A 250 mL flask contains air at 1.061 atm and 20.3 °C.
5 mL of ethanol is added, the flask is immediately sealed and then
warmed to 93.3 °C, during which time a small amount of the
ethanol vaporizes. The final pressure in the flask (stabilized at
93.3 °C) is 2.833 atm.
(Assume that the head space volume of gas in the flask
remains constant.)
a. What is the partial pressure of air, in the
flask at 93.3 °C?
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b. What is the partial pressure of the ethanol
vapour in the flask at 93.3 °C?
1homework pts
A 250 mL flask contains air at 1.061 atm and 20.3 °C. 5 mL of ethanol...
A 250 mL flask contains air at 0.9290 atm and 22.7 °C. 5 mL of ethanol is added, the flask is immediately sealed and then warmed to 91.5 °C, during which time a small amount of the ethanol vaporizes. The final pressure in the flask (stabilized at 91.5 °C) is 2.557 atm. (Assume that the head space volume of gas in the flask remains constant.) What is the partial pressure of air, in the flask at 91.5 °C? What is...
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A 250 mL flask, open to atmosphere, contains 0.0110 mol of air at 0 degrees C. Upon hearing, part of the air escapes, how much remains in the flask at 100 degrees Celsius? Can you please show me the steps to getting the correct answer? The correct answer is n=0.00817 moles remain.
A 5 mL sample of an unknown liquid is placed into a 250 mL flask that has had all of the air removed. The pressure measured in the flask is found to be 39.4 kPa. In a seperate trial the student puts 214 mL of the same liquid into the same flask at the same conditions. What will the pressure measure (in kPa) in the second trial? I have tried P1V1 = P2V2 and got the answer P2=268 kPa, but...
A 500.0-mL sealed flask contains 0.060 mol of neon and 0.050 mol of argon at 25°C. Select the correct partial pressures for each of the gases. R=0.0821 L.atm/K-mol Check all that apply. Pe= 5.4 atm P = 2.9 atm P.-2.9 atm = 5.4 atm Ar PA = 2.4 atm Do you know the answer?
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A sealed flask contains 100.0 mL H2O under 10.0 atm of total gas pressure. The gases above the water are N2 (5.1 atm), H2 (2.1 atm), and O2 (2.8 atm). If the mass of dissolved oxygen is 11.5 mg, what is the Henry’s law constant of O2? a. 1.14 x10-2 M/atm b. 1.28 x10-3 M/atm c. 4.11 x10-2 M/atm d. 3.59 x10-4 M/atm e. 1.78 x10-4 M/atm (answer is B) how do you solve?
(5 pts) A 5000. mL flask contains 600. mg of chlorine gas at a temperature of 22 °C. What is the pressure in atm inside the flask? (5 pts) A mixture of noble gases with a total pressure of 725 mm Hg at 25.0°C contains 8.86 x 103 mol Ar, 2.58 x 10-3 mol Ne and 5.84 x 10-mol Kr. Calculate the mole fraction of each gas in the mixture.
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1. Calculate what the pressure of the air is in the
flask at each of the temperatures for Trials 1 through 5. The
temperature is related to the temperature (at constant volume and
number of moles) through Gay-Lussac's Law.
2. Calculate the natural logarithm of the vapor
pressures, ln(P/P^-), where P^- is 100 kPa.
3. Calculate the inverse of the absolute
temperatures.
4. Prepare a graph (either by hand or in excel) with
ln(P/P^-) plotted on the y-axis and 1/T...