11 Gases. 760 torr a atm torr. 1 atm lb/in atm lb/in atm 760 mm Hg atm mm Hg. 1 atm. 101.

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1 47374_11_p qxd 2/9/07 9:09 AM Page Gases 11.1 a. Gaseous particles have greater kinetic energies at higher temperatures. Because kinetic energy is a measure of the energy of motion, the gaseous particles must be moving faster at higher temperatures than at lower values. Because particles in a gas are very far apart, gases can be easily compressed without the particles bumping into neighboring gas particles. Neighboring particles are much closer together in solids and liquids, and they will bump into each other and repel each other if the sample is compressed. Because the particles of a gas are very far apart, only a small amount of mass (due to the gas particles themselves) is found in a given volume of space a. At the higher temperature in the fire, the number of collisions against the container s walls increases because the gaseous particles have greater velocities. This increases the pressure in the can, and if that pressure exceeds what the container can endure, then the container will explode. The particles of a gas move faster at higher temperatures. This causes the particles to spread farther apart, reducing the gas sample s density. The density of air in the balloon is less, which causes it to rise until its density becomes equal to the surrounding atmosphere. Particles of a gas move rapidly and in straight lines until they reach a person who notices the odor a. temperature volume amount of gas d. pressure 11.4 a. temperature pressure volume d. amount of gas 11.5 Some units used to describe the pressure of a gas are pounds per square inch ( lb/in. 2, which is also abbreviated as psi), atmospheres (abbreviated atm), torr, mm Hg, in. Hg, pascals (Pa), and kilopascals (kpa) Statements a, d, and e describe the pressure of a gas a. d a atm 2.00 atm 2.00 atm 2.00 atm 467 mm Hg 760 torr 14.7 lb/in kpa 467 mm Hg 1 torr 1 mm Hg 467 mm Hg 1 cm 10 mm 203 kpa 1 in. Hg 2.54 cm d atm Pa 1520 torr 29.4 lb/in mm Hg atm 467 torr 18.4 in. Hg Pa 113

2 47374_11_p qxd 2/9/07 9:09 AM Page 114 Chapter The gases in the diver s lungs (and dissolved in the blood) will expand because pressure decreases as the diver ascends. Unless the diver exhales, the expanding gases could rupture the membranes in the lung tissues. In addition, the formation of gas bubbles in the bloodstream could cause the bends According to Boyle s law, gases expand as the pressure is decreased. Because atmospheric pressure decreases as altitude increases, the volume of gas sealed in the bag of chips will increase at higher altitudes During expiration, the volume (capacity) of the lungs decreases as pressure increases A respirator inflates the lungs with air that is rich in oxygen (and has a low concentration of carbon dioxide). This allows the blood to pick up more O 2 to be delivered to the body s cells as the blood circulates a. According to Boyle s law, for the pressure to increase while temperature and quantity of gas remain constant, the gas volume must decrease. Thus, cylinder A would represent the final volume. P mm Hg 0.86 atm P atm V ml V ml Because P 1 V 2 P 2 V 2, then V 2 P 1 V 1 /P atm V ml 1.2 atm 160 ml a. According to Boyle s law, the volume must increase when the pressure is decreased, so diagram C should represent the balloon s final volume at an increased altitude. When there is no change in the pressure, there is no change in the volume. Thus, diagram B would represent the final volume of the balloon. With an increase in pressure, there will be a decrease in volume. Thus, diagram A would describe the balloon s final volume a. The pressure doubles when the volume is halved. The pressure falls to one-third the initial pressure when the volume expands to three times its initial volume. 1 The pressure increases to ten times the original pressure when the volume decreases to 10 its initial volume a. The volume decreases to one-third the initial volume when pressure increases to three times its initial pressure. The volume doubles when pressure falls to one-half its initial pressure. The volume is five times greater when pressure is decreased to one-fifth its initial pressure From Boyle s law, we know that pressure is inversely related to volume. (For example, the pressure increases when the volume decreases.) a. Volume increases; pressure must decrease. 655 mm Hg 10.0 L 328 mm Hg 20.0 L Volume decreases; pressure must increase. 655 mm Hg 10.0 L 2620 mm Hg L The ml units must be converted to L for unit cancellation in the calculation, and because the volume decreases, pressure must increase. 655 mm Hg 10.0 L 1500 ml 1000 ml 1 L 4400 mm Hg

