Chapter 5 Gases. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–2 QUESTION.

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Presentation transcript:

Chapter 5 Gases

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–2 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–3 ANSWER, VT 4)All are the same Section 5.2 The Gas Laws (p. 181) Avogadro’s Law states that gas samples with the same P, and will have the same number of moles.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–4 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–5 ANSWER )flask C Section 5.6 Gases (p. 199) Ammonia has a lone pair that creates a polar molecule, while NO has an unpaired electron that leads to a bent polar molecule. N is a linear molecule with no dipole.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–6 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–7 ANSWER 1) Flask A Section 5.6 Gases (p. 199) Ammonia has the lowest molecular weight and will move faster than the two other molecules at the same temperature.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–8 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–9 QUESTION (continued)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–10 ANSWER 1) Your sample of gas has the higher pressure. Section 5.1 Pressure (p. 179) Mercury is much denser than water, so if both columns are the same height, more pressure is required to raise the mercury to that level than water.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–11 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–12 QUESTION (continued)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–13 ANSWER = / 3)1 : 4 Section5.3 The Ideal Gas Law (p. 186) A rearrangement of the Ideal Gas Law gives n PVRT. If the temperature of one vessel is four times that of the other then the number of moles must be ¼ of the other.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–14 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–15 QUESTION (continued)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–16 ANSWER u of time 5)1 : 2 Section 5.6 Gases (p. 199) The speed,, of molecules is proportional to the square root of the temperature. If the temperature increases by four times the speed of the molecules doubles, and thus the number of collisions with the wall per unit doubles.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–17 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–18 ANSWER 2)1.30 L Section 5.2 The Ideal Gas Laws (p. 186) Charles’ Law requires that the temperature be in units of Kelvins.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–19 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–20 ANSWER = / 3) 51.5 L Section 5.3 The Ideal Gas Law (p. 186) A rearrangement of the Ideal Gas Law gives V nRTP.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–21 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–22 QUESTION (continued)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–23 ANSWER 2 1 P 21 / 2 2 2)28 lb/in Section 5.3 The Ideal Gas Law (p. 186) P = TT is the equation used. The units of pressure can remain lb/in because they are on both sides.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–24 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–25 ANSWER 1 V / 1 P 2 / 2 3)1.88 atm Section 5.2 The Gas Laws (p. 181) Since the number of moles does not change, the equation P 1 T = 2 VT can be used.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–26 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–27 QUESTION (continued)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–28 ANSWER 2 2)Your friend should release the Cl first. Section 5.7 Effusion and Diffusion (p. 206) Chlorine is heavier than hydrogen. At the same temperature chlorine has the same kinetic energy as hydrogen, so it moves slower than hydrogen.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–29 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–30 ANSWER PTn n P T 5) low, high, low Section 5.8 Real Gases (p. 208) Ideal conditions are approached as the gas molecules interact the less and less. Low and low mean the number of collisions are low and high means the gas molecules will have very large kinetic energy vs. the potential energy of intermolecular forces when they do collide.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–31 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–32 ANSWER 2 2)Cl Section 5.4 Gas Stoichiometry (p. 190) Since all the gases have the same volume and number of molecules/atoms, the density is determined by the molecular mass.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–33 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–34 ANSWER 2 1)He Section 5.6 The Kinetic Molecular Theory of Gases (p. 199) ½ mv = kT. This is the relationship between kinetic energy and temperature. The smaller the mass of the gas molecule/atom the faster it moves at a given temperature.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–35 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–36 ANSWER, VT 5)All gases the same Section 5.4 Gas Stoichiometry (p. 190) P, and are the same, so the number of moles are the same.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–37 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–38 ANSWER 5)All gases the same Section 5.6 The Kinetic Molecular Theory of Gases (p. 199) At the same temperature, all the molecules have the same kinetic energy.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–39 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–40 ANSWER 3)2 Section 5.4 Gas Stoichiometry (p. 190) The larger the molecular mass the higher the density.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–41 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–42 ANSWER 4) 1.25 g/L Section 5.4 Gas Stoichiometry (p. 190) At STP, n = 1 mole (28.0 g). V = 22.4 L at STP.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–43 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–44 QUESTION (continued)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–45 ANSWER /. nP  V 1)22.4 L Section 5.4 Gas Stoichiometry (p. 190) V = nRTP = 2 and = mmHg (1 atm). The 2’s cancel, leaving = 22.4 L.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–46 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 5–47 ANSWER 22 3) 22.4 L Section 5.4 Gas Stoichiometry (p. 190) 28.0g of N is 1 mol of N. 1 mol of a gas at STP is the molar volume, or 22.4 L.