Using a spectrum of visible light, determine the approximate wavelength of each of these colors. What is the frequency and energy of a photon of each of these colors?

  Answers 1. Light has a wavelength and a frequency. 2. 4.09 × 10¹² s⁻¹ 3. 4.09 × 10¹⁴ s⁻¹ 4. 3.66 × 10⁻⁷ m 5. 297 m 6. 3.68 × 10⁻²⁰ J 7. 3.38 × 10⁻²² J 8. 3.15 m 9. E = 9.502 × 10⁻¹⁵ J; ν = 1.434 × 10¹⁹ s⁻¹ 10. Red: 660 nm; 4.54 × 10¹⁴ Hz; 3.01 × 10⁻¹⁹ J. Green: 520 nm; 5.77 × 10¹⁴ Hz; 3.82 × 10⁻¹⁹ J. Blue: 440 nm; 6.81 × 10¹⁴ Hz; 4.51 × 10⁻¹⁹ J. Somewhat different numbers are also possible.

Glossary

amplitude: extent of the displacement caused by a wave (for sinusoidal waves, it is one-half the difference from the peak height to the trough depth, and the intensity is proportional to the square of the amplitude) continuous spectrum: electromagnetic radiation given off in an unbroken series of wavelengths (e.g., white light from the sun) electromagnetic radiation: energy transmitted by waves that have an electric-field component and a magnetic-field component electromagnetic spectrum: range of energies that electromagnetic radiation can comprise, including radio, microwaves, infrared, visible, ultraviolet, X-rays, and gamma rays; since electromagnetic radiation energy is proportional to the frequency and inversely proportional to the wavelength, the spectrum can also be specified by ranges of frequencies or wavelengths frequency (ν): number of wave cycles (peaks or troughs) that pass a specified point in space per unit time hertz (Hz): the unit of frequency, which is the number of cycles per second, s⁻¹ intensity: property of wave-propagated energy related to the amplitude of the wave, such as brightness of light or loudness of sound photon: smallest possible packet of electromagnetic radiation, a particle of light wave: oscillation that can transport energy from one point to another in space wavelength (λ): distance between two consecutive peaks or troughs in a wave

]]> 2393 2362 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/periodic-trends/ Thu, 12 Apr 2018 03:52:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/periodic-trends/

Learning Objectives

By the end of this module, you will be able to:

• Be able to state how certain properties of atoms vary based on their relative position on the periodic table.

One of the reasons the periodic table is so useful is because its structure allows us to qualitatively determine how some properties of the elements vary versus their position on the periodic table. The variation of properties versus position on the periodic table is called periodic trends. There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row or down a whole column of the periodic table.

Atomic Radii

The atomic radius is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals.

The size of atoms vary and there are two periodic trends.  Atoms get smaller as you go from left to right across a period, and get larger as you go down a group.  Figure 1 "Atomic Radii Trends on the Periodic Table" shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius.

[caption id="attachment_4709" align="aligncenter" width="600"] Figure 1. Atomic Radii Trends on the Periodic Table.  Although there are some reversals in the trend (e.g., see Po in the bottom row), atoms generally get smaller as you go across the periodic table and larger as you go down any one column. Numbers are the radii in pm.[/caption]

The atomic size is easily explained when we examine how the electron configurations change as we move on the periodic table:

• As you go down a group, the valence electron configuration stays the same, but the number of shells is increasing. Each shell represents distance from the nucleus (as well as energy), thus we expect that ATOMIC SIZE INCREASES as you go down a row on the periodic table.
• As you go from left to right on the periodic table, you are adding electrons to the same shell, but, you are also adding protons (nuclear charge). These protons serve to pull the electrons closer to the nucleus. Thus, we expect that as you go from left to right along each period, ATOMIC SIZE DECREASES.

Example 1

Referring only to a periodic table and not to Figure 1 "Atomic Radii Trends on the Periodic Table", which atom is larger in each pair?

a) Si or S      b) S or Te

 

Solution

a) Si is to the left of S on the periodic table, so it is larger because as you go across the row, the atoms get smaller.

b) S is above Te on the periodic table, so Te is larger because as you go down the column, the atoms get larger.

 

Test Yourself

Referring only to a periodic table and not to Figure 1 "Atomic Radii Trends on the Periodic Table", which atom is smaller, Ca or Br?

 

Answer

Br

Example 2

For the following elements, write them in order of smallest to largest, using only the periodic table:

      K, As, F, N

 

Solution  

We use the periodic table and our knowledge of the trends in atomic size; further up and to the right are the smaller atoms. The order thus becomes:

Smallest     F, N, As, K       Largest

 

Test Yourself

For the following elements, write them in order of smallest to largest, using only the periodic table:

Rb, Si, Cl

  Answer Cl, Si, Rb

Ionization Energy

Ionization energy (IE) is the amount of energy required to remove an electron from an atom in the gas phase:

A(g) $latex \longrightarrow$ A⁺(g) + e⁻             ΔH ≡ IE

IE is usually expressed in kJ/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus.  However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases.

Figure 2 "Ionization Energy on the Periodic Table" shows values of IE versus position on the periodic table. Again, the trend isn’t absolute, but the general trends going across and down the periodic table should be obvious.

[caption id="attachment_4711" align="aligncenter" width="600"] Figure 2. Ionization Energy on the Periodic Table.  Values are in kJ/mol.[/caption]

IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with):

A(g) $latex \longrightarrow$ A⁺(g) + e⁻ IE₁ A⁺(g) $latex \longrightarrow$ A²⁺(g) + e⁻ IE₂ A²⁺(g) $latex \longrightarrow$ A³⁺(g) + e⁻ IE₃

and so forth.

Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration is 1s²2s²2p⁶3s²:

Mg(g) $latex \longrightarrow$ Mg⁺(g) + e⁻ IE₁ = 738 kJ/mol Mg⁺(g) $latex \longrightarrow$ Mg²⁺(g) + e⁻ IE₂ = 1,450 kJ/mol Mg²⁺(g) $latex \longrightarrow$ Mg³⁺(g) + e⁻ IE₃ = 7,734 kJ/mol

The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over five times the previous one. It suggests that there is more involved than simply overcoming a larger ionic charge. Why is it so much larger? Because the first two electrons are removed from the 3s subshell, but the third electron has to be removed from the n = 2 shell, specifically, the 2p subshell, which is lower in energy than the n = 3 shell.  It is evidence like this that demonstrate that electrons are organized in atoms in groups (shells and subshells).

Example 3

Which atom in each pair has the larger IE?

a) Ca or Sr      b) K or K⁺

 

Solution

a) Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE.

b) Because K⁺ has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K⁺ to be removed comes from another shell.

 

Test Yourself

Which atom has the lower ionization energy, C or F?

 

Answer

C

Electron Affinity

The opposite of IE is described by electron affinity (EA), which is the energy change when a gas-phase atom accepts an electron:

A(g) + e⁻$latex \longrightarrow$ A⁻(g)               ΔH ≡ EA

Electron affinity is also usually expressed in kJ/mol.

Electron affinity also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously. Figure 3 "Electron Affinity on the Periodic Table" shows EA values versus position on the periodic table for the s- and p-block elements. The trend isn’t absolute, especially considering the large positive EA values for the second column.

• Electron affinity generally increases in magnitude as we move to the right across a period (row) in the periodic table (across a period).
• There is not a definitive trend as you go down a group (column) on the periodic table; sometimes electron affinity increases, sometimes it decreases.

 

[caption id="attachment_4712" align="aligncenter" width="600"] Figure 3. Electron Affinity on the Periodic Table.  Values are in kJ/mol.[/caption]

Example 4

Predict which atom in each pair will have the highest magnitude of EA.

a) C or F      b) Na or S

 

Solution

a) C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA.

b) Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA.

 

Test Yourself

Predict which atom will have the highest magnitude of EA, As or Br.

 

Answer

Br

Key Concepts and Summary

Certain properties—notably effective atomic radius, IE, and EA—can be qualitatively understood by the positions of the elements on the periodic table.

Exercises

1. Write a chemical equation with an IE energy change.

2. State the trends in atomic radii as you go across and down the periodic table.

3. Which atom of each pair is larger?

a)  Na or Cs      b)  N or Bi 4.  Which atom of each pair is larger?

a)  K or Cl         b)  Ba or Bi

5.  Which atom has the higher IE?

a)  Na or S         b)  Ge or Br

6.  Which atom has the higher IE?

a)  Li or Cs         b)  Se or O

7.  A third-row element has the following successive IEs: 738; 1,450; 7,734; and 10,550 kJ/mol. Identify the element. 8.  For which successive IE is there a large jump in IE for Ca? 9.  Which atom has the greater magnitude of EA?

a)  C or F         b)  Al or Cl

Answers 1. Na(g) $latex \longrightarrow$ Na⁺(g) + e⁻ ΔH = IE (answers will vary) 2. As you go across, atomic radii decrease; as you go down, atomic radii increase. 3. a)  Cs         b) Bi 4. a)  K           b)  Ba 5. a)  S            b)  Br 6. a)  Li           b)  O 7. Mg 8. The third IE shows a large jump in Ca. 9. a)  F         b)  Cl

]]> 2398 2362 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/ Thu, 12 Apr 2018 03:52:48 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/end-of-chapter-material-7/

1. Calculate the frequency and wavelength of light that is emitted contains 3.45 x 10^-18J of energy.  What type of light is this?

2. Convert the following light wavelengths into frequencies (Hz). (The speed of light c=2.998 x 10⁸m/s)

a)  4.33 nm          b) 2.35 x 10^-10m          c)  735 nm          d) 4.57 mm

3. Convert the following light frequencies into wavelengths (expressing the result in the indicated units), assuming the light is moving at the speed of light (c=2.998 x 10⁸m/s)

a)  4.77 GHz (m)          b) 2.89 kHz (cm)          

c)  50. Hz (mm)          d) 2.88 MHz (mm)

4. How does the energy possessed by an emitted photon compare to the difference in energy levels that gave rise to the emission of the photon?

5. Explain the difference between continuous and discrete spectra.

6. Describe one aspect of Bohr’s model of the atom that “worked”. Describe one aspect of Bohr’s model of the atom that did not “work”.

7. Draw an electron shell model of the following

a)  chloride ion, Cl^1-          b) calcium ion, Ca²⁺

8. In what way are outer electron shells of oxygen and selenium similar? In what way are they different?

9. Why is the view of an electron being a particlein a fixed orbitunacceptable?

10. Explain the difference between an orbit and an orbital. Briefly describe the model that each refers to.

11. Which law(s) (e.g.: Aufbau principle, Pauli exclusion principle or Hund’s rule) is (are) violated in each of these ground state electron configurations. Write the correct configuration as well.

a)  B = 1s¹2s¹2p_x¹2p_y¹2p_z¹                                               

b)  Sc = [Ne] 3s²3p⁶3d³

c)  Na = 1s²1p⁷2s²3s¹                                                           

d)  C = 1s²2s²2p_x²

12. Write the electron configurations and show the orbital box diagram for each of the following. Show only the valence electrons with the box notation. a) Se      b) Cu     c) Fe    d) Si

13. Write the electron configuration, using core notation (condensed electron configuration), for the following. a) Pt      b) Zr      c) W

14. How many orbitals in an atom can have the designation: 5p, 4d, n=5, n=4?

15. The elements Si, Ga, As, Ge, Al, Cd, S and Se are all used in the manufacturing of various semiconductor devices. Write the expected electron configuration for each of these atoms

16. Write the expected ground-state electron configuration for the following:

a) The element(s) with one unpaired 5p electron.

b) The, as yet undiscovered alkaline earth metal after (i.e. below in the periodic table) radium.

c) The noble gas with electrons occupying 4f orbitals.

d) The first-row transition metal with the most UNPAIRED electrons (i.e. electrons singly in orbitals).

17. Which of the following electron configurations correspond to an excited state (i.e., is not a ground-state/lowest energy electron configuration)?  Identify the atoms and write the ground-state electron configurations where appropriate.

a)  1s²2s²3p¹                                                 b)  1s²2s²2p⁶ c)  1s²2s²2p⁴3s¹                                         d) [Ar]4s²3d⁵4p¹

18. Arrange the following in order of increasing atomic size:  N, O, S

19. Arrange the following in order of increasing ionization energy:  S, P, F

20. How many valence electrons do the following atoms have?

a)  Li           b) Al           c)  P

21. Arrange the following groups of atoms in order of increasing size:

a)  Be, Mg, Ca                   b)  Te, I, Xe                  c)  Ga, Ge, In

22. Arrange the atoms in the previous exercise in order of increasing first ionization energy.

23. In each of the following sets, which atom or ion has the smallest radius?

a)  Li, Na, K        b) P, As        c)  O⁺, O, O^-        d) S, Cl, Kr        e)  Pd, Ni, Cu

24. The first ionization energies of As and Se are 0.947 and 0.941 MJ/mol respectively.  Rationalize these values in terms of electron configurations.

25. For each of the pairs of elements  (O and F) and  (Ar and Br), pick the atom with:

a)  higher (more negative) electron affinity b)  higher ionization energy c)  larger size

26. What is the frequency of light if its wavelength is 1.00 m?

27. What is the energy of a photon if its wavelength is 1.00 meter?

28.  a)  Predict the electron configurations of Sc through Zn.

b)  From a source of actual electron configurations, determine how many exceptions there are from your predictions in part a.

29.  Recently, Russian chemists reported experimental evidence of element 117. Use the periodic table to predict its valence shell electron configuration. 30.  Which atom has a higher ionization energy (IE), O or P? 31.  Which atom has a smaller radius, As or Cl? 32.  How many IEs does an H atom have? Write the chemical reactions for the successive ionizations. 33.  Based on what you know of electrical charges, do you expect Na⁺ to be larger or smaller than Na?  

Answers

1.   5.21 x 10¹⁵s^-1;   5.76 x 10^-8m;   UV

2.   a) 6.92 x 10¹⁶Hz       b) 1.28 x 10¹⁸Hz      c) 4.08 x 10¹⁴Hz      d) 6.56 x 10¹³Hz

3.   a) 0.0629 m      b) 1.04 x 10⁷cm      c) 6.0 x 10⁹mm      d) 1.04 x 10⁸um

4.   equal

5.   continuous spectra contain all frequencies while discrete spectra only contain limited number of frequencies.

6.   The concept that the energy of the electron is quantized is true. The concept that the electron travels on a fixed path or “orbit” is not the case.

7.   a)                                              b)

 

8.   They are similar in that they both need 2 more electrons to have a full outer shell. They are different in that they have a different number of electrons in the outer shell. O has 6 but Se has16.

9.   Viewing the electron as a particle does not recognize that it has wave-like properties. Having a fixed orbit violates the Uncertainty Principle.

10.  orbit: 2-D circular path in which an electron can be found. Orbital: 3-D region of space in which there’s a high probability of finding the electron.

11.  a) Aufbau. 1s²2s²2p¹                                            b) Aufbau. [Ar]4s²3d¹ c) Aufbau & Pauli exclusion. 1s²2s²2p⁶3s¹         d) Hund’s rule. 1s²2s²2p²

12. Note: only the valence box diagram is shown in the following answers

a)  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴  

 

b)  1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

 

c)  1s²2s²2p⁶3s²3p⁶4s²3d⁶

d)  1s²2s²2p⁶3s²3p²

13.  a) [Xe]6s¹4f¹⁴5d⁹ b)  [Kr]5s²4d² c)  [Xe]6s²4f¹⁴5d⁴

14.  3; 5;  25;  16

15.  Si: 1s²2s²2p⁶3s²3p²    Ga: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p¹   As: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p³    Ge: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p²    Al: 1s²2s²2p⁶3s²3p¹    Cd: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰   S: 1s²2s²2p⁶3s²3p⁴    Se: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴   

16.  a) In: [Kr]5s²4d¹⁰5p¹or I: [Kr]5s²4d¹⁰5p⁵ b) Z = 120: [Rn]7s²5f¹⁴6d¹⁰7p⁶8s² c) Rn: [Xe]6s²4f¹⁴5d¹⁰6p⁶ d) Cr: [Ar]4s¹3d⁵

17.  a) excited state of B; 1s²2s²2p¹ b) ground state of Ne c) excited state of F: 1s²2s²2p⁵ d) excited state of Fe: [Ar]4s²3d⁶

18.  O < N < S

19.  P < S < F

20.  a) 1;    b) 3;    c) 5

21.  a) Be < Mg < Ca b) Xe < I < Te c) Ge < Ga < In

22.  a) Ca < Mg < Be b) Te < I < Xe c) In < Ga < Ge

23.  a) Li;  b) P;   c) O⁺;   d) Cl;  e) Cu

24.  As: [Ar]4s²3d¹⁰4p³and Se: [Ar]4s²3d¹⁰4p⁴. As has a half-filled 4p orbital which is more stable; Se has 2e in one of its 4p orbitals which gives rise to a larger electron-electron repulsion, making it easier for Se to lose an electron to reach a more stable half-filled 4p configuration, therefore Se has a smaller IE₁.

25.  a)  F, Br;    b) F, Ar;    c) O, Br

26. 3.00 × 10⁸ s⁻¹ 27. 1.99 × 10⁻²² J 28. a)  The electron configurations are predicted to end in 3d¹, 3d², 3d³, 3d⁴, 3d⁵, 3d⁶, 3d⁷, 3d⁸, 3d⁹, and 3d¹⁰. b)  Cr and Cu are exceptions. 29. Element 117’s valence shell electron configuration should be 7s²7p⁵. 30. O 31. Cl 32. H has only one IE: H → H⁺ + e⁻ 33. smaller

]]> 2399 2362 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/lewis-electron-dot-diagrams/ Thu, 12 Apr 2018 03:52:51 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/lewis-electron-dot-diagrams/

Learning Objective

By the end of this section, you will be able to:

• Draw a Lewis electron dot diagram for an atom or a monatomic ion.

In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful.

A Lewis electron dot diagram (or electron dot diagram or a Lewis diagram or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (It does not matter what order the positions are used.) For example, the Lewis electron dot diagram for hydrogen is simplyBecause the side is not important, the Lewis electron dot diagram could also be drawn as follows:

The electron dot diagram for helium, with two valence electrons, is as follows:

By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1s²2s¹, so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used:

Beryllium has two valence electrons in its 2s shell, so its electron dot diagram is like that of helium:

The next atom is boron. Its valence electron shell is 2s²2p¹, so it has three valence electrons. The third electron will go on another side of the symbol:

Again, it does not matter on which sides of the symbol the electron dots are positioned.

For carbon, there are four valence electrons, two in the 2s subshell and two in the 2p subshell. As usual, we will draw two dots together on one side, to represent the 2s electrons. However, conventionally, we draw the dots for the two p electrons on different sides. As such, the electron dot diagram for carbon is as follows:With nitrogen, which has three p electrons, we put a single dot on each of the three remaining sides:

For oxygen, which has four p electrons, we now have to start doubling up on the dots on one other side of the symbol. When doubling up electrons, make sure that a side has no more than two electrons.

Fluorine and neon have seven and eight dots, respectively:

With the next element, sodium, the process starts over with a single electron because sodium has a single electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we see that the Lewis electron dot diagrams of atoms will never have more than eight dots around the atomic symbol.

Example 1

What is the Lewis electron dot diagram for each element?      a) aluminum         b) selenium

 

Solution

a) The valence electron configuration for aluminum is 3s²3p¹. So it would have three dots around the symbol for aluminum, two of them paired to represent the 3s electrons:b) The valence electron configuration for selenium is 4s²4p⁴. In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows:

Test Yourself

What is the Lewis electron dot diagram for each element?  a) phosphorus          b) argon

 

Answer

For atoms with partially filled d or f subshells, these electrons are typically omitted from Lewis electron dot diagrams. For example, the electron dot diagram for iron (valence shell configuration 4s²3d⁶) is as follows:Elements in the same column of the periodic table have similar Lewis electron dot diagrams because they have the same valence shell electron configuration. Thus the electron dot diagrams for the first column of elements are as follows:

Monatomic ions are atoms that have either lost (for cations) or gained (for anions) electrons. Electron dot diagrams for ions are the same as for atoms, except that some electrons have been removed for cations, while some electrons have been added for anions. Thus in comparing the electron configurations and electron dot diagrams for the Na atom and the Na⁺ ion, we note that the Na atom has a single valence electron in its Lewis diagram, while the Na⁺ ion has lost that one valence electron:

Technically, the valence shell of the Na⁺ ion is now the n = 2 shell, which has eight electrons in it. So why do we not put eight dots around Na⁺? Conventionally, when we show electron dot diagrams for ions, we show the original valence shell of the atom, which in this case is the n = 3 shell and empty in the Na⁺ ion.

In making cations, electrons are first lost from the highest numbered shell, not necessarily the last subshell filled. For example, in going from the neutral Fe atom to the Fe²⁺ ion, the Fe atom loses its two 4s electrons first, not its 3d electrons, despite the fact that the 3d subshell is the last subshell being filled. Thus we haveAnions have extra electrons when compared to the original atom. Here is a comparison of the Cl atom with the Cl⁻ ion:

Example 2

What is the Lewis electron dot diagram for each ion?   a) Ca²⁺         b) O²⁻

 

Solution

a) Having lost its two original valence electrons, the Lewis electron dot diagram is just Ca²⁺.

Ca²⁺ b) The O²⁻ ion has gained two electrons in its valence shell, so its Lewis electron dot diagram is as follows:

Test Yourself

The valence electron configuration of thallium, whose symbol is Tl, is 6s²5d¹⁰6p¹. What is the Lewis electron dot diagram for the Tl⁺ ion?

 

Answer

Key Concepts and Summary

Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol.   Lewis electron dot diagrams for ions have fewer (for cations) or more (for anions) dots than the corresponding atom.

Exercises

1. Explain why the first two dots in a Lewis electron dot diagram are drawn on the same side of the atomic symbol. 2. What column of the periodic table has Lewis electron dot diagrams with two electrons? 3. Draw the Lewis electron dot diagram for each element.   a)  strontium      b)  silicon 4. Draw the Lewis electron dot diagram for each element.   a)  titanium        b)  phosphorus 5. Draw the Lewis electron dot diagram for each ion.  a)  Mg²⁺   b)  S²⁻ 6. Draw the Lewis electron dot diagram for each ion.  a)  Fe²⁺     b)  N³⁻

Answers 1. The first two electrons in a valence shell are s electrons, which are paired. 2. the second column of the periodic table 3. a)    b)    4. a)    b)    5. a)   Mg²⁺ b)    6. a)  Fe²⁺ b)   

]]> 2429 1538 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/electron-transfer-ionic-bonds/ Thu, 12 Apr 2018 03:52:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/electron-transfer-ionic-bonds/

Learning Objectives

By the end of this section, you will be able to:

• State the octet rule.
• Define ionic bond.
• Demonstrate electron transfer between atoms to form ionic bonds.

In Section 8.3 "Lewis Electron Dot Diagrams," we saw how ions are formed by losing electrons to make cations or by gaining electrons to form anions. The astute reader may have noticed something: Many of the ions that form have eight electrons in their valence shell. Either atoms gain enough electrons to have eight electrons in the valence shell and become the appropriately charged anion, or they lose the electrons in their original valence shell. The lower shell, now the valence shell, has eight electrons in it, so the atom becomes positively charged. For whatever reason, having eight electrons in a valence shell is a particularly energetically stable arrangement of electrons. The trend that atoms like to have eight electrons in their valence shell is called the octet rule. When atoms form compounds, the octet rule is not always satisfied for all atoms at all times, but it is a very good rule of thumb for understanding the kinds of bonding arrangements that atoms can make.

It is not impossible to violate the octet rule. Consider sodium: in its elemental form, it has one valence electron and is stable. It is rather reactive, however, and does not require a lot of energy to remove that electron to make the Na⁺ ion. We could remove another electron by adding even more energy to the ion, to make the Na²⁺ ion. However, that requires much more energy than is normally available in chemical reactions, so sodium stops at a 1+ charge after losing a single electron. It turns out that the Na⁺ ion has a complete octet in its new valence shell, the n = 2 shell, which satisfies the octet rule. The octet rule is a result of trends in energies and is useful in explaining why atoms form the ions that they do.

Now consider an Na atom in the presence of a Cl atom. The two atoms have these Lewis electron dot diagrams and electron configurations:For the Na atom to obtain an octet, it must lose an electron; for the Cl atom to gain an octet, it must gain an electron. An electron transfers from the Na atom to the Cl atom:

resulting in two ions—the Na⁺ ion and the Cl⁻ ion:

Both species now have complete octets, and the electron shells are energetically stable. From basic physics, we know that opposite charges attract. This is what happens to the Na⁺ and Cl⁻ ions:

where we have written the final formula (the formula for sodium chloride) as per the convention for ionic compounds, without listing the charges explicitly. The attraction between oppositely charged ions is called an ionic bond, and it is one of the main types of chemical bonds in chemistry. Ionic bonds are caused by electrons transferring from one atom to another.

In electron transfer, the number of electrons lost must equal the number of electrons gained. We saw this in the formation of NaCl. A similar process occurs between Mg atoms and O atoms, except in this case two electrons are transferred:The two ions each have octets as their valence shell, and the two oppositely charged particles attract, making an ionic bond:

Remember, in the final formula for the ionic compound, we do not write the charges on the ions.

What about when an Na atom interacts with an O atom? The O atom needs two electrons to complete its valence octet, but the Na atom supplies only one electron:The O atom still does not have an octet of electrons. What we need is a second Na atom to donate a second electron to the O atom:

These three ions attract each other to give an overall neutral-charged ionic compound, which we write as Na₂O. The need for the number of electrons lost being equal to the number of electrons gained explains why ionic compounds have the ratio of cations to anions that they do. This is required by the law of conservation of matter as well.

Example 1

With arrows, illustrate the transfer of electrons to form calcium chloride from Ca atoms and Cl atoms.

 

Solution

A Ca atom has two valence electrons, while a Cl atom has seven electrons. A Cl atom needs only one more to complete its octet, while Ca atoms have two electrons to lose. Thus we need two Cl atoms to accept the two electrons from one Ca atom. The transfer process looks like this:The oppositely charged ions attract each other to make CaCl₂.

 

Test Yourself

With arrows, illustrate the transfer of electrons to form potassium sulfide from K atoms and S atoms.

 

Answer

The strength of ionic bonding depends on two major characteristics: the magnitude of the charges and the size of the ion. The greater the magnitude of the charge, the stronger the ionic bond. The smaller the ion, the stronger the ionic bond (because a smaller ion size allows the ions to get closer together). The measured strength of ionic bonding is called the lattice energy. Some lattice energies are given in Table 1 "Lattice Energies of Some Ionic Compounds."

Compound	Lattice Energy (kJ/mol)
LiF	1,036
LiCl	853
NaCl	786
NaBr	747
MgF2	2,957
Na2O	2,481
MgO	3,791

Table 1. Lattice Energies of Some Ionic Compounds

Chemistry Is Everywhere: Salt

The element sodium (part [a] in the accompanying figure) is a very reactive metal; given the opportunity, it will react with the sweat on your hands and form sodium hydroxide, which is a very corrosive substance. The element chlorine (part [b] in the accompanying figure) is a pale yellow, corrosive gas that should not be inhaled due to its poisonous nature. Bring these two hazardous substances together, however, and they react to make the ionic compound sodium chloride (part [c] in the accompanying figure), known simply as salt.

