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General Chemistry, 10th Edition Solution Manual

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    1 CHAPTER 1 Chemistry and Measurement ■ SOLUTIONS TO EXERCISES Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of signi ficant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 1 . 1 . From the law of conservation of mass, Mass of wood + mass of air = mass of ash + mass of gases Substituting, you obtain 1.85 grams + 9 .45 grams = 0.28 grams + mass of gases or, Mass of gases = (1.85 + 9.45 − 0.28) grams = 11.02 grams Thus, the mass of gases in the vessel at the end of the experiment is 11.02 grams. 1 . 2 . Phy sical properties: soft, silvery - colored metal; melts at 64°C. Chemical properties: reacts vigorously with water , with oxygen , and with chlorine. 1 . 3 . a . The factor 9.1 has the fewest significant figures, so the answer should be reported to two significant figures. 5.61 7.891 9.1  = 4. 8 6 = 4.9 b . The number with the least number of decimal places is 8.91. Therefore, round the answer to two decimal places. 8.91 − 6.435 = 2.4 7 5 = 2.48 c . The number with the least number of decimal places is 6.81. Therefore, round the answer to two decimal place s. 6.81 − 6.730 = 0.0 8 0 = 0.08 d . You first do the subtraction within parentheses. In this step, the number with the least number of decimal places is 6.81, so the result of the subtraction has two decimal p laces. The least significant figure for this step is underlined. 38.91  (6.81 − 6.730) = 38.91  0.0 8 0 Next, perform the multiplication. In this step, the factor 0.0 8 0 has the fewest significant figures, so round the answer to one significant figure. 38.91  0.0 8 0 = 3 .11 = 3

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    2 Chapter 1 : Chemistry and Measurement 1 . 4 . a . 1.84 x 10 −9 m = 1.84 nm b . 5.67 x 10 −12 s = 5.67 ps c . 7.85 x 10 −3 g = 7.85 mg d . 9.7 x 10 3 m = 9.7 km e . 0.000732 s = 0.732 ms, or 732 μs f . 0.000000000154 m = 0.154 nm, or 154 pm 1 . 5 . a . Substituting, we find that t C = 5 C 9 F    ( t F − 32°F) = 5 C 9 F    (102.5°F − 32°F) = 39. 1 67°C = 39.2°C b . Substituting, we find that T K = c 1 K 1 C t         + 273.15 K = 1 K 78 C 1 C           + 273.15 K = 19 5 .15 K = 195 K 1 . 6 . Recall that density equals mass divided by volume. You substitute 159 g for the mass and 20.2 g/cm 3 for the volume. d = m V = 3 159 g 20.2 cm = 7.8 7 1 g/cm 3 = 7.87 g/cm 3 The density of the metal equals that of iron. 1 . 7 . Rearrange the formula defining the density to obtain the volume. V = m d Substitute 30.3 g for the mass and 0.789 g/cm 3 for the density. V = 3 30.3 g 0.789 g/cm = 38. 4 0 cm 3 = 38.4 cm 3 1 . 8 . Since one pm = 10 − 12 m, and the prefix milli - means 10 − 3 , you can write 121 pm  12 10 m 1 pm   3 1 mm 10 m  = 1.21  10 − 7 mm 1 . 9 . 67.6 Å 3  3 10 10 m 1 Å         3 1 1 dm 10 m        = 6.76  10 − 26 dm 3 1 . 10 . From the definitions, you obtain the following conversion factors: 1 = 36 in 1 yd 1 = 2.54 cm 1 in 1 = -2 10 m 1 cm The conversion factor for yards to meters is as follows:

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    Chapter 1 : Chemistry and Measurement 3 1.000 yd x 36 in 1 yd x 2.54 cm 1 in x -2 10 m 1 cm = 0.9144 m (exact) Finally, 3.5 4 yd x 0.9144 m 1 yd = 3.2 3 7 m = 3.24 m ■ ANSWERS TO CONCEPT CHECKS 1 . 1 . B ox A contains a collection of identical units; therefore, it must represent an element. Box B contains a compound b ecause a compound is the chemical combination of two or more elements (two elements in this case). Box C contains a mixture because it is made up of two different substances. 1 . 2 . a . For a person who weighs less than 100 pounds, two significant figures are typically used, although one significant figure is possible (for example, 60 pounds). For a person who weighs 100 pounds or more, three significant figures are typically used to report the weight (given to the whole pound), although people often round to the nearest unit of 10, which may result in reporting the weight with two significant figures (for example, 170 pounds). b . Assuming a weight of 165 pounds, rounded to two significant figures this would be reported as 1.7 x 10 2 pounds. c . For example, 165 lb weighed on a scale that can measure in 100 - lb increments would be 200 lb. Using the conversion factor 1 lb = 0.4536 kg, 165 lb is equivalent to 74.8 kg. Thus, on a scale that can measure in 50 - kg increments, 165 lb would be 50 kg. 1 . 3 . a . If your leg is approximately 32 inches long, this would be equivalent to 0.81 m, 8.1 dm, or 81 cm. b . One story is approximately 10 feet, so three stories is 30 feet. This would be equivalent to approximately 9 m. c . Normal body temperature is 98.6°F, or 37.0°C. Thus, if your body temperature were 39°C (102°F), you would feel as if you had a moderate fever. d . Room temperature is approximately 72°F, or 22°C. Thus, if you were sitting in a room at 23°C (73°F), you would be comfortable in a short - sleeve shirt. 1 . 4 . Gold is a very unreactive substance, so comparing physical properties is probably your best option. However, color is a physical property you cannot rely on in this case to get your answer. One experiment you could perform is to determine the densities of the metal and the chunk of gold. You could measure the mass of the nugget on a balance and the volume of the nugget by water displacement. Using this information, you could calculate the density of the nugget. Repeat the experiment and calculations for the sample of gold. If the nugget is gold, the two densities should be equal and be 19.3 g/cm 3 . Also, you could determine the melting points of the metal and the chunk of pure gold. The two melting points should be the same (1338 K) if the metal is gold.

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    4 Chapter 1 : Chemistry and Measurement ■ ANSWERS TO SELF - ASSESSMENT AND REVIEW QUESTIONS 1 . 1 . One area of technology that chemistry has changed is the characteristics of materials. The liquid - crystal displays (LCDs) in devices such as watches, cell phones, computer monitors, and televisions are materials made of molecules designed by chemists. Electronics and communications have been transformed by the development of optical fibers to replace copper wires. In biology, chemistry has changed the way scientists view life. Biochemists have found that all forms of life share many of the same molecules and molecular processes. 1 . 2 . An experiment is an observation of natural phenomena carried out in a controlled manner so that the results can be duplicated and rational conclusions obtained. A theory is a tested explanation of basic natural phenomena. They are related in that a theory is based on the results of many experiments and is fruitful in suggesting other, new experiments. Also, an experiment can disprove a theory but can never prove it absolutely. A hypothesis is a tentative explanation of some regularity of nature. 1 . 3 . Rosenberg conducted controlled experiments and noted a basic relationship that could be stated as a hypothesis — that is, that certain platinum compounds inhibit cell division. This led him to do new experiments on the anticancer activity of these compounds. 1 . 4 . Matter is the general term for the material things around us. It is whatever occupies space and can be perceived by our senses. Mass is the quantity of matter in a material. The difference between mass and weight is that mass remains the same wherever it is measured, but weight is proportional to the mass of the object divided by the square of the distance between the center of mass of the object and that of the earth. 1 . 5 . The law of conservation of mass states that the total mass remains constant during a chemical change (chemical reaction). To demonstrate this law, place a sample of wood in a sealed vessel with air, and weigh it. Heat the vessel to burn the wood, and weigh the vessel after the experiment. The weight before the experiment and that after it should be the same. 1 . 6 . Mercury metal, which is a liquid, reacts with oxygen gas to form solid mercury(II) oxide. The color changes from that of metallic mercury (silvery) to a color that varies from red to yellow depending on the particle size of the oxide. 1 . 7 . Gases are easily compressible and fluid. Liquids are relatively incompressible and fluid. Solids are relatively incompressible and rigid. 1 . 8 . An example of a substance is the element sodium. Among its physical properties: It is a solid, and it melts at 98°C. Among its chemical properties: It reacts vigorously with water, and it burns in chlorine gas to form sodium chloride. 1 . 9 . An example of an element: sodium; of a compound: sodium chloride, or table salt; of a heterogeneous mixture: salt and sugar; of a homogeneous mixture: sodium chloride dissolved in water to form a solution. 1 . 10 . A glass of bubbling carbonated beverage with ice cubes contains three phases: gas, liquid, and solid. 1 . 11 . A compound may be decomposed by chemical reactions into elements. An element cannot be decomposed by any chemical reaction. Thus, a compound cannot also be an element in any case.

