Top Class Kimia Tg 5

  Chemistry  Form 5  Chapter 5 Consumer and Industrial Chemistry

3. Diagram 3 show a part of graphene. Graphene is an allotrope of carbon consisting of a single layer
of atoms arranged in two-dimensional honeycomb lattice. Graphene is widely used in many fields
in the world.

Rajah 3 menunjukkan sebahagian grafen. Grafen ialah alotrop karbon yang terdiri daripada satu lapisan atom
yang disusun dalam kekisi sarang lebah dua dimensi. Grafen banyak digunakan dalam pelbagai bidang di seluruh
dunia.

Diagram 3 / Rajah 3

(a) (i) Define allotrope.

Takrifkan alotrop.

(ii) Suggest two examples of allotrope carbon. [1 mark / 1 markah]
[1 mark / 1 markah]
Cadangkan dua contoh alotrop karbon.

(b) (i) Suggest two applications of graphene in the industry.

Cadangkan dua aplikasi grafen dalam industri.

[2 mark / 2 markah]
(ii) State one physical property for each of the graphene application suggested in 1(b)(i).

Nyatakan satu sifat fizik bagi setiap aplikasi grafen yang dicadangkan di 1(b)(i).

[2 mark / 2 markah]

Section B / Bahagian B

4. (a) (i) A handkerchief was stained with grease. Explain how a soap can act as a cleansing agent

to remove the grease stain from the handkerchief.

Sapu tangan dikotori dengan gris. Terangkan bagaimana sabun boleh bertindak sebagai agen pembersih
untuk menghilangkan kesan gris daripada sapu tangan.

[8 marks / 8 markah]

(ii) State two advantages of using a soap instead of a detergent as a cleaning agent.

Nyatakan dua kelebihan menggunakan sabun untuk menggantikan detergen sebagai bahan pencuci.

[2 marks / 2 markah]

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Chemistry  Form 5  Chapter 5 Consumer and Industrial Chemistry 

(b) (i) Table 4 shows examples of food with the food additives.

Jadual 4 menunjukkan contoh makanan dengan bahan tambah makanan.

Examples of food Food additives

Contoh makanan Bahan tambahan makanan

Jam and lychee in syrup P
Q
Jem dan laici dalam sirap R

Pickled mango and chilli

Mangga dan cili jeruk

Burger and sausage

Burger dan sosej

Table 4 / Jadual 4

P, Q and R in Table 1 are food additives.

P, Q dan R dalam Jadual 1 ialah bahan tambah makanan.

• State the type of the food additives P, Q and R.

Nyatakan jenis bahan tambahan makanan P, Q dan R.

• Explain briefly how the food additives work in preserving food.

Terangkan secara ringkas bagaimana bahan tambahan makanan berfungsi dalam mengawet
makanan.

[9 marks / 9 markah]

(ii) State one disadvantage of using food additives.

Nyatakan satu kelemahan penggunaan bahan tambahan makanan.

[1 mark / 1 markah]

Section C / Bahagian C
5. (a) Diagram 5.1 shows a photo taken from Tanjung Basu beach, one of the tourist spots.

Rajah 5.1 menunjukkan gambar yang diambil dari pantai Tanjung Basu, salah satu tempat pelancongan.

Diagram 5.1 / Rajah 5.1

(i) State the type of pollution occured. [1 mark / 1 markah]

Nyatakan jenis pencemaran yang berlaku.

(ii) State two side effects caused by this pollution. [2 marks / 2 markah]

Nyatakan dua kesan negatif yang disebabkan oleh pencemaran ini.

(iii) Based on each ‘R’ in the 4R approach, explain how you can help to overcome this

pollution.
Berdasarkan setiap ‘R’ dalam pendekatan 4R, terangkan bagaimana anda dapat membantu mengatasi

pencemaran ini. [8 marks / 8 markah]

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  Chemistry  Form 5  Chapter 5 Consumer and Industrial Chemistry

(b) Diagram 5.2 shows a registered Malaysia’s wastewater and sanitation company. It is a
government-owned company which develops and maintains a modern and efficient
wastewater system in Peninsular Malaysia.

Rajah 5.2 menunjukkan syarikat air sisa dan sanitasi Malaysia yang berdaftar. Syarikat milik kerajaan ini
membangunkan dan mengekalkan sistem air sisa secara moden dan berkesan di Semenanjung Malaysia.

Diagram 5.2 / Rajah 5.2

(i) What is wastewater? [1 mark / 1 markah]

Apakah sisa air?

(ii) State three main source of wastewater. [3 marks / 3 markah]

Nyatakan tiga sumber utama sisa air.

