{"id":2179,"date":"2018-04-11T23:52:09","date_gmt":"2018-04-12T03:52:09","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/end-of-chapter-material-3\/"},"modified":"2019-05-14T16:25:19","modified_gmt":"2019-05-14T20:25:19","slug":"end-of-chapter-material-3","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/end-of-chapter-material-3\/","title":{"raw":"6.5 End of Chapter Problems","rendered":"6.5 End of Chapter Problems"},"content":{"raw":"1. Chemical equations can also be used to represent physical processes. Write a chemical equation for the boiling of water, including the proper phase labels.\r\n\r\n2. Chemical equations can also be used to represent physical processes. Write a chemical equation for the freezing of water, including the proper phase labels.\r\n\r\n3. Explain why the following chemical equation\u00a0should not be considered a proper chemical equation:\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">4 Na(s) +\u00a02 Cl<sub class=\"subscript\">2<\/sub>(g) $latex \\longrightarrow$\u00a04 NaCl(s)<\/span><\/span><\/p>\r\n<p class=\"para\">4. Does the following chemical reaction proceed as written? Why or why not?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">3 Zn(s) +\u00a02 Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>(aq) $latex \\longrightarrow$\u00a03 Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a02 Al(s)<\/span><\/span><\/p>\r\n<p class=\"para\">5. Explain what is wrong with this double-replacement reaction.<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">NaCl(aq) +\u00a0KBr(aq) $latex \\longrightarrow$\u00a0NaK(aq) +\u00a0ClBr(aq)<\/span><\/span><\/p>\r\n6. Predict the products of and balance this double-replacement reaction.\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$\u00a0?<\/span><\/span><\/p>\r\n7. Write the complete and net ionic equations for this double-replacement reaction.\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">BaCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) $latex \\longrightarrow$\u00a0?<\/span><\/span><\/p>\r\n8. Write the complete and net ionic equations for this double-replacement reaction.\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$\u00a0?<\/span><\/span><\/p>\r\n9. Identify the spectator ions in this reaction. What is the net ionic equation?\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">NaCl(aq) +\u00a0KBr(aq) $latex \\longrightarrow$\u00a0NaBr(aq) +\u00a0KCl(aq)<\/span><\/span><\/p>\r\n10. Can a reaction be a composition reaction and a redox reaction at the same time? Give an example to support your answer.\r\n\r\n11. Can a reaction be a decomposition reaction and a redox reaction at the same time? Give an example to support your answer.\r\n\r\n12. Why is CH<sub class=\"subscript\">4<\/sub> not normally considered an acid?\r\n\r\n13. What are the oxidation numbers of the nitrogen atoms in these substances?\r\n\r\n<\/div>\r\na) \u00a0N<sub class=\"subscript\">2 \u00a0 \u00a0 \u00a0<\/sub>b) \u00a0NH<sub class=\"subscript\">3 \u00a0 \u00a0 \u00a0<\/sub>c) \u00a0NO \u00a0 \u00a0 \u00a0d) \u00a0N<sub class=\"subscript\">2<\/sub>O\r\n\r\ne) \u00a0NO<sub class=\"subscript\">2 \u00a0 \u00a0 \u00a0<\/sub>f) \u00a0N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4 \u00a0 \u00a0 \u00a0<\/sub>g) \u00a0N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">5 \u00a0 \u00a0 \u00a0<\/sub>h) \u00a0NaNO<sub class=\"subscript\">3<\/sub>\r\n<div class=\"question\">14. \u00a0Disproportion 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.<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 CuCl(aq) $latex \\longrightarrow$\u00a0CuCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Cu(s)<\/span><\/span><\/p>\r\n\r\n<\/div>\r\n<p class=\"Questions\">15. Write the unbalanced chemical equation for each of the following reactions.<\/p>\r\n<p class=\"Indentpoints\">a) solid mercury(II) oxide decomposes to produce liquid mercury metal and gaseous oxygen.<\/p>\r\n<p class=\"Indentpoints\">b) Solid zinc metal reacts with hydrochloric acid to produce zinc chloride dissolved in water and hydrogen gas.<\/p>\r\n<p class=\"Indentpoints\">c) Propane (C<sub>3<\/sub>H<sub>8<\/sub>) gas reacts with oxygen in air to produce gaseous carbon dioxide and water vapor.<\/p>\r\n<p class=\"Indentpoints\">d) Solid ammonium nitrate can be produced by bubbling ammonia gas through nitric acid solution.<\/p>\r\n<p class=\"Indentpoints\">e) Elemental boron can be produced by heating solid diboron trioxide with magnesium metal, also producing solid magnesium oxide as a by-product.<\/p>\r\n<p class=\"Questions\">16. Beneath each word equation, write the formula equation and balance it:<\/p>\r\n<p class=\"Indent\">a)<span>\u00a0<\/span>zinc + sulfuric acid<span>\u00a0 $latex \\longrightarrow$<\/span><span>\u00a0\u00a0<\/span>zinc sulfate + hydrogen<\/p>\r\n<p class=\"Indent\">b)<span>\u00a0<\/span>carbon + oxygen<span>\u00a0 $latex \\longrightarrow$<\/span><span>\u00a0\u00a0<\/span>carbon dioxide<\/p>\r\n<p class=\"Indent\">c)<span>\u00a0<\/span>hydrogen + oxygen<span>\u00a0 $latex \\longrightarrow$<\/span><span>\u00a0\u00a0<\/span>water<\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>aluminum + hydrochloric acid<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>aluminum chloride + hydrogen<\/span><\/p>\r\n<p class=\"Indent\">e)<span>\u00a0<\/span>chromium + oxygen<span>\u00a0 $latex \\longrightarrow$<\/span><span>\u00a0<\/span>chromium(III) oxide<\/p>\r\n<p class=\"Indent\">f)<span>\u00a0<\/span>potassium + water<span>\u00a0 $latex \\longrightarrow$<\/span><span>\u00a0\u00a0<\/span>potassium hydroxide + hydrogen<\/p>\r\n<p class=\"Indent\">g)<span>\u00a0<\/span>copper(II) oxide + hydrochloric acid<span>\u00a0$latex \\longrightarrow$<\/span><span>\u00a0\u00a0<\/span>copper(II) chloride + water<span>\u00a0 <\/span><\/p>\r\n<p class=\"Indent\">h)<span>\u00a0<\/span>sodium hydrogen carbonate + nitric acid<span>\u00a0$latex \\longrightarrow$<\/span><span>\u00a0\u00a0<\/span>sodium nitrate + water + carbon dioxide<\/p>\r\n<p class=\"Questions\">17. Balance the following equations:<\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">a) H<sub>3<\/sub>PO<sub>4\u00a0 <\/sub>+\u00a0 CaO $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">b) NH<sub>3<\/sub>\u00a0 +\u00a0 O<sub>2<\/sub> $latex \\longrightarrow$<sub>\u00a0<\/sub><\/span><span lang=\"PT-BR\">NO<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">c) Cl<sub>2<\/sub>O<sub>7<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O $latex \\longrightarrow$<\/span><span>\u00a0<\/span><span lang=\"PT-BR\">HClO<sub>4<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">d) Mg<sub>3<\/sub>N<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Mg(OH)<sub>2<\/sub>\u00a0 +\u00a0 NH<sub>3<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">e) FeSO<sub>4\u00a0<\/sub><\/span><span>$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Fe<sub>2<\/sub>O<sub>3<\/sub>\u00a0 +\u00a0 SO<sub>2<\/sub>\u00a0 +\u00a0 O<sub>2<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<\/span><span lang=\"PL\">P<sub>4<\/sub>\u00a0 +\u00a0 Cl<sub>2<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PL\">PCl<sub>3<\/sub><\/span><span lang=\"PT-BR\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PL\">g) <\/span><span lang=\"PT-BR\">MnO<sub>2<\/sub>\u00a0 +\u00a0 C<\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Mn\u00a0 +\u00a0 CO<sub>2<\/sub><\/span><span lang=\"PL\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">h) Na<sub>2<\/sub>O<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">NaOH\u00a0 \u00a0+\u00a0 O<sub>2<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">i) CaH<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Ca(OH)<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">j)<\/span>NaHCO<sub>3<\/sub><span>\u00a0$latex \\longrightarrow$\u00a0<\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u00a0 +\u00a0 CO<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<span lang=\"PT-BR\"><\/span><\/p>\r\n<p class=\"Questions\">18. Balance the following equations:<\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">a)<span>\u00a0 <\/span>Mg<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>MgO<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>KClO<sub>3<\/sub><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>KCl<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">c)<span>\u00a0 <\/span>Fe<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>Mg<span>\u00a0<\/span>+<span>\u00a0 <\/span>HCl<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>MgCl<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>Na<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>NaOH<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<span>\u00a0 <\/span>N<sub>2<\/sub>+<span>\u00a0 <\/span>H<sub>2<\/sub><span>\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>NH<sub>3<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">g)<span>\u00a0 <\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u202210H<sub>2<\/sub>O<span>\u00a0\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">h)<span>\u00a0 <\/span>Fe<span>\u00a0<\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub>+<span>\u00a0 <\/span>H<sub>2<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">i)<span>\u00a0 <\/span>F<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>HF +<span>\u00a0 <\/span>O<sub>2<\/sub><\/span><\/p>\r\n<p class=\"Questions\">19. Balance the following chemical equations.<\/p>\r\n<p class=\"Indent\"><span lang=\"ES-MX\">a)<span>\u00a0\u00a0 <\/span>Ca<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">(PO<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">)<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0\u00a0<\/span>+<span>\u00a0 \u00a0<\/span>SiO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>C<span>\u00a0<\/span><span>\u00a0\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0\u00a0<\/span>P<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>CaSiO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0\u00a0<\/span>+<span>\u00a0 \u00a0<\/span>CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"ES-MX\">b)<span>\u00a0\u00a0 <\/span>C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">6<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>O<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0\u00a0<\/span>CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O <\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"ES-MX\">c)<span>\u00a0\u00a0 <\/span>Fe<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0\u00a0\u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0\u00a0<\/span>Fe<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>SO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"ES-MX\">d)<span>\u00a0\u00a0 <\/span>P<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">5<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>P<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"ES-MX\">e)<span>\u00a0\u00a0 <\/span>C<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>SO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span> $latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O<span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Questions\">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.<\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">a)<span>\u00a0 <\/span>NaCl(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>AgNO<sub>3<\/sub>(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>BaCl<sub>2<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>SO<sub>4<\/sub>(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">c)<span>\u00a0 <\/span>FeCl<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>NH<sub>4<\/sub>OH(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>K<sub>2<\/sub>CrO<sub>4<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0<\/span>Pb(NO<sub>3<\/sub>)<sub>2<\/sub>(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>KClO<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>MgCl<sub>2<\/sub>(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<span>\u00a0 <\/span>(NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>CaCl<sub>2<\/sub>(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">g)<span>\u00a0 <\/span>NaC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>BaCl<sub>2<\/sub>(aq)\u00a0$latex \\longrightarrow$<\/span><span lang=\"PT-BR\"><\/span><\/p>\r\n<p class=\"Questions\">21. Identify each of the following unbalanced reaction equations as belonging to one or more of the following categories: precipitation, acid-base, or redox.<\/p>\r\n&nbsp;\r\n<p class=\"Indent\">a)<span>\u00a0 <\/span>Fe<sub>(s)<\/sub>+ H<sub>2<\/sub>SO<sub>4(aq)<\/sub><span>\u00a0$latex \\longrightarrow$\u00a0<\/span>Fe<sub>3<\/sub>(SO<sub>4<\/sub>)<sub>2(aq)<\/sub>+ H<sub>2(g)<\/sub><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>HClO<sub>4(aq)<\/sub>+ RbOH<sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">RbClO<sub>4(aq)<\/sub>+ H<sub>2<\/sub>O<sub>(l)<\/sub><\/span><\/p>\r\n<p class=\"Indent\">c)<span>\u00a0 <\/span>Ca<sub>(s)<\/sub>+ O<sub>2(g)<\/sub><span>\u00a0$latex \\longrightarrow$\u00a0<\/span>CaO<sub>(s)<\/sub><\/p>\r\n<p class=\"Indent\"><span lang=\"ES-MX\">d)<span>\u00a0 <\/span>H<sub>2<\/sub><\/span><span lang=\"PT-BR\">SO<sub>4(aq)<\/sub>+ NaOH<sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Na<sub>2<\/sub>SO<sub>4(aq)<\/sub>+ H<sub>2<\/sub>O<sub>(l)<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>Pb(NO<sub>3<\/sub>)<sub>2(aq)<\/sub>+ Na<sub>2<\/sub>CO<sub>3(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">PbCO<sub>3(s)<\/sub>+ NaNO<sub>3(aq)<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<span>\u00a0 <\/span>K<sub>2<\/sub>SO<sub>4(aq)<\/sub>+ CaCl<sub>2(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">KCl<sub>(aq)<\/sub>+ CaSO<sub>4(s)<\/sub><\/span><\/p>\r\n<p class=\"Questions\">22. Classify the following unbalanced chemical reactions by as many methods as possible.<\/p>\r\n<p class=\"Indent\">a)<span>\u00a0 <\/span>I<sub>4<\/sub>O<sub>9(s)<\/sub><span>\u00a0$latex \\longrightarrow$\u00a0<\/span>I<sub>2<\/sub>O<sub>6(s)<\/sub>+ I<sub>2(s)<\/sub>+ O<sub>2(g)<\/sub><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>Mg<sub>(s)<\/sub>+ AgNO<sub>3(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Mg(NO<sub>3<\/sub>)<sub>2(aq)<\/sub>+ Ag<sub>(s)<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">c)<span>\u00a0 <\/span>SiCl<sub>4(l)<\/sub>+ Mg<sub>(s)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">MgCl<sub>2(aq)<\/sub>+ Si<sub>(s)<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>CuCl<sub>2(aq)<\/sub>+ AgNO<sub>3(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Cu(NO<sub>3<\/sub>)<sub>2(aq)<\/sub>+ AgCl<sub>(s)<\/sub><\/span><\/p>\r\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>Al<sub>(s)<\/sub>+ Br<sub>2(l)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">AlBr<sub>3(s)<\/sub><\/span><\/p>\r\n<span lang=\"ES-MX\">f)<span>\u00a0 <\/span>HBr<\/span><span lang=\"PT-BR\"><sub>(aq)<\/sub>+ NaOH<sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">NaBr<sub>(aq)<\/sub>+ H<sub>2<\/sub>O<sub>(l)<\/sub><\/span>\r\n<h2><\/h2>\r\n<h2>Answers<\/h2>\r\n1. H<sub class=\"subscript\">2<\/sub>O(\u2113) <span>$latex \\longrightarrow$<\/span>\u00a0H<sub class=\"subscript\">2<\/sub>O(g)\r\n\r\n2. H<sub class=\"subscript\">2<\/sub>O(\u2113) <span>$latex \\longrightarrow$<\/span>\u00a0H<sub class=\"subscript\">2<\/sub>O(s)\r\n\r\n3.