3 47374_11_p qxd 2/9/07 9:09 AM Page d. The ml units must be converted to L for unit cancellation in the calculation, and because the volume decreases, pressure must increase. 1 L 120 ml 0.12 L 1000 ml 655 mm Hg 10.0 L mm Hg 0.12 L From Boyle s law, we know that pressure is inversely related to volume. a atm 5.00 L 6.00 atm 1.00 L The ml units must be converted to L for unit cancellation in the calculation atm 5.00 L 1000 ml 2.4 atm 2500 ml 1 L d atm 5.00 L 750 ml 1.20 atm 5.00 L 8.00 L 4.5 L 2.0 L 15.0 atm 20.0 L L 1000 ml 1 L 1700 mm Hg 1.00 atm 8.0 atm atm From Boyle s law, we know that pressure is inversely related to volume. (For example, the pressure increases when the volume decreases.) a. Pressure increases; volume must decrease L 25 L 1500 mm Hg The mm Hg units must be converted to atm for unit cancellation in the calculation, and because the pressure increases, volume must decrease atm 1.00 atm 50.0 L 25 L 2.0 atm The mm Hg units must be converted to atm for unit cancellation in the calculation, and because the pressure decreases, volume must increase atm 1.00 atm 50.0 L 100 L atm d. The mm Hg units must be converted to torr for unit cancellation in the calculation, and because the pressure increases, volume must decrease L 1 torr 45 L 850 torr 1 mm Hg From Boyle s law, we know that pressure is inversely related to volume atm a. 25 ml 50. ml 0.40 atm 0.80 atm 25 ml 10. ml 2.00 atm Gases 115

4 47374_11_p qxd 2/9/07 9:09 AM Page 116 Chapter d According to Charles law, there is a direct relationship between temperature and volume. For example, volume increases when temperature increases, while the pressure and amount of gas remain constant. a. Diagram C describes an increased volume corresponding to an increased temperature. Diagram A describes a decreased volume corresponding to a decreased temperature. Diagram B shows no change in volume, which corresponds to no change in temperature According to Charles law, an increase in temperature will give an increase in volume. a. Because the gas warms, its volume must also increase. Because the warm gas will cool, its volume must decrease. Because the gas warms, its volume must increase Heating a gas in a hot air balloon increases the volume of gas, which reduces its density and allows the balloon to rise above the ground As the temperature decreases, the volume of the gas in the tire decreases, which makes the tire appear lower or flat in the morning According to Charles law, gas volume is directly proportional to Kelvin temperature when P and n are constant. In all gas law computations, temperatures must be in kelvin units. (Temperatures in C are converted to K by the addition of 273.) a. When temperature decreases, volume must also decrease. 75C K 55C K 2500 ml 328 K 2400 ml 348 K When temperature increases, volume must also increase ml 680 K 4900 ml 348 K 25C K 2500 ml 248 K 1800 ml 348 K d ml 240 K 1700 ml 348 K According to Charles law, a change in a gas s volume is directly proportional to the change in its Kelvin temperature. In all gas law computations, temperatures must be in Kelvin units. (Temperatures in C are converted to K by the addition of 273.) a. 0C K 273 K 10.0 L 683 K 683 K C 4.00 L d. 25 ml 25 ml 0.80 atm 2500 mm Hg 0.80 atm 80.0 torr 273 K 1.2 L 4.00 L 82 K 273 K 2.50 L 171 K 4.00 L 273 K L 3.4 K 4.00 L 760 torr 6.1 ml 190 ml 82 K C 171 K C 3.4 K C Because gas pressure increases with an increase in temperature, the gas pressure in an aerosol can may exceed the tolerance of the can when it is heated and cause it to explode.

