# Lesson Note For SS1 Chemistry (Second Term)

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**Chemistry Scheme of Work for SS1 Second Term**

### Week 1: Revision/Introduction to the mole concept: Molar volume of gases, Avogadro’s number, Percentage of element in a compound.

### Writing and Balancing Chemical Equations.

### Week 2 – 3: Stoichiometry of Reactions: Calculation of Masses of Reactants and Products, Calculation of Volume of Reacting Gases. Empirical and Molecular formulae.

### Week 4 – 5: Chemical Laws and their Verification: Law of Conservation of Mass, Law of Constant Composition, Law of Multiple Proportion.

### Week 6 – 7: Chemical Combinations: Electrovalent Bond: Properties of Electrovalent

### Compounds, Covalent Bond: Properties of Covalent Compounds. Other Types of Bonding

### Week 8 – 9: TheKinetic theoryof Matter and the Gas Laws: Boyle’s Law, Charles’ Law, Ideal Gas Equation, Dalton’s Law of Partial Pressure.

### Week 10: Avogadro’s law, Gay-Lussac’s Law of Combining Volumes, Graham’s Law of Diffusion.

**REFERENCE BOOKS**

- New Chemistry for Senior Secondary School by Osei Yaw Ababio; U.T.M.E Past Questions and Answers.
- Practical Chemistry for Senior Secondary Schools by Godwin Ojokuku
- Outline Chemistry for Schools & Colleges by Ojiodu C.C.
- Chemistry Pass Questions for S.S.C.E and UTME.

**Lesson Note on Chemistry for SS1 Second Term**

**WEEK 1 TOPIC: **Revision/Introduction to the mole concept: Molar volume of gases, Avogadro’s number, Percentage of element in a compound.

**Lesson Note On **Introduction to the mole concept

**CONTENT**

- Relative atomic mass
- Relative molecular mass
- Molar volume of gases
- Percentage of an element in a compound

**THE MOLE**

A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 10^{23} in magnitude and is known as Avogadro’s number of particles.

The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon-12.

**RELATIVE ATOMIC MASS**

The relative atomic mass of an element is the number of time the average mass of one atom of that element is heavier than one twelfth the mass of one atom of Carbon-12. It indicates the mass of an atom of an element. For e.g, the relative atomic mass of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23, and 40 respectively.

The atomic mass of an element contains the same number of atoms which is 6.02 x 10^{23}atoms; 1 mole of hydrogen having atomic mass of 2.0g contains 6.02 x 10^{23} atoms.

**EVALUATION**

- Define relative atomic mass of an element
- State the relative atomic mass of the following elements: potassium, chlorine, silver, lead, phosphorus and nitrogen

**RELATIVE MOLECULAR MASS**

The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of Carbon-12

It is the sum of the relative atomic masses of all atoms in one molecule of that substance. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also that of an ion or radical.

**CALCULATION**

Calculate the relative molecular mass of:

- Magnesium chloride
- Sodium hydroxide
- Calcium trioxocarbonate

**Solution:**

- MgCl
_{ 2}= 24 + 35.5×2 = 24 + 71 = 95gmol^{-1} - NaOH = 23 + 16 + 1 = 40gmol
^{-1} - CaCO
_{3}= 40 + 12 +16×3 = 100gmol^{-1}

**EVALUATION**

- What is relative molecular mass of a compound?
- Calculate the relative molecular mass of (a) NaNO
_{3}(b) CuSO_{4}.5H_{2}O

**MOLAR VOLUME OF GASES**

The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p) is 22.4 dm^{3}. Thus 1 mole of oxygen gas of molar mass 32.0gmol^{-1} occupies a volume of 22.4dm^{3} at s.t.p and 1 mole of helium gas of molar mass 4.0gmol^{-1} occupies a volume of 22.4 dm^{3} at s.t.p.

Note: When the conditions of temperature and pressure are altered, the molar volume will also change. Also, standard temperature = 273K and standard pressure = 760mmHg.

**RELATIONSHIP BETWEEN QUANTITIES**

Molar mass = __mass(g)__ i.e. M = __m__ gmol^{-1}

Amount (moles) n

Note: Amount = Number of moles

Molar volume of gas =__ volume ( cm ^{3} or dm^{3})__ i.e. Vm =

__v__dm

^{3}mol

^{-1}

Amount (mole) n

Amount = __Reacting mass (g)__

Molar mass (gmol^{-1})

Also, Amount of substance = __Number of particles__

Avogadro’s constant

But, Avogadro’s constant = 6.02 x 10^{23}

Combining the two expressions:

__Reacting mass__ = __Number of particles__

Molar mass 6.02 x 10^{23}

**CALCULATIONS**

- What is the mass of 2.7 mole of aluminium (Al=27)?

**Solution:**

Amount = __Reacting mass__

Molar mass

Reacting mass = Amount x Molar mass

= 2.7mole x 27 gmol^{-1} = 72.9g.

- What is the number of oxygen atoms in 32g of the gas (O=16, N
_{A}= 6.02 x 10^{23})?

**Solution:**

__Reacting mass__= __Number of atoms__

Molar mass 6.02 x 10^{23}

Number of atoms = __Reacting mass x 6.02 x 10 ^{23}__

^{ }Molar mass

Molar mass of O_{2 }= 16×2 =32gmol^{-1}

Number of atoms = __32g x 6.02 x 10 ^{23}__

32gmol^{-1}

= 6.02 x 10^{23}

The number of oxygen atoms is 6.02 x 10^{23}

**EVALUATION**

- Define the molar volume of a gas
- How many molecules are contained in 1.12dm
^{3}of hydrogen gas at s.t.p?

