Form 4 Chemistry Project. [Radioactive]

Tugasan Cuti Sekolah
2014
Nama kumpulan : Radioactive
Guru Mata Pelajaran : Chung Sze Meang
Ahli Kumpulan :
Jawatan Nama Kelas Tandatangan
Ketua Kumpulan Muhammad Firdaus
4Sains2
4Sains2
AhliKumpulan
Nur Faezah
Nur Zahidah
Juwita
Muhammed Azmirrudyn
Charles Hasley
Ian Bidick

Buku Log
Perjumpaan Tarikh Ahli yang terlibat Perbincangan topik Catatan
1. 21/10/14
Semua ahli
Pembahagian tugas Tugas telah
dibahagikan.2. 22/10/14
3. 4/11/14 Bab 5, 6 & 7 Bab 5 berjaya
disiapkan manakala
terdapat beberapa
masalah dengan Bab 6
& 7
4. 11/11/14 M. Firdaus, Juwita,
Charles
Bab 4 Dijalankan dengan
lancar.
5. 4/12/14 M. Firdaus, Juwita,
N. Zahidah, Ian B.,
N. Faezah
Bab 4, 5, 6. Bab 4 telah disiapkan.
Dijalankan degan
lancar.

Chapter 2: The Structure ofthe Atom
A. Knowledge ( Definition, Meaning and Facts )
1. State the kinetic theory of matter.
State one example to support the kinetic theory of matter.
-
-
-
-
-
matter is composed of a large number of small particles (individual atoms or molecules) that
are in constant motion.
are attracted together by an attractive force (intermolecular forces).
have kinetic energy (to moves faster at a higher temperature).
This theory is supported by the process of diffusion that takes place between particles of
matter.
Example :
When perfume is sprayed into one corner of a room, the diffusion process spreads out the
perfume particles until the whole room is filled with the smell of perfume.
2. What are atom, molecule and ion?
Atom Molecule Ion
-the smallest particle of an
element that can take part in
a chemical reaction.
-a group of two or more
atoms which chemically
bonded together.
-a negatively-charged or
positively-charged
particles.
3. What is melting point?
- Temperature at which the substance changes from solid state into a liquid state at a given
pressure.
4. Define proton number and nucleon number
- Proton number is the number of protons in an atom.
- Nucleon number is the total number of protons and neutrons in an atom.
5. State the meaning of isotopes.
- The atoms of the same element with the same proton number but different nucleon number.
6. State the uses of isotopes such as carbon-14 and cobalt-60
Isotopes Uses
Carbon-14 To determine the age of archeological artifacts
Cobalt-60 Used in radiotherapy for the treatment of cancer

B. Understanding / Application / Analysis
7. Explain why the temperature remains unchanged during the melting point.
-
-
-
At the melting point, the matter is in solid and liquid states
The latent heat of fusion is absorbed to overcome forces at attraction among solid particles
Therefore,the temperature remains unchanged at the melting point.
8. Explain why the temperature remains unchanged during the freezing point.
- At the freezing point, the states of matter is in liquid and solid.
The latent heat of fusion is released to allow forces of attraction to form among particles.
Therefore,the temperature remains unchanged at the freezing point.
9. A solid compound is heated until it converts into gas. Explain the changes in energy content,
forces of attraction between the particles, and arrangement of particles.
Changes in energy content Kinetic energy increases
Forces of attraction between
the particles
Very weak
Arrangement of particles Are very apart and randomly arranged
10. State the main subatomic particles of an atom.
Compare and contrast the relative atomic mass and the relative charge of the subatomic particles
of the atom.
- proton, neutron and electron
-
Particles Relative charge Relative mass
Proton +1 1
Neutron 0 1
Electron -1 1/1840
11. 𝐴𝑙13
27
is the symbol of aluminium.
(a) Determine the number of neutron of aluminium.
(b) Draw the electron arrangement of aluminium.
(a) 14 neutron ( 27-13 = 14)
(b)

C. Synthesis (Experiment)
12). Solid Z has a melting point of 65°C. Describe a labratory experiment to determine the melting
point of Z. Your answer should show how the melting point of Z is determined.
Materials & Apparatus :Boiling tube, Tap water,Solid Z, 250cm³ beaker, thermometer, tripod
stand, retord stand and clamp, Bunsen burner, stopwatch, and wire gauze.
Procedure :
1. A boiling tube is filled with solid Z and a thermometer is put into it.
2. The boiling tube is suspended in a beaker half filled with water using a retort stand and a
clamp as shown in the diagram. The level of substance Z in the boiling tube must be below
the level of the water in the beaker.
3. The water heated and the substance Z is stirred with the thermometer.
4. The temperature and the state(s) of the substance is recorded on half-minutes intervals by
using stopwatch.
** water bath method is used because the melting point of substance Z is below 100°C, the
maximum temperature that can be attained by the waterbath.
Thermometer
Substance
Z
Tap water

Result
Heating of Solid Z
Time (min) Temperature (°C) State
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
Observation
A) At temperature below 65 °C, Z remains as solid.
B) When it is heated, heat energy is absorbed. This cause the particles to gain kinetic energy
and vibrate faster. Temperature increase to 65°C.
C) At 65°C , solid Z begins to melt. During the melting process, the temperature of substance
Z did not rise even though heating continues.
D) The temperature remains constant because the heat energy is absorbed by the particles to
overcome the forces between particles so that the solid can turn into a liquid. At this
temperature,both solid and liquid are present.
E) At temperature higher than 65°C, the substance Z has melted.
F) The temperature will continues to rise and the particles in liquid Z absorb heat energy and
move faster.
Conclusion: The melting point of solid Z is 65°C.

13. Compound W has a freezing point of 82°C. Describe a laboratory experiment to determine the
freezing point of W.
Materials & Apparatus :Boiling tube, Compound W, Conical flask, Thermometer, Retord stand
and clamp, stopwatch.
Procedure :
1. The boiling tube that contain compound W is put in a conical flask as shown in thefigure
above.
2. The compound W is stirred with thermometer continuously.
3. The temperature and state(s) of compound W is recorded half-minute intervals until the
temperature drops to about 60°C.
Result
Cooling of compound W
Time (min) Temperature (°C) State
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
Compund W

Observation
A) At temperature above than 82°C , compound W exist as a liquid.
B) When the liquid is cooled, particles in the liquid lose their kinetic energy. They move
slower as the temperature decreases.
C) At temperature 82°C, the liquid W begins to freeze. During freezing, the temperature of
W remains constant because the heat loss to the surrounding is exactly balanced by the
heat energy liberated as the particles attract one another to form a solid. At this
temperature both solid and liquid are present.
D) At temperature 82°C, all the liquid W has frozen.
E) The particles in solid W release heat energy and vibrate slower. The decrease from 82°C
to lower point
Precautionary steps
1. The boiling tube containing liquid W is placed in a conical flask to ensure an even cooling
process and to minimize heat loss to the surroundings.
2. Liquid W is stirred continuously to avoid supercooling. Supercooling is a condition in which
the temperature of a cooling drops below its normal freezing point, without the appearance of
a solid.
Conclusion: The freezing point of compound W is 82°C

Chapter 3: Chemical Formulae and Equations
A. Knowledge ( Definition, Meaning and Facts )
1. State the meaning of relative atomic mass based on carbon-12 scale.
- the ratio of the average mass of one atom of an element to
1
12
of the mass of an atom
of carbon-12.
Relative Atomic Mass
(RAM) of an element
= The average mass of 1 atom of an element
1
12
× the mass of one carbon− 12
2. Define mole.
- One mole is defined as the number of particles equal to that in exactly 12.00g of carbon-12
isotope.
3. State the meaning of molar mass
- Molar mass is the mass of one mole of a substance. (unit g mol−1)
4. State the meaning of molar volume of gas.
- Molar volume of gas means the volume occupied by 1 mole of any gas. (unit dm3 mol−1)
[ i.e. 22.4 dm3 mol−1 at STP or 24 dm3 mol−1 at room conditions ]
5. State the meaning of empirical formula
- The chemical formula that shows the simplest whole number ratio of atoms of each element in
the compound.
6. State the meaning of molecular formula
- The chemical formula that shows the actual number of atoms of each element in the
compound.

