station #1 - uw madison astronomy department #1 . solve the following:

Station #1

Solve the following:

1. If a container has a volume of 14,810oz & is filled with 1.949lbs of O2, calculate the pressure of O2 in atm at 529.47 R.

a. 0.765 atm b. 1.53 atm c. 1.02 atm d. 2.30 atm e. None of the answers listed here f. All of the answers here

2. A fuel delivery system is filled with methane at 73.4oF, 100,413.075Pa, and then placed in a water bath at 212oF. What is the pressure of the system in atm?

a. 0.0781 atm b. 0.156 atm c. .313 atm d. 0.625 atm e. 1.25 atm f. None of the above

3. If the system in question #2 has a pressure relief valve set to open at 1.33322x105Pa, will the valve open based on your answer?

a. Yes b. No c. Maybe d. We’re not sure e. Possibly f. Google would know this

4. Given 491.67 R & 101325Pa, find the density in (g/L) of CO2. a. 1.96 g/L b. 3.92 g/L c. 7.84 g/L d. 0.980 g/L e. 0.490 g/L f. 1.47 g/L

5. With respect to question #4, how many molecules per liter is this? a. 1.34 x 1022 molecules CO2/L b. 5.36 x 1021 molecules CO2/L c. 2.68 x 1022 molecules CO2/L d. 1.34 x 1021 molecules CO2/L e. 5.36 x 1022 molecules CO2/L

f. 2.68 x 1021 molecules CO2/L 6. If the conditions for CO2 were changed to 68oF & 101.325kPa, what is the density in

(g/L) of CO2? a. 0.23 g/L b. 0.46 g/L c. 0.92 g/L d. 1.8 g/L e. 3.6 g/L f. 7.2 g/L

7. With respect to question #6, how many molecules per liter would this be? a. 7.8 x 1023 molecules CO2/L b. 1.6 x 1023 molecules CO2/L c. 3.1 x 1023 molecules CO2/L d. 6.2 x 1023 molecules CO2/L e. 1.3 x 1022 molecules CO2/L f. 2.5 x 1022 molecules CO2/L

8. A colorless liquid is isolated under the following conditions: measure the volume of container “J”; 212oF; 100,525Pa; flask + gas = 78.416g; flask = 77.834g. Calculate the molar mass of the liquid.

a. 42.2 g/mol b. 48.3 g/mol c. 84.4 g/mol d. 106 g/mol e. 127 g/mol f. 169 g/mol

9. At 509.67 R & 102,500Pa, the density of dry air is 2.7753304x10-3lbs/”G”. Measure the volume of container “G”. What is the average molar mass at these conditions?

a. 19 g/mol b. 29 g/mol c. 39 g/mol d. 49 g/mol e. 59 g/mol f. 69 g/mol

10. A plastic container holds 0.0771lbs of ethylene gas (C2H4) at a pressure of 105,724Pa. What is the pressure in torr if 0.011lbs of ethylene gas is removed at constant T?

a. 113 torr b. 136 torr c. 170 torr

d. 227 torr e. 340 torr f. 680 torr

11. If you have 6.999oz of chlorine tri-fluoride gas @ 699 mmHg & 113oF, what is the mass (in g) of the sample?

a. 0.169 g ClF3 b. 0.337 g ClF3 c. 0.539 g ClF3 d. 0.674 g ClF3 e. 0.843 g ClF3 f. 1.01 g ClF3

12. A sample of Freon-12 (CF2Cl2) occupies 862.26oz @ 536.67 R & 153.3kPa. Find its volume @ STP.

a. 35.3 L b. 22.4 L c. 17.7 L d. 11.2 L e. 53.0 L f. 70.6 L

13. Calculate the volume, in L, of a sample of carbon monoxide that is 6.8oF & 48,929.309Pa if it occupies 123.42oz @ 77oF & 99,325.16Pa.

a. 2.15 L b. 3.21 L c. 6.44 L d. 8.37 L e. 8.05 L f. 9.66 L

14. A sample of chlorine gas is confined in a 169.07oz container @ 30,397.5Pa & 540.27 R. How many moles of gas are in the sample?

a. 0.030 mol Cl2 b. 0.061 mol Cl2 c. 0.092 mol Cl2 d. 0.123 mol Cl2 e. 0.154 mol Cl2 f. 0.185 mol Cl2

