(Solution Manual) A Complete Solution Manual For Chemistry

Zumdahl,Zumdahl, 9

Chemistry

Solution Manual

9th Edition

• $15.00 $50.00

---

Description

Solution Manual for A Complete Solution Manual For Chemistry, The downloadable file consists of 872 pages, and the total size of the file is 13.82 MB. Purchase now only for highly discounted price.

Author(s): Zumdahl, Zumdahl

---

---

Study Note Preview

TABLE OF CONTENTS Chapter 1 Chemical Foundations ......................................................................................... 1 Chapter 2 Atoms, Molecules, and Ions ...............................................................................25 Chapter 3 Stoichiometry .....................................................................................................46 Chapter 4 Types of Chemical Reactio ns and Solution Stoichiometry ..................................93 Chapter 5 Gases ............................................................................................................... 139 Chapter 6 Thermochemistry ............................................................................................. 184 Chapter 7 Atom ic Structure and Periodicity ...................................................................... 215 Chapter 8 Bonding: General Concepts ............................................................................. 250 Chapter 9 Covalent Bonding: Orbitals ............................................................................. 304 Chapter 10 Liquids and Solids ............................................................................................ 341 Chapte r 11 Properties of Solutions ..................................................................................... 380 Chapte r 12 Chemical Kinetics ............................................................................................ 418 Chapter 13 Chem ical Equilibrium ...................................................................................... 458 Chapter 14 Acids and Bases ............................................................................................... 500 Chapter 15 Acid -Base Equilibria ........................................................................................ 563 Chapter 16 Solubility and Complex Ion Equilibria ............................................................. 621 Chapter 17 Spontanei ty, Entropy, and Free Energy ............................................................. 659 Chapter 18 Electrochemistry .............................................................................................. 688 Chapter 19 The Nucleus: A C hemist’s View ...................................................................... 741 Chapter 20 The Representative Ele ments ............................................................................ 760 Chapter 21 Transition Metals and Coordination Chemistry ................................................. 782 Chapter 22 Organic and Biological Molecules .................................................................... 