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Version 130 – Exam 3 – McCord – (53740)
1
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This is
Dr. McCord’s
MORNING course
that is MW± From 11:00  12 noon, unique
53740.
McCord
MWF : 11am
g
= 9
.
8 m
/
s
2
1 atm = 101325 Pa = 760 torr
R
= see back oF answer sheet
P
=
ρ
·
g
·
h
(
P
+
a
(
n/V
)
2
)(
V

nb
) =
nRT
001
10.0 points
IF the average speed oF a water molecule at
25
◦
C is 640 m
·
s
−
1
, what is the average speed
at 100
◦
C?
1.
801 m/s
2.
320 m/s
3.
572 m/s
4.
716 m/s
correct
5.
5120 m/s
Explanation:
T
1
= 100
◦
C + 273
.
15 = 373
.
15 K
T
2
= 25
◦
C + 273
.
15 = 298
.
15 K
±rom kinetic molecular theory, the temper
ature is
directly
proportional to mean KE.
KE
mean
=
1
2
(MW)(average molecular speed)
2
and knowing MW is constant (it’s the same
gas) we get average speed oF gas molecules
Avg speed
T
1
Avg speed
T
2
=
√
T
1
√
T
2
Avg speed
T
2
=
(Avg speed
T
1
)
√
T
2
√
T
1
=
(640 m
/
s)
√
373
.
15 k
√
298
.
15 K
= 715
.
986 m
/
s
i.e.
, average speed oF gas molecules at 100
◦
C
is 715
.
986 m
/
s.
002
10.0 points
Which oF the Following gases would you pre
dict to have the largest value oF the van der
Waals coe²cient
b
?
1.
CO
2
2.
C
2
±Cl
5
correct
3.
C
2
±
6
4.
Cl
2
5.
C
2
±
2
Cl
4
Explanation:
The larger the van der Waals coe²cient ‘
b
’,
the larger the molecule. The largest molecules
have 8 atoms. OF these, C
2
±Cl
5
is larger than
C
2
±
2
Cl
4
and C
2
±
6
, because Cl is larger than
±.
003
10.0 points
Which would you expect to be least viscous?
1.
C
4
H
8
at 30
◦
C
2.
C
8
H
18
at 50
◦
C
3.
C
8
H
18
at 30
◦
C
4.
C
4
H
8
at 50
◦
C
correct
Explanation:
004
10.0 points
Which oF the Following can Form intermolecu
lar hydrogen bonds?
1.
PH
3
2.
H
2
CO
3.
(CH
3
)
2
NH
correct
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2
4.
CH
3
COCH
3
Explanation:
Only molecules with H attached to the
electronegative atoms N, O, and F can hy
drogen bond. Of the molecules given only
CH
3
(CH
3
)
2
NH
2
has the H directly bonded to
N, so it can undergo hydrogen bonding.
005
10.0 points
If 250 mL of a gas at STP weighs 2 g, what is
the molar mass of the gas?
1.
56
.
0 g
·
mol
−
1
2.
44
.
8 g
·
mol
−
1
3.
179 g
·
mol
−
1
correct
4.
28
.
0 g
·
mol
−
1
5.
8
.
00 g
·
mol
−
1
Explanation:
V
= 250 mL
P
= 1 atm
T
= 0
◦
C = 273.15 K
m
= 2 g
The density of the sample is
ρ
=
m
V
=
2 g
0
.
25 L
= 8 g
/
L
The ideal gas law is
P V
=
nRT
n
V
=
P
RT
with unit of measure mol/L on each side.
Multiplying each by molar mass (MM) gives
n
V
·
MM =
P
RT
·
MM =
ρ,
with units of g/L.
MM =
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This note was uploaded on 09/13/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner

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