\u00a0The coefficients are not in their lowest whole-number ratio.\r\n\r\n4. No; zinc is lower in the activity series than aluminum.\r\n\r\n5. In the products, the cation is pairing with the cation, and the anion is pairing with the anion.\r\n<p id=\"ball-ch04_s07_qs01_p22_ans\" class=\"para\">6.\u00a0<span class=\"informalequation\"><span class=\"mathphrase\">Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) <span>$latex \\longrightarrow$<\/span>\u00a0SrSO<sub class=\"subscript\">4<\/sub>(s) + 2 AgCl(s)\u00a0<\/span><\/span><\/p>\r\n<p class=\"para\">7. Complete ionic equation: Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) <span>$latex \\longrightarrow$<\/span>\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 AgCl(s)<\/p>\r\nNet ionic equation: The net ionic equation is the same as the complete ionic equation.\r\n\r\n8. Complete ionic equation:\u00a0Sr<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) <span>$latex \\longrightarrow$<\/span>\u00a0SrSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 AgCl(s)\r\n\r\nNet ionic equation: The net ionic equation is the same as the complete ionic equation.\r\n\r\n9. Each ion is a spectator ion; there is no overall net ionic equation.\r\n\r\n10. Yes; H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub>\u00a0<span>$latex \\longrightarrow$<\/span> 2 HCl (answers will vary)\r\n\r\n11. Yes; 2 HCl <span>$latex \\longrightarrow$<\/span>\u00a0H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> (answers will vary)\r\n\r\n12. It does not increase the H<sup class=\"superscript\">+<\/sup> ion concentration; it is not a compound of H<sup class=\"superscript\">+<\/sup>.\r\n\r\n13. a) \u00a00 \u00a0 \u00a0 \u00a0b) \u00a0\u22123 \u00a0 \u00a0 \u00a0c) \u00a0+2 \u00a0 \u00a0 \u00a0d) \u00a0+1\r\n\r\ne) \u00a0+4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f) +4 \u00a0 \u00a0 \u00a0 \u00a0g) \u00a0+5 \u00a0 \u00a0 \u00a0h) \u00a0+5\r\n\r\n14. Copper is disproportionating. Initially, its oxidation number is +1; in the products, its oxidation numbers are +2 and 0, respectively.\r\n<p class=\"Answers\"><span lang=\"PT-BR\">15.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0<\/span>HgO(s) $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Hg(l) + O<sub>2<\/sub>(g)\r\nb)<span>\u00a0 <\/span>Zn(s) + HCl(aq) $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">ZnCl<sub>2<\/sub>(aq) + H<sub>2<\/sub>(g)\r\nc)<span>\u00a0 <\/span>C<sub>3<\/sub>H<sub>8<\/sub>(g) + O<sub>2<\/sub>(g) $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(g)\r\nd)<span>\u00a0 <\/span>NH<sub>3<\/sub>(g) + HNO<sub>3<\/sub>(aq) $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">NH<sub>4<\/sub>NO<sub>3<\/sub>(s)\r\ne)<span>\u00a0 <\/span>B<sub>2<\/sub>O<sub>3<\/sub>(s) + Mg(s) $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">B(s) + MgO(s)<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">16.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0<\/span>Zn(s) + H<sub>2<\/sub>SO<sub>4<\/sub>(aq)<span>\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>ZnSO<sub>4<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>(g)\r\nb)<span>\u00a0 <\/span>C(s) + O<sub>2<\/sub>(g)\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">CO<sub>2<\/sub>(g)\r\nc)<span>\u00a0 <\/span>2H<sub>2<\/sub>(g) + O<sub>2<\/sub>(g)<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>2H<sub>2<\/sub>O(l)\r\nd)<span>\u00a0 <\/span>2Al(s) + 6HCl(aq)<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>2AlCl<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>3H<sub>2<\/sub>(g)\r\ne)<span>\u00a0 <\/span>4Cr(s) + 3O<sub>2<\/sub>(g)<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>2Cr<sub>2<\/sub>O<sub>3<\/sub>(s)\r\nf)<span>\u00a0 <\/span>2K(s) + 2H<sub>2<\/sub>O(aq)<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2KOH(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>(g)\r\ng)<span>\u00a0 <\/span>CuO(s) + 2HCl(aq)<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>CuCl<sub>2<\/sub>(aq) + H<sub>2<\/sub>O(l)\r\nh)<span>\u00a0 <\/span>NaHCO<sub>3<\/sub>(aq) + HNO<sub>3<\/sub>(aq)<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>NaNO<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O(l)<span>\u00a0<\/span>+\u00a0CO<sub>2<\/sub>(g)<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">17.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0 <\/span>2H<sub>3<\/sub>PO<sub>4<\/sub>\u00a0 +\u00a0 3CaO $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0 +\u00a0 3H<sub>2<\/sub>O\r\nb)<span>\u00a0 <\/span>4NH<sub>3<\/sub>\u00a0 +\u00a0 7O<sub>2<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">4NO<sub>2<\/sub>\u00a0 +\u00a0 6H<sub>2<\/sub>O\r\nc)<span>\u00a0 <\/span>Cl<sub>2<\/sub>O<sub>7<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">2HClO<sub>4<\/sub>\r\nd)<span>\u00a0 <\/span>Mg<sub>3<\/sub>N<sub>2<\/sub>\u00a0 +\u00a0 6H<sub>2<\/sub>O\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">3Mg(OH)<sub>2<\/sub>\u00a0 +\u00a0 2NH<sub>3<\/sub>\r\ne)<span>\u00a0 <\/span>4FeSO<sub>4<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">2Fe<sub>2<\/sub>O<sub>3<\/sub>\u00a0 +\u00a0 4SO<sub>2<\/sub>\u00a0 +\u00a0 O<sub>2<\/sub>\r\nf<\/span><span lang=\"PL\">)<span>\u00a0 <\/span>P<sub>4<\/sub>+ 6Cl<sub>2<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PL\">4PCl<sub>3<\/sub>\r\ng)<span>\u00a0 <\/span>MnO<sub>2<\/sub>\u00a0 +\u00a0 C $latex \\longrightarrow$\u00a0<\/span><span lang=\"PL\">Mn\u00a0 +\u00a0 CO<sub>2<\/sub>\r\n<\/span><span lang=\"PT-BR\">h)<span>\u00a0 <\/span>2Na<sub>2<\/sub>O<sub>2<\/sub>\u00a0 +\u00a0 2H<sub>2<\/sub>O\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">4NaOH\u00a0 +\u00a0 O<sub>2<\/sub>\r\ni)<span>\u00a0 <\/span>CaH<sub>2<\/sub>\u00a0 +\u00a0 2H<sub>2<\/sub>O $latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Ca(OH)<sub>2<\/sub>\u00a0 +\u00a0 2H<sub>2<\/sub>\r\n<\/span><span lang=\"PL\">j)<span>\u00a0 <\/span>2NaHCO<sub>3<\/sub><\/span><span lang=\"PL\"><span>\u00a0$latex \\longrightarrow$ <\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u00a0 +\u00a0 CO<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">18.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0<\/span>2Mg<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><span>\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2MgO\r\nb)<span>\u00a0 <\/span>2KClO<sub>3<\/sub><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2KCl<span>\u00a0 <\/span>+<span>\u00a0 <\/span>3O<sub>2<\/sub>\r\nc)<span>\u00a0 <\/span>3Fe<span>\u00a0<\/span>+<span>\u00a0 <\/span>2O<sub>2<\/sub><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub>\r\nd)<span>\u00a0 <\/span>Mg<span>\u00a0<\/span>+<span>\u00a0 <\/span>2HCl<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>MgCl<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>\r\ne)<span>\u00a0 <\/span>2Na<span>\u00a0<\/span>+<span>\u00a0 <\/span>2H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2NaOH<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>\r\nf)<span>\u00a0 <\/span>N<sub>2<\/sub>+<span>\u00a0 <\/span>3H<sub>2<\/sub><span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2NH<sub>3<\/sub>\r\ng)<span>\u00a0 <\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u2022 10H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>+<span>\u00a0<\/span>10H<sub>2<\/sub>O\r\nh)<span>\u00a0 <\/span>3Fe<span>\u00a0\u00a0<\/span>+<span>\u00a0 <\/span>4H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub>+<span>\u00a0 <\/span>4H<sub>2<\/sub>\r\ni)<span>\u00a0 <\/span>2F<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>2H<sub>2<\/sub>O<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>4HF +<span>\u00a0 <\/span>O<sub>2<\/sub><\/span><\/p>\r\n<p class=\"Answers\">19. Balance the following chemical equations.