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Unformatted text preview: fernandez (jf23742) Homework 7 Brodbelt (53765) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Homework 7: Gases 001 10.0 points The same number of grams of NH 3 and O 2 are placed in separate bulbs of equal vol ume and temperature under conditions when both gases behave ideally. Which statement is true? 1. The pressure in the NH 3 bulb is greater than the pressure in the O 2 bulb. correct 2. The pressure in the O 2 bulb is greater than the pressure in the NH 3 bulb. 3. Both bulbs contain the same number of moles of gas. 4. The bulb containing O 2 contains more molecules of gas. 5. The pressures in the two bulbs are the same. Explanation: The molecular weight of NH 3 is less than that of O 2 , so in equal masses there are more moles of NH 3 than of O 2 . At the same volume and temperature, the larger number of moles of NH 3 would exert a higher pressure. 002 10.0 points How many molecules are in 1.00 liter of O 2 gas at 56 C and 821 torr. 1. 32 molec 2. 2 . 24 10 23 molec 3. 1 . 83 10 25 molec 4. 2 . 41 10 22 molec correct 5. 4 . 00 10 2 molec Explanation: V = 1 L T = 56 C + 273 = 329 K P = 821 torr atm 760 torr = 1 . 08 atm Applying the ideal gas law equation, P V = nRT n = P V RT n = (1 . 08 atm) (1 L) ( . 08206 L atm mol K ) (329 K) 6 . 02 10 23 molec 1 mol = 2 . 41 10 22 molec 003 10.0 points At STP, 6.0 grams of CO gas will occupy a volume of 1. 3.5 liters. 2. 4.8 liters. correct 3. 22.4 liters. 4. 5.6 liters. 5. 2.24 liters. Explanation: T = 0 C + 273 = 273 K P = 1 atm n = 6 g mol 28 g = 0 . 214 mol Applying the ideal gas law, P V = nRT V = nRT P V = (0 . 214 mol) ( . 08206 L atm mol K ) (273 K) 1 atm = 4 . 79411 L 004 10.0 points A flask contains 0.123 moles of an ideal gas that occupies 781 mL. A second flask at the same temperature contains 0.0712 moles of the same gas. The pressure is the same in both flasks. What is the volume of the second flask? 1. 452 mL correct fernandez (jf23742) Homework 7 Brodbelt (53765) 2 2. 89,180 mL 3. 1350 mL 4. 904 mL 5. 6.8 mL Explanation: n 1 = 0 . 123 mol n 2 = 0 . 0712 mol V 1 = 781 mL Applying the ideal gas law equation, P V = nRT V 1 V 2 = n 1 n 2 V 2 = V 1 n 2 n 1 V 2 = (781 mL) (0 . 0712 mol) . 123 mol = 452 mL 005 10.0 points What is the volume of 0.500 moles of an ideal gas at 273 K and 760 torr? 1. 11.2 liters correct 2. 22.4 liters 3. 4.56 cubic fathoms 4. 380 ml 5. 2.24 liters Explanation: n = 0 . 5 mol R = 0 . 08206 L atm mol K P = 760 torr = 1 atm T = 273 K P V = nRT V = nRT P = (0 . 5 mol) ( . 08206 L atm mol K ) (273 K) 1 atm = 11 . 2012 L 006 10.0 points A 2.00 mole sample of gas is at a temperature of 100.0 C and occupies 3.00 liters. What is its pressure?...
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This note was uploaded on 07/27/2009 for the course CH 89959 taught by Professor Cowley during the Spring '09 term at University of Texas at Austin.
 Spring '09
 COWLEY

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