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malfutka [58]
2 years ago
12

Consider the function f(x) =

Chemistry
1 answer:
Butoxors [25]2 years ago
8 0
I think it’s b so yea
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The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A
Mamont248 [21]

Answer:

The  K_c for the reaction B(g) = \frac{1}{2}A will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for B(g) = \frac{1}{2}A

            K_c=[B]^2[A]

                 = 0.045

The reverse will be 2B\leftrightharpoons A

Then,      K_c = \frac{[A]}{[B]^2}

                    =  \frac{1}{0.045}

                    = 22m^{-1}

The equilibrium constant for B(g) = \frac{1}{2}A will be

               K_c = \sqrt{K_c}

                    =\sqrt{22}

                    = 4.69

Therefore, K_c for the reaction B(g) = \frac{1}{2}A will be 4.69.

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For the reversible, one-step reaction A+B<---->C+D, the forward rate constant is 52.4 /mol*h and the rate constant for the
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Answer:

—-—1———————-s

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