[caption id="attachment_3230" align="aligncenter" width="400"] Figure 1. Sodium + Chlorine = Sodium Chloride  (a) Sodium is a very reactive metal. (b) Chlorine is a pale yellow, noxious gas. (c) Together, sodium and chlorine make sodium chloride—salt—which is necessary for our survival.[/caption]

Salt is necessary for life. Na⁺ ions are one of the main ions in the human body and are necessary to regulate the fluid balance in the body. Cl⁻ ions are necessary for proper nerve function and respiration. Both of these ions are supplied by salt. The taste of salt is one of the fundamental tastes; salt is probably the most ancient flavouring known, and one of the few rocks we eat.

The health effects of too much salt are still under debate, although a 2010 report by the US Department of Agriculture concluded that “excessive sodium intake…raises blood pressure, a well-accepted and extraordinarily common risk factor for stroke, coronary heart disease, and kidney disease.”[footnote]US Department of Agriculture Committee for Nutrition Policy and Promotion, “Report of the Dietary Guidelines Advisory Committee on the Dietary Guidelines for Americans,” accessed January 5, 2010, http://www.cnpp.usda.gov/DGAs2010-DGACReport.htm[/footnote] It is clear that most people ingest more salt than their bodies need, and most nutritionists recommend curbing salt intake. Curiously, people who suffer from low salt (called hyponatria) do so not because they ingest too little salt but because they drink too much water. Endurance athletes and others involved in extended strenuous exercise need to watch their water intake so their body’s salt content is not diluted to dangerous levels.

Key Concepts and Summary

The tendency to form species that have eight electrons in the valence shell is called the octet rule.  The attraction of oppositely charged ions caused by electron transfer is called an ionic bond.   The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions.

Exercises

1. Comment on the possible formation of the K²⁺ ion. Why is its formation unlikely? 2. How many electrons does a Ba atom have to lose to have a complete octet in its valence shell? 3. How many electrons does an Se atom have to gain to have a complete octet in its valence shell? 4. With arrows, illustrate the transfer of electrons to form potassium chloride from K atoms and Cl atoms. 5. With arrows, illustrate the transfer of electrons to form scandium fluoride from Sc atoms and F atoms. 6. Which ionic compound has the higher lattice energy—KI or MgO? Why?  7. Which ionic compound has the higher lattice energy—BaS or MgO? Why?

Answers 1. The K²⁺ ion is unlikely to form because the K⁺ ion already satisfies the octet rule and is rather stable. 2. two 3. two 4.5.6. MgO because the ions have a higher magnitude charge 7. MgO because the ions are smaller

]]> 2443 1538 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/covalent-bonds/ Thu, 12 Apr 2018 03:53:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/covalent-bonds/

Learning Objectives

By the end of this section, you will be able to:

• Define covalent bond.
• Illustrate covalent bond formation with Lewis electron dot diagrams.
• Draw Lewis structures depicting the bonding in simple molecules

Ionic bonding typically occurs when it is easy for one atom to lose one or more electrons and another atom to gain one or more electrons. However, some atoms won’t give up or gain electrons easily. Yet they still participate in compound formation. How?

There is another mechanism for obtaining a complete valence shell: sharing electrons. When electrons are shared between two atoms, they make a bond called a covalent bond.

Let us illustrate a covalent bond by using H atoms, with the understanding that H atoms need only two electrons to fill the 1s subshell. Each H atom starts with a single electron in its valence shell:The two H atoms can share their electrons:

We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom’s valence shell:

Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. (This explains why hydrogen is one of the diatomic elements.) For simplicity’s sake, it is not unusual to represent the covalent bond with a dash, instead of with two dots:

H–H

Because two atoms are sharing one pair of electrons, this covalent bond is called a single bond.

As another example, consider fluorine. F atoms have seven electrons in their valence shell:These two atoms can do the same thing that the H atoms did; they share their unpaired electrons to make a covalent bond.

Note that each F atom has a complete octet around it now:

We can also write this using a dash to represent the shared electron pair:

There are two different types of electrons in the fluorine diatomic molecule. The bonding electron pair makes the covalent bond. Each F atom has three other pairs of electrons that do not participate in the bonding; they are called lone electron pairs. Each F atom has one bonding pair and three lone pairs of electrons.

Covalent bonds can be made between different elements as well. One example is HF. Each atom starts out with an odd number of electrons in its valence shell:The two atoms can share their unpaired electrons to make a covalent bond:

We note that the H atom has a full valence shell with two electrons, while the F atom has a complete octet of electrons.

Example 1

Use Lewis electron dot diagrams to illustrate the covalent bond formation in HBr.

 

Solution

HBr is very similar to HF, except that it has Br instead of F. The atoms are as follows:The two atoms can share their unpaired electron:

Test Yourself

Use Lewis electron dot diagrams to illustrate the covalent bond formation in Cl₂.

 

Answer

More than two atoms can participate in covalent bonding, although any given covalent bond will be between two atoms only. Consider H and O atoms:The H and O atoms can share an electron to form a covalent bond:

The H atom has a complete valence shell. However, the O atom has only seven electrons around it, which is not a complete octet. We fix this by including a second H atom, whose single electron will make a second covalent bond with the O atom:

(It does not matter on what side the second H atom is positioned.) Now the O atom has a complete octet around it, and each H atom has two electrons, filling its valence shell. This is how a water molecule, H₂O, is made.

Example 2

Use a Lewis electron dot diagram to show the covalent bonding in NH₃.

 

Solution

The N atom has the following Lewis electron dot diagram:It has three unpaired electrons, each of which can make a covalent bond by sharing electrons with an H atom. The electron dot diagram of NH₃ is as follows:

Test Yourself

Use a Lewis electron dot diagram to show the covalent bonding in PCl₃.

 

Answer

Lewis Structures

We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:

The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:

A single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.

The Octet Rule

The other halogen molecules (F₂, Br₂, I₂, and At₂) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.

The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl₄ (carbon tetrachloride) and silicon in SiH₄ (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:

Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH₃ (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:

Double and Triple Bonds

As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH₂O (formaldehyde) and between the two carbon atoms in C₂H₄ (ethylene):

A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN^–):

Writing Lewis Structures with the Octet Rule

For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:

For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:

1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

Let us determine the Lewis structures of SiH₄, CHO₂−, NO⁺, and OF₂ as examples in following this procedure:

1. Determine the total number of valence (outer shell) electrons in the molecule or ion.
   • For a molecule, we add the number of valence electrons on each atom in the molecule:
     $latex \begin{array}{r r l} \text{SiH}_4 & & \\[1em] & \text{Si: 4 valence electrons/atom} \times 1 \;\text{atom} & = 4 \\[1em] \rule[-0.5ex]{21em}{0.1ex}\hspace{-21em} + & \text{H: 1 valence electron/atom} \times 4 \;\text{atoms} & = 4 \\[1em] & & = 8 \;\text{valence electrons} \end{array}$
   • For a negative ion, such as CHO₂⁻, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):
     $latex \begin{array}{r r l} {\text{CHO}_2}^{-} & & \\[1em] & \text{C: 4 valence electrons/atom} \times 1 \;\text{atom} & = 4 \\[1em] & \text{H: 1 valence electron/atom} \times 1 \;\text{atom} & = 1 \\[1em] & \text{O: 6 valence electrons/atom} \times 2 \;\text{atoms} & = 12 \\[1em] \rule[-0.5ex]{21.5em}{0.1ex}\hspace{-21.5em} + & 1\;\text{additional electron} & = 1 \\[1em] & & = 18 \;\text{valence electrons} \end{array}$
   • For a positive ion, such as NO⁺, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:
     $latex \begin{array}{r r l} \text{NO}^{+} & & \\[1em] & \text{N: 5 valence electrons/atom} \times 1 \;\text{atom} & = 5 \\[1em] & \text{O: 6 valence electrons/atom} \times 1 \;\text{atom} & = 6 \\[1em] \rule[-0.5ex]{21em}{0.1ex}\hspace{-21em} + & -1 \;\text{electron (positive charge)} & = -1 \\[1em] & & = 10 \;\text{valence electrons} \end{array}$
   • Since OF₂ is a neutral molecule, we simply add the number of valence electrons:
     $latex \begin{array}{r r l} \text{OF}_{2} & & \\[1em] & \text{O: 6 valence electrons/atom} \times 1 \;\text{atom} & = 6 \\[1em] \rule[-0.5ex]{21em}{0.1ex}\hspace{-21em} + & \text{F: 7 valence electrons/atom} \times 2 \;\text{atoms} & = 14 \\[1em] & & = 20 \;\text{valence electrons} \end{array}$
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)When several arrangements of atoms are possible, as for CHO₂⁻, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In CHO₂⁻, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl₃, S in SO₂, and Cl in ClO₄⁻. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
   • There are no remaining electrons on SiH₄, so it is unchanged:
4. Place all remaining electrons on the central atom.
   • For SiH₄, CHO₂⁻, and NO⁺, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
   • For OF₂, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
   • SiH₄: Si already has an octet, so nothing needs to be done.
   • CHO₂⁻: We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:
   • NO⁺: For this ion, we added eight valence electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:This still does not produce an octet, so we must move another pair, forming a triple bond:
   • In OF₂, each atom has an octet as drawn, so nothing changes.

Polyatomic ions are bonded together with covalent bonds, as seen in the example of CHO₂−.  Because they are ions, however, they participate in ionic bonding with other ions. So both major types of bonding can occur at the same time.

Example 3

NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H₃CCH₃), acetylene (HCCH), and ammonia (NH₃). What are the Lewis structures of these molecules?

 

Solution

1. Calculate the number of valence electrons.HCN: (1 × 1) + (4 × 1) + (5 × 1) = 10H₃CCH₃: (1 × 3) + (2 × 4) + (1 × 3) = 14HCCH: (1 × 1) + (2 × 4) + (1 × 1) = 10NH₃: (5 × 1) + (3 × 1) = 8
2. Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
3. Where needed, distribute electrons to the terminal atoms:HCN: six electrons placed on NH₃CCH₃: no electrons remainHCCH: no terminal atoms capable of accepting electronsNH₃: no terminal atoms capable of accepting electrons
4. Where needed, place remaining electrons on the central atom:HCN: no electrons remainH₃CCH₃: no electrons remainHCCH: four electrons placed on carbonNH₃: two electrons placed on nitrogen
5. Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:HCN: form two more C–N bondsH₃CCH₃: all atoms have the correct number of electronsHCCH: form a triple bond between the two carbon atomsNH₃: all atoms have the correct number of electrons

 

Test yourself Both carbon monoxide, CO, and carbon dioxide, CO₂, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO₂ has been implicated in global climate change. What are the Lewis structures of these two molecules?

  Answers

Example 4

What is the proper Lewis electron dot diagram for CO₂?

 

Solution

The central atom is a C atom, with O atoms as surrounding atoms. We have a total of 4 + 6 + 6 = 16 valence electrons. Following the rules for Lewis electron dot diagrams for compounds gives usThe O atoms have complete octets around them, but the C atom has only four electrons around it. The way to solve this dilemma is to make a double bond between carbon and each O atom:

Each O atom still has eight electrons around it, but now the C atom also has a complete octet. This is an acceptable Lewis electron dot diagram for CO₂.

 

Test Yourself

What is the proper Lewis electron dot diagram for carbonyl sulfide (COS)?

 

Answer

Fullerene Chemistry

Carbon soot has been known to man since prehistoric times, but it was not until fairly recently that the molecular structure of the main component of soot was discovered. In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley (Figure 1), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C₆₀ buckminsterfullerene molecule (Figure 1 in Chapter 8 Introduction). An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C_60. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.

[caption id="" align="aligncenter" width="650"] Figure 1. Richard Smalley (1943–2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. Upon his death in 2005, the US Senate honored him as the “Father of Nanotechnology.” (credit: United States Department of Energy)[/caption]

Exceptions to the Octet Rule

Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:

• Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
• Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
• Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.

Examples of these will be covered later chemistry courses.

Food and Drink App: Vitamins and Minerals

Vitamins are nutrients that our bodies need in small amounts but cannot synthesize; therefore, they must be obtained from the diet. The word vitamin comes from “vital amine” because it was once thought that all these compounds had an amine group (NH₂) in it. This is not actually true, but the name stuck anyway.

All vitamins are covalently bonded molecules. Most of them are commonly named with a letter, although all of them also have formal chemical names. Thus vitamin A is also called retinol, vitamin C is called ascorbic acid, and vitamin E is called tocopherol. There is no single vitamin B; there is a group of substances called the B complex vitamins that are all water soluble and participate in cell metabolism. If a diet is lacking in a vitamin, diseases such as scurvy or rickets develop. Luckily, all vitamins are available as supplements, so any dietary deficiency in a vitamin can be easily corrected.

A mineral is any chemical element other than carbon, hydrogen, oxygen, or nitrogen that is needed by the body. Minerals that the body needs in quantity include sodium, potassium, magnesium, calcium, phosphorus, sulfur, and chlorine. Essential minerals that the body needs in tiny quantities (so-called trace elements) include manganese, iron, cobalt, nickel, copper, zinc, molybdenum, selenium, and iodine. Minerals are also obtained from the diet. Interestingly, most minerals are consumed in ionic form, rather than as elements or from covalent molecules. Like vitamins, most minerals are available in pill form, so any deficiency can be compensated for by taking supplements.

Figure 2. Vitamin and Mineral Supplements

Every entry down through pantothenic acid is a vitamin, and everything from calcium and below is a mineral.

Key Concepts and Summary

Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient molecules, and hypervalent molecules.

Exercises

1. Write the Lewis symbols for each of the following ions:

a) As^3–         b) I^–         c) Be²⁺         d) O^2–          e) Ga³⁺          f) Li⁺         g) N^3–

2. Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed:

a) MgS         b) Al₂O₃            c) GaCl₃          d) K₂O        e) Li₃N         f) KF

3. Write the Lewis structure for the diatomic molecule P₂, an unstable form of phosphorus found in high-temperature phosphorus vapor. 4. Write Lewis structures for the following:

a) O₂             b) H₂CO         c) AsF₃         d) ClNO         e) SiCl₄

f) H₃O⁺        g) NH₄⁺            h) BF₄⁻         i) HCCH         j) ClCN          k) C₂²⁺

5. Write Lewis structures for the following:

a)  SeCl₃⁺          b) Cl₂BBCl₂ (contains a B–B bond)

6. Correct the following statement: “The bonds in solid PbCl₂ are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl₂ are located on the Cl^– ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.” 7. Methanol, H₃COH, is used as the fuel in some race cars. Ethanol, C₂H₅OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO₂ and H₂O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas. 8. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl₂CO. Write the Lewis structures for carbon tetrachloride and phosgene. 9. The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.

a) the amino acid serine:

b) urea:

c) pyruvic acid:

d) uracil:

e) carbonic acid:

10. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. 11. How are single, double, and triple bonds similar? How do they differ? 12. How many electrons will be in the valence shell of H atoms when it makes a covalent bond? 13. What is the Lewis electron dot diagram of I₂? Circle the electrons around each atom to verify that each valence shell is filled. 14. What is the Lewis electron dot diagram of NCl₃? Circle the electrons around each atom to verify that each valence shell is filled. 15. Draw the Lewis electron dot diagram for each substance.   a)  SF₂         b)  BH₄⁻ 16. Draw the Lewis electron dot diagram for each substance.   a)  GeH₄     b)  ClF 17. Draw the Lewis electron dot diagram for each substance. Double or triple bonds may be needed.              a)  SiO₂    b)  C₂H₄ (assume two central atoms) 18. Draw the Lewis electron dot diagram for each substance. Double or triple bonds may be needed.

a)  CS₂       b)  NH₂CONH₂ (assume that the N and C atoms are the central atoms)

  Answers

1. a) eight electrons:

b) eight electrons: c) no electrons:   Be²⁺ d) eight electrons: e) no electrons:  Ga³⁺ f) no electrons: Li⁺ g) eight electrons:

2. a)

b) c) d) e) f) 3.

4. a)

In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule. b) c) d) e) f) g) h) i) j) k)

5. a) SeCl₃⁺:

b) Cl₂BBCl₂:

6. Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb²⁺ ion has a 6s² valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.

7. 8.

9. a)

b) c) d) e) 10.

11. Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.

12. two 13. 14. 15.

a)   

b)   

16.

a)   

b)   

17.

a)   

b)   

18.

a)   

b)   

Glossary

double bond: covalent bond in which two pairs of electrons are shared between two atoms free radical: molecule that contains an odd number of electrons hypervalent molecule: molecule containing at least one main group element that has more than eight electrons in its valence shell Lewis structure: diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion Lewis symbol: symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion lone pair: two (a pair of) valence electrons that are not used to form a covalent bond octet rule: guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond single bond: bond in which a single pair of electrons is shared between two atoms triple bond: bond in which three pairs of electrons are shared between two atoms

]]> 2484 1538 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=28

Chapter 4. Reduction/Oxidation Reactions (Redox)

4.2 Reduction/Oxidation Reaction

Learning Objectives

By the end of this section, you will be able to:

• Compute the oxidation states for elements in compounds and identify reducing agents and oxidizing agents.

Oxidation-Reduction Reactions

Earth’s atmosphere contains about 20% molecular oxygen, O₂, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O₂, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.

Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:

These equations show that Na atoms lose electrons while Cl atoms (in the Cl₂ molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:

In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:

The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.

1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion’s charge.
3. Oxidation numbers for common nonmetals are usually assigned as follows:
   • Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
   • Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, O₂²⁻), very rarely (so-called superoxides, O₂⁻), positive values when combined with F (values vary)
   • Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.

Example 3

Assigning Oxidation Numbers
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:

(a) H₂S

(b) SO₃²⁻

(c) Na₂SO₄

Solution
(a) According to guideline 1, the oxidation number for H is +1.

Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

(b) Guideline 3 suggests the oxidation number for oxygen is −2.

Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

(c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately.

According to guideline 2, the oxidation number for sodium is +1.

Assuming the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:

Check Your Learning
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:

(a) KNO₃

(b) AlH₃

(c) NH₄⁺

(d) H₂PO₄⁻

Answer:

(a) N, +5; (b) Al, +3; (c) N, −3; (d) P, +5

Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 4.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H₂ to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in HCl).

Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted in Figure 1 in Chapter 4 Introduction are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:

Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at

3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.

Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:

Metallic elements may also be oxidized by solutions of other metal salts; for example:

This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu²⁺ ions dissolve in the solution to yield a characteristic blue color (Figure 5).

Figure 5. (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)

Example 4

Describing Redox Reactions
Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.

(a)

(b)

(c)

(d)

(e)

Solution
Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.

(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(l) to +3 in GaBr₃(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br₂(l) to −1 in GaBr₃(s). The oxidizing agent is Br₂(l).

(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H₂O₂(aq) to 0 in O₂(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H₂O₂(aq) to −2 in H₂O(l). For disproportionation reactions, the same substance functions as an oxidant and a reductant.

(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C₂H₄(g) to +4 in CO₂(g). The reducing agent (fuel) is C₂H₄(g). Oxygen is reduced, its oxidation number decreasing from 0 in O₂(g) to −2 in H₂O(l). The oxidizing agent is O₂(g).

Check Your Learning
This equation describes the production of tin(II) chloride:

Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.

Answer:

Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant.

Key Concepts and Summary

 Redox reactions involve a change in oxidation number for one or more reactant elements. The Reducing Agent cause another reagent to become reduced.  In turn the reducing agent becomes ‘oxidized’.  The Oxidizing agent causes another agent to become oxidized.  In turn the oxidizing agent becomes reduced. Analogy:  A cleaning agent cleans something , but in turn beocmes ‘dirty’ as it picks up dirt from the thing it cleans

Chemistry End of Chapter Exercises

1. Determine the oxidation states of the elements in the following compounds:

   (a) NaI

   (b) GdCl₃

   (c) LiNO₃

   (d) H₂Se

   (e) Mg₂Si

   (f) RbO₂, rubidium superoxide

   (g) HF

2. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

   (a) H₃PO₄

   (b) Al(OH)₃

   (c) SeO₂

   (d) KNO₂

   (e) In₂S₃

   (f) P₄O₆

3. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

   (a) H₂SO₄

   (b) Ca(OH)₂

   (c) BrOH

   (d) ClNO₂

   (e) TiCl₄

   (f) NaH

4. Identify the oxidation-reduction reactions:

   (a)

   (b)

   (c)

   (d)

   (e)

   (f)

5. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:

   (a)

   (b)

   (c)

   (d)

   (e)

   (f)

7. Great Lakes Chemical Company produces bromine, Br₂, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl₂.
8. In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of Mg₃N₂, a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.
9. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO₂. (Hint: Water is one of the products.)
10. Calcium propionate is sometimes added to bread to retard spoilage. This compound can be prepared by the reaction of calcium carbonate, CaCO₃, with propionic acid, C₂H₅CO₂H, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate.
11. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:

    (a)

    (b)

12. Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid hydrogen sulfate as the only product. Write the two equations which represent these reactions.
13. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.

    (a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)

    (b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction

    (c) gaseous H₂S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)

14. Calcium cyclamate Ca(C₆H₁₁NHSO₃)₂ is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid C₆H₁₁NHSO₃H with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions.
15. Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method):

    (a)

    (b)

    (c)

    (d)

    (e)

    (f)

    (g)

    (h)

16. Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method):

    (a)

    (b)

    (c)

    (d)

    (e)

    (f)

    (g)

    (h)

17. Balance each of the following equations according to the half-reaction method:

    (a)

    (b)

    (c)

    (d)

    (e)

18. Balance each of the following equations according to the half-reaction method:

    (a)

    (b)

    (c)

    (d)

    (e)

    (f)

    (g)

    (h)

19. Balance each of the following equations according to the half-reaction method:

    (a)

    (b)

    (c)

 

Glossary

half-reaction

an equation that shows whether each reactant loses or gains electrons in a reaction.

oxidation

process in which an element’s oxidation number is increased by loss of electrons

oxidation-reduction reaction

(also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements

oxidation number

(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic

oxidizing agent

(also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced

reduction

process in which an element’s oxidation number is decreased by gain of electrons

reducing agent

(also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized

s

Solutions

Answers to Chemistry End of Chapter Exercises

2. (a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)

4. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.

6. (a) H +1, P +5, O −2; (b) Al +3, H +1, O −2; (c) Se +4, O −2; (d) K +1, N +3, O −2; (e) In +3, S −2; (f) P +3, O −2

8. (a) acid-base; (b) oxidation-reduction: Na is oxidized, H⁺ is reduced; (c) oxidation-reduction: Mg is oxidized, Cl₂ is reduced; (d) acid-base; (e) oxidation-reduction: P³⁻ is oxidized, O₂ is reduced; (f) acid-base

10.
(a) ;
(b) ;

12.
(a) ;
(b) ;
(c) ;
(d) ;

14.
(a)
(b)
(c)

16.

18.

20.

22.
(a)
(b)

24.
(a) step 1: , step 2: ;
(b) ;
(c) ;

26.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)

28.
(a)
(b)
(c)
(d)
(e)

30.
(a)
(b)
(c)

License

4.2 Reduction/Oxidation Reaction by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 28 3 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=48

Chapter 8. Gases

8.1 Gas Pressure

Learning Objectives

By the end of this section, you will be able to:

• Define the property of pressure
• Define and convert among the units of pressure measurements
• Describe the operation of common tools for measuring gas pressure
• Calculate pressure from manometer data

The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure 1). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.

Figure 1. The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail.

A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.

A smaller scale demonstration of this phenomenon is briefly explained.

Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.

In general, pressure is defined as the force exerted on a given area: . Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area.

Let’s apply this concept to determine which would be more likely to fall through thin ice in Figure 2—the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in²), so the pressure exerted by each foot is about 14 lb/in²:

The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in², so the pressure exerted by each blade is about 30 lb/in²:

Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall though thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:

Figure 2. Although (a) an elephant’s weight is large, creating a very large force on the ground, (b) the figure skater exerts a much higher pressure on the ice due to the small surface area of her skates. (credit a: modification of work by Guido da Rozze; credit b: modification of work by Ryosuke Yagi)

The SI unit of pressure is the pascal (Pa), with 1 Pa = 1 N/m², where N is the newton, a unit of force defined as 1 kg m/s². One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch—pounds per square inch (psi)—for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table 1 provides some information on these and a few other common units for pressure measurements

Unit Name and Abbreviation	Definition or Relation to Other Unit
pascal (Pa)	1 Pa = 1 N/m2 recommended IUPAC unit
kilopascal (kPa)	1 kPa = 1000 Pa
pounds per square inch (psi)	air pressure at sea level is ~14.7 psi
atmosphere (atm)	1 atm = 101,325 Pa air pressure at sea level is ~1 atm
bar (bar, or b)	1 bar = 100,000 Pa (exactly) commonly used in meteorology
millibar (mbar, or mb)	1000 mbar = 1 bar
inches of mercury (in. Hg)	1 in. Hg = 3386 Pa used by aviation industry, also some weather reports
torr	named after Evangelista Torricelli, inventor of the barometer
millimeters of mercury (mm Hg)	1 mm Hg ~1 torr
Table 1. Pressure Units

Example 1

Conversion of Pressure Units
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:

(a) torr

(b) atm

(c) kPa

(d) mbar

Solution
This is a unit conversion problem. The relationships between the various pressure units are given in Table 1.

(a)

(b)

(c)

(d)

Check Your Learning
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?

Answer:

0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar

We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure 3). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere.

Figure 3. In a barometer, the height, h, of the column of liquid is used as a measurement of the air pressure. Using very dense liquid mercury (left) permits the construction of reasonably sized barometers, whereas using water (right) would require a barometer more than 30 feet tall.

If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, p:

where h is the height of the fluid, ρ is the density of the fluid, and g is acceleration due to gravity.

Example 2

Calculation of Barometric Pressure
Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = 13.6 g/cm³.

Solution
The hydrostatic pressure is given by p = hρg, with h = 760 mm, ρ = 13.6 g/cm³, and g = 9.81 m/s². Plugging these values into the equation and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:)

Check Your Learning
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm³.

Answer:

10.3 m

A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube (h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure 4) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.

Figure 4. A manometer can be used to measure the pressure of a gas. The (difference in) height between the liquid levels (h) is a measure of the pressure. Mercury is usually used because of its large density.

Example 3

Calculation of Pressure Using a Closed-End Manometer
The pressure of a sample of gas is measured with a closed-end manometer, as shown to the right. The liquid in the manometer is mercury. Determine the pressure of the gas in:

(a) torr

(b) Pa

(c) bar

Solution
The pressure of the gas is equal to a column of mercury of height 26.4 cm. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 26.4 cm Hg, or mercury.) We could use the equation p = hρg as in Example 2, but it is simpler to just convert between units using Table 1.

(a)

(b)

(c)

Check Your Learning
The pressure of a sample of gas is measured with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:

(a) torr

(b) Pa

(c) bar

Answer:

(a) ~150 torr; (b) ~20,000 Pa; (c) ~0.20 bar

Example 4

Calculation of Pressure Using an Open-End Manometer
The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown to the right. Determine the pressure of the gas in:

(a) mm Hg

(b) atm

(c) kPa

Solution
The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)

(a) In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg

(b)

(c)

Check Your Learning
The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown to the right. Determine the pressure of the gas in:

(a) mm Hg

(b) atm

(c) kPa

Answer:

(a) 642 mm Hg; (b) 0.845 atm; (c) 85.6 kPa

Measuring Blood Pressure

Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure 5). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure—the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure—the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg).

Figure 5. (a) A medical technician prepares to measure a patient’s blood pressure with a sphygmomanometer. (b) A typical sphygmomanometer uses a valved rubber bulb to inflate the cuff and a diaphragm gauge to measure pressure. (credit a: modification of work by Master Sgt. Jeffrey Allen)

Meteorology, Climatology, and Atmospheric Science

Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure 6) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.

