I apologize of I'm wrong but its probably OA
Sorry once again!
~wise rat
Answer:
4.6305 * 10^-6 mol^3.L^-3
Explanation:
Firstly, we write the value for the solubility of Ca(IO3)2 in pure water. This equals 0.0105mol/L.
We proceed to write the dissociation reaction equation for Ca(IO3)2
Ca(IO3)2(s) <——->Ca2+(aq) + 2IO3-(aq)
We set up an ICE table to calculate the Ksp. ICE stands for initial, change and equilibrium. Let the concentration of the Ca(IO3)2 be x. We write the values for the ICE table as follows:
Ca2+(aq). 2IO3-(aq)
I. 0. 0.
C. +x. +2x
E. x. 2x
The solubility product Ksp = [Ca2+][IO3-]^2
Ksp = x * (2x)^2
Ksp = 4x^3
Recall, the solubility value for Ca(IO3)2 in pure water is 0.0105mol/L
We substitute this value for x
Ksp = 4(0.0105)^2 = 4 * 0.000001157625 = 4.6305 * 10^-6
Answer:
A
Explanation:
I think this is true letter a
Answer:
TRY HARD!
Explanation:
DO UR WORK ANG GET IT DONE!
