{"id":1292,"date":"2018-04-11T22:51:09","date_gmt":"2018-04-12T02:51:09","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/1-6-mathematical-treatment-of-measurement-results\/"},"modified":"2018-06-22T22:57:57","modified_gmt":"2018-06-23T02:57:57","slug":"1-6-mathematical-treatment-of-measurement-results","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/1-6-mathematical-treatment-of-measurement-results\/","title":{"raw":"2.4 Mathematical Treatment of Measurement Results - Unit Conversions","rendered":"2.4 Mathematical Treatment of Measurement Results &#8211; Unit Conversions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities<\/li>\r\n \t<li>Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties<\/li>\r\n \t<li>Learn about the various temperature scales that are commonly used in chemistry and how to convert from one scale to another.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm319461744\">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 <em>time<\/em> required for the athlete to run from the starting line to the finish line, and the <em>distance<\/em> between these two lines, and then computing <em>speed<\/em> from the equation that relates these three properties:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm290867056\" style=\"text-align: center\">$latex \\text{speed}= \\frac{\\text{distance}}{\\text{time}} $<\/div>\r\n<p id=\"fs-idm350487392\">An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm257675424\" style=\"text-align: center\">$latex \\frac{\\text{100 m}}{\\text{10 s}} = \\text{10 m\/s} $<\/div>\r\n<p id=\"fs-idm308822992\">Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100\/10 = 10) <em>and likewise<\/em> dividing the units of each measured quantity to yield the unit of the computed quantity (m\/s = m\/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m\/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm219214640\" style=\"text-align: center\">$latex \\text{time} = \\frac{\\text{distance}}{\\text{speed}} $<\/div>\r\n<p id=\"fs-idm24572080\">The time can then be computed as:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp106016944\" style=\"text-align: center\">$latex \\frac{\\text{25 m}}{\\text{10 m\/s}} = \\text{2.5 s} $<\/div>\r\n<p id=\"fs-idm316688784\">Again, arithmetic on the numbers (25\/10 = 2.5) was accompanied by the same arithmetic on the units (m\/m\/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m\/m), the result is \u201c1\u201d\u2014or, as commonly phrased, the units \u201ccancel.\u201d<\/p>\r\n<p id=\"fs-idp44099792\">These calculations are examples of a versatile mathematical approach known as <strong>dimensional analysis<\/strong> (or the <strong>factor-label method<\/strong>). Dimensional analysis is based on this premise: <em>the units of quantities must be subjected to the same mathematical operations as their associated numbers<\/em>. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.<\/p>\r\n\r\n<section id=\"fs-idm285086480\">\r\n<h2>Conversion Factors and Dimensional Analysis<\/h2>\r\n<p id=\"fs-idm273312256\">A ratio of two equivalent quantities expressed with different measurement units can be used as a <strong>unit conversion factor<\/strong>. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm256748160\" style=\"text-align: center\">$latex \\frac{\\text{2.54 cm}}{\\text{1 in.}}\\;\\text{(2.54 cm = 1 in.) or 2.54}\\frac{\\text{cm}}{\\text{in}} $<\/div>\r\n<p id=\"fs-idm205801120\">Several other commonly used conversion factors are given in <a href=\"#fs-idm222237232\" class=\"autogenerated-content\">Table 1<\/a>.<\/p>\r\n\r\n<table id=\"fs-idm222237232\" class=\"span-all\" summary=\"This table is divided into 3 columns. They are titled length, volume, and mass. The following units are under the length column: 1 meter is equal to 1.0936 yards, 1 inch is equal to 2.54 cm 1 kilometer is equal to 0.62137 miles, 1 mile is equal to 1609.3 meters. The following units are under the volume column: 1 liter is equal to 1.0567 quarts, 1 quart is equal to 0.94635 meters, one cubic foot is equal to 28.317 liters, 1 tablespoon is equal to 14.787 milliliters. The following units are under the mass column: 1 kilogram is equal to 2.2046 pounds, 1 pound is equal to 453.59 grams, 1 avoirdupois ounce is equal to 28.349 grams, 1 troy ounce is equal to 31.103 grams.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Length<\/th>\r\n<th>Volume<\/th>\r\n<th>Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1 m = 1.0936 yd<\/td>\r\n<td>1 L = 1.0567 qt<\/td>\r\n<td>1 kg = 2.2046 lb<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1 in. = 2.54 cm (exact)<\/td>\r\n<td>1 qt = 0.94635 L<\/td>\r\n<td>1 lb = 453.59 g<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1 km = 0.62137 mi<\/td>\r\n<td>1 ft<sup>3<\/sup> = 28.317 L<\/td>\r\n<td>1 (avoirdupois) oz = 28.349 g<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1 mi = 1609.3 m<\/td>\r\n<td>1 tbsp = 14.787 mL<\/td>\r\n<td>1 (troy) oz = 31.103 g<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"3\"><strong>Table 1.<\/strong> Common Conversion Factors<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm221472400\">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\u2019s vertical jump of 34 inches can be converted to centimeters by:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm153590912\" style=\"text-align: center\">$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} $<\/div>\r\nSince this simple arithmetic involves <em>quantities<\/em>, the premise of dimensional analysis requires that we multiply both <em>numbers and units<\/em>. 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} $, and the unit product thus simplifies to <em>cm<\/em>. (When identical units divide to yield a factor of 1, they are said to \u201ccancel.\u201d) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.\r\n\r\n<\/section><section>\r\n<div class=\"textbox shaded\" id=\"fs-idm150235328\">\r\n<h3>Example 1<\/h3>\r\n<p id=\"fs-idp22709840\">The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (<a href=\"#fs-idm222237232\" class=\"autogenerated-content\">Table 1<\/a>).<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm290807904\"><strong>Solution<\/strong>\r\nIf we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm138056288\" style=\"text-align: center\">$latex x \\ \\text{oz} = \\text{125 \\text{g}} \\times \\text{unit conversion factor}$<\/div>\r\n<p id=\"fs-idm300877280\">We write the unit conversion factor in its two forms:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm233281680\" style=\"text-align: center\">$latex \\frac{1 \\text{oz}}{28.349 \\text{g}} \\;\\text{and}\\;\\frac{28.349 \\text{g}}{1 \\text{oz}}$<\/div>\r\n<p id=\"fs-idm369973600\">The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm222314304\">\r\n<p style=\"text-align: center\">$latex \\begin{array}{r @{{}={}} l} x\\;\\text{oz} &amp; 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] &amp; (\\frac{125}{\\text{28.349}})\\text{oz} \\\\[1em] &amp; \\text{4.41 oz (three significant figures)} \\end{array}$<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm224226288\"><em><strong>Test Yourself<\/strong><\/em>\r\nConvert a volume of 9.345 qt to liters.<\/p>\r\n&nbsp;\r\n\r\n<strong><em>Answer<\/em><\/strong>\r\n\r\n8.844 L\r\n\r\n<\/div>\r\n<p id=\"fs-idm262838208\">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\u2014all the <em>factors<\/em> involved in the calculation must be appropriately oriented to insure that their <em>labels<\/em> (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.<\/p>\r\n\r\n<div class=\"textbox shaded\" id=\"fs-idm305814320\">\r\n<h3>Example 2<\/h3>\r\n<p id=\"fs-idm280005808\">What is the density of common antifreeze in units of g\/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp14199296\"><strong>Solution<\/strong>\r\nSince density = $latex \\frac{\\text{mass}}{\\text{volume}} $, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A \u00d7 unit conversion factor. The necessary conversion factors are given in <a href=\"#fs-idm222237232\" class=\"autogenerated-content\">Table 1<\/a>: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm336821696\" style=\"text-align: center\">$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} $<\/div>\r\n<p id=\"fs-idm244171104\">We need to use two steps to convert volume from quarts to milliliters.<\/p>\r\n\r\n<ol id=\"fs-idm287602768\" class=\"stepwise\">\r\n \t<li><em>Convert quarts to liters.<\/em>\r\n<div class=\"equation\" id=\"fs-idm279137456\" style=\"text-align: center\">$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} $<\/div><\/li>\r\n \t<li><em>Convert liters to milliliters.<\/em>\r\n<div class=\"equation\" id=\"fs-idm289883088\" style=\"text-align: center\">$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} $<\/div><\/li>\r\n<\/ol>\r\n<p id=\"fs-idm301215296\">Then,<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm144640128\" style=\"text-align: center\">$latex \\text{density} = \\frac{4.20 \\times 10^3 \\;\\text{g}}{3.78 \\times 10^3 \\;\\text{mL}} = 1.11 \\;\\text{g\/mL} $<\/div>\r\n<p id=\"fs-idm292167648\">Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm9873904\" style=\"text-align: center\">$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} $<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm219351440\"><em><strong>Test Yourself<\/strong><\/em>\r\nWhat is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?<\/p>\r\n&nbsp;\r\n\r\n<strong><em>Answer<\/em><\/strong>\r\n\r\n$latex 2.956 \\times 10^{-2} \\text{L} $\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idm306560960\">\r\n<h3>Example 3<\/h3>\r\n<p id=\"fs-idm338996800\">While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.<\/p>\r\n<p id=\"fs-idm127359616\">a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?<\/p>\r\n<p id=\"fs-idm315428528\">b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm327742336\"><strong>Solution<\/strong>\r\na) We first convert distance from kilometers to miles:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm60362224\" style=\"text-align: center\">$latex 1250\\;\\text{km} \\times \\frac{0.62137\\;\\text{mi}}{1\\;\\text{km}} = 777\\;\\text{mi} $<\/div>\r\n<p id=\"fs-idm142282576\">and then convert volume from liters to gallons:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm214696672\" style=\"text-align: center\">$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} $<\/div>\r\n<p id=\"fs-idp1583872\">Then,<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm114266320\" style=\"text-align: center\">$latex \\text{(average) mileage} = \\frac{777 \\;\\text{mi}}{56.3\\;\\text{gal}} = 13.8\\;\\text{miles\/gallon} = 13.8\\;\\text{mpg} $<\/div>\r\n<p id=\"fs-idm311190128\">Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm171836160\" style=\"text-align: center\">$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} $<\/div>\r\n<p id=\"fs-idm268939440\">b) Using the previously calculated volume in gallons, we find:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm328491840\" style=\"text-align: center\">$latex \\text{56.3 gal} \\times \\frac{\\$3.80}{1\\;\\text{gal}} = \\$214 $<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm125734208\"><em><strong>Test Yourself<\/strong><\/em>\r\nA Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).<\/p>\r\n<p id=\"fs-idm247721312\">a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?<\/p>\r\n<p id=\"fs-idm292238912\">b) If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answers<\/strong><\/em>\r\n\r\na) 51 mpg \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) $62\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm206910464\">\r\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 4<\/h3>\r\na) Convert 35.9 kL to liters.\r\n\r\nb)Convert 555 nm to meters.\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">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 <span class=\"inlineequation\">1,000\u00a0L\/1\u00a0kL<\/span>. Applying this conversion factor, we get<\/p>\r\n<span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_11.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_11.png\" alt=\"35.9 kL x (1000 L\/1 kL) = 35900 L\" class=\" wp-image-4836 alignnone\" width=\"233\" height=\"63\" \/><\/a><\/span>\r\n\r\nb) 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 10<sup class=\"superscript\">9<\/sup> nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: <span class=\"inlineequation\">1\u00a0m\/10<sup>9<\/sup>\u00a0nm<\/span>. Applying this conversion factor, we get\r\n\r\n<span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_12.