V(rms) = √(3RT/M)
Where R is the molar gas constant with value 8.314 and T is the temperature in Kelvin and M is the molar mass
v(rms F₂) = √(3 x 8.314 x 304 / 38)
v(rms F₂) = 14.1 m/s
v(rms Cl₂) = √(3 x 8.314 x 304 / 71)
v(rms Cl₂) = 10.3 m/s
v(rms Br₂) = √(3 x 8.314 x 304 / 160)
v(rms Br₂) = 6.88 m/s
The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.
Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr
Vapor pressure of pure water at
=25.81mmHg
= 
According to Dalton's law of partial pressures,
Total pressure = Partial pressure of Oxygen gas + Partial pressure of water
749 Torr = Partial pressure of Oxygen gas + 25.81 Torr
Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr
Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr
The ball will float as it’s density is less than the salt water.
Hey!
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Steps To Solve:
~Substitute
3(2) - 2(3)
~Subtract
6 - 6
~Simplify
0
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Answer:

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Hope This Helped! Good Luck!