<\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">a)<span>\u00a0 <\/span>2 Ca<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">(PO<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">)<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0 <\/span>6 SiO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0 <\/span>5 C<span>\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>P<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0<\/span>6 CaSiO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0<\/span>5 CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">b)<span>\u00a0 <\/span>3 C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">6<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>5 O<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 $latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>6 CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>9 H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O <\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">c)<span>\u00a0 <\/span>48 Fe<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>35 S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>72 Fe<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>64 SO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">d)<span>\u00a0 <\/span>24 P<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">5<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0<\/span>64O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>15 S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>16 P<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"ES-MX\">e)<span>\u00a0 <\/span>16 C<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>56 SO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0$latex \\longrightarrow$<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0\u00a0<\/span>24 C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 <\/span>+<span>\u00a0 <\/span>16 H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O<span>\u00a0 <\/span>+<span>\u00a0 <\/span>7 S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">20.<span>\u00a0 <\/span>a) NaCl<sub>(aq)<\/sub>+ AgNO<sub>3(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">AgCl<sub>(s)<\/sub>+ NaNO<sub>3(aq)<\/sub>\r\nNa<sup>+<\/sup><sub>(aq)<\/sub>+ Cl<sup>-<\/sup><sub>(aq)<\/sub>+ Ag<sup>+<\/sup><sub>(aq)<\/sub>+ NO<sub>3<\/sub><sup>-<\/sup><sub>(aq)<\/sub>\r\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $latex \\longrightarrow$\u00a0<\/span><\/span><span lang=\"PT-BR\">AgCl<sub>(s)<\/sub>+ Na<sup>+<\/sup><sub>(aq)<\/sub>+ NO<sub>3<\/sub><sup>-<\/sup><sub>(aq)<\/sub>\r\nCl<sup>-<\/sup><sub>(aq)<\/sub>+ Ag<sup>+<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">AgCl<sub>(s)<\/sub><\/span><\/p>\r\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">b) BaCl<sub>2(aq)<\/sub>+ H<sub>2<\/sub>SO<sub>4(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">BaSO<sub>4(s)<\/sub>+ 2HCl<sub>(aq)<\/sub>\r\nBa<sup>2+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>-<\/sup><sub>(aq)<\/sub>+ 2H<sup>+<\/sup><sub>(aq)<\/sub>+ SO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>\r\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $latex \\longrightarrow$\u00a0<\/span><\/span><span lang=\"PT-BR\">BaSO<sub>4(s)<\/sub>+ 2H<sup>+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>-<\/sup><sub>(aq)<\/sub>\r\nBa<sup>2+<\/sup><sub>(aq)<\/sub>+ SO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">BaSO<sub>4(s)<\/sub><\/span><\/p>\r\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">c) FeCl<sub>3(aq)<\/sub>+ 3NH<sub>4<\/sub>OH<sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Fe(OH)<sub>3(s)<\/sub>+ 3NH<sub>4<\/sub>Cl<sub>(aq)<\/sub>\r\nFe<sup>3+<\/sup><sub>(aq)<\/sub>+ 3Cl<sup>-<\/sup><sub>(aq)<\/sub>+ 3 NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ 3OH<sup>-<\/sup><sub>(aq)<\/sub>\r\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $latex \\longrightarrow$\u00a0<\/span><\/span><span lang=\"PT-BR\">Fe(OH)<sub>3(s)<\/sub>+ 3 NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ 3Cl<sup>-<\/sup><sub>(aq)<\/sub>\r\nFe<sup>3+<\/sup><sub>(aq)<\/sub>+ 3OH<sup>-<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">Fe(OH)<sub>3(s)<\/sub><\/span><\/p>\r\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">d) K<sub>2<\/sub>CrO<sub>4(aq)<\/sub>+ Pb(NO<sub>3<\/sub>)<sub>2(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">PbCrO<sub>4(s)<\/sub>+ 2KNO<sub>3(aq)<\/sub>\r\n2K<sup>+<\/sup><sub>(aq)<\/sub>+ CrO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Pb<sup>2+<\/sup><sub>(aq)<\/sub>+ 2NO<sub>3<\/sub><sup>-<\/sup><sub>(aq)<\/sub>\r\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $latex \\longrightarrow$\u00a0<\/span><\/span><span lang=\"PT-BR\">PbCrO<sub>4(s)<\/sub>+ 2K<sup>+<\/sup><sub>(aq)<\/sub>+ 2NO<sub>3<\/sub><sup>-<\/sup><sub>(aq)<\/sub>\r\nCrO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Pb<sup>2+<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">PbCrO<sub>4(s)<\/sub><\/span><\/p>\r\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">e) KClO<sub>3(aq)<\/sub>+ MgCl<sub>2(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">N.R.<\/span><\/p>\r\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">f) (NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3(aq)<\/sub>+ CaCl<sub>2(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">CaCO<sub>3(s)<\/sub>+ 2NH<sub>4<\/sub>Cl<sub>(aq)<\/sub>\r\n2NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ CO<sub>3<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Ca<sup>2+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>-<\/sup><sub>(aq)<\/sub>\r\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $latex \\longrightarrow$\u00a0<\/span><\/span><span lang=\"PT-BR\">CaCO<sub>3(s)<\/sub>+ 2NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>-<\/sup><sub>(aq)<\/sub>\r\nCO<sub>3<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Ca<sup>2+<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">CaCO<sub>3(s)<\/sub><\/span><\/p>\r\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">g) NaC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2(aq)<\/sub>+ BaCl<sub>2(aq)<\/sub><\/span><span>\u00a0$latex \\longrightarrow$\u00a0<\/span><span lang=\"PT-BR\">N.R.<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">21.<span>\u00a0 <\/span>a) redox \u00a0 \u00a0 \u00a0b) acid-base \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0c) redox\r\nd) acid-base \u00a0 \u00a0 \u00a0\u00a0e) precipitation \u00a0 \u00a0 \u00a0f) precipitation<\/span><\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">22.<span>\u00a0 <\/span><\/span>a) decomposition, which is a type of redox<\/p>\r\n<p class=\"Answers\">b) single-replacement, which is a type of redox<\/p>\r\n<p class=\"Answers\">c) single-replacement, which is a type of redox<\/p>\r\n<p class=\"Answers\">d) precipitation, which is a type of double replacement (but not redox!)<\/p>\r\n<p class=\"Answers\"><span lang=\"PT-BR\">e) composition (sometimes also called a <em class=\"emphasis\">combination reaction<\/em> or a <em class=\"emphasis\">synthesis reaction<\/em>),\u00a0which is a type of redox<\/span><\/p>\r\nf) acid-base,\u00a0which is a type of double replacement (but not redox!)","rendered":"<p>1. Chemical equations can also be used to represent physical processes. Write a chemical equation for the boiling of water, including the proper phase labels.<\/p>\n<p>2. Chemical equations can also be used to represent physical processes. Write a chemical equation for the freezing of water, including the proper phase labels.