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Unformatted text preview: Version 172 – Exam 3 2008 CH301 – Brodbelt – (53765) 1 This printout should have 25 questions. Multiplechoice 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 nonbonding electrons, 6 antibonding electrons, and 10 bonding electrons? 1. 2 correct 2. 1 3. 3 4. Explanation: e − antibond = 10 e − bond = 6 BO = bonding e − antibonding 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, Hbonding, 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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 Fall '07
 Fakhreddine/Lyon
 Atom, Electron, Chemical bond

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