15. If 1.47 x 10 -3 mol of Ar occupies container “I” (Measure the volume of container “I”), @ 78.8oF, what is the pressure in torr?

a. 366 torr

b. 732 torr c. 182 torr d. 1,460 torr e. 549 torr f. 641 torr

16. A sample of sulfur hexafluoride gas has a volume equivalent to that of container “L” @ 848.07 R. What temperature in oC is needed to reduce the volume equivalent to container “K”? (Measure the volumes of containers “L” & “K”).

a. -84 oC b. -66 oC c. -120 oC d. 0 oC e. -42 oC f. -96 oC

17. A 3144.71oz sample of dry air is cooled from 293oF to -7.6oF. What is the final volume in L?

a. 5.0 L b. 6.8 L c. 56 L d. 0.8 L e. 28 L f. 22.4 L

18. An automobile engine has a fuel/air mixture in a cylinder compressed from 101.325kPa to 962.588kPa. If the uncompressed volume is equivalent to the volume of container “M” what is the volume when fully compressed in mL? (Measure the volume of container “M”).

a. 39.5 mL b. 79.0 mL c. 158 mL d. 237 mL e. 316 mL f. 395 mL

19. An average adult has a lung capacity equivalent to the volume of container “H” of air with each breath. When the air is warmed from room temperature @ 527.67 R, to body temperature @ 558.27 R, what is the volume of the air exhaled in L? (Measure the volume of container “H”).

a. 0.089 L b. 0.11 L

c. 0.27 L d. 0.53 L e. 0.66 L f. 0.80 L

20. What does the inside pressure of an aerosol can become if initially the aerosol is 4.559625x105Pa & is then heated from 527.67 – 1571.67 R, in atm?

a. 3.0 atm b. 5.0 atm c. 7.0 atm d. 9.0 atm e. 11. atm f. 13. atm

21. If 213.029oz of He gas is stored at 77oF & 101.325kPa, what temperature is needed to maintain the He gas @ 67.628oz & 283.71kPa?

a. 260 K b. 520 K c. 780 K d. 1040 K e. 1300 K f. 1560 K

22. Using the standard molar volume of a gas @ STP, find out how many moles of air @ STP would be contained in a room measuring 4.11 m wide by 5.36 m long by 2.58 m high.

a. 6.35 x 104 mol b. 1.27 x 103 mol c. 2.54 x 103 mol d. 1.01 x 105 mol e. 1.52 x 105 mol f. 2.03 x 105 mol

23. How many moles of methane gas (CH4), are in 3.38140565 x 106 oz storage tank at STP? a. 2.23 x 103 mol b. 4.46 x 103 mol c. 6.69 x 103 mol d. 8.92 x 103 mol e. 1.12 x 104 mol f. 1.34 x 104 mol

24. With respect to question 23, how many grams of methane is this? a. 1.79 x 104 g CH4 b. 3.57 x 104 g CH4

c. 5.36 x 104 g CH4 d. 7.14 x 104 g CH4 e. 8.93 x 104 g CH4 f. 1.07 x 105 g CH4

25. With respect to question 23, how many grams of carbon dioxide could the same tank hold?

a. 3.92 x 106 g CO2 b. 4.90 x 106 g CO2 c. 6.53 x 106 g CO2 d. 9.80 x 106 g CO2 e. 1.31 x 105 g CO2 f. 1.96 x 105 g CO2

Station #2

Write out the formula for the following ions. 26.) Nitrate 27.) Acetate 28.) Sulfite 29.) Ammonium 30.) Hydroxide Write out the names of the following ions. 31.) SO42- 32.) PO43- 33.) NO2- 34.) HCO3- 35.) CO32-

Use the equation below to answer the following questions.

NaHCO3 + HCl NaCl + CO2 + H2O 2.50g 0.600M

36.) If you start with 2.50g of NaHCO3, how much HCl at 0.600M is required to react with all the NaHCO3? a.) 84.0 mL b.) 126 mL c.) 17.9 mL d.) 49.6 mL 37.) Based on 2.50g of NaHCO3, how many moles of reactant is this?

a.) 2.98 x 10-3 b.) 1.49 x 10-4

c.) 2.98 x 10-2

d.) 1.49 x 10-2 38.) What is the theoretical yield of NaCl, if you start with 2.50g NaHCO3?

a.) 0.696 g b.) 1.74 g

c.) 0.870 g

d.) 1.39 g 39.) In front of you is a vial labeled “A”. Use the balance to weigh the product. How much was produced?

a.) 1.35 g b.) 1.61 g

c.) 1.55 g d.) 1.47 g 40.) What is the percent yield based on the amount you weighed?

a.) 84.5 % b.) 92.5 %

c.) 89.1 % d.) 77.6%

Use the equation below to answer the following questions.

