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Ni(OH)₂ ⇄ Ni⁺² + 2 OH⁻
Ksp = [Ni⁺²][OH⁻]² = S (2S)² = 4S³
where S is molar solubility.
at pH = 10
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
S =
= 6 x 10⁻⁸ M
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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Answer:
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Explanation:
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