Answer:
0.2193 μm
Explanation:
The reaction showing the Photodissociation of ozone (O3) is given below as:
O₃ + hv --------------------------> O₂ + O⁺
H° (142.9) (0) (438kJ/mol).
The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:
ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.
This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.
Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ = 0.2193 μm
We calculate first for the number of moles of gases in the sample through the ideal gas equation.
n = PV/RT
n = (725 mmHg/760 mmHg/atm)(0.255 L) / (0.0821 L.atm/mol.K)(65 + 273.15)
n = 8.76 x 10^-3 mol
Then, we calculate for the mol N2 using the ratio of the pressure.
n N2 = (8.76 x 10^-3 mols)(231 mmHg/725 mmHg)
n N2 = 2.79 x 10^-3 moles
Then, multiply the value with the molar mass of N2 which is 28 grams per mol giving us the answer of 0.078 grams.
Explanation:
As it is given that both the given containers are at same temperature and pressure, therefore they have the same density.
So, mass of 
in container- 1 is as follows.
5.35 mol x molar mass of
= 7.61 mol x 146.06 g/mol
= 1111.52 g
Therefore, density of will be calculated as follows.
Density = 
density = 
= 0.532 g/mL
Now, mass of in container- 2 is calculated as follows.
4.46 L x 1000 mL/L x 0.532 g/mL
= 2372.72 g
Hence, calculate the moles of moles present in container 2 as follows.
No. of moles = 
= 
= 16.24 mol
Since, 1 mol contains 6 moles F atoms
.
So, 16.24 mol contains following number of atoms.
= 
= 97.46 mol
Thus, we can conclude that moles of F atoms in container 2 are 97.46 mol.
