The electrons and the nuclei will settle into positions that minimize repulsion and maximize attraction.
Answer:
C.) HOCl Ka=3.5x10^-8
Explanation:
In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below
we Know that
pKa= -log(Ka)
therefore
A) pKa of HClO2 = -log(1.2 x 10^-2)
=1.9208
B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644
C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45
D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979
If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution
The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.
So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.
Hence, HOCl will be chosen for buffer construction.
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
The answer to the question is C