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Lera25 [3.4K]
3 years ago
14

The color change in the titration of KMnO4 with oxalic acid at the endpoint is:

Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0

Answer:

d. Colorless to pink

Explanation:

The color change in the titration of KMnO4 with oxalic acid at the endpoint is colorless to pink.

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Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi
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Answer : The value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

Explanation :

The following equilibrium reactions are :

(1) 2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2 K_1=5.40\times 10^{-16}

(2) 2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l) K_2=1.06\times 10^{10}

(3) CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g) K_3=2.68\times 10^{-9}

The final equilibrium reaction is :

CO_2(g)\rightleftharpoons C(s)+O_2(g) K_{goal}=?

Now we have to calculate the value of K_{goal} for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

K_{goal}=\sqrt{K_1\times K_2\times K_3}

Now put all the given values in this expression, we get :

K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}

K_{goal}=1.238\times 10^{-7}

Therefore, the value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

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