Answer:
I guess B They have a charge
T = 20 % : 20 / 100 = 0.2
m1 = solute
m2 = Solvent
T = m1 / m1 + m2
0.2 = 500 g / 500 g + m2
0.2 * ( 500 + m2 ) = 500
0.2 * 500 + 0.2 m2 = 500
100 + 0.2 m2 = 500
0.2 m2 = 500 - 100
0.2 m2 = 400
m2 = 400 / 0.2
m2 = 2000 g of water
hope this helps!
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:
Explanation:
heat released by the solution
= 120 x 3.18 x ( 25 - 14 )
= 4197.6 J
= 4.1976 kJ
This is the heat gain by the salt
so enthalpy change = + 4.1976 kJ
because there is increase in enthalpy
It is endothermic process .