Your answer is D predator population will increase greatly. That's because if there's more prey there will be more food and its good for the predators to produce more because they don't got to worry about food. Speaking how they preys population grew.
So your answer is D.
Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.
<span>By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". </span>
<span>"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form." </span>
Answer : The equilibrium concentration of 
will be, (C) 
Explanation : Given,
Equilibrium constant = 14.5
Concentration of 
at equilibrium = 0.15 M
Concentration of 
at equilibrium = 0.36 M
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the values in this expression, we get:
![14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%280.15%29%5Ctimes%20%280.36%29%5E2%7D)
![[CH_3OH]=2.82\times 10^{-1}M](https://tex.z-dn.net/?f=%5BCH_3OH%5D%3D2.82%5Ctimes%2010%5E%7B-1%7DM)
Therefore, the equilibrium concentration of will be, (C)
