`H_(2)S`, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of `H_(2)S` in water at STP is 0.195m, calculate Henry's law constant.

H_2S a toxic gas with rottenegg like smell is used for the qualetative analysis, if the solubility of H_2S in water, at STP is 0.195 mole. kg^(-1) the Henry's law constant is

Which question represents charge because equation for the solution of H_(2)S in water?

If CO_(2) gas having a partial pressure of 1.67 bar is bubbled through 1 L water at 298 K the amount of CO_(2) dissolved in water in g L^(-1) is approximately. (Henry's law constant of CO_(2) is 1.67 K bar at 298 K )

The volume of chlorine required for the complete reaction of 10 litres of H_(2)S at STP is [Cl_(2)+H_(2)Srarr2HCI+S]

O_2 is bubbled through water at 293K, assuming that O_2 exerts a partial pressure of 0.98 bar, the solubility of O_2 in gm. L^(-1) is (Henry's law constant = 34 k bar)

Why is H_(2)O a liquid while H_(2)S is a gas ?

H_(2)S is more volatile than water because

One mole of phenol is warmed with sodium metal . If we assume 100% yield , volume of H_2 gas liberated at S.T.P is

The first ionization constant of H_(2)S is 9.1xx10^(-8) . Calculate the concentration of HS^(-) ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also ? If the second dissociation constant of H_(2)S is 1.2xx10^(-13) , calculate the concentration of S^(2-) under both conditions.

For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-) ] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals Potussium chromate is slowly aded toa solution containing 0.20M Ag NO_(3) , and 0.20M Ba(NO_(3))_(2) . Describe what happensif the K_(sp) for Ag_(2),CrO_(4) , is 1.1 xx 10^(-12) and the K_(sp) of BaCiO_(4) , is 1.2 xx 10^(-10) ,