Respuesta :
Answer:
there are 0.00043484598 mol of ozone molecules
Explanation:
To find how many molecules of ozone there are in those conditions we have to apply the ideal gas law P×V = n×R×T
Where P stands for Pressure of the gas, V for volume, n is for the amount of gas in moles, R for the universal gas constant which is 0.082 atm×l\ K×mol and T stands for temperature in Kelvin.
if we replace what with the information:
P×V=n×R×T
1.79×10-3 ×5l = n × 0.082 atm×l ×251K
K×mol
0,00895 atm×l = n × 20.582 atm×l
mol
we pass 20.582 atm×l to the other side
mol
0.00895 atm×l = n
20.582 atm×l\mol
we divide and simplify atm ×l with atm× l and the result is:
0.00043484598 mol=n
Answer:
There are [tex]2.6172 \times 10^{20}[/tex] molecules of ozone present in 5.00 L of air under the stratospheric ozone conditions of 251 K temperature and [tex]1.79 \times 10^{-3}[/tex] atm pressure.
Explanation:
To find the number of molecules of ozone you need to:
- Use the Ideal Gas equation to find the moles of ozone and,
- Use the Avogadro number ([tex]6.02214\times 10^{23}[/tex]) to find the molecules of ozone.
Given
[tex]V= 5.00 \:L\\T=251 \:K\\P=1.79 \times 10^{-3} \:atm[/tex]
The Ideal Gas equation is
[tex]PV=nRT[/tex]
solving for n
[tex]n=\frac{PV}{RT}[/tex]
where
[tex]R=0.08205 \:\frac{L\cdot atm}{mol \cdot K}[/tex]
Substituting the values given into the expression we derived for n, we obtain
[tex]n=\frac{(1.79 \times 10^{-3} \:atm)\cdot (5.00 \:L)}{(0.08205 \:\frac{L\cdot atm}{mol \cdot K})\cdot (251 \:K)}= 4.346\times 10^{-4} \:mol[/tex]
Find the number of molecules of ozone
[tex]4.346\times 10^{-4} \:mol \cdot 6.02214\times 10^{23}\:\frac{molecules}{mol} =2.6172 \times 10^{20} \:molecules[/tex]
