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2 years ago

Please help i'm being timed

Chemistry

1 answer:

Bad White [126]2 years ago

5 0

a. KCl = strong electrolyte

b. CCl4 = non-electrolyte

c. LiCl = strong electrolyte

d. Na2SO4 = strong electrolyte

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In a chemical reaction, the difference between the potential energy of the products of the potential energy of the reactants is

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Your question looks a bit incomplete as you have the same contents in options a) and d). According to your list, I can't see the correct answer, but I can give you one.The difference between the potential energy of the products of the potential energy of the reactants is equal to the enthalpy of the reaction.

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Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

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Hello there,

Your correct answer would be "<span>the energy of a photon".

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</span>

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