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stira [4]
3 years ago
5

What minimum concentration of clâ is required to begin to precipitate pbcl2? for pbcl2, ksp=1.17Ã10â5?

Chemistry
1 answer:
Juliette [100K]3 years ago
5 0
Missing in your question : [Pb2+] = 0.085 M

So from this balanced equation:
PbCl2  ↔ Pb+2   +   2Cl-

when Ksp = [Pb+2][Cl]^2
when we have Ksp = 1.17x10^-5 & [Pb+2] = 0.085 M so by substitution:
1.17x10^-5 = 0.085 [Cl-]^2

∴[Cl]^2 = (1.17x10^-5) / 0.085 = 1.38x10^-4
∴[Cl-] = √0.0002
          = 0.0117 M
So 0.0117 M is the minimum concentration requires precipitating PbCl2
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Answer:

The number of moles of Sr in one mole of Sr(HCO₃)₂ = 1 mole

The number of moles of H in one mole of Sr(HCO₃)₂ = 2 moles

The number of moles of C in one mole of Sr(HCO₃)₂ = 2 moles

The number of moles of O in one mole of Sr(HCO₃)₂ = 6 moles

Explanation:

The given chemical formula of the compound is Sr(HCO₃)₂

The number of atoms of Sr in the compound = 1

The number of atoms of H in the compound = 2

The number of atoms of C in the compound = 2

The number of atoms of O in the compound = 6

The number of atoms of each element present in each formula unit of Sr(HCO₃)₂ is proportional to the number of moles of each atom in one mole of Sr(HCO₃)₂

Therefore;

The number of moles of Sr in one mole of Sr(HCO₃)₂ = 1 mole

The number of moles of H in one mole of Sr(HCO₃)₂ = 2 moles

The number of moles of C in one mole of Sr(HCO₃)₂ = 2 moles

The number of moles of O in one mole of Sr(HCO₃)₂ = 6 moles.

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Consider this equilibrium reaction at 400 K. Br2(g)+Cl2(g)↽−−⇀2BrCl(g)Kc=7.0 If the composition of the reaction mixture at 400 K
Genrish500 [490]

Answer:

Q = 7.0

Q = kc. The reaction is in equilibrium

Explanation:

Based on the reaction:

Br₂ + Cl₂ ⇄ 2BrCl

Equilibrium constant of the reaction, kc, is the ratio of <em>equilibrium concentrations</em> products over reactants powered to its reaction coefficient:

Kc = [BrCl]² / [Br₂] [Cl₂] = 7.0

Now, reaction quotient, Q, is write as the same Kc but the concentrations are actual concentrations:

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