C6H5CHCHCO2H + Br2 CO2HCHBrCHBrC6H5

3.79g 0.500M

41.) If you start with 3.79g of C6H5CHCHCO2H, how much Br2 at 0.500M is required to react with all the C6H5CHCHCO2H?

a.) 41.2 mL b.) 51.2 mL

c.) 61.2 mL d.) .0512 mL 42.) Based on 3.79g of C6H5CHCHCO2H, how many moles of reactant is this?

a.) 1.28x 10-3 b.) 0.000256

c.) 2.56 x 10-2

d.) 1.28 x 10-2 43.) What is the theoretical yield of CO2HCHBrCHBrC6H5, if you start with 3.79g C6H5CHCHCO2H?

a.) 0.787 g b.) 1.74 g

c.) 0.870 g d.) 7.87 g

44.) In front of you is a vial labeled “B”. This is CO2HCHBrCHBrC6H5. Use the balance to weigh the product. How much was produced?

a.) 5.20 g b.) 2.50 g

c.) 1.55 g d.) 1.47 g 45.) What is the percent yield based on the amount you weighed?

a.) 88.5 % b.) 59.9 %

c.) 66.1 % d.) 77.6 % Please use the following equation(s) for questions 46 through 51.

BaCO3(s) + CuO(s) + Y2O3(s) YBa2Cu3O6.5(s) + CO2(g) YBa2Cu3O6.5(s) + O2(g) YBa2Cu3O7(s) 46.) Balance the equation by writing ONLY the coefficients on your answer sheet. 47.) If you have 1.60g of BaCO3, 2.50g CuO, and 5.25g Y2O3, which is limiting the production of YBa2Cu3O7 and how many moles of product would you get? 48.) What is the theoretical yield? A) 5.60 g

B) 16.40 g C) 2.70 g

D) 13.8 g E) 8.22 g F) 10.0 g

49.) How much CuO remains, in grams?

A) 9.27 g B) 6.10 g C) 4.88 g

D) 7.45 g E) 1.53 g F) 3.00 g

50.) What is the percent yield if the product produced is in the vial labeled “C”? 51.) What percentage of Y2O3 remains? A) 8.76 %

B) 17.5 % C) 26.3 %

D) 91.2 % E) 82.5 % F) 73.7 %

Please use the following equation(s) for questions 52 through 55.

K2Zr(C2O4)3(H2C2O4)·H2O was synthesized from ZrOCl2·8H2O with H2C2O4·2H2O & excess KOH. ZrOCl2·8H2O(s) + H2C2O4·2H2O(s) + KOH(aq) K2Zr(C2O4)3(H2C2O4)·H2O(s) + KCl(aq) + H2O(l) 52.) Write the coefficients which would balance the equation on your answer sheet. 53.) If vial “D” is the Zr compound, vial “E” is H2C2O4·2H2O, and you have 4.95g of KOH, which is limiting? A) ZrOCl2·8H2O(s)

B) H2C2O4·2H2O(s) C) KOH

D) K2Zr(C2O4)3(H2C2O4)·H2O(s) E) KCl

F) H2O 54.) What is the theoretical amount possible?

A) 12.0 g

B) 6.17 g C) 5.58 g

D) 6.00 g E) 2.79 g F) 3.09 g

55.) If the amount produced was measured as 12.0 grams, then what would be the percent yield? A) 215 %

B) 111 % C) 100 %

D) 108 % E) 50.0 % F) 55.4 %

Station #3

56. If a person is at 101,325Pa & has a body temperature of 558.27 R, how many moles of air (NO2) are in the lungs if the average lung capacity of an adult is 128.49oz?

A. 1.88 x 10-3 mol

B. 3.75 x 10-3 mol

C. 7.50 x 10-2 mol

D. 1.50 x 10-2 mol

E. 3.00 x 10-1 mol

F. 6.01 x 10-1 mol

57. Methane is sold in cylinders which have a volume of 1.481x103oz & contain 12.20lbs of the gas. What is the pressure in the cylinder if the temperature is 68oF?

