Download Ex 3 Key Fa 03

Chem 152 Exam 3 Fall 2003
Name: KEY
R = 0.0821 L atm / mol  K
1 atm = 760 mm Hg
STP = standard temperature and pressure: 273 K and 1 atm
All questions are worth 4 points.
Multiple Choice, circle the correct answer:
1. When the equation
NaOH + H2SO4 
is completed and balanced, one of the terms in the balanced equation is:
a) NaSO4
b) 2Na2SO4
c) H2OH
d) 2H2O
2. The correct Lewis structure for N2 includes:
a) one single bond
b) one double bond c) one triple bond
e) none of these
d) none of these
3. The bonding in MgO is:
a) ionic
b) covalent
4. The bonding in HCl is:
a) ionic
b) covalent
5. Which compound has a trigonal pyramidal molecular structure?
a) NaCl
b) CO2
c) NH3
d) H2O
e) CH4
6. When the reaction
NaHCO3 + HNO3 
is completed and balanced, one of the terms in the balanced equation is:
a) CO2
b) H2
c) NaHNO3
d) H2CO3
e) none of these
7. Which of these is a strong electrolyte:
a) HC2H3O2
b) CCl4
c) NaOH
d) NH3
8. Which of these is a strong acid:
a) HC2H3O2
b) NH3
c) HNO3
d) HClO
True or False
9. The O2 molecule is polar.
______F_______
10. When the temperature of a gas is decreased, the pressure increases.
______F_______
11. Water is a polar molecule.
______T_______
Calculations: You must show your setup for credit on all problems (even if you can do the math in your
head). Please circle your final answers.
12. How many mL of 6.0 M HCl do you need to prepare 500.0 mL of 0.10 M HCl?
Dilution:
M1V1 = M2V2
(6.0 M) V1 = (0.10 M)( 500.0 mL)
V1 = 8.3 mL
13. How many grams of NaNO3 are need to make 750.0 mL of a 0.10 M solution? Describe in words
how you would make the solution.
M x L = moles
(0.10 mole/L) (0.7500 L) = 0.075 mole NaNO3
0.075 mole NaNO3 x 85.00 g = 6.4 g NaNO3
1 mole
Weigh 6.4 g NaNO3 , dissolve it in some water in a 750 mL volumetric flask, then bring the volume up to
the line with water.
14. When 20.0 g FeS2 are reacted with 10.0 g O2, how many grams of Fe2O3 are produced?
4FeS2 + 11O2
20.0 g FeS2 x 1 mole FeS2
1 mole O2
32.00 g O2
2Fe2O3
+
x 2 mole Fe2O3 x 159.7 g Fe2O3
119.99 g FeS2
10.0 g O2 x

4 mole FeS2
= 13.3 g Fe2O3
1 mole Fe2O3
x 2 mole Fe2O3 x 159.7 g Fe2O3
11 mole O2
8SO2
= 9.07 g Fe2O3
1 mole Fe2O3
ANSWER: 9.07 g Fe2O3
15. In problem #14, if you run the reaction and produce 5.3 g Fe2O3, what is your percent yield?
% Y = AY/TY x 100 = 5.3 g/ 9.07 g x 100 = 58 %
16. A gas occupies a volume of 205 mL at 27C and 0.974 atm pressure. Calculate the volume the gas
would occupy at STP.
P1V1/T1 = P2V2/T2
(0.974 atm)(205 mL) / 300 K = (1 atm) V2 / 273 K
V2 = 182 mL or 0.182 L
17. How many grams of O2 gas are in a sample occupying 0.500 L at 170.0 mmHg and 25C?
PV = nRT
P = 170.0 mm Hg x 1 atm / 760 mm Hg = 0.2237 atm
PV/RT = n
(0.2237 atm)( 0.500 L)
= 0.00457 mol O2
(0.0821 L atm / mol  K)(298 K)
0.00457 mol O2 x 32.00 g / 1 mole = 0.146 g O2
18. What is the density of He gas at STP?
Assume you have 1 mole of He, then mass = molar mass = 4.003 g. Since we are at STP, 1 mole has a
volume of 22.4 L.
d = mass/vol = 4.003 g / 22.4 L = 0.179 g/L
19. For the following reaction, label each reactant as an acid or a base (according to the Bronsted / Lowry
definition):
HCO3base
+
HF

