Calculate the pOH first :
pOH =14-8.57=5.43
(OH)=10^-5.47
(OH)=3.72 x 10^-6 mol/L
Water and salt are the products of a neutralization reaction.
Answer:
As the temperature of the water increases, the time needed for the dye to spread decreases. This is because the kinetic energy between the liquid particles increases, therefore helping the dye to dissolve and spread throughout the water.
Explanation:
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.
According to the <span>Le
Chatelier's Principle, if a condition of system is changed, then the system try
to become equilibrium again by correcting that change.
N2(g) + 3H2(g) ⇄<span> 2NH3(g)
If H2(g) is added, then the concentration of H2(g) is high in the system. So
the system tries to reduce it and become to equilibrium
state. Hence to reduce that H2(g) the forward reaction is promoted. Then
Concentration of product, NH3 gas
is increased.</span></span>