A. 190 atm

B. 265 atm

C. 200 atm

D. 450 atm

E. 366 atm

F. 509 atm

58. Which sample contains more molecules: 67.63oz of Cl2 at STP or 101.44oz of CH4 at 540.00 R & 153,320Pa?

A. 5.0 L of Cl2

B. 2.5 L of Cl2

C. 2.0 L of Cl2

D. 3.0 L of CH4

E. 4.5 L of CH4

F. 6.0 L of CH4

59. Calculate the volume of 1.8502x10-2lbs N2 at 671.67 R and 106657Pa.

A. 1.09 L

B. 2.18 L

C. 4.37 L

D. 8.73 L

E. 10.9 L

60. What volume will 2.678x10-3lbs of SO2 gas occupy at 64.4oF & 100658Pa?

A. 41.61 L

B. 22.5 L

C. 0.914 L

D. 1.37 L

E. 2.29 L

F. 0.457 L

61. Calculate the volume of 0.030837lbs of nitrous oxide (N2O) at STP.

A. 3.6 L

B. 7.2 L

C. 10.8 L

D. 14.4 L

E. 18.0 L

F. 21.6 L

62. An electronic vacuum tube was sealed off during manufacturing at a pressure of 0.0023998Pa at 80.6oF. Its volume is 100,000mm3. How many gas molecules remain in the tube?

A. 4.64 x 1014 molecules

B. 2.32 x 1014 molecules

C. 1.16 x 1014 molecules

D. 1.45 x 1013 molecules

E. 5.8 x 1013 molecules

F. 2.9 x 1013 molecules

63. What volume would 5.507x10-2lbs of Ar occupy at 653.67 R & 97,992Pa?

A. 14.4 L

B. 18.0 L

C. 19.3 L

D. 22.4 L

E. 24.6 L

F. 32.0 L

64. How much volume would the gas from question 63 occupy if Ar were at STP?

A. 32.0 L

B. 24.6 L

C. 22.4 L

D. 19.3 L

E. 18.0 L

F. 14.4 L

65. An iron meteorite was analyzed for its isotopic argon content. The amount of 36Ar was 1.22x10-5in3 (@ STP) per kg of meteorite. If each 36Ar atom had been formed by a single cosmic event, how many such events must have occurred per kg of meteorite?

A. 1.08 x 1016 cosmic events

B. 2.16 x 1016 cosmic events

C. 4.32 x 1016 cosmic events

D. 2.7 x 1015 cosmic events

E. 5.4 x 1015 cosmic events

F. 1.4 x 1015 cosmic events

66. The reaction is as follows : A + B C

Was studied kinetically and the below data was obtained.

Experiment A B Rate (Mol/gal∙min) 1 1.0 M 1.0 M 0.0397 2 2.0 M 1.0 M 0.0794 3 1.0 M 2.0M 0.0397

Determine the rate expression. (Hint: convert Mol/gal∙min to Mol/L∙min).

A. rate = k [A]

B. rate = k [A][B]2

C. rate = k [A]2[B]

D. rate = k [A][B]0

E. Both B & C

F. Both A & D

67. Two reactants were combined to yield the following reaction : A + B products

Experiment A B Rate (M2/gal2∙sec) 1 1.0 1.0 0.0035 2 1.0 2.0 0.0070 3 3.0 1.0 0.0315

Which of the following best describe the rate of reaction? (Hint: convert M2/gal2∙sec to M2/L2∙sec).

A. rate = k [A][B]0

B. rate = k [A]0[B]

C. rate = k [A][B]2

D. rate = k [A]2[B]

E. rate = k [A]2[B]3

F. rate = k [A]3[B]2

68. For the reaction : 2NO(g) + H2(g) N2O(g) + H2O(g) occurs @ 1,520.33oF generates data in the following table.

PNO (Pa) PH2 (Pa) Rate of Pdecrease (Pa/min) 15,199 40,530 2,027 7,599 40,530 507

15,199 20,265 1,013

Determine k. (Hint: convert data to atm).

A. 0.417 atm-2∙min-1

B. 2.22 atm-2∙min-1

C. 0.0667 atm-2∙min-1

D. 1.67 atm-2∙min-1

E. 4.44 atm-2∙min-1

F. None of the above

69. If an “A” molecule reacts with a “B” molecule to yield AB2, what would the rate law for this reaction be?

A. rate = k [A][B]

B. rate = k [A]2[B]

C. rate = k [A][B]2

D. rate = k [A]2[B]3

E. rate = k [A]3[B]4

F. rate = k [A][B]3

70. With respect to question #69, if the initial rate of formation of AB2 is 2.0 x 10-5 M/s and the initial [A] & [B]’s are 0.30M, what would the value of the rate constant be?

