When:
Fe(OH)3 ↔
Fe3+ + 3OH-so, each 1 mol Fe3+ → 3 mol OH-
and when [OH]
= 1.2 x 10^-5∴ [Fe3+] = (1.2 x 10^-5) / 3
= 4 x 10^-6when
Keq = [Fe3+][OH]^3 = (4 x 10^-6) (1.2 x 10^-5)^3
∴Keq = 6.9 x 10^-21
The answer would be 1952.5 grams of potassium sulfate solution. The way to get the answer is to use the equation, 617/x = 31.6%/1 and then cross multiply to get 31.6%x= 617grams and then 0.316x=617 grams so then x = 617/0.316= 1952.5 grams.
The value of x : 2
<h3>Further explanation</h3>
Semi-normal Hydrochloric acid solution = 0.5 N = 0.5 M
for titration :
M₁V₁n₁=M₂V₂n₂(1=HCl,2=Na₂CO₃)
MW Na₂CO₃.xH₂O=143
MW Na₂CO₃.xH₂O = MW Na₂CO₃+ MW xH₂O = 143
MW Na₂CO₃ = 106 g/mol
MW xH₂O = 18x
Equation :
Answer:
I'm sorry i do not speak spanish , so i dont understand.
Explanation:
do you know english
NaBr gives fully seperated ions as Na⁺(aq) and Br⁻(aq) while AgBr gives partially dissociated ions as Ag⁺(aq) and Br⁻(aq).
NaBr(aq) →Na⁺(aq) + Br⁻(aq)
0.200 0.200
AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X
Ksp = [Ag⁺(aq)] [ Br⁻(aq)]
7.7 × 10⁻¹³ = X * (0.200 + X)
7.7 × 10⁻¹³ = X² + 0.200X
0 = X² + 0.200X-7.7 × 10⁻¹³
X1 = 3.8 x 10⁻¹²
X2 = -0.2
X = molar solubility of AgBr.
hence, X cannot be a negative value.
molar solubility of AgBr is 3.8 x 10⁻¹² M.