Figure 6. Meteorologists use weather maps to describe and predict weather. Regions of high (H) and low (L) pressure have large effects on weather conditions. The gray lines represent locations of constant pressure known as isobars. (credit: modification of work by National Oceanic and Atmospheric Administration)

In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.

The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure 7: the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.

Figure 7. Earth’s atmosphere has five layers: the troposphere, the stratosphere, the mesosphere, the thermosphere, and the exosphere.

Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.

Key Concepts and Summary

Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.

Key Equations

• 
• 

Chemistry End of Chapter Exercises

1. Why are sharp knives more effective than dull knives (Hint: think about the definition of pressure)?
2. Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?
3. Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?
4. A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa.
5. A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals?
6. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?
7. Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?
8. During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr and kPa?
9. The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
10. A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. What is the pressure of this gas in atmospheres and torr?
11. Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in., 1013.9 mbar.

    (a) What was the pressure in kPa?

    (b) The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Calculate the drop in pressure in torr.

12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?
13. The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:

    (a) torr

    (b) Pa

    (c) bar

    

14. The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in. Hg, determine the pressure of the gas in:

    (a) torr

    (b) Pa

    (c) bar

    

15. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg, determine the pressure of the gas in:

    (a) mm Hg

    (b) atm

    (c) kPa

    

16. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg, determine the pressure of the gas in:

    (a) mm Hg

    (b) atm

    (c) kPa

    

17. How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?

Glossary

atmosphere (atm)

unit of pressure; 1 atm = 101,325 Pa

bar

(bar or b) unit of pressure; 1 bar = 100,000 Pa

barometer

device used to measure atmospheric pressure

hydrostatic pressure

pressure exerted by a fluid due to gravity

manometer

device used to measure the pressure of a gas trapped in a container

pascal (Pa)

SI unit of pressure; 1 Pa = 1 N/m²

pounds per square inch (psi)

unit of pressure common in the US

pressure

force exerted per unit area

torr

unit of pressure;

Solutions

Answers to Chemistry End of Chapter Exercises

1. The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively.

3. Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice.

5. 0.809 atm; 82.0 kPa

7. 2.2 × 10² kPa

9. Earth: 14.7 lb in^–2; Venus: 13.1 × 10³ lb in⁻²

11. (a) 101.5 kPa; (b) 51 torr drop

13. (a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar

15. (a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa

17. With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since P_gas = P_atm + P_{vol liquid}.

License

8.1 Gas Pressure by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 48 3 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=62

Chapter 8. Gases

8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law

Learning Objectives

By the end of this section, you will be able to:

• Identify the mathematical relationships between the various properties of gases
• Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions

During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 1), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas—a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation—the ideal gas law—that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Pressure and Temperature: Amontons’s Law

Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 2) and the pressure increases.

Figure 2. The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.

This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 3. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.

Figure 3. For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at –273 °C, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero.

Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P–T relationship for gases is known as either Amontons’s law or Gay-Lussac’s law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:

where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio is therefore constant (i.e., ). If the gas is initially in “Condition 1” (with P = P₁ and T = T₁), and then changes to “Condition 2” (with P = P₂ and T = T₂), we have that and , which reduces to . This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)

Example 1

Predicting Change in Pressure with Temperature
A can of hair spray is used until it is empty except for the propellant, isobutane gas.

(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?

(b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?

Solution
(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)

(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P₁ and T₁ as the initial values, T₂ as the temperature where the pressure is unknown and P₂ as the unknown pressure, and converting °C to K, we have:

Rearranging and solving gives:

Check Your Learning
A sample of nitrogen, N₂, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?

Answer:

400 torr

Volume and Temperature: Charles’s Law

If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.

This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.

These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 4.

Figure 4. The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph’s origin, representing a temperature of absolute zero.

The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.

Mathematically, this can be written as:

with k being a proportionality constant that depends on the amount and pressure of the gas.

For a confined, constant pressure gas sample, is constant (i.e., the ratio = k), and as seen with the P–T relationship, this leads to another form of Charles’s law: .

Example 2

Predicting Change in Volume with Temperature
A sample of carbon dioxide, CO₂, occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?

Solution
Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking V₁ and T₁ as the initial values, T₂ as the temperature at which the volume is unknown and V₂ as the unknown volume, and converting °C into K we have:

Rearranging and solving gives:

This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).

Check Your Learning
A sample of oxygen, O₂, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?

Answer:

21.6 mL

Example 3

Measuring Temperature with a Volume Change
Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm³ when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm³. Find the temperature of boiling ammonia on the kelvin and Celsius scales.

Solution
A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V₁ and T₁ as the initial values, T₂ as the temperature at which the volume is unknown and V₂ as the unknown volume, and converting °C into K we have:

Rearrangement gives 

Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.

Check Your Learning
What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?

Answer:

635 mL

Volume and Pressure: Boyle’s Law

If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 5.

Figure 5. When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of 1/P vs. V is linear.

Unlike the P–T and V–T relationships, pressure and volume are not directly proportional to each other. Instead, P and V exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:

with k being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure () versus the volume (V), or the inverse of volume () versus the pressure (P). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to “linearize” their data. If we plot P versus V, we obtain a hyperbola (see Figure 6).

Figure 6. The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a hyperbola, whereas (b) the graph of (1/P) vs. V is linear.

The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.

Example 4

Volume of a Gas Sample
The sample of gas in Figure 5 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:

(a) the P–V graph in Figure 5

(b) the vs. V graph in Figure 5

(c) the Boyle’s law equation

Comment on the likely accuracy of each method.

Solution
(a) Estimating from the P–V graph gives a value for P somewhere around 27 psi.

(b) Estimating from the versus V graph give a value of about 26 psi.

(c) From Boyle’s law, we know that the product of pressure and volume (PV) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P₁V₁ = k and P₂V₂ = k which means that P₁V₁ = P₂V₂.

Using P₁ and V₁ as the known values 13.0 psi and 15.0 mL, P₂ as the pressure at which the volume is unknown, and V₂ as the unknown volume, we have:

Solving:

It was more difficult to estimate well from the P–V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.

Check Your Learning
The sample of gas in Figure 5 has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using:

(a) the P–V graph in Figure 5

(b) the vs. V graph in Figure 5

(c) the Boyle’s law equation

Comment on the likely accuracy of each method.

Answer:

(a) about 17–18 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the P–V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow

Breathing and Boyle’s Law

What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure 7).

Figure 7. Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs.

Moles of Gas and Volume: Avogadro’s Law

The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.

In equation form, this is written as:

Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n versus T.

Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).

The Ideal Gas Law

To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:

• Boyle’s law: PV = constant at constant T and n
• Amontons’s law: = constant at constant V and n
• Charles’s law: = constant at constant P and n
• Avogadro’s law: = constant at constant P and T

Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:

where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol^–1 K^–1 and 8.314 kPa L mol^–1 K^–1.

Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.

The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.

Example 5

Using the Ideal Gas Law
Methane, CH₄, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH₄. What is the volume of this much methane at 25 °C and 745 torr?

Solution
We must rearrange PV = nRT to solve for V: 

If we choose to use R = 0.08206 L atm mol^–1 K^–1, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the “right” units:

It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.

Check Your Learning
Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.

Answer:

350 bar

If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: using units of atm, L, and K. Both sets of conditions are equal to the product of n ×R (where n = the number of moles of the gas and R is the ideal gas law constant).

Example 6

Using the Combined Gas Law
When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 8). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?

Figure 8. Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)

Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have:

Solving for V₂:

(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)

Check Your Learning
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.

Answer:

0.193 L

The Interdependence between Ocean Depth and Pressure in Scuba Diving

Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure 9) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.

Figure 9. Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)

Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.

Standard Conditions of Temperature and Pressure

We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP, an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure 10).

Figure 10. Since the number of moles in a given volume of gas varies with pressure and temperature changes, chemists use standard temperature and pressure (273.15 K and 1 atm or 101.325 kPa) to report properties of gases.

Key Concepts and Summary

The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law).

The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant.

Key Equations

• PV = nRT

Chemistry End of Chapter Exercises

1. Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
2. Explain how the volume of the bubbles exhausted by a scuba diver (Figure 8) change as they rise to the surface, assuming that they remain intact.
3. One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.” (a) What is the meaning of the term “inversely proportional?” (b) What are the “other things” that must be equal?
4. An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.” (a) What is the meaning of the term “directly proportional?” (b) What are the “other things” that must be equal?
5. How would the graph in Figure 4 change if the number of moles of gas in the sample used to determine the curve were doubled?
6. How would the graph in Figure 5 change if the number of moles of gas in the sample used to determine the curve were doubled?
7. In addition to the data found in Figure 5, what other information do we need to find the mass of the sample of air used to determine the graph?
8. Determine the volume of 1 mol of CH₄ gas at 150 K and 1 atm, using Figure 4.
9. Determine the pressure of the gas in the syringe shown in Figure 5 when its volume is 12.5 mL, using:

   (a) the appropriate graph

   (b) Boyle’s law

10. A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?
11. What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?
12. A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
13. A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
14. A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
    
15. The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa?
16. How many moles of gaseous boron trifluoride, BF₃, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF₃?
17. Iodine, I₂, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I₂ vapor at a pressure of 0.462 atm?
18. How many grams of gas are present in each of the following cases?

    (a) 0.100 L of CO₂ at 307 torr and 26 °C

    (b) 8.75 L of C₂H₄, at 378.3 kPa and 483 K

    (c) 221 mL of Ar at 0.23 torr and –54 °C

19. A high altitude balloon is filled with 1.41 × 10⁴ L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
20. A cylinder of medical oxygen has a volume of 35.4 L, and contains O₂ at a pressure of 151 atm and a temperature of 25 °C. What volume of O₂ does this correspond to at normal body conditions, that is, 1 atm and 37 °C?
21. A large scuba tank (Figure 8) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C?
22. A 20.0-L cylinder containing 11.34 kg of butane, C₄H₁₀, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C.
23. While resting, the average 70-kg human male consumes 14 L of pure O₂ per hour at 25 °C and 100 kPa. How many moles of O₂ are consumed by a 70 kg man while resting for 1.0 h?
24. For a given amount of gas showing ideal behavior, draw labeled graphs of:

    (a) the variation of P with V

    (b) the variation of V with T

    (c) the variation of P with T

    (d) the variation of with V

25. A liter of methane gas, CH₄, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H₂, at STP. Using Avogadro’s law as a starting point, explain why.
26. The effect of chlorofluorocarbons (such as CCl₂F₂) on the depletion of the ozone layer is well known. The use of substitutes, such as CH₃CH₂F(g), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:

    (a) CCl₂F₂(g)

    (b) CH₃CH₂F(g)

27. As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 × 10¹⁸ alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 °C?
28. A balloon that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mount Crumpet?
29. If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
30. If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?

Glossary

absolute zero

temperature at which the volume of a gas would be zero according to Charles’s law.

Amontons’s law

(also, Gay-Lussac’s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant

Avogadro’s law

volume of a gas at constant temperature and pressure is proportional to the number of gas molecules

Boyle’s law

volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured

Charles’s law

volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant

ideal gas

hypothetical gas whose physical properties are perfectly described by the gas laws

ideal gas constant (R)

constant derived from the ideal gas equation R = 0.08226 L atm mol^–1 K^–1 or 8.314 L kPa mol^–1 K^–1

ideal gas law

relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws

standard conditions of temperature and pressure (STP)

273.15 K (0 °C) and 1 atm (101.325 kPa)

standard molar volume

volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally

Solutions

Answers to Chemistry End of Chapter Exercises

2. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.

4. (a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure

6. The curve would be farther to the right and higher up, but the same basic shape.

8. 16.3 to 16.5 L

10. 3.40 × 10³ torr

12. 12.1 L

14. 217 L

16. 8.190 × 10^–2 mol; 5.553 g

18. (a) 7.24 × 10^–2 g; (b) 23.1 g; (c) 1.5 × 10^–4 g

20. 5561 L

22. 46.4 g

24. For a gas exhibiting ideal behavior:

26. (a) 1.85 L CCl₂F₂; (b) 4.66 L CH₃CH₂F

28. 0.644 atm

30. The pressure decreases by a factor of 3.

License

8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 62 3 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=73

Chapter 8. Gases

5.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions

Learning Objectives

By the end of this section, you will be able to:

• Use the ideal gas law to compute gas densities and molar masses
• Perform stoichiometric calculations involving gaseous substances
• State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures

The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”^[1]

As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.

Density of a Gas

Recall that the density of a gas is its mass to volume ratio, . Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example 1.

Example 1

Derivation of a Density Formula from the Ideal Gas Law
Use PV = nRT to derive a formula for the density of gas in g/L

Solution

1. PV = nRT
2. Rearrange to get (mol/L):
3. Multiply each side of the equation by the molar mass, . When moles are multiplied by in g/mol, g are obtained:
   
4. 

Check Your Learning
A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas?

Answer:

= 2.02 g/mol; therefore, the gas must be hydrogen (H₂, 2.02 g/mol)

We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP.

Example 2

Empirical/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas
Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?

Solution
Strategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:

Empirical formula is CH₂ [empirical mass (EM) of 14.03 g/empirical unit].

Next, use the density equation related to the ideal gas law to determine the molar mass:

= 42.0 g/mol, , so (3)(CH₂) = C₃H₆ (molecular formula)

Check Your LearningAcetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?

Answer:

Empirical formula, CH; Molecular formula, C₂H₂

Molar Mass of a Gas

Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:

The ideal gas equation can be rearranged to isolate n:

and then combined with the molar mass equation to yield:

This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.

Example 3

Determining the Molar Mass of a Volatile Liquid
The approximate molar mass of a volatile liquid can be determined by:

1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (see Figure 1)

Figure 1. When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At t_l⟶g, the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)

Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm³ at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?

Solution
Since and , substituting and rearranging gives ,

then

Check Your Learning
A sample of phosphorus that weighs 3.243 × 10⁻² g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?

Answer:

124 g/mol P₄

The Pressure of a Mixture of Gases: Dalton’s Law

Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 2). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:

In the equation P_Total is the total pressure of a mixture of gases, P_A is the partial pressure of gas A; P_B is the partial pressure of gas B; P_C is the partial pressure of gas C; and so on.

Figure 2. If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the mixture is 1350 kPa.

The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:

where P_A, X_A, and n_A are the partial pressure, mole fraction, and number of moles of gas A, respectively, and n_Total is the number of moles of all components in the mixture.

Example 4

The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 × 10⁻³ mol of H₂, 1.00 × 10⁻³ mol of He, and 3.00 × 10⁻⁴ mol of Ne at 35 °C.

(a) What are the partial pressures of each of the gases?

(b) What is the total pressure in atmospheres?

Solution
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using :

The total pressure is given by the sum of the partial pressures:

Check Your Learning
A 5.73-L flask at 25 °C contains 0.0388 mol of N₂, 0.147 mol of CO, and 0.0803 mol of H₂. What is the total pressure in the flask in atmospheres?

Answer:

1.137 atm

Here is another example of this concept, but dealing with mole fraction calculations.

Example 5

The Pressure of a Mixture of Gases
A gas mixture used for anesthesia contains 2.83 mol oxygen, O₂, and 8.41 mol nitrous oxide, N₂O. The total pressure of the mixture is 192 kPa.

(a) What are the mole fractions of O₂ and N₂O?

(b) What are the partial pressures of O₂ and N₂O?

Solution
The mole fraction is given by and the partial pressure is .

For O₂,

and

For N₂O,

and

Check Your Learning
What is the pressure of a mixture of 0.200 g of H₂, 1.00 g of N₂, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

Answer:

1.87 atm

Collection of Gases over Water

A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 3), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

Figure 3. When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).

However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 4); more detailed information on the temperature dependence of water vapor can be found in Table 2, and vapor pressure will be discussed in more detail in the next chapter on liquids.

Figure 4. This graph shows the vapor pressure of water at sea level as a function of temperature.

Temperature (°C)	Pressure (torr)		Temperature (°C)	Pressure (torr)		Temperature (°C)	Pressure (torr)
–10	1.95		18	15.5		30	31.8
–5	3.0		19	16.5		35	42.2
–2	3.9		20	17.5		40	55.3
0	4.6		21	18.7		50	92.5
2	5.3		22	19.8		60	149.4
4	6.1		23	21.1		70	233.7
6	7.0		24	22.4		80	355.1
8	8.0		25	23.8		90	525.8
10	9.2		26	25.2		95	633.9
12	10.5		27	26.7		99	733.2
14	12.0		28	28.3		100.0	760.0
16	13.6		29	30.0		101.0	787.6
Table 2. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level

Example 6

Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure 3, what is the partial pressure of argon?

Solution
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:

Rearranging this equation to solve for the pressure of argon gives:

The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so:

Check Your Learning
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?

Answer:

0.583 L

Chemical Stoichiometry and Gases

Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.

We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.

Avogadro’s Law Revisited

Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to , a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

The explanation for this is illustrated in Figure 5. According to Avogadro’s law, equal volumes of gaseous N₂, H₂, and NH₃, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N₂ reacts with three molecules of H₂ to produce two molecules of NH₃, the volume of H₂ required is three times the volume of N₂, and the volume of NH₃ produced is two times the volume of N₂.

Figure 5. One volume of N₂ combines with three volumes of H₂ to form two volumes of NH₃.

Example 7

Reaction of Gases
Propane, C₃H₈(g), is used in gas grills to provide the heat for cooking. What volume of O₂(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

Solution
The ratio of the volumes of C₃H₈ and O₂ will be equal to the ratio of their coefficients in the balanced equation for the reaction:

From the equation, we see that one volume of C₃H₈ will react with five volumes of O₂:

A volume of 13.5 L of O₂ will be required to react with 2.7 L of C₃H₈.

Check Your Learning
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C₂H₂, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 10³ L of O₂ at 0 °C and 1 atm, will be required to burn the acetylene?

Answer:

3.34 tanks (2.34 × 10⁴ L)

Example 8

Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H₂(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N₂?

Solution
Because equal volumes of H₂ and NH₃ contain equal numbers of molecules and each three molecules of H₂ that react produce two molecules of NH₃, the ratio of the volumes of H₂ and NH₃ will be equal to 3:2. Two volumes of NH₃, in this case in units of billion ft³, will be formed from three volumes of H₂:

The manufacture of 683 billion ft³ of NH₃ required 1020 billion ft³ of H₂. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)

Check Your Learning
What volume of O₂(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C₂H₄(g), measured under the same conditions of temperature and pressure? The products are CO₂ and water vapor.

Answer:

51.0 L

Example 9

Volume of Gaseous Product
What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

Solution
To convert from the mass of gallium to the volume of H₂(g), we need to do something like this:

The first two conversions are:

Finally, we can use the ideal gas law:

Check Your Learning
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO₂ at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?

Answer:

1.30 × 10³ L

Greenhouse Gases and Climate Change

The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost  is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure 6).

Figure 6. Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.

There is strong evidence from multiple sources that higher atmospheric levels of CO₂ are caused by human activity, with fossil fuel burning accounting for about of the recent increase in CO₂. Reliable data from ice cores reveals that CO₂ concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO₂ concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure 7).

Figure 7. CO₂ levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years.

Click here to see a 2-minute video explaining greenhouse gases and global warming.

Susan Solomon

Atmospheric and climate scientist Susan Solomon (Figure 8) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.

For more information, watch this video about Susan Solomon.

Figure 8. Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)

Key Concepts and Summary

The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.

Key Equations

• 
• 
• 

Chemistry End of Chapter Exercises

1. What is the density of laughing gas, dinitrogen monoxide, N₂O, at a temperature of 325 K and a pressure of 113.0 kPa?
2. Calculate the density of Freon 12, CF₂Cl₂, at 30.0 °C and 0.954 atm.
3. Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
4. A cylinder of O₂(g) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder?
5. What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr?
6. What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr?
7. How could you show experimentally that the molecular formula of propene is C₃H₆, not CH₂?
8. The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
9. Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 °C?

   (a) Outline the steps necessary to answer the question.

   (b) Answer the question.

10. A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO₂, 805 g O₂, and 4,880 g N₂. At 25 degrees C, what is the pressure in the cylinder in atmospheres?
11. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO₂, 12.0% O₂, and the remainder N₂ at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
12. A sample of gas isolated from unrefined petroleum contains 90.0% CH₄, 8.9% C₂H₆, and 1.1% C₃H₈ at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
13. A mixture of 0.200 g of H₂, 1.00 g of N₂, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.
14. Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O₂ is not. If enough O₂ is added to a cylinder of H₂ at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?
15. A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 × 10⁻⁶ mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
16. A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 °C. What is the pressure of the carbon monoxide? (See Table 2 for the vapor pressure of water.)
17. In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
18. Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO:
    

    (a) Outline the steps necessary to answer the following question: What volume of O₂ at 23 °C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?

    (b) Answer the question.

19. Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:
    

    (a) Outline the steps necessary to answer the following question: What volume of H₂ at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H₂O?

    (b) Answer the question.

20. The chlorofluorocarbon CCl₂F₂ can be recycled into a different compound by reaction with hydrogen to produce CH₂F₂(g), a compound useful in chemical manufacturing:
    
    (a) Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 °C would be required to react with 1 ton (1.000 × 10³ kg) of CCl₂F₂?

    (b) Answer the question.

21. Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN₃). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
22. Lime, CaO, is produced by heating calcium carbonate, CaCO₃; carbon dioxide is the other product.

    (a) Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875° and 0.966 atm is produced by the decomposition of 1 ton (1.000 × 10³ kg) of calcium carbonate?

    (b) Answer the question.

23. Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C₂H₂, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC₂, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.

    (a) Outline the steps necessary to answer the following question: What volume of C₂H₂ at 1.005 atm and 12.2 °C is formed by the reaction of 15.48 g of CaC₂ with water?

    (b) Answer the question.

24. Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C₂H₆, to produce carbon dioxide and water, if the volumes of C₂H₆ and O₂ are measured under the same conditions of temperature and pressure.
25. What volume of O₂ at STP is required to oxidize 8.0 L of NO at STP to NO₂? What volume of NO₂ is produced at STP?
26. Consider the following questions:

    (a) What is the total volume of the CO₂(g) and H₂O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C₂H₆(g) measured at STP?

    (b) What is the partial pressure of H₂O in the product gases?

27. Methanol, CH₃OH, is produced industrially by the following reaction:
    Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
28. What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO₂ to BaO and O₂?
29. A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N₂ and 1.25 L of O₂ at STP. What is the colorless gas?
30. Ethanol, C₂H₅OH, is produced industrially from ethylene, C₂H₄, by the following sequence of reactions:
    
    
    What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
31. One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin?
32. A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
33. One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (−NH₂) in protein material are allowed to react with nitrous acid, HNO₂, to form N₂ gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH₂(NH₂)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N₂ collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample?
    

Glossary

Dalton’s law of partial pressures

total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.

mole fraction (X)

concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components

partial pressure

pressure exerted by an individual gas in a mixture

vapor pressure of water

pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature

Solutions

Answers to Chemistry End of Chapter Exercises

2. 4.64 g L⁻¹

4. 38.8 g

6. 72.0 g mol⁻¹

8. 88.1 g mol⁻¹; PF₃

10. 141 atm

12. CH₄: 276 kPa; C₂H₆: 27 kPa; C₃H₈: 3.4 kPa

14. Yes

16. 740 torr

18. (a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O₂ produced by decomposition of this amount of HgO; and determine the volume of O₂ from the moles of O₂, temperature, and pressure. (b) 0.308 L

20. (a) Determine the molar mass of CCl₂F₂. From the balanced equation, calculate the moles of H₂ needed for the complete reaction. From the ideal gas law, convert moles of H₂ into volume. (b) 3.72 × 10³ L

22. (a) Balance the equation. Determine the grams of CO₂ produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 × 10⁵ L

24. 42.00 L

26. (a) 18.0 L; (b) 0.533 atm

28. 10.57 L O₂

30. 5.40 × 10⁵ L

32. XeF₂

---

1. “Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, http://www-history.mcs.st-andrews.ac.uk/Quotations/Lagrange.html ↵

License

5.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 73 3 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=79

Chapter 8. Gases

5.4 The Kinetic-Molecular Theory

Learning Objectives

By the end of this section, you will be able to:

• State the postulates of the kinetic-molecular theory
• Use this theory’s postulates to explain the gas laws

The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.

The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)

1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
2. The molecules composing the gas are negligibly small compared to the distances between them.
3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.

The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law.

The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I

Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:

• Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 1).
• Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.
• Boyle’s law. If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).
• Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 1).
• Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.

Figure 1. (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time.

Molecular Velocities and Kinetic Energy

The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample.

In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 2).

Figure 2. The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, ν_p, is a little less than 400 m/s, while the root mean square speed, u_rms, is closer to 500 m/s.

The kinetic energy (KE) of a particle of mass (m) and speed (u) is given by:

Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m² s^–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square velocity of a particle, u_rms, is defined as the square root of the average of the squares of the velocities with n = the number of particles:

The average kinetic energy, KE_avg, is then equal to:

The KE_avg of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:

where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J/K (8.314 kg m²s^–2K^–1). These two separate equations for KE_avg may be combined and rearranged to yield a relation between molecular speed and temperature:

Example 1

Calculation of u_rms
Calculate the root-mean-square velocity for a nitrogen molecule at 30 °C.

Solution
Convert the temperature into Kelvin:

Determine the mass of a nitrogen molecule in kilograms:

Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m²s^–2:

Check Your Learning
Calculate the root-mean-square velocity for an oxygen molecule at –23 °C.

Answer:

441 m/s

If the temperature of a gas increases, its KE_avg increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KE_avg decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 3.

Figure 3. The molecular speed distribution for nitrogen gas (N₂) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.

At a given temperature, all gases have the same KE_avg for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher u_rms, with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower u_rms, and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure 4.

Figure 4. Molecular velocity is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.

The gas simulator may be used to examine the effect of temperature on molecular velocities. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures.

The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II

According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.

The rate of effusion of a gas depends directly on the (average) speed of its molecules:

Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here:

The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law.

Key Concepts and Summary

The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.

Key Equations

• 
• 
• 

Chemistry End of Chapter Exercises

1. Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
2. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.
3. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:

   (a) The pressure of the gas is increased by reducing the volume at constant temperature.

   (b) The pressure of the gas is increased by increasing the temperature at constant volume.

   (c) The average velocity of the molecules is increased by a factor of 2.

4. The distribution of molecular velocities in a sample of helium is shown in Figure 4. If the sample is cooled, will the distribution of velocities look more like that of H₂ or of H₂O? Explain your answer.
5. What is the ratio of the average kinetic energy of a SO₂ molecule to that of an O₂ molecule in a mixture of two gases? What is the ratio of the root mean square speeds, u_rms, of the two gases?
6. A 1-L sample of CO initially at STP is heated to 546 °C, and its volume is increased to 2 L.

   (a) What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?

   (b) What is the effect on the average kinetic energy of the molecules?

   (c) What is the effect on the root mean square speed of the molecules?

7. The root mean square speed of H₂ molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N₂ molecule at 25 °C?
8. Answer the following questions:

   (a) Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?

   (b) Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?

   (c) At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?

   (d) The average temperature of the gas in a hot air balloon is 1.30 × 10² °F. Calculate its density, assuming the molar mass equals that of dry air.

   (e) The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?

   (f) An average balloon has a diameter of 60 feet and a volume of 1.1 × 10⁵ ft³. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?

   (g) A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO₂ and H₂O gas is produced by the combustion of this propane?

   (h) A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?

9. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, , is the same at 0 °C and 100 °C.

Glossary

kinetic molecular theory

theory based on simple principles and assumptions that effectively explains ideal gas behavior

root mean square velocity (u_rms)

measure of average velocity for a group of particles calculated as the square root of the average squared velocity

Solutions

2. Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature.

