The question is incomplete, complete question is;
A solution of 
is added dropwise to a solution that contains
of 
and 
and 
.
What concentration of 
is need to initiate precipitation? Neglect any volume changes during the addition.
value 
value 
What concentration of is need to initiate precipitation of the first ion.
Answer:
Cadmium carbonate will precipitate out first.
Concentration of is need to initiate precipitation of the cadmium (II) ion is 
.
Explanation:
1) 
The expression of an solubility product of iron(II) carbonate :
![K_{sp}=[Fe^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=2.10%5Ctimes%2010%5E%7B-11%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B2.10%5Ctimes%2010%5E%7B-11%7D%7D%7B1.15%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=1.826\times 10^{-9}M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D1.826%5Ctimes%2010%5E%7B-9%7DM)
2) 
The expression of an solubility product of cadmium(II) carbonate :
![K_{sp}=[Cd^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=1.80%5Ctimes%2010%5E%7B-14%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B1.80%5Ctimes%2010%5E%7B-14%7D%7D%7B0.58%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=3.103\times 10^{-12} M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D3.103%5Ctimes%2010%5E%7B-12%7D%20M)
On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.
So, cadmium carbonate will precipitate out first.
And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the concentration.
For Ar :
1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L
x = 7.6 * 22.4
x = 170.24 L
-----------------------------------------------------------------
For C2H3:
1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L
y = 0.44 * 22.4
y = 9.856 L
hope this helps !.
L = 0 to n-1. Hence, it can have values 0, 1, 2.
<span>m = -l to +l. Hence, it can have values -2, -1, 0, 1, 2 </span>
<span>s = +1/2 and -1/2</span>
The concept used here is the Le Chatelier's principle. When a disturbance is introduced to the system, it favors the direction of reaction that minimizes the disturbance to regain equilibrium.
In endothermic reactions, the forward reaction is favored when the temperature is low. Otherwise, the reverse reaction is favored. When you add the amounts of substances on the reactant side, more products would formed favoring the forward reaction. If you increase concentration on the product side, you form more reactants so it would favor the reverse reaction. Lastly, since 10 moles of gases are needed in the reactant side, it would be favored during high pressure reaction.
