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孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!
SPM CHEMISTRY SEMINAR 2015 1 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!MASTERING ESSAY QUESTION IN PAPER 3
Experiment 1 (Chapter 2) Experiment 2 (Chapter 4) SPM 2008 Experiment 3 (Chapter 4)
Aim:To investigate the effect of the presence of impurity to the melting point of naphthalene (becomes lower)
Aim:To investigate the reaction of alkali metals in water
Aim:To investigate the reactions of alkali metals with oxygen / chlorine
Problem statement:How does
Problem statement:How does reactivity of Group 1 elements change when they react with water?
Problem statement:How do alkali metal react with oxygen / chlorine?
Variables:Manipulated:Responding:Fixed (controlled):
Variables:Manipulated: types of alkali metalResponding: reactivity of alkali metalFixed (controlled): size of alkali metal
Variables:Manipulated: types of alkali metalsResponding: reactivity of alkali metalsFixed (controlled): oxygen gas
Hypothesis: Hypothesis: When going down Group 1, alkali metal becomes more reactive in water.
Hypothesis: When going down Group 1, alkali metals becomes more recative in their reaction with oxygen
Materials: Materials: red litmus paper, distilled water, paper, filter paper, small lithium,sodium and potassium
Materials: 3 gas jar filled with oxygen, red litmus paper, paper, filter paper, small lithium,sodium and potassium
Apparatus: Apparatus: water through, knife, forceps Apparatus: gas jar spoon, gas jar cover, small knife, Bunsen burner, pair of forceps
Procedure: Procedure:1) A piece of lithium removed from the bottle with apair of forceps.2) A small piece of lithium is cut using a knife.3) The paraffin oil on the surface of lithium is removed by using a
piece of filter paper.4) The piece of lithium is placed into a water through half-filled with
cold distilled water using pair of forceps.5) When the reaction stops, a piece of red litmus paper is put into the
solution formed.6) All changes occured are recorded.7) Steps 1 to 6 is repeated using potassium and sodium respectively
to remove lithium.
Procedure:1) A piece of lithium removed from the bottle with a pair of
forceps.2) A small piece of lithium is cut using a knife.3) The paraffin oil on the surface of lithium is removed by using a
piece of filter paper.4) The piece of lithium is placed in a gas jar spoon.5) The lithium is start to burn when heated.6) The burning lithium quickly lowered into a gas jar filled with
oxygen.7) When recations stops, 20cm3 of distilled water is poured into gas
jar and shaken well.8) A piece of red litmus paper is put into solution formed.9) All changes that occured are recorded.10) Steps 1 to 9 is repeated using potassium and sodium
respectively to remove lithium.
Tabulation of data: Tabulation of data:Metal Observation
LithiumSodiumPotassium
Tabulation of data:Metal Observation
Lithium
Sodium
Potassium
Experiment 4 (Chapter 4) Experiment 5 (Chapter 6) Experiment 6 (Chapter 6)
SPM CHEMISTRY SEMINAR 2015 2 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Aim:To investigate the reaction of halogens with iron
Aim:To investigate the effect of concentration of ions on the selective discharge of ions at the electrodes of aqueous hydrochloric acid
Aim:To investigate the effect of the type of electrodes on the products of electrolysis of aqueous copper(II) sulphate solution.
Problem statement:How does reactivity of halogens change when they react with iron
Problem statement:How does the concentration of ions in hydrochloric acid affect the discharge of ions at the anode?
Problem statement:How does the type of electrodes affect the type of products formed during the electrolysis
Variables:Manipulated: types of halogensResponding: reactivity of halogensFixed (controlled): iron
Variables:Manipulated: concentration of chloride ionsResponding: ion discharged at the anodeFixed (controlled): type of electrode
Variables:Manipulated: types of electrodesResponding: types of product at anodeFixed (controlled): type of electrolyte
Hypothesis: The reactivity of halogens decreases down the group from chlorine to bromine when they react with iron.
Hypothesis: When the concentration of chloride ion is higher, then the chloride ion will be preferentially discharged.
Hypothesis: When copper electrodes are used instead of carbon electrodes, the types of products formed at the anode are different
Materials: potassium manganate (VII) crystal, concentrated HCL, liquid bromine, iodine crystals, soda lime and iron wool
Materials: 1.0 moldm-3 hydrochloric acid, 0.001 moldm-3 hydrochloric acid Materials: 0.1 moldm-3 Copper (II) sulphate solution
Apparatus: test tubes, test tube holders, rubbber stoppers, conical flask, combustion tube, delivery tube, Bunsen burner, retort stand with clamp, thistle funnle.
Apparatus: batteries, carbon electrodes, connecting wires, ammeter, electrolytic cell, tet tube, blue litmus paper, wooden splinter and matches
Apparatus: batteries, carbon, copper electrodes, connecting wires, ammeter, electrolytic cells, sandpaper, wooden splinter and matches
Procedure:1) A little iron wool is placed in the middle of combustion tube
and heated strongly.2) Concentrated HCL/liquid bromine/iodine crystal is heated
gently in test tube.3) Chlorine gas/bromine gas/iodine vapour liberated is passed
through the heated iron wool in the combustion tube until no further changes occur.
4) The excess chlorine gas/bromine gas/iodine vapour is absorbed by soda lime.
5) All changes are observed and recorded.
