1.

The percentage of Fe​^( +3)​ ion present in Fe0.93O1.00 is :

Answer»

`15% `
`5.5% `
`10.0% `
`11.5% `

SOLUTION :
`X(+2)+3(0.93 -x )-2 =0`
`X=0.79`
`Fe ^(+2)= 0.79 Fe^(+3)= 0.14`
`% Fe ^(+3)=(0.14)/(0.93 )xx 100 =15 %`


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