Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole. - R= 0.082

- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082
× 353 K
Solving:
V = (4.745 moles× 0.082
× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Learn more:
The bond<span> length in a </span>molecule<span> of </span>KBr<span> is measured to be 282.1 pm, compared with a </span>nonpolar<span> covalent radius sum for K and Br of 310,4. </span>
Explanation:
The integrated rate law for the zeroth order reaction is:
![[A]=-kt+[A]_0](https://tex.z-dn.net/?f=%5BA%5D%3D-kt%2B%5BA%5D_0)
The integrated rate law for the first order reaction is:
![[A]=[A]_0e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_0e%5E%7B-kt%7D)
The integrated rate law for the second order reaction is:
![\frac{1}{[A]}=kt+\frac{1}{[A]_0}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA%5D_0%7D)
Where,
is the active concentration of A at time t
is the active initial concentration of A
t is the time
k is the rate constant