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_12.png\" alt=\"555 nm x (1 m\/ 10^9 nm) = 5.55 x 10^-7 m\" class=\"alignnone wp-image-4837\" width=\"458\" height=\"75\" \/><\/a><\/span>\r\n<p id=\"ball-ch02_s04_p20\" class=\"para\">In the final step, we expressed the answer in scientific notation.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p class=\"simpara\">a) Convert 67.08 \u03bcL to liters. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b)\u00a0Convert 56.8 m to kilometers.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answers<\/em><\/strong><\/p>\r\n<p class=\"simpara\">a) 6.708 \u00d7 10<sup class=\"superscript\">\u22125<\/sup> L \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b)\u00a05.68 \u00d7 10<sup class=\"superscript\">\u22122<\/sup> km<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 5<\/h3>\r\n<p class=\"Indent\">Complete the following conversions\r\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>(use table 1.1 for any metric conversions;<span>\u00a0 <\/span>Note: 1 L = 1.0567 qt )<\/p>\r\n<p class=\"IndentSub\">a) 125 m = ? mm<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>b) 2.3 x 10<sup>-6 <\/sup>Mg = ? g<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>c) 2.5 L = ? qt<\/p>\r\n&nbsp;\r\n<p class=\"Solution\"><strong>Solution<\/strong><span><strong>\u00a0<\/strong>\u00a0 <\/span><\/p>\r\n<p class=\"Indentpoints\">a)<span>\u00a0\u00a0 <\/span>$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} $<\/p>\r\n\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em;text-align: left\">Note that this equivalence statement comes from the literal \u201cmeaning\u201d of milli in Table 1.1. Many students prefer to use the equivalence statement 1000 mm = 1 m. That is fine too, as long as you always remember to put the appropriate unit and number.<\/span><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\">b)<span>\u00a0\u00a0<\/span>$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} $<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">c)<\/span><span style=\"font-size: 1em\">\u00a0\u00a0<\/span>$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} $<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span><\/span><em style=\"font-size: 1em\"><strong>Test Yourself<\/strong><\/em><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">Complete the following conversions<\/span><sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sup><span style=\"font-size: 1em\">(note: 1 in = 2.54 cm exactly)<\/span><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">a) 124 mL = ? L\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b) 256 days = ? hrs<\/span><sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sup><span style=\"font-size: 1em\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c) 63.2 cm = ? in<\/span><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><em style=\"font-size: 1em\"><strong>Answers<\/strong><\/em><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">a) 0.124 L \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 6.14 x 10<\/span><sup>3<\/sup><span style=\"font-size: 1em\"> hrs \u00a0 \u00a0 \u00a0 \u00a0 \u00a0c) 24.9 in<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 6<\/h3>\r\n<p class=\"Indent\">How many nm are in 26.5 feet? (12 in = 1 ft ,<span>\u00a0 <\/span>and<span>\u00a0<\/span>2.54 cm = 1 in exactly)<\/p>\r\n&nbsp;\r\n<p class=\"Solution\"><strong>Solution<\/strong><span><strong>\u00a0<\/strong>\u00a0 <\/span><\/p>\r\n<p class=\"Indent\">Likely, you do not have a direct conversion between feet and nm. So, instead, ask yourself \u201cwhere can I go from feet\u201d. The only possibility is inches. Then, where can you go from inches? \u2026and so on. The overall path becomes:<\/p>\r\n<p class=\"IndentSub\"><span>\u00a0<\/span>feet\u00a0$latex \\longrightarrow$\u00a0inches\u00a0$latex \\longrightarrow$\u00a0cm\u00a0$latex \\longrightarrow$\u00a0m\u00a0$latex \\longrightarrow$\u00a0<span style=\"font-size: 1em\">nm<\/span><\/p>\r\n<p class=\"IndentSub\"><span>\u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/p>\r\n$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} $\r\n\r\n&nbsp;\r\n<p class=\"SelfTest\"><em><strong>Test Yourself<\/strong><\/em><\/p>\r\n<p id=\"ball-ch02_s04_p22\" class=\"para\"><span>A marathon is 26.4 miles. If 1 mile = 1760 yards, and 1 m = 1.094 yards, how many km are in a marathon?<\/span><\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n42.5 km\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 7<\/h3>\r\n<p class=\"Indent\">a)<span>\u00a0\u00a0 <\/span>A car is moving at 35.2 km\/h. How many cm\/s is this speed?\r\nb)<span>\u00a0\u00a0 <\/span>How many cm<sup>3<\/sup>are in 5 x 10<sup>2<\/sup>m<sup>3<\/sup>?<\/p>\r\n&nbsp;\r\n<p class=\"Solution\"><strong>Solution<\/strong><span>\u00a0\u00a0 <\/span><\/p>\r\n<p class=\"Indentpoints\">a)<span>\u00a0 <\/span>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.<\/p>\r\n<p class=\"IndentSub\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/p>\r\n$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} $\r\n\r\n&nbsp;\r\n<p class=\"Indentpoints\">Remember\u2026\r\n1 m<sup>3<\/sup>= 1 m x 1 m x 1 m, so<span>\u00a0 <\/span>1 m<sup>3<\/sup>= 100 cm x 100 cm x 100 cm, NOT 100 cm<sup>3<\/sup><\/p>\r\n<p class=\"IndentSub\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\r\n<p class=\"IndentSub\">$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} $<\/p>\r\n&nbsp;\r\n<p class=\"SelfTest\"><em><strong>Test Yourself<\/strong><\/em><\/p>\r\n<p class=\"Indent\">Complete the following conversions<sup><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/sup><\/p>\r\n<p class=\"Indent\">a)<span>\u00a0 <\/span>75 mi\/h = ? m\/s<span>\u00a0 <\/span>(1760 yd = 1 mi and 1 m = 1.094 yd)\r\nb)<span>\u00a0 <\/span>4.1 g\/cm<sup>3<\/sup>= ? kg\/m<sup>3<\/sup><\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answers<\/strong><\/em>\r\n\r\na) 34 m\/s \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 4.1 x 10<sup>3<\/sup> kg\/m<sup>3<\/sup>\r\n\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-idm206910464\">\r\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 8<\/h3>\r\n<p id=\"ball-ch02_s04_p22\" class=\"para\">How many cubic centimeters are in 0.883 m<sup class=\"superscript\">3<\/sup>?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p23\" class=\"para\">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<\/p>\r\n<span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_14.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_14.png\" alt=\"0.883m^3 x (100cm\/1m) x (100cm\/1m) x (100cm\/1m) = 883000 cm^3\" class=\"alignnone wp-image-4839\" width=\"593\" height=\"56\" \/><\/a><\/span>\r\n<p id=\"ball-ch02_s04_p24\" class=\"para\">You should demonstrate to yourself that the three meter units do indeed cancel.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p25\" class=\"para\">How many cubic millimeters are present in 0.0923 m<sup class=\"superscript\">3<\/sup>?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p26\" class=\"para\">9.23 \u00d7 10<sup class=\"superscript\">7<\/sup> mm<sup class=\"superscript\">3<\/sup><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 9<\/h3>\r\n<p id=\"ball-ch02_s04_p28\" class=\"para\">Convert 88.4 m\/min to meters\/second.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p29\" class=\"para\">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: <span class=\"inlineequation\">1\u00a0min\/60\u00a0s<\/span>. Apply and perform the math:<\/p>\r\n<span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_15.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_15.png\" alt=\"88.4m\/m x 1min\/60s = 1.47 m\/s\" class=\"alignnone wp-image-4840\" width=\"225\" height=\"52\" \/><\/a><\/span>\r\n<p id=\"ball-ch02_s04_p30\" class=\"para\">Notice how the 88.4 automatically goes in the numerator. That\u2019s because any number can be thought of as being in the numerator of a fraction divided by 1.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p31\" class=\"para\">Convert 0.203 m\/min to meters\/second.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p32\" class=\"para\">0.00338 m\/s or 3.38 \u00d7 10<sup class=\"superscript\">\u22123<\/sup> m\/s<\/p>\r\n\r\n\r\n[caption id=\"attachment_2116\" align=\"alignnone\" width=\"300\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Grapevinesnail_01-1-e1528932055925.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Grapevinesnail_01-1-e1528932055925.jpg\" alt=\"\" width=\"300\" height=\"177\" class=\"wp-image-2116 size-full\" \/><\/a> <strong>Figure 1.<\/strong> How fast is fast?\u00a0A 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! \u00a0Source: \u201cGrapevine snail\u201dby J\u00fcrgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idm206910464\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 10<\/h3>\r\n<p id=\"ball-ch02_s04_p36\" class=\"para\">How many nanoseconds are in 368.09 \u03bcs?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p37\" class=\"para\">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 <em class=\"emphasis\">micro-<\/em> and <em class=\"emphasis\">nano-<\/em>,<\/p>\r\n<span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_21.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_21.png\" alt=\"368.09 us x 1s\/10^6us x 10^9ns \/1s = 368090 ns = 3.608 x 10^5 ns\" class=\"alignnone wp-image-4846\" width=\"544\" height=\"56\" \/><\/a><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p38\" class=\"para\">How many milliliters are in 607.8 kL?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p39\" class=\"para\">6.078 \u00d7 10<sup class=\"superscript\">8<\/sup> mL<\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm206910464\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 11<\/h3>\r\n<p id=\"ball-ch02_s04_p43\" class=\"para\">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.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p44\" class=\"para\">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.<\/p>\r\n<span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_22.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_22.png\" alt=\"36.7 cm x 128.8 cm x 1 m\/100cm x 1 m\/100 cm = 0.472696 m^2 = 0.473 m^2\" class=\"alignnone wp-image-4847\" width=\"575\" height=\"60\" \/><\/a><\/span>\r\n<p id=\"ball-ch02_s04_p45\" class=\"para\">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.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p46\" class=\"para\">What is the volume of a block in cubic meters whose dimensions are 2.1 cm \u00d7 34.0 cm \u00d7 118 cm?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s04_p47\" class=\"para\">0.0084 m<sup class=\"superscript\">3<\/sup><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Chemistry Is Everywhere: The Gimli Glider<\/h3>\r\n<p id=\"ball-ch02_s04_p48\" class=\"para\">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 \u201cthe Gimli Glider.\u201d<\/p>\r\n\r\n<div class=\"informalfigure large\" id=\"ball-ch02_s04_f02\">\r\n<div class=\"copyright\">\r\n\r\n[caption id=\"attachment_2124\" align=\"aligncenter\" width=\"300\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-300x219.jpg\" alt=\"\" width=\"300\" height=\"219\" class=\"wp-image-2124 size-medium\" \/><\/a> <strong>Figure 2.<\/strong> 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. \u201cAircanada.b767\u2032\u2032 is in the the public domain.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"ball-ch02_s04_p49\" class=\"para\">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.<\/p>\r\n<p id=\"ball-ch02_s04_p50\" class=\"para\">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 <em class=\"emphasis\">pounds<\/em> 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\u2014an incident that would not have occurred if people were watching their units.<\/p>\r\n\r\n<\/div>\r\n<h2>Conversion of Temperature Units<\/h2>\r\n<p id=\"fs-idm262720192\">We use the word <strong class=\"no-emphasis\">temperature<\/strong> 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.<\/p>\r\n<p id=\"fs-idm308860096\">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 \u00b0C is defined as the freezing temperature of water and 100 \u00b0C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the <strong>Fahrenheit<\/strong> scale, the freezing point of water is defined as 32 \u00b0F and the boiling temperature as 212 \u00b0F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).