**PERCENTAGE OF AN ELEMENT IN A COMPOUND**

The percentage composition of an atom in a compound is the amount of the atom expressed in percentage.

Percentage of an element in a compound = __Mass of element in the compound__ x __100__

Molar mass of compound 1

**CALCULATIONS**

- What is the percentage by mass of nitrogen in NH
_{4}NO_{3}( H=1, N=14, 0=16)?

**Solution:**

Molar mass of NH_{4}NO_{3} = 14×2 + 1×4 + 16×3 = 80gmol^{-1}

Percentage by mass of N_{2} =__ Mass of N _{2}__ x

__100__

Molar mass of NH_{4}NO_{3} 1

= __28__ x __100__= 35%

80 1

- Calculate the percentage by mass of water of crystallization in MgSO
_{4}.7H_{2}O

(Mg=24, S=32, 0=16, H=1)

**Solution:**

Molar mass of MgSO_{4}.7H_{2}0 = 24 + 32 + 16×4 + 9(2+16) = 246gmol^{-1}

7 moles of water of crystallization = 126g

Percentage by mass of water = __Mass of H _{2}O __ x

__100__

Molar mass of MgSO_{4}.7H_{2}O 1

= __126g __ x __100__

246gmol^{-1}1

= 51.2%

**GENERAL EVALUATION**

- What is the number of molecules in 6.4g of SO
_{2}(N_{A}= 6.02 X 10^{23})? - What is the volume in cm
^{3}of 2.2g of CO_{2}at s.t.p ( C=12, O=16)? - Determine the percentage by mass of oxygen in Al
_{2}(SO_{4}).2H_{2}( Al=27, S=32, O=16, H=1)

**READING ASSIGNMENT**

New School Chemistry for Senior Secondary Schools by O. Y Ababio, Pg 28-31

**WEEKEND ASSIGNMENT**

- What is the relative atomic mass of potassium A. 40 B. 39 C. 32 D. 24
- An element with relative atomic mass 108 is A. Ca B. Cl C. Ag D. Al
- Modern standard element with which chemist define relative atomic mass is A.
^{12}C B.C^{13}^{3}H D.^{16}O - Calculate the relative molecular mass of CH
_{3}A. 60gmol^{-1}B. 70gmol^{-1}C. 80gmol^{-1}D. 90gmol^{-1} - How many moles are there in 12g of CO
_{2}(C=12, 0=16)?A. 0.27 B. 0.47 C. 0.16 D. 0.32

**THEORY**

- Calculate the actual number of atoms contained in 2.8dm
^{3}of chlorine (Molar volume of gas = 22.4dm^{3}, N_{A}= 6.02 X 10^{23}) - How many moles are there in 10g of iron (II) tetraoxosulphate (VI)?

**Related Lesson Notes:**

Lesson Note For SS1 Chemistry (First Term)

Lesson Note For SS1 Chemistry (Third Term)

**Lesson Note on Chemistry for SS1 Second Term**

**WEEK 2 TOPIC: **Stoichiometry of Reactions: Calculation of Masses of Reactants and Products, Calculation of Volume of Reacting Gases. Empirical and Molecular formulae.

**Lesson Note On **Stoichiometry of Reactions: Calculation of Masses of Reactants and Products, Calculation of Volume of Reacting Gases. Empirical and Molecular formulae.

Chemical equations are representation of chemical reactions in terms of the symbols and formulae of the elements and compounds involved. In a chemical equation, the reactants are always written on the left hand side while the products are written on the right hand side. For instance, if A and B combines together to give C and D, the equation of the reaction is written as:

A + B → C + D

Reactants Products

**BALANCING CHEMICAL EQUATIONS**

All equations must be balanced in order to comply with the law of conservation of matter. Equations are balanced through the use of coefficients in front of the formula and not by changing the subscript numbers within the formulae of the products.

Example 1: Write a balanced equation for the combustion of ammonia gas in air.

**Solution:**

Step I: Write the reactants and predict the products

NH_{3(g)} + O_{2(g)} → NO_{(g)} + H_{2}O_{(g)}

Step II: The equation is not balanced. Therefore the equation can be balanced by placing the right coefficient in front of each molecule to balance the number of atoms. Thus, the balanced equation is:

4NH_{3(g)} + 5O_{2(g)} → 4NO_{(g)} + 6H_{2}O_{(g)}

Example 2: Write a balanced equation for the combustion of ethane in oxygen.

**Solution:**

The general formula for the combustion of Alkanes is

C_{x}H_{y} + (x + y/4) O_{2} → XCO_{2} + y/2 H_{2}O

The molecular formula for ethane is C_{2}H_{6}, so, x=2 and y=6

Substituting x and y into the formula above gives

C_{2}H_{6} + (2 + 6/4) O_{2} →2CO_{2} + 6/2 H_{2}O

C_{2}H_{6} + 7/2 O_{2} →2CO_{2} + 3H_{2}O

The equation is balanced. However, equations are written with whole number coefficients. By multiplying the entire equation by 2, we get

2C_{2}H_{6} + 7O_{2} →4CO_{2} + 6H_{2}O

**IMPORTANCE OF CHEMICAL EQUATIONS**

- It gives us information on the product that can be formed from the combination of two or more reactants in a particular reaction.
- It tells us the physical states of the reactants and products.
- It indicates the direction of the reaction and whether the reaction is reversible.
- It tells us the stoichiometry of the reaction (i.e. the relationship between the amount of reactants and products) in terms of mole ratio of the reactants and products involved.

Consider the table below: …………………..

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