Practice MC Test Unit D (Ch 10) Gas Laws (pg 1 Of 10) - Ms. Campbell's .

AP Chem Practice MC Test Gas Laws 37. A ballon occupies a volume of 1.0 liter when it contains 0.16 g of helium at 37ºC and 1 atm pressure. If helium is added to the balloon until it contains 0.80 grams while pressure and temperature are kept constant, what will be the new volume of the balloon? a. 0.50 liter b. 1.0 liter c. 2.0 liters d. 4.0 liters e. 5.0 liters 38. A gaseous mixture of oxygen and nitrogen is maintained at a constant temperature. Which of the following MUST be true regarding the two gases? a. Their average kinetic energies will be the same. b. Their average molecular speeds will be the same. c. Their partial pressures will be the same. d. Their total masses will be the same e. Their densities will be the same. 39. An ideal gas fills a ballon at a temperature of 27ºC and 1 atm pressure. By what factor will the volume of the balloon change if the gas in the balloon is heated to 127ºC at constant pressure? 27 a. 127 3 b. 4 4 c. 3 2 d. 1 127 e. 27 40. Which of the following assumption(s) is (are) valid based on kinetic molecular theory. I. Gas molecules have negligible volume. II. Gas molecules have no attractive forces on each other. III. The temperature of a gas is directly proportional to its kinetic energy. a. I only b. III only c. I and III only d. II and III only e. I, II, and III page 5 of 5 41. A gas sample with a mass of 10 grams occupies 6.0 liters and exerts a pressure of 2.0 atm at a temperature of 26ºC. Which of the following expressions is equal to the molecular mass of the gas? a. b. c. d. e. (10 ) ( 0.0821) ( 299 ) gimol 1 ( 2.0 ) ( 6.0 ) ( 0.0821) ( 299 ) gimol 1 (10 ) ( 2.0 ) ( 6.0 ) ( 299 ) ( 2.0 ) ( 6.0 ) gimol 1 (10 ) ( 0.0821) (10 ) ( 2.0 ) ( 6.0 ) gimol 1 ( 299 ) ( 0.0821) ( 2.0 ) ( 6.0 ) gimol 1 (10 ) ( 299 ) ( 0.0821) 42. Chlorine gas and fluorine gas will combine to form one gaseous product. One L of Cl2 reacts with 3 L of F2 to produce 2 L of product. assuming constant temperature and pressure condition, what is the formula of the product? a. ClF b. Cl2F2 c. ClF2 d. Cl2F e. ClF3 43. A sample of helium gas is collected in a container. The helium gas effuses out through a tiny hole at a rate of 0.2 moles per minute. If a sample of methane is allowed to effuse under the same conditions, the rate of effusion of CH4 will be a. 0.1 moles per minute b. 0.14 moles per minute c. 0.2 moles per minute d. 0.4 moles per minute e. 0.8 moles per minute 44. A 1.1 L container will hold about 1.6 g of which of the following gases at 0ºC and 1 atm? a. SO2 b. O2 c. NH3 d. CH4 e. F2

AP Chem Practice MC Test 45. Gas Laws A(g) B(g) C(g) In a system where the reaction shown above is taking place, which change below increases the concentration of A(g)? I. addition of A(g) molecules at constant temperature and volume II. addition of A(g) molecules at constant temperature and pressure a. b. c. d. e. III. addition on A(g) molecules at constant volume and pressure I only III only II and III only I and III only I, II, and III 46. A 7 Liter container will hold about 12 g of which of the following gases at 0ºC and 1 atm? a. F2 b. Cl2 c. Br2 d. N2 e. NO page 6 of 6 47. What is the total volume of products formed at STP when 1.2 g of carbon is burned? a. 0.1 L b. 2 L c. 12 L d. 22 L e. 44 L 48. A sample of 18.0 g of aluminum metal is added to excess hydrochloric acid. The volume of hydrogen gas produced at 0.0ºC and 1 atm pressure is approximately a. 67 L b. 45 L c. 22 L d. 11 L e. 7 L 49. What mass of iron is required to react with 7.0 L of oxygen at 1 atm and 0ºC in order to produce Fe3O4 according to the balanced equation below? 3Fe 2O2 Fe3O4 a. b. c. d. 12 g 26 g 35 g 52 g