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    Chapter 1 : Chemistry and Measurement 5 1 . 12 . The precision refers to the closeness of the set of values obtained from identical measurements of a quantity. The number of digits reported for the value of a measured or calculated quantity (significant figures) indicates the precision of the value. 1 . 13 . Multiplication and division rule: In performing the calculation 100.0 x 0.0634 ÷ 25.31, the calculator display shows 0.2504938. We would report the answer as 0.250 because the factor 0.0634 has the least number of significant figures (three). Addition and subtraction rule: In performing the calculation 184.2 + 2.324, the calculator display shows 186.524. Because the quantity 184.2 has the least number of decimal places (one), the answer is reported as 186.5. 1 . 14 . An exact number is a number that arises when you count items or sometimes when you define a unit. For example, a foot is defined to be 12 inches. A measured number is the result of a comparison of a physical quantity with a fixed standard of measurement. For example, a steel rod measures 9.12 centimeters, or 9.12 times the standard centimeter unit of measurement. 1 . 15 . For a given unit, the SI system uses prefixes to obtain units of different sizes. Units for all other possible quantities are obtained by deriving them from any of the seven base units. You do this by using the base units in equations that define other physical quantities. 1 . 16 . An absolute temperature scale is a scale in which the lowest temperature that can be attained theoretically is zero. Degrees Celsius and kelvins have units of equal size and are related by the formula t C = ( T K − 273.15 K)  1°C 1 K 1 . 17 . The density of an object is its mass per unit volume. Because the density is characteristic of a substance, it can be helpful in identifying it. Density can also be useful in determining whether a substance is pure. It also provides a useful relationship between mass and volume. 1 . 18 . Units should be carried along because (1) the units for the answers will come out in the calculations, and (2), if you make an error in arranging factors in the calculation, this will become apparent because the final units will be nonsense. 1 . 19 . The answer is c, three significant figures. 1 . 20 . The answer is a, 4.43 x 10 2 mm. 1 . 21 . The answer is e, 75 mL. 1 . 22 . The answer is c, 0.23 mg. ■ ANSWERS TO CONCEPT EXPLORATIONS 1 . 23 . a . First, check the physical appearance of each sample. Check the particles that make up each sample for consistency and h ardness. Also, note any odor. Then perform on each sample some experiments to measure physical properties such as melting point, density, and solubility in water. Compare all of these results and see if they match.

