Contents
PAK-21 Activities..................................................................................... iii 3.2 Rusting as a Redox Reaction 49
Aktiviti PAK-21
Pengaratan sebagai Tindak Balas Redoks
3.3 Reactivity Series of Metals and Its
PTER 50
Applications
1
CHA CHA Rate of Reaction 1 Siri Kereaktifan Logam dan Aplikasinya 51
3.4 Electrolytic and Chemical Cells
Kadar Tindak Balas
Sel Elektrolisis dan Sel Kimia
Smart Notes 1 SPM Practice 3 52
1.1 Rate of Reaction 2 HOTS Challange 61
Kadar Tindak Balas
QQ 61
1.2 Factors Affecting the Rate of 3 PTER
Reaction
CHA CHA 4 Thermochemistry 62
Faktor-faktor yang Mempengaruhi Kadar
Termokimia
Tindak Balas
1.3 Collision Theory 4 Smart Notes 62
Teori Perlanggaran 6
4.1 Energy Changes in Chemical
SPM Practice 1
HOTS Challange 20 Reactions 63
QQ 20 Perubahan Tenaga dalam Tindak Balas Kimia
PTER 4.2 Heat of Precipitation 64
Haba Pemendakan
2 Carbon Compounds 21
4.3 Heat of Displacement 65
Sebatian Karbon Haba Penyesaran
Smart Notes 21 4.4 Heat of Neutralisation 66
26 Haba Peneutralan
2.1 Carbon Compounds
Sebatian Karbon 26 4.5 Heat of Combustion 67
Haba Pembakaran 69
2.2 Alkanes 27
Alkana SPM Practice 4
28
2.3 Alkenes HOTS Challange 78
Alkena 29
QQ 78
2.4 Isomerism 30
Keisomeran PTER Chemicals for Consumers 79
31
2.5 Alcohols 5 Bahan Kimia untuk Pengguna 79
Alkohol 32 81
2.6 Carboxylic Acids 32 Smart Notes 82
Asid Karboksilik 34
44 5.1 Soap and Detergent 83
2.7 Esters QQ 44 Sabun dan Detergen
Ester 84
45 5.2 Food Additives 91
2.8 Fats Bahan Tambah Makanan QQ 91
Lemak
5.3 Medicine
2.9 Natural Rubber Ubat
Getah Asli
SPM Practice 5
SPM Practice 2
HOTS Challange
HOTS Challange
CHA PTER Oxidation and Reduction Written Practical 92
104
3 Pengoksidaan dan Penurunan SPM Forecast Paper
Answers A1
Smart Notes 45
3.1 Redox Reactions 48
Tindak Balas Redoks
ii
CHA PTER Carbon Compounds
2 Sebatian Karbon
SMART Notes
Comparison between alkanes and alkenes / Perbandingan antara alkana dengan alkena
Alkanes Similarities / Differences Alkenes
Alkana Persamaan / Perbezaan Alkena
Saturation
Saturated hydrocarbon Ketepuan Unsaturated hydrocarbon
Hidrokarbon tepu Hidrokarbon tak tepu
Chemical bond
Contains single bonds Ikatan kimia Contains at least one double bond (C = C)
only (C – C) Mengandungi sekurang-kurangnya satu ikatan
Mengandungi ikatan tunggal General formula ganda dua (C = C)
sahaja (C – C) Formula am
CnH2n, n = 2, 3, 4, …
CnH2n +2, n = 1, 2, 3, … Solubility in water
Keterlarutan dalam air Insoluble
Insoluble Tak larut
Tak larut Boiling point
Takat didih Low
Low Rendah
Rendah Melting point
Takat beku Low
Low Density Rendah
Rendah Ketumpatan
Less dense than water
Less dense than water Conductivity Kurang tumpat daripada air
Kurang tumpat daripada air Kekonduksian
Cannot conduct electricity
Cannot conduct electricity Combustion Tidak mengkonduksikan elektrik
Tidak mengkonduksikan Pembakaran
elektrik Undergoes complete combustion in excess
oxygen to produce carbon dioxide and water.
Undergoes complete Mengalami pembakaran lengkap dalam oksigen
combustion in excess berlebihan untuk menghasilkan karbon dioksida
oxygen to produce carbon dan air.
dioxide and water. C2H4 + 3O2 → 2CO2 + 2H2O
Mengalami pembakaran
lengkap dalam oksigen
berlebihan untuk
menghasilkan karbon dioksida
dan air.
CH4 + 2O2 → CO2 + 2H2O
21 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
Alkanes Similarities / Differences Alkenes
Alkana Persamaan / Perbezaan Alkena
Substitution reactions / Reactions Addition reactions / Tindak balas penambahan:
Tindak balas penukargantian Tindak balas
C C + XY CC
XY
• Hydrogenation / Penghidrogenan
• Halogenation / Penghalogenan
• Addition of hydrogen halides
Penambahan hidrogen halida
• Hydration / Penghidratan
• Addition of acidified KMnO4 solution
Penambahan larutan KMnO4 berasid
• Polymerisation / Pempolimeran
Alcohols / Alkohol
• Functional group: hydroxyl group (-OH).
Kumpulan berfungsi: kumpulan hidroksil (-OH).
• General formula: CnH2n+1OH, n = 1, 2, 3, …
Formula am: CnH2n + OH, n = 1, 2, 3, …
• Simple alcohols are soluble in water.
Alkohol ringkas adalah larut dalam air.
• Colourless volatile liquid that produces strong smell at room conditions.
Cecair mudah meruap yang tak berwarna yang menghasilkan bau yang kuat pada keadaan bilik.
Preparation of ethanol / Penyediaan etanol
1. Hydration of ethene with steam at 300oC and 60 atm with H3PO4 as catalyst.
Penghidratan etena dengan stim pada 300°C dan 60 atm dengan H3PO4 sebagai mangkin.
C2H4 + H2O H3PO→4 C2H5OH
Ethene 300°C, 60 atm
Etana
2. Fermentation of glucose with yeast (zymase enzyme).
Penapaian glukosa dengan yis (enzim zimase).
C6H12O6 yeast / yi→s 2C2H5O H + 2CO2
Glucose
Glukosa
© Penerbitan Pelangi Sdn. Bhd. 22
Chemistry Form 5 Chapter 2 Carbon Compounds
Chemical properties of ethanol / Sifat kimia etanol
1. Very flammable and burns easily in oxygen to produce carbon dioxide and water.
Sangat mudah terbakar dan membakar dengan mudah dalam oksigen untuk menghasilkan karbon dioksida dan air.
C2H5OH + 3O2 → 2CO2 + 3H2O
2. Ethanol is dehydrated by heating at 180oC with concentrated sulphuric acid, H2SO4.
Etanol terdehidrat oleh pemanasan pada 180°C dengan asid sulfurik pekat, H2SO4.
C2H5OH 1H82S0OoC→4 C2H4 + H2O
3. Ethanol is oxidised to ethanoic acid with acidified potassium dichromate(VI).
Etanol teroksida kepada asid etanoik dengan kalium dikromat(VI) berasid.
C2H5OH [O→] CH 3COOH + H2O
4. Forms ester when refluxed with carboxylic acid and several drops of concentrated sulphuric acid as
catalyst.
Membentuk ester apabila direfluks dengan asid karboksilik dan beberapa titis asid sulfurik pekat sebagai mangkin.
C2H5OH + CH3COOH → CH3COOC2H5 + H2O
Carboxylic acids / Asid karboksilik
The sting from the bite of ant is caused by methanoic acid
Sengatan daripada gigitan semut adalah disebabkan oleh asid metanoik
• Functional group: Carboxyl group (-COOH).
Kumpulan berfungsi: kumpulan karboksil (-COOH)
• General formula: CnH2n+1COOH, n = 0, 1, 2, 3, …
Formula am: CnH2n + 1COOH, n = 0, 1, 2, 3, …
• Simple carboxylic acids are soluble in water to form acidic
solution.
Asid karboksilik ringkas adalah larut dalam air untuk membentuk
larutan berasid.
• Colourless liquid that produces sour/vinegar like smell at
room conditions.
Cecair tak berwarna yang menghasilkan bau masam/cuka pada
keadaan bilik.
23 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
Preparation of ethanoic acid / Penyediaan asid etanoik
Oxidation of ethanol with acidified potassium dichromate(VI).
Pengoksidaan etanol dengan kalium dikromat(VI) berasid.
C2H5OH [O]→ CH3COOH + H2O
Chemical properties of ethanoic acid / Sifat kimia asid etanoik
1. Reacts with a base to form salt and water (neutralisation).
Bertindak balas dengan bes untuk membentuk garam dan air (peneutralan).
CH3COOH + NaOH → CH3COONa + H2O
2CH3COOH + MgO → (CH3COO)2Mg + H2O
2. Reacts with a reactive metal to form salt and hydrogen gas. Fruits contain natural esters
Bertindak balas dengan logam reaktif untuk membentuk garam dan Buah-buahan mengandungi ester
semula jadi
gas hidrogen.
2CH3COOH + Zn → (CH3COO)2Zn + H2
3. Reacts with a metal carbonate to form salt, water and carbon dioxide gas.
Bertindak balas dengan logam karbonat untuk membentuk garam, air dan gas karbon dioksida.
2CH3COOH + CaCO3 → (CH3COO)2Ca + H2O + CO2
4. Reacts with an alcohol to form ester and water (esterification).
Bertindak balas dengan alkohol untuk membentuk ester dan air (pengesteran).
CH3COOH + C2H5OH H→+ C H3COOC2H5 + H2O
Esters / Ester
• Functional group / Kumpulan berfungsi:
O
COC
• General formula: CmH2m+1COOCnH2n+1, m = 0, 1, 2,… and n = 1, 2, 3, …
Formula am: CmH2m+1COOCnH2n+1, m = 0, 1, 2, … dan n = 1, 2, 3, …
Preparation / Penyediaan
Refluxing a mixture of alcohol and carboxylic acid with concentrated
sulphuric acid as catalyst (esterification).
Reflukskan campuran alkohol dan asid karboksilik dengan asid sulfurik pekat sebagai
mangkin (pengesteran).
Physical properties of esters / Sifat fizikal ester
• Produces fragrant smell (fruity / flowery smell)
Menghasilkan bau wangi (bau buah-buahan / bunga)
• Usually insoluble in water
Biasanya tidak larut dalam air
• Simple esters are volatile
Ester ringkas adalah mudah meruap
© Penerbitan Pelangi Sdn. Bhd. 24
Chemistry Form 5 Chapter 2 Carbon Compounds
Oils / Minyak Both are naturally Fats / Lemak
• Found in plants. occurring esters found • Found in animals.
Dijumpai dalam tumbuh- Dijumpai dalam haiwan.
in living things.
tumbuhan. Kedua-duanya adalah ester • Exist as a solid or semi-solid at
semula jadi yang didapati room temperature.
• Exist as a liquid at room dalam benda-benda hidup.
temperature. Wujud sebagai pepejal atau separa
pepejal pada suhu bilik.
Wujud sebagai cecair pada suhu
bilik. • Melting point more than 20oC.
Takat lebur melebihi 20°C.
• Melting point less than 20oC.
Takat lebur kurang daripada 20°C • Saturated (do not have any
double bond between carbon
• Unsaturated (contains at atoms).
least 1 double bond between
carbon atoms). Tepu (tiada sebarang ikatan ganda
dua antara atom-atom karbon).
Tak tepu (mengandungi sekurang-
kurangnya satu ikatan ganda dua
antara atom-atom karbon).
Natural rubber / Getah Asli
Coagulation of latex / Penggumpalan lateks
Rubber IHoyn-diorongheidn++ro+i–o–g+– en+––ns+++––––+––+++++–+–––+++–––+–+––––+++–+––+++–+–++++
molecule – Acid is added Particles collide with
Molekul getah – – Asid ditambahkan each other
– – Zarah-zarah berlanggar
–– – – H+ ions from acid neutralise sesama sendiri
–– the negative charges
–– – – (protein membrane)
Ion H+ daripada asid
–
– meneutralkan cas-cas negatif
– – – Protein membrane (membran protein)
Membran protein
Negative
charges –– ––
Cas-cas –
negatif – ––
Protein membrane breaks
Membran protein pecah
Protein membrane breaks Rubber particles collide
and rubber molecules are Zarah-zarah getah berlanggar
Rubber molecules clump together now free
(latex has coagulated) Membran protein pecah dan
molekul getah menjadi bebas
Molekul getah berkumpul bersama
(lateks menggumpal)
25 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
2.1 Carbon Compounds
Sebatian Karbon
1. (a) What is a carbon compound?
Apakah sebatian karbon?
A compound that contains the element in its structure.
Sebatian yang mengandungi unsur dalam strukturnya.
(b) The following shows several compounds. Circle the carbon compounds.
Berikut menunjukkan beberapa sebatian. Bulatkan sebatian-sebatian karbon.
Sodium chloride Carbon dioxide Sulphuric acid Ethanoic acid
Natrium klorida Karbon dioksida Asid sulfurik Asid etanoik
2. (a) What is a hydrocarbon? and elements.
Apakah hidrokarbon? dan sahaja.
A organic compound that only contains
Sebatian organik yang mengandungi unsur
(b) Hydrocarbons are divided into two groups.
Hidrokarbon dibahagikan kepada dua kumpulan.
(i) Name the two groups of hydrocarbons.
Namakan dua kumpulan hidrokarbon itu.
(ii) State the difference between the two groups named.
Nyatakan perbezaan antara dua kumpulan yang dinamakan.
The carbon atoms are bonded by covalent bonds in saturated hydrocarbons
whereas in unsaturated hydrocarbons, there is at least one carbon-carbon
bond.
Atom-atom karbon terikat oleh ikatan kovalen dalam hidrokarbon tepu tetapi dalam
hidrokarbon tak tepu, terdapat sekurang-kurangnya satu ikatan .
(c) State two main sources of hydrocarbons.
Nyatakan dua sumber utama hidrokarbon.
2.2 Alkanes
Alkana
3. (a) What is an alkane? hydrocarbon.
.
Apakah alkana?
An alkane is a
Alkana adalah hidrokarbon
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Chemistry Form 5 Chapter 2 Carbon Compounds
(b) Give the general formula for alkanes.
Berikan formula umum bagi alkana.
(c) State three physical properties of alkanes.
Nyatakan tiga sifat fizik bagi alkana.
(i) in water
dalam air
(ii) conduct electricity in all states
mengkonduksikan elektrik dalam semua keadaan
(iii) melting and boiling points
Takat lebur dan takat didih yang
(d) Balance the chemical equations for the following reactions of alkanes. Name the reactions
that occur.
Tuliskan persamaan kimia bagi tindak balas alkana berikut. Namakan tindak balas yang berlaku.
(i) Reaction of propane with excess oxygen.
Tindak balas etana dengan oksigen berlebihan.
Equation:
Persamaan
Reaction:
Tindak balas
(ii) Reaction of methane with chlorine gas in the presence of ultraviolet light.
Tindak balas metana dengan gas klorin dalam kehadiran cahaya ultraungu.
Equation:
Persamaan
Reaction:
Tindak balas
2.3 Alkenes
Alkena
4. (a) What is an alkene? hydrocarbon with at least one carbon-carbon .
Apakah alkena? dengan sekurang-kurangnya satu ikatan
An alkene is an
bond.
Alkena adalah hidrokarbon
(b) Write the general formula of alkenes.
Tuliskan formula am alkena.
27 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
5. Name the main products of the reactions of ethene. Combustion Reactions of
Namakan hasil utama tindak balas etena. Carbon Compounds
+ excess O2 INFO
Combustion O2 berlebihan (a)
Pembakaran + limited O2
O2 terhad (b)
+ H2 (c)
Ethene Addition reactions + Br2 (d)
+ HCl (e)
Etena Tindak balas penambahan
+ H2O (f)
+ acidified KMnO4
KMnO4 berasid
(g)
Addition polymerisation (h)
Pempolimeran penambahan
2.4 Isomerism
Keisomeran
6. (a) What are structural isomers?
Apakah isomer struktur?
Structural isomers are molecules with the same formula but with different
formula.
Isomer struktur adalah molekul dengan formula yang sama tetapi dengan formula
yang berbeza.
(b) Draw the structural formula and name all the isomers of butene according to the IUPAC
nomenclature.
Lukis formula struktur dan namakan semua isomer butena berdasarkan penamaan IUPAC.
© Penerbitan Pelangi Sdn. Bhd. 28
Chemistry Form 5 Chapter 2 Carbon Compounds
2.5 Alcohols
Alkohol
7. (a) State the functional group of alcohols.
Nyatakan kumpulan berfungsi alkohol.
(b) Write the general formula of alcohols.
Tuliskan formula am alkohol.
8. (a) State two ways of preparing ethanol.
Nyatakan dua cara untuk menyediakan etanol.
(i) of glucose with yeast
glukosa dengan yis
(ii) of ethene
etena
(b) Give three physical properties of ethanol.
Berikan tiga sifat fizik bagi etanol.
(i) Colourless
tak berwarna
(ii) melting and boiling points
Takat lebur dan takat didih yang
(iii) in water
dalam air
(c) State three uses of alcohols.
Nyatakan tiga kegunaan alkohol.
(i) As a
Sebagai
(ii) As an
Sebagai
(iii) As an solvent
Sebagai pelarut
29 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
9. Complete the chart below to show the reactions of ethanol.
Lengkapkan carta di bawah untuk menunjukkan tindak balas etanol.
Complete combustion (a) + Water
Pembakaran lengkap Air
+ excess oxygen + Water
oksigen berlebihan Air
Pengoksidaan (b) + Water
Ethanol Oxidation Air
Etanol + acidified potassium manganate(VII) + Water
kalium manganate(VII) berasid Air
Pendehidratan (c)
Dehydration
hot porcelain chips
serpihan porselin panas
Esterification (d)
Pengesteran
ethanoic acid + concentrated
sulphuric acid, reflux
asid etanoik + asid sulfurik pekat, refluks
2.6 Carboxylic Acids
Asid Karboksilik
10. (a) State the functional group and general formula of carboxylic acids.
Nyatakan kumpulan berfungsi dan formula am asid karboksilik.
(b) How would you prepare a carboxylic acid in the school laboratory?
Bagaimanakah anda menyediakan suatu asid karboksilik dalam makmal sekolah?
By refluxing an alcohol with oxidising agent such as .
Dengan mereflukskan alkohol dengan agen pengoksidaan seperti .
(c) State two physical properties of ethanoic acid.
Nyatakan dua sifat fizik asid etanoik.
(i) Colourless liquid with smell.
Cecair tak berwarna dengan bau .
(ii) in water and organic solvents.
dalam air dan pelarut organik.
(d) By refluxing an alcohol with oxidising agent such as
Dengan mereflukskan alkohol dengan agen pengoksidaan seperti
(e) (i) Colourless liquid with smell.
Cecair tak berwarna dengan bau
(ii) in water and organic solvents.
dalam air dan pelarut organik.
© Penerbitan Pelangi Sdn. Bhd. 30
Chemistry Form 5 Chapter 2 Carbon Compounds
11. Complete the word equations below to show the four main reactions of ethanoic acid.
Lengkapkan persamaan perkataan di bawah untuk menunjukkan empat tindak balas asid etanoik yang utama.
(a) Ethanoic acid + → salt + water
Asid etanoik + → garam + air
(b) Ethanoic acid + → salt + carbon dioxide + water
Asid etanoik + → garam + karbon dioksida + air
(c) Ethanoic acid + → salt + hydrogen
Asid etanoik + → garam + hidrogen
(d) Ethanoic acid + → ester + water
Asid etanoik + → ester + air
2.7 Esters
Ester
12. (a) Name the process to produce an ester.
Namakan proses untuk menghasilkan ester.
(b) State the catalyst used in the process to produce an ester.
Nyatakan mangkin yang digunakan dalam proses yang menghasilkan ester.
(c) Give three physical properties of esters.
Berikan tiga sifat fizik ester.
(i) Produces smell. (iii) Usually in water.
Menghasilkan bau Biasanya dalam air.
(ii) melting and boiling points
Takat lebur dan takat didih yang
13. The structures of two esters are given below. Name each ester and identify the alcohol and
carboxylic acid needed to prepare the ester.
Struktur dua ester diberikan di bawah. Namakan setiap ester tersebut dan kenal pasti alkohol dan asid karboksilik
yang diperlukan untuk menyediakan ester tersebut.
