For reaction A(g) + B(g) we start with 2 moles of A and B each. At equilibrium `0.8` moles of AB is formed. Then how much of A changes to AB

A(g) + B(g) ⇌ AB(g) is areversible reaction. At equilibrium 0.4 mole of AB is formed whn each A and B are tsken on emole. How much of A change into AB ?

For reaction A+2BtoC . The amount of C formed by starting the reaction with 5 mole of A and 8 mole of B is :

For reaction A+2BtoC . The amount of C formed by starting the reaction with 5 mole of A and 8 mole of B is :

In the following reaction, 3A (g)+B(g) hArr 2C(g) +D(g) , Initial moles of B is double at A . At equilibrium, moles of A and C are equal. Hence % dissociation is :

In the following reaction, 3A (g)+B(g) hArr 2C(g) +D(g) , Initial moles of B is double at A . At equilibrium, moles of A and C are equal. Hence % dissociation is :

For the Chemical reaction A_(2(g)) + B_(2(g)) hArr 2 AB(g) the amount of AB at equilibrium is affected by

For the Chemical reaction A_(2(g)) + B_(2(g)) hArr 2 AB(g) the amount of AB at equilibrium is affected by

For a reaction A + 2B to C , the amount of C formed by starting the reaction with 5 moles of A and 8 moles of B is