A. 2.22 x 10-4 M-2∙s-1

B. 4.44 x 10-4 M-2∙s-1

C. 6.66 x 10-4 M-2∙s-1

D. 7.41 x 10-4 M-2∙s-1

E. 8.82 x 10-4 M-2∙s-1

F. 10.1 x 10-3 M-2∙s-1

71. “A” reacts with “B” to yield “C”. The rate constant k = 2.00 x 10-3 M-1s-1. If 0.500 mole of “A” and 0.300 mole “B” are placed in container “F”, what is the initial rate of reaction? (Measure the volume of container “F”).

A. rate = 1.20 x 10-3 M/s

B. rate = 2.00 x 10-3 M/s

C. rate = 4.00 x 10-3 M/s

D. rate = 6.67 x 10-3 M/s

E. rate = 0.250 x 10-3 M/s

F. rate = 0.150 x 10-3 M/s

72. The following reaction was studied : X(g) + Y(g) Z(g) @ 980.33oF & the data in the table were observed.

[X] (M) [Y] (M) [Z] (M/min) 0.10 0.10 0.030 0.20 0.20 0.240 0.20 0.10 0.120

What is the rate law for this reaction?

A. rate = k [X][Y]

B. rate = k [X]2[Y]

C. rate = k [X][Y]2

D. rate = k [X]2[Y]3

E. rate = k [X]3[Y]4

F. rate = k [X][Y]3

73. With respect to question #72, what is the numerical value of k?

A. 10 M-2∙min-1

B. 15 M-2∙min-1

C. 20 M-2∙min-1

D. 25 M-2∙min-1

E. 30 M-2∙min-1

F. 35 M-2∙min-1

74. With respect to question #72, what would be the initial rate of “Z” formation if you had started with 0.15M “X” & 0.15M “Y”?

A. 0.404 M/min

B. 0.303 M/min

C. 0.202 M/min

D. 0.101 M/min

E. 0.0505 M/min

F. 0.02525 M/min

75. With respect to question #72, how would the rate change if, after the reaction had just begun, the volume of the container were abruptly doubled?

A. rate = 2x initial rate

B. rate = ½ initial rate

C. rate = 4x initial rate

D. rate = ¼ initial rate

E. rate = 8x initial rate

F. rate = 1/8 initial rate

Station #4

Balance the following chemical equations 76. ___ Cu(s) + ___ S8(s) ___ Cu2S(s) 77. ___ C8H18(l) + ___ O2(g) ___ CO2(g) + ___ H2O(l) 78. ___ I-(aq) + ___ O2(g) + ___ H2O(l) ___ I2(s) + ___ OH-(aq)

79. __ KMnO4(aq) + __ NaNO2(aq) + __ H2O(l) __ MnO2(s) + __ NaNO3(aq) +

__ KOH(aq)

80. ___ KOH(aq) + ___ H2O2(aq) + ___ Cr(OH)3(s) ___ K2CrO4(aq) + ___ H2O(l)

81. __ Mn2+(aq) +__ Br2(l) + __H2O(l) __MnO4-(aq) + __Br-(aq) + __H+(aq)

82. __Cr2O72-(aq) + __Cd(s) + __H+(aq) __Cr3+(aq) + __Cd2+(aq) + __H2O(l)

83. __AgCl(s) + __NO(g) + __H2O(l) __Ag(s) + __Cl-(aq) +__NO3-(aq) + __H+(aq)

84. __H2O2(aq) + __Ni2+(aq) __H+(aq) + __O2(g) + __Ni(s)

85. __Cu+(aq) +__PbO2(s) +__H+(aq) +__SO42-(aq) __Cu2+(aq) +__PbSO4(s) +__H2O(l)