Procedure:1) An electrolytic cell filled with 1.0 moldm-3 hydrochloric acid until
it is half full.2) The circuit is completed by connecting the electrodes to the switch,
ammeter, and batteries.3) The switch is turned on.4) Any observation is recorded.5) The gas produced at anode and cathode are collectedand tested with
a moist blue litmus paper and a lighted wooden splinter,6) Step 1 to 4 is repeated using 0.001 moldm-3 hydrochloric acid to
replace 1.0 moldm-3 hydrochloric acid.7) The gas produced at the anode and cathode are collected and tested
with glowing wooden splinter and lighted wooden splinter.
Procedure:1) Two carbon electrodes are cleaned with sandpaper.2) The Copper (II) sulphate solution is poured into an electrolytic
cell with carbon electrodes until it is half full.3) The circuit is completed by connecting the electrodes to the
switch, ammeter, and batteries.4) The switch is turned on foor 15 min.5) All observation at the anode, cathode and electrolyte are
recorded.6) Gas produced at anode is collected and tested with glowing
wooden splinter.7) Step 1 to 5 is repeated using copper electrodes
Tabulation of data:Halogens ObservationChlorineBromineIodine
Tabulation of data:Electrolyte Observation
Anode Cathode1.0 HCL
0.001 HCL
Tabulation of data:Electrode Observation
Anode Cathode ElectrolyteCarbonCopper
Experiment 7 (Chapter 6) SPM 2007 Experiment 8 (Chapter 6) Experiment 9 (Chapter 7)
SPM CHEMISTRY SEMINAR 2015 3 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Aim:To construct the electrochemical series based on potential differences between metals
Aim:To construct the electrochemical series using the principle of displacement of metals
Aim:To investigate the role of water in showing their properties of acids
Problem statement:Can an electrochemical series of metals be constructed based on the potential difference between two metals?
Problem statement:How can the electrochemical series of metals be constructed based on the displacement metals?
Problem statement:Is water neede for an acid to show its acidic properties?
Variables:Manipulated: pairs of metalsResponding: potential difference producedFixed (controlled): type and concentration of electrolye
Variables:Manipulated: metal strips/salt solution usedResponding: deposition of metalsFixed (controlled): concentration of salt solution
Variables:Manipulated: types of solventResponding: change in colour of blue litmus paperFixed (controlled): type of acid
Hypothesis: The further apart the two metals in the electrochemical series, the greater the potential difference produced/voltage produced.
Hypothesis: The greater the number of metals that can be displaced by a metal from their solutions, the higher is its position in the electrochemical series.
Hypothesis: an acid will show its acidic properties when dissolves in water.
Materials: 0.1 moldm-3 sulphuric acid, copper strip, lead, iron, zinc, aluminium, magnesium strip, sand paper
Materials: 1.0 mol dm-3 of magnesium nitrate, zinc nitrate, lead nitrate, copper nitrate, magnesium, zinc, lead and copper strips, sandpaper
Materials: glacial ethanoic acid, distilled water, dry propanone, blue litmus paper
Apparatus: 250 cm3 beaker, crocodile clips, voltmeter, and connecting wires.
Apparatus: test tubes, test tube rack Apparatus: test tube, droppers, tets tube rack
Procedure:1) The metal strips are celaned with sanpaper.2) A beaker filled with dilute sulphuric acid until two-thirds full.3) Magnesium strip as electrode A and copper strip as electrode B
are dipped into dilute sulphuric acid.4) The circuit is completed by connecting the metals to a
voltmeter.5) The reading of voltmeter produced is recorded.6) The metal strip that acts as the negative terminal is determined
and recorded.7) Step 1 to 6 are repeated using aluminium, zinc, lead, iron, and
copper to replace magnesium as electrode A.
Procedure:1) The strips are clened with sandpaper.2) 5cm3 og magnesium nitrate, zinc nitrate, lead nitrate, copper nitrate
solution are poured into 4 separate tubes.3) A strip of magnesium is placed into each of the solutions.4) Observation on the deposition of metals are recorded.5) Steps 1 to 4 are repeated using zinc, lead and copper to replace
magnesium strip.
Procedure:1) Three test tubes are labelled as A, B, and C. They are placed in
test tube racks.2) 1cm3 of glacial ethanoic acid is placed into each test tubes usind
a dropper.3) 2cm3 of distilled water is added to test tube B.4) 2cm3 of dry propanone added to test tube C.5) A piece of dry blue litmus paper is placed into each test tube.6) Any changes that occur are observed and recorded.
Tabulation of data:Pair of metal + terminal -terminal Voltage
Zn/CuMg/CuAl/CuAg/CuFe/Cu
Tabulation of data:Metal strip Mg(NO3) Zn(NO3)2 Pb(NO3)2 Cu(NO3)2
MagnesiumZincLead
Copper
Tabulation of data:Test tube Observation
ABC
Experiment 10 (Chapter 2) Experiment 11 (Chapter 7) SPM 2009 Experiment 12 (Chapter 8)
SPM CHEMISTRY SEMINAR 2015 4 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Aim:To investigate the role of water in showing their properties of alkalis
Aim:To investigate the relationship between pH values with molarity of acid
Aim:To construct the ionic equation for the concentration for the formation of lead(II) chromate(VI)
Problem statement:Is water needed for an alkali to show its alkali properties?
