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
Answer:
none of the above is the answer
Ionic compounds are formed when the transfer of electrons is a metal and non metal.<span />
<h3>Answer:</h3>
The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.
<h3>Explanation:</h3>
Lets take start with the melting point of both compounds.
n-Butane = - 140 °C
Trimethylamine = - 117 °C
Intermolecular Forces in n-Butane:
As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.
Intermolecular Forces in Trimethylamine:
Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.
How to change an emerson thermostat from celsius to fahrenheit.
Answer at the picture above.