5 47374_11_p qxd 2/9/07 9:09 AM Page The pressure in a tire is lower at the typical temperatures encountered on a winter morning, but it increases as the temperature increases. If the pressure increases beyond the tire s pressure rating, then the tire may suffer a blowout According to Gay-Lussac s law, temperature is directly related to pressure. For example, the temperature increases when the pressure increases. In all gas law computations, temperatures must be in kelvin units. (Temperatures in C are converted to K by the addition of 273.) a. When temperature decreases, pressure must also decrease. P torr P 2? T 1 155C K T 2 0C K P torr 273 K 770 torr 428 K When pressure decreases, temperature must also increase. 12C K 35C K 1.40 atm 308 K K According to Gay-Lussac s law, pressure is directly related to temperature. For example, pressure increases when the temperature increases. In all gas law computations, temperatures must be in Kelvin units. (Temperatures in C are converted to K by the addition of 273.) a. At constant volume, P torr P torr T 1 0C K T 2? 1500 torr 273 K 1600 K C 250 torr At constant volume, P mm Hg P mm Hg T 1 40.C K T 2? 680 torr 313 K 288 K C 740 torr a. boiling point vapor pressure atmospheric pressure d. boiling point Boiling would occur in b and c because the vapor pressure equals the atmospheric (external) pressure in both a. Water boils at temperatures less than 100 C because the atmospheric pressure is less than on top of a mountain. The boiling point is the temperature at which the vapor pressure of a liquid becomes equal to the external (in this case, atmospheric) pressure. The pressure inside a pressure cooker is greater than ; therefore, water boils above 100C. Foods cook faster at higher temperatures a. At a higher altitude, the external pressure is less, which means that water boils at a lower temperature. At an external pressure of 2.0 atm, water will boil at 120C When Boyle s, Charles, and Gay-Lussac s laws are brought together, they form the combined gas law. P 1 V 1 P 2V 2 T 1 T a. P 2 P 1V 1 T V 2 2 P 1V 1 T 2 P 2 T 1 V 2 T 1 Gases 117

6 47374_11_p qxd 2/9/07 9:09 AM Page 118 Chapter T K; V L; P mm Hg (1.1) a. T K; V L L 325 K 4.25 atm 1.85 L 298 K T 2 12C K; V L L 285 K 3.07 atm 2.25 L 298 K T 2 47C K; V L L 320 K atm 12.8 L 298 K T 1 112C K; P atm; V ml a. T K; P mm Hg (0.866 atm); V 2? ml 1.20 atm 281 K V ml 743 ml atm 385 K T 2 75C K; P atm; V 2? ml 1.20 atm 348 K V ml 1400 ml 0.55 atm 385 K T 2 15C K; P atm; V 2? ml 1.20 atm 258 K V ml 3.84 ml 15.4 atm 385 K T 1 225C K; P atm; V ml T 2 25C K; P atm; V 2? V ml 1.80 atm 0.80 atm T 1 8C K; V ml; P atm T 2 37C K; V ml; P 2? 50.0 ml 310 K P atm 1.10 atm ml 281 K 248 K 498 K 110 ml Addition of more air molecules to a tire or basketball will increase its volume because more moles of gas are added Air molecules are escaping from the balloon (it is deflating) as it flies around the room According to Avogadro s law, the change in the gas volume is directly proportional to the change in the number of moles of gas. a. When moles decrease, volume must also decrease. (One-half of 4.00 mol 2.00 mol. ) 2.00 mol 8.00 L 4.00 L 4.00 mol When moles increase, volume must also increase. 1 mol Ne 25.0 g Ne 1.24 mol Ne added g Ne (1.50 mol 1.24 mol 2.74 mol) 2.74 mol 8.00 L 14.6 L 1.50 mol 1.50 mol 3.50 mol 5.00 mol 5.00 mol 8.00 L 26.7 L 1.50 mol 118

