Solute=potassium chloride, solvent=water
Answer:
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No, because the appearance has nothing to do with system as well as the fact that appearance is only how it looks
(Btw, that toga picture is like... ughhhh!) hope this helps you ♥︎☀︎☁︎♨︎
<u>Answer:</u> Copper (I) iodide will precipitate first.
<u>Explanation:</u>
We are given:
of CuCl = 
of CuI = 
Concentration of 
Concentration of 
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
![K_{sp}=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
Putting values in above equation, we get:
![1.0\times 10^{-6}=[Cu^+]\times 0.021](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-6%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.021)
![[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-6%7D%7D%7B0.021%7D%3D4.76%5Ctimes%2010%5E%7B-5%7DM)
Concentration of copper (I) ion = 
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
Putting values in above equation, we get:
![5.1\times 10^{-12}=[Cu^+]\times 0.017](https://tex.z-dn.net/?f=5.1%5Ctimes%2010%5E%7B-12%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.017)
![[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B5.1%5Ctimes%2010%5E%7B-12%7D%7D%7B0.017%7D%3D3.00%5Ctimes%2010%5E%7B-10%7DM)
Concentration of copper (I) ion = 
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.