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Chapter 12 Gases

Chapter 12 Gases. What is the pressure of the gas in the bulb?. P gas = P h P gas = P atm P gas = P h + P atm P gas = P h - P atm P gas = P atm - P h. What is the pressure of the gas in the bulb?. P gas = P h P gas = P atm P gas = P h + P atm P gas = P h - P atm

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Chapter 12 Gases

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  1. Chapter 12Gases

  2. What is the pressure of the gas in the bulb? • Pgas = Ph • Pgas = Patm • Pgas = Ph+ Patm • Pgas = Ph-Patm • Pgas = Patm - Ph

  3. What is the pressure of the gas in the bulb? • Pgas = Ph • Pgas = Patm • Pgas = Ph+ Patm • Pgas = Ph-Patm • Pgas = Patm - Ph

  4. 2 NO(g) + O2 (g) 2 NO2 (g) If 250 mL of NO is placed in a flask with O2, what volume of O2 is needed for complete reaction? • 100 mL • 125 mL • 200 mL • 250 mL • Cannot be determined from the given information.

  5. 2 NO(g) + O2 (g) 2 NO2 (g) If 250 mL of NO is placed in a flask with O2, what volume of O2 is needed for complete reaction? • 100 mL • 125 mL • 200 mL • 250 mL • Cannot be determined from the given information.

  6. If an equal mass of each gas is put into a separate balloon, which will have the greatest volume?Assume that they are all the same temperatureand pressure. • He • H2 • N2 • Ne • O2

  7. If an equal mass of each gas is put into a separate balloon, which will have the greatest volume?Assume that they are all the same temperatureand pressure. • He • H2 • N2 • Ne • O2

  8. If equal masses of CH4, C2H6, and C3H8 are placedin a flask, which of the following is true? • PCH4 = PC2H6 = PC3H8 • PCH4 ~ PC2H6 ~ PC3H8 • PCH4 > PC2H6 > PC3H8 • PCH4 < PC2H6< PC3H8 • None of the above

  9. If equal masses of CH4, C2H6, and C3H8 are placedin a flask, which of the following is true? • PCH4 = PC2H6 = PC3H8 • PCH4 ~ PC2H6 ~ PC3H8 • PCH4 > PC2H6 > PC3H8 • PCH4 < PC2H6< PC3H8 • None of the above

  10. Arrange the gases according to increasing molecular speed. He (25°C) He (100°C) Ne (25°C) Ne (0°C) • He (25) < He (100) < Ne (25) < Ne (0) • He (25) < He (100) < Ne (0) < Ne (25) • Ne (0) < Ne (25) < He (25) < He (100) • Ne (25) < Ne (0) < He (100) < He (25) • Ne (0) < He (25) < Ne (25) < He (100)

  11. Arrange the gases according to increasing molecular speed. He (25°C) He (100°C) Ne (25°C) Ne (0°C) • He (25) < He (100) < Ne (25) < Ne (0) • He (25) < He (100) < Ne (0) < Ne (25) • Ne (0) < Ne (25) < He (25) < He (100) • Ne (25) < Ne (0) < He (100) < He (25) • Ne (0) < He (25) < Ne (25) < He (100)

  12. flask 1 flask 2 If a mixture of gas A and gas B is moved from flask 1 to flask 2, which of the following is true: • PA, PB, and Ptot decrease • PA, PB, and Ptot increase • PA and PB decrease, Ptot remains the same • PA, PB, and Ptot remain the same • PA and PB remain the same, Ptot decreases

  13. flask 1 flask 2 If a mixture of gas A and gas B is moved from flask 1 to flask 2, which of the following is true: • PA, PB, and Ptot decrease • PA, PB, and Ptot increase • PA and PB decrease, Ptot remains the same • PA, PB, and Ptot remain the same • PA and PB remain the same, Ptot decreases

  14. A gas initially at 2.0 atm is in an adjustable volume container of 10. L in volume. If the pressure is decreased to 0.50 atm, what is the new volume? 40. L 20. L 10. L 5.0 L

