Sulphur and oxygen are known to form two compounds The sulphur content in one of these is 51% while in the other is 41% Show that these data is in agreement with the law of multiple proportions
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therefore if % of S = 1% , the % of O will be = 49 /51= 0.96. This discussion on Sulphur and oxygen are known to form two compounds. The Sulphur content in one of these is 51% and in other is 41%.
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In the first compound % of S = 51% , hence % of O will be = 49%
therefore if % of S = 1% , the % of O will be = 49 /51= 0.96.
In the second compound
the % of S = 41% , hence % of O will be = 59%
if % of S = 1% , % of O = 59/ 41 = 1.439
therefore the simple ratio between O: O in these oxide = 1.439/0.96 : 0.96/ 0.96 = 1.5 : 1 = 3:2 hence the law
Ratio of oxygen in oxides of sulphur is 3:2
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