Question

1 mole of N2H4 lost 10 moles of electrons and formed a new compound,Y in which all hydrogen's are present with out change in the oxidation state. The oxidation number of nitrogen in the new compound Y is
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Answer 1

$\huge\underline\mathfrak\purple{Answer}$

1 mole N2H4 loses 10mole e-

So, 1 N2H4 loses 10e-

So,  new compound Y=(N2H4)10+

Charge on y is 10+

Let charge on N be x.

2x+4=10

x=+3

$\huge\underline\mathfrak\black{Thanks}$

Answer 2

Answer:

N2H4 here the oxidation state of 2N is -4

it loses its 10e- which means that in the product side 2N is oxidised by +10e-

oxidation state of 2N in product side = -4+10 e- = +6e-

then for 1N += +6/2 = +3 is the oxidation state of nitrogen in the new compound .

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Explanation: