Question

solve this I need the answer plz​

Answer 1

Answer:

Explanation:

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