last name first name signature
1
2
3 4 5 6 7 8 9 10 11 12
13 14 15 16 17
18
1 2
3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
1.008 4.003
6.941 9.012 10.81 12.01 14.01 16.00 19.00 20.18
22.99 24.31 26.98 28.09 30.97 32.07 35.45 39.95
39.10 40.08 44.96 47.87 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 72.64 74.92 78.96 79.90 83.80
85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.76 127.60 126.90 131.29
132.91 137.33 138.91 178.49 180.95 183.84 186.21 190.23 192.22 195.08 196.97 200.59 204.38 207.20 208.98 (209) (210) (222)
(223) (226) (227) (267) (268) (269) (270) (270) (278) (281) (282) (285) (286) (289) (290) (293) (294) (294)
H He
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og
58 59 60 61 62 63 64 65 66 67 68 69 70 71
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
140.12 140.91 144.24 (145) 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97
232.04 231.04 238.03 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (266)
constants
R = 0.08206 L atm/mol K R = 8.314 J/mol K NA= 6.022× 1023/mol h = 6.626× 10−34 J·s c = 3.00× 108 m/s g = 9.81 m/s2
conversions
1 atm = 760 torr 1 atm = 101325 Pa 1 atm = 1.01325 bar 1 bar = 105Pa
◦F =◦C(1.8) + 32 K =◦C + 273.15
conversions
1 in = 2.54 cm 1 ft = 12 in 1 yd = 3 ft 1 mi = 5280 ft 1 lb = 453.6 g 1 ton = 2000 lbs 1 tonne = 1000 kg 1 gal = 3.785 L 1 gal = 231 in3 1 gal = 128 fl oz 1 fl oz = 29.57 mL
water data
Cs,ice= 2.09 J/g◦C Cs,water= 4.184 J/g◦C Cs,steam= 2.03 J/g◦C ρwater= 1.00 g/mL ρice= 0.9167 g/mL ρseawater= 1.024 g/mL
∆Hfus= 334 J/g
∆Hvap= 2260 J/g Kw= 1.0× 10−14
This exam should have exactly 20 questions. Each question is equally weighted at 5 points each. You will enter your answer choices on the virtual bubblehseet after you have finished. Your score is based on what you submit on the vir- tual bubblesheet and not what is circled on the exam.
1.
Three gases (H2, O2 and Ar) are each in their own separate containers. Each container is at the same temperature and pressure. Which container of gas has the highest number of particles per liter? Which gas has the most energy?a. Ar; Ar
b. All are the same; H2
c. H2; all are the same d. Ar; all are the same e. O2; O2
f. H2; H2
●
g. All are the same; All are the same h. All are the same; ArExplanation: In an ideal gas several assumptions have to be made. Most importantly, the identity of the gas does not matter and all gasses are treated the same.
The number of particles are constant because they take up no volume in each of these containers. Also the energy is directly dependent on temperature of the gases and in this case the temperature is the same therefore the energy is the same.
2.
Which of the following gases do you predict to have the largest value of the Van der Waals coefficient b?a. C2H6
b. H2
●
c. C4H10 d. CH4e. C3H8
Explanation: C4H10 is the largest molecule and therefore will need the largest correction due to volume of the gas.
3.
Consider the following combustion reaction:C5H12+ 8O2−→ 5CO2+ 6H2O
How many mols of CO2 are produced if 13 mols of pentane are reacted with excess oxygen with a 91%
yield?
a. 65 mols
●
b. 59 mols c. 71 mols d. 12 mols e. 44 molsExplanation: Moles of CO2 is 5 times the pentane or 65 mole if 100% complete. Percent yield scales that down (91% of 65) to give 59 mols CO2 of actual yield.
4.
Properly balance the following chemical equation:KI + Pb(NO3)2−→ KNO3 + PbI2
What is the sum of the coefficients after balancing?
a. 4 b. 9
●
c. 6 d. 3 e. 10Explanation: The coefficients are as follows:
2KI + Pb(NO3)2−→ 2KNO3+ PbI2
Therefore, when summing them: 2 + 1 + 2 + 1 = 6
5.
What is the molar mass of C18H11Br?a. 109 g/mol b. 216 g/mol
6.
There is a gas canister with 4 mols of He gas. If a valve is opened and 2 mols of He gas escape will the pressure increase or decrease and by what factor?●
a. decrease, 2× b. increase, 2× c. increase, 4× d. decrease, 4× e. decrease, 1× Explanation: Pn11 = Pn2
2 ; P41 = P22 ; P21 = P2 ; The pressure decreases by a factor of 2.
7.
The average speed of a gas at 300 K is 410 m/s.What will the speed be of that gas when heated up to 675 K ?
a. 525 m/s b. 923 m/s c. 273 m/s d. 182 m/s
●
e. 615 m/sExplanation: Use the following ratio: vv2
1 =
√T2 T1, so
v2
400 =
√675
300 = 1.5 v2= 400(1.5) = 675
8.
What is the percent composition of earths air?●
a. 78% N2, 21% O2, 1% Ar b. 78% N2, 21% O2, 1% CO2c. 70% N2, 20% O2, 10% CO2
d. 70% N2, 30% O2
e. 70% N2, 20% O2, 10% Ar
Explanation: This answer can be found in the class notes.
9.
