Conjugate base of hydrazoic acid is:

{N H}_{3}^{-}

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N_{2}^{-}

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N_{3}^{-}

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N^{3 -}

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Solution

The correct option is C $N_{3}^{-}$

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Similar questions

N_{3}^{-}

lim n \to \infty [\frac{1}{n^{3}} + \frac{2^{2}}{n^{3}} + \frac{3^{2}}{n^{3}} + \dots + \frac{n^{2}}{n^{3}}]

l i m x \to \infty (\frac{1}{n^{3} + 1} + \frac{4}{n^{3} + 1} + \frac{9}{n^{3} + 1} + \dots \dots + \frac{n^{2}}{n^{3} + 1})

{lim}_{n \to \infty} (\frac{1^{2}}{1 - n^{3}} + \frac{2^{2}}{1 - n^{3}} + . . . . + \frac{n^{2}}{1 - n^{3}})

5 < n_{1} < n_{2} < n_{3} < n_{4}

n_{1} + n_{2} + n_{3} + n_{4} = 35

(n_{1}, n_{2}, n_{3}, n_{4})

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