In a reaction one mole of $N_{2} H_{4}$ loses ten moles of electrons to form a new compound Y. Assuming that all nitrogen appears in the new compound. What is the oxidation number of N in Y?

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Solution

The oxidation state of Nitrogen in $N_{2} H_{4}$ is $- 2$
Therefore,
Two $N$ makes it $- 4$ ; Hence they loose $10 e^{-}$
Total charge $+ 6$ and since $H$ is still $+ 1$
Two $N$ divides the $+ 6$
Then oxidation state of $N$ is $+ 3$ in the new compound.

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mole of

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Q. A mole of

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N_{2} H_{4}

loses

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A mole of $N_{2} H_{4}$ loses 10 mol of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation number of hydrogen.)

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In a reaction one mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all nitrogen appears in the new compound. What is the oxidation number of N in Y?

The oxidation state of Nitrogen in N2H4 is −2 Therefore,Two N makes it −4 ; Hence they loose 10e−Total charge +6 and since H is still +1 Two N divides the +6Then oxidation state of N is +3 in the new compound.

In a reaction one mole of $N_{2} H_{4}$ loses ten moles of electrons to form a new compound Y. Assuming that all nitrogen appears in the new compound. What is the oxidation number of N in Y?

The oxidation state of Nitrogen in $N_{2} H_{4}$ is $- 2$
Therefore,
Two $N$ makes it $- 4$ ; Hence they loose $10 e^{-}$
Total charge $+ 6$ and since $H$ is still $+ 1$
Two $N$ divides the $+ 6$
Then oxidation state of $N$ is $+ 3$ in the new compound.