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211chp5 - Chapter 5 Homework Solutions 12 a Decreases b...

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Chapter 5 Homework Solutions 9/26/11 12. a) Decreases. b) Increases. As the ball rises in the air, its kinetic energy is converted to potential energy. This is seen as the velocity of the ball decreasing and eventually reaching a value of zero, at which point the ball will begin to fall back to Earth. c) The ball with the heavier mass will travel at a slower velocity (KE = ½mv 2 ) and so will reach zero velocity sooner. Hence, it won’t travel as far (half as far to be exact). 14. a) KE = ½(850 lb) 2 2 2 /s m kg 1 J 1 km 1 m 1000 mi 1 km .609 1 s 3600 hr 1 hr mi 66 lb kg 454 0 . = 1.7 x 10 5 J b) Since KE = ½mv 2 , halving the velocity, quarters the kinetic energy. (i.e. KE 34 mph = ¼KE 68 mph ) c) The kinetic energy is lost as friction of the brakes on the wheels, tires on the road, and (minimally) movement through the air. 22. a) electrostatic attraction; no work is done (the particles do not move over a distance) b) magnetic attraction; work is done 26. a) E = q + w = 850 J + (-382 J) = 468 J, endothermic b) E = -3140 J + 0 J = -3140 J, exothermic c) E = -6.47 kJ + 0 J = -6.47 kJ, exothermic 28. a) The potential energy of the system has decreased. b) E is positive. c) q = 0 and w > 0. 40. In each part, the answer assumes all other things are the same. a) CO 2 (g) > CO 2 (s) since the heat of sublimation must be added to the solid to make the gas b) 2 mol H 2 (g) > 1 mol H 2 (g) because there are twice as many particles to hold heat. c) H 2 (g) + ½ O 2 (g) > H 2 O (g) because breaking apart a water molecule requires the input of energy. d) 300 ºC > 100 ºC the higher temperature transfers extra heat to the nitrogen molecules 42. a) Since H is positive, the reaction is endothermic. b) heat transfer = (45.0 g CH3OH ) CH3OH CH3OH CH3OH mol 1 kJ 0.7 9 g 2.04 3 mol 1 = 127 kJ (heat will be consumed, H = 127 kJ) c) mass H2 = (25.8 kJ) H2 H2 H2 mol g 02 . 2 kJ 0.7 9 mol 2 = 1.15 g H = -25.8 kJ (The sign changes because the reaction is reversed.)
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d) heat transfer = (50.9 g CO ) CO CO CO mol 1 kJ 0.7 9 g 8.01 2 mol 1 = 165 kJ
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