General Chemistry, 10th Edition Solution Manual

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1CHAPTER1Chemistry and Measurement■SOLUTIONS TO EXERCISESNote on significant figures:If the final answer to a solution needs to be rounded off, it is given firstwithone nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded tothe correct number of significant figures. In multistepproblems, intermediate answers are given with atleast 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 gasesSubstituting, you obtain1.85 grams+9.45 grams=0.28 grams+mass of gasesor,Mass of gases=(1.85+9.45−0.28) grams=11.02 gramsThus, the mass of gases in the vessel at the end of the experiment is 11.02 grams.1.2.Physical properties: soft, silvery-colored metal; melts at 64°C.Chemical properties: reacts vigorously with water,withoxygen, and withchlorine.1.3.a.The factor 9.1 has the fewest significant figures, so the answer should be reported to twosignificant figures.5.617.8919.1= 4.86 = 4.9b.The number with the least number of decimalplaces is 8.91. Therefore, round the answer totwo decimal places.8.91−6.435=2.475=2.48c.The number with the least number of decimal places is 6.81. Therefore, round the answer totwo decimal places.6.81−6.730=0.080=0.08d.You first do the subtraction within parentheses. In this step, the number with the leastnumber of decimal places is 6.81, so the result of the subtraction has two decimal places.The least significant figure for this step is underlined.38.91(6.81−6.730)=38.910.080Next, perform the multiplication. In this step, the factor 0.080 has the fewest significantfigures, so round the answer to one significant figure.38.910.080=3.11=3

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2Chapter1:Chemistry and Measurement1.4.a.1.84 x 10−9m = 1.84 nmb.5.67 x 10−12s = 5.67 psc.7.85 x 10−3g = 7.85 mgd.9.7 x 103m = 9.7 kme.0.000732 s = 0.732 ms, or 732 μsf.0.000000000154 m = 0.154 nm, or 154 pm1.5.a.Substituting, we find thattC=5 C9 F(tF− 32°F) =5 C9 F(102.5°F − 32°F) = 39.167°C= 39.2°Cb.Substituting, we find thatTK=c1 K1 Ct+ 273.15 K =1 K78 C1 C+ 273.15 K = 195.15 K= 195 K1.6.Recall that density equals mass divided by volume. You substitute 159 g for the mass and 20.2g/cm3for the volume.d=mV=3159 g20.2 cm=7.871 g/cm3=7.87 g/cm3The density of the metal equals that of iron.1.7.Rearrange the formula defining the density to obtain the volume.V=mdSubstitute 30.3 g for the massand 0.789 g/cm3for the density.V=330.3 g0.789 g/cm=38.40 cm3=38.4 cm31.8.Since one pm=10−12m, and the prefix milli-means 10−3, you can write121 pm1210m1 pm31 mm10m=1.2110−7mm1.9.67.6 Å331010m1 Å311 dm10m=6.7610−26dm31.10.From the definitions, you obtain the following conversion factors:1=36 in1 yd1=2.54 cm1 in1=-210m1 cmThe conversion factor for yards to meters is as follows:

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Chapter1:Chemistry and Measurement31.000ydx36 in1 ydx2.54 cm1 inx-210m1 cm=0.9144m(exact)Finally,3.54ydx0.9144 m1 yd=3.237 m=3.24 m■ANSWERS TO CONCEPT CHECKS1.1.Box A contains a collection of identical units; therefore, it must represent an element. Box Bcontains a compound because 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 differentsubstances.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 whoweighs 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, whichmay 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 bereported as 1.7 x 102pounds.c.For example, 165 lb weighed on a scale that can measure in 100-lb increments would be200 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, or81 cm.b.One story is approximately 10 feet, so three stories is 30 feet. This would be equivalent toapproximately 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 at23°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 bestoption. 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 ofgold. You could measure the mass of the nugget on a balance and the volume of the nugget bywater displacement. Using this information, you could calculate the density of the nugget. Repeatthe experiment and calculations for the sample of gold. If the nugget is gold, the two densitiesshould be equal and be 19.3 g/cm3.Also, you could determine the melting points of the metal and the chunk of pure gold. The twomelting points should be the same (1338 K) if the metal is gold.

