8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013 http://slidepdf.com/reader/full/skema-pecutan-akhir-chemistry-p2-p3-spm-2013 1/18 SECTIO 1. (a) an (b) Zn (c) copp (d) To e (e) Repe (f) Mass No of m Ratio Empirical 2. (a)(i) (ii) U (iii) W (b) (i) 2. (ii) X + (c)(i) Th (ii) X, Y, (d) (i) 4 (ii) 2X + 3. (a)(i) (ii) Grou (iii) Ato (b)(i) 2. (ii) (iii) Q →(c) P an (d)(i) ion (ii) (e)(i) Co N A hydrous cal 2HCl →r(II) oxide sure that t at the heati X 5.32 l 0.76 1 formula = .1 y have the Z + O 2 →2H 2 O →.4 14 and pe P has 4 val Q 2+ + 2e R. They ha ic bond alent bond ium chlorid ZnCl 2 + H 2 e combusti g, cooling – 4.560 = / 64 = 0.0 XO same 3 she 2X 2 O 2XOH + iod 2 lence electr ve the sam n tube is t nd weighin 0.768 g 12 lls occupied 2 ns and has proton nu tally filled process u O 5.520 – 5 0.192 / 1 1 with electro 2 shells occ ber but dif ith hydroge til a consta .328 = 0.19 = 0.012 ns upied with ferent nucle n gas nt mass is p 2 g lectrons on number. roduced
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013
Zinc is more electropositive than copper in electrochemical series
Zinc becomes negative terminal and copper becomes positive terminal
Negative terminal:
Zinc release electron to form zinc ion. Oxidation occur. Half equation: Zn → Zn2+ + 2e
Observation: Zinc becomes thinner
The electron move from zinc electrode to copper electrode through external circuit /
external wire
Positive terminal:
Copper(II) ion accept electron to form copper atom. Reduction occur.
Half equation: Cu2+ + 2e → Cu
Observation: Copper becomes thicker
15. (a)(i) Lead(II) nitrate: neutralization reaction through the reaction between lead(II) oxide and
nitric acid
Lead(II) sulphate: Double decomposition through the reaction between lead(II) nitrate
solution and sodium sulphate solution
(ii) lead(II) nitrate solution and sodium sulphate solution
(b) Preparation of lead(II) nitrate crystal:
1. Pour 100 cm3 of 1 mol dm-3 nitric acid into a beaker and the solution is heated gently.
2. Add solid lead(II) oxide little by little until in excess.
3. Stir the mixture using glass rod.4. The heating is stopped when lead(II) oxide is no longer dissolve in nitric acid.
5. Filter the mixture
6. The filtrate is heated/evaporated until 1/3 its original volume
7. The saturated lead(II) nitrate solution is cooled to room temperature.
8. The lead(II) nitrate crystals are filtered
9. The lead(II) nitrate crystals are pressed between two filter paper
10. Chemical equation: PbO + 2HNO3 → Pb(NO3)2 + H2O
(c) Test for chloride ion:
1. Pour 2 cm3 sodium chloride solution into a test tube2. Add 2 cm3 dilute hydrochloric acid followed by 2 cm3 silver nitrate solution into the test tube
3. Shake the mixture
4. White precipitate is formed
5. Chloride ion is confirmed present
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013
1. Pour 2 cm3 iron(II) nitrate solution into a test tube
2. Add 2 cm3 of potassium hexacyanoferrate (III) solution into the test tube
3. Shake the mixture
4. Dark blue colouration is formed
5. Iron(II) ion is confirmed present
PAPER 3
16. (a)
Pink colouration Iron does not rust Very high intensity of dark blue colouration Iron rust very quicklyLow intensity of dark blue colouration Iron rust quickly
(b) (i) Type of metal in contact with iron
(ii) Rusting of iron / Intensity of dark blue colouration
(iii) Iron nails
(c) Fe → Fe2+ + 2e
(d) The further the distance of metal from iron in electrochemical series, the iron rust very
quickly
(e)
Metal more electropositive than iron Metal less electropositive than ironZinc Copper
tin
(f) When the iron nail coiled with less electropositive metal is immersed in jelly solutioncontaining phenolphthalein and potassium hexacyanoferrate(III) solutions, dark bluecolouration is formed.
17. (a) Reading in 2 d.p with correct unit e.g 0.80 cm
(b)
Type of metal Diameter of dent (cm) Average diameter of dent(cm)1 2 3
Pure copper Alloy X(c) Alloy X is harder than pure copper // The smaller the diameter of dent, the metal is harder
(d) Bronze // Brass
18. (a)
2.8 V 0.8 V1.4 V 0.4 V
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013
19. (a) How to identify the solubility of salt in water? // How to identify the soluble salt and
insoluble salt when dissolve in water?
