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Homework 07

# Homework 07 - Su Yung Homework 7 Due Nov 8 2007 11:00 pm...

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Su, Yung – Homework 7 – Due: Nov 8 2007, 11:00 pm – Inst: FAKHREDDINE 1 This print-out should have 47 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Each question is worth 1 point. 001 (part 1 of 1) 1 points A gas has a volume of 2.0 L at a pressure of 3.00 atm. What pressure is needed to change the volume to 6.00 L at constant temperature? 1. 4.0 atm 2. 9.0 atm 3. 2.0 atm 4. 1.0 atm correct Explanation: V 1 = 2.00 L P 1 = 3 atm V 2 = 6.00 L Boyle’s law, P 1 V 1 = P 2 V 2 , relates the vol- ume and pressure of a sample of gas. P 2 = P 1 V 1 V 2 = (3.00 atm) (2.00 L) 6.00 L = 1.0 atm 002 (part 1 of 1) 1 points A balloon is inflated outdoors on a cold day in North Dakota at a temperature of - 40 C to a volume of 2.00 L. The pressure remains constant. What is the volume of the balloon indoors at a temperature of 25 C? 1. 2.0 L 2. - 3 . 2 L 3. 1.6 L 4. 2.6 L correct Explanation: T 1 = - 40 C + 273 = 233 K V 1 = 9 . 0 L T 2 = 25 C + 273 = 298 K Charles’s 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)(298 K) 233 K = 2 . 6 L 003 (part 1 of 1) 1 points A sample of a gas contains 0 . 305 g of carbon, 0 . 407 g of oxygen and 1 . 805 g of chlorine. A different sample of the same gas having a mass of 8 . 84 g occupies 2 L at STP. What is the molecular formula for the compound? 1. C 2 O 2 Cl 4 2. COCl 2 correct 3. COCl 4. C 2 O 2 Cl 2 5. CO 2 Cl Explanation: V = 2 L P = 1 atm T = 0 C + 273 = 273 K Applying the ideal gas law equation, P V = n R T n = P V R T = (1 atm) (2 L) 273 K · mol · K 0 . 08206 L · atm = 0 . 0892762 mol MW = 8 . 84 g 0 . 0892762 mol = 99 . 0185 g / mol Total mass = 0 . 305 g C + 0 . 407 g O +1 . 805 g Cl = 2 . 517 g C : % C = 0 . 305 g C 2 . 517 g × 100% = 12 . 1176% C mol C = 0 . 121176 99 . 0185 g mol · mol 12 g = 0 . 999889 mol

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Su, Yung – Homework 7 – Due: Nov 8 2007, 11:00 pm – Inst: FAKHREDDINE 2 O : % O = 0 . 407 g O 2 . 517 g × 100% = 16 . 17% O mol O = 0 . 1617 99 . 0185 g mol · mol 16 g = 1 . 00071 mol Cl : % Cl = 1 . 805 g Cl 2 . 517 g × 100% = 71 . 7124% Cl mol Cl = 0 . 717124 99 . 0185 g mol · mol 35 . 5 g = 2 . 00024 mol Therefore the formula is COCl 2 . 004 (part 1 of 1) 1 points A mixture of 0 . 75 mol H 2 gas and 0 . 75 mol N 2 gas is introduced into a 15 liters container having a pinhole leak at 30 C. After a period of time, which of the following is true? 1. The partial pressure of H 2 exceeds that of N 2 in the container. 2. The partial pressure of N 2 exceeds that of H 2 in the container. correct 3. The partial pressures of the two gases change, but they remain equal. 4. The partial pressures of both gases in- crease above their initial values. Explanation: X N 2 = 0 . 79 P total = 1 atm Because of the pinhole leak, the con- tainer will become equilibrated with the at- mosphere. Air is composed of 79% N 2 , 20% O 2 , and 1% other gases. Eventually the gases in the container will have the same ratios as air and the pressure will be 1 atm. X N 2 = 0 . 79 P N 2 = X N 2 P total = 0 . 79 (1 atm) = 0 . 79 atm .
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Homework 07 - Su Yung Homework 7 Due Nov 8 2007 11:00 pm...

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