The following table shows an actual
example:--
+--------------+---------------+-----------+
| Chalk taken. | Gas obtained. | Standard. |
+--------------+---------------+-----------+
| 0.0873 gram | 17.8 c.c. | 0.4904 |
| 0.1305 " | 27.3 " | 0.4780 |
| 0.1690 " | 35.8 " | 0.4721 |
| 0.1905 " | 40.4 " | 0.4715 |
| 0.2460 " | 52.5 " | 0.4686 |
| 0.3000 " | 64.0 " | 0.4687 |
+--------------+---------------+-----------+
It will be seen that the standard decreases as the quantity of chalk
increases; this points to a deficiency in the quantity of gas evolved.
Then
0.3000 = 64.0 c.c.
0.0873 = 17.8 "
------ = ----
0.2127 = 46.2 "
and 0.2127?100/46.2 = 0.4604. Then, multiplying the weight of chalk
taken by 100, and dividing by 0.4604, we get the calculated results of
the following table:--
+--------------+------------+-----------------+-------------+
| Chalk taken. | Gas found. | Gas calculated. | Difference. |
+--------------+------------+-----------------+-------------+
| 0.0873 gram | 17.8 c.c. | 18.9 c.c. | -1.1 c.c. |
| 0.1305 " | 27.3 " | 28.3 " | -1.0 " |
| 0.1690 " | 35.8 " | 36.7 " | -0.9 " |
| 0.
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