Answer :
AgI should precipitate first.
The concentration of 
when CuI just begins to precipitate is, 
Percent of remains is, 0.0076 %
Explanation :
for CuI is 
for AgI is 
As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.
Now we have to calculate the concentration of iodide ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
![1\times 10^{-12}=0.0079\times [I^-]](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-12%7D%3D0.0079%5Ctimes%20%5BI%5E-%5D)
![[I^-]=1.25\times 10^{-10}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.25%5Ctimes%2010%5E%7B-10%7DM)
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ag^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
![8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M](https://tex.z-dn.net/?f=8.3%5Ctimes%2010%5E%7B-17%7D%3D%5BAg%5E%2B%5D%5Ctimes%201.25%5Ctimes%2010%5E%7B-10%7DM)
![[Ag^+]=6.64\times 10^{-7}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D6.64%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the percent of remains in solution at this point.
Percent of remains = 
Percent of remains = 0.0076 %
A solid, a liquid or a gas.
Answer:
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Explanation:
Step 1: Data given
Kp = 4.7 x 10^3 at 400K
Pressure of CH3OH = 0.250 atm
Pressure of HCl = 0.600 atm
Volume = 10.00 L
Step 2: The balanced equation
CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)
Step 3: The initial pressure
p(CH3OH) = 0.250atm
p(HCl) = 0.600 atm
p(CH3Cl)= 0 atm
p(H2O) = 0 atm
Step 3: Calculate the pressure at the equilibrium
p(CH3OH) = 0.250 - X atm
p(HCl) = 0.600 - X atm
p(CH3Cl)= X atm
p(H2O) = X atm
Step 4: Calculate Kp
Kp = (pHO * pCH3Cl) / (pCH3* pHCl)
4.7 * 10³ = X² /(0.250-X)(0.600-X)
X = 0.249962
p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm
p(HCl) = 0.600 - 0.249962 = 0.350038 atm
p(CH3Cl)= 0.249962 atm
p(H2O) = 0.249962 atm
Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)
Kp = 4.7 *10³
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Answer:
molecular so number 3. ...