<\/p>\n<p>3. Explain why the following chemical equation\u00a0should not be considered a proper chemical equation:<\/p>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">4 Na(s) +\u00a02 Cl<sub class=\"subscript\">2<\/sub>(g) [latex]\\longrightarrow[\/latex]\u00a04 NaCl(s)<\/span><\/span><\/p>\n<p class=\"para\">4. Does the following chemical reaction proceed as written? Why or why not?<\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">3 Zn(s) +\u00a02 Al(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>(aq) [latex]\\longrightarrow[\/latex]\u00a03 Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a02 Al(s)<\/span><\/span><\/p>\n<p class=\"para\">5. Explain what is wrong with this double-replacement reaction.<\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">NaCl(aq) +\u00a0KBr(aq) [latex]\\longrightarrow[\/latex]\u00a0NaK(aq) +\u00a0ClBr(aq)<\/span><\/span><\/p>\n<p>6. Predict the products of and balance this double-replacement reaction.<\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex]\u00a0?<\/span><\/span><\/p>\n<p>7. Write the complete and net ionic equations for this double-replacement reaction.<\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">BaCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) [latex]\\longrightarrow[\/latex]\u00a0?<\/span><\/span><\/p>\n<p>8. Write the complete and net ionic equations for this double-replacement reaction.<\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex]\u00a0?<\/span><\/span><\/p>\n<p>9. Identify the spectator ions in this reaction. What is the net ionic equation?<\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">NaCl(aq) +\u00a0KBr(aq) [latex]\\longrightarrow[\/latex]\u00a0NaBr(aq) +\u00a0KCl(aq)<\/span><\/span><\/p>\n<p>10. Can a reaction be a composition reaction and a redox reaction at the same time? Give an example to support your answer.<\/p>\n<p>11. Can a reaction be a decomposition reaction and a redox reaction at the same time? Give an example to support your answer.<\/p>\n<p>12. Why is CH<sub class=\"subscript\">4<\/sub> not normally considered an acid?<\/p>\n<p>13. What are the oxidation numbers of the nitrogen atoms in these substances?<\/p>\n<\/div>\n<p>a) \u00a0N<sub class=\"subscript\">2 \u00a0 \u00a0 \u00a0<\/sub>b) \u00a0NH<sub class=\"subscript\">3 \u00a0 \u00a0 \u00a0<\/sub>c) \u00a0NO \u00a0 \u00a0 \u00a0d) \u00a0N<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>e) \u00a0NO<sub class=\"subscript\">2 \u00a0 \u00a0 \u00a0<\/sub>f) \u00a0N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4 \u00a0 \u00a0 \u00a0<\/sub>g) \u00a0N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">5 \u00a0 \u00a0 \u00a0<\/sub>h) \u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\n<div class=\"question\">14. \u00a0Disproportion 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.<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 CuCl(aq) [latex]\\longrightarrow[\/latex]\u00a0CuCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Cu(s)<\/span><\/span><\/p>\n<\/div>\n<p class=\"Questions\">15. Write the unbalanced chemical equation for each of the following reactions.<\/p>\n<p class=\"Indentpoints\">a) solid mercury(II) oxide decomposes to produce liquid mercury metal and gaseous oxygen.<\/p>\n<p class=\"Indentpoints\">b) Solid zinc metal reacts with hydrochloric acid to produce zinc chloride dissolved in water and hydrogen gas.<\/p>\n<p class=\"Indentpoints\">c) Propane (C<sub>3<\/sub>H<sub>8<\/sub>) gas reacts with oxygen in air to produce gaseous carbon dioxide and water vapor.<\/p>\n<p class=\"Indentpoints\">d) Solid ammonium nitrate can be produced by bubbling ammonia gas through nitric acid solution.<\/p>\n<p class=\"Indentpoints\">e) Elemental boron can be produced by heating solid diboron trioxide with magnesium metal, also producing solid magnesium oxide as a by-product.<\/p>\n<p class=\"Questions\">16. Beneath each word equation, write the formula equation and balance it:<\/p>\n<p class=\"Indent\">a)<span>\u00a0<\/span>zinc + sulfuric acid<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><span>\u00a0\u00a0<\/span>zinc sulfate + hydrogen<\/p>\n<p class=\"Indent\">b)<span>\u00a0<\/span>carbon + oxygen<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><span>\u00a0\u00a0<\/span>carbon dioxide<\/p>\n<p class=\"Indent\">c)<span>\u00a0<\/span>hydrogen + oxygen<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><span>\u00a0\u00a0<\/span>water<\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>aluminum + hydrochloric acid<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>aluminum chloride + hydrogen<\/span><\/p>\n<p class=\"Indent\">e)<span>\u00a0<\/span>chromium + oxygen<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><span>\u00a0<\/span>chromium(III) oxide<\/p>\n<p class=\"Indent\">f)<span>\u00a0<\/span>potassium + water<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><span>\u00a0\u00a0<\/span>potassium hydroxide + hydrogen<\/p>\n<p class=\"Indent\">g)<span>\u00a0<\/span>copper(II) oxide + hydrochloric acid<span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><span>\u00a0\u00a0<\/span>copper(II) chloride + water<span>\u00a0 <\/span><\/p>\n<p class=\"Indent\">h)<span>\u00a0<\/span>sodium hydrogen carbonate + nitric acid<span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><span>\u00a0\u00a0<\/span>sodium nitrate + water + carbon dioxide<\/p>\n<p class=\"Questions\">17. Balance the following equations:<\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">a) H<sub>3<\/sub>PO<sub>4\u00a0 <\/sub>+\u00a0 CaO [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">b) NH<sub>3<\/sub>\u00a0 +\u00a0 O<sub>2<\/sub> [latex]\\longrightarrow[\/latex]<sub>\u00a0<\/sub><\/span><span lang=\"PT-BR\">NO<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">c) Cl<sub>2<\/sub>O<sub>7<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O [latex]\\longrightarrow[\/latex]<\/span><span>\u00a0<\/span><span lang=\"PT-BR\">HClO<sub>4<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">d) Mg<sub>3<\/sub>N<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Mg(OH)<sub>2<\/sub>\u00a0 +\u00a0 NH<sub>3<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">e) FeSO<sub>4\u00a0<\/sub><\/span><span>[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Fe<sub>2<\/sub>O<sub>3<\/sub>\u00a0 +\u00a0 SO<sub>2<\/sub>\u00a0 +\u00a0 O<sub>2<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<\/span><span lang=\"PL\">P<sub>4<\/sub>\u00a0 +\u00a0 Cl<sub>2<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PL\">PCl<sub>3<\/sub><\/span><span lang=\"PT-BR\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PL\">g) <\/span><span lang=\"PT-BR\">MnO<sub>2<\/sub>\u00a0 +\u00a0 C<\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Mn\u00a0 +\u00a0 CO<sub>2<\/sub><\/span><span lang=\"PL\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">h) Na<sub>2<\/sub>O<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">NaOH\u00a0 \u00a0+\u00a0 O<sub>2<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">i) CaH<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Ca(OH)<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">j)<\/span>NaHCO<sub>3<\/sub><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u00a0 +\u00a0 CO<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<span lang=\"PT-BR\"><\/span><\/p>\n<p class=\"Questions\">18. Balance the following equations:<\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">a)<span>\u00a0 <\/span>Mg<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>MgO<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>KClO<sub>3<\/sub><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>KCl<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">c)<span>\u00a0 <\/span>Fe<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>Mg<span>\u00a0<\/span>+<span>\u00a0 <\/span>HCl<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>MgCl<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>Na<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>NaOH<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<span>\u00a0 <\/span>N<sub>2<\/sub>+<span>\u00a0 <\/span>H<sub>2<\/sub><span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>NH<sub>3<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">g)<span>\u00a0 <\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u202210H<sub>2<\/sub>O<span>\u00a0\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">h)<span>\u00a0 <\/span>Fe<span>\u00a0<\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub>+<span>\u00a0 <\/span>H<sub>2<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">i)<span>\u00a0 <\/span>F<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>HF +<span>\u00a0 <\/span>O<sub>2<\/sub><\/span><\/p>\n<p class=\"Questions\">19. Balance the following chemical equations.