\(AO_{2}\) disproportionates in to \(AO_{4}^{-1}\) and \(A^{ n+}\) ion. If the mole ratio of \(AO_{2}\) undergone oxidation and reduction is 2:3, the value of "n" is

AO_(2) disproportionates into AO_(4)^(-1) and A^(n+) ion. If the mole ratio of AO_(2) undergone oxidation and reduction is 2:3 , the value of n is

3MnO_2 + 4Al to 3Mn + 2Al_2O_3 + Heat Identify the substances undergone oxidation and reduction in this reaction

One mole of MnO_(4)^(2-) disproportionates to yield :

In the disproportionation reaction ( unbalanced ) , Br_(2) + OH^(-) rarr Br^(-) + BrO_(3)^(-) + H_(2)O The ratio of Br_(2) molecules undergoing oxidation and reduction is "P/Q" then "P+Q" is

n' arithmetic means are there between 4 and 36. If the ratio of 3rd and (n-2)th mean is 2:3, find the value of n.

3MnO_(2) + 4Al rarr 3 Mn + 2Al_(2)O_(3)+ heat. Identify the substances undergone oxidation and reduction reactions.

In the reaction, A^(n+)+MnO_(4)^(-) rarr A^(5) +Mn^(2+) if 0 0.05 mole of A^(n+) is oxidized by 0.02 mole of MnO_(4)^(-) , the value of 'n' is

One mole of \(Fe^{2+}\) ions are oxidized to \(Fe^{3+}\) ion with the help of "x" moles of \(KMnO_{4}\).The value of "5x" is

If (n!)/(2!(n-2!)) and (n!)/(4!(n-4!)) are in the ratio 2:1, find the value of n.

Br_2 undergoes disproportionation reaction in basic medium to give Br^(ɵ) ion and BrO_3^(ɵ) (bromate) ion in reduction and oxidation reaction.