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Unformatted text preview: Su, Yung Exam 3 Due: Nov 16 2007, 9:00 am Inst: FAKHREDDINE 1 This printout should have 19 questions. Multiplechoice 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 2KNO 3 (s) 2KNO 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 . 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(bymass). Whatisthemo 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 . Youve 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: . 75 mol NaOH . 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 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....
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This note was uploaded on 04/15/2009 for the course CH 54740 taught by Professor Fakhreddine during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Fakhreddine
 Chemistry

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