<\/p>\n<p class=\"Indent\"><span lang=\"ES-MX\">a)<span>\u00a0\u00a0 <\/span>Ca<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">(PO<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">)<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0\u00a0<\/span>+<span>\u00a0 \u00a0<\/span>SiO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>C<span>\u00a0<\/span><span>\u00a0\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0\u00a0<\/span>P<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>CaSiO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0\u00a0<\/span>+<span>\u00a0 \u00a0<\/span>CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"ES-MX\">b)<span>\u00a0\u00a0 <\/span>C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">6<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>O<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0\u00a0<\/span>CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O <\/span><\/p>\n<p class=\"Indent\"><span lang=\"ES-MX\">c)<span>\u00a0\u00a0 <\/span>Fe<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0\u00a0\u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0\u00a0<\/span>Fe<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>SO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"ES-MX\">d)<span>\u00a0\u00a0 <\/span>P<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">5<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>P<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"ES-MX\">e)<span>\u00a0\u00a0 <\/span>C<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0\u00a0<\/span>SO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span> [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O<span>\u00a0 \u00a0<\/span>+<span>\u00a0 \u00a0<\/span>S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Questions\">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.<\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">a)<span>\u00a0 <\/span>NaCl(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>AgNO<sub>3<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>BaCl<sub>2<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>SO<sub>4<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">c)<span>\u00a0 <\/span>FeCl<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>NH<sub>4<\/sub>OH(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>K<sub>2<\/sub>CrO<sub>4<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0<\/span>Pb(NO<sub>3<\/sub>)<sub>2<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>KClO<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>MgCl<sub>2<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<span>\u00a0 <\/span>(NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>CaCl<sub>2<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">g)<span>\u00a0 <\/span>NaC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>BaCl<sub>2<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex]<\/span><span lang=\"PT-BR\"><\/span><\/p>\n<p class=\"Questions\">21. Identify each of the following unbalanced reaction equations as belonging to one or more of the following categories: precipitation, acid-base, or redox.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"Indent\">a)<span>\u00a0 <\/span>Fe<sub>(s)<\/sub>+ H<sub>2<\/sub>SO<sub>4(aq)<\/sub><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span>Fe<sub>3<\/sub>(SO<sub>4<\/sub>)<sub>2(aq)<\/sub>+ H<sub>2(g)<\/sub><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>HClO<sub>4(aq)<\/sub>+ RbOH<sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">RbClO<sub>4(aq)<\/sub>+ H<sub>2<\/sub>O<sub>(l)<\/sub><\/span><\/p>\n<p class=\"Indent\">c)<span>\u00a0 <\/span>Ca<sub>(s)<\/sub>+ O<sub>2(g)<\/sub><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span>CaO<sub>(s)<\/sub><\/p>\n<p class=\"Indent\"><span lang=\"ES-MX\">d)<span>\u00a0 <\/span>H<sub>2<\/sub><\/span><span lang=\"PT-BR\">SO<sub>4(aq)<\/sub>+ NaOH<sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Na<sub>2<\/sub>SO<sub>4(aq)<\/sub>+ H<sub>2<\/sub>O<sub>(l)<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>Pb(NO<sub>3<\/sub>)<sub>2(aq)<\/sub>+ Na<sub>2<\/sub>CO<sub>3(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">PbCO<sub>3(s)<\/sub>+ NaNO<sub>3(aq)<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">f)<span>\u00a0 <\/span>K<sub>2<\/sub>SO<sub>4(aq)<\/sub>+ CaCl<sub>2(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">KCl<sub>(aq)<\/sub>+ CaSO<sub>4(s)<\/sub><\/span><\/p>\n<p class=\"Questions\">22. Classify the following unbalanced chemical reactions by as many methods as possible.<\/p>\n<p class=\"Indent\">a)<span>\u00a0 <\/span>I<sub>4<\/sub>O<sub>9(s)<\/sub><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span>I<sub>2<\/sub>O<sub>6(s)<\/sub>+ I<sub>2(s)<\/sub>+ O<sub>2(g)<\/sub><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">b)<span>\u00a0 <\/span>Mg<sub>(s)<\/sub>+ AgNO<sub>3(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Mg(NO<sub>3<\/sub>)<sub>2(aq)<\/sub>+ Ag<sub>(s)<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">c)<span>\u00a0 <\/span>SiCl<sub>4(l)<\/sub>+ Mg<sub>(s)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">MgCl<sub>2(aq)<\/sub>+ Si<sub>(s)<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">d)<span>\u00a0 <\/span>CuCl<sub>2(aq)<\/sub>+ AgNO<sub>3(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Cu(NO<sub>3<\/sub>)<sub>2(aq)<\/sub>+ AgCl<sub>(s)<\/sub><\/span><\/p>\n<p class=\"Indent\"><span lang=\"PT-BR\">e)<span>\u00a0 <\/span>Al<sub>(s)<\/sub>+ Br<sub>2(l)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">AlBr<sub>3(s)<\/sub><\/span><\/p>\n<p><span lang=\"ES-MX\">f)<span>\u00a0 <\/span>HBr<\/span><span lang=\"PT-BR\"><sub>(aq)<\/sub>+ NaOH<sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">NaBr<sub>(aq)<\/sub>+ H<sub>2<\/sub>O<sub>(l)<\/sub><\/span><\/p>\n<h2><\/h2>\n<h2>Answers<\/h2>\n<p>1. H<sub class=\"subscript\">2<\/sub>O(\u2113) <span>[latex]\\longrightarrow[\/latex]<\/span>\u00a0H<sub class=\"subscript\">2<\/sub>O(g)<\/p>\n<p>2. H<sub class=\"subscript\">2<\/sub>O(\u2113) <span>[latex]\\longrightarrow[\/latex]<\/span>\u00a0H<sub class=\"subscript\">2<\/sub>O(s)<\/p>\n<p>3.\u00a0The coefficients are not in their lowest whole-number ratio.<\/p>\n<p>4. No; zinc is lower in the activity series than aluminum.<\/p>\n<p>5. In the products, the cation is pairing with the cation, and the anion is pairing with the anion.<\/p>\n<p id=\"ball-ch04_s07_qs01_p22_ans\" class=\"para\">6.