Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 12 - Chemical Equilibrium - Exercise 12.4 - Manipulating Equilibrium Constants - Page 532: a

Answer

$$K_c = 2.50 \times 10^{-29}$$

Work Step by Step

1. The coefficients are half as big (divided by 2). Therefore: $$K_{c2} = (K_{c1})^{1/2} = (6.25 \times 10^{-58})^{1/2} = 2.50 \times 10^{-29}$$
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