(iii) Suggest three benefits of efficient wastewater treatment. [3 marks / 3 markah]

Cadangkan tiga kelebihan rawatan air sisa secara berkesan.

(iv) State two impacts of improper wastewater treatment on human health.

Nyatakan dua kesan rawatan air sisa yang tidak terurus terhadap kesihatan manusia.

[2 marks / 2 markah]

HOTS Challenge

Diagram below shows a bottle of mineral supplement given to Darren who had a bone fracture.
Explain how this supplement can help him to resolve the health problem faster and better.

Rajah di bawah menunjukkan sebotol mineral tambahan yang diberikan kepada Darren yang mengalami patah
tulang. Terangkan bagaimana makanan tambahan ini dapat membantu menyelesaikan masalah kesihatan Darren
dengan lebih cepat dan baik.

Quiz 5

21PAK- 96

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SPM Model Paper

Paper 1 / Kertas 1
[1 hour 15 minutes / 1 jam 15 minit]

[40 marks / 40 markah]

Answer all questions in this section.

Jawab semua soalan dalam bahagian ini.

1. Which of the following is a A Molten silver iodide A 2.8
natural polymer? consists of silver atoms B 2.8.1
and iodine atoms. C 2.8.2
Antara berikut, yang manakah D 2.8.8.5
polimer semula jadi? Leburan argentum iodida
terdiri daripada atom 6. Which of the following
A Nylon / Nilon argentum dan atom iodin. is correctly matched the
B Polystyrene / Polistirena isotopes and its use?
C Starch / Kanji B Molten silver iodide
consists of silver ions Antara berikut, yang manakah
2. Which of the following and iodide ions. pasangan isotop dan
are the correct safety kegunaannya yang betul?
precautions when Leburan argentum iodida
handling bromine in an terdiri daripada ion argentum A Iodine-131: To treat
experiment? dan ion iodida. thyroid cancer

Antara berikut, yang manakah C Molten silver iodide Iodin-131: Untuk merawat
langkah keselamatan yang consists of silver atoms kanser tiroid
betul apabila mengendalikan and iodine ions.
bromin di dalam eksperimen? B Gamma rays: To detect
Leburan argentum iodida leaks in underground
I Wear goggles terdiri daripada atom pipes
Pakai kaca mata argentum dan ion iodin.
Sinaran gamma: Untuk
keselamatan D Molten silver iodide mengesan kebocoran paip
consists of silver iodide di bawah tanah
II Wear a mask molecules that move
freely. C Carbon-13: To estimate
Pakai topeng muka the age of artefacts
Leburan argentum iodida
III Conduct the terdiri daripada molekul Karbon-13: Untuk
experiment on a argentum iodida yang menganggar usia artifak
workbench bergerak bebas.
D Sodium-24: To
Menjalankan eksperimen di 4. Who discovered protons? generate nuclear
atas meja kerja energy
Siapakah yang menemui
IV Conduct the proton? Natrium-24: Untuk
experiment on a table menghasilkan tenaga nuklear
A J.J. Thomson
Menjalankan eksperimen di B James Chadwick 7. Which of the following
atas meja C Ernest Rutherford device can be used to
detect isotopes?
A I and II C II and IV 5. The nucleon number of
atom X is 23 and it has 12 Antara berikut, alat yang
I dan II II dan IV neutrons in its nucleus. manakah boleh digunakan
What is the electron untuk mengesan isotop?
B I and III D III and IV arrangement of ion X?
A Geiger counter
I dan III IIII dan IV Nombor nukleon atom X ialah
23 dan mempunyai 12 neutron Kaunter Geiger
dalam nukleusnya. Apakah
3. Which of the following susunan elektron ion X? B X-ray machine

statement is true about Mesin sinar-X
molten silver iodide?
C Nuclear reactor
Antara berikut, pernyataan
yang manakah benar tentang Reaktor nuklear
leburan argentum iodida?

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  Chemistry  Form 5  SPM Model Paper

8. Which of the following chemical name and molar mass are of Elements. What is
correctly matched?
the possible electron
[Relative atomic mass: K = 39, O = 16, H = 1, Mg = 24,
arrangement of ion N?
Cl = 35.5, Ag = 108, N = 14, Pb = 207, I = 127] Atom M dan atom N
merupakan dua unsur dari
Antara berikut, yang manakah pasangan nama kimia dan jisim molar kumpulan yang sama. Diberi
yang dipadankan dengan betul? nombor proton atom M ialah 8
[Jisim atom relatif: K = 39, O = 16, H = 1, Mg = 24, Cl = 35.5, Ag = 108, dan atom N terletak di bawah
N = 14, Pb = 207, I = 127] atom M dalam Jadual Berkala

Chemical name Molar mass (g mol–1) Unsur. Apakah susunan
elektron yang mungkin bagi
Nama kimia Jisim molar (g mol–1)

A Potassium oxide 55 ion N?