  15. Correct Answer: 40. L 20. L 10. L 5.0 L 1 =  V constant P Thus, 2.00 atm(10. L) = 0.50 atm (Vfinal) Vfinal = 2.00 atm(10. L)/0.50 atm = 40. L

  16. Assuming pressure is held constant, to what volume will a balloon initially at 1.0 L change if its temperature is decreased from 300 K to 75 K? 1.0 L 2.0 L 0.25 L 4.0 L

  17. Correct Answer: 1.0 L 2.0 L 0.25 L 4.0 L =  V constant T Thus, 1.0 L/300 K = (Vfinal)/75 K Vfinal = 75 K/(1.0 L)300 K = 0.25 L

  18. At standard temperature and pressure, how many moles of gas are present in a box with a volume of 112 L? 1.00 moles 2.00 moles 5.00 moles 0.200 moles

  19. Correct Answer: 1.00 moles 2.00 moles 5.00 moles 0.200 moles = PV nRT ( ) ( ) ( ) nRT 1 mol 0.08206 L·atm/mol· K 273.15 K = = = V 22.41 L P 1.000 atm Thus, at STP 22.41 L = 112 L 1.00 mol n n = 5.00 moles

  20. At standard temperature and pressure, a hot-air balloon is filled with helium only to a volume of 4480 L. How many grams of helium are needed to fill the balloon? 200. g 400. g 800. g 50.0 g

  21. Correct Answer: 200. g 400. g 800. g 50.0 g At STP 22.41 L = 4480 L 1.00 mol n n = 200. moles Since MW(He) is 4.0 g/mol Mass = (200. g)(4.0 g/mol) = 800. g

  22. A gas sample occupies a volume of 4.00 L at 20°C. The temperature at which the gas would double its volume is 10°C 40°C 288°C 313°C

  23. Correct Answer: 10°C 40°C 288°C 313°C The temperature scale is absolute, however; Vinitial/(Vfinal) = Tfinal/(Tinitial) 2 = Tfinal/293 K Tfinal = 586 K, or 313°C

  24. N2(g) + 3 H2(g) 2 NH3(g) At STP, 16 L of N2 and 48 L of H2 are mixed. Assuming all the reactants are consumed, how many L of NH3 will be produced? 8.0 L 16 L 24 L 32 L 64 L

  25. Correct Answer: 8.0 L 16 L 24 L 32 L 64 L  V n (constant P, T ) According to Avogadro’s law, mole ratios in the chemical equation will be volume ratios under identical conditions. Because the reactants are in a stoichiometric 3:1 volume ratio, the product will have stoichiometric equivalence. Thus, (16 L N2)(2 mol NH3/1 mol N2) = 32 L NH3

  26. A container holds a mixture of oxygen, neon, and helium gases whose partial pressures are 150 torr, 300 torr, and 450 torr, respectively. The mole fraction of neon is • 0.17 • 0.33 • 0.50 • 0.67  = P P i i total

  27. Correct Answer: • 0.17 • 0.33 • 0.50 • 0.67  = P P i i total Xi = Pi/Ptotal Xi = (300 torr)/(150 + 300 + 450) torr Xi = 300 torr/900 torr = 0.33

  28. A sample of He gas initially at STP is compressed to a smaller volume at constant temperature. What effect does this have on the rms speed of the atoms? Increases Decreases No effect

  29. Correct Answer: Increases Decreases No effect The rms speed is directly proportional to the square root of the temperature, which does not change in this example.

  30. An unknown gas effuses at half the rate of helium. This gas is likely to be which of the following? H2 CH4 Ne O2 Ar

  31. Correct Answer: H2 CH4 Ne O2 Ar (r1/2)2 =M2/M1 M2= (r1/r2)2M1 M2= (2/1)2(4.0 g/mol) = 16.0 g/mol Therefore it could be CH4

  32. Real gases deviate from ideal behavior at __________ and _________. High temperature; low pressure Low temperature; high pressure High temperature; high pressure Low temperature; low pressure

  33. Correct Answer: High temperature; low pressure Low temperature; high pressure High temperature; high pressure Low temperature; low pressure At low temperature and high pressure, intermolecular forces increase as the molecules get closer together.

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