Chapters5and6Key - Chemistry 161 Study Guide Chapters 5...

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Unformatted text preview: Chemistry 161 Study Guide Chapters 5 & 6 Harold 1. The valve between a 2.00-L bulb, in which the gas pressure is1.00 atm, and a 3.00-L bulb, in which the gas pressure is 1.50 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant and no reaction occurring? Let the gases be called A and B P A V A = n A RT n A = (P A V A )/(RT) AND P B V B = n b RT n B = (P B V B )/(RT) ALSO P T V T = n T RT where n T = n A + n B and V T = V A + V B P T = n T RT = (n A + n B )RT = {(P A V A )/(RT) + (P B V B )/(RT)} RT = (P A V A + P B V B ) ----------------- ------------------------------------- --------------------------------------------------------------------------------------------------- ------------------------------------------------ V T V A + V B V A + V B V A + V B THUS P T = 1.00atm x 2.00L + 1.50atm x 3.00L = 6.50 Latm = 1.30 atm ------------------------------------------------------------------------------------------------------------- -------------------------------- 2.00 L + 3.00 L 5.00 L 2. For an ideal gas, which pairs of variables are directly proportional to each other (if all other factors remain constant)? 1. P, T 2. P, V 3. V, T 4. n, V a. 1 and 2 only b. 3 and 4 only c. 2 only d. 1 and 3 only e. 1, 3, and 4 only 3. The temperature of a gas in a sealed container changes from 20.0 o C to 40.0 o C. If the volume remains constant, the pressure will change from 740 mmHg to? (Convert temp to Kelvin first) Since V is constant then P 1 /T 1 = P 2 /T 2 or P 2 = P 1 x T 2 /T 1 so ... P 2 = 740 x 313.15/293.15mm Hg P 2 = 790. mm Hg 4. At standard conditions, 4.16 L of a gas weighed 6.32 g. The gas is a. C 2 H 2 . b. NO. c. PH 3 . d. HCl. e. CH 3 SH. In standard conditions 1 mole fills 22.41 L of space, so 4.16 L is 4.16 L/22.41 L/mole = 0.18563 moles of gas and the molecular weight of the gas is 6.32 g/ 0.18563 moles = 34.0 g/mole PH 3 is the only one listed with this molar weight. 5. The density of a gas is 3.48 g/L at STP. What is its molecular weight? d = m/V for one mole of gas at STP mass (m) is the molar weight. The volume , V = 22.41 L/mole so... m = d x V or M.W. = d x V or 3.48g/L x 22.41 L/mole = M.W. = 78.0 g/mol 6. At 27 o C and 275 mmHg, an unknown pure gas has a density of 0.647 g/L. Which of the following gases could be the unknown gas? PV = nRT put in the molar weight form is M.W. = (mRT)/PV and m = dV so... M.W. = (dVRT)/PV Volume cancels out M.W. = (dRT)/P = (0.647g/Lx0.0821Latm/molKx300K)/(275 mmHg/760mmHg/atm) = 44.0 g/mol a. N 2 O b. Ne c. C 2 H 6 d. CO e. F 2 7. A mixture consisting of 0.100 mol N7....
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Chapters5and6Key - Chemistry 161 Study Guide Chapters 5...

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