810 iii 1 CHAPTER 1 CHEMICAL FOUNDATIONS Questions 17. A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction or the ideal gas law, PV = nRT. A theory (model) is an attempt to explain why something happens. Dalton ’s atomic t heory explains why mass is conserved in a chemical reaction. The kinetic molecular theory explains why pressure and volume are inversely related at constant temperature and moles of gas present , as well as explaining the other mathematical relationships summarized in PV = nRT. 18. A dynamic process is one that is active as opposed to static. In terms of the scientific method, scientists are always performing experiments to prove or disprove a hypothesis or a law or a theory. Scientists do not stop asking questions just because a given theory seems to account satisfactorily for some aspect of natural behavior. The key to the scientific method is to continually ask quest ions and perform experiments. Science is an active process , not a static one. 19. The fundamental steps are (1) making observations ; (2) formulating hypotheses ; (3) performing experiment s to test the hypothese s. The key to the scientific method is performi ng experiments to test hypotheses . If after the test of time the hypothese s seem to account satisfactorily for some aspect of natural behavior, then the set of tested hypoth eses turn s into a theory (model). However , scientists continue to perform experiments to refine or replace existing theories. 20. A random error has equal probability of being too high or too low. This type of error occurs when estimating the value of the last digit of a measurement. A systematic er ror is one that always occurs in the same direction, either too high or too low. For example, t his type of error would occur if the balance you were using weighed all objects 0.20 g too high, that is , if the balance wasn’t calibrated correctly. A random error is an indeterminate error , whereas a systematic error is a determinate error. 21. A qualitative observation expresses what makes something what it is; it does not involve a number; e.g., the air we breathe is a mixture of gases, ice is less dense than water, rotten milk stinks. The SI units are mass in kilograms, length in meters, and volume in the derived units of m 3. The assumed uncertainty in a number is ± 1 in the last signific ant figure of the number. The precision of an instrument is related to the number of significant figures associated with an 2 CHAPTER 1 CHEMICAL FOUNDATIONS experimental reading on that instrument. Different instruments for measuring mass, length, or volume have varying degrees of precisi on. Some instruments only give a few significant figures for a measurement , whereas others will give more significant figures. 22. Precision: reproducibility; a ccuracy: the agreement of a measurement with the true value. a. Imprecise and inaccurate data: 12.32 cm, 9.63 cm , 11.98 cm, 13.34 cm b. Precise but inaccurate data: 8.76 cm, 8.79 cm, 8.72 cm, 8.75 cm c. Precise and accurate data: 10.60 cm, 10.65 cm, 10.63 cm, 10.64 cm Data can be imprecise if the measuring device is imprecise as well as if the user of the measuring device has poor skills. Data can be inaccurate due to a systematic error in the measuring device or with the user. For example, a balance may read all masses as weighing 0.2500 g too high or the user of a graduated cylinder may read all measurements 0.05 mL too low. A set of measurements that are imprecise implies that all the numbers are not close to each other. If the numbers aren’t reproducible, then all the numbers can’t be very close to the true value. Some say that if the aver age of imprecise data gives the true value, then the data are accurate ; a better description is that the data takers are extremely luck y. 23. Significant figures are the digits we associate with a