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110

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Chemistry

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Dec 6, 2023

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CHEM 110Gases (Key)Ch. 9 Gas Laws 1. A balloon has a volume of 2.50 L indoors at 22 o C. If the balloon is taken outdoors on a cold day in Escanaba where the air temperature is -15 o C (5 o F), what will its volume be in liters? V 1 T 1 = V 2 T 2 V 1 T 2 = V 2 T 1 V 2 = V 1 T 2 T 1 = ( 2.5 L )( 258 K ) ( 295 K ) = 2.18 L V = 2.18 L 2. A sample of helium at a pressure of 745 torr and in a volume of 2.58 L was heated from 24.0 to 75.0 o C. What was the final pressure (in atm) of the helium? P 1 T 1 = P 2 T 2 P 1 T 2 = P 2 T 1 P 2 = P 1 T 2 T 1 = ( 745 torr )( 348 K ) ( 297 K ) = 873 torr P = 873 torr 3. There is an unknown volume of gas at a pressure of 0.5 atm and a temperature of 325 K. If the pressure is raised to 1.2 atm, temperature is lowered to 320 K, and the final volume is measured to be 48 liters, what was the initial volume of the gas (in liters)? Initial Final P (atm) 0.5 1.2 T (K) 325 320 V (L) ? 48 P 1 V 1 T 1 = P 2 V 2 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 1 = P 2 V 2 T 1 P 1 T 2 = ( 1.2 atm )( 48 L )( 325 K ) ( 0.5 atm ) ( 320 K ) = 112 L
V = 112 L Ideal Gas Law 4. An unknown gas is placed in a 1.500 L bulb at a pressure of 356 mmHg and a temperature of 22.5 o C, and is found to weigh 0.9847 g. What is the molecular mass of the unknown gas? M.M. = g/mol n ( mol ) = PV RT = ( 356 mmHg × 1 torr 1 mmHg × 1 atm 760 torr ) ( 1.500 L ) ( 0.0821 Latm mol K ) ( 295.5 K ) = 0.0290 mol M . M . = 0.9847 g 0.0290 mol = 34 g mol M.M. = 34 g/mol Stoichiometry and Gas Laws 5. Hydrogen gas can be prepared by reaction of zinc metal with hydrochloric acid: Zn (s) + 2 HCl (aq) ? ZnCl 2 (aq) + H 2 (g) a. How many liters of H 2 would be formed at 742 mm Hg and 15 o C if 25.5 g of Zn was allowed to react? 25.5 g Zn× 1 mol Zn 65.4 g × 1 mol H 2 1 mol Zn = 0.389 mol H 2 V = nRT P = ( 0.389 mol ) ( 0.0821 Latm mol K ) ( 288 K ) ( 742 mmHg × 1 torr 1 mmHg × 1 atm 760 torr ) = 9.44 L V = 9.44 L b. How many grams of Zn would you start with if you wanted to prepare 5.00 L of H 2 at 350 mmHg and 30.0 o C? n = PV RT = ( 350 mmHg × 1 torr 1 mmHg × 1 atm 760 torr ) ( 5.0 L ) ( 0.0821 Latm mol K ) ( 303 K ) = 0.0925 mol H 2
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