Answer:
To the right
Explanation:
Step 1: Given data
- Partial pressure of PCl₅ (pPCl₅) = 0.548 atm
- Partial pressure of PCl₃ (pCl₃) = 0.780 atm
- Partial pressure of Cl₂ (pCl₂) = 0.780 atm
Step 2: Write the balanced equation
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Step 3: Calculate the pressure reaction quotient

Step 4: Determine whether the reaction proceeds to the right or to the left as equilibrium is approached
Since <em>Qp < Kp</em>, the reaction will proceed to the right to attain the equilibrium.
So it’s OKAYYYYY ITS OKAYYYY
The elements in group 7, which fluorine is in also, would have similar properties due to the valence electrons and the configuration. <span />
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
Learn more here:brainly.com/question/15962928