(a) (i) Name of ester:
HO HHH Nama ester
&' &&&
(ii) Alcohol:
H ! C ! C !O! C ! C ! C ! H
& &&& Alkohol
H HHH (iii) Carboxylic acid:
Asid karboksilik
(b) (i) Name of ester:
HH O H HH Nama ester
&& ' &&&
(ii) Alcohol:
H ! C ! C !O! C ! C ! C ! C ! H
Alkohol
&& &&&
(iii) Carboxylic acid:
HH HHH
Asid karboksilik
31 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
2.8 Fats
Lemak
14. (a) What are fats and oils?
Apakah lemak dan minyak?
(b) Name the products obtained from the hydrolysis of fat or oil. dan
Namakan hasil yang diperoleh daripada hidrolisis lemak atau minyak. Oils / Minyak
and /
15. Distinguish between fats and oils.
Bezakan antara lemak dengan minyak.
Fats / Lemak
(a) Source (b)
Sumber
Physical state at room
(c) temperature (d)
Keadaan fizik pada
suhu bilik
Content proportion of
(e) sturated fatty acid (f)
Nisbah kandungan
asid lemak tepu
2.9 Natural Rubber
Getah Asli
16. Draw the structure and name the monomer of natural rubber according to the IUPAC nomenclature.
Lukis struktur dan namakan monomer getah asli berdasarkan penamaan IUPAC.
17. (a) What is coagulation of latex?
Apakah penggumpalan lateks?
The process of of latex into rubber.
.
Proses lateks kepada getah
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Chemistry Form 5 Chapter 2 Carbon Compounds
(b) Explain why a sample of latex coagulates after being exposed to air for several hours.
Jelaskan mengapa suatu sampel lateks menggumpal selepas didedahkan kepada udara selama beberapa
jam.
This is because the in air react with latex to produce acid. The acid
the negative charges on the protein membrane of latex particles and
causes the coagulation to occur.
Hal ini kerana di udara bertindak balas dengan lateks untuk menghasilkan asid. Asid
itu cas-cas negatif pada membran protein zarah-zarah lateks dan menyebabkan
penggumpalan berlaku.
(c) Name one substance that
Namakan satu bahan yang
(i) causes coagulation of latex to occur faster.
menyebabkan penggumpalan lateks berlaku lebih cepat.
(ii) causes coagulation of latex to occur slower.
menyebabkan penggumpalan lateks berlaku lebih lambat.
18. (a) What is vulcanisation? , and more
Apakah pemvulkanan? , Vulcanisation
of Rubber
The process to modify natural rubber into a
material.
Proses untuk mengubah suai getah asli kepada bahan yang lebih
dan .
(b) Compare the properties of vulcanised rubber and unvulcanised rubber. INFO
Bandingkan sifat-sifat getah tervulkan dengan getah tak tervulkan.
Unvulcanised rubber Properties Vulcanised rubber
Getah tak tervulkan Sifat Getah tervulkan
(i) Hardness (ii)
Kekerasan
(iii) Elasticity (iv)
Kekenyalan
(v) Resistance to oxidation (vi)
Ketahanan terhadap pengoksidaan
(vii) Toughness (viii)
Kekuatan
(ix) Durability (x)
Tahan lama
33 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
SPM Practice 2
SPM Paper 11 SPM Year ‘13 ‘14 ‘15 ‘16 ‘17
Paper 65555
Analysis
1
2.1 Carbon Compounds 2.2 Alkanes 2.3 Alkenes
Sebatian Karbon Alkana Alkena
1. Which of the following 3. One mole of an alkane is 5. Propene reacts with steam in
is correct about saturated 2S0P1M6 completely combusted in air. 2S0P1M7 the presence of concentrated
hydrocarbons? Which alkane produces
phosphoric acid as catalyst
Antara yang berikut, yang carbon dioxide and water in a at temperature 300oC and
manakah betul tentang mol ratio of 2:3? pressure 60 atm to produce
hidrokarbon tepu? compound X.
Satu mol alkana dibakar dengan What is compound X?
A Contains carbon atoms lengkap dalam udara.
and hydrogen atoms only. Propena bertindak balas dengan
Alkana yang manakah stim dengan kehadiran asid
Mengandungi atom karbon menghasilkan karbon dioksida fosforik pekat sebagai mangkin
dan atom hidrogen sahaja. dan air dalam nisbah mol 2:3? pada suhu 300°C dan tekanan
60 atm untuk menghasilkan
B Contains carbon atoms A Methane sebatian X.
and hydrogen atoms with
single bonds only. Metana Apakah sebatian X?
Mengandungi atom karbon B Ethane A C3H6
dan atom hidrogen dengan B C3H8
ikatan tunggal sahaja. Etana C C3H7OH
D C3H7COOH
C Contains carbon atoms C Propane
and hydrogen atoms with 6. Diagram 1 shows the
one double bond only. Propana 2S0P1M6 structural formula of
Mengandungi atom karbon D Butane compound K.
dan atom hidrogen dengan
satu ikatan ganda dua Butana Rajah 1 menunjukkan formula
sahaja. struktur sebatian K.
4. Which of the following is
D Contains carbon atoms 2S0P1M4 produced when butane is H HH
and hydrogen atoms with & &&
one triple bond only. burnt in excess oxygen? H!C"C!C!C!H
Mengandungi atom karbon Yang manakah antara berikut &&
dan atom hidrogen dengan dihasilkan apabila butana dibakar HH
satu ikatan ganda tiga dalam oksigen berlebihan? H!C!H
sahaja. &
I Carbon dioxide H
2. Which of the following is
an example of a saturated Karbon dioksida Diagram 1 / Rajah 1
hydrocarbon?
II Carbon monoxide What is the percentage of
Antara yang berikut, yang carbon by mass in compound
manakah ialah contoh Karbon monoksida K?
hidrokarbon tepu?
III Water vapour Apakah peratusan karbon
A Ethene mengikut jisim dalam sebatian K?
Wap air
Etena [Relative atomic mass / Jisim
IV Carbon atom relatif: H = 1; C = 12]
B Ethane A 16.67%
Karbon B 17.14%
Etana C 83.33%
A I and II D 85.71%
C Ethanol I dan II
B I and III
Etanol I dan III
C II and IV
D Ethanoic acid
II dan IV
Asid etanoik
D III and IV
III dan IV
© Penerbitan Pelangi Sdn. Bhd. 34
Chemistry Form 5 Chapter 2 Carbon Compounds
7. Which substance can be 10. What is produced when A Similar chemical
2S0P1M5 used to differentiate between 2S0P1M5 hydrogen and butene are properties but different
molecular formulae
propane and propene? passed over nickel catalyst at
temperature 180oC? Sifat kimia yang sama
Bahan yang manakah dapat tetapi formula molekul yang
digunakan untuk membezakan Apakah yang terhasil apabila berbeza
propana dan propena? hidrogen dan butena dilakukan
ke atas mangkin nikel pada suhu B Similar in physical
A Tetrachloromethane 180oC? properties but different
molecular formulae
Tetraklorometana A Butane
Sifat fizik yang sama tetapi
B Liquid bromine Butana formula molekul yang
berbeza
Cecair bromin B Butanol
C Similar structure but
C Sulphuric acid Butanol different molecular
formulae
Asid sulfurik C Butanoic acid
Struktur yang sama tetapi
D Potassium hydroxide Asid butanoik formula molekul yang
berbeza
Kalium hidroksida D Butyl butanoate
D Similar molecular
8. Which structure shows an Butil butanoat formula but different
2S0P 1M4 structural formulae
unsaturated hydrocarbon? 11. The equation below shows
the conversion of butene Formula molekul yang sama
Struktur yang manakah to compound P through tetapi formula struktur yang
menunjukkan hidrokarbon tak hydration process. berbeza
tepu?
Persamaan di bawah
A (CH3)2CHCH2CH3 menunjukkan pertukaran
B CH3CH2CH2CH2CH3 butena kepada sebatian P
C CH3CHCHCH2CH2CH3 melalui proses penghidratan.
D CH3CH2C(CH3)3
9. Diagram 2 shows the Hydration 2.5 Alcohols
2S0P1M3 structural formulae of two BBuutteennae Pen ghidratan
CSeobmatpiaonuPnd P Alkohol
organic compounds.
What is compound P? 13. Diagram 3 shows the set up
Rajah 2 menunjukkan formula of apparatus for a reaction.
struktur bagi dua sebatian Apakah sebatian P?
organik. Rajah 3 menunjukkan susunan
A Butane radas bagi suatu tindak balas.
H H H Butana Glass wool
& & & soaked with
H!C"C!C!H B Butanol butanol Porcelain chips
& Wul kaca yang Serpihan porselin
H Butanol
dibasahi butanol Butene gas
H H H H C Butanoic acid
& & & & Gas butena
H !C"C!C!C!H Asid butanoik
& &
H H D Butyl butanoate
Diagram 2 / Rajah 2 Butil butanoate
2.4 Isomerism Heat
Panaskan
Keisomeran
Water
Which property is the same 12. Which of the following best Air
for both compounds? describes the similarities
and differences between Diagram 3 / Rajah 3
Sifat yang manakah adalah sama compounds which exist as
bagi kedua-dua sebatian itu? isomers? What is the reaction?
A Boiling point Antara yang berikut, yang Apakah tindak balas ini?
manakah paling baik menerangkan
Takat didih persamaan dan perbezaan antara A Oxidation
sebatian-sebatian yang wujud
B Relative molecular mass sebagai isomer? Pengoksidaan
Jisim molekul relatif B Reduction
C Number of isomers Penurunan
Bilangan isomer C Hydrogenation
D Solubility in water Penghidrogenan
Keterlarutan dalam air D Dehydration
Pendehidratan
35 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
14. Below are the names of two Carboxylic Acids Homologous Functional
M 2R01S M7 series group
organic compounds. 2.6 Asid Karboksilik
Siri homolog Kumpulan
Di bawah adalah nama bagi dua berfungsi
sebatian organik.
Methanol Ethanol 16. A liquid produced A Alkene CC
Metanol effervescence when reacted
Etanol with potassium carbonate Alkena
solution. What is the
Which of the following molecular formula of the B Carboxylic O
statements is true about both liquid? acid C OH
compounds?
Suatu cecair menghasilkan Asid OH
Antara pernyataan berikut, yang karboksilik O
manakah benar tentang kedua- pembuakan apabila bertindak CO
dua sebatian itu? balas dengan larutan kalium C Ester
karbonat. Apakah formula
A Both have the same molekul cecair ini? Ester
functional group.
A C2H5COOH D Alcohol
Kedua-duanya mempunyai B CH3CH2CH2OH
kumpulan berfungsi yang C CH3COOK Alkohol
sama. D C2H5COOCH3
19. Diagram 4 shows the
B Both have the same 17. A compound has the structure of an organic
melting and boiling 2S0P 1M5 compound that produces
points. following properties. fragrant smell.
Kedua-duanya mempunyai Suatu sebatian mempunyai sifat- Rajah 4 menunjukkan struktur
takat lebur dan takat didih sifat berikut. sebatian organik yang
yang sama. menghasilkan bau harum.
• Changes the colour of
C Both react differently blue litmus paper to red H HH OH
with acidified potassium
manganate(VII) solution. Menukarkan warna kertas HC CCOCCH
litmus biru kepada merah
Kedua-duanya bertindak HHH H
balas dengan berbeza • Tastes sour
dengan larutan kalium Diagram 4 / Rajah 4
manganat(VII) berasid. Berasa masam
Which of the following can be
D Both have different • Produces gas bubbles used to make the compound
general formulae. when reacting with zinc shown?
Kedua-duanya mempunyai Menghasilkan gelembung Antara yang berikut, yang
formula am yang berbeza. gas apabila bertindak balas manakah boleh digunakan
dengan zink untuk membuat sebatian yang
ditunjukkan?
15. Dehydration of alcohol What is the molecular
2S0P1M4 produces alkene and water. formula of the compound? A Ethanol and ethanoic acid
Which alkene is formed
Apakah formula molekul sebatian Etanol dan asid etanoik
if pentan-1-ol undergoes itu?
dehydration? B Ethanol and propanoic
A C3H8 acid
Pendehidratan alkohol B C3H7OH
menghasilkan alkena dan air. C CH3COOH Etanol dan asid propanoik
D CH3COOC2H5
Alkena yang manakah terbentuk C Propanol and ethanoic
apabila pentan-1-ol mengalami 2.7 Esters acid
pendehidratan?
Ester Propanol dan asid etanoik
A CH3CH2CH2CH=CH2
18. Which of the following D Propanol and propanoic
B CH3CH=CHCH2CH3 shows the correct functional acid
group?
C CH3 Propanol dan asid propanoik
| Yang manakah antara berikut
CH2=CCH2CH3 menunjukkan kumpulan berfungsi
yang betul?
D CH3
|
CH3CH2C=CH2
© Penerbitan Pelangi Sdn. Bhd. 36
Chemistry Form 5 Chapter 2 Carbon Compounds
20. Diagram 5 shows the Process Q A I and II
structural formula of an V egetable oil Proses Q Margarine
organic compound. Minyak sayuran Marjerin I dan II
Rajah 5 menunjukkan formula Diagram 7 / Rajah 7 B I and III
struktur suatu sebatian organik.
What is process Q? I dan III
O
Apakah proses Q? C II and IV
CH3 CH2 CH2 C O CH2 CH3
A Hydration II dan IV
Diagram 5 / Rajah 5
Penghidratan D III and IV
What is the name of the
organic compound? B Fermentation III dan IV
Apakah nama sebatian organik ini? Penapaian 24. A lorry wants to transfer
2S0P1M5 latex from a rubber estate to
A Ethyl propanoate C Hydrogenation
a rubber factory.
Etil propanoat Penghidrogenan Which substance can be
B Ethyl butanoate D Halogenation added to the latex to prevent
it from coagulating?
Etil butanoat Penghalogenan
Sebuah lori hendak
C Propyl ethanoate 2.9 Natural Rubber memindahkan lateks dari ladang
getah ke sebuah kilang getah.
Propil etanoat Getah Asli
Bahan yang manakah boleh
D Butyl ethanoate 23. Which of the following ditambahkan kepada lateks
2S0P1M5 shows the correct comparison untuk mencegahnya daripada
Butil etanoat menggumpal?
between vulcanised rubber
21. Diagram 6 shows oranges and unvulcanised rubber? A Tetrachloromethane
M 2R01S M6
that produce sweet smell. Antara yang berikut, yang Tetraklorometana
manakah menunjukkan
Rajah 6 menunjukkan buah oren perbandingan yang betul antara B Ethanoic acid
yang menghasilkan bau wangi. getah tervulkan dengan getah tak
tervulkan? Asid etanoik
Diagram 6 / Rajah 6 Vulcanised Unvulcanised C Distilled water
rubber rubber
Air suling
Getah Getah tak
tervulkan tervulkan D Ammonia solution
What is the name of the I Lower Higher Larutan ammonia
elasticity elasticity
substance that gives the 25. Substance P coagulates latex.
Rendah Tinggi What is substance P?
sweet smell? Evaluating keelastikan keelastikan
Bahan P menggumpalkan lateks.
Apakah nama bahan yang II Harder Softer Apakah bahan P?
memberikan bau wangi itu?
Lebih keras Lebih lembut A Ammonia solution
A Octyl ethanoate
III Cannot Can withstand Larutan ammonia
Oktil etanoat withstand heat
heat B Ethanoic acid
B Propan-2-ol Tahan haba
Tidak tahan Asid etanoik
Propan-2-ol haba
C Potassium hydroxide
C Linolenic acid
Kalium hidroksida
Asid linolenik
D Potassium chloride
Kalium klorida
D 2-methylpentane IV Cannot Oxidises
2-metilpentana oxidise easily easily
2.8 Fats Tidak teroksida Teroksida
dengan mudah dengan mudah
Lemak
22. Diagram 7 shows the process
2S0P1M3 of manufacturing margarine
from vegetable oil.
Rajah 7 menunjukkan proses
pembuatan marjerin daripada
minyak sayuran.
37 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds SPM Year ‘13 ‘14 ‘15 ‘16 ‘17
Paper
SPM Paper 12 Analysis 1 1 2 1 1
2 3
Section A / Bahagian A
1. Diagram 1 shows the conversion of alkene X, C3H6 into compound Y by process I. Compound Y can be
2S0P 1M3
converted back into alkene X by process II. Subtopic 2.3, 2.4
Rajah 1 menunjukkan pertukaran alkena X, C3H6 kepada sebatian Y melalui proses I. Sebatian Y dapat ditukar balik kepada
alkena X melalui proses II.
Process I / Proses I
Alkene X, C3H6 H2O (H3PO4, 300°C, 60 atm) Compound Y
Process II / Proses II
Alkena X, C3H6 Sebatian Y
Diagram 1 / Rajah 1
(a) Name alkene X.
Namakan alkena X.
[1 mark / 1 markah]
(b) Compare the solubility of alkene X and compound Y in water.
Bandingkan keterlarutan alkena X dan sebatian Y dalam air.
[2 marks / 2 markah]
(c) (i) What is observed if alkene X is passed into bromine water?
Apakah yang diperhatikan apabila alkena X dilalukan ke dalam air bromin?
[1 mark / 1 markah]
(ii) Write the balanced chemical equation for the reaction in (c)(i).
Tuliskan persamaan kimia seimbang bagi tindak balas dalam (c)(i).
[2 marks / 2 markah]
(d) Based on Diagram 1,
Berdasarkan pada Rajah 1,
(i) name compound Y.
namakan sebatian Y.
[1 mark / 1 markah]
(ii) write the balanced chemical equation for the conversion of alkene X into compound Y in process I.
tuliskan persamaan kimia seimbang bagi pertukaran alkena X kepada sebatian Y dalam proses I.
[2 marks / 2 markah]
© Penerbitan Pelangi Sdn. Bhd. 38
Chemistry Form 5 Chapter 2 Carbon Compounds
(e) Draw a labelled diagram to show how process II can be carried out in the school laboratory.
Lukiskan rajah berlabel untuk menunjukkan bagaimana proses II dapat dilakukan di dalam makmal sekolah.
[2 marks / 2 markah]
2. Diagram 2 shows the structural formulae of compound P and compound Q. Subtopic 2.2, 2.3, 2.4
Both compounds P and Q are isomers of a carbon compound.
Rajah 2 menunjukkan formula struktur bagi sebatian P dan sebatian Q.
Kedua-dua sebatian P dan Q adalah isomer satu sebatian karbon.
H H H H H H CH3 H H
& & & & & & & & &
H !C!C!C!C!C!H H ! C ! C ! C ! C ! H
& & & & & & & & &
H H H H H H H H H
Compound P Compound Q
Sebatian P Sebatian P
Diagram 2 / Rajah 2
(a) (i) Give the meaning of isomer.
Berikan maksud isomer.
[1 mark / 1 markah]
(ii) What is the molecular formula of the carbon compound in Diagram 4?
Apakah formula molekul bagi sebatian karbon dalam Rajah 4?
[1 mark / 1 markah]
(iii) Name compound P and compound Q using IUPAC nomenclature.
Namakan sebatian P dan sebatian Q dengan menggunakan penamaan IUPAC.
P: Q:
[2 marks / 2 markah]
(iv) Draw the structural formula for another isomer of the carbon compound in Diagram 4.
Lukiskan formula struktur bagi satu lagi isomer bagi sebatian karbon dalam Rajah 4.
[1 mark / 1 markah]
39 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
(b) Compound P can be prepared from an alkene.
Sebatian P dapat disediakan daripada suatu alkena.
(i) State the general formula of an alkene.
Nyatakan formula am bagi alkena.
[1 mark / 1 markah]
(ii) What is the name of the process for the conversion of the alkene into compound P?
Apakah nama proses bagi penukaran alkena tersebut kepada sebatian P?
(iii) State the temperature and catalyst for the conversion. [1 mark / 1 markah]
[2 marks / 2 markah]
Nyatakan suhu dan mangkin bagi penukaran tersebut.
Temperature / Suhu:
Catalyst / Mangkin:
3. The chemical equation below shows the combustion of alcohol W in excess oxygen. Alcohol W contains 4
2S0P 1M7
carbon atoms per unit molecule. Subtopic 2.4, 2.5, 2.7
Persamaan kimia di bawah menunjukkan pembakaran alkohol W dalam oksigen berlebihan. Alkohol W mengandungi 4
atom karbon per unit molekul.
Alcohol W + Oxygen gas : Gas K + Water vapour
Alkohol W Gas oksigen Gas K Wap air
(a) (i) What is the general formula of alcohol?
Apakah formula am alkohol?
[1 mark / 1 markah]
(ii) What is the molecular formula of alcohol W?
Apakah formula molekul alkohol W?