86. __HClO4 + __P4O10 __H3PO4 + __Cl2O7

87.__C6H5Cl + __SiCl4 + __Na __(C6H5)4Si + __NaCl

88.__K4Fe(CN)6 + __H2SO4 + __H2O __K2SO4 + __FeSO4 + __(NH4)2SO4 + __CO

89.__H3PO4 + __(NH4)2MoO4 + __HNO3 __(NH4)3PO4∙12MoO3 + __NH4NO3 + __H2O

90.__CuSCN + __KIO3 + __HCl __CuSO4 + __KCl + __HCN + __ICl + __H2O

91.__KMnO4 + __FeSO4 + __H2SO4 __MnSO4 + __K2SO4 + __Fe2(SO4)3 + __H2O

92.__H3PO4 + __Mg(OH)2 __Mg3(PO4)2 + __H2O

93.__Cu2S + __HNO3 __Cu(NO3)2 + __CuSO4 + __NO2 + __H2O

94.__Al2(SO4)3 + __Ca(OH)2 __CaSO4 + __Al(OH)3

95.__Ca3(PO4)2 + __H2SO4 __CaSO4 + __H3PO4

96.__FeSO4 + __K3[Fe(CN)6] __Fe3[Fe(CN)6]2 + __K2SO4

97.__CaHPO4∙2H2O + __NaOH + __H2O __Na2HPO4∙12H2O + __Ca(OH)2

98.__Ca3(PO4)2 + __SiO2 + __C __CaSiO3 + __P4 + __ CO

99.__V2O5 + __Al __Al2O3 + __V

100.__MgNH4PO4 __Mg2P2O7 + __NH3 + __H2O

101.__VO2Cl + __NH4OH __NH4VO3 + __NH4Cl + __H2O

102.__Na2SnO3 + __ H2S __SnS2 + __NaOH + __H2O

103.__Ca(OH)2 + __P4O10 + __H2O __Ca(H2PO4)2

104.__H2S + __Cl2 __S8 + __HCl

105.__C10H16 + __Cl2 __C + __HCl

106.__Ca3P2 + __H2O __Ca(OH)2 + __PH3

107.__V2O5 + __HCl VOCl3 + __H2O

108.__K4Fe(CN)6 + __KMnO4 + __H2SO4 __KHSO4 + __Fe2(SO4)3 + __MnSO4 + __HNO3 + __CO2 + __H2O

109.__K3[Fe(SCN)6] + __Na2Cr2O7 + __H2SO4 __Fe(NO3)3 + Cr2(SO4)3 + __CO2 + __H2O + __Na2SO4 + __KNO3

110.__K4[Fe(SCN)6] + __K2Cr2O7 + __H2SO4 __Fe2(SO4)3 + __Cr2(SO4)3 + __CO2 + __H2O + __K2SO4 + __KNO3

Station #5

111. Which of the postulates of kinetic molecular theory can be used to justify Dalton’s Law of Partial Pressures?

a. Postulate 1: Particle Volume

b. Postulate 2: Particle Motion

c. Postulate 3: Particle Collisions

d. Postulate 1 & no exertion of force on varied components

e. Postulate 2 & no exertion of force on varied components

f. Postulate 3 & no exertion of force on varied components

112. Why are spark plugs not necessary in a diesel engine, based on kinetic molecular theory?

a. The volume of the cylinders is different from gasoline engines

b. It’s the kinetics

c. The type of fuel is different

d. The shape of the cylinders is different from gasoline engines

e. The fuel gets heated above the ignition temperature of the gas

f. The gas gets heated above the ignition temperature of the fuel

113. When a tire is pumped up rapidly, its temperature rises. Would this effect be expected if air were an ideal gas, even though it is not?

a. Ideally, yes

b. Energy expended on the gas is converted to static energy, so there is minimal temperature rise

c. Energy expended on the tire in turn acts on the gas, so yes

d. Energy is transferred to gas molecules as kinetic energy, so heating is possible

e. Both a & d

f. We are sure glad we brought our calculators

114. If a real gas is changed from its initial state to its final state by adiabatic expansion, why can Boyle’s law not be used to calculate the volume?

a. The temperature would be raised by the expansion

b. The temperature would be lowered by the expansion

c. Boyle’s law only deals with pressure and volume

d. Boyle’s law application suggests temperature must be constant

e. Both b & d

f. b, c, & d

115. Which of the following represent vector quantities with respect to kinetic molecular theory? Circle or shade in (however you have answered other questions on your answer sheet) all that apply!