number. They contain all of the certain digits and the fir st uncertain digit (the first estimated digit). What follows is one thousand indicated to varying numbers of significant figures: 1000 or 1 × 10 3 (1 S.F.); 1.0 × 103 (2 S.F.); 1.00 × 103 (3 S.F.); 1000. or 1.000 × 103 (4 S.F.). To perfor m the calculation, the addition /subtraction significant figure rule is applied to 1.5 − 1.0. The result of this is the one -significant -figure answer of 0.5. Next, the multi- plication/division rule is applied to 0.5/0.50. A one -significant -figure number divided b y a two-significant -figure number yields an answer with one significant figure (answer = 1). 24. From Figure 1.9 of the text, a change in temperature of 180° F is equal to a change in temperature of 100 °C and 100 K. A degree unit on the Fahrenheit scale is not a large as a degree unit on the Celsius or Kelvin scales. Therefore, a 20° change in the Celsius or Kelvin temperature would correspond to a larger temperature change than a 20 ° change in the Fahrenheit scale . The 20° temperature change on the Celsius and Kelvin scales are equal to each other. 25. Straight line equation: y = mx + b, where m is the slope of the line and b is the y -intercept. For the T F vs. T C plot: TF = (9/5)T C + 32 y = m x + b The slope of the plot is 1.8 ( = 9/5) and the y-intercept is 32 °F. For the TC vs. T K plot: TC = T K − 273 y = m x + b The slope of the plot is 1, and the y-intercept is − 273° C. CHAPTER 1 CHEMICAL FOUNDATIONS 3 26. a. coffee; saltwater; the air we breath e (N2 + O 2 + others); brass (Cu + Zn) b. book; human being; tree; desk c. sodium chloride (NaCl); water (H 2O); glucose (C6H12O6); carbon dioxide (CO 2) d. nitrogen (N 2); oxygen (O 2); copper (Cu) ; zinc (Zn) e. boiling water; freezing water; meltin g a popsic le; dry ice subliming f. Elecrolysis of molten sodium chloride to produce sodium and chlorine gas; the explosive reaction between oxygen and hydrogen to produce water; photosynthesis , which co nverts H2O and CO 2 into C 6H12O6 and O 2; the combustion of gasoline in our car to produce CO 2 and H 2O Exercises Significant Figures and Unit Conversions 27. a. exact b. inexact c. exact d. inexact (π has an infinite number of decimal places.) 28. a. one significant figure (S.F.). The implied uncertainty is ± 1000 pages. More significant figures should be added if a more precise number is known. b. two S.F. c. four S.F. d. two S.F. e. infinite number of S.F. (exact number) f. one S.F. 29. a. 6.07 × 1510−; 3 S.F. b. 0.003840 ; 4 S.F. c. 17.00 ; 4 S.F. d. 8 × 108; 1 S.F. e. 463.8052 ; 7 S.F. f. 300; 1 S.F. g. 301; 3 S.F. h. 300.; 3 S.F. 30. a. 1 00; 1 S.F. b. 1.0 × 102; 2 S.F. c. 1.00 × 103; 3 S.F. d. 100.; 3 S.F. e. 0.0048 ; 2 S.F. f. 0.00480 ; 3 S.F. g. 4.80 × 310−; 3 S.F. h. 4.800 × 310−; 4 S.F. 31. When rounding, the last significant figure stays the same if the number after this significant figure is less than 5 and i ncreases by one if the number is greater than or equal to 5. a. 3.42 × 410− b. 1.034 × 104 c. 1.7992 × 101 d. 3.37 × 105 32. a. 4 × 105 b. 3.9 × 105 c. 3.86 × 105 d. 3.8550 × 105 33. Volume measurements are estimated to one place past the markings on the glassware. The first graduated cylinder is labeled to 0.2 mL volume increments, so we estimate volumes to 4 CHAPTER 1 CHEMICAL FOUNDATIONS the hundredths place. Realistically, the uncertainty in this graduated cylinde r is ±0.05 mL. The second cylinder, with 0.02 mL volume increments, will