Final Exam-solutions - Version 237 Final Exam vanden bout...

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Version 237 – Final Exam – vanden bout – (51640) 1 This print-out should have 44 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points What is the Δ S fus ±or mercury i± it has a ±reezing point o± - 38 . 8 C and a Δ H fus = 2 . 292 kJ / mol? 1. - 0 . 06 J / mol · K 2. - 59 J / mol · K 3. 9 . 78 J / mol · K correct 4. 59 J / mol · K 5. 0 . 06 J / mol · K Explanation: FP = - 38 . 8 C = 234 . 35 K Δ H = 2 . 292 kJ / mol Δ S fus = Δ H fus T = 2 . 292 kJ / mol 234 . 35 K = 0 . 00978024 J / mol · K = 9 . 78024 kJ / mol · K 002 10.0 points The decomposition o± hydrogen peroxide to ±orm water is a frst order process. I± it takes 20 minutes ±or the initial concentration to ±all ±rom 1.6 M to 0.8 M, how much time has passed when only 0.05 M o± the initial 1.6 M remains? 1. 120 minutes 2. 100 minutes correct 3. 160 minutes 4. 80 minutes 5. 40 minutes Explanation: 003 10.0 points Name the molecule C C O C C C 1. pentanal 2. 4-pentanol 3. pentanone 4. pentanoic acid 5. pentanol 6. pentanoic ether 7. ethyl propyl ether correct 8. hexanoic ether 9. ethyl propyl amine 10. ethyl propyl ester Explanation: 004 10.0 points In the determination o± iron in vitamins, Fe 2+ is titrated with permanganate (MnO 4 ) in acidic solution. The products o± the reaction are Fe 3+ and Mn 2+ . In the balanced equation, how many electrons are trans±erred? 1. 10 2. 1 3. 7 4. 5 correct Explanation: 005 10.0 points Over what temperature range would a 4 m solution o± NaCl remain a liquid? (±or water
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Version 237 – Final Exam – vanden bout – (51640) 2 k b = 0.512 C kg mol 1 and k f = 1.86 C kg mol 1 ) 1. 114 . 9 C 2. 95 . 9 C 3. 85 . 1 C 4. 81 C 5. 119 C correct 6. 104 . 1 C Explanation: Pure water will remain a liquid between 0 C and 100 C. A 4 m NaCl solution would have a freezing point depression of 2 · 4 · 1 . 86 = 14 . 88 C, and a new freezing point of - 14 . 88 C. It would have a boiling point elevation of 2 · 4 · 0 . 512 = 4 . 096 C, and a new boiling point of 104 . 096 C. The so- lution would thus be stable from 104.096 to - 14 . 88 C, a range of 119 C. 006 10.0 points The equilibrium constant K p for the reaction I 2 (g) + Br 2 (g) 2 IBr(g) + 11 . 7 kJ is 280 at 150 C. Suppose that a quantity of IBr is placed in a closed reaction vessel and the system is allowed to come to equilibrium at 150 C. When equilibrium is established, the pressure of IBr is 0.200 atm. What is the pressure of I 2 at equilibrium? 1. 0.067 atm 2. 0.168 atm 3. None of these 4. 0.096 atm 5. 0.012 atm correct Explanation: At equilibrium, P IBr = 0 . 200 atm K p = 280 I 2 (g) + Br 2 (g) 2 IBr(g) + 11.7 kJ 0 0 y x x - 2 x x x y - 2 x y - 2 x = 0 . 2 K p = P 2 IBr P I 2 · P Br 2 = 280 0 . 2 2 x 2 = 280 x = 0 . 2 280 = 0 . 0119523 P I 2 = 0 . 0119523 atm 007 10.0 points ? is the nuclear process by which heavier
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This note was uploaded on 05/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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Final Exam-solutions - Version 237 Final Exam vanden bout...

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