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Unformatted text preview: lastname firstname signature McCord CH301 TTh Please read the following!! You will have to turn in ALL of your exam materials at the end of the exam. This includes your exam copy, your bubblesheet, and your scratch paper. Be sure and sign your exam copy as well as your bubblesheet. Your exam copy can be picked up at the window of WEL 2.212 beginning Monday, 12/14 at 9 AM. Your scores for the final should be available on Quest sometime later this afternoon (after 2:30 PM). R = 0 . 08206 L atm/mol · K R = 8 . 314 J/mol · K R = 62 . 36 L · torr/mol · K 1 L · atm = 101.325 J 1 eV = 1.602 × 10 − 19 J c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10 − 34 J · s m e = 9 . 11 × 10 − 31 kg R = 3 . 29 × 10 15 s − 1 N A = 6 . 022 × 10 23 mol − 1 k = 1 . 381 × 10 − 23 J/K g = 9 . 81 m/s 2 d water = 1 . 00 g/mL 1 atm = 1 . 01325 × 10 5 Pa 1 in = 2.54 cm Version 073 – Final – McCord – (53130) 2 This printout should have 55 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When 4 . 1 g of phosphorus was burned in chlo rine, the product was a phosphorus chloride. Its vapor took 1 . 77 times as long to effuse as the same amount of CO 2 under the same con ditions of temperature and pressure. What is the molar mass of the phosphorus chloride? 1. 102 g / mol 2. 156 g / mol 3. 87 . 7 g / mol 4. 138 g / mol correct Explanation: Because the phosphorus chloride took 1 . 77 times as long to effuse, the rate of effusion of CO 2 must be 1 . 77 times that of the phospho rus compound: Eff CO 2 = 1 . 77 Eff PCl 3 MW CO 2 = 12 . 01 g / mol + 2(16 g / mol) = 44 . 01 g / mol Eff ∝ radicalbigg 1 MW , so Eff CO 2 Eff PCl 3 = radicalBigg MW PCl 3 MW CO 2 MW PCl 3 = MW CO 2 parenleftbigg Eff CO 2 Eff PCl 3 parenrightbigg 2 = 44 . 01 g / mol parenleftbigg 1 . 77 Eff PCl 3 Eff PCl 3 parenrightbigg 2 = 137 . 879 g / mol . 002 10.0 points What is the volume occupied by 0.170 grams of gaseous H 2 S at 27 ◦ C and 380 torr? 1. 2.24 liters 2. 0.25 liter correct 3. 50.0 liters 4. 1.00 liter 5. 0.020 liter 6. 4.00 liters 7. None of these 8. 0.446 liter 9. 2.46 liters 10. 0.170 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 = nRT V = nRT P V = (0 . 005 mol) ( . 08206 L · atm mol · K ) (300 K) . 5 atm = 0 . 24618 L 003 10.0 points Which of the following best describes the range of atomic radii? 1. 1 to 100 ˚ A 2. 5 to 10 ˚ A 3. 0.4 to 3 ˚ A correct 4. 10 to 30 ˚ A 5. 1 to 15 ˚ A Version 073 – Final – McCord – (53130) 3 6. 0.05 to 1 ˚ A Explanation: Atomic radii vary from 30 pm (He) to 300 pm (Fr). 1 ˚ A= 10 − 10 m = 100 pm, so the range of atomic radii is 0.3 to 3 ˚ A....
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 Spring '09
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 Atom, Enthalpy, Chemical bond

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