<\/p>\r\n<p id=\"fs-idm288396336\">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:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm6121760\" style=\"text-align: center\">$latex \\text{length in feet} = \\left( \\frac {1\\;\\text{ft}}{12\\;\\text{in.}}\\right) \\times \\text{length in inches} $<\/div>\r\n<p id=\"fs-idm161552768\">where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (y = mx + b). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales\u2019 zero points (b).<\/p>\r\n<p id=\"fs-idm292928800\">The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as <em>x<\/em> and the Fahrenheit temperature as <em>y<\/em>, the slope, <em>m<\/em>, is computed to be:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm229969840\">\r\n<p style=\"text-align: center\">$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}} $<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idm206340880\">The y-intercept of the equation, <em>b<\/em>, is then calculated using either of the equivalent temperature pairs, (100 \u00b0C, 212 \u00b0F) or (0 \u00b0C, 32 \u00b0F), as:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm234430272\" style=\"text-align: center\">$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} $<\/div>\r\n<p id=\"fs-idm218561008\">The equation relating the temperature scales is then:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm305465424\">\r\n<p style=\"text-align: center\">$latex T_{^\\circ\\text{F}} = (\\frac{9 \\;^\\circ\\text{F}}{5 \\;^\\circ\\text{C}} \\times T_{^\\circ\\text{C}}) + 32\\;^\\circ\\text{C} $<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idm208304512\">An abbreviated form of this equation that omits the measurement units is:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm226315088\" style=\"text-align: center\">$latex T_{^\\circ\\text{F}} = \\frac{9}{5} \\times T_{^\\circ\\text{C}} + 32 $<\/div>\r\n<p id=\"fs-idm356038704\">Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm138168256\" style=\"text-align: center\">$latex T_{^\\circ\\text{C}} = \\frac{5}{9} (T_{^\\circ\\text{F}} - 32) $<\/div>\r\n<p id=\"fs-idm131802496\">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 \u2212273.15 \u00b0C. 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\u2019s chapter on gases).<\/p>\r\n<p id=\"fs-idm309875456\">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:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp289344\" style=\"text-align: center\">$latex T_{\\text{K}} = T_{^\\circ\\text{C}} + 273.15 $<\/div>\r\n<div class=\"equation\" id=\"fs-idm303916848\" style=\"text-align: center\">$latex T_{^\\circ\\text{C}} = T_{\\text{K}} - 273.15 $<\/div>\r\n<p id=\"fs-idm288599552\">The 273.15 in these equations has been determined experimentally, so it is not exact. <a href=\"#CNX_Chem_01_06_TempScales\" class=\"autogenerated-content\">Figure 3<\/a> shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.<\/p>\r\n\r\n<figure id=\"CNX_Chem_01_06_TempScales\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1300\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_01_06_TempScales-2.jpg\" alt=\"A thermometer is shown for the Fahrenheit, Celsius and Kelvin scales. Under the Fahrenheit scale, the boiling point of water is 212 degrees while the freezing point of water is 32 degrees. Therefore, there are 180 Fahrenheit degrees between the boiling point of water and the freezing point of water. Under the Celsius scale, the boiling point of water is 100 degrees while the freezing point of water is 0 degrees. Therefore, there are 100 Celsius degrees between the boiling point and freezing point of water. Under the kelvin scale, the boiling point of water is 373.15 K, while the freezing point of water is 273.15 K. 233.15 K is equal to negative 40 degrees Celsius, which is also equal to negative 40 degrees Fahrenheit.\" width=\"1300\" height=\"911\" \/> <strong>Figure 3.<\/strong> The Fahrenheit, Celsius, and kelvin temperature scales are compared.[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<p id=\"fs-idm296725200\">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.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<div class=\"example\" id=\"fs-idm75569040\">\r\n<h3>Example\u00a012<\/h3>\r\n<p id=\"fs-idm290946544\">Normal body temperature has been commonly accepted as 37.0 \u00b0C (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?<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm288550064\"><strong>Solution<\/strong><\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm223031696\" style=\"text-align: left\">$latex \\text{K} = ^\\circ\\text{C} + \\text{273.15} = \\text{37.0} + \\text{273.2} = \\text{310.2 K} $<\/div>\r\n<div class=\"equation\" id=\"fs-idp114907616\" style=\"text-align: left\">$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} $<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp16310096\"><em><strong>Test Yourself<\/strong><\/em>\r\nConvert 80.92 \u00b0C to K and \u00b0F.<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answers<\/strong><\/em>\r\n\r\n354.07 K, 177.7 \u00b0F\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idm309453552\">\r\n<h3>Example 13<\/h3>\r\n<p id=\"fs-idm421846336\">Baking a ready-made pizza calls for an oven temperature of 450 \u00b0F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm36231872\"><strong>Solution<\/strong><\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm32912\" style=\"text-align: left\">$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)} $<\/div>\r\n<div class=\"equation\" id=\"fs-idm144212448\" style=\"text-align: left\">$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)}$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm296519728\"><em><strong>Test Yourself<\/strong><\/em>\r\nConvert 50 \u00b0F to \u00b0C and K.<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answers<\/strong><\/em>\r\n\r\n10 \u00b0C, 280 K\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm292051392\" class=\"summary\">\r\n<div class=\"callout block\" id=\"ball-ch02_s04_n07\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 14<\/h3>\r\na) What is 98.6 \u00b0F in degrees Celsius?\r\n\r\nb) What is 25.0 \u00b0C in degrees Fahrenheit?\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) Using the first formula from above, we have<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm32912\" style=\"text-align: left\">$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} $<\/div>\r\n&nbsp;\r\n\r\nb) Using the second formula from above, we have\r\n<div class=\"equation\" id=\"fs-idp114907616\" style=\"text-align: left\">$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} $<\/div>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p class=\"simpara\">a) Convert 0 \u00b0F to degrees Celsius.<\/p>\r\n<p class=\"simpara\">b) Convert 212 \u00b0C to degrees Fahrenheit.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answers<\/em><\/strong><\/p>\r\n<p class=\"simpara\">a) \u221217.8 \u00b0C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b)\u00a0414 \u00b0F<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 15<\/h3>\r\n<p id=\"ball-ch02_s05_p07\" class=\"para\">If normal room temperature is 72.0 \u00b0F, what is room temperature in degrees Celsius and kelvins?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch02_s05_p08\" class=\"para\">First, we use the formula to determine the temperature in degrees Celsius:<\/p>\r\n$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} $\r\n<p id=\"ball-ch02_s05_p09\" class=\"para\">Then we use the appropriate formula above to determine the temperature in the Kelvin scale:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">K = 22.2 \u00b0C +\u00a0273.15 = 295.4 K<\/span><\/span>\r\n<p id=\"ball-ch02_s05_p10\" class=\"para\">So, room temperature is about 295 K.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s05_p11\" class=\"para\">What is 98.6 \u00b0F on the Kelvin scale?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch02_s05_p12\" class=\"para\">310.2 K<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Food and Drink App: Cooking Temperatures<\/h3>\r\n<p id=\"ball-ch02_s05_p78\" class=\"para\">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\u00b0F, while a cake may be baked at 350\u00b0F and a chicken roasted at 400\u00b0F. The broil setting on many ovens is 500\u00b0F, which is typically the highest temperature setting on a household oven.<\/p>\r\n<p id=\"ball-ch02_s05_p79\" class=\"para\">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\u00b0F (100\u00b0C), but in Denver, the Mile-High City, water boils at about 200\u00b0F (93.3\u00b0C), which can significantly lengthen cooking times. Good cooks need to be aware of this.<\/p>\r\n<p id=\"ball-ch02_s05_p80\" class=\"para\">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\u00b0F (122\u00b0C); 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 <em class=\"emphasis\">autoclave<\/em>.)<\/p>\r\n<p id=\"ball-ch02_s05_p81\" class=\"para\">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\u2014a 425\u00b0F oven in the United States is equivalent to a 220\u00b0C oven in other countries. These days, many oven thermometers are marked with both temperature scales.<\/p>\r\n\r\n<\/div>\r\n<section id=\"fs-idm292051392\" class=\"summary\">\r\n<div id=\"fs-idp86805728\" class=\"textbox shaded\">\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-1-2.png\" alt=\"\" width=\"129\" height=\"80\" class=\"alignleft\" \/>\r\n<p id=\"fs-idm169361696\">Need a refresher or more practice with unit conversion? Visit this site (<a href=\"https:\/\/viuvideos.viu.ca\/media\/Unit+Conversion\/0_o671v9j6\">https:\/\/viuvideos.viu.ca\/media\/Unit+Conversion\/0_o671v9j6<\/a>) to go over the basics of unit conversions.<\/p>\r\nVideo source: Unit conversion by keyj\r\n\r\n<\/div>\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idm126307616\">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.<\/p>\r\n\r\n<\/section><section id=\"fs-idm299998176\" class=\"key-equations\">\r\n<h2>Key Equations<\/h2>\r\n<ul id=\"fs-idm309482144\">\r\n \t<li>$latex T_{^\\circ\\text{C}} = \\frac{5}{9} \\times T_{^\\circ\\text{F}} - 32 $<\/li>\r\n \t<li>$latex T_{^\\circ\\text{F}} = \\frac{9}{5} \\times T_{^\\circ\\text{C}} + 32 $<\/li>\r\n \t<li>$latex T_\\text{K} = {^\\circ\\text{C}} + 273.15 $<\/li>\r\n \t<li>$latex T_{^\\circ\\text{C}} = \\text{K} - 273.15 $<\/li>\r\n<\/ul>\r\n<\/section>\r\n<div class=\"callout block\" id=\"ball-ch02_s05_n06\">\r\n<div class=\"qandaset block\" id=\"ball-ch02_s05_qs01\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div class=\"question\">\r\n<p id=\"ball-ch02_s05_qs01_p1\" class=\"para\">1. \u00a0Perform the following conversions.<\/p>\r\n\r\n<\/div>\r\na) \u00a0255\u00b0F to degrees Celsius \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0\u2212255\u00b0F to degrees Celsius\r\n\r\nc) \u00a050.0\u00b0C to degrees Fahrenheit \u00a0 \u00a0 \u00a0 \u00a0 \u00a0d) \u00a0\u221250.0\u00b0C to degrees Fahrenheit\r\n<div class=\"question\">\r\n<p id=\"ball-ch02_s05_qs01_p3\" class=\"para\">2. \u00a0Perform the following conversions.<\/p>\r\na) \u00a0100.0\u00b0C to kelvins \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0\u2212100.0\u00b0C to kelvins\r\n\r\nc) \u00a0100 K to degrees Celsius \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0d) \u00a0300 K to degrees Celsius\r\n\r\n<\/div>\r\n<div class=\"question\"><\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch02_s05_qs01_p5\" class=\"para\">3. \u00a0Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch02_s05_qs01_p9\" class=\"para\">4. \u00a0The hottest temperature ever recorded on the surface of the earth was 136\u00b0F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins?<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">5. \u00a0Write the two conversion factors that exist between the two given units.<\/span><\/p>\r\n\r\n<\/div>\r\na) \u00a0milliliters and laters \u00a0 \u00a0 \u00a0b) \u00a0microseconds and seconds \u00a0 \u00a0 \u00a0\u00a0c) \u00a0kilometers and meters\r\n<div class=\"question\"><span style=\"font-size: 1em\">6. \u00a0Perform the following conversions.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\na) \u00a05.4 km to meters \u00a0 \u00a0 \u00a0b) \u00a00.665 m to millimeters \u00a0 \u00a0 \u00a0c) \u00a00.665 m to kilometers\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">7. \u00a0Perform the following conversions.