HEBAT SPM 2019 - CHEMISTRY

Practice Rate of Reaction
1 Kadar Tindak Balas 1






Practice Carbon Compounds
2 Sebatian Karbon 15






Practice Oxidation and Reduction
3 Pengoksidaan dan Penurunan 27






Practice Thermochemistry
4 Termokimia 38







Practice Chemicals for Consumers
5 Bahan Kimia untuk Pengguna
50





Written Practical 58




Forecast Paper 64




Answers A1














00 CONTENT.indd 1 14/02/2019 5:06 PM

PRACTICE Carbon Compounds
2 Sebatian Karbon





A nalysis Paper ’11 ’12 ’13 ’14 ’15 ’16 ’17 ’18

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Objective Questions PAPER 1 SPM 2 3        
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Carbon Compounds formula molekul bagi hidrokarbon B Ethane undergoes
2.1 Sebatian Karbon itu? addition polymerisation
[Jisim atom relatif: C = 12, H = 1, reaction.
O = 16]
1. Which of the following is an Etana mengalami tindak balas
unsaturated hydrocarbon? A C 3 H 4 pempolimeran penambahan.
Antara yang berikut, yang B C 2 H 6 C Ethane undergoes
manakah merupakan hidrokarbon C C 3 H 8 substitution reaction
tak tepu? when exposed to the
A H H H H D C 4 H 8 sunlight.
H H H
H H H
H H H
H H H
3
H C C C C H 3. 10 cm of hydrocarbon X was Etana mengalami tindak
H H H C C C C C C C C C C C C H H H balas penukargantian
completely burnt in oxygen. apabila terdedah kepada
H H The total volume of the cahaya matahari.
H H H
H H H
3
B H products was 30 cm . Which D Ethane undergoes
H H H
of the following chemical
dehydration when it
H C H equations represents the is passed through hot
H H H C C C H H H
combustion of hydrocarbon X? porcelain chips.
H
H H H 10 cm hidrokarbon X dibakar Etana mengalami
3
dengan lengkap dalam oksigen. pendehidratan apabila
C H H H Jumlah isi padu hasil tindak melalui serpihan porselin
H H H
H H H
H H H
balas ialah 30 cm . Persamaan panas.
3
H C C C H
H H H C C C C C C C C C H H H kimia yang manakah mewakili
pembakaran hidrokarbon X? 5. The following chemical
H H
H H H H H H equation represents the
H C H A CH 4 (g) + 2O 2 (g) →
H H H
H H H
C C C
CO 2 (g) + 2H 2 O(g) reaction between butane and
H oxygen.
H H H
B C 2 H 4 (g) + 3O 2 (g) → Persamaan kimia yang berikut
D H H H H 2CO 2 (g) + 2H 2 O(g) mewakili tindak balas antara
H H H H H H H H H H H H butana dengan oksigen.
C C 3 H 8 (g) + 5O 2 (g) →
H C C C C H
H H H C C C C C C C C C C C C H H H 3CO 2 (g) + 4H 2 O(g)
H H H OH D 2C 2 H 6 (g) + 7O 2 (g) → 2C 4H 10(g) + 13O 2(g) →
8CO 2 (g) + 10H 2 O(l)
H H H
OH
OH
H H H
H H H
OH
4CO 2 (g) + 6H 2 O(g) 2C 4 H 10 (g) + 13O 2 (g) →
2. A hydrocarbon is burnt 8CO 2 (g) + 10H 2 O(ce)
completely in air to form Alkanes
13.2 g of carbon dioxide gas 2.2 Alkana
and 7.2 g of water. What is What is the minimum
the molecular formula of the 4. Which statement is correct volume of oxygen required
hydrocarbon? about ethane? at room temperature and
[Relative atomic mass: Pernyataan yang manakah betul pressure for the complete
C = 12, H = 1, O = 16] tentang etana? combustion of 1 mol of
Satu hidrokarbon dibakar dengan A Ethane is the smallest butane?
lengkap dalam udara untuk member of alkane. [1 mol of gas occupies the
3
membentuk 13.2 g gas karbon Etana merupakan ahli volume of 24 dm at room
dioksida dan 7.2 g air. Apakah terkecil bagi alkana. temperature and pressure]
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Chemistry Form 5 Practice 2 Carbon Compounds
Berapakah isi padu minimum Alkenes D Both burn completely in
oksigen yang diperlukan pada 2.3 Alkena the air to produce carbon
suhu dan tekanan bilik bagi dioxide and water.
pembakaran lengkap 1 mol Kedua-duanya terbakar
butana? 7. What is the general formula dengan lengkap dalam
[1 mol gas menempati isi padu of alkenes? udara menghasilkan karbon
24 dm pada suhu dan tekanan Apakah formula am bagi alkena? dioksida dan air.
3
bilik] A C n H 2n
A 78 dm 10. The following reactions
3
B 156 dm B C n H 2n + 2 involve propene.
3
C C n H 2n + 1 OH
C 312 dm Tindak balas yang berikut
3
D 360 dm 3 D C n H 2n + 1 COOH melibatkan propena.
8. The following is the equation
6. The following chemical that represents the reaction H H H H H H
equation shows the reaction between pentene and
between hydrocarbon X and chlorine. H C C C H H C C C H
chlorine in the presence of Persamaan yang berikut mewakili H H H Process X H
sunlight. tindak balas antara pentena Proses X
Persamaan kimia yang berikut dengan klorin. Process Y
menunjukkan tindak balas Proses Y
antara hidrokarbon X dengan Pentene + Cl 2 → Q H H H
klorin dalam kehadiran cahaya Pentena + Cl 2 → Q
matahari. H C C C H
Which of the following is the H Cl H
H H H
| | | name of Q?
Hydrocarbon X + Cl 2 → H – C – C – C – H + HCl Antara nama yang berikut, yang
Hidrokarbon X | | | manakah Q? What substances can be used
H Cl H as the reagents in Process X
A 2-chloropentane and Process Y?
Which of the following is 2-kloropentana Apakah bahan yang boleh
hydrocarbon X ? B 1,2-dichloropentane digunakan dalam Proses X dan
Antara yang berikut, yang 1,2-dikloropentana Proses Y?
manakah hidrokarbon X? C 2,2-dichloropentane Process X Process Y
A 2,2-dikloropentana Proses X Proses Y
H H H D 2,3-dichloropentane
2,3-dikloropentana
H C C C H A Hydrogen Steam
H H H 9. Which of the following is chloride Wap air
Hidrogen
true about both ethane and klorida
B ethene?
H
H
H
Antara yang berikut, yang B Steam Acidified
manakah benar tentang kedua-
H C C C H Wap air potassium
dua etana dan etena?
A Both are unsaturated (VII)
H manganate
hydrocarbons. Kalium
C H H H H Kedua-duanya adalah manganat(VII)
hidrokarbon tak tepu. berasid
H C C C C H B Both can undergo
polymerisation. C Hydrogen Hydrogen
H H H H
Kedua-duanya boleh Hidrogen chloride
mengalami pempolimeran. Hidrogen
D H H H H C Both can decolourise the klorida
colour of bromine water.
H C C C C H Kedua-duanya boleh D Hydrogen Chlorine
melunturkan warna air bromin. Hidrogen Klorin
H H



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Chemistry Form 5 Practice 2 Carbon Compounds
11. Butane is formed from the 16. Diagram 2 shows the apparatus
reaction of Compound X H H H H H set-up for a reaction.
with hydrogen at 180°C in CH 3 C C C C C H Rajah 2 menunjukkan susunan
the presence of platinum as a radas bagi satu tindak balas.
catalyst. CH 3 CH 3 H CH 3 H Porcelain chips
What is X? Serpihan porselin
Butana terhasil daripada tindak What is the IUPAC name of Boiling tube
balas antara sebatian X dengan the organic compound? Tabung didih Gas
hidrogen pada suhu 180°C Apakah nama IUPAC bagi Gas
dengan kehadiran platinum sebatian organik itu?
sebagai mangkin. A 2,2-dimethyl-3- Glass wool
Heat
Apakah X? propylpentane soaked in Panaskan
A C 4 H 8 2,2-dimetil-3-propilpentana ethanol
Kapas kaca
B C 4 H 9 OH B 3-propyl-2,2- direndam Water
C C 3 H 7 COOH dimethylpentane dalam etanol Air
D C 4 H 9 COOH 3-propil-2, 2-dimetilpentana
C 2, 3, 5, 5-tetramethylpentane Diagram 2 / Rajah 2
2, 3, 5, 5-tetrametilpentana Which of the following is the
Isomerism
2.4 Keisomeran D 2, 3, 5-trimethyhexane property of the gas formed?
2, 3, 5-trimetilheksana
Antara yang berikut, yang manakah
adalah sifat gas yang terbentuk?
12. Diagram 1 shows a Alcohols A Has the smell of vinegar.
conversion of butanol. 2.5 Alkohol Mempunyai bau cuka.
Rajah 1 menunjukkan satu siri B Burns with a non-smoky
perubahan butanol.
14. What is the functional group blue flame.
Dehydration of alcohols? Terbakar dengan nyalaan
Pendehidratan Compound X
C 4 H 9 OH Apakah kumpulan berfungsi biru tidak berasap.
Sebatian X
alkohol? C Decolourises
Diagram 1 / Rajah 1 A Hydroxyl group acidified potassium
Kumpulan hidroksil manganate(VII) solution.
Which of the following is the B Carbon-carbon double Melunturkan warna kalium
isomer of Compound X? bond manganat(VII) berasid.
Antara yang berikut, yang Ikatan ganda dua karbon- D Turns red litmus paper
manakah isomer bagi Sebatian X? karbon blue.
I But-1-ene C Carboxyl group Menukarkan kertas litmus
But-1-ena Kumpulan karboksil merah kepada biru.
II 2-methylprop-1-ene D Hydroxide ion
2-metilprop-1-ena Ion hidroksida 17. The following flow chart
III 2-methylpropane shows the conversion of
2-metilpropana 15. A mixture of grape juice ethanol to ethene and ethanol
IV 2,2-dimethylbutane and yeast is kept at 35°C to ethanoic acid.
2,2-dimetilbutana for three days. What are the Carta alir yang berikut
A I and II products formed? menunjukkan pertukaran etanol
I dan II Satu campuran jus anggur dan kepada etena dan etanol kepada
B I and IV yis disimpan pada 35°C selama asid etanoik.
I dan IV tiga hari. Apakah hasil yang
C II and III terbentuk? Process Process
I
II
II dan III A Ethanol and water Ethene ← Ethanol → Ethanoic
D III and IV Etanol dan air Proses Proses acid
III dan IV B Ethanoic acid and oxygen I II
Asid etanoik dan oksigen Etena ← Etanol → Asid
13. The following is the C Ethanol and carbon etanoik
structural formula of an dioxide What are Process I and
organic compound. Etanol dan karbon dioksida Process II?
Berikut adalah formula struktur D Starch and carbon dioxide Apakah Proses I dan Proses II?
bagi satu sebatian organik.
Kanji dan karbon dioksida


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02[Hebat Chem F5].indd 17 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
19. What is the structural
Process I Process II Esters
Proses I Proses II formula of the organic 2.7 Ester
compound formed when
A Oxidation Esterification ethanol reacts with propanoic 20. What is the catalyst used
Pengoksidaan Pengesteran acid?
Apakah formula struktur bagi in the preparation of ethyl
B Dehydration Oxidation sebatian organik yang terbentuk ethanoate from ethanol and
Pendehidratan Pengoksidaan apabila etanol bertindak balas ethanoic acid?
dengan asid propanoik? Apakah mangkin yang digunakan
C Fermentation Dehydration A O dalam penyediaan etil etanoat
Penapaian Pendehidratan daripada etanol dan asid etanoik?
CH 3 CH 2 C O CH 3 A Yeast
D Esterification Fermentation Yis
Pengesteran Penapaian B O B Nickel
CH 3 C O CH 2 CH 2 CH 3 Nikel
C Manganese(IV) oxide
C
Carboxylic Acids O Mangan(IV) oksida
2.6 Asid Karboksilik CH 3 CH 2 C O CH 2 CH 3 D Concentrated sulphuric
D acid
18. Organic compound X reacts O Asid sulfurik pekat
separately with sodium, CH CH CH C O CH CH
sodium hydroxide and 3 2 2 2 3
sodium carbonate. What is
compound X? 21. Which structural formula does not correspond to the name of the
Sebatian organik X bertindak ester?
balas secara berasingan dengan Formula struktur yang manakah tidak sepadan dengan nama ester?
natrium, natrium hidroksida
dan natrium karbonat. Apakah Structural formula Compound name
sebatian X? Formula struktur Nama sebatian
A H H H H
A O H H Ethyl methanoate
H C C C C H ' & & Etil metanoat
H!C!O!C!C!H
H H &
H
B H H H H
B H H O H H Propyl propanoate
& & ' & &
H C C C C H Propil propanoat
H!C!C!C!O!C!C!H
O H H H & & & &
H H H H
H
C H O H H H Propyl ethanoate
& ' & & & Propil etanoat
C O H!C!C!O!C!C!C!H
& & & &
CH C O CH CH CH
3 2 2 3 H H H H
D H O H Methyl ethanoate
D & ' & Metil etanoat
O H!C!C!O!C!H
& &
CH 3 CH 2 CH 2 C O H H H











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Chemistry Form 5 Practice 2 Carbon Compounds
22. Diagram 3 shows the Fats C Carbohydrate
changes of carbon 2.8 Lemak Karbohidrat
compounds involving a series D Protein
of reactants. 23. What is the process that Protein
Rajah 3 menunjukkan penukaran converts unsaturated fat to
sebatian karbon yang melibatkan saturated fat? Natural Rubber
beberapa siri tindak balas. 2.9 Getah Semula Jadi
Apakah proses yang menukarkan
Substance A KMnO 4 /H + Butanoic acid lemak tak tepu kepada lemak tepu?
Bahan A Asid butanoik A Fermentation 25. Which of the following is
Penapaian the property of rubber after
B Hydrogenation undergoing vulcanisation?
Esterification Penghidrogenan Antara yang berikut, yang
Pengesteran C Hydrolysis manakah sifat getah selepas
Substance O Hidrolisis mengalami pemvulkanan?
Bahan O D Polymerisation A Less resistant to heat
Diagram 3 / Rajah 3 Pempolimeran Kurang rintangan terhadap
haba
24. The following shows the
Name substances A and O. formula of a compound. B Molecules of rubber slide
Namakan bahan A dan O. Berikut menunjukkan formula more easily over one
another
Substance A Substance O bagi satu sebatian. Molekul getah menggelongsor
Bahan A Bahan O
C H 31 COO CH 2 lebih mudah antara satu
15
A Butene Butanol sama lain
Butena Butanol C H COO CH C Can return to its original
15 31
B Butanol Butene shape after it is stretched
Butanol Butena C H 31 COO CH 2 Boleh kembali kepada bentuk
15
asal apabila diregangkan
C Butyl What is the compound? D The melting point decreases
Butene butanoate
Butena Apakah sebatian itu? Takat lebur berkurang
Butil butanoat A Saturated fat
D Butyl Lemak tepu
Butanol B Unsaturated fat
Butanol butanoate
Butil butanoat Lemak tak tepu

Subjective Questions PAPER 2

SECTION A (a) (i) State the name of Process I.
1. Diagram 1 shows the reactions of but-2-ene, Nyatakan nama Proses I.
C 4 H 8 .
Rajah 1 menunjukkan tindak balas but-2-ena, C 4 H 8 . [1 mark / 1 markah]
(ii) Describe briefly how Process I can be
C 4H 10
carried out.
Process I Huraikan dengan ringkas bagaimana Proses
Proses I I boleh dijalankan.
Bromine in
tetrachloromethane
Compound X Bromin dalam
Sebatian X tetraklorometana
But-2-ene Compound Y [2 marks / 2 markah]
C 4 H 8 (OH) 2
But-2-ena Sebatian Y
(iii) Write the chemical equation for
Diagram 1 / Rajah 1 Process I.
Tuliskan persamaan kimia bagi Proses I.

[1 mark / 1 markah]


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02[Hebat Chem F5].indd 19 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
(b) (i) Suggest a suitable chemical for (a) Name the homologous series for:
Compound X. Namakan siri homolog bagi:
Cadangkan bahan kimia yang sesuai bagi Compound A :
Sebatian X. Sebatian A
Compound B :
[1 mark / 1 markah] Sebatian B
(ii) Draw the structural formula of [2 marks / 2 markah]
C 4 H 8 (OH) 2 . (b) Compound A reacts with water to produce
Lukis formula struktur C 4 H 8 (OH) 2 . Compound B.
Sebatian A bertindak balas dengan air
menghasilkan Sebatian B.
(i) Write the chemical equation for the
reaction.
[1 mark / 1 markah] Tuliskan persamaan kimia bagi tindak balas
(c) (i) State the observation that occurs in tersebut.
the reaction between but-2-ene and
bromine in tetrachloromethane.
Nyatakan pemerhatian yang berlaku dalam [1 mark / 1 markah]
tindak balas antara but-2-ena dengan (ii) State the temperature and pressure
bromin dalam tetraklorometana. required.
Nyatakan suhu dan tekanan yang
diperlukan.
[1 mark / 1 markah]
(ii) State the name of Compound Y. Temperature / Suhu:
Nyatakan nama Sebatian Y.
Pressure / Tekanan:
[1 mark / 1 markah]
(d) Draw the structural formulae of two [2 marks / 2 markah]
isomers of but-2-ene. (c) Compound B undergoes dehydration
Lukis formula struktur bagi dua isomer process to form Compound A.
but-2-ena.
Draw the set-up of apparatus for the
reaction.
Sebatian B mengalami proses pendehidratan
untuk membentuk Sebatian A.
Lukis susunan alat radas bagi tindak balas
tersebut.

[2 marks / 2 markah]

2. Table 2 shows the molecular formulae of three
compounds.
Jadual 2 menunjukkan formula molekul bagi tiga
sebatian.

Compound Molecular [2 marks / 2 markah]
formula
Sebatian (d) Compound B reacts with Compound C to
Formula molekul produce a compound with a sweet smell.
Sebatian B bertindak balas dengan Sebatian C
A C 3 H 6 untuk menghasilkan satu sebatian yang berbau
harum.
B C 3 H 7 OH (i) Name the product formed.

C C 2 H 5 COOH Namakan hasil yang terbentuk.

Table 2 / Jadual 2 [1 mark / 1 markah]


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02[Hebat Chem F5].indd 20 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
(ii) Draw the structural formula of the (vi) Hydrocarbon X reacts with bromine in
compound formed. tetrachloromethane to form a compound
Lukis formula struktur bagi sebatian yang with the structural formula shown below.
terbentuk. Hidrokarbon X bertindak balas dengan
bromin dalam tetraklorometana untuk
membentuk satu sebatian dengan formula
struktur di bawah.

