Version 001 HW02-gas laws vandenbout (52130) 1

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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 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? mm Hg correct mm Hg mm Hg mm Hg V L mm Hg V 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 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: atm. T 1 75 C K 1 5 atm T C K 2? Applying the Gay-Lussac law, 1 T 1 2 T 2 2 1T 2 T atm (5 atm)(566 K) 348 K Holt da 10 rev 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: ka. V cm 3 V 2 93 cm ka 2? Applying Boyle s law, 1 V 1 2 V 2 2 1V 1 (27.4 ka)(130 cm3 ) V 2 93 cm ka Gas Vol 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? L correct L L L L ( ) 1 atm 1 (1040 torr) 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 ( atm)(5 L) 2 1 atm L

2 Version 001 HW02-gas laws vandenbout (52130) 2 Brodbelt 430abc 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? mol mol mol mol mol mol mol correct V 14 L T 351 K 1.11 atm p 82.5% V n n V (1.11 atm)(14 L) ( mol K L atm )(351 K) mol H 2 For 82.5% yield n H mol H n H2 ( mol H 2 ) mol HCl mol H 2 Holt da 10 3 sample 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: C. V ml V ml T 1 0 C K T 2? Applying Charles Law, V 1 T 1 V 2 T 2 T 2 T 1V 2 V K (273 K)(501 ml) 317 ml C K C Mlib points IfIhave44.8litersofnitrogengasatstandard temperature and pressure, how much will it weigh? g correct kg g g u 1 atm T 273 K V n n V n (1 atm)(44.8 L) ) (273 K) ( L atm K mol g Chemrin3e T points V 44.8 L 28 g 1 mol

3 Version 001 HW02-gas laws vandenbout (52130) 3 At 80.0 C and 12.0 Torr, the density of camphor vapor is g L 1. What is the molar mass of camphor? g mol g mol g mol 1 correct g mol g mol 1 T 80 C K K 1 atm (12 Torr) atm 760 Torr ρ 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, ( g/l)( mol K L atm atm ( K) g/mol Lyon e5 q4a points What is the density of nitrogen gas at ST? g/l g/l correct g/l g/l ) 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 mm Hg. What is the molar mass of the compound? g/mol g/mol g/mol g/mol correct g/mol T 36 C K atm mm Hg 2.88 atm 760 mm Hg V n n V mol L MW (2.88 atm) (309 K)( L atm molk ) 7.71 g/l mol L 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

4 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? moles moles moles correct moles moles AtST,1moleofanidealgashasavolume of 22.4 L; i.e., 22.4 L. mol (15 L) mol 22.4 L 0.67 mol Mlib 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? L L L 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 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 L L L L L correct n 0.65 kg C 6 H 12 O g 1 kg mol C 6 H 12 O 6 3 mol CH g C 6 H 12 O 6 1 mol C 6 H 12 O mol CH 4 T 295 K 1 atm 800 torr atm 760 torr

5 Version 001 HW02-gas laws vandenbout (52130) 5 Applying the ideal gas law equation, V n V n V ( mol)( L atm mol K) (295 K) atm L Msci 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? L L correct L L L correct L L 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)( )(546 K) L T 546 K 4. Cannot be determined 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 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, increase, 18 V CO2 V CH 4 n CO2 n CH4 (6.1 L N 2)(3 mol H 2 ) 1 mol N L H 2 Msci 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 points You have a sample of H 2 gas and Ar gas at

6 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 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 g mol 1 : 32.0 g mol Holt da 11 5 practice points What is the mass of oxygen gas in a 15.2 L container at 46.0 C and 4.25 atm? Correct answer: g. T 46.0 C K 4.25 atm V 15.2 L m? n V (4.25 atm)(15.2 L) ( L atm mol K) (319 K) mol O 2 ( ) 32 g m ( mol) g O 2 mol Holt da 11 review 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 atm, calculate the temperature. Correct answer: C. d 1.4 g/cm atm M 2.00 g/mol T? M d R T

7 T M dr Version 001 HW02-gas laws vandenbout (52130) 7 (2.00 g/mol)( atm ) (1.4 g/cm 3 ) ( mol K L atm ) 1 ml 1 cm 3 1 L 1000 ml C Holt da 11 6 practice points What is the molar mass of a gas if g of the gas occupies a volume of 493 ml at 9.00 C and 1.76 atm? Correct answer: g/mol atm V L M? T 9.00 C K m g V CO L At ST we can use the standard molar volume, 22.4 L/mol mol CO L 22.4 L/mol 8.00 mol CO 2 2 mol NaCl 1 mol CO mol NaCl Chemrin3e points The analysis of a hydrocarbon revealed that it was % C and % H by mass. When 3.47 g of the gas was stored in a 1.1 L flask at 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)( L atm mol K) (282 K) (1.761 atm)(0.493 L) g/mol Brodbelt 12 04a points For the reaction 2 HCl+Na 2 CO 3 2 NaCl+H 2 O+CO liters of CO 2 is collected at ST. How many moles of NaCl are also formed? moles moles moles moles moles correct moles 2. C 3 H 6 correct 3. C 2 H 4 4. C 8 H C 7 H C 4 H C 3 H 5 8. C 3 H 8 9. C 5 H 10 m 3.47 g V 1.1 L T K 490 Torr R 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

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