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The % of Fe^(+2) in Fe(0.93)O(1.00) is ....

The `%` of `Fe^(+2)` in `Fe_(0.93)O_(1.00)` is .

A

`15%`

B

`85%`

C

`93%`

D

`7%`

Text Solution

Verified by Experts

The correct Answer is:
B
`(10)/(12)xx2xx6.022xx10^(23)`= total neutrons
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