Hydrazinc reacts with `KIO_(3)` in pressence of HCl as
`N_(2)H_(4) +IO_(3) + 2H^(-)+Cl^(-) to ICI+N_(2)+3H_(2)O`
The equivalent masses of `N_(2)H_(4) and KIO_(3)` respectively are:

Hydrazine reacts with KIO_(3) in presence of HCl as : N_(2)H_(4)+IO_(3)^(-)+2H^(+)+Cl^(-) rarrICI+N_(2)+3H_(2)O The equivalent masses of N_(2)H_(4) and KIO_(3) respectively are :

Give the empirical formula of N2H4

What is the ratio by mass of N and H in NH_3

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) If excess water is added into a bottle sample labelled as "112% H_(2)SO_(4) " and is reacted with 5.3 g NaCO_(3) then find the volume of CO_(2) evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :

N_(2) (g) reacts with H_(2) (g) in either of the following ways depending upon supply of H_(2) (g) : N_(2)(g)+H_(2)(g) to N_(2)H_(2)(l) N_(2)(g)+2H_(2)(g) to N_(2)H_(4) (g) If 5 L N_(2) (g) and 3 L H_(2) (g) are taken initially (at same temperature and pressure ), calculate the contraction in valume after the reaction (in L ).

In the following reaction, As_2S_3 + H^+ +NO_3^- rarr NO + H_2O+As O_4^(3-) + SO_4^(2-) the equivalent mass of As_2 S_3 is related to its molecular mass by :

For the reaction : C_(3)H_(8)(g) + 5O_(2) (g) to 3 CO_(2) (g) + 4H_(2) O (l) at constant temperature Delta H - Delta U

Calculate the enthalpy change in the reaction, 4NH_(3) + 3O_(2) to 2N_(2) + 6H_(2) O at 298 K,given that the enthalpy of formation at 298 K for NH_(3) and H_(2)O are - 46.0 and - 286.0 KJ mol^(-1) respectively.

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 Consider the following reaction. H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O What is the equivalent mass of H_(3)PO_(2) ?(mol.Wt.is M)