During the electrolysis of the molten lithium chloride, the Lithium ions (Li⁺) at the cathode undergoes reduction, and the electron configuration of lithium becomes 1s²2s¹.
<h3>What is electrolysis?</h3>
Electrolysis can be described as the process in which the electric current is passed through the chemical compound to break them. In this process, the atoms and ions are interchanged by the addition or removal of electrons.
The ions are allowed to move freely in this process. When an ionic compound is melted or dissolved in water then ions are produced which can move freely.
During the electrolysis of molten lithium chloride, the lithium ions reach the cathode and accept the electrons while chloride ions reach at anode and loss electrons to become chlorine gas.
At anode : 2 Cl⁻ → Cl₂ + 2e⁻
At cathode: 2 Li⁺ + 2e⁻ → Li
Learn more about electrolysis, here:
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In photosynthesis, light energy is used to convert CO2 into carbohydrates.
Answer:
-162,5 kJ/mol
Explanation:
Cl(g) + 2O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ/mol (as we used the reaction in the opposite direction, it will turn the enthalpy from exothermic to endothermic)
2O3(g) --> 3O2(g) ΔH = -285.3 kJ/mol
Cl(g) + O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ
+ 2O3 (g) --> 3O2(g) ΔH = - 285.3 kJ
O3(g) + Cl(g) --> ClO(g) + 2O2(g) ΔH = 122.8 + (-285.3) = -162,5 kJ
Answer:
Initially the function is symmetric with respect to the axis of the one dimensional box. In the final state it is also symmetrical, however you can envision a snapshot of the system as the light field is interacting with the wave-function wherein a node begins to develop as is shown in the middle and the wave function is evolving from the initial to final state. Now consider that the electron density during process is the square of the wave function:
Electron density during transition
As can be seen in the initial and final states the electron density is symmetrically distributed with respect to the axis of the box. However with the field on, the electron density is not symmetrically distributed and a transitory dipole moment can be present. To relate back to real molecules think of each of those orbitals as a linear combination of atomic orbitals. One important factor is the symmetry. But there may be one other factor that will be just as important as symmetry. If you treat orbital 1 as a linear combination over n orbitals and orbital 2 as a linear combinations of orbitals as well, there will be a spatial over lap between the orbital in the ground state and the orbital in the excited state. If there is no spatial overlap between the ground state and excited state orbitals there will be no transition dipole moment. However, if the electrons are in the same place spatially, a large transition dipole moment will result.
Explanation:
