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exam 3 spring 2007 - Kolar Leah Exam 3 Due noon Inst McCord...

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Kolar, Leah – Exam 3 – Due: Apr 11 2007, noon – Inst: McCord 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. R = 0 . 08206 L · atm/mol · K R = 8 . 314 Pa · m 3 /mol · K R = 62 . 36 L · torr/mol · K 1 atm = 101325 Pa Assume balloons X and Y are regular latax rubber balloons. 001 (part 1 of 1) 10 points Which would you expect to be least viscous? 1. C 8 H 18 at 30 C 2. C 4 H 8 at 30 C 3. C 8 H 18 at 50 C 4. C 4 H 8 at 50 C correct Explanation: 002 (part 1 of 1) 10 points At STP, 7.0 grams of CO gas will occupy a volume of 1. 11.2 liters. 2. 5.6 liters. correct 3. 3.5 liters. 4. 4.8 liters. 5. 22.4 liters. Explanation: n = 7 g · mol CO 28 g CO = 0 . 25 mol CO P = 1 atm T = 273 . 15 K Applying the ideal gas law, P V = n R T V = n R T P V = (0 . 25)(0 . 08206)(273 . 15) 1 = 5 . 60367 L 003 (part 1 of 1) 10 points Consider four molecules I) CHCl 3 II) CH 4 III) CH 3 Cl IV) CCl 4 Which of these exhibit permanent dipole- dipole interactions? 1. None of these 2. I and III only correct 3. I only 4. I, III, and IV only 5. III only Explanation: Dipole-dipole interactions occur in polar molecules. CHCl 3 and CH 3 Cl are polar be- cause their dipole moments do not cancel. CH 4 and CCl 4 are symmetric; their dipole moments cancel and the overall molecule is non-polar. 004 (part 1 of 1) 10 points Calculate the ratio of the rate of effusion of CO 2 to that of He (at the same temperatures). 1. 11 : 1 2. 1 : 1 3. 11 : 1 4. 11 2 : 1 5. 1 : 11 2
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Kolar, Leah – Exam 3 – Due: Apr 11 2007, noon – Inst: McCord 2 6. 1 : 11 7. 1 : 11 correct Explanation: Eff CO 2 Eff He = MW HC p MW CO 2 = r 4 44 = r 1 11 005 (part 1 of 1) 10 points What volume of pure oxygen gas (O 2 ) mea- sured at 546 K and 1.00 atm is formed by complete dissociation of 0 . 5 mol of Ag 2 O? 2 Ag 2 O(s) 4 Ag(s) + O 2 (g) 1. 11 . 2 L correct 2. 16 . 8 L 3. 22 . 4 L 4. 5 . 6 L 5. 33 . 6 L Explanation: n O 2 = 0 . 5 mol Ag 2 O · 1 mol O 2 2 mol Ag 2 O = 0 . 25 mol O 2 P = 1 atm T = 546 K PV = n R T V = n R T P = (0 . 25 mol) (0 . 0826 L · atm mol · K ) (546 K) 1 atm = 11 . 2012 L 006 (part 1 of 1) 10 points A mixture of CO, CO 2 and O 2 is contained
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