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Unformatted text preview: Version 373 Exam 3 mccord (50970) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. McCord CH301 This exam is only for Dr. McCords CH301 classes. 001 10.0 points A mixture of oxygen and helium is 92.3% by mass oxygen. It is collected at atmospheric pressure (687 torr). What is the partial pressure of oxygen in this mixture? 1. 333 Torr 2. 446 Torr 3. 299 Torr 4. 414 Torr correct 5. 688 Torr Explanation: Assume you have 100 g of this mixture; calculate the number of moles: n O 2 = (92 . 3 g O 2 ) 1 mol O 2 31 . 9988 g O 2 = 2 . 91261 mol O 2 . n He = (7 . 7 g He) 1 mol He 4 . 0026 g He = 1 . 92375 mol He . n tot = n O 2 + n He = 2 . 91261 mol O 2 + 1 . 92375 mol He = 4 . 83636 mol gas Daltons Law: P O 2 = P tot O 2 = P tot n O 2 n tot = (687 Torr) 2 . 91261 mol O 2 4 . 83636 mol gas = 413 . 733 Torr 002 10.0 points A gas has a volume of 2.00 liters at a tempera ture of 127 C. What will be the volume of the gas if the temperature is increased to 327 C? (Assume the pressure remains constant.) 1. 6.00 liters 2. 3.00 liters correct 3. 2.00 liters 4. 4.00 liters Explanation: T 1 = 127 C + 273 = 400 K V 1 = 2 . 00 L T 2 = 327 C + 273 = 600 K Charles Law relates the volume and ab solute (Kelvin) temperature of a sample of gas: V 1 T 1 = V 2 T 2 V 2 = V 1 T 2 T 1 = (2 . 00 L) (600 K) 400 K = 3 . 00 L 003 10.0 points A mixture of CO, CO 2 and O 2 is contained within a 275 mL flask at 0 C. If the total pres sure is 780 torr, the CO has a partial pressure of 330 torr and the CO 2 has a partial pressure of 330 torr, what is the partial pressure of O 2 ? 1. 120 torr correct 2. 780 torr 3. 900 torr 4. 660 torr Version 373 Exam 3 mccord (50970) 2 5. 330 torr Explanation: P total = 780 torr P CO = 330 torr P CO 2 = 330 torr P total = P CO + P CO 2 + P O 2 P O 2 = P total P CO P CO 2 = 780 torr 330 torr 330 torr = 120 torr 004 10.0 points If a real gas has a volume that is larger than you would predict based on the ideal gas law, then the intermolecular forces for that gas are dominated by 1. attractive forces 2. repulsive forces correct 3. gravitational forces 4. neither attractive nor repulsive forces 5. compression forces Explanation: If the volume of the real gas is larger than you would predict based on the ideal gas law, then the compression factor Z > 1. This means the intermolecular forces are domi nated by repulsions between the gas particles (atoms or molecules). 005 10.0 points The gauge in this problem is an absolute pres sure gauge so assume the pressure is exactly as stated in this problem (do not adjust it)....
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This note was uploaded on 04/30/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Fakhreddine/Lyon
 Chemistry

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