7 47374_11_p qxd 2/9/07 9:09 AM Page 119 Gases a g O 2 1 mol O g O mol O 2 New moles mol O 2 n mol 0.50 mol 0.65 mol At STP, the molar volume of any gas is 22.4 L/mol. a. d At STP, the molar volume of any gas is 22.4 L/mol. a. 2.5 mol N L 56 L N 1 mol N 2 2 d L 0.65 mol O mol O 2 65 L mol 10.0 L 15.0 L mol O 2 remain 1 mol He 4.00 g He 4.00 g He 1.00 mol He New moles 1.00 mol 0.15 mol 1.15 mol gas 1.15 mol gas 15.0 L 0.15 mol O L 44.8 L O 2 1 mol O L 2.00 mol O L CO 2 1 mol CO L 6.40 g O 2 1 mol O L O L O g O 2 1 mol O g Ne 1 mol Ne g Ne mol CO L Ne 1 mol Ne mol He 22.4 L 1000 ml 9410 ml 1 mol He 1 L 1 mol Ne g Ne 11.2 L 10.1 g Ne 22.4 L 1 mol Ne 1 L 1600 ml 1000 ml 1 mol H L mol H At STP, the volume of any gas is 22.4 L/mol. a. For F 2, the molar mass is g/mol g F 2 1 mol F 2 1 mol F L 1.70 g F 2/L For CH 4, the molar mass is g/mol g CH 4 1 mol CH g CH 1 mol CH L 4 /L For Ne, the molar mass is g Ne/mol g Ne 1 mol Ne g Ne/L 1 mol Ne 22.4 L d. For SO 2, the molar mass is g SO 2 /mol g SO 2 1 mol SO g SO 1 mol SO L 2 /L At STP, the volume of any gas is 22.4 L/mol. a. For C 3 H 8, the molar mass is g/mol g C 3 H 8 1 mol C 3H g C 1 mol C 3 H L 3 H 8 /L 1000 ml 1 L Ne ml Ne 119

8 47374_11_p qxd 2/9/07 9:09 AM Page 120 Chapter For NH 3, the molar mass is g/mol g NH 3 1 mol NH g NH 1 mol NH L 3 /L For Cl 2, the molar mass is g Cl 2 /mol g Cl 2 1 mol Cl 2 1 mol Cl L 3.17 g Cl 2/L d. For Ar, the molar mass is g Ar/mol g Ar 1 mol Ar 1.78 g Ar/L 1 mol Ar 22.4 L P n V PV n V n P n PV 10.0 g Kr V n P (2.00 mol)( Latm)(300 K) (10.0 L)(molK) (4.0 mol)( Latm)(291 K) (1.40 atm)(molk) (845 mm Hg)(20.0 L) (62.4 Lmm Hg)(295 K) molk 1 mol Kr g Kr mol Kr (0.119 mol)(62.4 Lmm Hg)(298 K) (575 mm Hg)(molK) 4.93 atm 68 L g O 2 1 mol O g O L, or 3850 ml g N 2 1 mol N g mol T PV nr (630. mm Hg)(50.0 L) (0.892 mol)(62.4 Lmm Hg) molk 566 K C a. n g CO 2 1 mol CO g CO mol T PV nr (455 mm Hg)(0.525 L) ( mol)(62.4 Lmm Hg) molk L 1 mol mol 22.4 L 0.84 g 42 g/mol mol 1.28 g 22.4 L 28.7 g/mol 1 L 1 mol T 295 K V 1.00 L P 685 mm Hg n PV (685 mm Hg)(1.00 L) mol (62.4 Lmm Hg)(295 K) molk Molar mass 1.48 g 39.8 g/mol mol 745 K C 120

9 47374_11_p qxd 2/9/07 9:09 AM Page 121 Gases d. n PV (0.95 atm)(2.30 L) mol ( Latm)(297 K) molk Molar mass 2.96 g 33 g/mol mol a L 1 mol mol 22.4 L 11.6 g 130. g/mol mol g 22.4 L 16.0 g/mol 1 L 1 mol n PV (1.20 atm)(1.00 L) mol ( Latm)(295 K) d a. Molar mass n PV molk Molar mass 2.32 g 51.2 g/mol mol 1 mol Mg 8.25 g Mg mol Mg g Mg 1 mol Mg 1 mol H 2 1 mol Mg n PV mol H 2 molk g 16.9 g/mol mol (685 mm Hg)(1.23 L) mol (62.4 Lmm Hg)(298 K) (735 mm Hg)(5.00 L) (62.4 Lmm Hg)(291 K) molk 1 mol Mg 1 mol H L 1 mol H L H 2 released at STP g Mg 1 mol Mg mol H g Mg a g NH 4 NO 3 1 mol NH 4NO g NH 4 NO mol NH 4 NO mol NH 4 NO 3 4 mol H 2O 2 mol NH 4 NO mol H 2 O V n P n PV (0.644 mol)( Latm)(623 K) (0.950 atm)(molk) (0.950 atm)(10.0 L) ( Latm)(623 K) molk 34.7 L mol NH 4 NO g NH 4NO 3 1 mol NH 4 NO g NH 4 NO mol O 2 2 mol NH 4NO 3 1 mol O mol g C 4 H 10 1 mol C 4H g C 4 H mol O 2 2 mol C 4 H mol O 2 V n P (6.17 mol)( Latm)(298 K) (0.850 atm)(molk) 178 L 121