Practice Test D.1 (Ch 10) 1. a 2. e Gas Laws ANSWERS (pg 7 of 10) It is important to convert to moles to consider the situation presented. 14 g of N2 is 0.5 mol and 22 g of CO2 is 0.5 mol. You are meant to realize that two containers with the same number of moles at the same temperature and same pressure will have to be the same volume. PV PV This problem is a “before and after” problem, which means you need to break out the combined gas law. 1 1 2 2 . n1T1 n2T2 2.5 P 2 . With a bit of algebra, 400 300 you should realize you are looking for ¾ of 2.5, half of 2.5 is 1.25, and it must be larger than this, thus the only option is (e). In this problem, we need a molar mass to determine the molecular formula. Notice that the conditions in the problem, 0ºC 1.3g 22.4L and 1 atm are STP conditions, and this allows us to use 22.4 L/mol. , which means you have to do a roush 1L 1mol estimate of this calculation. To make the math easy, consider 22.4 to be 21, and ⅓ of 21 is 7, thus a MM of 29. Of the options given, the only formula with a molar mass of 29 is (b), C2H4 with a molar mass of 28. Gases follow the gas laws best (most mathematically correct) when they are not be pushed to turn in to a liquid. Gases will liquify at high pressure and low temperature, thus gases behave most ideally at lower pressure and higher temperatures. Set up a RICE box. When the temp and volume are constant, the pressure of a gas is proportional to its number of moles, so you can treat the pressure values, the same way you would treat mole values. Set up a rice box as shown, which allows you to calculate the pressure at the end using Dalton’s Law of Partial Pressures; Ptotal P1 P2 P3 . In the RCE box shown below, you can see the pressure of the nitrogen and oxygen given in the I row, and the problem tells us there is no product. The problem also tells us there is 0.75 of the product at the end. This lets us calculate how much oxygen was used up to 2O2 produce the 0.75 atm of product: 0.75atm 1.5atm for O 2 and how much nitrogen was used to produce the 0.75 1N 2O4 Reread the problem and realize that n and V remain constant, thus the problem reduces to 3. b 4. d 5. c atm of product: 0.75atm 6. 1N 2 0.75atm for N 2 . Put these in the change row and then do the subtraction to find the End 1N 2O4 row. R N2 2 O2 N2O4 I 1 1.5 0 C -0.75 -1.5 0.75 E 0.25 0 0.75 c The word “evacuated in the problem means emptied out. Thus there is no carbon dioxide at the start. The simple stoichiometry tells us 0.33 mol of calcium carbonate will form 0.33 mol of carbon dioxide. Then use the ideal gas law to 0.082atmiL ( 0.33mol ) ( 300K ) nRT moliK calculate pressure. P Substitute P This looks really bad to try and solve without V (1L ) a calculator. Sometimes you will have an easier time using fractions to find the common factors. Look what happens when 1 (0.082)(300) you change 0.33 to ⅓, and I will leave off all the units. P 3 1 7. a 8. c 9. 10. 11. 12. b a a d As the gases effuse out of the vessel, the gases that have a slowest velocity will leak out slower, thus carbon dioxide with the largest molar mass will go out the slowest, as the other gases move out faster, and the carbon dioxide’s moles proportional to the other gases will be higher than at the start. The other statements are flat out false. The should appear to be a “before and after” problem, thus use the combined gas law without n, since the container is sealed, P P 0.5 P and without volume since you are told that there is a constant volume. 1 2 Substitute 2 and solve. T1 T2 200 400 K ºC 273. atm is changed to mmHg by multiplying by 760. This is Boyle’s Law, part of the combined gas law. When you cool a substance down, the molecules slow down, and molecular motion would stop at absolute zero, 0 K or 273ºC. When this occurs, there is no way to slow the molecules further, thus there is no way to cool the temperature further, making absolute zero the coldest possible temperature.