[1 mark / 1 markah]
(iii) Alcohol W has isomers.
Write the name of any two isomers of alcohol W.
Alkohol W mempunyai isomer.
Tuliskan nama bagi mana-mana dua isomer alkohol W.
[2 marks / 2 markah]
(b) Based on the equation above,
Berdasarkan pada persamaan di atas,
(i) identify gas K
kenal pasti gas K
[1 mark / 1 markah]
© Penerbitan Pelangi Sdn. Bhd. 40
Chemistry Form 5 Chapter 2 Carbon Compounds
(ii) suggest a simple chemical test to identify the presence of gas K. State the observation also.
cadangkan satu ujian kimia ringkas untuk mengenal pasti kehadiran gas K. Nyatakan pemerhatiannya juga.
[2 marks / 2 markah]
(c) When alcohol W is refluxed with compound V in the presence of concentrated sulphuric acid, a compound
with fragrant smell is produced. The structural formula of the compound produced is shown below.
Apabila alkohol W direfluks dengan sebatian V dalam kehadiran asid sulfurik pekat, satu sebatian dengan bau wangi
dihasilkan. Formula struktur bagi sebatian yang terhasil ditunjukkan di bawah.
H HO HHHH
&&' &&&&
H!C!C!C!O!C!C!C!C!H
&& &&&&
HH HHHH
(i) By referring to the structure of the compound shown above, name compound V.
Dengan merujuk kepada struktur sebatian yang ditunjukkan di atas, namakan sebatian V.
[1 mark / 1 markah]
(ii) Name the process that produces the compound shown above.
Namakan proses yang menghasilkan sebatian yang ditunjukkan di atas.
[1 mark / 1 markah]
(iii) The compound shown above floats on top of cold water to form two layers of colourless liquid.
Explain why.
Sebatian yang ditunjukkan di atas terapung di atas permukaan air sejuk untuk membentuk dua lapisan cecair tak
berwarna. Terangkan mengapa.
[1 mark / 1 markah]
41 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
Section B / Bahagian B
4. (a) Table 4 shows the information on three members of a homologous series. Subtopic 2.2, 2.3, 2.5, 2.6, 2.10
Jadual 4 menunjukkan maklumat tentang tiga ahli satu siri homolog.
Member of homologous series Boiling point (°C) Preparation Hydrogenation product
Ahli siri homolog Takat didih (°C)
Penyediaan Hasil penghidrogenan
Ethene -102
C2H5OH → C2H4 + H2O Ethane
Etena
Etana
Propene - 48 C3H7OH → C3H6 + H2O Propane
Propena Propana
Butene -7 C4H9OH → C4H8 + H2O Butane
Butena Butana
Table 4 / Jadual 4
Based on Table 4, state and explain five characteristics of a homologous series.
Berdasarkan Jadual 4, nyatakan dan terangkan lima ciri siri homolog.
[10 marks / 10 markah]
(b) The following information is about organic compound Z.
Maklumat yang berikut adalah tentang sebatian organik Z.
• Empirical formula is CH2
Formula empirik ialah CH2
• Relative molecular mass is 42
Jisim molekul relatif ialah 42
• Decolourises liquid bromine in the dark
Menyahwarnakan cecair bromin dalam keadaan yang gelap
Based on the information given,
Berdasarkan pada maklumat yang diberi,
(i) determine the molecular formula of Z. [Relative atomic mass: H = 1; C = 12]
tentukan formula molekul bagi Z. [Jisim atom relatif: H = 1; C = 12]
[2 marks / 2 markah]
(ii) state the name of the homologous series for Z and explain your answer. [2 marks / 2 markah]
nyatakan nama siri homolog bagi Z dan terangkan jawapan anda.
(iii) write a balanced chemical equation for the reaction of compound Z with liquid bromine in the dark.
tulis satu persamaan kimia yang seimbang bagi tindak balas sebatian Z dengan cecair bromin dalam keadaan
yang gelap.
[2 marks / 2 markah]
(c) Diagram 4 shows the structural formulae of two organic compounds, A and B.
Rajah 4 menunjukkan formula struktur bagi dua sebatian organik, A dan B.
HH HO
H C C OH H C C OH
HH H
Compound A Compound B
Sebatian A Sebatian B
Diagram 4 / Rajah 4
Briefly describe two simple chemical tests to differentiate the two compounds given.
Terangkan secara ringkas dua ujian kimia ringkas untuk membezakan kedua-dua sebatian yang diberi.
[4 marks / 4 markah]
© Penerbitan Pelangi Sdn. Bhd. 42
Chemistry Form 5 Chapter 2 Carbon Compounds
Section C / Bahagian C
5. (a) Compound R is a hydrocarbon with the molecular formula of C4H10.
2S0P1M4 Sebatian R adalah hidrokarbon dengan formula molekul C4H10.
(i) Compound R has two isomers.
Draw the structural formulae for the isomers of R and name them.
Sebatian R mempunyai dua isomer.
Lukiskan formula struktur bagi isomer-isomer R dan namakannya.
[4 marks / 4 markah]
(ii) Compound R burns completely in excess oxygen to produce carbon dioxide and water.
Write the balanced chemical equation for the combustion of R and calculate the volume of carbon
dioxide gas produced when 11.6 g of R is combusted completely.
[Relative atomic mass: C = 12; O = 16; H = 1; Molar volume of gas at room conditions =
24.0 dm3 mol–1]
Sebatian R membakar lengkap dalam oksigen berlebihan untuk menghasilkan karbon dioksida dan air.
Tuliskan persamaan kimia seimbang bagi pembakaran R dan kirakan isi padu gas karbon dioksida yang terhasil
apabila 11.6 g R dibakar selengkapnya.
[Jisim atom relatif: C = 12; O = 16; H = 1; Isi padu molar gas pada keadaan bilik = 24.0 dm3 mol–1]
[6 marks / 6 markah]
(b) Table 5 shows the properties of four organic compounds. Each compound has two carbon atoms per
molecule.
Jadual 5 menunjukkan sifat-sifat empat sebatian organik. Setiap sebatian mempunyai dua atom karbon per molekul.
Organic compound Properties
Sebatian organik Sifat-sifat
K A colourless, volatile liquid that produces fragrant smell.
Cecair mudah meruap tak berwarna yang menghasilkan bau wangi.
Cannot dissolve in water and decolourises the reddish-brown colour of liquid bromine
L in the dark condition.
Tak larut dalam air dan menyahwarnakan warna perang kemerahan cecair bromin dalam
keadaan gelap.
Dissolves in water and reacts with magnesium ribbon to produce a colourless and
M flammable gas.
Larut dalam air dan bertindak balas dengan pita magnesium untuk menghasilkan gas tak
berwarna yang mudah terbakar.
Dissolves very easily in water and burns with blue flame to produce carbon dioxide and water.
N Larut dengan sangat mudah dalam air dan membakar dengan nyalaan biru untuk menghasilkan
karbon dioksida dan air.
Table 5 / Jadual 5
Based on Table 5, draw the structural formulae for compounds K, L, M and N.
Berdasarkan pada Jadual 5, lukiskan formula struktur bagi sebatian K, L, M dan N.
[4 marks / 4 markah]
(c) Diagram 5 shows how compound N can be converted into compound L and compound M respectively.
Rajah 5 menunjukkan bagaimana sebatian N boleh ditukarkan kepada sebatian L dan sebatian M masing-masing.
Compound L Oxidation Compound N Dehydration Compound M
Sebatian L Pengoksidaan Sebatian N Pendehidratan Sebatian M
Diagram 5 / Rajah 5
43 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Chapter 2 Carbon Compounds
(i) Draw a labelled diagram to show how compound L is prepared and collected from compound N.
Lukiskan rajah berlabel untuk menunjukkan bagaimana sebatian L disediakan dan dikumpul daripada sebatian
N.
[2 marks / 2 markah]
(ii) State the condition and reagent used to convert compound N into compound M and write the
chemical equation for the reaction.
Nyatakan keadaan dan reagen yang digunakan untuk menukarkan sebatian N kepada sebatian M dan tulis
persamaan kimia bagi tindak balas itu.
[4 marks / 4 markah]
Challenge Applying
TBK glove factory received multiple complains that the gloves they produced tear off easily and not durable.
The glove factory produces gloves directly from natural rubber latex.
Kilang sarung tangan TBK menerima banyak aduan bahawa sarung tangan yang dihasilkannya mudah koyak dan tidak
tahan lama. Kilang sarung tangan ini menghasilkan sarung tangan terus daripada lateks getah semula jadi.
Give suggestion to TBK glove factory so that they can overcome the defects of their gloves. Briefly describe
how your suggestion can be carried out.
Berikan cadangan kepada kilang sarung tangan TBK supaya mereka dapat mengatasi kelemahan pada sarung tangan
mereka. Terangkan secara ringkas bagaimana cadangan anda dapat dilakukan.
© Penerbitan Pelangi Sdn. Bhd. 44
Answers
CHA1PTER Rate of Reaction 1.3 Collision Theory (ii) At higher concentration,
Kadar Tindak Balas Teori Perlanggaran there are more particles of
Volume of H2 (cm3) reactants per unit volume.
Isi padu H2 (cm3)1.1Rate of Reaction 5. (a) collide, energy / berlanggar, tenaga This increases the frequency
Kadar Tindak Balas for the particles of reactants
(b) minimum / minimum to collide. Therefore, the
1. quantity, time frequency of effective collisions
kuantiti, masa 6. (a) larger, increases, increases, increases. The rate of reaction
increases increases.
lebih besar, meningkatkan, bertambah, Pada kepekatan yang lebih tinggi,
bertambah terdapat lebih zarah bahan tindak
2. (a) beginning, highest balas per unit isi padu. Hal ini
awal, tertinggi (b) more, faster, increases, increases, meningkatkan frekuensi bagi zarah-
increases zarah bahan tindak balas untuk
(b) decreases, decreases berlanggaran. Oleh itu, frekuensi
berkurang, berkurang lebih tinggi, pantas, meningkatkan, perlanggaran berkesan bertambah.
bertambah, bertambah Kadar tindak balas bertambah.
(c) (i) Overall average rate of reation
(c) more, increases, increases,
Kadar purata tindak balas increases
keseluruhan lebih banyak, meningkatkan, bertambah, (e)
bertambah
= 48 cm3 Volume of CO2 (cm3)
180 s (d) higher, increases, increases, Isi padu CO2 (cm3)
increases, increases
= 0.2667 cm3 s–1
lebih tinggi, bertambah, meningkatkan,
(ii) Average rate of reaction from bertambah, bertambah 180
90 s to 150 s 7. (a) rate, chemically / kadar, kimia 90
Purata kadar tindak balas dari 90 s
(b) activation / pengaktifan
hingga 150 s
8. (a) Ammonia / Ammonia
= (47 – 36) cm3
(150 – 90) s (b) Vanadium(V) oxide
Vanadium(V) oksida 0 Time (min)
= 0.1833 cm3 s–1 2 Masa (min)
(c) Nitric acid / Asid nitrik
(iii) Average rate of reaction in the 2. (a) CaCO3 + 2HCl → CaCl2 + H2O +
(d) Nickel / Nikel CO2
second minute
Purata kadar tindak balas dalam (b) (i) Because the reaction has
completed
minit kedua
Kerana tindak balas sudah selesai
= (42 – 27) cm3 SPM Practice 1
(120 – 60) s (ii) 2.50 g
= 0.25 cm3 s–1 (c)
Factors Affecting the Rate of Reaction SPM Paper 1
1.2 Faktor-faktor yang Mempengaruhi Kadar 1. C 2. C 3. B 4. C 5. D Delivery tube
Tindak Balas 6. A 7. C 8. D 9. C 10. C Salur penghantar
11. A 12. B 13. C 14. B 15. D
3. (a) Total surface area – smaller, higher 16. C 17. D 18. D 19. D 20. C Hydrochloric Burette
Jumlah luas permukaan – kecil, tinggi 21. A acid, 1.0 mol dm–3 Buret
Asid hidroklorik,
(b) Concentration of reactant – increases 1.0 mol dm–3
Kepekatan bahan tindak balas –
SPM Paper 2
meningkat
Section A / Bahagian A Water
(c) Temperature – increases Air
Suhu – meningkat
1. (a) ZnCO3 + H2SO4 → Calcium carbonate chips
(d) Catalyst – changes ZnSO4 + H2O + CO2 Serpihan kalsium karbonat
Mangkin – mengubah
(b) The volume of carbon dioxide gas (d) (i)
(e) Pressure – increases produced
Tekanan – meningkat
Isi padu gas karbon dioksida yang
4. dihasilkan Mass of calcium carbonate (g)
Jisim kalsium karbonat (g)
(a) (c) Average rate of reaction
Kadar tindak balas purata
4.00
= —18—0 —cm––3
(b) 120 s Curve at temperature 30°C
0 Lengkung pada suhu 30°C
= 1.5 cm3 s–1
(d) (i) Set 2. The acid used has a 2.50
Time (s) higher concentration.
Masa (s) Set 2. Asid yang digunakan t1 Time (s) / Masa (s)
mempunyai kepekatan yang lebih
tinggi.
A1 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
(ii) When the temperature 4. (a) A catalyst is a chemical substance (ii) Mg + 2HCl → MgCl2 + H2
increases, hydrogen ions and that changes the rate of a reaction
carbonate ions receive more (b) (i) Number of moles of HCl
kinetic energy and move by altering the activation energy of Bilangan mol HCl
faster. This increases the
chances for the ions to collide the reaction. = –0—.1—× —25– = 0.0025 mol [1]
with one another effectively. Mangkin adalah bahan kimia yang 1000
The frequency of effective
collision increases and the mengubah kadar tindak balas dengan From the equation,
rate becomes higher. mengubah tenaga pengaktifan tindak
balas. Number of moles of H2
Apabila suhu bertambah, ion
hidrogen dan ion karbonat Daripada persamaan,
menerima lebih tenaga kinetik dan
bergerak lebih pantas. (b) Copper(II) sulphate solution Bilangan mol H2
Hal ini meningkatkan peluang bagi Larutan kuprum(II) sulfat = —12
ion-ion tersebut berlanggar sesama × mol of HCl
sendiri dengan berkesan. Frekuensi mol HCl
perlanggaran berkesan meningkat (c) (i) Set I:
dan kadar menjadi lebih tinggi.
Average rate / Kadar purata = —21 × 0.0025 = 0.00125 mol
(e) Total surface area of calcium [1]
carbonate / Concentration of = 24 cm3 = 0.40 cm3 s–1.
hydrochloric acid / Presence of a 60 s
suitable catalyst ∴ volume of H2 / isi padu H2
Jumlah luas permukaan kalsium Set II: = 0.00125 × 24 dm3
karbonat / Kepekatan asid hidroklorik /
Kehadiran mangkin yang sesuai Average rate / Kadar purata = 0.03 dm3 [1]
3. (a) (i) Na2S2O3 + H2SO4 → = 15 cm3 = 0.25 cm3 s–1. (c) (i) • Rate of reaction in
Na2SO4 + SO2 + S + H2O 60 s Experiment II is higher than
in Experiment I.
(ii) The formation of sulphur (ii) Rate of reaction for Set I is
causes the '×' mark to Kadar tindak balas dalam
disappear from sight. higher than Set II. Eksperimen II lebih tinggi
Kadar tindak balas bagi Set I daripada dalam Eksperimen I.
Pembentukan sulfur menyebabkan
tanda ‘×’ hilang daripada adalah lebih tinggi daripada Set II. [1]
penglihatan.
(iii) The catalyst added in Set I • In Experiment II, the
(b) 0.017, 0.048, 0.077, 0.111, 0.143,
0.167 lowers the activation energy of hydrochloric acid with higher
(c) (i) the reaction. As a result, the concentration is used.
Temperature (°C) frequency of effective collision Dalam Eksperimen II, asid
Suhu (°C) hidroklorik dengan kepekatan
between zinc atoms and yang tinggi digunakan. [1]
hydrogen ions increases and • There are more particles of
the rate becomes higher. hydrochloric acid per unit
Mangkin yang ditambahkan
volume used in
dalam Set I merendahkan tenaga
pengaktifan tindak balas. Akibatnya, Experiment II.
frekuensi perlanggaran berkesan
antara atom zink dengan ion Terdapat lebih banyak zarah
hidrogen bertambah dan kadar asid hidroklorik per unit isi padu
meningkat. digunakan dalam Eksperimen II.
(d) (i)(ii) [1]
Energy • This increases the chance
Tenaga
for the particles of reactants
70 EII EI Set II
60 Zn + H2SO4 Set I to collide effectively.
50 Hal ini meningkatkan peluang
40
30 bagi zarah bahan tindak balas
20 untuk berlanggar dengan lebih
10 efektif. [1]
0 –Ti1me (s–1) ZnSO + H2 • The frequency of effective
M–a1sa (s–1) collisions increases, causing
0.02 Reaction path the rate to increase.
0.04 Laluan tindak balas
0.06 Frekuensi perlanggaran yang
0.08 Section B / Bahagian B efektif menyebabkan kadarnya
0.10 meningkat. [1]
0.12
0.14 (ii) • By adding a suitable
0.16 catalyst, such as copper(II)
0.18 sulphate solution.
(ii) Rate of reaction 5. (a) (i) • Rate of reaction is defined Dengan menambah mangkin
Kadar tindak balas yang sesuai, seperti larutan
as the change in the kuprum(II) sulfat. [1]
(d) (i) The rate of reaction increases quantity of magnesium • When a catalyst is added,
the catalyst provides an
with temperature. that dissolved in the acid alternative pathway for the
Kadar tindak balas bertambah reaction to occur with a
dengan suhu. against time taken. lower activation energy.
Kadar tindak balas ditakrifkan
Apabila mangkin ditambahkan,
sebagai perubahan kuantiti mangkin akan membekalkan
laluan alternatif bagi tindak
(ii) At higher temperature, the magnesium yang larut dalam balas berlaku dengan tenaga
pengaktifan yang rendah. [2]
reactant particles possess more asid melawan masa yang
• The particles of reactants
kinetic energy. The frequency diambil. [1] are able to collide
effectively even though they
of effective collisions increases. • The factors affecting the
Hence, this increases the rate rate of reaction are:
Faktor-faktor yang
of reaction.
Pada suhu yang lebih tinggi, mempengaruhi kadar tindak
zarah-zarah bahan tindak balas ialah:
balas mempunyai lebih tenaga
kinetik. Frekuensi perlanggaran – Temperature
berkesan bertambah. Oleh itu, ini Suhu [1]
meningkatkan kadar tindak balas.
– Concentration
Kepekatan [1]
© Penerbitan Pelangi Sdn. Bhd. A2
Chemistry Form 5 Answers
contain lesser energy. (iii) Average rate of reaction for Terdapat lebih banyak
Kadar tindak balas purata bagi: perlanggaran antara zarah-
Zarah bahan tindak balas boleh zarah bahan tindak balas.
berlanggar dengan efektif Experiment I / Eksperimen I Frekuensi perlanggaran
walaupun mempunyai tenaga berkesan bertambah.
yang rendah. [1] = –1—20—c—m–3 = 2 cm3 s–1 [1]
60 s
• The frequency of effective Section C / Bahagian C
collisions increases, resulting Experiment II / Eksperimen II
rate to increase also. 7. (a) Factor 1: Concentration of
= –1—20—c—m–3
Frekuensi perlanggaran 42 s reactants [1]
yang berkesan meningkat Faktor 1: Kepekatan bahan tindak balas
menyebabkan kadarnya = 2.86 cm3 s–1 [1]
meningkat. [1] • The rate of reaction increases
Experiment III / Eksperimen III
(iii) • The acid used in with the concentration of the
Experiment II has double –1—20—c—m–3
the concentration of the = 30 s reactants. [1]
acid in Experiment I. Kadar tindak balas bertambah
= 4 cm3 s–1 [1]
Asid yang digunakan dalam dengan kepekatan bahan tindak
Eksperimen II mempunyai (iv) • Rate of reaction of
kepekatan berganda berbanding balas.
asid di dalam Eksperimen I. [1] Experiment II is higher than
Factor 2: Total surface area of the
• As the number of moles that of Experiment I. [1]
of reactant has doubled, Kadar tindak balas Eksperimen reactants [1]
the quantity of product Faktor 2: Jumlah luas permukaan bahan
(hydrogen gas) becomes II lebih tinggi daripada
double also. Eksperimen I. tindak balas
Apabila bilangan mol bahan • The rate of reaction increases
tindak balas digandakan, kuantiti
hasil tindak balas (gas hidrogen) • Experiment II uses calcium with the total surface area of the
juga digandakan. [1]
carbonate powder which is reactants. [1]
Kadar tindak balas bertambah
smaller size than calcium
dengan jumlah luas permukaan
carbonate chips used in
bahan tindak balas.