10 47374_11_p qxd 2/9/07 9:09 AM Page 122 Chapter g Al a a g KNO 3 1 mol KNO g KNO 3 1 mol O 2 2 mol KNO mol O 2 V n P (0.247 mol O 2)( Latm)(308 K) (1.19 atm)(molk) n PV 1 mol Al g Al 3 mol O 2 4 mol Al 0.15 mol O L 1 mol O L O 2 (STP) (725 mm Hg)(4.00 L) (62.4 Lmm Hg)(688 K) molk mol NO 2 4 mol NH 3 4 mol NO g NH 3 1 mol NH g NH 3 P O2 765 mm Hg 22 mm Hg 743 mm Hg n PV (743 mm Hg)(0.256 L) mol O (62.4 Lmm Hg)(297 K) 2 molk P O mm Hg 751 mm Hg n PV (751 mm Hg)(0.425 L) (62.4 Lmm Hg)(289 K) molk 0.15 mol O mol NO L O mol NO Each gas particle in a gas mixture exerts a pressure as it strikes the walls of the container. The total gas pressure for any gaseous sample is thus a sum of all the individual pressures. When the portion of the pressure due to a particular type of gaseous particle is discussed, it is only part of the total. Accordingly, these portions are referred to as partial pressures Because the helium and oxygen particles in the sample must have the same average kinetic energies, each particle will exert the same average pressure against the container s walls. For the partial pressures for each gas to be the same, there must be equal numbers of these two types of gaseous particles To obtain the total pressure in a gaseous mixture, add the partial pressures (provided each carries the same pressure unit). P total P Nitrogen P Oxygen P Helium 425 torr 115 torr 225 torr 765 torr To obtain the total pressure in a gaseous mixture, add the partial pressures (provided each carries the same pressure unit). P total P Argon P Neon P Nitrogen 415 mm Hg 75 mm Hg 125 mm Hg 615 mm Hg 615 mm Hg atm Because the total pressure in a gaseous mixture is the sum of the partial pressures (same pressure unit), addition and subtraction are used to obtain the missing partial pressure. P Nitrogen P total (P Oxygen P Helium ) 925 torr (425 torr 75 torr) 425 torr 122

11 47374_11_p qxd 2/9/07 9:09 AM Page Because the total pressure in a gaseous mixture is the sum of the partial pressures, addition and subtraction are used to obtain the missing partial pressure. P Helium P total P Mixture of Oxygen, Nitrogen, and Neon 1.50 atm 1.20 atm 0.30 atm a. 2; fewest number of gas particles exerts the lowest pressure. 1; greatest number of gas particles exerts the highest pressure a. 3; volume increases. 1; volume decreases. 3; volume decreases. d. 3; volume increases. e. 2; volume stays the same a. A; volume decreases when temperature decreases. C; volume increases when pressure decreases. A; volume decreases when the moles of gas decrease. d. B; doubling temperature doubles the volume, but losing half the gas particles decreases the volume by half. The two effects cancel and no change in volume occurs. e. C; increasing the moles increases the volume to keep T and P constant a. Volume decreases; pressure increases. Increase particles; pressure increases. Decrease temperature; pressure decreases V ml T K P mm Hg a. The volume of the chest and lungs will decrease when compressed during the Heimlich maneuver. A decrease in volume causes the pressure to increase. A piece of food would be dislodged with a sufficiently high pressure T 1 15C K; P mm Hg; V ml 1000 ml P atm 912 mm Hg; V L 1 L 2500 ml 912 mm Hg 288 K 207 K C 4250 ml 745 mm Hg Remember to convert temperatures to Kelvin units and solve for new volume, V torr 228 K 750 L 1500 L 0.20 atm 760 torr 281 K V 2? L T K P atm (425 ml)(0.980 atm)(178 K) V ml (297 K)(0.115 atm) n PV (12 atm)(35.0 L) ( Latm)(278 K) molk 1.8 mol CO mol CO molecules 1 mol CO molecules CO atm ml Gases g N 2 1 mol N g N mol N 2 P n V (1.78 mol N 2)( Latm)(298 K) 15.0 L(molK) 2.92 atm 123

