note chemistry form 4 english

CHEMISTRY 2012 HJH ROSSITA RADZAK

1

CHAPTER 2: MATTER

1. To determine melting point

Water bath – to ensure uniform heating

Plot graph – y – axis (Start at 50 o C).

2. The kinetic theory of matter (solid, solid –liquid, liquid, liquid-gas, gas)

Matter is made up of tiny and discrete particles (atoms, ions or molecules), there is space between these

particles.

a) arrangement of particles-pack closely together, orderly arrangement, far apart from each other

b) Movement of particles-vibrates, rotate, slowly, rapidly, vigorously, randomly.

c) kinetic energy content-

d) change in energy content-

e) force of attraction-strong, weak, weaker

Example:

Graph shows the heating curve of element X.

Describe the graph in terms of states of matter, particle arrangements and changes in energy.

Sample answer:

Stage State of matter Particles arrangement Changes in energy

to – t1 Liquid The particles are close to each other.

The particles arrangement is not orderly.

The kinetic energy increases

t1 - t2 Liquid –

gaseous

Some of particles are close to each other

and some far apart.


The kinetic energy is constant

t2 – t3 Gaseous The particles are far away from each other.


The kinetic energy increases

3. Diffusion – the movement of particles (atom/ions/molecule) of substance in between the particles of

another substance / from highly concentrated area to less concentrated area. Gas > liquid > solid

4. Isotope – are atoms of the same element (same number of proton) with different number of

neutrons/nucleon number.

Temperature,

oC

Time, s to t1 t2 t3

Boiling point

Cooling Heating


2

Example: Compare atom 12

X and atom 14

Y

6 6

Atom X Y

Proton number 6 6

No. of electron 6 6

Valence electron 4 4

Number of neutron / nucleon number 6/ 12 8 /14

Chemical properties similar

Physical properties different

No. of occupied shell similar

CHAPTER 3: CHEMICAL FORMULA

1. Empirical formula: chemical formula that shows the simplest whole number ratio of atoms of each

element in a compound, CH2

2. Molecular formula: a chemical formula that shows the actual number of atoms of each element that is

present in a molecule of the compound, C2H4

a) Unreactive metal – reaction oxide metal with

hydrogen gas, (CuO, PbO ,SnO )

b) Diagram

Reactive metal ( Mg, Zn – burn in excess oxygen /

air ) – more reactive than H2

c) Procedure

- Weigh and record the mass of combustion tube

with porcelain dish

- Add a spatula of copper (II) oxide on the

porcelain dish. Weigh the tube again.

- Allow hydrogen gas flow into the tube for 5 – 10

minutes.

- Burn the excess hydrogen.

- Heat copper (II) oxide strongly.

- Turn off the flame when black solid turns brown

completely.

- Continue the flow of hydrogen until the set of

apparatus cool down to room temperature.

- Weigh the combustion tube with its content.

- -Repeat the process heating, cooling and

weighing until a constant mass is obtained and

record.

- Weigh and record a crucible with its lid

- Clean Mg ribbon with sand paper then coil the

Mg ribbon and place into the crucible. Weigh and

record.

- Heat strongly

- When Mg ribbon start to burn, cover the crucible

with lid.

- Lift / raise the lid at intervals.

- When the burning is complete, remove the lid and

heat strongly.

-Allow the crucible to cool down.

-Weigh and record the crucible with content and

lid.

-Repeat the process heating, cooling and weighing

until a constant mass is obtained and record.

- Observation : White fume / solid formed

Result :

- combustion tube with porcelain dish = a g

- combustion tube with porcelain dish + copper (II)

oxide = b g

-combustion tube with porcelain dish + copper = c g

- mass of copper = ( c- a) g , Mass of oxygen = ( b- c ) g

- mass of crucible + lid = a g

- mass of crucible + lid + Mg = b g

- mass of crucible + lid + magnesium oxide = c g

- mass of Mg = ( b – a ) g

- mass of oxygen = ( c – b) g

6p

8n

6p

6n

X

Y

Hydrogen


3

Calculation: Element / atom Cu O

Mass (g) x y

Number of mole x / 64 y / 16

Simplest ratio of mole

Element / atom Mg O

Mass (g) x y

Number of mole x / 24 y / 16

Simplest ratio of mole

Precaution :

1. The flow of H2 must be continuous during

cooling – to prevent hot copper metal from

oxidized.

2. Allow hydrogen gas flow into the tube for 5 –

10 minutes to unsure air totally removed. The

mixture H2 and air may cause an explosion. 3. To determine all air totally removed, collect the

air and place lighted splinter, the gas burn quietly.

[To prepare H2]

4. Zn + 2HCl ZnCl2 + H2

5. Anhydrous calcium chloride – to dry the H2 gas.

6. CuO + H2 Cu + H2O

Precaution :

1. Clean Mg ribbon with sand paper to remove the

layer of oxide on its surface.

2. Lift / raise the lid at intervals to allow air in

3. When Mg ribbon start to burn, cover the crucible

with lid to avoid the white fume produced from

being escape to the air.

4. Repeat the process heating, cooling and

weighing to make sure all magnesium is

completely reacted with oxygen.

5. 2Mg + O2 2MgO

3. Based on the two formulae Na2O, CuI

(a) State the oxidation number for sodium, and copper.

(b) Name both the compound based on IUPAC nomenclature system.

(c) Explain the difference between the names of the two compounds based on IUPAC nomenclature

system.

Sample answer:

Na2O CuI

Oxidation number for

sodium and copper

+1 +1

IUPAC Nomenclature Sodium oxide Copper (I) iodide

Reason Does not have roman number

because sodium has only one

oxidation number

Has roman number because

copper has more than one

oxidation number

CHAPTER 4: PERIODIC TABLE

1. Explain the following statements, referring to the electron arrangement of the elements.

(a) The elements of Group 18 are unreative and exist as monoatomic particles (3 marks)

The points are:

(Duplet /octet) electron arrangement.

No tendency to donate, accept, share electrons

Remain as individual particles


4

(b) The reactivity of Group I elements increases down the group, whereas the reactivity of Group

17 elements decreases down the group. (12 marks)

The points are:

Explanation Group I Group 17

1 Change in proton number Increases Increases

2 Change in number of electrons and electron filled shells Increases Increases

3 Change in atomic size/radius/diameter Increases Increases

4 Strength of electrostatic attraction between nucleus and

valence electron

Decreases /weaker Decreases /weaker

5 Tendency to Releases electron

increases

Attract t/ accept

electron decreases

6 To become Positive ion Negative ion

7 Reactivity Increases Decreases

(2) Explain how the melting point of Group 1 elements change down the group (4 marks)

decreases gradually

Reason

atomic size increases

metallic bonding between the atoms become weaker

Less energy is required / needed to overcome this metallic bonding.

(3) Chemical properties of element in group 17

I Reaction with water Cl2 + H2O HCl + HOCl

II Reaction with sodium hydroxide Cl2 + 2NaOH NaCl + NaOCl + H2O

III Reaction with iron 3Cl2 + 2Fe 2FeCl3 (brown solid)

Example: Compare the reactivity of reactions between chlorine and bromine with iron. [Diagram III]

Reaction Reactants Observation

A Iron + chlorine gas The hot iron wool ignites rapidly with a bright flame.

A brown solid is formed.

B Iron + bromine gas The hot iron wool glows moderately with fast.

A brown solid is formed.

Sample answer:

a) Chemical equation: 3Cl2 + 2Fe 2FeCl3

b) The reactivity of reaction A is higher than reaction B.

c) The atomic size of chlorine is smaller than bromine.

d) The forces of attraction of the nucleus toward the electrons are stronger.

e) It is easier for chlorine atom to attract/receive electron.

Chlorine gas

Sodium hydroxide solution

Soda lime

Hot iron wool

HCl (cons) +

KMnO4(s)

To

produce

Cl2

III II


5

(4) Across period 3, atomic radius (atomic size) decreases / electronegativity increases. Explain.

a) Proton number increases by one unit.

b) The number of valence electrons in each atom increases.

c) Positive charge of the nucleus increases, thus

d) Nuclei attraction on valence electron increases.

e) Atomic radius (atomic size) decreases

f) Tendency to receive electron increases (to form negative ion) thus electronegativity increases.

(5) Chemical properties of the oxide of element across Period 3 changes from basic oxide to amphoteric

oxide to acidic oxide.

Basic oxide – sodium oxide (Na2O)

Amphoteric oxide – Aluminium oxide (Al2O3)

Acidic oxide – sulphur dioxide, SO2

CHEPTER 5: CHEMICAL BOND

(a) Group 1 elements react with Group 17 elements to produce compounds that have high melting points.

