Exam 3 - Version 172 Exam 3 2008 CH301 Brodbelt (53765) 1...

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Unformatted text preview: Version 172 Exam 3 2008 CH301 Brodbelt (53765) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The use of programmable calculators, cell phones, and other electronic devices is pro- hibited on this exam. Assume all numerical values are given with three significant dig- its (even if some of the digits are missing). GOOD LUCK! 001 10.0 points What is the bond order of two atoms that have 4 non-bonding electrons, 6 anti-bonding electrons, and 10 bonding electrons? 1. 2 correct 2. 1 3. 3 4. Explanation: e antibond = 10 e bond = 6 BO = bonding e - anti-bonding e 2 = 10- 6 2 = 2 002 10.0 points A sample of N 2 gas occupies a volume of 746 mL at STP. What volume would N 2 gas occupy at 155 C at a pressure of 368 torr? 1. 588 mL 2. 983 mL 3. 323 mL 4. 2415 mL correct 5. 312 mL 6. 1792 mL 7. 566 mL 8. 3295 mL Explanation: V 1 = 746 mL T 1 = 273 . 15 K P 1 = 1atm = 760torr T 2 = 155 C = 428 K P 2 = 368torr Combined Gas Law: P 1 V 1 T 1 = P 2 V 2 T 2 V 2 = P 1 V 1 T 2 T 1 P 2 = (760torr)(746mL)(428 . 15K) (273 . 15 K)(368torr) = 2414 . 9mL 003 10.0 points 0.850 moles of N 2 originally at 85 C is cooled such that it now occupies 17.55 L at 1.25 atm. What is the final temperature of the gas? 1. 85.0 C 2. 65.4 C 3. Not enough information is given. 4. 31.4 C 5. 17.2 C 6. 41.4 C correct 7. 314.5 C Explanation: The initial temperature is not needed to solve the problem. Use PV = nRT PV/ ( nR ) = T 1 . 25(17 . 55) / . 850(0 . 08206) = 314 . 5 K 314 . 5 K- 273 . 15 = 41 . 4 C 004 10.0 points What is the volume occupied by 0.170 grams of gaseous H 2 S at 27 C and 380 torr? 1. None of these 2. 2.24 liters Version 172 Exam 3 2008 CH301 Brodbelt (53765) 2 3. 0.446 liter 4. 0.25 liter correct 5. 0.020 liter 6. 0.170 liter 7. 2.46 liters 8. 50.0 liters 9. 4.00 liters 10. 1.00 liter Explanation: n = 17 g mol 34 g = 0 . 005 mol T = 27 C + 273 = 300 K P = 380 torr 1 atm 760 torr = 0 . 5 atm Applying the ideal gas law, P V = n RT V = n RT P V = (0 . 005 mol) ( . 08206 L atm mol K ) (300 K) . 5 atm = 0 . 24618 L 005 10.0 points Based on the strength of intermolecular forces, which would you expect to have the lowest boiling point? 1. CaI 2 2. SrI 2 3. C 2 H 6 correct 4. NH 3 5. C 4 H 8 Explanation: The intermolecular forces, from strongest to weakest, are interionic, H-bonding, dipole- dipole, and dispersion forces. The stronger the force holding molecules together, the more energy is required to break those attractions, so molecules with strong intermolecular forces have higher boiling points....
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This note was uploaded on 09/13/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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Exam 3 - Version 172 Exam 3 2008 CH301 Brodbelt (53765) 1...

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