AP Chem Practice Multiple Choice 13. b page 8 of 8 Whenever you need to compare the rate or ratio of the speed of one gas to another gas, you should use this equation rate1 MM 2 . While this equation is on the formula sheet, remember that you only get a periodic table for multiple rate2 MM 1 choice. Substitute 14. c Chapter 10 CH 4 64 and solve. SO2 16 Use the combined gas law with n removed because without any other information, you can assume the container is sealed. 1 V 1) ( 4L ) 3 2 P1V1 P2V2 ( Substitute with pressure and temp with values of 1 , though you may find it far less 1 T1 T2 (1) 2 complicated by picking the original pressure as 3 and temperature as 2 15. c Because of the combined gas law, and then solve. P1 P2 TP , which you can rearrange, 2 1 P2 , thus you may be tempted to say the T1 T2 T1 pressure would be triple, but you must change to Kelvin temperature 16. c ( 3) ( 4L ) (1) V2 (2) (1) ( 573) P1 ( 373) which does not triple the pressure. Rearrange the combined gas law and change to Kelvin, again assuming the container is sealed and moles remains constant. P1V1 P2V2 T P , thus V1 2 1 V2 T1 T2 T1 P2 17. d Use a RICE box to solve this problem. The volume is constant, and we can assume that at equilibrium (the end) that the temperature will return to the starting temperature, thus T is constant, which allows us to substitute with pressure values. R NH4NO2 N2 2 H2O I 6 0 0 C x x 2x E 6 x x 2x We know that the pressure at the ned must add up to 14. Thus (6 x) x x 14 and solve for x 4 but the question asks for the pressure of water vapor, which is 2x. PV PV 18. e Use the combined gas law 1 1 2 2 , remember to convert to Kelvin, and assume moles are constant. T1 T2 ( 400 ) ( 5 ) (V2 ) ( 700 ) (1000 ) ( 350 ) 19. d 20. d 21. a 22. c Rearrange and look for common factors The helium must be ¾ of the total pressure. Remember that the gas laws “work best” and a gas behaves most ideally when their molecules are far apart and moving very fast. Gases vary from the ideal when their molecules begin to slow down enough, lower temps, and get pushed close together enough, high pressure, that they start to feel their intermolecular attractions for one another. This makes them “sticky” to each other and eventually they will turn to a liquid. Even before condensation can occur, the gas law proportions will not mathematically work out so well. When temperature increases, kinetic energy increases, which in turn causes pressure to increase since the flask has constant volume. The number of moles is constant because the flask is sealed. We know that each gas causes a portion of the total pressure. Thus we can make a proportion using moles and pressures. ( 2000 ) ( 6 ) N Thus the total number of moles is 8, and n1 ntotal P n to solve for the total number of moles. total 1 ntotal total P1 Ptotal P1 (1500 ) ¼ of that is 2 moles of hydrogen with a molar mass of 2 g/mol. 23. b 24. b ( 350 ) ( 400 ) ( 5 ) V 2 ( 700 ) (1000 ) Using the combined gas law with moles and volume constant. ( 350 ) (1200 ) P and look for common P1 P2 Substitute 2 T1 T2 ( 300 ) factors. When give masses it is important to convert to moles so you can compare the quantities of gas molecules. 8 g of O2 is 0.25 mol and 7 g of N2 is also 0.25 mol, Thus at constant temperature and pressure, for equal moles of gas, the volume would remain constant.

AP Chem Practice Multiple Choice 25. d page 9 of 9 The mole fraction (symbolized N1) of a gas causes that same fraction of the total pressure. N1 Ptotal P1 The mole fraction of the nitrogen is 26. e Chapter 10 ( 0.4 ) ( 0.5 ) Thus (4) 1 8 (5) The is the same problem as the previous problem, solve for the mole fraction of nitrogen, easiest to divide the 5 into the 700 to get 140, then triple that to get 420. ( 3) 700 . You will likely find it (5) (1) 1.2 0.4 ( 3) 27. c The information tells us that the neon is ⅓ of the total amount of gas, thus 28. e Just like our lab work, the pressure for any gas collected over water is caused by the gas and the water vapor pressure. A simple subtraction allows the calculation of the gas itself. 781 23 758. The term, isothermally, means occurs at the same temperature. Thus this is a combined gas law problem with equal moles ( 6 ) ( 24 ) P PV and equal temperature. P1V1 P2V2 Solve for P2 and substitute and look for common factors. 1 1 P2 2 V2 ( 36 ) 29. c 30. c 31. c 32. d 33. e 34. a 35. e 36. d When you increase the temperature of a sealed rigid container, the mass nor the volume will change, thus the density will not change. Kinetic energy will increase, which in turn increased the pressure. Take the easy route by realizing this problem is at STP, and you know that the volume of a mole of any gas at STP is 22.4 L/ (14g ) 28g / mol mol. Thus 11.2 L is a ½ mol of gas. ( 0.5mol ) This is a stoichiometry problem that requires a balanced equation. Ethane is C2H6 thus C2H6 7/2O2 2CO2 3H2O 1 2CO2 Convert the ethane to moles, but keep it as a fraction, ⅕ Thus 22.4 While this math may not look so smooth, 5 1C2 H 6 consider the 22.4 to be 20, to get an answer of 8 and survey the answers to see that a value of 9 L compared to the rest of the choices, is close enough to select and move on. As you must know by now, molecular speed of gases is inversely proportional to their molar mass, thus list the gases from largest MM to smallest MM. Gas laws don’t work out as well when the molecules are pushed too close together, causing them “notice” each other and “feel the stickiness.” PV Use the ideal gas law and solve for n. n Substitute in the values given in the problem. RT rate1 rate2 MM 2 . While this equation is on the formula sheet, remember that you only get a periodic table for multiple MM 1 choice. Substitute what is given in the problem 37. e gas H2 2 , and look for a gas that would give 1/9th inside the square root. x This would of course be water with its molar mass of 18. Since the gas in this problem is all one gas, helium, the masses given will be proportional to the number of moles present. V V This allows us to use the combined gas law with pressure and temperature dropping out. 1 2 . Rearrange and substitute n1 n2 in the masses, since in this problem it is all the same gas, helium. like ½ but it actually solves to 1 (1) V ( 0.2 ) 2 which is n2V1 V2 n1 ( 0.8 ) (1) V ( 0.16 ) 2 If you are not careful, this looks 1 (1) V2 and this of course cleans up to 5 L. Again, if you are confused 2 10 0.8 (1) 4 as to why we can just enter the masses in for the moles, consider this: V2 and you can see the molar masses drop 0.16 4 38. a out of the equation. If substances are at the same temperature, they must have the same amount of kinetic energy.

