Su, Yung – Homework 7 – Due: Nov 8 2007, 11:00 pm – Inst: FAKHREDDINE
1
This
printout
should
have
47
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
Each question is worth 1 point.
001
(part 1 of 1) 1 points
A gas has a volume of 2.0 L at a pressure of
3.00 atm. What pressure is needed to change
the volume to 6.00 L at constant temperature?
1.
4.0 atm
2.
9.0 atm
3.
2.0 atm
4.
1.0 atm
correct
Explanation:
V
1
= 2.00 L
P
1
= 3 atm
V
2
= 6.00 L
Boyle’s law,
P
1
V
1
=
P
2
V
2
, relates the vol
ume and pressure of a sample of gas.
P
2
=
P
1
V
1
V
2
=
(3.00 atm) (2.00 L)
6.00 L
= 1.0 atm
002
(part 1 of 1) 1 points
A balloon is inflated outdoors on a cold day
in North Dakota at a temperature of

40
◦
C
to a volume of 2.00 L. The pressure remains
constant. What is the volume of the balloon
indoors at a temperature of 25
◦
C?
1.
2.0 L
2.

3
.
2 L
3.
1.6 L
4.
2.6 L
correct
Explanation:
T
1
=

40
◦
C + 273 = 233 K
V
1
= 9
.
0 L
T
2
= 25
◦
C + 273 = 298 K
Charles’s law relates the volume and ab
solute (Kelvin) temperature of a sample of
gas:
V
1
T
1
=
V
2
T
2
V
2
=
V
1
T
2
T
1
=
(2
.
00 L)(298 K)
233 K
= 2
.
6 L
003
(part 1 of 1) 1 points
A sample of a gas contains 0
.
305 g of carbon,
0
.
407 g of oxygen and 1
.
805 g of chlorine. A
different sample of the same gas having a mass
of 8
.
84 g occupies 2 L at STP. What is the
molecular formula for the compound?
1.
C
2
O
2
Cl
4
2.
COCl
2
correct
3.
COCl
4.
C
2
O
2
Cl
2
5.
CO
2
Cl
Explanation:
V
= 2 L
P
= 1 atm
T
= 0
◦
C + 273 = 273 K
Applying the ideal gas law equation,
P V
=
n R T
n
=
P V
R T
=
(1 atm) (2 L)
273 K
·
mol
·
K
0
.
08206 L
·
atm
= 0
.
0892762 mol
MW =
8
.
84 g
0
.
0892762 mol
= 99
.
0185 g
/
mol
Total mass = 0
.
305 g C + 0
.
407 g O
+1
.
805 g Cl
= 2
.
517 g
C :
% C =
0
.
305 g C
2
.
517 g
×
100%
= 12
.
1176% C
mol C = 0
.
121176
99
.
0185 g
mol
¶
·
mol
12 g
= 0
.
999889 mol
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Su, Yung – Homework 7 – Due: Nov 8 2007, 11:00 pm – Inst: FAKHREDDINE
2
O :
% O =
0
.
407 g O
2
.
517 g
×
100%
= 16
.
17% O
mol O = 0
.
1617
99
.
0185 g
mol
¶
·
mol
16 g
= 1
.
00071 mol
Cl :
% Cl =
1
.
805 g Cl
2
.
517 g
×
100%
= 71
.
7124% Cl
mol Cl = 0
.
717124
99
.
0185 g
mol
¶
·
mol
35
.
5 g
= 2
.
00024 mol
Therefore the formula is COCl
2
.
004
(part 1 of 1) 1 points
A mixture of 0
.
75 mol H
2
gas and 0
.
75 mol
N
2
gas is introduced into a 15 liters container
having a pinhole leak at 30
◦
C. After a period
of time, which of the following is true?
1.
The partial pressure of H
2
exceeds that of
N
2
in the container.
2.
The partial pressure of N
2
exceeds that of
H
2
in the container.
correct
3.
The partial pressures of the two gases
change, but they remain equal.
4.
The partial pressures of both gases in
crease above their initial values.
Explanation:
X
N
2
= 0
.
79
P
total
= 1 atm
Because
of
the
pinhole
leak,
the
con
tainer will become equilibrated with the at
mosphere.
Air is composed of 79% N
2
, 20%
O
2
, and 1% other gases. Eventually the gases
in the container will have the same ratios as
air and the pressure will be 1 atm.
X
N
2
= 0
.
79
P
N
2
=
X
N
2
P
total
= 0
.
79 (1 atm)
= 0
.
79 atm
.
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 Spring '09
 Fakhreddine
 Chemistry, mol

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