`A_(2)O_(X)` is oxidised to `AO_(3)^(-)` by `MnO_(4)^(-)` in acidic medium. If `1.5times10^(-3)` mole of `A_(2)O_(X)` requires 40ml of `0.03M-KMnO_(4)` solution in acidic medium. The value of `x`

An oxide X_(2)O_(n) is oxidised to XO_(3)^(-) by MnO_(4)^(-) in acidic medium.If 1.5times10^(-3) moles of X_(2)O_(n) requires 1.8times10^(-3) moles of MnO_(4)^(-) for complete oxidation.Assuming that oxidation state of oxygen is -2 in all compounds.The value of

The ion A^(n+) is oxidised to AO_(3)^(-) by MnO_(4)^(-) changing to Mn^(2+) in acid medium. Given that 2.68 xx10^(-3) mole of A^(n+) required 1.61 xx10^(-3) mole of MnO_(4)^(-) . What is the value of n.

The volume of 10 volume H_(2)O_(2) solution that decolourises 200 mL of 2 N KMnO_(4) solution in acidic medium is

In a titration, H_(2)O_(2) is oxidised to O_(2) by MnO_(4)^(-) . 24 mL of 0.1M H_(2)O_(2) requires 16 mL of 0.1M MnO_(4)^(-) solution. Hence MnO_(4)^(-) changes to :

10 moles of ferric oxalate is oxidised by x mole of MnO_(4)^(-) in acidic medium. The value of 'x' is :

A 3 mole mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) required 100 mL of 2M KMnO_(4) solution in acidic medium. Find the mole of FeSO_(4) in the mixture.