Version 001 HW02-gas laws vandenbout (52130) 1

Version 001 HW02-gas laws vandenbout (52130) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Mlib 76 2075 001 10.0 points A gas is enclosed in a 10.0 L tank at 1200 mm Hg pressure. Which of the following is a reasonable value for the pressure when the gas is pumped into a 5.00 L vessel? 1. 2400 mm Hg correct 2. 24 mm Hg 3. 0.042 mm Hg 4. 600 mm Hg V 1 10.0 L 1 1200 mm Hg V 2 5.0 L Boyle s law relates the volume and pressure of a sample of gas: 1 V 1 2 V 2 2 1V 1 (1200 mm Hg)(10.0 L) V 2 5 L 2400 mm Hg Holt da 10 4 sample 3 002 10.0 points A sample of gas in a closed container at a temperature of 75 C and a pressure of 5 atm is heated to 293 C. What pressure does the gas exert at the higher temperature? Correct answer: 8.13218 atm. T 1 75 C+273 348 K 1 5 atm T 2 293 C+273 566 K 2? Applying the Gay-Lussac law, 1 T 1 2 T 2 2 1T 2 T 1 8.13218 atm (5 atm)(566 K) 348 K Holt da 10 rev 22 003 10.0 points A flask containing 130 cm 3 of hydrogen was collected under a pressure of 27.4 ka. What pressure would have been required for the volume ofthe gastohave been 93 cm 3, assuming the same temperature? Correct answer: 38.3011 ka. V 1 130 cm 3 V 2 93 cm 3 1 27.4 ka 2? Applying Boyle s law, 1 V 1 2 V 2 2 1V 1 (27.4 ka)(130 cm3 ) V 2 93 cm 3 38.3011 ka Gas Vol 004 10.0 points A 5.00-L sample of a gas exerts a pressure of 1040 torr at 50.0 C. In what volume would the same sample exert a pressure of 1.00 atm at 50.0 C? 1. 6.84 L correct 2. 1.95 L 3. 0.581 L 4. 3.33 L 5. 10.5 L ( ) 1 atm 1 (1040 torr) 1.36842 atm 760 torr 2 1 atm V 1 5 L Since the number of moles and the temperature of the gas sample are held constant, the change in the volume is determined by Boyle s Law: 1 V 1 2 V 2 V 2 1V 1 (1.36842 atm)(5 L) 2 1 atm 6.84211 L

Version 001 HW02-gas laws vandenbout (52130) 2 Brodbelt 430abc 005 10.0 points Consider the following reaction of Al with HCl to produce hydrogen gas. 2Al+6HCl 2AlCl 3 +3H 2 Thisreactionhasayieldof82.5percent. How manymolesofhclareneededtoproduce14.0 L of H 2 at 351 K and 1.11 atm? 1. 1.08 mol 2. 0.446 mol 3. 0.655 mol 4. 0.540 mol 5. 0.890 mol 6. 0.223 mol 7. 1.31 mol correct V 14 L T 351 K 1.11 atm p 82.5% V n n V (1.11 atm)(14 L) (0.0821 mol K L atm )(351 K) 0.539263 mol H 2 For 82.5% yield n H2 0.539263 mol H 2 0.825 n H2 (0.653652 mol H 2 ) 1.3073 mol HCl 0.653652 mol H 2 Holt da 10 3 sample 2 006 10.0 points 6 mol HCl 3 mol H 2 A sample of nitrogen gas is contained in a pistonwithafreely movingcylinder. At0 C, the volume of the gas is 317 ml. To what temperature must the gas be heated to occupy a volume of 501 ml? Correct answer: 158.461 C. V 1 317 ml V 2 501 ml T 1 0 C+273 273 K T 2? Applying Charles Law, V 1 T 1 V 2 T 2 T 2 T 1V 2 V 1 431.461 K (273 K)(501 ml) 317 ml C K 273 158.461 C Mlib 04 1093 007 10.0 points IfIhave44.8litersofnitrogengasatstandard temperature and pressure, how much will it weigh? 1. 56 g correct 2. 28 kg 3. 28 g 4. 44.8 g 5. 56 u 1 atm T 273 K V n n V n (1 atm)(44.8 L) ) (273 K) ( 0.08206 L atm K mol 55.9941 g Chemrin3e T04 33 008 10.0 points V 44.8 L 28 g 1 mol