(4 marks)

The points are:

Ionic compound produced

Because involve transfer of electrons between metal atom and non metal atom.

Metal atom donates valence electron to form positive ion, non metal atom accepts electron to

negative ion.

The oppositely charged ions are held together by strong electrostatic force.

More heat energy is needed to overcome the strong force of attraction.

Formation of ionic compound (metal [Group 1,2 & 13] and non metal [ Group 14, 15, 16& 17])

Sample answer:

1. Electron arrangement of atoms ( Na , 28.1 ; Cl 2.8.7 ) // valence electrons

2. To achieve stable / octet electron arrangement

3. Atom ( Na) releases one / valence electron to form sodium ion, Na+

4. Half equation ( Na Na+ + e)

5. Atom (Cl) gain / accept electron to form chloride ion, Cl-

6. Half equation ( Cl + e Cl- )

7. Oppositely charged ion, Na+ & Cl

- are attracted to one another by strong electrostatic force of

attraction to form ionic compound, NaCl

8. Diagram

Formation of covalent compound (nonmetal)

1. electron arrangement of the atom /valence electrons

2. to achieve duplet /octet electron arrangement

3. Atom (Carbon) contributes 4 electrons while (H) atom contributes 1 electron (for sharing).

4. one ( Carbon ) atom share 4 pairs of electrons with 4 (H) atoms to form covalent compound , CH4 /

ratio

5. diagram


6

Compare the physical properties of covalent and ionic compound

Properties Covalent compound ( naphthalene) Ionic compound ( sodium chloride)

Melting and

boiling

- low

- consist of molecules

- weak inter molecular forces

between molecules - less energy needed to overcome the

weak forces

- high

- consist of oppositely charged ions

- the ions are held together by strong

electrostatic forces . - more heat energy needed to overcome the

strong forces

Electrical

conductivity

- consist of molecules

- does not conduct electricity in any

state (molten or aqueous).

- consist of oppositely charged ions

- conduct electricity in molten or aqueous

solution.

- in molten or aqueous solution, ions can move

freely.

CHAPTER 6: ELECTROCHEMISTRY

1. Factor that affect the electrolysis of an aqueous solution

(a) position of ions in the electrochemical series (cathode)

(b) concentration of ions in the electrolyte - halide ( Chloride, bromide and iodide)

(c) type of electrodes used in the electrolysis – ( anode – metal )

Application

(i) Electroplating

anode – electroplating metal ( less electropositive metal / Cu, Ag, Ni )

cathode – metal /object to be electroplated

electrolyte - solution that contains the metal ions of electroplating metal

(ii) Purification

anode – impure metal ( Cu Cu2+

+ 2e )

cathode –pure metal ( Cu2+

+ 2e Cu )

electrolyte - solution that contains the metal ions ( Cu2+

)

(iii) Extraction of metal (reactive metal, sodium, aluminium)

– Down`s Process – extraction of sodium from molten sodium chloride.

– Extraction of aluminium from molten aluminium oxide ( bauxite)


7

2. To construct the electrochemistry based on tendency to release electron /potential differences - voltaic

cell/ Electrochemical cell.

3. To construct the electrochemistry based on ability / tendency of metal

to displace another metal from it salts solution.

Displacement reaction: a metal which is higher in the electrochemical

series is able to displace a metal below it in a series from its salt solution.

Example: Zn + CuSO4 ZnSO4 + Cu // Zn Zn2+

+ 2e / Cu2+

+ 2e Cu

Cell P Cell Q

4. Compare and contrast cell P and Q. Include in your answer the observation and half equation for the

reaction of the electrodes in both cells.

Cell P Characteristics Cell Q

Electrical chemical Energy change Chemical electrical

+ve / anode: copper (OXIDATION)

-ve / cathode: copper

Electrode +ve/cathode: copper

-ve/ anode: lead (OXIDATION)

Cu2+

, H+

OH- , SO4

2- Ions present in

the electrolyte

Cu2+

, H+

OH- , SO4

2-

Anode :Cu Cu2+

+ 2e

(type of electrode)

Cathode: Cu2+

+ 2e Cu ( ECS)

Half equation

Anode: Pb Pb2+

+ 2e

Cathode: Cu2+

+ 2e Cu (ECS)

Anode: copper electrode become thinner

Cathode: brown solid formed/ becomes

thicker.

Electrolyte: intensity blue solution /

concentration of Cu2+

solution remain.

Rate of ionized of copper atom to form

copper (II) ion at the anode same as rate of

discharged copper (II) ion at the cathode.

Observation

Anode: becomes thinner

Cathode: becomes thicker / brown solid

formed

Electrolyte: intensity blue solution

decrease / blue becomes paler

copper

Copper(II)

sulphate solution

lead


8

CHAPTER 7: ACID AND BASE

An acid is chemical substance which ionizes in water to produce hydrogen ion, H

+

A base is a chemical substance which ionizes in water to produce hydroxide ions, OH-

Alkali is a soluble base.

Basicity is the number of ionisable hydrogen atoms per molecule of an acid.

1. Explain why these two solutions have different pH values

identify strong acid , weak acid

definition strong acid

definition weak acid

concentration of H+

relationship between pH value and concentration of hydrogen ions, H+

Sample answer:

1. Hydrochloric acid is a strong acid while methanoic acid is a weak acid.

2. Hydrochloric acid completely ionizes in water to form higher concentration of hydrogen ions.

HCl + H2O H3O+ + Cl

- // HCl H

+ + Cl

- , H3O

+ , hydroxonium ion

3. Methanoic acid ionizes partially in water to form lower concentration hydrogen ions

CH3COOH CH3COO- + H

+

4. The higher the concentration of hydrogen ions the lower the pH value.

2. Aim: To determine the end point during the neutralization of potassium hydroxide

and hydrochloric acid

Apparatus: 25 cm3 pipette, burette , 250 cm

3 conical flask, retort stand, white tile

Material: potassium hydroxide and hydrochloric acid 0.1 mol dm-3

, phenolphathalein.

Procedure:

1. Rinse a burette with a small amount hydrochloric acid 0.1 mol dm-3

.

2. Clamp the burette on retort stand.

3. Fill the burette with hydrochloric acid 0.1 mol dm-3

.Adjust the meniscus level of acid to a reading at 0.

4. Record the initial burette reading.

5. Pipette 25.0 cm3 of potassium hydroxide 0.1 mol dm

-3 into conical flask.

6. Add two drop of phenolphathalein.

7. Add hydrochloric acid 0.1 mol dm-3

carefully. Swirl the conical flask during the process.

8. When the colour of the mixture turn paler, add hydrochloric acid drop by drop.

9. Stop adding the hydrochloric acid as soon as the solution turns colourless.

10. Record the final burette reading.

11. Repeat steps 1-10 twice. Tabulate your reading.

The pH value of 1.0 mol dm-3

hydrochloric acid is 1

The pH value of 1.0 mol dm-3

methanoic acid is 4

MAVA = a

MBVB b


9

Result :

Titration 1 2 3

Final burette reading, cm3

Initial burette reading, cm3

Volume of hydrochloric acid 0.1 mol dm-3

, cm3

3. Preparation Standard solution ( 0.1 mol dm-3

NaOH, 100 cm3)

1. calculate the mass of solute ( mole = 0.1 x 100/1000 , 0.01 = mass/ 40)

2. weigh 0.4g of NaOH in weighing bottle using digital balance / electronic balance

3. pour into a beaker, rinse the bottle with distilled water.

4. dissolve NaOH with a little ( 10 – 20 cm3 )distilled water.

5. transfer the mixture into volumetric flask 100 cm3 rinse the beaker with distilled water.

6. pour the washings into volumetric flask 100 cm3

7. add distilled water, shake well

8. add distilled water drop by drop to finally bring the volume of solution to the 100 cm3 mark /

calibration mark.

Preparation of a standard solution by dilution method

M1V1 = M2V2

M1 – initial molarity

V1 - initial volume

M2 – final molarity

V2 – final volume

NOTE : CONCENTRATION – 1. MOLARITY - mol dm-3

2. g dm-3

Neutralization in our daily lives

Agriculture Powdered lime (CaO) , limestone (CaCO3), ashes of burnt wood

Used to treat acidic soil.

Industries 1. Powdered lime (CaO)

Used to treat acidic effluent from factories, acidic gas SO2 emitted by power station and

industries.

2. Ammonia prevent the coagulation of latex by neutralizing the acid produced by

bacteria in the latex.

Health 1. Anti-acids contain bases such as aluminium hydroxide and magnesium hydroxide to

neutralize the excess acid in the stomach.

2. Vinegar (citric acid) is used to cure wasp stings that are alkaline in nature.

3. Baking powder (NaHCO3) is used to cure bee stings and ant bites that are acidic in

nature.

