-~
6
cost'
p13
3J
3 3
log
(cos u)
=
log
6
+
log
3
-
2
log
14
=
=
0.767185
+
0.71568
-
1.71919
=
=
9.773474
-
10
u =
53°
5428' 14"
The other
3 roots
are
-p
3
-p
3
2J
24
2J
cos
cos
cos
U
3
`U
`3
`U
`3
p
-.5
+
120°)
+
240°)
Only
positive
roots
are
usable
in
the
problem.
Since
2
J -
£
is
positive,
the other factor
must
also be
positive
if the
product
is to
be
positive.
^
is
an
acute
angle,
hence its cosine
positive,
and the first
root is usable. The cosine of the second
angle
(situated in the
second
quadrant)
is
negative,
thus the root is
unusable.
The cosine
of the third
~
+
240° is smaller than 260°, hence it
still
lies in
the third
quadrant,
and hence the 3d root is unusable.
log
x
=
log
2
+
2
cos
17°
49'
21"
log
x
=
0.61420
log
10
p
=
0.38580
p
=
0.243
27.
MATURA
EXAMINATION (G) CHEMISTRY
[21
September
1896, 2-4 P.M.]
Albert Einstein
PROBLEM.
How
many
liters of 30%
hydrochloric acid,
whose
specific
gravity
is 1.15,
are
obtained
from 39.5 kg table salt when neutral sodium
sulfate forms
as
the reaction side
product.
Describe the substances
formed
during
the
reaction.
23