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Unformatted text preview: Version 100 – Exam 3 – McCord – (53110) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. McCord CH301mwf This exam is only for McCord’s MWF CH301 class. PLEASE carefully bubble in your UTEID and Version Number! 001 10.0 points The process by which a gas is converted to a liquid is called 1. sublimation. 2. vaporization. 3. ionization. 4. condensation. correct Explanation: Liquid Vapor Evaporation Condensation 002 10.0 points Calcium cyanamide reacts with H 2 O(g) CaNCN(s) + 3 H 2 O(g) → CaCO 3 (s) + 2 NH 3 (g) at 150 ◦ C and 1 atm pressure. If 80.0 L of NH 3 (g) are produced, what volume of steam is required? 1. 26.7 L 2. 40.0 L 3. 240.0 L 4. 80.0 L 5. 120.0 L correct Explanation: V NH 3 = 80 L P = 1 atm T = 150 ◦ C + 273 . 15 = 423 . 15 K For the NH 3 , apply the ideal gas law: P V = nRT n = P V RT = (1 atm)(80 L) ( . 08206 L · atm mol · K ) (423 . 15 K) = 2 . 3039 mol NH 3 Thus for the H 2 O, n = (2 . 3039 mol NH 3 ) 3 mol H 2 O 2 mol NH 3 = 3 . 45585 mol H 2 O and V = nRT P = (3 . 45585 mol H 2 O)(0 . 08206 L · atm mol · K ) 1 atm × (423 . 15 K) = 120 L H 2 O 003 10.0 points Hydrogen bonds are really just a very strong form of dipoledipole interaction. 1. False 2. True correct Explanation: Permanent dipoledipole interactions are stronger than London forces and occur be tween polar covalent molecules due to charge separation. Hbonds are a special case of very strong dipoledipole interactions. They only occur when H is bonded to small, highly electroneg ative atoms – F, O or N only. 004 10.0 points A badly tuned automobile engine can release about 50 moles of carbon monoxide per hour. At 35 ◦ C, what volume of carbon monoxide Version 100 – Exam 3 – McCord – (53110) 2 is released in a sixhour period if the atmo spheric pressure is 740 Torr? 1. 7 . 9 × 10 5 L 2. 1 . 3 × 10 5 L 3. 7 . 8 × 10 3 L correct 4. 1 . 3 × 10 3 L 5. 10 L Explanation: n = (50 mol CO / h)(6 h) = 300 mol CO T = 35 ◦ C + 273 = 308 K P = (740 Torr) 1 atm 760 Torr = 0 . 973684 atm P V = nRT V = nRT P = (300 mol)(0 . 08206 L · atm / mol / K) . 973684 atm × (308 K) = 7787 . 15 L 005 10.0 points A 750.0 mL metal bulb is filled with 0.421 g of CH 4 and an unknown mass of N 2 . If the total pressure in the bulb is 3.77 atm at 315 ◦ C, then how much N 2 is also in the bulb? 1. 14.5 grams 2. 0.549 grams 3. 0.0323 grams 4. 0.904 grams correct 5. 1.17 grams 6. 2.39 grams 7. 0.00713 grams Explanation: n CH 4 = (0 . 421 g) · mol 16 g = 0 . 0263 mol V = 750 mL = 0 . 75 L T = 315 ◦ C + 273 = 588 K Applying the ideal gas law equation, P V = nRT n tot = P V RT = (3 . 77 atm) (0 . 75 L) (0 . 08206 L · atm mol · K ) (588 K) = 0 . 0585995 mol n tot = n N 2 + n CH 4 n N 2 = n tot n CH 4 = 0 . 0585995 mol . 0263 mol = 0 . 0322995 mol m N 2 = n N 2 MM N 2 = (0 . 0322995 mol) 28 g N 2 1 mol N 2 = 0 . 904385 g N 2 006 10.0 points If the compression factor...
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This note was uploaded on 11/09/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
 Fall '07
 Fakhreddine/Lyon

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