4. Toothpaste contains bases that neutralize the acid produces by bacteria in our mouth


10

CHAPTER 8 SALT

A salt is a compound formed when the hydrogen ion, H+ from an acid is replaced by a metal ion or an

ammonium ion, NH4+

Preparation of soluble salt

– acid + reactive metal(Zn / Mg) salt + H2 / 2H+ + Mg Mg

2+ + H2

– acid + base ( metal oxide) salt + water

– acid + alkali salt + water / H+ + OH

- H2O ( NaOH, KOH, NH4OH)

– acid + carbonate metal salt + CO2 + H2O / 2H+ + CO3

2+ CO2 + H2O

Procedure:

1. pour ( 25 – 100cm3) acid ( 0.5 – 2.0 mol dm

-3) into a beaker

2. heat slowly

3. add solid (metal / base/ carbonate ) a little until excess / no more dissolve

4. stir

5. filter the mixture into evaporating dish

6. heat (slowly) the filtrate until 1/3 from original volume / saturated solution formed

7. cool down the saturated solution (until crystallized )

8. filter (to separate the crystals)

9. dry / transfer onto filter paper / dry between sheets of filter paper

Observation

Chemical equation

Preparation of insoluble salt – precipitation reaction / double decomposition reaction

Pb2+

+ SO42-

PbSO4

Example : Preparation of lead(II)sulphate.

Procedure

1. pour ( 25 – 50cm3) of soluble salt Pb(NO3)2 into a beaker

2. add ( 25 – 50cm3) of soluble salt (Na2SO4)

3. stir

4. filter the mixture

5. rinse residue / solid / precipitate

6. dry between sheets of filter paper

Observation

Chemical equation

Ionic equation

Action of heat on salt

Carbonate oxide metal (base) + CO2 except Na, K and NH4+

Example: CuCO3 CuO + CO2

Nitrate oxide metal + nitrogen oxide + oxygen except Na, K, (2NaNO3 2NaNO2 + O2 )

Example : 2Mg(NO3)2 2MgO + 4NO2 + O2

(Brown gas)

Ammonium chloride ammonia gas + hydrogen chloride gas, (NH4Cl NH3 + HCl )


11

Confirmatory test for cation and anion

1. State the material / chemical / reagent

2. procedure

3. observation

4. conclusion

Example: You are given a bottle of ammonium chloride solution. Describe chemical test to

verify the cation and anion.

(a) test for cation (NH4+)

1. pour 2 cm3 the solutions into a test tube

2. add 1 cm3 copper (II) sulphate solution

3. blue precipitate soluble in excess to form dark blue solution.

OR

4. add 2 to 3 drops of Nessler reagent into the test tube

5. brown precipitate.

6. Ammonium ions (NH4+) present.

(b) test for anion (Cl-)

1. pour 2 cm3 the solution into a test tube

2. add 1 cm3 of dilute nitric acid and silver nitrate solution.

3. white precipitate formed

4. confirm the presence of chloride ions

Example: You are given lead (II) nitrate and aluminium nitrate solution. Describe chemical test to

verify the cation and anion.

(c) test for cation

1. pour 2 cm3 the solutions into different test tubes

2. add 1 cm3 potassium iodide solution into the test tubes

3. yellow precipitate formed

4. lead (II) ion present

(d) test for anion

1. pour 2 cm3 of lead (II) nitrate solution into a test tube

2. add 1 cm3 of dilute sulphuric acid

3. add 1 cm3 of iron (II) sulphate solution

4. shake the mixure

5. tilt the test tube, add concentrated sulphuric acid carefully // drop by drop down the side

of the test tube

6. the brown ring formed

7. nitrate ion, NO3- present.


12

Aim : To construct the ionic equation for the formation of lead (II) chromate(VI)

[Continuous variation method]

Apparatus : Test tubes of the same size, test tube rack, burette, retort stand with

clamp, ruler, glass rod, dropper.

Material : 0.5 mol dm-3

potassium chromate (VI) solution, 0.5 mol dm-3

lead (II)

nitrate solution.

Procedure : 1. Seven test tubes of the same size were labelled from number 1 to 7. They were placed in a test tube

rack.

2. A burette was filled 0.5 mol dm-3

lead (II) nitrate solution, 5.00 cm3 of the lead (II) nitrate

solution was run into each the seven tubes.

3. Another burette was filled with 0.5 mol dm-3

potassium chromate (VI) solution.

4. Potassium chromate (VI) solution from the burette was added into each of the seven test tubes

according to the volumes specified in the table.

5. The mixture in each test tube was stirred with a clean glass rod.

6. The test tubes were left aside for about an hour.

7. The height of the precipitate in each test tube was measured. The colour of the solution above the

precipitate in each test tube was observed and recorded.

Results:

Test tube 1 2 3 4 5 6 7

Volume of 0.5 mol

dm-3

Pb(NO3)2 /cm3

5.00 5.00 5.00 5.00 5.00 5.00 5.00

Volume of 0.5 mol

dm-3

K2Cr O4 /cm3

1.00 2.00 3.00 4.00 5.00 6.00 7.00

Height of precipitate

(cm)

0.60 1.20 1.80 2.40 3.00 3.00 3.00

Colour of solution

above the precipitate

colourless colourless colourless colourless colourless yellow yellow

Paper 2

Discussion

The volume of 0.5 mol dm-3

potassium chromate (VI), solution required to exactly react with 5.00 cm3 of 0.5

mol dm-3

lead (II) nitrate solution is 5.00 cm3.

Calculation:

Number of moles lead (II) ions = MV

= 0.5 x 5.00/1000 = 0.0025 mol.

Number of moles chromate (VI) ions = MV

= 0.5 x 5.00/1000 = 0.0025 mol.

Simplest mole ratio of lead (II) ions : chromate (VI) ions

0.0025 : 0.0025

1 : 1


13

Discussions:

1. A yellow precipitate of lead (II) chromate (VI) is formed in each of the seven test tubes.

2. The height of the precipitate increases gradually from test tubes 1 to 5 because more and more lead (II)

chromate (VI) is formed due to the increasing amount of potassium chromate (VI) added to the test tubes.

3. The colour of solution above the precipitate in test tubes 1 to 4 are colourless due to the excess lead (II)

nitrate.

4. The colour of solution above the precipitate in test tubes 6 to 7 is yellow due to the excess potassium

chromate (VI).

5. Ionic equation: Pb2+

+ Cr2O72-

PbCr2O7

Conclusion:

As / when the volume of potassium chromate (VI) solution used increases, the height of the precipitate

increases until it achieves a maximum height.

CHAPTER 9 : MANUFACTURED SUBSTANCES IN INDUSTRY

1. Contact process: manufactured sulphuric acid

Stage Equation Explanation

1 S + O2 SO2 Sulphur is burned in the excess of oxygen gas to produce sulphur

dioxide gas.

2 2SO2 + O2 2SO3 SO2 is then heated in excess oxygen gas, catalyst Vanadium (V)

oxide, 1 atm and 450 – 550 o C , to produce sulphur trioxide gas.

3 SO3 + H2SO4 H2S2O7 Gas sulfur trioxide dissolve in sulphuric acid to produce oleum

4 H2S2O7 + H2O 2H2SO4 Oleum is added to water to produce sulfuric acid

Gas SO3 is not dissolve in water to produce H2SO4 straight away because the reaction will produce a

lot of heat which is dangerous( cause the forming of acid fumes)

Usage of sulphuric acid:

To manufacture fertilizer, soap and detergent

To make explosive material, paint / pigment, polymer

As metal cleaner and electrolyte in car battery.

2. Haber Process

N2 + 3H2 2NH3

Condition: Catalyst: iron, temperature: 450 – 550 oC, Pressure 200 – 500 atm

Usage: to manufacture fertilizer

2NH3 + H2SO4 (NH4 )2 SO4

3NH3 + H3PO4 (NH4 )3 PO4

NH3 + HNO3 NH4NO3


14

3. High percentage of nitrogen is a good fertilser for plants. How to calculate %N in fertiliser?

urea CO(NH2)2 and ammonium nitrate (NH4NO3), which one is a better fertiliser?

[ RAM : N,14; C,12 ;O,16; H,1]

Sample answer:

% N in Urea = mass of nitrogen / RMM urea x 100

= 2x14 / 60 x 100 = 46.67%

% N in NH4NO3 = 2x14 / 80 x 100 = 35.00 %

Urea is a good fertilizer than ammonium nitrate, because the percentage of nitrogen in urea higher than

ammonium nitrate.

4. Describe how toxic waste product from factory affects the quality of the environment. Your description

should include the following aspects. Source, process and effect.

