Answer:
Concentration AgBr at saturation = 7.07 x 10⁻⁷M
Explanation:
Given AgBr(s) => Ag⁺(aq) + Br⁻(aq) ; Ksp = 5 x 10⁻¹³ = [Ag⁺][Br⁻]
I --- 0 0
C --- +x +x
E --- x x
[Ag⁺][Br⁻] = (x)(x) = x² = 5 x 10⁻¹³ => x = SqrRt(5 x 10⁻¹³) = 7.07 x 10⁻⁷M
No of mole = reacting mass/ molar mass
No of mole = 0.06/56
No of mole = 0.00107mol....
Or...
1mole of iron contain 56gmol¯¹
:-x mole is contained in 0.06g
Porata
And you'll have..
x = 0.06/56
x = 0.00107mol.....
C I think it’s C I’m semi guessing