CHM
1845
07102
Chapter
4
Reducing
Reduction
oxydation
Agent
⑦
2
-
27
2
-
0
↓
·
3
-
7
F2s
+
Bi
*
i
Bi
+
15
Cr0+
NHy
:(u
+
vz
+
it,12
-
oxydazing
7(s)
Agent
oxydation
Reduction
Molarity:
#
of
moles
of
solute
pen
x
0.42
moles
of
NaOH
M
=
1
solution
<-
->
iL
->
Molanity:?
103
1
HCl
+
Na
OH
->
Nacd
+
2
r)
&60mL
->
20.260L
(9)
(af)
Molarity
as
convertion
factor.
:2
=
1615
=
7.6M
1.6M
7.6
moles
L
-
is
les
#iten
23.99
3
moles.
S.2X
=
76.72
moles
7.07
How
many
moles
of
NaOH
are
these
in
6.7
Lof7.6
rsolutions,
S
=
16.72
moles
40
+600
Ind
Ind
6.37.
mole
=
9.8
moles
[
/
=
644.8grans
=
640
How
many
of
1.67
M
HC
is
required
to
completely
neutralize
NaOH?
1,
=
3.11
M.V.
M2a
W
=
5.2
L
M
=
7.6 7 M
V
=?
2
NaOH
+THzSOy
->
iNa,
SO,
+
2
HO
5.2LxsmbratH
nol
ASOy
=
8.0s
moe
e
x
8.06
nols
=
b.09L
.
Y
moses
1