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\na) \u00a017.8 \u03bcg to grams \u00a0 \u00a0 \u00a0b) \u00a07.22 \u00d7 10<sup class=\"superscript\">2<\/sup> kg to grams \u00a0 \u00a0 \u00a0c) \u00a00.00118 g to nanograms\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">8. \u00a0Perform the following conversions.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\na) \u00a09.44 m<sup class=\"superscript\">2<\/sup> to square centimetres \u00a0 \u00a0 \u00a0b) \u00a03.44 \u00d7 10<sup class=\"superscript\">8<\/sup> mm<sup class=\"superscript\">3<\/sup> to cubic meters\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">9. \u00a0Why would it be inappropriate to convert square centimeters to cubic meters?<\/span><\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch02_s04_qs01_p13\" class=\"para\">10. \u00a0Perform the following conversions.<\/p>\r\na) \u00a045.0 m\/min to meters\/second \u00a0 \u00a0 \u00a0 b) \u00a00.000444 m\/s to micrometers\/second\r\n\r\nc) \u00a060.0 km\/h to kilometers\/second\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">11. \u00a0Perform the following conversions.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\na) \u00a00.674 kL to milliliters \u00a0 \u00a0 \u00a0b) \u00a02.81 \u00d7 10<sup class=\"superscript\">12<\/sup> mm to kilometers \u00a0 \u00a0 \u00a0c) \u00a094.5 kg to milligrams\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">12. \u00a0Perform the following conversions.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\na) \u00a06.77 \u00d7 10<sup class=\"superscript\">14<\/sup> ms to kilo seconds \u00a0 \u00a0 \u00a0\u00a0b) \u00a034,550,000 cm to kilometers\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">13. \u00a0Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\na) \u00a088 ft\/s to miles\/hour (Hint: use 5,280 ft = 1 mi.) \u00a0 \u00a0 \u00a0b) \u00a00.00667 km\/h to meters\/second\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">14. \u00a0What is the area in square millimeters of a rectangle whose sides are 2.44 cm \u00d7 6.077 cm? Express the answer to the proper number of significant figures.<\/span><\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">15. \u00a0The formula for the area of a triangle is 1\/2 \u00d7 base \u00d7 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.<\/span><\/div>\r\n<div class=\"question\">\r\n\r\n16. \u00a0<span style=\"font-size: 1em\">Write conversion factors (as ratios) for the number of:<\/span>\r\n<p id=\"fs-idm279869696\">a) yards in 1 meter \u00a0 \u00a0 \u00a0b) liters in 1 liquid quart \u00a0 \u00a0 \u00a0c) pounds in 1 kilogram<\/p>\r\n17. \u00a0The 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?\r\n\r\n18. \u00a0Soccer 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?\r\n\r\n19. \u00a0How many milliliters of a soft drink are contained in a 12.0-oz can?\r\n\r\n20. \u00a0The diameter of a red blood cell is about 3 \u00d7 10<sup>\u22124<\/sup> in. What is its diameter in centimeters?\r\n\r\n21. \u00a0Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?\r\n\r\n22. \u00a0Many medical laboratory tests are run using 5.0 \u03bcL blood serum. What is this volume in milliliters?\r\n\r\n23. \u00a0Use scientific notation to express the following quantities in terms of the SI base units:\r\n<p id=\"fs-idm287857616\">a) 0.13 g \u00a0 \u00a0 \u00a0b) 232 Gg \u00a0 \u00a0 \u00a0c) 5.23 pm \u00a0 \u00a0 \u00a0d) 86.3 mg \u00a0 \u00a0 \u00a0e) 37.6 cm \u00a0 \u00a0 \u00a0f) 54 \u03bcm \u00a0 \u00a0 \u00a0g) 1 Ts \u00a0 \u00a0 \u00a0h) 27 ps \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0i) 0.15 mK<\/p>\r\n24. \u00a0Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank?\r\n\r\n25. \u00a0A long ton is defined as exactly 2240 lb. What is this mass in kilograms?\r\n\r\n26. \u00a0Make the conversion indicated in each of the following:\r\n<p id=\"fs-idm367197520\">a) the length of a soccer field, 120 m (three significant figures), to feet<\/p>\r\n<p id=\"fs-idp51878496\">b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers<\/p>\r\n<p id=\"fs-idm307271088\">c) the area of an 8.5 t 11-inch sheet of paper in cm<sup>2<\/sup><\/p>\r\n<p id=\"fs-idm127105888\">d) the displacement volume of an automobile engine, 161 in.<sup>3<\/sup>, to liters<\/p>\r\n<p id=\"fs-idm218455584\">e) the estimated mass of the atmosphere, 5.6 t 10<sup>15<\/sup> tons, to kilograms<\/p>\r\n<p id=\"fs-idm98022832\">f) the mass of a bushel of rye, 32.0 lb, to kilograms<\/p>\r\n<p id=\"fs-idm162390144\">g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)<\/p>\r\n27. \u00a0A chemist\u2019s 50-Trillion Angstrom Run\u00a0would be an archeologist\u2019s 10,900 cubit run. How long is one cubit in meters and in feet? (1 \u00c5 = 1 \u00d7 10<sup>\u22128<\/sup> cm)\r\n\r\n28. \u00a0As 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?\r\n\r\n29. \u00a0A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds?\r\n\r\n30. \u00a0Solve these problems about lumber dimensions.\r\n<p id=\"fs-idm124373120\">a) To describe to a European how houses are constructed in the US, the dimensions of \u201ctwo-by-four\u201d lumber must be converted into metric units. The thickness \u00d7 width \u00d7 length dimensions are 1.50 in. \u00d7 3.50 in. \u00d7 8.00 ft in the US. What are the dimensions in cm \u00d7 cm \u00d7 m?<\/p>\r\n<p id=\"fs-idm97712544\">b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?<\/p>\r\n31. \u00a0Calculate the density of aluminum if 27.6 cm<sup>3<\/sup> has a mass of 74.6 g.\r\n\r\n32. \u00a0Calculate these masses.\r\n<p id=\"fs-idp56604960\">a) What is the mass of 6.00 cm<sup>3<\/sup> of mercury, density = 13.5939 g\/cm<sup>3<\/sup>?<\/p>\r\n<p id=\"fs-idm84745936\">b) What is the mass of 25.0 mL octane, density = 0.702 g\/cm<sup>3<\/sup>?<\/p>\r\n33. \u00a0Calculate these volumes.\r\n<p id=\"fs-idm290040480\">a) What is the volume of 25 g iodine, density = 4.93 g\/cm<sup>3<\/sup>?<\/p>\r\n<p id=\"fs-idm307942992\">b) What is the volume of 3.28 g gaseous hydrogen, density = 0.089 g\/L?<\/p>\r\n34. \u00a0Convert the boiling temperature of gold, 2966 \u00b0C, into degrees Fahrenheit and kelvin.\r\n\r\n35. \u00a0Convert the temperature of the coldest area in a freezer, \u221210 \u00b0F, to degrees Celsius and kelvin.\r\n\r\n36. \u00a0Convert the boiling temperature of liquid ammonia, \u221228.1 \u00b0F, into degrees Celsius and kelvin.\r\n\r\n37. \u00a0The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 \u00b0C. What was the temperature on the Fahrenheit scale?\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<b>Answers<\/b>\r\n\r\n1. \u00a0a) \u00a0124\u00b0C \u00a0 \u00a0 \u00a0b) \u00a0\u2212159\u00b0C \u00a0 \u00a0 \u00a0c) \u00a0122\u00b0F \u00a0 \u00a0 \u00a0d) \u00a0\u221258\u00b0F\r\n\r\n2. \u00a0a) \u00a0373 K \u00a0 \u00a0 \u00a0b) \u00a0173 K \u00a0 \u00a0 \u00a0c) \u00a0\u2212173\u00b0C \u00a0 \u00a0 \u00a0d) \u00a027\u00b0C\r\n\r\n3.<b>\u00a0 \u00a0<\/b>\u2212273\u00b0C. This is the lowest possible temperature in degrees Celsius.\r\n\r\n4. \u00a057.8\u00b0C; \u00a0331 K\r\n\r\n5. \u00a0<span class=\"inlineequation\">a) \u00a01,000\u00a0mL\/1\u00a0L<\/span> and <span class=\"inlineequation\">1\u00a0L\/1,000\u00a0mL \u00a0 \u00a0 \u00a0<\/span><span class=\"inlineequation\">b) \u00a01,000,000\u00a0\u03bcs\/1\u00a0s<\/span> and <span class=\"inlineequation\">1\u00a0s\/1,000,000\u00a0\u03bcs<\/span>\r\n\r\n<span class=\"inlineequation\">c) \u00a01,000\u00a0m\/1\u00a0km<\/span> and <span class=\"inlineequation\">1\u00a0km1,000\u00a0m<\/span>\r\n\r\n6. \u00a0a) \u00a05,400 m \u00a0 \u00a0 \u00a0b) \u00a0665 mm \u00a0 \u00a0 \u00a0c) \u00a06.65 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> km\r\n\r\n7. \u00a0a) \u00a01.78 \u00d7 10<sup class=\"superscript\">\u22125<\/sup> g \u00a0 \u00a0 \u00a0b) \u00a07.22 \u00d7 10<sup class=\"superscript\">5<\/sup> g \u00a0 \u00a0 \u00a0c) 1.18 \u00d7 10<sup class=\"superscript\">6<\/sup> ng\r\n\r\n8. \u00a0a) \u00a094,400 cm<sup class=\"superscript\">2 \u00a0 \u00a0 \u00a0<\/sup>b) \u00a00.344 m<sup class=\"superscript\">3<\/sup>\r\n\r\n9. \u00a0One is a unit of area, and the other is a unit of volume.\r\n\r\n10. \u00a0a) \u00a00.75 m\/s \u00a0 \u00a0 \u00a0\u00a0b) \u00a0444 \u00b5m\/s \u00a0 \u00a0 \u00a0c) \u00a01.666 \u00d7 10<sup class=\"superscript\">\u22122<\/sup> km\/s\r\n\r\n11. \u00a0a) \u00a0674,000 mL \u00a0 \u00a0 \u00a0b) \u00a02.81 \u00d7 10<sup class=\"superscript\">6<\/sup> km \u00a0 \u00a0 \u00a0c) \u00a09.45 \u00d7 10<sup class=\"superscript\">7<\/sup> mg\r\n\r\n12. \u00a0a) \u00a06.77 \u00d7 10<sup class=\"superscript\">8<\/sup> ks \u00a0 \u00a0 \u00a0b) \u00a0345.5 km\r\n\r\n13. \u00a0a) \u00a06.0 \u00d7 10<sup class=\"superscript\">1<\/sup> mi\/h \u00a0 \u00a0 \u00a0b) \u00a00.00185 m\/s\r\n\r\n14. \u00a01.48 \u00d7 10<sup class=\"superscript\">3<\/sup> mm<sup class=\"superscript\">2<\/sup>\r\n\r\n15. \u00a0<span style=\"font-size: 11px\"><\/span>3.35 \u00d7 10<sup class=\"superscript\">3<\/sup> cm<sup class=\"superscript\">2<\/sup>\r\n<p id=\"fs-idp10346048\">16. \u00a0a) $latex \\frac{\\text{1.0936 yd}}{\\text{1 m}} $ \u00a0 b) $latex \\frac{\\text{0.94635 L}}{\\text{1 qt}} $ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0c) $latex \\frac{\\text{2.2046 lb}}{\\text{1 kg}} $<\/p>\r\n<p id=\"fs-idm205490112\">17. $latex \\frac{\\text{2.0 L}}{\\text{67.6 fl oz}} = \\frac{\\text{0.030 L}}{\\text{1 fl oz}}$\r\nOnly two significant figures are justified.<\/p>\r\n<p id=\"fs-idm184575152\">18. 68\u201371 cm; 400\u2013450 g<\/p>\r\n<p id=\"fs-idm208576112\">19. 355 mL<\/p>\r\n<p id=\"fs-idp32913072\">20. 8 \u00d7 10<sup>\u22124<\/sup> cm<\/p>\r\n<p id=\"fs-idm196114640\">21. yes; weight = 89.4 kg<\/p>\r\n<p id=\"fs-idm137236672\">22. 5.0 \u00d7 10<sup>\u22123<\/sup> mL<\/p>\r\n<p id=\"fs-idm211581456\">23. a) 1.3 \u00d7 10<sup>\u22124<\/sup> kg \u00a0 \u00a0 \u00a0b) 2.32 \u00d7 10<sup>8<\/sup> kg \u00a0 \u00a0 \u00a0c) 5.23 \u00d7 10<sup>\u221212<\/sup> m \u00a0 \u00a0 \u00a0d) 8.63 \u00d7 10<sup>\u22125<\/sup> kg \u00a0 \u00a0 \u00a0e) 3.76 \u00d7 10<sup>\u22121<\/sup> m \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f) 5.4 \u00d7 10<sup>\u22125<\/sup> m \u00a0 \u00a0 \u00a0g) 1 \u00d7 10<sup>12<\/sup> s \u00a0 \u00a0 \u00a0h) 2.7 \u00d7 10<sup>\u221211<\/sup> s \u00a0 \u00a0 \u00a0i) 1.5 \u00d7 10<sup>\u22124<\/sup> K<\/p>\r\n<p id=\"fs-idm268990752\">24. 45.4 L<\/p>\r\n<p id=\"fs-idp176467360\">25. 1.0160 \u00d7 10<sup>3<\/sup> kg<\/p>\r\n<p id=\"fs-idm134479920\">26. \u00a0a) 394 ft \u00a0 \u00a0 \u00a0b) 5.9634 km \u00a0 \u00a0 \u00a0c) 6.0 \u00d7 10<sup>2 \u00a0 \u00a0 \u00a0<\/sup>d) 2.64 L \u00a0 \u00a0 \u00a0e) 5.1 \u00d7 10<sup>18<\/sup> kg \u00a0 \u00a0 \u00a0f) 14.5 kg \u00a0 \u00a0 \u00a0g) 324 mg<\/p>\r\n<p id=\"fs-idp194388912\">27. 0.46 m; 1.5 ft\/cubit<\/p>\r\n<p id=\"fs-idm126197920\">28. Yes, the acid's volume is 123 mL.<\/p>\r\n<p id=\"fs-idm321078368\">29. 62.6 in (about 5 ft 3 in.) and 101 lb<\/p>\r\n<p id=\"fs-idm182624144\">30. (a) 3.81 cm \u00d7 8.89 cm \u00d7 2.44 m; (b) 40.6 cm<\/p>\r\n<p id=\"fs-idm209433984\">31. 2.70 g\/cm<sup>3<\/sup><\/p>\r\n<p id=\"fs-idp10159024\">32. (a) 81.6 g; (b) 17.6 g<\/p>\r\n<p id=\"fs-idm203289008\">33. (a) 5.1 mL; (b) 37 L<\/p>\r\n<p id=\"fs-idm24577456\">34. 5371 \u00b0F, 3239 K<\/p>\r\n<p id=\"fs-idm143339312\">35. \u221223 \u00b0C, 250 K<\/p>\r\n<p id=\"fs-idp51877792\">36. \u221233.4 \u00b0C, 239.8 K<\/p>\r\n<p id=\"fs-idm367616944\">37. 113 \u00b0F<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2><span style=\"font-family: Roboto, Helvetica, Arial, sans-serif\">Glossary<\/span><\/h2>\r\n<strong>dimensional analysis:\u00a0<\/strong>(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\r\n\r\n<strong>Fahrenheit:\u00a0<\/strong>unit of temperature; water freezes at 32 \u00b0F and boils at 212 \u00b0F on this scale\r\n\r\n<strong>unit conversion factor:\u00a0<\/strong>ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities<\/li>\n<li>Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties<\/li>\n<li>Learn about the various temperature scales that are commonly used in chemistry and how to convert from one scale to another.