This is an incomplete question, the image for the given question is attached below.
Answer : The wavelength of photon would be absorbed, 
Explanation :
From the given diagram of energy we conclude that,
Energy at ground state, A = 400 zJ
Energy of 2nd excited state, C = 1050 zJ
Now we have to calculate the energy of the photon.


Now we have to calculate the wavelength of the photon.
Formula used :

As, 
So, 
where,
E = energy of photon = 
= frequency of photon
h = Planck's constant = 
= wavelength of photon = ?
c = speed of light = 
Now put all the given values in the above formula, we get:


Therefore, the wavelength of photon would be absorbed, 
Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>
Explanation:
Given: The base dissociation constant:
= 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant:
= 1 × 10⁻¹⁴
<em><u>The acid dissociation constant </u></em>(
)<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
<u>The acid dissociation constant: </u>![K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Cleft%20%5BB%20%5Cright%20%5D%20%5Cleft%20%5BH_%7B3%7DO%5E%7B%2B%7D%5Cright%20%5D%7D%7B%5Cleft%20%5BBH%5E%7B%2B%7D%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B%28x%29%28x%29%7D%7B%280.1%20-%20x%29%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B0.1%20-%20x%7D)





<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>
See the sketch attached.
<h3>Explanation</h3>
The Lewis structure of a molecule describes
- the number of bonds it has,
- the source of electrons in each bond, and
- the position of any lone pairs of electrons.
Atoms are most stable when they have eight or no electrons in their valence shell (or two, in case of hydrogen.)
- Each oxygen atom contains six valence electrons. It demands <em>two</em> extra electrons to be chemically stable.
- Each sulfur atom contains six valence electrons. It demands <em>two </em> extra electrons to be chemically stable.
- Each hydrogen atom demands <em>one</em> extra electron to be stable.
H₂O contains two hydrogen atoms and one oxygen atom. It would take an extra 2 + 2 × 1 = 4 electrons for all its three atoms are stable. Atoms in an H₂O would achieve that need by sharing electrons. It would form a total of 4 / 2 = 2 O-H bonds.
Each O-H bond contains one electron from oxygen and one from hydrogen. Hydrogen has no electron left. Oxygen has six electrons. Two of them have went to the two O-H bonds. The remaining four become 4 / 2 = 2 lone pairs. The lone pairs repel the O-H bonds. By convention, they are placed on top of the two H atoms.
Similarly, atoms in a SO₂ molecule demands an extra 2 × 2 + 2 = 6 electrons for its three atoms to become chemically stable. It would form 6 / 2 = 3 chemical bonds. Loops are unlikely in molecules without carbon. As a result, one of the two O atoms would form two bonds with the S atom while the other form only one.
Atoms are unstable with an odd number of valence electrons. The S atom in SO₂ would have become unstable if it contribute one electron to each of the three bond. It would end up with 3 × 2 + 3 = 9 valence electrons. One possible solution is that it contributes two electrons in one particular bond. One of the three bonds would be a coordinate covalent bond, with both electrons in that bond from the S atom. In some textbooks this type of bonds are also known as dative bonds.
Dots and crosses denotes the origin of electrons in a bond. Use the same symbol for electrons from the same atom. Electrons from the oxygen atoms O are shown in blue in the sketch. They don't have to be colored.
If more reactant is added, the equation will shift to the right in order to make more product (which will increase the products)