MOOREHW3 - Chapter L0 Gases and The Atmosphere 1a Answer(a...

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Unformatted text preview: Chapter L0: Gases and The Atmosphere 1a Answer: (a) 0.9-1? atm {h} 950. mm Hg {c} 5421o1't' {d} .93.? lsPn (e) 16.91 atm .S‘rt'otegi' and Explanation: Convert a series of pressure quantities into other pressure units. T_.'se Table 10.2 to design conversion factors to achieve the conversions. a 1 atm a {a} 120. inran >< q— = 0.94 r atm :60. mini-lg (b) 1.35 atmxm= 950.1111an 1 atm , 7’60 torr (c) :42 nmngx_—= 542 torr 360 nunl-Ig 1 rt 101.325 RP (a) T401111an >< i >< —a = 93.?- kPa ‘ 160. mmJ-Ig 1 atm (e) T00. kPa x A = 6.91atm 101.325 kPa v“ Reasonable Answer Check: The unit of atm represents a lot more pressure than the units okaa. torr. and mm Hg. so it makes sense that the numbers of atmospheres are always much smaller than the pressure expressed in these other units. The unit of kPa represents more pressure than the units of torr and mm Hg. so it makes sense that the numbers of kPa are always smaller than the pressure expressed in the other units of torr and nun Ha. Torr and nun H; are the same size. so their quantities should be identical. rJI 'J'l Answer: 21 mm Hg .S‘n'ottgr and £.'tp.-Tomtfon.' Given the balanced chemical equation for a chemical reaction. the mass of one reactant. and excess other reactant. determine the pressure of the product produced at a specified volume and temperature. Convert front grams to rnoles. Use the stoichiometric relationship from the balanced equation to determine moles of product. Use the ideal gas law to determine the pressure of the product. 0050 g 134me 1rnolB41-110 X 10 mol 1-110 _ —= 0.004? mol 1-110 33.32 g B_—H—_G 2 mol 341-110 ' T=30 3C' - 214.15 = 303 K " L anti 1 (0.004?- mol [—110le 0.08206 x603 K) nH , 0 RT ' 1,_ mol - K _.|' _ P— - = ' _ ' — 0.02 atm V (4.2: L) 260 mmHa 0.02? atmx—“= 2l 1111an 1 atm v’ Reasonable Answer Check: The relative quantities of B 4l-I10 and 1-130 seem sensible. All units cancel properly in the calculation of pressure. This is a relatively low pressure for water but the sample is also small. 1'5. Answer: (a) 1.98 atm [1)me = 0.438 arm, p3} = 0.182 atm,pA1. = 1.36 atm 8-1. 116. 1-10. Snotegv and Explanation: Given three containers of gases. with known volume and pressure. determine the total pressure and partial pressures of the gases when the three containers are opened to each other. L'se Boyle s law to determine how the pressure ofeach gas changes with the increase in total volume and then use Dalton's law to determine the total pressures after mixing. Use the definition of partial pressure to show that the pressures calculated are the partial pressure. The gas in one chamber is allowed to diffuse into all three chambers. so its final vohune increases to the total of the three volumes: V = 3.00 L — 2.00 — 5.00 L = 10.00 L TOT ..v . [1.45 atmlx(3.00 L) Bovle's law: Pf ,M = Lg“ Lg“ ForOs. Pfo. = P?” = ' I ' _ 0:138 atm ' I“ v1.1... - ' — vf 1.10.00 L} {0.903 atmill x (2.00- Lj [in atm-Il x (5.00 L) For 31;. Pig-1 =4: 0.182 atm For PflAl. =+=136 atm (10.00 L) (a) Pf_ .3. =pf_01 -pf_:‘-1 + pigr = 0.438 atm — 0.182 atm + 1.36 atm = 1.98 atm (10.00 L} (b) Partial pressure is the pressure each gas would cause if it were alone in the container. Pm” = p0. = 0.438 atm. Pix. = px. = 0.182 atm. Piss... = pAZ. =136 atm v“ Reasouobfe Airsu‘w' (knack: All the individual final pressures are less than the initial pressures. which makes sense because the volume is larger. The total pressure is a weighted average of the three pressures. and is influenced most by the gas present in largest quantity (Ar). i"ii351l'éFZ-"E'ipifl'flflfiflfl.' The behavior of real gases is discussed in Section 10.9. At low temperatures. the molecules are moving relatively slowly; however. when the pressure is very low. they are still quite far apart. As external pressures increases. the gas volume decreases. the slow molecules are squeezed closer together. and the attractions among the molecules get stronger. Figure 10.16 shows that a gas molecule strikes the walls of the container with less force due to the attractive forces between it and its neighbors. This makes the mathematical product PV smaller than the mathematical product nRT. Ami-tar: Box (13) Strategy and Basil-arteriole: The