Problem statement:What is the relation between pH values with molarity of an acid?
Problem statement:How to construct an ionic equation for the formation of lead(II) chromate(VI)?
Variables:Manipulated: types of solventsResponding: chnage in colour of red litmus paperFixed (controlled): type of alkali and litmus paper
Variables:Manipulated: molarity of acidResponding: pH valuesFixed (controlled): type of acid
Variables:Manipulated: volumes of lead(II) nitrate solutionResponding: height of yellow precipitateFixed (controlled): volumes of potassium chromate(VI) soution
Hypothesis: an alkali will only show its alkaline properties when dissolve in water
Hypothesis: When the molarity of acid increases, its pH values increases Hypothesis: as the volume of potassium chromate(VI) solution increases, the height of the yellow precipitate increases untill all the lead (II) nitrate has reacted.
Materials: dry ammonia gas stoppered in atest tube, ammonia gas dissolved in propanone, aqueous ammonia solution and red litmus paper.
Materials: hydrocloric acids of 1.0 mol dm-3, 0.1 mol dm-3, 0.01 mol dm-3, 0.001 mol dm-3
Materials: 0.5 mol dm-3 lead (II) nitrate solution and o.5 mol dm-3 potassium chromate (VI) solution
Apparatus: test tubes and droppers Apparatus: pH meter, 100cm3 beakers, and 100cm measuring cylinder Apparatus: test tubes of the same size, test tube rack, 50cm3 burette, retort stand with clamp and ruler
Procedure:1) A piece of dry litmus paper is put into a stoppered test tube of
dry ammonia gas and the test tube is then stoppered back immediately.
2) The effect of the dry ammonia gas on the red litmus paper is recorded.
3) Another piece of dry red litmus aper is put in 5cm3 of aqueous ammonia solution in aseparate test tube.
4) Step 3 of the experiment is repeated using ammonia dissolved in propanone to replace aqueous ammonia solution.
Procedure:1) 30cm3 of 1.0 mol dm-3 hydrochloric acid is put in dry beaker.2) The probe of a pH meter that has been washed with distilled water
is immersed in 30 cm3 of the 1.0 mol dm-3 hydrochloric acid.3) The pH values shown on the pH meter is recorded.4) The pH values of hydric acid solutions with different molarities
are measured one by one in dry beakers as in steps 1 to 3.
Procedure:1. 7 test tubes of same size are labelled from 1 to 7. They are
placed on test tube rack.2. A burette is filled with 0.5 moldm-3 potassium chromate (VI)
solution. 5.0 moldm-3 of the potassium chromate (VI) solution is placed into each test tube.
3. Another burette is filled with 0.5 moldm-3 lead (II) nitrate solution.
4. lead (II) nitrate solution from the burette is added into each of the 7 test tubes.
5. The mixture in each test tube is stirred using glass rod. A yellow precipitate is formed in each test tube.
6. The test tubes are left aside for about an hour.7. The height of precipitate in each test tube is measured using
metre rule. The colour above the precipitate in each test tube is observed and recorded.
Tabulation of data:Condition of ammonia Observation Inference
Tabulation of data:Molarities (mol dm-3) pH values
1.00.10.010.001
Tabulation of data:Test tube 1 2 3 4 5 6 7Volume of 0.5 moldm-3 potassium chromate (VI) solutionVolume of 0.5 moldm-3 lead (II) nitrate solution.Height of precipitateColour of solution above precipitate
Experiment 13 (Chapter 9) SPM 2003/05 Experiment 14 (Chapter 9) SPM 2003 Experiment 15 (Chapter 10)
SPM CHEMISTRY SEMINAR 2015 5 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Aim:To compare the hardness of pure metal and its alloy
Aim:To compare the rate of rusting between iron, steel and stainless steel
Aim:To investigate the effect of total surface area of the reactant on the rate of reaction
Problem statement:Are alloys harder than pure metal?( is bronze harder than copper )
Problem statement:How does the rate of rusting between iron, steel and stainless steel differ?
Problem statement:How does the total exposed surface area of a solid reactant affect the rate of reaction?
Variables:Manipulated: types of materials to make the metal blockResponding: diameter of the dent made by a steel ball bearingFixed (controlled): size of steel ball bearing
Variables:Manipulated: types of nails(iron, steel, stainless steel)Responding: rate of rustingFixed (controlled): size of nails
Variables:Manipulated: size of marble chipsResponding: rate of reactionFixed (controlled): temperature
Hypothesis: Bronze is harder than copper Hypothesis: pure iron rusts faster than steel while stainless steel does not rust easily
Hypothesis: When the total surface area of marble chips increases, the rate of reaction increases
Materials: copper block, bronze block, ball beraing, cellophane tape and thread
Materials: iron nail, steel nail, stainless steel nail, 5% jelly solution and potassium hexacyanoferrate(III) solution and sandpaper
Materials: 80cm3 0.1moldm-3 hydrochloric acid, 2g large marble chip, 2g small marble chip, water
Apparatus: metre rule, retort stand with clamp, 1 kg weight Apparatus:test tubes
Apparatus: 50cm3 measuring cylinder, 150cm3 conical flask, stopper with delivery tube, basin, burette, electronic balance, stopwatch
Procedure:1) A metre rule is clamped to a retort stand, and a piece of copper
block is placed on the based of the retort stand.2) A steel ball bearing is placed on the copper block and a piece of
cellophane tape is used to hold the ball bearing in place.3) A 1 kg weight is hung at a height of 50 cm above the copper
block.4) The weight is dropped onto the ball bearing placed on the
copper block.5) The diameter of the dent made by the ball bearing is measured
to the nearest 0.5mm.6) Step 1 to 5 is repeated using br onze block to replace copper
block.
