Answer:
Mg^2+ and OH- are the chemical species present at the equilibrium. Mg(OH)2 will not affect the equilibrium.
Explanation:
Step 1: data given
Reactants are Solid Mg(OH)2 and H2O(l)
Kc1 = 1.8 * 10^-11
Step 2: The balanced equation
Mg(OH)2(s) ⇄ Mg2+(aq) + 2OH-(aq)
Step 3: Define the equilibrium constant Kc
Kc = [OH-]²[Mg^2+]
Pure solids and liquids do not have any effect or influence on the equilibrium in the reaction. So they are not included in the equilibrium constant expression.
This means Mg^2+ and OH- are the chemical species present at the equilibrium. Mg(OH)2 will not affect the equilibrium.
You can boil or evaporate the water and the salt will be left behind as a solid. If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid. So, I would say maybe oil.
The compound is sodium chloride
If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation
Molarity after dilution : 0.0058 M
<h3>Further explanation
</h3>
The number of moles before and after dilution is the same
The dilution formula
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
M₁=0.1 M
V₁=6.11
V₂=105.12