12 47374_11_p qxd 2/9/07 9:09 AM Page 124 Chapter n PV (2500 mm Hg)(2.00 L) (62.4 Lmm Hg)(291 K) molk 0.28 mol CH g CH 4 1 mol CH g CH mol CH molecules T 297 K V 4.60 L a V n P D n PV ml mol N g N 2 1 mol N g N 2 Mass Volume g O 2 1 mol Molar mass Using ideal gas law: 1 mol O molecules mol O 2 (0.066 mol)(62.4 Lmm Hg)(278 K) 845 mm Hg(molK) 1.00 g CO 2 1 mol CO g CO mol CO 2 P n ( mol)( Latm)(297 K) atm V (4.60 L)(molK) 91.5 mm Hg P total P Nitrogen P water P Nitrogen P total P water 745 mm Hg 32 mm Hg 713 mm Hg (713 mm Hg)(0.25 L) (62.4 Lmm Hg)(303 K) molk 1 L 1 mol gas 1000 ml 22.4 L (STP) g mol 1.15 g gas mol gas 1 mol 22.4 L (STP) mol N mol gas 115 g/mol 1.4 L 1.43 g/l 1000 ml 1 L 1400 ml n g 1 molar mass (MM) PV n PV g g MM MM PV (1.15 g)( Latm)(273 K) MM 115 g/mol (1.0 atm)(0.225 L)(molK) 1 torr 1 mm Hg Molar mass g of gas/mol of gas n PV (748 torr)(0.941 L) mol (62.4 Lmm Hg)(293 K) molk Molar mass 1.62 g 42.1 g/mol mol 124

13 47374_11_p qxd 2/9/07 9:09 AM Page 125 Gases 1 L 1 mol gas ml mol gas 1000 ml 22.4 L (STP) Molar mass g/mol 1.02 g gas/ mol gas 30.0 g/mol Using ideal gas law: Because CH 3 has a molar mass of 15.0, there must be two CH 3 units in a molecular formula of C 2 H g/mol 15.0 g/ef 2.00 (CH 3) 2 C 2 H g Zn g Mg a a PV n PV (g/mm) MM g PV (1.02 g)( Latm)(273 K) MM 30.0 g/mol (1.0 atm)(0.762 L)(molK) V n P molecules NO mol NO 2 7 mol O 2 4 mol NO L O 2 1 mol O 2 16 L O 2 n PV mol H 2 O 4 mol NH 3 6 mol H 2 O g NH 3 1 mol NH g NH 3 n PV (748 torr)(1.88 L) mol (62.4 Lmm Hg)(293 K) Molar mass 3.24 molk g 42.1 g/mol mol 2.78 g C 1 mol C mol C mol C g C 0.46 g H 1 mol H 0.46 mol H mol H g H Empirical formula CH g/ef 42.1 g/mol n 14.0 g/ef 3 (CH 2) 3 C 3 H 6 molecular formula n PV 1 mol Zn g Zn 1 mol H 2 1 mol Zn 22.4 L H 2 (STP) 1 mol H L H 2 (STP) 1 mol Mg g Mg 1 mol H 2 1 mol Mg mol H 2 (0.494 mol)(62.4 Lmm Hg)(297 K) (835 mm Hg)(molK) (725 mm Hg)(5.00 L) (62.4 Lmm Hg)(648 K) molk (752 mm Hg) (0.782 L) (62.4 Lmm Hg)(296 K) mol molk Molar mass g/mol 2.23 g 70.1 g/mol mol Empirical formula mass of CH g/mol 70.1 g/mol 14.0 g/mol 5.01 CH 2 5 C 5 H L 1 mol molecules NO mol NO mol H 2 O 125

