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6Chapter 2: Measurements and Calculations1.6093 km 716.9 km1 mif.445.5 mi g.36.2 m h.0.501 km a.5.25 oz b.125 g 1 lb 0.276 lb453.59 gc.125 g 1 lb16 oz 4.41 oz1 lb453.59 gd.125 mL 1L 0.125 L1000 mLe.125 mL 1.057 qt 2 pt 0.264 pt 1000 mL 1 qtf.2.5 mi 1.6093 km 4.0 km1 mig.2.5 mi 1.6093 km 1000 m 4.0 103 m1 mi1 kmh.2.5 mi 1.6093 km 1000 m 100 cm 4.0 105 cm1 mi1 km1m1 km 0.0362 km1000 m1000 m100 cm 5.01 104 cm1 km1m66.1 lb 0.328 lb16 oz68. 190 mi 1.9 102 mi to two significant figures1.9 102 mi 1 km 3.1 102 km0.62137 mi3.1 102 km 1000 m 3.1 105 m1 km1.9 102 mi 5280 ft 1.0 106 ft1 mi70. 1 10–10 m 100 cm 1 10–8 cm1m1 10–8 cm 1 in 4 10–9 in.2.54 cm1 10–8 cm 1m109 nm 0.1 nm100 cm1mCopyright Houghton Mifflin Company. All rights reserved.

Chapter 2: Measurements and Calculations72. freezing74. 27376. Fahrenheit (F)78.80.82.TC TK 273a.–210 273 63 Kb.275 K – 273 2ºCc.778 K – 273 505ºCd.778 273 1051 KTF 1.80(TC ) 32a.1.80(78.1) 32 173ºFb.1.80(40.) 32 104ºFc.1.80(–273) 32 –459ºFd.1.80(32) 32 90.ºFTF 1.80(TC ) 32TC (TF 32)/1.80a.275 – 273 2ºCb.(82 – 32)/1.80 28ºCc.1.80(–21) 32 –5.8ºF (–6ºF)d.(–40 – 32)/1.80 –40 ºC (the Celsius and Fahrenheit temperatures are the same).84. g/cm3 (g/mL)86. 100 in388. Density is a characteristic property that is always the same for a pure substance.90. copper92. density massvolume122.4 g 22 g/cm335.5 cma.d b. 100 cm 53v 0.57 m3 5.7 10 cm 1m 3d c.1.9302 104 g 0.034 g/cm355.7 10 cmm 0.0175 kg 1000 g 17.5 g1 kgCopyright Houghton Mifflin Company. All rights reserved.7

8Chapter 2: Measurements and Calculationsd 17.5 g 0.962 g/mL 0.962 g/cm318.2 mL3d. 100 cm 53v 0.12 m3 1.2 10 cm 1m d 2.49 g 2.1 10–5 g/cm31.2 105 cm 394. 1.0 kg 1.0 103 g1.0 103 g 96. m 3.5 lb 1 mL 0.91 L (two significant figures)1.097 g453.59 g 1.59 103 g1 lb3 2.54 cm 53v 1.2 104 in.3 1.97 10 cm1in d 1.59 103 g 8.1 10–3 g/cm3531.97 10 cmThe material will float.98. 5.25 g 1 cm3 0.500 cm3 0.500 mL10.5 g11.2 mL 0.500 mL 11.7 mL100.a.50.0 cm3 19.32 g 966 g1 cm3b.50.0 cm3 7.87 g 394 g1 cm3c.50.0 cm3 11.34 g 567 g1 cm3d.50.0 cm3 2.70 g 135 g1 cm 3a.3.011 1023 301,100,000,000,000,000,000,000b.5.091 109 5,091,000,000c.7.2 102 720d.1.234 105 123,400e.4.32002 10–4 0.000432002f.3.001 10–2 0.03001102.Copyright Houghton Mifflin Company. All rights reserved.