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4Chapter1:Chemistry and Measurement■ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS1.1.One area of technology that chemistry has changed is the characteristics of materials. The liquid-crystal displays (LCDs) in devices such as watches, cellphones, computer monitors, andtelevisions are materials made of molecules designed by chemists. Electronics andcommunications have been transformed by the development of optical fibers to replace copperwires. In biology, chemistry has changed the way scientists view life. Biochemists have foundthat 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 thatthe results can be duplicated and rational conclusions obtained. A theory is a tested explanation ofbasic natural phenomena. They are related in that a theory is based on the results of manyexperiments and is fruitful in suggesting other, new experiments. Also, an experiment candisprove a theory but can never prove it absolutely. A hypothesis is a tentative explanation ofsome regularity of nature.1.3.Rosenberg conducted controlled experiments and noted a basic relationship that could be statedas a hypothesis—that is, that certain platinum compounds inhibit cell division. This led him to donew 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 canbe perceived by our senses. Mass is the quantity of matter in a material. The difference betweenmass and weight is that mass remains the same wherever it is measured, but weight isproportional to the mass of the object divided by the square of the distance between the center ofmass 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 chemicalchange (chemical reaction). To demonstrate this law, place a sample of wood in a sealed vesselwith air, and weigh it. Heat the vessel to burn the wood, and weigh the vessel after theexperiment. 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. Thecolor changes from that of metallic mercury (silvery) to a color that varies from red to yellowdepending on the particle size of the oxide.1.7.Gases are easily compressible and fluid. Liquids are relatively incompressible and fluid. Solidsare relatively incompressible and rigid.1.8.An example of a substance is the element sodium. Among its physical properties: It is a solid, andit melts at 98°C. Among its chemical properties: It reacts vigorously with water, and it burns inchlorine gas to form sodium chloride.1.9.An example of an element: sodium; of a compound: sodium chloride, or table salt; of aheterogeneous mixture: salt and sugar; of a homogeneous mixture: sodium chloride dissolved inwater to form a solution.1.10.A glass of bubbling carbonated beverage with ice cubes contains three phases: gas, liquid, andsolid.1.11.A compound may be decomposed by chemical reactions into elements. An element cannot bedecomposed by any chemical reaction. Thus, a compound cannot also be an element in any case.

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Chapter1:Chemistry and Measurement51.12.The precision refers to the closeness of the set of values obtained from identical measurements ofa 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, thecalculator display shows 0.2504938. We would report the answer as 0.250 because the factor0.0634 has the least number of significant figures (three).Addition and subtraction rule: In performing the calculation 184.2 + 2.324, the calculator displayshows 186.524. Because the quantity 184.2 has the least number of decimal places (one), theanswer is reported as 186.5.1.14.An exact number is a number that arises when you count items or sometimes when you define aunit. For example, a foot is defined to be 12 inches. A measured number is the result of acomparison of a physical quantity with a fixed standard of measurement. For example, a steel rodmeasures 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 otherpossible quantities are obtained by deriving them from any of the seven base units. You do this byusing 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 attainedtheoretically is zero. Degrees Celsius and kelvins have units of equalsizeand are related by theformulatC=(TK−273.15 K)1°C1 K1.17.The density of an object is its mass per unit volume. Because the density is characteristic of asubstance, it can be helpful in identifying it. Density can also be useful in determining whether asubstance 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 thecalculations, and (2), if you make an error in arranging factors in the calculation, this will becomeapparent 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 102mm.1.21.The answer is e, 75 mL.1.22.The answer is c, 0.23 mg.■ANSWERS TO CONCEPT EXPLORATIONS1.23.a.First, check the physical appearance of each sample. Check the particles that make up eachsample for consistency and hardness. Also, note any odor. Then perform on each samplesome experiments to measure physical properties such as melting point, density, andsolubility in water. Compare all of these results and see if they match.

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6Chapter1:Chemistry and Measurementb.It is easier to prove that the compounds were different by finding one physical property thatis different, say different melting points. To prove the two compounds were the samewould 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 notdifficult to measure, and it is relatively accurate. The density of a powder is not as easy todetermine 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 tellif there was a difference in behavior. Many white powders dissolve in water. Theirchemical compositions are not the same.1.24.Part 1a.3 g+1.4 g+3.3 g=7.7 g=8 gb.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, youaccumulate 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 101with only one significant figure in the answer.The student probably applied the rule for addition (instead of for multiplication) after thefirst step.f.The answer 28.5 g is correct.g.Don’t round off intermediate answers. Indicate the round-off position after each step byunderlining the least significant digit.Part 2a.The calculated answer is incorrect. It should be 11 cm3. The answer given has too manysignificant figures. There is also a small round off error due to using a rounded-off valuefor 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 bearings11.234 g1 ball bearing31 cm3.1569 g=3.90889=3.909 cm3d.There was no rounding off of intermediate steps; all the factors are as accurate as possible.■ANSWERS TO CONCEPTUAL PROBLEMS1.25.a.Two phases: liquid and solid.b.Three phases: liquid water, solid quartz, and solid seashells.1.26.If the material is a pure compound, all samples should have the same melting point, the samecolor, and the same elemental composition. If it is a mixture, these properties should differdepending on the composition.

