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Number of electrons present in 3.6 mg of...

Number of electrons present in 3.6 mg of `NH_(4)^(+)` are : `(N_(A) = 6 xx 10^(23))`

A

`1.2 xx 10^(21)`

B

`1.2 xx 10^(20)`

C

`1.2 xx 10^(22)`

D

`2 xx 10^(-3)`

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