B. Understanding / Application / Analysis
7. Explain why we could not determine the empirical formula of copper (ll) oxide by heating
copper powder in a crucible.
- Copper is less almost not a reactive metal
8. Compare and contrast empirical formula with molecular formula using ethane as an example.
Empirical formula Molecular formula
C2H6 C2H6
9. Vinegar is a dilute ethanoic acid with a molecular formula of CH3COOH.
(a) Find the empirical formula of ethanoic acid.
(b) Find the percentage composition by mass of carbon in ethanoic acid.
(a) CH2O
(b) 40 %
10. 3.6 g of carbon reacted with 0.8 g of hydrogen to form a compound.
(a) Determine the empirical formula of the compound formed.
(b) Given that the relative molecular mass of the compound is 88 g, find its molecular formula.
[Relative atomic mass : C, 12 ; H, 1 ]
(a)
(b)
Element Carbon , C Hydrogen , H
Mass (g) 3.6 0.8
Number of moles 0.3 0.8
Most simplest ratio 1 3
Empirical formula : CH3
n (CH3) = 88
n (12+ 3) = 88
15n = 88
n = 6
Molecular formula : C6H18
11. Hydrogen gas is reacted with 20 g of hot copper (ll) oxide powder to produce solid copper and
water.
(a) Write the chemical equation for the reaction.
(b) Calculate the maximum mass of solid copper formed.
[Relative atomic mass : Cu, 64 ; O, 16 ; H, 1]
(a) Hydrogen (g) + copper (ll) oxide (s) copper (s) + water (l)
(b) H2 + CuO CuO + H2O

i. No. of moles of CuO
=
mass
RMM
=
20 g
80
= 0.25 mol
ii. 1 mol of CuO 1 mol of Cu
0.25 mol of CuO 0.25 mol of Cu
iii. Mass of solid copper
= mol × RAM
= 0.25 × 64
= 16 g
12. Describe a laboratory experiment to determine the empirical formula of magnesium oxide. Your
answer should include all the precautions and calculations involved.
[Relative atomic mass : Mg, 24; O, 16 ]
Material : Magnesium ribbon, sandpaper
Apparatus : Crucible with lid, Bunsen burner, pipe-clay triangle, chemical balance, tripod stand.
Procedure :
1. A crucible and its lid are weighed.
2. A 10cm length of magnesium ribbon is cleaned with sandpaper to remove the oxide layer on
its surface.
3. The ribbon is coiled loosely and placed in the crucible. The crucible with its lid and content
are weighed.
4. The apparatus is set up as shown in the diagram.

5. The crucible is heated strongly without its lid.
6. When the magnesium starts to burn, the crucible is covered with its lid.
7. The lid is carefully raised slightly, using tongs, at intervals.
8. When the burning is complete, the lid is removed and the crucible is heated strongly for 1 to 2
minutes.
9. The crucible with its lid still on is allowed to cool to room temperature.
10. The crucible with its lid and content are weighed again.
11. The processes of heating, cooling and weighing are repeated until a constant mass is obtained.
The constant mass obtained is recorded.
Result :
Description Mass(g)
Crucible + lid x
Crucible + lid + magnesium y
Crucible + lid + magnesium oxide z
Magnesium y - x
Oxygen z - y
Calculation :
Element Magnesium, Mg Oxygen, O
Mass y - x z - y
No of moles =
Mass(g)
Molar mass (gmol )
y – x
24
z – y
16
Simplest ratio (whole no) r v
Thus, the emperical formula for Magnesium oxide is MgᵣOᵥ
Precautionary steps Explanation
Magnesium ribbon must be cleaned with
sandpaper.
To remove the oxide formed on the magnesium
ribbbon.
Crucible must be closed with a lid while the
magnesium is being heated.
To prevent particles of magnesium oxide from
escaping.
Lid of the crucible must be opened once in a
while during the heating process of magnesium.
To let it the oxygen gas to ensure there is
sufficient/enough oxygen gas for complete
reaction.
Steps of heating, cooling and weighing repeated
until a fixed mass is obtained.
To ensure all magnesium as fully reacted with
oxygen to form oxygen oxide.

Discussion :
1. Magnesium reacts with oxygen in the air to form white fumes, magnesium oxide.
Magnesium + Oxygen  magnesium oxide
Conclusion :
The emperical formulae of magnesium oxide is MgO.
13. Describe a laboratory experiment to determine the emperical formula of copper(ll) oxide.Your
answer should include all the precautions and calculations involved.
[Relative atomic mass: Cu, 64; O, 16]
Material : Copper oxide powder, zinc pieces, dilute hydrochloric acid, anhydrous calcium chloride.
Apparatus : Round bottomed flask, combustion tube with small opening at its end, stopper with
delivery tube, chemical balance, retort stand with clamp, thistle funnel,U-tube.
Procedure :
1. Combustion tube mass with the porcelain dish in it is weighed.
2. A spatula full of copper(ll) oxide is added to the porcelain dish. The tube is weighed again.
3. The apparatus is set up as shown in the diagram.
4. Hydrogen gas is allowed to flow into the set of apparatus for 5 to 10 minutes to remove all the
air in the tube.
5. To determine whether all the air has been removed from the tube, the gas that has comes out
from the small hole is collected in the a test tube. Then ,the gas is tested with a lighted

wooden splinter. If the gas burns with a ‘pop’ sound, then all the air has been totally removed
from the combustion tube.
6. The excess hydrogen gas that flows out from the small hole of combustion tube is burnt and
copper(ll) oxide is heated strongly.
7. Flame is turned off when copper(ll) oxide turn completely brown.
8. The flow of hydrogen gas is continued until the set of apparatus cools down to room
temperature.
9. Combustion tube mass with its content is weighed again.
10. The heating, cooling and weighing are repeated until a constant mass is obtained. The
constant mass is recorded.
Result :
Description Mass(g)
Combustion tube + porcelain dish x
Combustion tube + porcelain dish + copper(ll) oxide y
Combustion tube + porcelain dish + copper z
Copper z - y
Oxygen y - z
Calculation :
Element Copper, Cu Oxygen, O
Mass z – x y – z
No of moles =
mass(g)
Molar mass(gmol )
z – x
64
y – z
16
Simplest ration (whole no) v r
Thus, emperical formula for copper oxide is CuᵥOᵣ
Precautionary steps Explanation
Combustion tube must be flled with hydrogen
before the hydrogen gas is ignited.
A mixture of oxygen and hydrogen gas can cause
explosion when ignited.
Hydrogen gas must be continous throughout the
experiment.
Unstable flow of hydrogen gas can cause oxygen
from air to flow into combustion tube, explosion
may occur.
Steps of heating, cooling and weighing repeated
until a fixed mass is obtained.
To ensure that copper(ll) oxide is fully reacted to
form copper.
Hydrogen gas is still passed through apparatus
during the cooling until room temperature is
reached.
To prevent the hot copper from reacting with
oxygen in the air to form copper oxide again.
Discussion :
1. The function of anhydrous calcium chloride is to dry the hydrogen gas.
2. Copper(ll) oxide is black in colour. It reacts with hydrogen gas to produce brown copper metal.
Hydrogen gas + Copper(ll) oxide  copper + water
Conclusion :
The emperical formulae of copper(ll) oxide is C

