Calculate the average speed of an oxygen gas molecule, O2, at 21.0∘C .
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How much faster do ammonia (NH3) molecules effuse than carbon monoxide (CO) molecules?
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Highest rate of effusion to lowest rate
He, H2S, C3H8, Ar, H2
highest rate is lowest molar mass
lowest rate is highest molar mass
Which of these FOLLOWING gases in decreasing order of effusion rate has the fastest to lowest ?
NH3
SO2
UF6
PH3
CL2
H2,Ne,PH3,F2,C4H8, highest effusion is H2 and lowest effusion is C4H8.
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To calculate the average speed of an oxygen gas molecule at 21.0∘C, we can use the root mean square (rms) formula for the average speed of a gas molecule:
v_avg = √(3RT/M)
where:
v_avg is the average speed of the gas molecule
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (21.0 + 273.15 = 294.15 K)
M is the molar mass of the gas (for oxygen, the molar mass of O2 is 32.00 g/mol)
Substituting these values into the formula:
v_avg = √(3 * 8.314 J/(mol·K) * 294.15 K / 32.00 g/mol)
v_avg = √(760.57 J/mol) ≈ 27.6 m/s (rounded to two decimal places)
Therefore, the average speed of an oxygen gas molecule at 21.0∘C is approximately 27.6 m/s.
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To compare the effusion rates of ammonia (NH3) and carbon monoxide (CO) molecules, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:
Rate1 / Rate2 = √(M2 / M1)
where:
Rate1 is the effusion rate of gas 1
Rate2 is the effusion rate of gas 2
M1 is the molar mass of gas 1 (ammonia, NH3)
M2 is the molar mass of gas 2 (carbon monoxide, CO)
The question asks for the difference in rate, which means we need to calculate the ratio of the rates:
Rate1 / Rate2 = √(M2 / M1)
Let's calculate this ratio:
Molar mass of NH3 (M1) = 14.01 g/mol
Molar mass of CO (M2) = 28.01 g/mol
Rate1 / Rate2 = √(28.01 g/mol / 14.01 g/mol)
= √2
≈ 1.41
Therefore, ammonia (NH3) molecules effuse approximately 1.41 times faster than carbon monoxide (CO) molecules.
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To compare the rates of effusion for helium (He), hydrogen sulfide (H2S), propane (C3H8), argon (Ar), and hydrogen (H2), we can use Graham's law of effusion as mentioned earlier.
From Graham's law, we know that the rate of effusion is inversely proportional to the square root of the molar mass. Higher rates of effusion correspond to lower molar masses.
Therefore, in order from highest rate to lowest rate of effusion, we can arrange the gases as follows:
Highest rate of effusion: H2
2nd highest rate of effusion: He
3rd highest rate of effusion: H2S
4th highest rate of effusion: Ar
Lowest rate of effusion: C3H8