How many significant figures are in 1.2 x 10^5
1 answer:
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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
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 =
Answer:
Explanation:
Hello!
In this case, since the Gibbs free energy of any process is related with the enthalpy change, temperature and entropy change as shown below:
For a chemical reaction it is simply modified to:
Thus, since the enthalpy of reaction is given as -304.2 kJ and the entropy as -414.2 J/K (-0.4142 kJ/K), at 775 K the Gibbs free energy of reaction turns out:
Whose result means this is a nonspontaneous reaction.
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