_When substances are mixed by volume_, the sp. g. of the mixture is the mean of those of its constituents, and may be calculated in the usual way for obtaining averages. 1 c.c. of a substance having a sp. g. of 1.4 mixed with 1 c.c. of another having a sp. g. of 1.0 will yield 2 c.c. of a substance having a sp. g. of 1.2. If, however, we write gram instead of c.c. in the above statement, the resulting sp. g. will be 1.16. The simplest plan is to remember that the sp. g. is the weight divided by the volume (sp. g. = w/v) and the sp. g. of a mixture is the sum of the weights divided by the sum of the volumes (sp. g. = (w + w' + w", &c.)/(v + v' + v", &c.)). In the above example the sum of the volumes is 2 c.c.; the weights (got by multiplying each volume by its corresponding sp. g.) are 1.4 gram and 1 gram. The sum of the weights divided by the sum of the volumes is 2.4/2 or 1.2. The sp. g. of a mixture of 10 c.c. of a substance having a sp. g. of 1.2, with 15 c.c. of another having a sp. g. of 1.5 may be thus found:-- sp. g. = (12+22.5)/(10+15) = 1.38 multiply each volume by its sp. g. to get its weight: 10×1.2 = 12 15×1.5 = 22.5 add these together (12+22.5 = 34.5) and divide by the sum of the volumes (10+15 = 25): 25)34.5(1.38 25 -- 95, &c. The sp. g. will be 1.38, provided the mixture is not accompanied by any change of volume. The same formula will serve when the proportion of the ingredients is given by weight. A mixture of 4 parts by weight of galena (sp. g. 7.5) with 5 parts of blende (sp. g. 4) will have a sp. g. of 5.06: sp. g. = (4+5)/(0.53+1.25) = 9/1.78 = 5.06 It is necessary in this case to calculate the volumes of the galena and of the blende, which is done by dividing the weights by the sp. gravities: thus, 4 divided by 7.5 gives 0.53 and 5 divided by 4 gives 1.25. The converse problem is a little more difficult. Given the sp. g. of a mixture and of each of the two ingredients, the percentage by weight of the heavier ingredient may be ascertained by the following rule, which is best expressed as a formula. There are three sp. gravities given; if the highest be written H, the lowest L and that of the mixture M, then: Percentage of heavier mineral = (100×H×(M-L))/(M×(H-L)) Suppose a sample of tailings has a sp. g. of 3.0, and is made up of quartz (sp. g. 2.6) and pyrites (sp. g. 5.1): then the percentage of pyrites is 27: (100×5.1×(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