Answer:
2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)
0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x
i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L
Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)
10.5 = x*x/(0.205-2x)^2
=> 10.5(4x^2-0.82x+0.042) = x^2
=>42x^2-8.61x+0.441=x^2
=>41x^2-8.61x+0.441 = 0
This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121
The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.
Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L
Answer:
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Explanation:
Answer:
(c) only Ca2+(aq) and Hg2+(aq)
Explanation:
- In the first step, hydrochloric acid (HCl) is added to the solution. In this case the equilibrium that could take place is:
Ag⁺(aq) + Cl⁻(aq) ↔ AgCl(s)
But no precipitate was formed, so Ag⁺(aq) is absent.
- By adding H₂SO₄(aq) the next equilibrium that could take place is:
Ca⁺²(aq) + SO₄⁻²(aq) ↔ CaSO₄(s)
A white precipitate was formed, so Ca⁺² is present in the solution.
- The following could take place after adding H₂S(aq):
Hg²⁺(aq) + S⁻² ↔ HgS(s)
A black precipitate formed, so Hg⁺² is present as well.
