# Chapter 12 - Chemical Equilibrium - Conceptual Exercise 12.3 - Properties of Equilibrium - Page 529: d

$[trans-2-butene] = 0.15 \space mol/L$

#### Work Step by Step

$$K_c = \frac{[trans]}{[cis]}$$ $$[trans] = K_c \times [cis] = (1.47) \times (0.10 \space mol/L) = 0.15 \space mol/L$$

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