14 47374_11_p qxd 2/9/07 9:09 AM Page 126 Chapter The partial pressure of each gas is proportional to the number of particles of each type of gas that is present. Thus, a ratio of partial pressure to total pressure is equal to the ratio of moles of that gas to the total number of moles of gases that are present: 126 P Helium /P total n Helium /n total Solving the equation for the partial pressure of helium yields: 2.0 mol P Helium 2400 torr 600 torr 8.0 mol 6.0 mol P Oxygen 2400 torr 1800 torr 8.0 mol Because the total pressure is to be reported in mm Hg, the atm and torr units (for argon and nitrogen, respectively) must be converted to mm Hg, as follows: 0.25 atm 190 mm Hg 1 mm Hg 360 torr 360 mm Hg 1 torr and P total P Argon P Helium P Nitrogen 190 mm Hg 350 mm Hg 360 mm Hg 900 mm Hg ( mm Hg) Because the partial pressure of nitrogen is to be reported in torr, the atm and mm Hg units (for oxygen and argon, respectively) must be converted to torr, as follows: 760 torr 0.60 atm 460 torr (O 2) 425 mm Hg 1 torr 425 torr (Ar) 1 mm Hg and P Nitrogen P total (P Oxygen P Argon ) 1250 torr (460 torr 425 torr) 370 torr Because we need the answer in mm Hg, we must include a pressure unit conversion factor for oxygen. P Oxygen atm 342 mm Hg 1 mm Hg 255 torr 255 mm Hg 1 torr P total 255 mm Hg 342 mm Hg 597 mm Hg a. n(n 2 ) PV (840 mm Hg)(2.0 L) (62.4 Lmm Hg)(297 K) molk P Hydrogen P total P water vapor 755 mm Hg 21 mm Hg 734 mm Hg n PV (734 mm Hg) (0.415 L) mol H (62.4 L mm Hg)(296 K) mol H 2 mol K 2 mol Al 3 mol H g Al 1 mol Al g Al a. PO mm Hg 25 mm Hg 719 mm Hg n PV (719 mm Hg) (0.226 L) mol O (62.4 L mm Hg)(299 K) 2 mol K 2 mol NO 2 1 mol N g NO 2 1 mol NO g NO 2

15 47374_11_p qxd 2/9/07 9:09 AM Page 127 Gases d a. False. The flask containing helium has more moles and thus more atoms of helium. False. There are different numbers of moles in the flasks, which means the pressures are different. True. There are more moles of helium, which makes the pressure of helium greater than that of neon. d. False. Density is molar mass divided by molar volume. Because the molar masses are different, their densities are different a mol O 2 2 mol KClO 3 3 mol O mol KClO mol KClO g KClO 3 1 mol KClO g KClO 3 reacted V 25.0 L T 413 K P atm n PV (0.900 atm)(25.0 L) mol ( Latm)(413 K) molk Molar mass 92.9 g 140. g/mol mol T PV nr (1.30 atm)(25.0 L) C (0.664 mol)( Latm) molk L 1 mol mol 22.4 L 1.02 g 30.0 g/mol mol 9.60 g C 1 mol C mol C/0.799 mol 1.00 mol C g C 2.42 g H 1 mol H 2.40 mol H/0.799 mol 3.00 mol H g H Empirical formula CH 3 Empirical mass (1.008) g/ef 30.0 g/mol g/ef 2 (CH 3) 2 C 2 H g NaN 3 1 mol NaN g NaN mol NaN 3 3 mol N 2 2 mol NaN mol N 2 At STP 3.04 mol N L 1 mol N L n PV (1.00 atm)(7.50 L) ( Latm)(310 K) molk mol O 2 6 mol H 2O 6 mol O mol H 2 O 18.0 g C 6 H 12 O 6 1 mol C 6H 12 O mol C 6 H 12 O 6 6 mol H 2O g C 6 H 12 O 6 1 mol C 6 H 12 O mol H 2 O mol H 2 O is the smaller number of moles of product. Thus, O 2 is the limiting reactant mol H 2 O g H 2O 1 mol H 2 O 5.32 g H 2O 127

16 47374_11_p qxd 2/9/07 9:09 AM Page 128 Chapter n N2 PV (1.08 atm)(2.00 L) ( Latm)(298 K) molk mol N 2 n O2 PV (0.118 atm)(4.00 L) mol O 2 (smaller amount of reactant) ( Latm)(298 K) molk 2 mol NO g NO mol O g NO 1 mol O 2 1 mol NO If the initial volume is 1 V, the final volume is 1 2 V, or V. (Temperature remains constant). V P mm Hg 1190 mm Hg V 2 128