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Chapter1:Chemistry and Measurement71.27.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 firstimmerse 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. Aslong as the two points are far enough apart to obtain readings of the desired accuracy, thethermometer can be used in experiments.b.You could make 19 evenly spaced marks on the thermometer between the two originalpoints, each representing a difference of 5°C. You may divide the space between the twooriginal points into fewer spaces as long as you can read the thermometer to obtain thedesired 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 Scale does notcorrespond to one degree on the Fahrenheit scale and that −100F corresponds to 0onYour Scale (different “zero” points). As stated in the problem, in the desired range of 100Your Scale degrees, there are 120 Fahrenheit degrees. Therefore,the relationship can beexpressed as 120F = 100YS, since it covers the same temperature range. Now you needto “scale” the two systems so that they correctly convert from one scale to the other. Youcould set up an equation with the known data points and then employ the information fromthe relationship above.For example, to construct the conversion betweenYS andF, you could perform thefollowing steps:Step 1:F=YSNot a true statement, but one you would like to make true.Step 2:°F=°YS120°F100°YSThis equation takes into account the difference in the size between the temperature unit onthe two scales but will not give you the correct answer because it doesn’t take into accountthe different zero points.Step 3: By subtracting 100F from your equation from Step 2, you now have the completeequation that converts betweenF andYS.F=(YS120°F100°YS)−100Fb.Using the relationship from part a, 66°YS is equivalent to(66YS120°F100°YS)−100F=−20.8F=−21F

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8Chapter1:Chemistry and Measurement1.30.Some physical properties you could measure are density, hardness, color, and conductivity.Chemical properties of sodium would include reaction with air, reaction with water, reaction withchlorine, reaction with acids, bases, etc.1.31.The empty boxes are identical, so they do not contribute to any mass or density difference. Sincethe edge of the cube and the diameter of the sphere are identical, they will occupy the samevolume in each of the boxes; therefore, each box will contain the same number of cubes orspheres. If you view the spheres as cubes that have been rounded by removing wood, you canconclude that the box containing the cubes must have a greater mass of wood; hence, it must havea greater density.1.32.a.Since the bead is less dense than any of the liquids in the container, the bead will float ontop of all the liquids.b.First, determine the density of the plastic bead. Since density is mass divided by volume,you getd=mV=23.9210g0.043 mL= 0.911 g/mL = 0.91 g/mLThus, the glass bead will pass through the top three layers and float on the ethylene glycollayer, 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 unitfor mass is kg, the approximate weights for the items presented in the problem are asfollows: a grain of sand, 110−5kg; a paper clip, 1 x 10−3kg; a nickel, 510−3kg; a 5.0-gallon bucket of water, 2.0101kg; a brick, 3 kg; a car, 1103kg.1.34.When taking measurements, never throw away meaningful information even if there is someuncertainty in the final digit. In this case, you are certain that the nail is between 5 and 6 cm. Theuncertain, yet still important, digit is between the 5 and 6 cm measurements. You can estimatewith reasonable precision that it is about 0.7 cm from the 5 cm mark, so an acceptable answerwould be 5.7 cm. Another person might argue that the length of the nail is closer to 5.8 cm, whichis also acceptable given the precision of the ruler. In any case, an answer of 5.7 or 5.8 shouldprovide useful information about the length of the nail. If you were to report the length of the nailas 6 cm, you would be discarding potentially useful length information provided by themeasuring instrument. If a higher degree of measurement precision were needed (more significantfigures), you would need to switch to a more precise ruler—for example, one that had mmmarkings.1.35.a.The number of significant figures in this answer follows the rules for multiplication anddivision. Here, the measurement with the fewest significant figures is the reported volume0.310 m3, which has three. Therefore, the answer will have three significant figures. SinceVolume =LxWxH, you can rearrange and solve for one of the measurements, say thelength.L=VWH=30.310 m(0.7120 m) (0.52145 m)= 0.83496 m = 0.835 m