H2CO3
+
F-
acid
(reasoning: acids are H+ donors, bases are H+ acceptors)
20. A HCl solution is used to titrate a Ca(OH)2 solution. What is the Molarity of the HCl if 50.0 mL of it
is required to titrate 50.0 mL of 0.500 M Ca(OH)2 ?
2 HCl(aq) + Ca(OH)2 (aq) 
CaCl2 (aq) + 2 H2O(l)
(0.0500 L)(0.0500 mol/L) = 0.0250 mole Ca(OH)2
MHCl = moles HCl / L HCl =
x
2 mole HCl
1 mole Ca(OH)2
0.0500 mole HCl / 0.0500L = 1.00 M HCl
= 0.0500 mole HCl
Draw Lewis structures for the following molecules or ions:
21) NH4+
22) SO2
Note: these are too hard to type so I will describe the # of bonds and lone pairs. You should draw them.
4 single bonds, no lone pairs
1 single bond, 1 double bond.
3 lone pairs on O with single, 2 lone pairs on
O with double, one lone pair on S.
23) NO3-
24) HNO2
one double bond, two single bonds
3 lone pairs on O's with single,
2 lone pairs on O with double, none on N.
O=N–O–H
with 2 lone pairs on each O, one on N and
none on H.
25) H2O
2 single bonds, 2 lone pairs on O, none on H's
Group
Period
1
IA
1A
1
H
1
1.008
2
3
4
5
7
2
IIA
2A
13
14
IIIA IVA
3A 4A
15
VA
5A
16
17
VIA VIIA
6A 7A
2
He
4.003
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.941
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
12
17
18
Na
Mg
22.99
19
24.31
3
IIIB
3B
4
IVB
4B
5
VB
5B
20
21
22
23
6
7
VIB VIIB
6B
7B
24
25
8
9
10
------- VIII ------------- 8 -------
11
IB
1B
12
IIB
2B
13
14
15
16
Al
Si
P
S
26.98
28.09
30.97
32.07
35.45
39.95
26
29
30
31
32
33
34
35
36
27
28
Cl Ar
K
Ca
Sc
Ti
V
Cr Mn Fe Co
Ni
Cu
Zn Ga Ge As Se Br Kr
39.10
40.08
44.96
47.88
50.94
52.00
54.94
55.85
58.47
58.69
63.55
65.39
40
41
42
43
44
45
37
38
39
Rb
Sr
Y
85.47
87.62
56
55
6
18
VIIIA
8A
46
47
48
49
Zr Nb Mo Tc Ru Rh
Pd
Ag
Cd
In
88.91
91.22
92.91
106.4
107.9
112.4
114.8
57
72
73
80
81
95.94
(98)
101.1
74
75
76
Cs
Ba La* Hf Ta
132.9
137.3
138.9
178.5
180.9
183.9
186.2
87
88
89
104
105
106
107
Fr
(223)
102.9
Lanthanide Series*
Actinide Series~
72.59
74.92
78.96
79.90
50
51
52
53
54
I
Xe
126.9
131.3
85
86
Sn Sb Te
118.7
127.6
83
84
77
78
79
Pt
Au
Hg Tl
Pb
190.2
190.2
195.1
197.0
200.5
207.2
108
109
110
111
112
114
116
118
---
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---
---
---
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()
()
()
()
()
()
(227)
(257)
(260)
(263)
(262)
(265)
(266)
58
59
60
61
62
63
64
66
Dy
Ho Er Tm Yb Lu
140.1
140.9
144.2
(147)
150.4
152.0
157.3
158.9
162.5
164.9
167.3
168.9
173.0
175.0
90
91
92
93
94
95
96
97
98
99
100
101
102
103
U
(238)
70
(210)
65
(231)
69
209.0
Tb
Th Pa
68
Bi Po
Ce Pr Nd Pm Sm Eu Gd
232.0
67
204.4
82
121.8
71
Np Pu Am Cm Bk
Cf
Es Fm Md No Lr
(237)
(249)
(254)
(242)
83.80
Ir
W Re Os
Ra Ac~ Rf Db Sg Bh Hs Mt
(226)
69.72
(243)
(247)
(247)
(253)
(256)
(254)
(257)
At Rn
(210)
(222)