Neutral oxides show neither basic nor acidic properties and hence do not form salts when reacted with acids or bases, e.g., carbon monoxide (CO); nitrous oxide (N₂O); nitric oxide (NO)

Fractional distillation can be used to separate ethanol and water. These two liquids are miscible with each other. This means that they can be mixed together to form a single uniform liquid. Their different boiling points allow them to be separated, as water boils at 100℃ and ethanol boils at 78℃.

When the mixture is heated, the ethanol will evaporate first. Some of the water may also evaporate with the ethanol so the temperature of the column must be controlled so that any water vapour formed condenses in the column and drops into the flask. This leads to the alcohol vaporising at the top and out into the container. This means that the ethanol produced first will be pure and should not contain any water.

Once all of the ethanol has been removed, the solution is heated further and the water will evaporate off once its boiling temperature is reached. The water obtained should be pure as all of the ethanol has been removed. This method allows both the ethanol and the water to be collected separately.

Electronic configuration = 2, 8, 8, 2

The outermost shell is the last shell hence the last shell has 2 electrons as shown in the electronic configuration.

Lime water is calcium hydroxide [Ca(OH)₂]

Carbon (IV) oxide is carbon dioxide (CO₂)

When carbon dioxide is passed through lime water, it turns milky due to formation of white precipitate in form of calcium carbonate (CaCO₃).

The reaction is shown below.

Ca(OH)₂+CO₂→CaCO₃+H₂O

Mass = 5.3 g
Molar mass = 106

Mole = $\frac{5.3}{106}$ = 0.05 mol

Per dm³ means volume = 1 dm³

Molarity = $\frac{\mathrm{0.05\; mol}}{1{\mathrm{dm}}^3}$ = 0.05 mol dm^-3

Convert the cm³ to dm³ by dividing it by 1000

100 cm³ = 0.1 dm³

Volume = 0.1 dm³

Molarity = 0.5 moldm^-3

Molar mass = 58.5

0.5 = $\frac{\mathrm{Mass}}{\mathrm{58.5\; x\; 0.1}}$

Mass = 0.5 x 58.5 x 0.1 g = 2.93 g

Method I

Using the formula

Mole = 0.001 mol

Avogadro's constant = 6.02 x 10²³ per mol

Number of atoms = 6.02 x 10²³ x 0.001

Number of atoms = 6.02 x 10²⁰

Method II

Using proportionality

1 mole contains 6.02 x 10²³ atoms

0.001 mol contains 6.02 x 10²³ x 0.001 atoms

0.001 mol contains 6.02 x 10²⁰ atoms

Example of a trioxocarbonate (IV) salt is calcium carbonate.

The reaction with hydrochloric acid is shown below:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Note: CO₂ is carbon (IV) oxide.

Differences between evaporation and boiling

Boiling occurs at a particular temperature but evaporation occurs at any temperature

Electronic configuration = 2, 8, 8, 2

The outermost shell is the last shell hence the last shell has 2 electrons as shown in the electronic configuration.

Nucleon is the collective term for protons and neutrons. Nucleons are the particles found in the nucleus of atoms.

Mass = 5.3 g
Molar mass = 106

Mole = $\frac{5.3}{106}$ = 0.05 mol

Per dm³ means volume = 1 dm³

Molarity = $\frac{\mathrm{0.05\; mol}}{1{\mathrm{dm}}^3}$ = 0.05 mol dm^-3

Convert the cm³ to dm³ by dividing it by 1000

100 cm³ = 0.1 dm³

Volume = 0.1 dm³

Molarity = 0.5 moldm^-3

Molar mass = 58.5

0.5 = $\frac{\mathrm{Mass}}{\mathrm{58.5\; x\; 0.1}}$

Mass = 0.5 x 58.5 x 0.1 g = 2.93 g

Method I

Using the formula

Mole = 0.001 mol

Avogadro's constant = 6.02 x 10²³ per mol

Number of atoms = 6.02 x 10²³ x 0.001

Number of atoms = 6.02 x 10²⁰

Method II

Using proportionality

1 mole contains 6.02 x 10²³ atoms

0.001 mol contains 6.02 x 10²³ x 0.001 atoms

0.001 mol contains 6.02 x 10²⁰ atoms

Fats consist of the monomers or smaller molecules, glycerol and fatty acid. These molecules are composed of the chemical elements hydrogen, carbon, and oxygen.

In a neutral atom, the number of electrons equals the number of protons. Since the protons (11) are more than the electrons (10), means one (1) electron has been lost/transferred to another to form a cation (an ion).

Mass = 4.0 g
Molar mass = 16

Mole = $\frac{4}{16}$ = 0.25

The volume of an ideal gas at STP = 22.4 dm³/mol

If 1 mol = 22.4 dm³

0.02 mol = $\frac{22.4{\mathrm{dm}}^3\mathrm{x\; 0.02\; mol}}{\mathrm{1\; mol}}$ = 0.448dm³

The boiling point of a liquid depends on pressure, vapour pressure and molecular weight.

Factors affecting boiling point

Pressure: Boiling point increases as the pressure increases. If external pressure is lower than the atmospheric pressure then the boiling point is less than the normal boiling point. It increases when the pressure is more than the atmospheric pressure.

Effect of impurities: The boiling point increases with the increase in impurities.

Intermolecular/Cohesive forces: If the forces are weak, then the boiling point is also low and if the forces are strong then the boiling is high.

Molecular weight: Boiling increases if the chain length of the molecule increases.

Branching: With the increase in branching, the boiling point decreases.

The atomic masses of some elements are in fraction instead of whole numbers. This is due to the existence of isotopes having different mass number.

The isotopes of an element have different atomic mass and the percentage of these isotopes vary in nature.

In a neutral atom, the number of protons equals the number of electrons.

Number of protons = 13 = Number electrons

Mass = 5.85

Na = 23
Cl = 35.5

Molar mass = 23 + 35.5 = 58.5

Mole = $\frac{5.85}{58.5}$ = 0.10

Since the concentration is in dm³, convert the cm³ volume to dm³ by dividing by 1000

250 cm³ = $\frac{250}{1000}$ = 0.25 dm³

Mole = Concentration x Volume

Mole = 2 mol dm^-3 x 0.25 dm³ = 0.5 moles

The initial and final moles are the same. Thus the mole doesn't change but only the concentration changes.

Initial molarity = Final molarity

Moles = Concentration x Volume

C₁ x V₁ = C₂ x V₂

C₁ = 0.5 mol dm^-3

V₁ = 100 cm³

C₂ = 0.1 mol dm^-3

0.5 mol dm^-3 x 100 cm³ = 0.1 mol dm^-3 x V₂

Divide both sides by 0.1

V₂ = $\frac{\mathrm{0.5\; x\; 100}}{0.1}$ = 500 cm³

The final volume must be 500 cm³

Volume of HCl + Water = 500 cm³

Volume of HCl = 100 cm³

100 cm³ + Water = 500 cm³

Water = 500 cm³ - 100 cm³

Water = 400 cm³

Since the concentration is in per dm³, convert the cm³ to dm³ by dividing by 1000

250 cm³ = $\frac{250}{1000}$ dm³ = 0.25 dm³

Concentration = 0.10 mol dm^-3

Molar mass = 39 + 16 + 1 = 56 g mol^-1

Note: unit for molar mass is grams per mole

Volume = 0.25 dm³

Mass = 0.10 mol dm^-3 x 56 g mol^-1 x 0.25 dm³ = 1.4 g