CH 3 CH CH CH 3
Br Br
[1 mark / 1 markah] Based on your answer in (a)(v),
identify the possible hydrocarbon X.
SECTION B Berdasarkan jawapan anda di (a)(v), kenal
pasti hidrokarbon X yang mungkin.
3. (a) When 0.02 mol of hydrocarbon X was [1 mark / 1 markah]
completely burnt in oxygen, 1920 cm (b) (i) Explain the natural coagulation of latex.
3
of carbon dioxide was formed at room Terangkan penggumpalan lateks secara
conditions. Hydrocarbon X decolourises the semula jadi.
bromine in tetrachloromethane. [3 marks / 3 markah]
Apabila 0.02 mol hidrokarbon X terbakar (ii) State the name of one chemical that
dengan lengkap dalam oksigen, 1920 cm can be used to prevent the coagulation
3
karbon dioksida terbentuk pada suhu bilik. of latex. Explain how the chemical
Hidrokarbon X melunturkan warna bromin dalam prevents the coagulation of latex.
tetraklorometana. Nyatakan satu bahan kimia yang
(i) Calculate the number of carbon atoms boleh digunakan untuk menghalang
in one molecule of hydrocarbon X. penggumpalan lateks. Terangkan
[1 mol of gas occupies the volume bagaimana bahan kimia itu menghalang
of 24 dm at room temperature and penggumpalan lateks.
3
pressure] [3 marks / 3 markah]
Hitung bilangan atom karbon dalam satu
molekul hidrokarbon X. SECTION C
[1 mol gas menempati isi padu 24 dm pada 4. (a) (i) State the name of one alkane and one
3
suhu dan tekanan bilik] alkene with the same number of carbon
[4 marks / 4 markah] atoms. Write their molecular formulae.
(ii) To which homologous series does X Nyatakan nama satu alkana dan satu
belong? alkena dengan bilangan atom karbon yang
Dalam kumpulan siri homolog manakah X sama. Tuliskan formula molekulnya.
berada? [1 mark / 1 markah] [4 marks / 4 markah]
(iii) Suggest the molecular formula of (ii) Based on the answer in (a)(i), explain
hydrocarbon X. the statements below.
Cadangkan formula molekul bagi hidrokarbon Berdasarkan jawapan anda di (a)(i),
X. [1 mark / 1 markah] terangkan pernyataan di bawah.
(iv) Write the chemical equation for the I: Alkenes burn with sootier flames as
complete combustion of hydrocarbon compared to alkanes.
X in oxygen. [Relative atomic mass: C = 12, H = 1]
Tuliskan persamaan kimia bagi pembakaran I: Alkena terbakar dengan lebih banyak
lengkap hidrokarbon X dalam oksigen. jelaga berbanding alkana.
[1 mark / 1 markah] [Jisim atom relatif: C = 12, H = 1]
(v) Draw all the possible structural [3 marks / 3 markah]
formulae of the isomers of II: Alkanes undergo substitution
hydrocarbon X. State the name for all reaction while alkenes undergo
the isomers. addition reaction.
Lukis semua formula molekul yang mungkin II: Alkana mengalami tindak balas
bagi isomer hidrokarbon X. Nyatakan nama penukargantian manakala alkena mengalami
bagi semua isomer tersebut. tindak balas penambahan.
[6 marks / 6 markah] [6 marks / 6 markah]


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02[Hebat Chem F5].indd 21 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
(b) Diagram 4 shows the conversions of Nyatakan nama satu alkohol yang
alcohol to alkene and alcohol to carboxylic mempunyai lebih daripada satu atom
acid. karbon. Lukis formula strukturnya.
Rajah 4 menunjukkan penukaran alkohol kepada [2 marks / 2 markah]
alkena dan alkohol kepada asid karboksilik. (ii) Draw a flow chart to show how the
alcohol mentioned in (b)(i) is converted
Alkene Alcohol Carboxylic acid to an alkene and a carboxylic acid.
Alkena Alkohol Asid karboksilik Lukis carta alir untuk menunjukkan
bagaimana alkohol yang dinyatakan dalam
Diagram 4 / Rajah 4 (b)(i) ditukarkan kepada alkena dan asid
(i) State the name of one alcohol that has karboksilik. [4 marks / 4 markah]
more than one carbon atom. Draw its (iii) State one use of alcohol.
structural formula. Nyatakan satu kegunaan alkohol.
[1 mark / 1 markah]







OBJECTIVE QUESTIONS A I and II Rubber A Rubber B
I dan II Getah A Getah B
1. Diagram 1 shows a process B I and IV A Less elastic More
of preparing margarine from I dan IV
SPM elastic
CLONE Kurang
2013 palm oil through Process X. C II and III kenyal Lebih kenyal
Rajah 1 menunjukkan proses II dan III
menyediakan marjerin daripada D II and IV B Stronger Weaker and
minyak kelapa sawit melalui II dan IV and harder softer
Proses X. Kuat dan Lemah dan
Process X 3. Which substance is used to keras lembut
Palm oil Proses X Margarine SPM differentiate hexane and hexene? C High Low
Minyak sawit Marjerin CLONE Bahan yang manakah digunakan melting melting
2015
untuk membezakan heksana dan point point
Diagram 1 / Rajah 1 heksena? Takat lebur Takat lebur
What is process X? A Limewater tinggi rendah
Apakah proses X? Air kapur D Difficult to Easy to
A Halogenation B Bromine water oxidise oxidise
Penghalogenan Air bromin Sukar Mudah
B Esterification C Hydrochloric acid teroksida teroksida
Pengesteran Asid hidroklorik
C Hydrogenation D Ammonia 5. What is the product formed
Penghidrogenan Ammonia SPM when hexene reacts with
D Oxidation CLONE steam at 300°C and 60 atm
2015
Pengoksidaan 4. Which of the following is the in the presence of phosphoric
2. What are the products formed SPM correct match of properties of acid as a catalyst?
CLONE
Rubber A and Rubber B?
2015
SPM when butane is burnt in Apakah hasil yang terbentuk
CLONE Antara yang berikut, yang apabila heksena bertindak balas
2013 excess oxygen? manakah padanan sifat yang dengan stim pada 300°C dan
Apakah hasil yang terbentuk betul bagi Getah A dan Getah B? 60 atm dalam kehadiran asid
apabila butana dibakar dalam fosforik sebagai mangkin?
oksigen berlebihan?
I Carbon dioxide c c c c A Hexane
Karbon dioksida c = c c = c c = c c = c s s s s Heksana
s s
s s
II Carbon monoxide c = c c = c c = c c = c c = c c c c c c c B Hexanol
s s
c = c
Karbon monoksida s s Heksanol
III Carbon c = c c = c c = c c c c c c c C Hexanoic acid
c = c
Karbon A B Asid heksanoik
IV Water D Hexyl hexanoate
Air Heksil heksanoat
© Penerbitan Pelangi Sdn. Bhd. 22




02[Hebat Chem F5].indd 22 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
6. Latex is a milk-like liquid C Add hydrochloric acid to What is X?
SPM obtained from a tapped latex Butanol bertindak balas dengan
CLONE
2015 rubber tree. Latex coagulates Menambahkan asid larutan kalium manganat(VII)
easily due to the bacterial hidroklorik pada lateks berasid yang bertindak sebagai
action on latex. What is the D Add zinc chloride to agen pengoksidaan untuk
most suitable way to prevent latex membentuk Sebatian X.
coagulation of latex? Menambahkan zink klorida Apakah X?
Lateks ialah cecair seperti susu pada lateks A C 3 H 6
yang diperoleh daripada pokok B C 4 H 8
getah yang ditoreh. Lateks 7. 1 mol of alkene is burnt in C C 3 H 7 COOH
senang tergumpal hasil daripada SPM excess oxygen. Which alkene D C 4 H 9 COOH
tindakan bakteria. Apakah CLONE produces carbon dioxide and
2016
cara yang paling sesuai untuk water in a mol ratio of 2:2? 11. Diagram 2 shows the
menghalang lateks daripada 1 mol alkena terbakar dalam SPM structural formula of an ester.
menggumpal? oksigen berlebihan. Alkena yang CLONE Rajah 2 menunjukkan formula
2017
A Add ammonia solution to manakah menghasilkan karbon struktur bagi suatu ester.
latex dioksida dan air dalam nisbah
Menambahkan larutan mol 2:2? H O H H H
ammonia pada lateks A Ethene C Butane
B Add ethanoic acid to Etena Butena H C C O C C C H
latex B Propene D Pentene H H H H
Menambahkan asid etanoik Propena Pentena
pada lateks Diagram 2 / Rajah 2
8. The following information shows the properties of organic Name the alcohol that reacts
with ethanoic acid to produce
SPM compound X.
CLONE the ester?
2016 Maklumat berikut menunjukkan sifat-sifat bahan organik X.
Namakan alkohol yang bertindak
• Has a carboxyl group. balas dengan asid etanoik untuk
Mempunyai kumpulan karboksil. menghasilkan ester itu?
• Releases a gas which turns limewater chalky when it is added A Ethanol C Butanol
with calcium carbonate. Etanol Butanol
Membebaskan gas yang mengeruhkan air kapur apabila dicampurkan B Propanol D Pentanol
dengan kalsium karbonat. Propanol Pentanol
• Produces a substance which has a sweet smell when it is
reacted with butanol. 12. Diagram 3 shows a series of
Menghasilkan bahan yang berbau wangi apabila bertindak balas dengan SPM conversion of a compound.
CLONE
butanol. 2017 Rajah 3 menunjukkan siri
perubahan bagi suatu sebatian.
Which substance is X? Compound A Water Carbon dioxide gas and water
Bahan yang manakah adalah X?. C 3 H 7 OH + 2[O] Sebatian A + Air Gas karbon dioksida dan air
A Ethane
Etana What is Compound A? Polymerisation Combustion
B Ethene Apakah Sebatian A? Pempolimeran Pembakaran
Etena A Propane Polyethene Compound X Ethanol
C Ethanol Propane Polietena Sebatian X Etanol
Etanol B Propene Diagram 3 / Rajah 3
D Ethanoic acid Propena
Asid etanoik C Propanoic acid What is the general formula
Asid propanoik of Compound X?
9. The following equation D Butyl propanoate Apakah formula am bagi
SPM represents the conversion of Butyl propanoat Sebatian X?
CLONE
2017 propanol to Compound A A C n H 2n
through an oxidation process. 10. Butanol reacts with acidified B C n H 2n+2
Persamaan berikut mewakili SPM potassium manganate(VII) C C nH 2n+1 OH
CLONE
tindak balas penukaran propanol 2017 solution which acts as an
kepada Sebatian A melalui oxidising agent to form D C n H 2n+1 COOH
proses pengoksidaan. Compound X.

23 © Penerbitan Pelangi Sdn. Bhd.






02[Hebat Chem F5].indd 23 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
13. Used in the preparation of Formula struktur yang manakah C H H
SPM cosmetics and perfumes. mempunyai ciri yang sama
CLONE seperti pernyataan di atas?
2017 H C C H
Digunakan dalam penyediaan
barang kosmetik dan minyak A H H H OH
wangi.
H C C H D
Which of the following H O H H
structural formula has the B H O H C C O C C H
same characteristic as the H C C OH
above statement? H H H
H
(c) Draw one isomer of Alcohol A.
Subjective Questions PAPER 2 Lukis satu isomer bagi Alkohol A.
SECTION A
1. Diagram 1.1 shows the combustion of Alcohol A. [1 mark / 1 markah]
SPM Alcohol A contains 4 carbon atoms per unit molecule.
CLONE (d) Diagram 1.2 shows the set-up of apparatus
2017 Rajah 1.1 menunjukkan pembakaran Alkohol A.
Alkohol A mempunyai 4 atom karbon per unit molekul. to preparean ester in the laboratory.
Rajah 1.2 menunjukkan susunan radas untuk
menyediakan ester di dalam makmal.


Filter funnel Concentrated
Corong turas sulphuric acid O
Limewater Asid sulfurik pekat
Spirit lamp Air kapur H C O C C C C H
Lampu pelita Alcohol A +
Alcohol A Carboxylic acid B
Alkohol A Alkohol A +
Asid karboksilik B
Diagram 1.1 / Rajah 1.1 Heat
Panaskan
(a) (i) State the general formula of alcohol.
Nyatakan formula am bagi alkohol Diagram 1.2 / Rajah 1.2
(i) By referring to the structural
[1 mark / 1 markah] formula of the ester formed, identify
Carboxylic acid B.
(ii) State the molecular formula of Alcohol A. Dengan merujuk kepada formula struktur
Nyatakan formula molekul bagi Alkohol A. ester yang terbentuk, kenal pasti Asid
karboksilik B.
[1 mark / 1 markah]
(iii) State two products of the combustion [1 mark / 1 markah]
of Alcohol A. (ii) Write a balanced chemical equation
Nyatakan dua hasil tindak balas for the esterification of Alcohol A and
pembakaran Alkohol A. Carboxylic acid B.
Tuliskan persamaan kimia seimbang bagi
pengesteran antara Alkohol A dengan Asid
karboksilik B.
[2 marks / 2 markah]
(b) Combustion of Alcohol A releases a gas. [2 marks / 2 markah]
Describe a chemical test to identify the gas.
Pembakaran Alkohol A membebaskan sejenis gas. (iii) State the observation when the ester
Huraikan satu ujian kimia untuk mengenalpasti formed is poured into a basin of water.
gas tersebut. Nyatakan pemerhatian apabila ester yang
terhasil dituang ke dalam satu besen air.

[1 mark / 1 markah]
[2 marks / 2 markah]
© Penerbitan Pelangi Sdn. Bhd. 24






02[Hebat Chem F5].indd 24 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds

SECTION B
2. Table 2 shows the information of four organic compounds A, B, C, and D.
Jadual 2 menunjukkan maklumat bagi empat sebatian organik A, B, C, dan D.
SPM
CLONE
2015
Organic Type of carbon Number of
compound compound carbon atoms Specific reaction
Sebatian Jenis sebatian Bilangan Tindak balas khusus
organik organik karbon atom
A Hydrocarbon 6 Decolourises acidified potassium manganate(VII) solution
Hidrokarbon Menyahwarnakan larutan kalium manganat(VII) berasid
B Hydrocarbon 6 Does not decolourise acidified potassium manganate(VII)
Hidrokarbon solution
Tidak menyahwarnakan larutan kalium manganat(VII) berasid
C Non- 3 Reacts with acidified potassium manganate(VII) solution
hydrocarbon to produce an acidic solution
Bukan hidrokarbon Bertindak balas dengan larutan kalium manganat(VII) berasid
untuk menghasilkan larutan berasid
D Non- 3 Reacts with magnesium to produce salt and hydrogen gas
hydrocarbon Bertindak balas dengan magnesium untuk menghasilkan garam
Bukan hidrokarbon dan gas hidrogen
Table 2 / Jadual 2
(a) Based on the information in Table 2: (ii) State four physical properties of the
Berdasarkan maklumat dalam Jadual 2: compound formed.
(i) State the name of each of the compounds. Nyatakan empat ciri fizik bagi sebatian yang
Nyatakan nama bagi setiap sebatian itu. terhasil.
(ii) Draw the structural formula for each of [6 marks / 6 markah]
the compounds.
Lukis formula struktur bagi setiap sebatian SECTION C
itu. 3. (a) The following information is about carbon
(iii) State the functional group for compound X.
compounds C and D. Maklumat yang berikut adalah tentang sebatian
Nyatakan kumpulan berfungsi bagi sebatian karbon X.
C dan D.
[10 marks / 10 markah] • Empirical formula = C 2 H 4 O
(b) Compound A and B both burn to produce Formula empirik = C 2 H 4 O
sooty flame. • Relative molecular mass = 88
Determine which compound produces more Jisim molekul relatif = 88
soot. Explain your answer.
Sebatian A dan B kedua-duanya terbakar (i) Determine the molecular formula of
menghasilkan nyalaan berjelaga. Compound X.
Tentukan sebatian yang manakah menghasilkan Tentukan formula molekul bagi Sebatian X.
lebih jelaga. Terangkan jawapan anda. [2 marks / 2 markah]
[4 marks / 4 markah] (ii) Describe briefly about Compound X.
(c) Compound C reacts with Compound D to Your answer should include the
form a compound. following aspects:
Sebatian C bertindak balas dengan Sebatian D Huraikan dengan ringkas tentang Sebatian X.
untuk menghasilkan satu sebatian. Jawapan anda hendaklah mengandungi
(i) Write the chemical equation for the aspek-aspek yang berikut:
reaction.
Tuliskan persamaan kimia untuk tindak
balas tersebut.




25 © Penerbitan Pelangi Sdn. Bhd.






02[Hebat Chem F5].indd 25 14/02/2019 5:13 PM

Chemistry Form 5 Practice 2 Carbon Compounds
• Homologous series (ii) State the name of substance Y.
Siri homolog Nyatakan nama sebatian Y.
• Functional group [1 mark / 1 markah]
Kumpulan berfungsi (iii) Compound Y can be used as a food additive.
• General formula State the type of food additive and
Formula am give your reason.
• Structural formula Sebatian Y boleh digunakan sebagai bahan
Formula struktur tambah makanan.
• Two chemical reactions and Nyatakan jenis bahan tambah makanan dan
examples berikan alasan anda.
Dua tindak balas kimia dan contoh- [2 marks / 2 markah]
contohnya (iv) State one example of food additive, Q
[8 marks / 8 markah] which is used to enhance the taste of
food.
(b) Compound X reacts with butanol to produce Nyatakan satu contoh bahan tambah
Compound Y. makanan, Q yang digunakan untuk
Sebatian X bertindak balas dengan butanol meningkatkan rasa makanan.
menghasilkan Sebatian Y. [1 mark / 1 markah]
(i) Write the chemical equation for the (v) State three side effects of taking
reaction. In your answer, include the excessive food additive, Q for a long
condition for the reaction. period of time.
Tuliskan persamaan kimia bagi tindak balas Nyatakan tiga kesan sampingan
itu. Dalam jawapan anda, nyatakan keadaan pengambilan secara berlebihan bahan
bagi tindak balas itu. tambah makanan, Q dalam jangka masa
[3 marks / 3 markah] yang panjang. [3 marks / 3 markah]







Pancake syrup
Ingredients: High fructose corn syrup, corn syrup, water, maple syrup, cellulose gum, natural flavours,
artificial flavours, caramel colour, salt, sodium hexametaphosphate, sodium benzoate and sorbic acid.
Sirap pankek
Kandungan: Sirap jagung tinggi fruktosa, sirap jagung, air, sirap mapel, gam selulosa, perisa semula jadi, perasa buatan,
warna karamel, garam, natrium heksametafosfat, natrium benzoat dan asid sorbik.