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    6 Chapter 1 : Chemistry and Measurement b . It is easier to prove that the compounds were different by finding one physical property that is different, say different melting points. To prove the two compounds were the same would require showing that every physical property was the same. c . Of the properties listed in part a, the melting point would be most convincing. It is not difficult to measure, and it is relatively accurate. The density of a powder is not as easy to determine as the melting point, and solubility is not reliable enough on its own. d . No. Since neither solution reached a saturation point, there is not enough information to tell if there was a difference in behavior. Many white powders dissolve in water. Their chemical compositions are not the same. 1 . 24 . Part 1 a . 3 g + 1.4 g + 3.3 g = 7 .7 g = 8 g b . First, 3 g + 1.4 g = 4 .4 g = 4 g. Then, 4 g + 3.3 g = 7 .3 g = 7 g. c . Yes, the answer in part a is more accurate. When you round off intermediate steps, you accumulate small errors and your answer is not as accurate. d . The answer 29 g is correct. e . This answer is incorrect. It should be 3 x 10 1 with only one significant figure in the answer. The student probably applied the rule for addition (instead of for multiplication) after the first step. f . The answer 28.5 g is correct. g . Don’t round off intermediate answers. Indicate the round - off position after each step by underlining the least significant digit. Part 2 a . The calculated answer is incorrect. It should be 11 cm 3 . The answer given has too many significant figures. There is also a small round off error due to using a rounded - off value for the density. b . This is a better answer. It is reported with the correct number of significant figures (three). It can be improved by using all of the digits given for the density. c . V = 10 ball bearings 1  1.234 g 1 ball bearing  3 1 cm 3.1569 g = 3.90 8 89 = 3.909 cm 3 d . There was no rounding off of intermediate steps; all the factors are as accurate as possible. ■ ANSWERS TO CONCEPTUAL PROBLEMS 1 . 25 . a . Two phases: liquid and solid. b . Three phases: liquid water, solid quart z, and solid seashells. 1 . 26 . If the material is a pure compound, all samples should have the same melting point, the same color, and the same elemental composition. If it is a mixture, these properties should differ depending on the composition.

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    Chapter 1 : Chemistry and Measurement 7 1 . 27 . a . You need to establish two points on the thermometer with known (defined) temperatures — for example, the freezing point (0  C) and boiling point (100  C) of water. You could first immerse the thermometer in an ice - water bath and mark the level at this point as 0  C. Then, immerse the thermometer in boiling water, and mark the level at this point as 100  C. As long as the two points are far enough apart to obtain readings of the desired accuracy, the thermometer can be used in experiments. b . You could make 19 evenly spaced marks on the thermometer between the two original points, each representing a difference of 5°C. You may divide the space between the two original points into fewer spaces as long as you can read the thermometer to obtain the desired accuracy. 1 . 28 . a. b. c. 1 . 29 . a . To answer this question, you need to develop an equation that converts between  F and  YS. To do so, you need to recognize that one degree on the Your Sc ale does not correspond to one degree on the Fahrenheit scale and that −100  F corresponds to 0  on Your Scale (different “zero” points). As stated in the problem, in the desired range of 100 Your Scale degrees, there are 120 Fahrenheit degrees. Therefore, the relationship can be expressed as 120  F = 100  YS, since it covers the same temperature range. Now you need to “scale” the two systems so that they correctly convert from one scale to the other. You could set up an equation with the known data points and then employ the information from the relationship above. For example, to construct the conversion between  YS and  F, you could perform the following steps: Step 1:  F =  YS Not a true statement, but one you would like to make true. Step 2: °F = °YS  120°F 100°YS This equation takes into account the difference in the size between the temperature unit on the two scales but will not give you the correct answer because it doesn’t take into account the different zero points. Step 3: By subtracting 100  F from your equation from Step 2, you now have the complete equation that converts between  F and  YS.  F = (  YS  120°F 100°YS ) − 100  F b . Using the relationship from part a, 66°YS is equivalent to (66  YS  120°F 100°YS ) − 100  F = − 2 0 .8  F = − 21  F