Experiment I. [1]
Eksperimen II menggunakan Factor 3: Temperature of the
serbuk kalsium karbonat yang reaction [1]
Faktor 3: Suhu tindak balas
bersaiz lebih kecil daripada
serpihan kalsium karbonat yang • The rate of reaction increases
6. (a) • Particles of smaller size have digunakan dalam Eksperimen I. with the temperature of the
larger surface area. [1] • The smaller the particle reaction. [1]
Kadar tindak balas bertambah
Zarah yang bersaiz lebih kecil size of the reactant, the
dengan suhu tindak balas.
mempunyai luas permukaan yang larger the total surface area
lebih besar. exposed for reaction. [1] Factor 4: Presence of a catalyst [1]
Semakin kecil saiz zarah bahan Faktor 4: Kehadiran mangkin
• Rate of reaction is higher if the
tindak balas, semakin besar
surface area is larger. [1] jumlah luas permukaan yang • The presence of a positive
terdedah untuk tindak balas
Kadar tindak balas adalah lebih tinggi berlaku. catalyst increases the rate of a
jika luas permukaan lebih besar. reaction. [1]
Kehadiran mangkin positif
(b) (i) CaCO3 + 2HCl → • As a result, the frequency
meningkatkan kadar suatu tindak
CaCl2 + H2O + CO2 [1] of effective collisions
balas.
From the equation, 1 mol of increases. [1]
Akibatnya, frekuensi (b)
CaCO3 produces 1 mol of
perlanggaran berkesan Quantity of product
CO2. [1] bertambah. Kuantiti hasil tindak balas
Daripada persamaan, 1 mol CaCO3
menghasilkan 1 mol CO2. • The rate of reaction of
The number of moles CO2 Experiment III is higher
Bilangan mol CO2
than that of Experiment II.
= number of moles CaCO3
bilangan mol CaCO3 [1]
Kadar tindak balas Eksperimen
= –10—0.50– = 0.005 mol [1] 0 Time
III lebih tinggi daripada Masa
Eksperimen II.
∴ The volume of CO2 • The temperature of reaction [2]
Isi padu CO2
for Experiment III is higher • At the beginning, the rate of
= 0.005 × 24 = 0.12 dm3 [1]
than that of Experiment II. reaction is high. [1]
(ii)
[1] Pada mulanya, kadar tindak balas
Suhu tindak balas Eksperimen
Volume of gas (cm3) adalah tinggi.
Isi padu gas (cm3) III lebih tinggi daripada
Eksperimen II. • This is shown by the gradient or
III II I • Higher temperature tangent of the curve. The tangent
120 increases the kinetic energy
of the particles of reactants. is very steep at the beginning. [1]
[1]
Ini dapat ditunjukkan dengan
Suhu yang lebih tinggi
meningkatkan tenaga kinetik kecerunan atau tangen lengkung.
zarah-zarah bahan tindak balas.
Tangen itu sangat curam pada
• There will be more
collisions between the mulanya.
particles of reactants.
The frequency of effective • It is because at the beginning,
collisions increases. [1]
the concentration of the
0 Time (s) reactants is high. [1]
Masa (s)
30 42 60 Ini disebabkan kepekatan bahan
[3] tindak balas adalah tinggi pada
mulanya.
A3 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
• Towards the end, the rate of (b) • Carbon dioxide H
Karbon dioksida &
reaction decreases and lastly H!C!H
• Ethanoic acid H&H
becomes zero. [1] Asid etanoik &&&
H!C"C!C!H
Pada akhir tindak balas, kadar
2-Methylpropene / 2-Metilpropena
tindak balas berkurang dan akhirnya 2. (a) carbon, hydrogen
karbon, hidrogen
menjadi sifar.
• The gradient or tangent is flat at
the end of the reaction, showing (b) (i) Saturated hydrocarbons and
the rate is zero. [1] unsaturated hydrocarbons. 2.5 Alcohols
Hidrokarbon tepu dan hidrokarbon Alkohol
Kecerunan atau tangen adalah tak tepu.
mendatar pada akhir tindak balas
menunjukkan kadar adalah sifar. (ii) single, double 7. (a) Hydroxyl group, –OH
tunggal, ganda dua Kumpulan hidroksil, –OH
• Towards the end, the
(b) CnH2n + 1OH, n = 1, 2, 3, …
concentration of the reactant (c) Petroleum and natural gas
Petroleum dan gas asli 8. (a) (i) Fermentation
decreases and eventually Penapaian
becomes zero when the reaction (ii) Hydration
Penghidratan
stops. [1] 2.2 Alkanes (b) (i) liquid / Cecair
Pada akhir tindak balas, kepekatan Alkana
(ii) Low / rendah
bahan tindak balas berkurang dan 3. (a) saturated / tepu (iii) Soluble / Larut
(c) (i) fuel / bahan api
akhirnya menjadi sifar apabila tindak (b) CnH2n + 2, n = 1, 2, 3, … (ii) antiseptic / antiseptik
(c) (i) Insoluble / Tak larut (iii) organic / organic
balas selesai. (ii) Cannot / Tidak boleh
(iii) Low / rendah
(c) • A pressure cooker cooks food
(d) (i) C3H8 + 5O2 → 3CO2 + 4H2O
under high pressure. [1]
Complete combustion
Periuk tekanan memasak makanan di Pembakaran lengkap
bawah tekanan yang tinggi. (ii) CH4 + Cl2 → CH3Cl + HCl
• The higher pressure causes the Substitution reaction
Tindak balas penukargantian
water in the cooker to boil at a
higher temperature. [1] 9. (a) Carbon dioxide / Karbon dioksida
(b) Ethanoic acid / Asid etanoik
Tekanan yang lebih tinggi (c) Ethene / Etena
(d) Ethyl propanoate / Etil propanoat
menyebabkan air dalam periuk
tekanan mendidih pada suhu yang
lebih tinggi.
• At a higher temperature, food Alkenes
Alkena
particles have more kinetic 2.3 2.6 Carboxylic Acids
Asid Karboksilik
energy. The frequency of
collisions between food particles 4. (a) unsaturated, double
tak tepu, ganda dua.
increases. [1] 10. (a) Carboxyl group
Pada suhu yang lebih tinggi, (b) CnH2n, n = 2, 3, 4, … Kumpulan karboksil
zarah-zarah makanan mempunyai CnH2n + 1COOH, n = 0, 1, 2, …
tenaga kinetik yang tinggi. Frekuensi 5. (a) CO2 and H2O / CO2 dan H2O
perlanggaran antara zarah-zarah (b) acidified potassium manganate(VII)
makanan bertambah. (b) C and H2O or CO and H2O kalium manganat(VII) berasid
C dan H2O atau CO dan H2O
• The frequency of effective (c) (i) sour / masam
(c) Ethane / Etana (ii) Soluble / Larut
collisions increases and the rate (d) 1,2-dibromoethane
1,2-dibromoetana
of reaction increases. Food is 11. (a) base / bes
(e) 1-chloroethane / 1-kloroetana (b) metal carbonate / karbonat logam
cooked faster. [1] (f) Ethanol / Etanol (c) metal / logam
(g) 1,2-ethanediol / 1,2-etanadiol (d) alcohol / alkohol
Frekuensi perlanggaran berkesan (h) Polyethene / Polietena
bertambah dan kadar tindak balas
bertambah. Makanan dimasak
dengan lebih cepat.
Challenge 2.7 Esters
Ester
When cut into smaller pieces, the total 2.4 Isomerism 12. (a) Esterification / Pengesteran
Keisomeran (b) Concentrated sulphuric acid
surface area of the chicken meat exposed to Asid sulfurik pekat
6. (a) molecular, structural (c) (i) fragrant / wangi
fire is increased. This causes the process to molekul, struktur (ii) Low / rendah
(iii) insoluble / tidak larut
cook the chicken meat to complete faster. (b) H H H H
Apabila dipotong kepada kepingan yang lebih kecil, &&&& 13. (a) (i) Propyl ethanoate
jumlah luas permukaan daging ayam yang terdedah Propil etanoat
kepada api bertambah. Ini menyebabkan proses H!C"C!C!C!H (ii) Propanol / Propanol
untuk selesai memasak daging ayam menjadi lebih && (iii) Ethanoic acid / Asid etanoik
cepat. HH
(b) (i) Ethyl butanoate / Etil butanoat
2PTER Carbon Compounds 1-Butene / 1- Butena (ii) Ethanol / Etanol
Sebatian Karbon (iii) Butanoic acid / Asid butanoik
CHA HHHH
&&&&
H!C!C"C!C!H
&&
2.1 Carbon Compounds
Sebatian Karbon H H
1. (a) carbon / karbon 2-Butene / 2-Butena
© Penerbitan Pelangi Sdn. Bhd. A4
Chemistry Form 5 Answers
2.8 Fats (b) Alkene X is insoluble in water, but (ii) Esterification
Lemak compound Y is soluble in water. Pengesteran
14. (a) Natural esters Alkena X tidak larut di dalam air, tetapi (iii) This because the compound
Ester semula jadi sebatian Y larut di dalam air. is less dense than water and
insoluble in water.
(b) Fatty acids, glycerol (c) (i) The reddish-brown colour of
Asid lemak, gliserol Hal ini kerana sebatian itu kurang
bromine decolourises. tumpat daripada air dan tidak larut
15. (a) Animals / Haiwan Warna perang kemerahan bromin dalam air.
(b) Plants / Tumbuhan dinyahwarnakan
(c) Solid or semi-solid (ii) C3H6 + Br2 → C3H6Br2 Section B / Bahagian B
Pepejal atau separa pepejal (d) (i) Propanol
4. (a) • All members of the same
(d) Liquid / Cecair (ii) C3H6 + H2O → C3H7OH homologous series can be
(e) Higher / Lebih tinggi (e)
(f) Lower / Lebih rendah represented by the same
Glass wool soaked general formula. For example,
with propanol
Wul kaca yang Porcelain chips ethene, propene and butene are
dibasahi propanol Serpihan porselin
members of alkene homologous
Propene gas
2.9 Natural Rubber Gas propena series and can be represented
Getah Asli
by the general formula CnH2n,
where n = 2, 3, 4, …
[2]
16. H CH3 H H Semua ahli siri homolog dapat
diwakili oleh formula am yang sama.
HCCCCH Heat Sebagai contohnya, etena, propena
Panaskan
dan butena adalah ahli siri homolog
2-methylbut-1,3-diene Water alkena dan boleh diwakili oleh
2-metilbut-1,3-diena Air
formula am CnH2n, dengan n = 2, 3, 4,
…
17. (a) hardening, solid 2. (a) (i) Isomers are molecules with • The members of a homologous
pengerasan, pepejal series show a gradual change
the same molecular formula in physical properties such as
(b) bacteria, neutralises the boiling points as the number
bakteria, meneutralkan but with different structural of carbon atoms per molecule
increases. For example, the
(c) (i) Methanoic acid formula.
Asid metanoik Isomer adalah molekul dengan boiling points of alkenes
formula molekul yang sama tetapi
(ii) Ammonia solution formula struktur yang berbeza.
Larutan ammonia
(ii) C5H12 increase from ethene to butene.
18. (a) harder, stronger, elastic (iii) P: Pentane [2]
keras, kuat, elastik Pentana Ahli siri homolog menunjukkan suatu
perubahan sifat fizikal secara beransur-
(b) (i) Softer / Lebih lembut Q: 2-methylbutane ansur seperti takat didih apabila
(ii) Harder / Lebih keras 2-metilbutana bilangan atom karbon per molekul
(iii) Less elastic / Kurang kenyal bertambah. Sebagai contohnya, takat
(iv) More elastic / Lebih kenyal (iv) H CH3 H didih alkena bertambah dari etena ke
& & & butena.
(v) Lower resistance H ! C ! C ! C ! H
Kurang ketahanan & & & • All members of the same
H CH3 H
(vi) Higher resistance homologous series can be
Lebih ketahanan (b) (i) CnH2n, n = 2, 3, 4, …
prepared using the same
(vii) Less / Kurang (ii) Hydrogenation
(viii) More / Lebih Penghidrogenan method. For example, alkenes
(ix) Less durable (iii) 180°C, Nickel / Nikel can be prepared from the
Kurang tahan lama
3. (a) (i) CnH2n+1OH, n = 1, 2, 3, … dehydration of a corresponding
(x) More durable
Lebih tahan lama (ii) C4H9OH alcohol. [2]
Semua ahli siri homolog dapat
(iii) Butan-1-ol / Butan-1-ol
disediakan dengan kaedah
Butan-2-ol / Butan-2-ol yang sama. Sebagai contohnya,
alkena dapat disediakan melalui
2-methylpropan-1-ol pendehidratan alkohol yang berpadan
2-metilpropan-1-ol dengannya.
SPM Practice 2 2-methylpropan-2-ol • The members of a homologous
SPM Paper 1 2-metilpropan-2-ol
series undergo the same
1. B 2. B 3. B 4. B 5. C (any two / mana-mana dua)
6. D 7. B 8. C 9. D 10. A reaction and yield similar
11. B 12. D 13. D 14. A 15. A (b) (i) Carbon dioxide
16. A 17. C 18. B 19. C 20. B Karbon dioksida products. For example, all
21. A 22. C 23. B 24. D 25. B
(ii) Flow the gas into a test tube alkenes undergo hydrogenation
SPM Paper 2 containing lime water. The
Section A / Bahagian A lime water turns chalky. to produce corresponding
1. (a) Propene / Propena Alirkan gas itu ke dalam tabung uji alkanes. [2]
yang mengandungi air kapur. Air Ahli-ahli siri homolog yang sama
kapur menjadi keruh.
mengalami tindak balas yang
(c) (i) Propanoic acid sama dan menghasilkan hasil yang
Asid propanoik serupa. Sebagai contohnya, semua
alkena mengalami penghidrogenan
untuk menghasilkan alkana yang
berkenaan.
A5 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
• Each subsequent member of a H Reagent: Acidified potassium
& manganate(VII) solution [1]
homologous series is different HH ! C ! HH
& & & Reagen: Larutan kalium
from the previous member by H ! C ! C ! C ! H manganat(VII) berasid
& & &
a –CH2 group. For example, H H H Equation / Persamaan:
ethene (C2H4), propene, (C3H6)
and butene (C4H8) successively CH3CH2OH + 2[O] KMnO4/H+ CH3COOH + H2O
differ from one another by a – Reflux
CH2 group. [2] 2-methylpropane
2-metilpropana
Setiap ahli yang seterusnya dalam Refluks [2]
siri homolog adalah berbeza dengan [4]
ahli yang sebelumnya dari segi satu (ii) 2C4H10 + 13O2 → 8CO2 + Challenge
10H2O + 10H2O [2]
kumpulan –CH2. Sebagai contohnya,
etena (C2H4), propena (C3H6) dan Number of mol C4H10 The gloves should be vulcanised.
butena (C4H8) adalah berbeza dari Bilangan mol C4H10
satu dengan yang lain dari segi satu Vulcanisation can be done by immersing the
kumpulan –CH2. = 11.6 gloves into disulphur dichloride solution for
[(4 × 12) + (10 × 1)]
(b) (i) n = ——4—2 –– several minutes.
(12 + 2) = 0.2 mol [1] Sarung tangan tersebut perlu divulkankan.
Pemvulkanan dapat dilakukan dengan
= 3 [1] From the equation, merendamkan sarung tangan ke dalam larutan
Daripada persamaan, disulfur diklorida selama beberapa minit.
Molecular formula
Formula molekul 2 mole of C4H10 produces
= (CH2)3 8 mol of CO2. [1] 3PTER
= C3H6 [1] 2 mol C4H10 menghasilkan 8 mol CHA Oxidation and Reduction
[1] Pengoksidaan dan Penurunan
(ii) Z is an alkene. CO2
Z ialah suatu alkena.
∴ 0.2 mole of C4H10 produces
This is because Z has the 3.1 Redox Reactions
0.8 mol CO2. Tindak Balas Redoks
general formula of CnH2n, 0.2 mol C4H10 menghasilkan
where n = 3 / It decolourises
0.8 mol CO2.
liquid bromine in the dark. [1]
Ini adalah kerana Z mempunyai Volume of CO2 / Isi padu CO2 1. (a) Losing / Kehilangan
formula am CnH2n, dengan n = 3 / = 0.8 × 24 dm3 = 19.2 dm3 (b) Gaining / Penerimaan
Z menyahwarnakan cecair bromin
dalam keadaan yang gelap. [1] (c) Gaining / Penerimaan
(iii) C3H6 + Br2 → C3H6Br2 [2] (b) K: O H (d) Losing / Kehilangan
' & (e) Elimination / Penyingkiran
H ! C ! O ! C ! H (f) Receiving / Penerimaan
& (g) Increase / Peningkatan
(c) Test 1: Mix both compounds H [1] (h) Decrease / Pengurangan
with calcium carbonate chips 2. (a) (i) oxidises, reduction
[1] L: H H
separately. & & mengoksidakan, penurunan
H ! C " C ! H
Ujian 1: Campuran kedua-dua sebatian (ii) chlorine water, acidified
dengan cebisan kalsium karbonat secara potassium manganate(VII)
[1]
berasingan. air klorin, kalium manganat(VII)
M: H O berasid
Results: Compound B gives & '
H ! C ! C ! O H (b) (i) reduces, oxidation
effervescence but compound & menurunkan, pengoksidaan
A does not react with calcium
H (ii) electropositive metals,
carbonate chips. [1] potassium iodide solution
Keputusan: Sebatian B membuak tetapi [1] logam elektropositif, larutan kalium
sebatian A tidak bertindak balas dengan [1] iodida
N: H H
cebisan kalsium karbonat. & &
H ! C ! C ! O H
Test 2: Reflux both compounds & &
separately with acidified potassium H H
manganate(VII). [1]
Ujian 2: Refluks kedua-dua sebatian (c) (i) 3. (a) +3 (b) +5 (c) +6
dengan kalium manganat(VII) berasid Porcelain chips (d) +7 (e) –3 (f) +4
secara berasingan. Serpihan porselin
Results: Compound A Rusting as a Redox Reaction
Pengaratan sebagai Tindak Balas Redoks
decolourises acidified potassium Compound M 3.2
manganate(VII) but compound B Sebatian M
does not. [1] 4. (a) redox, oxidised, ions
redoks, dioksidakan, ion
Keputusan: Sebatian A Heat
menyahwarnakan kalium manganat(VII) Panaskan (b) rusting / pengaratan
berasid tetapi sebatian B tidak. Glass wool soaked Water (c) (i) Oxygen, water, oxidising,
with compound N Air
Section C / Bahagian C Wul kaca yang reducing
dibasahkan dengan Oksigen, air, pengoksidaan,
5. (a) (i) H H H H sebatian N penurunan
& & & &
H ! C ! C ! C ! C ! H [2] (ii) anode, oxidised,
& & & & [1]
H H H H (ii) Condition: Under reflux Fe → Fe2+ + 2e–
Keadaan: Di bawah refluks
anod, dioksidakan,
Butane Fe → Fe2+ + 2e–
Butana
© Penerbitan Pelangi Sdn. Bhd. A6
Chemistry Form 5 Answers
(iii) cathode, reduction, SPM Paper 2 (ii) G
Carbon
2H2O + O2 + 4e– → 4OH– Section A / Bahagian A electrode P Carbon
katod, penurunan, Elektrod electrode Q
1. (a) (i) Presence of oxygen and water karbon P Elektrod
2H2O + O2 + 4e– → 4OH– Kehadiran oksigen dan air karbon Q
(iv) iron(II) hydroxide, green,
2Fe + O2 + 2H2O → 2Fe(OH)2 (ii) • Fe(s) → Fe2+(aq) + 2e–
ferum(II) hidroksida, hijau, Fe(p) → Fe2+(ak) + 2e–
2Fe + O2 + 2H2O → 2Fe(OH)2 • 2H2O(l) + O2(g) + 4e– →
(v) hydrated iron(III) oxide, reddish- 4OH–(aq)
brown 2H2O(ce) + O2(g) + 4e– → (d)
ferum(III) oksida terhidrat, perang- G
4OH–(ak)
kemerahan
(b) (i) After Fe2+ and OH– ions are
formed, both combine to form
Reactivity Series of Metals and Its green precipitate of Fe(OH)2.