4. H₂O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier.

6. (a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to times its initial value; u_rms is proportional to .

8. (a) equal; (b) less than; (c) 29.48 g mol⁻¹; (d) 1.0966 g L⁻¹; (e) 0.129 g/L; (f) 4.01 × 10⁵ g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min⁻¹

License

5.4 The Kinetic-Molecular Theory by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 79 3 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/1-3-laboratory-techniques-for-separation-of-mixtures/ Wed, 25 Apr 2018 20:03:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3184

Learning Objectives

By the end of this section, you will be able to:

• Describe different methods of separation.
• Identify which separation method is most suited for a given mixture.
• Identify what physical change occurs during the separation process.

A mixture is composed of two or more types of matter that can be present in varying amounts and can be physically separated by using methods that use physical properties to separate the components of the mixture, such as evaporation, distillation, filtration and chromatography. Evaporation can be used as a separation method to separate components of a mixture with a dissolved solid in a liquid.  The liquid is evaporated, meaning it is convert from its liquid state to gaseous state.  This often requires heat.  Once the liquid is completely evaporated, the solid is all that is left behind. [caption id="attachment_3243" align="aligncenter" width="300"] Figure 1. Evaporation can be used as a separation technique.[/caption] Distillation is a separation technique used to separate components of a liquid mixture by a process of heating and cooling, which exploits the differences in the volatility of each of the components. [caption id="attachment_3242" align="aligncenter" width="300"] Figure 2. Distillation apparatus.[/caption]

Distillation procedure: 1) the round bottom flask contains the liquid mixture which must be heated to a vigorous boil, 2) the component with the lower boiling point will change into its gaseous state, 3) upon contact with the water-cooled condenser, the gas will condense, 4) trickle down into the graduated cylinder where the chemist can them recuperate the final distilled liquid, and  5) the other liquid component remains in the round bottom flask.    

Filtration is a separation technique used to separate the components of a mixture containing an undissolved solid in a liquid.  Filtration may be done cold or hot, using gravity or applying vacuum, using a Buchner or Hirsch funnel or a simple glass funnel .  The exact method used depends on the purpose of the filtration, whether it is for the isolation of a solid from a mixture or removal of impurities from a mixture.

[caption id="attachment_3244" align="aligncenter" width="267"] Figure 3.  Filtration apparatus.[/caption] Filtration procedure: 1) the mixture is pored through a funnel lined with a filter paper, 2) the filtrate (liquid) drips through to the filter flask, 3) the solid remains in the funnel. Though chromatography is a simple technique in principle, it remains the most important method for the separation of mixtures into its components. It is quite versatile for it can be used to separate mixtures of solids, or of liquids, or mixtures of solids and liquids combined, or in the case of gas chromatography, can separate mixtures of gases.  The two elements of chromatography are the stationary phase and the mobile phase.  There are many choices of stationary phases, some being alumina, silica, and even paper.  The mobile phase, in liquid chromatography, can also vary.  It is often either a solvent or a mixture of solvents and is often referred to as the eluant..  A careful choice of eluting solvent helps to make the separation more successful.  The mixture is placed on the stationary phase.  The eluant passes over the mixture and continues to pass through the stationary phase carrying along the components of the mixture.  If a component in the mixture has greater affinity for the mobile phase (eluant) than the stationary phase, it will tend to be carried along easily with the eluant. If another component in the mixture has a greater affinity for the stationary phase than the mobile phase then it will not be carried along so easily.  A separation is thus obtained when the different components in a mixture have different affinity for the stationary and mobile phase.  Three important types of chromatography based on the principles discussed above are: 1) thin layer chromatography (TLC), 2) column chromatography, and 3) gas chromatography. [caption id="attachment_4632" align="aligncenter" width="300"] Figure 4. Thin layer chromatography is a one type of chromatography. a) The stationary phase can be a thin film of alumina or silica on glass or even paper.  The plate is placed in a developing tank which contains the mobile phase (eluant) which travels up the plate by capillary action.  b)  A separation is obtained because the component of the mixture that has a stronger affinity for the eland (compound 2) travels faster up the plate, than the component that has a strong affinity to the stationary phase (compound 1).[/caption]

Example 1

Identify which separation method is most suited for the following mixtures:  

Separation methods:	A mixture of solids	A mixture of liquids	A mixture of a solid dissolved in a liquid	A mixture of solid and liquid
Evaporation
Distillation
Filtration
Chromatography

 

Solution  

Separation methods:	A mixture of solids	A mixture of liquids	A mixture of a solid dissolved in a liquid	A mixture of solid and liquid
Evaporation	NO	NO	YES*	YES**
Distillation	NO	YES*	YES*	NO
Filtration	NO	NO	NO	YES
Chromatography	YES*	YES*	YES*	YES*

* Success depends on the physical properties of the components in the mixture. ** Would work but filtration is so much faster.  

Test Yourself

What method of separation would be most effective on the following mixtures:

a)  Sea water b)  Gold nuggets in water. c)  A solution of alcohol (liquid) and water.

  Answers

a) evaporation or distillation (chromatography not effective here)

b) filtration

c) distillation

Key Concepts and Summary

Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as evaporation, distillation, filtration and chromatography.  Which separation method used when attempting to separate a mixture depends on what kind of mixture it is (what states of matter are present) and on the physical properties of the components.

Exercises

1.  What method of separation would be most effective on the following mixtures:

a)  Vinegar (a solution of acetic acid (liquid) in water) b)  Loose tea leaves in tea. c)  Copper sulfate (solid) in water.

 

2.  Identify what physical change occurs during the following separation processes.

a)  Distillation of a solution comprising of 50:50 acetone and water b)  Filtration to remove tea leaves from tea. c)  Evaporation for water from a sugar solution to obtain sugar crystals.

d) Taking a sand and salt mixing, mixing it with water, followed by filtration to remove the sand, then evaporating the salt solution to retrieve salt crystals.

 

3.  Propose a method of separate the following complex mixtures:

a) A mixture of sand, sea water (water and salt)

b) A mixture of  marbles, small gold nuggets, and sugar

  Answers

1. a) distillation;   b) filtration;   c) evaporation

2. a)  The lower boiling liquid (acetone) would undergo a phase change (evaporation) upon heating, then once the gaseous acetone comes in contact with the condenser it would under another phase change (condensation).   b)  No phase changes, this simply involves physical removal of the leaves via filtration.  c) Water would undergo a phase change (evaporation) upon heating. c) The salt would dissolve in the water, then during the evaporation step, water would undergo a phase change (evaporation) upon heating. 3. a) Filtration to remove the sand, then evaporating the salt solution to retrieve salt crystals.    b) Manual separation of the marbles (removing by using your fingers), dissolve the rest in water, then filtration to remove the gold nuggets, and then evaporation of the water to retrieve sugar crystals.  

Glossary

chromatography: is a separation technique based on how the different components in a mixture have different affinity for the stationary and mobile phase distillation: is a separation technique used to separate components of a liquid mixture by a process of heating and cooling evaporation: is a separation method used to separate of a mixture of a liquid with a dissolved solid, involving removal of a   liquid by evaporating it and leaving behind a solid filtration: is a separation technique used to separate the components of a mixture containing an undissolved solid in a liquid by using a funnel lined with filter paper to retain the solids while letting the liquid through.

]]> 3184 1247 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/langara-end-of-chapter-problems/ Tue, 01 May 2018 02:47:31 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3263

2.  Consider three samples of H₂O: 1g of ice, 1g of water and 1g of vapor. How do the volumes of these samples compare with one another?  How is the volume related to the physical state?

3. Classify the following either a physical change, physical property, chemical change or chemical property: a)  Oxygen is a gas b)  Dry ice sublimes (goes directly from a solid to a gas) c)  A bottle of wine turning to vinegar d)  Sugar dissolving in water e)  Acid produced by bacteria can cause tooth decay

4. Correct the following statements: a)  Elements in a period have similar chemical properties. b)  In the modern periodic table, the elements are arranged in order of increasing atomic mass. c)  Elements can be classified as either metalloid or nonmetal/

5. Clearly explain the similarities and differences between molecules and mixtures.

6.  How are elements and compounds similar?  How are they different?

7.  How are compounds and molecules similar?  How are they different?

8.  How are compounds and mixtures similar?  How are they different?

9.  Classify each of the following using as many terms as are applicable from this list: element, compound, molecule, homogeneous mixture, heterogeneous mixture, pure substance. a)  sugar               b)  pure apple juice                 c)  oxygen gas in air               d)  distilled water

 Answers:

1.

2.  The volume of the vapor will be the largest, as it contains the most empty space. The volume of the liquid will be the next largest, and the volume of the solid will be the smallest because atoms are tightly packed in a solid (thus very little empty space).

3.   a) physical property     b) physical change     c) chemical change     d) physical change     e) chemical property

4.   a) Elements in a periodgrouphave similar chemical properties. b) In the modern periodic table, the elements are arranged in order of increasing atomic massnumber. c) Elements can be classified as either metalloid or nonmetal or metal

5.   Both molecules and mixtures contain more than one thing. Molecules contain more than one atom, whereas mixtures contain more than one substance. A molecule is a pure substance, whereas a mixture is not. In a molecule, the “things” (atoms) are bonded, where as in a mixture they are not. A molecule has a fixed ratio of “things” (atoms), whereas a mixture has a variable ratio.

6.   Both are pure substances (hence have fixed composition) but compounds contain more than one element bonded together, whereas elements contain only one type of atom.

7.   Both contain fixed ratios of bonded atoms (at least 2 atoms), but compounds must contain >1 type of element, whereas molecules CAN be 2 or more of the same element. All compounds are molecules but not all molecules are compounds.

8.   Both contain more than 1 substance, but compounds are bonded, whereas mixtures are not AND compounds must have a FIXED RATIO (fixed composition, as they are a pure substance) whereas mixtures can have variable composition (are NOT pure substances!)

9.   a) pure substance, compound, exists as molecules b) homogeneous mixture c) element, exists as molecules (O₂), pure substance d) molecule, compound, pure substance

]]> 3263 1247 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/2-6-langara-end-of-chapter-problems/ Tue, 01 May 2018 17:39:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3291 2 mL of water         b) 2.0x10² mL of water         c) 2.00x10² mL of water d) 2.000x10² mL of water         e) 200 mL of water         f) 200. mL of water 2. Write the following in correct scientific notation and list the number of significant figures in each: a) 0.00406      b)  3,500,000 c) 154 x 10^-2      d)  0.00001256 x 10⁶ 3. State the number of significant figures in each of the following: a) 275       b)  2.75         c) 0.275 d)  0.00275         e) 2750.0          f) 2.75 x 10⁵ 4. Write the following numbers in scientific notation: a) 426.7      b) 337300.0 c)  0.000003       d)  0.02003 5. Express the results of each of the following calculations to the appropriate number of significant figures: a) 13.196 + 0.0825 + 2.32 + 0.0013 = b) 721.56 – 0.394 = c)  (5.23x10^-2)  + (6.01x10^-3)  +  (8x10^-3)  + (3.273x10^-2)   = d)   (3.21 x 432 x 65)/563  = e)   (8.57x10^-2  x  6.02x10²³  x  2.543) / (361  x  907)  = f)   (2.01 x 43.9 x 67.0) / (23.9 x 0.016)  =

6. Express the results of each of the following calculations to the appropriate number of significant figures, and in scientific notation: a) (4.9x10³)² = b)  (3.8x10^-6)  x  (2.1x10²)  = c)  (4.4x10^-7)  ÷  (1.2x10⁵)  = d)  (7.86x10^-2)  – (18.6x10^-3)  = e)  (1.6x10³)  – (8.53x10²)  + 7.5  =

7. For the following calculations, determine the answer using the appropriate number of significant figure, units and using proper scientific notation: a) (1.93 x 10² g)(44.7 m/s)² / 2 (where 2 is an exact number) b)  (8.334 x 10⁷g) / (1.95 x 10²cm)³ c)  4.20 m x 1.1 m – (4.5 x 10³cm²)

8. Solve the following, expressing the answers in scientific notation: a) (0.101 cm)(0.15 cm) + (10.50 cm)(0.105 cm)  = b)   (0.27104 m / 0.0150 sec)  -  ( 38.171 m / 3.022 sec)  =

9. Express the results of each of the following calculations to the appropriate number of significant figures: a) (1.00x10¹⁸) +  (5.6x10¹⁷) = b)   (317  -  314.35)  ÷  4.0  = c)  30.5 + 3.05 + 0.305 + 0.0305 = d)   (3.03  +  8.14)  ÷  (427.78 - 362.060)  = e)   (6.5 g + 9.5 g)  ÷  25.00 mL = 10.  The volume of a container is 5982 mm³. a) What is the volume in cubic feet, ft³? (1 ft = 0.3048 m) b) If it takes 0.23 seconds to add 1.00 mL of water to the container, how many seconds does it take to fill the entire container? 11.  One mile equals 1.609 kilometers. If you are going 110 km/h, how many minutes will it take you to travel 254 miles? 12.  On July 23^rd, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel is 0.803 g/mL, or 1.77 lb/L. The plane had 7682 L of fuel on board in Montreal. The ground crew there multiplied the 7682 L by the factor 1.77 and concluded that they had 13,597 kg of fuel on board and needed an additional 8703 kg for the trip. They divided 8703 kg by the factor 1.77 and concluded that they needed to add 4916 L of fuel. They added 5000 L. On its flight, the plane ran out of fuel and crashed near Winnipeg, hundred of kilometers short of its destination (there were few injuries and luckily, no fatalities). What mistake did the ground crew make? How much fuel SHOULD they have added before take off? 13.  Perform the following calculation and report the answer in cm, with proper significant figures:   13.25 cm + 26 mm – 7.8 cm + 0.186 m 14.  Complete the following conversions: 498 cm/s           =   ________ km/h 63 km/h               =   ________ cm/s 89^oC                 =   ________ K 821 K                  =   ________ ^oC 802 kg/m³          =   ________ g/mL 10024 g/mL        =   ________ kg/L 38 mL                =   ________ cm3 0.00924 km         =  ________ mm 7098 mm           =   ________ dm 2987 µm             =   ________ mm 0.78 L                =   ________ cm3 908 mg               =   ________ kg 87.8 mm            =   ________ nm 89.2 m³               =   ________ mm³ 3 x 10⁸m/s        =   ________ km/year 15.  An Erlenmeyer flask has a mass of 392.6 g when empty. When filled with water (density = 1.00 g/cm³), the total mass is 503.5 g. If the flask is emptied and then filled with chloroform (density = 1.48 g/cm³), what will the total mass (container + chloroform) be? 16.  Alcohol has a density of 789 g/L. If you need 85 g of alcohol, what volume of alcohol would you need? 17.  Nickel has a density of 8.90 x 10³g/L and mercury has a density of 13.6 x 10³g/L. a) What volume of mercury has the same mass as a 40.0 cm³ piece of nickel? b) What mass of nickel occupies the same volume as 200.0 g of mercury? 18.  Gold has a density of 19.3 g/mL. If 5.79 mg of gold is hammered into a gold leaf of uniform thickness with an area of 4.46 x 10³ mm², what is the thickness of the gold leaf? 19.  The square nut pictured below is 14.00 mm on edge, 6.00 mm thick, and has a 7.0 mm diameter hole. The density of the metal used in the nuts is 7.87 g/cm³. Approximately how many of these nuts are present in a 1.00 lb package? (Note: 1 lb = 453.6 g) 20.  A container of unknown volume has a mass of 32.105 g. When filled completely with a fluid of density 0.9982 g/mL, the container and contents have a mass of 42.062 g (at 20 ^oC). When filled with benzene at 20 ^oC, the container and contents have a mass of 40.873 g. What is the density of benzene at 20 ^oC? 21.  A 3.50 mL piece of boron has a mass of 8.19 g. What is the density of boron? 22. An object made of iron is immersed in water. The object has a mass of 250 g. If the density of iron is 7.86 g/cm³, what is the volume of the water displaced by the iron object? 23.  Evaluate 0.00000000552 × 0.0000000006188 and express the answer in scientific notation. You may have to rewrite the original numbers in scientific notation first. 24.  Express the number 6.022 × 10²³ in standard notation. 25.  When powers of 10 are multiplied together, the powers are added together. For example, 10² × 10³ = 10²⁺³ = 10⁵. With this in mind, can you evaluate (4.506 × 10⁴) × (1.003 × 10²) without entering scientific notation into your calculator? 26.  Consider the quantity two dozen eggs. Is the number in this quantity “two” or “two dozen”? Justify your choice. 27.  Fill in the blank: 1 km = ______________ μm. 28.  Fill in the blank: 1 cL = ______________ ML. 29.  Express 67.3 km/h in meters/second. 30.  Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 mi/h into kilometers/hour. 31.  Convert 52.09 km/h into meters/second. 32.  Use the formulas for converting degrees Fahrenheit into degrees Celsius to determine the relative size of the Fahrenheit degree over the Celsius degree. 33. What is the mass of 12.67 L of mercury? 34.  What is the volume of 2.884 kg of gold?

Answers

1.   a) ± 100 mL of water         b) ± 10 mL of water         c) ± 1 mL of water         

d) ± 0.1 mL of water         e) ± 100 mL of water         f) ± 1 mL of water   

And the most precise measurement is (d) because it has the least amount of uncertainty. b)  ± 1 mL or so, 438 mL c)  ± 0.001 °C or so, 10.060 ^oC

2.   a) 4.06 x 10^-3= 3 significant figures b)  3.5 x 10⁶= 2 significant figures c) 1.54 or 1.54 x 10⁰= 3 significant figures d)  1.256 x 10¹= 4 significant figures

3.   a) 3    b) 3   c) 3    d) 3    e) 5   f) 3

4.   a) 4.267x10²       b) 3.373000x10⁵      c) 3x10^-6      d) 2.003x10^-2

5.   a) 15.60      b) 721.17      c) 9.9x10^-2      d) 1.6x10²      e) 4.01x10¹⁷      f) 1.5x10⁴

6.   a) 2.4x10⁷      b) 8.0x10^-4        c) 3.7x10^-12      d) 6.00x10^-2       e) 8x10²

7.   a) 1.93 x 10⁵g·m²/s²      b)  11.2 g/cm³      c) 4.2 m²or 4.2 x 10⁴cm²

8.   a) 1.12 cm²       b) 5.4 m/s

9.   a) 1.56x10¹⁸      b) 0.7      c) 33.9      d) 0.1700      e) 0.640 g/mL

10.  a) 2.113 x 10^-4ft³      b) 1.4 s

11. 220 minutes (2.2 x 10²minutes)

12. The crew used the wrong conversion factor (using 1.77 lb/L as if it were 1.77 kg/L) and didn’t pay attention to proper cancellation of units (to catch the mistake).  They should have added 2.01 x 10⁴L of fuel (at least).

13.  26.6 cm (or 26.7 cm OK)

14.  17.9;   1.8 x 10³;   362;   548;   0.802;   10024;   38;

9.24 x 10³;   70.98;   2.987;   7.8 x 10²;    9.08 x 10^-4;

8.78 x 10⁷;    8.92 x 10¹⁰;   9 x 10¹²

15.  557 g

16.  0.11 L

17.  a) 0.0262 L or 26.2 cm³       b)  131 g

18.  6.73 x 10^-5mm

19.  approximately 61 nuts

20.  0.8790 g/mL

21.  2.34 g/mL

22.  32 mL

23.  3.42 × 10⁻¹⁸

24.  602,200,000,000,000,000,000,000

25.  4.520 × 10⁶ 26.  The quantity is two; dozen is the unit. 27.  1,000,000,000 28.  1/100,000,000 29.  18.7 m/s 30.  96.1 km/h 31.  14.47 m/s 32.  One Fahrenheit degree is nine-fifths the size of a Celsius degree. 33.  1.72 × 10⁵ g 34.  149 mL

]]> 3291 2084 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/introduction-13/ Mon, 07 May 2018 15:33:21 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3402

]]> 3402 2141 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-1-names-of-elements/ Mon, 07 May 2018 16:39:33 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3410

Learning Objectives

By the end of this section, you will be able to:

• Recognize the symbols and knowing the names the elements.

To be able to name compounds, we need to start with knowing the names of the elements.  Table 1 lists the names of the elements, those in bold are the 51 elements that students in an introductory chemistry course should know.

1 - H - Hydrogen 2 - He - Helium 3 - Li - Lithium 4 - Be - Beryllium 5 - B - Boron 6 - C - Carbon 7 - N - Nitrogen 8 - O - Oxygen 9 - F - Fluorine 10 - Ne - Neon 11 - Na - Sodium 12 - Mg - Magnesium 13 - Al - Aluminum 14 - Si - Silicon 15 - P - Phosphorus 16 - S - Sulfur 17 - Cl - Chlorine 18 - Ar - Argon 19 - K - Potassium 20 - Ca - Calcium 21 - Sc - Scandium 22 - Ti - Titanium 23 - V - Vanadium 24 - Cr - Chromium 25 - Mn - Manganese 26 - Fe - Iron 27 - Co - Cobalt 28 - Ni - Nickel 29 - Cu - Copper 30 - Zn - Zinc

31 - Ga - Gallium 32 - Ge - Germanium 33 - As - Arsenic 34 - Se - Selenium 35 - Br – Bromine 36 - Kr - Krypton 37 - Rb - Rubidium 38 - Sr - Strontium 39 - Y - Yttrium 40 - Zr – Zirconium 41 - Nb - Niobium 42 - Mo - Molybdenum 43 - Tc - Technetium 44 - Ru - Ruthenium 45 - Rh - Rhodium 46 - Pd - Palladium 47 - Ag - Silver 48 - Cd - Cadmium 49 - In - Indium 50 - Sn - Tin 51 - Sb - Antimony 52 - Te - Tellurium 53 - I - Iodine 54 - Xe - Xenon 55 - Cs - Cesium 56 - Ba - Barium 57 - La - Lanthanum 58 - Ce - Cerium 59 - Pr - Praseodymium 60 - Nd - Neodymium

61 - Pm - Promethium 62 - Sm - Samarium 63 - Eu - Europium 64 - Gd - Gadolinium 65 - Tb - Terbium 66 - Dy - Dysprosium 67 - Ho - Holmium 68 - Er - Erbium 69 - Tm - Thulium 70 - Yb – Ytterbium 71 - Lu - Lutetium 72 - Hf - Hafnium 73 - Ta - Tantalum 74 - W - Tungsten 75 - Re - Rhenium 76 - Os - Osmium 77 - Ir - Iridium 78 - Pt - Platinum 79 - Au - Gold 80 - Hg - Mercury 81 - Tl - Thallium 82 - Pb - Lead 83 - Bi - Bismuth 84 - Po - Polonium 85 - At - Astatine 86 - Rn - Radon 87 - Fr - Francium 88 - Ra - Radium 89 - Ac - Actinium 90 - Th - Thorium

91 - Pa - Protactinium 92 - U - Uranium 93 - Np - Neptunium 94 - Pu - Plutonium 95 - Am - Americium 96 - Cm - Curium 97 - Bk - Berkelium 98 - Cf - Californium 99 - Es - Einsteinium 100 - Fm - Fermium 101 - Md - Mendelevium 102 - No - Nobelium 103 - Lr - Lawrencium 104 - Rf - Rutherfordium 105 - Db - Dubnium 106 - Sg - Seaborgium 107 - Bh - Bohrium 108 - Hs - Hassium 109 - Mt - Meitnerium 110 - Ds - Darmstadtium 111 - Rg – Roentgenium 112 - Cn - Copernicium 113 - Uut - Ununtrium 114 - Fl - Flerovium 115 - Uup - Ununpentium 116 - Lv - Livermorium 117 - Uus - Ununseptium 118 - Uuo - Ununoctium

Table 1. The names of the elements.  Those in bold are the elements that students in an introductory chemistry course should know.

The chemical elements are named by the International Union of Pure and Applied Chemistry (IUPAC), which generally adopts the name chosen by the discoverer of the element. Often the name refers to a place, a property of the element or a scientist.   At times, there has been some controversy of which research group actually discovered the element, and therefore which group gets the privilege of naming the element. This delayed the naming of the elements for a considerable amount of time.  Checkout element naming controversy to learn more about contention that existed in naming certain elements.

Key Concepts and Summary

Knowing the names and recognizing the symbol of the 51 elements (bolded in Table 1) that students in an introductory chemistry course will not only make it feasible for students to be able to name compounds but will help them be familiar with the common elements and compounds they may encounter in their daily lives.

Activity

Make yourself a stack of small sized Qcards.  On one side have the name of the element (e.g. hydrogen) and on the other side have its symbol (e.g. H).  Make a complete set of all the elements you should know (see bolded elements in Table 1).  Then use these Qcards to quiz yourself.

Glossary

International Union of Pure and Applied Chemistry (IUPAC): is an international federation that represents chemists in individual countries, which has several responsibilities, one being the standardization of chemical nomenclature including the naming of new elements in the periodic table.  

