In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M acetone 5 mL 1.0 M HCl 5 mL 0.0050 M I2 10 mL H2O While keeping the total volume at 25 mL and keeping the concentration of H ion and I2 as in the original mixture, how could you double the molarity of the acetone in the reaction mixture

Explanation:

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