AP Chem Practice Multiple Choice Chapter 10 page 10 of 10 39. c (The original answer sheet said the answer is b, it is actually c) This is a before and after problem. Use the combined gas law V1 V2 TV 400V1 and be sure and convert to Kelvin. Solve 2 1 V2 and substitute V2 T1 T2 T1 300 40. e These are the three most important assumptions of the KMT. 41. a You can revise the ideal gas law PV nRT substituting n 42. e 43. a 44. b 45. d 46. a 47. b 48. c 49. b m DRT , OR use the Molar Mass Kitty Kat! MM , but MM P m mRT remember that D , so you can substitute it back in MM . Then substitute in the data. V VP Since the reaction is at constant temp and pressure, you can write an equation using the volumes as moles. Knowing the stoichiometry of the reactants and products will tell you what the formula must be. Cl2 3F2 2 Cl?F? thus ClF3 rate1 MM 2 methane 4 1 Use and substitute , thus the methane is the rate of the helium, thus 0.1 moles/min rate2 MM 1 helium 16 2 1.6g 22.4L . 1.1L 1mol This may look “messy” but the 1.1 divide into the 22.4 20 times, and thus 1.6 20 is just over 30, thus oxygen. moles Remember that concentration is concentration . In option (I.) adding more molecules to a constant volume will Volume increase concentration as well as pressure. In option (II.) the container must have a flexible volume if the temp and pressure are to stay constant, thus the addition of more molecules, with a proportional increase in volume will mean that the concentration stays constant. And in option (III.) the addition of more molecules with a constant volume means that the concentration will go up, and since the pressure is being held constant, the temperature would be decreasing Don’t miss the fact that this problem is also at STP, thus 7 L is ⅓ of a liter. Calculate molar mass 12g MM 36g / mol ⅓mol 1mol 1C 22.4L Write a balanced equation: C O2 CO2 1.2g 2L 12g 1CO2 1mol Don’t miss the fact that this problem is at STP. To figure out which gas, you need to calculate a molar mass. Again, you must write a balanced equation: 2Al 6HCl 2AlCl3 3H2 Convert the 18 g of Al to moles 18g 2 27g / mol 3 2 3H 2 22.4L mol 22L 3 2Al 1mol Skip this problem. It will show up in the next unit. To convert an empirical formula to a molecular formula, you must know the molar mass of the molecular compound. The good news in this problem is that the gas is collected at standard conditions, 0ºC and 1 atm. At these conditions, 1 mole of any gas occupies 22.4 L. This allows the mass given to be converted to a molar 1.3g 22.4L 30g mass: If this math without a calculator is getting you down, remember that you only need an 1L 1mol 1mol approximate value, ⅓ of 22.4 is approximately 7, add that to 22.4 to get 30. When you survey the answers, it appears that two answers are two close to distinguish without a calculator, after all, b) C2H4 is only 28 g/mol and c) C2H6 is 30g/mol, so how should you easily distinguish without a calculator? Remember that molecular formulas are always whole number multiples of the empirical formula, thus c) is not even an option as it is not a multiple of the empirical formula.

Practice MC Test unit D (Ch 10)Gas Laws (pg 1 of 10) 8. When a sample of carbon dioxide gas in a closed container of constant volume at 0.5 atm and 200 K is heated until its temperature reaches 400 K, its new pressure is closest to a. 0.25 atm b. 0.50 atm c. 1.0 atm d. 1.5 atm e. 2.0 atm 9. Liquid nitrogen has a boiling point of -196ºC.