Winter 2008
February 28,2008
Chemistry 122
Exam 2
Oregon State University
Dr. Richard Nafshun
Instructions: You should have with you several number two pencils, an eraser, your 3" x 5" note card, a
calculator, and your University ID Card. If you have notes with you, place them in a sealed backpack
and place the backpack OUT OF SIGHT or place the notes directly on the table at the front of the room.
Fill in the front page of the Scantron answer sheet with your last name, first name, middle initial, and
student identification number. Leave the class section number and the test form number blank.
This exam consists of 25 multiple-choice questions. Each question has four points associated with it.
Select the best multiple-choice answer by filling in the corresponding circle on the rear page of the
answer sheet. If you have any questions before the exam, please ask. If you have any questions during
the exam, please ask the proctor. Open and start this exam when instructed. When finished, place your
Scantron form in the appropriate stack and present your University ID Card to the proctor. You may
keep the exam packet, so please show your work and mark the answers you selected on it.
R = 0.0821 Leatm/mol*K
M = mol/L
= nRT
SC: 2r = s
Na
Sodium
I
m = mollkg
ATb= i d b
kb(HZO)= 0.5 12 OCIm
760 mm Hg = 760 torr = 1 atm
ATf = imkf
kf (HzO) = 1.86 "Clm
BCC: 4r = sd3
I
Mg
Mamerium
FCC: 4r = sd2
IVA
VA
VIA
5
6
7
8
9
10
B
C
N
0
F
Ne
Bomn
Carbon
Nimgen
Oxygen
Pluorine ,
Neon
10.81
12.01 1
14.0067
15.9994
18.9984
20.179
13
Calcium
1
40.08
Co
Ni
Cu
16
17
18
S
Cl
Ar
Silicon
Phosphorus
Sulfur
Chlodne
Argon
28.0855
30.97376
32.06
35.453
39.948
Si.
Aluminum
Sc
15
P
14
A1
.
26.9815
Ca
31
32
33
34
35
36
Zn
Ga
Ge
As
Se
Br
Kr
Scandium
Cobalt
Nickel
copper
Zinc
Gallium
Gcmanium
Anenic
Selenium
Bmminc
Krypton
44.9559
58.9332
58.70
63.546
65.38
69.72
72.59
74.9216
78.96
79.904
83.80
47
48
49
50
51
52
53
54
'
37
38
40
41
42
43
44
45
46
Rb
Sr
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Rubidium
Strontium
Zirconium
Niobium
Ruthenium
Rhodium
Palladium
Silver
Tin
85.4678
87.62
91.22
92.9064
101.07
102.9055
106d
107.868
118.69
Molybdenum Technetium
95.94
98.906
Sn
56
77
78
79
82
Ba
Ir
Pt
Au
Pb
Cesium
Barium
Iridium
Platinum
Gold
Lead
132.9054
137.33
192.22
195.09
196.9665
109
110
Francium
I
I
Radium
VLIA
IlJA
1
1 Scaborgium l~cilsbahrium Hasrium
'~etinides I ~ u h ~ f o r d i u m Hahnium
57
58
La
Ce
Lanthanium
Cerium
138.9055
140.12
'
I
1
I
1
61
62
63
64
Prn
Sm
Eu
Gd
Tb
Samarium
Europium
Gadolinium
Terbium
150.4
151.96
157.25
158.9254
162.50
164.9304
144.24
145
65
!ismu*
(
-1
60
Rawodymium Neodymium hmerhium
140.9077
Mcimerium
Nd
59 ,
Pr
Mt
I
Polonium
I
Aslatine
stable region?
67
68
69
70
71
Dy
Ho
Er
Tm
Yb
Lu
Dysprosium
Holmium
Erbium
Thulium
Yucrbium
Lulctium
167,26
168.9342
173.M
174.967
66
,
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Ac
Th
Pa
U
Np
.
.
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Nptunium
1
Plutonium
Anx2ri um
Actinium
2270278
1
Thorium
232.038i
3
3
1
Uranium
0
2
I
m W 2
(244)
C"um
(247)
1
"lium
(247)
Californium Einsteinium
(251)
1
(254)
1
Fermium
(2.57)
Mmdel~vium Noklium
1
17.581
I
7.59
(
Radon
1.
Consider the phase diagrams below. Which diagram could correctly describe C02?
