When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)
So, according to the balanced equation of the reaction:
and by using ICE table:
Ag2CrO4(s) → 2Ag+ (Aq) + CrO4^2-(aq)
initial 0 0
change +2X +X
Equ 2X X
∴ Ksp = [Ag+]^2[CrO42-]
so by substitution:
∴ 3.83 x 10^-11 = (2X)^2* X
3.83 x 10^-11 = 4 X^3
∴X = 2.1 x 10^-4
∴[CrO42-] = X = 2.1 x 10^-4 M
[Ag+] = 2X = 2 * (2.1 x 10^-4)
= 4.2 x 10^-4 M
when we comparing with the actual concentration of [Ag+] and [CrO42-]
when moles Ag+ = molarity * volume
= 0.004 m * 0.005L
= 2 x 10^-5 moles
[Ag+] = moles / total volume
= 2 x 10^-5 / 0.01L
= 0.002 M
moles CrO42- = molarity * volume
= 0.0024 m * 0.005 L
= 1.2 x 10^-5 mol
∴[CrO42-] = moles / total volume
= (1.2 x 10^-5)mol / 0.01 L
= 0.0012 M
by comparing this values with the max concentration that is saturation in the solution
and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.
∴ the excess will precipitate out
Answer:
The density of solution is 1.283 g/mL.
Explanation:
Molarity of the KOH before dilution = 
Volume of the solution before dilution = 
Molarity of the KOH after dilution = 
Volume of the solution after dilution = 
(1 mL = 0.001 L)
Mass of 0.2076 moles of KOH:
0.2076 mol × 56 g/mol = 11.6256 g
Mass of KOH is solution = 11.6265 g
Mass of the solution = M
Mass percentage of solution = 30.0% of KOH
M = 38.755 g
Density of the solution , d= 
The density of solution is 1.283 g/mL.
Answer:
i can't understand your question haha
Answer:
oh it's easy
Explanation:
Take the hydrate
N
a
2
S
2
O
3
∙
5
H
2
O
. Are there ionic forces between the
N
a
+
and the
S
2
O
2
−
3
and ion-dipole forces between the cation/anions and the water?
