Answer:
hope this image is helpful for you
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:
Explanation:
i notice that theri is no image so i wont be able to help you sorry
Answer: (D) it is not spontaneous.
Answer:its the thrid one i think
Explanation:
