Chemistry Reference
In-Depth Information
Because 0.618
8
mol of solute is 24.4
4
g, the molar mass is
(b)
Assume 100.00 g of water in each solution.
a
12.00 g solute
88.00 g H
2
O
24.4
4
g
0.618
8
mol
39.5 g/mol
100.00 g H
2
O
b
13.64 g solute
a
2.00 g solute
98.00 g H
2
O
15.110
P
P
°
X
231 torr
(268 torr)
X
100.00 g H
2
O
b
2.041 g solute
231 torr
268 torr
X
0.861
9
There is a total of 15.68 g solute in 215.68 g solution, so the
percent by mass is
Let
y
number of moles of solute per mole of benzene.
15.68 g solute
215.68 g solution
100.0%
Then
7.270% solute
1.000 mol benzene
1.000 mol
y
X
0.861
9
0.861
9
(1.000
y
)
1.000 mol
16 Oxidation Numbers
16.1
0.861
9
y
0.138
1
mol
(a)
MA
3
(b)
YZ
3
y
0.160 mol
16.2
Electrons are present, and at least one type of ion must be
present to balance the charge.
Check:
1.000 mol
1.160 mol
0.862
16.3
The sum of the oxidation numbers of all the atoms is equal to
the charge on the species (rule 1).
(a)
15.113
Assume any mass of
CH
3
OH,
such as 70.0 g. (Then there
is 30.0 g
CH
2
O.
) Calculate the number of moles of each
12
(b)
2
(c)
4
substance:
16.4
(a)
3
(b)
1
(c)
2
a
1 mol CH
3
OH
32.0 g CH
3
OH
16.5
(a)
The oxidation number of copper is
2;
that of the sulfur
70.0 g CH
3
OH
b
2.18
8
mol CH
3
OH
atom is
(b)
The oxidation number of copper is
2.
1;
that of the sulfur
a
1 mol CH
2
O
30.0 g CH
2
O
30.0 g CH
2
O
b
1.00 mol CH
2
O
atom is
(c)
Copper(II) sulfide and copper(I) sulfide, respectively.
2.
The total number of moles is 3.18
8
mol, and the mole frac-
tion of
16.6
The elemental silver (on the left) has an oxidation number of
0 (rule 2), and the silver in
CH
3
OH
is
Ag
2
O
has an oxidation number of
2.18
8
mol
3.18
8
mol
0.686
1.
Note that silver always has
1
as its oxidation number
in its compounds.
16.7
(a)
The oxidation number of the nitrogen atom is
3;
that of
15.120
The number of moles of gas in a liter is given by the ideal gas
law:
each hydrogen atom is
1,
and that of the chlorine atom
is
(b)
The oxidation number of sulfur is
1.
(1.00 atm) (1.00 L)
(0.0821 L
#
atm/mol
#
K) (298 K)
0.0409 mol
PV
RT
4;
that of each
n
fluorine atom is
1.
The number of moles of ammonia in a liter of 15 M solution
is 15 mol. The ratio of numbers of moles is
16.9
Right
16.10
(a)
Two electrons for each of three A's, or six electrons in all
(b)
Six electrons
(c)
3 : 1
15 mol
0.0409 mol
370
1
16.11
The element that changed oxidation number has already
been balanced, so oxygen needs to have an oxidation number
of
15.121
For 1.00 kg of water:
a
32.0 g CH
3
OH
1 mol CH
3
OH
1 kg
1000 g
2.
55.2 mol CH
3
OH
b
a
b
16.12
Six electrons must be added to the left side.
1.76
6
kg CH
3
OH
16.14
(a)
Double substitution is one reaction type described in
Chapter 8 in which oxidation numbers do not change.
Combinations and decompositions of
compounds
to
form new compounds (for example, calcium oxide plus
carbon dioxide yielding calcium carbonate, or vice
versa) also have no changes of oxidation numbers.
(b)
No. (We balanced simple oxidation-reduction equations
in Chapter 8.)
a
1 mol H
2
O
18.0 g H
2
O
1000 g H
2
O
b
55.5
6
mol H
2
O
55.5
6
mol H
2
O
1.76
6
kg CH
3
OH
31.5 m H
2
O
Molality
15.123
The mole fraction of water must be 0.809 because the total
of both mole fractions must equal 1.000. In a solution con-
taining 1.000 mol total, there are 0.191 mol of methyl alco-
hol and 0.809 mol of water. The mass of the water is
. .
16.15
. .
. .
. .
a
18.0 g H
2
O
1 mol H
2
O
Cl
P
Cl
. .
. .
. .
0.809 mol H
2
O
b
14.5
6
g H
2
O
Cl
. .
Thus,
The oxidation number of phosphorus is that of
the sulfur atom is and that of each of the chlo-
rine atoms is Thus, the sulfur in is like
the oxygen in and the phosphorus has the same oxi-
dation number in both compounds.
5
0
5;
6
8
2;
0.191 mol alcohol
0.0145
6
kg water
13.1 m
Molality
7
8
1.
PSCl
3
POCl
3
,
15.124 (a)
Assume 100.0 g of each solution. Then we have
12.00 g solute
2.00 g solute
200.0 g solution
16.17
(a)
4
(b)
6
(c)
4
7.00% solution
(d)
2
(e)
4