Sample answer:

1. [Source] sulphur dioxide gas produced by factory or burning of fossil fuels

2. [process ] sulphur dioxide gas dissolves in rain water / water to form acid rain,

2SO2 +O2 + 2H2O 2H2SO4]

3. [effect ] toxic waste / acid flows to into lakes and rivers, acid rain lowers the pH value of water, soil

and air.

4. Fish and other aquatic organisms die.

5. acid rain corrodes concrete buildings and metal structures

6. acid destroys trees in forest

7. Acid rain reacts with minerals in soil to produces salt which are leached out the top soil.

8. Plants die of malnutrition and diseases.

9. Soil becomes acidic, unsuitable for growth of plants and destroys the roots of plants.

10. sulphur dioxide causes respiratory problems in humans.

5. POLIMER: - large molecules made up of identical repeating sub-units of monomers which are

joined together by covalent bonds.

Synthetic polymer Monomer Uses

Polythene Ethene Plastic bags, plastic container

Polypropene Propene Piping, car batteries

Polyvinyl chloride, PVC Chloroethene Artificial leather, water pipe

Perspex Methylmethacrylate Safety glass, reflectors

monomer ( ethene) polymer (polyethene)


15

ALLOY

An alloy is a mixture of two or more elements with a certain fixed composition in which the major

component is a metal.

1. The composition , properties and uses of some alloy

Alloy Composition Properties Uses

Bronze Cu

Tin

-Hard and strong

-does not corrode easily

-has shiny surface

-in building of statue or monuments.

-in making of medals

-swords and artistic material

Brass Cu

Zinc

-harder than copper -in making of musical instruments and

kitchenware

Steel Iron

Carbon

Hard and strong -in construction of buildings and bridges

-in building of the body of cars and railway

tracks

Stainless

steel Iron

Carbon

Chromium

-shiny

-strong

-does not rust

-in making of cutlery

-in making of surgical

instrument

Duralumin Aluminium

Copper

Magnesium

manganese

-light

-strong

-in building of the body of aeroplane and

bullet trains

Pewter Tin, Copper

antimony

-lustre, shiny

-strong

In making of souvenirs

Bronze is harder than pure copper. Explain.

Reason:

1. The presence of atoms of other metals / tin that are different sizes

2. Disrupt the orderly arrangement of copper atoms

3. Tin atoms reduce the layers of copper atoms from sliding

4. Alloy is stronger and harder than pure metal

2. You have learnt the steel is an alloy of iron. Steel is harder than pure iron. Both iron and steel can rust

when exposed to air and water. Do they rust at the same rate?

Aim : To compare the rate of rusting between iron, steel and stainless steel

Problem Statement

How does the rate of rusting between iron, steel and stainless steel differ?

Hypothesis

Iron rust faster than steel and steel rust faster than stainless steel.

Variables

Manipulated : Iron, steel and stainless steel.

Responding : intensity / amount of dark blue colour / rate of rusting

Fixed : size of nail, concentration of solution, duration of rusting

Copper atom

Stanum atom


16

Procedure:

1. Clean the nails with sand paper (to removed the rust from all the nails)

2. Place the iron nail, steel nail and stainless steel nail into the test tube A, B and C respectively.

3. Prepare a 5 % jelly solution by adding 5 g jelly to 100 cm3 of boiling water. Add a few drop of potassium

hexacyanoferrate (III) solution.

4. Pour the hot jelly into the test tubes until all the nails are fully immersed.

5. Leave the nails for 3 days.

6. Observe and record the intensity of the dark blue colour.

Tabulation of data

Paper 2

Conclusion

1. The concentration of Fe2+

ions in the test tube A is higher than in test tube B. No Fe2+

ions are present in

test tube C.

2. The rate of rusting in test tube A is higher than that in test tube B. No rusting takes place in test tube C.

Alloy slow down the rate of rusting.

Properties, composition and uses different type of glass

Type Properties Chemical

composition

Uses

Fused glass -Very high softening point

-Highly heat resistant

-Does not crack when temperature changes

-very resistant to chemical reactions

-difficult to be shaped

SiO2 Lenses, telescope

mirrors, optical fibres,

Laboratory glassware.

Soda lime

glass

-low softening point

-does not withstand heating

-break easily

-less resistant to chemical reactions

-easy to be shaped

- cracks easily with sudden change in temperature

SiO2

CaCO3 /

Na2CO3

Flat glass, light bulb,

mirrors, glass

containers.

Borosilicate -lower thermal coefficient

-heat resistant

- Does not crack when temperature changes

-very resistant to chemical reactions

-does not break easily

SiO2

B2O3

Na2O

Laboratory glassware,

cooking utensils.

Automobile

headlights.

Lead glass -low softening point

-high density

-High refractive index

SiO2

PbO

CaO

Decorative items,

crystal glass ware,

lens, prism, chandelier

Test tube The intensity of the dark blue colour //

rate of rusting

A

B

C


17

Composite Materials is a structural material that is formed by combining two or more different

substances such as metal, alloys, glass, ceramics and polymers.

Composite

material

Component Properties of

component

Properties of

composite

Uses of

components

Reinforced

concrete

Concrete Hard but brittle, low

tensile strength

Stronger, high tensile

strength does not

corrode easily, can

withstand higher

applied forces and

loads, cheaper.

Construction of

framework for

highway, bridges

and high-rise

building Steel Hard with high tensile

strength but expensive

and can corrode.

Super-

conductor

Copper(II)oxide,

barium oxide

Insulators of electricity Conducts electricity Generators,

transformers,

electric cable,

amplifiers,

computer parts

MRI

Fibre optics Glass of low

refractive index

Transparent, does reflect

light rays.

reflect light rays and

allow light rays to

travel along the fiber

Transmit data in the

form of light in

telecommunications

Glass of high

refractive index

Fibre glass Glass Heavy, strong but brittle

and non-flexible

Light, strong, tough,

resilient and flexible

wit high tensile

strength not

inflammable, low

density, easily

coloured, shaped and

moulded.

Water storage

tanks, small boat,

helmet

Polyester plastic Light, flexible, elastic

but weak and

inflammable

Photo-

chromic

glass

Glass Transparent, does reflect

light rays.

Sensitive to light :

darkens when light

intensity is high,

becomes clear when

light intensity is low.

Photochromic

optical lens, camera

lens, car

windshields, optical

switches, light

intensity meters.

Silver chloride

or silver

bromide

Sensitive to light


18

CHAPTER 10: RATE OF REACTION

Rate of reaction is the change in selected quantity of reactants or products per time taken.

Aplication

1. Explain why potatoes fried in boiling oil cook faster than potatoes boiled in boiling water?

Answer:

- Boiling point of oil is higher than boiling point of water

- At higher temperature potatoes is faster to cook

2. Based on the collision theory, explain why we need to store fresh milk in refrigerator.

Answer:

(i) the temperature inside the refrigerator is lower

(ii) bacteria are not active at low temperature

(iii) decomposition of milk caused by bacteria will slow down

(iv) this will keep the milk fresh for along time

Collision theory

Effective collision: Collision which achieve activation energy (minimum amount) and with correct

orientation.

Temperature

1. As temperature increases, the kinetic energy of the particles ( H+, S2O3

2- ) increases /

2. Frequency of collision between particles ( H+, S2O3

2- ) increases

3. Frequency of effective collision increases

4. Rate of reaction increases

Size of particles (total surface area) 1. The smaller the size of particles,

2. The larger the total surface area exposed to the collision

3. Frequency of collision between particles increases



Concentration of the solution

1. The higher the concentration of the solution,

2. The greater the number of particles per volume

3. Frequency of collision between particles increases



Catalyst

1. The presence of catalyst provide an alternative pathway / route

2. with lower activation energy

3. Frequency of effective collision between particles increases

4. Rate of reaction increases.


19

Note:

1. Catalyst – a substance which alters the rate of chemical reaction while remains chemically

unchanged at the end of reaction.

2. Observable changes for measuring the rate of reaction.

(a) volume of gas liberated

(b) precipitate formation

(c) change in mass during reaction, colour ,temperature, pressure

1. Catalyst (Manganese (IV) oxide)

a) Decomposition of sodium chlorate (V), 2NaClO3 2NaCl + 3O2

b) Decomposition hydrogen peroxide , 2H2O2 2H2O + O2

2. Catalytic converters in the car exhaust system contain rhodium, platinum or chromium (III) oxide Cr2O3.

Example:

1. Aim: To investigate the effect of temperature of sodium thiosulphate Na2S2O3 solution on the rate of

reaction

Problem Statement:

How does temperature of sodium thiosulphate Na2S2O3 solution affect the rate of reaction?