Procedure:1) Three test tubes are half filled with jelly solution and are labelled
as A, B and C.2) 1 cm3 of potassium hexacyanoferrate (III) solution is added to
every test tube.3) An iron nail, a steel nail and a stainless steel are polished with
sandpaper to remove any rust formed. The nails are then placed in the three test tubes labelled A, B and C respectively.
4) All test tubes are allowed to stand for 2 days before they are examiined.
5) Changes occurred on the nail are observed and recorded.
Procedure:1) Fill a burette with water and invert it over water in a basin.2) Clamp the burette vertically using a retort stand. Records its
initial reading.3) Measure 40cm3 of 0.1 moldm-3 hydrochloric acid, HCL using a
measuring cylinder. Transfer the acid into a conical flask.4) Weigh 2g of large marble chips. Put them into the conical flask.
Close the conical flask immediately with astopper which is joined to the delivery tube and start the stopwatch.
5) Record the burette reading at intervals of 30 seconds until the reaction stops.
6) Repeate the steps 1 to 6 using 2g of small marble chips to replace 2g of large marble chips.
Tabulation of data:MetalBlock
Diameter of dent (mm)I II II average
CopperBronze
Tabulation of data:Test tube Type of nail Observation
A Iron nailB Steel nailC Stainless steel nail
Tabulation of data:Times
Burette reading
Volume gas
Experiment 16 (Chapter 10) Experiment 17 (Chapter 10) Experiment 18 (Chapter 10) SPM 2010
Aim:To investigate the effect of concentration on the rate of reaction
Aim:To investigate the effect of temperature on the rate of reaction
Aim:To investigate the effect of catalyst on the rate of recation
SPM CHEMISTRY SEMINAR 2015 6 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Problem statement:How does the concentration of reactant affect the rate of reaction?
Problem statement:How does the temperature affect the rate of reaction?
Problem statement:How does catalyst affect the rate of reaction?
Variables:Manipulated: concentration of sodium thiosulphate solutionResponding: time taken for the cross ‘X’ to disappearFixed (controlled): concentration dilute sulphuric acid
Variables:Manipulated: temperature of sodium thiosulphate solutionResponding: time taken for the cross ‘X’ to diasappearFixed (controlled): concentration of sodium thiosulphate solution
Variables:Manipulated: amount of catalyst usedResponding: the volume of oxygen given off at half-minute intervalsFixed (controlled): volume of hydrogen peroxide
Hypothesis: When the concentration of sodium thiosulphate is higher, the higher the rate of reaction
Hypothesis: The higher the temperature of the reactant, the higher the rate of reaction
Hypothesis: The rate of decomposition of hydrogen peroxide increases when the amount of catalyst used is increased
Materials: 0.2 mol dm-3 sodium thiosulphate solution, 1.0 mol dm-3
sulphuric acid and distilled waterMaterials: 0.1 mol dm-3 sodium thiosulphate solution and 1.0 mol dm-3 sulphuric acid
Materials: 0.2 mol dm-3 hydrogen peroxide nad manganese(VI) oxide
Apparatus: 10cm3 and 100cm3 measuring cylinders, 100cm3 conical flask, white paper marked with cross ‘X’ and stopwatch
Apparatus: conical flask, 10cm3 measuring cylinder, thermometer, stopwatch, white paper marked with cross ‘X’, wire gauze, tripod stand, and Bunsen burner.
Apparatus: measuring cylinder, conical flask, delivery tube, rubber stopper, retort stand, clamp and burette.
Procedure:1) Using ameasuring cylinder, measure and pour 45cm3 of 0.2
moldm-3 sodium thiosulphate, Na2S2O3 solution into a conical flask.
2) Place the conical flask on top of a piece of white paper with mark ‘X’.
3) Using another measuring cylinder, measure 5cm3 of 0.1 mol dm-3 sulphuric acid.
4) Pour the sulphuric acid quickly and carefully into the conical flask and start the stop watch immediately.
5) Swirl the mixture in the conical flask and start the stopwatch immediately.
6) Swirl the mixture in the conical flask a few times and place in back on the white paper.
7) Obeserve the mark ‘X’ vertically from the top through the solution.
8) Record the time t required for the mark ‘X’ to disappear from sight.
9) Repeat the experiment four more times using different volumes of 0.2 mol dm-3 sodium thiosulphate of different concentration.
Procedure:1)
Procedure:1) Using a measuring cylinder, 25 cm3 of 0.2 mol dm-3 hydrogen
peroxide is measured into a conical flask and 0.5 g of manganese(VI) oxide is added to the hydrogen peroxide.
2) The conical flask is immediately closed with a stopper fitted with delivery tube and the stopwatch is started simultaneously. The conical flask is swirled gently.
3) The total volume of oxygen gas given off is determined from
the burette reading at intervals of 12
minute for 4 minutes.
4) The experiment is repeated using 0.20 g of manganese(VI) oxide instead of 0.50 g of manganese(VI) oxide.