Chapter 2: Measurements and Calculationsg.2.9901 10–7 0.00000029901h.4.2 10–1 e.millimetersa.36.2 blim 1400 kryll 5.07 104 kryll1 blimb.170 kryll 1 blim 0.12 blim1400 kryllc. 1 blim –5272.5 kryll 3.70 10 blim 1400 kryll 104.106.22108. 52 cm 110. 1 lb 1 in 20. in2.54 cm1 kg 1.20 2.45euro 1.33 per pound2.2 lb 1 euro1 kg112. X 1.26 C 14114. d 36.8 g 3.50 g/L (3.50 10–3 g/cm3)10.5 L116. For ethanol, 100. mL 0.785 g 78.5 g1 mLFor benzene, 1000 mL 0.880 g 880. g1 mLtotal mass, 78.5 880. 959 eCopyright Houghton Mifflin Company. All rights reserved.9

10Chapter 2: Measurements and Calculations120.a.2; positiveb.11; negativec.3; positived.5; negativee.5; positivef.0; zerog.1; negativeh.7; negativea.1; positiveb.3; negativec.0; zerod.3; positivee.9; negativea.The decimal point must be moved five places to the left; 2.98 10–5 0.0000298.b.The decimal point must be moved nine places to the right; 4.358 109 4,358,000,000.c.The decimal point must be moved six places to the left; 1.9928 10–6 0.0000019928.d.The decimal point must be moved 23 places to the right; 6.02 1023 602,000,000,000,000,000,000,000.e.The decimal point must be moved one place to the left; 1.01 10–1 0.101.f.The decimal point must be moved three places to the left; 7.87 10–3 0.00787.g.The decimal point must be moved seven places to the right; 9.87 107 98,700,000.h.The decimal point must be moved two places to the right; 3.7899 102 378.99.i.The decimal point must be moved one place to the left; 1.093 10–1 0.1093.j.The decimal point must be moved zero places; 2.9004 100 2.9004.k.The decimal point must be moved four places to the left; 3.9 10–4 0.00039.l.The decimal point must be moved eight places to the left; 1.904 10–8 0.00000001904.a.1/102 1 10–2b.1/10–2 1 102c.55/103 122.124.126.5.5 101 5.5 10–21 103Copyright Houghton Mifflin Company. All rights reserved.

Chapter 2: Measurements and Calculationsd.(3.1 106)/10–3 e.(106)1/2 1 103f.(106)(104)/(102) g.1/0.0034 h.3.453/10–4 113.1 106 3.1 1091 10 3(1 106 )(1 104 ) 1 108(1 102 )1 2.9 102-33.4 103.453 3.453 1041 10-4128. Kelvin, K130. centimeter132. 0.105 m134. 1 kg (100 g 0.1 kg)136. 10 cm (1 cm 10 mm)138. 2.8 (the hundredths place is estimated)140.a.0.000426b.4.02 10–5c.5.99 106d.400.e.0.00600a.2149.6 (The answer can be given only to the first decimal place because 149.2 is knownonly to the first decimal place)b.5.37 103 (The answer can be given only to two decimal places because 4.34 is known onlyto two decimal places. Moreover, since the power of ten is the same for each number, thecalculation can be performed directly.)c.Before performing the calculation, the numbers have to be converted so that they contain thesame power of ten.142.4.03 10–2 – 2.044 10–3 4.03 10–2 – 0.2044 10–2 3.83 10–2 (The answer can begiven only to the second decimal place because 4.03 10–2 is known only to the seconddecimal place.)d.Before performing the calculation, the numbers have to be converted so that they contain thesame power of ten.2.094 105 – 1.073 106 2.094 105 – 10.73 105 –8.64 105Copyright Houghton Mifflin Company. All rights reserved.

