The mass of gas particle affects the rate of effusion by decreasing the rate as the mass increases. This describe by the Graham's Law of Effusion where the rate of effusion is inversely proportional to the square root of the molar mass of a substance. Hope this helps.
Answer:
Element
Explanation:
The symbol for gold is Au which only has one capital letter. This means that there is only one thing that makes up gold which would make it an element.
hope this helps :)
From the calculations, we can see that, the change in the freezing point is -0.634°C.
<h3>What is freezing point?</h3>
The term freezing point refers to the temperature at which a liquid is changed to solid.
Given that;
ΔT = K m i
Number of moles sucrose = 35.0 g/ 342.30 g/mol = 0.1 moles
molality = 0.1 moles/ 300.0 * 10^-3 Kg
= 0.33 m
Thus;
ΔT = -1.86°C/mol * 0.33 m * 1
= -0.634°C
Learn more about freezing point:brainly.com/question/3121416
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Answer:
V₂ = 530.5 mL
Explanation:
Given data:
Initial temperature = 20.0°C
Final temperature = 40.0 °C
Final volume = 585 mL
Initial volume = ?
Solution:
Initial temperature = 20.0°C (20+273 = 293 K)
Final temperature = 40.0 °C (40+273 = 323 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 585 mL × 293 K / 323 K
V₂ = 171405 mL.K / 323 K
V₂ = 530.5 mL
Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:
If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):
6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
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I hope it helps!
