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CHM1040
Answer to Exercise Set 2
FUNDAMENTAL LAWS OF CHEMISTRY
1.
Based on the Law of Conservation of Mass, if 16.0 g of oxygen react completely with 2.0 g hydrogen then
18.0 g of water are produced. Based on the Law of Definite Composition, if 16.0 g of oxygen react with 2.0
g of hydrogen then 4.0 g of oxygen will react with 0.5 g of hydrogen. (Both cases have an oxygen to
hydrogen mass ratio of 8:1.) Therefore based on the Law of Conservation of Mass, there will be 1.5 g of
hydrogen left and will produce 4.5 g of water.
2.
Answer
i. We can use the Law of Multiple Proportions.
ii. For simplicity take 1.00 g G to be 1.00 part of G and 3.00 g M to be 1.00 part of M.
Note: Two different elements cannot both have the masses equal to one.
Therefore, in compound I there is one part G and one part M or a formula of GM
Likewise, in compound II there is one part G and two parts M or a formula of GM
2
Finally, in compound III there is one part G and three parts M or a formula of GM
3
.
iii. Now 1.00 g G is proportional to 1 units of G and 3.00 g M is proportional to 2 units of M.
Therefore,
compound I is GM
2
, compound II is GM
4
, and compound III is GM
6
.
iv. Now 1.00 g G is proportional to 2 units of G and 3.00 g M is proportional to 5 units of M.
Therefore,
compound I is G
2
M
5
, compound II is G
2
M
10
or GM
5
, and compound III is G
2
M
15
.
v. Now 1.00 g G is proportional to 1 unit of G and 6.00 g M is proportional to 1 unit of M.
Therefore,
compound II is GM, compound I is GM
0.5
or G
2
M, and compound III is GM
1.5
or G
2
M
3
.
QUESTIONS ON FUNDAMENTAL SUBATOMIC PARTICLE
3.
His experiments showed that regardless of the material used for the cathode the particles produced all had
the same properties (e/m) and therefore were in fact the same particles.
Since all materials produced the
same particles they must in fact be present in all materials and hence are fundamental to all matter. The same
could not be said for the canal rays produced because they had different properties (e/m) which varied
depending on the gas left in the tube.
4.
Neutrons could not be isolated and characterized by the same methods as the proton and electron because the
neutron has no charge which is required of a particle which is to be affected by a magnetic or electric field.
5.
a.The droplets carry different charges because there may be 1, 2, 3 or more excess electrons on each droplet.
The electron charge then is likely to be the greatest common factor in all the observed charges. Assuming
this to be so, we calculate the apparent electronic charge from each droplet as follows:
A:
9.6132 x 10
19
C / 2 e

= 4.8066 x 10
19
C/e

B:
14.4198 x 10
19
C / 3 e

= 4.8066 x 10
19
C/e

C:
4.8066 x 10
19
C / 1 e

= 4.8066 x 10
19
C/e

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D:
19.2264 x 10
19
C / 4 e

= 4.8066 x 10
19
C/e

The value 4.8066 x 10
19
C could be taken as the unit charge on a single electron as this value may
be multiplied by integers (whole numbers) to yield the other charges observed.
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 Spring '10
 Staff
 Chemistry

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