The diagram above shows the food label of a pancake syrup. Sorbic acid or 2,4-hexadienoic acid, is an antimicrobial
agent often used as a food preservative. The E number for sorbic acid is E200.
What is the function of food preservative?
Rajah di atas menunjukkan satu label makanan pada sirap penkek. Asid sorbik atau asid 2,4-heksadienoik adalah sejenis agen
antimikrobial yang selalu digunakan sebagai pengawet makanan. Nombor E bagi asid sorbik ialah E200.
Apakah fungsi pengawet makanan?

















© Penerbitan Pelangi Sdn. Bhd. 26






02[Hebat Chem F5].indd 26 14/02/2019 5:13 PM

Frekuensi perlanggaran kadar tindak balas yang tinggi.
PRACTICE Rate of Reaction
1 Kadar Tindak Balas antara ion hidrogen SECTION
dengan kalsium karbonat B
dalam Set II lebih tinggi 3. (a) (i) The three factors are:
Objective Questions PAPER 1 daripada dalam Set I. Tiga faktor tersebut adalah:
1. D 2. A 3. B 4. C 5. A 3. The frequency of • Surface area of reactant
6. B 7. A 8. B 9. B 10. B effective collision Luas permukaan bahan
11. D 12. B 13. A 14. C 15. D between hydrogen ions tindak balas [1]
16. C 17. B 18. B 19. D 20. C and calcium carbonate • Concentration of solution
21. C 22. C 23. A 24. C 25. D is higher in Set II than in Kepekatan larutan [1]
Set I.
Subjective Questions PAPER 2 Frekuensi perlanggaran • Temperature of the
reaction
SECTION A berkesan antara ion Suhu bahan tindak balas
hidrogen dengan kalsium
1. (a) Concentration of hydrogen ion karbonat dalam Set II [1]
per unit volume lebih tinggi daripada (ii)
Eye
Kepekatan ion hidrogen per isi dalam Set I. Mata
padu unit 2. (a) Size of reactant particles Conical
(b) (i) CaCO 3 (s)+ 2HCl(aq) → Saiz zarah bahan tindak balas flask
CaCl 2 (aq) + H 2 O(l) + (b) (i) Change in the volume of Kelalang kon
CO 2 (g) CO 2 gas released with time. Na S O + HCl
2
3
2
CaCO 3 (p)+ 2HCl(ak) → Perubahan isi padu gas White paper with
CaCl 2 (ak) + H 2 O(ce) + CO 2 terbebas dengan masa. cross ‘×’
CO 2 (g) (ii) Change in the volume of Kertas putih dengan
(ii) Number of moles of HCl CO 2 gas with time can tanda ‘×’ [2]
Bilangan mol HCl be recorded easily and • A white piece of paper with a
0.1 × 100 accurately. –3
= 1 000 Perubahan isi padu gas cross 'X' on it, 0.2 mol dm
= 0.01 mol CO 2 terbebas dengan sodium thiosulphate solution
and 1.0 mol dm hydrochloric
–3
2 mol of HCl produces masa boleh direkodkan acid are prepared.
dengan mudah dan tepat.
mol of CO 2 Sekeping kertas putih dengan
2 mol HCl menghasilkan (c) 1. Concentration of tanda ‘X’ di atasnya, larutan
hydrochloric acid used –3
1 mol CO 2 natrium tiosulfat 0.2 mol dm
0.01 mol of HCl produces Kepekatan asid hidroklorik dan asid hidroklorik
yang digunakan 0.1 mol dm disediakan. [1]
–3
0.005 mol of CO 2
0.01 mol menghasilkan 2. Temperature of the reaction • Using a measuring cylinder,
Suhu tindak balas 50 cm of sodium
3
0.005 mol CO 2
Volume of CO 2 gas = (d) (i) The gradient of graph for thiosulphate is measured and
Experiment II is steeper
poured into a conical flask.
0.005 × 24 = 0.12 dm 3 than that of Experiment I. 50 cm larutan natrium
3
Isi padu gas CO 2 Kecerunan graf bagi tiosulfat disukat dan dituang
(c) (i) The rate of reaction in Set Eksperimen II lebih curam ke dalam kelalang kon
II is higher than in Set I. daripada kecerunan graf dengan menggunakan
Kadar tindak balas dalam bagi Eksperimen I. silinder penyukat. [1]
Set II lebih tinggi daripada (ii) The reactants have • The conical flask containing
dalam Set I. completely reacted. sodium thiosulphate is placed
(ii) 1. The concentration of Bahan tindak balas habis on the white paper marked
hydrogen ion per unit bertindak balas. with cross ‘X’.
volume in Set II is higher (iii) The rate of reaction is zero Kelalang kon yang
than in Set I. for both reactions after mengandungi larutan natrium
Kepekatan ion hidrogen time x and y. tiosulfat diletakkan di atas
per isi padu unit dalam Kadar tindak balas adalah kertas putih yang bertanda ‘x’.
Set II lebih tinggi kosong bagi kedua-dua [1]
daripada dalam Set I. tindak balas selepas • Using a small measuring
2. The frequency of masa x dan y. cylinder, 5 cm of 1.0 mol dm
–3
3
collision between (e) Reactants with smaller particle hydrochloric acid is measured.
hydrogen ions and size produce a higher rate of 5 cm asid hidroklorik
3
calcium carbonate is reaction. 1.0 mol dm disukat dengan
–3
higher in Set II than in Bahan tindak balas dengan saiz menggunakan silinder
Set I. zarah yang kecil menghasilkan penyukat yang kecil. [1]
A1 © Penerbitan Pelangi Sdn. Bhd.
08 Ans[Pra A++ Chem F5].indd 1 14/02/2019 5:25 PM

Chemistry Form 5 Answers
• The hydrochloric acid Concentration of Na 2 S 2 O 3 after mixing (b) • A catalyst is a substance
measured is quickly Kepekatan Na 2 S 2 O 3 selepas dicampur that changes the rate of
poured into the conical reaction but it is chemically
flask containing sodium = M 1 V 1 unchanged at the end of the
thiosulphate solution. V 2 reaction.
The stopwatch is started 0.2 × volume of Na 2 S 2 O 3 used Mangkin ialah bahan yang
immediately. = / mengubah kadar tindak balas
Asid hidroklorik yang disukat 50 tetapi tidak akan berubah
dituang ke dalam kelalang 0.2 × Isi padu Na 2 S 2 O 3 yang digunakan secara kimia pada akhir
kon yang mengandungi 50 tindak balas. [1]
larutan natrium tiosulfat [1]
dengan cepat. Jam randik • Activation energy is the
dihidupkan dengan serta- A graph of concentration of sodium minimum energy that the
1
–3
–1
merta. [1] thiosulphate (mol dm ) against Time (s ) reactant particles must
• The time is recorded as soon is plotted. possess during collision in
as cross 'x' disappears from Graf kepekatan natrium tiosulfat (mol dm ) order for a chemical reaction
–3
view. melawan 1 (s ) diplotkan. to take place.
–1
Masa direkodkan sebaik Masa Tenaga pengaktifan ialah
sahaja tanda 'x' hilang tenaga minimum yang
daripada pandangan. [1] Concentration of diperlukan oleh zarah-zarah
• The steps are repeated with Na S O (mol dm ) bahan tindak balas semasa
–3
2
3
2
different mixtures of sodium Kepekatan –3 perlanggaran supaya tindak
thiosulphate solution and Na S O (mol dm ) balas kimia boleh berlaku. [1]
2
2
3
distilled water as shown in
the table below. SECTION C
Langkah ini diulang dengan 4. (a) • The experiment is carried out
menggunakan campuran in a conical flask. The flask
larutan natrium tiosulfat 1 is fitted with a stopper and a
–1
dengan air suling yang – (s ) delivery tube is connected to
Time
berbeza seperti yang – (s ) an inverted burette filled with
1
–1
ditunjukkan dalam jadual di Masa
water.
bawah. [1] [1] Eksperimen ini dijalankan
Experiment 1 2 3 4 5 Based on the graph, the concentration di dalam kelalang kon yang
–3
Eksperimen of Na 2 S 2 O 3 (mol dm ) is directly ditutupi oleh penutup dan
1
–1
Volume of Na 2 S 2 O 3 proportional to Time (s ) sebatang tiub penghantar
(cm ) 3 menghubungkan dengan
Isi padu Na 2 S 2 O 3 50 40 30 20 10 Berdasarkan graf, kepekatan natrium buret yang terbalik dalam
(cm ) 3 tiosulfat (mol dm ) berkadar terus bekas yang berisi air suling.
–3
1
–1
Volume of distilled dengan Masa (s ) 3 –3 [1]
• 50 cm of 1.0 mol dm
water (cm ) 3 0 10 20 30 40
Isi padu air suling Na 2 S 2 O 3 (aq) + 2HCl(aq) → hydrochloric acid is pipetted
(cm ) 3 2NaCl(aq) + H 2 O(l) + SO 2 (g) + S(s) into the conical flask.
3
Total volume of Na 2 S 2 O 3 (ak) + 2HCl(ak) → 50 cm asid hidroklorik
1.0 mol dm dipipetkan ke
–3
solution (cm ) 3 50 50 50 50 50 2NaCl(ak) + H 2 O(ce) + SO 2 (g) + S(p) dalam kelalang kon. [1]
Jumlah isi padu [1]
larutan (cm ) 3 The higher the concentration of sodium • Lumps of calcium carbonate
thiosulphate, the shorter the time taken are placed in the flask and
Concentration of for a certain mass of sulphur to be immediately the conical flask
Na 2 S 2 O 3 after
mixing (mol dm ) 0.2 0.16 0.12 0.08 0.04 precipitated that is for the cross ‘×’ to is fitted with the stopper.
–3
Kepekatan Na 2 S 2 O 3 disappear from view. Therefore, the Ketulan kalsium karbonat
selepas dicampur higher the concentration of sodium dimasukkan ke dalam
–3
(mol dm ) thiosulphate, the higher the rate of kelalang dan ditutup dengan
reaction. penutup serta-merta. [1]
Time taken (s) Semakin tinggi kepekatan larutan
Masa yang diambil 24 30 42 62 111 natrium tiosulfat, semakin singkat masa • The stopwatch is started.
(s) Jam randik dimulakan. [1]
yang diambil untuk jisim tertentu sulfur • The volume of gas is
1 –1 termendak iaitu masa yang diambil untuk measured at regular intervals.
Time (s ) 0.042 0.0033 0.024 0.016 0.009 tanda ’x’ hilang daripada pandangan.
1 –1 Oleh itu, semakin tinggi kepekatan Isi padu gas disukat pada
Masa (s ) natrium tiosulfat, semakin tinggi kadar selang masa yang tetap. [1]
[2] tindak balas. [1]
© Penerbitan Pelangi Sdn. Bhd. A2
08 Ans[Pra A++ Chem F5].indd 2 14/02/2019 5:25 PM

Chemistry Form 5 Answers
• The set-up of the apparatus hidroklorik akan (iii) • Powder has a higher
is as shown: meningkatkan bilangan surface area than chips
Susunan radas ditunjukkan zarah per unit isi padu. or lumps
seperti di bawah: [1] [1] Serbuk mempunyai luas
• The frequency of permukaan yang lebih
Delivery tube Gas collisions between tinggi berbanding ketulan
Tiub penghantar Gas [1]
Burette particles is higher
Buret as there are more and therefore increases
the rate of collisions.
Conical flask molecules present in a dan menyebabkan kadar
Kelalang kon certain volume. perlanggaran meningkat.
Frekuensi perlanggaran
Hydrochloric [1]
acid antara zarah meningkat • Higher rate of collision
Asid hidroklorik Calcium carbonate disebabkan bilangan leads to higher
Kalsium karbonat molekul yang ada dalam frequency of effective
• The readings of the volume kepekatan yang tertentu. collisions and thus
of gas at fixed intervals are [1] higher rate of reaction.
recorded in the table below. Hence, more effective Kadar perlanggaran
Bacaan isi padu pada sela collisions are produced yang tinggi akan
masa yang tetap direkodkan and the rate of reaction meningkatkan frekuensi
dalam jadual di bawah. increases. perlanggaran berkesan
yang meningkatkan
Time / Oleh itu, lebih banyak kadar tindak balas. [1]
Masa 0 1 2 3 4 5 6 7 perlanggaran berkesan
(min) yang terhasil dan kadar
Volume tindak balas meningkat.
of gas / (ii) • Increasing the OBJECTIVE QUESTIONS
Isi temperature will increase
padu the rate of reaction. 1. D 2. C 3. C 4. D 5. A
gas Meningkatkan suhu akan 6. A 7. A 8. A 9. C 10. B
(cm ) meningkatkan kadar SUBBJECTIVE QUESTIONS
3
[2] tindak balas. [1]
• A graph of volume against • As the temperature SECTION A
+
2+
time is plotted. increases, the average 1. (a) Zn(s) + 2H (aq) → Zn (aq) +
Graf isi padu melawan masa kinetic energy possessed H 2 (g)
+
2+
diplotkan. [2] by the molecules Zn(p) + 2H (ak) → Zn (ak) +
• From the graph, the increases and the H 2 (g)
instantaneous rate of particles will move faster. (b) Catalyst / Mangkin
reaction at t minute can be Semakin tinggi suhu, (c) (i) Number of mol of Zn
determined from the gradient semakin tinggi tenaga Bilangan mol Zn
of the graph at t minute. kinetik purata yang Mass / Jisim
Daripada graf, kadar dimiliki oleh molekul dan = RAM / JAR
tindak balas pada minit t zarah akan bergerak = 2.0
boleh ditentukan daripada dengan lebih cepat. [1] 65
kecerunan graf pada minit t. • The frequency of = 0.0308 mol
[1] collisions is higher Number of mol of HCl
Max: 10 and more molecules Bilangan mol HCl
(b) (i) • The rate of reaction have energy greater = MV
increases when a than or equal to 1000
25(0.1)
more concentrated activation energy, E a . = 1000
hydrochloric acid is Frekuensi perlanggaran
used. lebih tinggi dan lebih = 0.0025 mol
Kadar tindak balas banyak molekul yang (ii) From the chemical
bertambah apabila mempunyai tenaga equation,
kepekatan asid yang lebih besar atau Daripada persamaan kimia,
hidroklorik yang lebih sama dengan tenaga 2 mol HCl : 1 mol Zn
tinggi digunakan. [1] pengaktifan, E a . [2] 0.0025 mol HCl: 0.00125
• An increase in This leads to an mol Zn
concentration will increase in the number Excess number of mol of
increase the number of of effective collisions. zinc / Bilangan mol zink
berlebihan
particles per unit volume. Ini akan meningkatkan = 0.0308 – 0.00125 mol
Peningkatan dalam bilangan perlanggaran = 0.02955 mol
kepekatan asid berkesan.


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08 Ans[Pra A++ Chem F5].indd 3 14/02/2019 5:25 PM

Chemistry Form 5 Answers
Mass of excess zinc for Experiment II. This Semakin tinggi kepekatan
Jisim zink berlebihan is because Experiment I asid hidroklorik, semakin
= Number of mole × RAM uses powdered calcium bertambah bilangan ion
+
Bilangan mol × JAR carbonate. The smaller the H per isi padu. Frekuensi
= 0.02955 × 65 size of calcium carbonate, perlanggaran antara
= 1.92 g the larger the total surface karbonat ion dengan ion
+
(iii) area of calcium carbonate H bertambah. Frekuensi
perlanggaran berkesan
Mass of zinc (g) that the exposed to the antara ion karbonat dan
reaction. The frequency
Jisim zink (g) +
I of collision between ion H juga bertambah.
+
carbonate ion with H ion Kadar tindak balas
increases. The frequency semakin bertambah.
II of effective collision
1.92 between carbonate ions SECTION C
and H ions also increases.
+
Time (s) 3. (a) (i) I : 420 / 100 [1]
Masa (s) The rate of reaction
increases. = 4.2 cm s [1]
3 -1
B Kadar tindak balas bagi II : 420 / 140 [1]
SECTION
3 -1
2. (a) (i) P Eksperimen I lebih tinggi = 3.0 cm s [1]
berbanding kadar tindak
(ii) Copper(II) sulphate acts as (ii) 1. Factor: Catalyst
a catalyst. The presence balas bagi Eksperimen II. 2. A basin is filled with
Hal ini kerana Eksperimen
of catalyst provides an I menggunakan serbuk water until half full.
alternative pathway with kalsium karbonat. 3. A burette full with water
lower activation energy. Semakin kecil saiz kalsium is inverted into the basin.
The frequency of effective karbonat yang digunakan, 4. The burette is clamped
collision increases. The semakin luas jumlah vertically using a retort
rate of reaction increases. luas permukaan kalsium stand.
Kuprum(II) sulfat bertindak karbonat yang terdedah 5. 100 cm of 1.0 mol dm
3
-3
sebagai mangkin. kepada tindak balas. hydrochloric acid is
Kehadiran mangkin Frekuensi perlanggaran measured and poured
menyediakan lintasan antara ion karbonat into the conical flask.
alternatif dengan tenaga dengan ion H bertambah. 6. 5 g of magnesium
+
pengaktifan yang lebih Frekuensi perlanggaran powder is weighed and
rendah. Frekuensi berkesan antara ion placed into the conical
perlanggaran berkesan karbonat dengan ion H + flask.
bertambah. Kadar tindak juga bertambah. Kadar 7. 1 cm of 1.0 mol dm -3
3
balas meningkat. tindak balas semakin copper(II) sulphate
bertambah.
(iii) Zn + 2HCl → ZnCl 2 + H 2 solution is measured and
Number of mol of HCl (ii) The rate of reaction for poured into the conical
Bilangan mol HCl Experiment III is higher flask.
MV than the rate of reaction 8. The conical flask is
= for Experiment I. This is
1000 because, Experiment III closed immediately with
= 1.0(100) useds higher concentration a stopper fitted with
1000 of hydrochloric acid. delivery tube.
= 0.1 mol As the concentration 9. A stopwatch is started
From the chemical of hydrochloric acid immediately.
equation, increases, the number 10. The volume of gas
Daripada persamaan kimia, of H ions per volume released is recorded
+
2 mol HCl → 1 mol H 2 increases. The frequency every 30 seconds
interval.
0.1 mol HCl → 0.1 of collision between
+
2 carbonate ions and H ions 1. Faktor: Mangkin
= 0.05 mol increases. The frequency 2. Satu besen diisi dengan
Volume of H 2 gas of effective collision air sehingga separuh
Isi padu gas H 2 between carbonate ions penuh.
+
= Number of mol × molar and H ions also increases. 3. Sebuah buret yang telah
volume The rate of reaction diisi penuh dengan air
Bilangan mol × isi padu increases. diterbalikkan di dalam
molar Kadar tindak balas bagi besen tersebut.
Eksperimen III lebih tinggi
= 0.05 mol × 24 dm × 1000 berbanding kadar tindak 4. Buret tersebut
3
= 1200 cm 3 balas Eksperimen I. Ini kemudiannya diapitkan
menegak menggunakan
(b) (i) The rate of reaction for kerana Eksperimen III kaki retort.
Experiment I is higher menggunakan kepekatan 5. 100 cm asid hidroklorik
3
than the rate of reaction asid hidroklorik yang tinggi. 1.0 mol dm disukat
-3
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08 Ans[Pra A++ Chem F5].indd 4 14/02/2019 5:25 PM