/_-_-----7
(C)
2.
Which of the following is false?
Graphite is a network covalent compound. T ~ U L
(A)
Diamond is a network covalent compound. T.\*-b
(B)
Network covalent compounds typically melt at higher temperatures than molecules. ~ r % o (C)
."4..,0
(D) CPotassmm chroride^i3a neWdre
(E)
Carbon d i o x i d e - b n z KC4 ; a
.7cL"i"+.-~a.T
q
; B ~ ; LC @ M ~ O * A ~
3.
Consider the phase diagram below for compound foolsgoldium. The temperature of sublimation
at 1 atm is:
it; &
(C)
(D)
(E)
4.
200 "C
212°C
Sodium chloride melts at 804 "C. Sodium iodide melts at 65 1 "C. The difference in melting
points can be attributed to:
(A)
Different intermolecular forces (dispersion, dipole-dipole, hydrogen bonding)
(D)
Network covalent co
One is a molecule (attractions by intermolecular forces), one is an ionic compound
(attractions by charges)
5.
MgO melts near 2800 "C. KC1 melts near 776 "C. The difference in melting points can be
attributed to:
(A)
(B)
(C)
(D)
(E)
dipole, hydrogen bonding)
Network covalent compounds
The sheet-like structure
J
i o n k L?
c.S
(k:?L
6.
MP)
1-propanol, CH3CH2CH20H,boils at +97.1 "C. Butane, CH3CH2CH2CH3boils at -0.5 "C. The
difference in melting points can be attributed to:
I -prop an01
(A)
(B)
(c)
(D)
(E)
butane
don-
po\ar
Different distances between nuclei. (ionic size)
Network covalent compounds
One is a molecule (attractions by intermolecular forces), one is an ionic compound
(attractions by charges)
7.
Consider dimethyl ether, CH30CH3. The intermolecular forces present in CH30CH3are:
4
%"
fl"
3?
-%?
8.
*%Q
%@4
6"@
Dispersion forces only.
d
P
.&Q*%
&T@?@P
P'%
&kN
(A)
(B)
(c>
(D)
(E)
boo
&"qJ'
'on
*LO@%'
c"' 'vo6
*&
&k*fi
orq9'n
ydrogen bonding.
Hydrogen bonding only.
Consider He, CH30CH3,CH3CH20H,diamond, LiF, CsF, MgF2, and AlN. Arranged in
increasing melting point, these are:
Lowest r n ~
(D)
(E)
Hiehest m ~
~ < diamond
He < CH3CH20H < CH30CH3<LiF < CsF < M ~ <FA1N
He < CH30CH3 < CH3CH20H < A1N <LiF < CsF < MgF2 < diamond
t-\ ;?hn$ b.
M a \ & \~ r Qe:&
)
9.
Which of the following has a hydrophilic end (polar, water-loving end) and a hydrophobic end
(non-polar, water-fearing end) and has the ability to bridge water molecules to non-polar
molecules?
(A)
(B)
(C)
The polymer Teflon (-CF2CF2-)
Graphite
Ethane, CH3CH3
pat
on-Qc\ce~
10.
Which of the following cannot undergo free radical polymerization?
11.
Which of the following molecules will not form hydrogen bonds?
H
H
H
H
I I
I HI
H
I
I
H
H
Q
H
H
H
H
I
1
H
I
1
H
I
1
H
I
1
H
H-C-C-C-
C
A"
12.
13.
The intermolecular forces that are most significant in accounting for the high boiling point of
liquid water relative to other substances of similar molecular weight islare the:
(A)
(B)
(C)
(Dl
Dispersion forces
ole interactions
(E)
Ionic charges
Which has a higher melting point, sodium fluoride or aluminum fluoride? Why?
(A)
sodium fluoride has a higher melting point because it has weaker dispersion forces than
aluminum fluoride.
(B)
sodium fluoride has a higher melting point because it has stronger dispersion forces than
aluminum fluoride.
(C)
aluminum fluoride has a higher melting point because it has stronger dispersion forces
than sodium fluoride.
(D)
aluminum fluoride has a higher melting point because it has a greater mass than sodium
fluoride.
,.. ..,-:.,.,.:,..
.;,<,-.<
,';.,'+h
7
.+--has a higher melting point because it has greater ionic charges than
sodium fluoride.