]]> 3410 2141 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/4-4-end-of-chapter-problems/ Tue, 08 May 2018 18:37:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3475 1. Name the following: a)  CaC₂H₃O₂         b) Fe₂O₃            c)  S₂F₅               d)  H₂SO₃            e) CuNO₃ f)  PbCO₃               g)  PF₄                h)  KMnO₄           i) B₂O₃               j)  HNO₂

2. Name the following: a)  CoCl₂                        b) Cr(OH)₃         c)  ICl                  d)  Mg₃P₂            e) Ag₃N f)  NH₄F                 g)  BaO               h)  Na₃As           i)  ZnBr₂              j) SF₆ k)  Ba(NO₃)₂          l) Ni(ClO₄)₂         m)  Zn(ClO₂)₂      n) HNO₃             o)  Ca(MnO₄)₂     p)  CuHCO₃            q) HF                 r)  NH₄HSO₄        s) HgO

3. Name the following: a)  Cd₃P₂                        b)  H₂SO₃            c) (NH₄)₂S          d) MnF₃              e)  Al₂O₃ f)   HCN                    g)  OF₂                h)  CrN                i)  Co₂S₃              j)  P₂O₃ k)   CS₂                      l)  CdO                m)  AsH₃             n)  IF₃                  o)  FePO₄ p)  H₃PO₄                 q)  TiO₂               r)  HClO              s) HI                  

4. Determine the formula for the following compounds: a)  aluminum sulfide                b)  ammonium iodide                c)  nickel(II) iodide d)  lithium phosphide               e)  gold(III) phosphide              f)  phosphoric acid g)  lithium oxide                       h) nickel(II) phosphate            i)  bromic acid j)  xenon dioxide                      k) bromine monofluoride         l)  aluminum bisulfate m)  cobalt(II) carbonate           n) potassium phosphate        o)  lithium nitrite

5. Determine the formula for the following compounds: a)  gold(III) chloride                  b)  iron(III) oxide                      c) chloric acid d)  manganese(II) bromide      e) cobalt(II) hydroxide            f)  arsenic dioxide g)  copper(II) hydroxide          h) sulfurous acid                    i)  calcium iodide j)  cadmium oxide                    k) nitrogen triiodide                 l)  sulfur monoxide m) iron(III) chromate               n)  boron trihydride                 o) manganese(III) oxide p)  sulfur hexafluoride            q) strontium hypochlorite       r)  cobalt(II) phosphate s)  magnesium bicarbonate

6. Write the formula for the following: a)  chloric acid                        b) xenon hexafluoride                  c)  iron(III) nitrate d)  acetic acid                         e) chromium(III) sulfide                 f)  lithium hydrogen carbonate g)  aluminum oxide                  h) hydrofluoric acid                      i)  copper(II) permanganate

7. Name each molecule: a) PF₃         b) CO         c) Se₂Br₂       d) SF₄         e) P₂S₅        f) N₂O₃      g) PCl₅      h) O₂F₂   i) CO₂      j) SeF₆      k) NO 8. Name the following ionic compounds: a)  CuSO₄                    b)  FeBr₃                      c)  CsClO₄ d)  (NH₄)₂Cr₂O₇            e) KHSO₄                    d)  CrF₂ 9. Determine the formula for the following ionic compounds:

a)  manganese(IV) oxide        b)  magnesium perchlorate c) antimony(III) nitrate          d)  sodium iodide

10. Determine the formula for the following ionic compounds: a)  potassium permanganate              b) iron(II) sulfate c)  sodium phosphate                         d)  copper(II) nitride e)  aluminum sulfide                            f)  chromium(III) oxide 11. Determine the chemical formula for each of the following:

 a)  dinitrogen pentoxide          b)  hydrochloric acid          c) hypochlorous acid             d)  nitric acid

12. Determine the chemical formulas of the following:

      a)  boron trifluoride     b)  bromic acid            c) sulfurous acid

Answers

1.  a) calcium acetate              b) iron(III) oxide        c) disulfur pentafluoride      d) sulfurous acid  e) copper(I) nitrate             f) lead(II) carbonate      g) phosphorous tetrafluoride      h) potassium permanganate  i) diboron trioxide               j) nitrous acid

2.  a) cobalt(II) chloride     b) chromium(III) hydroxide   c) iodine monochloride  d) magnesium phosphide e) silver nitride            f) ammonium fluoride    g) barium oxide            h) sodium arsenide i) zinc bromide             j) sulfur hexafluoride   k) barium nitrate          l) nickel(II) perchlorate m) zinc chlorite            n) nitric acid    o) calcium permanganate      p) copper(I) hydrogen carbonate  

q) hydrofluoric acid    r) ammonium hydrogen sulfate   s) mercury(II) oxide

3.  a) cadmium phosphide  b) sulphurous acid   c) ammonium sulfide    d) manganese(III) fluoride e) aluminum oxide        f) hydrocyanic acid    g) oxygen difluoride     h) chromium(III) nitride i) cobalt(III) sulfide       j) diphosphorous trioxide    k) carbon disulfide       l) cadmium oxide m) arsenic trihydride    n) iodine trifluoride    o) iron(III) phosphate    p) phosphoric acid q) titanium(IV) oxide     r) hypochlorous acid    s) hydroiodic acid

4.  a) Al₂S₃         b) NH₄I        c) NiI₂         d) Li₃P    e) AuP          f) H₃PO₄      g) Li₂O         h) Ni₃(PO₄)₂ i) HBrO₃        j) XeO₂         k) BrF         l) Al(HSO₄)₃    m) CoCO₃     n) K₃PO₄     o) LiNO₂

5.  a) AuCl₃        b) Fe₂O₃      c) HClO₃      d) MnBr₂      e) Co(OH)₂    f) AsO₂        g) Cu(OH)₂   h) H₂SO₃ i) CaI₂           j) CdO         k) NI₃          l) SO     m) Fe(CrO₄)₃  n) BH₃         o) Mn₂O₃      p) SF₆ q) SrClO        r) Co₃(PO₄)₂  s) Mg(HCO₃)₂

6.  a) HClO₃         b) XeF₆        c) Fe(NO₃)₃   d) CH₃COOH  e) Cr₂S₃       f) LiHCO₃ g) Al₂O₃          h) HF          i) Cu(MnO₄)₂

7.  a) phosphorus trifluoride  b) carbon monoxide   c) diselenium dibromide

d) sulfur tetrafluoride       e) diphosphorus pentasulfide         f) dinitrogen trioxide        

g) phosphorous pentachloride        h) dioxin difluoride          i) carbon dioxide 

j) selenium hexafluoride              k) nitrogen monoxide

8. a) copper(II) sulfate      b) iron(III) bromide       c) cesium perchlorate        d) ammonium dichromate e) potassium hydrogen sulfate        f) chromium(II) fluoride 9.   a)  First, identify the anion and cation and their charges. Manganese is Mn, and we are told from the name (type II) that it has a charge of +4. Oxide is O^2-.  Now, do the “cross” method of the charges, and reduce: manganese(IV) oxide = MnO₂

b)  Magnesium is Mg²⁺(type I), and perchlorate is ClO₄^-;  magnesium perchlorate = Mg(ClO₄)₂

c)  Antimony(III) is Sb with a charge of +3, according to the type II name. Nitrate is NO₃^-

      antimony(III) nitrate = Sb(NO₃)₃

d)  Sodium is Na⁺and iodide is I^-   ;  sodium iodide = NaI

10.  a) KMnO₄    b) FeSO₄    c) Na₃PO₄     d) Cu₃N₂    e) Al₂S₃    f) Cr₂O₃ 11.a)  Dinitrogen means 2 nitrogen atoms and pentoxide is five oxygen atoms = N₂O₅

b)  Hydro means that it must be a general acid. “Chlor” is the root for Cl, which has a charge of –1. Thus H⁺and Cl^-together make HCl (note that we have to treat this as an ionic compound, the charges must balance).

c)  Be careful!  Hypo does NOT mean it is a general acid. It is simply the prefix for the polyatomic oxyanion. Because the ending is “ous”, we know that the ion ending must be “ite”. Thus, this originates from the hypochlorite ion = ClO^-. Together with H⁺this makes HClO.

d)  “Nitric” tells us that this originates from the nitrate ion = NO₃^-. With H⁺, this makes HNO₃

12. a) BF₃;  b) HBrO₃;  c) H₂SO₃]]> 3475 2141 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-2-mass-terminology/ Fri, 11 May 2018 18:51:05 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3495

Learning Objectives

By the end of this section, you will be able to:

• Calculate formula masses for covalent and ionic compounds

We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.

Formula Mass

In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance’s formula.

Formula Mass for Covalent Substances

For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl₃), a covalent compound once used as a surgical anesthetic and now primarily used in the production of the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 1 outlines the calculations used to derive the molecular mass of chloroform, which is 119.377 amu.

1 C mass	= 12.011 amu
1 H masses	= 1.00794 amu
3 Cl masses = 3 x 35.4527 amu	= 106.3581
Total	= 119.377 amu = the molecular mass of CHCl3

Likewise, the molecular mass of an aspirin molecule, C₉H₈O₄, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure 2).

9 C mass = 9 x 12.011 amu	= 108.099 amu
8 H masses = 8 x 1.00794 amu	= 8.06352 amu
4 O masses = 4 x 15.9994 amu	= 63.9976 amu
Total	= 180.160 amu = the molecular mass of C9H8O4

Example 1

Ibuprofen, C₁₃H₁₈O₂, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass for this compound?

 

Solution Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:

13 C mass = 13 x 12.011 amu	= 156.143 amu
18 H masses = 18 x 1.00794 amu	= 18.14292 amu
2 O masses = 2 x 15.9994 amu	= 31.9988 amu
Total	= 206.285 amu = the molecular mass of C13H18O2

 

Test Yourself Acetaminophen, C₈H₉NO₂, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass for this compound?

  Answer 151.16 amu

Example 2

What is the molecular mass of each substance?

a) NBr₃       b) C₂H₆

 

Solution

a) Add one atomic mass of nitrogen and three atomic masses of bromine:

1 N mass	= 14.0067 amu
3 Br masses = 3 × 79.904 amu	= 239.712 amu
Total	= 253.719 amu = the molecular mass of NBr3

  b) Add two atomic masses of carbon and six atomic masses of hydrogen:

2 C masses = 2 × 12.011 amu	= 24.022 amu
6 H masses = 6 × 1.00794 amu	= 6.04764 amu
Total	= 30.070 amu = the molecular mass of C2H6

The compound C₂H₆ also has a common name—ethane.

 

Test Yourself

What is the molecular mass of each substance?

a) SO₂         b) PF₃

 

Answers

a) 64.065 amu         b) 87.969 amu

Formula Mass for Ionic Compounds

Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.”

As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na⁺, and chloride anions, Cl⁻, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (Figure 3).

1 Na mass	= 22.9898 amu
1 Cl masses	= 35.4527 amu
Total	= 58.4425 amu = the molecular mass of NaCl

Figure 3. Table salt, NaCl, contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu.

Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.

Example 3

Aluminum sulfate, Al₂(SO₄)₃, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?

  Solution The formula for this compound indicates it contains Al³⁺ and SO₄²⁻ ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al₂S₃O₁₂. Following the approach outlined above, the formula mass for this compound is calculated as follows:

2 Al mass = 2 x 26.9815 amu	= 53.9630 amu
3 S masses = 3 x 32.066 amu	= 96.198 amu
12 O masses = 12 x 15.9994 amu	= 191.9928 amu
Total	= 342.154 amu = the molecular mass of Al2S3O12

 

Test Yourself Calcium phosphate, Ca₃(PO₄)₂, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?

  Answer 310.18 amu

Example 4

What is the molecular mass of Fe(NO₃)₃?

 

Solution  

There is 1 atom of Fe, 3 atoms of N and 3(3) = 9 atoms of O

Thus, the molecular mass:

1(55.847amu) + 3(14.0067amu) + 9(15.9994amu) = 241.862 amu

 

Test Yourself

What is the mass, in amu’s, of 5.292 x 10²¹ molecules of Ni(NO₃)₂?    Answer 9.669 x 10²³ amu

Chemistry Is Everywhere: Sulfur Hexafluoride

On March 20, 1995, the Japanese terrorist group Aum Shinrikyo (Sanskrit for “Supreme Truth”) released some sarin gas in the Tokyo subway system; twelve people were killed, and thousands were injured. Sarin (molecular formula C₄H₁₀FPO₂) is a nerve toxin that was first synthesized in 1938 (Figure 4). It is regarded as one of the most deadly toxins known, estimated to be about 500 times more potent than cyanide. Scientists and engineers who study the spread of chemical weapons such as sarin (yes, there are such scientists) would like to have a less dangerous chemical, indeed one that is nontoxic, so they are not at risk themselves.

[caption id="attachment_4854" align="aligncenter" width="499"] Figure 4.  The nerve toxin Sarin (molecular formula C₄H₁₀FPO₂).[/caption]

Sulfur hexafluoride is used as a model compound for sarin. SF₆ (Figure 5) has a similar molecular mass (about 146 amu) as sarin (about 140 amu), so it has similar physical properties in the vapour phase. Sulfur hexafluoride is also very easy to accurately detect, even at low levels, and it is not a normal part of the atmosphere, so there is little potential for contamination from natural sources. Consequently, SF₆ is also used as an aerial tracer for ventilation systems in buildings. It is nontoxic and very chemically inert, so workers do not have to take special precautions other than watching for asphyxiation.

[caption id="attachment_4855" align="aligncenter" width="251"] Figure 5. Sulfur hexafluoride[/caption]

Sulfur hexafluoride also has another interesting use: a spark suppressant in high-voltage electrical equipment. High-pressure SF₆ gas is used in place of older oils that may have contaminants that are environmentally unfriendly (part (c) in the accompanying figure).

Key Concepts and Summary

Isotopic mass (the mass of an isotope of an element, expressed in amu) of naturally occurring isotopes of a given element, is used to calculate the atomic mass of that element (expressed in amu).  The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass.

Exercises

1. a) What is the atomic mass of an oxygen atom?

    b) What is the molecular mass of oxygen in its elemental form (meaning in the form it naturally occurs in)?

2. a) What is the atomic mass of bromine?

    b) What is the molecular mass of bromine in its elemental form?

3. Determine the molecular mass of each substance.

a)  F₂          b)  CO         c)  CO₂

4.  Determine the mass of each substance.

a)  Na         b)  B₂O₃         c)  S₂Cl₂

5. Determine the formula mass of each substance.

a)  GeO₂         b)  IF₃         c)  XeF₆ 6. What is the total mass (amu) of carbon in each of the following molecules?

a) CH₄         b) CHCl₃         c) C₁₂H₁₀O₆         d) CH₃CH₂CH₂CH₂CH₃

7. Calculate the molecular or formula mass of each of the following:

a) P₄         b) H₂O         c) Ca(NO₃)₂         d) CH₃CO₂H (acetic acid)

e) C₁₂H₂₂O₁₁ (sucrose, cane sugar) 8. Determine the molecular mass of the following compounds:

a)

b)

c)

d)

9. Which molecule has a molecular mass of 28.05 amu?

a)

b)

c)

Answers 1. a)  15.999 amu         b)  The elemental for of oxygen is O₂.  Its moleculass mass is 31.998 amu. 2.  a) 79.904 amu         b) 159.808 amu 3. a)  37.997 amu         b)  28.010 amu         c)  44.010 amu 4. a)  22.9898 amu         b)  69.620 amu         c)  135.037 amu 5. a)  104.59 amu         b)  183.900 amu         c)  245.280 amu 6. a) 12.011 amu            b) 12.011 amu         c) 144.132 amu         d) 60.055 amu

7. a) 123.895 amu         b) 18.015 amu         c) 164.088 amu         d) 60.052 amu       e) 342.300 amu

8. a) 56.108 amu         b) 54.092 amu         c) 199.9977 amu         d) 97.9952 amu

9. B

Glossary

formula mass: sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass

]]> 3495 1354 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-4-percent-composition-langara/ Mon, 14 May 2018 20:03:39 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3510

Learning Objectives

By the end of this section, you will be able to:

• Compute the percent composition of a compound from experimental mass measurements
• Compute the percent composition of a compound from its chemical formula

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will begin exploring how a chemist may go about determining the identity of a compound from experimental mass measurements and calculating percent composition.

Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

$latex \% \;\text{H} = \frac{\text{mass H}}{\text{mass compound}} \times 100 \% $

$latex \% \;\text{C} = \frac{\text{mass C}}{\text{mass compound}} \times 100 \% $

If analysis of a 10.0-g sample of this gas showed that it contains 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

$latex \%\;\text{H} = \frac{2.5 \;\text{g H}}{10.0 \;\text{g compound}} \times 100 \% = 25 \% $

$latex \%\;\text{C} = \frac{7.5 \;\text{g C}}{10.0 \;\text{g compound}} \times 100 \% = 75 \% $

Example 1

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed that it contains 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

 

Solution To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

$latex \%\;\text{C} = \frac{7.34 \;\text{g C}}{12.04 \;\text{g compound}} \times 100\% = 61.0\% $ $latex \%\;\text{H} = \frac{1.85 \;\text{g H}}{12.04 \;\text{g compound}} \times 100\% = 15.4\% $ $latex \%\;\text{N} = \frac{2.85 \;\text{g N}}{12.04 \;\text{g compound}} \times 100\% = 23.7\% $

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

 

Test Yourself A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

  Answer 12.1% C, 16.1% O, 71.8% Cl

Determining Percent Composition from The Chemical Formula

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH₃), ammonium nitrate (NH₄NO₃), and urea (CH₄N₂O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH₃ contains one N atom weighing 14.0067 amu and three H atoms weighing a total of (3 × 1.00794 amu) = 3.02382 amu. The formula mass of ammonia is therefore (14.0067 amu + 3.02382 amu) = 17.0305 amu, and its percent composition is:

$latex \%\;\text{N} = \frac{14.0067 \;\text{amu N}}{17.0305 \;\text{amu NH}_3} \times 100\% = 82.2448\% $ $latex \%\;\text{H} = \frac{3.02382 \;\text{amu N}}{17.0305 \;\text{amu NH}_3} \times 100\% = 17.7553\% $

This same approach may be taken for a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated in Example 2. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.

Example 2

Aspirin is a compound with the molecular formula C₉H₈O₄. What is its percent composition?

 

Solution To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C₉H₈O₄. It is convenient to consider 1 mol of C₉H₈O₄ and use its molar mass (180.160 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

$latex \begin{array}{r @{{}={}} l} \%\text{C} & \frac{9 \;\text{mol C} \;\times\; \text{molar mass C}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{9 \times 12.011 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{108.099 \;\text{g/mol}}{180.160 \;\text{g/mol}} \times 100 \\[1em] \%\text{C} & 60.002\%\;\text{C} \end{array}$

$latex \begin{array}{r @{{}={}} l} \%\text{H} & \frac{8 \;\text{mol H} \;\times\; \text{molar mass H}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{8 \times 1.00794 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{8.06352 \;\text{g/mol}}{180.160 \;\text{g/mol}} \times 100 \\[1em] \%\text{H} & 4.47575\%\;\text{H} \end{array}$

$latex \begin{array}{r @{{}={}} l} \%\text{O} & \frac{4 \;\text{mol O} \;\times\; \text{molar mass O}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{4 \times 15.994 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{63.9976 \;\text{g/mol}}{180.160 \;\text{g/mol}} \times 100 \\[1em] \%\text{O} & 35.5226\%\;\text{O} \end{array}$

Note that these percentages sum to equal 100.00% when appropriately rounded.

 

Test Yourself To three significant digits, what is the mass percentage of iron in the compound Fe₂O₃?

  Answer  69.9% Fe

Example 3

What is the percent by mass of H in 5.00g of H₂O

 

Solution  

We are given the total mass of the compound (5.00g). To solve for percent composition, we need to determine the mass of H in the sample. Remember that ONLY the chemical formula gives us the ability to go from H₂O to H, and it only refers to particles or moles. Thus, our pathway:

      grams H₂O $latex \longrightarrow$ mol H₂O $latex \longrightarrow$ mol H $latex \longrightarrow$ grams H

Then we will apply the percent composition formula.

    $latex 5.00\; \rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g of H}_2\text{O}\; \times \frac{1\; \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}_2\text{O}} {18.0153\; \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g H}_2\text{O}} \times \frac{2\; \rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}_2\text{O}} \times \frac{1.01\; \text{g H}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}_2\text{O}}$ $latex = 0.561\; \text{g of H}$

  $latex \%\;\text{H} = \frac{0.561 \;\text{g H}}{5.00 \;\text{g water}} \times 100 \% = 11.2 \% $  

Test Yourself

What is the percent by mass of O in 10.0 g of Cu(NO₃)₂?

  Answer 51.18%

Example 4

Determine the percent composition of H in propanol (C₃H₇OH)

 

Solution  

Considering one mol of C₃H₇OH, % of H in C₃H₇OH  =

$latex \%\;\text{H} = \frac{\text{mass of 8 mol H}}{\text{mass of 1 mol propanol}} \times 100 \% =$ $latex \frac{8.06352\; \text{g H}}{60.096\; \text{g propanol}} \times 100 \% = 13.418 \% $  

Test Yourself

What is the percent composition of each element in H₂SO₄?

  Answer H = 2.06%, S = 32.69%, O = 65.25%

Example 5

Hemoglobin contains 0.33% Fe by mass. The molar mass of hemoglobin is 6.8 x 10⁴g/mol. How many grams of Fe are in 0.020 mol of hemoglobin?

 

Solution  

The percent composition essentially refers to the mass of element in 100 g of compound. Thus 0.33% = 0.33g of Fe in 100 g of hemoglobin. We can use this as a conversion factor (but don’t forget that it’s based on 100 g of compound!)

Mol hemoglobin $latex \longrightarrow$ mass hemoglobin $latex \longrightarrow$ mass of Fe

$latex 0.020\; \rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol of hemoglobin}\; \times \frac{6.8 \times 10^4\; \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g hemoglobin}} {1\; \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol hemoglobin}} \times \frac{0.33\; \text{g Fe}}{100\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g hemoglobin}} $ $latex = 4.5\; \text{g of Fe}$

 

Test Yourself

How many grams of Ba₃P₂ would contain 4.23 g of Ba, if Ba₃P₂ contains 86.9% Ba by mass?   Answer 4.87 g Ba₃P₂

Key Concepts and Summary

The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound.

Key Equations

• $latex \%\text{X} = \frac{\text{mass X}}{\text{mass commpound}} \times 100\% $

Exercises

1. Calculate the following to four significant figures:

a) the percent composition of ammonia, NH₃

b) the percent composition of photographic “hypo,” Na₂S₂O₃

c) the percent of calcium ion in Ca₃(PO₄)₂

2. Determine the percent ammonia, NH₃, in Co(NH₃)₆Cl₃, to three significant figures.   Answers

1. a) % N = 82.24%, % H = 17.76%

b) % Na = 29.08%, % S = 40.56%, % O = 30.36% c) % Ca²⁺ = 38.76%

2. % NH₃ = 38.2%

Glossary

percent composition: percentage by mass of the various elements in a compound

]]> 3510 1354 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/5-6-end-of-chapter-problems/ Mon, 14 May 2018 20:04:35 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3512 72atoms of Fe? 3. a) What is the mass (in grams) of 9.01 mol of Ca? b) How many mol are present in 21.2 g of Zn? 4. a) How many mol are present in 10.421 g of Fe? b) What is the mass (in kilograms) of 905.25 mol of lead? 5. a) How many atoms are present in a 25.0 g sample of Na? b) What is the mass (in grams) of 5.00 x 10²⁵atoms of Cr? 6. How many atoms are present in 21.2 mg of Ag? 7. How many atoms are in a piece of gold measuring 1 cm x 2 cm x 0.5 cm? (Density of gold is 19.32 g/mL) 8. The density of mercury (Hg) is 13.53 g/mL. How many litres will 4.2 x 10²¹atoms of Hg occupy? 9. What is the molar mass of Cu(SO₄)₂? 10. What is the molar mass of:  a)  H₂O          b) Ca₃(PO₄)₂          c) C₂H₅OH 11. How many mol of O are present in 4.32 x 10³² molecules of Cu(SO₄)₂? 12. How many atoms of H are present in 4.2 mol of H₂O? 13. How many grams of S are present in 32.1 grams of Cu(SO₄)₂? 14. How many grams of Ni are present in 50.0 g of NiNO₃? 15. Complete the following conversions: 9.024 x 10²³ Hg atoms = ____ g Hg 96.35 g NO = ____ molecules NO 253.52 g CO₂= ____ TOTAL atoms 521.2 g (NH₄)₂Cr₂O₇  =  _____ g H 1.371 g C₂H₅OH = ____ atoms of H 16. The hemoglobin content of blood is about 15.5 g/100 mL of blood. The molar mass of hemoglobin is about 64,500 g/mol and there are 4 iron atoms in a hemoglobin molecule. Approximately how many Fe atoms are present in 6 L of blood in a typical adult? 17. List the conversion factors, or combination of conversion factors that you would use in the following problem situations (no calculations required): a) You need to go from g H₂O to mol H₂O b) You need to go from mol H₂SO₄ to molecules H₂SO₄ c)  From g HCl to molecules HCl d)  From mol KMnO₄ to atoms O e)  From g KCN to g C f)  From mL CH₃OH to mol CH₃OH g)  From molecules NaCl to atoms Cl h)  From kg KBr to number of Br-ions i)  From mol H₂SO₄ to mL of H₂SO₄ j)  From mol MgO to g O k)  From mm³ NaOH to mol NaOH l)  From molecules HCl to g H m)  From mol C in C₆H₆ to g C₆H₆ 18. For all the situations in #17, calculate the final answer based on a starting unit of 5. E.g.:  for (a)  How many mol H₂O are in 5 g of H₂O? for (b)  How many molecules of H₂SO₄ are in 5 mol H₂SO₄?   etc… NOTE:  for f, i and k, the following densities are needed: CH₃OH  d = 0.791 g/mL           H₂SO₄   d = 1.45 g/mL        NaOH   d = 1.22 g/mL 19. Calculate the following: a) Total number of ions in 38.1 g of CaF₂ (hint… how many ions TOTAL are in one molecule of CaF₂?) b) Mass in g of 2.04 x 10²¹ molecules of N₂O₅ c)  Mass in mg of 3.58 mol CuCl₂ d)  Mass in g of 9.64 x 10²⁴ molecules of Cl₂O₇ 20. Oxygen is required for metabolic combustion of foods. Calculate the number of atoms in 38.0 g of oxygen gas, the amount absorbed from the lungs at rest in about 15 minutes. 21. An imperial quart of oil is spilled on a lake. If the molecules were to spread out in an film one molecule deep, what would be the area of the oil slick, in square miles? The density of oil = 0.8g/mL ; area covered by one molecule = 0.5 nm². Although oil is really a mixture of compounds of C & H, assume the oil is simply C₁₆H₃₄ and 1 qt = 1.1L and 1 mile = 1.6 km. 22. Why would it be necessary to clarify what you mean when you say “1 mole of nitrogen” or “1 mole of hydrogen”? Why is this clarification not necessary when referring to “1 mole of lead” or “1 mole of water”? 23. How many magnesium ions and how many nitride ions are in 4.75 mol of Mg₃N₂? 24. What mass of O₂ contains the same number of moles of molecules as 52.0 g of N₂? 25. What mass of sodium hydroxide (NaOH) contains the same number of moles as 126 g of nitric acid (HNO₃)? 26. Ringer’s lactate is an aqueous (in water) physiological solution used for intraveneous fluid therapy. A 1.00 L sample of the solution contains 5.96 g of NaCl, 3.1 g of NaC₃H₅O₃(sodium lactate), 0.3 g of KCl and 0.2 g CaCl₂. a) How many moles of each compound are in 1.00 L of solution? b) The ingredients dissolve to form Na⁺, K⁺, Ca²⁺, Cl^- and C₃H₅O₃^- ions.  How many moles of each ion are in 1.00 L of solution.  Note: there are several sources of Na⁺and Cl^-. 27. The density of liquid benzene, C₆H₆ is 0.879 g/mL at 15^oC. What is the volume in milliliters of 1.00 mol of benzene at this temperature? 28. A sample of 0.370 mol of a metal oxide (M₂O₃) weighs 55.4 g. a) How many moles of O are in the sample? b) How many grams of M are in the sample? c)  What element is represented by M? 29. a) How many grams of H₂S are there in 0.400 moles of H₂S? b) How many grams of hydrogen and sulfur are contained in 0.400 moles of H₂S? c)  How many molecules of H₂S are contained in 0.400 moles of H₂S? d)  How many atoms of H and S are contained in 0.400 moles of H₂S? 30. a) How many moles of SO₂ are represented by 9.54 g of SO₂? b) How many atoms of oxygen does this represent? 31. What is the average mass (in grams) of one oxygen atom? 32. What is the mass in grams of one molecule of CH₃OH? 33. The density of gold is approx. 2 x 10¹ g/cm³. What is the volume of one gold atom? 34. How many moles are there in one atom? 35. What is the mass percent of each atom in (NH₄)₂Cr₂O₇? 36. Calculate the following mass percents: a) Mass % of H in NH₄HCO₃       b) Mass % of Mn in KMnO₄ 37. Calculate the following mass fractions: a) Mass fraction of Cl in CaCl₂     b) Mass fraction of P in P₄O₇ 38. The effectiveness of nitrogen fertilizer is determined mainly by its mass % N. Rank the following in terms of their effectiveness: KNO₃; NH₄NO₃; Co(NH₂)₂; NH₄SO₄. Show your calculations. 39. A compound of iodine and cesium contains 63.94 g of metal and 61.06 g of nonmetal. How many grams of cesium are in 38.77 g of the compound? How many grams of iodine? 40. What is the empirical formula and the empirical formula molar mass of each of the following:  a) C₂H₄   b) C₂H₆O₂   c) N₂O₅    d) Ba₃(PO₄)₂     e) Te₄I₁₆ 41. What is the molecular formula of each of the following: a) Empirical formula = CH₂; molecular molar mass = 42.08 g/mol b)  EF = NH₂; MMM = 32.05 g/mol c)  EF = NO₂; MMM = 92.02 g/mol d)  EF = CHN; MMM = 135.14 g/mol e)  EF = CH;  MMM = 78.11 g/mol f)  EF = C₃H₆O₂; MMM = 74.08 g/mol g)  EF = C₇H₄O₂; MMM = 240.02 g/mol 42. A 3.450 g sample of nitrogen reacts with 1.970 g of oxygen to form a compound. Determine the empirical formula for this compound. 43. A compound containing carbon, hydrogen and oxygen is found to be 40.00% carbon and 6.700% hydrogen by mass. The molar mass of the compound is between 115 g/mol and 125 g/mol. Determine the empirical formula and molecular formula of this compound. 44. Determine the empirical formula for the following situations: a) 0.0630 mol of chlorine atom combined with 0.220 mol of oxygen atoms b) 2.45 g silicon combined with 12.4 g of chlorine c)  a compound with 27.3% carbon and 72.7% oxygen by mass d)  a hydrocarbon (containing only C and H) which has 79.9% by mass C 45. A compound containing only silicon and chlorine contains 79.1% chlorine by mass and has a molar mass of 269 g/mol. What is the molecular formula? 46. A sample of 0.600 mole of metal reacts completely with fluorine to form 46.8 g of MF₂. a) How moles of F are in the sample of MF₂ that formed? b) How many grams of M are in this sample of MF₂? c)  What element is represented by M? 47. A sample of nicotine contains 6.16 mmol of C, 8.56 mmol of H and 1.23 mmol of N. What is its empirical formula? (note: 1000 mmol = 1 mol) 48. Cortisol (molar mass = 362.47 g/mol), one of the major steroid hormones, is a key factor in the synthesis of protein. Its profound effect on the reduction of inflammation explains its use in the treatment of rheumatoid arthritis. Cortisol is 69.6% C, 8.34% H and 22.1% O by mass. What is its molecular formula? 49. 3.00g of molybdenum (Mo) combines with sulfur to produce 5.50 g of a compound. What is the empirical formula of this compound? 50. A 5.70 g sample of an iron oxide compound was heated in a stream of hydrogen gas, which reacted with the oxygen in the compound to form water. The water vapor was carried away in the stream of gas, leaving only pure metallic iron weighing 4.20g. Calculate the simplest possible formula for the original iron oxide. 51. When 2.31g of a carboxylic acid compound is burned in O₂, the only products are 1.33g of H₂O and 3.38g of CO₂. Calculate the empirical (simplest possible) formula for this compound. 52. A compound was analyzed and found to have the following percentage composition: aluminum = 15.7%, sulfur = 28.11%, oxygen = 56.12%. The molar mass of the compound is known to be approximately 684 g/mol. What is the molecular formula?  