Chemistry Form 5 Answers
dan dituang ke dalam Jumlah luas permukaan bagi setiap tuala (c) Porcelaine chips
kelalang kon. adalah berbeza. Semakin besar jumlah Serpihan porselin
6. 5 g serbuk magnesium luas permukaan, semakin cepat tuala itu Propene gas
ditimbang dan kering. Gas propena
dimasukkan ke dalam
kelalang kon. Heat
7. 1 cm larutan kuprum(II) PRACTICE Carbon Compounds Glass wool Panaskan
3
soaked in
2
sulfat 1.0 mol dm -3 Sebatian Karbon propanol
disukat dan dituang ke PAPER 1 Wul kaca yang
dibasahi propanol
dalam kelalang kon. Objective Questions Water
8. Kelalang kon ditutup 1. A 2. C 3. A 4. C 5. B Air
dengan serta merta 6. A 7. A 8. B 9. D 10. C (d) (i) Propyl propanoate
menggunakan 11. A 12. A 13. D 14. A 15. C Propil propanoat
penyumbat bersama 16. C 17. B 18. D 19. C 20. D (ii)
salur penghantar. 21. B 22. D 23. B 24. A 25. C H H O H H H
9. Jam randik dimulakan
dengan serta merta. Subjective Questions PAPER 2 H C C C O C C C H
10. Isi padu gas yang SECTION A H H H H H
dibebaskan direkodkan
selang setiap 30 saat. 1. (a) (i) Hydrogenation SECTION B
(b) (i) x = 30 C [1] Penghidrogenan 3. (a) (i) Volume of carbon dioxide
o
y = 40 C [1] (ii) A mixture of but-2-ene and formed from one mole of
o
hydrogen is passed over a
(ii) 1. The higher the heated nickel at 180°C. hydrocarbon X
temperature of HCl, the Campuran but-2-ena dan Isi padu karbon dioksida
higher the kinetic energy hidrogen dilalukan ke atas yang terbentuk daripada
of H ion. satu mol hidrokarbon X
+
2. The frequency of nikel yang dipanaskan pada = 1920 [1]
suhu 180°C.
collision between Ni 0.02
3
S 2 O 3 ion and H ion in (iii) C 4 H 8 (g) + H 2 (g) ⎯→ C 4 H 10 (g) = 96 000 cm [1]
+
2-
Experiment II is higher (b) (i) Acidified potassium Number of moles of carbon
than Experiment I. manganate(VII) solution dioxide formed
3. The frequency of Larutan kalium Bilangan mol karbon
effective collision manganat(VII) berasid dioksida yang terbentuk
between S 2 O 3 ion and 96 000
2-
H ion in Experiment II is (ii) CH 3 CH CH CH 3 = 24 000
+
higher than Experiment I. = 4 mol [1]
4. The rate of reaction in OH OH Hence, the number of
Experiment II is higher (c) (i) The brown colour of carbon atoms in one
than Experiment I. bromine turns colourless molecule of hydrocarbon X
1. Semakin tinggi suhu Warna perang bromin Oleh itu, bilangan atom
HCl, semakin tinggi bertukar menjadi tidak karbon dalam satu molekul
tenaga kinetik ion H berwarna hidrokarbon X
+
yang bertindak balas. (ii) 2,3-dibromobutane
2. Frekuensi perlanggaran 2,3-dibromobutana = 4 atoms / 4 atom [1]
antara ion S 2 O 3 ion (d) H H H H (ii) Alkene / Alkena [1]
2-
dengan ion H lebih tinggi (iii) C 4 H 8 [1]
+
dalam Eksperimen II H C C C C H (iv) C 4 H 8 (g) + 6O 2 (g) →
berbanding Eksperimen I. H H 4CO 2 (g) + 4H 2 O(l)
3. Frekuensi perlanggaran C 4 H 8 (g) + 6O 2 (g) →
berkesan antara ion H 4CO 2 (g) + 4H 2 O(ce)
S 2 O 3 ion dengan H C H [1]
2-
ion H lebih tinggi H H
+
dalam Eksperimen II H C C C H (v) H H H H
berbanding Eksperimen I. H C C C C H
4. Kadar tindak balas lebih H H H
tinggi dalam Eksperimen 2. (a) (i) Alkene
II berbanding dengan Alkena But-1-ene / But-1-ena [2]
Eksperimen I. Compound B – Alcohol H H H H
Sebatian B – Alkohol H C C C C H
(b) (i) C 3 H 6 + H 2 O → C 3 H 7 OH H H
Total surface area of each towel is (ii) Temperature / Suhu : 300°C But-2-ene / But-2-ena [2]
different. The bigger the total surface Pressure / Tekanan : 60 atm
area, the faster it dries.

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08 Ans[Pra A++ Chem F5].indd 5 14/02/2019 5:25 PM

Chemistry Form 5 Answers
(ii) I : • Percentage of tetrachloromethane
H H
carbon by mass in through addition reaction.
H C C C H Butena bertindak
butane, C 4 H 10
CH 3 H Peratus karbon balas dengan bromin
dalam tetraklorometana
berdasarkan jisim
2-methylprop-1-ene
2-metilprop-1-ena [2] dalam butana, C 4 H 10 melalui tindak balas [1]
penambahan.
(12)(4)
(vi) But-2-ene / But-2-ena [1] = [12(4) + 1(10)] × 100 • C 4 H 8 (g) + Br 2 (l) →
(b) (i) • Bacteria in latex produce C 4 H 8 Br 2 (l)

acids. = 82.8% [1] C 4 H 8 (g) + Br 2 (ce) →
Bakteria di dalam lateks • Percentage of carbon
menghasilkan asid. [1] by mass in butene, C 4 H 8 C 4 H 8 Br 2 (ce)
• Hydrogen ions in the Peratus karbon [1]
acid neutralise negative berdasarkan jisim • Bromine is added to
charges on the particles dalam butena, C 4 H 8 C=C to form a saturated
of rubber. = (12)(4) × 100 compound, C 4 H 8 Br 2 .
Ion hidrogen dalam asid [12(4) + 1(8)] Bromin ditambahkan
meneutralkan cas negatif = 85.7% [1] ke dalam C=C untuk
di sekeliling zarah getah. The percentage of carbon membentuk sebatian
[1] by mass in butene is higher tepu, C 4 H 8 Br 2 . [1]
• This will enable the than in butane. Hence, Max: 6
particles to collide butene burns with a more (b) (i) Ethanol / Etanol
among each other, sooty flame than butane. CH 3 – CH 2 – O – H [2]
causing coagulation. Peratus karbon berdasarkan (ii)
Hal ini menyebabkan jisim dalam butena adalah Porcelain chips
zarah berlanggar lebih tinggi berbanding Serpihan porselin
antara satu sama lain butana. Oleh itu, butena CH 2 = CH 2 ←⎯⎯⎯⎯⎯⎯⎯ CH 3 –CH 2 – OH
Heat
dan menyebabkan terbakar dengan lebih Panaskan
penggumpalan. [1] berjelaga berbanding Acidified potassium
(ii) • Ammonia / Ammonia [1] butana. [1] manganate(VII)
• When ammonia is added II : Kalium manganat(VII)
to latex, the ionisation of • Butane is a saturated berasid
ammonia produces OH compound with only ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ CH 3 COOH + H 2 O
–

Reflux
ions. carbon–carbon single Refluks
Apabila ammonia bonds, C–C. [4]
ditambah kepada lateks, Butana adalah sebatian
pengionan ammonia tepu dengan ikatan (iii) Ethanol is used as a
menghasilkan ion OH . [1] tunggal antara atom solvent for varnishes.
–
• The OH ions neutralise karbon-karbon, C–C. [1] Etanol digunakan sebagai
–
pelarut untuk varnis.
[1]
the acid produced by • Hydrogen in butane is
bacteria. substituted by bromine in
Ion OH meneutralkan the presence of UV light.
–
asid yang dihasilkan Hidrogen dalam
oleh bakteria. [1] butana digantikan OBJECTIVE QUESTIONS
• The rubber particles dengan bromin dengan 1. C 2. B 3. B 4. A 5. B
remain negatively kehadiran cahaya UV. [1] 6. A 7. A 8. D 9. C 10. C
charged and keep on • C 4 H 10 (g) + Br 2 (g) → 11. B 12. A 13. D
repelling and stay in C 4 H 9 Br(l) + HBr(g)
liquid state. C 4 H 10 (g) + Br 2 (g) → SUBJECTIVE QUESTIONS
Zarah getah C 4 H 9 Br(ce) + HBr(g)
mengekalkan cas negatif [1] SECTION A
dan kekal menolak 1. (a) (i) C n H 2n+1 OH n = 1, 2, 3….
antara satu sama • Butene is an unsaturated (ii) C 4 H 9 OH
lain dan kekal dalam compound with at (iii) Carbon dioxide gas and water
keadaan cecair. [1] least one double bond Gas karbon dioksida dan air
Max: 3 between C–C atoms.
Butena adalah sebatian (b) 1. Flow the gas through
limewater.
SECTION C tak tepu dengan 2. Limewater turns chalky.
4. (a) (i) Alkane: Butane, C 4 H 10 sekurang-kurangnya 1. Salurkan gas ke dalam air kapur.
satu ikatan ganda dua
Alkana: Butana, C 4 H 10 [2] antara atom C–C. [1] 2. Air kapur menjadi keruh.
Alkene: Butene, C 4 H 8 • Butene reacts
Alkena: Butena, C 4 H 8 [2]
with bromine in
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08 Ans[Pra A++ Chem F5].indd 6 14/02/2019 5:25 PM

Chemistry Form 5 Answers
(c) 6(12) (ii) • Sweet-smelling
H H H H H = 6(12) + 14(1) × 100% Berbau wangi
H C C C C H H C H = 83.72% • Less dense than water
H H Kurang tumpat
H H H OH Percentage of carbon by mass berbanding air
H C C C OH in hexene, C 6 H 12 • Colourless liquid at room
Peratusan karbon dalam
H H H temperature
heksena, C 6 H 12 Cecair tanpa warna
H H H H H = 6(12) × 100% pada suhu bilik
H C C C C H H C H 6(12) + 12(1) • Very volatile
H H = 85.71% Mudah meruap
H H OH H Hexene contains higher
H C C C H percentage of carbon by mass • Insoluble in water
Tidak larut dalam air
H OHH than in hexane. SECTION C
(Any one/ Mana-mana satu) Hexene produces more sooty 3. (a) (i) (C 2 H 4 O)n = 88
flame than hexane.
(d) (i) Methanoic acid Heksena mengandungi 24n + 4n + 16n = 88
Asid metanoik peratusan karbon yang lebih 44n = 88
(ii) HCOOH + C 4 H 9 OH → tinggi. n = 2
HCOOC 4 H 9 + H 2 O Heksena menghasilkan lebih Molecular formula
(iii) Ester floats on water jelaga berbanding heksana. of Compound X is
Ester terapung di atas (c) (i) C 3 H 7 OH + C 2 H 5 COOH → C 3 H 7 COOH.
permukaan air C 2 H 5 COOC 3 H 7 + H 2 O Formula molekul
bagi Sebatian X ialah
SECTION B C 3 H 7 COOH.
2. (a) (i) A – Hexene / Heksena
B – Hexane / Heksana (ii) Homologous series of Compound X is carboxylic acid.
C – Propanol / Propanol Siri homolog bagi Sebatian X adalah asid karboksilik.
D – Propanoic acid / Asid
propanoik The functional group is carboxyl group.
A – H H H H H H Kumpulan berfungsi adalah kumpulan karboksilik.
The general formula of Compound X is C 2n H 2n+1 COOH, where n = 1, 2, ...
=
H C C C C C C H
Formula am bagi Sebatian X ialah C 2n H 2n+1 COOH,
H H H H n = 1, 2, ...
B – H H H H H H Structural formula
Formula struktur,
H C C C C C C H H H H O
& & & '
H H H H H H H!C!C!C!C!OH
& & &
H H H
C – H H H or / atau
Compound X undergoes neutralisation reaction by reacting with alkali solution
H C C C OH to produce salt and water.
H H H Example: C 3 H 7 COOH + NaOH → C 3 H 7 COONa + H 2 O
H H H Sebatian X mengalami tindak balas peneutralan iaitu bertindak balas
dengan larutan alkali untuk menghasilkan garam dan air.
H C C C H Contoh: C 3H 7COOH + NaOH → C 3H 7COONa + H 2O
H OH H Compound X reacts with metal carbonate to produce salt, water and
carbon dioxide gas.
D – H H O Example: 2C 3 H 7 COOH + CaCO 3 → (C 3 H 7 COO) 2 Ca + H 2 O + CO 2
Sebatian X bertindak balas dengan logam karbonat membentuk garam, air
H C C C OH dan gas karbon dioksida.
H H Contoh: 2C 3 H 7 COOH + CaCO 3 → (C 3 H 7 COO) 2 Ca + H 2 O + CO 2
(ii) C – Hydroxyl group (b) (i) C 3 H 7 COOH + C 4 H 9 OH → (iii) As a flavouring agent
Kumpulan hidroksil C 3 H 7 COOC 4 H 9 + H 2 O because it has a fruity smell
D – Carboxyl group Sebagai agen perisa
Kumpulan karboksil Condition: Concentrated kerana mempunyai bau
sulphuric acid
(b) Percentage of carbon by mass Keadaan: Asid sulfurik pekat yang harum
in hexane C 6 H 14 (ii) Butyl butanoate (iv) Monosodium glutamate
Peratusan karbon dalam Mononatrium glutamat
Butil butanoat
heksana, C 6 H 14

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08 Ans[Pra A++ Chem F5].indd 7 14/02/2019 5:25 PM

Chemistry Form 5 Answers
(v) Cancer, hypertension, brain SECTION B • So, hydrogen acts as the
damage reducing agent.
Kanser, tekanan darah 2. (a) (i) • Air in the combustion Oleh itu, hidrogen
tinggi, kerosakan otak tube must be totally bertindak sebagai agen
removed before the penurunan. [1]
metal oxide is heated.
Udara di dalam tiub (b) • Air and water are needed for
pembakaran mesti rusting.
To prevent the growth of fungi. disingkirkan sepenuhya Air dan udara diperlukan
Untuk menghalang pertumbuhan fungi. sebelum oksida logam untuk pengaratan. [1]
dipanaskan. [1] • When the surface of iron is in
• The flow of hydrogen contact with water, the middle
PRACTICE Oxidation and Reduction gas must be continuous region of the water droplet
3 Pengoksidaan dan Penurunan throughout the acts as the anode.
experiment. Apabila permukaan ferum
bersentuhan dengan air,
Objective Questions PAPER 1 Pengaliran gas hidrogen kawasan tengah titisan air
hendaklah berterusan
1. C 2. D 3. C 4. B 5. D sepanjang eksperimen. bertindak sebagai anod. [1]
6. A 7. C 8. B 9. D 10. B [1] • At the anode, iron is oxidised
2+
11. B 12. B 13. C 14. A 15. A (ii) Zinc, hydrogen, iron. to Fe ion.
16. D 17. A 18. D 19. A 20. B Zink, hidrogen, ferum [1] Pada anod, ferum dioksidakan
[1]
kepada ion ferum(II).
21. C 22. B 23. C 24. D 25. B
(iii) Fe 2 O 3 (s) + 3H 2 (g) → • Fe(s) → Fe (aq) + 2e –
2+
2+
Subjective Questions PAPER 2 2Fe(s) + 3H 2 O(l) Fe(p) → Fe (ak) + 2e – [1]
Fe 2 O 3 (p) + 3H 2 (g) → • The region at the edge of
SECTION A 2Fe(p) + 3H 2 O(ce) the water droplet acts as the
1. (a) (i) The pale green solution [1] cathode.
turns yellow. (vi) • Hydrogen gains oxygen Kawasan pinggir titisan air
Larutan hijau pudar to form water. bertindak sebagai katod. [1]
menjadi kuning. Hidrogen menerima • The electrons released flow
(ii) The orange solution turns oksigen untuk through the metal to the
green. membentuk air. [1] cathode where dissolved
Larutan jingga menjadi hijau. • Hydrogen undergoes oxygen is reduced.
(b) To complete the circuit by allowing oxidation. Elektron yang dilepaskan
the transfer of ions to occur. Hidrogen mengalami mengalir melalui logam
Untuk melengkapkan litar pengoksidaan. [1] kepada katod di mana
dengan membenarkan ion • Iron(III) oxide loses its oksigen diturunkan. [1]
mengalir melaluinya. oxygen to form iron. • O 2 (g) + 2H 2 O(l) + 4e →
–
(c) (i) +6 Ferum(III) oksida 4OH (aq)
–
(ii) +6 to +3 kehilangan oksigen O 2 (g) + 2H 2 O(ce) + 4e →
–
–
(d) untuk membentuk ferum. 4OH (ak)
e – G e – [1] [1]
• Iron(III) oxide undergoes • The Fe ions formed
2+
–
Graphite Graphite reduction. combine with the OH ions to
electrode X electrode Y Ferum(III) oksida form iron(II) hydroxide.
2+
Elektrod Elektrod mengalami penurunan. Ion Fe terbentuk bergabung
grafit X grafit Y dengan ion OH membentuk
-
[1]
• Iron(III) oxide causes ferum(II) hidroksida. [1]
(e) (i) Potassium dichromate(VI) hydrogen to be oxidised. • The iron(II) hydroxide is then
Kalium dikromat(VI) Ferum(III) oksida further oxidised by oxygen to
(ii) Potassium dichromate(VI) menyebabkan hidrogen form hydrated iron(III) oxide.
oxidises Fe ions to Fe ions. dioksidakan. [1] This hydrated oxide is called
3+
2+
Kalium dikromat(VI) • So, iron(III) oxide acts as rust.
mengoksidakan ion Fe the oxidising agent. Ferum(II) hidroksida
2+
kepada Fe . Oleh itu, ferum(III) kemudiannya dioksidakan oleh
3+
oksigen membentuk ferum(III)
(f) (i) The brown colour of oksida bertindak sebagai oksida terhidrat yang disebut
bromine decolourises. agen pengoksidaan. [1] sebagai karat. [1]
Warna perang bromin • Hydrogen causes
dilunturkan. iron(III) oxide to be Max: 8
2+
(ii) 2Fe (aq) + Br 2 (aq) → reduced. SECTION C
2Fe (aq) + 2Br (aq) Hidrogen menyebabkan
3+
–
2Fe (ak) + Br 2 (ak) → ferum(III) oksida 3. (a) (i) Zn(s) + CuSO 4 (aq) →
2+
2Fe ak) + 2Br (ak) diturunkan. [1] ZnSO 4 (aq) + Cu(s)
3+(
-
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08 Ans[Pra A++ Chem F5].indd 8 14/02/2019 5:25 PM