Applications The green precipitate, Fe(OH)2
3.3 Siri Kereaktifan Logam dan Aplikasinya will further oxidise into +–
Fe(OH)3 in the presence of
5. (a) oxygen / oksigen oxygen and water. Copper Magnesium
(b) decreasing / menurun Kuprum Magnesium
(c) higher, lower 4Fe(OH)2(s) + 2H2O(l) + O2(g) →
lebih tinggi, lebih rendah 4Fe(OH)3(s) Copper(II) Magnesium
(d) Carbon / Karbon sulphate sulphate
(e) electrolysis / elektrolisis Fe(OH)3 will then decompose solution Porous pot solution
into Fe2O3.xH2O, which is the Larutan Pasu berliang Larutan
6. Copper, Tin, Iron, Zinc, Magnesium, kuprum(II) magnesium
Sodium reddish-brown solid called sulfat sulfat
Kuprum, Stanum, Ferum, Zink, Magnesium, rust. 3. (a) To provide oxygen for the reaction
Natrium Selepas ion-ion Fe2+ dan Untuk membekalkan oksigen bagi tindak
3.4 Electrolytic and Chemical Cells OH– terbentuk, kedua-duanya balas
Sel Elektrolisis dan Sel Kimia bergabung untuk membentuk
mendakan hijau, Fe(OH)2. (b) Heat up the metal powder until it is
7. (a) Cathode / Katod Mendakan hijau, Fe(OH)2, akan hot before heating the potassium
dioksidakan dengan selanjutnya chlorate(V) to supply the oxygen
(b) Negative terminal / Terminal negatif kepada Fe(OH)3 dalam kehadiran for the reaction.
oksigen dan air.
(c) Anode / Anod Panaskan serbuk logam sehingga
4Fe(OH)2(p) + 2H2O(ce) + O2(g) menjadi panas sebelum memanaskan
(d) Positive terminal / Terminal positif kalium klorat(V) untuk membekalkan
→ 4Fe(OH)3(p) oksigen bagi tindak balas itu.
(e) Electrical energy → chemical
energy Fe(OH)3 akan mengurai kepada (c) Magnesium / Magnesium
Fe2O3.xH2O, iaitu suatu pepejal
Tenaga elektrik → tenaga kimia perang-kemerahan yang disebut (d) (i) The brown powder changed to
karat. black.
(f) Chemical energy → electrical
energy (ii) +2 to +3 / +2 kepada +3 Serbuk perang bertukar menjadi
hitam.
Tenaga kimia → tenaga elektrik (c) (i) Magnesium is more
(ii) 2Cu(s) + O2(g) → 2CuO(s)
(g) Electrons flow from anode to electropositive than iron. 2Cu(p) + O2(g) → 2CuO(p)
cathode / Elektron mengalir dari anod
ke katod Therefore, it oxidises more (iii) 0 to +2 / 0 kepada +2
(h) Electrons flow from negative easily than iron. Magnesium (e) In this reaction, metal P displaces
terminal to positive terminal copper from copper(II) oxide to
oxidises itself by losing produce P oxide. In this reaction,
Elektron mengalir dari terminal negatif ke metal P is oxidised whereas
terminal positif electrons to iron to prevent copper(II) ions are reduced.
Oxidation and reduction occurs
(i) Occurs at the anode iron from oxidation. simultaneously. Therefore, a redox
Berlaku di anod Magnesium lebih elektropositif reaction occurred.
daripada besi. Oleh itu, magnesium
(j) Occurs at the negative terminal teroksida lebih mudah daripada Dalam tindak balas ini, logam P
Berlaku di terminal negatif besi. Magnesium mengoksidakan menyesarkan kuprum daripada
dirinya dengan melepaskan elektron kuprum(II) oksida untuk menghasilkan
(k) Occurs at the cathode kepada besi untuk mengelakkannya oksida P. Dalam tindak balas ini,
Berlaku di katod daripada teroksida. logam P dioksidakan manakala ion
kuprum(II) diturunkan. Pengoksidaan dan
(l) Occurs at the positive terminal (ii) Mg(s) → Mg2+(aq) + 2e– penurunan berlaku secara serentak. Oleh
Berlaku di terminal positif itu, tindak balas redoks berlaku.
Mg(p) → Mg2+(ak) + 2e–
(f) (i) Carbon / Karbon
SPM Practice 3 2. (a) Purple
(ii) Oxidation of Q produces oxide
SPM Paper 1 Ungu of Q which exists in gas state
at room conditions.
1. D 2. A 3. C 4. D 5. C (b) (i) 2I– → I2 + 2e-
6. B 7. D 8. D 9. C 10. A Pengoksidaan Q menghasilkan
11. B 12. A 13. B 14. D 15. C (ii) Oxidation / Pengoksidaan oksida Q yang wujud dalam
16. B 17. A 18. B 19. B 20. B keadaan gas pada keadaan bilik.
(iii) Test the solution with some
starch solution. A blue black
colourisation is produced.
Uji larutan tersebut dengan sedikit
larutan kanji. Warna biru tua akan
terhasil.
(c) (i) MnO4– + 8H+ + 5e- → Mn2+ +
4H2O
A7 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
Section B / Bahagian B disebabkan oleh pembentukan • In the experiment, iron(II)
4. (a) (i) Potassium: +1 / Kalium: +1 [1] iodin.
sulphate is used as the reducing
Iron: +3 / Ferum: +3 [1] • Acidic potassium
agent. [1]
(ii) K2O: Potassium oxide [1] dichromate(VI) is orange in Dalam eksperimen ini, ferum(II) sulfat
Kalium oksida colour at the beginning. [1] digunakan sebagai agen penurunan.
Kalium dikromat(VI) berasid
Fe2O3: Iron(III) oxide [1] • The manganate(VII) ions that
berwarna jingga pada asalnya.
Ferum(III) oksida act as an oxidising agent will
(iii) • The roman number is not • After 10 minutes, the help to oxidise iron(II) ions into
necessary for potassium solution turns green iron(III) ions. [1]
oxide because potassium because of the formation of Ion manganat(VII) yang bertindak
only exhibits one stable chromium(III) ions. [1] sebagai agen pengoksidaan akan
oxidation number. [1] Selepas 10 minit, larutan membantu dalam mengoksidakan ion
Nombor roman tidak diperlukan bertukar menjadi hijau ferum(II) kepada ion ferum(III).
bagi kalium oksida kerana disebabkan oleh pembentukan • The half-equation for the
kalium hanya menunjukkan satu ion kromium(III). oxidation process is:
Persamaan setengah bagi proses
nombor pengoksidaan yang Section C / Bahagian C
pengoksidaan ialah:
stabil. 5. (a) • The suitable metal is zinc. [1]
Logam yang sesuai ialah zink. Fe2+ → Fe3+ + e–. [1]
• Roman number is used
• The colourless solution contains • The electrons will flow from
for iron(III) oxide because carbon electrode P to carbon
zinc ions, Zn2+ and sulphate electrode Q through the external
iron exhibit more than one
ions, SO42–. [1] circuit across the galvanometer.
stable oxidation number. [1] Larutan tidak berwarna mengandungi
Nombor roman digunakan untuk [1]
ion zink, Zn2+ dan ion sulfat, SO42–. Elektron akan mengalir dari
ferum(III) oksida kerana ferum
menunjukkan lebih daripada • During the reaction, the elektrod karbon P ke elektrod
satu nombor pengoksidaan karbon Q menerusi litar luar melalui
yang stabil. oxidation number of zinc galvanometer.
• At carbon electrode Q, the
(b) (i) Potassium iodide [1] increases from 0 to +2. [1]
Kalium iodida manganate(VII) ions and
Semasa tindak balas berlaku, nombor
hydrogen ions will receive
pengoksidaan zink bertambah dari 0
electrons and undergo
(ii) • At the negative terminal, ke +2.
Di terminal negatif, reduction. [1]
• Oxidation occurs on zinc. [1] Di elektrod karbon Q, ion
2I– → I2 + 2e– Pengoksidaan berlaku pada zink.
manganat(VII) dan ion hidrogen akan
(Writing the equation [1], • Zinc acts as a reducing agent in menerima elektron dan mengalami
penurunan.
balancing the number of the reaction. [1]
atoms and charges [1]) Zink bertindak sebagai agen
(Tulis persamaan [1],
penurunan dalam tindak balas.
seimbangkan bilangan atom dan
cas [1]) • Half-equation for the reaction on • The half-equation for the
zinc: Zn → Zn2+ + 2e– [1]
reduction process is:
Setengah persamaan bagi tindak
• At the positive terminal, Persamaan setengah bagi proses
Di terminal positif, balas pada zink: Zn → Zn2+ + 2e-
penurunan ialah:
Cr2O72– + 14H+ + 6e– →
• On the other hand, the oxidation MnO4– + 8H+ + 5e– →
2Cr3+ + 7H2O number of copper decreases Mn2+ + 4H2O. [1]
(Writing the equation [1], from +2 to 0. [1] • The overall equation for the
balancing the number Sementara itu, nombor pengoksidaan redox reaction is:
of atoms [1], balancing kuprum berkurang dari +2 ke 0. Persamaan keseluruhan bagi tindak
the number of charges • Reduction occurs on copper(II) balas redoks ialah:
[1]) / (Tulis persamaan [1], ions. [1] 5Fe2+ + MnO4– + 8H+
Penurunan berlaku pada ion kuprum(II). → 5Fe3+ + Mn2+ + 4H2O. [1]
seimbangkan bilangan atom [1],
• The iron(II) sulphate solution
seimbangkan bilangan cas [1]) • Copper(II) ions act as an oxidising
slowly changes its colour
(iii) • At the negative terminal, agent in the reaction. [1]
from green to yellow, while
oxidation occurred because Ion kuprum(II) bertindak sebagai agen
pengoksidaan dalam tindak balas. the acidified potassium
electrons are donated. [2]
Di terminal negatif, • Half-equation for the reaction on manganate(VII) solution
copper: Cu2+ + 2e– → Cu [1]
pengoksidaan berlaku kerana changes from purple to
elektron didermakan. Persamaan setengah bagi tindak
colourless. [2]
• At the positive terminal, balas pada kuprum: Larutan ferum(II) sulfat bertukar
reduction occurred because
Cu2+ + 2e– → Cu perlahan-lahan dari warna hijau ke
kuning manakala larutan kalium
electrons are accepted. [2] (b) • Apparatus set up: / Susunan radas: manganat(VII) berasid bertukar dari
Di terminal positif, penurunan ungu ke tidak berwarna.
Carbon G Carbon
berlaku kerana elektron electrode P electrode Q
Elektrod Elektrod Maximum / Maksimum: 10
diterima. karbon P karbon Q
• Potassium iodide solution is Iron(II) Acidified
sulphate potassium
colourless at the beginning. solution manganate(VII)
Larutan ferum(II) solution
[1] sulfat Larutan kalium Challenge
Larutan kalium iodida adalah manganat(VII)
berasid (a) (i) A
tidak berwarna pada asalnya.
[2] (ii) Iron rusts faster in the presence of
• After 10 minutes, the salt solution.
solution turns yellow Besi berkarat dengan lebih cepat dalam
kehadiran larutan garam.
because of the formation of Dilute sulphuric
acid
iodine. [1] Asid sulfurik cair
Selepas 10 minit, larutan
bertukar menjadi kuning
© Penerbitan Pelangi Sdn. Bhd. A8
Chemistry Form 5 Answers
(b) (i) The iron rusts slower. Number of moles of lead(II) sulphate Heat change / Perubahan haba, Q
Besi berkarat dengan lebih perlahan. = mcθ
produced = 200 × 4.2 × 6.75 = 5670 J
(ii) Zinc is more electropositive than Bilangan mol plumbum(II) sulfat yang terhasil
iron and protects iron from rusting.
= 0.5 mol dm–3 × —5—0 – dm–3 Number of moles of water
Zink lebih elektropositif daripada besi dan 1000
melindungi besi daripada pengaratan. = 0.025 mol produced
Bilangan mol air yang dihasilkan
CHA 4PTER Thermochemistry ∴ Heat of precipitation, = 1.0 × —10—0– = 0.10 mol
Termokimia Haba pemendakan, 1000
∆H = − —01.2—062—05 = −50.4 kJ mol–1 ∴ Heat of neutralisation
Haba peneutralan, ∆H
4.1 Energy Changes in Chemical Reactions Energy = − —56—7—0
Perubahan Tenaga dalam Tindak Balas Kimia Tenaga 0.10
Pb2+ + SO42–
= −56.7 kJ mol–1
1. (a) releases / membebaskan
(b) Increases / Bertambah ⌬H = –50.4 kJ mol–1 Energy H+ + OH–
PbSO4 Tenaga
(c)
Energy Reactants ⌬H = –56.7 kJ mol–1
Tenaga Bahan tindak balas H2O
4.3 Heat of Displacement
Haba Penyesaran
Products 5. metal, salt, electropositive (b) less, Less, absorbed, ionise
Hasil tindak balas logam, garamnya, elektropositif kurang, Kurang, diserap, mengion
(d) Reactants, products 6. Heat change 4.5 Heat of Combustion
Bahan tindak balas, hasil tindak balas. Perubahan haba, Q Haba Pembakaran
(e) Formation of chemical bonds = mcθ 9. (a) burned, oxygen
Pembentukan ikatan kimia = 50 × 4.2 × 3.5 dibakar, oksigen
= 735 J (b) exothermic
eksotermik
2. (a) absorbs / menyerap Number of moles of zinc displaced
(b) Decreases / Berkurang Bilangan mol zink yang disesarkan 10. (a) Mass of water / Jisim air = 500 g
Temperature change
(c) = 0.1 × —5—0 – = 0.005 mol Perubahan suhu
1000
Energy Products = (42.0 – 28.0)°C
Tenaga Hasil tindak balas ∴ Heat of displacement,
Haba penyesaran, = 14.0°C
∴ Heat change / Perubahan haba, Q
∆H = − —07.—0305—5 = 500 × 4.2 × 14.0
Reactants = −147 kJ mol−1 = 29 400 J
Bahan tindak balas
(d) Reactants, products Energy Mg + Zn2+ Mass of ethanol combusted
Bahan tindak balas, hasil tindak balas Tenaga Jisim etanol yang terbakar
(e) Breaking of chemical bonds ⌬H = –147 kJ mol–1 = (67.28 – 66.13) g = 1.15 g
Pemecahan ikatan kimia Mg2+ + Zn
∴ Number of moles of ethanol
Heat of Precipition burned
Haba Pemendakan
4.2 Bilangan mol etanol yang terbakar
4.4 Heat of Neutralisation = —1.–1—5 = 0.025 mol
Haba Peneutralan 46
3. precipitate, ions
mendakan, ion-ionnya 7. water, hydrogen, hydroxide Heat of combustion / Haba
air, hidrogen, hidroksida pembakaran, ∆H
4. Average initial temperature
Purata suhu awal 8. (a) Average initial temperature = – —209—.042—050– = –1176 kJ mol–1
Purata suhu awal
= —28—.0—+2—29—.0– = 28.5°C Energy
Temperature change, θ = —28—.5—+—29–.—0 Tenaga C2H5OH + 3O2
Perubahan suhu, θ 2
= (31.5 – 28.5)°C
= 3.0°C = 28.75°C ⌬H = –1176 kJ mol–1
2CO2 + 3H2O
Total volume of solution Temperature change
Jumlah isi padu campuran Perubahan suhu, θ (b) (i) greater / lebih besar
= (50 + 50) cm3 = 100 cm3
= (35.5 – 28.75)°C (ii) heat / haba
∴ Total mass of solution
Jumlah jisim larutan = 6.75°C (iii) 1. windshield
= 100 g pengadang angin
Total volume of solution
Heat change / Perubahan haba, Q Jumlah isi padu larutan 2. thin
= mcθ nipis
= 100 × 4.2 × 3.0 = 1260 J = (100 + 100) cm3 = 200 cm3
∴ Total mass of solution / Jumlah
jisim larutan = 200 g
A9 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
SPM Practice 4 alkana dibakar dengan lengkap dalam (ii) Exothermic / Eksotermik
SPM Paper 1 oksigen berlebihan di bawah keadaan
piawai. • The total energy absorbed
(b) (i) to break the chemical
Heat of combustion (kJ mol–1) bonds in the reactants
Haba pembakaran (kJ mol–1)
1. C 2. C 3. A 4. C 5. C is more than the energy
6. C 7. A 8. D 9. A 10. B
11. C 12. D 13. D 14. B 15. B released when the chemical
16. D 17. B 18. A 19. D 20. B
bonds are formed in the
SPM Paper 2
4000 products. [2]
Jumlah tenaga yang diserap
Section A / Bahagian A 3500 untuk memecahkan ikatan
3000 kimia dalam bahan tindak balas
1. (a) Mg + Cu2+ → Mg2+ + Cu adalah lebih daripada tenaga
2500 yang dibebaskan apabila ikatan
(b) (i) Heat change Nu2m00b0er0of 1carb2on a3tom4s p5er m6olecule kimia terbentuk dalam hasil
Perubahan haba tindak balas.
= mcθ Bilangan atom karbon per molekul
= 100(4.2)(38.0 – 29.0) • As a result, heat energy
= 3780 J
is released during the
(ii) Number of moles of copper(II)
sulphate (ii) The more the number of reaction. [1]
Akibatnya, tenaga haba
Bilangan mol kuprum(II) sulfat carbon atoms per molecule,
= —10—100—×00—0.2– = 0.02 mol dibebaskan semasa tindak
the more products will be
balas.
produced. Therefore, more
Endothermic / Endotermik
heat is released during the
• The total energy absorbed
formation of the bonds in the
to break the chemical
products.
Heat of displacement of Semakin banyak bilangan atom bonds in the reactants
copper by magnesium
karbon per molekul, semakin is less than the energy
banyak hasil akan dihasilkan. Oleh
Haba penyesaran kuprum oleh itu, lebih banyak haba dibebaskan released when the chemical
magnesium semasa pembentukan ikatan pada
hasil. bonds are formed in the
= – —307—.082—0 products. [2]
= –189 kJ mol–1 Jumlah tenaga yang diserap
(iii) –2650 kJ mol–1 untuk memecahkan ikatan
kimia dalam bahan tindak balas
(c) (iv) Number of moles of hexane adalah kurang daripada tenaga
Energy Bilangan mol heksana yang dibebaskan apabila ikatan
Tenaga kimia terbentuk dalam hasil
= —4.—3 = 0.05 mol tindak balas.
86
Mg + Cu2+
Heat released • As a result, heat energy
⌬H = –189 kJ mol–1 Haba yang terbebas
Mg2+ + Cu is absorbed during the
= heat of combustion
× number of moles of hexane reaction. [1]
haba pembakaran Akibatnya, tenaga haba diserap
× bilangan mol heksana semasa tindak balas.
(d) Use two layers of polystyrene cups = 3849 × 0.05 (b) (i) The heat change when one
= 192.45 kJ
instead of one. mole of hexane is completely
Gunakan dua lapisan cawan polistirena (c) No, because the combustion of
methane is incomplete and the burnt in air under standard
berbanding dengan hanya menggunakan heat is lost to the surroundings.
satu lapisan. conditions. [2]
Tidak, kerana pembakaran metana Perubahan haba apabila satu mol
(e) The heat change that occurred adalah tidak lengkap dan haba hilang ke heksana dibakar dengan lengkap
persekitaran. dalam udara di bawah keadaan
when one mole of copper is piawai.
displaced from its salt solution (ii) Heat absorbed by water
Haba yang diserap oleh air
using magnesium.
Perubahan haba yang berlaku apabila Section B / Bahagian B = mcθ [1]
= 400 × 4.2 × (59 – 27) [1]
satu mol kuprum disesarkan daripada 3. (a) (i) Exothermic / Eksotermik
larutan garamnya dengan menggunakan = 53.76 kJ
magnesium. • A reaction that releases
Mass of hexane burnt
(f) (i) The temperature change heat to the surroundings. [1] Jisim heksana yang dibakar
Tindak balas yang
in this experiment is lower = (138.06 – 136.34) g
membebaskan haba ke
compared to that of using = 1.72 g
persekitaran.
magnesium powder. Relative molecular mass of
Perubahan suhu dalam eksperimen • Example: Neutralisation of hexane
ini adalah lebih rendah berbanding
dengan menggunakan serbuk acid and alkali [1] Jisim molekul relatif heksana
magnesium.
