In a reaction involving the lodination of acetone, the following volumes were used to make up the reaction mixture: 5.00 mL 2.15 M acetone + 10.0 mL 1.95 M HCI+ 10.0 mL 0.00450 Mly +25.0 mL H₂O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A moles of A = [(A) x V where [A] is the molarity of A and V is the volume in liters of the solution of A that was used. moles b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50.0 ml, 0.0500 L. and the number of moles of acetone was found in Part (a). moles of A V of soln in liters Macetone (A) c. How could you double the molarity of acetone in the reaction mixture, keeping the total volume at 50 ml. and keeping the same concentrations of Hion and Is as in the original mixture? ml acetone and Use HCI and Is the same. ml water, keeping the volumes of

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In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:
5.00 mL 2.15 M acetone + 10.0 mL 1.95 M HCI+ 10.0 mL 0.00450 M l₂ + 25.0 mL H₂O
a. How many moles of acetone were in the reaction mixture? Recall that, for a component A
moles of A = [A] x V
where [A] is the molarity of A and V is the volume in liters of the solution of A that was used.
moles
b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50.0 ml, 0.0500 L, and the number of moles of acetone was found in Part (a).
moles of A
V of soln in liters
Macetone
(A)
c. How could you double the molarity of acetone in the reaction mixture, keeping the total volume at
50 ml and keeping the same concentrations of
Hion and
la as in the original mixture?
ml acetone and
Use
HCI and
I, the same.
ml water, keeping the volumes of
Transcribed Image Text:In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5.00 mL 2.15 M acetone + 10.0 mL 1.95 M HCI+ 10.0 mL 0.00450 M l₂ + 25.0 mL H₂O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A moles of A = [A] x V where [A] is the molarity of A and V is the volume in liters of the solution of A that was used. moles b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50.0 ml, 0.0500 L, and the number of moles of acetone was found in Part (a). moles of A V of soln in liters Macetone (A) c. How could you double the molarity of acetone in the reaction mixture, keeping the total volume at 50 ml and keeping the same concentrations of Hion and la as in the original mixture? ml acetone and Use HCI and I, the same. ml water, keeping the volumes of
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