Chemistry Form 5 Answers
Zn(p) + CuSO 4 (ak) → Agen pengoksidaan: Ion (iv) Zinc ion / Ion zink
ZnSO 4 (ak) + Cu(p) kuprum(II) [1] (b) In set II, carbon is more
[1] • Copper(II) ions causes reactive than metal X and zinc
• The oxidation number of zinc to be oxidised to is more reactive than metal X in
zinc increases from 0 to zinc ions. reactivity series.
+2. Ion kuprum(II) Dalam set II, karbon lebih
Nombor pengoksidaan menyebabkan zink reaktif berbanding logam X
zink bertambah daripada dioksidakan kepada ion dan zink lebih reaktif daripada
0 kepada +2. [1] zink. [1] logam X dalam siri kereaktifan.
• The oxidation number • Reducing agent: Zinc (c) (i) Mg, C, Zn, X
of copper in copper(II) Agen penurunan: Zink (ii) Copper / Kuprum
sulphate decreases from [1]
+2 to 0. • Zinc causes copper(II)
Nombor pengoksidaan ions to be reduced to SECTION B
kuprum dalam larutan copper. 2. (a) (i) • Iron atom, Fe undergoes
kuprum(II) sulfat Zink menyebabkan oxidation and oxygen
berkurang daripada +2 ion kuprum diturunkan molecules undergo
kepada 0. [1] kepada kuprum. [1] reduction.
(ii) • Oxidising agent: Atom ferum,
Copper(II) ions Fe mengalami
pengoksidaan dan
(iii) NaOH(aq/ak) + HCl(aq/ak) → NaCl(aq/ak) + H 2 O(l/ce) [1] molekul oksigen
| | | | | | | | |
Oxidation number: mengalami penurunan.
Nombor • At the anode, iron atom
+1–2+1 +1–1 +1 –1 +1–2
pengoksidaan: [1] loses two electrons to
2+
• All the elements in the • The electrons then flow to form iron(II) ion, Fe .
reaction do not undergo graphite electrode Y. Pada anod, atom ferum
any change in oxidation Elektron mengalir ke elektrod melepaskan dua elektron
numbers. grafit Y. [1] untuk membentuk ion
2+
Semua unsur tidak At graphite electrode Y: ferum(II), Fe .
mengalami sebarang Pada elektrod grafit Y: • The electrons flow
perubahan nombor • The bromine receives electrons through iron to the edge
pengoksidaan. [1] and are reduced to Br ions. of the water droplet.
–
(b) Bromin menerima elektron Elektron mengalir
G dan diturunkan kepada ion melalui ferum kepada
Br . – [1] hujung titisan air.
• Reduction process occurs. • At the cathode, oxygen
Graphite Graphite Proses penurunan berlaku. [1] molecules gain electrons
electrode X electrode Y • Br 2 (aq) + 2e → 2Br (aq) to form hydroxide ions,
–
–
Elektrod grafit X Elektrod grafit Y Br 2 (ak) + 2e → 2Br (ak) [1] OH . –
–
–
Potassium Bromine Pada katod, molekul
iodide solution water oksigen menerima
Larutan kalium iodida Air bromin elektron dan membentuk
Dilute sulphuric acid OBJECTIVE QUESTIONS ion hidroksida, OH . –
Asid sulfurik cair
[2] 1. D 2. A 3. B 4. C 5. C • Iron(II) ion combines
At graphite electrode X: 6. D 7. C 8. A 9. C with hydroxide ions to
form iron(II) hydroxide.
Pada elektrod grafit X: SUBJECTIVE QUESTIONS Ion ferum(II) bergabung
• The potassium iodide solution dengan ion hidroksida
contains I ions. SECTION A untuk membentuk
–
Larutan kalium iodida 1. (a) (i) Redox reaction is a ferum(II) hidroksida.
mengandungi ion I . – [1] reaction whereby oxidation
–
• The I ions lose electrons and and reduction occur • Iron(II) hydroxide
undergoes further
are oxidised to iodine. simultaneously.
oxidation to form rust.
Ion iodida kehilangan Tindak balas redoks Ferum(II) hidroksida
elektron dan dioksidakan adalah tindak balas yang melalui pengoksidaan
kepada iodin. [1] mana pengoksidaan dan selanjutnya untuk
• Oxidation process occurs. penurunan berlaku secara membentuk karat
Proses pengoksidaan serentak.
berlaku. [1] (ii) C + ZnO → Zn + CO 2 (ii) • Alloying / Pengaloian
• 2I (aq) → I 2 (aq) + 2e (iii) +2 to 0 • Using protective coating
–
–
2I (ak) → I 2 (ak) + 2e – [1] +2 kepada 0 for example greasing,
–
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08 Ans[Pra A++ Chem F5].indd 9 14/02/2019 5:25 PM

Chemistry Form 5 Answers
painting, electroplating or segi perubahan nombor 4. Bromine water is added
galvanising. pengoksidaan carefully to the other arm of
Menggunakan lapisan • Zinc undergoes oxidation the U-tube.
pelindung, contohnya as its oxidation number 5. A carbon electrode is dipped
dengan menggunakan increases from 0 to +2. in each of the solutions.
gris, mengecat, Zink mengalami 6. The electrodes are connected
penyaduran elektrik dan pengoksidaan apabila to a galvanometer by using
penggalvanian nombor pengoksidaannya connecting wires.
(b) 2Mg + O 2 → 2MgO [1] bertambah dari 0 kepada +2. 1. Tiub-U diapitkan pada kaki
2+
Number of mole of Mg • Copper(II) ion, Cu retort.
Bilangan mol Mg undergoes reduction as 2. Asid sulfurik cair dituang ke
7.2 oxidation number of copper dalam tiub-U.
=
24 in copper(II) ion decreases 3. Larutan ferum(II) sulfat
= 0.3 mol [1] from +2 to 0. ditambah ke dalam salah
2 moles of Mg produces 2 Ion kuprum(II), Cu 2+ satu lengan tiub-U secara
moles of MgO mengalami penurunan berhati-hati.
2 mol Mg menghasilkan 2 mol apabila nombor 4. Air bromin ditambahkan
MgO [1] pengoksidaannya ke dalam satu lagi lengan
0.3 mole of Mg produces 0.3 berkurang daripada +2 tiub-U.
mole of MgO kepada 0. 5. Elektrod karbon dicelupkan
0.3 mol Mg menghasilkan 0.3 SECTION C ke dalam kedua-dua larutan.
mol MgO [1] 3. (a) 1. Reaction I is not a redox 6. Elektrod disambungkan
Mass of MgO / Jisim MgO reaction. kepada galvanometer
= 0.3 x [24 + 16] = 12.0 g [1] 2. No change in the oxidation menggunakan wayar
(c) Redox reaction in terms of numbers for all the elements penyambung.
electron gain/loss: before and after the reaction. Ionic equation / Persamaan ion
Tindak balas redoks dari segi 3. Reaction II is a redox 2Fe 2+ + Br 2 → 2Fe + 2Br -
3+
penerimaan/kehilangan elektron reaction.
• Zinc is above copper in the 4. The oxidation number of Mg equation – 1
Electrochemical Series. increases from 0 to +2 while balancing – 1
Zink berada di atas kuprum the oxidation number of Ag Labeled diagram
dalam Siri Elektrokimia. decreases from +1 to 0. Gambar rajah berlabel
• Zinc undergoes oxidation 1. Tindak balas I bukan tindak e –
as zinc atom releases balas redoks. G
electrons to form zinc ion, 2. Tiada perubahan pada
Zn . nombor pengoksidaan setiap – + Carbon
2+
Zink mengalami unsur sebelum dan selepas electrodes
pengoksidaan di mana tindak balas. Elektrod
atom zink melepaskan 3. Tindak balas II ialah tindak karbon
elektron untuk membentuk balas redoks.
ion zink, Zn . 4. Nombor pengoksidaan bagi FeSO (aq / ak) Br (aq / ak)
2+
2
4
• Copper(II) ion undergoes Mg meningkat daripada 0 Dilute sulphuric acid
reduction as copper(II) ion, kepada +2 manakala nombor Asid sulfurik cair
2+
Cu receives electrons to pengoksidaan bagi Ag diagram – 1
form copper atom. menurun daripada +1 kepada label – 1
Ion kuprum(II) mengalami 0.
penurunan. Ion kuprum(II), (b) A : Mg / Al / Zn
Cu menerima elektron B : Cu
2+
untuk membentuk atom Set I :
kuprum. Mg + CuSO 4 → Cu + MgSO 4 To prevent rusting
• Oxidation half equation: (reactant-1, product-1) Untuk mengelakkan pengaratan
Zn → Zn + 2e - Set II :
2+
Persamaan setengah Cu + Ag 2SO 4 → 2Ag + CuSO 4 PRACTICE Thermochemistry
pengoksidaan: Zn → Zn + 2e - (equation -1, balancing-1) 4 Termokimia
2+
• Reduction half equation: (c) Procedure / Prosedur
-
Cu + 2e → Cu 1. A U-tube is connected to a
2+
Persamaan setengah retort stand. Objective Questions PAPER 1
-
penurunan: Cu + 2e → Cu 2. Dilute sulphuric acid is 1. D 2. B 3. A 4. B 5. B
2+
Redox reaction in terms of poured into the U-tube. 6. B 7. D 8. A 9. A 10. A
change in the oxidation number: 3. Iron(II) sulphate solution is 11. A 12. C 13. B 14. B 15. C
Tindak balas redoks dari added carefully to one of the 16. C 17. D 18. D 19. D 20. B
arms of the U-tube. 21. A 22. C 23. A
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08 Ans[Pra A++ Chem F5].indd 10 14/02/2019 5:25 PM

Chemistry Form 5 Answers

Subjective Questions PAPER 2 (f) Suhu campuran menurun
2.8
Energy / Tenaga sebanyak = 1.4°C
SECTION A CaCO 3 (s) + 2NaNO 3 (aq) 2
1. (a) To let the solutions achieve a CaCO 3 (p) + 2NaNO 3 (ak) 2. (a) The heat produced when one
uniform temperature. mole of ethanol is completely
Supaya larutan mencapai suhu Ca(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) ∆H = +11.76 kJ mol –1 burnt in oxygen under standard
yang sekata. Ca(NO 3 ) 2 (ak) + Na 2 CO 3 (ak) conditions.
(b) Ca(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) → Haba yang dihasilkan apabila
CaCO 3 (s) + 2NaNO 3 (aq) (g) The temperature of the mixture satu mol etanol terbakar
2.8
Ca(NO 3 ) 2 (ak) + Na 2 CO 3 (ak) → will drop by 2 = 1.4°C dengan lengkap dalam oksigen
CaCO 3 (p) + 2NaNO 3 (ak) pada keadaan piawai.
(c) A white precipitate is formed. (b)
The temperature of the mixture Thermometer / Termometer
drops. Wind shield / Pengadang angin
Mendakan putih terbentuk. Water Copper can / Bekas kuprum
Suhu campuran menurun. Air Tripod stand / Tungku kaki tiga
(d) Heat change = mcθ Ethanol Spirit lamp / Lampu spirit
Perubahan haba = mcθ Etanol Wooden block / Bongkah kayu
= (40 + 40) × 4.2 × 2.8
= 940.8 J (c) Copper is a good conductor Jadi, jumlah tenaga haba
(e) Number of moles of calcium of heat, thus it can transfer yang terbebas semasa
nitrate the heat released during the pembakaran 1 mol etanol
Bilangan mol kalsium nitrat combustion of ethanol to the = 1 × 25.2
= 2 x 40 water. 0.01848
1000 Kuprum adalah konduktor haba = 1364 kJ
= 0.08 mol yang baik, oleh itu kuprum ∴ Heat of combustion of
boleh memindahkan haba yang ethanol
Number of moles of sodium terbebas kepada air semasa
carbonate Haba pembakaran etanol
Bilangan mol natrium karbonat pembakaran etanol. = –1364 kJ mol –1
2 x 40 (d) C 2 H 5 OH(l) + 3O 2 (g) → (f) Heat is released to the
= 1000 2CO 2 (g) + 3H 2 O(l)
= 0.08 mol C 2 H 5 OH(ce) + 3O 2 (g) → surrounding.
Haba terbebas ke persekitaran.
2CO 2 (g) + 3H 2 O(ce)
Number of moles of calcium (e) (i) Mass of ethanol burnt 3. (a) • The intensity of the blue
carbonate precipitate = 0.08 mol Jisim etanol yang terbakar colour of the mixture
Bilangan mol mendakan decreases.
kalsium karbonat = 0.08 mol = 142.78 – 141.93 Keamatan warna biru
= 0.85 g campuran itu berkurang.
Precipitation of 0.08 mol of Number of moles of ethanol
CaCO 3 absorbs 940.8 J of heat burnt • A brown solid is formed.
energy. Bilangan mol etanol yang Pepejal berwarna perang
Pemendakan 0.08 mol CaCO 3 terbakar terbentuk.
menyerap 940.8 J tenaga haba. 0.85 (b) Zn(s) + CuSO 4 (aq) →
=
46
∴ The total heat energy = 0.01848 mol ZnSO 4 (aq) + Cu(s)
absorbed during the Zn(p) + CuSO 4 (ak) →
precipitation of one mole of (ii) Heat change = mcθ ZnSO 4 (ak) + Cu(p)
CaCO 3 Perubahan haba = mcθ (c) (i) 0.65 g
Jumlah tenaga haba yang = 200 × 4.2 × (57.0 – 27.0) (ii) 8°C
diserap semasa pemendakan = 25 200 J (iii) Number of moles of copper
= 25.2 kJ displaced
satu mol CaCO 3
1 (iii) Combustion of 0.01848 = Number of moles of
= 0.08 × 940.8 mol of ethanol gives out copper(II) sulphate
= 11760 J 25.2 kJ of heat energy. Bilangan mol kuprum
= 11.76 kJ So, the total heat energy disesarkan
given out during the = Bilangan mol kuprum(II)
∴ Heat of precipitation of combustion of one mole of
CaCO 3 = +11.76 kJ mol –1 ethanol sulfat
50 x 0.20
Haba pemendakan CaCO 3 Pembakaran 0.01848 mol = 1000
= +11.76 kJ mol –1 etanol membebaskan = 0.01 mol
25.2 kJ tenaga haba.




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08 Ans[Pra A++ Chem F5].indd 11 14/02/2019 5:25 PM

Chemistry Form 5 Answers
(iv) Heat change = mcθ Kemudian, atom hidrogen Number of moles of
Perubahan haba = mcθ dan atom klorin hydrochloric acid, HCl used
= 50 × 4.2 × (8.0) bergabung untuk Bilangan mol asid hidroklorik
= 1680 J membentuk molekul yang digunakan
= 1.68 kJ hidrogen klorida. [1] 1 x 50
(v) Displacement of 0.01 mol • Energy is given out during = 1000
of copper by zinc gives out the formation of bonds. = 0.05 mol [1]
1.68 kJ of heat energy. Tenaga dibebaskan
–
+
Penyesaran 0.01 mol semasa pembentukan • H (aq) + OH (aq) → H 2 O(l)
–
kuprum oleh zink ikatan. [1] H (ak) + OH (ak) → H 2 O(ce)
+
membebaskan 1.68 kJ ∆H = –57 kJ mol
–1
tenaga haba. • The total energy given The reaction between one
∴ The total heat energy out during the formation mole of H ions and one mole
+
given out during the of bonds is greater of OH ions to form one mole
–
displacement of one mole than the total energy of water releases 57 kJ of
of copper absorbed during the heat energy.
Jumlah tenaga haba breaking of bonds.
yang terbebas semasa Jumlah tenaga yang Tindak balas antara satu mol
+
penyesaran satu mol terbebas semasa ion H dengan satu mol ion
–
kuprum pembentukan ikatan OH membentuk satu mol air
yang membebaskan 57 kJ
= 1 × 1.68 = 168 kJ lebih besar berbanding tenaga haba. [1]
0.01 jumlah tenaga yang
The heat of displacement diserap semasa • Hydrochloric acid is a strong
of copper by zinc pemutusan ikatan. [1] monoprotic acid.
Haba penyesaran kuprum (iii) Asid hidroklorik adalah asid
oleh zink Energy / Tenaga monoprotik yang kuat. [1]
= –168 kJ mol –1 • 0.05 mol of hydrochloric acid
SECTION B H 2 (g) + CI 2 (g) dissociates completely to
form 0.05 mole of H ions.
+
4. (a) (i) A chemical reaction that ∆H = –184 kJ mol –1 0.05 mol asid hidroklorik
gives out heat to the 2HCI(g) mengion sepenuhnya
surrounding. membentuk 0.05 mol ion H .
+
Suatu tindak balas kimia [2] [1]
yang membebaskan haba (iv) Number of moles of • The neutralisation involves
ke persekitaran. [1] chlorine reacted the reaction between 0.05
(ii) • When hydrogen and Bilangan mol klorin yang mol of H ions and
+
chlorine combine to bertindak balas 0.05 mol of OH ions (from
–
form hydrogen chloride, = 14.2 excess sodium hydroxide
the bonds between the 2(35.5) solution) to form 0.05 mol of
hydrogen atoms in each = 0.2 mol [1] water.
molecule and the bonds When 1 mole of chlorine Peneutralan ini melibatkan
between the chlorine reacts completely, the heat tindak balas antara 0.05
atoms in each molecule given out = 184 kJ mol ion H dengan 0.05 mol
+
are broken. Apabila 1 mol klorin ion OH (daripada natrium
–
Apabila hidrogen bertindak balas lengkap, hidroksida berlebihan) untuk
dan klorin bergabung haba yang terbebas membentuk 0.1 mol air. [1]
membentuk hidrogen = 184 kJ [1] • Sulphuric acid is a strong
klorida, ikatan antara diprotic acid.
atom hidrogen dalam ∴ When 0.2 mole of Asid sulfurik merupakan asid
setiap molekul dan chlorine reacts completely, diprotik yang kuat. [1]
ikatan antara atom klorin the heat given out • 0.05 mol of sulphuric acid
dalam setiap molekul Apabila 0.2 mol klorin dissociates completely to
berpecah. [2] bertindak balas lengkap, form 0.1 mole of H ions.
+
• Energy is absorbed for haba yang terbebas 0.05 mol asid sulfurik
the breaking of these = 0.2 × 184 mengion sepenuhnya
bonds. = 36.8 kJ [1] membentuk 0.1 mol ion H [1]
+
Tenaga diserap untuk (b) • Number of moles of sulphuric • The neutralisation involves
memutuskan ikatan ini. acid, H 2 SO 4 used the reaction between 0.1 mol
[1] Bilangan mol asid sulfurik, of H ions and 0.1 mol of OH
–
+
• Then the hydrogen H 2 SO 4 yang digunakan ions (from excess sodium
atoms and the chlorine 1 x 50 hydroxide solution) to form
atoms combine to form = 1000 0.1 mol of water. Thus, the
hydrogen chloride = 0.05 mol heat released is doubled.
molecules.
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08 Ans[Pra A++ Chem F5].indd 12 14/02/2019 5:25 PM

