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During preparation of triphenylmethanol (MW= 323), 646 mg of triphenylbromolmethane, was added to a 36 mL...

During preparation of triphenylmethanol (MW= 323), 646 mg of triphenylbromolmethane, was added to a 36 mL of 50% water (excess). This yielded 130 mg of triphenylmethanol (MW=260). What is the theoretical yield of the triphenylmethanol? What is the percent yield of the reaction? Please show all work in explanation.

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