a. force

b. momentum

c. energy

d. speed

e. velocity

f. temperature

g. pressure

116. Is the rate of effusion of a gas when compared to its rate of diffusion,

a. Higher

b. Lower

c. Same

d. Differ in sign but quantity is the same

e. Too bad we cannot use google

f. Wiki-leaks would know this

117. If two gases have molecules of approximately equal size, and if the ratio of effusion is compared to the ratio of diffusion, is the ratio

a. Higher

b. Lower

c. Same

d. Good thing we brought our calculators

e. Is there a computer in the lab which might know

f. Maybe someone else knows this

118. Under standard temperature and pressure conditions, what would the ratio of relative rates at which inert gases would diffuse through a common orifice. Compare Ar, He, & Kr. [Hint: Use Graham’s Law of Diffusion where rate α 1/(mass)^(1/2)]

a. 0.9492 : 2.999 : 0.6552

b. 0.7910 : 2.499 : 0.5460

c. 0.6328 : 1.999 : 0.4368

d. 0.4746 : 1.499 : 0.3276

e. 0.3164 : 1.000 : 0.2184

f. 0.1582 : 0.4998 : 0.1092

Given the following data for a series of experiments between O2 & NO, answer the proceeding questions.

Experiment Init. [O2] (mol/L) Init. [NO] (mol/L) Initial Rate (mol/L∙s) 1 1.10 x 10-2 1.30 x 10-2 3.21 x 10-3

2 2.20 x 10-2 1.30 x 10-2 6.40 x 10-3

3 1.10 x 10-2 2.60 x 10-2 12.8 x 10-3 4 3.30 x 10-2 1.30 x 10-2 9.60 x 10-3 5 1.10 x 10-2 3.90 x 10-2 28.8 x 10-3

119. What order is the reaction with respect to NO?

120. What is the order with respect to O2?

121. From the above data determine the rate expression.

Given the following reaction and experimental data, answer the proceeding questions.

NO2(g) + CO(g) NO(g) + CO2(g)

Experiment Init. Rate (mol/L∙s) Initial [NO2] (mol/L) Initial [CO] (mol/L) 1 0.0050 0.10 0.10 2 0.080 0.40 0.10 3 0.0050 0.10 0.20

122. What is the reaction order with respect to NO2?

123. What is the reaction order with respect to CO?

124. What is the overall reaction order?

125. What is the rate expression for this reaction?


4A(g) + 3B(g) 2C(g)

- The following data were obtained at constant T.

Experiment Init. [A] (mol/L) Init. [B] (mol/L) Initial Rate (mol/L∙min)

1 0.100 0.100 5.00 2 0.300 0.100 45.0 3 0.100 0.200 10.0 4 0.300 0.200 90.0

126. What is the order with respect to A?

127. What is the order with respect to B?

128. Write the rate expression.

129. Calculate k.


A(g) + B(g) + C(g) D(g)

- The following data were obtained at constant T.

Experiment Init [A] (mol/L) Init [B] (mol/L) Init [C] (mol/L) Init Rate (mol/L∙s)