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm319461744\">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 <em>time<\/em> required for the athlete to run from the starting line to the finish line, and the <em>distance<\/em> between these two lines, and then computing <em>speed<\/em> from the equation that relates these three properties:<\/p>\n<div class=\"equation\" id=\"fs-idm290867056\" style=\"text-align: center\">[latex]\\text{speed}= \\frac{\\text{distance}}{\\text{time}}[\/latex]<\/div>\n<p id=\"fs-idm350487392\">An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of<\/p>\n<div class=\"equation\" id=\"fs-idm257675424\" style=\"text-align: center\">[latex]\\frac{\\text{100 m}}{\\text{10 s}} = \\text{10 m\/s}[\/latex]<\/div>\n<p id=\"fs-idm308822992\">Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100\/10 = 10) <em>and likewise<\/em> 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:<\/p>\n<div class=\"equation\" id=\"fs-idm219214640\" style=\"text-align: center\">[latex]\\text{time} = \\frac{\\text{distance}}{\\text{speed}}[\/latex]<\/div>\n<p id=\"fs-idm24572080\">The time can then be computed as:<\/p>\n<div class=\"equation\" id=\"fs-idp106016944\" style=\"text-align: center\">[latex]\\frac{\\text{25 m}}{\\text{10 m\/s}} = \\text{2.5 s}[\/latex]<\/div>\n<p id=\"fs-idm316688784\">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 \u201c1\u201d\u2014or, as commonly phrased, the units \u201ccancel.\u201d<\/p>\n<p id=\"fs-idp44099792\">These calculations are examples of a versatile mathematical approach known as <strong>dimensional analysis<\/strong> (or the <strong>factor-label method<\/strong>). Dimensional analysis is based on this premise: <em>the units of quantities must be subjected to the same mathematical operations as their associated numbers<\/em>. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.<\/p>\n<section id=\"fs-idm285086480\">\n<h2>Conversion Factors and Dimensional Analysis<\/h2>\n<p id=\"fs-idm273312256\">A ratio of two equivalent quantities expressed with different measurement units can be used as a <strong>unit conversion factor<\/strong>. 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,<\/p>\n<div class=\"equation\" id=\"fs-idm256748160\" style=\"text-align: center\">[latex]\\frac{\\text{2.54 cm}}{\\text{1 in.}}\\;\\text{(2.54 cm = 1 in.) or 2.54}\\frac{\\text{cm}}{\\text{in}}[\/latex]<\/div>\n<p id=\"fs-idm205801120\">Several other commonly used conversion factors are given in <a href=\"#fs-idm222237232\" class=\"autogenerated-content\">Table 1<\/a>.<\/p>\n<table id=\"fs-idm222237232\" class=\"span-all\" summary=\"This table is divided into 3 columns. They are titled length, volume, and mass. The following units are under the length column: 1 meter is equal to 1.0936 yards, 1 inch is equal to 2.54 cm 1 kilometer is equal to 0.62137 miles, 1 mile is equal to 1609.3 meters. The following units are under the volume column: 1 liter is equal to 1.0567 quarts, 1 quart is equal to 0.94635 meters, one cubic foot is equal to 28.317 liters, 1 tablespoon is equal to 14.787 milliliters. The following units are under the mass column: 1 kilogram is equal to 2.2046 pounds, 1 pound is equal to 453.59 grams, 1 avoirdupois ounce is equal to 28.349 grams, 1 troy ounce is equal to 31.103 grams.\">\n<thead>\n<tr valign=\"top\">\n<th>Length<\/th>\n<th>Volume<\/th>\n<th>Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1 m = 1.0936 yd<\/td>\n<td>1 L = 1.0567 qt<\/td>\n<td>1 kg = 2.2046 lb<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1 in. = 2.54 cm (exact)<\/td>\n<td>1 qt = 0.94635 L<\/td>\n<td>1 lb = 453.59 g<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1 km = 0.62137 mi<\/td>\n<td>1 ft<sup>3<\/sup> = 28.317 L<\/td>\n<td>1 (avoirdupois) oz = 28.349 g<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1 mi = 1609.3 m<\/td>\n<td>1 tbsp = 14.787 mL<\/td>\n<td>1 (troy) oz = 31.103 g<\/td>\n<\/tr>\n<tr>\n<td colspan=\"3\"><strong>Table 1.<\/strong> Common Conversion Factors<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm221472400\">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\u2019s vertical jump of 34 inches can be converted to centimeters by:<\/p>\n<div class=\"equation\" id=\"fs-idm153590912\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<p>Since this simple arithmetic involves <em>quantities<\/em>, the premise of dimensional analysis requires that we multiply both <em>numbers and units<\/em>. 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.}}[\/latex]. Just as for numbers, a ratio of identical units is also numerically equal to one, [latex]\\frac{\\text{in.}}{\\text{in.}}=\\text{1}[\/latex], and the unit product thus simplifies to <em>cm<\/em>. (When identical units divide to yield a factor of 1, they are said to \u201ccancel.\u201d) 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.<\/p>\n<\/section>\n<section>\n<div class=\"textbox shaded\" id=\"fs-idm150235328\">\n<h3>Example 1<\/h3>\n<p id=\"fs-idp22709840\">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 (<a href=\"#fs-idm222237232\" class=\"autogenerated-content\">Table 1<\/a>).<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm290807904\"><strong>Solution<\/strong><br \/>\nIf 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.<\/p>\n<div class=\"equation\" id=\"fs-idm138056288\" style=\"text-align: center\">[latex]x \\ \\text{oz} = \\text{125 \\text{g}} \\times \\text{unit conversion factor}[\/latex]<\/div>\n<p id=\"fs-idm300877280\">We write the unit conversion factor in its two forms:<\/p>\n<div class=\"equation\" id=\"fs-idm233281680\" style=\"text-align: center\">[latex]\\frac{1 \\text{oz}}{28.349 \\text{g}} \\;\\text{and}\\;\\frac{28.349 \\text{g}}{1 \\text{oz}}[\/latex]<\/div>\n<p id=\"fs-idm369973600\">The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.<\/p>\n<div class=\"equation\" id=\"fs-idm222314304\">\n<p style=\"text-align: center\">[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}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm224226288\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nConvert a volume of 9.345 qt to liters.<\/p>\n<p>&nbsp;<\/p>\n<p><strong><em>Answer<\/em><\/strong><\/p>\n<p>8.844 L<\/p>\n<\/div>\n<p id=\"fs-idm262838208\">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\u2014all the <em>factors<\/em> involved in the calculation must be appropriately oriented to insure that their <em>labels<\/em> (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.<\/p>\n<div class=\"textbox shaded\" id=\"fs-idm305814320\">\n<h3>Example 2<\/h3>\n<p id=\"fs-idm280005808\">What is the density of common antifreeze in units of g\/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp14199296\"><strong>Solution<\/strong><br \/>\nSince density = [latex]\\frac{\\text{mass}}{\\text{volume}}[\/latex], 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 \u00d7 unit conversion factor. The necessary conversion factors are given in <a href=\"#fs-idm222237232\" class=\"autogenerated-content\">Table 1<\/a>: 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:<\/p>\n<div class=\"equation\" id=\"fs-idm336821696\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<p id=\"fs-idm244171104\">We need to use two steps to convert volume from quarts to milliliters.<\/p>\n<ol id=\"fs-idm287602768\" class=\"stepwise\">\n<li><em>Convert quarts to liters.<\/em>\n<div class=\"equation\" id=\"fs-idm279137456\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<\/li>\n<li><em>Convert liters to milliliters.<\/em>\n<div class=\"equation\" id=\"fs-idm289883088\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-idm301215296\">Then,<\/p>\n<div class=\"equation\" id=\"fs-idm144640128\" style=\"text-align: center\">[latex]\\text{density} = \\frac{4.20 \\times 10^3 \\;\\text{g}}{3.78 \\times 10^3 \\;\\text{mL}} = 1.11 \\;\\text{g\/mL}[\/latex]<\/div>\n<p id=\"fs-idm292167648\">Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:<\/p>\n<div class=\"equation\" id=\"fs-idm9873904\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm219351440\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWhat is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?<\/p>\n<p>&nbsp;<\/p>\n<p><strong><em>Answer<\/em><\/strong><\/p>\n<p>[latex]2.956 \\times 10^{-2} \\text{L}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\" id=\"fs-idm306560960\">\n<h3>Example 3<\/h3>\n<p id=\"fs-idm338996800\">While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.<\/p>\n<p id=\"fs-idm127359616\">a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?<\/p>\n<p id=\"fs-idm315428528\">b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm327742336\"><strong>Solution<\/strong><br \/>\na) We first convert distance from kilometers to miles:<\/p>\n<div class=\"equation\" id=\"fs-idm60362224\" style=\"text-align: center\">[latex]1250\\;\\text{km} \\times \\frac{0.62137\\;\\text{mi}}{1\\;\\text{km}} = 777\\;\\text{mi}[\/latex]<\/div>\n<p id=\"fs-idm142282576\">and then convert volume from liters to gallons:<\/p>\n<div class=\"equation\" id=\"fs-idm214696672\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<p id=\"fs-idp1583872\">Then,<\/p>\n<div class=\"equation\" id=\"fs-idm114266320\" style=\"text-align: center\">[latex]\\text{(average) mileage} = \\frac{777 \\;\\text{mi}}{56.3\\;\\text{gal}} = 13.8\\;\\text{miles\/gallon} = 13.8\\;\\text{mpg}[\/latex]<\/div>\n<p id=\"fs-idm311190128\">Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:<\/p>\n<div class=\"equation\" id=\"fs-idm171836160\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<p id=\"fs-idm268939440\">b) Using the previously calculated volume in gallons, we find:<\/p>\n<div class=\"equation\" id=\"fs-idm328491840\" style=\"text-align: center\">[latex]\\text{56.3 gal} \\times \\frac{\\$3.80}{1\\;\\text{gal}} = \\$214[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm125734208\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nA Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).<\/p>\n<p id=\"fs-idm247721312\">a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?<\/p>\n<p id=\"fs-idm292238912\">b) If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answers<\/strong><\/em><\/p>\n<p>a) 51 mpg \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) $62<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm206910464\">\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p>a) Convert 35.9 kL to liters.<\/p>\n<p>b)Convert 555 nm to meters.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">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 <span class=\"inlineequation\">1,000\u00a0L\/1\u00a0kL<\/span>. Applying this conversion factor, we get<\/p>\n<p><span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_11.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_11.png\" alt=\"35.9 kL x (1000 L\/1 kL) = 35900 L\" class=\"wp-image-4836 alignnone\" width=\"233\" height=\"63\" \/><\/a><\/span><\/p>\n<p>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 10<sup class=\"superscript\">9<\/sup> nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: <span class=\"inlineequation\">1\u00a0m\/10<sup>9<\/sup>\u00a0nm<\/span>. Applying this conversion factor, we get<\/p>\n<p><span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_12.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_12.png\" alt=\"555 nm x (1 m\/ 10^9 nm) = 5.55 x 10^-7 m\" class=\"alignnone wp-image-4837\" width=\"458\" height=\"75\" \/><\/a><\/span><\/p>\n<p id=\"ball-ch02_s04_p20\" class=\"para\">In the final step, we expressed the answer in scientific notation.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p class=\"simpara\">a) Convert 67.08 \u03bcL to liters. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b)\u00a0Convert 56.8 m to kilometers.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answers<\/em><\/strong><\/p>\n<p class=\"simpara\">a) 6.708 \u00d7 10<sup class=\"superscript\">\u22125<\/sup> L \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b)\u00a05.68 \u00d7 10<sup class=\"superscript\">\u22122<\/sup> km<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 5<\/h3>\n<p class=\"Indent\">Complete the following conversions<br \/>\n<span>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>(use table 1.1 for any metric conversions;<span>\u00a0 <\/span>Note: 1 L = 1.0567 qt )<\/p>\n<p class=\"IndentSub\">a) 125 m = ? mm<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>b) 2.3 x 10<sup>-6 <\/sup>Mg = ? g<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>c) 2.5 L = ? qt<\/p>\n<p>&nbsp;<\/p>\n<p class=\"Solution\"><strong>Solution<\/strong><span><strong>\u00a0<\/strong>\u00a0 <\/span><\/p>\n<p class=\"Indentpoints\">a)<span>\u00a0\u00a0 <\/span>[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}[\/latex]<\/p>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em;text-align: left\">Note that this equivalence statement comes from the literal \u201cmeaning\u201d 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.