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When 100 " mL of " 0.06 M Fe (NO3)3,50mL...

When 100 " mL of " 0.06 M Fe `(NO_3)_3,50mL ` of `0.2M FeCl_3` and `100mL` of `0.26M` Mg `(NO_3)_2`, are mixed In the final solution…..
`[Fe^(3+)]=` ….
`[NO_3^(ɵ)]=`…..
`[Cl^(ɵ)]=`……
`[Mg^(2+)]=`……

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