`HNO_(3)` oxidies `NH_(4)^(+)` ions to nitrogen and itself gets reduced to `NO_(2)`. The moles of `HNO_(3)` required by 1 mole of `(NH_(4))_(2)SO_(4)` is:

HNO_(3) oxidised NH_(4)^(+) ions to nitrogen and itself gets reduced to NO_(2) . The moles of HNO_(3) required by 1 mole of (NH_(4))_(2)SO_(4) is

HNO_(3) oxidises NH_(4)^(+) to nitrogen and gets reduced to NO_(2) . Mols of HNO_(3) reduced by one mol NH_(4)Cl

[Co(NH_(3))_(5)NO_(2)]SO_(4) shows

Calculate the number of moles of ammonia required to produce 2.5 moles of [Cu(NH_(3))_(4)]SO_(4).

Number of moles of electrons take up when 1 mole of NO_(3)^(-) ions is reduced to 1 mole of NH_(2)OH is:

A 2L solution was prepared by mixing of NH_(3) and (NH_(4))_(2)SO_(4) . If pH of solution was found 8 and solution contains one mole of NH_(3) then moles of (NH_(4))_(2)SO_(4) was taken will be [pKb of NH_(3)=5] 2 4 5 1

A 2L solution was prepared by mixing of NH_(3) and (NH_(4))_(2)SO_(4) . If pH of solution was found 8 and solution contains one mole of NH_(3) then moles of (NH_(4))_(2)SO_(4) was taken will be [pKb of NH_(3)=5] 2 4 5 1

By the reduction of HNO_(3) to NO_(2) the number of moles of electrons involved per mole of HNO_(3) is

How many moles of NO_(2) are produced when 1 mole of B reacts with HNO_(3) ?