Chapter 4 : Periodic Table ofElements.
A. Knowledge (Definition, meaning and facts)
1. State the basic principle of arranging the elements in the periodic table from its proton number.
The elements are arranged according to the ascending proton numbers of the
electrons.
2. State the physical properties of Group 1.
 good conductor of heat & electricity.
 soft & cut easily with knife.
 are grey in colour (silvery & shiny surfaces)
 low melting/boiling points
 low densities, float on water.
3. State the physical properties of Group 17.
 non-metal, non-conductors of heat & electricity.
 the colour becomes darker when going down the group.
 low boiling points.
4. State the changes in the atomic size and electronegativity of elements across the Period 3.
 Atomic size decreases across Period 3.
 Electronegativity increases across Period 3.
5. State three special properties of transition elements?
a) Form coloured compounds :
 Fe2+
(green solution)
 Fe3+
(brown solution)
 Ni2+
(green solution)
 Mn2+
(pink solution)
 Cu2+
(blue solution)
b) Shows different oxidation number in their compounds.
 Iron(II) chloride, FeCl2
 Iron(III) chloride, FeCl3
 Copper(I) oxide, Cu2O
 Copper(II) oxide, CuO
c) Are useful catalysts
 fine iron powder – haber process(ammonia)
 platinum, Pt – Ostwald process(nitric acid)
 nickel, Ni – manufacture margarine
 vanadium (V) oxide, V2O5 – contact process (sulphuric acid)
d) Form complex ions
 hexamine chromium (III) ion [Cr(NH3)6]3+
 hexacyanoferrate(II) ion [Fe(CN)6]4-
 hexacyanoferrate(III) ion [Fe(CN)

B. Understanding / Application / Analysis.
6. State the position of element X39
20 in Periodic Table. Explain your answer.
- Group 2, Period 4. This is because calcium is an alkali earth metal. Furthermore, calcium has
4 shells of electrons of 2.8.8.2 has 2 valence electrons.
7. Going across Period 3 from sodium to argon, the atomic size decreases. Explain.
When going across Period 3 from sodium to argon, the atomic size decreases. This is
because :
- The number of proton in the nucleus increases.
- This increases the positive charge of the nucleus.
- The attractive force between nucleus and the electrons become stronger.
- The electrons are pulled closer to the nucleus causing the atomic size to decreases.
8. Going across Period 3 from sodium to argon, the electronegativity increases. Explain.
When going across Period 3 from sodium to argon, the electronegativity increases. This is
because :
- The number of proton in the nucleus increases.
- This increases the positive charge or the nucleus.
- The attractive force between the nucleus and the electrons in the outermost shell become
stronger.
The increases the tendency to attract electron across the period.
9. The reactivity of Group 1 increases when going down the group. Explain why.
The reactivity of Group 1 increases when going down the group because :
a) The distance between outermost shell occupied with electrons and the nucleus becomes
further.
b) The force of attraction between nucleus and the valence electron becomes weaker.
c) The atom becomes easier to lose its valence electron to achieve the stable duplet/octet
electron arrangement.
10. The reactivity of Group 17 increases when going down the group. Explain why.
The reactivity of Group 17 increases when going down the group because :
a) The distance between outermost shell occupied with electrons and the nucleus become
further.
b) The force of attraction between nucleus and the valence electron becomes weaker.
c) The atom becomes more difficult to attract electron to its outermost shell.
11. Why helium gas is not reactive?
- Because helium is duplet electron arrangement. These electron arrangement are very stable.
They do not accept, donate or share electrons with other elements.
12. X is an element from Group 1. X is burnt in oxygen and the product is dissolved in water. What is
the property of the solution formed? Explain why.
- X burns slowly with a red flame and liberates white fumes which become a white solid on
cooling.

- The white solid dissolves in water to produce a colourless solution, which turns red litmus
paper blue.
13. Chlorine gas is dissolved in water. What can you observe if a piece of blue litmus paper is
immersed into the solution formed? Explain why.
- This solution turns blue litmus paper to red, then white. This is because chlorine dissolves rapidly in
water forming acidic solutions which turn blue litmus paper to red. The solutions formed are also
bleaching agents which then turn the litmus paper to white due to the presence of hypochlorous acid
or hypobromous acid.
14. W is an element from Group 1. Predict the chemical reaction of W with:
a) water,
W moves slowly on the water surface with a soft ‘hisss’ sound. A colourless solution that turns
red litmus paper to blue is formed.
Chemical equation : 2Li + 2H2O  2LiOH + H2
b) oxygen,
W burns slowly with a red flame and liberates white fumes which become a white solid on
cooling. The white solid dissolves in water to produce a colourless solution, which turns red
litmus paper to blue.
Chemical equation : 4Li + O2  2Li2O
15. Describe a laboratory experiment to compare the reactivity of elements in Group 1 : lithium,
sodium and potassium.
Apparatus & materials : water toughs, small knife, forceps, small pieces of lithium, sodium and
potassium, distilled water,red litmus paper and filter paper.
Procedure :
1. A small piece of lithium is cut using a knife.
2. The oil on the surface of lithium is removed by rolling it on a piece of filter paper.
3. The lithium is then placed slowly onto the water surface in a water tough with the help of forceps,
as shown in diagram.
4. All changes that occur are recorded.
5. When the reaction stops, the solution formed is tested with a piece of red litmus paper.
6. Steps 1 to 5 are repeated using sodium and potassium respectively to replace lithium.
Observation :
Lithium : lithium moves slowly on the surface with a ‘hiss’ sound. A colourless solution that turns red
litmus paper to blue is formed.
Sodium : sodium melts to become a small sphere, moves rapidly and randomly on the surface with
a ’hiss’ sound. A colourless solution that turns red litmus paper to blue is formed.
Potassium : potassium melts to become a small sphere, burns with a lilac flame, moves very rapidly
and randomly on the water surface with a ‘hiss’ and ‘pop’ sounds. A colourless solution that turns red
litmus paper to blue is formed.

16. Describe a laboratory experiment to compare the reactivity of chlorine, bromine and iodine in the
reaction with iron wool. State the observation and the write chemical equations involved in
reactions.
Apparatus & materials :
combustion tubes, boiling tubes, conical flask, retort stand and clamp, Bunsen burner, thistle funnel,
stoppers, delivery tubes, potassium manganite (VII) crystals, concentrated hydrochloric acid, liquid
bromine, solid iodine, iron wool and soda-lime.
Procedure :
1. The arrangement of apparatus as shown in it set up.
2. The iron wool is heated strongly until is it red-hot .
3. Chlorine gas is passed over the red-hot iron wool in the combustion tube until no further change
occurs.
4. All the changes are recorded.
5. Step 1- 4 is repeated by replacing chlorine gas with bromine and iodine vapour ( heating is
necessary to evaporate bromine and iodine)
Observation:
Chlorine : The hot iron wool ignites rapidly with a bright flame .A brown solid is formed.
Bromine : The hot iron wool glows moderately bright, moderately fast and less vigorously. A brown
solid is formed.
Iodine : The hot iron wool glows dimly and slowly. A brown solid is formed.

