22. Figure 5.10 p. 195 Balloons Holding 1.0 L of Gas at 25º C and 1 atm . Each contains 0.041 mol of gas, or 2.5 x 10 22 molecules, even though their masses are different ( equal numbers, different masses ).
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35. Figure 5.11 A Mole of Any Gas Occupies a Volume of Approximately 22.4 L at STP How many particles are in each box? 6.022 x 10 23 particles
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52. Figure 5.12 p. 206 Partial Pressure of Each Gas in a Mixture
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62. Figure 5.13 The Production of Oxygen by Thermal Decomposition of KCIO 3
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73. Figure 5.18 Path of One Particle in a Gas. Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of its container.
80. Figure 5.21 p. 219The Effusion of a Gas into an Evacuated Chamber Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.
92. Figure 5.24 p. 222 Plots of PV/nRT Versus P for Several Gases (200 K) Behavior is close to ideal ONLY at low pressures (less than 1 atm)
93. Figure 5.25 Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures Deviations are smaller at higher temperatures.
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Hinweis der Redaktion
Z5e 188 Section 5.1 Pressure
Z5e 189 Fig. 5.2
Z5e 189 Fig. 5.3
Z5e 189 Units of Pressure 96.5 kPa 724 torr 0.952 atm
Z5e 190 Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro
Hrw 314 ref. Fig 10-10
Z5e 192 Fig. 5.6 Plot of PV vs P at P below 1 atm Ideal gas expected to have constant value of PV Boyle’s law is good approximation at relatively low pressures
(20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x); so x = 2.04 L (30.6)(740) = (x)(750); x = 30.2 torr = 402 kPa
(20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x); so x = 2.04 L (30.6)(740) = (x)(750); x = 30.2 torr = 402 kPa
Z5e 193 Charles’ Law
Hrw 316 and rf hrw 318 fig. 10-12.
Z5e 194 Fig. 5.8 Plot V vs. T Gas samples contain different numbers of moles
(247)/(22 + 273) = (x)(273 + 98); x = 311 ml
(40.5)/(26.4 + 273) = (81.0)/x; x = 593 K = 320. Celsius
Z5e 195 Avogadro’s Law
Z5e 195 Fig. 5.10
V is constant, so (3.8)/(295) = (x)/373; x = 4.8 atm (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22); x = 680 ml
V is constant, so (3.8)/(295) = (x)/373; x = 4.8 atm (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22); x = 680 ml
Z5e 196 Section 5.3 The Ideal Gas Law Also 62.396 l torr/k mol = 8.3145 J/k mol
PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
PV = nRT T = (1.85)(47.3)/(.0821 * 1.62) = 658 K n = PV/RT = 6.51 mol; n * M = 651 * 83.8 = 546 g Use combined law; x = 3.99 L (can convert torr to atm, but they cancel out so don’t need to for this problem).
Z5e Section 5.4 Gas Stoichiometry
Z5e 202
Balance 1st! Stoich MW HgO 216.59; Hg 200.59; O2 32; ans. 0.212 L Use ideal gas law; ans. 0.537 L
Balance 1st! Stoich MW HgO 216.59; Hg 200.59; O2 32; ans. 0.212 L Use ideal gas law; ans. 0.537 L
MW NaHCO 3 84 Not at STP so use PV=nRT; ans. 21.5 g Use stoich to get moles HCl, then mole ratio --> moles CO 2 , then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = 1.08 mole CO 2 ; so concentration is 0.415 M
MW NaHCO 3 84 Not at STP so use PV=nRT; ans. 21.5 g Use stoich to get moles HCl, then mole ratio --> moles CO 2 , then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = 1.08 mole CO 2 ; so concentration is 0.415 M
T & P constant so cancel out. L proportional to moles, so 10 L watch Kg! Ans. Rounded to 16 500 (from calc ans of 16 470)
T & P constant so cancel out. L proportional to moles, so 10 L watch Kg! Ans. Rounded to 16 500 (from calc ans of 16 470)
LR = O 2 , ans. 313 g ans. 52 L correct to STP = 783 L; ans. 895 g.
Z5e 205
Hrw 345
Hrw additional sample problem 11-5
Hrw additional sample problem 11-5
Hrw additional sample problem 11-6, modified so that density is used, where density is rounded down to 3.36 g/L from 3.365 g/L.
D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
D = MP/RT = 0.68 g/L n = 0.0083; M = 97; EF = 49 so molecular formula = C 2 H 2 Cl 2
Z5e 205 Section 5.5 Dalton’s Law of Partial Pressures
Z5e 206 Fig 5.12
Z5e 205 Section 5.5 Dalton’s Law of Partial Pressures
Z5e 207
592/752 = 0.787 0.787 x 5.25 = 4.13 atm Note: can use mole fraction regardless of units (don’t have to convert). Also, Air pressure is 752 (so this is the total pressure in the denominator)
592/752 = 0.787 0.787 x 5.25 = 4.13 atm Note: can use mole fraction regardless of units (don’t have to convert). Also, Air pressure is 752 (so this is the total pressure in the denominator)
P1V1 = P2V2; total V = 9 L CH4 (4)(2.70) = (9)(x); x = 1.2 atm N2 ans. 0.763 O2 = 0.294 Sum = 2.127 = ans.
Z5e 209
Z5e 209 Fig. 5.13
Make sure reaction is balanced MW of ammonium nitrate = 80. G Use spider to get moles of N20 = .0325, then use PV = nRT, but first have to: Convert 21 torr to 2.8 kPa 94.0 - 2.8 = 91.2 kPa N2O convert P to atm d/t R ( or use 62.4)!!! Convert C to K Ans. 6.6 L
Make sure reaction is balanced MW of ammonium nitrate = 80. G Use spider to get moles of N20 = .0325, then use PV = nRT, but first have to: Convert 21 torr to 2.8 kPa 94.0 - 2.8 = 91.2 kPa N2O convert P to atm d/t R ( or use 62.4)!!! Convert C to K Ans. 6.6 L
Z5e 211 Section 5.6 The Kinetic Molecular Theory of Gases
Z5e 216, 217
M rms = sq root 3RT/u Must express R using joules and M = Kg/mol (not g, d/t R) R = 8.3145 J/K*Mol 411 m/s 1928 m/s 324 m/s
Z5e 218 Fig. 5.18: Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of its container.
Z5e Section 5.7 Effusion and Diffusion
Z5e 219 Fig 5.21: Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.
Z5e 220
Z5e 221 Diffusion
a. ans. 40.96 b. rate NH3/rate HCl = 2.47/2.47 = 1/1 = sq rt (.0025*17/x*36.5) = .00177 mol HCl. Alternate method: (0.251/2.47 divided by (x)/2.47) = (sq rt 36.5)/(sq rt 17) = 0.00177 mol HCl. c. Rate N2is faster than rate x by 55/38 times = 1.45 times faster. Then, 1.45/1 = Sq Rt M x divided by sq rt N 2 and answer is 59 g/mol (sig figs)
Z5e Section 5.8 Real Gases
Z5e 224
Z5e 222: Behavior is close to ideal ONLY at low pressures (less than 1 atm)
Z5e 222
a. P = nRT/V = 12.2 atm b. P = nRT/V - nb - a (n/v) 2 = 11.0 atm
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