Hypothesis:

When the temperature of sodium thiosulphate Na2S2O3 solution increases, the rate of reaction increases.// the

higher the temperature of sodium thiosulphate solution, the higher the rate of reaction.

Variables:

Manipulated :Temperature of sodium thiosulphate solution.

Responding :Rate of reaction/ Time taken for the cross „X‟ to disappear from the sight.

Fixed : Concentration and volume of sulphuric acid, concentration and volume of sodium

thiosulphate solution.

Apparatus : 150 cm3 connical flask, 50 cm

3 measuring cylinder,10cm

3 measuring cylinder, stopwatch,

thermometer, Bunsen burner, tripod stand, wire gauze.

Materials: 0.2 mol dm-3

sodium thioulphate solution, 1.0 mol dm-3

sulphuric acid, white paper marked “X”

at the centre.

Procedure:

1. 50 cm3 of 0.2 mol dm

-3 sodium thiosulphate solution is measured using measuring cylinder and poured

into a conical flask.

2. The temperature of the solution is measured with a thermometer.

3. The conical flask is placed on a white paper marked`X`.

4. 5 cm3 of 1 mol dm

-3 sulphuric acid is measured and then poured quickly and

carefully into the sodium thiosulphate solution.

5. The stopwatch is started immediately and the conical flask is swirled.

6. The mark `X` is viewed / observed vertically from above.

7.The stopwatch is stopped as soon as the mark disappear from sight.

8.Time taken is recorded.

9. Steps 1 to 9 are repeated by using the different temperature of sodium thiosulphate solution.


20

Data and Observation

Experiment Temperature ,

(oC)

Time taken for the “X” mark to

disappear from view, t (s)

1/ time taken ,

1/t ( s-1

)

1 28

2 35

3 40

4 45

5 50

Discussion

Based on plotted graph: [ calculation ]

The higher the temperature of sodium thiosulphate, the shorter the time taken for cross„X‟ to disappear

from the sight.

The rate of reaction directly proportional to the temperature of sodium thiosulphate solution used. //

As the temperature sodium thiosulphate solution increases, the time taken decreases. Therefore the rate of

reaction increases.

Conclusion :

The rate of reaction increases as the temperature sodium thioulphate solution increases.

Energy profile diagram

1. Ea – activation energy without catalyst

2. Ea‟ - activation energy with catalyst

3. Exothermic reaction – heat released /given out

4. Energy content in reactants higher than products

5. ^ H is the energy difference in reactants and products

6. Heat given out during bond formation is greater than

heat absorbed during bond breaking

7. Exothermic reacton.

- ^ H


21

2. Aim: to investigate effect of catalyst on the rate of decomposition hydrogen peroxide.

Problem statement: how does a catalyst affect the rate of decomposition hydrogen peroxide?

Hypothesis: manganese (IV) oxide, MnO2 increases the rate of decomposition of hydrogen peroxide

Variables:

Manipulated : presence of manganese (IV) oxide (MnO2)

Responding : rate of reaction

Fixed : concentration of H2O2, initial temperature of H2O2 solution.

Apparatus: test tube, 10 cm3 measuring cylinder, test tube rack, spatula.

Materials: (5-10) – volume of H2O2 solution, manganese (IV) oxide (MnO2) powder, wooden splinter

Procedures:

1. label two test tube as A and B

2. Using a measuring cylinder measure 5 cm3 of 20 – volume of H2O2 solution and pour into test

tube A.

3. Add ½ spatula of manganese (IV) oxide powder into test tube A.

4. Shake the test tube.

5. Immediately place a glowing splinter into the test tube.

6. Observe and record the changes.

7. Repeat the same procedure for test tube B without MnO2

Observation: [Paper 2]

Test tube Observation

A Effervescence occurred. The glowing wooden splinter relight.

B No effervescence. The glowing wooden splinter did not relight.

Discussion:

Manganese (IV) oxide (MnO2) increases the rate of decomposition of hydrogen peroxide. Decomposition

of hydrogen peroxide produces oxygen gas. 2H2O2 2H2O + O2


22

CHAPTER 11: CARBON COMPOUND

1. Hydrocarbon – chemical compound containing carbon and hydrogen atom only.

2. Alkene – chemical compound containing carbon and hydrogen atom and at least

one carbon-carbon double bond ( C = C )

3. Isomers are molecules with the same molecular formula, but with different structural formula.

Example: C4H10 – butane

1. C2H4 + [O] + H2O C2H4(OH)2 [ purple turns colourless] //[ orange turns green]

2. CH3COOH + C2H5OH CH3COO C2H5 + H2O

3. C2H4 + H2O C2H5OH

4. C6H12O6 2C2H5OH + 2CO2

H 2SO4, cons

H 3 PO4, 60 atm, 300 oC

CH3COO C2H5

Ethyl ethanoate

CH3COOH

Carboxyl

-COOH

C2H4 Double bond between

C atoms, C=C

C6H12O6

KMnO4/ H+,

K2Cr2O7/ H+

C2H6

C2H5Br

C2H4(OH)2

C2H4Br2

C2H5OH

Esterification

H2SO4

- CH2- CH2-

A

d

d

i

t

i

o

n

Oxidation

Fermention

Br2

H2

KMnO4/ H+,

K2Cr2O7/ H+

H2O

HX

CnH 2n+ 2 , n = 1,2 alkane

CnH2n , n = 2, 3 alkene

CnH 2n+ 1 OH, n = 1, 2 alcohol

CnH 2n+1 COOH , n=0,1.. Carboxylic

acid

C2H5OH

Hydroxyl

-OH

KMnO4/H+ / K2Cr2O7/ H

+

Yeast

n-butane

2-methylpropane


23

Homologous

series

General formula Functional group Member , example

Alkane CnH2n + 2 , n = 1,2.. Single covalent bond

between carbon atoms. C- C

Ethane

Alkene CnH2n , n = 2.. Double covalent bond

between carbon atoms. C=C

Ethene

Alcohols CnH2n + 1 OH, n = 1,2.. Hydroxyl group / - OH Ethanol

Carboxylic

acid

CnH2n + 1 COOH, n =

0,1,2..

Carboxyl group , -COOH

Ethanoic acid CH3COOH


24

4. Your are required to prepare one namely ester by using ethanoic acid is one of the reactants. By using a

namely alcohol, describe one experiment to prepare the ester. In your description include the chemical

equation and observation involved.

Ester: ethylethanoate

Material: ethanol, etahanoic acid, water, concentrated sulphuric acid

Apparatus: Boiling tube / test tube, Bunsen burner, test tube holder, beaker

Procedure:

1. Pour 2 cm3 of ethanol into a boiling tube / test tube

2. Add 1 cm3 of ethanoic acid

3. Add 2 to 4 drops of concentrated sulphuric acid

4. Heat the mixture gently for about two minutes

5. Pour the mixture into a beaker containing water.

Observation: Sweet/ pleasant / fruity smell // insoluble in water

Chemical equation: CH3COOH + C2H5OH CH3COO C2H5 + H2O

4. Dehydration of alcohol

Diagram of set up of apparatus

1. Complete and functional

2. Labels of set up of apparatus correct

Procedure: a) Place some glass wool in a boiling tube

b) Use a dropper to add propan-1-ol to wet the glass wool.

c) Clamp the boiling tube horizontally and placed unglazed porcelain chips in the mid section of

the boiling tube.

d) Heat the unglazed porcelain chips strongly.

e) Then heat the glass wool gently to vaporize the propanol.


25

f) [Description of the chemical test to the gas collected in the test tube.]

Add 1 cm3

of bromine water and shake well.

[Observation]:

Reddish brown colour of bromine decolourised Or,

Add 1 cm3

of acidified potassium manganate(VII) solution and shake well.

