Answer: the pH of the solution is 4.52
Explanation:
Consider the weak acid as Ha, it is dissociated as expressed below
HA H⁺ + A⁻
the Henderson -Haselbach equation can be expressed as;
pH = pKa + log( [A⁻] / [HA])
the weak acid is dissociated into H⁺ and A⁻ ions in the solution.
now the conjugate base of the weak acid HA is
HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}
so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;
pKₐ = -logKₐ
pKₐ = -log ( 7.4×10⁻⁵ )
pKₐ = 4.13
now thw pH is
pH = pKₐ + log( [A⁻] / [HA])
pH = 4.13 + log( [0.540] / [0.220])
pH = 4.13 + 0.3899
pH = 4.5199 = 4.52
Therefore the pH of the solution is 4.52
<u>Answer:</u> The mole ratio of H : O in ammonium nitrate is 4 : 3.
<u>Explanation:</u>
We are given a compound named ammonium nitrate having formula 
There are 3 elements in this compound which are nitrogen, hydrogen and oxygen.
To calculate the mole ratio, we write the ratio of their subscripts. For this compound, it is:
The mole ratio of H and O for this compound is 4 : 3.
Answer:
97 J
Explanation:
Step 1: Given data
- Mass of the sample (m): 12 kg
- Specific heat capacity (c): 0.231 J/kg.°C (this can also be expressed as 0.231 J/kg.K)
- Initial temperature: 45 K
Step 2: Calculate the temperature change
ΔT = 80 K - 45 K = 35 K
Step 3: Calculate the heat required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.231 J/kg.K × 12 kg × 35 K = 97 J