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Chapter1:Chemistry and Measurement9b.The number of significant figures in this answer follows the rules for addition andsubtraction. The measurement with the least number of decimal places is the result 1.509m, which has three. Therefore, the answer will have three decimal places. Since the result isthe sum of the three measurements, the third length is obtained by subtracting the other twomeasurements from the total.Length=1.509 m−0.7120 m−0.52145 m=0.27555 m=0.276 m1.36.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 themass of a specific amount(volume) of an item, material, substance, or collection of things. You could use 1 kg of feathersand 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 PROBLEMSNote on significant figures:If the final answer to a solution needs to be rounded off, it is given first withone nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded tothe correct number of significant figures. In multistep problems, intermediate answers are given with atleast 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 dioxidePlugging in gives15.9 g+20.0 g=29.3 g+mass of carbon dioxideMass of carbon dioxide=15.9 g+20.0 g−29.3 g=6.6 g1.38.By the law of conservation of mass:Mass of iron+mass of acid=mass of contents of beaker+mass of hydrogenPlugging in gives5.6 g+15.0=20.4 g+mass of hydrogenMass of hydrogen=5.6 g+15.0 g−20.4 g=0.2 g1.39.By the law of conservation of mass:Mass of zinc+mass of sulfur=mass of zinc sulfideRearranging and plugging in giveMass of zinc sulfide=65.4 g+32.1 g=97.5 gFor the second part, letx= mass of zinc sulfide that could be produced. By the law ofconservation of mass:20.0 g+mass of sulfur=x

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10Chapter1:Chemistry and MeasurementWrite a proportion that relates the mass of zinc reacted to the mass of zinc sulfide formed, whichshould be the same for both cases.mass zincmass zinc sulfide=65.4 g97.5 g=20.0 gxSolving givesx=29.81 g=29.8 g1.40.By the law of conservation of mass:Mass of aluminum+mass of bromine=mass of aluminum bromidePlugging in and solving give27.0 g+Mass of bromine=266.7 gMass of bromine=266.7 g−27.0 g=239.7 gFor the second part, letx= mass of bromine that reacts. By the law of conservation of mass:18.1g+x=mass of aluminum bromideWrite a proportion that relates the mass of aluminum reacted to the mass of bromine reacted,which should be the same for both cases.mass aluminummass bromine=27.0 g239.7 g=18.1 gxSolving givesx=160.7g=161g1.41.a. Solidb. Liquidc. Gasd. Solid1.42.a. Solidb. Solidc. Solidd. Liquid1.43.a.Physical changeb.Physical changec.Chemical changed.Physical change1.44.a.Physical changeb.Chemical changec.Chemical changed.Physical change1.45.Physical change: Liquid mercury is cooled to solid mercury.Chemical changes: (1) Solid mercury oxide forms liquid mercury metaland 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 toform solid iodine.Chemical change: Solid iodine and zinc metal are ignited to form a white powder.

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Chapter1:Chemistry and Measurement111.47.a.Physical propertyb.Chemical propertyc.Physical propertyd.Physical propertye.Chemical property1.48.a.Physical propertyb.Chemical propertyc.Physical propertyd.Chemical propertye.Physical property1.49.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 givealuminum iodide.1.50.Physical properties: (1) is a solid; (2) has an orange-red color; (3) has a density of11.1 g/cm3; (4) is insoluble in water.Chemical property: Mercury(II) oxide decomposes when heated to give mercury and oxygen.1.51.a.Physical processb.Chemical reactionc.Physical processd.Chemical reactione.Physical process1.52.a.Chemical reactionb.Physical processc.Physical processd.Physical processe.Chemical reaction1.53.a.Solutionb.Substancec.Substanced.Heterogeneous mixture

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12Chapter1:Chemistry and Measurement1.54.a.Homogeneous mixture, if fresh; heterogeneous mixture, if spoiledb.Substancec.Solutiond.Substance1.55.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.fourb.threec.fourd.fivee.threef.four1.58.a.fiveb.fourc.twod.foure.threef.four1.59.40,000 km=4.0 x 104km1.60.150,000,000 km=1.50108km1.61.a.8.710.03010.031= 8.457 = 8.5b.0.71 + 89.3 = 90.01 = 90.0c.9340.00435 + 107 = 4.0629 + 107 = 111.06 = 111d.(847.89− 847.73)14673 = 0.1614673 = 2347 = 2.3103

