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3 years ago

from the equation 2PbO + C = 2Pb + CO2, detemine which one is reactant is oxidised and which is reduced.

Chemistry

2 answers:

ss7ja [257]3 years ago

6 0

Answer:

Explanation:

C is oxidised because C up to C+4

PbO is reduced because PbO from Pb2+ down to Pb0

KatRina [158]3 years ago

5 0

Answer:

lead (Pb) is reduced carbon is oxidized

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4 years ago

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the

kramer

Answer:

The equilibrium constant of the given reaction is 0.01351.

Explanation:

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

The equilibrium constant of the reaction = $K_3=1.85\times 10^{-10}$

$K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}$ ...[1]

$AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$

The equilibrium constant of the reaction = $K_4=1.17\times 10^{-4}$

$K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}$ ..[2]

$[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}$

$PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$

The expression of equilibrium constant of the creation is ;

$K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}$

Dividing [1] by [2]

$\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}$

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}$

Substituting the value of $[Cl^-]$ from [2] :

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}$

$\frac{K_3}{K_4}=K_4\times K$

$K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}$

$K=0.01351$

The equilibrium constant of the given reaction is 0.01351.

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from the equation 2PbO + C = 2Pb + CO2, detemine which one is reactant is oxidised and which is reduced.​

from the equation 2PbO + C = 2Pb + CO2, detemine which one is reactant is oxidised and which is reduced.