Contoh: Peneutralan asid dan = (12 × 6) + (1 × 14) = 86 [1]
alkali
(ii) Zinc is less electropositive Endothermic / Endotermik Number of moles of hexane
than magnesium. • Reaction that absorbs heat burnt
Zink adalah kurang elektropositif from the surroundings. [1] Bilangan mol heksana yang terbakar
daripada magnesium. Tindak balas yang menyerap —1.—72–
haba dari persekitaran. 86
2. (a) The heat released when one mole = = 0.02 mol [1]
of an alkane is completely burnt • Example: Dissolving
ammonium salts in water [1] Combustion of 0.02 moles of
in excess oxygen under standard
Contoh: Melarutkan garam hexane produces 53.76 kJ heat.
conditions. ammonium dalam air Pembakaran 0.02 mol heksana
Haba yang dibebaskan apabila satu mol
menghasilkan 53.76 kJ haba.
© Penerbitan Pelangi Sdn. Bhd. A10
Chemistry Form 5 Answers
∴ Combustion of 1 mole of (b) (i) The heat change when Fuel value / Nilai bahan api
hexane produces 2688 kJ lead(II) ions and sulphate ions = —–5—4—62–
114
heat. [1] combined to form one mole of
Pembakaran 1 mol heksana lead(II) sulphate precipitate. = –47.91 kJ g–1
menghasilkan 2688 kJ haba. [2]
Perubahan haba apabila ion
Heat of combustion of hexane plumbum(II) dan ion sulfat (b) Thrust power of the cars in Malaysia is
Haba pembakaran heksana bergabung untuk membentuk satu expected to be more powerful because
mol mendakan plumbum(II) sulfat. the petrol we use in Malaysia has a
= –2688 kJ mol–1 [1] higher fuel value and generates more
(ii) energy per gram of fuel for the car.
(iii) Heat is lost to the
surroundings. Energy Kuasa tujahan kereta di Malaysia dijangka lebih
Tenaga Pb(NO3)2 + Na2SO4 berkuasa kerana petrol yang kita gunakan di
Haba hilang ke persekitaran. [1] Malaysia mempunyai nilai bahan api yang lebih
⌬H = –50 kJ mol–1 tinggi dan menjana tenaga yang lebih banyak
Combustion of hexane is PbSO4 + 2NaNO3 per gram bahan api untuk kereta.
incomplete.
5PTER
Pembakaran heksana adalah tidak
lengkap. [1]
Section C / Bahagian C
4. (a) (i) The heat change when 1 [2] CHA Chemicals for Consumers
Bahan Kimia untuk Pengguna
mole of copper(II) ions are
displaced by zinc from its salt (iii) • Sodium sulphate solution,
solution. [2] lead(II) nitrate solution and Soap and Detergent
Sabun dan Detergen
Perubahan haba apabila 1 mol ion sulphuric acid are strong 5.1
kuprum(II) disesarkan oleh zink electrolytes that dissociate
daripada larutan garamnya. completely in water. [1] 1. (a) Saponification / Saponifikasi
Larutan natrium sulfat, larutan
(ii) • Zinc powder dissolves in plumbum(II) nitrat dan asid (b) concentrated sodium hydroxide,
sulfurik adalah elektrolit kuat heated, sodium chloride, soap,
the solution/brown copper yang mencerai sepenuhnya di filtered
dalam air.
powder is formed. [1] natrium hidroksida pekat, dipanaskan,
natrium klorida, sabun, dituraskan
Serbuk zink larut dalam larutan/
serbuk perang kuprum terhasil.
• The intensity of the blue • Both sodium sulphate
copper(II) sulphate solution solution and sulphuric acid 2. (a) (i) Hydrophobic / Hidrofobik
(ii) Hydrophilic / Hidrofilik
decreases. [1] dissociate to give sulphate Hydrophobic / Hidrofobik
(b) (i) Hydrophilic / Hidrofilik
Keamatan warna biru larutan ions. [1] (ii)
Kedua-dua larutan natrium
kuprum(II) sulfat berkurangan.
sulfat dan asid sulfurik mencerai
• The thermometer records 3. (a) Vegetable oils or animal fats Minyak
untuk memberikan ion sulfat. sayuran atau lemak haiwan
an increase in temperature.
• Precipitation of lead(II) (b) Petroleum or long-chain alcohols
[1] Petroleum atau alkohol berantai panjang
Termometer merekodkan suatu sulphate involves the
(c) Effective
kenaikan suhu. reaction of free moving Pb2+ Berkesan
(d) Effective
(iii) Zn(s) + CuSO4(aq) → and SO42– ions. [1] Berkesan
(e) Ineffective
ZnSO4(aq) + Cu(s) [1] Pemendakan plumbum(II) sulfat Tidak berkesan
(f) Effective
Zn(p) + CuSO4(ak) → melibatkan tindak balas ion-ion Berkesan
ZnSO4(ak) + Cu(p) Pb2+ dan SO42– yang bergerak (g) Biodegradable
bebas. Terbiodegradasi
Number of moles of copper(II)
sulphate used • The ionic equation for both (h) Non-biodegradable
Tidak terbiodegradasi
reactions is the same, (i) Less soluble
Kurang larut
Bilangan mol kuprum(II) sulfat yang which is Pb2+(aq) + SO42– (j) More soluble
digunakan (aq) → PbSO4(s). [1] Lebih larut
Persamaan ion bagi kedua-dua
= —501—0×0—00—.5
tindak balas adalah sama,
= 0.025 mol [1]
iaitu Pb2+(ak) + SO42–(ak) →
Heat released in the reaction PbSO4(p).
Haba yang dibebaskan dalam
Challenge
tindak balas
= 0.025 × 210 000 = 5250 J [1] (a) For ethanol / Bagi etanol:
Heat released [1] Relative molecular mass 5.2 Food Additives
Haba yang terbebas [1] Jisim molekul relatif Bahan Tambah Makanan
= mcθ = 5250
= (2 × 12) + (5 × 1) + 16 + 1 4. food, improve
50 × 4.2 × θ = 5250 = 46 makanan, memperbaiki
∴ θ = 25°C Fuel value / Nilai bahan api 5. (a) (i) last longer
= —–1—436—78– tahan lebih lama
(iv) • A higher temperature will = –29.96 kJ g–1
(ii) Sodium nitrite
be recorded. [1] For isooctane / Bagi isooktana, Natrium nitrit
(b) (i) oxidation
Suatu suhu yang lebih tinggi Relative molecular mass pengoksidaan
Jisim molekul relatif
akan direkodkan. (ii) Ascorbic acid
= (8 × 12) + (18 × 1) Asid askorbik
• Magnesium is more = 114
electropositive than zinc. [1]
Magnesium adalah lebih
elektropositif berbanding zink.
A11 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
(c) (i) attractive / menarik (iii) As an artificial flavouring solution. [1]
which gives pineapple flavour Tambahkan 50 cm3 larutan natrium
(ii) Azo dyes / Pewarna azo Sebagai perisa tiruan yang
(d) (i) taste / rasa memberikan rasa nanas hidroksida 5 mol dm-3 dengan
(ii) Monosodium glutamate (f) Aspartame / Aspartame perlahan-lahan dan cermat.
Mononatrium glutamat
2. (a) (i) M • Boil the mixture with low heat
(e) (i) separating / diasingkan
(ii) Skin disease / Penyakit kulit and stir it with a glass rod until
(ii) Acacia gum / Gam acacia
(f) (i) texture / tekstur (iii) The extract of its leaves is the oil layer becomes invisible
applied on the affected area.
(ii) Starch / Kanji after approximately 20 minutes.
Ekstrak daripada daunnya disapu
5.3 Medicine pada kawasan yang dijangkiti. [1]
Ubat Didihkan campuran dengan api yang
(b) (i) Antibiotics – to kill and prevent
kecil dan kacaunya dengan rod
the growth of bacteria kaca sehingga lapisan minyak tidak
Antibiotik – untuk membunuh dan kelihatan selepas kira-kira 20 minit.
mencegah pertumbuhan bakteria
6. (a) For treating stomach discomfort • After all of the oil has reacted,
due to gas Psychotherapeutic medicine
add some water and stir the
Untuk mengubati ketidakselesaan perut – to change the emotions and
akibat angin mixture. [1]
behaviour of the patient Selepas semua minyak telah
(b) For treating skin problems Ubat psikoterapeutik – untuk
Untuk mengubati masalah kulit mengubah emosi dan kelakuan bertindak balas, tambahkan sedikit air
(c) For treating malaria pesakit dan kacau campuran itu.
Untuk mengubati malaria
(d) For reducing high blood pressure • Cool the mixture and add 15 g
Untuk merendahkan tekanan darah tinggi of sodium chloride to 50 cm3
(e) For treating respiratory problems
of water. Pour the cooled
such as asthma and skin problems
Untuk mengubati masalah pernafasan mixture into the sodium chloride
seperti asma dan masalah kulit
Analgesics – to relieve pain solution. [1]
7. (a) (i) pain / kesakitan
(ii) Paracetamol / Parasetamol without causing numbness or Sejukkan campuran dan tambahkan
(iii) Aspirin / Aspirin
(b) (i) bacteria / bakteria affecting consciousness 15 g natrium klorida kepada
(ii) Streptomycin / Streptomisin Analgesik – melegakan kesakitan
(iii) Penicillin / Penisilin tanpa menyebabkan kebas atau 50 cm3 air. Tuangkan campuran
(c) (i) emotion / emosi mempengaruhi kesedaran
(ii) Barbiturate / Barbiturat yang telah disejukkan kepada larutan
(iii) Tranquiliser / Penenang
natrium klorida.
SPM Practice 5
(ii) May cause bleeding in the • Let this new mixture cool
SPM Paper 1
stomach completely. The solid that forms
1. B 2. C 3. B 4. C 5. B Mungkin menyebabkan pendarahan
6. B 7. C 8. C 9. A 10. D is the soap. [1]
11. B 12. D 13. B 14. B 15. B dalam perut
16. D 17. C 18. C Biarkan campuran baru ini sejuk
SPM Paper 2 (iii) So that all the bacteria that dengan sepenuhnya. Pepejal yang
Section A / Bahagian A cause the disease are killed terhasil ialah sabun.
Supaya semua bakteria yang
1. (a) Azo compounds / Sebatian azo (b) • When soap is added to water,
(b) As a preservative and as a food menyebabkan penyakit dapat
flavouring dibunuh soap molecules ionise into
Sebagai pengawet dan perisa makanan
(iv) Calms down the patient’s positive and negative ions. [1]
(c) As a preservative Apabila sabun ditambahkan kepada
Sebagai pengawet emotion so that the patient
air, molekul sabun mengion kepada
(d) (i) As an antioxidant and as a can sleep properly ion-ion positif dan negatif.
food flavouring Melegakan emosi pesakit supaya
• The head (hydrophilic end) of
Sebagai pengantioksida dan perisa pesakit dapat tidur dengan baik
makanan the soap dissolves in water but
3. (a) (i) Saponification / Saponifikasi
(ii) Ascorbic acid / Asid askorbik the tail (hydrophobic end) of the
(ii) Glycerol / Gliserol
(e) (i) Carboxylate group soap goes into the grease. [2]
Kumpulan karboksilat (iii) To help precipitate the solid Kepala (hujung hidrofilik) sabun
soap
(ii) Ethanol and butanoic acid larut dalam air tetapi ekor (hujung
Etanol dan asid butanoik Untuk membantu dalam hidrofobik) sabun masuk ke dalam
memendakkan pepejal sabun gris.
(b) (i) P – soap / sabun, • When the cleaning process is
Q – detergent / detergen
conducted, the agitation causes
(ii) The white precipitate that is
formed when soap reacts with the grease stains to be removed
certain mineral ions that are
present in hard water. from the cloth. [1]
Mendakan putih yang terbentuk Apabila proses pencucian dilakukan,
apabila sabun bertindak balas
dengan sesetengah ion mineral tindakan pengocakan menyebabkan
yang hadir dalam air liat.
kesan gris ditanggalkan daripada
(iii) Mg2+ and Ca2+ / Mg2+ dan Ca2+
kain.
(iv) Easily biodegradable
Mudah terbiodegradasi • When removed, the grease will
be broken into tiny droplets
called an emulsion. [1]
Apabila ditanggalkan, gris akan
dipecahkan kepada titisan-titisan kecil
yang disebut emulsi.
Section B / Bahagian B • The emulsion particles are
surrounded by negative ions
4. (a) • Place 20 g of vegetable oil in a which cause the repulsion
beaker. [1] among the emulsion particles.
[1]
Masukkan 20 g minyak sayuran
dalam sebuah bikar. Zarah-zarah emulsi dikelilingi oleh ion
negatif yang menyebabkan penolakan
• Slowly and carefully add 50 cm3 antara zarah-zarah emulsi.
of 5 mol dm–3 sodium hydroxide
© Penerbitan Pelangi Sdn. Bhd. A12
Chemistry Form 5 Answers
• Therefore, the emulsion • Example: Penicillin, From the equation, 2 moles
particles do not merge again. [1] streptomycin [1]
of aspirin react with excess
Oleh itu, zarah-zarah emulsi tidak Contoh: Penisilin,
bergabung semula. streptomisin sodium carbonate to produce
• When rinsed with water, the dirt (3) Psychotherapeutic 1 mole of CO2. [1]
and emulsion will be removed medicine Daripada persamaan, 2 mol
and the cloth is cleansed. [1] aspirin bertindak balas dengan
Ubat psikoterapeutik [1] natrium karbonat berlebihan untuk
Apabila dibilas dengan air, kotoran menghasilkan 1 mol CO2.
dan emulsi akan ditanggalkan dan • Change the emotions
kain dibersihkan. Number of moles of aspirin
and behaviour of a Bilangan mol aspirin
(c) • When soap is dissolved in patient [1] = (—9 —× —12—) +—(—8 3—× 1—) —+ —(4—× –1—6)
Mengubah emosi dan = —183—0
hard water, the soap molecules
kelakuan seseorang
combine with the Mg2+ and Ca2+ pesakit
ions. [1] • Example: Barbiturate, = 0.017 mol [1]
Apabila sabun larut dalam air liat,
tranquiliser, ∴ Number of moles of CO2
molekul sabun bergabung dengan produced / Bilangan mol CO2
ion-ion Mg2+ dan Ca2+. amphetamine [1]
Contoh: Barbiturat, ubat yang terhasil
• The combination produces
penenang, amfetamin
scum, an insoluble solid. [1]
Penggabungan ini menghasilkan (b) (i) • Chemicals that are required: = —1 × 0.017 mol
2
kekat, suatu pepejal tak terlarutkan. Salicylic acid and ethanoic
• As a result, the hydrophilic anhydride [2] = 0.0085 mol [1]
end of soap cannot dissolve in Bahan kimia yang diperlukan: Volume of CO2 at s.t.p.
Asid salisilik dan etanoik Isi padu CO2 pada s.t.p.
water causing the soap to be anhidrida = 0.0085 × 22.4
ineffective in hard water. [1] • Chemical equation and the = 0.190 dm3 [1]
Oleh itu, hujung hidrofilik sabun tidak structure of aspirin:
dapat larut dalam air menyebabkan
sabun menjadi kurang berkesan Persamaan kimia dan struktur Challenge
dalam air liat.
aspirin: O
• When detergent is dissolved COOH C CH3 Ashley is a small child and consuming
OH + O
in hard water, the detergent C CH3 aspirin may lead to stomach bleeding due
Salicylic acid EthanOoic anhydride
molecules do not form insoluble Asid salisilik to its acidic nature. Instead of aspirin,
solid. [1] Etanoik anhidrida paracetamol can be given to Ashley because
Apabila detergen larut dalam air liat,
paracetamol is not acidic and does not upset
molekul detergen tidak membentuk
pepejal tak terlarut. the stomach.
Ashley adalah seorang kanak-kanak kecil dan
• Both the hydrophilic and pengambilan aspirin mungkin menyebabkan
pendarahan dalam perut disebabkan oleh sifat
hydrophobic ends of the keasidannya. Sebaliknya, parasetamol boleh
diberikan kepada Ashley kerana parasetamol tidak
detergent are still effective. COOH berasid dan tidak memudaratkan perut.
Therefore, the detergent is still
effective in hard water. [1] O C CH3 + CH3COOH
Kedua-dua hujung hidrofilik dan
hidrofobik detergen masih berkesan. O Written Practical
Aspirin
Oleh itu, detergen masih berkesan di Aspirin
dalam air liat. • Other name of aspirin: Chapter 1
Acetylsalicylic acid
• Detergent is a better cleaning [2] 1. (a) (i) The ‘X’ mark on the paper
Nama lain aspirin: Asid [1]
agent compared to soap. [1] asetilsalisilik disappears due to the
Detergen adalah bahan pencuci yang
formation of yellow precipitate.
lebih baik daripada sabun. Tanda ‘X’ pada kertas tidak
kelihatan lagi disebabkan oleh
Section C / Bahagian C (ii) Chemical equation for the pembentukan mendakan kuning.
reaction between aspirin and
5. (a) (i) A medicine is a substance sodium carbonate: (ii) The yellow precipitate formed
that helps to prevent or cure is sulphur.
Mendakan kuning yang terbentuk
diseases. [1] Persamaan kimia bagi tindak balas
Ubat ialah suatu bahan yang antara aspirin dengan natrium ialah sulfur.
karbonat:
membantu dalam mencegah atau
mengubati penyakit. COOH (b) (i) Concentration of sodium
(ii) (1) Analgesics / Analgesik [1] thiosulphate
Kepekatan natrium tiosulfat
• Relieve pain 2 O C CH3 + Na2CO3
Melegakan kesakitan [1] O (ii) Rate of reaction
Kadar tindak balas
• Example: Aspirin,
paracetamol, codeine COONa (iii) Concentration of sulphuric
[1]
2 O C CH3 + CO2 + H2O acid / Size of conical flask
Contoh: Aspirin, O Kepekatan asid sulfurik / Saiz
parasetamol, kodeina [1]
kelalang kon
(2) Antibiotics / Antibiotik [1]
• Kill and prevent the (c) The higher the concentration of
growth of bacteria [1]
sodium thiosulphate, the higher the
Membunuh dan
mencegah pertumbuhan rate of reaction.
bakteria Semakin tinggi kepekatan natrium
tiosulfat, semakin tinggi kadar tindak
balas tersebut.
A13 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
(d) Rate of reaction is operationally 2. (a) At t = 60 s / Pada t = 60 s, (e) Procedure / Prosedur:
defined as the time taken for the
‘X’ mark to disappear. Reading of burette Delivery tube Hgaysdrogen
Bacaan buret = 22.30 cm3 Salur penghantar Gas hidrogen
Kadar tindak balas didefinisi secara
operasi sebagai masa yang diambil bagi Volume of gas collected / Isi padu aHcyiddrochloric Burette
tanda ‘X’ tidak kelihatan lagi. gas yang dikutip = 27.70 cm3 Asid hidroklorik Buret
cMoailgnesium
(e) (i) At t = 120 s / Pada t = 120 s, Pita magnesium Water
Air
Set I II III IV Reading of burette / Bacaan buret =
9.90 cm3
Concentration
Volume of gas collected
of sodium Isi padu gas yang dikutip
thiosulphate = 40.10 cm3 1. A burette is filled with water
until it is full.
(mol dm–3) 0.5 0.4 0.3 0.2 (b) As the reaction between zinc
granules and hydrochloric acid Sebuah buret diisikan dengan air
Kepekatan proceeds, the total volume of gas sehingga penuh.
evolved increases.
natrium tiosulfat 2. It is then inverted over water
Apabila tindak balas antara butiran zink in a basin and clamped
(mol dm–3) dengan asid hidroklorik berlaku, jumlah isi vertically using a retort stand.
padu gas yang terbebas bertambah.
Time (s) 10 13 17 25 Kemudian, buret itu ditelangkupkan
Masa (s) (c) The rate of reaction is defined as ke dalam air dalam sebuah besen
the change of the volume of gas dan diapitkan tegak dengan
—138—0 (s–1) 0.10 0.08 0.06 0.04 produced against the time taken for menggunakan kaki retort.
—183—0 (s–1) the reaction to occur.
3. The water level in the burette
Kadar tindak balas ditakrifkan sebagai is adjusted and the initial
(ii) perubahan isi padu gas yang terhasil burette reading is recorded.
lawan masa yang diambil bagi tindak
Concentration (mol dm–3) balas berlaku. Aras air dalam buret diselaraskan
Kepekatan (mol dm–3) dan bacaan awal buret direkodkan.
3. (a) Problem statement:
0.5 Pernyataan masalah: 4. 25 cm3 of 0.1 mol dm–3
0.4 hydrochloric acid is measured
0.3 How does the size of magnesium using a measuring cylinder and
0.2 increase the rate of magnesium poured into a conical flask.