\u00a0<span class=\"informalequation\"><span class=\"mathphrase\">Ag<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0SrCl<sub class=\"subscript\">2<\/sub>(aq) <span>[latex]\\longrightarrow[\/latex]<\/span>\u00a0SrSO<sub class=\"subscript\">4<\/sub>(s) + 2 AgCl(s)\u00a0<\/span><\/span><\/p>\n<p class=\"para\">7. Complete ionic equation: Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) <span>[latex]\\longrightarrow[\/latex]<\/span>\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 AgCl(s)<\/p>\n<p>Net ionic equation: The net ionic equation is the same as the complete ionic equation.<\/p>\n<p>8. Complete ionic equation:\u00a0Sr<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) <span>[latex]\\longrightarrow[\/latex]<\/span>\u00a0SrSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 AgCl(s)<\/p>\n<p>Net ionic equation: The net ionic equation is the same as the complete ionic equation.<\/p>\n<p>9. Each ion is a spectator ion; there is no overall net ionic equation.<\/p>\n<p>10. Yes; H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub>\u00a0<span>[latex]\\longrightarrow[\/latex]<\/span> 2 HCl (answers will vary)<\/p>\n<p>11. Yes; 2 HCl <span>[latex]\\longrightarrow[\/latex]<\/span>\u00a0H<sub class=\"subscript\">2<\/sub> +\u00a0Cl<sub class=\"subscript\">2<\/sub> (answers will vary)<\/p>\n<p>12. It does not increase the H<sup class=\"superscript\">+<\/sup> ion concentration; it is not a compound of H<sup class=\"superscript\">+<\/sup>.<\/p>\n<p>13. a) \u00a00 \u00a0 \u00a0 \u00a0b) \u00a0\u22123 \u00a0 \u00a0 \u00a0c) \u00a0+2 \u00a0 \u00a0 \u00a0d) \u00a0+1<\/p>\n<p>e) \u00a0+4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f) +4 \u00a0 \u00a0 \u00a0 \u00a0g) \u00a0+5 \u00a0 \u00a0 \u00a0h) \u00a0+5<\/p>\n<p>14. Copper is disproportionating. Initially, its oxidation number is +1; in the products, its oxidation numbers are +2 and 0, respectively.<\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">15.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0<\/span>HgO(s) [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Hg(l) + O<sub>2<\/sub>(g)<br \/>\nb)<span>\u00a0 <\/span>Zn(s) + HCl(aq) [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">ZnCl<sub>2<\/sub>(aq) + H<sub>2<\/sub>(g)<br \/>\nc)<span>\u00a0 <\/span>C<sub>3<\/sub>H<sub>8<\/sub>(g) + O<sub>2<\/sub>(g) [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(g)<br \/>\nd)<span>\u00a0 <\/span>NH<sub>3<\/sub>(g) + HNO<sub>3<\/sub>(aq) [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">NH<sub>4<\/sub>NO<sub>3<\/sub>(s)<br \/>\ne)<span>\u00a0 <\/span>B<sub>2<\/sub>O<sub>3<\/sub>(s) + Mg(s) [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">B(s) + MgO(s)<\/span><\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">16.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0<\/span>Zn(s) + H<sub>2<\/sub>SO<sub>4<\/sub>(aq)<span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>ZnSO<sub>4<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>(g)<br \/>\nb)<span>\u00a0 <\/span>C(s) + O<sub>2<\/sub>(g)\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">CO<sub>2<\/sub>(g)<br \/>\nc)<span>\u00a0 <\/span>2H<sub>2<\/sub>(g) + O<sub>2<\/sub>(g)<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>2H<sub>2<\/sub>O(l)<br \/>\nd)<span>\u00a0 <\/span>2Al(s) + 6HCl(aq)<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>2AlCl<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>3H<sub>2<\/sub>(g)<br \/>\ne)<span>\u00a0 <\/span>4Cr(s) + 3O<sub>2<\/sub>(g)<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>2Cr<sub>2<\/sub>O<sub>3<\/sub>(s)<br \/>\nf)<span>\u00a0 <\/span>2K(s) + 2H<sub>2<\/sub>O(aq)<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2KOH(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>(g)<br \/>\ng)<span>\u00a0 <\/span>CuO(s) + 2HCl(aq)<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>CuCl<sub>2<\/sub>(aq) + H<sub>2<\/sub>O(l)<br \/>\nh)<span>\u00a0 <\/span>NaHCO<sub>3<\/sub>(aq) + HNO<sub>3<\/sub>(aq)<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0\u00a0<\/span>NaNO<sub>3<\/sub>(aq)<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub>O(l)<span>\u00a0<\/span>+\u00a0CO<sub>2<\/sub>(g)<\/span><\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">17.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0 <\/span>2H<sub>3<\/sub>PO<sub>4<\/sub>\u00a0 +\u00a0 3CaO [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>\u00a0 +\u00a0 3H<sub>2<\/sub>O<br \/>\nb)<span>\u00a0 <\/span>4NH<sub>3<\/sub>\u00a0 +\u00a0 7O<sub>2<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">4NO<sub>2<\/sub>\u00a0 +\u00a0 6H<sub>2<\/sub>O<br \/>\nc)<span>\u00a0 <\/span>Cl<sub>2<\/sub>O<sub>7<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">2HClO<sub>4<\/sub><br \/>\nd)<span>\u00a0 <\/span>Mg<sub>3<\/sub>N<sub>2<\/sub>\u00a0 +\u00a0 6H<sub>2<\/sub>O\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">3Mg(OH)<sub>2<\/sub>\u00a0 +\u00a0 2NH<sub>3<\/sub><br \/>\ne)<span>\u00a0 <\/span>4FeSO<sub>4<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">2Fe<sub>2<\/sub>O<sub>3<\/sub>\u00a0 +\u00a0 4SO<sub>2<\/sub>\u00a0 +\u00a0 O<sub>2<\/sub><br \/>\nf<\/span><span lang=\"PL\">)<span>\u00a0 <\/span>P<sub>4<\/sub>+ 6Cl<sub>2<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PL\">4PCl<sub>3<\/sub><br \/>\ng)<span>\u00a0 <\/span>MnO<sub>2<\/sub>\u00a0 +\u00a0 C [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PL\">Mn\u00a0 +\u00a0 CO<sub>2<\/sub><br \/>\n<\/span><span lang=\"PT-BR\">h)<span>\u00a0 <\/span>2Na<sub>2<\/sub>O<sub>2<\/sub>\u00a0 +\u00a0 2H<sub>2<\/sub>O\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">4NaOH\u00a0 +\u00a0 O<sub>2<\/sub><br \/>\ni)<span>\u00a0 <\/span>CaH<sub>2<\/sub>\u00a0 +\u00a0 2H<sub>2<\/sub>O [latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Ca(OH)<sub>2<\/sub>\u00a0 +\u00a0 2H<sub>2<\/sub><br \/>\n<\/span><span lang=\"PL\">j)<span>\u00a0 <\/span>2NaHCO<sub>3<\/sub><\/span><span lang=\"PL\"><span>\u00a0[latex]\\longrightarrow[\/latex] <\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u00a0 +\u00a0 CO<sub>2<\/sub>\u00a0 +\u00a0 H<sub>2<\/sub>O<\/span><\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">18.<span>\u00a0\u00a0 <\/span>a)<span>\u00a0<\/span>2Mg<span>\u00a0 <\/span>+<span>\u00a0 <\/span>O<sub>2<\/sub><span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2MgO<br \/>\nb)<span>\u00a0 <\/span>2KClO<sub>3<\/sub><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2KCl<span>\u00a0 <\/span>+<span>\u00a0 <\/span>3O<sub>2<\/sub><br \/>\nc)<span>\u00a0 <\/span>3Fe<span>\u00a0<\/span>+<span>\u00a0 <\/span>2O<sub>2<\/sub><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub><br \/>\nd)<span>\u00a0 <\/span>Mg<span>\u00a0<\/span>+<span>\u00a0 <\/span>2HCl<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>MgCl<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub><br \/>\ne)<span>\u00a0 <\/span>2Na<span>\u00a0<\/span>+<span>\u00a0 <\/span>2H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2NaOH<span>\u00a0 <\/span>+<span>\u00a0 <\/span>H<sub>2<\/sub><br \/>\nf)<span>\u00a0 <\/span>N<sub>2<\/sub>+<span>\u00a0 <\/span>3H<sub>2<\/sub><span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>2NH<sub>3<\/sub><br \/>\ng)<span>\u00a0 <\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>\u2022 10H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Na<sub>2<\/sub>CO<sub>3<\/sub>+<span>\u00a0<\/span>10H<sub>2<\/sub>O<br \/>\nh)<span>\u00a0 <\/span>3Fe<span>\u00a0\u00a0<\/span>+<span>\u00a0 <\/span>4H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>Fe<sub>3<\/sub>O<sub>4<\/sub>+<span>\u00a0 <\/span>4H<sub>2<\/sub><br \/>\ni)<span>\u00a0 <\/span>2F<sub>2<\/sub><span>\u00a0 <\/span>+<span>\u00a0 <\/span>2H<sub>2<\/sub>O<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"PT-BR\"><span>\u00a0<\/span>4HF +<span>\u00a0 <\/span>O<sub>2<\/sub><\/span><\/p>\n<p class=\"Answers\">19. Balance the following chemical equations.