.<<'
.L-,,,.,
_-_U"*.'Ul.'-"l...--,"
:htS.:~>,;r
i.c!':><.>-:a%"-
= s " , % ~ ? h ; . a * w d ~ u * & ~ : ~ - ? b ~ ~ ~ "
&
,*,,,
"\.-
'w,r+,*,-#,
w-s**
14.
The equivalent number of atoms in the FCC unit cell is:
I
~
3(,, ,n 4
Cr.)
~
bx t = 3
(;~lcr%!
.
9
15.
16.
The structure below [from a ~ @ s e Worksheet] represents:
(A)
An SC unit cell
(E)
A prokaryotic cell
The cubic form for the fictitious element Nathanium (named for a CH 122 TA) is FCC. The
atomic radius is 132.0 pm and the molar mass is 267.4 glmol. The density of Nathanium is:
[I m = 1 x 1012pm
1 m = 100 cm]
ti=
(C)
(D)
(E)
8.55 g/cm3
49.4 g / ~ m 3
2.03g/cm3
hZ
3
0 Ma$'?,
+
Z67'Y
.
1
*o\
"A(),:
)=
(&r%uloa3r+.~
....................
1
.- .
................
.......
,
,
......
-z1
\ . 7 7 & ~ , 09
17.
The freezing point of 0.500 m aqueous AlC13 is:
(A)
(B)
(C)
(D)
18.
-2.79 "C
+2.79 "C
+5.58 "C
-5.58 "C
AT+=
;m k C i
( ~ [ o . s u o ~ \ , % br~ 3\ , 7 z Q ~
7 5 = 0%- 3 . 7 "~ - ~3 .~ 7 ' ~ ~ ~
Whch of the following sets of compounds are expected to be soluble in water?
'l--.---v""--/
(A)
(B)
(C)
(Dl
\ ~ . ~ \ #an&
.r
; 0 ~ 9 ; &
19.
A student dissolves 12.000 g of an unknown polymer in 800 mL of water at 320 K. She
measures the osmotic pressure to be 0.0677 mm Hg. What is the molar mass of the polymer?
(A)
(B)
(C)
(D)
(E)
20.
2.71x106glmo
@ E x lo6 qim
1.73 x 10' g/mol
1.73x106glmol
2.26 x lo6 glmol
ttd= nRT
n=
-
nq
RT 3L o c h 7 7
w*
wq f 7 6 0
d , _ _ - j . - - - - - = - * _ _
( a . o e z \ PI^^.
b C'lb-.Xazc u)
Generally, which of the following generally does not increase the reaction rate?
(A)
Increase the temperature
T~-J-
- Eq
-..__Y
/\.~+m)?(O.%oo
___________
L)
2 1.
The half-life is:
(C)
(D)
The amount of time required for the entire sample to decay
42 years
.+
22.
A student ( 9 ) obtains a 100.0 gram sample of 1311(tli2= 8.00 days). How many grams of "'I
will remain after 16.00 days?
---
(A)
(B)
(C)
(D)
(El
8.0grams
\00,03
'50.07
8
b&.)9
23.
0
5
A student ( 9 .) obtains a 100.0 gram sample of 13'1(tin = 8.00 days). How long will it take so
that only 10.0 grams of l3 '1 remain?
(A)
(B)
8.2 days
16.4 days
J0-
Car, k
24.
The following are initial rate data for:
1
( 4
(B)
(C)
(D)
(E)
25.
Experiment
1
Initial [A1
A + B + C
1
Initial IB1
00ubtc
1
Initial Rate
161 -4
+Lr p k ~
The rate law is Rate = ~ [ A ] ~ [ B ] ' .
The rate law is Rate = ~[A]'[B]~.
%&FU
~ p r h (A>
' ~
)20
* C
cL\*115c
r9,f 4-
The Chemistry 122 final exam is Monday, March 17,2008 at 4:OOpm; yes, this is better than
early in the morning. After the chemistry final I will be ...
(A)
Watching the C-SPAN baseball steroid hearings
(B)
Making a poster for my dorm room wall that has the daily countdown to the new Indiana
Jones movie (In theaters May 22)
(C)
Two words: Tom and Gisele
(E)
Re-watching The Oscars
[Any response will receive full credit; even no response]
ih