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The number of electron in `3.1 mg NO_(3)^(-)` is (`N_(A)` = `6 xx 10^(23)` )

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The number of electron in `3.1 mg NO_(3)^(-)` is (`N_(A)` = `6 xx 10^(23)` )
A. 32
B. `1.6xx10^(-3)`
C. `9.6xx 10^(20)`
D. `9.6xx10^(23)`
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Correct Answer - C
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