book even asnwer ch5-6

book even asnwer ch5-6 - CHAPTER 5-6 Chapter 5 14. a. The...

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Unformatted text preview: CHAPTER 5-6 Chapter 5 14. a. The volume will decrease. b. The volume will decrease. c. The volume will increase. 22. a. A; A higher pressure in container A means more molecules and thus more mass and a higher density. b. They are equal. Equal pressures and temperatures mean the same number of molecules per volume. c. B; A higher temperature means that container A has fewer molecules and thus less mass and a lower density. 36. a. 0.160 atm b. 0.305 atm 38. 726 Torr 46. 69 Torr extra 48. 4.4 atm 50. 3.85 L 52. 546 °C 64. mol released = 8.00 - 4.69 = 3.31 moles 70. Ptota1= 1.09 x 103 mmHg P1 V1 P2V2 V1P1T2 _ = = 3 3.00 atm (25 + 273)K 72' T1 T2 ’ V2 P2T1 4'20 "10 L x 151 atm x (17 + 273)K V2 = 85.7 L 76. Molecular mass = 111 u 84. Calculate the empirical formula. 8563 c £41293 1c —1——=100 ' g x12.011gc‘ ' “1° x7.1293m01C ' mol H 1 __ 14.37 g H x — 14.2574 mol H x m — 2.00 14.027 CH2 12.011 + 2(1.0079) = empirical formula 1.69 g X 0.08206 L atm dRT L K mol x (24 + 273)K 9W=— = ——-—-—-—-——-—-——- =42.13 g/mol P 743 H a—tm— m g x 760 mmHg 42.13 u formula unit = 3 00 formula unit molecule x 14.027 u ' molecule C3H6 émol 02 88. ?L02=3.06LCOx—m=1.53L02 02 is limiting H l-mm‘wymvwwvmw v mm" - m . WW 4 W""“‘"“““““ ‘"“"1W‘W"‘W'WWW"I*‘WW * W”"‘"’""”“""""'" 90. 92. 100. 102. 108. 110. 120. 124. 26. 28. 30. 32. 38. lmolC02 ?LC02=1.48L02x1— 51110102 = 2.96 L coz 7.42 L 02 5.998 g H202 7 % yield = 6.00% 2.78 x 10-2 2.82 L gas (VP of H20 at 24°C is 22 Torr.) Rearranging the equaton. 1 3.1%.,“2 ll V viz 1 2 N ._ 2=_._.. 2mu 3 P ek < II who viz V II WIN ' constant T Thus, at the same T and P, two equal volumes of gas should be comprised of the same number of molecules. 972.0 km/hr. a. 35.0 L Hz b. 38.6 L H2 c. 35.0 L H2 d. 35.2 L H2124. 342 m/s CHAPTER 6 110 J = - 557 J 557 J of work done by system a. Yes; an example would be an exothermic reaction that does pressure-volume work. b. No; if q and w are both negative, ?E cannot be zero, it must be negative. - 5.66 x 103 kJ mol ?H and ?E are very nearly equal because the P?V~ 0 for this reaction. The number of moles of gases is unchanged in the reaction. -1.03 x 103 kJ 42. -4.44x103kJ 50. 6.34J/°C 54. 329 °C 58. mSS = 2.3 x 102 g 66. -44.7 k] /g gasoline 17.9 kJ 68. 0C 70. 3.29x1o4kg 74. ?H°=-818.2kJ 76. ?H° = 219.0 kJ 82. -984.9 kJ /mol 90. ?H = ?U + P?V The P?Vvalue will be negative, as Vf(l) is less than Vinifis), so ?U will be larger by a very small value. ...
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book even asnwer ch5-6 - CHAPTER 5-6 Chapter 5 14. a. The...

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