Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.

Temperature = 
As molar mass of
is 32 g/mol. Hence, the number of moles of
are calculated as follows.

Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Im pretty sure a cinder-cone volcano
In the first distillation this week, Hexane from the original solvent makes a larger contribution to the vapor pressure of the mixture.
In between hexane and toluene, the hexane will have more vapor pressure contribution in the solution. The boiling point of hexane is much lower than toluene. Therefore, it will evaporate easily at low temperatures and start exerting pressure on the solution.
Hence between hexane and toluene, because of more vapor pressure of hexane and lower boiling point, it will easily evaporate and exerts pressure.
Therefore, from the original solvent, hexane makes a larger contribution to the vapor pressure of the mixture.
To learn more about vapor pressure and hexane, visit: brainly.com/question/28206662
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