Chapter 5: Chemical Bonds
A.Knowledge [Definition, Meaning and Facts]
1. What is anion?
- Anion is a negatively charged ion that would be attracted to anode in electrolysis.
- Anion is formed when an atom receives electron.
- Anion are attracted to cation by strong electrostatic force of attraction in ionic bond.
2. What is cation?
- Cation is a postively charged ion that would be attracted to cathode in electrolysis.
- Cation is formed when an atom releases electron.
- Cation are attracted to anion by strong electrostatic force of attraction in ionic bond.
-
3. State two physical properties of ionic compound :
- High melting or boiling point
- Conducts electricity in liquid and in aqueous solution.
4. State two physical properties of covalent compound:
- Low melting or boiling point
- Does not conduct electricity in any state
B. Understanding, Aplication, Analysis
5. Exlain why Sodium Chloride can conduct electricity in aqueous state but cannot conduct
electricity in solid state.
- In aqueous state, sodium chloride dissolvesin water and dissociates to produce Na 
and Cl
ions.
These ions can move freely in the solution and conducts electricity.
- In solid state,sodium chloride’s ion are bond together tightly and cannot move freely to conduct
electricity
6. Magnesium chloride and hydrogen chloride are two compunds of chlorine. At room temperature,
magnesium chloride existed as solid but hydrogen chloride exist as a gas. Explain why.
Magnesium Chloride Hydrogen Chloride (HCl)
- Ionic bond are formed between Mg 2
and
Cl
ions.
(ionic compound)
- Magnesium is a metal and chlorine is non
metal.
- Has high melting and boiling point.
- Hydrogen and chlorine atoms share
electrons through covalent bonding.
(Covalent compound)
- Both elements are non metals
- Has low melting and boiling point.

7. Describe the formation of ionic bond in sodium chloride.
 The electron arrangement of a sodium atom is 2.8.1.
 Sodium atom releases one valence electron to form sodium ion, Na+
Na → Na+
+ e-
 The electron arrangement of sodium ion is 2.8.
 Sodium ion achieves a stable octet electron arrangement.
 The electron arrangement of a chlorine atom is 2.8.7.
 Chlorine atom receives one electron to form chloride ion, Cl-
Cl + e-
→ Cl-
 The electron arrangement of chloride ion is 2.8.
 Chloride ion achieves a stable octet electron arrangement.
 Strong electrostatic force pulls the sodium ion and chloride iontogether.
 An ionic bond is formed.
 Sodium chloride is an ionic compound.
Na+
+ Cl-
→ NaCl
8. By using examples, describe the formation of covalent bond between element from Group 14 and
element from Group 17.
 Formation of covalent bond of Methane, CH4
 A carbon atom has an electron arrangement of 2.4.
 Carbon atom has 4 valence electrons.
 A carbon atom needs 4 more electrons to achieve the stable octet electron arrangement.
 A hydrogen atom has 1 valence electron.
 A hydrogen needs 1 more electron to achieve the stable duplet electron arrangement.
 Each of the 4 hydrogen atoms contributes 1 electron and a carbon atom contributes 4 electron
for sharing to form the single bond covalent compound, methane, CH4.

C. Synthesis
9. Draw electron arrangement of the compound formed from the following elements.
a) Nitrogen and hydrogen
Ammonia, NH3
b) Carbon and oxygen
Carbon Dioxide, CO2

c) Magnesium and chlorine
Magnesium Chloride, MgCl2
d) Carbon and hydrogen
Ethane, C2H6
e) Hydrogen and Chlorine
Hydrogen Chloride, HCl
f) Sodium and oxygen
Sodium Oxide, Na2O

Chapter 6: Electrochemistry
1. State the meaning of electrolyte.
Electrolytes are substances that can conduct electricity in the molten state or aqueous solution.
2. State the meaning of electrolysis.
Electrolysis is the decomposition of a substance (electrolyte) in the molten state or aqueous
solution to its elements by electric current.
3. State three factors affecting electrolysis of an aqueous solution.
i. Position of ions in the electrochemical series.
ii. The concentration of ions in the electrolyte solution.
iii. Type of electrode.
B. Understanding/Application/Analysis
4. Explain why solution of hydrogen chloride in water can conduct electricity but hydrogen chloride
in methylbenzene cannot conduct electricity.
a. Water molecules, H2O, in aqueous solution ionise or dissociate slightly to form hydrogen ions,
H+
and hydroxide ions, OH-
.
H2O (l)  H+
(aq) + OH-
(aq)
Therefore,aqueos solutions containts hydrogen ions, H+
and hydroxide ions, OH-
, in addition
to electrolyte ions.
b. Methylbenzene or toluene, C7H8, cannot conduct electricity because it is a non-electrolyte
substance which does not undergo any chemical change when an electric current passes through.
Methylbenzene consists of molecules only. The absence of ions that can carry electrical charges
makes it unable to conduct electricity.
5. By using example, explain how the following factors can determine the selective discharge of
ions at the electrodes.
ii. Types of electrodes.
If the type of metal used as the anode (positive electron) is the same as the type of metal ion
present in the electrolyte solution, then the metal atom (anode) will release electrons to form
metal ions at the anode. For example :-
Electrolysis of copper(II) chloride solution using copper electrodes.
Half equation:
Cu (s)  Cu2+
(aq) + 2e-

At the cathode, the metal ions receive electrons and are deposited as metal atoms.
Half equation:
Cu2+
(aq) + 2e-
 Cu (s)
iii. Concentration of the ions.
The concentration of ions in the electrolyte solution also influences the selection of ions for
discharge.
The more concentrated ions will be selected for discharge at the anode.
The type of ion chosen for discharge at the cathode is determined by the position of the ion in
the electrochemical series. For example :-
Electrolysis of 1.0 mol dm-3
sodium chloride, NaCl, solution (concentrated solution)
Ions present: Cl, OH-
, Na+
, H+
At the anode, Cl-
ions will be discharged at the anode because of its concentration.
Half equation:
2Cl-
(aq)  Cl2 (g) + 2e-
At the cathode, H+
ions will be discharge because it is located at lower position of the
electrochemical series than Na+
ions.
Half equation:
2H+
(aq) + 2e-
 H2 (g)
iv. Position of ions in the Electrochemical series.
The positions of ions in the electrochemical series influences the selection of ions discharged
in an aqueous solution.
Cations: K+
Na+
Ca2+
Mg2+
Al3+
Zn2+
Fe2+
Sn2+
Pb2+
H+
Cu2+
Ag+
F-
SO4
2-
NO3
-
Cl-
Br-
I-
OH-
There is greater tendency for ions located in the electrochemical series to be selected for
discharge compared to ions located higher in the series. For example:-
Electrolysis of sodium sulphate solution, Na2SO4
Ions present: Cathode: Na+
,H+
Anode: SO4
2-
,OH-
More easily discharged

At the anode, OH-
is selectively discharged because it is located lower than SO4
2-
ion in the
electrochemical series.
Half equation:
4OH-
(aq)  2H2O (l) + O2 + 4e-
At the cathode, H+
is selectively discharged because it is located lower than Na+
ion in the
electrochemical series.
Half equation:
2H+
(aq) + 2e-
 H2 (g)
6. Describe the electrolysis of molten lead(II) bromide.
At the anode (+)
- Negative ions or anions (Br-) are attracted to the
anode (+)
- Negative ions will be discharged by releasing
electrons to the anode. These electrons will flow
to the positive terminal of the battery.
2Br- (l)  Br2 (g) + 2e
At the cathode (-)
- Positive ions or cations (Pb2+) are attracted to the
cathode (-).
- Positive ions will be discharged by accepting
eletrons at the cathode to form neutral lead
atoms.
Pb2+ (l) + 2e-  Pb (s)

7. Describe the extraction of aluminium by electrolysis.
Extraction of aluminium from aluminium ore (bauxite) by industrial electrolysis. The main
content of bauxite are aluminium oxide, Al2O3.
8. Draw the structure of Daniell cell and explain how it can produce electricity.
- The Daniell cell consists of two metal electrodes that are connected through the external
cricuit with the electrolyte solution, either separated by a porous pot or a salt bridge.
- The porous pot or salt bridge functions to separate the two electrolyte solutions, but allows
ions to pass through to complete the circuit.
- A metal which is located higher in the electrochemical series (the more electropositive metal)
will act as a negative terminal of the cell. These metals tend to donate electrons.
- A metal which is located lower in the electrochemical series (the less electropositive metal)
will act as a positive terminal of the cell. Thus, converting chemical energy to electrical
energy.
At the anode:
- Anions O2- are attracted to the anode.
- O2- ions will be discharged by
releasing electron at the anode
producing colourless gas bubbles of
oxygen gas, O2.
- Half equation:
2O2- (l)  O2 (g) + 4e-
At the cathode:
- Al3+ ions are attracted to the cathode.
- Al3+ will be discharged by deposing a
grey solid of aluminium metal, Al.

