Answer
$[trans-2-butene] = 0.15 \space mol/L$
Work Step by Step
$$K_c = \frac{[trans]}{[cis]}$$ $$[trans] = K_c \times [cis] = (1.47) \times (0.10 \space mol/L) = 0.15 \space mol/L$$
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 - Conceptual Exercise 12.3 - Properties of Equilibrium - Page 529: d
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