Tabulation of data: Tabulation of data: Tabulation of data:
Experiment 19 (Chapter 11) SPM 2006 Experiment 20 (Chapter 12) Experiment 21 (Chapter 13)
Aim:To compare the elasticity of vulcanized rubber and unvulcanized rubber
Aim:To investigate the effect of other metals on rusting
Aim:To determine the heat of combustion of various liquid alcohol
SPM CHEMISTRY SEMINAR 2015 7 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Problem statement:Is vulcanized rubber more elastic than unvulcanized rubber?
Problem statement:How do different types of metals in contact with iron affect rusting?
Problem statement:Does alcohol with more carbon atoms per molecule have a higher heat of combustion?
Variables:Manipulated: type of rubberResponding: lenght of rubber stripFixed (controlled): size of rubber strip, mass of weight
Variables:Manipulated: different metal in contact with ironResponding: presence of blue colouration//rustingFixed (controlled): clean iron nails
Variables:Manipulated: different types of alcoholResponding: heat oof combustionFixed (controlled): volume of water, copper can
Hypothesis: vulcanized rubber is more elastic than unvulcanized rubber Hypothesis: when a more electropositive metals is in contact with iron, the metal inhibits rusting. When a less electropositive is in contact with iron, the metal speeds up rusting
Hypothesis: the higher the number of carbon atoms in the molecules, the higher the heat of combustion
Materials: vulcanized rubber strip and unvulcanized rubber strip Materials: 5 iron nails, 3cm magnesium ribbon, 3 cm copper strip, 3cm zinc strip, 3cm tin strip, 50cm3 hot jelly solution with a little potassium hexacyanoferrate(III) solution and phenolphthalein, sandpaper
Materials: 50cm3 of Methanol, Ethanol, Propanol, Butanol and 800cm3
water
Apparatus: retort stand and clamp, bulldog clip, metre rule, weights Apparatus: 5 test tubes, test tube rack Apparatus: copper can, tripod stand, thermometer, 100cm3 measuring cylinder, spirit lamp, weighing balance, wooden block, wind shield.
Procedure:1) Hang both rubber strips using bulldog clips.2) Measure the initial lenght of both rubber strips.3) Hang 10g weight to the end of each rubber strip.4) Remove the weight and measure the lenght of both rubber strips.5) Repeat steps 2 to 4 using 20g, 30g, 40g and 50g weights to
replace the 10g weights.6) Record the data.
Procedure:1) Clean all the five iron nails, magnesium ribbon, strips of copper,
zinc and tin with sandpaper.2) Coil four nails tightly with magnesium ribbon, strips of cooper,
zinc and tin each.3) Place all the iron nails in separate test tubes.4) Pour the same volume of hot jelly containing potassium
hexacyanoferrate(III) solution and phenolphthalein indicator int the test tubes in a test tube rack and leave them aside for a day.
5) Record the observation.
Procedure:1) Using a measuring cylinder, measure 200cm3 of water and pour
it into a copper can.2) Place the copper can on atripod stand.3) Measure the initial temperature of the water and record the
reading.4) Pour 50cm3 of methanol into a spirit lamp and then weight the
lamp and its content. Record the mass.5) Put the lamp and light up the wick of the lamp immediately.6) Stir the water continuously until the temperature of water
increase about 30°C.7) Put off the flame and record the highest temperature reached by
the water.8) Weight the lamp and its content immediately and record its
reading.9) Repeat steps 1 to 8 using other alcohol like Ethanol, Propanol,
Butanol.
Tabulation of data: Tabulation of data: Tabulation of data:
Tabulation of data for experiment 21
Alcohol Methanol Ethanol Propan-1-ol Butan-1-olInitial temperature (°C)Highest temperature (°C)Rise in temperature of water (°C)Mass of spirit lamp before burning (g)Mass of spirit lamp after burning (g)
SPM CHEMISTRY SEMINAR 2015 8 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!Mass of alcohol burnt (g)
Tabulation of data for experiment 20
Test tube Pair of metals Intensity of the blue colour in the solution Intensity of pink colour in the solution inferenceA FeB Fe + ZnC Fe + CuD Fe + MgE Fe + Sn
Tabulation of data for experiment 19
Type of rubber Unvulcanised rubber Vulcanised rubberInitial length (cm)Stretched length (cm)Increase in length (cm)Length after removal of weight (cm)Difference in length (cm)
Tabulation of data for experiment 18
Time (s)Burette reading (cm3)Volume of hydrogen gas (cm3)
SPM CHEMISTRY SEMINAR 2015 9 PREPARED BY: MS PHUAN
GUIDELINE ON ANSWERING QUESTIONS IN CHEMISTRY PAPER 3 2015
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!
FORMAT (4541/3)TIME TYPE OF INSTRUMENT NUMBER OF
QUESTIONCONSTRUCT MARKS
1 ½ HOURS
STRUCTURE ITEM OPEN RESPONSE ITEM REPORT ITEM
2 OR 3 COMPULSORY QUESTION 1 & 2 33QUESTION 3 17
QUESTION 1 & 2 : SOME COMMAND WORDS 1. To differentiate between products and observationObservation: describe what you observed Products: names in full , not formula.