[Observation]:

Purple colour of potassium manganate(VII) solution decolourised

Chemical equation: C3H7OH C3H6 + H2O

Industrial extraction of palm oil


26

5. Table shows results of latex coagulation

Procedure Observation

Propanoic acid (weak acid) is added to latex Latex coagulates immediately

Latex is left under natural conditions Latex coagulates slowly

Explain why there is a difference in these observations

Answer:

1. Acid ionizes in water to produce high concentration of / a lot of hydrogen ions

2. Hydrogen ions, H+ neutralize the negative charges on the protein membranes

3. The rubber particles collide and the protein membranes break

4. Rubber molecules are released and combine with one another and entangle.

5. The existence of bacteria in natural conditions

6. The growth of bacteria produce / lactic acid /weak acid / low concentration of H+ ions.

7. Due to the slow bacterial action, the coagulation of latex takes a longer time to occur.

[Monomer of natural rubber: 2 – methylbuta-1,3- diene , C5H8 / isoprene ]

Explain how to prevent coagulation of latex

1. Add ammonia solution

2. Ammonia solution contains / ionized to produce hydroxide ions , OH-

3. Hydroxide ions, OH- neutralized the hydrogen ions, H

+ / acid produced by the bacteria

4. The rubber particles remain negatively charged and coagulation is prevented.

6. [Paper 3]

Aim: To compare the elasticity / strength of vulcanised and unvulcanised rubber

Problem statement: Does vulcanised rubber more elastic than unvulcanised rubber

Hypothesis: Vulcanised rubber is more elastic than unvulcanised rubber

Variable: Manipulated : vulcanised rubber and unvulcanised rubber

Responding : length of rubber strip / elasticity

Fixed : mass of weight, size of rubber

Material and apparatus:

Retort stand, bulldog clip, meter ruler, weight, vulcanised and unvulcanised rubber

Procedure:

1. Hang both rubber strips to the retort stand with bulldog clip.

2. Measure the initial length of both rubber strips and record.

3. Hang 50 g weight to the end of each rubber using bulldog clip.

4. Remove the weight and measure the length of both rubber strips and record.//

5. Record all the data obtained.

Rubber particles

Rubber molecules -

-

-

-

- Protein

membranes


27

Unvulcanised rubber Vulcanised rubber

Result / Data

Type of rubber Initial length , cm Length after removal of weight , cm

vulcanised

unvulcanised

Compares and contrasts the properties of vulcanized rubber

Vulcanized rubber Elasticity Unvulcanised rubber

Harder Hardness Less harder

More elastic Elasticity Less elastic

Stronger Tensile strength Weaker

Can withstand higher temperature Resistance to heat Cannot withstand higher

temperature

Less easily oxidized Resistance to oxidation More easily oxidized

Does not become soft and sticky easily Effect of organic solvent Become soft and sticky easily

Conclusion:

1. Vulcanised rubber is more elastic than unvulcanised rubber due to the presence of cross-linkage of

sulfur atoms between the rubber molecules. Vulcanised rubber could return to its original length

after removal of the weight.

To prepare vulcanised rubber

Rubber can be vulcanized by dipping natural rubber sheets into disulphur dichloride solution

in methylbenzene or heated with sulphur.

Note:

Vulcanised rubber is more heat resistance due to the presence of cross-linkage of sulfur atoms

increases the size of rubber molecules. Force of attraction between molecules will increase.


28

7. Compare and differentiate between namely alkene and alkane

Alkane ( hexane ) Alkene ( hexene )

1 Hydrocarbon ( contain C and H atom)

2 Low melting and boiling point

3 Insoluble in water, soluble in organic solvent

4 Cannot conduct electricity

5 Density less than water

6 Completely combustion produce CO2 + H2O

7 Saturated , single covalent bond, C-C Unsaturated , contain at least one double bond C=C

8 Unreactive – undergo substitution with

halogen in the presence of sunlight / UV ray

Reactive – undergo addition reaction( hydrogenation,

halogenations, oxidation, polymerization, with halide,

steam(hydration)

9 General formula , CnH2n+2 , n = 1,2 … , CnH2n , n= 2 …

10 Identify test

1. Combustion, burn less soot flame.

(% of carbon per molecule is lower)

1. More soot flame.

( % of carbon per molecule is higher).

Chemical tests

2. add bromine water , brown colour

remains

3. add acidified KMnO4 , purple colour

remains

2. decolorized brown bromine water

3. purple colour is decolourized


29

CHAPTER 12: REDOX

Redox reactions are chemical reactions involving oxidation and reduction occurring simultaneously.

1. Transfer of electron, Mg Mg2+

+ 2e // Cu2+

+ 2e Cu

2. Loss or gain oxygen, C + 2CuO 2Cu + CO2

3. Loss or gain hydrogen, H2S + Cl2 2HCl + S

4. changes in oxidation number

Rusting of iron

1. When iron exposed to water and oxygen

2. Iron atom releases 2 electrons to form iron (II) ion, Fe2+

/ is oxidized to form iron (II) ion, Fe2+

3. Fe Fe2+

+ 2e // (anode) [ oxidation]

4. Iron acts as reducing agent

5. Oxygen and water receives /gain electrons to form hydroxide ions.

6. O2 + 2H2O + 4e 4OH- (cathode) [reduction]

7. Oxygen acts as oxidizing agent.

8. Iron (II) ion, Fe2+

combine with hydroxide ion, OH- to form iron (II) hydroxide, Fe(OH)2.

9. Iron (II) hydroxide, Fe(OH)2 oxidized by oxygen to form iron (III) oxide, brown solid/precipitate,

Fe2O3.x H2O. // Fe2+

Fe3+

+ e

Effect of the contact of other metals on the rusting of iron.

Aim : To investigate the effect of in contact of other metals on the

rusting of iron.

Problem statement:

How does the effect on rusting when iron is in contact with another metal? //

How does different type of metal in contact with affect the rusting of iron?

Hypothesis :

When a more electropositive metal is in contact with iron, the metal inhibits rusting.

When a less electropositive metal is in contact with iron, the metal speeds up the rusting.

Variable:

Manipulated : Type of metal that in contact with iron.

Responding : Rusting of iron

Fixed : Iron nails, temperature, medium in which iron nails are kept.


30

Apparatus : Test tube, test tube rack

Materials : iron nails, magnesium ribbon, copper strip, zinc strip, tin strip, hot jelly solution, potassium

hexacyanoferat (III) , K3Fe(CN)6 solution, phenolphthalein indicator, sand paper.

Procedure:

1. Five iron nails, magnesium ribbon, copper strip, zinc strip and tin strip were cleaned with sand paper.

2. Four iron nails were coiled tightly with the magnesium ribbon, copper strip, zinc strip and tin strip

respectively.

3. All five iron nails were placed in separate test tube.

4. The volume of hot jelly solution that was mixed with a little K 3Fe(CN)6 solution and phenolphthalein

indicator was poured into the each test tube to completely cover all the nails.

5. The test tubes were kept in a test tube rack and were aside for a day.

6. All observations were recorded.

Observation

Metal Observation

Intensity of dark

blue colouration

Intensity of pink

colouration

Condition of nail

Fe Low The surface of the nail was partially covered with

reddish brown solid

Fe-

Mg

High No reddish brown solid was found on the surface of

the nail.

Fe-Zn High No reddish brown solid was found on the surface of

the nail.

Fe-Sn Moderate Low The whole surface of the nail was covered with

reddish brown solid

Fe-Cu High Low The whole surface of the nail was heavily covered

with reddish brown solid

The nail in test tube A rusted a little. No rusting occurred to the nails in test tubes B and C .The nail in test

tube D rusted but the nail in test tube E rusted even more.

Discussion

1. Based on the observations magnesium and zinc metals inhibit rusting of iron, while copper and tin

metals speed up rusting of iron.

2. This is because magnesium and zinc are more electropositive than iron. Magnesium atom or zinc

atom releases its electron more easily than iron.

Mg Mg2+

+ 2e

O2 + 2H2O + 4e 4OH-

3. Copper and tin are less electropositive than iron. Iron atom releases its electrons more easily than

copper atom or tin atom.

4. Fe Fe2+

+ 2e

5. The less electropositive metals that in contact with iron, the faster the rusting of iron occurs.

6. The more electropositive metals that in contact with iron prevent iron from rusting.

Conclusion:

Rusting can be prevented when iron is in contact with a more electropositive metal. Rusting occurs faster

when iron is in contact with a less electropositive metal.


31

1. Displacement reaction Metal:

Example: Zn + CuSO4 ZnSO4 + Cu // Zn + Cu2+

Cu + Zn2+

a) Zn atom oxidized to Zn

2+ , Zn Zn

2+ + 2e

b) Oxidation number of Zn changes / increase from 0 to +2,

c) Zn acts as reducing agent.

d) Copper (II) ion reduced to Cu, Cu2+

+ 2e Cu

e) Oxidation number of copper changes / decrease from +2 to 0

f) Cu2+

ion acts as oxidizing agent

Example:

An experiment is carried out to determine the relative position of three metals, silver, L and M, in the

electrochemical series.

Experiment

Observation grey deposit

colourless solution

grey deposit

light blue solution

no change

Based on results, arrange the three metals in order of increasing electropositivity. Explain you answer.

Sample answer:

1. Silver, M and L

2. L can displace silver from silver nitrate solution.

3. L is more electropositive than silver // L is higher than silver in electrochemical series.

4. M metal can displace silver from silver nitrate solution.

5. M is more electropositive than silver // M is higher than silver in the electrochemical series.

6. M cannot displace L from L nitrate solution.

7. M is less electropositive than L // L is higher than M in the electrochemical series.

silver nitrate

solution

silver nitrate

solution

L nitrate

solution

L M M


32

2. Displacement of Halogen:

Aim: To investigate oxidation and reduction in the displacement of halogen from its halide solution.