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    8 Chapter 1 : Chemistry and Measurement 1 . 30 . Some physical propert ies you could measure are density, hardness, color, and conductivity. Chemical properties of sodium would include reaction with air, reaction with water, reaction with chlorine, reaction with acids, bases, etc. 1 . 31 . The empty boxes are identical, so they do not contribute to any mass or density difference. Since the edge of the cube and the diameter of the sphere are identical, they will occupy the same volume in each of the boxes; therefore, each box will contain the same number of cubes or spheres. If you view the spheres as cubes that have been rounded by removing wood, you can conclude that the box containing the cubes must have a greater mass of wood; hence, it must have a greater density. 1 . 32 . a . Since the bead is less dense than any of the liquids in the container, the bead will float on top of all the liquids. b . First, determine the density of the plastic bead. Since density is mass divided by volume, you get d = m V = 2 3.92 10 g 0.043 mL   = 0.9 1 1 g/mL = 0.91 g/mL Thus, the glass bead will pass through the top three layers and float on the ethylene glycol layer, which is more dense. c . Since the bead sinks all the way to the bottom, it must be more dense than 1.114 g/mL. 1 . 33 . a . A paper clip has a mass of about 1 g. b . Answers will vary depending on your particular sample. Keeping in mind that the SI unit for mass is kg, the approximate weights for the items presented in the problem are as follows: a grain of sand, 1  10 −5 kg; a paper clip, 1 x 10 −3 kg; a nickel, 5  10 −3 kg; a 5.0 - gallon bucket of water, 2.0  10 1 kg; a brick, 3 kg; a car, 1  10 3 kg. 1 . 34 . When taking measurements, never throw away meaningful information eve n if there is some uncertainty in the final digit. In this case, you are certain that the nail is between 5 and 6 cm. The uncertain, yet still important, digit is between the 5 and 6 cm measurements. You can estimate with reasonable precision that it is about 0.7 cm from the 5 cm mark, so an acceptable answer would be 5.7 cm. Another person might argue that the length of the nail is closer to 5.8 cm, which is also acceptable given the precision of the ruler. In any case, an answer of 5.7 or 5.8 should provi de useful information about the length of the nail. If you were to report the length of the nail as 6 cm, you would be discarding potentially useful length information provided by the measuring instrument. If a higher degree of measurement precision were needed (more significant figures), you would need to switch to a more precise ruler — for example, one that had mm markings. 1 . 35 . a . The number of significant figures in this answer follows the rules for multiplication and division. Here, the measurement with the fewest significant figures is the reported volume 0.310 m 3 , which has three. Therefore, the answer will have three significant figures. Since Volume = L x W x H , you can rearrange and solve for one of the measurements, say the length. L =  V W H = 3 0.310 m (0.7120 m) (0.52145 m) = 0.83 4 96 m = 0.835 m

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    Chapter 1 : Chemistry and Measurement 9 b . The number of significant figures in this answer follows the rules for addition and subtraction. The measurement with the least number of decimal places is the result 1.509 m, which has three. Therefore, the answer will have three decimal places. Since the result is the sum of the three measurements, the third length is obtained by subtracting the other two measurements from the total. Length = 1.509 m − 0.7120 m − 0.52145 m = 0.27 5 55 m = 0.276 m 1 . 36 . The mass of something (how heavy it is) depends on how much of the item, material, substance, or collection of things you have. The density of something is the mass of a specific amount (volume) of an item, material, substance, or collection of things. You could use 1 kg of feathers and 1 kg of water to illustrate that they have the same mass yet have very different volumes; therefore, they have different densities. ■ SOLUTIONS TO PRACTICE PROBLEMS Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to t he correct number of s ignificant figures. In multi step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 1 . 37 . By the law of conservation of mass: Mass of sodium carbonate + mass of acetic acid solution = mass of contents of reaction vessel + mass of carbon dioxide Plugging in gives 15.9 g + 20.0 g = 29.3 g + mass of carbon dioxide Mass of carbon dioxide = 15.9 g + 20.0 g − 29.3 g = 6.6 g 1 . 38 . By the law of conservation of mass: Mass of iron + mass of acid = mass of contents of beaker + mass of hydrogen Plugging in gives 5.6 g + 15.0 = 20.4 g + mass of hydrogen Mass of hyd rogen = 5.6 g + 15.0 g − 20.4 g = 0.2 g 1 . 39 . By the law of conservation of mass: Mass of zinc + mass of sulfur = mass of zinc sulfide Rearranging and plugging in give Mass of zinc sulfide = 65.4 g + 32.1 g = 97.5 g For the second part, let x = mass of zinc sulfide that could be produced. By the law of conservation of mass: 20.0 g + mass of sulfur = x

