If "66%" of "N_(2)O_(4)(g)" is dissociated into "NO_(2)(g)" at "27^(@)C" and "1" atm pressure,the volume occupied by 11.5g" of "N_(2)O_(4)" would be about

N_(2)O_(4) is 60% dissociated into NO_(2) at 340K and 1 atm pressure. Find the volume of 10 g N_(2)O_(4) occupy under these condition.

Influence of pressure, temperature, concentration and addition of inert gas on a reversible chemical reaction in equilibrium can be explained by formulating the expression for equilibrium constant K_(c) or K_(p) for the equilibrium. On the other hand Le Chatelier principle can be theoretically used to explain the effect of P , T or concentration on the physical or chemical equilibrium both. N_(2)O_(4) is 66% dissociated into NO_(2) at 340K and 1 atmospheric pressure. The volume occupied by 10gN_(2)O_(4) under these conditions is:

At 60^(@) and 1 atm, N_(2)O_(4) is 50% dissociated into NO_(2) then K_(p) is

At 77^(@)C and one atmospheric pressure , N_(2)O_(4) is 70% dissociated into NO_(2) What will be the volume occupied by the mixture under these conditions if we start with 10 g of N_(2)O_(4) ?

2 mole of N_(2)O_(4) (g) is kept in a closed container at 298 K and 1 atmosphere pressure. It is heated to 596 K when 20% by mass of N_(2)O_(4) (g) decomposes to NO_(2) . The resulting pressure is

At 27^(@)C and 1 atm pressure , N_(2)O_(4) is 20% dissociation into NO_(2) . What is the density of equilibrium mixture of N_(2)O_(4) and NO_(2) at 27^(@)C and 1 atm ?

At 27^(@)C and 1 atm pressure , N_(2)O_(4) is 20% dissociation into NO_(@) .What is the density of equilibrium mixture of N_(2)O_(4) and NO_(2) at 27^(@)C and 1 atm?

One "mole" of N_(2)O_(4)(g) at 100 K is kept in a closed container at 1.0 atm pressure. It is heated to 400 K , where 30% by mass of N_(2)O_(4)(g) decomposes to NO_(2)(g) . The resultant pressure will be