Select the answer which shows the correct order, slowest to fastest, of the relative rate of diffusion of the gases.
1 answer:
Diffusion is the process of a substance spreading out to evenly fill its container or environment. Rate of diffusion of a gas is inversely proportional to the molar mass of the gas.
Lighter(lower) the molar mass of the gas , faster will be its rate of diffusion and heavier (higher) the molar mass of the gas , slower will be its rate of diffusion.
We have to arrange the given gases from slowest rate of diffusion to fastest rate of diffusion that means we need to arrange gases from higher molar mass to lower molar mass.
Molar mass of given gases are :
Cl = 35.5 g/mol
Xe = 131.29 g/mol
He = 4.00 g/mol
N = 14.00 g/mol
So correct order for slowest rate of diffusion (highest molar mass) to fastest rate of diffusion (lowest molar mass) is :
Xe , Cl , N , He
Xe having the highest molar mass will have the slowest rate of diffusion and He with lowest molar mass will have the fastest rate of diffusion, so option 'c' is correct.
Note : Slowest rate of diffusion = High Molar Mass
Fastest rate of diffusion = Low Molar Mass
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Answer: Option (a) is the correct answer.
Explanation:
Entropy means the degree of randomness present within the molecules of an object.
When a substance is present in a solid state then its molecules are held together because of strong intermolecular forces of attraction.
As a result, molecules are not able to move freely and therefore, there will be no randomness.
But when a substance changes into liquid state then its molecules mover further apart from each other due to gain in kinetic energy. Hence, randomness increases.
On the other hand, when a substance moves from liquid to gaseous state then its molecules gain more kinetic energy due to which there will be high degree of randomness.
Therefore, as the products in the given reaction are present in liquid and gaseous state. So, this means that there has been an increase in entropy.
A decrease in entropy is represented by negative sign and an increase in entropy is represented by a positive sign.
Thus, we can conclude that the sign of for the given reaction is
> 0.