(b) Manipulated : Type of salts // Salt A and salt BResponding : Solubility of salt in waterFixed : Type of solvent // distilled water // water(c) When the salt dissolves in water, it is soluble salt whereas when the salt not dissolves inwater, it is insoluble salt(d) Apparatus : Beaker, glass rod, spatula, measuring cylinder.Materials : Salt A, salt B, distilled water
(e) 1. Measure 50 cm3 of distilled water (using measuring cylinder) and pour into a beaker.2. Add one spatula of salt A into the beaker.3. Stir the mixture (using glass rod)4. Record the solubility of salt in water.5. Repeat steps 1 to 4 using salt B.
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013
Aim of experiment To investigate the effect of rusting of iron when in contact with metal
X[Magnesium] and metal Y[copper]
All variablesinvolved
Manipulated variable: Types of metal in contact with iron//Magnesium and
copper
Responding variable: Rusting or iron
Controlled variable: Iron nails// jelly solution
Statement ofhypothesis
When iron in contact with magnesium, iron does not rust /
When iron in contact with copper, the iron rust
List of materialsand apparatus:
Apparatus : Test tube, test tube rack, dropper, glass rodMaterials : Iron nails, sandpaper, magnesium ribbon, copper strip, hot jelly,
potassium hexacyanoferrate (III) and phenolphthalein
Experimentalprocedure:
1. Clean three iron nails, Magnesium ribbon and Copper strip with
sandpaper.
2. Two iron nails are coiled with magnesium and copper each
3. Place the three nails into three different test tubes
4. Add 4 drops of potassium hexacyanoferrate (III) solution followed by 4drops of phenolphthalein into hot jelly solution. Stir the mixture.5. Pour hot jelly solutions into all the test tubes until it covers the entire nail
6. Left the test tube aside for one day7. Any observation are recordedTabulation of data Pairs of metal Intensity of blue
colour
Intensity of pink
colour
Iron Nail
Iron Nail +
magnesium
Iron Nail +copper
21.
Problem statement How does the concentration of ions affect the selective discharge of ions at theanode?
All variables involved Manipulated: concentration of chloride ionsResponding: Ions discharged at anode/product at anodeConstant: carbon electrodes
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013
When the concentration of chloride ions is higher, chloride ions will be selectivelydischarged at the anode.When the concentration of chloride ions is lower, hydroxide ions will be selectivelydischarged at the anode. //When the concentration of chloride ions is higher, the product at anode is chlorinegas.
When the concentration of chloride ions is lower, the product at anode is oxygengas
1. Measure 100 cm3 of 0.001 mol dm-3 hydrochloric acid solution using measuringcylinder and pour into an electrolytic cell.2. Dip two carbon electrodes into the solution. Invert the two test tubes containinghydrochloric acid solution onto both electrodes.3. Connect the electrodes to a switch, ammeter and two dry cells by usingconnecting wires.4. Close the switch and the observation at anode is recorded.
5. Collect the gas in a test tube at anode. Place a glowing wooden splinter into themouth of the test tube. Record the observation.6. Repeat steps 1 to 4 using 1 mol dm-3 hydrochloric acid solution to replace 0.001mol dm-3 hydrochloric acid solution.7. Collect the gas in a test tube at anode. Place a moist blue litmus paper into themouth of the test tube. Record the observation.
Tabulation of data Concentration of hydrochloric acid(mol dm-3)
Observation at Anode
0.0011.0
22.
Aim of experiment To investigate the cleansing action of soap and detergent in hard water
All variables involved Manipulated variable: Soap and detergentResponding variable: The presence of greasy stain on clothFixed variable: Volume of hard water// volume of cleaning agent
Statement ofhypothesis
The cleansing action of a detergent is more effective than soap in hard water.
List of materials andapparatus:
Materials: 5% soap solution, 5% detergent solution, hard water, two small pieces
of cloth with greasy stains.
Apparatus: 100 cm3 beaker, 50 cm3 measuring cylinder, glass rod
Experimentalprocedure:
1. Measure 50 cm3 of hard water using measuring cylinder and pour into a beaker.2. Add 50 cm3 of soap solution into the beaker. Stir the mixture.
3. Immerse a small piece of cloth with greasy stains into the mixture.4. Rub the cloth gently.5. Record the presence of greasy stain on the cloth.6. Repeat steps 1 to 5 using detergent solution.
Tabulation of data Type of cleaning agent ObservationSoap + hard waterDetergent + hard water
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013
Problem statement How to compare the electrical conductivity between molten naphthalene andmolten lead(II) bromide?
All variables involved Manipulated: naphthalene and lead(II) bromide // Types of compoundResponding: Reading of ammeter/electrical conductivityConstant: mass of compound// carbon electrodes
Materials: naphthalene and lead(II) bromide Apparatus: crucible, carbon electrodes, two dry cells, connecting wire, ammeter,switch, Bunsen burner, electronic balance
Experimentalprocedure:
1. Measure 50 g of naphthalene and pour into a crucible.2. Dip two carbon electrodes into the naphthalene.3. Connect the electrodes to a switch, two dry cells and ammeter usingconnecting wires.4. Heat the naphthalene until its melt.5. Close the switch and the observation is recorded.6. Repeat steps 1 to 5 using lead(II) bromide to replace naphthalene.
Tabulation of data Type of compound ObservationNaphthalene moltenLead(II) bromide molten