How much faster does helium escape through a porous container than ozone?
1 answer:
According to Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".
r₁ / r₂ =
---- (1)
r₁ = Rate of effusion of He
r₂ = Rate of Effusion of O₃
M₁ = Molecular Mass of He = 4 g/mol
M₂ = Molecular Mass of O₃ = 48 g/mol
Putting values in eq. 1,
r₁ / r₂ =
r₁ / r₂ =
r₁ / r₂ = 3.46
Result:
Therefore, Helium will effuse 3.46 times more faster than Ozone.
r₁ / r₂ =
r₁ = Rate of effusion of He
r₂ = Rate of Effusion of O₃
M₁ = Molecular Mass of He = 4 g/mol
M₂ = Molecular Mass of O₃ = 48 g/mol
Putting values in eq. 1,
r₁ / r₂ =
r₁ / r₂ =
r₁ / r₂ = 3.46
Result:
Therefore, Helium will effuse 3.46 times more faster than Ozone.
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<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%
<u>Explanation:</u>
- To calculate the number of moles, we use the equation:
.....(1)
<u>For lead (II) bromide:</u>
Given mass of lead (II) bromide = 0.7822 g
Molar mass of lead (II) bromide = 367 g/mol
Putting values in equation 1, we get:
- The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:
By Stoichiometry of the reaction:
1 mole of lead (II) bromide is produced from 2 moles of potassium bromide
So, 0.0021 moles of lead (II) bromide will be produced from = of potassium bromide
- Now, calculating the mass of potassium bromide by using equation 1, we get:
Molar mass of KBr = 119 g/mol
Moles of KBr = 0.0042 moles
Putting values in equation 1, we get:
- To calculate the percentage composition of KBr in the mixture, we use the equation:
Mass of mixture = 5.000 g
Mass of KBr = 0.4998 g
Putting values in above equation, we get:
Hence, the percent by mass of KBr in the mixture is 9.996 %