Assume all four of the gases below are at the same temperature. Now rank them in order of decreasing average velocities.a. CO2> O2> NO2> H2
b. H2> CO2> O2> NO2
●
c. H2> O2> CO2> NO2d. CO2> NO2> H2> O2
e. H2> O2> NO2> CO2
Explanation: Kinetic speed is Vrms is proportional to
√1
M. Least mass is the fastest and most mass is slowest.
10.
How many grams of NaCl (58.44 g/mol) do you need to dissolve into 3.80 liters of water too make a solution that has a concentration of 0.0500 M ? a. 4.42 gb. 5.23 g
●
c. 11.1 g d. 0.772 g e. 22.3 gExplanation: Molarity is defined as moles/liter.
0.0500 mol/L× 3.8 L/gal × 58.44 g/mol = 11.1 g
11.
A gas mixture is made up of 5 moles of H2, 3 moles of N2, and 9 moles of O2. The partial pressure of the nitrogen is known to be 108 Torr. What is the total pressure of this mixture?●
a. 612 Torr b. 720 Torr c. 180 Torr d. 324 Torr e. 432 TorrExplanation: The mole fraction of nitrogen is 3/17.
Divide the partial of nitrogen by its mole fraction:
108/(3/17) = 612 Torr.
12.
An series of ideal gas law problems are all given with volume given in gallons, and pressure in torr.Temperature and moles remain in Kelvin and moles.
What would the working value of R be using these units? Specifically, what is R in galmol·Torr·K ?
a. 62.36 b. 82.06
●
c. 16.48 d. 0.08206 e. 236.1 Explanation:(0.08206 L· atm mol· K
) ( 1 gal 3.785 L
) (760 Torr 1 atm
)
= 16.48
13.
Properly balance the following chemical equation:2C6H14+ 19O2−→ 12CO2+ 14H2O Now we react 2.74 mols of C6H14 with 11.02 mol of oxygen. Assuming the reaction goes to completion, how many moles of H2O are produced ?
a. 14.0 mol b. 8.73 mol c. 19.18 mol d. 11.12 mol
●
e. 8.12 molExplanation: You can run 1.37 mol of rxn with 2.74 mol of hexane. You can run 0.58 mol of rxn with 11.02 mol oxygen. The oxygen is limiting then. Take the 0.58 mol rxn and multiply by 14 mol water per rxn and get 8.12 mol water.
14.
What is the density of oxygen gas at STP?a. 0.058 g/L
●
b. 1.43 g/L15.
There are two tanks of gas connected by a valve.Gas A is in a 10 L tank with a pressure of 6 atm. Gas B is in a 3 L tank with a pressure of 12 atm. What is partial pressure of gas A after the valve is opened and the two gases have been allowed to mix?
●
a. 4.61 atm b. 6 atm c. 15.23 atm d. 2.76 atm e. 12 atmExplanation: Use the volumes to determine the fraction of the total container each gas will have and multiply the fraction by the pressure associated with the gas. The new pressure will represent the partial pressure gas A has in the container.
16.
Consider a Maxwell-Boltzmann distribution plot of gas velocities vs number of particles (the classic plot).Assuming all the gases listed (Noble Gases) are at the same temperature, which one will have the broadest distribution of velocities in a given container?
a. Ar b. Kr c. Xe d. Rn
●
e. NeExplanation: The lightest gas will have the greatest range of velocities - the distribution is broader. The lightest gas listed is neon, Ne.
17.
Define the Van der Waals coefficient b.a. corrective factor to adjust for attractive sources between the molecules
●
b. corrective factor to adjust the gas particle volume c. corrective factor to adjust for inelastic collisions d. corrective factor to adjust for the temperature of the18.
What is the identity of a gas if 57 g of it has a pressure of 518 torr, a volume of 50 L, and a temperature of 277 K?a. CO2
b. N2 c. O2
d. Cl2
●
e. F2Explanation: Using the Ideal Gas Law, solve for n and get 1.5 moles of gas. 57g/1.5mol = 38 g/mol which corresponds to fluorine gas.
19.
Pentel mechanical pencil lead is composed of 61%graphite, 14% synthetic resin and 25% carbon black by mass. How many total moles of carbon black are present in a package of pencil lead if the package had a net weight of 50 grams? Note that carbon black is just carbon (C).
a. 0.52 mol b. 0.58 mol c. 104 mol
●
d. 1.04 mol e. 61.4 molExplanation: Use the percent composition to deter- mine the grams of carbon in the pencil lead. 0.25 × 50 g = 12.5 g Next, Use divide by the molecular weight of carbon 12 g/mol to convert the grams to mols.
20.
At what temperature and pressure are gases least ideal?●
a. Low temperature and high pressure.b. High temperature and low pressure.
c. Constant temperature and constant pressure.
d. Constant temperature and high pressure.
e. Low temperature and constant pressure.
Explanation: Gases do not behave ideally at low temperatures because low temperatures mean lower kinetic energies, which translates to having lower velocities. The particles move slowly enough that attraction can occur between molecules, violating one of the ideal gas assumptions. High pressure means that the particles are crammed into a small space, which violates the assumption that the volume of the gas is negligible.
After you are finished, then, and only then, can you get out your phone (or internet device) and submit your answers via the virutal bubblesheet. Use the QR code below or type in the URL to open the bubblesheet page.
https://mccord.cm.utexas.edu/neon