<\/span><\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\">b)<span>\u00a0\u00a0<\/span>[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}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">c)<\/span><span style=\"font-size: 1em\">\u00a0\u00a0<\/span>[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}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span><\/span><em style=\"font-size: 1em\"><strong>Test Yourself<\/strong><\/em><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">Complete the following conversions<\/span><sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sup><span style=\"font-size: 1em\">(note: 1 in = 2.54 cm exactly)<\/span><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">a) 124 mL = ? L\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b) 256 days = ? hrs<\/span><sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sup><span style=\"font-size: 1em\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c) 63.2 cm = ? in<\/span><\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\"><em style=\"font-size: 1em\"><strong>Answers<\/strong><\/em><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span style=\"font-size: 1em\">a) 0.124 L \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 6.14 x 10<\/span><sup>3<\/sup><span style=\"font-size: 1em\"> hrs \u00a0 \u00a0 \u00a0 \u00a0 \u00a0c) 24.9 in<\/span><\/div>\n<\/div>\n<\/div>\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 6<\/h3>\n<p class=\"Indent\">How many nm are in 26.5 feet? (12 in = 1 ft ,<span>\u00a0 <\/span>and<span>\u00a0<\/span>2.54 cm = 1 in exactly)<\/p>\n<p>&nbsp;<\/p>\n<p class=\"Solution\"><strong>Solution<\/strong><span><strong>\u00a0<\/strong>\u00a0 <\/span><\/p>\n<p class=\"Indent\">Likely, you do not have a direct conversion between feet and nm. So, instead, ask yourself \u201cwhere can I go from feet\u201d. The only possibility is inches. Then, where can you go from inches? \u2026and so on. The overall path becomes:<\/p>\n<p class=\"IndentSub\"><span>\u00a0<\/span>feet\u00a0[latex]\\longrightarrow[\/latex]\u00a0inches\u00a0[latex]\\longrightarrow[\/latex]\u00a0cm\u00a0[latex]\\longrightarrow[\/latex]\u00a0m\u00a0[latex]\\longrightarrow[\/latex]\u00a0<span style=\"font-size: 1em\">nm<\/span><\/p>\n<p class=\"IndentSub\"><span>\u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/span><\/p>\n<p>[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}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p class=\"SelfTest\"><em><strong>Test Yourself<\/strong><\/em><\/p>\n<p id=\"ball-ch02_s04_p22\" class=\"para\"><span>A marathon is 26.4 miles. If 1 mile = 1760 yards, and 1 m = 1.094 yards, how many km are in a marathon?<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>42.5 km<\/p>\n<\/div>\n<\/div>\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 7<\/h3>\n<p class=\"Indent\">a)<span>\u00a0\u00a0 <\/span>A car is moving at 35.2 km\/h. How many cm\/s is this speed?<br \/>\nb)<span>\u00a0\u00a0 <\/span>How many cm<sup>3<\/sup>are in 5 x 10<sup>2<\/sup>m<sup>3<\/sup>?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"Solution\"><strong>Solution<\/strong><span>\u00a0\u00a0 <\/span><\/p>\n<p class=\"Indentpoints\">a)<span>\u00a0 <\/span>We must convert km [latex]\\longrightarrow[\/latex] m [latex]\\longrightarrow[\/latex] cm AND convert h [latex]\\longrightarrow[\/latex] min [latex]\\longrightarrow[\/latex] 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.<\/p>\n<p class=\"IndentSub\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/p>\n<p>[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}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p class=\"Indentpoints\">Remember\u2026<br \/>\n1 m<sup>3<\/sup>= 1 m x 1 m x 1 m, so<span>\u00a0 <\/span>1 m<sup>3<\/sup>= 100 cm x 100 cm x 100 cm, NOT 100 cm<sup>3<\/sup><\/p>\n<p class=\"IndentSub\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n<p class=\"IndentSub\">[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}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p class=\"SelfTest\"><em><strong>Test Yourself<\/strong><\/em><\/p>\n<p class=\"Indent\">Complete the following conversions<sup><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span><\/sup><\/p>\n<p class=\"Indent\">a)<span>\u00a0 <\/span>75 mi\/h = ? m\/s<span>\u00a0 <\/span>(1760 yd = 1 mi and 1 m = 1.094 yd)<br \/>\nb)<span>\u00a0 <\/span>4.1 g\/cm<sup>3<\/sup>= ? kg\/m<sup>3<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answers<\/strong><\/em><\/p>\n<p>a) 34 m\/s \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) 4.1 x 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/p>\n<\/div>\n<\/div>\n<section id=\"fs-idm206910464\">\n<div class=\"section\" id=\"ball-ch02_s04\" lang=\"en\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 8<\/h3>\n<p id=\"ball-ch02_s04_p22\" class=\"para\">How many cubic centimeters are in 0.883 m<sup class=\"superscript\">3<\/sup>?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch02_s04_p23\" class=\"para\">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<\/p>\n<p><span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_14.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_14.png\" alt=\"0.883m^3 x (100cm\/1m) x (100cm\/1m) x (100cm\/1m) = 883000 cm^3\" class=\"alignnone wp-image-4839\" width=\"593\" height=\"56\" \/><\/a><\/span><\/p>\n<p id=\"ball-ch02_s04_p24\" class=\"para\">You should demonstrate to yourself that the three meter units do indeed cancel.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p25\" class=\"para\">How many cubic millimeters are present in 0.0923 m<sup class=\"superscript\">3<\/sup>?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p26\" class=\"para\">9.23 \u00d7 10<sup class=\"superscript\">7<\/sup> mm<sup class=\"superscript\">3<\/sup><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 9<\/h3>\n<p id=\"ball-ch02_s04_p28\" class=\"para\">Convert 88.4 m\/min to meters\/second.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch02_s04_p29\" class=\"para\">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: <span class=\"inlineequation\">1\u00a0min\/60\u00a0s<\/span>. Apply and perform the math:<\/p>\n<p><span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_15.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_15.png\" alt=\"88.4m\/m x 1min\/60s = 1.47 m\/s\" class=\"alignnone wp-image-4840\" width=\"225\" height=\"52\" \/><\/a><\/span><\/p>\n<p id=\"ball-ch02_s04_p30\" class=\"para\">Notice how the 88.4 automatically goes in the numerator. That\u2019s because any number can be thought of as being in the numerator of a fraction divided by 1.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p31\" class=\"para\">Convert 0.203 m\/min to meters\/second.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p32\" class=\"para\">0.00338 m\/s or 3.38 \u00d7 10<sup class=\"superscript\">\u22123<\/sup> m\/s<\/p>\n<figure id=\"attachment_2116\" aria-describedby=\"caption-attachment-2116\" style=\"width: 300px\" class=\"wp-caption alignnone\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Grapevinesnail_01-1-e1528932055925.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Grapevinesnail_01-1-e1528932055925.jpg\" alt=\"\" width=\"300\" height=\"177\" class=\"wp-image-2116 size-full\" \/><\/a><figcaption id=\"caption-attachment-2116\" class=\"wp-caption-text\"><strong>Figure 1.<\/strong> How fast is fast?\u00a0A 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! \u00a0Source: \u201cGrapevine snail\u201dby J\u00fcrgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idm206910464\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 10<\/h3>\n<p id=\"ball-ch02_s04_p36\" class=\"para\">How many nanoseconds are in 368.09 \u03bcs?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch02_s04_p37\" class=\"para\">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 <em class=\"emphasis\">micro-<\/em> and <em class=\"emphasis\">nano-<\/em>,<\/p>\n<p><span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_21.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_21.png\" alt=\"368.09 us x 1s\/10^6us x 10^9ns \/1s = 368090 ns = 3.608 x 10^5 ns\" class=\"alignnone wp-image-4846\" width=\"544\" height=\"56\" \/><\/a><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p38\" class=\"para\">How many milliliters are in 607.8 kL?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p39\" class=\"para\">6.078 \u00d7 10<sup class=\"superscript\">8<\/sup> mL<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm206910464\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 11<\/h3>\n<p id=\"ball-ch02_s04_p43\" class=\"para\">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.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch02_s04_p44\" class=\"para\">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.<\/p>\n<p><span class=\"informalequation\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2015\/11\/converting_units_22.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/converting_units_22.png\" alt=\"36.7 cm x 128.8 cm x 1 m\/100cm x 1 m\/100 cm = 0.472696 m^2 = 0.473 m^2\" class=\"alignnone wp-image-4847\" width=\"575\" height=\"60\" \/><\/a><\/span><\/p>\n<p id=\"ball-ch02_s04_p45\" class=\"para\">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.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p46\" class=\"para\">What is the volume of a block in cubic meters whose dimensions are 2.1 cm \u00d7 34.0 cm \u00d7 118 cm?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s04_p47\" class=\"para\">0.0084 m<sup class=\"superscript\">3<\/sup><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Chemistry Is Everywhere: The Gimli Glider<\/h3>\n<p id=\"ball-ch02_s04_p48\" class=\"para\">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 \u201cthe Gimli Glider.\u201d<\/p>\n<div class=\"informalfigure large\" id=\"ball-ch02_s04_f02\">\n<div class=\"copyright\">\n<figure id=\"attachment_2124\" aria-describedby=\"caption-attachment-2124\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-300x219.jpg\" alt=\"\" width=\"300\" height=\"219\" class=\"wp-image-2124 size-medium\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-300x219.jpg 300w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-768x562.jpg 768w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-65x48.jpg 65w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-225x165.jpg 225w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1-350x256.jpg 350w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Aircanada.b767-300er.c-ggmx.arp_-1.jpg 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-2124\" class=\"wp-caption-text\"><strong>Figure 2.<\/strong> 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. \u201cAircanada.b767\u2032\u2032 is in the the public domain.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<p id=\"ball-ch02_s04_p49\" class=\"para\">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.<\/p>\n<p id=\"ball-ch02_s04_p50\" class=\"para\">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 <em class=\"emphasis\">pounds<\/em> 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\u2014an incident that would not have occurred if people were watching their units.<\/p>\n<\/div>\n<h2>Conversion of Temperature Units<\/h2>\n<p id=\"fs-idm262720192\">We use the word <strong class=\"no-emphasis\">temperature<\/strong> 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.<\/p>\n<p id=\"fs-idm308860096\">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 \u00b0C is defined as the freezing temperature of water and 100 \u00b0C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the <strong>Fahrenheit<\/strong> scale, the freezing point of water is defined as 32 \u00b0F and the boiling temperature as 212 \u00b0F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).<\/p>\n<p id=\"fs-idm288396336\">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:<\/p>\n<div class=\"equation\" id=\"fs-idm6121760\" style=\"text-align: center\">[latex]\\text{length in feet} = \\left( \\frac {1\\;\\text{ft}}{12\\;\\text{in.}}\\right) \\times \\text{length in inches}[\/latex]<\/div>\n<p id=\"fs-idm161552768\">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\u2019 zero points (b).