Name: _________KEY________________
CHEM&141 – Final Exam – F08 - 200 Points
*Valuable values and equations on last page
1. All the following compounds are insoluble in water except: (4 points)
a. BaF2
b. CaF2
c. MgF2
d. CsF
2. C2H2 has ____ pi-bonds. (4 points)
a. 0
b. 1
c. 2
d. 4
3. The net dipole moment of NF3 is ________. (4 points)
a. Towards the N
b. Towards the F on the left
c. Towards the F on the right
d. None of the above
Give the name/molecular formulae for the following: (3 points each – 12 points total)
4. As2O3
Diarsenic trioxide
5. TiO2
Titanium (IV) oxide
6. H3PO4(aq)
Phosphoric acid
7. Mo2(SO3)7
Molybdenum (VII) sulfite
8. Give the short-hand electron configuration for: (5 points each – 10 points total)
a. Ni2+
[Ar]3d8
b. Pb2+
[Xe]6s24f145d10
___________/34 points total
Fall 2008
CHEM&141 Final Exam
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Give the: a) Lewis dot structures (including resonance, if necessary), b) hybridizations of
the central atom, c) electron and molecular geometries of the central atom, and d) bond
angles for the central atom for the following species: (30 points total)
9. OF2
a. Lewis structure:
F
O
F
b. Hybridization:
sp3
c. Electron/molecular geometries:
Tetrahedral/bent
d. Bond angle:
<<109.5°
10. HOSO3a. Lewis structure:
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CHEM&141 Final Exam
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__
__
O
H
O
O
S
O
H
O
O
S
O
O
__
O
H
O
S
O
O
b. Hybridization: sp3
__________/14 points total
c. Electron/molecular geometries:
Tetrahedral/tetrahedral
d. Bond angle: 109.5°
11. XeI2
a. Lewis structure:
I
Xe
I
b. Hybridization:
Sp3d
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CHEM&141 Final Exam
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c. Electron/molecular geometries:
Trigonal bipyramid/linear
d. Bond angle:
180°/120°
_______________/16 points total
For the following:
i) State the molecular and net ionic equations for each (include phase
labels!) If there is no reaction write “NR” (8 points – 16 points total)
ii) Circle the type of reaction. For non-reactions just circle No Reaction
(NR). (4 points – 8 points total)
12. A solution of mercury (II) nitrate is mixed with a solution of potassium sulfate
yielding a precipitate.
Acid-Base
Decomposition
Double Displacement
Single Displacement
Redox
No Reaction
Molecular: Hg(NO3)2(aq) + K2SO4(aq)  HgSO4(s) + 2KNO3(aq)
Ionic: Hg2+(aq) + 2NO3-(aq) + 2K+(aq) + SO42-(aq)  2K+(aq) + NO3-(aq) + HgSO4(s)
Net-ionic: Hg2+(aq) + SO42-(aq)  HgSO4(s)
13. Mercury reacts with oxygen in air to produce solid mercury (II) oxide.
Acid-Base
Decomposition
Double Displacement
Single Displacement
Redox
No Reaction
Molecular: 2Hg(l) + O2(g)  2HgO(s)
Ionic: ditto
Net-ionic: ditto
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CHEM&141 Final Exam
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_______________/24 points total
14.
8 58407346
(12.0   ) 8. __

 0.1 (5 points)
(80)
80
__
15. The melting point of radon is -71.15 °C. What is this in Kelvin and degrees
Fahrenheit? (6 points)
K = 273.15 -71.15 = 202.00
°F = 1.80 x (-71.15) + 32 = -96.07
16. If the RDA for vitamin C is 60. mg per day and there are 70. mg of vitamin C per
100. g of orange, how many 3.0 oz. oranges would you have to eat each week to
meet this requirement? (454 g = 1.0 lb & 16 oz. = 1.0 lb) (10 points)
1orange 16oz 1lb
100.g orange
60.mg Vitamin C 7days