2.Examples of observation:- solid or precipitated formed. ( colur must be mentioned)-solid is soluble / insoluble in named reagent-gas released ( gas colour must be stated )-change in colour : state the initial and final colour
incorrect correctclear solution
no reaction seen
hydrogen gas released
movement shown by ammeter
purple colour solution disappear
product of electrolysis at cathode:
brown precipitate seen
litmus paper becomes blue
the solution becomes colourless
SPM CHEMISTRY SEMINAR 2015 10 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!
OPERATIONAL DEFINITION : What you do ….. What you observe .
Example:
Tabulation of data for experiment 17
Set Temperature (°C) Time, t(s) 1/time (s-1)I.
II.III.
SPM CHEMISTRY SEMINAR 2015 11 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!IV.V.
SPM CHEMISTRY SEMINAR 2015 12 PREPARED BY: MS PHUAN
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!
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 otherb) 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
SPM CHEMISTRY SEMINAR 2015 13 PREPARED BY: MS PHUAN
Temperature,oC
Time, s to t1 t2 t3
Boiling point
CoolingHeating
GUIDELINE ON ANSWERING QUESTIONS IN CHEMISTRY PAPER 2 2015
孔子补习学院 CONFUCIUS TUTION CENTRE 1 对 1 补习,全 A+的保证,成绩只有上上上!
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 energyto – 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.
Example: Compare atom 12X and atom 14Y 6 6
Atom X YProton number 6 6No. of electron 6 6Valence electron 4 4Number of neutron / nucleon number 6/ 12 8 /14Chemical properties similarPhysical properties differentNo. 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
- 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.
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6p
6p
X
Y
Hydrogen
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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.
- 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
Calculation:Element / atom Cu O
Mass (g) x yNumber of mole x / 64 y / 16Simplest ratio of mole
Element / atom Mg OMass (g) x yNumber of mole x / 24 y / 16Simplest 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 in3. 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 CuIOxidation number for sodium and copper
+1 +1
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IUPAC Nomenclature Sodium oxide Copper (I) iodideReason 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 171 Change in proton number Increases Increases2 Change in number of electrons and electron filled shells Increases Increases3 Change in atomic size/radius/diameter Increases Increases4 Strength of electrostatic attraction between nucleus and
valence electronDecreases /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 + HOClII Reaction with sodium hydroxide Cl2 + 2NaOH à NaCl + NaOCl + H2OIII Reaction with iron 3Cl2 + 2Fe à 2FeCl3 (brown solid)
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Chlorine gas
Soda lime
Hot iron wool
HCl (cons) +
To
produce Cl2II
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Example: Compare the reactivity of reactions between chlorine and bromine with iron. [Diagram III]
Reaction Reactants ObservationA 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.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, thusd) Nuclei attraction on valence electron increases.e) Atomic radius (atomic size) decreasesf) 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 electrons2. To achieve stable / octet electron arrangement3. 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- )
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III
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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 electrons2. to achieve duplet /octet electron arrangement3. 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
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 electroplatedelectrolyte - 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)
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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 Q4. 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 QElectrical 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
Cu2+ , H+
OH- , SO42-
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 Anode: becomes thinner
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copper
Copper(II)
lead
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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.
ObservationCathode: becomes thicker / brown solid formedElectrolyte: intensity blue solution decrease / blue becomes paler
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 ion3. 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 tileMaterial: 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.
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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
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11. Repeat steps 1-10 twice. Tabulate your reading.
Result :
Titration 1 2 3Final 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 balance3. 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 well8. 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 molarityV1 - initial volumeM2 – final molarityV2 – 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 woodUsed 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
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MAVA = a
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nature.4. Toothpaste contains bases that neutralize the acid produces by bacteria in our mouth
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+ + H2
– acid + base ( metal oxide) à salt + water – acid + alkali à salt + water / H+ + OH- à H2O ( NaOH, KOH, NH4OH)– acid + carbonate metal à salt + CO2 + H2O / 2H+ + CO3
2+ à CO2 + H2OProcedure:
1. pour ( 25 – 100cm3) acid ( 0.5 – 2.0 mol dm-3) into a beaker2. heat slowly3. add solid (metal / base/ carbonate ) a little until excess / no more dissolve4. stir5. filter the mixture into evaporating dish6. heat (slowly) the filtrate until 1/3 from original volume / saturated solution formed7. 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
ObservationChemical equation
Preparation of insoluble salt – precipitation reaction / double decomposition reactionPb2+ + SO4
2- à PbSO4
Example : Preparation of lead(II)sulphate.Procedure
1. pour ( 25 – 50cm3) of soluble salt Pb(NO3)2 into a beaker2. add ( 25 – 50cm3) of soluble salt (Na2SO4)
3. stir4. filter the mixture5. rinse residue / solid / precipitate6. dry between sheets of filter paper
ObservationChemical equationIonic 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)
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Ammonium chloride à ammonia gas + hydrogen chloride gas, (NH4Cl à NH3 + HCl )Confirmatory test for cation and anion
1. State the material / chemical / reagent2. procedure3. 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 tube2. add 1 cm3 copper (II) sulphate solution3. blue precipitate soluble in excess to form dark blue solution.
OR
4. add 2 to 3 drops of Nessler reagent into the test tube5. brown precipitate.6. Ammonium ions (NH4
+) present.