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Chapter1:Chemistry and Measurement131.62.a.8.710.575.871= 0.8456 = 0.85b.8.937 − 8.930 = 0.007c.8.937 + 8.930 = 17.867d.0.0001554.6 + 1.002 = 0.00819 + 1.002 = 1.0101 = 1.0101.63.Thevolume of the first sphere isV1=(4/3)r3=(4/3)(5.15cm)3=572.15cm3The volume of the second sphere isV2=(4/3)r3=(4/3)(5.00 cm)3=523.60 cm3The difference in volume isV1−V2=572.15cm3−523.60 cm3=48.55cm3=49cm31.64.The length of the cylinder between the two marks isl=3.50 cm−3.20 cm=0.30cmThe volume of iron contained between the marks isV=r2l=(1.500 cm)20.30cm=2.12 cm3=2.1cm31.65.a.5.8910−12s = 5.89 psb.0.2010 m = 2.01 dmc.2.56010−9g =2.560 ngd.6.05103m = 6.05 km1.66.a.4.85110−6g = 4.851 μgb.3.1610−2m = 3.16 cmc.2.59110−9s = 2.591 nsd.8.9310−12g = 8.93 pg1.67.a.6.15 ps = 6.1510−12sb.3.781 μm = 3.78110−6mc.1.546 Å = 1.54610−10md.9.7 mg = 9.710−3g1.68.a.6.20 km = 6.20103mb.1.98 ns = 1.9810−9sc.2.54 cm = 2.5410−2md.5.23 μg = 5.2310−6g

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14Chapter1:Chemistry and Measurement1.69.a.tC=5°C9°F(tF− 32°F) =5°C9°F(68°F − 32°F) = 20.0°C = 20.°Cb.tC=5°C9°F(tF− 32°F) =5°C9°F(−23°F − 32°F) = −30.56°C = −31°Cc.tF= (tC9°F5°C) + 32°F = (26°C9°F5°C) + 32°F = 78.8°F = 79°Fd.tF= (tC9°F5°C) + 32°F = (−81°C9°F5°C) + 32°F = −113.8°F = −114°F1.70.a.tC=5°C9°F(tF− 32°F) =5°C9°F(51°F − 32°F) =10.556C = 11Cb.tC=5°C9°F(tF− 32°F) =5°C9°F(−11°F − 32°F) = −23.9°C = −24°Cc.tF= (tC9°F5°C) + 32°F = (−41°C9°F5°C) + 32°F = −41.8°F = −42°Fd.tF= (tC9°F5°C) + 32°F = (22°C9°F5°C) + 32°F = 71.6°F = 72°F1.71.tF=(tC9°F5°C)+32°F=(−20.0°C9°F5°C)+32°F=−4.0°F=−4.0°F1.72.tF=(tC9°F5°C)+32°F=(−222.7°C9°F5°C)+32°F=−368.86°F=−368.9°F1.73.d=mV=312.4 g1.64 cm=7.560 g/cm3=7.56 g/cm31.74.d=mV=23.6 g30.0 mL=0.7867g/mL=0.787g/mL1.75.First, determine the density of the liquid.d=mV=6.71 g8.5 mL=0.7894=0.79 g/mLThe density is closest to ethanol (0.789 g/cm3).1.76.First, determine the density of the mineral sample.d=mV=35.94 g0.73 cm=8.137=8.1g/cm3The density is closest tocinnabar(8.10g/cm3).1.77.The mass of platinum is obtained as follows.Mass=dV=21.4 g/cm35.9 cm3=126 g=1.3102g

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Chapter1:Chemistry and Measurement151.78.The mass of gasoline is obtained as follows.Mass=dV=0.70 g/mL43.8 mL=30.66g=31 g1.79.The volume of ethanol is obtained as follows. Recall that 1 mL = 1 cm3.Volume=md=319.8 g0.789 g/cm=25.09 cm3=25.1 cm3=25.1 mL1.80.The volume of bromine is obtained as follows.Volume=md=88.5 g3.10 g/mL=28.54 mL=28.5 mL1.81.Since 1 kg=103g, and1 mg=10−3g, you can write0.480 kgx310g1 kg31 mg10g=4.80105mg1.82.Since 1 mg=10−3g, and 1 μg=10−6g, you can write611mgx-310g1 mg-61 μg10g=6.11105μg1.83.Since 1 nm=10−9m, and 1 cm=10−2m, you can write555 nm910m1 nm21 cm10m=5.5510−5cm1.84.Since 1 Å=10−10m,and 1 mm = 10−3m,you can write0.96 Å1010m1 Å31 mm10m=9.610−8mm1.85.Since 1 km=103m, you can write3.73108km33310m1 km=3.731017m3Now, 1 dm=10−1m. Also, note that 1 dm3=1 L. Therefore, you can write3.731017m33-11 dm10m=3.731020dm3=3.731020L1.86.1 μm=10−6m, and 1 dm=10−1m. Also, note that 1 dm3=1 L. Therefore, you can write1.3 μm33-610m1 μm3-11 dm10m=1.310−15dm3=1.310−15L1.87.3.58 short tonx2000 lb1 short ton16 oz1 lb1 g0.03527 oz=3.248106g=3.25106g

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