 7.0 oranges
week
3.0oz
1lb 454 g 70.mg Vitamin C
1day
1week
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CHEM&141 Final Exam
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_______________/21 points total
17. Find the empirical and molecular formulae of heroine, given the following mass
percent composition: C 68.22%, H 6.29%, N 3.81%, O 21.66%. Its molecular
weight is: 369.41 g/mol. (15 points)
1mol
 5.680molC  0.272mol  20.88C  21C
12.01g
1mol
H : 6.29 g 
 6.23molH  0.272mol  22.9 H  23H
1.01g
1mol
N : 3.81g 
 0.272molN  0.272mol  1.00 N  1N
14.01g
1mol
O : 21.66 gO 
 1.354molO  0.272mol  4.978O  5O
16.00 g
Thus, empirical formula = C 21H 23 NO5 & empirical formula weight = 369.45g/mol
C : 68.22 g 
369.41g/mol
1
369.45g/mol
Therefore, molecular formula is the same as the empirical formula; i.e., C 21H 23 NO5
n=
18. 0.74 L of 0.0470 M BeCl2 is thoroughly mixed with 1.08 L of 0.0484 M AgNO3.
What is the mass (in grams) of the precipitate produced? BeCl2 = 79.91 g/mol,
AgNO3 = 169.94 g/mol, AgCl = 143.32 g/mol, Be(NO3)2 = 133.15 g/mol (15
points)
BeCl2(aq) + 2AgNO3(aq)  2AgCl(s) + Be(NO3)2(aq)
0.047mol 2molAgCl 143.32 g


 10.gAgCl
1L
1mol
1mol
0.0484mol 2molAgCl 143.32 g
AgNO3 :1.08L 


 7.49 gAgCl
1L
2mol
1mol
Thus, mass produced is 7.49 g
BeCl2 : 0.74 L 
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CHEM&141 Final Exam
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_____________/30 points total
If only 6.98 grams of precipitate is produced, what is the percent yield? (4 points)
6.98 g
100%  93.2%
7.49 g
19. In the following reaction:
Sc(ClO4)3(s)  ScCl3(s) + 6O2(g)
What is the pressure, in atmospheres, in which oxygen gas was collected from the
2.48 g sample of reactant given a flask with a volume of 1.00L and a temperature
of 625.0°C? MW Sc(ClO4)3 = 385.97 g/mol and MW O2 = 32.00 g/mol (15
points)
6molO2
1mol reactant

 0.0386molO2
385.97 g reactant 1molreactant
L  atm
(0.0386mol )(0.08206
)(898.2 K )
mol  K
P
 2.85atm
1.00 L
2.48 g reactant 
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CHEM&141 Final Exam
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_________/19 points total
20. 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to
1.80 mol, what new volume will result, given a constant temperature and
pressure? (10 points)
V2
5.00 L

0.965mol 1.80mol
V2  9.33L
21. How many kJ of energy are required to raise the temperature of 3.80 grams of
radon gas from 25.00°C to 100.00°C? Specific heat capacity = 0.094 J/gK. (12
points)
Q  mcT  (3.80 g )(0.094 J
gK
)(75.00 K )  26.8 J 
1kJ
 2.68 10 2 kJ
1000 J
__________/22 points total
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CHEM&141 Final Exam
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22. Photochromic sunglasses, which darken when exposed to light, contain a small
amount of colorless AgCl(s) embedded in the glass. When irradiated with light,
metallic silver atoms are produced and the glass darkens: AgCl(s)
Ag(s) + Cl(g).
Escape of the chlorine atoms is prevented by the rigid structure of the glass, and
the reaction therefore reverses as soon as the light is removed. (10 points)
a. If 310 kJ/mol of energy is required to make the reaction proceed, what
wavelength of light (in nm) is necessary?
E photon  h  h
c

3
 (6.626  10 J  s )
2.998  108 m

1mol
1000 J
s  310 kJ 

23
mol 6.022 10 photons 1kJ
110 nm
 390nm  3.9  10 2 nm
1m
Is this visible light?
  3.9 107 m 
9
No, it’s in the ultraviolet or UV range.
23. Oxygen gas reacts with solid iron to generate iron(III) oxide, as per the
following equation: (10 points)
4Fe(s) + 3O2(g) → 2Fe2O3(s)
a. If 65.1 g of iron is allowed to mix with 30.0 g of oxygen gas, which
is the limiting reagent?
1molFe 2molproduct
Fe : 65.1gFe 

 0.583mol product
55.85 gFe
4molFe
1molO2
2molproduct
O2 : 30.0 g 

 0.625mol product
32.00 gO2
3molO2
Thus, the limiting reagent is iron.
b. What mass of iron(III) oxide is formed?
0.583mol product 
159.70 g
 93.1g product
mol
____________/20 points total
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CHEM&141 Final Exam
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1 point extra credit: please circle where I have used alliteration.
“Valuable values” from the first page right below the title of the exam.
Q = mcΔT
PV = nRT; R = 0.08206 Latm/molK
°F = 1.80°C + 32
K = 273.15 + °C
FC = VE – [LE + ½ BE]
Ephoton = h; h = 6.626 x 10-34 Js
C = ; c = 2.998 x 108 m/s
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CHEM&141 Final Exam
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