<\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">a)<span>\u00a0 <\/span>2 Ca<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">(PO<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">)<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0 <\/span>6 SiO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0 <\/span>5 C<span>\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>P<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0<\/span>6 CaSiO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">+<span>\u00a0<\/span>5 CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">b)<span>\u00a0 <\/span>3 C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">6<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>5 O<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 [latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>6 CO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>9 H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O <\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">c)<span>\u00a0 <\/span>48 Fe<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>35 S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0 \u00a0<\/span>72 Fe<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>64 SO<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">d)<span>\u00a0 <\/span>24 P<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">S<\/span><sub><span lang=\"PT-BR\">5<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0<\/span>64O<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0<\/span>15 S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>16 P<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Answers\"><span lang=\"ES-MX\">e)<span>\u00a0 <\/span>16 C<\/span><sub><span lang=\"PT-BR\">3<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 <\/span>+<span>\u00a0\u00a0 <\/span>56 SO<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0[latex]\\longrightarrow[\/latex]<\/span><\/span><span lang=\"ES-MX\"><span>\u00a0\u00a0<\/span>24 C<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">H<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\">O<\/span><sub><span lang=\"PT-BR\">4<\/span><\/sub><span lang=\"ES-MX\"><span>\u00a0 <\/span>+<span>\u00a0 <\/span>16 H<\/span><sub><span lang=\"PT-BR\">2<\/span><\/sub><span lang=\"ES-MX\">O<span>\u00a0 <\/span>+<span>\u00a0 <\/span>7 S<\/span><sub><span lang=\"PT-BR\">8<\/span><\/sub><span lang=\"ES-MX\"><\/span><\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">20.<span>\u00a0 <\/span>a) NaCl<sub>(aq)<\/sub>+ AgNO<sub>3(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">AgCl<sub>(s)<\/sub>+ NaNO<sub>3(aq)<\/sub><br \/>\nNa<sup>+<\/sup><sub>(aq)<\/sub>+ Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub>+ Ag<sup>+<\/sup><sub>(aq)<\/sub>+ NO<sub>3<\/sub><sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\longrightarrow[\/latex]\u00a0<\/span><\/span><span lang=\"PT-BR\">AgCl<sub>(s)<\/sub>+ Na<sup>+<\/sup><sub>(aq)<\/sub>+ NO<sub>3<\/sub><sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\nCl<sup>&#8211;<\/sup><sub>(aq)<\/sub>+ Ag<sup>+<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">AgCl<sub>(s)<\/sub><\/span><\/p>\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">b) BaCl<sub>2(aq)<\/sub>+ H<sub>2<\/sub>SO<sub>4(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">BaSO<sub>4(s)<\/sub>+ 2HCl<sub>(aq)<\/sub><br \/>\nBa<sup>2+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub>+ 2H<sup>+<\/sup><sub>(aq)<\/sub>+ SO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub><br \/>\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\longrightarrow[\/latex]\u00a0<\/span><\/span><span lang=\"PT-BR\">BaSO<sub>4(s)<\/sub>+ 2H<sup>+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\nBa<sup>2+<\/sup><sub>(aq)<\/sub>+ SO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">BaSO<sub>4(s)<\/sub><\/span><\/p>\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">c) FeCl<sub>3(aq)<\/sub>+ 3NH<sub>4<\/sub>OH<sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Fe(OH)<sub>3(s)<\/sub>+ 3NH<sub>4<\/sub>Cl<sub>(aq)<\/sub><br \/>\nFe<sup>3+<\/sup><sub>(aq)<\/sub>+ 3Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub>+ 3 NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ 3OH<sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\longrightarrow[\/latex]\u00a0<\/span><\/span><span lang=\"PT-BR\">Fe(OH)<sub>3(s)<\/sub>+ 3 NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ 3Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\nFe<sup>3+<\/sup><sub>(aq)<\/sub>+ 3OH<sup>&#8211;<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">Fe(OH)<sub>3(s)<\/sub><\/span><\/p>\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">d) K<sub>2<\/sub>CrO<sub>4(aq)<\/sub>+ Pb(NO<sub>3<\/sub>)<sub>2(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">PbCrO<sub>4(s)<\/sub>+ 2KNO<sub>3(aq)<\/sub><br \/>\n2K<sup>+<\/sup><sub>(aq)<\/sub>+ CrO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Pb<sup>2+<\/sup><sub>(aq)<\/sub>+ 2NO<sub>3<\/sub><sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\longrightarrow[\/latex]\u00a0<\/span><\/span><span lang=\"PT-BR\">PbCrO<sub>4(s)<\/sub>+ 2K<sup>+<\/sup><sub>(aq)<\/sub>+ 2NO<sub>3<\/sub><sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\nCrO<sub>4<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Pb<sup>2+<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">PbCrO<sub>4(s)<\/sub><\/span><\/p>\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">e) KClO<sub>3(aq)<\/sub>+ MgCl<sub>2(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">N.R.<\/span><\/p>\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">f) (NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3(aq)<\/sub>+ CaCl<sub>2(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">CaCO<sub>3(s)<\/sub>+ 2NH<sub>4<\/sub>Cl<sub>(aq)<\/sub><br \/>\n2NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ CO<sub>3<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Ca<sup>2+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\longrightarrow[\/latex]\u00a0<\/span><\/span><span lang=\"PT-BR\">CaCO<sub>3(s)<\/sub>+ 2NH<sub>4<\/sub><sup>+<\/sup><sub>(aq)<\/sub>+ 2Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\nCO<sub>3<\/sub><sup>2-<\/sup><sub>(aq)<\/sub>+ Ca<sup>2+<\/sup><sub>(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">CaCO<sub>3(s)<\/sub><\/span><\/p>\n<p class=\"AnswersSub\"><span lang=\"PT-BR\">g) NaC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2(aq)<\/sub>+ BaCl<sub>2(aq)<\/sub><\/span><span>\u00a0[latex]\\longrightarrow[\/latex]\u00a0<\/span><span lang=\"PT-BR\">N.R.<\/span><\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">21.<span>\u00a0 <\/span>a) redox \u00a0 \u00a0 \u00a0b) acid-base \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0c) redox<br \/>\nd) acid-base \u00a0 \u00a0 \u00a0\u00a0e) precipitation \u00a0 \u00a0 \u00a0f) precipitation<\/span><\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">22.<span>\u00a0 <\/span><\/span>a) decomposition, which is a type of redox<\/p>\n<p class=\"Answers\">b) single-replacement, which is a type of redox<\/p>\n<p class=\"Answers\">c) single-replacement, which is a type of redox<\/p>\n<p class=\"Answers\">d) precipitation, which is a type of double replacement (but not redox!)<\/p>\n<p class=\"Answers\"><span lang=\"PT-BR\">e) composition (sometimes also called a <em class=\"emphasis\">combination reaction<\/em> or a <em class=\"emphasis\">synthesis reaction<\/em>),\u00a0which is a type of redox<\/span><\/p>\n<p>f) acid-base,\u00a0which is a type of double replacement (but not redox!)<\/p>\n","protected":false},"author":330,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"6.5 End of Chapter 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