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part 2 says: ____ M The dissociation of molecular lodine into lodine atoms is represented as 1,6)=216) At 1000.0 K, the equilibrium constant for the

part 2 says: ____ M
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The dissociation of molecular lodine into lodine atoms is represented as 1,6)=216) At 1000.0 K, the equilibrium constant for the reaction is 3.80 x 10. Suppose you start with 0.0459 mot of 1, in a 2.29 1. Pusk at 1000.0 K What are the concentrations of the gases at equilibrium? Part 1 of 2 What is the equilibrium concentration of 12? Be sure your answer has the correct number of significant digits. X Part 2 of 2 What is the equilibrium concentration of I? Be sure your answer has the correct number of significant digits

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