Just post the question on here
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:
is reduced in the reaction
Explanation:
The given reaction is
The oxidation number of 
is changed from 
And The oxidation number of 
is changed from 
Hence, 
is oxidized and is reduced
The subscriot 2 means that in the formula there are two parts of K, and the subscript 1 (implicit) for S, indicates that there is one part of S.
This is, the formula gives the ratio of the elements K and S in the compound, which is:
2 atoms of K : 1 atom of S.
Answer: there are 2 atoms of K and 1 atom of S in a molecule of K2S.