0.1 dissolving in dilute hydrochloric
acid? 25 cm3 asid hidroklorik
0 0.02 0.04 0.06 0.08 0.10 1 (s–1)/ 0.1 mol dm-3 disukat dengan
Time Bagaimanakah saiz magnesium menggunakan silinder penyukat
0.05 mempengaruhi kadar magnesium larut dan dituang ke dalam sebuah
1 (s –1) dalam asid hidroklorik cair? kelalang kon.
Masa
(b) Variables / Pemboleh ubah: 5. The apparatus is set up as
shown in the diagram.
Manipulated: Size of magnesium
(f) (i) Directly proportional Dimanipulasikan: Saiz magnesium Radas disediakan seperti yang
Berkadar terus ditunjukkan dalam rajah.
Responding: Rate of reaction
(i i) From gr aph, —118—30 = 0.05 s–1 Bergerak balas: Kadar tindak balas 6. 2 g of magnesium coil is
\ Time = 20 s weighed using an electronic
Constant: Concentration and balance and added to the acid
Daripada graf, —1—3 = 0.05 s–1 volume of hydrochloric acid, in the conical flask.
∴ Masa = 20 s 180 temperature, mass of magnesium
2 g pita magnesium ditimbang
(g) A smaller conical flask has a Dimalarkan: Kepekatan dan isi padu asid dengan penimbang elektronik dan
hidroklorik, suhu, jisim magnesium ditambahkan kepada asid dalam
smaller base and requires lesser kelalang kon.
(c) Hypothesis / Hipotesis:
sulphur precipitate to cause the ‘X’ 7. The conical flask is closed
The smaller the particle size of immediately with a stopper
mark to disappear. magnesium, the higher the rate of fitted with a delivery tube. At
Kelalang kon yang lebih kecil mempunyai reaction. the same time, the stopwatch
is started.
dasar yang lebih kecil dan memerlukan Semakin kecil saiz zarah magnesium,
kurang mendakan sulfur untuk semakin tinggi kadar tindak balas. Kelalang kon ditutup serta-merta
menyebabkan tanda ‘X’ tidak kelihatan dengan penyumbat yang dipasang
lagi hilang. (d) Substances / Bahan: dengan salur penghantar. Pada
masa yang sama, jam randik
(h) Anions / Anion: 0.1 mol dm–3 hydrochloric acid, dimulakan.
Sulfat, magnesium powder, magnesium
Sulphate / SO 2- coil, water 8. The gas released is collected
4 in the burette by downward
Asid hidroklorik 0.1 mol dm-3, serbuk displacement of water as
Cations / Kation: magnesium, pita magnesium, air shown in the diagram.
Hydrogen / Hidrogen, H+
Apparatus / Radas: Gas yang dibebaskan dikutip dalam
Cations / Kation: buret melalui penyesaran ke bawah
Sodium / Natrium, Na+ 150 cm3 conical flask, burette, air seperti yang ditunjukkan dalam
50 cm3 measuring cylinder, rajah.
basin, stopper with delivery tube,
electronic balance, stopwatch 9. The conical flask is shaken
slowly throughout the
Kelalang kon 150 cm3, buret, silinder experiment.
penyukat 50 cm3, besen, penyumbat
dengan salur penghantar, penimbang
elektronik, jam randik
© Penerbitan Pelangi Sdn. Bhd. A14
Kelalang kon digoncangkan 11. Steps 1 to 10 are repeated Chemistry Form 5 Answers
dengan perlahan-lahan sepanjang
eksperimen. using 2 g of magnesium (c) Hypothesis / Hipotesis:
Vulcanised rubber is more elastic
10. The volume of gas collected powder to replace the 2 g of
in the burette is recorded at than unvulcanised rubber.
regular intervals of 30 seconds magnesium coil. All the other Getah tervulkan adalah lebih kenyal
until no more gas is liberated.
conditions remain unchanged. daripada getah tak tervulkan.
Isi padu gas yang terkutip dalam Langkah 1 hingga 10 diulang
buret direkodkan pada sela (d) Materials / Bahan:
masa 30 saat sehingga tiada gas dengan menggunakan 2 g serbuk Vulcanised and unvulcanised
dibebaskan. magnesium untuk menggantikan 2
g pita magnesium. Semua keadaan rubber strands
yang lain dikekalkan tidak berubah. Jalur getah tervulkan dan jalur getah tak
(f) Tabulation of data: tervulkan
Penjadualan data:
Using magnesium coil: Apparatus / Radas:
Menggunakan pita magnesium: Retort stand and clamp, bulldog
Time(s) Burette reading (cm3) Volume of gas liberated (cm3) clips, 50 g weight, ruler
Masa (s) Bacaan buret (cm3) Isi padu gas yang terbebas (cm3) Kaki retot dan penyepit, klip, pemberat
Volume of gas liberated (cm3)
0 Isi padu gas yang terbebas (cm3) 50 g, pembaris
30
60 (e) Procedure / Prosedur:
90
120 Clips
150 Klip
180
210 Rubber strand
240 Jalur getah
270 Clips
300 Klip
50 g weight
Using magnesium powder: Pemberat 50 g
Menggunakan serbuk magnesium:
Retort stand
Time(s) Burette reading (cm3) Kaki retort
Masa (s) Bacaan buret (cm3)
1. Using clips, a strand of
0 vulcanised rubber is attached
30 to a retort stand as shown in
60 the diagram.
90
120 Dengan menggunakan klip, sejalur
150 getah tervulkan dipasangkan pada
180 kaki retort seperti yang ditunjukkan
210 dalam rajah.
240
270 2. The original length of the
300 vulcanised rubber strand is
measured.
Chapter 2
Panjang asal jalur getah tervulkan
1. (a) Aim / Tujuan: diukur.
To compare the elasticity of vulcanised and unvulcanised rubber 3. A 50 g weight is hung on the
Untuk membandingkan kekenyalan getah tervulkan dengan getah tak tervulkan vulcanised rubber strand for
several minutes.
(b) Variables / Pemboleh ubah:
Suatu pemberat 50 g digantungkan
Manipulated: Type of rubber strands pada jalur getah tervulkan selama
Dimanipulasikan: Jenis jalur getah beberapa minit.
Responding: Length of rubber strands when stretched 4. The weight is removed
Bergerak balas: Panjang jalur getah apabila diregang and the final length of the
vulcanised rubber strand is
Constant: Initial length of rubber strands, mass of weight applied measured.
Dimalarkan: Panjang asal jalur getah, jisim pemberat yang digunakan
Pemberat itu dialihkan dan panjang
A15 akhir jalur getah tervulkan diukur.
5. Steps 1 to 4 are repeated
using unvulcanised rubber
strand.
Langkah 1 hingga 4 diulang
dengan menggunakan jalur getah
tak tervulkan.
6. The data collected is recorded
in the table.
Data yang dikumpul direkodkan
dalam jadual.
© Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers Unvulcanised rubber Vulcanised rubber (d) Materials / Bahan:
Getah tak tervulkan Getah tervulkan
(f) Tabulation of data: 8.0 8.0 Chlorine water, bromine water, 0.5
Penjadualan data: mol dm–3 potassium chloride solution,
9.2 8.0 0.5 mol dm–3 potassium bromide
Type of rubber strand solution, tetrachloromethane
Jenis jalur getah 1.2 0.0
Initial length (cm) Air klorin, air bromin, larutan kalium
Panjang asal (cm) klorida 0.5 mol dm-3, larutan kalium
Final length (cm) bromida 0.5 mol dm-3, tetraklorometana
Panjang akhir (cm)
Increase in length Apparatus / Radas:
Penambahan dalam panjang (cm)
Test tubes, test tube rack
Chapter 3
Tabung uji, rak tabung uji
1. (a) Metal Observation
Logam Pemerhatian (e) Procedure / Prosedur:
Iron Faint flame with bright glow 1. A test tube is filled with 2 cm3
Ferum Nyalaan kecil dengan baraan yang cerah of 0.5 mol dm–3 potassium
chloride solution.
Aluminium Very bright flame
Aluminium Nyalaan yang sangat cerah Sebuah tabung uji diisi dengan
2 cm3 larutan kalium klorida
Lead Bright glow
Plumbum Baraan yang cerah 0.5 mol dm-3.
(b) Variable Action to be taken 2. 2 cm3 of bromine water is
Pemboleh ubah Tindakan yang diambil added to the test tube and the
mixture is shaken thoroughly.
(i) Types of metals (i) Replace the metal with different
Jenis logam 2 cm3 air bromin ditambahkan
metals kepada tabung uji dan campuran
Gantikan logam dengan logam-logam yang itu digoncang dengan sekata.
berlainan 3. 2 cm3 of tetrachloromethane is
added to the mixture and the
(ii) Vigour of the flame or glow (ii) Observe the intensity of the flame or mixture produced is shaken
Kecergasan nyalaan atau baraan thoroughly.
glow
(iii) Quantity of metal /quantity of KClO3 Perhatikan keamatan nyalaan atau baraan 2 cm3 tetraklorometana
Kuantiti logam /kuantiti KClO3 ditambahkan kepada campuran
(iii) Use a constant mass of metal/KClO3 itu dan campuran yang terhasil
Gunakan jisim logam / KClO3 yang tetap digoncang dengan sekata.
(c) The metal that is located higher in the reactivity series gives a brighter intensity of 4. After a few seconds, the
flame or glow. colour of the aqueous and the
tetrachloromethane layers are
Logam yang terletak lebih tinggi dalam siri kereaktifan memberikan keamatan nyalaan atau baraan observed.
yang lebih terang.
Selepas beberapa saat,
(d) (i) Aluminium, iron, lead, copper warna lapisan akueus dan
Aluminium, ferum, plumbum, kuprum tetraklorometana diperhatikan.
(ii) Al, Zn, Fe, Pb, Cu 5. Another test tube is filled
with 2 cm3 of 0.5 mol dm–3
(e) Metals that are less reactive than potassium bromide solution.
Metals that are more reactive than carbon
carbon Tabung uji yang satu lagi diisi
Logam yang lebih reaktif daripada karbon Logam yang kurang reaktif daripada dengan 2 cm3 larutan kalium
karbon bromida 0.5 mol dm-3.
Aluminium / Aluminium
Potassium / Kalium Iron / Ferum 6. 2 cm3 of chlorine water is
Copper / Kuprum added to the test tube and the
mixture is shaken thoroughly.
2. (a) Aim / Tujuan:
2 cm3 air klorin ditambahkan
To investigate the displacement of halogens from its halide solutions kepada tabung uji dan campuran
Untuk mengkaji penyesaran halogen daripada larutan halidanya itu digoncang dengan sekata.
(b) Hypothesis / Hipotesis: 7. 2 cm3 of tetrachloromethane
is added to the mixture. The
A more reactive halogen displaces a less reactive halogen from its halide solution. mixture produced is shaken
Halogen yang lebih reaktif menyesarkan halogen yang kurang reaktif daripada larutan halidanya. thoroughly.
(c) Variables / Pemboleh ubah: 2 cm3 tetraklorometana ditambahkan
kepada campuran itu. Campuran
Manipulated: Type of mixture yang terhasil digoncang dengan
Dimanipulasikan: Jenis campuran sekata.
Responding: Colour change of the solution 8. After a few seconds, the
Bergerak balas: Perubahan warna larutan colour of the aqueous and the
tetrachloromethane layers are
Constant: Volume of the solution observed.
Dimalarkan: Isi padu larutan
Selepas beberapa saat,
warna lapisan akueus dan
tetraklorometana diperhatikan.
© Penerbitan Pelangi Sdn. Bhd. A16
Chemistry Form 5 Answers
(f) Tabulation of data: (i) (i) The heat of neutralisation is
Penjadualan data: defined as the amount of heat
released when one mole of
Mixture Colour of Colour of Inference water is produced from the
Campuran Inferens reaction between an acid
aqueous layer tetrachloro-methane and an alkali under standard
Bromine + Warna lapisan conditions.
potassium layer
chloride akueus Warna lapisan Haba peneutralan ditakrifkan
Bromin + kalium tetrakloro-metana sebagai jumlah haba yang
klorida dibebaskan apabila satu mol air
Yellow Red Bromine is present. No dihasilkan daripada tindak balas
Chlorine + Kuning Merah antara suatu asid dengan suatu
potassium displacement reaction alkali di bawah keadaan piawai.
bromide
Klorin + kalium has occurred. (ii) Experiment II uses a strong
bromida Bromin hadir. Tiada tindak acid whereas Experiment I
balas penyesaran berlaku. uses a weak acid. The heat
released in Experiment I is
Yellow Red Bromine is present. less than that in Experiment
Kuning Merah II because a small portion of
The displacement of heat released in Experiment
I is absorbed to help the
bromine has occurred. dissociation of ethanoic acid.
Bromin hadir. Penyesaran
bromin telah berlaku. Eksperimen II menggunakan
asid kuat sementara Eksperimen
Chapter 4 I menggunakan asid lemah.
Haba yang dibebaskan dalam
1. (a) Initial temperature of mixture / Suhu awal campuran: 27.0°C Eksperimen I kurang daripada
Final temperature of mixture / Suhu akhir campuran: 33.0°C Eksperimen II kerana sebahagian
Change in temperature / Perubahan suhu: 6.0°C kecil haba yang dibebaskan
dalam Eksperimen I diserap untuk
(b) Experiment I Experiment II membantu dalam penceraian asid
Experiment Eksperimen I Eksperimen II etanoik.
Eksperimen
T1 (j)
Initial temperature of mixture (°C) 27.0
Suhu awal campuran (°C) T2
Final temperature of mixture (°C) 33.0 T3 Name of Heat of Type of
Suhu akhir campuran (°C) acid neutralisation acid
Change in temperature (°C) 6.0 Nama asid (kJ mol–1) Jenis asid
Perubahan suhu (°C) Haba peneutralan
(kJ mol–1)
Ethanoic –50.4 Weak acid
Asid lemah
(c) The neutralisation between a (f) Change in temperature acid
strong acid and a strong alkali = final temperature of mixture Asid etanoik
releases more heat than the
neutralisation between a weak acid – initial temperature of mixture Hydrochloric –57.3 Strong acid
and a strong alkali. Perubahan suhu = suhu akhir campuran Asid kuat
acid
Peneutralan antara asid kuat dengan – suhu awal campuran Asid
alkali kuat membebaskan lebih banyak hidroklorik
haba berbanding dengan peneutralan (g) 1. A colourless mixture of
antara asid lemah dengan alkali kuat. solution is obtained. Propanoic –50.1 Weak acid
Asid lemah
(d) The change in temperature in Suatu campuran larutan tak acid
Experiment II is greater than berwarna diperoleh. Asid propanoik
the change in temperature in
Experiment I, whereby T3 > 6.0°C, 2. The vinegar smell of ethanoic Chapter 5
approximately 7.0°C. acid disappears.
1. (a) Aim / Tujuan:
Perubahan suhu dalam Eksperimen II Bau cuka asid etanoik hilang.
lebih besar daripada perubahan suhu To investigate the effectiveness of
dalam Eksperimen I, dengan T3 > 3. The polystyrene cup becomes soap in soft water and hard water
6.0°C, kira-kira 7.0°C. warmer.
Untuk mengkaji keberkesanan sabun
(e) The initial temperature and final Cawan polistirena menjadi lebih dalam air lembut dan air liat
temperature of both experiments panas.
must be recorded so that the (b) Variables / Pemboleh ubah:
change in temperature for both (h) 1. The volumes of the acid and
experiments can be obtained. the alkali. Manipulated: Type of water
Dimanipulasikan: Jenis air
Suhu awal dan suhu akhir kedua- Isi padu asid dan alkali.
dua eksperimen mestilah direkodkan Responding: Presence of oily stain
untuk membolehkan kita memperoleh 2. The concentrations of the acid on cloth
perubahan suhu bagi kedua-dua and alkali.
eksperimen. Bergerak balas: Kehadiran kesan tompok
Kepekatan asid dan alkali. minyak pada kain
3. The type of cup used in the Constant: Soap used
experiment. Dimalarkan: Sabun yang digunakan
Jenis cawan yang digunakan dalam
eksperimen.
A17 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
(c) Hypothesis / Hipotesis: Kira-kira 1 g serbuk sabun (b) (i) Mass of 0.06 mol K
ditambahkan ke dalam kedua-dua Jisim 0.06 mol K
Soap is able to remove the oily bikar masing-masing. = 0.06 × 62
stain on cloth in soft water but is = 3.72 g
unable to remove the oily stain in 3. The two cloths with the oily
hard water. stains are immersed into the Mass of O / Jisim O
beakers and the cloths are = (3.72 – 1.44 – 0.36) g
Sabun dapat menanggalkan kesan agitated using the glass rod. = 1.92 g
tompok minyak pada kain dalam air
lembut tetapi tidak dapat menanggalkan Dua helai kain dengan kesan
kesan tompok minyak dalam air liat. tompok minyak direndamkan dalam (ii)
bikar dan kain dikocakkan dengan
(d) Materials / Bahan: menggunakan rod kaca. Element C H O
Unsur
Soap, distilled water, sea water 4. After that, the two beakers 1.44 0.36 1.92
and its content are left for Mass
Sabun, air suling, air laut 5 minutes. Jisim (g) 1.44 0.36 1.92
12 1 16
Apparatus / Radas: Selepas itu, kedua-dua bikar dan Number of = 0.12 = 0.12
kandungannya dibiarkan selama 5 mol = 0.36
Two 500 cm3 beakers, two pieces minit. Bilangan mol 0.12 0.12
of cloths stained with the same oily 0.12 0.36 0.12
stain, glass rod 5. Then, the cloths in both Simplest =1 0.12 =1
beakers are agitated again mol ratio =3
Dua bikar 500 cm3, dua helai kain yang using the glass rod before Nisbah mol
dikotori dengan kesan tompok minyak the water in both beakers is teringkas
yang sama, rod kaca poured away.
∴ Empirical formula
(e) Procedure / Prosedur: Kemudian, kain-kain dalam kedua-
dua bikar dikocak lagi dengan Formula empirik = CH3O
1. Two 500 cm3 beakers are menggunakan rod kaca sebelum
filled with approximately air dalam kedua-dua bikar itu (iii) n = [12 + 62 16]
400 cm3 of distilled water and dituangkan. (3 × 1) +
sea water respectively.
6. The cloths are rinsed with = 2
Dua bikar 500 cm3 diisi dengan water and observed.
kira-kira 400 cm3 air suling dan air ∴ Molecular formula
laut masing-masing. Kain-kain itu dibilas dengan air dan Formula molekul
diperhatikan.
2. About 1 g of soap powder = (CH3O)2= C2H6O2
is added into both beakers
respectively. (c) One molecule of K contains 2
carbon atoms, 6 hydrogen atoms
(f) Tabulation of data: and 2 oxygen atoms.
Penjadualan data:
Satu molekul K mengandungi 2 atom
Beaker Presence of soap bubbles Presence of oily stain after cleaning karbon, 6 atom hidrogen dan 2 atom
oksigen.
Bikar Kehadiran buih sabun Kehadiran kesan tompok minyak selepas dicuci 3. (a) Group 17 / Kumpulan 17
(b) (i) 2Na + X2 → 2NaX
Distilled A lot of soap bubbles are Oily stain disappears after cleaning.
Kesan tompok minyak hilang selepas dicuci. (ii) (8.775 – 3.450) g = 5.325 g
water formed.
Air suling Banyak buih sabun terhasil.
Sea Very little soap bubbles are Oily stain remains after cleaning. (iii) Assume the relative atomic
Kesan tompok minyak kekal selepas dicuci. mass of X = m
water formed
Air laut Sangat sedikit buih sabun Andaikan jisim atom relatif X = m
terhasil.
