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Equal weights of methane and oxygen and mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is:
- a)1/3
- b)½
- c)2/3
- d)(1/3) × (273/298)
Correct answer is 'A'. Can you explain this answer?
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Equal weights of methane and oxygen and mixed in an empty container at...
Molar mass of methane = 16.042g mol-1 Molar mass of oxygen = 32.00 g mol-1 Therefore if say, 32g of methane and 32g of oxygen mixed,there is 2 moles of methane and 1 moles of oxygen. n= PV/RT pressure is directly proportional to the number of moles Oxygen is 1/3 of total number of moles and hence it exerts 1/3 of the total pressure The answer is a.
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Equal weights of methane and oxygen and mixed in an empty container at...
°C and 1 atm pressure. Assuming ideal gas behavior, what is the mole fraction of each gas in the mixture?
First, we need to calculate the number of moles of each gas in the mixture. Since equal weights are used, we can assume that we have the same mass of each gas. The molar mass of methane is 16 g/mol and the molar mass of oxygen is 32 g/mol. Therefore, we have:
Number of moles of methane (nCH4) = Mass of methane / Molar mass of methane
nCH4 = 16 g / 16 g/mol = 1 mol
Number of moles of oxygen (nO2) = Mass of oxygen / Molar mass of oxygen
nO2 = 16 g / 32 g/mol = 0.5 mol
The total number of moles in the mixture is therefore:
nTotal = nCH4 + nO2 = 1.5 mol
Now we can calculate the mole fraction of each gas in the mixture:
Mole fraction of methane = nCH4 / nTotal = 1 / 1.5 = 0.67
Mole fraction of oxygen = nO2 / nTotal = 0.5 / 1.5 = 0.33
Therefore, the mole fraction of methane in the mixture is 0.67 and the mole fraction of oxygen is 0.33.
First, we need to calculate the number of moles of each gas in the mixture. Since equal weights are used, we can assume that we have the same mass of each gas. The molar mass of methane is 16 g/mol and the molar mass of oxygen is 32 g/mol. Therefore, we have:
Number of moles of methane (nCH4) = Mass of methane / Molar mass of methane
nCH4 = 16 g / 16 g/mol = 1 mol
Number of moles of oxygen (nO2) = Mass of oxygen / Molar mass of oxygen
nO2 = 16 g / 32 g/mol = 0.5 mol
The total number of moles in the mixture is therefore:
nTotal = nCH4 + nO2 = 1.5 mol
Now we can calculate the mole fraction of each gas in the mixture:
Mole fraction of methane = nCH4 / nTotal = 1 / 1.5 = 0.67
Mole fraction of oxygen = nO2 / nTotal = 0.5 / 1.5 = 0.33
Therefore, the mole fraction of methane in the mixture is 0.67 and the mole fraction of oxygen is 0.33.
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Equal weights of methane and oxygen and mixed in an empty container at...
Let us consider both oxygen and methane are taken x gm. Then no of moles of oxygen is (x/32) and no of moles of methane is (x/16). So no of mole fraction of oxygen is (x/32)/(x/16+x/32)= 1/3. So the fraction of total pressure exerted by oxygen is 1/3.
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Equal weights of methane and oxygen and mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is:a)1/3b)½c)2/3d)(1/3) × (273/298)Correct answer is 'A'. Can you explain this answer?
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