Answer:
I am confused as to what you're asking.
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 ∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
= 2x10^-8 Mc) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M
Answer:
5
Explanation:
Balance the equation in order of C,H,O and then you should be able to find the coefficient
Answer:
The gas occupy 2406.4 mL at 80 K.
Explanation:
Given data:
Initial volume of gas = 752 mL
Initial temperature = 25 K
Final temperature = 80 K
Final volume = ?
Solution:
The given problem is solved by using charle's law.
V₁/T₁ = V₂/T₂
V₂ = V₁. T₂ /T₁
V₂ = 752 mL × 80 k / 25 K
V₂ = 60160 mL. k/25 K
V₂ = 2406.4 mL
They both break down and and erode