When $KMnO_4$ is reduced with oxalic acid in acidic solution,the oxidation number of $Mn$ changes from

Iodine reacts with concentrated $HNO_3$ to yield $Y$ along with other products. The oxidation state of iodine in $Y$ is

What are the oxidation states of the two sulfur atoms in thiosulfuric acid $(H_2S_2O_3)$?

The sum of the oxidation numbers of all the carbons in $C_6H_5CHO$ is

In the reaction,$MnO_{4(aq)}^{-} + Br_{(aq)}^{-} \rightarrow MnO_{2(s)} + BrO_{3(aq)}^{-}$,the correct change in oxidation number of the species involved is

One mole of $N_2H_4$ loses $10 \ mol$ of electrons to form a new compound $Y$. Assuming that all nitrogen atoms appear in the new compound,what is the oxidation state of $N$ in $Y$? (There is no change in the oxidation state of hydrogen)