Answer:
0.225M
Explanation:
Step 1:
Data obtained from the question. This includes:
Volume of the stock solution (V1) = 125mL
Molarity of the stock solution (M1) = 0.90 M
Volume of the diluted solution (V2) = 125 + 375 = 500mL
Molarity of the diluted solution (M2) =..?
Step 2:
Determination of the molarity of the diluted solution.
This is obtained by using the dilution formula as follow:
M1V1 = M2V2
0.9 x 125 = M2 x 500
Divide both side by 500
M2 = (0.9 x 125) /500
M2 = 0.225M
Therefore, the molarity of the diluted solution is 0.225M
I think it is True because when the plates collide then a volcano forms. (Subduction Zone)
Answer:
Molality = 1.428m
Explanation:
Molality, m, is an unit of concentration defined as the ratio between moles of solute and kg of solvent:
Molality: Moles solute / kg solvent.
In the problem, water is the solvent (Is the compound in the higher quantity) whereas NaNO₃ is the solute.
Moles of solute: 0.3355moles NaNO₃.
Kg solvent:
Thus, molality of the solution is:
Molality = 0.3355 moles NaNO₃ / 0.2350kg
<h3>Molality = 1.428m</h3>
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