have an uncertainty of ±0.005 mL. The approximate volume in the first graduated cylinder is 2.85 mL, and the volume in the other graduated cylinder is approximately 0.280 mL. The t otal volume would be: 2.85 mL +0.280 mL 3.13 mL We should report the total volume to the hundredths place because the volume from the first graduated cylinder is only read to the hundredths (read to two decimal places). The first graduated cylinder is the least precise volume measurement because the uncertainty of this instrument is in the hundredths place, while the uncertainty of the second graduated cylinder is to the thousandths place. It is always the lease precise measurement that lim its the precision of a calculation. 34. a. Volumes are always estimated to one position past the marked volume increments. The estimated volume of the first beaker is 32.7 mL, the estimated volume of the middle beaker is 33 mL, and the estimated volume in the last beaker is 32.73 mL. b. Yes, all volumes could be identical to each other because the more precise volume readings can be rounded to the other volume readings. But because the volumes are in three different measuring devices, each with its own unique uncertainty, we cannot say with certainty that all three beakers contain the same amount of water. c. 32.7 mL 33 mL 32.73 mL 98.43 mL = 98 mL The volume in the middle beaker can only be estimated to the ones place, which dictates that the sum of the volume should be reported to the ones place. As is always the case, the least precise measurement determines the precision of a calculation. 35. For addition and/or subtraction, the result has the same number of decimal places as the number in the calculation with the fewest decimal places. When the result is rounded to the correct number of significant figures, the last significant figure stays the same if the number after this significant figure is less than 5 and increases by one if the number i s greater than or equal to 5. The underline shows the last significant figure in the intermediate answers. a. 212.2 + 26.7 + 402.09 = 640. 99 = 641.0 b. 1.0028 + 0.221 + 0.10337 = 1.32 717 = 1.327 c. 52.331 + 26.01 − 0.9981 = 77.34 29 = 77.34 d. 2.01 × 102 + 3.014 × 103 = 2.01 × 102 + 30.14 × 102 = 32.1 5 × 102 = 3215 When the exponents are different, it is easiest to apply the addition/subtraction rule when all numbers are based on the same power of 10. e. 7.255 − 6.8350 = 0.42 = 0.420 (first uncertain digit is in the third decimal place). CHAPTER 1 CHEMICAL FOUNDATIONS 5 36. For multiplication and/or division, the result has the same number of significant figures as the number in the calculation with the fewest significant figures. a. 2.26 352.261.01273 0.0821 0.102= =× × b. 0.14 × 6.022 × 1023 = 8.431 × 1022 = 8.4 × 1022; since 0.14 only has two significant figures, the result should only have two significant figures. c. 4.0 × 104 × 5.021 × 310− × 7.34993 × 102 = 1.476 × 105 = 1.5 × 105 d. 12 12 76 10 67.6 10 6766.6 10 00.310 00.2× = × = ×× − 37. a. Here, apply the multiplication/division rule first; then apply the addition/subtraction rule to arrive at the one-decimal -place answer. We will generally round off at intermediate steps in order to show the correct number of significant figures. However, you should round off at the end of all the mathematical operations in order to avoid round- off error. The best way to do calcul ations is to keep track of the correct number of significant figures during intermediate steps, but round off at the end. For this problem, we underlined the last significant figure in the intermediate steps. 