Fall 07 HW6 solution

Fall 07 HW6 solution - Homework Assignment #6 Name_...

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Homework Assignment #6 Name____________________________ Laboratory Section (circle one) MW1.30 MW6.30 TTh7 1. Chapter 5, Problem 45, 68, 112 5.45 Total moles = n T = PV/RT = ( )( ) ( ) ( ) 3 626 mmHg 355 mL 1 atm 10 L L • atm 760 mmHg 1 mL 0.0821 273 35 K mol • K - + = 0.011563657 mol (unrounded) Moles Ne = n Ne = (0.146 g Ne) (1 mol Ne / 20.18 g Ne) = 0.007234886 mol Ne (unrounded) Moles Ar = n T - n Ne = (0.011563657 - 0.007234886) mol = 0.004328771 = 0.0043 mol Ar 5.68 To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (equation 5.14). 2 Rate O Rate Kr = 2 Molar Mass Kr Molar Mass O = 83.80 g/mol 32.00 g/mol = 1.618255 = 1.618 5.112 Molar volume uses exactly one mole of gas along with the temperature and pressure given in the problem. V / n = RT / P = ( ) ( ) L • atm 0.0821 730. K mol • K 90 atm = 0.66592 = 0.67 L/mol 2. Consider two 1.0L samples of gas: one is He and the other is N 2. Both are at 1.00 atm and 25°C. For each quantity given below, indicate which has: a. larger mass b for same number of moles nitrogen will have larger mass as the molar mass of nitrogen is greater than helium b. greater density b nitrogen, as density is proportional to molar mass. c. faster average molecular speed b helium, as molecular speed is faster for gas with lower molar mass d. slower rate of effusion b nitrogen, as effusion rate is inversely proportional to the square root of the molar mass e. 3. Chapter 6, Problem 9, 11, 18, 19, 25, 27, 30, 34, 38, 42, 45, 51, 59, 64, 68, 78, 85 (a, b only), 111 6.9 Δ E = q + w = -255 cal + (-428 cal) = -683 cal 6.11 Δ E = q + w = ( ) 3 10 J 0.615 kJ 1 kJ + ( ) 3 10 cal 4.184 J 0.247 kcal 1 kcal 1 cal  = 1648.4 =
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This note was uploaded on 02/12/2008 for the course CHEM 400 taught by Professor Gupta during the Fall '07 term at American River.

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Fall 07 HW6 solution - Homework Assignment #6 Name_...

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