The answer to the question is "B. Roman Numerals"
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
For the answer to the questions above,
a) Ag2CO3(s) => Ag2O(s)+CO2(g)
<span>b) Cl2(g)+2(KI)(aq) => I2(s)+2(KCl)(aq) (coefficients are for balanced equation) </span>
<span>net ionic is Cl2(g)+2I- => I2(s)+2Cl-(aq) </span>
<span>c) I2(s)+3(Cl2)(g)=>2(ICl3)
</span>I hope I helped you with your problem