From the chemical equation,
SPM Forecast Paper (iii) Temperature of 450 – 550oC Dari persamaan kimia,
SPM Paper 1 and at a pressure. Bilangan mol Na = Bilangan mol X2
Suhu 450 – 550oC dan pada ×2
1. A 2. B 3. D 4. B 5. A
6. C 7. B 8. C 9. C 10. D tekanan atmosfera. = 3.450 = 5.325 ×2
11. C 12. C 13. D 14. C 15. A 23 2m
16. B 17. B 18. B 19. A 20. D
21. C 22. B 23. D 24. B 25. D (c) (i) Dissolve the sulphur trioxide ∴ m = 35.5
26. C 27. D 28. B 29. D 30. A
31. D 32. D 33. D 34. C 35. D gas into concentrated (iv) X is chlorine / X ialah klorin.
36. C 37. C 38. B 39. A 40. A
41. C 42. C 43. A 44. A 45. C sulphuric acid (c) (i) Bromine, X, fluorine
46. D 47. C 48. D 49. D 50. C Melarutkan gas sulfur trioksida Bromin, X, fluorin
kepada asid sulfurik
SPM Paper 2 (ii) Bromine, X, fluorine
(ii) SO3 + H2SO4 → H2S2O7 Bromin, X, fluorin
Section A / Bahagian A
(iii) By dissolving oleum into water 4. (a) Organic compounds that contain
1. (a) Contact process / Proses Sentuh Dengan melarutkan oleum ke elements carbon and hydrogen
(b) (i) II only.
dalam air
(ii) Vanadium(V) oxide Sebatian organik yang mengandungi
Vanadium(V) oksida 2. (a) (i) Chemical formula that shows unsur karbon dan hidrogen sahaja.
the simplest ratio of the (b) (i) Alkene / Alkena
number of atoms of every (ii) Alkane / Alkane
(c) (i) Because Q is a saturated
element that form a molecule
hydrocarbon
of the substance. Kerana Q ialah hidrokarbon tepu
Formula kimia yang menunjukkan
nisbah teringkas bilangan atom (ii) C3H6 + Br2 → C3H6Br2
setiap unsur yang membentuk satu
molekul sesuatu bahan.
© Penerbitan Pelangi Sdn. Bhd. A18
Chemistry Form 5 Answers
(d) Acidified potassium manganate(VII) (iii) Heat of neutralisation • Each aluminium atom will
solution. When added to P, the Haba peneutralan lose 3 electrons to achieve
stable electron arrangement
purple colour of the solution = – 1417.5 J of 2.8 and form Al3+ ion. [2]
0.025 mol
decolourises but when added to
= –56.7 kJ mol–1 Setiap atom aluminium akan
Q, the purple colour of the solution menghilangkan 3 elektron untuk
(d) mencapai susunan elektron
remains. Energy / Tenaga stabil 2.8 dan membentuk ion
Larutan kalium manganat(VII) berasid. Al3+.
Apabila ditambahkan kepada P, warna • Each oxygen atom will
ungu larutan dilunturkan tetapi apabila
ditambahkan kepada Q, warna ungu HCl + NaOH receive 2 electrons to
larutan kekal. ∆H = –56.7 kJ mol–1
NaCl + H2O achieve stable electron
(e) (i) P
arrangement of 2.8 and
(ii) Because P has higher
percentage of carbon than Q form O2- ion. [2]
Kerana P mempunyai peratusan Setiap atom oksigen akan
karbon yang lebih tinggi daripada Q menerima 2 elektron untuk
mencapai susunan elektron stabil
(f) Hydrogenation / Penghidrogenan Section B / Bahagian B 2.8 dan membentuk ion O2-.
5. (a) (i) Yellow / Kuning 7. (a) (i) Duralumin [1] • The electron arrangement
of aluminium oxide is: [2]
(ii) Na2S2O3 + 2HCl → 2NaCl +
H2O + SO2 + S (ii) Duralumin alloy is harder than Susunan elektron bagi
aluminium oksida adalah:
(b) (i) Rate of reaction pure aluminium. [1]
Kadar tindak balas
Duralumin adalah lebih keras 3+ 2–
(ii) When concentration of sodium daripada aluminium tulen.
thiosulphate increases, the • Addition of foreign atoms 2 AI 3O
rate of reaction increases also. into pure aluminium causes
Apabila kepekatan natrium tiosulfat
irregular arrangement of
bertambah, kadar tindak balas
bertambah juga. atoms in duralumin due to Total / Jumlah: 8
different sizes of atoms. [2]
Penambahan atom asing (c) • The extraction of aluminium
(iii) When the concentration of ke dalam aluminium tulen from aluminium oxide is
sodium thiosulphate increases, menyebabkan susunan atom tak conducted using carbon
there are more thiosulphate sekata dalam duralumin yang electrodes. [1]
ions per unit volume. This disebabkan oleh saiz atom yang Pengekstrakkan aluminium daripada
increases the chances for berbeza. aluminium oksida dilakukan dengan
thiosulphate and hydrogen • This gives greater menggunakan elektrod-elektrod
ions to collide effectively. The obstruction for the layers of karbon.
frequency of effective collision atoms in duralumin to slide • Molten aluminium oxide is
increases and the rate on one another, increasing added with some cryolite to
becomes higher. its hardness. [1] reduce the melting point of
Apabila kepekatan natrium tiosulfat
Ini memberikan lebih halangan aluminium oxide. [1]
bertambah, terdapat lebih banyak
ion tiosulfat per unit isi padu. Ini yang kepada lapisan atom Leburan aluminium oksida
meningkatkan peluang bagi ion-
ion tiosulfat dan hidrogen untuk dalam duralumin untuk ditambahkan dengan sedikit kriolit
berlanggar dengan berkesan.
Frekuensi perlanggaran berkesan menggelonsor pada satu untuk merendahkan takat lebur
bertambah dan kadar tindak balas
menjadi lebih tinggi. sama lain, menambahkan aluminium oksida.
kekerasannya.
• The Al3+ ions will be attracted
Total / Jumlah: 4 to cathode and the O2- ions
will be attracted to anode for
(b) (i) • Aluminium atoms do discharging. [1]
(iv) From graph, / Dari graf, not have octet electron Ion Al3+ akan tertarik ke katod dan
ion O2- akan tertarik ke anod untuk
When concentration arrangement. [1] nyahcas.
Atom aluminium tidak
= 0.40 mol dm–3, 1 = 0.02 s-1 • Molten aluminium will be produced
time mempunyai susunan elektron when Al3+ ions discharges. [1]
Apabila kepekatan oktet. Leburan aluminium akan dihasilkan
apabila ion Al3+ dinyahcaskan.
1 • Aluminium reacts with
masa Al3+ → Al + 3e-
= 0.40 mol dm–3, = 0.02 s-1 oxygen to achieve octet
1 electron arrangement to
0.02
∴ Time / Masa = = 50 s attain higher stability. [1]
Aluminium bertindak balas
6. (a) Because it is a good insulator of • Oxygen gas will be produced
dengan oksigen untuk mencapai when O2- ions discharges. [1]
heat
Kerana cawan polistirena adalah penebat susunan elektron oktet untuk Gas oksigen akan dihasilkan apabila
ion O2- dinyahcaskan.
haba yang baik memperoleh kestabilan yang
2O2- → O2 + 4e-
(b) HCl + NaOH → NaCl + H2O lebih tinggi.
• The overall equation for the
Total / Jumlah: 2 extraction of aluminium is: [1]
(c) (i) Number of mol water (ii) • The electron arrangements Persamaan keseluruhan bagi
Bilangan mol air pengekstrakkan aluminium ialah:
of aluminium and oxygen
4Al3+ + 6O2– → 4Al + 3O2
= 1.0 × 25 or / atau 0.5 × 50 are 2.8.3 and 2.6
1000 1000 Maximum / Maksimum: 5
respectively. [2]
= 0.025 mol Susunan elektron bagi
aluminium dan oksigen adalah
(ii) Heat change / Perubahan haba 2.8.3 dan 2.6 masing-masing.
= (25 + 50) × 4.2 × 4.5
= 1417.5 J
A19 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
8. (a) (i) Functional group 3. • Ethanol undergoes dehydration (b) 1. Volumetric flask used is 250 cm3.
Kumpulan berfungsi:
when its vapour is passed over [1]
Hydroxyl, -OH / Hidroksil, -OH. hot porcelain chips. [1] Kelalang volumetrik 250 cm3
[1]
etanol mengalami pendehidratan digunakan.
Other example apabila wapnya dilalukan pada 2. Number of moles of KOH
Contoh lain: Bilangan mol KOH
serpihan porselin panas.
Methanol, propanol • To produce ethene and water = 1 × 250 = 0.25 mol [1]
Metanol, propanol vapour [1] 1000
(any one / mana-mana satu) [1] Untuk menghasilkan etena dan wap 3. Mass of potassium hydroxide
air.
needed / Jisim kalium hidroksida
(ii) As fuel and as antiseptics. [2] Total / Jumlah: 6 diperlukan
Sebagai bahan api dan sebagai
Section C / Bahagian C = 0.25 × 56
antiseptik.
9. (a) (i) • S: Water / Air [1] = 14 g [1]
(b) Method 1: • T: Tetrachloromethane / 4. Weigh 14 g of KOH solid in a
beaker. [1]
Cara 1: Propanone / Methylbenzene
Timbang 14 g pepejal KOH dalam
• Fermentation / Penapaian [1] [1] bikar.
• Glucose and yeast (zymase) Tetraklorometana / Propanon / 5. Add distilled water. [1]
Metilbenzena Tambahkan air suling.
are mixed together in a closed
container and stored in a cool • Experiment I / Eksperimen I: 6. Stir until all the KOH solid
dissolves. [1]
and dark place. [2] Ethanoic acid ionises in
Glukosa dan yis (zimase) Kacau sehingga semua pepejal KOH
water to produce H+ ions. melarut.
dicampurkan bersama dalam bekas
[1]
tertutup dan disimpan dalam tempat
Asid etanoik mengion dalam air
yang sejuk dan gelap. 7. Pour the solution into the
untuk menghasilkan ion H+.
• The chemical equation for the volumetric flask. [1]
• Experiment II / Eksperimen II:
reaction [2] Tuangkan larutan ke dalam kelalang
Persamaan kimia bagi tindak balas Ethanoic acid cannot ionise volumetrik.
tersebut: in tetrachloromethane / 8. Rinse the beaker, glass rod and
C6H12O6 → 2C2H5OH + 2CO2 propanone / methylbenzene filter funnel. [1]
and is in the form of Bilas bikar, rod kaca dan corong
turas.
Method 2: molecules. [1]
Asid etanoik tidak mengion 9. Add distilled water until near
Cara 2:
dalam tetraklorometana / the graduation mark, add
• Hydration of ethene [1] propanon / metilbenzena dan water drop by drop until
Penghidratan etena berada dalam bentuk molekul. the meniscus reaches the
• Ethene and water vapour are (ii) • Ethanoic acid is a weak graduation mark. [1]
acid. [1]
flown over phosphoric acid Tambahkan air suling sehingga
Asid etanoik adalah asid lemah.
catalyst at the temperature of menghampiri garis penanda,
300oC and pressure of 60 atm • Hydrochloric acid is a tambahkan air titis demi titis
to produce ethanol. [2] strong acid. [1] sehingga aras meniskus mencapai
Etena dan wap air dilalukan pada Asid hidroklorik adalah asid garis penanda.
mangkin asid fosforik pada suhu
300oC dan tekanan 60 atm untuk kuat. 10. Stopper the volumetric flask
menghasilkan etanol.
• Ethanoic acid ionises and shake the solution. [1]
• The chemical equation for the partially in water to produce Tutup kelalang volumetrik dengan
menggunakan penutup dan
low concentration of goncangkan larutan.
reaction: [2] hydrogen ions. [1]
Persamaan kimia bagi tindak balas Asid etanoik mengion separa 10. (a) • Reaction I is not a redox
reaction. No change in the
tersebut: H3PO4 dalam air untuk menghasilkan
oxidation number for all the
kepekatan ion hidrogen yang
elements before and after the
C2H4 + H2O 300°C, 60 atm C2H5OH rendah.
reaction. [2]
Total / Jumlah: 10 • The lower the concentration Tindak balas I bukan tindak balas
of hydrogen ions, the
(c) 1. • Etanol is completely higher the pH value. [1] redoks. Tiada perubahan pada
nombor pengoksidaan setiap unsur
combusted in excess oxygen. Semakin rendah kepekatan ion sebelum dan selepas tindak balas.
hidrogen, semakin tinggi nilai
[1] pH.
Etanol dibakar lengkap dalam • Hydrochloric acid ionises • Reaction II is a redox reaction.
oksigen berlebihan. The oxidation number of zinc
increases (from 0 to +2).
• To produce carbon dioxide completely in water to
and water [1] produce high concentration
Untuk menghasilkan karbon
of hydrogen ions. [1] [2]
dioksida dan air. Asid hidroklorik mengion Tindak balas II ialah tindak balas
2. • Ethanol can be oxidised lengkap dalam air untuk redoks. Nombor penoksidaan bagi
by acidified potassium menghasilkan kepekatan ion zink meningkat (daripada 0 kepada
manganate(VII) under reflux [1] hidrogen yang tinggi. +2).
etanol boleh dioksidakan oleh • The higher the (b) • P: Bromine / Chlorine water /
kalium manganat(VII) berasid di
concentration of hydrogen Acidified KMnO4 solution [1]
bawah refluks. ions, the lower the pH Bromin / Air klorin / Larutan kalium
• To produce ethanoic acid. [1] value. [1] manganat(VII) berasid
untuk menghasilkan asid etanoik. Semakin tinggi kepekatan ion
• Q: Zinc / Zink [1]
hidrogen, semakin rendah nilai
pH.
© Penerbitan Pelangi Sdn. Bhd. A20
Chemistry Form 5 Answers
• Fe2+ undergoes oxidation • Bromine water undergoes (b)
reaction because it releases reduction because it gains
electron. [1] electron. [1] Diagram Observation Inference
Rajah Pemerhatian Inferens
Fe2+ mengalami pengoksidaan kerana Air bromin mengalami penurunan
melepaskan elektron. kerana menerima elektron. 1.1 The bulb Aqueous sodium
• Fe3+ + e– → Fe2+ [1] lights up chloride conducts
Mentol
• Zn → Zn2+ + 2e– [1] menyala electricity
Akueus
(c) • Oxidising agent: Acidified potassium manganate(VII) solution (or any other natrium klorida
suitable oxidising agent) [1] mengkonduksikan
elektrik
Agen pengoksidaan: Larutan kalium manganat(VII) berasid (atau mana-mana agen
pengosidaan lain yang sesuai) 1.2 The bulb Liquid
• Reducing agent: Iron(II) solution (or any other suitable reducing agent) [1] cannot light tetrachloromethana
Agen penurunan: Larutan ferum(II) (atau mana-mana agen penurunan lain yang sesuai)
• Labelled diagram / Rajah berlabel: up cannot conduct
Mentol tidak electricity
menyala Cecair
G tetraklorometana tidak
mengkonduksikan
elektrik
Carbon Carbon 1.3 The bulb Molten lead(II)
electrode electrode
Elektrod lights up bromide conducts
Elektrod karbon Mentol
karbon menyala electricity
Leburan
Acidified potassium Iron(II) sulphate plumbum(II) bromida
manganate(VII) Lsaorluuttaionnferum(II) mengkonduksikan
Larutasnokluatliiuomn sulfat elektrik
manganat(VII) DAsiliudtseulsfuurlipkhcuairric acid
berasid [6]
[2] (c) (i) Substances that conducts
Procedure / Prosedur:
electricity in molten or
1. Pour dilute sulphuric acid into a U-tube.
aqueous state using its free
Tuangkan asid sulfurik cair ke dalam tiub-U.
ions. [3]
2. Using a dropper / Add slowly iron(II) sulphate solution into one arm of the Bahan yang mengkonduksikan
elektrik dalam keadaan leburan
U-tube and acidified potassoium manganate(VII) solution into another arm. [1] atau akueus dengan menggunakan
ion-ion bebasnya.
Dengan menggunakan penitis / Tambah perlahan-lahan larutan ferum(II) sulfat ke dalam (ii)
satu lengan tiub-U dan larutan kalium manganat(VII) berasid ke dalam satu lagi lengan. Electrolyte Non-electrolyte
Elektrolit Bukan elektrolit
3. Place carbon electrodes in each side of the arm of the U-tube. [1]
Letakkan elektrod karbon ke dalam setiap sisi lengan tiub-U. Aqueous sodium Liquid
4. Connect the electrodes to a galvanometer by using connecting wires / chloride
Akueus natrium klorida tetrachloromethane
Complete the external circuit. [1] Cecair
Molten lead(II) tetraklorometana
Sambungkan elektrod kepada galvanometer dengan menggunakan wayar penyambung / bromide
Leburan plumbum(II)
Lengkapkan litar luar. bromida
5. Record the observations. [1]
Rekodkan pemerhatian.
Observations / Pemerhatian: [3]
• The green colour of iron(II) sulphate solution turns brown. [1]
Warna hijau larutan ferum(II) sulfat bertukar kepada perang. (d) (i) The bulb cannot light up. [3]
(ii) Mentol tidak menyala.
• The purple colour of acidified potassium manganate(VII) solution turns
There is no free ions in
colourless. [1] sodium chloride powder
because all the ions are
Warna ungu larutan kalium manganat(VII) berasid berubah kepada tidak berwarna. trapped in the ionic lattice. [3]
Tiada ion bebas dalam serbuk
SPM Paper 3 natrium klorida kerana semua ion
adalah terperangkap dalam kekisi
1. (a) (i) Type of compound used ionik.
Jenis sebatian yang digunakan
2. (a) (i) Hexane / Heksana [3]
(ii) Electrical conductivity of the compound (ii) Hexene / Heksena
Kekonduksian elektrik sebatian
(iii) Voltage of battery
Voltan bateri [3]
A21 © Penerbitan Pelangi Sdn. Bhd.
Chemistry Form 5 Answers
(b) 2. The temperature of the
Diagram Observation hydrochloric acid is measured
Rajah Pemerhatian
and recorded.
2.1 The purple colour of acidified potassium manganate(VII) remains. Suhu asid hidroklorik disukat dan
Warna ungu kalium manganat(VII) berasid kekal. direkodkan.
2.2 The purple colour of acidified potassium manganate(VII) decolourises. 3. A small piece of magnesium
Warna ungu kalium manganat(VII) berasid dilunturkan.
ribbon is placed into the
[3]
boiling tube above and
(c) (i) % Carbon in hexane • Constant variable:
Pemboleh ubah dimalarkan: the stopwatch is started
% Karbon dalam heksana
Volume and concentration of immediately.
= 6 × 12 × 100% hydrochloric acid Secebis kecil pita magnesium
86 Isi padu dan kepekatan asid diletakkan ke dalam tabung didih
hidroklorik di atas dan jam randik dimulakan
serta-merta.
= 83.72% 4. The time taken for the reaction
% Carbon in hexene to complete is recorded.
Masa yang diambil untuk tindak
% Karbon dalam heksena (c) Hypothesis: balas selesai direkodkan.
Hipotesis:
= 6 × 12 × 100% 5. Steps 1 to 4 are repeated by
84 When the temperature of a
heating the hydrochloric acid
= 85.71% [3] reaction increases, the rate of
to 40oC before placing an
(ii) Hexene gives sootier reaction also increases.
Apabila suhu tindak balas meningkat, identical magnesium ribbon
combustion because hexene
kadar tindak balas juga meningkat. into the acid.
has higher percentage of Langkah 1 hingga 4 diulang
(d) Substances / Bahan: dengan memanaskan asid
carbon. [3] hidroklorik kepada 40oC sebelum
Heksena memberikan pembakaran 0.1 mol dm-3 hydrochloric acid, meletakkan pita magnesium yang
magnesium ribbon. serupa ke dalam asid tersebut.
yang lebih berjelaga kerana
heksena mempunyai peratusan 0.1 mol dm-3 asid hidroklorik, pita (f) Tabulation of data:
karbon yang lebih tinggi. magnesium. Penjadulan data:
3. (a) Problem statement: Apparatus / Radas:
Pernyataan masalah:
How the temperature affects the 10 cm3 measuring cylinder, boiling Temperature Time taken
Suhu
rate of a reaction? tube, stopwatch, thermometer, for reaction to
Bagaimanakah suhu mempengaruhi
Bunsen burner. complete
kadar suatu tindak balas? 10 cm3 silinder penyukat, tabung didih, Masa yang diambil
untuk tindak balas
(b) Variables: jam randik, termometer, penunu Bunsen.
Pemboleh ubah: selesai
(e) Procedure / Prosedur:
• Manipulated variable:
Pemboleh ubah dimanipulasikan: 1. 5 cm3 of hydrochloric acid is Room
Temperature / Suhu measured and poured into a temperature
Suhu bilik
• Responding variable: boiling tube.
Pemboleh ubah bergerak balas: 5 cm3 asid hidroklorik disukat dan 40°C
Rate of reaction / Kadar tindak balas dituang ke dalam sebuah tabung
didih.
© Penerbitan Pelangi Sdn. Bhd. A22
Notes
A23 © Penerbitan Pelangi Sdn. Bhd.
Notes
© Penerbitan Pelangi Sdn. Bhd. A24
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