4326.0705.80 623.0470.0 1.3526.2+ + = 0.8 148 + 0.754 4 + 186.558 = 188.1 b. Here, the mathematical operation requires that we apply the addition/subtraction rule first, then apply the multiplication/division rule. 6.191.2 404.6 1.177.1891.2 404.6 × −×= = 12 c. 6.071 × 510− − 8.2 × 610− − 0.521 × 410− = 60.71 × 610−− 8.2 × 610− − 52.1 × 610− = 0. 41 × 610− = 4 × 710− d. 26 1213 12 1213 13 13 1213 12 103.6 10761024 10 63 104100.4 10 38 103.6 104100.4 108.3 −− − − − − × ×× × + ×× + × × + ×× + ×= = = e. 4755.19 4175.38.21.45.9=+ + + = 4.89 = 4.9 Uncertainty appears in the first decimal place. The average of several numbers can only be as precise as the least precise number. Averages can be exceptions to the significant figure rules. f. 925.8002.0100925.8905.8 925.8= ×− × 100 = 0.22 38. a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025 6 CHAPTER 1 CHEMICAL FOUNDATIONS b. 17 98 34 10 82.7 10 54.210 998.2 10 6262.6 − −− × = ×× × × c. 1.285 × 210− + 1.24 × 310− + 1.879 × 110− = 0.1285 × 110− + 0.0124 × 110−+ 1.879 × 110− = 2.020 × 110− When the exponents are different, it is easiest to apply the addition/subtraction rule when all numbers are based on the same power of 10. d. 27 23 2310 29.2 10 02205.600138.0 10 02205.6) 00728.1 00866.1( −× × ×−= = e. 1 22 22 2 101.8 100 10 875.910 080.0100 10 875.910 795.9 10 875.9 −× × ××× ×× − ×= = f. 310 625.1 10 824.0 10 942.0 310 625.1 10 234.8 10 42.93 3 3 3 2 2× + × + × × + × + ×= = 1.130 × 103 39. a. 8.43 cm × mm3.84mmm 1000 cm100m1= × b. 2.41 × 102 cm × cm100m1 = 2.41 m c. 294.5 nm cm 10 945.2mcm100 nm101m1 5 9−× × ×× = d. 1.445 × 104 m × mkm 10001= 14.45 km e. 235.3 m × mmm 1000 = 2.353 × 105 mm f. 903.3 nm mμ 9033.0mmμ101 nm101m16 9=×× ×× 40. a. 1 Tg kg101g 1000kg1 Tgg 101 912 × ××× = b. 6.50 × 102 Tm nm 10 50.6mnm101 Tmm 101 239 12 ××××× = c. 25 fg kg 105.2 kg 10 25g 1000kg1 fg 101g1 17 18 15− −× × × ×× = = d. 8.0 dm3 × 31 dmL = 8.0 L (1 L = 1 dm3 = 1000 cm3 = 1000 mL) e. 1 mL Lμ101LLμ101 mL0001L1 36 ××× × = f. 1 μg pg101gpg 101 gμ101g1 612 6××× ×× = CHAPTER 1 CHEMICAL FOUNDATIONS 7 41. a. Appropriate conversion fa ctors are found in Appendix 6. In general, the number of significant figures we use in the conversion factors will be one more than the number of significant figures from the numbers given in the problem. This is usually sufficient to avoid round- off error. 3.91 kg × kg 4536.0lb1= 8.62 lb; 0.62 lb × lboz16 = 9.9 oz Baby’s weight = 8 lb and 9.9 oz or , to the nearest ounce, 8 lb and 10 . oz. 51.4 cm × cm54.2in1 = 20.2 in ≈ 20 1/4 in = baby’s height b. 25,000 mi × mikm61.1 = 4.0 × 104 km; 4.0 × 104 km × kmm 1000 = 4.0 × 107 m c. V = 1 × w × h = 1.0 m ×    × ×   ×dm10m1dm1.2cm100m1cm6.5 = 1.2 × 210− m3 1.2 × 210−m3 × 33 dmL1 mdm01×   = 12 L 12 L × 33 cm54.2in1 Lcm 1000    × = 730 in3; 730 in3 × 3 in12ft1    = 0.42 ft3 42. a. 908 oz × lbkg 4536.0 oz16lb1× = 25.7 kg b. 12.8 L × qt4gal1 L 9463.0qt1× = 3.38 gal c. 125 mL × L 9463.0qt1 mL 1000L1× = 0.132 qt d. 2.89 gal × L1mL 1000 qt057.1L1 gal1qt4× × = 1.09 × 104 mL e. 4.48 lb × lb1g6.453 = 2.03 × 103 g f. 550 mL × Lqt06.1 mL 1000L1× = 0.58 qt 43. a. 1.25 mi × mifurlongs8 = 10.0 furlongs; 10.0 furlongs × furlongrods40 = 4.00 × 102 rods 4.00 × 102 rods × cm100m1 incm54.2 ydin36 rodyd5.5× × × = 2.01 × 103 m 8 CHAPTER 1 CHEMICAL FOUNDATIONS 2.01 × 103 m × m 1000km1 = 2.01 km b. Let's as sume we know this distance to ± 1 yard. First , convert 26 miles to yards. 