9. Draw the strucuture of a dry cell and explain how it can produce electricity.
- A dry cells converts chemical energy into electrical energy. The chemical reaction produces
electrons, which then collect at the negative terminal of the battery. When there is a
connection between the positive and negative terminal, the electrons flow to the positive
terminal. This electron flow is what we experience as an electric current.
10. Describe a laboratory experiment to extract lead from lead(II) oxide using electrolysis.
Carbon reduction method
- On heating with carbon, the carbon (charcoal) is oxidized to carbon dioxide (or carbon
monoxide) while lead is reduced from a positive “ion” in lead(II) oxide to the neutral metallic
element:
2PbO + C  2Pb + CO2
11. Describe a laboratory experiment to show that types of electrodes affect the seletive discharge of
ions in electrolysis of copper(II) sulphate solution.

Problem statement: Does the type of electrode influence the selective discharge of ion?
Hypothesis: The type of electrode will affect the selective discharge of ion.
Variables
a) Manipulated: Type of electrode.
b) Responding: Product at the electrodes.
c) Constant: Type of electrolyte.
Conclusion: The type of electrode will influence the type of ion discharge
12. Describe a laboratory experiment to show that concentration of ions affect the selective discharge
of ions in electrolysis of hydrchloric acid solution.
Problem statement: How does the concentration of ion affect its selective discharge?
Hypothesis: The more concentrated ion will be selected for discharge at the anode.
Variable:
a) Manipulated: concentration of ion
b) Responding: product at the electrodes.
c) Constant: type of electrode/type of electrolyte.
Electrode Observation Test for gas Inference
Anode Greenish yellow gas bubbles
are released
Gas produces bleach
moist red litmus
paper
Chlorine gas is produce
2Cl-
(g)  Cl2 (g) + 2e-
Cathode Colourless gas bubbles are
released
Gas produces a ‘pop’
sound with a lighted
wooden splinter
Hydrogen gas is produced
2H+
(aq) + 2e-  H2 (g)
Electrode Observation Inference
Copper Copper electrode dissolves/becomes thinner Copper(II) ion is formed Cu(s) 
Cu2+
(aq) + 2e-
Carbon A brown solid is deposited Copper is formed
Cu2+
(aq) + 2e-
 Cu (s)

Conclusion:
a. In a concentrated electrolyte solution, ions that are more concentrated,that is chloride ions, are
selected for discharge at the anode to form chlorine gas.
b. At the cathode, ions that are lower in the electrochemical series are selected for discharge to form
hydrogen gas.
13. You are given magnesium ribbon, copper plate, magnesium nitrate solution, copper(II) sulphate
solution, connecting wires with crocodile clips, 250cm3
beaker, voltmeter and porous pot.
Construct a voltaic cell by using the above materials. Explain how the voltaic cell can produce
electricity. Your answer must include observations and half equations for reaction at anode and
cathode.
- A porous pot has fine pores that allowions to flow through but can prevent the two different
aqueous solutions from mixing.
- As magnesium is more reactive than copper, magnesium becomes the negative terminal. It
release 2 electrons to come Mg2+
ions.
Anode[Oxidation]:
Mg  Mg2+
+ 2e-
-
At the positive terminal, Cu2+
ions in the copper(II) sulphate solution accept 2 electrons to
form copper.
Cathode[Reduction]:
Cu2+
+ 2e-  Cu
Overall equation:
Mg + Cu2+  Mg2+
+ Cu
14. Describe a laboratory experiment to construct the electrochemical series of magnesium, copper,
zinc and lead based on the potential difference in a voltaic cell.

Problem statement: How does the position of two metals in the electrochemical series affect the
voltmeter reading?
Hypothesis: The further apart the position of the two metals in the electrochemical series, the higher
the value of the voltmeter reading.
Variables:
a) Manipulated: metal pairs.
b) Responding: The voltage value/voltmeter reading
c) Constant: The concentration of copper(II) sulphate solution (electrolyte)
Result:
Metal pairs Voltage/v
Copper and zinc 1.1
Copper and lead 0.5
Copper and magnesium 1.8
Conclusion
The electrochemical series can be constructed by measuring the voltage value of the different pairs of
metals in the voltaic cell.

Chapter 7: Acids and Bases
1. State the meaning of acid and alkali.
- Acid is a chemical compound which ionises or undergoes dissociation of water to produce
hydrogen ions, H+
(or hydroxonium ions, H3O+
) which moves freely.
- Alkali is an aqueous solution of a base containing hydroxide ions, OH-
, which moves freely.
2. What is the meaning of strong acid and weak alkali?
- Strong acid is an acid that dissociates or ionises completely in water to produce a high
concentration of hydrogen ion, H+
.
- Weak alkali is a base that ionises partially in water to produce a solution with low concentration
of hydroxide ion.
3. What is neutralisation?
Neutralisation is a reaction that takes place between an acid with a base to form salt and water.
B. Understanding/Application/Analysis
4. The pH value of ammonia solution in water is 9 but the pH value of ammonia solution in
trichloromethane is 7. Explain why the pH values of the two solutions are different?
- In water,ammonia solution can ionise or undergo dissociation to produce hydroxide ions.
- Ammonia solution in trichloromethane (organic solvent) is dry and does not show alkaline
properties.
-
5. 80cm3
of distilled water is added to 20cm3
of 2.0 mol dm-3
solution of HCl. Find the molarity of
solution.
Molarity of the solution in mol dm-3
M1V1/1000 = M2V2/1000
(2.0 mol dm-3
)(20cm3
)/1000 = (M2)(80cm3
)/1000
M2 = 0.05 mol dm-3
6. In a tiltration, 40cm3
of 0.25 mol dm-3
solution of potassium hydroxide, KOH solution is need to
neutralise 20 cm3
of nitric acid, HNO3. Calculate the molarity of the nitric acid, HNO3.
Molarity of the nitric acid, HNO3.
M1V1/M2V2 = a/b
(M1)(20 cm3
)/(0.25 mol dm-3
)(40cm3
) = 1/1
M1 = 0.5 mol dm-3

7. Given dilute nitric acid and dilute sulphuric acid have the same concentration of 0.5 mol dm-3
. In
a neutralization experiment, 20 cm3
of nitric acid is required to neutralize 20 cm3
of sodium
hydroxide solution but only 10cm3
of sulphuric acid is required to neutralize 20cm3
of sodium
hydroxide solution. Explain why.
Given the formula for neutralization, M1V1/M2V2 = a/b
Molarity of sodium hydroxide when neutralizing with sulphuric acid.
2NaOH + H2SO4  Na2SO4 + 2H2O
(0.5)(10)/(M2)(20) = 1/2
M2 = 0.5 mol dm-3
Molarity of sodium hydroxide when neutralizing with nitric acid.
NaOH + HNO3  NaNO3 + H2O
(0.5)(20)/(M2)(20) = 1/1
M2 = 0.5 mol dm-3
Based on the calculation, the reason why sulphuric acid needed less volume to neutralize the
sodium hydroxide solution because sodium hydroxide has higher number of moles needed to react
than the reaction with nitric acid. Hence,the number of moles of alkali or acid affects the volume
needed to neutralize.
8. Describe a chemical test to show that a give solution is an acid.
Test with litmus paper.
- To test whether the chemical is an acid, a blue litmus paper is used to determined it.
- The blue litmus paper is immersed in a beaker of 50 cm3
of 0.5 mol dm-3
of hydrchloric acid.
- The blue litmus paper turns to red and indicates the solution as an acid.
9. Describe an experiment to determine the concentration of sodium hydroxide by using tiltration
process. You are give 0.2 mol dm-3
of dilute sulphuric acid, phenolphthalein, burrette, pipette and
a conical flask.

