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 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 particles arrangement is not orderly.
The kinetic energy is constant
t2 – t3 Gaseous The particles are far away from each other. The particles arrangement is not orderly.
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
Example: Compare atom 12X and atom 14Y 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 2 6p 8n 6p 6n X Y Hydrogen
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 à ZnCl2 + H2
5. Anhydrous calcium chloride – to dry the H2 gas. 6. CuO + H2 à 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 à 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
(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 à HCl + HOCl
II Reaction with sodium hydroxide Cl2 + 2NaOH à NaCl + NaOCl + H2O III Reaction with iron 3Cl2 + 2Fe à 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 à 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. 4
Chlorine gas
Sodium hydroxide solution Soda lime
Hot iron wool HCl (cons) + KMnO4(s) To produce Cl2 III II
e) It is easier for chlorine atom to attract/receive electron.
(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 à Na+ + e)
5. Atom (Cl) gain / accept electron to form chloride ion, Cl -6. Half equation ( Cl + e à 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
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 à Cu2+ + 2e ) cathode –pure metal ( Cu2+ + 2e à 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)
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 à ZnSO4 + Cu // Zn à Zn2+ + 2e / Cu2+ + 2e à 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 àchemical Energy change Chemical à electrical
+ve / anode: copper (OXIDATION)
-ve / cathode: copper Electrode +ve/cathode: copper-ve/ anode: lead (OXIDATION) Cu2+ , H+
OH- , SO42- Ions present in the electrolyte Cu 2+ , H+ OH- , SO4 2-Anode :Cu à Cu2+ + 2e (type of electrode) Cathode: Cu2+ + 2e à Cu ( ECS) Half equation Anode: Pb à Pb2+ + 2e Cathode: Cu2+ + 2e à 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
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 à H3O+ + Cl- // HCl à H+ + Cl- , H3O+ , hydroxonium ion
3. Methanoic acid ionizes partially in water to form lower concentration hydrogen ions CH3COOH à 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 cm3 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.
8
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
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.
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) à salt + H2 / 2H+ + Mg à Mg2+ + H 2
– acid + base ( metal oxide) à salt + water – acid + alkali à salt + water / H+ + OH- à H
2O ( NaOH, KOH, NH4OH)
– acid + carbonate metal à salt + CO2 + H2O / 2H+ + CO32+ à 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- à 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 à oxide metal (base) + CO2 except Na, K and NH4+ Example: CuCO3 à CuO + CO2
Nitrate à oxide metal + nitrogen oxide + oxygen except Na, K, (2NaNO3 à 2NaNO2 + O2 ) Example : 2Mg(NO3)2 à 2MgO + 4NO2 + O2
(Brown gas)
Ammonium chloride à ammonia gas + hydrogen chloride gas, (NH4Cl à NH3 + HCl ) 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-)
7. pour 2 cm3 the solution into a test tube
8. add 1 cm3 of dilute nitric acid and silver nitrate solution. 9. white precipitate formed
10. 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
11. pour 2 cm3 the solutions into different test tubes
12. add 1 cm3 potassium iodide solution into the test tubes 13. yellow precipitate formed
14. lead (II) ion present (d) test for anion
15. pour 2 cm3 of lead (II) nitrate solution into a test tube 16. add 1 cm3 of dilute sulphuric acid
17. add 1 cm3 of iron (II) sulphate solution 18. shake the mixure
19. tilt the test tube, add concentrated sulphuric acid carefully // drop by drop down the side of the test tube
20. the brown ring formed 21. nitrate ion, NO3- present.
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
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- à 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 à SO2 Sulphur is burned in the excess of oxygen gas to produce sulphur dioxide gas.
2 2SO2 + O2 à 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 à H2S2O7 Gas sulfur trioxide dissolve in sulphuric acid to produce oleum 4 H2S2O7 + H2O à2H2SO4 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 à 2NH3
Condition: Catalyst: iron, temperature: 450 – 550 oC, Pressure 200 – 500 atm Usage: to manufacture fertilizer
2NH3 + H2SO4 à (NH4 )2 SO4 3NH3 + H3PO4 à (NH4 )3 PO4
NH3 + HNO3 à NH4NO3
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
14
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.
Copper atom
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 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
SiO2 B2O3 Na2O Laboratory glassware, cooking utensils. Automobile Test tube The intensity of the dark blue colour //
rate of rusting A
B C
-very resistant to chemical reactions -does not break easily
headlights. Lead glass -low softening point
-high density
-High refractive index
SiO2 PbO CaO
Decorative items, crystal glass ware, lens, prism, chandelier
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
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+, S
2O32- ) increases /
2. Frequency of collision between particles ( H+, S
2O32- ) 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 4. Frequency of effective collision increases
5. Rate of reaction 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 4. Frequency of effective collision increases
5. Rate of reaction 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.
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 à 2NaCl + 3O2 b) Decomposition hydrogen peroxide , 2H2O2 à 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 cm3 measuring cylinder,10cm3 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. 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
20
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
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 à 2H2O + O2
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] 22 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-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+ n-butane 2-methylpropane
2. CH3COOH + C2H5OH CH3COO C2H5 + H2O
3. C2H4 + H2O C2H5OH 4. C6H12O6 2C2H5OH + 2CO2
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 H 2SO4, cons
H 3 PO4, 60 atm, 300 oC
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 à 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.
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 à C3H6 + H2O
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 26 Rubber particles Rubber molecules -Protein membranes
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.
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.
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
CHAPTER 12: REDOX
Redox reactions are chemical reactions involving oxidation and reduction occurring simultaneously. 1. Transfer of electron, Mg à Mg2+ + 2e // Cu2+ + 2e à Cu
2. Loss or gain oxygen, C + 2CuO à 2Cu + CO2 3. Loss or gain hydrogen, H2S + Cl2 à 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 à 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 à 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+ à Fe3+ + e
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.
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 à Mg2+ + 2e
O2 + 2H2O + 4e à 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 à 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.
1. Displacement reaction Metal:
Example: Zn + CuSO4 à ZnSO4 + Cu // Zn + Cu2+ à Cu + Zn2+
a) Zn atom oxidized to Zn2+ , Zn à Zn2+ + 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 à 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 depositcolourless solution •• grey depositlight blue solution • no change Based on results, arrange the three metals in order of increasing electropositivity. Explain you answer. Sample answer: silver nitrate solution silver nitrate solution L nitrate solution L M M
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.
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 +
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
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)
potassium iodide solution bromine water
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.
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 à 2MgO 34 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
2. hydrogen gas with oxide of less reactive metal
H2 + PbO à Pb + H2O
3. carbon with oxide metal C + 2CuO à 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.
4. Carbon dioxide with metal CO2 + 2Mg à 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 à CO2 C + CO2 à 2CO
C, CO2 , 2CO reduced the iron oxides to iron 2 Fe2O3 + 3C à 4Fe + 3CO2
Oxidizing reducing agent agent
Fe2O3 + 3CO à 2Fe + 2CO2
CaCO3 à CaO + CO2 ( lime stone decomposed) CaO + SiO2 à CaSiO3 ( impurities )
Redox reaction in various chemical cells
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- à H 2O
4. Heat of combustion – heat releases when 1 mole of alcohol burnt completely in excess oxygen. C2H5OH + 3O2 à 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
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:
38
Heat of combustion of butanol = a J (z/74) mol
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.
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
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
(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 cm distilled water and solid NaCl3 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:
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 ) cm liquid soap into the beaker.3
