Chemistry, asked by veeraranjane38, 23 days ago

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​

Answers

Answered by dhruvpratapsinghh
0

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%.

Answered by grbh23
6

Answer:

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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