The sp of strontium carbonate, SrCO3, is 5.60×10−10 . Calculate the solubility of this compound in g/L.

3.50 × 10⁻³ g/L

Explanation:

Step 1: Make an ICE chart for the solution of SrCO₃

"S" represents the molar solubility.

         SrCO₃(s) ⇄ Sr²⁺(aq) + CO₃²⁻(aq)

I                             0               0

C                           +S              +S

E                             S               S

The solubility product constant (Ksp) is:

Ksp = [Sr²⁺] [CO₃²⁻] = S²

S = √Ksp = √5.60 × 10⁻¹⁰ = 2.37 × 10⁻⁵ M

Step 2: Convert "S" to g/L

The molar mass of SrCO₃ is 147.63 g/mol.

2.37 × 10⁻⁵ mol/L × 147.63 g/mol = 3.50 × 10⁻³ g/L

just did the assignment and the answer was #1 +2 n

i think its 5049.

i hope this !