<\/p>\n<p id=\"fs-idm292928800\">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 <em>x<\/em> and the Fahrenheit temperature as <em>y<\/em>, the slope, <em>m<\/em>, is computed to be:<\/p>\n<div class=\"equation\" id=\"fs-idm229969840\">\n<p style=\"text-align: center\">[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}}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-idm206340880\">The y-intercept of the equation, <em>b<\/em>, is then calculated using either of the equivalent temperature pairs, (100 \u00b0C, 212 \u00b0F) or (0 \u00b0C, 32 \u00b0F), as:<\/p>\n<div class=\"equation\" id=\"fs-idm234430272\" style=\"text-align: center\">[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}[\/latex]<\/div>\n<p id=\"fs-idm218561008\">The equation relating the temperature scales is then:<\/p>\n<div class=\"equation\" id=\"fs-idm305465424\">\n<p style=\"text-align: center\">[latex]T_{^\\circ\\text{F}} = (\\frac{9 \\;^\\circ\\text{F}}{5 \\;^\\circ\\text{C}} \\times T_{^\\circ\\text{C}}) + 32\\;^\\circ\\text{C}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-idm208304512\">An abbreviated form of this equation that omits the measurement units is:<\/p>\n<div class=\"equation\" id=\"fs-idm226315088\" style=\"text-align: center\">[latex]T_{^\\circ\\text{F}} = \\frac{9}{5} \\times T_{^\\circ\\text{C}} + 32[\/latex]<\/div>\n<p id=\"fs-idm356038704\">Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:<\/p>\n<div class=\"equation\" id=\"fs-idm138168256\" style=\"text-align: center\">[latex]T_{^\\circ\\text{C}} = \\frac{5}{9} (T_{^\\circ\\text{F}} - 32)[\/latex]<\/div>\n<p id=\"fs-idm131802496\">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&#8217;s volume and temperature suggested that the volume of a gas would be zero at \u2212273.15 \u00b0C. 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\u2019s chapter on gases).<\/p>\n<p id=\"fs-idm309875456\">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}}[\/latex]. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:<\/p>\n<div class=\"equation\" id=\"fs-idp289344\" style=\"text-align: center\">[latex]T_{\\text{K}} = T_{^\\circ\\text{C}} + 273.15[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idm303916848\" style=\"text-align: center\">[latex]T_{^\\circ\\text{C}} = T_{\\text{K}} - 273.15[\/latex]<\/div>\n<p id=\"fs-idm288599552\">The 273.15 in these equations has been determined experimentally, so it is not exact. <a href=\"#CNX_Chem_01_06_TempScales\" class=\"autogenerated-content\">Figure 3<\/a> shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.<\/p>\n<figure id=\"CNX_Chem_01_06_TempScales\"><figcaption>\n<figure style=\"width: 1300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_01_06_TempScales-2.jpg\" alt=\"A thermometer is shown for the Fahrenheit, Celsius and Kelvin scales. Under the Fahrenheit scale, the boiling point of water is 212 degrees while the freezing point of water is 32 degrees. Therefore, there are 180 Fahrenheit degrees between the boiling point of water and the freezing point of water. Under the Celsius scale, the boiling point of water is 100 degrees while the freezing point of water is 0 degrees. Therefore, there are 100 Celsius degrees between the boiling point and freezing point of water. Under the kelvin scale, the boiling point of water is 373.15 K, while the freezing point of water is 273.15 K. 233.15 K is equal to negative 40 degrees Celsius, which is also equal to negative 40 degrees Fahrenheit.\" width=\"1300\" height=\"911\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> The Fahrenheit, Celsius, and kelvin temperature scales are compared.<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<p id=\"fs-idm296725200\">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.<\/p>\n<div class=\"textbox shaded\">\n<div class=\"example\" id=\"fs-idm75569040\">\n<h3>Example\u00a012<\/h3>\n<p id=\"fs-idm290946544\">Normal body temperature has been commonly accepted as 37.0 \u00b0C (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?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm288550064\"><strong>Solution<\/strong><\/p>\n<div class=\"equation\" id=\"fs-idm223031696\" style=\"text-align: left\">[latex]\\text{K} = ^\\circ\\text{C} + \\text{273.15} = \\text{37.0} + \\text{273.2} = \\text{310.2 K}[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idp114907616\" style=\"text-align: left\">[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}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp16310096\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nConvert 80.92 \u00b0C to K and \u00b0F.<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answers<\/strong><\/em><\/p>\n<p>354.07 K, 177.7 \u00b0F<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" id=\"fs-idm309453552\">\n<h3>Example 13<\/h3>\n<p id=\"fs-idm421846336\">Baking a ready-made pizza calls for an oven temperature of 450 \u00b0F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm36231872\"><strong>Solution<\/strong><\/p>\n<div class=\"equation\" id=\"fs-idm32912\" style=\"text-align: left\">[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]<\/div>\n<div class=\"equation\" id=\"fs-idm144212448\" style=\"text-align: left\">[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)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm296519728\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nConvert 50 \u00b0F to \u00b0C and K.<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answers<\/strong><\/em><\/p>\n<p>10 \u00b0C, 280 K<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm292051392\" class=\"summary\">\n<div class=\"callout block\" id=\"ball-ch02_s04_n07\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 14<\/h3>\n<p>a) What is 98.6 \u00b0F in degrees Celsius?<\/p>\n<p>b) What is 25.0 \u00b0C in degrees Fahrenheit?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) Using the first formula from above, we have<\/p>\n<div class=\"equation\" id=\"fs-idm32912\" style=\"text-align: left\">[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}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>b) Using the second formula from above, we have<\/p>\n<div class=\"equation\" id=\"fs-idp114907616\" style=\"text-align: left\">[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}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p class=\"simpara\">a) Convert 0 \u00b0F to degrees Celsius.<\/p>\n<p class=\"simpara\">b) Convert 212 \u00b0C to degrees Fahrenheit.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answers<\/em><\/strong><\/p>\n<p class=\"simpara\">a) \u221217.8 \u00b0C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b)\u00a0414 \u00b0F<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 15<\/h3>\n<p id=\"ball-ch02_s05_p07\" class=\"para\">If normal room temperature is 72.0 \u00b0F, what is room temperature in degrees Celsius and kelvins?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch02_s05_p08\" class=\"para\">First, we use the formula to determine the temperature in degrees Celsius:<\/p>\n<p>[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}[\/latex]<\/p>\n<p id=\"ball-ch02_s05_p09\" class=\"para\">Then we use the appropriate formula above to determine the temperature in the Kelvin scale:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">K = 22.2 \u00b0C +\u00a0273.15 = 295.4 K<\/span><\/span><\/p>\n<p id=\"ball-ch02_s05_p10\" class=\"para\">So, room temperature is about 295 K.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s05_p11\" class=\"para\">What is 98.6 \u00b0F on the Kelvin scale?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch02_s05_p12\" class=\"para\">310.2 K<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Food and Drink App: Cooking Temperatures<\/h3>\n<p id=\"ball-ch02_s05_p78\" class=\"para\">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\u00b0F, while a cake may be baked at 350\u00b0F and a chicken roasted at 400\u00b0F. The broil setting on many ovens is 500\u00b0F, which is typically the highest temperature setting on a household oven.<\/p>\n<p id=\"ball-ch02_s05_p79\" class=\"para\">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\u00b0F (100\u00b0C), but in Denver, the Mile-High City, water boils at about 200\u00b0F (93.3\u00b0C), which can significantly lengthen cooking times. Good cooks need to be aware of this.<\/p>\n<p id=\"ball-ch02_s05_p80\" class=\"para\">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\u00b0F (122\u00b0C); 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 <em class=\"emphasis\">autoclave<\/em>.)<\/p>\n<p id=\"ball-ch02_s05_p81\" class=\"para\">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\u2014a 425\u00b0F oven in the United States is equivalent to a 220\u00b0C oven in other countries. These days, many oven thermometers are marked with both temperature scales.<\/p>\n<\/div>\n<section id=\"fs-idm292051392\" class=\"summary\">\n<div id=\"fs-idp86805728\" class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-1-2.png\" alt=\"\" width=\"129\" height=\"80\" class=\"alignleft\" \/><\/p>\n<p id=\"fs-idm169361696\">Need a refresher or more practice with unit conversion? Visit this site (<a href=\"https:\/\/viuvideos.viu.ca\/media\/Unit+Conversion\/0_o671v9j6\">https:\/\/viuvideos.viu.ca\/media\/Unit+Conversion\/0_o671v9j6<\/a>) to go over the basics of unit conversions.<\/p>\n<p>Video source: Unit conversion by keyj<\/p>\n<\/div>\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idm126307616\">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.<\/p>\n<\/section>\n<section id=\"fs-idm299998176\" class=\"key-equations\">\n<h2>Key Equations<\/h2>\n<ul id=\"fs-idm309482144\">\n<li>[latex]T_{^\\circ\\text{C}} = \\frac{5}{9} \\times T_{^\\circ\\text{F}} - 32[\/latex]<\/li>\n<li>[latex]T_{^\\circ\\text{F}} = \\frac{9}{5} \\times T_{^\\circ\\text{C}} + 32[\/latex]<\/li>\n<li>[latex]T_\\text{K} = {^\\circ\\text{C}} + 273.15[\/latex]<\/li>\n<li>[latex]T_{^\\circ\\text{C}} = \\text{K} - 273.15[\/latex]<\/li>\n<\/ul>\n<\/section>\n<div class=\"callout block\" id=\"ball-ch02_s05_n06\">\n<div class=\"qandaset block\" id=\"ball-ch02_s05_qs01\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"question\">\n<p id=\"ball-ch02_s05_qs01_p1\" class=\"para\">1. \u00a0Perform the following conversions.<\/p>\n<\/div>\n<p>a) \u00a0255\u00b0F to degrees Celsius \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0\u2212255\u00b0F to degrees Celsius<\/p>\n<p>c) \u00a050.0\u00b0C to degrees Fahrenheit \u00a0 \u00a0 \u00a0 \u00a0 \u00a0d) \u00a0\u221250.0\u00b0C to degrees Fahrenheit<\/p>\n<div class=\"question\">\n<p id=\"ball-ch02_s05_qs01_p3\" class=\"para\">2. \u00a0Perform the following conversions.<\/p>\n<p>a) \u00a0100.0\u00b0C to kelvins \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0\u2212100.0\u00b0C to kelvins<\/p>\n<p>c) \u00a0100 K to degrees Celsius \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0d) \u00a0300 K to degrees Celsius<\/p>\n<\/div>\n<div class=\"question\"><\/div>\n<div class=\"question\">\n<p id=\"ball-ch02_s05_qs01_p5\" class=\"para\">3. \u00a0Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch02_s05_qs01_p9\" class=\"para\">4. \u00a0The hottest temperature ever recorded on the surface of the earth was 136\u00b0F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins?<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">5. \u00a0Write the two conversion factors that exist between the two given units.<\/span><\/p>\n<\/div>\n<p>a) \u00a0milliliters and laters \u00a0 \u00a0 \u00a0b) \u00a0microseconds and seconds \u00a0 \u00a0 \u00a0\u00a0c) \u00a0kilometers and meters<\/p>\n<div class=\"question\"><span style=\"font-size: 1em\">6. \u00a0Perform the following conversions.<\/span><\/div>\n<div class=\"question\">\n<p>a) \u00a05.4 km to meters \u00a0 \u00a0 \u00a0b) \u00a00.665 m to millimeters \u00a0 \u00a0 \u00a0c) \u00a00.665 m to kilometers<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">7. \u00a0Perform the following conversions.<\/span><\/div>\n<div class=\"question\">\n<p>a) \u00a017.8 \u03bcg to grams \u00a0 \u00a0 \u00a0b) \u00a07.22 \u00d7 10<sup class=\"superscript\">2<\/sup> kg to grams \u00a0 \u00a0 \u00a0c) \u00a00.00118 g to nanograms<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">8. \u00a0Perform the following conversions.