g
= 9
.
81 m/s
2
c
= 3
×
10
8
m/s
d
water
= 1
.
00 g/mL
1 atm = 1.01325
×
10
5
Pa
N
A
= 6
.
022
×
10
23
mol

1
k
= 1
.
381
×
10

23
J/K
R
= 0
.
08206 L
·
atm/mol
·
K
R
= 8
.
314 J/mol
·
K
R
= 62
.
36 L
·
torr/mol
·
K
1 L
·
atm = 101.325 J
E
=
hν
c
=
λ/ν
λ
=
h
p
=
h
mv
1
2
mv
2
=
hν

Φ
Tλ
max
=
1
5
c
2
c
2
= 0
.
0144 K
·
m
ν
=
R
1
n
2
1

1
n
2
2
¶
R
= 3
.
29
×
10
15
Hz
E
n
=
n
2
h
2
8
mL
2
n
= 1
,
2
,
3
,
· · ·
P
=
dhg
PV
=
nRT
P
total
=
P
A
+
P
B
+
P
C
+
· · ·
x
A
=
P
A
/P
total
P
+
a
n
2
V
2
¶
(
V

nb
) =
nRT
v
rms
=
3
RT
M
¶
1
/
2
rate of effusion (or speed)
∝
r
T
M
Δ
U
=
q
+
w
H
=
U
+
PV
Δ
U
=
q
V
=
C
V
Δ
T
=
n C
V,
m
Δ
T
Δ
H
=
q
P
=
C
P
Δ
T
=
n C
P,
m
Δ
T
w
=

P
Δ
V
w
=

Δ
nRT
Δ
U
= Δ
H

P
Δ
V
Δ
U
= Δ
H

Δ
nRT
Δ
H
rxn
= Δ
H
1
+ Δ
H
2
+ Δ
H
3
+
...
Δ
H
◦
rxn
=
X
nB.E.
(react)

X
nB.E.
(prod)
Δ
H
◦
rxn
=
X
n
Δ
H
◦
f
(prod)

X
n
Δ
H
◦
f
(react)
Δ
S
◦
rxn
=
X
nS
◦
(prod)

X
nS
◦
(react)
Δ
G
◦
rxn
=
X
n
Δ
G
◦
f
(prod)

X
n
Δ
G
◦
f
(react)
Δ
H
◦
r
(
T
2
) = Δ
H
◦
r
(
T
1
) + Δ
C
P
Δ
T
Δ
C
P
=
X
nC
P,
m
(prod)

X
nC
P,
m
(react)
w
=

nRT
ln
V
2
V
1
Δ
S
=
nR
ln
V
2
V
1
Δ
S
=
C
V
ln
T
2
T
1
Δ
S
=
C
P
ln
T
2
T
1
Δ
H
vaporization
≈
85 J/mol K
d
S
= d
q
rev
/T
S
=
k
ln
W
G
=
H

TS
Δ
G
= Δ
H

T
Δ
S
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Moore, Lauren – Final 1 – Due: May 10 2007, 1:00 pm – Inst: McCord
2
This
printout
should
have
56
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
How many sigma and pi bonds are in the
molecule CH
2
CCH
2
?
1.
4 sigma and 2 pi
2.
6 sigma and 2 pi
correct
3.
2 sigma and 2 pi
4.
6 sigma and 0 pi
5.
8 sigma and 0 pi
Explanation:
002
(part 1 of 1) 10 points
What volume of hydrogen at STP would be
required to produce 1.000 mole of HCl?
H
2
(g) + Cl
2
(g)
→
2 HCl(g)
1.
6.6 L
2.
22.4 L
3.
5.3 L
4.
11.2 L
correct
5.
44.8 L
Explanation:
At STP, 1 mole of an ideal gas has a volume
of 22.4 L;
i.e.
,
22
.
4 L
mol
.
V
H
2
= 1 mol HCl
×
1 mol H
2
2 mol HCl
×
22
.
4 L H
2
mol H
2
= 11
.
2 L
003
(part 1 of 1) 10 points
Which is the weakest type of attractive force
between particles?
1.
hydrogen bond
2.
dispersion forces
correct
3.
ionic bond
4.
covalent bond
Explanation:
London forces, dispersion forces, or induced
dipoles all describe the same intermolecular
force. London forces are induced, shortlived,
and very weak.
Molecules and atoms can
experience London forces because they have
electron clouds. London forces result from the
distortion of the electron cloud of an atom or
molecule by the presence of nearby atoms or
molecules.
Permanent dipoledipole interactions are
stronger than London forces and occur be
tween polar covalent molecules due to charge
separation.
Hbonds are a special case of very strong
dipoledipole interactions.
They only occur
when H is bonded to small, highly electro
magnetic atoms – F, O or N only.
Ionion interactions are the strongest due to
extreme charge separation and occur between
ionic molecules.
They can be thought of as
both inter and intramolecular bonding.
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 Spring '08
 McCord
 Chemistry, Atom, Chemical bond, Lauren – Final

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