Exam 03 - Su Yung Exam 3 Due 9:00 am Inst FAKHREDDINE This...

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Su, Yung – Exam 3 – Due: Nov 16 2007, 9:00 am – Inst: FAKHREDDINE 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Each question is worth as assigned. 001 (part 1 of 1) 10 points What is the density of fluorine gas at STP? 1. 0.620 g/L 2. 1.70 g/L correct 3. 4.00 g/L 4. 2.50 g/L 5. 0.925 g/L Explanation: 002 (part 1 of 1) 12 points A 42 gram sample of impure potassium ni- trate (KNO 3 ) was heated to complete decom- position according to the equation 2 KNO 3 (s) 2 KNO 2 (s) + O 2 (g) After the reaction was complete, the volume of the dry gas produced was 2 liters at 61 . 8 C and 625 torr. How many grams of KNO 3 were present in the original sample? (Assume that only the potassium nitrate had decomposed.) Correct answer: 12 . 1049 grams. Explanation: V = 2 liters T = 61 . 8 C = 334 . 8 K P = 625 torr · 1 atm 760 torr = 0 . 822368 atm P V = n R T n = P V R T = (0 . 822368 atm)(2 liters) (0 . 08206 L · atm mol · K ) (334 . 8 K) = 0 . 0598659 mol 0 . 0598659 mol O 2 · 2 mol KNO 3 1 mol O 2 · 101 . 1 g mol = 12 . 1049 grams KNO 3 003 (part 1 of 1) 12 points The information provided on a bottle contain- ing a NaOH solution reads density = 1.330 g/mL; 30% NaOH (by mass). What is the mo- larity of the solution? (NaOH = 40.0 g/mol) 1. 14.25 M 2. 0.30 M 3. 1.42 M 4. 0.56 M 5. 9.98 M correct 6. 6.32 M Explanation: density soln = 1 . 330 g / mL % NaOH = 30% MW NaOH = 40 . 0 g / mol Assume 100 g of solution (as you should for any percent problem); this means that 30 g (30%) is NaOH and 70 g (the remainder) is H 2 O. The number of moles of NaOH is 30 g NaOH 1 mol NaOH 40 g NaOH = 0 . 75 mol NaOH . You’ve assumed that you have 100 g of solution, so use the density to convert the mass of this solution into its volume: 100 g soln 1 mL 1 . 330 g = 75 . 2 mL soln = 0 . 0752 L soln The molarity is then calculated in the usual way: 0 . 75 mol NaOH 0 . 0752 L soln = 9 . 98 M 004 (part 1 of 1) 10 points If the temperature of an ideal gas is raised from 100 C to 200 C, while the pressure re- mains constant, the volume 1. None of these correct
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Su, Yung – Exam 3 – Due: Nov 16 2007, 9:00 am – Inst: FAKHREDDINE 2 2. remains the same. 3. doubles. 4. increases by a factor of 100. 5. goes to 1 2 of the original volume. Explanation: T 1 = 100 C + 273 = 373 K T 2 = 200 C + 273 = 473 K Volume is directly proportional to temper- ature expressed on the Kelvin scale: V 1 T 1 = V 2 T 2 V 2 V 1 = T 2 T 1 The temperature changes by a factor of T 2 T 1 = 473 K 373 K = 1 . 27 , so the volume increases by the same factor 1.27. None of the specific answers gives this factor. 005 (part 1 of 1) 10 points For a substance that remains a gas under the conditions listed, deviation from the Ideal Gas Law would be most pronounced at 1. - 100 C and 2.0 atm.
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