Ksp for potassium chloride, KCl, is 21.7. A 5.0-liter container filled with otherwise-pure water, at standard temperature and pressure, contains 21.5 moles of dissolved KCl. (A) Write a chemical reaction describing the dissolution of KCl. (B) Write an equation for Ksp in terms of the reactants and products from part (A). (C) Find the solubility of KCl, in moles/liter. (D) Is the above solution undersaturated, supersaturated, or at equilibrium with respect to KCl

See explanation below

Explanation:

a) First, as the KCl is a salt, this will ve completely dissociated by water in solution and the reaction taking place would be:

KCl <> K⁺ + Cl⁻

b) According to the above reaction, the Ksp or constant of solubility product will be just the concentration of the products. This is because solid compounds do not contribute in the equilibrium expression, cause they have a constant concentration near to 1. Same thing happens with liquid compounds. So, based on this the Ksp expression for the above reaction is:

KCl(s) <> K⁺(aq) + Cl⁻(aq)

Ksp = [K⁺] [Cl⁻]

c) To find this solubility we need to write the Ksp expression from above, use the given Ksp value and then solve for the solubility:

KCl(s) <> K⁺(aq) + Cl⁻(aq)

Ksp = [K⁺] [Cl⁻]

21.7 = s * s

21.7 = s²

s = 4.66 mol/L

d) To know this, we need to see the concentration of the KCl in solution with the given data of 21.5 moles and the 5 liters of solution:

[K⁺] = [Cl⁻] = 21.5 / 5 = 4.3 mol/L

If we want to calculate the Ksp at this conditions we have:

Ksp2 = (4.3)²

Ksp2 = 18.49

As this value is lower than the actual Ksp value, we can conclude that this solution is undersaturated.

Hope this helps

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.