Chemistry Form 5 Answers
Peneutralan ini melibatkan disukat menggunakan Initial temperature of
tindak balas antara 0.1 mol silinder penyukat. Larutan hydrochloric acid = t 1
ion H dengan 0.1 mol ion itu kemudian dituang ke Suhu awal asid
+
OH (daripada larutan natrium dalam cawan polistirena.
–
hidroksida berlebihan) [1] hidroklorik = t 1
Initial temperature of
membentuk 0.1 mol air. Oleh 2. 50 cm of 2.0 mol dm sodium hydroxide
3
–3
itu, haba terbebas adalah hydrochloric acid is
dua kali ganda. [1] measured using another solution = t 2
measuring cylinder. It is Suhu awal larutan natrium
SECTION C then poured into another hidroksida = t 2 14444444444244444444443 [1]
polystyrene cup. Average initial temperature
5. (a) (i) HCl(aq) + NaOH(aq) → 50 cm asid hidroklorik Purata suhu awal
3
NaCl(aq) + H 2 O(l) 2.0 mol dm disukat t 1 + t 2
–3
HCl(ak) + NaOH(ak) → menggunakan silinder = 2 = T 1
NaCl(ak) + H 2 O(ce) penyukat yang berlainan. Highest temperature of
∆H = –x kJ mol [2] Larutan itu kemudian mixture = T 2
–1
(ii) CH 3 COOH(aq) + NaOH(aq) dituang ke dalam cawan Suhu tertinggi
→ CH 3 COONa(aq) + H 2 O(l) polistirena yang lain. [1] campuran = T 2
CH 3 COOH(ak) + NaOH(ak) 3. The initial temperatures
→ CH 3 COONa(ak) + H 2 O(ce) of both solutions are Number of moles of
∆ H = –y kJ mol –1 [2] measured and recorded. hydrochloric acid
(b) • The value of x is greater than Suhu awal kedua-dua Bilangan mol asid hidroklorik
the value of y. larutan disukat dan 2.0 x 50
Nilai x lebih tinggi berbanding direkodkan. [1] = 1000 = 0.1 mol 144444244443 [1]
nilai y. [1] 4. The hydrochloric acid Number of moles of sodium
• Hydrochloric acid is a strong is poured quickly hydroxide = 0.1 mol
acid while ethanoic acid is a and carefully into the Bilangan mol natrium
weak acid. polystyrene cup containing hidroksida = 0.1 mol
Asid hidroklorik merupakan sodium hydroxide solution. ∴ Number of moles of water
asid kuat manakala asid Asid hidroklorik dituang formed / Bilangan mol air
etanoik merupakan asid dengan cepat dan terbentuk = 0.1 mol
lemah. [1] berhati-hati ke dalam Heat produced / Haba terbentuk
• Hydrochloric acid dissociates cawan polistirena yang = 100 × 4.2 × (T 2 – T 1 )
completely in water. mengandungi larutan = x J [1]
Asid hidroklorik mengion natrium hidroksida. [1] Heat of neutralisation
lengkap di dalam air. [1] 5. The mixture is stirred using Haba peneutralan
• Ethanoic acid dissociates the thermometer. –x
partially in water. Campuran dikacau dengan = —————– kJ mol –1 [1]


Asid etanoik mengion separa menggunakan termometer. 0.1 × 1000
di dalam air. [1] [1]
• Hence, some of the heat 6. The highest temperature
energy given out during the of the reaction mixture is OBJECTIVE QUESTIONS
neutralisation reaction is used recorded.
to dissociate the ethanoic Suhu tertinggi campuran 1. D 2. C 3. B 4. B 5. B
acid completely. tindak balas direkodkan. [1] 6. A 7. B 8. A 9. B 10. D
Oleh itu, sesetengah tenaga 7. Steps 1 to 6 are repeated SUBJECTIVE QUESTIONS
3
yang terbebas semasa using 50 cm of 2.0 mol
–3
tindak balas peneutralan dm ethanoic acid to SECTION A
digunakan semula untuk asid replace the hydrochloric 1. (a) (i) Heat of neutralisation is the
etanoik mengion dengan acid. heat released when 1 mole
sepenuhnya. [1] Langkah 1 hingga of water is formed from
(c) Procedure / Prosedur: 6 diulang dengan 3 neutralisation of acid with
1. 50 cm of 2.0 mol dm menggunakan 50 cm –3 an alkali.
3
–3
asid etanoik 2.0 mol dm
sodium hydroxide solution untuk menggantikan asid Haba peneutralan ialah
is measured using a hidroklorik. [1] perubahan haba yang
measuring cylinder. It dibebaskan apabila 1 mol
is then poured into a Calculation / Pengiraan: air terbentuk daripada
polystyrene cup. Hydrochloric acid + sodium peneutralan antara asid
50 cm larutan natrium hydroxide solution dengan alkali. [1]
3
–3
hidroksida 2.0 mol dm Asid hidroklorik + larutan
natrium hidroksida
A13 © Penerbitan Pelangi Sdn. Bhd.






08 Ans[Pra A++ Chem F5].indd 13 14/02/2019 5:25 PM

Chemistry Form 5 Answers
(ii) CH 3 COOH / Ethanoic acid SECTION C Displacement of 1 mole of Cu
Asid etanoik 2. (a) Number of moles of copper(II) releases 26 250 J
(iii) 1. The heat of sulphate Penyesaran 1 mol Cu
neutralisation for Bilangan mol kuprum(II) sulfat membebaskan 26 250 J
Experiment I is higher MV 0.0125 mole of Cu releases
than Experiment II. = 1000 26 250 x 0.0125 = 328.13 J
2. HCl ionises completely 0.5(25) 0.0125 mol Cu membebaskan
in water whereas = 1000 = 0.0125 mol 2650 x 0.0125 = 328.13 J
CH 3 COOH ionises Δ H = mcθ
partially in water. 328.13 J = (25)(4.2)(θ)
3. Some of the heat θ = 3.1°C
released in Experiment
II is absorbed to further (b)
ionise CH 3 COOH Reaction I II
molecules. Tindak balas
1. Haba peneutralan dalam Type of reaction Exothermic Endothermic
Eksperimen I lebih tinggi Jenis tindak balas Eksotermik Endotermik
daripada Eksperimen II.
2. HCl mengion lengkap Temperature change Increase Decrease
Perubahan suhu
Berkurang
Bertambah
dalam air manakala
CH 3 COOH mengion Change in total energy Total energy content of Total energy content of
separa dalam air. content of reactants and reactants is higher than reactants is lower than
3. Sebahagian haba total energy content of total energy content of total energy content of
products
products
products
yang dibebaskan Perubahan jumlah Kandungan tenaga Kandungan tenaga
dalam Eksperimen II
diserap semula untuk kandungan tenaga bahan tindak balas bahan tindak balas
bahan tindak balas
lebih rendah daripada
lebih tinggi daripada
mengionkan molekul
CH 3 COOH yang belum dan jumlah kandungan kandungan tenaga hasil kandungan tenaga hasil
mengion. tenaga hasil tindak balas tindak balas tindak balas
(b) (i) Number of mol of propanol
3
Bilangan mol propanol (c) 1. 100 cm of water is measured Sumbu pelita mesti menyentuh
= 0.6 / 3(12) + 7(1) + 16 + 1 and poured into a copper can. dasar bekas kuprum dan
3
= 0.01 mol [1] 100 cm air disukat dan sumbu dinyalakan.
(ii) H = mc dituangkan ke dalam bekas 7. The water is stirred
kuprum.
= 160 × 4.2 × 30 continuously with the
= 20 160 J [1] 2. A thermometer is placed thermometer until its
(iii) 0.01 mol of propanol into the water and the initial temperature increases by 30°C,
temperature, T 1 is recorded.
the flame is put off.
releases 20 160 J of heat Termometer diletakkan di Air dikacau secara berterusan
energy
1 mol of propanol releases dalam air dan suhu awal, T 1 dengan termometer sehingga
direkodkan.
suhu naik sebanyak 30°C,
2 016 000 J of heat energy
0.01 mol propanol 3. The metal can is placed on a nyalaan dipadamkan.
membebaskan 20 160 J tripod stand. 8. The highest temperature, T 2 of
tenaga haba Bekas kuprum diletakkan di water is recorded.
1 mol propanol membebaskan atas tungku kaki tiga. Suhu tertinggi T 2 dicapai oleh
2 016 000 J tenaga haba 4. A lamp is filled with propanol. air dicatatkan.
∆H = –2016 kJ mol –1 The lamp is weighed and the 9. The final mass of the lamp,
(c) initial mass, m 1 is recorded. m 2 and its content is weighed
Thermometer Sebuah pelita diisi dengan immediately and recorded.
Termometer propanol. Pelita itu ditimbang Jisim akhir lampu dan
Windshield dan jisim awal, m 1 dicatatkan. kandungannya, m 2 ditimbang
Pengadang angin 5. A windshield is placed to dengan serta merta dan
Water reduce heat loss to the moving dicatatkan.
Air Metal can
Tin logam air in the surrounding. 10. Steps 1 to 9 are repeated with
Tripod stand Pengadang angin diletakkan butanol to replace propanol.
Tungku kaki tiga untuk mengurangkan Langkah 1 hingga 9
kehilangan haba disebabkan diulangi menggunakan
Propanol Spirit lamp pergerakan udara di butanol untuk menggantikan
Propanol Lampu spirit persekitaran. propanol.
Diagram – 1 6. The wick of the lamp must
Label – 1 touch the base of the copper
Windshield – 1 can and the wick is lighted.
© Penerbitan Pelangi Sdn. Bhd. A14


08 Ans[Pra A++ Chem F5].indd 14 14/02/2019 5:25 PM

Chemistry Form 5 Answers

Thermometer 2. (a) To relieve pain and to bring down Persamaan – Kedua-duanya
Termometer the temperature during fever. mengandungi dua bahagian:
Windshield Untuk mengurangkan sakit ekor hidrofobik hidrokarbon dan
[2]
Pengadang angin
Copper container dan menurunkan suhu semasa kepala hidrofilik berion.
demam.
Difference – The ionic head of
Water Bekas kuprum
Air Tripod stand (b) Aspirin / Aspirin soap is carboxylate ion whereas
Fuel Tungku kaki tiga (c) Liver and kidney damage the ionic head of detergent is a
Spirit lamp
Bahan api Lampu spirit Rosak hati dan ginjal sulphonate ion.
Wooden block (d) Antibiotic /Antibiotik Perbezaan – Kepala berion
Bongkah kayu sabun ialah ion karboksilat
(e) Streptomycin / Streptomisin manakala kepala berion
(f) To kill all the bacteria so that no detergen ialah ion sulfonat. [2]
bacteria can develop resistance
to the antibiotic (c) Cleaning agent B [1]

This is because it does not form
Serve in polystyrene container. Untuk membunuh semua scum in hard water,
Foamed polystyrene contains millions bakteria supaya tiada bakteria Bahan pencuci B. Hal ini
of tiny air bubbles trapped in the yang boleh bertahan terhadap kerana bahan pencuci B tidak
foam. Air is a poor heat conductor antibiotik tersebut membentuk kekat di dalam air
and since polystyrene has a high (g) Barbiturate or tranquiliser liat, [1]
thermal resistance, it is very effective in Barbiturat atau penenang whereas cleaning agent A
preventing heat transfer. (h) To reduce depression, tension forms scum in hard water.
Menghidang dalam bekas polistirena. and anxiety manakala, bahan pencuci A
Polistirena mengandungi berjuta- Untuk megurangkan kemurungan, membentuk kekat di dalam air
juta gelembung udara halus yang ketegangan dan kebimbangan liat. [2]
boleh memerangkap haba. Udara (i) Addiction and drowsiness (d) Cleaning agent A [1]
adalah pengalir haba yang lemah, Ketagihan dan mengantuk This is because it is
dan disebabkan polisterina bersifat 3. (a) Natural food preservatives are biodegradable or can be
pemerangkap haba, maka suhu daripada substances derived from natural decomposed by bacteria.
sup cendawan sukar untuk dibebaskan. resources used to prevent Bahan pencuci A. Hal ini
spoilage of food. kerana bahan pencuci A
PRACTICE Chemicals for Consumers Bahan pengawet semula jadi terbiodegradasi atau boleh
5 Bahan Kimia untuk Pengguna adalah bahan yang diterbitkan diurai oleh bakteria. [1]
daripada sumber semula It is also not toxic to aquatic
Objective Questions PAPER 1 jadi yang digunakan untuk animals.
1. A 2. D 3. A 4. A 5. C menghalang kerosakan makanan. Bahan pencuci A juga tidak
6. B 7. A 8. A 9. A 10. A (b) (i) Vinegar / Cuka toksik kepada hidupan akuatik.
11. D 12. C 13. C 14. A 15. A (ii) Sodium chloride [2]
16. D 17. C 18. A 19. C 20. B Natrium klorida (e)
21. B (iii) Sugar / Gula Additive Function
(c) Salt draws water out of the cell of Bahan tambah Fungsi
Subjective Questions PAPER 2 the microorganisms by osmosis. Sodium To eliminate mud
This will kill the microorganisms
SECTION A through dehydration. tripolyphosphate stain and soften the
Natrium tripolifosfat water
1. (a) Alkaline hydrolysis Garam mengeluarkan air [1] Untuk
Hidrolisis beralkali daripada sel mikroorganisma menyingkirkan
(b) C 17 H 35 COONa melalui osmosis. Ini akan kotoran lumpur dan
(c) Ionic head, COO – membunuh mikroorganisma melembutkan air [1]
Bahagian kepala berion, COO – melalui pendehidratan. Sodium perborate As a bleaching
(d) Saponification (d) (i) Flavourings / Perisa Natrium perborat agent to remove
Saponifikasi (ii) To increase the taste and [1] dirty stain
(e) Ester / Ester smell of the food Sebagai agen
(f) Glycerol / Gliserol Untuk meningkatkan rasa peluntur untuk
(g) Alcohol / Alkohol dan bau makanan menyingkirkan
(h) • Soap is not effective in hard (e) Food dye / Pewarna makanan kotoran [1]
water. SECTION B Biological enzyme To remove blood
Sabun tidak berkesan dalam Enzim biologi [1] and food stains
air liat. 4. (a) A – Soap / Sabun [1] Untuk
• Soap cannot clean effectively B – Detergent / Detergen [1] menyingkirkan
in acidic solution. (b) Similarity – Both contain two kotoran darah dan
Sabun tidak boleh mencuci parts: hydrophobic hydrocarbon makanan [1]
dengan berkesan dalam tail and hydrophilic ionic head.
larutan berasid.



A15 © Penerbitan Pelangi Sdn. Bhd.






08 Ans[Pra A++ Chem F5].indd 15 14/02/2019 5:25 PM

Chemistry Form 5 Answers

Sodium silicate and As a drying agent (e) • Sugar or salt – Smoking can partially
sodium sulphate to absorb moisture Gula atau garam remove the water.
Natrium silikat dan so that the – Addition of sugar and Menyalai boleh
natrium sulfat [1] detergent remains salt has the effect of menyingkirkan sebahagian
dry reducing the activity of air [1]
Sebagai agen microorganisms. – Smoke also contains
pengering Penambahan gula dan formaldehyde and phenols
untuk menyerap garam mempunyai that kill the bacteria.
lembapan bagi kesan penurunan aktiviti Asap juga mengandungi
mengekalkan mikroorganisma. [1] formaldehid dan fenol yang
kekeringan – Sugar or salt has the ability membunuh bakteria. [1]
detergen [1] to draw out water from the – Food that is preserved using
smoke is smoked meat.
Fluorescent To absorb cell of the microorganisms Makanan yang diawet
through osmosis. This will
substance ultraviolet light so kill the microorganisms. menggunakan asap ialah
Bahan pendarflour that white cloth Gula atau garam daging salai. [1]
[1] becomes whiter mempunyai keupayaan
Untuk menyerap untuk menyingkirkan
cahaya ultraungu air daripada sel
supaya pakaian mikroorganisma OBJECTIVE QUESTIONS
putih menjadi melalui osmosis. Hal 1. A 2. B 3. C 4. C 5. C
semakin putih [1] ini akan membunuh 6. A
[Any three / Mana-mana tiga] mikroorganisma itu. [1]
Max: 6 – Foods that are preserved SUBJECTIVE QUESTIONS
SECTION C using sugar are condensed SECTION A
sweetened milk and jam.
5. (a) Food additives are chemicals Makanan yang diawetkan 1. (a) (i) Saponification
that are added in small menggunakan gula adalah Saponifikasi
quantities to the food for susu pekat dan jem. [1] (ii) Sodium chloride
various functions. – Foods that are preserved Natrium klorida
Bahan tambah makanan using salt are salted fish, (iii) To precipitate the soap
adalah bahan kimia yang salted meat or salted Untuk menghasilkan
ditambah dalam kuantiti yang vegetable. mendakan sabun
kecil kepada makanan untuk Makan yang diawetkan (iv) Glycerol
pelbagai fungsi. [2] menggunakan garam ialah Gliserol
(b) • Growth of bacteria or fungi ikan masin, daging masin (b) (i) Set 1 : Oily stain remains
Kotoran kekal
Pertumbuhan bakteria atau atau sayuran masin. [1] Set 2 : Oily stain is
kulat [1] • Acid / Asid removed
• Oxidation of food – Lowering the pH values Kotoran hilang
Pengoksidaan makanan [1] would make bacteria less (ii) • Hard water contains Ca
2+
active.
(c) Antioxidant. To prevent [1] Merendahkan pH akan and Mg ion.
2+
oxidation of food which causes menyebabkan bakteria Air liat mempunyai ion
rancidity in food. [1] Ca dan Mg .
2+
2+
kurang aktif.
Pengantioksida. Untuk – Example of acid commonly • Soap reacts with Ca
2+
menghalang pengoksidaan used is vinegar (ethanoic and Mg ion to form
2+
makanan yang menyebabkan acid) which is used in pickling scum.
rasa tengik pada makanan. [2] or lactic acid in yoghurt. Sabun bertindak
(d) (i) • To inhibit the growth of Contoh asid yang biasa balas dengan ion
microorganisms digunakan ialah cuka (asid Ca dan Mg untuk
2+
2+
Untuk menghalang etanoik) yang digunakan menghasilkan kekat.
pertumbuhan dalam penjerukan
mikroorganisma [1] atau asid laktik dalam (iii) Bleach // Sodium perborate
• To give pink or red pembuatan yogurt. [1] Peluntur // Natrium perborat
colour to the meat • Smoking / Menyalai
Untuk memberikan – Smoking is a process of
warna merah jambu atau allowing warm air containing River water is hard water which contains
merah kepada daging [1] smoke to pass over the food. Mg and Ca ions. Soap will produce
2+
2+
(ii) It may cause stomach cancer Menyalai adalah proses scum in hard water.
Boleh menyebabkan yang membenarkan udara Air sungai ialah air liat yang
kanser perut [1] suam yang mengandungi mengandungi ion Mg dan ion Ca .
2+
2+
asap melalui makanan. [1]
Sabun menghasilkan kekat di dalam air
liat.
© Penerbitan Pelangi Sdn. Bhd. A16
08 Ans[Pra A++ Chem F5].indd 16 14/02/2019 5:25 PM

