A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7
Answer: O 10.1 is your answer! Please mark me as brainlyest!?
Answer:
See below
Explanation:
ΔQ = m c T ΔQ = heat required(J) m = mass (g) T = C° temp change
c = heat capacity in J/g-C
Answer:
Explanation: By adding base to a solution increases concentration of OH ions.
Answer:
the answer is C hope I helped
