Answer:
![[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Explanation:
Hello there!
In this case, for the ionization of silver iodide we have:
![AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]](https://tex.z-dn.net/?f=AgI%28s%29%5Crightleftharpoons%20Ag%5E%2B%28aq%29%2BI%5E-%28aq%29%5C%5C%5C%5CKsp%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:
![[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D%5BH%5E%2B%5D%3D10%5E%7B-3.55%7D%3D2.82x10%5E%7B-4%7DM)
Now, we can set up the equilibrium expression as shown below:

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:
![x=[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=x%3D%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Best regards!
Explanation:
In first one it should be 2Mg
and in second = 2Mgo
hope it helps....
I’m guessing it would be A because the river discharge tends to gather near coastlines where the river ends
You may want to ask the questions seperatly more likely for someone to answer
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