Answers

1. 2.55 x 10²⁴ gidgets 2. 8.6 x 10⁴⁸ mol of Fe 3. a) 361 g of Ca         b) 0.324 mol of Zn 4.  a) 0.18660 mol of Fe      b) 187.57 kg of Pb 5. a) 6.55 x 10²³ atoms of Na    b) 4.32 x 10³ g of Cr 6. 1.18 x 10²⁰ atoms 7. 6 x 10²² Au atoms 8. 1.0 x 10^-4 L 9. 255.673 g/mol of Cu(SO₄)₂ 10. a) 18.0152 g/mol     b) 310.177 g/mol     c)  46.069 g/mol 11. 5.74 x 10⁹ mol of O 12. 5.1 x 10²⁴ atoms of H 13. 8.05 g of S 14. 24.3 g of Ni

15.  a) 300.6 g Hg b) 1.934 x 10²⁴ molecules of NO c) 1.0407 x 10²⁵ atoms d) 16.67 g H e) 1.075 x 10²³ atoms

16.  3 x 10²² iron atoms

17.   a) g H₂O $latex \longrightarrow$ mol H₂O using molar mass of H₂O

b) mol H₂SO₄ $latex \longrightarrow$ molecules H₂SO₄ using Avogadro’s number

c) g HCl $latex \longrightarrow$ mol HCl using molar mass HCl $latex \longrightarrow$ molecules HCl using Avogadro’s number

d) mol KMnO₄ $latex \longrightarrow$ mol O using chemical formula $latex \longrightarrow$ atoms O using Avogadro’s number

e) g KCN $latex \longrightarrow$ mol KCN using molar mass of KCN $latex \longrightarrow$ mol C using chemical formula $latex \longrightarrow$ g C using molar mass of C

f) mL CH₃OH $latex \longrightarrow$ g CH₃OH using density $latex \longrightarrow$ mol CH₃OH using molar mass of CH₃OH

g) molecules NaCl $latex \longrightarrow$ atoms Cl using formula ratio  OR molecules NaCl $latex \longrightarrow$ moles NaCl using Avogadro’s number $latex \longrightarrow$ moles Cl using the chemical formula $latex \longrightarrow$ atoms Cl using Avogadro’s number

h) kg KBr $latex \longrightarrow$ g KBr using metric conversions $latex \longrightarrow$ mol KBr using the molar mass of KBr $latex \longrightarrow$ moles Br^- ions using chemical formula $latex \longrightarrow$ number of Br^- ions using Avogadro’s number

i) mol H₂SO₄ $latex \longrightarrow$ g H₂SO₄ using molar mass of H₂SO₄ $latex \longrightarrow$ mL H₂SO₄ using density of H₂SO₄

j) mol MgO $latex \longrightarrow$ mol O using the chemical formula $latex \longrightarrow$ g O using molar mass of O

k) mm³ NaOH $latex \longrightarrow$ cm³ NaOH using metric conversions $latex \longrightarrow$ g NaOH using the density $latex \longrightarrow$ mol NaOH using the molar mass of NaOH

l) molecules HCl $latex \longrightarrow$ moles HCl using Avogadro’s number $latex \longrightarrow$ moles H using the chemical formula $latex \longrightarrow$ g H using the molar mass of H

m) mol C $latex \longrightarrow$ mol C₆H₆ using the chemical formula $latex \longrightarrow$ g C₆H₆ using the molar mass of C₆H₆

18.   a) 0.3 mol                b) 3 x 10²⁴ molecules        c) 8 x 10²² molecules d) 1 x 10²⁵ atoms          e) 0.9 g                                f) 0.1 mol g) 5 ions (or 5 atoms)    h) 3 x 10²⁵ ions                   i) 3 x 10² mL j) 8 x 10¹ g                     k) 2 x 10^-4 mol                    l) 8 x 10^-24 g m) 7 x 10¹ g

19.   a) 8.82 x 10²³ ions         b) 0.366 g        c) 4.81 x 10⁵ mg d)  2.93 x 10³ g

20.   1.43 x 10²⁴ atoms of O (7.15 x 10²³ molecules O₂)

21.  0.5 mile²

22.  Because 1 mole of nitrogen may be 1 mole of N or 1 mole of N₂, etc. Nitrogen and hydrogen (but not lead and water) are commonly found as diatomic molecules.

23   8.58 x 10²⁴Mg²⁺ions and 5.72 x 10²⁴N^3-ions

24.  59.4 g of O₂ (note: the nitrogen is given as N₂)

25.  80.0 g of NaOH

26.  a) 0.102 mol NaCl, 0.028 mol NaC₃H₅O₃, 0.004mol KCl     and 0.002 mol CaCl₂ b) Na⁺ = 0.130 mol,  K⁺ = 0.004 mol, Ca²⁺ = 0.002,     Cl^- = 0.110 mol; C₃H₅O₃^- = 0.028 mol

27.  88.9 mL

28.  a) 1.11 mol O    b) 37.6 g M    c) V

29.  a) 13.6 g H₂S                          b) 0.806 g H, 12.8 g S c) 2.41 x 10²³ molecules H₂S        d)  4.82 x 10²³H atoms and 2.41 x 10²³ S atoms

30.  a) 0.149 mol SO₂       b) 1.79 x 10²³atoms O

31. 2.65682 x 10^-23 g

32.  5.3208 x 10^-23 g

33.  2 x 10^-23 cm³

34.  1.6605388 x 10^-24 mol

35.  N= 11.114%, H= 3.1990%, Cr= 41.256%, O= 44.431%

36.  a) 6.3749%     b) 34.7634%

37.  a)  0.638883    b) 0.525222

38. NH₄NO₃> Co(NH₂)₂> KNO₃ > NH₄SO₄

39. 19.83 g Cs, 18.94 g I

40.  a) C₂H₄= CH₂; 14.027 g/mol b) C₂H₆O₂= CH₃O; 31.034 g/mol c) N₂O₅= N₂O₅; 108.0104 g/mol d) Ba₃(PO₄)₂= Ba₃(PO₄)₂; 601.924 g/mol e) Te₄I₁₆= TeI₄; 635.22 g/mol

41.  a) C₃H₆             b) N₂H₄             c) N₂O₄ d) C₅H₅N₅                e) C₆H₆              f) C₃H₆O₂ g) C₁₄H₈O₄

42. N₂O

43.  EF = CH₂O, MF = C₄H₈O₄

44.  a) Cl₂O₇            b) SiCl₄            c) CO₂            d) CH₃

45.  Si₂Cl₆

46.  a) 1.20 mol F     b) 24.0 g M     c) Ca

47.  C₅H₇N

48.  C₂₁H₃₀O₅

49.  Mo₂S₅ 

50.  Fe₄O₅(Fe₈O₁₀etc. are also possible, but not the simplest)

51.  CH₂O   (Note that, based on the products, the unknown compound could only contain C, H and O, and note that all of the C in the original compound ends up in the CO₂and all the H ends up in the H₂O.)

52.  Al₄S₆O₂₄

   ]]> 3512 1354 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/development-of-quantum-theory/ Fri, 18 May 2018 19:12:56 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3668

Learning Objectives

By the end of this module, you will be able to:

• Extend the concept of wave–particle duality that was observed in electromagnetic radiation to matter as well
• Understand the general idea of the quantum mechanical description of electrons in an atom, and that it uses the notion of three-dimensional wave functions, or orbitals, that define the distribution of probability to find an electron in a particular part of space

The electron shell model gives us a satisfactory picture of the electrons in atom because it accounts for the quantization of the electron energy and it does not constrain the electrons to a fixed orbit.  It also allows us to begin to explain why elements behave the way they do. However many fundamental aspects of the atom are not addressed by this simple model. How exactly do the electrons move? What controls the maximum number of electrons in the shell? And, there are also a number of chemical and physical properties of the elements that remain unexplained by the electron shell model. A more advanced model is required to approach these issues.

The Wave Mechanics Model of the Atom

Bohr’s ideas were useful but were applied only to the hydrogen atom. However, later researchers generalized Bohr’s ideas into a new theory called quantum mechanics, which explains the behaviour of electrons as if they were acting as a wave, not as particles. Quantum mechanics predicts two major things: quantized energies for electrons of all atoms (not just hydrogen) and an organization of electrons within atoms. Electrons are no longer thought of as being randomly distributed around a nucleus (from Electron Shell Model) or restricted to certain orbits (from Bohr's Model). Instead, electrons are collected into groups and subgroups that explain much about the chemical behaviour of the atom.

Wave/Particle Duality and the Uncertainty Principle

We’ve seen that electromagnetic radiation (a form of energy) can be viewed as a wave, or as a particle (a photon, or “packet”). That is, light has a dual nature. If energy can behave like a particle and a wave, can matter also exhibit this duality? It turns out that it does! In the 1920’s Louis Victor de Broglie showed that all moving particles can be described as wave-like. We call this general phenomenon wave/particle duality. While it may seem odd to think of a physical object, such as a baseball, behaving like a wave, that is just what the theory suggests. However, for any object that is large, the particle nature is dominant and the wave nature is insignificant. But for tiny objects, such as electrons, the wave-like nature is physically meaningful and can be observed. Around the same time, Werner Heisenberg proposed that the more accurately you know the position of a particle, the less you know about its momentum, and vice versa. This is now called the Heisenberg Uncertainty Principle. This means we can no longer talk about precise trajectories or locations of electrons. Instead, there is always some uncertainty and we can only talk about the PROBABILITY of finding an electron in a certain time and place.

The Schrödinger Wave Equation and the Wave Mechanics Model

Together, the quantization of energy, wave / particle duality and the Uncertainty Principle all led to Wave Mechanics, developed by Erwin Schrödinger in the 1920’s. Schrödinger presented his Wave Mechanics in the form of a very complicated mathematical equation which, when solved gives us a series of WAVE FUNCTIONS, each having a specific energy. A wave function is a mathematical function (like f(x) = x²), and we can plot a graph of the function (Figure 1) to get a picture of the ORBITAL—the space in which the electron is most likely to be found. Depending on the shape of the space, the orbitals have been assigned a name such as: s, p, d, and f. Note that an orbital is much different than an orbit!  An orbit is a set path on which an electron travels. We can’t talk about a set path because of the uncertaintly principle. An orbital is more like a probability map: the region in space where the electron is most likely to be over time.  The differing intensity of the shaded areas is an indication of electron density, which is a measure of the probability of locating an electron in a particular region of space (Figure 1 and 2). [caption id="" align="aligncenter" width="1300"] Figure 1. This illustrates graphs showing the probability (y axis) of finding an electron for the 1s, 2s, 3s orbitals as a function of distance from the nucleus and how it relates to the space around the nucleus.[/caption]

  [caption id="attachment_3660" align="aligncenter" width="573"] Figure 2.  Electron probability maps of lower and higher energy orbitals for the electron in a H atom.[/caption] Looking at the probability maps of the electron in hydrogen (Figure 2), we can perhaps best picture an electron as a “cloud”. To simplify the picture of an orbital, however, we generally limit or define the orbital to be the zone within which the electron has 90% probability of being contained, and we represent this space with a solid. Figure 3 shows the many different complex shapes that the electron movement traces out. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found.  Keep in mind the “cloud” image when viewing these representations however. [caption id="" align="aligncenter" width="1300"] Figure 3. Shapes of s, p, d, and f orbitals.[/caption]

Describing atoms using the Wave Mechanics Model

Properties of electrons in orbitals

Orbitals are labeled with a number (known as the “shell”) followed by a letter (s, p, d, or f, known as the “subshell”), for example 1s, 2s, 2p, 3s. The subshells occur with a certain multiplicity; there are always seven f orbitals, five d orbitals, three p orbitals, and one s orbital (these could be called the “sub-subshells”). Any atom has all of these orbitals though they may or may not be occupied by electrons; those orbitals that are not occupied are available for possible excited states. Each orbital has a relative energy as shown in Figure 4, where the orbitals are represented as lines. [caption id="" align="aligncenter" width="1300"] Figure 4. The chart shows the energies of electron orbitals in a multi-electron atom (up to 4p).[/caption] Electrons and other subatomic particles behave as if they are spinning (we cannot tell if they really are, but they behave as if they are). Electrons themselves have two possible spin states, spin up (graphically represented as an upward pointing arrow, ˦ ), and spin down (graphically represented as a downward pointing arrow, ˨ ). An orbital can hold a maximum of two electrons; when an orbital has two electrons in it, one electron will be spin up, and one will be spin down. This last aspect is an outcome of the Pauli Exclusion Principle, which states that no two electrons in the same atom can behave identically.

[caption id="" align="aligncenter" width="472"] Figure 5. Electrons with spin values ±1/2 in an external magnetic field.[/caption]

Key Concepts and Summary

The quantum mechanical model of atoms describes the three-dimensional position of the electron in a probabilistic manner. Therefore, atomic orbitals describe the areas in an atom where electrons are most likely to be found.

Exercises

1. How are the Bohr model and the quantum mechanical model of the hydrogen atom similar? How are they different? 2. Draw out the 1s, 2s and 3s orbitals and describe the differences and similarities between them. 3. Draw out the 2p_x, 2p_y and 2p_z orbitals and describe the differences and similarities between them. 4. Draw out the 3p and 3s orbitals and describe the differences and similarities between them. 5. The following represents examples of what type of orbital? Answers 1. Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model assumes that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, n = 1, 2, 3, …, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons do not move in precise orbits (such orbits violate the Heisenberg uncertainty principle) and, instead, a probabilistic interpretation of the electron’s position at any given instant is used, with a mathematical function ψ called a wavefunction that can be used to determine the electron’s spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, n (the same one used by Bohr), which specifies shells such that orbitals having the same n all have the same energy and approximately the same spatial extent; the angular momentum quantum number l, which is a measure of the orbital’s angular momentum and corresponds to the orbitals’ general shapes, as well as specifying subshells such that orbitals having the same l (and n) all have the same energy; and the orientation quantum number m, which is a measure of the z component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen’s discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [s orbitals] can have zero angular momentum). 2. Differences: n values are different therefore the energies of the orbitals are different (1s is lower in energy, 2s higher in energy and 3s is highest in energy) and size of the orbitals are different (1s smaller, 2s bigger, 3s biggest). Similarities: they are all s-orbitals, therefore their shape are the same (spherical), they can only contain two electrons with opposite spins 3. Differences: the orientation of the orbitals is different Similarities: n values are the same therefore their energies of the orbitals and size of the orbitals are the same; they are all p-orbitals, therefore their shape are the same (dumbbell); they can only contain two electrons with opposite spins 4. Differences: one is a p-orbital and the other is an s-orbital, therefore their shapes are different (dumbbell and spherical), p-orbitals can have different orientations, but s-orbitals don't Similarities: n values are the same therefore their energies of the orbitals and size of the orbitals are the same; they can only contain two electrons with opposite spins 5. These are two examples of d-orbitals.

Glossary

atomic orbital: mathematical function that describes the behavior of an electron in an atom (also called the wavefunction), it can be used to find the probability of locating an electron in a specific region around the nucleus, as well as other dynamical variables d orbital: region of space with high electron density that is either four lobed or contains a dumbbell and torus shape. An electron in this orbital is called a d electron electron density: a measure of the probability of locating an electron in a particular region of space, it is equal to the squared absolute value of the wave function ψ f orbital: multi-lobed region of space with high electron density. An electron in this orbital is called an f electron Heisenberg uncertainty principle: rule stating that it is impossible to exactly determine both certain conjugate dynamical properties such as the momentum and the position of a particle at the same time. The uncertainty principle is a consequence of quantum particles exhibiting wave–particle duality p orbital: dumbbell-shaped region of space with high electron density. An electron in this orbital is called a p electron Pauli exclusion principle: specifies that no two electrons in an atom can have the same value for all four quantum numbers quantum mechanics: field of study that includes quantization of energy, wave-particle duality, and the Heisenberg uncertainty principle to describe matter s orbital: spherical region of space with high electron density. An electron in this orbital is called an s electron wavefunction (ψ): mathematical description of an atomic orbital that describes the shape of the orbital; it can be used to calculate the probability of finding the electron at any given location in the orbital, as well as dynamical variables such as the energy and the angular momentum

]]> 3668 2362 4 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/8-7-end-of-chapter-problems/ Tue, 22 May 2018 19:16:49 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3792 1. List 5 ions that are isoelectronic with Kr.

2. Determine the electron configuration for the following ions: a)  Pd²⁺     b) Rh³⁺     c)  Ca²⁺     d) S^2-

3. If carbon formed a stable ion, what charge would you expect it to have?  Explain.

4. Explain the differences and similarities between covalent bonds, ionic bonds and polar covalent bonds.

5. Given that ions in LiF and MgO are of similar size, which compound has stronger ionic bonding?  Use Couloumb’s law to explain your answer.

6. How can ionic compounds be neutral if they contain positive and negative ions?

7. Are ions present in a sample of NH₃?  Explain.

8. Are molecules present in a sample of KBr? Explain.

9. The radii of sodium and potassium ions are 102 pm and 138 pm respectively. Which compound has stronger ionic attractions, sodium chloride or potassium chloride?

10. An ionic compound forms when lithium (Z = 3) reacts with oxygen (Z = 8). If a sample of the compound contains 5.3 x 10²⁰lithium ions, how many oxide ions does it contain?

11. Explain the difference between electronegativity and ionization energy. What is the general periodic table trend for each? How do we use that trend to predict bond characteristics?

12. For each of the following bonds, draw a dipole moment (if it is present) and show the partial charges. Hint: Just focus on the periodic table trend.

      a) C-F      b)   N-O    c)   Si-F     d)  Si-N

13. Without using actual electronegativity values, predict which bond in #12 above will have the most ionic character. Do the electronegativity values support your statement?  Explain.

14. Write the Lewis dot diagram for the following:

      a)  SiF₄            b) NBr₃           c)  PO₄^3-         d) CO               e)  NO₃^- 

15. Give Lewis electron dot symbols for:

      a)  Mg              b)  Mg²⁺           c) I^-                 d) Te                e)  Ga

16. Draw electron dot diagrams for the following compounds. If the compound is ionic, give the dot symbols for the ions separately, so as not to imply that the ions are sharing electrons.  (In a group of atoms, the central atom is underlined.)

      a)  OCl₂           b) MgCl₂          c)  CF₄             d)  HCN            e)SO₂             f)  BaO            g)  H₂S

17. Give electron dot diagrams for: 

      a)  PO₄^3-          b)  NH₄⁺          c) ClO₂^-           d)  SO₃^2-

18. Give electron dot diagrams for the following. If more than one possible structure satisfies the rules, write ALL possible structures:

      a)  N₂               b)  HCN            c)  NCO^-

 

 Answers

1.   Se^2-,  Br^-,  Rb⁺,  Sr²⁺,  Y³⁺

2.   a) [Kr]4d⁸;   b) [Kr]4d⁶;   c) [Ar];  d) [Ar]

3.   C⁴⁺or C^4-. To reach the next noble gas configuration, which is very stable.

4.   All bonds are a force that holds atoms together that results from the attraction of ions to the nuclei of atoms. In ionic bonds, electrons are donated from the metal to the nonmetal. In a covalent bond, electrons are shared between two identical non metals. In a polar covalent bond, electrons are unequally shared between two different non metals.

5.   If the sizes are the same, the bond strength will depend only on the charges (strength is proportional to q₁x q₂). Mg and O both have charges with the magnitude of 2 (+2 and -2), whereas Li and F have charges of 1 (+1, -1). Therefore the strength of the MgO bond will be stronger than the LiF bond.

6.   Ionic compounds are neutral overall. The total negative charge of the anions is balanced by the total positive charge of the cations.

7.   No. NH₃is a covalent compound made of N and H (both non-metal). Even though it would have polar covalent bonds (and thus partial positive and negative charges) this does not make them ions.

8.   No. KBr is an ionic compound and exists as an ionic solid, not as individual molecules. KBr is better described as a formula unit that represents the overall composition of the ionic solid.

9.   According to Couloumb’s Law, the strength of a bond is inversely proportional to the distance between atoms. If sodium is smaller, the distance between the centre of sodium and the centre of chloride will be smaller, therefore the bond strength will be greater as compared to KCl.

10.  Li⁺and O^2-therefore compound formed would be Li₂O. Thus there are half as many oxide ions compared to Li ions, therefore 2.7 x 10²⁰ions.

11.  Ionization energy is the ability of an atom’s nucleus to attract one of the atoms own electrons, whereas electronegativity (EN) is the relativity ability of an atom in a molecule to attract the shared electrons in a chemical bond. The general trend for both IE and EN are the same: they increase bottom to top and right to left. The difference in EN values determines the polarity of a bond, therefore the further apart the two atoms are on the periodic table, the more polar the bond will be.

12.             

13.  Si—F is the most polar bond. ΔEN (C—F) = 4.0 – 2.5 = 1.5 ΔEN (N—O) = 3.5 – 3.0 = 0.5 ΔEN (Si—F) = 4.0 – 1.8 = 2.2 ΔEN (Si—N) = 3.0 – 1.8 = 1.2 The larger the the difference in EN values, the more polar the bond.

14. 15. 16. 17. 18. ]]> 3792 1538 8 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-functional-groups-cw/ Fri, 25 May 2018 19:59:22 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3836

Learning Objectives

By the end of this section, you will be able to:

• Define functional group.
• Identify and name the functional groups mentioned here.
• Draw the functional groups mentioned here.
• Know the difference between saturated and unsaturated hydrocarbons.

Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms.  The structure of capsaicin, the compound which is the source of the heat in hot chili peppers, incorporates several functional groups, labeled in the figure below and explained throughout this section. [caption id="attachment_3838" align="aligncenter" width="621"] Figure 1. Functional groups in capsaicin.[/caption] As one progresses their study of organic chemistry, it becomes extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures, and the nomenclature of simple organic compounds.  Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups behave in organic reactions. The 'default' in organic chemistry, essentially, the lack of any functional groups, is given the term alkane, characterized by single bonds between carbon and carbon, or between carbon and hydrogen.  Methane, CH₄, is the natural gas you may burn in your furnace.  Octane, C₈H₁₈, is a component of gasoline.

Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:

$latex \text{CH}_4(g)\;+\;2\text{O}_2(g)\;{\longrightarrow}\;\text{CO}_2(g)\;+\;2\text{H}_2\text{O}(g)$

Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH₄, is the principal component of natural gas. Butane, C₄H₁₀, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (see Figure 2).

[caption id="" align="aligncenter" width="1300"] Figure 2. In a column for the fractional distillation of crude oil, oil heated to about 425 °C in the furnace vaporizes when it enters the base of the tower. The vapors rise through bubble caps in a series of trays in the tower. As the vapors gradually cool, fractions of higher, then of lower, boiling points condense to liquids and are drawn off. (credit left: modification of work by Luigi Chiesa)[/caption]

Alkenes, sometimes called olefins, have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds.  Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening.  If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas.  Alkynes burn readily. Ethyne, commonly called acetylene, is used as a fuel in welding blow torches. Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens  - in other words, they are saturated with hydrogen atoms.  The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. An arene, or an aromatic hydrocarbon is a hydrocarbon with alternative double and single bonds between carbon atoms forming ring.  This functional group is exemplified by benzene, which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic, and naphthalene, a compound with a distinctive 'mothball' smell.  Arenes are planar ring structures, and are widespread in nature. When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or a haloalkane.  Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules. In the alcohol functional group, a carbon is single-bonded to an OH group.  The OH group, by itself, is referred to as a hydroxyl. Ethanol, CH₃CH₂OH, also called ethyl alcohol, is a particularly important alcohol for human use. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars: Except for methanol, all alcohols can be classified as primary, secondary, or tertiary.  In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon.  In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol. The sulfur analog of an alcohol is called a thiol, from the Greek thio, for sulfur. Note that the definition of a phenol states that the hydroxyl oxygen must be directly attached to one of the carbons of the aromatic ring. The compound below, therefore, is not a phenol - it is a primary alcohol.  The distinction is important, because there is a significant difference in the reactivity of alcohols and phenols. In an ether functional group, a central oxygen is bonded to two carbons. Below is the structure of diethyl ether, a common laboratory solvent and also one of the first compounds to be used as an anesthetic during operations. Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins.  The sulfur analog of an ether is called a thioether or sulfide. Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.  One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations when reacted with an acid. In the case where a nitrogen has four bonds to carbon, which is somewhat unusual in biomolecules, it is called a quaternary ammonium ion. Note: Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols and amines - the definitions are different.  In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.

In some amines, the nitrogen atom replaces a carbon atom in an aromatic hydrocarbon. Pyridine is one such heterocyclic amine. A heterocyclic compound contains atoms of two or more different elements in its ring structure.

[caption id="" align="aligncenter" width="248"] Pyridine[/caption]

DNA in Forensics and Paternity

The genetic material for all living things is a polymer of four different molecules, which are themselves a combination of three subunits. The genetic information, the code for developing an organism, is contained in the specific sequence of the four molecules, similar to the way the letters of the alphabet can be sequenced to form words that convey information. The information in a DNA sequence is used to form two other types of polymers, one of which are proteins. The proteins interact to form a specific type of organism with individual characteristics.

