The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
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The answer to this question is A- evaporation
The chemical element of atomic number 7, a colorless, odorless unreactive gas that forms
about 78 percent of the earth's atmosphere.
With all of the information given (pressure, volume, temperature, and the molar mass), we need a formula that relates this all together. The formula we need is the ideal gas law, PV=nRT. Since the pressure is defined in millimeters of mercury, we need the R value that correlates with this, which is 62.4; on top of this, we need the temperature in Kelvin - simply add 273.15 to convert from Celsius. With all of this information, simply plug-and-chug:
PV=nRT
(800)(3.7) = n(62.4)(37 + 273.15)
n = 0.1529 moles
Finally, the problem is asking the amount of air in grams. We have moles, so all we need to do is multiply that value by the molar mass.
0.1529 moles x 29 grams per mole =
4.435 grams of air
The balloon has 4.435 grams of air inside it.
Hope this helps!
