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:
Indeed, the chemist is capable of identifying the compound present in the sample.
Explanation:
In one mole of K₂O, potassium has a mass of 2 × 39.1 g = 78.2 g, while the total mass of K₂O is 94.2 g. The mass ratio of K compared to K₂O is calculated as 78.2 g / 94.2 g = 0.830.
For 1 mole of K₂O₂, potassium's mass remains the same at 78.2 g, but the total mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ then equates to 78.2 g / 110.2 g = 0.710.
When the chemist measures the mass of K in relation to the overall sample, the mass ratio can be computed.
- If the mass ratio is 0.830, then it indicates a pure K₂O compound.
- If the mass ratio is 0.710, it indicates a pure K₂O₂ compound.
- If the mass ratio falls outside of 0.830 or 0.710, the sample is assessed to be a mixture.
The resulting temperature, following the change in volume and pressure, is -27.26°C. To find this temperature, we apply the combined gas law equation—a formulation where initial and final pressures, volumes, and temperatures are compared. Given the initial conditions and transformations, when we input the stipulated values, we reach the conclusion that the resultant temperature is -27.26°C.
