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Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).
[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]
[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]
Correct answer is between '1.29,1.31'. Can you explain this answer?
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Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 at...
The density of O2 gas at 300 K and 1.0 atm can be calculated using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Rearranging the equation to solve for n/V (which is the molar density), we get:
n/V = P/RT
The molar mass of O2 is 32.00 g/mol, so the mass density can be found by multiplying the molar density by the molar mass:
ρ = (n/V) x M
where M is the molar mass.
Substituting the given values, we get:
n/V = (1.0 atm)/(0.08206 L atm/mol K x 300 K) = 0.0407 mol/L
ρ = (0.0407 mol/L) x (32.00 g/mol) = 1.30 g/L
Therefore, the density of O2 gas at 300 K and 1.0 atm is 1.30 g/L.
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Rearranging the equation to solve for n/V (which is the molar density), we get:
n/V = P/RT
The molar mass of O2 is 32.00 g/mol, so the mass density can be found by multiplying the molar density by the molar mass:
ρ = (n/V) x M
where M is the molar mass.
Substituting the given values, we get:
n/V = (1.0 atm)/(0.08206 L atm/mol K x 300 K) = 0.0407 mol/L
ρ = (0.0407 mol/L) x (32.00 g/mol) = 1.30 g/L
Therefore, the density of O2 gas at 300 K and 1.0 atm is 1.30 g/L.
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Community Answer
Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 at...
V = RT/P
= 0.082×300/1.0
= 24.6
So, density = mass/ volume
= 32/24.6
= 1.31
= 0.082×300/1.0
= 24.6
So, density = mass/ volume
= 32/24.6
= 1.31
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Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer?
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Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer?.
Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming ideal gas behavior, the density of O2 gas at 300 K and 1.0 atm is ______ g L–1 (rounded up to two decimal places).[R = 0.082 L atm mol–1 K–1, molar mass of O2 = 32]Correct answer is between '1.29,1.31'. Can you explain this answer?.
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