Answer
$$K_c = 2.50 \times 10^{-29}$$
Work Step by Step
1. The coefficients are half as big (divided by 2). Therefore:
$$K_{c2} = (K_{c1})^{1/2} = (6.25 \times 10^{-58})^{1/2} = 2.50 \times 10^{-29}$$
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
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
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
