1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Bas_tet [7]
4 years ago
12

Write a chemical equation representing the first ionization energy for lithium. use e− as the symbol for an electron.

Chemistry
2 answers:
horsena [70]4 years ago
7 0

\boxed{{\text{Li}} \to {\text{L}}{{\text{i}}^ + } + {{\text{e}}^ - }} is the chemical equation that represents the first ionization energy of lithium.

Further Explanation:

Ionization energy is the amount of energy that is required to remove the most loosely bound valence electrons from the isolated neutral gaseous atom. It is denoted by IE. The value of IE is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.

Ionization energy is further represented as first ionization, second ionization and so on. When the first electron is removed from a neutral, isolated gaseous atom, the energy needed for the purpose is known as the first ionization energy, written as  {\text{I}}{{\text{E}}_{\text{1}}}. Similarly, when the second electron is removed from the positively charged species (cation), the ionization energy is called the second ionization energy \left( {{\text{I}}{{\text{E}}_2}} \right) and so on.

The atomic number of lithium is 3. So its electronic configuration is 1{s^2}2{s^1}. Its outermost electron is present in 2<em>s</em> orbital. The amount of energy required to remove this electron of lithium atom is known as the first ionization energy of lithium.

The equation that represents the first ionization of lithium is {\text{Li}} \to {\text{L}}{{\text{i}}^ + } + {{\text{e}}^ - }.

Learn more:

1. The subatomic particle with the least mass: brainly.com/question/2224691

2. Number of carbon atoms in 1.3-carat diamond: brainly.com/question/4235993

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: first ionization energy, lithium, electronic configuration, atomic number, electron, neutral, isolated, gaseous atom, IE1., equation, size of the atom.

just olya [345]4 years ago
4 0

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>
  • Ionization energy: brainly.com/question/1971327
  • Trend in ionization energy: brainly.com/question/1971327
  • First ionization energy: brainly.com/question/1971327

Level: High school  

Subject: Chemistry  

Topic: Periodic table and chemical families  

Sub-topic: Ionization energy

You might be interested in
What is the molality of an aqueous solution containing FeCl3 (MM = 162.2 g/mol) with a mole fraction of FeCl3 of 0.15?
Natalka [10]

Answer:

10 m

Explanation:

The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.

The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:

0.85 mol × 18.02 g/mol = 15 g = 0.015 kg

The molality of FeCl₃ is:

m = moles of solute / kilogram of solvent

m = 0.15 mol / 0.015 kg

m = 10 m

3 0
4 years ago
How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc
Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

c_{1} v_{1} =c_{2} v_{2}

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

v_{1} =\frac{c_{2} v_{2} }{c_{1} }

in this case that would be:

v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL

5 0
3 years ago
How many amino acids can compose a protein molecule?
Vladimir79 [104]
The answer is A: Between 50 and 5,000 amino acids
3 0
3 years ago
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

8 0
3 years ago
A container is filled to a volume of 55.2 L at 61 °C. While keeping the
babymother [125]

Answer:

4.45 atm

Explanation:

Applying,

PV = P'V'............ Equation 1

Where P = Initial pressure of the container, V = Initial volume of the container, P' = Final pressure of the container, V' = Final volume of the container.

make P the subject of the equation

P = P'V'/V........... Equation 2

From the question,

Given: V = 55.2 L, P' = 8.53 atm, V' = 28.8 L

Substitute these values into equation 2

P = (8.53×28.8)/55.2

P = 4.45 atm

5 0
4 years ago
Read 2 more answers
Other questions:
  • What mass of AgBr can be produced starting with 34.3G of NaBr? (Stoichiometry method)
    6·1 answer
  • Balancing Chemical Equations.
    7·1 answer
  • Give an example of a physical property that is also a characteristic property
    14·1 answer
  • How does the polar jet stream influence the weather on the southern tip of soth america?
    8·1 answer
  • Discuss the relationship between amount of energy and the color of light produced by excited electrons
    10·1 answer
  • Which one of the following statements about mixtures is correct [2] [3] The different substances in a mixture are always in a de
    14·1 answer
  • Which of the following is the correct word equation for the reaction described below?
    8·1 answer
  • Please help
    14·1 answer
  • How many grams of oxygen gas are there in a 2.3 L tank at 7.5 atm and 24 C
    9·1 answer
  • How many moles are present in 3.4 x 1023 atoms of Na?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!