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Metal oxide have 72.4% and70% metal second oxide of metal is M2O3 find first

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3 years ago

Ishika dammani
44 Points
```							Ratio of metal and oxygen in second oxide, M2O3 = 70:30Let molecular mass of metal = MTherefore, the percentage by weight of the metal in the oxide(2×M×100)/(2×m+3×16)=70200M=140M+336060M=3360M=56kg Now,Moles of metal in first oxide = 72.4 / 56 = 1.29Moles of oxygen in second oxide = 27.6 / 16 = 1.72Ration of moles of metal and oxygen = 1.29 : 1.72  = 1 : 1.33  = 3 : 4Hence, formula of the first oxide = M3O4
```
3 years ago
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