Procedure:

1. Pour 2m cm3 of potassium bromide solution into a test tube.

2. Add 2 cm3 of chlorine water to the test tube and shake the mixture.

3. Add 2 cm3 of 1,1,1-trichloroethane / tetrachlorometane to the test tube and shake the mixture and

leave it on the test tube rack

4. Record theobservation.

5. Repeat steps 1 to 4 using another halogens and halide solutions.

Tabulation of data:

Halogen

Halide

solution

Chlorine Bromine Iodine

Potassium chloride X X

Potassium bromine / X

Potassium iodide / /

Example: Cl2 + 2KI 2KCl + I2 // Cl2 + 2I- I2 + 2Cl

-

Cl2 + 2e 2Cl- ( reduction) 2I

- I2 + 2e (oxidation

3. Transfer of electron at a distance – U-tube

Procedure:

1. clamp a U-tube to a retort stand

2. pour dilute sulphuric acid

3. add solution (oxidizing agent) into one end of the arm of the U-tube

4. Add solution (reducing agent) into the other end.

5. place / dip carbon electrodes into each arm of the U-tube

6. connect the electrodes to a voltmeter/ galvanometer using connecting wire

7. leave the apparatus for 30minutes

8. record the observation

potassium iodide solution bromine water


33

4. Based on electron transfer, EXPLAIN the oxidation and reduction reaction in

(i) Changing of Fe2+

ions to Fe3+

ions

(ii) Changing of Fe3+

ions to Fe2+

ions

Use a suitable example for each of the reaction. Include half equations in your answer.

Sample answer:

(i)

a. Fe+2

Fe+3

+ e

b. Br2 + 2e 2Br –

2. Iron (II) ions releases / donates electron to become iron(III) ions. Iron(II) ions are oxidized.

3. Bromine molecules receive/ gain electrons to form bromide ions. Bromine molecules are reduced.

(any suitable oxidizing agent, Cl2, KMnO4/H+ )

(ii)

1. Fe+3

+ e Fe+2

2. Zn Zn+2

+ 2e

3. Iron(III) ions gain electron to become iron(II) ions. Iron(III) ions are reduced.

4. Zinc atoms releases/ donates electrons to form zinc ions. Zinc atoms are oxidized.

(a: any suitable reducing agent)

5. Describe an experiment to investigate oxidation and reduction in the change of iron(II) ions to iron(III)

ions and vice versa.

(i) Changing of Fe2+

ions to Fe3+

ions

Procedure:

1. Pour 2 cm3 of freshly prepared iron(II)sulphate solution into a test tube.

2. Using dropper, add bromine water drop by drop until no further changes are observed.

3. Heat slowly / gently

4. Add 3 drops of potassium hexacyanoferrate (II) solution / sodium hydroxide solution.

5. Dark blue precipitate // brown precipitate formed.

(ii) Changing of Fe3+

ions to Fe2+

ions

Procedure:

1. Pour 2 cm3 of iron(III)sulphate solution into a test tube.

2. Add half spatula of zinc / Mg powder to the solution.

3. Shake the mixture until no further changes are observed.

4. Filter the mixture.

5. Add 3 drops of potassium hexacyanoferrate (III) solution / sodium hydroxide solution into the filtrate.

6. Dark blue precipitate // green precipitate formed.


34

Reactivity series

1. reactive metal with oxygen

Aim: 1. to investigate the reactivity of metal with oxygen

2. To arrange metals in term of their reactivity with oxygen

Procedure: 1. Put one spatula of potassium manganate(VII), KMnO4 ,

into a boiling tube.

2. Push some glass wool into the boiling tube and clamp horizontally.

3. Place one spatula magnesium powder on a piece of asbestos paper

and put into the boiling tube.

4. Heat magnesium powder strongly and then heat the solid KMnO4.

5. Observe and record how vigorous the reaction and colour of

the residue when it is hot and when it is cold.

2Mg + O2 2MgO

2. hydrogen gas with oxide of less

reactive metal

H2 + PbO Pb + H2O

3. carbon with oxide metal

C + 2CuO 2Cu + CO2

Aim:

To determine the position of carbon in the

reactivity series of metals

Procedure:

1. Mix thoroughly a spatula of carbon powder and

a spatula of copper(II)oxide in a crucible.

2. Heat the mixture strongly.

3. Record the observation.

4. Repeat steps 1 to 3, using magnesium oxide, aluminium oxide and zinc oxide to replace

copper(II)oxide.

K

Na

Ca

Mg

Al

C

Zn

H

Fe

Sn

Pb

Cu

Hg

Ag

Au

Positions of

carbon and

hydrogen in the

reacting series

of metal

Produce oxygen


35

4. Carbon dioxide with metal

CO2 + 2Mg 2MgO + C

Application of reactivity series in the extraction of metals

Extraction of iron from its ores, hematite, Fe2O3

Extraction of tin from its ores, cassiterite, SnO2

- in blast furnace , carbon / coke as a reducing agent.

Example:

C + O2 CO2

C + CO2 2CO

C, CO2 , 2CO reduced the iron oxides to iron

2 Fe2O3 + 3C 4Fe + 3CO2

Fe2O3 + 3CO 2Fe + 2CO2

CaCO3 CaO + CO2 ( lime stone decomposed)

CaO + SiO2 CaSiO3 ( impurities )

Oxidizing reducing

agent agent


36

Redox reaction in various chemical cells


37

CHAPTER 13: THERMOCHEMISTRY

1. Exothermic

– A chemical reaction that gives out heat to the surroundings

- The reactants lose heat energy to form the products

- The energy content of reactants is higher than products

- ΔH negative

2. Energy level diagram (label energy, reactants and product with correct chemical / ionic formula, heat of

reaction with unit.

3. Heat of reaction – heat change/releases when 1 mole of product formed. [ kJmol- ]

= mCǾ / mole

Heat of neutralization – heat releases when 1 mole of H+ combines with 1 mol of OH

-

to form

1 mole of water. H+

+ OH- H2O

4. Heat of combustion – heat releases when 1 mole of alcohol burnt completely in excess oxygen.

C2H5OH + 3O2 2CO2 + 3H2O

5. As the number of carbon atom per molecule increases, the heat of combustion increases, due to more

products formed (CO2 & H2O) . Therefore more heat released when more bonds are formed.

6. To determine heat of combustion (material and apparatus, procedure, tabulation of data, calculation,

observations, precautions).

Procedure:

1. (100 – 200) cm3 of water is measured using a measuring cylinder

2. and poured into a copper tin.

3. The initial temperature of water is measured and recorded, θ 1

4. A spirit lamp is filled with butanol/ other alcohol and weighed, x gram

5.The spirit lamp is light and put under the copper can.

6.The water is stirred continuously with a thermometer.

7.When the temperature of water increased by 30oC, the flame is put off.

8.The spirit lamp is weighed again, y gram

9.The highest temperature is recorded, θ2


38

Results:

Mass of weight of spirit lamp + butanol /g x

Final mass of spirit lamp + butanol /g y

Mass of butanol used/g (x-y) // z

Highest temperature of water /oC θ1

Initial temperature of water /oC θ2

Increased in temperature /oC (θ1 - θ2 ) // θ3

Calculation:

Heat change = mcθ

= 100 x 4.2 x (θ2 – θ1)

= a J

Precautions: 1. Make sure the flame from the combustion of ethanol touches the bottom of the copper can // The

spirit lamp is placed very close or just beneath the bottom of the copper can.

2. Stir the water in the copper can continuously.

3. The spirit lamp must be weighed immediately (because the ethanol is very volatile).

4. A wind shield must be used during experiment.

Heat of displacement

Aim: To determine the heat of displacement of copper by zinc and iron

Procedure:

1. Measure 25 cm3 of 0.2 mol dm

-3 of copper(II)sulphate solution and pour into a plastic cup / polystrene cup.

2. Record the initial temperature of the solution.

3. Pour 0.5g of zinc powder into the solution.

4. Stir the mixture with thermometer

5. Measure and record the highest temperature of the reacting mixutre.

Tabulation of data:

Metal Initial temperature, oC Highest temperature,

oC

Zinc

Iron

Heat of precipitation

Aim: To determine the heat of precipitaion of silver chloride, AgCl

Apparatus: plastic cup, thermometer, measuring cylinder

Material : silver nitrate solution , 0.5 mol dm-3

, sodium chloride solution, 0.5 mol dm-3

Procedure:

1. Measure 20 cm3 0.5 mol dm

-3 of silver nitrate solution and pour into plastic cup.

