Version 643 – Exam 1 – Mccord – (50970)
1
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printout
should
have
31
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
It takes light with a wavelength of 261 nm,
or less,
to break the O
H bond in wa
ter.
What energy does this correspond to
(in kJ/photon) and what is the O
H bond
strength in kJ/mol?
1.
7
.
6
×
10

22
kJ/photon, 458 kJ/mol
cor
rect
2.
7
.
6
×
10

22
kJ/photon, 458,000 kJ/mol
3.
6
.
6
×
10

25
kJ/photon, 4
×
10

4
kJ/mol
4.
6
.
6
×
10

25
kJ/photon, 0.40 kJ/mol
Explanation:
002
10.0 points
If a particle is confined to a onedimensional
box of length 300 pm, for Ψ
3
the particle is
most likely to be found at
1.
50, 150, and 250 pm, respectively.
cor
rect
2.
17.3 pm.
3.
100 and 200 pm, respectively.
4.
0 pm.
5.
300 pm.
Explanation:
003
10.0 points
C
6
H
6
= 78.11 g/mol
Cl
2
= 70.91 g/mol
C
6
H
5
Cl = 112.55 g/mol HCl = 36.46 g/mol
If 39.0 g of C
6
H
6
reacts with excess chlorine
and produces 30.0 g of C
6
H
5
Cl in the reaction
C
6
H
6
+ Cl
2
→
C
6
H
5
Cl + HCl
,
what is the percent yield of C
6
H
5
Cl?
1.
13.2%
2.
50.0%
3.
53.4%
correct
4.
76.9%
5.
69.4%
Explanation:
m
C
6
H
6
= 39.0 g
m
C
6
H
5
Cl
= 30.0 g
Our first step is to determine the theoretical
yield of the reaction.
The reaction started
with 39.0 g of C
6
H
6
. We convert from grams
to moles using the molar mass:
? mol C
6
H
6
= 39
.
0 g C
6
H
6
×
1 mol C
6
H
6
78
.
11 g C
6
H
6
= 0
.
4993 mol C
6
H
6
From the balanced equation we see that
1 mole C
6
H
5
Cl is produced for each mole
of C
6
H
6
reacted.
Therefore, if 0.4993 mol
C
6
H
6
were reacted we would expect to pro
duce 0.4993 mol C
6
H
5
Cl.
We use the molar mass of C
6
H
5
Cl to covert
from moles to grams:
? g C
6
H
5
Cl = 0
.
4993 mol C
6
H
5
Cl
×
112
.
55 g C
6
H
5
Cl
1 mol C
6
H
5
Cl
= 56
.
196 g C
6
H
5
Cl
This is the theoretical yield; the maximum
number of grams of C
6
H
5
Cl of that could
be poduced from 39.0 g C
6
H
6
.
To find the
percent yield, we divide the actual yield by
the theoretical yield:
? percent yield =
30
.
0 g C
6
H
5
Cl
56
.
196 g C
6
H
5
Cl
×
100%
= 53
.
4%
004
10.0 points
How many electrons could be described by
the quantum numbers
n
= 5 and
m
s
= +
1
2
?
1.
50
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Version 643 – Exam 1 – Mccord – (50970)
2
2.
1
3.
2
4.
25
correct
5.
5
Explanation:
For a given principle energy level,
n
, a total
of 2
n
2
electrons are described. Exactly half of
those,
n
2
, will have a +
1
2
spin.
005
10.0 points
How many electrons can possess this set of
quantum numbers: principal quantum num
ber
n
= 4, angular quantum number
ℓ
= 2?
1.
10
correct
2.
2
3.
8
4.
14
5.
18
6.
0
7.
16
8.
4
9.
12
10.
6
Explanation:
Use the rules for the quantum numbers:
If
n
= 4 and
ℓ
= 2 (
i.e.
, 4
d
), then
m
ℓ
=

2
,

1
,
0
,
+1
,
+2 are permitted; there are
five different orbitals and
m
s
=
±
1
2
, each
holding two electrons.
006
10.0 points
What is the de Broglie wavelength of a bowl
ing ball rolling down a bowling alley lane?
Assume the mass of the ball is 4500 g and it
is moving at 4.12 m/s.
1.
1
.
229
×
10

32
m
2.
3
.
57389
×
10

35
m
correct
3.
1
.
4725
×
10

37
m
4.
3
.
574
×
10

38
m
Explanation:
λ
=
h
p
=
h
m
·
v
=
6
.
626
×
10

34
kg
·
m
2
s
(4
.
5 kg
×
4
.
12 m
/
s)
= 3
.
57389
×
10

35
m
007
10.0 points
Calculate the longestwavelength line in the
Balmer series for hydrogen.
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 Fall '07
 Fakhreddine/Lyon
 Atomic orbital, C6 H5 Cl

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