In acidic medium,one mole of $MnO_4^-$ accepts how many moles of electrons in a redox process?

In the half-reaction $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,the equivalent weight of $Cr_2O_7^{2-}$ is obtained by dividing its molecular weight by:

In which reaction is there a change in the oxidation state?

For the redox reaction $MnO_4^- + C_2O_4^{2-} + H^{+} \to Mn^{2+} + CO_2 + H_2O$,the correct coefficients of the reactants for the balanced reaction are:
$MnO_4^-$ : $C_2O_4^{2-}$ : $H^{+}$

The equivalent weight of $MnSO_4$ is half of its molecular weight when it is converted to

$KMnO_4$ oxidises oxalic acid in acidic medium. The number of $CO_2$ molecules produced per mole of $KMnO_4$ is