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>
You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25
That should be the right answer but I'm not sure. It has been awhile since I have done this.
When two atoms of this element move towards each other, they will combine in a covalent bond to form a diatomic molecule.
Looking at the electron configuration of the atoms; 1s2 2s2 2p4, we can see that these are atoms of elements in group 16. The elements in group 16 has a general outer electronic configuration of ns2np4. They have a valency of 2.
When two atoms of this element approach each other, they will combine in a covalent bond to form a compound. If this element is depicted as X, the compound formed is X2.
Learn more: brainly.com/question/1527403
The half life for radioactive can be calculated as:
N /N0 = (1 /2) ^ n
n = T /T half
According to question there are n number of half life are present which would result in remaining amount of element as n.