RECITATION – CHAPTER 22
1. The mass of an electron is
m
= 9.11
×
10
–31
kg, so the number of electrons in a collection
with total mass
M
= 75.0 kg is
31
31
75.0 kg
8.23
10
electrons.
9.11 10
kg
M
n
m

=
=
=
×
×
The total charge of the collection is
(
29
(
29
31
19
13
8.23
10
1.60
10
C
1.32
10
C.
q
ne

= 
= 
×
×
= 
×
2. There are two protons (each with charge
e
) in each molecule, so there are
2
A
N
protons in a
mole of hydrogen.
The total quantity of positive charge is
(
29
(
29
(
29
23
19
5
2
2
6.02
10
1.60
10
C
1.93
10
C
0.193 MC.
A
q
N e

=
=
×
×
=
×
=
3. Each electron has charge –
e
.
Removing one electron leaves the coin with charge
e
.
The
number of electrons that must be removed to leave a net charge of
7
1.0
10
C
q

=
×
is
7
11
19
1.0
10
C
6.3
10 .
1.6
10
C
q
n
e


×
=
=
=
×
×
4. 250 cm
3
of water has a mass of 250 g since the density of water is 1.0 g/cm
3
.
Since one
mole of water has a mass of 18 g, there are 250/18 = 14 moles of water in this volume.
There
are ten protons (each with charge
e
) in each molecule of H
2
O, so the total quantity of positive
charge is
(
29
(
29
(
29
(
29
(
29
23
19
7
14
10
140
6.02
10
1.60
10
C
1.3
10 C
13MC.
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