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!
it small because that bond is not very polar notice that all of those arrows they cancel. because they point in opposite directions. that's why BH 3 is nonpolar if we were to draw the Lewis structure of bf3. ... so even though the boron fluorine bond is polar the molecule as a whole is nonpolar.
Answer:
(C) The recrystallization solvent should be nonvolatile.
Explanation:
Recrystallization is the separation technique which is used to purify the solid compounds in their crystal or amorphous form.
Some properties follow the recrystallization process as:
The solids are more soluble in hot solvent as compared to the solubility in the cold solvent.
The solvent must have lower boiling point and can be volatile easily.
The solvent should not react with the compound.
The impurities must be soluble in the cool solvent, so that only the pure product crystallizes.
Hence, Answer - C which is not an ideal characteristic.
Aye you have the same class as me bruh I need help on some chemistry qustions
