Предмет: Химия, автор: ost2234

. Яка масова частка Феруму у сполуці: FePO4

Ответы

Автор ответа: jinsun20

Ответ:

W(Fe)=M(Fe)/M(FeO)=56/72≈0.78≈78%

w(Fe)=M(Fe)*2/M(Fe2O3)=56*2/160=0.7=70%

w(Fe)=M(Fe)*3/M(Fe3O4)=56*3/232≈0.724≈72.4%

Объяснение:

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