CHEMISTRY WKST KEY: Ch. 6 Review
p.1
23
1)
6.022 x 10
2)
a) 24.31 g Mg
3)
a) ratio of the atoms
4)
a) BaCl2
137.33 g + 2(35.45 g) = 208.23 g
b) P4O10
4(30.97g) + 10(16.00 g) = 283.88 g
c) Pb(C2H3O2)4
207.2 g + 8(12.01 g) + 12(1.01 g) + 8(16.00 g) = 443.40 g
d) Ba(OH)2•8H2O
137.33 g + 2(16.00 g) + 2(1.01 g) + 8(18.02 g) = 315.51 g
5)
b) 6.022 x 10
)
(
a)
)(
(
(
a) NaCl
)(
)(
b) H2O
)( 8
8
)
8
)
)
)(
9)
(
)
H
(
c)
7)
H
)(
)
8
b)
8)
H
8
)
H
(
c)
6)
H
(
H
b)
Mg atoms
b) mole ratio of the elements
(
a)
23
)
CHEMISTRY WKST KEY: Ch. 6 Review
10)
g solute = (2.25 M)(40.00 g/mol)(1.750 L) = 158 g NaOH
11)
H
( 8
)(
(
12)
a)
CuSO4
)
63.55 g + 32.07 g + 4(16.00 g) = 159.62 g
(
b)
Ba(NO3)2
)
137.33 g + 2(14.01 g) + 6(16.00 g) = 261.35 g
(
13)
8
)
H
(
(
)
)
)
14)
H
100x = 22.25x + 11125 g
77.75x = 11125 g
x = 143.1 g Mg(NO3)2
p.2
CHEMISTRY WKST KEY: Ch. 6 Review
p.3
15)
ZnSO4
65.38 g + 32.07 g + 4(16.00 g) = 161.45 g
8
161.45x = 3269 g
x = 20.25 g Zn
16)
8
0.342 x = 22340 g
x = 65321.64 g = 65300 g
17)
a) C3H4O3
18)
b) CH
8
8
8
8
S1.00Cl2.01
SCl2
- assume 100 g sample
60.8 % Na
60.8 g Na
28.5 % B
28.5 g B
10.5 % H
20)
8
2.62 g S
8.44 g – 2.62 g = 5.82 g Cl
19)
c) C14H18N2O5
C2HCl
8
8
10.5 g H
8
H
8
8
Na1.00B1.00H3.94
H
H
H
60.48
C2x3H1x3Cl1x3
C6H3Cl3
NaBH4
CHEMISTRY WKST KEY: Ch. 6 Review
21)
p.4
- assume 100 g sample
42.87 % C
42.87 g C
3.598 % H
3.598 g H
28.55 % O
28.55 g O
25.00 % N
25.00 g N
8
8
H
H
H
H
8
8
8
8
8
8
C2.001H2.00O1.000N1.000
8
8
C2H2ON  empirical formula
8
C2H2ON ≈ 56
8
22)
C2x3H2x3O1x3N1x3
(
a)
)
(
b)
)(
)(
(
c)
d)
8
C6H6O3N3  molecular formula
)(
8
)
)
8
K = °C + 273 = 12.9°C + 273.15 = 286.05 K = 286.0 K
e)
°F = 1.8(°C) + 32 = 1.8(−44°C) + 32 = −79 + 32 = −47°F
“ z ”
23)
A Kelvin
24)
The atomic mass for hydrogen (as well as for all elements) is the average of ALL the hydrogen atoms in the
universe.
25)
The purpose of distilling a solution is to separate the liquid from the solutes dissolved in the liquid.
26)
This was a physical change because the liquid, which was H2O, never changed—it was still H2O as steam and H2O
when converted back into a liquid. The color seen in the solution was from something that was dissolved in the
liquid.
27)
When steam is converted back into liquid water, the water loses energy. This is an exothermic process.
CHEMISTRY WKST KEY: Ch. 6 Review
p.5
28)
When distilling the colored substance in the solution is left behind in the boiler—only the pure liquid (H2O in our
case) was distilled. When filtering, the filter paper does NOT stop any substance still dissolved in the liquid.
Therefore, the colored substance passes through the filter paper since it was still dissolved in the H 2O.
29)
Solubility of the solids in the liquid determines whether or not they can be separated by filtering. One of the solids in
the mixture is soluble in H2O while the other substance was insoluble in H2O. The insoluble solid will be stopped by
the filter paper while the soluble solid will pass through.
30)
a)
1 mol Fe(s) : 1 mol Cu(s)
b)
Since the ratio between Fe(s) and FeSO4(aq) is 1:1, there will be 0.75 mol FeSO4 produced.
c)
CuSO4 when dissolved in H2O gives a blue color to the solution. The filtrate was still blue after filtering,
indicating that not all the CuSO4 reacted. Thus, there was enough CuSO4 to make sure all the Fe reacted.
31)
You will find the mass of the object by weighing it on a balance. There are two ways to find the volume:
 If the object is small enough, place an known amount of water in a graduated cylinder. Place the object into
the water and see how much the volume of the water in the graduated cylinder increases—this will be the
volume of the object.
 If the object is too large for a graduated cylinder, put water into a container with an overflow spout—fill it with
enough water to have some come out of the spout. After the water quits coming out of the spout, place a
graduated cylinder under the spout, place the object into the water, and catch the water that comes out of
the spout. The volume of water that comes out will be the volume of the object.
32) 1.00
33)