Answer:
The amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.
Explanation:
The chemical reaction between an acid and a base to form salt and water only is called a Neutralization reaction. Chemically
H⁺ + OH⁻ = H₂0
Hence, one mole of hydroxonium ion (H⁺) will combine with one mole of hydroxide ion (OH⁻) to give salt and water only.
In a completely neutralized reaction, the resulting salt is formed when there is complete dissociation of the acid and base to give salt and water with a pH of 7.
In the given question, the stated pH of between 8-9 tells us that the salt produced in this particular neutralization reaction is basic or alkaline. This usually occurs when a strong base reacts with a weak acid, producing a higher concentration of the hydroxide ion at equilibrium.
Hence the amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.
If the concentration or molarity of the acid is known, then the exact amount of base required to neutralize it can be calculated. This is usually done via titrating the acid against drop wise solution of the base. Neutralization usually occurs when there is a change in colour of the resulting solution. The pH of the resulting solution can be determined using a litmus paper.
A blue litmus paper is indicative of a basic solution while a red litmus paper is indicative of an acidic solution.
Ultraviolet rays has the most energy
Answer:
If it served you, give me 5 stars please, thank you!
<h3><u>c) 13.29 mL</u></h3>
Answer:
Explanation:
M=D times V
Answer-3,633.84g
Rounded Answer (correct sig figs)- 3600g
Answer:
12.50g
Explanation:
T½ = 2.5years
No = 100g
N = ?
Time (T) = 7.5 years
To solve this question, we'll have to find the disintegration constant λ first
T½ = In2 / λ
T½ = 0.693 / λ
λ = 0.693 / 2.5
λ = 0.2772
In(N/No) = -λt
N = No* e^-λt
N = 100 * e^-(0.2772*7.5)
N = 100*e^-2.079
N = 100 * 0.125
N = 12.50g
The sample remaining after 7.5 years is 12.50g