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Version 083 – Make up 1 – laude – (51635)
1
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Constants
R
= 8
.
314 JK
−
1
mol
−
1
Defnitions
Q
≡
[C]
c
[D]
d
[A]
a
[B]
b
G
≡
H

TS
K
w
≡
[H
+
][OH
−
]
pH
≡ 
log [H
+
]
pOH
≡ 
log [OH
−
]
Equations
Δ
G
= Δ
H

T
Δ
S
ln
p
P
2
P
1
P
=
Δ
H
◦
vap
R
p
1
T
1

1
T
2
P
Π =
MRT
Δ
T
b
=
iK
b
m
Δ
T
f
=
iK
f
m
P
i
=
P
◦
i
χ
i
P
=
s
i
P
i
q
=
mc
Δ
T
q
=
m
Δ
H
Δ
G
◦
=

RT
ln
K
K
=
e
−
Δ
G
◦
RT
ln
K
2
K
1
=
Δ
H
◦
R
p
1
T
1

1
T
2
P
Q
=
K
[at equil.]
[H
+
] = (
K
a
C
a
)
1
/
2
[OH
−
] = (
K
b
C
b
)
1
/
2
[H
+
] =
K
a
(
C
a
/C
b
)
pH = p
K
a
+ log(
C
b
/C
a
)
[OH
−
] =
K
b
(
C
b
/C
a
)
pOH = p
K
b
+ log(
C
a
/C
b
)
001
6.0 points
Write the equilibrium expression For the Fol
lowing reaction:
2
NO
(
aq
) + 4 H
2
O(
ℓ
)
←→
3
H
2
(
aq
) + 2
H
+
(
aq
) + 2
NO
−
3
(
aq
)
1.
K
=
[[
H
+
]
2
·
[
NO
−
3
]
2
[
NO
]
2
2.
K
=
[
NO
]
2
[
H
+
]
2
·
[
NO
−
3
]
2
3.
K
=
[
NO
]
2
[
H
2
]
3
·
[
H
+
]
2
·
[
NO
−
3
]
2
4.
K
=
[
H
2
]
3
·
[
H
+
]
2
·
[
NO
−
3
]
2
[
NO
]
2
·
[
H
2
O
]
4
5.
K
=
[
H
2
]
3
·
[
H
+
]
2
·
[
NO
−
3
]
2
[
NO
]
2
correct
Explanation:
Set up
K
, products in the numerator, reac
tants in the denominator, all raised to respec
tive stoichiometric coe±cients.
002
6.0 points
30
.
2 g oF glycerine (C
3
H
8
O
3
, with MW
92
.
1 g
/
mol) are dissolved in 150 g oF wa
ter. What is the boiling point oF the solution?
(
K
b
oF water = 0
.
515
◦
C
/
m
)
1.
0.104
◦
C
2.
101.13
◦
C
correct
3.
101.52
◦
C
4.
100.10
◦
C
5.
1.13
◦
C
Explanation:
m
glyc
= 30
.
2 g
MW
glyc
= 92
.
1 g
/
mol
m
water
=150 g
K
b
= 0
.
515
◦
C
/
m
Δ
T
b
=
K
b
m
=
K
b
mol glycerol
kg water
= (0
.
515
◦
C
/m
)
30
.
2
92
.
1
mol C
3
H
8
O
3
0
.
150 kg water
= 1
.
13
◦
C
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View Full DocumentVersion 083 – Make up 1 – laude – (51635)
2
T
b
=
T
0
b
+ Δ
T
b
= 101
.
13
◦
C
003
6.0 points
What would be the pH of a 0.563 M solution of
ammonia (NH
3
) at room temperature? The
K
b
of ammonia is 1
.
78
×
10
−
5
.
1.
11.5
correct
2.
9
3.
2.5
4.
4.5
5.
7
Explanation:
[OH
−
] = (K
b
·
C
b
)
1
/
2
= (1
.
78
×
10
−
5
·
0
.
563)
1
/
2
= (10
−
5
)
1
/
2
= 10
−
2
.
5
pOH =

log[OH
−
] =

log(10
−
2
.
5
) = 2
.
5
pH = pK
w

pOH = 14

2
.
5 = 11
.
5
004
6.0 points
What would be molar solubility of a very ex
pensive salt, awesomium, which is composed
of 3 cations and 2 anion (X
3
Y
2
) and whose
K
sp
is 1
.
08
×
10
−
18
?
1.
1 M
2.
1
×
10
−
9
M
3.
0
.
0001 M
correct
4.
6
.
1
×
10
−
7
M
Explanation:
For a salt of the form X
3
Y
2
molar solubility = x =
p
K
sp
27
·
4
P
0
.
2
x =
p
1
.
08
×
10
−
18
27
·
4
P
0
.
2
= 0
.
0001 M
005
6.0 points
Which of the following phase changes is not
correctly paired with the sign of its change in
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 Spring '07
 Holcombe

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