26 mi × ft3yd1 mift 5280× = 45,760. yd 26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards 46,145 yard × yd5.5rod1 = 8390.0 rods; 8390.0 rods × rods40furlong1 = 209.75 furlongs 46,145 yard ×cm100m1 incm54.2 ydin36× × = 42,195 m; 42,195 m × m 1000km1 = 42.195 km 44. a. 1 ha × 22 m 1000km1 ham000,10    × = 1 2 2km 10−× b. 5.5 acre × 22 cm100m1 incm54.2 ydin36 rodyd5.5 acrerod160    × × × × = 2.2 × 104 m2 2.2 × 104 m2 × 2 4m101ha1 × = 2.2 ha; 2.2 × 104 m2 × 2 m 1000km1    = 0.022 km2 c. Area of lot = 120 ft × 75 ft = 9.0 × 103 ft2 9.0 × 103 ft2 × 22 rod160acre1 yd5.5rod1 ft3yd1×   × = 0.21 acre; acre000,31$ acre21.0500,6$= We can use our result from (b) to get the conver sion factor between acres and hectares (5.5 a cre = 2.2 ha.). Thus 1 ha = 2.5 acre. 0.21 acre × acre5.2ha1 = 0.084 ha; the price is: ha000,77$ ha084.0500,6$= 45. a. 1 troy lb × g 1000kg1 graing 0648.0 pwgrains24 oztroypw20 lbtroyoztroy12× × × × = 0.373 kg 1 troy lb = 0.373 kg × kglb205.2 = 0.822 lb b. 1 troy oz ×graing 0648.0 pwgrains24 oztroypw20× × = 31.1 g 1 troy oz = 31.1 g × g200.0carat1 = 156 carats CHAPTER 1 CHEMICAL FOUNDATIONS 9 c. 1 troy lb = 0.373 kg; 0.373 kg × g3.19cm1 kgg 10003 × = 19.3 cm3 46. a. 1 grain ap × ap dramg888.3 scruples3ap dram1 ap grain20scruple1× × = 0.06480 g From the previous question, we are given that 1 grain troy = 0.0648 g = 1 grain ap. So the two are the same. b. 1 oz ap × g1.31*troyoz1 ap dramg888.3 apozap dram8× × = 1.00 oz troy; *see Exercise 45 b. c. 5.00 × 102 mg × ap dramscruples3 g888.3ap dram1 mg 1000g1× × = 0.386 scruple 0.386 scruple × scrupleap grains20 = 7.72 grains ap d. 1 scruple × ap dramg888.3 scruples3ap dram1× = 1.296 g 47. 15.6 g × g65.0capsule1 = 24 capsules 48. 1.5 teaspoons × teaspoon50.0acetmg.80 = 240 mg acetaminophen kg454.0lb1 lb24acetmg240× = 22 mg acetaminophen/kg kg454.0lb1 lb35acetmg240× = 15 mg acetaminophen/kg The range is from 15 to 22 mg acetaminophen per kg of body weight. 49. warp 1.71 = yd/h 2030knot1 hmin60 mins60 myd 1.094 sm10 3.005.008 × × × ×    ×× = 2.9 1 × 109 knots hmin60 mins60 km609.1mi1 m 1000km1 sm10 00.300.58 × × × ×    ×× = 3.36 × 109 mi/h 50. s58.9m.100 = 10.4 m/s; hmin60 mins60 m 1000km1 s58.9m.100× × × = 37.6 km/h 10 CHAPTER 1 CHEMICAL FOUNDATIONS ydft3 myd 0936.1 s58.9m.100× × = 34.2 ft/s; hmin60 mins60 ft 5280mi1 sft2.34× × × = 23.3 mi/h 1.00 × 102 yd × m.100s58.9 yd 0936.1m1× = 8.76 s 51. kmmi 6214.0 hkm65× = 40.4 = 40. mi/h To the correct number of significant figures (2), 65 km/h does not violate a 40 m i/h speed limit. 52. 112 km × mi65h1 kmmi 6214.0× = 1.1 h = 1 h and 6 min 112 km × galL785.3 mi28gal1 kmmi 6214.0× × = 9.4 L of gasoline 53. euro$1.32 lb 2.2046kg1 kgeuros 2.45× × = $1.47/lb One pound of peaches costs $1. 47. 54. For the gasoline car: 500. mi gal50.3$ mi0.28gal1× × = $62.5 For the E85 car: 500. mi gal85.2$ mi5.22gal1× × = $63.3 The E85 vehicle would cost slightly more to drive 500. miles as compared to the gasoline vehicle ($63.3 versus $62.5). 55. Volume of lake = 100 mi2 × 2 mift 5280  × 20 ft = 6 × 1010 ft3 6 × 1010 ft3 × mLμg0.4 cmmL1 incm 2.54 ftin12 33 × ×  × = 7 × 1014 μg mercury 7 × 1014 μg g101kg1 gμ101g1 3 6×× ×× = 7 × 105 kg of mercury 56. Volume of room = 18 ft × 12 ft × 8 ft = 1700 ft3 (carrying one extra significant figure) CHAPTER 1 CHEMICAL FOUNDATIONS 11 1700 ft3 33 3 3 m48cm100m1 incm54.2 ftin12=   ×  ×  × 48 m3 COgμ101COg1 mCOgμ000,400 6 3×× × = 19 g = 20 g CO (to 1 sig. fig.) Temperature 57. a. TC = 95(TF − 32) = 95(−459°F − 32) = − 273°C ; TK = T C + 273 = −273°C + 273 = 0 K b. TC = 95(−40.