initial volume is 1.8 L and the final volume is 0.9 L. This 2:1 ratio in the gas volumes means for every two molecules of gas reactants there must be one molecule of gas products. The reactant count is six. so the box that has three product molecules fits these observations. The correct box is Box (b): 6 AB:(g-_l —'l" 3 211341;?) Answer: [a] More significant. because of more collisions 1111 More significant. because of more collisions tcl Less significant. because the molecules will move faster .S‘n'orrgr and Eipfdxrition: (a) ‘When the gas is compressed to a smaller volume. the molecules will be closer together and they will collide with each other more often. The temperature is fixed. so the molecules will hit each other at the same average speed; however. with more collisions the interactions between the molecules will be more significant. (b) V's-lien more molecules of the same gas are added to the container. the molecules will be closer together and they will collide with each other more often. The temperature is fixed. so the molecules will hit each other at the same average speed; however. with more collisions the interactions between the molecules will be more signficant. V's-lien the temperature is increased. the average kinetic energy of the molecules is increased and the molecules will move faster. That means they will collide with each other more often. Because the speed has increased the time these molecules spend in proximity to each other will decrease. so. compared to the situation in (a) and (hi. the effect of intermolecular interactions will less significant. Chapter 11: Liquids. Solids. and Materials 'JI on F0 The boiling point is when tlte yapor pressure of the liquid is equal to the atmospheric pressure acting on the liquid. When the atmospheric pressure is 1 atm. the boiling point is called the normal boiling point. (Section 11.2) Sublimation is the direct cotu‘ersion of a solid to a gas. {Section 11.3) The ttnit cell of a crystal is a small portion of the crystal that. when repeated along the directions defined by its edges. represents the entire structure of the crystal. Translation of the atom' s and-"or ion's position by one unit cell dintension in any direction along its coordinate axes generates the entire structure. (Section 11.6) .ltrsu'er: Substance D; it has lower vapor pressure at any given temperature than A. B._ or C. Strategy and Eftpfarmtr'ou: Vapor pressure increases as more molecules are able to escape the liquid state. so the substance with the greatest interntolecular attractive forces at a given temperature is the substance with the lowest vapor pressure. Looking at the graph at 25 3C1 C‘Lll‘R'E D has the lowest vapor presstu'e1 so substance D has the greatest intermolecular attractive forces. Answer: (a) molecular solid (b) ionic solid (c) metallic solid or network solid (d) amorphous solid Strategy and Eitp-Tonarfotr: Use the characterization conclusions as described in the solution to Question 56: (a) A solid that melts below 100 3C' and is insoluble in water is probably a nonpolar molecular solid. (b) An ionic solid will conduct electricity only when melted. (c) A solid that is insoluble iii water and conducts electricity is probably a metallic solid: though it might be t network solid like graphite. (d) A noncrystalline solid that has a wide melting point range is an amorphous solid. C'hapte: 12: l-".'.els. Organic C'hertticals. and Polymers 15. 16. In order to undergo addition polymerization. a molecule must have one or more double bonds. Condensation polymers are made from linking two different functional groups on the monomer molecules. Both ester and amide linkages contain a carbonyl [C'=O)._ since they are both made by reacting carboxylic acids All condensation polymers have water as a byproduct. '. (a) Neoprene is an example of a synthetic addition polymer. (b) Nylon is an example of a synthetic condensation polymer. (c) Polyisostyrene is a natural addition polymer. ...
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This note was uploaded on 06/03/2008 for the course CHEM 1A taught by Professor Nitsche during the Fall '08 term at Berkeley.

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MOOREHW3 - Chapter L0 Gases and The Atmosphere 1a Answer(a...

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