What is the concentration of `O_(2)`in a freshwater stream in equilibrium with air at `30^@`and `2.0`bar? Given,`K_(H)`(Henry's law constant )of`O_(2)=2.0xx10^(-3)mol//kg`bar at `30^@C`.

What is the concentration of dissolved oxygen at 25^@C at 1 bar if partial pressure of oxygen is 0.22 bar? The Henry's law constant for oxygen is 1.3 xx 10^(-3) mol dm 3 bar1.

Calculate the proporation of O _(2) N_(2) dissolved in water at 298K. When air containing 20% O _(2)and 80% N_(2) by volume is in equilibrium with water at 1 atm pressure. Henry's law constants for two gases are K_(H) (O_(2)) =4.6 xx 10^(4)atm and K_(H) (N_(2)) = 8.5 xx 10 ^(4)atm.

CO(g) is dissolved in H_(2)O at 30^@ C and 0.020 atm.Henry's law constant for this system is 6.20xx10^(4) atm.Thus mole fraction CO(g) is

If O_(2) gas is bubbled through water at 293 K, how many millimoles of O_(2) gas would dissolve in 1 litre of water? Assume that the partial pressure of O_(2) is 0.987 bar. Given that Henry's law constant for O_(2) at 293K is 34.86 kbar.

The Henry's law constant for O_(2) , dissolved in water is 4.34 xx 10^(4) atm at certain temperature. If the partial pressure of O_(2) , in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O_(2) , in the solution?

For the reaction : N_(2)O_(4)(g)hArr 2NO_(2)(g) the concentration of the equilibrium mixture at 300 K are [N_(2)O_(4)]=4.8xx10^(-2)mol L^(-1) and [NO_(2)]=1.2xx10^(-2)mol L^(-1). K_(c ) for the reaction is :

Calculate the equilibrium constant (K) for the formation of NH^ in the following reaction: N_2(g) + 3H_2(g) At equilibrium, the concentration of NH_3 , H_2 and N_2 are 1.2 xx 10^(-2) ,3.0 xx 10^(-2) and 1.5 xx 10^(-2) M respectively.

For an equilibrium reaction, N_(2)O_(4)(g) hArr 2NO_(2)(g) , the concentrations of N_(2)O_(4) and NO_(2) at equilibrium are 4.8 xx 10^(-2) and 1.2 xx 10^(-2) mol//L respectively. The value of K_(c) for the reaction is