Calculate the solubility at of AgCl in pure water and in a ).0010 M solution. You'll find data in the ALEKS Data tab. Round both of your answers to significant digits.

1.34x10⁻⁵ mol / L is the solubility of AgCl

Explanation:

Ksp of AgCl is defined as:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:

Ksp = [Ag⁺] [Cl⁻]

Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):

Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:

[Ag⁺] = X

[Cl⁻] = X

Where X is the amount of AgCl that dissolvesin water. X = solubility:

Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]

1.8x10⁻¹⁰ = [X] [X]

1.8x10⁻¹⁰ = X²

X =

its going to be mutualism cuz they both benefit from the relationship.

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a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.