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lesantik [10]
2 years ago
8

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M ac

etone 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
Chemistry
1 answer:
Slav-nsk [51]2 years ago
7 0

Answer:

Explanation:

pls give me point so i can ask questions

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Sveta_85 [38]

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

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