exam three ch - Version 052 Exam Three: Shear Shear (52375)...

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Version 052 – Exam Three: Shear – Shear – (52375) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. DrRuth says: In all cases, select the answer that BEST answers the question asked. The total number oF points on this exam is 350. Read the whole thing frst and plan which to answer frst because you Feel confdent about them. Notice the harder ones and save them up till last so you don’t use up all your time working on them. Unless otherwise stated, assume the gases in all questions are ideal. Shear 52375 g = 9 . 8 m / s 2 1 atm = 101325 Pa = 760 torr R = 0.08206 L atm K 1 mol 1 R = 0.08314 L bar K 1 mol 1 R = 8.314 J K 1 mol 1 E±usion rate oF gas 1 E±usion rate oF gas 2 = r M 2 M 1 E±usion rate at T 1 E±usion rate at T 2 = r T 1 T 2 KE = 2 / 3 RT u rms = r 3 RT M 001 10.0 points IF it takes 15 s For a certain sample oF neon to e±use through a porous barrier, how much time will it take For the same amount oF ni- trogen gas to e±use through the barrier under the same conditions? 1. 15 s 2. less than 15 s 3. greater than 15 s correct Explanation: Rate oF e±usion 1 MW MWs For each gas: Ne : 20 . 179 g / mol N 2 : 28 . 0134 g / mol The heaviest gas will have particles moving at the lowest average speed so it will e±use more slowly. 002 10.0 points A 6.5 L sample oF nitrogen at 25 C and 1.5 atm is allowed to expand to 13.0 L. The temperature remains constant. What is the fnal pressure? 1. 0.12 atm 2. 0.38 atm 3. 3.0 atm 4. 0.75 atm correct 5. 0.063 atm Explanation: V 1 = 6.5 L V 2 = 13.0 L P 1 = 1.5 atm Boyle’s law relates the volume and pressure oF a sample oF gas: P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 = (1 . 5 atm) (6 . 5 L) 13 . 0 L = 0 . 75 atm 003 20.0 points At STP a gas occupies 121 mL. How many milliliters will this gas occupy at - 52 C and 1.18 atm? 1. - 19.5 mL 2. 115 mL 3. 83.0 mL correct 4. 127 mL
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Version 052 – Exam Three: Shear – Shear – (52375) 2 Explanation: P 1 = 1 atm P 2 = 1 . 18 atm T 1 = 273 . 15 K V 1 = 121 mL T 2 = - 52 C + 273 . 15 = 221 . 15 K P 1 V 1 T 1 = P 2 V 2 T 2 V 2 = P 1 V 1 T 2 T 1 P 2 = (1 atm) (121 mL) (221 . 15 K) (273 . 15 K) (1 . 18 atm) = 83 . 0212 mL 004 15.0 points Rank the molecules CH 3 F , C 2 H 6 , H 2 O , H 2 , He in terms of increasing viscosity. 1. He , H 2 , C 2 H 6 , CH 3 F , H 2 O correct 2. C 2 H 6 , H 2 O , CH 3 F , He , H 2 3. H 2 , He , CH 3 F , H 2 O , C 2 H 6 4. None of the answers is correct. 5. H 2 O , CH 3 F , C 2 H 6 , H 2 , He Explanation: The intermolecular attractions, ranked from weakest to strongest, are London forces (also called van der Waals forces or induced dipoles), dipole-dipole interactions, hydro- gen bonding interactions, and ion-ion interac- tions. In general, the larger the molecule, the greater the total intermolecular forces. Strong intermolecular interactions cause molecules to “stick” to one another, and are thus more
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exam three ch - Version 052 Exam Three: Shear Shear (52375)...

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