For each chemical reaction, fill in the exponents in the formula for Ks
BaCro (s) + Ba?- (aq) + Cro?- (aq)
K., =[Ba?"1" [CRO-
Mg(OH)2 + Mg²+(aq) + 20H (aq)
Ks = [Mg²+1"[OH]
SIF (S)
Sr- (aq) + 2F"(aq)
Cuz (PO4)2(s) + 3Cu?- (aq) + 2PO, 2 (aq)
Kg = [Cu²+ ][PO, 1
K., = [Sr2-1 [F-1

The answer to your question is given below

Explanation:

BaCrO4(s) <=> Ba^2+(aq) + CrO4^2-(aq)

Ksp = [Ba^2+]^A [CrO4^2-]^B

From the equation above,

Ba^2+ and CrO4^2- has 1 coefficient each. Therefore, the Ksp is given as:

Ksp = [Ba^2+] [CrO4^2-]

A = 1

B = 1

Mg(OH)2 <=> Mg^2+(aq) + 2OH^-(aq)

Ksp = [Mg^2+]^D [OH^-]^E

From the above equation,

Mg^2+ has a coefficient of 1

OH^- has a coefficient of 2

Therefore, the Ksp can be written as:

Ksp = [Mg^2+] [OH^-]^2

D = 1

E = 2

SrF2(s) <=> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+]^C [F^-]^D

From the equation above,

Sr^2+ has a coefficient of 1.

F^- has a coefficient of 2.

Ksp = [Sr^2+] [F^-]^2

C = 1

D = 2

Cu3(PO4)2(s) <=> 3Cu^2+(aq) + 2PO4^3-(aq)

Ksp = [Cu^2+]^F [PO4^3-]^G

From the equation above,

Cu^2+ has a coefficient of 3

PO4^3- has a coefficient of 2

Ksp = [Cu^2+]^3 [PO4^3-]^2

F = 3

G = 2

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