A genetic molecule is called DNA, which stands for deoxyribonucleic acid. The four molecules that make up DNA are called nucleotides. Each nucleotide consists of a single- or double-ringed molecule containing nitrogen, carbon, oxygen, and hydrogen called a nitrogenous base. Each base is bonded to a five-carbon sugar called deoxyribose. The sugar is in turn bonded to a phosphate group $latex (-\text{PO}_4^{\;\;3-})$ When new DNA is made, a polymerization reaction occurs that binds the phosphate group of one nucleotide to the sugar group of a second nucleotide. The nitrogenous bases of each nucleotide stick out from this sugar-phosphate backbone. DNA is actually formed from two such polymers coiled around each other and held together by hydrogen bonds between the nitrogenous bases. Thus, the two backbones are on the outside of the coiled pair of strands, and the bases are on the inside. The shape of the two strands wound around each other is called a double helix (see Figure 3).

It probably makes sense that the sequence of nucleotides in the DNA of a cat differs from those of a dog. But it is also true that the sequences of the DNA in the cells of two individual pugs differ. Likewise, the sequences of DNA in you and a sibling differ (unless your sibling is an identical twin), as do those between you and an unrelated individual. However, the DNA sequences of two related individuals are more similar than the sequences of two unrelated individuals, and these similarities in sequence can be observed in various ways. This is the principle behind DNA fingerprinting, which is a method used to determine whether two DNA samples came from related (or the same) individuals or unrelated individuals.

Using similarities in sequences, technicians can determine whether a man is the father of a child (the identity of the mother is rarely in doubt, except in the case of an adopted child and a potential birth mother). Likewise, forensic geneticists can determine whether a crime scene sample of human tissue, such as blood or skin cells, contains DNA that matches exactly the DNA of a suspect.

Watch this video animation of how DNA is packaged for a visual lesson in its structure.

 

The basicity of an amine’s nitrogen atom plays an important role in much of the compound’s chemistry. Amine functional groups are found in a wide variety of compounds, including natural and synthetic dyes, polymers, vitamins, and medications such as penicillin and codeine. They are also found in many molecules essential to life, such as amino acids, hormones, neurotransmitters, and DNA.

Addictive Alkaloids

Since ancient times, plants have been used for medicinal purposes. One class of substances, called alkaloids, found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with H₃O⁺ in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant:

$latex \text{R}_3\text{N}\;+\;\text{H}_3\text{O}^{+}\;+\;\text{Cl}^{-}\;{\longrightarrow}\;[\text{R}_3\text{NH}^{+}]\;\text{Cl}^{-}\;+\;\text{H}_2\text{O}$

The name alkaloid means “like an alkali.” Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base:

$latex [\text{R}_3\text{NH}^{+}]\;\text{Cl}^{-}\;+\;\text{OH}^{-}\;{\longrightarrow}\;\text{R}_3\text{N}\;+\;\text{H}_2\text{O}\;+\;\text{Cl}^{-}$

The structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant:

In these diagrams, as is common in representing structures of large organic compounds, carbon atoms in the rings and the hydrogen atoms bonded to them have been omitted for clarity. The solid wedges indicate bonds that extend out of the page. The dashed wedges indicate bonds that extend into the page. Notice that small changes to a part of the molecule change the properties of morphine, codeine, and heroin. Morphine, a strong narcotic used to relieve pain, contains two hydroxyl functional groups, located at the bottom of the molecule in this structural formula. Changing one of these hydroxyl groups to a methyl ether group forms codeine, a less potent drug used as a local anesthetic. If both hydroxyl groups are converted to esters of acetic acid, the powerfully addictive drug heroin results (Figure 4).

[caption id="" align="aligncenter" width="349"] Figure 4. Poppies can be used in the production of opium, a plant latex that contains morphine from which other opiates, such as heroin, can be synthesized. (credit: Karen Roe)[/caption]

There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl.   Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways.  In a ketone, the carbon atom of a carbonyl is bonded to two other carbons.  In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon.  The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.

Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.

Dimethyl ketone, CH₃COCH₃, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals.

When a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to an oxygen, nitrogen, or sulfur, the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a set of related functional groups.  The eponymous member of this family is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl group.   The conjugate base of a carboxylic acid is a carboxylate.  Carboxylic esters, usually just called 'esters' and amides are some of the other carboxylic acid derivatives.

The simplest carboxylic acid is formic acid, HCO₂H, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests.

Acetic acid, CH₃CO₂H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon.

The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure 5). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, $latex \text{C}_3\text{H}_5\text{(OH)}_3$, with large carboxylic acids, such as palmitic acid, $latex \text{CH}_3\text{(CH}_2)_{14}\text{CO}_2\text{H}$, stearic acid, $latex \text{CH}_3\text{(CH}_2)_{16}\text{CO}_2\text{H}$, and oleic acid, $latex \text{CH}_3\text{(CH}_2)_7\text{CH} = \text{CH(CH}_2)_7\text{CO}_2\text{H}$. Oleic acid is an unsaturated acid; it contains a $latex \text{C} = \text{C}$ double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds.

[caption id="" align="aligncenter" width="430"] Figure 5. Over 350 different volatile molecules (many members of the ester family) have been identified in strawberries. (credit: Rebecca Siegel)[/caption]

Esters and amides are very common in biological molecules and/or metabolic pathways.  Amides can be produced when carboxylic acids react with amines or ammonia in a process called amidation. A water molecule is eliminated from the reaction, and the amide is formed from the remaining pieces of the carboxylic acid and the amine:

The reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both amine and carboxylic acid substituents) link together in a polymer to form proteins.

Proteins and Enzymes

Proteins are large biological molecules made up of long chains of smaller molecules called amino acids. Organisms rely on proteins for a variety of functions—proteins transport molecules across cell membranes, replicate DNA, and catalyze metabolic reactions, to name only a few of their functions. The properties of proteins are functions of the combination of amino acids that compose them and can vary greatly. Interactions between amino acid sequences in the chains of proteins result in the folding of the chain into specific, three-dimensional structures that determine the protein’s activity.

Amino acids are organic molecules that contain an amine functional group (–NH₂), a carboxylic acid functional group (–COOH), and a side chain (that is specific to each individual amino acid). Most living things build proteins from the same 20 different amino acids. Amino acids connect by the formation of a peptide bond, which is a covalent bond formed between two amino acids when the carboxylic acid group of one amino acid reacts with the amine group of the other amino acid. The formation of the bond results in the production of a molecule of water (in general, reactions that result in the production of water when two other molecules combine are referred to as condensation reactions). The resulting bond—between the carbonyl group carbon atom and the amine nitrogen atom is called a peptide link or peptide bond. Since each of the original amino acids has an unreacted group (one has an unreacted amine and the other an unreacted carboxylic acid), more peptide bonds can form to other amino acids, extending the structure. (Figure 6) A chain of connected amino acids is called a polypeptide. Proteins contain at least one long polypeptide chain.

[caption id="" align="aligncenter" width="448"] Figure 6. This condensation reaction forms a dipeptide from two amino acids and leads to the formation of water.[/caption]

Enzymes are large biological molecules, mostly composed of proteins, which are responsible for the thousands of metabolic processes that occur in living organisms. Enzymes are highly specific catalysts; they speed up the rates of certain reactions. Enzymes function by lowering the activation energy of the reaction they are catalyzing, which can dramatically increase the rate of the reaction. Most reactions catalyzed by enzymes have rates that are millions of times faster than the noncatalyzed version. Like all catalysts, enzymes are not consumed during the reactions that they catalyze. Enzymes do differ from other catalysts in how specific they are for their substrates (the molecules that an enzyme will convert into a different product). Each enzyme is only capable of speeding up one or a few very specific reactions or types of reactions. Since the function of enzymes is so specific, the lack or malfunctioning of an enzyme can lead to serious health consequences. One disease that is the result of an enzyme malfunction is phenylketonuria. In this disease, the enzyme that catalyzes the first step in the degradation of the amino acid phenylalanine is not functional (Figure 7). Untreated, this can lead to an accumulation of phenylalanine, which can lead to intellectual disabilities.

[caption id="" align="aligncenter" width="405"] Figure 7. A computer rendering shows the three-dimensional structure of the enzyme phenylalanine hydroxylase. In the disease phenylketonuria, a defect in the shape of phenylalanine hydroxylase causes it to lose its function in breaking down phenylalanine.[/caption]

Finally, a nitrile group is characterized by a carbon triple-bonded to a nitrogen. A single compound often contains several functional groups, particularly in biological organic chemistry.  The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups, a compound with several alcohol groups is often referred to as a ‘polyol’. The hormone testosterone, the amino acid phenylalanine, and the glycolysis metabolite dihydroxyacetone phosphate all contain multiple functional groups, as labeled below. [caption id="attachment_3890" align="aligncenter" width="542"] Figure 7. Natural products are compounds produced by a living organism.  Here are three examples of natural products, which contain a variety of functional groups.[/caption] While not in any way a complete list, this section has covered many of the important functional groups that are encountered in organic chemistry. Table 1 provides a summary of all of the functional groups listed in this section, plus a few more that may be encountered in a second year organic chemistry course. [caption id="attachment_3858" align="aligncenter" width="903"] Table 1. Examples of some common functional groups in organic chemistry.[/caption]

Key Concepts and Summary

Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms.  Organic chemist learn to correlate functional groups to the chemistry that they do.  For example, any molecule that contains a carboxylic acid, will be able to do an acid-base reaction when mixed with a base.  This allows the organic chemist to have a sense of the various chemistry a compound can or cannot do based on the functional groups that are present in the structure.

Activity

Make yourself a stack of small sized Qcards.  On one side have the name of the functional group (e.g. alcohol) and on the other side have its structure (see Table 1).  Make a complete set of all the functional groups you should know (see Table 1).  You can even include some compounds like those below in exercise 1 - on one side of the card have the compound, on the other the names of the functional groups.  Then use these Qcards to quiz yourself.  This will help you recognize the functional groups in larger compounds.

Exercises

1. Answer the following questions for each of these compounds. a) Name the circled functional groups: A = ?,  B =?,  C = ? b) What is the chemical formula of the compound? c) How many lone pairs are there in the compound?   Phenylpropanolamine is a psychoactive drug which is used as a stimulant and decongestant in prescription and over-the-counter cough and cold medicines.   Triiodothyronine is a thyroid hormone which affects many physiological processes in the body, such as growth, metabolism and heart rate.   Aldosterone is a hormone which is involved in the function of the kidneys.   Ephedrine is a drug commonly used as a stimulant, decongestant and also as a concentration aid.   Clomifene is mainly used for ovarian stimulation in female infertility which is due to anomulation. 2. Among the five compounds listed in question 1, which would do an acid-base reaction when mixed with sodium hydroxide (which is a base).   Answers   Phenylpropanolamine: a) A = Primary Amine, B = Secondary Alcohol, C = Arene b) C₉H₁₃NO c) 3 lone pairs – one on the nitrogen and two on the oxygen   Triiodothyronine: a) A = Carboxylic Acid, B = Ether, C = Primary Amine b) C₁₅H₁₂I₃NO₄ c) 18 lone pairs – one on each nitrogen, two on each oxygen, and three on each iodine   Aldosterone: a) A = Secondary Alcohol, B = Aldehyde, C = Ketone b) C₂₁H₂₈O₅ c) 10 lone pairs – two on each oxygen   Ephedrine: a) A = Secondary Amine, B = Secondary Alcohol, C = Arene b) C₁₀H₁₅NO c) 3 lone pairs – one on the nitrogen and two on the oxygen   Clomifene: a) A = Arene, B = Ether, C = Tertiary Amine b) C₂₆H₂₈ClNO c) 6 lone pairs – one on the nitrogen, two on the oxygen, and three on the chlorine   2. Any molecule that contains a carboxylic acid, will be able to do an acid-base reaction when mixed with a base.  Therefore among the five compounds in question 1, the only one that has a carboxylic acid is Triiodothyronine.

 ]]> 3836 1829 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-1-condensed-structure-and-line-structure-cw/ Fri, 25 May 2018 20:36:14 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3849

Learning Objectives

By the end of this section, you will be able to:

• Interpret condense and line structures.
• Draw the condensed structure of a given Lewis Structure or line structure.
• Draw the line structure of a given Lewis Structure or condense structure.

If you look ahead in this chapter and in other resources at the way organic compounds are drawn, you will see that the figures are somewhat different from the Lewis structures you are used to seeing in your general chemistry book. In some sources,  you will see condensed structures for smaller molecules instead of full Lewis structures:   [caption id="attachment_3851" align="aligncenter" width="463"] Figure 1. Comparison between Lewis structures and condensed structures.[/caption]

Example 1

Determine the Lewis Structure of the following condensed structure of oleic acid, a fatty acid that is found naturally in various animal and vegetable fats and oils.

CH₃(CH₂)₇CH=CH(CH₂)₇COOH

Solution

Start by drawing the CH₃. The (CH₂)₇ represents a repeating unit, meaning you must draw seven CH₂'s one after another, which are bonded to a CH which is bonded to a CH, and then another seven CH₂'s.  The COOH represent a carboxylic acid, which means you have a C=O connected to an O-H.  Always double check your structure to ensure every carbon is making four bonds.  When you do this, you will see the two CH must be double bonded.  

Test Yourself Common organic compounds that you likely have at home are: acetone (CH₃COCH₃) found in nail polish remover, acetic acid (CH₃COOH) found in vinegar, and isopropanol ((CH₃)₂CHOH) found in rubbing alcohol. Determine the Lewis Structure for each of these household chemicals.

  Answer

More commonly, organic and biological chemists use an abbreviated drawing convention called line structures, also known as skeletal structures or line bond structures.  The convention is quite simple and makes it easier to draw molecules, but line structures do take a little bit of getting used to. Carbon atoms are depicted not by a capital C, but by a ‘corner’ between two bonds, or a free end of a bond. Open-chain molecules are usually drawn out in a 'zig-zig' shape. Hydrogens attached to carbons are generally not shown: rather, like lone pairs, they are simply implied (unless a positive formal charge is shown, all carbons are assumed to have a full octet of valence electrons). Hydrogens bonded to nitrogen, oxygen, sulfur, or anything other than carbon are shown, but are usually drawn without showing the bond. The following examples illustrate the convention. [caption id="attachment_3853" align="aligncenter" width="482"] Figure 2. Comparison between Lewis structure and line structure.[/caption] As you can see, the 'pared down' line structure makes it much easier to see the basic structure of the molecule and the locations where there is something other than C-C and C-H single bonds.  For larger, more complex biological molecules, it becomes impractical to use full Lewis structures.  Conversely, very small molecules such as ethane should be drawn with their full Lewis or condensed structures. Sometimes, one or more carbon atoms in a line structure will be depicted with a capital C, if doing so makes an explanation easier to follow. If you label a carbon with a C, you also must draw in the hydrogens for that carbon.

Example 2

Draw the line structures for these two molecules:

 

Solution Each carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):

 

Test Yourself Draw the line structures for these two molecules:

  Answers

Example 3

Identify the chemical formula of the molecule represented here:

 

Solution There are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of C₈H₁₆.

Location of the hydrogen atoms:

 

Test Yourself Identify the chemical formula of the molecule represented here:

  Answer C₉H₂₀

Example 4

Determine the Lewis Structure of the following line structure of L-ascorbic acid (vitamin C):

Solution

At each corner or intersection of lines or end of a line, add a C:

Each C makes fours bonds.  If some are not visible, then add a bond to H for each bonds missing. Also show the bond between the O's and H's.

 

Test Yourself Determine the Lewis Structure of the following line structure of acetaminophen, the pain and fever medicine found in Tylenol.

  Answer

Key Concepts and Summary

Condensed structures and line structures are a way organic structures can be represented in a very concise manner.

Exercises

1. Draw the line bond structure for the following compounds: a) (HO)₃C(CH₂)₂N(CH₂CHO)CH(CH₂CH₃)₂ b) CH₃(CH₂)₃CH(CH₃)(CH₂)₂OCH₂CH₃ c) HOOCCH₂O(CH₂)₂N(CH₂CH₃)₂ d) HOCCH₂CH(CH₃)CH₂CH(CH₃)(CH₂)₂COOCH₃ e) H₂NCH₂[CH(CH₃)]₂(CH)₄CO(CH₂)₂CH₃ f) ClCH₂CH(CH₃)COCH₂OCH₂CO(CH₂)₂CH₃ 2. For the following compounds,give the chemical formula and the condensed structure:   Answers 1. 2. a) C₁₃H₂₆O;   CH₃(CH₂)₃C(CH₃)₂C(CH₃)₂CH₂COCH₃ b) C₈H₁₄O₂;   CHOCH₂CO(CH₂)₄CH₃ c) C₅H₇O₄N;   COOH(CH₂)₂(CO)₂NH₂ d) C₁₂H₂₅O₂N;   CH₃OCO(CH₂)₂C(CH₃)₂C(CH₃)₂CH(CH₃)NH₂ e) C₁₂H₂₅ON;   CH₃(CH₂)₄CON(CH₂CH₂CH₃)₂ f) C₈H₁₆O₃;   CH₃CH₂CH(OCH₃)(CH₂)₃COOH

]]> 3849 1829 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-x-end-of-chapter-problems-langara/ Fri, 25 May 2018 23:04:45 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=3870 1. Draw the line bond structure for the following compounds: a) Br[CH(CH₃)]₄COCH(CH₃)CH₂CH₃ b) CH₃(CH)₈CH₂COOH c) CH₃CH₂CH(CHO)(CH₂)₂CH(CH₃)CH₂Br d) (HO)₂CH(CH₂)₂O(CH₂)₃COOH 2. For the following compounds, give the chemical formula and the condensed structure: 3. Answer the following questions for each of these compounds. a) Name the circled functional groups: A = ?,  B = ?, C = ? b) What is the chemical formula of the compound? c) How many lone pairs are there in the compound? Phenylalanine is an α-amino acid. Aspirin, also known as acetylsalicylic acid, is often used to relieve minor aches and pains, and to reduce fever. Tetracycline is an antibiotic used against many bacterial infections. Ampicillin is an antibiotic used against many bacterial infections since 1961. Forskolin is a natural product produced by the Indian Coleus plant (Coleus forskohlii).

4. Cycloalkanes are named based on the number of C atoms in them, just like regular alkanes, but with the prefix cyclo- on the name. What are the names of the three smallest cycloalkanes?

5. Draw the bond-line structure of all noncyclic alkanes with only four C atoms.

6. Cyclic alkanes can also have substituent groups on the ring. Draw the bond-line structure of all cyclic alkanes with only four C atoms.

7. What is the maximum number of methyl groups that can be on a propane backbone before the molecule cannot be named as a propane compound?

8. In the gasoline industry, what is called isooctane is actually 2,2,4-trimethylpentane. Draw the structure of isooctane.

9. The actual name for the explosive TNT is 2,4,6-trinitrotoluene. If the structure of TNT is as shown below, propose the structure of the parent compound toluene.

10. Draw the smallest molecule that can have a separate aldehyde and carboxylic acid group.

11. Name the functional group(s) in the following structure:

12. Ethyl acetate is a common ingredient in nail-polish remover because it is a good solvent. Its IUPAC name is ethyl ethanoate.  Draw the structure of ethyl acetate.

13. Draw the structure of diethyl ether, once used as an anesthetic.

14. Write the chemical reaction of HCl with trimethylamine.

15.Give the IUPAC name the following organic compounds.

Answers

1. 2.  a) C₈H₁₂O₄      HOOCCH₂COCH₂CO(CH₂)₂CH₃ b) C₈H₁₀O₃         HOC(CH)₄(CH₂)₂COOH c) C₁₄H₂₈              CH₃(CH)₂CH(CH[CH₃]CH₂CH₃)(CH₂)₅CH₃ d) C₉H₁₇Cl        Cl(CH₂)₂C(CHCH₂CH₃)(CH₂)₂CH₃ 3. Phenylalanine a) A = Arene, B = Carboxylic Acid, C = Primary Amine b) C₉H₁₁NO₂ c) 5 lone pairs – one on the nitrogen and two on each oxygen, Aspirin a) A = Ester, B = Arene, C = Carboxylic Acid b) C₉H₈O₄ c) 8 lone pairs – two on each oxygen, Tetracycline a) A = Tertiary Alcohol, B = Tertiary Amine, C = Ketone b) C₂₂H₂₄N₂O₈ c) 18 lone pairs – one on each nitrogen and two on each oxygen Ampicillin  a) A = Tertiary Amide, B = Carboxylic Acid, C = Arene b) C₁₆H₁₉N₃O₄S c) 13 lone pairs – one on each nitrogen, two on each oxygen and two on the sulfur Forskolin a) A = Secondary Alcohol, B = Ether, C = Ketone b) C₂₂H₃₄O₇ c) 14 lone pairs –two on each oxygen, 4. cyclopropane, cyclobutane, and cyclopentane 5. 6. 7. two 8.         9. 10. 11. alcohol (or more specifically phenol), arene, ketone, primary amine 12. 13. 14. Triethylamine is a base and HCl is an acid, therefore you get an acid-base reaction. (CH₃)₃N + HCl $latex \longrightarrow$ (CH₃)₃NHCl 15. a) 4-ethyloctane          b) 3-ethyl-2,4-dimethylhexane c) 2,4-dibromo-1-chloro-3-methylpentane        d) 5,5-dimethyl-3-propyl-1-heptene e) 4-methyl-4-octene      f)  2-ethyl-1,6-heptadiene       g) 3-ethyl-1,4-hexadiene h) 4-ethyl-2,5-octadiene     i) 3,3-dimethyl-4-octyne       j) 4-methyl-2-hexyne]]> 3870 1829 9 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/9-7-summary-of-nomenclature-rules-me/ Thu, 31 May 2018 03:12:04 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=4017 Figure 1. IUPAC Nomenclature Guide[/caption] The names are made of three main parts: 1) specifying the information about the substituents; 2) specifying the information about the parent chain (or ring); and 3) the ending which specifies what functional group is present in the structure being named.

	Two Rules
Alkanes:  #-substituents-PREFIX+ANE	Rule 1. Identify the longest chain of carbon atoms. Rule 2. Names and position of the substituents. ***
Alkenes: #-substituents-PREFIX-#-ENE	Rule 1. Identify the longest chain of carbons which contains the double bond and its position.  And when numbering the main chain, the double gets the lowest possible number.Rule 2. Names and position of the substituents.
Alkynes: #-substituents-PREFIX-#-YNE	Rule 1. Identify the longest chain of carbons which contains the triple bond and its position.  And when numbering the main chain, the triple gets the lowest possible number.Rule 2. Names and position of the substituents.
Arenes (specifically benzene derivatives): #-substituents-BENZENE	Rule 1. Identify the arene ring (BENZENE). Rule 2. Names and position (if more than one) of the substituents: If there are two or more substituents on a benzene molecule, the relative positions must be numbered. The substituent that is first alphabetically is assigned position 1, and the ring is numbered in a circle to give the other substituents the lowest possible number(s).
Alcohols: #-substituents-PREFIX-#-ANE+OL	Rule 1. Identify the longest chain of carbons which contains the OH group and its position.  And when numbering the parent chain, the hydroxyl group gets the lowest possible number. Rule 2. Names and position of the substituents.***
Ethers: PREFIX+OXY-PREFIX-ANE	Rule 1. Identify the longest carbon branch  Rule 2. Names of the substituent, the other carbon branch (PREFIX+OXY)
Aldehydes: #-substituents-PREFIX-ANE+AL	Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-ANE+AL).  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name. Rule 2. Names and position of the substituents.***
Ketones: #-substituents-PREFIX-ANE+ONE	Rule 1. Identify the longest chain of carbons which contains the carbonyl group.  And when numbering the parent chain, the carbonyl group gets the lowest possible number.  In the smaller ketones (propanone and butanone), the locant number is not used because there is no alternative placement in these smaller ketones. Rule 2. Names and position of the substituents.***
Carboxylic Acids: #-substituents-PREFIX-ANE+OIC ACID	Rule 1. Identify the longest chain of carbons which contains the carbonyl group.  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name. Rule 2. Names and position of the substituents.***
Esters: ALKYL #-substituents-PREFIX-ANE+OATE	Rule 1. Identify the longest chain of carbons which contains the carbonyl group.  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name.  AND then name the other carbon chain (PREFIX+YL). Rule 2. Names and position of the substituents.***
Amides: #-substituents-PREFIX-ANE+AMIDE	Rule 1. Identify the longest chain of carbons which contains the carbonyl group (PREFIX-ANE+AMIDE).  And when numbering the parent chain, the carbonyl group gets the lowest possible number, therefore it is always 1 and therefore is not included in the name. Rule 2. Names and position of the substituents.***

***If there are substituents on the parent chain, their names and position on the chain must be included at the front of the name. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. Multiple substituents are named individually and placed in alphabetical order at the front of the name. It is helpful to recognize the similarities between the rules of alkenes and alkynes, and between the rules of alcohols  and the carbonyl functional groups.  Ethers and amines have their own unique naming procedures.

The systematic methods of naming amines follows a different format:

primary amines:

ALKYLamine  

______________

secondary amines:

ALKYLALKYLamine    or   diALKYLamine

______________

tertiary amines:

ALKYLALKYLALKYLamine    or   triALKYLamine

______________

Remember: there are two ways that are commonly used to name ethers.  The common method is also important to know.

Common method of naming ethers:

ALKYLALKYL ether    or   diALKYL ether 

]]> 4017 1829 8 0 https://pressbooks.bccampus.ca/chem1114langaracollege/authors/ Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/authors/ ]]> 7 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/ Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/cover/ ]]> 8 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/table-of-contents/ Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/table-of-contents/ ]]> 9 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/about/ Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/about/ ]]> 10 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/buy/ Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/buy/ ]]> 11 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/access-denied/ Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/access-denied/ ]]> 12 0 0 0 https://pressbooks.bccampus.ca/chem1114langaracollege/?metadata=book-information Tue, 10 Apr 2018 21:33:55 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/2018/04/10/book-information/ 16 0 1 0 © Jun 25, 2018 is licensed under a Creative Commons Attribution Non-Commercial ShareAlike License 4.0 license. CHEM 1114 - Introduction to Chemistry was adapted by Shirley Wacowich-Sgarbi by combining a courseware package from the Langara College Chemistry Department and a combination of OpenStax textbooks, including:  Paul Flowers, Klaus Theopold and Richard Langley's textbook Chemistry {Download for free at http://cnx.org/contents/85abf193-2bd2-4908-8563-90b8a7ac8df6@9.311.}; Jessie A. Key's textbook Introductory Chemistry- 1st Canadian Edition {Download for free at https://open.bccampus.ca/find-open-textbooks/?uuid=c7025f6b-f32b-4d0a-865e-f473d9f98fb6&contributor=&keyword=&subject=Chemistry};  and portions of chapter 1 from Tim Soderberg's textbook Organic Chemistry with a Biological Emphasis {Download for free at https://open.bccampus.ca/find-open-textbooks/?uuid=f01f98b4-925d-4e94-bc82-c1d707e6280e&contributor=&keyword=&subject=Chemistry}.]]> https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=84

Chapter 8. Gases

5.5 Effusion and Diffusion of Gases

Learning Objectives

By the end of this section, you will be able to:

• Define and explain effusion and diffusion
• State Graham’s law and use it to compute relevant gas properties

If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule

In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 1). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 1. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).

Figure 1. (a) Two gases, H₂ and O₂, are initially separated. (b) When the stopcock is opened, they mix together. The lighter gas, H₂, passes through the opening faster than O₂, so just after the stopcock is opened, more H₂ molecules move to the O₂ side than O₂ molecules move to the H₂ side. (c) After a short time, both the slower-moving O₂ molecules and the faster-moving H₂ molecules have distributed themselves evenly on both sides of the vessel.

