Chechnya. 345. Gcbjshjkfs
1,000 mL/ 1 L
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One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.
Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
The reaction between mercury (Hg) and sulfur (S) to form HgS is:
Hg + S ------------- HgS
Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS
The given mass of Hg = 246 g
Atomic mass of Hg = 200.59 g/mol
# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles
Based on the reaction stoichiometry,
# moles of S that would react = 1.226 moles
Atomic mass of S = 32 g/mol
Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g
39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS
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
