First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
= - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts
Answer:
D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.
Explanation:
- It is a redox reaction that is consisted of two half-reactions:
Oxidation reaction:
Zn losses 2 electrons and is oxidized to Zn²⁺:
<em>Zn → Zn²⁺ + 2e⁻.</em>
<em></em>
Reduction reaction:
H⁺ gains 1 electron and is reduced to H:
<em>2H⁺ + 2e⁻ → H₂.</em>
<em></em>
<em>So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.</em>
<em></em>
Explanation:
<em><u>Solutions. 1. If 47 g of KCl dissolved in enough water to give 375 mL of soloution, what is the molarity ... vo volume of solute . ... v/v ethanol, how much 95% v/v ethanol ... prepare 200. mL ...</u></em>
The SI unit for distance is metres - m.
The SI unit for speed or velocity is metres per second - m/s.
The SI unit for acceleration is metres per second squared. - m/s^2.
The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.
As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.
The Interactions found in these compounds are London Dispersion Forces.
And among several factors at which London Dispersion Forces depends, one is the size of molecule.
Size of Molecule:
There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