1 0.0500 0.0500 0.0100 6.25 x 10-3

2 0.1000 0.0500 0.0100 1.25 x 10-2

3 0.1000 0.1000 0.0100 5.00 x 10-2

4 0.0500 0.0500 0.0200 6.25 x 10-3

130. What is the reaction order with respect to A?

131. What is the reaction order with respect to B?

132. What is the reaction order with respect to C?

133. Write the rate law.

134. Calculate k.

Answer Sheet

School Name

Team Number

Team Member

Team Member

KEY

1 A B C D E F

2 A B C D E F

3 A B C D E F

4 A B C D E F

5 A B C D E F

6 A B C D E F

7 A B C D E F

8 A B C D E F

9 A B C D E F

10 A B C D E F

11 A B C D E F

12 A B C D E F

13 A B C D E F

14 A B C D E F

15 A B C D E F

16 A B C D E F

17 A B C D E F

18 A B C D E F

19 A B C D E F

20 A B C D E F

21 A B C D E F

22 A B C D E F

23 A B C D E F

24 A B C D E F

25 A B C D E F

Station #1

Total Points (5 each) = 125 points

26 52

27 53 A B C D E F

28 54 A B C D E F

29 55 A B C D E F

30

31

32

33

34

35

36 A B C D E F

37 A B C D E F

38 A B C D E F

39 A B C D E F

40 A B C D E F

41 A B C D E F

42 A B C D E F

43 A B C D E F

44 A B C D E F

45 A B C D E F

46

47 BaCO3 0.00405 moles // 4.05 x 10-3 moles

48 A B C D E F

49 A B C D E F

50

51 A B C D E F

Station #2

CH3COO-

NO3-

SO32-

NH4+

63.00%

1, 4, 4, 1, 2, 1, 4

sulfate

phosphate

nitrite

bicarbonate

carbonate

8, 12, 2, 4, 8, 4, 1, 4 // 4, 6, 1, 2, 4, 2, 1/2, 2

OH-

Total points (5 each) = 150 points

56 A B C D E F

57 A B C D E F

58 A B C D E F

59 A B C D E F

60 A B C D E F

61 A B C D E F

62 A B C D E F

63 A B C D E F

64 A B C D E F

65 A B C D E F

66 A B C D E F

67 A B C D E F

68 A B C D E F

69 A B C D E F

70 A B C D E F

71 A B C D E F

72 A B C D E F

73 A B C D E F

74 A B C D E F

75 A B C D E F

Station #3


Points

76 16 1 8 377 2 25 16 18 478 4 1 2 2 4 579 2 3 1 2 3 2 680 4 3 2 2 8 581 2 5 8 2 10 16 682 1 3 14 2 3 7 683 3 1 2 3 3 1 4 784 1 1 2 1 1 585 2 1 4 1 2 1 2 786 12 1 4 6 487 4 1 8 1 8 588 1 6 6 2 13 6 689 1 12 21 1 21 12 690 4 7 14 4 7 4 7 5 891 2 10 8 2 1 5 8 792 2 3 1 6 493 1 12 1 1 10 6 694 1 3 3 2 495 1 3 3 2 496 3 2 1 3 497 1 2 10 1 1 598 2 6 10 6 1 10 699 3 10 5 6 4

100 2 1 2 1 4101 1 2 1 1 1 5102 1 2 1 2 1 5103 2 1 2 2 4104 8 8 1 16 4105 1 8 10 16 4106 1 6 3 2 4107 1 6 2 3 4108 10 122 299 162 5 122 60 60 188 9109 1 16 58 1 16 6 58 16 3 9110 6 97 355 3 97 36 355 91 36 9

188

Station #4* Write ONLY the coefficients which balance the equation

Total points

111 A B C D E F

112 A B C D E F

113 A B C D E F

114 A B C D E F

115 A B C D E F

116 A B C D E F

117 A B C D E F

118 A B C D E F

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134 k = 50.0 L2/mol2∙sec

second order overall

Rate = k [ NO2 ]2

second order

Station #5

second order

first order

Rate = k [ O2 ] [ NO ]2

second order

zero order


first order

Rate = k [ A ]2 [ B ]

first order

second order

zero order

Rate = k [ A ] [ B ]2 [ C ]0 // Rate = k [ A ] [ B ]2

k = 5.00 x 103 L2/mol2∙min

References

Temperature Pressure

o F = (9/5) o C + 32 1 atm = 760mmHg = 760 torr

o C = (5/9) (o F – 32) 1 atm = 1.01325 bar = 101325 Pa = 101.325 kPa

K = o C + 273.15 1 atm = 14.7 psi

R = [(K - 273.15) (1.8000)] + 491.67 STP = 1 atm & 0 o C

STP = 22.4 L for 6.022 x 1023 particles/atoms/molecules

Miscellaneous

Mass Avogadro’s Number

1 lb = 454 g 1 mole = 6.022 x 1023 particles/atoms/molecules

Length

1 in = 2.54 cm

Volume

1 oz = 29.5735 mL 1 cm3 = 1,000 mm3

1 cm3 = 1 mL

1 in3 = 16.387 cm3

1 gal = 3.78 L

K.M.T (Kinetic Molecular Theory) – Postulates

Postulate 1 (Particle volume) – A gas consists of a large collection of individual particles. The volume of an individual particle is extremely small compared with the volume of the container. In essence, the model pictures gas particles as having mass but no volume.

Postulate 2 (Particle Motion) – Gas particles are in constant, random, straight-line motion, except when they collide with the container walls or with each other.

Postulate 3 (Particle Collisions) – Collisions are elastic, which means that, somewhat like minute billiard balls, the colliding molecules exchange energy but they do not lose any energy through friction. Thus, their total kinetic energy (Ek) is constant. Between collisions, the molecules do not influence each other at all.i

station #1 - uw madison astronomy department #1 . solve the following:

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