Question

A metal form two oxides higher oxide is 80perce metal and0.72g lower oxide is oxidied . So that thisdata is in the law of multiple proportion

Answer 1

Explanation:

80% of higher oxide 0.80 gms : 0.64 gm of metal in the higher oxide. Thus before oxidation, there is 0.64 gm of metal in the lower oxide. So there is 0.72 gms - 0.64 gms = 0.08 gms of oxygen in lower oxide. and 0.80 - 0.64 = 0.16 gms of oxygen in lower oxide.