Recall that when a reaction is at equilibrium, the forward and reverse reactions occur at the same rate. To illustrate this idea, consider the reaction of A (small, red spheres) and B (large, blue spheres) to form AB. A+B ⇌ AB

Notice that the reaction never stops. Even after several minutes, there is A and B left unreacted, and the forward and reverse reactions continue to occur. Also note that amounts of each species (i. e., their concentrations) stay the same.

Required:
What is the value of the equilibrium constant for this reaction?
Assume each atom or molecule represents a 1 M concentration of that substance.

Equilibrium constant Kc = [x]² / [A - x] [B - x]

Explanation:

The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium

ie Kc = [products] / [reactants].

The balanced equation of the reaction is given as : A + B ⇄ AB

At the beginning of the reaction,

Initial concentration I = A = 1M

                                       B = 1M

                                      AB = 0M

After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become

                                         C = reactant A = [A - x] M

                                                 rectant B =   [B - x] M

                                              Product AB =  [x] [x] M

At equilibrium, the concentrations are,

                                            E  = rectant A = [A - x] M

                                                   reactant B = [B - x] M

                                                   product AB = [x]² M

therefore , the equilibrium constant, Kc  = [products]/[reactants]

                                                                   = [x]² / [A - x] [B - x]

you need to give more information

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%