Answer
$2.6\times10^{-9} M$
Work Step by Step
$K_{c} = \frac{[Br_{2}][F_{2}]^{5}}{[BrF_{5}]^{2}} = \frac{(0.0018)(0.0090)^{5}}{(0.0064)^{2}} = 2.6\times10^{-9} M$
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 - Questions for Review and Thought - Topical Questions - Page 523b: 34
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