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    10 Chapter 1 : Chemistry and Measurement Write a proportion that relates the mass of zinc reacted to the mass of zinc sulfide formed, which should be the same for both cases. mass zinc mass zinc sulfide = 65.4 g 97.5 g = 20.0 g x Solving gives x = 29. 8 1 g = 29.8 g 1 . 40 . By the law of conservation of mass: Mass of al uminum + mass of bromine = mass of aluminum bromide Plugging in and solving give 27.0 g + Mass of bromine = 266.7 g Mass of bromine = 266.7 g − 27.0 g = 239.7 g For the second part, let x = mass of bromine that reacts. By the law of conservation of ma ss: 1 8.1 g + x = mass of aluminum bromide Write a proportion that relates the mass of aluminum reacted to the mass of bromine reacted, which should be the same for both cases. mass aluminum mass bromine = 27.0 g 239.7 g = 18.1 g x Solving gives x = 1 6 0 .7 g = 1 61 g 1 . 41 . a . Solid b . Liquid c . Gas d . Solid 1 . 42 . a . Solid b . Solid c . Solid d . Liquid 1 . 43 . a . Physical change b . Physical change c . Chemical change d . Physical change 1 . 44 . a . Physical change b . Chemical change c . Chemical change d . Physical change 1 . 45 . Physical change: Liquid mercury is cooled to solid mercury. Chemical changes: (1) Solid mercury oxide forms liquid mercury metal and gaseous oxygen; (2) glowing wood and oxygen form burning wood (form ash and gaseous products). 1 . 46 . Physical changes: (1) Solid iodine is heated to gaseous iodine; (2) gaseous iodine is cooled to form solid iodine. Chemical change: Solid iodine and zinc metal are ignited to form a white powder.

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    Chapter 1 : Chemistry and Measurement 11 1 . 47 . a . Physical property b . Chemical property c . Physical property d . Physical property e . Chemical property 1 . 48 . a . Physical property b . Chemical property c . Physical property d . Chemical property e . Physical property 1 . 49 . Physical properties: (1) Iodine is solid; (2) the solid has lustrous blue - black crystals; (3) the crystals vaporize readily to a violet - colored gas. Chemical properties: (1) Iodine combines with many metals, such as with aluminum to give aluminum iodide. 1 . 50 . Physical properties: (1) is a solid; (2) has an orange - red color; (3) has a density of 11.1 g/cm 3 ; (4) is insoluble in water. Chemical property: Mercury(II) oxide decomposes when heated to give mercury and oxygen. 1 . 51 . a . Physical process b . Chemical reaction c . Physical process d . Chemical reaction e . Physical process 1 . 52 . a . Chemical reaction b . Physical process c . Physical process d . Physical process e . Chemical reaction 1 . 53 . a . Solution b . Substance c . Substance d . Heterogeneous mixture

