When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
Answer:
The answer to your question is
1.- 1.686 x 10²⁴ atoms
2.- 0.25 moles
Explanation:
1.-
1 mol ---------------- 6.023 x 10²³ atoms
2.8 moles ---------- x
x = (2.8 x 6.023 x 10²³) / 1
x = 1.686 x 10²⁴ atoms
2.- 1 mol ------------------ 6.023 x 10 ²³ molecules
x moles ------------- 1.50 x 10²³ molecules
x = (1.50 x 10²³ x 1) / 6.023 x 10²³
x = 0.25 moles
Answer:
Explanation:
All of above except carbon dioxide