Explanation:
The rate at which gases effuse is inversely related to the square root of their molar masses.
In this case, half of the helium (1.5 L) passed through the membrane in 24 hours. Therefore, we can calculate the effusion rate of He gas as follows.
= 0.0625 L/hr
Given that the molar mass of He is 4 g/mol and for 
it is 32 g/mol.
Now,
= 2.83
Thus, the effusion rate of 
= 
Rate of = 0.022 L/hr.
This implies that 0.022 L of gas will effuse in one hour.
Consequently, to find the duration needed for 1.5 L of gas to effuse, we calculate as follows.
= 68.18 hours
Thus, we can conclude that it will require 68.18 hours for half of the oxygen to effuse through the membrane.
Answer: Option (a) is the correct answer.
Explanation:
Under conditions of low pressure and high temperature, gas molecules exhibit negligible attractions or repulsions among themselves. Hence, gases behave ideally in these scenarios.
Conversely, at low temperatures, there is a reduction in the kinetic energy of gas molecules, while high pressure compels the molecules to be closer together.
Thus, attractive forces emerge between molecules in conditions of low temperature and high pressure, causing gases to be termed real gases.
Therefore, we conclude that the ideal gas law becomes less accurate when pressure increases and temperature decreases.
Ca3(PO4)2 is the correct formula.
Answer: The mole fraction of hydrogen gas at 20°C is 0.975
Explanation:
The information provided includes:
Water vapor pressure at 20°C is 17.5 torr
Total pressure at 20°C = 700.0 torr
Hydrogen gas vapor pressure at 20°C = (700.0 - 17.5) torr = 682.5 torr
To find hydrogen gas's mole fraction at 20°C, we utilize Raoult's law, represented by:
where,
= pressure of hydrogen gas = 682.5 torr
= total pressure = 700.0 torr
= mole fraction of hydrogen gas =?
Substituting the values into the equation yields:
Thus, the mole fraction of hydrogen gas at 20°C equals 0.975