Version 001 HW02-gas laws vandenbout (52130) 3 At 80.0 C and 12.0 Torr, the density of camphor vapor is 0.0829 g L 1. What is the molar mass of camphor? 1. 243 g mol 1 2. 3490 g mol 1 3. 152 g mol 1 correct 4. 20.3 g mol 1 5. 34.5 g mol 1 T 80 C + 273.15 K 353.15 K 1 atm (12 Torr) 0.0157895 atm 760 Torr ρ 0.0829 g/l The ideal gas law is V n n V with unit of measure mol/l on each side. Multiplying each by molar mass (MM) gives n V MM with units of g/l. MM ρ MM ρ, (0.0829 g/l)( 0.08206 mol K L atm 0.0157895 atm (353.15 K) 152.152 g/mol Lyon e5 q4a 009 10.0 points What is the density of nitrogen gas at ST? 1. 2.50 g/l 2. 1.25 g/l correct 3. 0.500 g/l 4. 0.625 g/l ) 5. 4.00 g/l Lyon end quiz 010 (part 1 of 3) 10.0 points A chemist has synthesized a greenish-yellow gaseous compound that contains only chlorine and oxygen and has a density of 7.71 g/l at 36.0 degrees Celsius and 2188.8 mm Hg. What is the molar mass of the compound? 1. 25.8 g/mol 2. 86.9 g/mol 3. 51.5 g/mol 4. 67.9 g/mol correct 5. 109 g/mol T 36 C+273 309 K atm 2188.8 mm Hg 2.88 atm 760 mm Hg V n n V 0.1136 mol L MW (2.88 atm) (309 K)(0.08206 L atm molk ) 7.71 g/l 0.1136 mol L 67.87 g/mol 011 (part 2 of 3) 10.0 points What is the molecular formula of the compound? 1. ClO 4 2. ClO 3 3. ClO 4. ClO 2 correct

Version 001 HW02-gas laws vandenbout (52130) 4 5. ClO 5 MW ClO2 (35.5)+2(16) 67.5 g/mol 012 (part 3 of 3) 10.0 points How many moles of this gaseous compound are there in 15 L at ST? 1. 1.0 moles 2. 5.0 moles 3. 0.67 moles correct 4. 0.50 moles 5. 3.0 moles AtST,1moleofanidealgashasavolume of 22.4 L; i.e., 22.4 L. mol (15 L) mol 22.4 L 0.67 mol Mlib 04 1147 013 10.0 points What is the final volume if 10 L methane reacts completely with 20 L oxygen CH 4 (g)+2o 2 (g) CO 2 (g)+2h 2 O(l) at 100 C and 2 atmospheres? 1. 30 L 2. 20 L 3. 15 L 4. 10 L correct 5. Cannot be determined from the information given. V CH4 10 L V O2 20 L Avogadro s rinciple tells us that because and T are the same, V n, so we can work with the volume instead of the number of moles. From the equation above, we need 2 times more L of O 2 than that of CH 4, and we DO have this! Both reagents are present in stoichiometric amounts. We will have no left over reagents. Find out how much CO 2 is made based on the L of O 2 : ( ) 1 L CO2 V (20 L O 2 ) 10 L CO 2 2 L O 2 Note we could have just as easily done this calculation with the volume of CH 4, because bothreagentsarefullyusedupinthereaction. Wealsoassumethevolumeoftheliquidwater formed is negligible. Msci 02 1216 014 10.0 points Calculate the volume of methane (CH 4 ) produced by the bacterial breakdown of 0.65 kg of sugar (C 6 H 12 O 6 ) C 6 H 12 O 6 3CH 4 +3CO 2 at 295 K and 800 torr. 1. 455 L 2. 149 L 3. 252 L 4. 3036 L 5. 249 L correct n 0.65 kg C 6 H 12 O 6 1000 g 1 kg mol C 6 H 12 O 6 3 mol CH 4 180 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 10.8333 mol CH 4 T 295 K 1 atm 800 torr 1.05263 atm 760 torr

Version 001 HW02-gas laws vandenbout (52130) 5 Applying the ideal gas law equation, V n V n V (10.8333 mol)( 0.08206 L atm mol K) (295 K) 1.05263 atm 249.138 L Msci 12 1505 015 10.0 points If the reaction N 2 (g)+3h 2 (g) 2NH 3 (g) is carried out at constant temperature and pressure, how much H 2 is required to react with 6.1 L liters of N 2? 1. 9.15 L 2. 18.3 L correct 3. 24.4 L 1. 4.48 L 2. 2.24 L correct 3. 3.36 L 4. 6.72 L 5. 1.12 L n O2 0.1 mol Ag 2 O 0.05 mol O 2 1 atm 1 mol O 2 2 mol Ag 2 O V n V n (0.05)(0.08206)(546 K) 1 2.24024 L T 546 K 4. Cannot be determined. 5. 12.2 L V N2 6.1 L n N2 1 mol V n, so n V, and n CH4 n CO2 V CH 4 V CO2 n H2 3 mol LDE Ideal Gas Calculation 001 017 10.0 points If we increase the volume of a gaseous system by a factor of 3 and raise the temperature by a factor of 6, then the pressure of the system will (increase/decrease) by a factor of (2/18): 1. decrease, 18 2. increase, 18 V CO2 V CH 4 n CO2 n CH4 (6.1 L N 2)(3 mol H 2 ) 1 mol N 2 18.3 L H 2 Msci 12 1514 016 10.0 points What volume of pure oxygen gas (O 2 ) measured at 546 K and 1.00 atm is formed by complete dissociation of 0.1 mol of Ag 2 O? 2Ag 2 O(s) 4Ag(s)+O 2 (g) 3. increase, 2 correct 4. decrease, 2 Tripling the volume will decrease the pressure by a factor of 3 and sextupling the temperature will increase the pressure by a factor of 6, resulting a double the original pressure. DVB Avagadro s Gas Law 018 10.0 points You have a sample of H 2 gas and Ar gas at