Aim: To determine the end point of a tiltration between sodium hydroxide solution and sulphuric acid
and hence calculate the concentration of the sulphuric acid.
Problem statement: Can sulphuric acid concentration be determined by tiltration?
Hypothesis: The concentration of sulphuric acid can be determine b tiltrating with a standard solution
of sodium hydroxide.
Variables:
a) Manipulated: concentration of sulphuric acid
b) Responding: volume of sulphuric acid required
c) Constant: concentration of sodium hydroxide, volume of sodium hydroxide.
Material:
0.4 mol dm-3
sodium hydroxide solution, 0.2 mol dm-3
of sulphuric acid, phenolphthalein.
Apparatus: Burette, pipette, pipette filler, beaker, conical flask, burette stand and clamp, white tile,
filter funnel.
Procedure:
1. A clean burette is rinsed with a little sulphuric acid. The burette is clampted to a retort stand.
2. The burette is filled with sulphuric acid. The initial burette reading is recorded.
3. A pipette is rinsed with sodium hydroxide solution.
4. 25.0 cm3
of the aqueos sodium hydroxide solution is pipetted into a conical flask.
5. Two drops of phenolphthalein are added to the conical flask. The alkali solution turns pink.
6. The flask is placed on top of a white tile as shown in the diagram.
7. The sulphuric acid is added slowly into the conical flask. The solution in the flask is swirled
continuosly.
8. When the solution in the conical flas starts to become pale pink, the acid is added drop by drop.
The conical flask is shaken after each drop of acid is added.
9. The addition of the acid is stopped when the pink colour just disappears.
10. The final burette reading is recorded.
11. The above tiltration is repeated a few times to obtain accurate and consistent results.
Result:
Tiltration number 1 2 3
Final burette reading
(cm3
)
20.40 21.00 28.00
Initial burette reading
(cm3
)
0.5 1.00 8.00
Volume of sulphuric acid
(cm3
)
19.90 20.00 20.00

Average volume of sulphuric acid used
= 19.90 + 20.00 + 20.00/3
= 20.00 cm3
Calculation:
H2SO4 (aq) + 2NaOH (aq)  Na2SO4 + 2H2O
Using the formula, MaVa/MbVb = a/b
Ma x 20.00 / 0.40 x 25.00 = ½
Ma = 0.40 x 25.00 /2 x 20.00 = 0.25 mol dm-3
Conclusion:
The concentration of sulphuric acid determine by tiltrating with a standard solution of sodium
hydroxide is 0.25 mol dm-3
.
10. You are given solid sodium chloride. Describe how you can prepare sodium chloride solution of
0.2 mol dm-3
in a laboratory using 250 cm3
volumetric flask.
Aim: To prepare 250cm3
of a standard solution of 0.2 mol dm-3
sodium chloride, NaCl.
Materials: Solid sodium chloride, NaCl, distilled water.
Apparatus: 250cm3
volumetric flask, beaker, chemical balance, spatula, filter funnel.

Procedure:
1. Calculate the required mass of sodium chloride
Number of moles of NaCl = MV/1000 = 0.2 x 250/1000 = 0.05 mol.
Mass of sodium chloride = 58.4 x 0.05 = 2.92 g
2. An empty beaker is weighed.
3. 2.92 g of solid sodium chloride is weighed.
4. Distilled water is added to the solid sodium chloride in the beaker. The mixture is stirred until
all the solid dissolve.
5. Distilled water is added into volumetric flask up to one-third full.
6. Sodium chloride solution is poured into the volumetric flask using filter funnel.
7. The beaker and the filter funnel is rinsed severaltimes with a little distilled water and the
washings are transferred into the volumetric flask.
8. Distilled water is added to the volumetric flask slowly until the meniscus is exactly at 250cm3
graduation.
9. The volumetric flask is stoppered and inverted severaltimes until the solution mixes well.
Conclusion:
250cm3
of standard solution of 0.2 mol dm-3
sodium chloride can be prepared by dissolving 2.92g of
solid sodium chloride in 250cm3
volumetric flask.

Chapter 8 : Salts
A. Knowledge [Definition, meaning and facts]
1. What is salt?
A salt is a compound formed from the replacement of hydrogen ions, H+
, in an acid by a metal ion or
ammonium ion, NH4
+
.
2. What is double decomposition reaction?
3. A double decompostion reaction is a reaction involving ion exchange to produce insoluble salts
(precipitate).
4. State one uses of salts in agriculture, medical field and food preparation.
Salts Uses
Food
11. Table salt, (NaCl) Monosodium glutamate,
C5H8NO4Na
12. Sodium benzoate, (NaC7H5O2) & sodium
nitrite, NaNO2
Food flavouring
Food preservatives
Medicine
 Calcium sulphate sesquihydrate,
2CaSO4.H2O
 Milk of Magnesia (magnesium hyrdroxide),
Mg(OH)2
 Barium sulpate, BaSO4
Plaster of Paris
Antacid
Barium meal for patients
Agriculture
 Ammonium sulphate, (NH4)2SO4
Ammonium nitrate, NH4NO3
Ammonium phosphate, (NH4)3PO4
Fertilisers

B. Synthesis
1. Describe the preparation of zinc sulphate.
- 50cm of 1 mol of dm-3
zinc(II) nitrate solution is added to 50cm3
of 1 mol dm-3
sodium
sulphate solution in a test tube.
Zn(NO3)2 (aq)+ Na2SO4 (aq)  ZnSO4 (s) + 2NaNO3 (aq)
- The resulting white precipitate, zinc(II) sulphate is formed and filtered.
- The precipitate is rinsed with a little distilled water.
- The white precipitate of zinc(II) obtained was dried with filter paper.
-
2. Describe the preparation of lead(II) chloride.
lead(II) nitrate solution is added to 50cm3
of 1 mol dm-3
sodium
chloride solution in a beaker.
Pb(NO3)2 (aq) + 2NaCl (aq)  PbCl2 (s) + 2NaNO3 (aq)
- The mixture in the beaker is stirred.
- The resulting white precipitate, lead(II) chloride is formed and filtered.
- The precipitate is rinsed with a little distlled water
- The white precipitate of lead(II) chloride is obtained was dried with filter paper.
3. Describe the formation of potassium nitrate.
silver nitrate solution is added to 50cm3
of 1 mol dm-3
potassium
iodide solution in a beaker.
AgNO3 (aq) + KI (aq)  AgI (s)+ KNO3 (aq)
- The mixture formed bright yellow precipitate of silver iodide and soluble potassium nitrate
solutions.
4. Describe an experiment to construct the ionic equation for formation of lead(II) sulphate.
Aim
To construct an ionic equation for formation of lead(II) sulphate.
Problem statement
How can the ionic equation for the reaction between lead(II) nitrate, Pb(NO3)2 with potassium
sulphate, K2SO4,be constructed?