We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:

The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.

A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 2). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.

Figure 2. Diffusion occurs when gas molecules disperse throughout a container. Effusion occurs when a gas passes through an opening that is smaller than the mean free path of the particles, that is, the average distance traveled between collisions. Effectively, this means that only one particle passes through at a time.

If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 3). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:

This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:

Figure 3. A balloon filled with air (the blue one) remains full overnight. A balloon filled with helium (the green one) partially deflates because the smaller, light helium atoms effuse through small holes in the rubber much more readily than the heavier molecules of nitrogen and oxygen found in air. (credit: modification of work by Mark Ott)

Example 1

Applying Graham’s Law to Rates of Effusion
Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.

Solution
From Graham’s law, we have:

Using molar masses:

Hydrogen effuses four times as rapidly as oxygen.

Check Your Learning
At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide effuse?

Answer:

52 mL/s

Here’s another example, making the point about how determining times differs from determining rates.

Example 2

Effusion Time Calculations
It takes 243 s for 4.46 × 10⁻⁵ mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10⁻⁵ mol Ne to effuse?

Solution
It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:

and combine it with Graham’s law:

To get:

Noting that amount of A = amount of B, and solving for time for Ne:

and substitute values:

Finally, solve for the desired quantity:

Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.

Check Your Learning
A party balloon filled with helium deflates to of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air ( = 28.2 g/mol) to deflate to of its original volume?

Answer:

32 h

Finally, here is one more example showing how to calculate molar mass from effusion rate data.

Example 3

Determining Molar Mass Using Graham’s Law
An unknown gas effuses 1.66 times more rapidly than CO₂. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?

Solution
From Graham’s law, we have:

Plug in known data:

Solve:

The gas could well be CH₄, the only gas with this molar mass.

Check Your Learning
Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Answer:

163 g/mol

Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment

Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of ²³⁵U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% ²³⁵U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF₆, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The ²³⁵UF₆ molecules have a higher average speed and diffuse through the barrier a little faster than the heavier ²³⁸UF₆ molecules. The gas that has passed through the barrier is slightly enriched in ²³⁵UF₆ and the residual gas is slightly depleted. The small difference in molecular weights between ²³⁵UF₆ and ²³⁸UF₆ only about 0.4% enrichment, is achieved in one diffuser (Figure 4). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.

Figure 4. In a diffuser, gaseous UF₆ is pumped through a porous barrier, which partially separates ²³⁵UF₆ from ²³⁸UF₆ The UF₆ must pass through many large diffuser units to achieve sufficient enrichment in ²³⁵U.

The large scale separation of gaseous ²³⁵UF₆ from ²³⁸UF₆ was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10^–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF₆.

Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.

Key Concepts and Summary

Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).

Key Equations

• 
• 

Chemistry End of Chapter Exercises

1. A balloon filled with helium gas is found to take 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?
2. Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure 1.
3. Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.
4. Heavy water, D₂O (molar mass = 20.03 g mol^–1), can be separated from ordinary water, H₂O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H₂O and D₂O.
5. Which of the following gases diffuse more slowly than oxygen? F₂, Ne, N₂O, C₂H₂, NO, Cl₂, H₂S
6. During the discussion of gaseous diffusion for enriching uranium, it was claimed that
   ²³⁵UF₆ diffuses 0.4% faster than ²³⁸UF₆. Show the calculation that supports this value. The molar mass of ²³⁵UF₆ = 235.043930 + 6 × 18.998403 = 349.034348 g/mol, and the molar mass of ²³⁸UF₆ = 238.050788 + 6 × 18.998403 = 352.041206 g/mol.
7. Calculate the relative rate of diffusion of ¹H₂ (molar mass 2.0 g/mol) compared to that of ²H₂ (molar mass 4.0 g/mol) and the relative rate of diffusion of O₂ (molar mass 32 g/mol) compared to that of O₃ (molar mass 48 g/mol).
8. A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
9. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH₄Cl forms where gaseous NH₃ and gaseous HCl first come into contact. (Hint: Calculate the rates of diffusion for both NH₃ and HCl, and find out how much faster NH₃ diffuses than HCl.)

   At approximately what distance from the ammonia moistened plug does this occur?

Glossary

diffusion

movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)

effusion

transfer of gaseous atoms or molecules from a container to a vacuum through very small openings

Graham’s law of effusion

rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses

mean free path

average distance a molecule travels between collisions

rate of diffusion

amount of gas diffusing through a given area over a given time

Solutions

Answers to Chemistry End of Chapter Exercises

1. 4.2 hours

3. Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: . Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy:

Therefore,

5. F₂, N₂O, Cl₂, H₂S

7. 1.4; 1.2

9. 51.7 cm

License

5.5 Effusion and Diffusion of Gases by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 84 3 7 0 https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/ Mon, 30 Nov -0001 00:00:00 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=chapter&p=91

Chapter 8. Gases

5.6 Non-Ideal Gas Behavior

Learning Objectives

By the end of this section, you will be able to:

• Describe the physical factors that lead to deviations from ideal gas behavior
• Explain how these factors are represented in the van der Waals equation
• Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
• Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation

Thus far, the ideal gas law, PV = nRT, has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.

One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, V_m) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with:

Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. Figure 1 shows plots of Z over a large pressure range for several common gases.

Figure 1. A graph of the compressibility factor (Z) vs. pressure shows that gases can exhibit significant deviations from the behavior predicted by the ideal gas law.

As is apparent from Figure 1, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.

Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas (Figure 2). The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law.

Figure 2. Raising the pressure of a gas increases the fraction of its volume that is occupied by the gas molecules and makes the gas less compressible.

At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure 3). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.

Figure 3. (a) Attractions between gas molecules serve to decrease the gas volume at constant pressure compared to an ideal gas whose molecules experience no attractive forces. (b) These attractive forces will decrease the force of collisions between the molecules and container walls, therefore reducing the pressure exerted compared to an ideal gas.

There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.

The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in the ideal gas law is , and the “correction” to the volume is nb. Note that when V is relatively large and n is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, PV = nRT. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table 3.

Gas	a (L2 atm/mol2)	b (L/mol)
N2	1.39	0.0391
O2	1.36	0.0318
CO2	3.59	0.0427
H2O	5.46	0.0305
He	0.0342	0.0237
CCl4	20.4	0.1383
Table 3. Values of van der Waals Constants for Some Common Gases

At low pressures, the correction for intermolecular attraction, a, is more important than the one for molecular volume, b. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by PV = nRT over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in Figure 1. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing P). At very high pressures, the gas becomes less compressible (Z increases with P), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.

Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case.

Example 1

Comparison of Ideal Gas Law and van der Waals Equation
A 4.25-L flask contains 3.46 mol CO₂ at 229 °C. Calculate the pressure of this sample of CO₂:

(a) from the ideal gas law

(b) from the van der Waals equation

(c) Explain the reason(s) for the difference.

Solution
(a) From the ideal gas law:

(b) From the van der Waals equation:

This finally yields P = 32.4 atm.

(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO₂ molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.

Check your Learning
A 560-mL flask contains 21.3 g N₂ at 145 °C. Calculate the pressure of N₂:

(a) from the ideal gas law

(b) from the van der Waals equation

(c) Explain the reason(s) for the difference.

Answer:

(a) 46.562 atm; (b) 46.594 atm; (c) The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions.

Key Concepts and Summary

Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.

Key Equations

• 
• 

Chemistry End of Chapter Exercises

1. Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?
   
2. Explain why the plot of PV for CO₂ differs from that of an ideal gas.
   
3. Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.

   (a) high pressure, small volume

   (b) high temperature, low pressure

   (c) low temperature, high pressure

4. Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas.
5. For which of the following gases should the correction for the molecular volume be largest:

   CO, CO₂, H₂, He, NH₃, SF₆?

6. A 0.245-L flask contains 0.467 mol CO₂ at 159 °C. Calculate the pressure:

   (a) using the ideal gas law

   (b) using the van der Waals equation

   (c) Explain the reason for the difference.

   (d) Identify which correction (that for P or V) is dominant and why.

7. Answer the following questions:

   (a) If XX behaved as an ideal gas, what would its graph of Z vs. P look like?

   (b) For most of this chapter, we performed calculations treating gases as ideal. Was this justified?

   (c) What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.

   (d) What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.

   (e) In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?

Glossary

compressibility factor (Z)

ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation

van der Waals equation

modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior

Solutions

1. Gases C, E, and F

3. The gas behavior most like an ideal gas will occur under the conditions in (b). Molecules have high speeds and move through greater distances between collision; they also have shorter contact times and interactions are less likely. Deviations occur with the conditions described in (a) and (c). Under conditions of (a), some gases may liquefy. Under conditions of (c), most gases will liquefy.

5. SF₆

7. (a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures they behave close enough to ideal gases that they are approximated as such, however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the ideal gas equation (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure 1) (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure 1. (e) low temperatures

License

5.6 Non-Ideal Gas Behavior by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

]]> 91 3 8 0 https://pressbooks.bccampus.ca/chem1114langaracollege/front-matter/preface/ Thu, 12 Apr 2018 02:50:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/front-matter/preface/

About Chem 1114 - Introduction to Chemistry

CHEM 1114 - Introduction to Chemistry is designed for a one-semester introductory chemistry course. For many students, this course provides their first introduction to chemistry. As such, this textbook provides an important opportunity for students to learn some of the core concepts of chemistry and understand how those concepts apply to their lives and the world around them. The text has been developed to meet the scope and sequence of most introductory chemistry courses, including an initial emphasis on the skills required (chapter 1 and 2) for the laboratory portion of the course. A strength of CHEM 1114 - Introduction to Chemistry is that instructors can customize the book, adapting it to the approach that works best in their classroom.

Coverage and Scope Our CHEM 1114 - Introduction to Chemistry textbook adheres to the scope and sequence of most introductory chemistry courses nationwide. We strive to make chemistry, as a discipline, interesting and accessible to students. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression of the fundamental concepts of chemical science. Topics are introduced within the context of familiar experiences whenever possible, treated with an appropriate rigor to satisfy the intellect of the learner, and reinforced in subsequent discussions of related content. The organization and pedagogical features were developed and vetted with feedback from chemistry educators dedicated to the project.

• Chapter 1: Chemistry: An Experimental Science
• Chapter 2: Making Measurements
• Chapter 3: Atoms, Molecules, and Ions
• Chapter 4: Chemical Nomenclature
• Chapter 5: Chemical Composition
• Chapter 6: Chemical Reactions and Equations
• Chapter 7: Stoichiometry of Chemical Reactions
• Chapter 8: Electronic Structure of Atoms
• Chapter 9: Chemical Bonding and Lewis Structures
• Chapter 10: Organic Chemistry

Pedagogical Foundation Throughout CHEM 1114 - Introduction to Chemistry, you will find features that draw the students into scientific inquiry by taking selected topics a step further. Students and educators alike will appreciate discussions in these feature boxes.

• Chemistry in Everyday Life ties chemistry concepts to everyday issues and real-world applications of science that students encounter in their lives. Topics include cell phones, solar thermal energy power plants, plastics recycling, and measuring blood pressure.
• How Sciences Interconnect feature boxes discuss chemistry in context of its interconnectedness with other scientific disciplines. Topics include neurotransmitters, greenhouse gases and climate change, and proteins and enzymes.
• Portrait of a Chemist features present a short bio and an introduction to the work of prominent figures from history and present day so that students can see the “face” of contributors in this field as well as science in action.

Comprehensive Art Program Our art program is designed to enhance students’ understanding of concepts through clear, effective illustrations, diagrams, and photographs.

 

Interactives That Engage CHEM 1114 - Introduction to Chemistry incorporates links to relevant interactive exercises and animations that help bring topics to life through our Link to Learning feature.

Assessments That Reinforce Key Concepts In-chapter Examples walk students through problems by posing a question, stepping out a solution, and then asking students to practice the skill with a “Test Yourself” component. The book also includes assessments at the end of each chapter so students can apply what they’ve learned through practice problems and easily verify their solutions since all answers have been included.

Curation To broaden access and encourage community curation, CHEM 1114 - Introduction to Chemistry is “open source” licensed under a Creative Commons Attribution Non-Commercial Share Alike (CC-BYNCSA) license. The academic science community is invited to submit examples, emerging research, and other feedback to enhance and strengthen the material and keep it current and relevant for today’s students.

]]> 1246 0 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-1-essential-ideas/ Thu, 12 Apr 2018 02:50:47 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-1-essential-ideas/

Chapter Topics

• Chemistry in Context
• Phases and Classification of Matter
• Physical and Chemical Properties
• Laboratory Techniques for Separation of Mixtures

]]> 1247 0 1 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-2-atoms-molecules-and-ions/ Thu, 12 Apr 2018 02:51:09 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-2-atoms-molecules-and-ions/

Chapter Topics

• Early Ideas in Atomic Theory
• Evolution of Atomic Theory
• Atomic Structure and Symbolism
• Chemical Formulas
• The Periodic Table

]]> 1293 0 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-3-composition-of-substances-and-solutions/ Thu, 12 Apr 2018 02:51:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-3-composition-of-substances-and-solutions/

Chapter Topics

• Mass Terminology
• The Mole
• Percent Composition
• Determining the Empirical and Molecular Formulas

 ]]> 1354 0 5 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-4-stoichiometry-of-chemical-reactions/ Thu, 12 Apr 2018 02:51:40 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-4-stoichiometry-of-chemical-reactions/

Chapter Topics

• Reaction Stoichiometry
• Limiting Reagent and Reaction Yields
• Molarity
• Other Units for Solution Concentrations
• Quantitative Chemical Analysis

]]> 1405 0 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-7-chemical-bonding-and-molecular-geometry/ Thu, 12 Apr 2018 02:52:26 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-7-chemical-bonding-and-molecular-geometry/

Chapter Topics

• Ionic Bonding
• Covalent Bonding
• Lewis Electron Dot Diagrams
• Electron Transfer: Ionic Bonds
• Covalent Bonds and Lewis Structures
• Formal Charges and Resonance

]]> 1538 0 9 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-20-organic-chemistry/ Thu, 12 Apr 2018 03:47:13 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-20-organic-chemistry/

Chapter Topics

• Condensed Structure and Line Structure
• Functional Groups
• Nomenclature of Hydrocarbons and Alkyl Halides
• Nomenclature of Alcohols and Ethers
• Nomenclature of Amines
• Nomenclature of Aldehydes, Ketones, Carboxylic Acids, Esters and Amides
• Summary of Nomenclature Rules

]]> 1829 0 10 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-2-measurements/ Thu, 12 Apr 2018 03:51:54 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-2-measurements/

Chapter Topics

• Expressing Numbers
• Measurements and Units
• Measurement Uncertainty, Accuracy, and Precision
• Mathematical Treatment of Measurement Results - Unit Conversions
• Density - Just Another Conversion Factor

]]> 2084 0 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-3-atoms-molecules-and-ions/ Thu, 12 Apr 2018 03:52:02 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-3-atoms-molecules-and-ions/

Chapter Topics

• Names of Elements
• Ionic and Molecular Compounds
• Nomenclature of Simple Ionic and Molecular Compounds

 ]]> 2141 0 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/front-matter/acknowledgement/ Thu, 21 Jun 2018 19:23:30 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=front-matter&p=4773 CHEM 1114 - Introduction to Chemistry was adapted by Shirley Wacowich-Sgarbi by combining a courseware package from the Langara College Chemistry Department and a combination of OpenStax textbooks, including Paul Flowers, Klaus Theopold and Richard Langley's textbook Chemistry, Jessie A. Key's textbook Introductory Chemistry- 1st Canadian Edition and portions of chapter 1 from Tim Soderberg's textbook Organic Chemistry with a Biological Emphasis. Shirley Wacowich-Sgarbi, June 2018

From Paul Flowers, Klaus Theopold and Richard Langley's textbook Chemistry: Content Leads

Paul Flowers, PhD, University of North Carolina - Pembroke

Klaus Theopold, PhD, University of Delaware Richard Langley, PhD, Stephen F. Austin State University

Senior Contributing Author

William R. Robinson, PhD

Contributing Authors

Mark Blaser, Shasta College; Simon Bott, University of Houston; Donald Carpenetti, Craven Community College; Andrew Eklund, Alfred University; Emad El-Giar, University of Louisiana at Monroe; Don Frantz, Wilfrid Laurier University; Paul Hooker, Westminster College; Jennifer Look, Mercer University; George Kaminski, Worcester Polytechnic Institute; Carol Martinez, Central New Mexico Community College; Troy Milliken, Jackson State University; Vicki Moravec, Trine University; Jason Powell, Ferrum College; Thomas Sorensen, University of Wisconsin–Milwaukee; Allison Soult, University of Kentucky

Contributing Reviewers Casey Akin, College Station Independent School District; Lara AL-Hariri, University of Massachusetts–Amherst; Sahar Atwa, University of Louisiana at Monroe; Todd Austell, University of North Carolina–Chapel Hill; Bobby Bailey, University of Maryland–University College; Robert Baker, Trinity College; Jeffrey Bartz, Kalamazoo College; Greg Baxley, Cuesta College; Ashley Beasley Green, National Institute of Standards and Technology; Patricia Bianconi, University of Massachusetts; Lisa Blank, Lyme Central School District; Daniel Branan, Colorado Community College System; Dorian Canelas, Duke University; Emmanuel Chang, York College; Carolyn Collins, College of Southern Nevada; Colleen Craig, University of Washington; Yasmine Daniels, Montgomery College–Germantown; Patricia Dockham, Grand Rapids Community College; Erick Fuoco, Richard J. Daley College; Andrea Geyer, University of Saint Francis; Daniel Goebbert, University of Alabama; John Goodwin, Coastal Carolina University; Stephanie Gould, Austin College; Patrick Holt, Bellarmine University; Kevin Kolack, Queensborough Community College; Amy Kovach, Roberts Wesleyan College; Judit Kovacs Beagle, University of Dayton; Krzysztof Kuczera, University of Kansas; Marcus Lay, University of Georgia; Pamela Lord, University of Saint Francis; Oleg Maksimov, Excelsior College; John Matson, Virginia Tech; Katrina Miranda, University of Arizona; Douglas Mulford, Emory University; Mark Ott, Jackson College; Adrienne Oxley, Columbia College; Richard Pennington, Georgia Gwinnett College; Rodney Powell, Coastal Carolina Community College; Jeanita Pritchett, Montgomery College–Rockville; Aheda Saber, University of Illinois at Chicago; Raymond Sadeghi, University of Texas at San Antonio; Nirmala Shankar, Rutgers University; Jonathan Smith, Temple University; Bryan Spiegelberg, Rider University; Ron Sternfels, Roane State Community College; Cynthia Strong, Cornell College; Kris Varazo, Francis Marion University; Victor Vilchiz, Virginia State University; Alex Waterson, Vanderbilt University; JuchaoYan, Eastern New Mexico University; Mustafa Yatin, Salem State University; Kazushige Yokoyama, State University of New York at Geneseo; Curtis Zaleski, Shippensburg University; Wei Zhang, University of Colorado–Boulder

From Jessie A. Key's textbook Introductory Chemistry - 1st Canadian Edition [an adaption from David W. Ball's textbook Introductory Chemistry]: I would like to acknowledge the team at BCcampus for all their hard work on this project. Project managers Amanda Coolidge and Clint Lalonde, and the entire editorial team were instrumental to the success of this work. As well, I would like to thank my colleagues at Vancouver Island University for their support. Jessie A. Key, September 2014

From David W. Ball's textbook Introductory Chemistry: The decision to write a new textbook from scratch is not one to be taken lightly. The author becomes a saint to some and a sinner to others—and the feedback from the “others” is felt more acutely than the feedback from the “some”! Ultimately, the decision to write a new book comes from the deep feeling that an author can make a positive contribution to the field, and that it is ultimately time well invested. It also helps that there are people supporting the author both personally and professionally. The first person to thank must be Jennifer Welchans of Flat World Knowledge. I have known Jen for years; indeed, she was instrumental in getting me to write my first academic book, a math review book that is still available through another publisher. We reconnected recently, and I learned that she was working for a new publisher with some interesting publishing ideas. With her urging, the editorial director and I got together, first by phone and then in person, to discuss this project. With all the enthusiasm and ideas that Flat World Knowledge brought to the table, it was difficult not to sign on and write this book. So thanks, Jen—again. Hopefully this won’t be the last book we do together. Thanks also to Michael Boezi, editorial director at Flat World Knowledge, for his enthusiastic support. Jenn Yee, project manager at Flat World Knowledge, did a great job of managing the project and all of its pieces—manuscript, answers to exercises, art, reviews, revisions, and all the other things required to put a project like this together. Vanessa Gennarelli did a great job of filling in when necessary (although Jenn should know better than to take a vacation during a project). Kudos to the technology team at Flat World Knowledge, who had the ultimate job of getting this book out: Brian Brennan, David Link, Christopher Loncar, Jessica Carey, Jon Gottfried, Jon Williams, Katie Damo, Keith Avery, Mike Shnaydman, Po Ki Chui, and Ryan Lowe. I would also like to thank the production team at Scribe Inc., including Stacy Claxton, Chrissy Chimi, Melissa Tarrao, and Kevin McDermott. This book would not exist without any of these people. Thanks to Mary Grodek and Bill Reiter of Cleveland State University’s Marketing Department for assistance in obtaining a needed photograph. A project like this benefits from the expertise of external reviewers. I would like to thank the following people for their very thoughtful evaluation of the manuscript at several stages: Sam Abbas, Palomar College; Bal Barot, Lake Michigan College; Sherri Borowicz, Dakota College of Bottineau; Ken Capps, Central Florida Community College; Troy Cayou, Coconino Community College; Robert Clark, Lourdes College; Daniel Cole, Central Piedmont Community College; Jo Conceicao, Metropolitan Community College; Bernadette Corbett, Metropolitan Community College; James Fisher, Imperial Valley College; Julie Klare, Gwinnett Technical College; Karen Marshall, Bridgewater College; Tchao Podona, Miami-Dade College; Kenneth Rodriguez, California State University–Dominguez Hills; Mary Sohn, Florida Institute of Technology; Angie Spencer, Greenville Technical College; Charles Taylor, Pomona College; Susan T. Thomas, The University of Texas at San Antonio; Linda Waldman, Cerritos College Thanks especially to ANSR Source, who performed accuracy checks on various parts of the text. Should any inaccuracies remain, they are the responsibility of the author. I hope that readers will let me know if they find any; one of the beauties of the Flat World process is the ability to update the textbook quickly, so that it will be an even better book tomorrow. I am looking forward to seeing how the Flat World Knowledge model works with this book, and I thank all the adopters and users in advance for their help in making it a better text. David W. Ball, February 2011

]]> 4773 0 2 0 https://pressbooks.bccampus.ca/chem1114langaracollege/front-matter/about-the-authors/ Fri, 22 Jun 2018 18:10:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/?post_type=front-matter&p=4812 Shirley Wacowich-Sgarbi, PhD, Langara College - Vancouver, BC Dr. Shirley Wacowich-Sgarbi is a chemistry instructor at Langara College in Vancouver, British Columbia.  She received her B.Sc. (Honours) from McGill University majoring in chemistry with the bio-organic option.  She then earned her PhD in organic chemistry from the University of Alberta doing carbohydrate research while mentored by Dr. Dave Bundle.  After doing postdoctoral research on cancer vaccines with Dr. Chi-Huey Wong at the Scripps Research Institute in San Diego, she worked a few years in a biotech company in San Diego and then a few years in a biotech company in Vancouver, BC.  In 2005, she made the leap into teaching chemistry at Langara College, where she currently teaches introductory chemistry, general chemistry and organic chemistry.

Paul Flowers, PhD, University of North Carolina - Pembroke

Dr. Paul Flowers earned a BS in Chemistry from St. Andrews Presbyterian College in 1983 and a PhD in Analytical Chemistry from the University of Tennessee in 1988. After a one-year postdoctoral appointment at Los Alamos National Laboratory, he joined the University of North Carolina–Pembroke in the fall of 1989. Dr. Flowers teaches courses in general and analytical chemistry, and conducts experimental research involving the development of new devices and methods for microscale chemical analysis. Klaus Theopold, PhD, University of Delaware Dr. Klaus Theopold (born in Berlin, Germany) received his Vordiplom from the Universität Hamburg in 1977. He then decided to pursue his graduate studies in the United States, where he received his PhD in inorganic chemistry from UC Berkeley in 1982. After a year of postdoctoral research at MIT, he joined the faculty at Cornell University. In 1990, he moved to the University of Delaware, where he is a Professor in the Department of Chemistry and Biochemistry and serves as an Associate Director of the University’s Center for Catalytic Science and Technology. Dr. Theopold regularly teaches graduate courses in inorganic and organometallic chemistry as well as General Chemistry. Richard Langley, PhD, Stephen F. Austin State University Dr. Richard Langley earned BS degrees in Chemistry and Mineralogy from Miami University of Ohio in the early 1970s and went on to receive his PhD in Chemistry from the University of Nebraska in 1977. After a postdoctoral fellowship at the Arizona State University Center for Solid State Studies, Dr. Langley taught in the University of Wisconsin system and participated in research at Argonne National Laboratory. Moving to Stephen F. Austin State University in 1982, Dr. Langley today serves as Professor of Chemistry. His areas of specialization are solid state chemistry, synthetic inorganic chemistry, fluorine chemistry, and chemical education. Jessie Key, PhD, Vancouver Island University - Nanaimo, BC  Dr. Jessie Key is a professor of chemistry at Vancouver Island University in Nanaimo, British Columbia. He received his Ph.D from the University of Alberta in Edmonton, Alberta, and his B.Sc (Hons.) from Thompson Rivers University in Kamloops, British Columbia. Jessie's main area of research expertise is chemical biology; with a focus on fluorophore synthesis, cellular labelling and bioassays. He currently teaches general chemistry and organic chemistry at Vancouver Island University, and does research on the use of technology in chemical education. David Ball, PhD, Cleveland State University  Dr. Ball is a professor of chemistry at Cleveland State University in Ohio. He earned his PhD from Rice University in Houston, Texas. His specialty is physical chemistry, which he teaches at the undergraduate and graduate levels. About 50 percent of his teaching is in general chemistry: chemistry for nonscience majors, GOB, and general chemistry for science and engineering majors. In addition to this text, he is the author of a math review book for general chemistry students, a physical chemistry textbook with accompanying student and instructor solutions manuals, and two books on spectroscopy (published by SPIE Press). He is coauthor of a general chemistry textbook (with Dan Reger and Scott Goode), whose third edition was published in January 2009. His publication list has over 180 items, roughly evenly distributed between research papers and articles of educational interest. Timothy Soderberg, PhD, University of Minnesota

Dr. Soderbergh is an associate professor at the University of Minnesota.  He earned his PhD from the University of Utah.  His main area of research expertise include bioorganic chemistry, cloning and genetic engineering, protein expression and purification and enzymatic assays.

]]> 4812 0 3 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-4-chemical-reactions-and-equations/ Thu, 12 Apr 2018 03:52:06 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-4-chemical-reactions-and-equations/

Chapter Topics

• Writing and Balancing Chemical Equations
• Precipitation Reactions
• Acid-Base Reactions
• Oxidation-Reduction Reactions

]]> 2162 0 6 0 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-8-electronic-structure/ Thu, 12 Apr 2018 03:52:41 +0000 https://pressbooks.bccampus.ca/chem1114langaracollege/part/chapter-8-electronic-structure/

Chapter Topics

• Electromagnetic Energy
• Quantization of the Energy of Electrons
• Development of Quantum Theory
• Electronic Structure of Atoms
• Periodic Trends

]]> 2362 0 7 0