(b) test for anion (Cl-)1. pour 2 cm3 the solution into a test tube2. add 1 cm3 of dilute nitric acid and silver nitrate solution.3. white precipitate formed4. 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 cation1. pour 2 cm3 the solutions into different test tubes2. add 1 cm3 potassium iodide solution into the test tubes3. yellow precipitate formed 4. lead (II) ion present
(d) test for anion1. pour 2 cm3 of lead (II) nitrate solution into a test tube2. add 1 cm3 of dilute sulphuric acid3. add 1 cm3 of iron (II) sulphate solution4. shake the mixure5. 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.
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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
DiscussionThe 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) ions0.0025 : 0.0025 1 : 1
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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+ + Cr2O7
2- 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 Explanation1 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 oleum4 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 detergentTo make explosive material, paint / pigment, polymerAs 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]
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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 à 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 structures6. 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 UsesPolythene Ethene Plastic bags, plastic containerPolypropene Propene Piping, car batteriesPolyvinyl chloride, PVC Chloroethene Artificial leather, water pipePerspex Methylmethacrylate Safety glass, reflectors
ALLOYAn 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
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polymer (polyethene)monomer ( ethene)
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Alloy Composition Properties UsesBronze 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 CuZinc
-harder than copper -in making of musical instruments and kitchenware
Steel IronCarbon
Hard and strong -in construction of buildings and bridges-in building of the body of cars and railway tracks
Stainless steel
IronCarbonChromium
-shiny-strong-does not rust
-in making of cutlery-in making of surgicalinstrument
Duralumin AluminiumCopperMagnesiummanganese
-light-strong
-in building of the body of aeroplane and bullet trains
Pewter Tin, Copperantimony
-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 sizes2. Disrupt the orderly arrangement of copper atoms3. 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 StatementHow 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.
VariablesManipulated : Iron, steel and stainless steel.Responding : intensity / amount of dark blue colour / rate of rusting Fixed : size of nail, concentration of solution, duration of rustingProcedure:
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.
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Copper atom
Stanum atom
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5. Leave the nails for 3 days.6. Observe and record the intensity of the dark blue colour.
Tabulation of data
Paper 2Conclusion 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
PbOCaO
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
Construction of framework for
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Test tube The intensity of the dark blue colour // rate of rusting
ABC
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corrode easily, can withstand higher applied forces and loads, cheaper.
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 partsMRI
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
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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.
Temperature1. As temperature increases, the kinetic energy of the particles ( H+, S2O3
2- ) increases /2. Frequency of collision between particles ( H+, S2O3
2- ) increases3. Frequency of effective collision increases4. 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 collision3. Frequency of collision between particles increases4. Frequency of effective collision increases5. Rate of reaction increases
Concentration of the solution1. The higher the concentration of the solution,2. The greater the number of particles per volume3. Frequency of collision between particles increases4. Frequency of effective collision increases5. 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 increases4. 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
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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)
1 282 353 404 455 50
Discussion Based on plotted graph: [ calculation ]
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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
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 reactionFixed : concentration of H2O2, initial temperature of H2O2 solution.
Apparatus: test tube, 10 cm3 measuring cylinder, test tube rack, spatula.
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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
- ^ H
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Materials: (5-10) – volume of H2O2 solution, manganese (IV) oxide (MnO2) powder, wooden splinter
Procedures:1. label two test tube as A and B2. 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 ObservationA 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
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n-butane
2-methylpropane
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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
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
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H 2SO4, cons
H 3 PO4, 60 atm, 300 oC
CH3COO C2H5
Ethyl ethanoate
CH3COOH
Carboxyl
C2H4
Double bond between C atoms,
C6H12O6
C2H6
C2H5Br
C2H4(OH)2
C2H4Br2
C2H5OH
Esterification
H2SO4
- CH2- CH2-
A
d
d
i
t
i
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
C2H5OH
Hydroxyl
KMnO4/H+ / K2Cr2O7/ H+
Yeast
KMnO4/ H+,
K2Cr2O7/ H+
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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 acidApparatus: Boiling tube / test tube, Bunsen burner, test tube holder, beakerProcedure:1. Pour 2 cm3 of ethanol into a boiling tube / test tube2. Add 1 cm3 of ethanoic acid3. Add 2 to 4 drops of concentrated sulphuric acid 4. Heat the mixture gently for about two minutes5. Pour the mixture into a beaker containing water.
Observation: Sweet/ pleasant / fruity smell // insoluble in water
Chemical equation: CH3COOH + C2H5OH à CH3COO C2H5 + H2O
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4. Dehydration of alcohol Diagram of set up of apparatus1. Complete and functional2. 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 decolourisedOr,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
Industrial extraction of palm oil
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5. Table shows results of latex coagulation
Procedure ObservationPropanoic acid (weak acid) is added to latex Latex coagulates immediatelyLatex is left under natural conditions Latex coagulates slowly
Explain why there is a difference in these observationsAnswer:1. Acid ionizes in water to produce high concentration of / a lot of hydrogen ions2. Hydrogen ions, H+ neutralize the negative charges on the protein membranes 3. The rubber particles collide and the protein membranes break4. Rubber molecules are released and combine with one another and entangle.5. The existence of bacteria in natural conditions6. 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
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Rubber particles--
-
-
-Protein membranes
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2. Ammonia solution contains / ionized to produce hydroxide ions , OH-
3. Hydroxide ions, OH- neutralized the hydrogen ions, H+ / acid produced by the bacteria4. 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 rubberVariable:Manipulated : vulcanised rubber and unvulcanised rubberResponding : length of rubber strip / elasticityFixed : 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.