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    12 Chapter 1 : Chemistry and Measurement 1 . 54 . a . Homogeneous mixture, if fresh; heterogeneous mixture, if spoiled b . Substance c . Solution d . Substance 1 . 55 . a . A pure substance with two phases present, liquid and gas. b . A mixture with two phases present, solid and liquid. c . A pure substance with two phases present, solid and liquid. d . A mixture with two phases present, solid and solid. 1 . 56 . a . A mixture with two phases present, solid and liquid. b . A mixture with two phases present, solid and liquid. c . A mixture with two phases present, solid and solid. d . A pure substance with two phases present, liquid and gas. 1 . 57 . a . four b . three c . four d . five e . three f . four 1 . 58 . a . five b . four c . two d . four e . three f . four 1 . 59 . 4 0 ,000 km = 4.0 x 10 4 km 1 . 60 . 15 0 ,000,000 km = 1.50  10 8 km 1 . 61 . a . 8.71 0.0301 0.031  = 8. 4 57 = 8.5 b . 0.71 + 89.3 = 90. 0 1 = 90.0 c . 934  0.00435 + 107 = 4.0 6 29 + 107 = 11 1 .06 = 111 d . (847.89 − 847.73)  14673 = 0.1 6  14673 = 2 3 47 = 2.3  10 3

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    Chapter 1 : Chemistry and Measurement 13 1 . 62 . a . 8.71 0.57 5.871  = 0.8 4 56 = 0.85 b . 8.937 − 8.930 = 0.007 c . 8.937 + 8.930 = 17.867 d . 0.00015  54.6 + 1.002 = 0.008 1 9 + 1.002 = 1.01 0 1 = 1.010 1 . 63 . The volume of the first sphere is V 1 = (4/3)  r 3 = (4/3)   (5.1 5 cm) 3 = 5 7 2 . 15 cm 3 The volume of the second sphere is V 2 = (4/3)  r 3 = (4/3)   (5.00 cm) 3 = 52 3 .60 cm 3 The difference in volume is V 1 − V 2 = 57 2 .15 cm 3 − 52 3 .60 cm 3 = 4 8 . 55 cm 3 = 49 cm 3 1 . 64 . The length of the cylinder between the two marks is l = 3.50 cm − 3. 2 0 cm = 0. 3 0 cm The volume of iron contained between the marks is V =  r 2 l =   (1.500 cm) 2  0. 3 0 cm = 2. 1 2 cm 3 = 2. 1 cm 3 1 . 65 . a . 5.89  10 −12 s = 5.89 ps b . 0.2010 m = 2.01 dm c . 2.560  10 −9 g = 2.560 ng d . 6.05  10 3 m = 6.05 km 1 . 66 . a . 4.851  10 −6 g = 4.851 μg b . 3.16  10 −2 m = 3.16 cm c . 2.591  10 −9 s = 2.591 ns d . 8.93  10 −12 g = 8.93 pg 1 . 67 . a . 6.15 ps = 6.15  10 −12 s b . 3.781 μm = 3.781  10 −6 m c . 1.546 Å = 1.546  10 −10 m d . 9.7 mg = 9.7  10 −3 g 1 . 68 . a . 6.20 km = 6.20  10 3 m b . 1.98 ns = 1.98  10 −9 s c . 2.54 cm = 2.54  10 −2 m d . 5.23 μg = 5.23  10 −6 g

