1?(3-2.6))/(3?(5.1-2.6)) = (510?0.4)/(3?2.5) = 204/7.5 = 27.2
The same problem could be solved with the help of a little algebra by
the rule already given, as thus: the sp. g. of a mixture equals the sum
of the _weights_ of the constituents divided by the sum of the
_volumes_. Then 100 grams of the tailings with _x_ per cent. of pyrites
contain 100-_x_ per cent. of quartz. The sum of the weights is 100. The
volume of the pyrites is _x_/5.1 and of the quartz (100-_x_)/2.6.
Then we have by the rule
3 = 100/((_x_/5.1)+(100-_x_)/2.6)
3 = 1326/(510-2.5_x_)
204 = 7.5_x_
and _x_ = 27.2
If the percentage (P) and sp. g. (H) of one constituent and the sp. g.
(M) of the mixture are known, the sp. g. of the other constituent may be
calculated by the following formula, in which _x_ is the required sp.
g.:
_x_ = ((100-P)?M?H)/((100?H)-(P?M))
For example, "tailings" (sp. g. 3.0) containing 27.2 per cent. of
pyrites (sp. g. 5.1) will contain (100-27.2), 72.8 per cent. of earthy
matter having a mean sp. g. of _x_:
_x_ = ((100-27.2)?3?5.1)/((100?5.1)-(27.2?3))
= 1113.84/428.4 = 2.6
The differences in sp. g. corresponding to differences in strength have
been carefully determined and tabulated in the case of the stronger
acids and of many other liquids. Such tables are given at the end of
this book.
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