Thomas, Phillip – Homework 7 – Due: Oct 31 2007, 11:00 pm – Inst: Brodbelt
1
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Brodbelt CH301
Homework 7
Due October 31, 2007 at 11:00 PM
001
(part 1 oF 1) 10 points
The same number oF grams oF NH
3
and O
2
are placed in separate bulbs oF equal vol
ume and temperature under conditions when
both gases behave ideally. Which statement
is true?
1.
The bulb containing O
2
contains more
molecules oF gas.
2.
The pressure in the O
2
bulb is greater
than the pressure in the NH
3
bulb.
3.
The pressures in the two bulbs are the
same.
4.
The pressure in the NH
3
bulb is greater
than the pressure in the O
2
bulb.
correct
5.
Both bulbs contain the same number oF
moles oF gas.
Explanation:
The molecular weight oF NH
3
is less than
that oF O
2
, so in equal masses there are more
moles oF NH
3
than oF O
2
. At the same volume
and temperature, the larger number oF moles
oF NH
3
would exert a higher pressure.
002
(part 1 oF 1) 10 points
How many molecules are in 1.00 liter oF O
2
gas at 56
◦
C and 821 torr.
1.
2
.
24
×
10
23
molec
2.
4
.
00
×
10

2
molec
3.
32 molec
4.
1
.
83
×
10
25
molec
5.
2
.
41
×
10
22
molec
correct
Explanation:
V
= 1 L
T
= 56
◦
C + 273 = 329 K
P
= 821 torr
·
atm
760 torr
= 1
.
08 atm
Applying the ideal gas law equation,
P V
=
n R T
n
=
P V
R T
n
=
(1
.
08 atm)(1 L)
(
0
.
08206
L
·
atm
mol
·
K
)
(329 K)
·
6
.
02
×
10
23
molec
1 mol
= 2
.
41
×
10
22
molec
003
(part 1 oF 1) 10 points
At STP, 6.0 grams oF CO gas will occupy a
volume oF
1.
4.8 liters.
correct
2.
5.6 liters.
3.
22.4 liters.
4.
2.24 liters.
5.
3.5 liters.
Explanation:
T
= 0
◦
C + 273 = 273 K
P
= 1 atm
n
= 6 g
·
mol
28 g
= 0
.
214 mol
Applying the ideal gas law,
P V
=
n R T
V
=
n R T
P
V
=
(0
.
214 mol)
(
0
.
08206
L
·
atm
mol
·
K
)
(273 K)
1 atm
= 4
.
79411 L
004
(part 1 oF 1) 10 points
A ±ask contains 0.123 moles oF an ideal gas
that occupies 781 mL. A second ±ask at the
same temperature contains 0.0712 moles oF
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View Full DocumentThomas, Phillip – Homework 7 – Due: Oct 31 2007, 11:00 pm – Inst: Brodbelt
2
the same gas. The pressure is the same in
both fasks. What is the volume oF the second
fask?
1.
1350 mL
2.
904 mL
3.
452 mL
correct
4.
89,180 mL
5.
6.8 mL
Explanation:
n
1
= 0
.
123 mol
n
2
= 0
.
0712 mol
V
1
= 781 mL
Applying the ideal gas law equation,
P V
=
n R T
V
1
V
2
=
n
1
n
2
V
2
=
V
1
n
2
n
1
V
2
=
(781 mL)(0
.
0712 mol)
0
.
123 mol
= 452 mL
005
(part 1 oF 1) 10 points
What is the volume oF 0.500 moles oF an ideal
gas at 273 K and 760 torr?
1.
11.2 liters
correct
2.
4.56 cubic Fathoms
3.
2.24 liters
4.
380 ml
5.
22.4 liters
Explanation:
006
(part 1 oF 1) 10 points
A 2.00 mole sample oF gas is at a temperature
oF 100.0
◦
C and occupies 3.00 liters. What is
its pressure?
1.
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 Fall '07
 Fakhreddine/Lyon
 Chemistry, pH

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