Answer:
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Explanation:
Hello,
In this case, by considering the dissolution of silver bromide:
![AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}](https://tex.z-dn.net/?f=AgBr%28s%29%5Crightleftharpoons%20Ag%5E%2B%28aq%29%2BBr%5E-%28aq%29%20%5C%20%5C%20%5C%20Ksp%3D%5BAg%5E%2B%5D%5BBr%5E-%5D%3D7.7x10%5E%7B-13%7D)
And the formation of the complex:
![Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7](https://tex.z-dn.net/?f=Ag%5E%2B%28aq%29%2B2NH_3%28aq%29%5Crightleftharpoons%20Ag%28NH_3%29_2%5E%2B%28aq%29%5C%20%5C%20%5C%20Kf%3D%5Cfrac%7B%5BAg%28NH_3%29_2%5E%2B%5D%7D%7B%5BAg%5E%2B%5D%5BNH_3%5D%5E2%7D%3D1.6x10%5E7)
We obtain the balanced net ionic equation by adding the aforementioned equations:
Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:
![AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}](https://tex.z-dn.net/?f=AgBr%28s%29%2BAg%5E%2B%28aq%29%2B2NH_3%28aq%29%5Crightleftharpoons%20Ag%28NH_3%29_2%5E%2B%28aq%29%2BBr%5E-%2BAg%5E%2B%5C%5C%5C%5CK%3D%5BAg%5E%2B%5D%5BBr%5E-%5D%2A%5Cfrac%7B%5BAg%28NH_3%29_2%5E%2B%5D%7D%7B%5BAg%5E%2B%5D%5BNH_3%5D%5E2%7D)
So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:
Best regards.
A valley but I'm not 100% sure
I believe the answer is Zinc because it has 30protona
Answer:
d. CH3CH2OH
Explanation:
Molecular solution are solutions when a molecular compound is dissolved in them.
CH3CH2OH represents Ethanol or alcohol carrying "OH" group and is a molecule. CH3CH2OH or Ethanol forms a molecular solution in water.
The equation for Ethanol dissolving in water as follows:
CH3CH2OH(l) => CH3CH2OH(aq)
CH3CH2OH srays together as molecules. During dissolving, ethanol molecule fit into spaces between water molecules and completely mix with water.
While HCN, CH3COOH and Ba(OH)2 form ionic solution.
Hence, the correct option is d. CH3CH2OH.