12Chapter 2: Measurements and Calculations144.a.(2.9932 104)(2.4443 102 1.0032 101) (2.9932 104)(24.443 101 1.0032 101) (2.9932 104)(25.446 101) 7.6166 106b.(2.34 102 2.443 10–1)/(0.0323) (2.34 102 0.002443 102)/(0.0323) (2.34 102)/(0.0323) 7.24 103146.c.(4.38 10–3)2 1.92 10–5d.(5.9938 10–6)1/2 2.4482 10–31 year12 months;12 months1 year148.a.908 oz 1 lb1 kg 25.7 kg16 oz 2.2046 lbb.12.8 L 1 qt1 gal 3.38 gal 0.94633 L 4 qtc.125 mL 1L1 qt 0.132 qt1000 mL 0.94633 Ld.2.89 gal 4 qt1L1000 mL 1.09 104 mL 1 gal 1.0567 qt1Le.4.48 lb 453.59 g 2.03 103 g1 lbf.550 mL 1L1.0567 qt 0.58 qt1000 mL1L150. Assuming exactly 6 gross, 864 pencils152.a.Celsius temperature (175 – 32)/1.80 79.4 CKelvin temperature 79.4 273 352 Kb.255 – 273 –18 Cc.(–45 – 32)/1.80 –43 Cd.1.80(125) 32 257 F154. 85.5 mL 0.915 g 78.2 g1 mLCopyright Houghton Mifflin Company. All rights reserved.

Chapter 2: Measurements and Calculations156. m 155 lb 453.59 g 7.031 104 g1 lb33 12 in 2.54 cm 53v 4.2 ft3 1.189 10 cm 1 ft 1 in d 7.031 104 g 0.59 g/cm31.189 105 cm3158. TF 1.80(TC ) 32a.23 Fb.32 Fc.–321 Fd.–459 Fe.187 Ff.–459 FCopyright Houghton Mifflin Company. All rights reserved.13

CHAPTER 3MatterCHAPTER ANSWERS2.forces4.liquids6.gaseous8.The stronger the interparticle forces, the more rigid is the sample overall.10. Gases are easily compressed into smaller volumes whereas solids and liquids are not. Because agaseous sample consists mostly of empty space, it is the empty space that is compressed whenpressure is applied to a gas.12. This is a chemical change; the mercury disappears and is replaced by an orange solid.14. Magnesium is malleable and ductile.16. The most common physical changes are changes in state: solid to liquid, liquid to gaseous, solid togaseous (and their opposites).18.a.physical: The iron is only being heated.b.chemical: The sugars in the marshmallow are being reduced to carbon.c.chemical: Most strips contain a peroxide that decomposes.d.chemical: The bleach oxidizes dyes in the fabric.e.physical: Evaporation is only a change of state.f.physical: The salt is modifying only the physical properties of the solution, not undergoing achemical reaction.g.chemical: The drain cleaner breaks bonds in the hair.h.physical: Students will most likely reply that this is a physical change since the perfume isevaporating; the sensation of smell, however, depends on chemical processes.i.physical: The sublimation is only a change of state.j.physical: The wood is only being physically divided into smaller pieces.k.chemical: The cellulose in the wood is reacting with oxygen gas.20. Compounds consist of two or more elements combined together chemically in a fixedcomposition, no matter what their source may be. For example, water on earth consists ofmolecules containing one oxygen atom and two hydrogen atoms. Water on Mars (or any otherplanet) has the same composition.22. compoundsCopyright Houghton Mifflin Company. All rights reserved.

Chapter 3: Matter1524. Typically, the properties of a compound and the elements that constitute it are very different.Consider the properties of liquid water and the hydrogen and oxygen gases from which the waterwas prepared. Consider the properties of sodium chloride (table salt) and the sodium metal andchlorine gas from which it might have been prepared.26. Assuming the magnesium and sulfur had been measured out in exactly the correct ratio forcomplete reaction, what would remain after heating would be a pure compound. If there were anexcess of either magnesium or sulfur, however, the material left after reaction would be a mixtureof the compound and the excess reagent.28. solutions: window cleaner, shampoo, rubbing alcoholmixtures: salad dressing, jelly beans, the change in my pocket30.a.primarily a pure compound, but fillers and anti-caking agents may have been addedb.mixturec.mixtured.pure usd.heterogeneouse.homogeneous32.34. Consider a mixture of salt (sodium chloride) and sand. Salt is soluble in water, sand is not. Themixture is added to water, stirred to dissolve the salt and then filtered. The salt solution passesthrough the filter; the sand remains on the filter. The water can then be evaporated from the salt.36. The solution is heated to vaporize (boil) the water. The water vapor is then cooled so that itcondenses back to the liquid state, and the liquid is collected. After all the water is vaporized fromthe original sample, pure sodium chloride will remain. The process consists of physical changes.38. Since X is a pure substance, the fact that two different solids form when electrical current ispassed indicates that X must be a compound.40. Because vaporized water is still the same substance as solid water, no chemical reaction hasoccurred. Sublimation is a physical change.42. far apart44. chemical46. chemical48. ight Houghton Mifflin Company. All rights reserved.

