The chemical equation representing the first ionization energy for lithium is given by;
Li → Li + e-
<h2>Further Explanation; </h2><h3>Ionization energy</h3>
- Ionization energy is the energy required to remove outermost electrons from the outermost energy level. Energy is required to remove an electron from an atom.
- The closer an electron is to the nucleus the more energy is required, since the electron is more tightly bound to the atom thus making it more difficult to remove, hence higher ionization energy.
- Ionization energy increases across the periods and decreases down the group from top to bottom.
- Additionally, the ionization energy increases with subsequent removal of a second or a third electron.
<h3>First ionization energy </h3>
- This is the energy required to remove the first electron from the outermost energy level of an atom.
- Energy needed to remove the second electron to form a divalent cation is called the second ionization energy.
<h3>Trends in ionization energy </h3><h3>1. Down the group(top to bottom)</h3>
- Ionization energy decreases down the groups in the periodic table from top to bottom.
- It is because as you move down the group the number of energy levels increases making the outermost electrons get further from the nucleus reducing the strength of attraction to the nucleus.
- This means less energy will be required compared to an atoms of elements at the top of the groups.
<h3>2. Across the period (left to right)</h3>
- Ionization energy increases across the period from left to right.
- This can be explained by an increase in nuclear energy as extra protons are added to the nucleus across the period increasing the strength of attraction of electrons to the nucleus.
- Consequently, more energy is needed to remove electrons from the nucleus.
Keywords: Ionization energy, periodic table, energy levels, electrons
<h3>Learn more about</h3>
Level: High school
Subject: Chemistry
Topic: Periodic table and chemical families
Sub-topic: Ionization energy
5.167g of calcium chloride is dissolved in 101.0mL of water in a calorimeter whose calorimeter constant is 15.3J/°C. The temperature rises from 18.4°C to 27.2 ...
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al 
is required.
Therefore for 2 moles of Al,
Total 
F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
brainly.com/question/27985929
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