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Question 2 2 pts You add 0.58 mL of n-propanol (0.803 g/mL, 60.1 g/mol) to an excess of acetic acid to create the pear ester
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OH OH Weight of propanol (0.58 mL) x (0.803 g/mol) 0.46 grams moles of propanol (0.46 g) (60.1 g/mol) 0.00775 mol moles of pr

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Question 2 2 pts You add 0.58 mL of n-propanol (0.803 g/mL, 60.1 g/mol) to an excess of acetic acid to create the p...
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