Chemistry Form 5 Answers
Pembentukan mendakan sulfur 4. The two beakers are left
mengaburkan pangkah. overnight.
Chapter 1 (e) A more accurate results can be Kedua-dua bikar dibiarkan
obtained. It can also change semalaman.
1. (a) Concentration the concentration of sodium 5. Any changes that occur
of sodium Rate of thiosulphate when the volume are recorded.
thiosulphate reaction, of sodium thiosulphate solution Sebarang perubahan yang
solution, A , 1 (s ) used is varied. berlaku direkodkan.
–1
50 D Hasil yang lebih tepat boleh (e) Tabulation of data
(mol dm ) Kadar diperoleh. Jumlah keseluruhan Penjadualan data
–3
Kepekatan larutan tindak 1 yang dimalarkan juga akan
natrium tiosulfat, balas, mengubah kepekatan natrium Beaker Observation
A (mol dm ) (s ) D sulfat apabila isi padu larutan Bikar Pemerhatian
–1
–3
50 natrium tiosulfat berubah. A (latex + ammonia
1.0 0.0333 (f) First: Sodium thiosulpate solution solution)
0.8 0.0263 Pertama: Larutan narium tiosulfat A (lateks + larutan
ammonia)
Second: Water
0.6 0.0200 Kedua: Air B (latex only)
0.4 0.0135 Third: Hydrochloric acid B (lateks sahaja)
0.2 0.0065 Ketiga: Asid hidroklorik
(g) The rate is directly proportional
(b) to the concentration of sodium 3. (a) Beaker Observation
Rate of reaction (s ) thiosulphate solution. Bikar Pemerhatian
–1
Kadar tindak balas (s –1 ) Kadar adalah berkadar A The latex coagulates
langsung dengan kepekatan
0.03 larutan natrium tiosulfat. within 5 minutes.
Lateks menggumpal
Chapter 2 dalam masa 5 minit.
0.02 2. (a) Aim: To investigate the effect of B The latex does not
alkali on the coagulation of latex. coagulate after 6
Tujuan: Menyiasat kesan alkali hours.
0.01 terhadap penggumpalan lateks Lateks tidak
menggumpal selepas
(b) Hypothesis: Ammonia prevents 6 jam.
coagulation of latex
Hipotesis: Ammonia mencegah C The latex coagulates
0.2 0.4 0.6 0.8 1.0 penggumpalan lateks after 6 hours.
–3
Concentration of Na S O (mol dm ) (c) Materials: Latex, 2 mol dm Lateks menggumpal
-3
2
3
2
Kepekatan Na 2 S 2 O 3 (mol dm –3 ) ammonia solution selepas 6 jam.
(c) (i) Concentration of sodium Bahan: Lateks, larutan (b) Ethanoic acid speeds up
thiosulphate solution ammonia 2 mol dm -3 the coagulation of latex but
Kepekatan larutan natrium Apparatus: Beakers, glass rod, ammonia solution prevents the
3
tiosulfat 50 cm measuring cylinder coagulation of latex.
(ii) Changing the volume Radas: Bikar, rod kaca, silinder Asid etanoik mempercepatkan
3
of sodium thiosulphate penyukat 50 cm penggumpalan lateks tetapi
solution but keeping the (d) Procedure: / Prosedur: larutan ammonia mencegah
total volume constant. 1. 20 cm of latex is poured penggumpalan lateks.
3
Menukar isi padu larutan into each of the two (c) (i) Presence of ethanoic acid
natrium tiosulfat tetapi beakers labelled A and B. or ammonia.
memalarkan jumlah isi 20 cm lateks dituang ke Kehadiran asid etanoik
3
padu. dalam setiap bikar berlabel atau ammonia.
(iii) Concentration of A dan B. (ii) Coagulation of latex
hydrochloric acid 2. About 5 cm of ammonia Penggumpalan lateks
3
Kepekatan asid hidoklorik solution is added into (iii) Volume of latex
(iv) Use the same volume and beaker A. Isi padu lateks
3
concentration of HCl for 5 cm larutan ammonia (d) (i) The latex coagulates within
the repeated experiments. ditambahkan ke dalam 5 minutes.
Gunakan isi padu dan bikar A. Lateks menggumpal dalam
kepekatan HCl yang sama 3. The mixture is stirred well masa 5 minit.
untuk eksperimen yang with a glass rod.
diulang. Campuran dikacau dengan (ii) The nitric acid added in
(d) The formation of sulphur baik dengan menggunakan the beginning is used to
neutralise the ammonia
precipitate obscures the cross. rod kaca.
A17 © Penerbitan Pelangi Sdn. Bhd.



08 Ans[Pra A++ Chem F5].indd 17 14/02/2019 5:25 PM

Chemistry Form 5 Answers
solution. After all the Pemboleh ubah mendatar seperti yang
ammonia solution is dimanipulasikan: Jenis logam ditunjukkan dalam rajah.
neutralised, the excess Responding variable: Intensity 3. One spatula of copper
acid is used to coagulate of flame or glow powder is placed on a
the latex. Pemboleh ubah bergerak piece of asbestos paper
Asid nitrik yang ditambah balas: Keamatan nyalaan atau and is put into the tube.
pada awal eksperimen baraan Satu spatula serbuk
digunakan untuk Fixed variable: Amount of kuprum diletak pada kertas
meneutralkan larutan metal powder asbestos dan diletakkan
ammonia. Pemboleh ubah dimalarkan: dalam tabung didih.
Selepas semua larutan Jumlah serbuk logam 4. The copper powder is
ammonia dineutralkan, asid (c) Hypothesis: The more reactive heated strongly. Then,
yang berlebihan digunakan a metal, the more vigorous the the solid potassium
untuk menggumpalkan metal burns in oxygen manganate(VII) is heated.
lateks. Hipotesis: Semakin reaktif How vigorously copper
(e) (i) Bacteria in the air attacks suatu logam, semakin kuat reacts with oxygen is
the latex and produces logam tersebut terbakar dalam observed.
acid. This acid causes oksigen. Serbuk kuprum
coagulation of latex. (d) Materials: Copper powder, zinc dipanaskan dengan kuat.
Bakteria dalam udara powder, magnesium powder, Kemudian, pepejal kalium
menyerang lateks dan lead powder, solid potassium manganat(VII) dipanaskan.
menghasilkan asid. manganate(VII), asbestos Kekuatan tindak balas
Asid ini menyebabkan paper, glass wool kuprum dengan oksigen
penggumpalan lateks. Bahan: Serbuk kuprum, diperhatikan.
(ii) The quantity of acid in serbuk zink, serbuk 5. Steps 1 to 4 are repeated
beaker A is more than that magnesium, serbuk plumbum, using zinc powder,
in beaker C. pepejal kalium manganat(VII), magnesium powder and
Kuantiti asid dalam bikar kertas asbestos, wul kaca lead powder to replace
A adalah lebih tinggi Apparatus: Boiling tube, retort copper powder.
daripada bikar C. stand and clamp, Bunsen Langkah 1 hingga 4 diulang
(iii) Beaker C acts as a control. burner, spatula and forceps menggunakan serbuk zink,
serbuk magnesium dan
Bikar C bertindak sebagai Radas: Tabung didih, kaki retort
kawalan. dan pengapit, penunu Bunsen, serbuk plumbum untuk
menggantikan serbuk
(f) Subtance that can Subtance that spatula dan forseps kuprum.
coagulate latex cannot coagulate (e) Procedure / Prosedur: (f) Tabulation of data:
Bahan yang latex Metal powder Penjadualan data:
boleh Bahan yang Serbuk logam Metal powder Observation
menggumpalkan tidak boleh Solid potassium Serbuk logam Pemerhatian
lateks menggumpalkan Glass wool manganate(VII)
Pepejal kalium
lateks Wul kaca manganat(VII) Copper
Hydrochoric acid Sodium Kuprum
Asid hidroklorik hydroxide Zinc
solution Zink
Larutan natrium Heat Heat
hidroksida Panaskan Panaskan Magnesium
Methanoic acid Potassium Asbestos paper Magnesium
Asid metanoik hydroxide Kertas asbestos Lead
solution Plumbum
Larutan kalium 1. One spatula of solid
hidrokisda potassium manganate(VII) Chapter 4
is put into a boiling tube.
Satu spatula pepejal 5. (a) Aim: To determine the heat of
Chapter 3 kalium manganat(VII) combustion of various liquid
4. (a) Aim of the experiment: dimasukkan ke dalam alcohols
Tujuan: Menentukan haba
To determine the order of tabung didih. pembakaran bagi pelbagai
metals in the Reactivity Series 2. Some glass wool is pushed cecair alkohol
Tujuan eksperimen: into the tube. The tube is (b) Manipulated variable:
Menentukan susunan logam clamped horizontally as Type of liquid alcohol
dalam Siri Kereaktifan shown in the diagram. Pemboleh ubah
(b) Manipulated variable: Type of Wul kaca dimasukkan dimanipulasikan:
metal ke dalam tabung didih. Jenis cecair alkohol
Tabung didih diapit secara



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08 Ans[Pra A++ Chem F5].indd 18 14/02/2019 5:25 PM

Chemistry Form 5 Answers
Responding variable: Value of under the copper tin as is removed. The highest
heat of combustion shown in the diagram. The temperature of water is
Pemboleh ubah bergerak apparatus is then shielded recorded.
balas: Nilai haba pembakaran using windshield. Apabila suhu meningkat
Fixed variable: Volume of Pelita tersebut diletakkan dalam kira-kira 30°C, api
water, copper can di atas bongkah kayu di dipadamkan. Suhu tertinggi
Pemboleh ubah dimalarkan: bawah tin kuprum seperti air direkodkan.
Isi padu air, tin kuprum yang ditunjukkan dalam 9. The spirit lamp containing
(c) Hypothesis: Different rajah. Radas dilindungi the methanol is weighed
alocohols have different heat of dengan pengadang angin. again immediately.
combustion. 6. The wick of the lamp is Pelita yang mengandungi
Hipotesis: Alkohol yang lighted immediately. metanol ditimbang semula
serta-merta.
Sumbu pelita dinyalakan
berlainan mempunyai haba serta-merta. 10. Steps 1 to 9 are repeated
pembakaran yang berbeza. 7. The water is stirred using ethanol, propan-1-ol
(d) Materials: Methanol, ethanol, continuously using the and butan-1-ol to replace
propan-1-ol, butan-1-ol, water. thermometer. methanol.
Bahan: Metanol, etanol, Air dikacau dengan Langkah 1 hingga 9 diulang
propan-1-ol, butan-1-ol, air menggunakan termometer. menggunakan etanol,
Apparatus: 250 cm copper 8. Once the temperature of propan-1-ol dan butan-
3
can, tripod stand, thermometer, the water is increased 1-ol untuk menggantikan
100 cm measuring cylinder, by about 30°C, the flame metanol.
3
spirit lamp, electronic balance,
wooden block, windshield (f) Tabulation of data / Penjadualan data:
3
Radas: 250 cm tin kuprum,
tungku kaki tiga, termometer, Alcohol Methanol Ethanol Propan-1-ol Butan-1-ol
3
100 cm silinder penyukat, Alkohol Metanol Etanol Propan-1-ol Butan-1-ol
pelita, neraca elektronik,
bongkah kayu, pengadang Mass of spirit lamp
angin + alcohol before
(e) Procedure / Prosedur: combustion (g)
Thermometer Windshield Jisim pelita + alkohol
Termometer Pengadang angin sebelum pembakaran (g)
Mass of spirit lamp +
Copper can
Water alcohol after combustion
Air Tin kuprum (g)
Spirit lamp Tripod stand Jisim pelita + alkohol
Pelita Tungku kaki tiga selepas pembakaran (g)
Methanol Wooden block Initial temperature of
Metanol Bongkah kayu water ( C)
o
o
1. 200 cm of water is Suhu awal air ( C)
3
measured and poured into Highest temperature of
a copper can. o
water ( C)
200 cm air diukur dan Suhu tertinggi air
3
dituang ke dalam tin kuprum.
2. The copper can is placed
on a tripod stand without a Chapter 5 (e) Filter the mixture using filter
wire gauze. 6. (a) To hydrolyse the fats funnel and filter paper.
Tin kuprum diletakkan di Untuk menghidrolisis lemak Tapis campuran menggunakan
atas tungku kaki tiga tanpa (b) To reduce the solubility of soap corong turas dan kertas
kasa dawai. in the water. penapis.
3. The initial temperature of Untuk mengurangkan (f) (i) Foam Scum
the water is measured after keterlarutan sabun di dalam air. Buih Kekat
five minutes. (c) White flakes of soap formed
Suhu awal air diukur float on the surface of the
selepas lima minit. solution.
4. 50 cm of methanol is Emping putih sabun yang Pipe water + soap Sea water + soap
3
poured into a spirit lamp. terbentuk terapung pada Air paip+ sabun Air laut + sabun
The spirit lamp containing permukaan larutan.
the methanol is weighed. (d) When the white flakes of soap Foam (ii) Scum
50 cm metanol dituang ke are stirred with water, bubbles Buih Kekat
3
dalam pelita. Pelita tersebut
ditimbang. or foams are formed.
5. The spirit lamp is then Apabila emping putih sabun
Pipe water + soap
placed on a wooden block digaul dengan air, gelembung Sea water + soap
atau buih terhasil. Air paip+ sabun Air laut + sabun
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08 Ans[Pra A++ Chem F5].indd 19 14/02/2019 5:25 PM

Chemistry Form 5 Answers
(d) 1. Labelled axis with units: 5. (a) Electrolysis is a process
temperature (°C), time (s) whereby an electrolyte is
Paper 1 Label paksi dan unit: decomposed to its constituent
1. B 2. C 3. B 4. C 5. B suhu (°C), masa (s) elements when electric current
6. C 7. C 8. C 9. D 10. C 2. Suitable scale: covers 50% passes through it.
11. B 12. B 13. B 14. B 15. D of graph paper Elektrolisis adalah proses
16. B 17. B 18. D 19. D 20. A Skala yang sesuai: meliputi penguraian elektrolit kepada
21. B 22. B 23. B 24. D 25. C 50% kertas graf unsur juzuknya apabila arus
26. A 27. B 28. C 29. D 30. D 3. All points are plotted correctly. elektrik dialirkan melaluinya.
+
2+
-
31. B 32. A 33. C 34. C 35. C Semua titik diplot dengan (b) H , Cu , OH , SO 4 2-
36. C 37. B 38. B 39. B 40. A betul. (c) Anode / Anod: 4OH → 2H 2 O +
-
41. B 42. B 43. C 44. A 45. A 4. Correct shape and smooth O 2 + 4e -
46. D 47. C 48. A 49. C 50. C curve. Cathode / Katod: Cu + 2e →
-
2+
Bentuk graf yang betul dan
Paper 2 lengkung yang licin. Cu
Section A Temperature (°C) (d) (i) Brown solid is formed
Pepejal perang terbentuk


Suhu (°C)
1. (a) (i) Type of medicine / Jenis ubat: (ii) Copper metal
Analgesic / Analgesik 60 Logam kuprum
Example / Contoh: 50 x
Paracetamol / Parasetamol 40 x x (e) 1. Insert a glowing wooden
(ii) P: Analgesic / Analgesik 30 x x splinter into the test tube.
Q: Antidepressant 2. The splinter lights up.
Antidepresen 20 1. Masukkan kayu uji berbara
R: Streptomycin / Penicillin 10 ke dalam tabung uji.
Streptomisin / Penisilin Time (s) 2. Kayu uji menyala.
Suhu (s)
(b) Name / Nama: 10 20 30 40 50 60 70 80 90 (f) 1. Blue to pale blue
Lecithin / Lesitin (e) The higher the temperature of 2. Copper(II) ions are
Type / Jenis: sodium thiosulphate solution, discharged to form copper
Stabiliser / Penstabil the less the time taken for 'X' atoms.
(c) (i) Sausage / burger mark to disappear from sight. 3. Concentration of copper(II)
Sosej / burger Semakin tinggi suhu larutan ions decreases.
(ii) Prevents food from being natrium tiosulfat, semakin 1. Biru kepada biru cair
spoilt by slowing down the berkurang masa yang 2. Ion kuprum(II) dinyahcas
growth of microorganisms diperlukan untuk tanda ‘X’ menjadi atom kuprum
Mengelakkan makanan hilang dari pandangan. 3. Kepekatan ion kuprum(II)
daripada rosak dengan 4. (a) (i) Draw on the diagram, from berkurang
melambatkan pertumbuhan P to Q. 6. (a) Reaction I: Neutralisation
mikroorganisma Lukis pada rajah, dari P ke Q. Peneutralan
2. (a) (i) Alkene / Alkena (ii) Dilute sulphuric acid Reaction II: Double
(ii) C 3 H 6 Asid sulfurik cair decomposition / Precipitation
9
(b) C 3 H 6 + O 2 ➝ 3CO 2 + 3H 2 O (accept other suitable Penguraian ganda dua /
2
Pemendakan
(c) (i) Hydroxyl group answer / Terima jawapan (b) (i) Copper(II) sulphate
Kumpulan hidroksil lain yang sesuai) Kuprum(II) sulfat
(ii) 300°C, 60 atm (iii) Oxidising agent: (ii) Carbon dioxide. Pass the
(d) (i) 1,2-dibromopropane Agen pengoksidaan gas through limewater.
1,2-dibromopropana Bromine water / Air bromin Limewater turns chalky.
(ii) Brown solution becomes Reducing agent: Gas karbon dioksida.
colourless Agen penurunan: Lalukan gas pada air
Larutan perang menjadi Iron(II) sulphate solution kapur. Air kapur menjadi
tidak berwarna Larutan ferum(II) sulfat keruh.
(e) Propyl ethanoate (iv) Chlorine water (c) (i) CuSO 4 + Pb(NO 3 ) 2 →
Propil etanoat Air klorin
PbSO 4 + Cu(NO 3 ) 2
3. (a) Sulphur / Sulfur (b) (i) The colourless potassium (ii) 2 cm solution of salt C is
3
(b) 2HCl + Na 2 S 2 O 3 ➝ 2NaCl + iodide turns brown. poured into a test tube. 2
H 2 O + SO 2 + S Kalium iodida yang tidak cm of dilute sulphuric acid
3
(c) Eye berwarna berubah kepada is added followed by 2 cm
3
warna perang.
Mata (ii) Potassium iodide solution of iron(II) sulphate solution.
Sodium thiosulphate solution The mixture is shaken.
Larutan natrium tiosulfat Larutan kalium iodida The test tube is slanted
+ (iii) 2I ➝ I 2 + 2e –
-
hydrochloric acid and few drops of
asid hidroklorik (iv) 0 to -1 concentrated H 2 SO 4 acid is
0 kepada -1 added slowly. Brown ring is
White paper marked with ‘’ formed.
Kertas putih yang bertanda ‘’
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08 Ans[Pra A++ Chem F5].indd 20 14/02/2019 5:25 PM