<\/span><\/div>\n<div class=\"question\">\n<p>a) \u00a09.44 m<sup class=\"superscript\">2<\/sup> to square centimetres \u00a0 \u00a0 \u00a0b) \u00a03.44 \u00d7 10<sup class=\"superscript\">8<\/sup> mm<sup class=\"superscript\">3<\/sup> to cubic meters<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">9. \u00a0Why would it be inappropriate to convert square centimeters to cubic meters?<\/span><\/div>\n<div class=\"question\">\n<p id=\"ball-ch02_s04_qs01_p13\" class=\"para\">10. \u00a0Perform the following conversions.<\/p>\n<p>a) \u00a045.0 m\/min to meters\/second \u00a0 \u00a0 \u00a0 b) \u00a00.000444 m\/s to micrometers\/second<\/p>\n<p>c) \u00a060.0 km\/h to kilometers\/second<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">11. \u00a0Perform the following conversions.<\/span><\/div>\n<div class=\"question\">\n<p>a) \u00a00.674 kL to milliliters \u00a0 \u00a0 \u00a0b) \u00a02.81 \u00d7 10<sup class=\"superscript\">12<\/sup> mm to kilometers \u00a0 \u00a0 \u00a0c) \u00a094.5 kg to milligrams<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">12. \u00a0Perform the following conversions.<\/span><\/div>\n<div class=\"question\">\n<p>a) \u00a06.77 \u00d7 10<sup class=\"superscript\">14<\/sup> ms to kilo seconds \u00a0 \u00a0 \u00a0\u00a0b) \u00a034,550,000 cm to kilometers<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">13. \u00a0Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.<\/span><\/div>\n<div class=\"question\">\n<p>a) \u00a088 ft\/s to miles\/hour (Hint: use 5,280 ft = 1 mi.) \u00a0 \u00a0 \u00a0b) \u00a00.00667 km\/h to meters\/second<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">14. \u00a0What is the area in square millimeters of a rectangle whose sides are 2.44 cm \u00d7 6.077 cm? Express the answer to the proper number of significant figures.<\/span><\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">15. \u00a0The formula for the area of a triangle is 1\/2 \u00d7 base \u00d7 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.<\/span><\/div>\n<div class=\"question\">\n<p>16. \u00a0<span style=\"font-size: 1em\">Write conversion factors (as ratios) for the number of:<\/span><\/p>\n<p id=\"fs-idm279869696\">a) yards in 1 meter \u00a0 \u00a0 \u00a0b) liters in 1 liquid quart \u00a0 \u00a0 \u00a0c) pounds in 1 kilogram<\/p>\n<p>17. \u00a0The 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?<\/p>\n<p>18. \u00a0Soccer 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?<\/p>\n<p>19. \u00a0How many milliliters of a soft drink are contained in a 12.0-oz can?<\/p>\n<p>20. \u00a0The diameter of a red blood cell is about 3 \u00d7 10<sup>\u22124<\/sup> in. What is its diameter in centimeters?<\/p>\n<p>21. \u00a0Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?<\/p>\n<p>22. \u00a0Many medical laboratory tests are run using 5.0 \u03bcL blood serum. What is this volume in milliliters?<\/p>\n<p>23. \u00a0Use scientific notation to express the following quantities in terms of the SI base units:<\/p>\n<p id=\"fs-idm287857616\">a) 0.13 g \u00a0 \u00a0 \u00a0b) 232 Gg \u00a0 \u00a0 \u00a0c) 5.23 pm \u00a0 \u00a0 \u00a0d) 86.3 mg \u00a0 \u00a0 \u00a0e) 37.6 cm \u00a0 \u00a0 \u00a0f) 54 \u03bcm \u00a0 \u00a0 \u00a0g) 1 Ts \u00a0 \u00a0 \u00a0h) 27 ps \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0i) 0.15 mK<\/p>\n<p>24. \u00a0Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank?<\/p>\n<p>25. \u00a0A long ton is defined as exactly 2240 lb. What is this mass in kilograms?<\/p>\n<p>26. \u00a0Make the conversion indicated in each of the following:<\/p>\n<p id=\"fs-idm367197520\">a) the length of a soccer field, 120 m (three significant figures), to feet<\/p>\n<p id=\"fs-idp51878496\">b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers<\/p>\n<p id=\"fs-idm307271088\">c) the area of an 8.5 t 11-inch sheet of paper in cm<sup>2<\/sup><\/p>\n<p id=\"fs-idm127105888\">d) the displacement volume of an automobile engine, 161 in.<sup>3<\/sup>, to liters<\/p>\n<p id=\"fs-idm218455584\">e) the estimated mass of the atmosphere, 5.6 t 10<sup>15<\/sup> tons, to kilograms<\/p>\n<p id=\"fs-idm98022832\">f) the mass of a bushel of rye, 32.0 lb, to kilograms<\/p>\n<p id=\"fs-idm162390144\">g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)<\/p>\n<p>27. \u00a0A chemist\u2019s 50-Trillion Angstrom Run\u00a0would be an archeologist\u2019s 10,900 cubit run. How long is one cubit in meters and in feet? (1 \u00c5 = 1 \u00d7 10<sup>\u22128<\/sup> cm)<\/p>\n<p>28. \u00a0As 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?<\/p>\n<p>29. \u00a0A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds?<\/p>\n<p>30. \u00a0Solve these problems about lumber dimensions.<\/p>\n<p id=\"fs-idm124373120\">a) To describe to a European how houses are constructed in the US, the dimensions of \u201ctwo-by-four\u201d lumber must be converted into metric units. The thickness \u00d7 width \u00d7 length dimensions are 1.50 in. \u00d7 3.50 in. \u00d7 8.00 ft in the US. What are the dimensions in cm \u00d7 cm \u00d7 m?<\/p>\n<p id=\"fs-idm97712544\">b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?<\/p>\n<p>31. \u00a0Calculate the density of aluminum if 27.6 cm<sup>3<\/sup> has a mass of 74.6 g.<\/p>\n<p>32. \u00a0Calculate these masses.<\/p>\n<p id=\"fs-idp56604960\">a) What is the mass of 6.00 cm<sup>3<\/sup> of mercury, density = 13.5939 g\/cm<sup>3<\/sup>?<\/p>\n<p id=\"fs-idm84745936\">b) What is the mass of 25.0 mL octane, density = 0.702 g\/cm<sup>3<\/sup>?<\/p>\n<p>33. \u00a0Calculate these volumes.<\/p>\n<p id=\"fs-idm290040480\">a) What is the volume of 25 g iodine, density = 4.93 g\/cm<sup>3<\/sup>?<\/p>\n<p id=\"fs-idm307942992\">b) What is the volume of 3.28 g gaseous hydrogen, density = 0.089 g\/L?<\/p>\n<p>34. \u00a0Convert the boiling temperature of gold, 2966 \u00b0C, into degrees Fahrenheit and kelvin.<\/p>\n<p>35. \u00a0Convert the temperature of the coldest area in a freezer, \u221210 \u00b0F, to degrees Celsius and kelvin.<\/p>\n<p>36. \u00a0Convert the boiling temperature of liquid ammonia, \u221228.1 \u00b0F, into degrees Celsius and kelvin.<\/p>\n<p>37. \u00a0The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 \u00b0C. What was the temperature on the Fahrenheit scale?<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><b>Answers<\/b><\/p>\n<p>1. \u00a0a) \u00a0124\u00b0C \u00a0 \u00a0 \u00a0b) \u00a0\u2212159\u00b0C \u00a0 \u00a0 \u00a0c) \u00a0122\u00b0F \u00a0 \u00a0 \u00a0d) \u00a0\u221258\u00b0F<\/p>\n<p>2. \u00a0a) \u00a0373 K \u00a0 \u00a0 \u00a0b) \u00a0173 K \u00a0 \u00a0 \u00a0c) \u00a0\u2212173\u00b0C \u00a0 \u00a0 \u00a0d) \u00a027\u00b0C<\/p>\n<p>3.<b>\u00a0 \u00a0<\/b>\u2212273\u00b0C. This is the lowest possible temperature in degrees Celsius.<\/p>\n<p>4. \u00a057.8\u00b0C; \u00a0331 K<\/p>\n<p>5. \u00a0<span class=\"inlineequation\">a) \u00a01,000\u00a0mL\/1\u00a0L<\/span> and <span class=\"inlineequation\">1\u00a0L\/1,000\u00a0mL \u00a0 \u00a0 \u00a0<\/span><span class=\"inlineequation\">b) \u00a01,000,000\u00a0\u03bcs\/1\u00a0s<\/span> and <span class=\"inlineequation\">1\u00a0s\/1,000,000\u00a0\u03bcs<\/span><\/p>\n<p><span class=\"inlineequation\">c) \u00a01,000\u00a0m\/1\u00a0km<\/span> and <span class=\"inlineequation\">1\u00a0km1,000\u00a0m<\/span><\/p>\n<p>6. \u00a0a) \u00a05,400 m \u00a0 \u00a0 \u00a0b) \u00a0665 mm \u00a0 \u00a0 \u00a0c) \u00a06.65 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> km<\/p>\n<p>7. \u00a0a) \u00a01.78 \u00d7 10<sup class=\"superscript\">\u22125<\/sup> g \u00a0 \u00a0 \u00a0b) \u00a07.22 \u00d7 10<sup class=\"superscript\">5<\/sup> g \u00a0 \u00a0 \u00a0c) 1.18 \u00d7 10<sup class=\"superscript\">6<\/sup> ng<\/p>\n<p>8. \u00a0a) \u00a094,400 cm<sup class=\"superscript\">2 \u00a0 \u00a0 \u00a0<\/sup>b) \u00a00.344 m<sup class=\"superscript\">3<\/sup><\/p>\n<p>9. \u00a0One is a unit of area, and the other is a unit of volume.<\/p>\n<p>10. \u00a0a) \u00a00.75 m\/s \u00a0 \u00a0 \u00a0\u00a0b) \u00a0444 \u00b5m\/s \u00a0 \u00a0 \u00a0c) \u00a01.666 \u00d7 10<sup class=\"superscript\">\u22122<\/sup> km\/s<\/p>\n<p>11. \u00a0a) \u00a0674,000 mL \u00a0 \u00a0 \u00a0b) \u00a02.81 \u00d7 10<sup class=\"superscript\">6<\/sup> km \u00a0 \u00a0 \u00a0c) \u00a09.45 \u00d7 10<sup class=\"superscript\">7<\/sup> mg<\/p>\n<p>12. \u00a0a) \u00a06.77 \u00d7 10<sup class=\"superscript\">8<\/sup> ks \u00a0 \u00a0 \u00a0b) \u00a0345.5 km<\/p>\n<p>13. \u00a0a) \u00a06.0 \u00d7 10<sup class=\"superscript\">1<\/sup> mi\/h \u00a0 \u00a0 \u00a0b) \u00a00.00185 m\/s<\/p>\n<p>14. \u00a01.48 \u00d7 10<sup class=\"superscript\">3<\/sup> mm<sup class=\"superscript\">2<\/sup><\/p>\n<p>15. \u00a0<span style=\"font-size: 11px\"><\/span>3.35 \u00d7 10<sup class=\"superscript\">3<\/sup> cm<sup class=\"superscript\">2<\/sup><\/p>\n<p id=\"fs-idp10346048\">16. \u00a0a) [latex]\\frac{\\text{1.0936 yd}}{\\text{1 m}}[\/latex] \u00a0 b) [latex]\\frac{\\text{0.94635 L}}{\\text{1 qt}}[\/latex] \u00a0 \u00a0 \u00a0 \u00a0 \u00a0c) [latex]\\frac{\\text{2.2046 lb}}{\\text{1 kg}}[\/latex]<\/p>\n<p id=\"fs-idm205490112\">17. [latex]\\frac{\\text{2.0 L}}{\\text{67.6 fl oz}} = \\frac{\\text{0.030 L}}{\\text{1 fl oz}}[\/latex]<br \/>\nOnly two significant figures are justified.<\/p>\n<p id=\"fs-idm184575152\">18. 68\u201371 cm; 400\u2013450 g<\/p>\n<p id=\"fs-idm208576112\">19. 355 mL<\/p>\n<p id=\"fs-idp32913072\">20. 8 \u00d7 10<sup>\u22124<\/sup> cm<\/p>\n<p id=\"fs-idm196114640\">21. yes; weight = 89.4 kg<\/p>\n<p id=\"fs-idm137236672\">22. 5.0 \u00d7 10<sup>\u22123<\/sup> mL<\/p>\n<p id=\"fs-idm211581456\">23. a) 1.3 \u00d7 10<sup>\u22124<\/sup> kg \u00a0 \u00a0 \u00a0b) 2.32 \u00d7 10<sup>8<\/sup> kg \u00a0 \u00a0 \u00a0c) 5.23 \u00d7 10<sup>\u221212<\/sup> m \u00a0 \u00a0 \u00a0d) 8.63 \u00d7 10<sup>\u22125<\/sup> kg \u00a0 \u00a0 \u00a0e) 3.76 \u00d7 10<sup>\u22121<\/sup> m \u00a0 \u00a0 \u00a0 \u00a0 \u00a0f) 5.4 \u00d7 10<sup>\u22125<\/sup> m \u00a0 \u00a0 \u00a0g) 1 \u00d7 10<sup>12<\/sup> s \u00a0 \u00a0 \u00a0h) 2.7 \u00d7 10<sup>\u221211<\/sup> s \u00a0 \u00a0 \u00a0i) 1.5 \u00d7 10<sup>\u22124<\/sup> K<\/p>\n<p id=\"fs-idm268990752\">24. 45.4 L<\/p>\n<p id=\"fs-idp176467360\">25. 1.0160 \u00d7 10<sup>3<\/sup> kg<\/p>\n<p id=\"fs-idm134479920\">26. \u00a0a) 394 ft \u00a0 \u00a0 \u00a0b) 5.9634 km \u00a0 \u00a0 \u00a0c) 6.0 \u00d7 10<sup>2 \u00a0 \u00a0 \u00a0<\/sup>d) 2.64 L \u00a0 \u00a0 \u00a0e) 5.1 \u00d7 10<sup>18<\/sup> kg \u00a0 \u00a0 \u00a0f) 14.5 kg \u00a0 \u00a0 \u00a0g) 324 mg<\/p>\n<p id=\"fs-idp194388912\">27. 0.46 m; 1.5 ft\/cubit<\/p>\n<p id=\"fs-idm126197920\">28. Yes, the acid&#8217;s volume is 123 mL.<\/p>\n<p id=\"fs-idm321078368\">29. 62.6 in (about 5 ft 3 in.) and 101 lb<\/p>\n<p id=\"fs-idm182624144\">30. (a) 3.81 cm \u00d7 8.89 cm \u00d7 2.44 m; (b) 40.6 cm<\/p>\n<p id=\"fs-idm209433984\">31. 2.70 g\/cm<sup>3<\/sup><\/p>\n<p id=\"fs-idp10159024\">32. (a) 81.6 g; (b) 17.6 g<\/p>\n<p id=\"fs-idm203289008\">33. (a) 5.1 mL; (b) 37 L<\/p>\n<p id=\"fs-idm24577456\">34. 5371 \u00b0F, 3239 K<\/p>\n<p id=\"fs-idm143339312\">35. \u221223 \u00b0C, 250 K<\/p>\n<p id=\"fs-idp51877792\">36. \u221233.4 \u00b0C, 239.8 K<\/p>\n<p id=\"fs-idm367616944\">37. 113 \u00b0F<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2><span style=\"font-family: Roboto, Helvetica, Arial, sans-serif\">Glossary<\/span><\/h2>\n<p><strong>dimensional analysis:\u00a0<\/strong>(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<\/p>\n<p><strong>Fahrenheit:\u00a0<\/strong>unit of temperature; water freezes at 32 \u00b0F and boils at 212 \u00b0F on this scale<\/p>\n<p><strong>unit conversion factor:\u00a0<\/strong>ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit<\/p>\n","protected":false},"author":330,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"2.4 Mathematical Treatment of Measurement Results - Unit Conversions","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[54],"class_list":["post-1292","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":2084,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1292","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/users\/330"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1292\/revisions"}],"predecessor-version":[{"id":4802,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1292\/revisions\/4802"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/2084"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1292\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=1292"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=1292"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=1292"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=1292"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}