The number of electrons in 3.1 mg `NO_(3^(-))` is:

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

The oxidation number of nitrogen in NO_(3)^(-) is:

If mass of an electron is 9.1xx10^(-31) kg, the number of electrons in 1 mg is

The total number of electrons presnt in 1 mL Mg Given density of ._(12) Mg^(24) = 1. 2 g//mL .

Nitrogen has an atomic number of 7 and oxygen has an atomic number of 8. The total number of electron in the nitrate ion (NO_(3)^(-)) is :

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