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# Chem 101 Book Answers (Ch.7) - Chapter 7 States of Matter...

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Chapter 7 States of Matter and Changes in State 7-1 1. A weather report indicates a barometric pressure of 29.2 in. What is this pressure expressed in torr? 29.2 in x 25.4 mm 1 in = 742 mm Hg = 742 torr 3. If the gas in Exercise 2 is heated to 35 o C, what would be the separation between the two mercury levels? 11 2 2 2 Write the gas law for the two experiments: P V = nRT & P V = nRT V,n, and R are not subscripted because they are unchanged. PT Dividing the two equations eliminates V, n, and R to yield = 2 21 1 2 T , so P = P T T = 273 + 35 = 308, T = 273 + 20 = 293, and P = 758 + 423 = 1181 mm from Exercise 2, so 308 P = 1183 mm = 1241 mm, and the height of the column = 1241 - 758 = 483 mm 293 × × 5. Convert the following temperatures to the Celsius scale: Use K = ° C + 273.15 a) 4 K = -269 ° C b) 350 K = 77 ° C c) 186.4 K = -86.8 ° C d) 657 K = 384 ° C 7. What is the pressure inside a 7.20-L container filled with 0.254 moles of CO 2 at 35 o C? P = nRT V = (0.254)(0.0821)(308) 7.20 = 0.892 atm 9. The pressure of air in a tire at 10 o C is 22 psi. After several miles of driving at high speed, the pressure is 28 psi. Assume the volume of the tire is unchanged and calculate the temperature of the air in the tire. The number of moles of gas, its volume and the ideal gas law constant are all unchanged so we can write nR V = P 1 T 1 = P 2 T 2 or T 2 = T 1 x P 2 P 1 , T 1 = 10 + 273 = 283 K, P 1 = 22 psi, and P 2 = 28 psi. Note we can use any units for the two pressures as long as they are the same because the ratio of pressures is unitless. T 2 = 283 x 28 22 = 360 K which is 360 - 273 = 87 o C 11. Determine the concentrations of the following gases in moles/L: a) 3.0 moles of gas at 900 K and 2.6 atm First, determine the volume: V = nRT P = (3.0 mol)(0.0821 L atm K -1 mol -1 )(900 K) 2.6 atm = 85 L Then determine the concentration: [gas] = n V = 3.0 mol 85 L = 0.035 mol/L = 0.035 M b) 6.2 L of a gas at 250 o C and 800 torr First, determine the number of moles: n = PV RT = ( 800 760 atm)(6.2 L) (0.0821 L atm K -1 mol -1 )(523 K) = 0.15 mol Then determine the concentration: [gas] = n V = 0.15 mol 6.2 L = 0.025 mol/L = 0.025 M c) 4.0 g of Ne in a 0.56 L container at 200 o C. First determine the number of moles: 4.0 g Ne × 1 mole 20.18 g = 0.20 mol Ne Then determine the concentration: [Ne] = n V = 0.20 mol 0.56 L = 0.35 mol/L = 0.35 M

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States of Matter and Changes in State 7-2 13. What are the partial pressures of O 2 and N 2 and the total pressure in a 3.5-L flask at 77 o C that contains 4.0 g O 2 and 7.0 g N
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Chem 101 Book Answers (Ch.7) - Chapter 7 States of Matter...

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