Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
Answer:
if the oil is already 60 c and you heat up the hot plate to the same degree you are not changing anything
hope this helps :)
Answer:
18.3 kilopascals
Explanation:
We are given that the volume of this container is 0.0372 meters^3, that the mass of water is 4.65 grams, and that the temperature of this water vapor ( over time ) is 368 degrees Kelvins. This is a problem where the ideal gas law is an " ideal " application.
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First calculate the number of moles present in the water ( H2O ). Water has a mass of 18, so it should be that n, in the ideal gas law - PV = nRT, is equal to 4 / 18. It is the amount of the substance.
We now have enough information to solve for P in PV = nRT,
P( 0.0372 ) = 4 / 18( 8.314 )( 368 ),
P ≈ 18,276.9
Pressure ≈ 18.3 kilopascals
<u><em>Hope that helps!</em></u>
Answer:
SO < CO2 < C3H8
Explanation:
Entropy refers to the degree of disorderliness of a system. The standard molar entropy of a substance refers to the entropy of 1 mole of the substance vunder standard conditions.
The molar entropy depends on the number of microstates in the system which in turn depends on the number of atoms in the molecule.
C3H8 has 11 atoms and hence the highest number of microstates followed by CO2 having three atoms and least of all SO having only two atoms.
