When the first reaction equation is:
AgI(S) ↔ Ag+(Aq) + I-(Aq)
So, the Ksp expression = [Ag+][I-]
∴Ksp = [Ag+][I-] = 8.3 x 10^-17
Then the second reaction equation is:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+
So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2
∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7
by combining the two equations and solve for Ag+:
and by using ICE table:
AgI(aq) + 2NH3 ↔ Ag(NH3)2+ + I-
initial 2.5 0 0
change -2X +X +X
Equ (2.5-2X) X X
so K = [Ag(NH3)2+] [I-] / [NH3]^2
Kf * Ksp = X^2 / (2.5-2X)
8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X
∴ X = 5.9 x 10^-5
∴ the solubility of AgI = X = 5.9 x 10^-5 M
The third class lever s<span>have </span>the effort<span> placed amongst </span>load<span> and the fulcrum.</span>
PH= log[H3O+]
10.25=log [H3O+]
[H3O+] = 10^10.25
[H3O+]= 1.778 ×10^10
According to the glossary in the back of 'Chemistry: A Central Science - Twelfth Edition', a precipitate is "An insoluble substance that forms in, and separates from, a solution."
the atomic number determines the tyoe of element an atom is.
the atomic number means the number of protons in an atom.
atoms of different elements have different numbers of protons and different atomic numbers so you can identify the element based on the atomic number