16Chapter 3: Matterc.heterogeneous (Unless you work hard to get all the lumps out!)d.Although strictly heterogeneous, it may appear homogeneous.e.heterogeneous52. Answer depends on student’s response54. physical, chemical56. O2 and P4 are both still elements, even though the ordinary forms of these elements consist ofmolecules containing more than one atom (but all atoms in each respective molecule are thesame). P2O5 is a compound because it is made up of two or more different elements (not all theatoms in the P2O5 molecule are the same).58. Assuming there is enough water present in the mixture to have dissolved all the salt, filter themixture to separate out the sand from the mixture. Then distill the filtrate (consisting of salt andwater), which will boil off the water, leaving the salt.60. The most obvious difference is the physical states: Water is a liquid under room conditions;hydrogen and oxygen are both gases. Hydrogen is flammable. Oxygen supports combustion.Water does neither.Copyright Houghton Mifflin Company. All rights reserved.

Cumulative Review Chapters 1–3ANSWERS2.By now, after having covered three chapters in this book, it is hoped that you have adopted an“active” approach to your study of chemistry. You may have discovered, perhaps through adisappointing grade on a quiz (though we hope not), that you have to get really involved withchemistry. You can’t just sit and take notes or just look over the solved examples in the textbook.You have to learn to solve problems. You have to learn how to interpret problems and how toreduce them to the simple mathematical relationships you have studied. Whereas in some coursesyou might get by on just giving back on exams the facts or ideas presented in class, in chemistryyou have to be able to extend and synthesize what has been discussed and to apply the material tonew situations. Don’t get discouraged if this is difficult at first: it’s difficult for everyone at first.4.It is difficult sometimes for students (especially beginning students) to understand why certainsubjects are required for a given college major. The faculty of your major department, however,have collectively many years of experience in the subject in which you have chosen to specialize.They really do know which courses will be helpful to you in the future. They may have hadtrouble with the same courses that now give you trouble, but they realize that all the work will beworth it in the end. Some courses you take, particularly in your major field itself, have obviousand immediate utility. Other courses, oftentimes chemistry included, are provided to give you ageneral background knowledge that may prove useful in understanding your own major or othersubjects related to your major. In perhaps a burst of bravado, chemistry has been called “thecentral science” by one team of textbook authors. This moniker is very true; however, in order tounderstand biology, physics, nutrition, farming, home economics, or whatever, it helps to have ageneral background in chemistry.Commonly Used Prefixes in the Metric SystemPrefixMeaningPower of The metric system is in use in most of the world because its system of units and multiples issimple to remember, and the system permits easy conversion between units. The various multiplesand subdivisions of the basic units are based on factors of ten, which are also the basis for ournumber system. The United States uses a historical system in which there is no simplerelationship among most units. Although several attempts have been made to gradually change theUnited States over to the metric system, no widespread support for the program has beenachieved. Although 2-liter soda bottles were accepted without too much complaint (since theyCopyright Houghton Mifflin Company. All rights reserved.

18Cumulative Review Chapters 1–3replaced a similar-sized 2-quart bottle), the thought of having a size 85 waist (in cm) may berepugnant to too many Americans! There obviously also would be a great cost to industry toretool all machinery and measurement devices in metric units.6.Whenever a scientific measurement is made, we always employ the instrument or measuringdevice we are using t

of Introductory Chemistry, Introductory Chemistry: A Foundation, and Basic Chemistry by Steven S. Zumdahl. Several hundred new problems and questions have been prepared for the new editions of the text, which we hope will be of even greater help to your students in gaining an