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    14 Chapter 1 : Chemistry and Measurement 1 . 69 . a . t C = 5°C 9°F  ( t F − 32°F) = 5°C 9°F  (68°F − 32°F) = 2 0 .0°C = 20.°C b . t C = 5°C 9°F  ( t F − 32°F) = 5°C 9°F  (−23°F − 32°F) = −3 0 .56°C = −31°C c . t F = ( t C  9°F 5°C ) + 32°F = (26°C  9°F 5°C ) + 32°F = 7 8 .8°F = 79°F d . t F = ( t C  9°F 5°C ) + 32°F = (−81°C  9°F 5°C ) + 32°F = −11 3 .8°F = −114°F 1 . 70 . a . t C = 5°C 9°F  ( t F − 32°F) = 5°C 9°F  (51°F − 32°F) = 1 0 .556  C = 11  C b . t C = 5°C 9°F  ( t F − 32°F) = 5°C 9°F  (−11°F − 32°F) = −2 3 .9°C = −24°C c . t F = ( t C  9°F 5°C ) + 32°F = (−41°C  9°F 5°C ) + 32°F = −4 1 .8°F = −42°F d . t F = ( t C  9°F 5°C ) + 32°F = (22°C  9°F 5°C ) + 32°F = 7 1 .6°F = 72°F 1 . 71 . t F = ( t C  9°F 5°C ) + 32°F = ( − 2 0 . 0 °C  9°F 5°C ) + 32°F = − 4 . 0 °F = − 4 .0°F 1 . 72 . t F = ( t C  9°F 5°C ) + 32°F = ( − 222.7 °C  9°F 5°C ) + 32°F = − 3 68 . 8 6 °F = − 3 68 .9 °F 1 . 73 . d = m V = 3 12.4 g 1.64 cm = 7.5 6 0 g/cm 3 = 7.56 g/cm 3 1 . 74 . d = m V = 23.6 g 30.0 mL = 0. 78 6 7 g/mL = 0. 787 g/mL 1 . 75 . First, determine the density of the liquid. d = m V = 6.71 g 8.5 mL = 0.7 8 94 = 0.79 g/mL The density is closest to ethanol (0.789 g/cm 3 ). 1 . 76 . First, determine the density of the mineral sample. d = m V = 3 5.94 g 0.73 cm = 8. 1 37 = 8.1 g/cm 3 The density is closest to cinnabar ( 8.10 g/cm 3 ). 1 . 77 . The mass of platinum is obtained as follows. Mass = d  V = 21.4 g/cm 3  5.9 cm 3 = 1 2 6 g = 1.3  10 2 g

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    Chapter 1 : Chemistry and Measurement 15 1 . 78 . The mass of gasoline is obtained as follows. Mass = d  V = 0.70 g/mL  43.8 mL = 3 0 .66 g = 31 g 1 . 79 . The volume of ethanol is obtained as follows. Recall that 1 mL = 1 cm 3 . Volume = m d = 3 19.8 g 0.789 g/cm = 25. 0 9 cm 3 = 25.1 cm 3 = 25.1 mL 1 . 80 . The volume of bromine is obtained as follows. Volume = m d = 88.5 g 3.10 g/mL = 28. 5 4 mL = 28.5 mL 1 . 81 . Since 1 kg = 10 3 g, and 1 mg = 10 − 3 g, you can write 0.480 kg x 3 10 g 1 kg  3 1 mg 10 g  = 4.80  10 5 mg 1 . 82 . Since 1 mg = 10 − 3 g, and 1 μg = 10 − 6 g, you can write 611 mg x -3 10 g 1 mg  -6 1 μg 10 g = 6.11  10 5 μg 1 . 83 . Since 1 nm = 10 − 9 m, and 1 cm = 10 − 2 m, you can write 555 nm  9 10 m 1 nm   2 1 cm 10 m  = 5.55  10 − 5 cm 1 . 84 . Since 1 Å = 10 − 10 m, and 1 mm = 10 − 3 m, you can write 0.96 Å  10 10 m 1 Å   3 1 mm 10 m  = 9.6  10 − 8 m m 1 . 85 . Since 1 km = 10 3 m, you can write 3.7 3  10 8 km 3  3 3 10 m 1 km       = 3.73  10 17 m 3 Now, 1 dm = 10 − 1 m. Also, note that 1 dm 3 = 1 L. Therefore, you can write 3.73  10 17 m 3  3 -1 1 dm 10 m       = 3.73  10 20 dm 3 = 3.73  10 20 L 1 . 86 . 1 μm = 10 − 6 m, and 1 dm = 10 − 1 m. Also, note that 1 dm 3 = 1 L. Therefore, you can write 1.3 μm 3  3 -6 10 m 1 μm        3 -1 1 dm 10 m       = 1.3  10 − 15 dm 3 = 1.3  10 − 15 L 1 . 87 . 3. 58 short ton x 2000 lb 1 short ton  16 oz 1 lb  1 g 0.03527 oz = 3.2 4 8  10 6 g = 3.25  10 6 g

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