Hypothesis
The ionic equation can be constructed by determining the volume of potassium sulphate, K2SO4,
which is added to the lead(II) nitrate, Pb(NO3)2, to produce a specific height of precipitate of
lead(II) sulphate, PbSO4.
Variables
b) Manipulated: The volume of lead(II) nitrate solution.
c) Responding: The height of the precipitate.
d) Constant: Volume and concentration of potassium sulphate.
Materials
Solutions of 1.0 mol dm-3
lead(II) nitrate solution, Pb(NO3)2,1.0 mol dm-3
potassium sulphate,
K2SO4.
Apparatus
Two burettes,retort clamp and stand, eight test tubes, eight test tubes rack,beakers,ruler
Procedure
12. Using a burette, 5.0cm3
of 1.0 mol dm-3
potassium sulphate solution is filled into 8 test tubes.
13. By using the second burette, 0.5cm3
of 1.0 mol dm-3
lead(II) nitrate is added to the first test tube,
1.0 cm3
in the second, 1.5cm3
in the third and so on until 4.0 cm3
is added to the 8th
test tube.
14. Each test tube is shaken and left for about 30 minutes
15. The height of the precipitate in each test tube is measure with a ruler.
K2SO4

Results
Test tube number 1 2 3 4 5 6 7 8
Volume of lead (II) nitrate
solution (cm3
)
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Height of precipitate (cm) 1 1 1.5 2 2 2 2 2
Analysis
From the table, 2.0 cm3
of 1.0 mol dm-3
lead(II) nitrate solution reacts with exactly 2.0 cm3
of
1.0 mol dm-3
potassium sulphate solution, K2SO4.
Calculation
Number of moles of Pb(NO3)2/Pb2+
MV/100 = 1.0 x 2/1000 = 0.002 mol
Number of moles of K2SO4/ SO4
2+
= MV/1000 = 1.0 x 2/1000 = 0.002 mol
0.00025 mol of ion Pb2+
reacts completely with 0.005 mol of ion SO4
2+
1 mol of ion Pb2+
reacts completely with 0.002/0.002 = 1 mol of ion SO4
2-
Therefore the ionic equation for the reaction is
Pb2+
(aq) + SO4
2-
(aq) PbSO4 (s)
Conclusion
The ionic for the double decomposition reaction between the solution of lead (II) nitrate and
potassium sulphate is Pb2+
(aq) + SO4
2-
(aq)  PbSO4 (s)

16. Solid W is a salt. Describe the tests you would carry out to confirm the presence of zinc ions and
17. nitrate ions in the salt.
Test for presence ofzinc ions, Zn2+
Test for presence ofnitrate ions, NO3
-
18. Solid X is a metal carbonate. Describe the tests you would carry out to confirm that X consists of
carbonate ions besides heating.
Tests ofcarbonate ions, CO3
2-
Sodium
hydroxide
solution
Solid W
- Add sodium hydroxide solution
little by little until it excess.
- White soluble precipitate in
excess sodium hydroxide
solutions.
- Zinc ions are confirmed if white
soluble precipitate are formed.
Dilute sulphuric
acid + iron (II)
sulphate
solution +
concentrate
sulphuric acid
Solid W
- Put solid W in a test tube.
- Dilute sulphuric acid
followed by iron(II)
sulphaye solution.
- Tilt the test tube and pour
slowly concentrated
sulphuric acid
- A brown ring is formed.
- Nitrate ions are confirmed to
be present.
- Pour sodium carbonate solution
in a test tube
- Add dilute hydrochloric acid
(any acid).
- Pass the gas through the lime
water.
- Lime water becomes cloudy.
- Carbonates ion is confirmed to
be present.
Solid X

19. You are given four test tubes with solution consist of zinc ions, lead(II) ions, aluminium ions and
magnesium ions respectively. Describe the test you would carry out to confirm the ion that are
present in each test tube.
20. You are given potassium chloride solution, lead(II) oxide powder and dilute nitric acid. Describe
how you would prepare lead(II) chloride salt from the given materials.
- 50cm3
of 1 mol dm-3
dilute nitric acid is measure and poured into a beaker and is heated.
- Lead(II) oxide powder is added to the nitric acid and the mixture is stirred until lead oxide is
in excess.
- The mixture then triggers a reaction and produced a solution of lead(II) nitrate.
PbO (s) + 2HNO3 (aq) Pb(NO3)2 (aq) + H2O (s)
- The product of lead(II) nitrate solution is then poured into a beaker of 50cm3
of 1 mol dm-3
potassium chloride solution and the mixture is stirred.
- Salt crystals formed are filtered and rinsed with a little distiller water and dried with filter
paper.
Pb(NO3)2 (aq) + 2KCl (aq)  PbCl2 (s) + 2KNO3 (aq)
Pb2+
(aq) + 2I-
(aq)  PbI2 (s)
Ammoniasolution
Aqueoussolutionof
cations(Pb2+
,Zn2+
,
Al3+
,Mg2+
)
- Aqueoussolutionof Pb2+
,Zn2+
,Al3+
and Mg2+
are addedinto
the testtubes.
- Ammoniasolutionsare addedlittle by little until inexcess.
- White precipitationof Pb2+
,Zn2+
, Al3+
andMg2+
are formedin
the excessaqueousammonia.

21. You are given dilute sulphuric acid, copper(II) nitrate solution and sodium carbonate solution.
Describe how you would prepare copper(II) sulphate salt from the given materials.
- Dilute sulphuric acid is poured into sodium carbonate to produce sodium sulphate, carbon
dioxide and water.
H2SO4 (aq) + Na2CO3 (aq) Na2SO4 (aq) + CO2 (g) + H2O (l)
- The produced sodium sulphate solution is the poured into a test tube filled with copper(II)
nitrate solution to produce blue solution of copper(II) sulphate salt and sodium nitrate salt
through double decomposition.
Na2SO4 (aq) + Cu(NO3)2 (aq)  CuSO4 (aq) + NaNO3 (aq)
Chapter 9: Manufactured Substances In Industry
1. What is the meaning of alloy?
An alloy is a substance formed from a mixture of metals with other elements.
2. What is the meaning of composite materials?
Composite materials are made from two or more constituent materials with significantly different
physical or chemical properties, that when combined, produce a material with characteristics
different from the individual components.
State the components of the following composite materials:
a) Reinforced concrete
Made from a mixture of cement,gravel, sand, water,iron or steel.
b) Superconductor
Materials that has no electrical resistance (zero resistance) such yttrium oxide, Y2O3, barium
carbonate, BaCO3, and copper oxide, CuO.
c) Optical fibre
Fine transparent glass tube that is made of molten glass.
d) Fiber glass
Consists of silica, SiO2, sodium carbonate, Na2CO3 and calcium carbonate, CaCO3.
e) Photochromic glass
Made from molten silica mixed with a little silver chloride.
3. State the catalyst, temperature and pressure of the following process:
a) Contact process
Catalyst: Vanadium(V) oxide, V2O5