2. Measure and record the initial temperature of silver nitrate solution.

3. Measure 20 cm3 0.5 mol dm

-3 of sodium chloride solution and pour into plastic cup.

4. Measure and record the initial temperature of sodium chloride solution.

Heat of combustion of butanol = a J

(z/74) mol


39

5. Add the sodium chloride soltuions into the silver nitrate solution quickly and stir the mixture.

6. Measure and record the highest temperature of the reacting mixture.

Tabulation of data:

initial temperature of silver nitrate solution, oC

initial temperature of sodium chloride solution, oC

Average temperature of both solutions, oC

highest temperature of the reacting mixture, oC

Heat of precipitation is the heat released / heat change when one mole of precipitate is formed from their

ions in aqueous solution.

Aplication of exothermic and endothermic reaction

ammonium nitrate

(NH4NO3) Calcium chloride or

magnesium sulphate

sodium acetate crystals


40

CHAPTER 14: CHEMICALS FOR CONSUMERS

Example:

1. (a) A student washed his socks which had oily stains. Explain the cleansing action of soap on the oily

stains.

In water soap ionizes to form ions/anion CH3(CH2)x COO- and cation, sodium ions, Na

+

The anions consists of hydrophilic part ( -COO -) and hydrophobic part (hydrocarbon)

Hydrophilic part dissolve in water only but hydrophobic part dissolve in grease only.

The anions reduce surface tension of water, causing wetting of greasy surface.

During washing and scrubbing, the anions pull the grease and lifted it off the surface and break it into a small droplets (Emulsifying agent)

Rinsing away the dirty water removes the grease (the dirt) and excess soap and the surface is clean.

(b) Another student carried out four experiments to investigate the cleansing effect of soap and detergent on

oily stains in soft water and hard water respectively.

Compare the cleansing effect between

(i) Experiments I and II


41

(ii) Experiment II and IV

Explain the differences in the observation

Exp. I and II

The oily stain disappears in Experiment I but remains oily in Experiment II.

Hard water contains Ca2+

and Mg2+

ions which reacts with soap ions to form

scum (insoluble salt)

The formation of scum makes anions less efficient for cleaning the oily stain on the sock

In soft water, all anions are used to clean the oily stain

Thus, soap is only effective as a cleansing agent in soft water and ineffective in hard water.

Exp. II and IV

The sock in Experiment II remains oily but is clean in experiment IV.

The soap anions form scum when reacts with Ca2+

and Mg2+

ions in hard water.

The formation of scum makes anions less efficient for cleaning

The detergent anions CH3(CH2)x OSO3- / CH3(CH2)x SO3

- do not form a precipitate with

Ca2+

and Mg2+

in hard water.

Hence, detergent cleans effectively in hard water but soap does not clean effectively in hard water.

2. Preparation of soap

Procedure

1. pour 10 cm3 palm oil ( vegetable oil ) into a beaker

2. add 50 cm3 of 5.0 mol dm

-3 NaOH / KOH solution

3. heat the mixture for (10 minutes)

4. stir

5. stop heating, add 50 cm3 distilled water and solid NaCl

6. boil the mixture for 5 minutes

7. cool

8. filter, wash / rinse

9. dry ( press the residue between filter papers

Test

10. Place a small amount of the residue into a test tube add distilled water, shake it well.

produce a lot of lather ( very foamy)

Observation : white solid, slippery and produce a lot of lather ( very foamy).

Chemical equation:

O


42

3. You are given liquid soap, sample of hard water, sample of soft water and other materials.

Describe an experiment to investigate the effect of cleaning action of the soap in different types of water.

You description must include example of hard water and soft water, observation and conclusion.

[10 marks]

Sample answer:

1. hard water : sea water

2. soft water : distilled water

Materials: liquid soap, sea water, distilled water, pieces of cloth with oil stain.

Apparatus: beaker (suitable container), glass rod, measuring cylinder

Procedure:

1. pour (100 – 200) cm3 sea water into a beaker/ suitable container

2. Add (10 – 20 ) cm3 liquid soap into the beaker.

3. stir the mixture

4. Place a piece of cloth with oil stain into the beaker.

5. Record the observation.

6. Repeat step 1 – 4 using distilled water.

Observation:

1. The oil stain in hard water remained but removed in soft water.

Conclusion:

1. Hard water contains Mg2+

or Ca2+

. Soap anion formed scum (insoluble salt) when react with Mg2+

or

Ca2+

.

2. Soap is not an effective cleansing agent in hard water but only effective in soft water.


43

Compare and contrast soap and detergent

Soap Detergent

Sodium carboxylate Sodium alkyl sulphate Sodium alkylbenzene sulphonate

Material : fat, vegetable oil,

NaOH / KOH, mol dm-3

Petroleum fractions : long chain

alcohol, NaOH / KOH,

5 mol dm-3

, H2SO4

Petroleum fractions : long chain

alkene, NaOH / KOH,

5 mol dm-3

, H2SO4

Preparation

Saponification

1. sulphonation

2. neutralization

1. alkylation

2. sulphonation

3. neutralization

The additives in detergent

Type Function Example

Fragrances To add fragrance to both the detergent and

fabrics

Biological

enzymes

To remove protein stains such as blood Amylases, proteases, celluloses, lipases

Whitening agents To convert stains into colourless

substances

Sodium perborate

Suspension agents To prevent the dirt particles removed from

redepositing onto cleaned fabrics

Carboxymethylcellulose (CMC)

Fillers To add to the bulk of the detergent and

enable it to be pour easily

Sodium sulphate, sodium silicate

Optical whitening To add brightness and whiteness to white

fabrics.

Fluorescent dyes

Builder To enhance the cleaning efficiency of

detergent by softening the water

Sodium tripolyphosphate

CH3 (CH2)15 COO- Na

+


44

Food additive

Type Function Examples

Preservatives To slow down/ prevent the

growth of microorganism,

therefore food can kept for

longer periods of time

Salts/sugar: draws the water out of the cells of

microorganism and retards the growth of

microorganism.

Vinegar: provides an acidic condition that inhibits the

growth of microorganism.

NaNO3 (Burger)

Benzoic acid / sodium benzoate: to slow down the

growth of microorganism.

Antioxidants To prevent oxidation that can

causes rancid fats and brown

fruits

Ascorbic acid and vitamin E (Tocopherol)

Flavorings To improve the taste of food

and restore taste loss because

of processing.

Sugar , salt, MSG, vinegar, aspartame and synthetic

essences (ester)

Stabilizers To prevent emulsion from

separating out.

Lecithin, fatty acid

Thickeners Its use to thicken foods Pectin, acacia gum, gelatin

Dyes To add or restore the colour in

food in order to enhance its

visual appeal and match

consumers expectations.

Natural dyes and artificial dyes: Azo compounds or

triphenyl compound.

Medicine

Type Function Example Effect on health

Analgesic To relieve pain without affected

consciousness

Aspirin

-Internal bleeding and

ulceration

-can cause brain and liver

damage to children

Paracetamol Over dose can cause brain and

liver damage

Codeine Addiction, depression and

nausea

Antibiotics To treat infections cause by

bacteria (tuberculosis, TB) and

pneumonia. Can kill or slow

down the growth of bacteria.

Penicillin

(Penicillium

notatum)

Can cause allergic reaction.

Streptomycin Can cause nausea, vomiting,

dizziness, rashes, fever

Psychotherapeutic To alter the abnormal thinking,

feelings and behaviors. Divide

into 3 categories :

a) stimulant: to reduce fatigue

Amphetamines

-High dose can lead to anxiety,

hallucinations, severe

depression, and psychological

dependence.


45

b) antidepressant:

to reduce tension and anxiety

c) antipsychotic: to treat

psychiatric illness

Barbiturate /

tranquilizer

chlorpromazine

haloperidol,

clozapine

Overdose can lead to

respiratory difficulties,

sleeplessness, come, death.

Dizziness, drowsiness, rapid

heartbeat.

The existence of Chemicals 1. Detergent:

* wear gloves when working with strong detergents to protect your hands

* use biodegradable detergent

* use appropriate amounts of detergents

2. Food additives

* Be wise consumer. Read the label to know what you are eating.

* Avoid consuming too much salts and sugar

* avoid foodstuff with additives which are you sensitive to.

* avoid rewarding children with junk food.

3. Medicine:

* do not store up medicines.

* no self medication

* do not take medicine prescribe for someone else

* check for expiry date

* follow your doctor`s instructions for taking medicine.

* keep away from children

* do not overdose


46

Some common medical plant and their functions

note chemistry form 4 english

Documents

arrangement of particles

discrete particles atoms

b movement of particles

solid liquid

number of proton

gas liquid solid

gas matter

orderly arrangement