<u>Solution and Explanation:</u>
[La3+] = 0.1 M
<u>At the equilibrium:
</u>
La(IO3)3 <----> La3+ + 3 IO3-
0.1 +s 3s
![\mathrm{Ksp}=[\mathrm{La} 3+][\mathrm{IO} 3-]^{\wedge} 3](https://tex.z-dn.net/?f=%5Cmathrm%7BKsp%7D%3D%5B%5Cmathrm%7BLa%7D%203%2B%5D%5B%5Cmathrm%7BIO%7D%203-%5D%5E%7B%5Cwedge%7D%203)

Since Ksp is small, s can be ignored as compared to the 0.1
The above shown expression thus becomes:



<u>Answer: 1.4*10^-4 M
</u>
Answer is in picture below.
Use 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; mass percentage of the chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; amount of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; amount of carbon.
n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.
This compound is dichlorocarbene CCl₂.
The answer is Q because the red it's absorbed by object Q meaning only black will show