°F − 32) = − 40.°C; TK = −40.°C + 273 = 233 K c. TC = 95(68°F − 32) = 20. °C; TK = 20.°C + 273 = 293 K d. TC = 95(7 × 107°F − 32) = 4 × 107°C; TK = 4 × 107°C + 273 = 4 × 107 K 58. 96.1°F ± 0.2°F; first, convert 96.1°F to °C. T C = 95(TF − 32) = 95(96.1 − 32) = 35.6°C A change in temperature of 9°F is equal to a change in temperature of 5°C. So the uncertainty is: ±0.2°F × F9C5 °° = ±0.1°C. Thus 96.1 ±0.2°F = 35.6 ± 0.1°C . 59. a. TF = 59 × T C + 32 = 59 × 39.2°C + 32 = 102.6°F (Note: 32 is exact.) TK = T C + 273.2 = 39.2 + 273.2 = 312.4 K b. TF = 59 × (−25) + 32 = − 13°F; TK = −25 + 273 = 248 K c. TF = 59 × (−273) + 32 = − 459°F; TK = −273 + 273 = 0 K d. TF = 59 × 801 + 32 = 1470°F; TK = 801 + 273 = 1074 K 60. a. TC = T K − 273 = 233 − 273 = - 40.°C TF = 59× T C + 32 = 59 × (−40.) + 32 = − 40.°F b. TC = 4 − 273 = − 269°C; T F = 59 × (−269) + 32 = − 452°F 12 CHAPTER 1 CHEMICAL FOUNDATIONS c. TC = 298 − 273 = 25°C; T F = 59 × 25 + 32 = 77°F d. TC = 3680 − 273 = 3410°C; T F = 59 × 3410 + 32 = 6170°F 61. TF = 59× T C + 32; f rom the problem, we want the temperature where T F = 2TC. Substituting: 2TC = 59× T C + 32, (0.2)TC = 32, TC = 2.032 = 160° C TF = 2TC when the temperature in Fahrenheit is 2(160) = 320° F. Because all numbers when solving the equation are exact numbers, the calculated temperatures are also exact numbers. 62. TC = 95(TF – 32) = 95(72 – 32) = 22° C TC = T K – 273 = 313 – 273 = 40. °C The difference in temperature between Jup iter at 313 K and Earth at 72° F is 40 .°C – 22 °C = 18°C. 63. a. A change in temperature of 140° C is equal to 50 °X. Therefore, X50C 140 oo is the unit con- version between a degree on the X scale to a degree on the Celsius scale. To account for the different zero points, − 10° must be subtracted from the temperature on the X scale to get to the Celsius scale. The conversion between ° X to ° C is: TC = T X × X50C 140 oo − 10°C, T C = T X × X5C14 oo − 10°C The conversion between °C to ° X would be: TX = (T C + 10°C) C14X5 oo b. Assuming 10° C and C14X5 oo are exact numbers: TX = (22.0 °C + 10° C) C14X5 oo = 11.4 °X c. Assuming exact numbers in the temperature conversion formulas: TC = 58.0 °X × X5C14 oo − 10°C = 152° C CHAPTER 1 CHEMICAL FOUNDATIONS 13 TK = 152° C + 273 = 425 K TF = C5F9 oo × 152° C + 32 °F = 306° F 64. a. A change in temperature of 160°C equals a change in temperature of 100°A. So A 100C 160 °° is our unit conversion for a degree change in temperature. At the freezing point: 0°A = − 45°C Combining these two pieces of information: TA = (T C + 45°C) × C 160A 100 °°= (T C + 45°C) × C8A5 °° or T C = T A × A5C8 °° − 45°C b. TC = (T F − 32) × 95; T C = T A × 58 − 45 = (T F − 32) × 95 TF − 32 = 59   − × × 4558TA = 2572TA× − 81, T F = T A × A25F72 °° − 49°F c. TC = T A × 58 − 45 and T C = T A; so T C = T C × 58 − 45, 5T3c= 45, T C = 75°C = 75°A d. TC = 86°A × A5C8 °° − 45°C = 93°C; T F = 86°A × A25F72 °° − 49°F = 199°F = 2.0 × 102°F e. TA = (45°C + 45°C) × C8A5 °° = 56°A Density 65. Mass = 350 lb lbg6.453× = 1.6 × 105 g; V = 1.2 × 104 in3 3 incm54.2  × = 2.0 × 105 cm3 Density = 3 3 55 g/cm80.0 cm100.2g101 volumemass= ××= Because the material has a density less than water, it will float in water. 66. 33 3 3 cm52.0g0.2d;cm52.0 )cm50.0( 14.334r34V = × × == =π = 3.8 g/cm3 The ball will sink. 100oA 115oC 160oC -45oC100oA 0oA