1.

NaHC_(2)O_(4) is 0.1 M when neutralised with NaOH. Hence, it is ...... when oxidised with MnO_(4-)//H^(+) .

Answer»


Solution : `NaHC_(2)O_(4) to n=2`(as a REDUCING agent)
So, `0.1xx2=0.2N NaHC_(2)O_(4)` as a RA.


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