Matter doesn't have to take up space as long as it has mass.
2 answers:
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The question is incomplete, here is the complete question:
Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?
A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.
<u>Answer:</u> The mole fraction of gas B (hydrogen gas) is 0.557
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For nitrogen gas:</u>
Given mass of nitrogen gas = 10.25 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
- <u>For hydrogen gas:</u>
Given mass of hydrogen gas = 2.05 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
- <u>For ammonia gas:</u>
Given mass of ammonia gas = 7.63 g
Molar mass of ammonia gas = 17 g/mol
Putting values in equation 1, we get:
Mole fraction of a substance is given by:
Moles of gas B (hydrogen gas) = 1.025 moles
Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles
Putting values in above equation, we get:
Hence, the mole fraction of gas B (hydrogen gas) is 0.557
Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.
Explanation:
Given: = 178 K,
= 0.3 atm
= 278 K,
= ?
According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.
Formula used to calculate the pressure is as follows.
Substitute the values into above formula is as follows.
Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.