Unvulcanised rubber Vulcanised rubber
Result / Data
Type of rubber Initial length , cm Length after removal of weight , cmvulcanisedunvulcanised
Compares and contrasts the properties of vulcanized rubber
Vulcanized rubber Elasticity Unvulcanised rubberHarder Hardness Less harderMore elastic Elasticity Less elasticStronger Tensile strength WeakerCan withstand higher temperature Resistance to heat Cannot withstand higher
temperatureLess easily oxidized Resistance to oxidation More easily oxidized
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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 point3 Insoluble in water, soluble in organic solvent4 Cannot conduct electricity5 Density less than water6 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 test1. Combustion, burn less soot flame. (% of carbon per molecule is lower)
1. More soot flame. ( % of carbon per molecule is higher).
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Chemical tests2. add bromine water , brown colour remains3. 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 Cu2. Loss or gain oxygen, C + 2CuO à 2Cu + CO2
3. Loss or gain hydrogen, H2S + Cl2 à 2HCl + S4. 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 agent5. 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,
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Fe2O3.x H2O. // Fe2+ à 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 ironFixed : Iron nails, temperature, medium in which iron nails are kept.
Apparatus : Test tube, test tube rackMaterials : 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 ObservationIntensity 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
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reddish brown solidFe-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+ + 2eO2 + 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+ + 2e5. 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+ + 2eb) 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 Cue) Oxidation number of copper changes / decrease from +2 to 0f) 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.
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silver nitrate
silver nitrate
L nitrate
solutionL MM
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Sample answer:
1. Silver, M and L2. 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 rack4. Record theobservation. 5. Repeat steps 1 to 4 using another halogens and halide solutions.
Tabulation of data:
HalogenHalide solution
Chlorine Bromine Iodine
Potassium chloride X XPotassium bromine / XPotassium iodide / /
Example: Cl2 + 2KI à 2KCl + I2 // Cl2 + 2I- à I2 + 2Cl-
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Cl2 + 2e à 2Cl- ( reduction) 2I- à I2 + 2e (oxidation
3. Transfer of electron at a distance – U-tube
Procedure:1. clamp a U-tube to a retort stand2. pour dilute sulphuric acid3. add solution (oxidizing agent) into one end of the arm of the U-tube4. Add solution (reducing agent) into the other end.5. place / dip carbon electrodes into each arm of the U-tube6. connect the electrodes to a voltmeter/ galvanometer using connecting wire7. leave the apparatus for 30minutes8. 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 + eb. 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 + 2e3. 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
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bromine water potassium iodide solution
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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 / gently4. 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 series1. 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
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K
Na
Ca
Mg
Al
C
Zn
H
Fe
Positions of carbon and hydrogen in the reacting series of metal
Produce oxygen
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2. hydrogen gas with oxide of less reactive metalH2 + PbO à Pb + H2O
3. carbon with oxide metalC + 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 metalCO2 + 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 2COC, CO2 , 2CO reduced the iron oxides to iron
2 Fe2O3 + 3C à 4Fe + 3CO2
Fe2O3 + 3CO à 2Fe + 2CO2
CaCO3 à CaO + CO2 ( lime stone decomposed) CaO + SiO2 à CaSiO3 ( impurities )
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Oxidizing reducing
agent agent
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Redox reaction in various chemical cells
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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 negative2. 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 H2O
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.
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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 gram5.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 gram9.The highest temperature is recorded, θ2
Results:Mass of weight of spirit lamp + butanol /g xFinal mass of spirit lamp + butanol /g yMass of butanol used/g (x-y) // zHighest 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 displacementAim: 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, oCZinc
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Heat of combustion of butanol = a J
(z/74) mol
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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, oCinitial temperature of sodium chloride solution, oCAverage temperature of both solutions, oChighest 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
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ammonium nitrate (NH4NO3) Calcium chloride or
magnesium sulphatesodium acetate crystals
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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.
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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 cm 3 palm oil ( vegetable oil ) into a beaker2. add 50 cm 3 of 5.0 mol dm-3 NaOH / KOH solution3. heat the mixture for (10 minutes)4. stir 5. stop heating, add 50 cm 3 distilled water and solid NaCl6. boil the mixture for 5 minutes7. cool8. filter, wash / rinse9. 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:
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O
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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 water2. 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) cm 3 sea water into a beaker/ suitable container2. Add (10 – 20 ) cm 3 liquid soap into the beaker.3. stir the mixture4. 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.
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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
PreparationSaponification
1. sulphonation2. neutralization
1. alkylation2. sulphonation3. neutralization
The additives in detergent
Type Function ExampleFragrances To add fragrance to both the detergent and
fabricsBiological 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
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CH3 (CH2)15 COO- Na+
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Food additive
Type Function ExamplesPreservatives 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.
MedicineType 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.
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b) antidepressant:to reduce tension and anxiety
c) antipsychotic: to treat psychiatric illness
Barbiturate / tranquilizer
chlorpromazinehaloperidol, clozapine
Overdose can lead to respiratory difficulties, sleeplessness, come, death.
Dizziness, drowsiness, rapid heartbeat.
The existence of Chemicals1. 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
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Some common medical plant and their
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