When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M
Answer: mmmmmm asking for mrs.howard work I see lol good luck grace
Explanation:
Answer: A, 1.3*1020, 1326
A. 1326
B. 2960.9
C. 9804
D. 8559.6
Answer:
Explanation:
formula of osmotic pressure is as follows
p= n RT
n is mole of solute per unit volume
If m be the grams of solute needed
m gram = m / 227.1 moles
m / 227.1 moles dissolved in .279 litres
n = m / (227.1 x .279 )
= m / 63.36
substituting the values in the osmotic pressure formula
5.14 = (m / 63.36) x .082 x 298
m / 63.36 = .21
m = 13.32 grams .
The answer is false because it goes in a full 360 degree circle