Chemistry Form 5 Answers
3
2 cm larutan garam C (ii) • The boiling point of 8. (a) • Acidic soil is treated with
terlarut dituang ke dalam chlorine is higher than soda lime powder.
tabung uji. 2 cm asid fluorine. Tanah berasid dirawat dengan
3
sulfurik cair diikuti dengan Takat didih klorin lebih serbuk kapur tohor. [1]
2 cm larutan ferum(II) tinggi daripada flourin. [1] • Basic soil is treated with
3
sulfat ditambah. Campuran • The size of chlorine compost.
digoncang. Tabung molecule is bigger than Tanah berbes dirawat
uji dicondongkan dan fluorine molecule. dengan kompos. [1]
beberapa titis H 2 SO 4 pekat Saiz molekul klorin lebih • Acidic gases emitted by
dititiskan perlahan-lahan. besar daripada molekul industries are neutralised
Cincin perang terbentuk. flourin. [1] with soda lime.
Section B • The force of attraction Gas-gas berasid yang
7. (a) (i) • Electron arrangement of between chlorine dibebaskan oleh kilang
fluorine atom is 2.7 molecule is stronger. dineutralkan dengan kapur
Susunan elektron bagi Daya tarikan antara tohor. [1]
atom fluorin ialah 2.7 [1] molekul klorin lebih kuat. • Baking powder cures bee
• Flourine atom has seven [1] stings.
valence electrons • More heat energy is Serbuk penaik merawat
Atom fluorin mempunyai needed sengatan lebah. [1]
tujuh elektron valens [1] Lebih banyak tenaga (b) (i) Acid X: Nitric acid
• Flourine is located in haba diperlukan [1] / hydrochloric acid
Group 17 • to overcome the (any strong acid except
Flourin berada dalam stronger forces between sulphuric acid)
kumpulan 17 [1] molecules. Asid X: Asid nitrik / asid
• Flourine atom has untuk mengatasi daya hidroklorik (mana-mana
two shells filled with yang kuat antara asid kuat kecuali asid
electrons molekul klorin. [1]
Atom flourin mempunyai (b) (i) Ionic compound / Sebatian sulfurik) [1]
dua petala yang terisi ion: Magnesium nitrate / Acid Y: Ethanoic acid (any
elektron [1] Magnesium nitrat weak acid)
• Hence, fluorine is Covalent compound Asid Y: Asid etanoik
located in Period 2 Sebatian kovalen: (mana-mana asid lemah)
Maka, flourin berada Tetrachloromethane [1]
dalam Kala 2 [1] Tetraklorometana (ii) Carbon dioxide
Karbon dioksida [1]
(ii) Ionic compound Characteristic Covalent compound (iii) HCl + CaCO 3 ➝ CaCl 2 +
Sebatian ionik Ciri-ciri Sebatian kovalen CO 2 + H 2 O [2]
• Mg(NO 3 ) 2 is high • CCl 4 is low (iv) • Acid X is a strong acid
Mg(NO 3 ) 2 lebih tinggi. CCl 4 lebih rendah while acid Y is a weak
• There are strong • The forces between acid.
electrostatic forces CCl 4 molecules are Asid X ialah asid kuat
2+ - weak.
between Mg and NO 3 manakala asid Y ialah
Terdapat daya Melting point / Daya di antara asid lemah. [1]
elektrostatik yang kuat Boiling point molekul CCl 4 adalah
antara Mg dengan Takat lebur / lemah • Acid X ionises
2+
-
NO 3. takat didih • Less heat energy is completely in water to
• More heat energy is needed to overcome form high concentration
needed to overcome the forces. of hydrogen ion.
the forces. Kurang tenaga haba Asid X mengion dengan
Lebih banyak tenaga yang diperlukan lengkap dalam air
haba yang diperlukan untuk mengatasi daya dan menghasilkan
untuk mengatasi daya tersebut. ion hidrogen yang
tersebut. berkepekatan tinggi. [1]
• Conducts electricity • Does not conduct • Acid Y ionises partially
in molten or aqueous electricity because in water to form low
state because there Electrical there are no free concentration of
are free moving ions. moving ions / exist as hydrogen ion.
Mengalirkan arus Conductivity molecules. Asid Y mengion
Kebolehan
elektrik dalam mengalir arus Tidak mengalirkan separa dalam air
keadaan leburan arus elektrik kerana
atau akueus kerana elektrik tiada ion yang bebas dan menghasilkan
terdapat ion yang bergerak / wujud ion hidrogen yang
bebas bergerak. dalam keadaan molekul. berkepekatan rendah. [1]
[8]
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08 Ans[Pra A++ Chem F5].indd 21 14/02/2019 5:25 PM

Chemistry Form 5 Answers
• The higher the bilangan sebenar atom 9. The crucible, lid and its
concentration of hidrogen dan atom karbon content are cooled at room
hydrogen ion, the lower dalam pentena. temperature and then
the pH value. 4. 1 molekul pentena weighed again.
Semakin tinggi mempunyai 5 atom karbon 10. The process of heating,
kepekatan ion hidrogen, dan 10 atom hidrogen. cooling and weighing are
semakin rendah nilai pH. (b) 1. Relative formula mass of repeated until a constant
[1] ammonium sulphate = 2[14 + mass is obtained.
• HCl ➝ H + Cl - [1] 4(1)] + 32 +4(16) = 132 1. 10 cm pita magnesium
+
• CH 3 COOH ➝ CH 3 COO - 2. Relative molecular mass of dibersihkan dengan
+ H [1] urea = 12 + 16 + 2[14 + 2(1)] menggunakan kertas pasir.
+
(c) 2KOH + H 2 SO 4 ➝ K 2 SO 4 + = 60 2. Mangkuk pijar dengan
2H 2 O [1] 3. Percentage of nitrogen by penutup ditimbang.
Number of mole KOH mass in ammonium sulphate 3. Pita magnesium digulung dan
Bilangan mol KOH = 14/132 x100% = 10.61% diletakkan di dalam mangkuk
= 1 x 50 = 0.05 mol [1] 4. Percentage of nitrogen by pijar.
1000 mass in urea = 14/60 x 100% 4. Mangkuk pijar, penutup dan
From the equation, 2 moles = 23.3% pita magnesium ditimbang.
of KOH reacts with 1 mole of 5. Urea is the better fertiliser. 5. Mangkuk pijar dipanaskan
H 2 SO 4 . dengan kuat tanpa penutup.
Daripada persamaaan, 2 mol 6. It is because urea has higher
KOH bertindak balas dengan 1 percentage of nitrogen by 6. Apabila pita magnesium
mass.
mula terbakar, mangkuk pijar
mol H 2 SO 4 . [1] ditutup dengan penutup.
Therefore, 0.05 mole of KOH 1. Jisim molekul relatif bagi 7. Penutup dibuka sekali sekala
ammonium sulfat = 2[14 +
reacts with 0.025 mole of H 2 SO 4 4(1)] + 32 +4(16) = 132 dengan menggunakan
Jadi, 0.05 mol KOH bertindak 2. Jisim molekul relatif bagi urea penyepit.
balas dengan 0.025 mol H 2 SO 4 = 12 + 16 + 2[14 + 2(1)] = 60 8. Apabila pita magnesium
[1] 3. Peratus nitrogen mengikut berhenti terbakar, penutup
Volume of H 2 SO 4 jisim dalam ammonium sulfat dibuka dan mangkuk pijar
Isi padu H 2 SO 4 = 14/132 x100% = 10.61% dipanaskan dengan kuat
0.025 x 1000 selama 2 minit lagi.
= 4. Peratus nitrogen mengikut
0.5 jisim dalam urea = 14/60 x 9. Mangkuk pijar, penutup dan
= 0.050 dm 100% = 23.3% kandungannya dibiarkan
3
= 50 cm [1] sejuk pada suhu bilik dan
3
5. Urea merupakan baja yang ditimbang sekali lagi.
Section C lebih baik. 10. Proses pemanasan,
9. (a) 1. Empirical formula is the 6. Ini kerana urea mempunyai penyejukan dan
formula that shows the peratus nitrogen mengikut penimbangan diulang
simplest ratio of hydrogen jisim yang lebih tinggi. beberapa kali sehingga jisim
atoms and carbon atoms in (c) 1. 10 cm of magnesium ribbon tetap diperoleh.
pentene. is cleaned with sand paper.
2. The simplest ratio of carbon 2. A crucible and its lid are 10. (a)
atom to hydrogen atom is weighed.
1:2. 3. The magnesium ribbon is Energy
Tenaga
3. Molecular formula is the coiled and placed in the
formula that shows the crucible. Zn + CuSO
actual number of hydrogen 4. The crucible, lid and 4
atoms and carbon atoms in magnesium ribbon are ∆H = –217 kJ mol –1
pentene. weighed again.
4. 1 molecule of pentene 5. The crucible is heated ZnSO + Cu
4
contains 5 carbon atoms and strongly without its lid.
10 hydrogen atoms. 6. When the magnesium starts [2]
1. Formula empirik ialah formula to burn, the crucible is 1. Some heat is lost to the
yang menunjukkan nisbah covered with its lid. surrounding.
teringkas bagi atom hidrogen 7. The lid of the crucible is lifted Sebahagian haba hilang ke
dan atom karbon dalam pentena. from time to time using tongs. persekitaran. [1]
2. Nisbah teringkas bagi 8. When the magnesium ribbon 2. Heat is absorbed by the
atom karbon kepada atom stops burning, the lid is apparatus.
hidrogen ialah 1:2. removed and the crucible is Haba diserap oleh alat radas.
3. Formula molekul ialah heated strongly for another 2 [1]
formula yang menunjukkan minutes.



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08 Ans[Pra A++ Chem F5].indd 22 14/02/2019 5:25 PM

Chemistry Form 5 Answers

(b) Reaction I II (c) Test tube A:
Tindak balas Tabung uji A:
Type of reaction Exothermic Endothermic Iron rusts // Iron atom releases
Jenis tindak balas Eksotermik Endotermik electrons // Iron(II) ions are
formed.
Temperature change Increases Decreases Besi berkarat // Atom ferum
Perubahan suhu Bertambah Berkurang membebaskan elektron // Ion
Formation and Weak bonds are Strong bonds are ferum(II) terbentuk.
breaking of bond broken and strong broken and weak Test tube B:
Pembentukan dan bonds are formed. bonds are formed. Tabung uji B:
pemecahan ikatan Ikatan lemah Ikatan kuat dipecahkan Iron does not rust // Iron(II) ions
dipecahkan dan ikatan dan ikatan lemah are not present.
kuat dibentuk. dibentuk. Besi tidak berkarat // Ion
ferum(II) tidak terbentuk.
(c) 1. Apparatus: Polystyrene cup, 8. Larutan argentum nitrat
thermometer, measuring dituang dengan cepat dan Test tube C:
cylinder. cermat ke dalam larutan Tabung uji C:
2. Materials: [Suitable salt natrium klorida. Iron rusts // Iron atom releases
solution to produce a 9. Campuran tindak balas dikacau. electrons // Iron(II) ions are
precipitate / any insoluble 10. Suhu tertinggi dicatatkan. formed.
salt] Besi berkarat // atom ferum
1. Alat radas: Cawan Paper 3 membebaskan elektron // Ion
polisterina, termometer, 1. (a) • Manipulated variable ferum(II) terbentuk.
silinder penyukat Pemboleh ubah dimanipulasi: (d) Sample answer
2. Bahan: [Garam bersesuaian Metal coiled to iron nail Sampel jawapan
untuk menghasilkan Logam yang dililit pada paku • When a more
mendakan / garam tak • Responding variable electropositive metal is in
terlarut] Pemboleh ubah bergerak contact with iron, the metal
Procedure: balas: inhibits rusting.
3. [25-200] cm of [0.1-2.0] mol Rusting // Presence of blue Apabila besi menyentuh
3
dm silver(II) nitrate solution / pink colour logam yang lebih
-3
is measured and Pengaratan besi // elektropositif, logam
4. poured into a polystyrene Kehadiran warna biru / tersebut menghalang
cup. merah jambu pengaratan //
5. [25-200] cm of [0.1-2.0] mol • Constant variable • If the metal in contact with
3
dm sodium chloride solution Pemboleh ubah iron is higher than iron in
-3
is measured and dimalarkan: the Electrochemical Series,
6. poured into a different Size of iron nail // Type of rusting does not occur.
polystyrene cup. nail // Medium in which the Apabila besi menyentuh
7. The initial temperature of iron nails are kept logam yang berada lebih
Saiz paku // Jenis paku //
both solutions is measured. Medium yang digunakan tinggi daripada ferum
8. Silver(II) nitrate solution is untuk menyimpan paku dalam Siri Elektrokimia,
poured quickly into sodium pengaratan tidak berlaku.
chloride solution. (b) Test tube A: (e) Sample answer

Tabung uji A:
9. The mixture is stirred. Intensity of blue colour Sampel jawapan
10. The highest temperature is produced is very high • When iron is in contact
recorded. Keamatan warna biru yang with a metal lower than
Prosedur: terhasil sangat tinggi iron in the Electrochemical
3. [25-200] cm larutan Test tube B: Series, blue colour
3
argentum nitrat [0.1-2.0] mol Tabung uji B: indicates rusting
dm disukat dan Apabila besi menyentuh
-3
4. dituang ke dalam cawan Intensity of pink colour logam yang berada di
produced is high
polisterina. Keamatan warna merah jambu bawah ferum dalam Siri
5. [25-200] cm larutan natrium yang terhasil tinggi Elektrokimia, warna biru
3
klorida [0.1-2.0] mol dm Test tube C: menunjukkan pengaratan
-3
disukat dan Tabung uji C: besi berlaku //
6. dituang dalam cawan Intensity of blue colour • Rusting occurs when iron
polisterina yang lain. produced is low nail is in contact with a
7. Suhu awal kedua-dua larutan Keamatan warna biru yang less electropositive metal
dicatatkan. terhasil rendah and form blue colour.
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08 Ans[Pra A++ Chem F5].indd 23 14/02/2019 5:25 PM

Chemistry Form 5 Answers
Pengaratan berlaku apabila Untuk mengkaji jenis elektrod (e) Procedure
paku menyentuh logam memberi kesan terhadap jenis Prosedur:
yang kurang elektropositif hasil pada anod semasa proses • Carbon electrodes are
dan warna biru terhasil // elektrolisis cleaned with sand paper.
• Blue colour is formed when (b) Variables: Elektrod karbon
iron is in contact with a Pemboleh ubah: dibersihkan menggunakan
metal lower than iron in the Manipulated variable: kertas pasir.
3
Electrochemical Series. Pemboleh ubah • 50.0 cm of copper(II)
Warna biru terhasil apabila dimanipulasikan: sulphate solution is
paku menyentuh logam Type of anode // Carbon and measured and poured
yang berada lebih rendah copper electrodes into the electrolytic cell /
daripada ferum dalam Siri Jenis anod / Elektrod karbon beaker.
3
Elektrokimia. dan elektrod kuprum 50.0 cm larutan kuprum(II)
(f) Sample answer: Responding variable: sulfat disukat dan dituang
ke dalam sel elektrolisis /
Sampel jawapan: Pemboleh ubah bergerak balas: bikar.
• The higher the intensity of Products formed at the anode • A test tube filled with
the blue colour, the higher Hasil pada anod copper(II) sulphate solution
the rate of rusting. is inverted at the carbon
Semakin tinggi keamatan Fixed variable: anode.
warna biru, semakin tinggi Pemboleh ubah dimalarkan: Tabung uji yang berisi
kadar pengaratan besi. Concentration and volume of larutan kuprum(II) sulfat
• When the intensity of blue copper(ll) sulphate solution ditelangkupkan pada anod
Kepekatan dan isi padu larutan
colour is high, the rate of karbon.
rusting is high. kuprum(II) sulfat • The switch is turned on.
Apabila keamatan (c) Hypothesis: Suis dihidupkan.
warna biru tinggi, kadar Hipotesis: • The gas collected is tested
pengaratan besi juga When a copper electrode is and the observation is
tinggi. used as the anode, it becomes recorded.
(g) • Fe ion / Iron(II) ion thinner whereas when a carbon Gas yang terhasil diuji dan
2+
pemerhatian dicatatkan.
electrode is used as the anode,
2+
Ion Fe / Ion ferum(II) • Steps 1 – 4 are repeated
• Fe ➝ Fe + 2e – bubble gas is produced. using copper electrodes to
2+
Apabila elektrod kuprum
(h) Cu, Fe, Zn, Mg digunakan sebagai anod, anod replace carbon electrodes.
(i) The intensity of blue colour is semakin nipis manakala apabila Langkah 1 – 4 diulang
higher in day 4 compared to day 1. elektron karbon digunakan dengan menggunakan
Keamatan warna biru lebih sebagai anod, gelembung gas elektrod kuprum untuk
tinggi pada hari keempat terhasil. menggantikan elektrod
berbanding hari pertama. (d) Materials karbon.
(j) Metals that slow down rusting: Bahan: (f) Tabulation of data:
Logam yang melambatkan • 0.1 mol dm copper(II) Penjadualan data:
-3
pengaratan: sulphate solution
Magnesium, Zinc Larutan kuprum(II) sulfat Type of electrodes Observation at the
anode
Magnesium, Zink 0.1 mol dm -3 Jenis elektrod Pemerhatian pada
• Carbon electrodes
Metals that speed up rusting: Elektrod karbon anod
Logam yang mempercepatkan • Copper electrodes
pengaratan: Elektrod kuprum
Copper / Kuprum
(k) Intensity of blue colour Apparatus Carbon
Radas:

produced is the highest. • Batteries / Bateri Karbon
Keamatan warna biru yang • Connecting wire with
terhasil paling tinggi. crocodile clips
2. (a) Aim of the experiment: Wayar penyambung
Tujuan eksperimen: dengan klip buaya
To investigate the type of • Electrolytic cell // Beaker
electrodes affect the type of Sel elektrolisis / Bikar
product formed at the anode • Wooden splinter Copper
Kuprum
during electrolysis Kayu uji
• Test tube
Tabung uji

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08 Ans[Pra A++ Chem F5].indd 24 14/02/2019 5:25 PM