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What gas effuses four times as fast as oxygen?
Helium is a gas that effuses four times faster than oxygen. This is because helium has a lower molar mass compared to oxygen, leading to faster effusion rates as per Graham's law of effusion.
Compare the rate of effusion of hydrogen and helium?
The rate of effusion of helium is higher than hydrogen because helium has a lower molar mass. The rate of effusion is inversely proportional to the square root of the molar mass, meaning lighter gases effuse faster. Helium, being lighter than hydrogen, effuses faster.
How much faster does UF 235 effuse compared with UF 238?
Uranium-235 (UF₃₅) effuses faster than uranium-238 (UF₂₃₈) because of its lower molecular weight. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass. Since UF₃₅ has a molar mass of approximately 235 g/mol and UF₂₃₈ has a molar mass of about 238 g/mol, UF₃₅ effuses roughly 1.01 times faster than UF₂₃₈.
How much faster does nitrogen monoxide effuse in comparison to dinitrogen tetroxide(N2O4)?
To compare the effusion rates of nitrogen monoxide (NO) and dinitrogen tetroxide (N2O4), we can use Graham's law of effusion, which states that the rate of effusion is inversely proportional to the square root of the molar masses of the gases. The molar mass of NO is approximately 30 g/mol, while that of N2O4 is about 92 g/mol. Therefore, nitrogen monoxide effuses faster than dinitrogen tetroxide, specifically, it effuses approximately 1.73 times faster (√(92/30) ≈ 1.73).
What is the ratio of effusion rates between helium and radon gas?
The ratio of effusion rates between helium and radon gas is approximately √(Molar mass of gas 2 / Molar mass of gas 1), which in this case would be √(222 / 4) = √55.5 ≈ 7.46. This means that radon gas effuses approximately 7.46 times slower than helium gas under the same conditions.
How do you calculate the ratio of effusion rates for nitrogen times two and neon?
To calculate the ratio of effusion rates for nitrogen (N2) and neon (Ne), use Graham's law of effusion: Ratio = (Molar mass of neon / Molar mass of nitrogen)^(1/2) For neon (Ne) with a molar mass of 20.18 g/mol and nitrogen (N2) with a molar mass of 28.02 g/mol, the ratio of their effusion rates would be approximately √(20.18 / 28.02) ≈ 0.75.
Does oxygen effuse 1.07 times faster than nitrogen?
No, oxygen does not effuse 1.07 times faster than nitrogen. The effusion rate of a gas is inversely proportional to the square root of its molar mass, so the effusion rate of oxygen would be √(Molar mass of nitrogen / Molar mass of oxygen) ≈ √(28.02 / 32) ≈ 0.91 times faster than nitrogen.
During an effusion experiment oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles of another gas under the same conditions. What is the molar mass of the unknown gas (?
According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. If oxygen gas (O₂) effuses 2.5 times faster than the unknown gas, we can express this relationship as: ( \frac{r_{O_2}}{r_{unknown}} = \sqrt{\frac{M_{unknown}}{M_{O_2}}} ) Given that ( r_{O_2} = 2.5 \times r_{unknown} ) and the molar mass of O₂ is approximately 32 g/mol, we can rearrange the equation to find the molar mass of the unknown gas: ( 2.5 = \sqrt{\frac{M_{unknown}}{32}} ) Squaring both sides gives ( 6.25 = \frac{M_{unknown}}{32} ), leading to ( M_{unknown} = 6.25 \times 32 = 200 , g/mol ). Thus, the molar mass of the unknown gas is approximately 200 g/mol.
How does the rate of effusion of sulfur dioxide SO2 compare to that of helium (He) (Note the molar masses are SO2 64 gmol He 4.0 gmol.)?
The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Helium has a molar mass of 4.0 g/mol, while sulfur dioxide has a molar mass of 64 g/mol. Therefore, the rate of effusion of SO2 will be √(4.0/64) = 1/4 times that of helium. In other words, sulfur dioxide will effuse more slowly than helium.
What is the molar mass of a gas that takes three times longer to effuse than helium?
You want Graham's law of effusion, which is:rate1/rate2 = square root of (M2/M1). You're trying to find M2, because you know that M1 = 4 and rate 1 = 1 for He. I just set rate1 to 1 because it's a question that involves relative amounts). Rate 2 = 3xrate1 = 3. Plug in and solve.After a little rearranging, you should get:3^2 = M2 /4, or 36 g/mole.Something like O2 would fit the bill.ahaha, im smart !!
The molar mass of argon is 40 gmol. What is the molar mass of a gas if it effuses at 0.91 times the speed of argon gas?
graham's law (Rate 1 / rate 2) ^2 = MM2 / MM1 (Rate x / rate Ar) ^2 = MM Ar / MM x (.907)^2 = 39.9481 / MMx MMx = 39.9481 / 0.8226 Molar mass of unknown gas = 48.56 g / mole
An unknown gas is found to diffuse through a porous membrane 2.92 times more slowly than rm H 2.?
The molar mass of the unknown gas is 2.92 times greater than that of hydrogen gas, because the rate of diffusion is inversely proportional to the square root of the molar mass of the gas. This relationship is described by Graham's Law of Effusion.
