*I-2
**II-1
C
C
NC
NC
***II-2
C
NC
°VII-1
°°VII-2
C
C
NC
NC
CHEMISTRY 130-01
Quiz 3 Form A Fall 2010
Chapter 5
Chemical Calculations
Use your Periodic Table for molar masses, and calculate them to the nearest 0.01 g. Show
your work, with units, for all calculations.
NAME:
KEY
A. (1.5 Points) Carry out the following conversions:
*1. Find the number of moles of S (sulfur) in 50.0 g of S. Show your work.
Answer:
1.56 mol S
1 mol S x 50.0 g S = 1.56 mol S
32.06 g S
*2. Find the number of grams of Se (selenium) in 5.00 moles of Se. Show your work.
Answer:
395 g Se
78.96 g Se x 5.00 mol Se = 395 g Se
1 mol Se
3. Find the number of atoms of Te (tellurium) in 5.00 g of Te. Show your work.
Answer:
2.36 x 1022 atoms Te
6.022 x 1023 atoms Te x 1 mol Te x 5.00 g Te = 2.36 x 1022 atoms Te
1 mol Te
127.60 g Te
B.
(1 Point) Determine the molar mass of K2SO4 (potassium sulfate). Show your work.
Answer:
174.27 g K2SO4/mol
2 mol K x 39.10 g K = 78.20 g K
mol K
1 mol S x 32.07 g P = 32.07 g S
mol P
4 mol O x 16.00 g O = 64.00 g O
mol O
174.27 g K2SO4/mol
**C.
(1.5 Points) Balance each of the following equations:
1
C5H12 +
1
CH4 +
1
SiO2
8
4
+
O2
Cl2
3
C
→
→
→
5
CO2 +
6
H2O
4
HCl +
1
CCl4
2
CO
+
1
SiC
***D. (2 Points) Use the chemical equation shown below to calculate the quantities
requested. Show your work.
2 H2 + O2 →
2 H2O
1. What mass of water can be formed from 5.00 g of O2 and excess H2?
Answer:
5.63 g H2O
5.00 g O2 x
1 mol O2 x 2 mol H2O x 18.02 g H2O = 5.63 g H2O
32.00 g O2
1 mol O2
mol H2O
2. What mass of H2 would be required to react with 5.00 g of O2 to produce water?
Answer:
0.631 g H2
5.00 g O2 x
1 mol O2 x 2 mol H2 x 2.02 g H2 = 0.631 g H2
32.00 g O2
1 mol O2
mol H2
Alternately, (5.63 g H2O – 5.00 O) g = 0.63 g H2
Chapter 6a Gas Laws
A.
(1 point) You finish your Diet Coke in the car and cap the empty bottle. It’s a hot day,
the temperature is 35°C (95°F). The air pressure is 763 mm Hg. The next morning, it’s
much cooler, 18°C (65°F). What is the pressure of the air in the bottle? Assume that its
volume is constant.
Answer:
P1
T1
P2
T2
721 mm Hg
= 763 mm Hg
= 35°C + 273 = 308
= x
= 18°C + 273 = 291 K
P1 = P2
T1
T2
P1 x T2 = P2
T1
763 mm Hg x 291 K
308 K
=
721 mm Hg
°B. (1 point) The bottle in the previous problem contains 0.0233 moles of air when it is
closed. What is the volume of the bottle?
Answer:
0.589 L
P = 763 mm Hg x
1.00 atm = 1.00 atm
760 mm Hg
V = x
T = 308 K
n = 0.0233 mol
C.
PV = nRT
V = nRT
P
0.0233 mol x 308 K x 0.0821 L • atm = 0.589 L
1.00 atm
mol • K
(0.5 point) A flask similar to the one you used in the Charles’ Law experiment had a mass
of 256.401 g. The students filled it with water and weighed it again. The filled flask had a
mass of 408.003 g. What volume of water did the flask contain?
Answer:
152 mL
408.003 g
–256.401 g
151.602 g
density of water = 1.00 g/mL
1 mL x 151.602 g = 152 mL
1.00 g
Chapter 6b Phases and Intermolecular Forces
A.
°°B.
(1 point) Determine whether each statement is true (T) or false (F), and mark your
answer in the space provided.
T
1. Only molecules with N-H, O-H, and F-H bonds undergo hydrogen bonding.
F
2. Dipole - dipole forces only act between nonpolar molecules.
T
3. When a compound melts, it releases energy.
F
4. While a compound is boiling, its temperature increases dramatically.
F
5. Dispersion forces only occur in nonpolar compounds.
(2 point) Consider the following molecules, then answer the questions below.
H
H
C O
H
H
C C
H
A. Formaldehyde
H
B. Ethylene
1. Give the name of the compound that will have the higher boiling point.
Formaldehyde
2. What forces are present in that molecule but not present in the other one?
Dipole-dipole forces
Consider the following molecules, then answer the questions below.
H
H H
H C O
H
H C C O
H
A. Methanol
H H
H
B. Ethanol
1. Give the name of the compound that will have the higher boiling point.
Ethanol
2. What forces are stronger in that molecule than in the other one?
Dispersion Forces