Version 001 HW02-gas laws vandenbout (52130) 6 the same temperature and pressure, but the H 2 gas has twice the volume of the Ar gas. Assuming the gases behave ideally, which gas has the larger number density (gas particles per volume)? 1.itdependsonthevalueofthetemperature and pressure 2. the H 2 gas 3. they are the same correct 4. the Ar gas At the same temperature and pressure the numberdensityofthegasesmustbethesame. V n The H 2 gas must have twice the twice the number of moles since it hastwice the volume at the same T and. DVB gas density 019 10.0 points Which has the higher mass density: a sample of O 2 with a volume of 10 L, or a sample of Cl 2 with a volume of 3 L. Bothsamples are at the same temperature and pressure. 1. they are the same 2. the Cl 2 correct 3. the O 2 4.itdependsonthevalueofthetemperature and pressure At the same temperature and pressure the molar volume or number density of the two gases will be identical (assuming they are ideal). However, the mass of Cl 2 is more than two times as large as O 2. Therefore the density of the Cl 2 sample will be about twice as high. density mass volume n M.W. volume M.W. The only difference between the two samples at the same temperature and pressure is their molecular weight(m.w.). Therefore the ratio of the densities will be the ratio of their molecular weights. density of Cl 2 : density of O 2 70.91 g mol 1 : 32.0 g mol 1 2.2 Holt da 11 5 practice 2 020 10.0 points What is the mass of oxygen gas in a 15.2 L container at 46.0 C and 4.25 atm? Correct answer: 78.9312 g. T 46.0 C+273 319 K 4.25 atm V 15.2 L m? n V (4.25 atm)(15.2 L) ( 0.0821 L atm mol K) (319 K) 2.4666 mol O 2 ( ) 32 g m (2.4666 mol) 78.9312 g O 2 mol Holt da 11 review 22 021 10.0 points One method of estimating the temperature of the center of the sun is based on the assumptionthatthecenterconsistsofgasesthathave an average molar mass of 2.00 g/mol. If the density of the center of the sun is 1.40 g/cm 3 at a pressure of 1.30 10 9 atm, calculate the temperature. Correct answer: 2.26205 10 7 C. d 1.4 g/cm 3 1.3 10 9 atm M 2.00 g/mol T? M d R T

T M dr Version 001 HW02-gas laws vandenbout (52130) 7 (2.00 g/mol)( 1.3 10 9 atm ) (1.4 g/cm 3 ) ( 0.0821 mol K L atm ) 1 ml 1 cm 3 1 L 1000 ml 2.26205 10 7 C Holt da 11 6 practice 1 022 10.0 points What is the molar mass of a gas if 0.483 g of the gas occupies a volume of 493 ml at 9.00 C and 1.76 atm? Correct answer: 12.8805 g/mol. 1.761 atm V 0.493 L M? T 9.00 C+273 282 K m 0.483 g V CO2 179.2 L At ST we can use the standard molar volume, 22.4 L/mol. 8.00 mol CO 2 179.2 L 22.4 L/mol 8.00 mol CO 2 2 mol NaCl 1 mol CO 2 16.0 mol NaCl Chemrin3e 04 42 024 10.0 points The analysis of a hydrocarbon revealed that it was 85.6281% C and 14.3719% H by mass. When 3.47 g of the gas was stored in a 1.1 L flask at 168.344 C, it exerted a pressure of 490 Torr. What is the molecular formula of the hydrocarbon? 1. C 4 H 6 M m V (0.483 g)( 0.0821 L atm mol K) (282 K) (1.761 atm)(0.493 L) 12.8805 g/mol Brodbelt 12 04a 023 10.0 points For the reaction 2 HCl+Na 2 CO 3 2 NaCl+H 2 O+CO 2 179.2 liters of CO 2 is collected at ST. How many moles of NaCl are also formed? 1. 6.0 moles 2. 32.0 moles 3. 12.5 moles 4. 8.0 moles 5. 16.0 moles correct 6. 4.0 moles 2. C 3 H 6 correct 3. C 2 H 4 4. C 8 H 16 5. C 7 H 14 6. C 4 H 10 7. C 3 H 5 8. C 3 H 8 9. C 5 H 10 m 3.47 g V 1.1 L T 104.806 K 490 Torr R 0.08206 L atm/k/mol The empirical formula derived from the elemental analyses is (CH 2 ) n. Applying the ideal gas law, V n m MW MW mt R V

Version 001 HW02-gas laws vandenbout (52130) 8 ( ) (3.47 g)(104.806 K) 760 Torr (490 Torr)(1.1 L) 1 atm 0.08206 L atm/k/mol 42.0797 g/mol. The empirical formula mass is 14.0266 g/mol. The value of nintheformula (CH 2 ) n is, therefore, equalto 3.