Liquid `NH_(3)` dissociation to a slight extent, At a certain temp. its self dissociation constant `K_(SDC(NH_(3))=10^(-30))`. The number of `NH_(4)^(+)` ions are present per 100 cm^(3)` of pure liquid are :

Liquid ammonia ionises to a slight extent. At -50^(@)C , its self ionisation constant, K_(NH_(3))=[NH_(4)^(+)][NH_(2)^(-)]=10^(-30) . How many amide ions are present per cm^(3) of pure liquid ammonia ? ("Assume" N=6.0xx10^(23))

Liquid ammonia ionises to a slight extent. At -50^@C its self ionisation constant K_(NH_3) = [NH_4^+] [NH_2^-] = 10^(-30) How many amide ions are present per cm^3 of pure liquid ammonia . (Assume N= 6.0xx10^(23) )?

Liquid ammonia ionises to a slight extent. Its self ionisation constant is 10^(-30) at - 50^(@) C , then the number of amide iona present per 1 cc of it at -50^(@) C is

0.00135 mole of NH_(3) dissociates in 1.0 litre solution of 0.10M . Calculate the dissociation constant of NH_(3) .

Liquid NH_(3) ionises to a slight extent. At -50^(@)C , its ionic product K_(NH_(3)) = [overset(Θ)Nh_(4)] [overset(Θ)NH_(2)] is 10^(-30) . How many amide ions, overset(o+)NH_(2) are present per mm^(3) of pure liquid NH_(3) ?

Dissociation constant of NH_(4)OH is 1.8 xx 10^(-5) . The hydrolysis constant of NH_(4)Cl would be

At - 50^(@) C , the self - ionization constant (ion product ) of NH_(3)" is " K_(NH_(3)) = [NH_(4)^(+)] [NH_(2)^(-)] = 10^(-30) M^(2) . How many amide ions are present per mm^(3) of pure liquid ammonia ?