Temperature: 450ºC
Pressure:One atmosphere
b) Haber Process
Catalyst: Iron fillings
Temperature: 450ºC
Pressure:200 atmosphere
4. What is the meaning of polymers?
Name the monomer of polythene and polyvinyl chloride.
Polymers are long chain molecules that consist of a combination of many small molecules.
Monomer of polythene: Ethene
Monomer of Polyvinyl chloride: Chloroethane
5. State four types of glass and their compositions. List the uses of each glass.
Type of glass Compositions Uses
Fused silica glass Silica, or silicon dioxide, SiO2
Lenses,glasses of spectacles,
glass tubes, ultraviolet column
Soda lime glass
Sodium silicate, Na2SiO3, and
calcium silicate, CaSiO3
Bottles, glass containers, glass
tubes, window glass
Borosilicate glass
Silica, SiO2, boron oxide, B2O3 Bowls, plates, dishes, pots and
glass apparatus in the
laboratory such as a test tubes,
beakers and flasks
Lead glass
Silica, SiO2, lead(II) oxide,
PbO,and sodium oxide, Na2O
Lenses,prisms, glass and
ornamets (crystal)

6. What is ceramics? State the properties and list the uses of ceramics.
Ceramics are substances made from clay that has been heated to high temperatures.
Main content of ceramics is a silicate, SiO2.
Properties of ceramics include:
a. Brittle
b. Very hard
c. Corrosion resistant
d. Very high melting point
e. Cracks with extreme temperature changes
f. Heat insulator
g. Withstands compression
h. Non-conductor of electricity
B. Understand/Application/Analysis
7. Bronze is an alloy consisting of copper and tin. Explain why bronze is harder than copper.
- Copper atoms have the same size and shape,are closely and orderly arranged but there is still
space between atoms.
- Under an applied force,the copper atoms easily slide over each other causing the copper
metal to be ductile and malleable.
- When beaten,metal atoms slide to fill the empty spaces between the metalatoms. The
situation causes the copper metal to be soft and not strong.
- The foreign atoms in the bronze alloy, which is the tin atoms, are larger or smaller than the
copper atoms in size, and will fill the empty spaces in the copper metal.
- The tin atoms prevent layers of copper atoms from sliding over one another easily.
- These increases the strength and hardness of metals, prevent corrosion of metals and improve
the appearance of the metal to be more attractive or easily.
i. Resistant to the chemicals
(withstands corrosion)
Tin atoms
Copper
atoms

8. Explain how acid rain formed.
Describe how acid rain causes environmental pollution.
- Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide
(SO2) and nitrogen oxides (NO2) are released into the air. These substances can rise very high
into the atmosphere, where they mix and react with water, oxygen and other chemicals to
form more acidic pollutants, known as acid rain.
- Sulfur dioxide and nitrogen oxides dissolve very easily in water and can be carried very far by
the wind. As a result, the two compounds can travel long distances where they become part of
the rain, sleet, snow, and fog that we experience on certain days.
- Acid rain causes many environmental pollution. For instance, acid rain decreases the pH
levels and increase acidity of water sources such as lakes and rivers, which in turn will not be
suitable for organisms such as plants, animals and even humans. Moreover, acid rain causes
corrodes building, statues, damage to crops, catastrophe such as landslide and in some rare
cases causes the corrosion of the skin and affects health.
9. Explain the industrial process involved in the manufacture of sulphuric acid.
Write all chemicals equation involved.
Sulphuric acid, H2SO4,is manufactured in the industry through Contact Process
i. Stage I
Sulphur dioxide gas, SO2,can be produced through two ways:
a) The combustion of sulphur in air.
S (s) + O2 (g)  SO2 (g)
b) Roasting of sulphide ores such zinc sulphide, ZnS, in air.
2ZnS (s) + 3O2 (g)  2ZnO (s) + 2SO2 (g)
v. Stage II
Mixture of sulphure dioxide gas, SO2, and air is passed over vanadium(V) oxide, V2O5, (acting
as catalyst) at a temperature of 450ºC and the pressure at one atmosphere to produce sulphur
trioxide gas, SO3.
2SO2 (g) + O2 (g) 2SO3 (g)
vi. Stage III
Sulphur trioxide gas, SO3, is dissolved in concentrated sulphuric acid, H2SO4, to produce oleum,
H2S2O7.
SO3 (g) + H2SO4 (l) H2S2O7 (l)
vii. Stage IV

Water was added to oleum, H2S2O7 to dilute it to produce sulphuric acid
H2S2O7 (l) + H2O  2H2SO4 (l)
22. Explain the industrial process involved in the manufacture of ammonia gas.
Write all the chemical equations involved.
Ammonia gas is produced through the Haber Process.
1. In the Haber Process,dry nitrogen gas, N2,and dry hydrogen gas, H2,are mixed in the mol or
volume ratio to 1 to 3 (1:3)
2. The gas mixture is then passed over iron fillings (catalyst) at temperatures from 450°C
and compressed under a pressure of 200 atmosphere to produce ammonia gas, NH3.
N2 (g) + 3H2 (g) 2NH3 (g)
3. Ammonia gas, NH3,obtained is cooled and condesnsed to form liquid ammonia, NH3 (aq).
Diagram shows a summary of the production of ammonia.
N2 (g) + 3H2 (g) 2NH3 (g)
Nitrogen is obtained
by fractional
distillitation of
liquid air
Mole ratio (volume)
of nitrogen to
hydrogen is 1:3.
Hydrogen is
obtained from
natural gas
methane, CH4.
Conditions:
Temperature :
450°C
Pressure: 200
atmosphere
Catalyst: Iron
fillings

C. Synthesis
1. Describe the laboratory experiment to prepare ammonium sulphate (ammonium fertilizer).
Neutralisations reactions between aqueous ammonia , NH3 (ak), (alkali) and an acid solution to
produce ammonium salts, NH4
+
,which can be used in fertilisers.
Example:
3NH3 + H3PO4  (NH4)3PO4 (Ammonium Phosphate)
NH3 + HNO3  NH4NO3 (Ammonium Nitrate)
2NH3 + H2SO4  (NH4)2SO4 (Ammonium sulphate)
The apparatus set-up below can be used to prepare ammonium sulphate salt, (NH4)2SO4
2. Describe a laboratory experiment to compare the hardness of brass and copper.
Aim
To study the hardness and strength of a copper metal compared to its alloy, brass.
Problem statement
Is brass (copper alloy) harder than pure copper?
Hypothesis
Diameter of the dent on brass is smaller than the diameter of the dent on pure copper.
Variables
a) Manipulated: Type of metal block.
b) Responding: Diameter of the dent.
c) Constant: Height of weight/mass of weight.
Materials
Copper block, brass block, steelball bearings, cellophane tape, string.
- 1 mol of dm-3
ammonia solution, is drippled drop
by drop into the sulphuric acid while stirring until
the solution smells of ammonia (or red litmus
paper turns blue).
- The solution is poured into the evaporating dish
and the following steps are conducted.
1. Heat the solution until saturated/concentrated
2. Cool down the solution until salt crystals form
3. Filter salt crystals
4. Rinse with a little distilled water’
5. Dry with filter paper
H2SO4 (aq) + 2NH3 (aq)  (NH4)2SO4 (aq)

Apparatus
Metre rule, a weight of 1 kg, retort stand.
Procedure
1. A steel ball is stuck on the surface of the brass block using cellophane tape.
2. A weight of 1 kg is hung at a height of 50cm from the surface of the brass block.
3. The weight is released to fall on the steelball bearing
4. Diameter of the dent made in the brass block surface is measured using a metre rule and
recorded in the following table.
5. Step 1to step are repeated twice to obtain the average diameter of the dents.
6. The experiment is repeated using a pure copper block.
Results
Type of metal
block
Diameter of dent (mm) Average
diameter (mm)1 2 3
Brass 3.0 3.2 3.1 3.1
Copper 4.3 4.4 4.2 4.3
Analysis
The dent on the brass block is smaller than the dent on pure copper block.
Conclusion
Brass is harder than copper.

Form 4 Chemistry Project. [Radioactive]

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Form 4 Chemistry Project. [Radioactive]