Kalium oksida A 2.6 C 2.8.6

B Magnesium chloride 59 B 2.8 D 2.8.8

Magnesium klorida 13. Molecule P is usually used
as a disinfectant. It also
C Silver nitrate 232 eliminates bacteria, molds
and algae that commonly
Argentum nitrat grow in water supply
reservoirs. Which of the
D Lead(II) iodide 461 following is the correct
drawing of electron
Plumbum(II) iodida arrangement for atom P?

9. The relative molecular 11. A compound has the Molekul P biasanya digunakan
following percentage of sebagai pembasmi kuman.
mass of substance X2Y2Z3 is composition, 49.48% Molekul P juga menghilangkan
160. If the relative atomic carbon, 5.15% hydrogen, bakteria, kulat dan alga yang
28.87% nitrogen and biasanya tumbuh di takungan
mass of atom X is 24 and the remaining is oxygen. air. Antara berikut, yang
Given that the relative manakah lukisan susunan
atom Y is 32, calculate the molecular mass of the elektron yang betul untuk atom
compound is 194. What is P?
relative atomic mass of the molecular formula of
the compound? A
atom Z.
Jisim molekul relatif suatu [Relative atomic mass: H =
jbisaihmanatXom2Y2rZe3laitaiflaahto1m60X. Jika 1, C = 12, N = 14, O = 16]
ialah
24 dan atom Y ialah 32, hitung Suatu sebatian mempunyai
jisim atom relatif atom Z. peratus komposisi berikut,
49.48% karbon, 5.15% hidrogen,
A 16 C 39 28.87% nitrogen dan selebihnya
ialah oksigen. Diberi jisim molekul
B 32 D 64 relatif sebatian tersebut ialah 194. P
Apakah formula molekul sebatian
1 0. A pack of instant coffee tersebut? B

contains 125 mg of caffeine, [Jisim atom relatif: H = 1, C = 12, P
N = 14, O = 16]
C8H10N4O2. Calculate C –
the number of caffeine A C4H5N2
B C8H10N4 P
molecules in the coffee. C C4H5N2O
[Relative atomic mass: D C8H10N4O2 D –

H=1, C=12, N=14, O=16; 1 2. Atom M and atom N are P
two elements from the
Avogadro constant = same group. Given that
the proton number of
6.02 × 1023 mol–1] atom M is 8 and atom
N is located below atom
Satu paket kopi segera M in the Periodic Table
mengandungi 125 mg kafein,
mC8oHle10kNu4lOka2.feHinitudni gdablialamngkaonpi itu. 98
[Jisim atom relatif: H = 1, C = 12,
N = 14, O = 16; Pemalar
Avogadro = 6.02 × 1023 mol–1]

A 3.88 × 1020
B 3.88 × 1021
C 3.88 × 1022
D 3.88 × 1023

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Chemistry  Form 5  SPM Model Paper 

1 4. Diagram 1 shows an apparatus set-up to investigate the 16. Amethyst is a purple
reaction between chlorine gas and iron wool. germstone that consists
of silicone dioxide and
Rajah 1 menunjukkan susunan radas untuk mengkaji tindak balas antara manganese.
gas klorin dengan wul besi.
Which of the following
Combustion tube Soda lime statement best explains
Tiub pembakaran Soda kapur the colour formed in
amethyst?
Iron wool
Wul besi Batu kecubung ialah batu
permata berwarna ungu yang
Conical Heat Retort stand terdiri daripada silikon dioksida
flask Panaskan Kaki retort dan mangan.
Kelalang kon Antara berikut, pernyataan
yang manakah paling
Concentrated Potassium tepat menerangkan tentang
hydrochloric acid manganate(VII) pembentukan warna pada batu
Asid hidroklorik crystals kecubung?
pekat Hablur kalium manganat(VII)
A Germstone normally
Diagram 1 / Rajah 1 contains colour.

Which of the following 1 5. Chlorine gas is channelled Batu permata biasanya
statements are incorrect into a pail of water. Then, mengandungi warna.

about the reaction? a rusty nail is immersed B Presence of impurities
causes the formation of
Antara berikut, pernyataan yang into the pail and left for a colour.
manakah tidak benar tentang
tindak balas tersebut? day. The rusty nail turns Kehadiran bendasing

I Hot iron wool burns shiny grey the next day. menghasilkan warna.

very brightly. Explain this situation. C Manganese is a
coloured metal.
Wul besi panas terbakar Gas klorin disalurkan ke
dengan sangat terang. dalam sebuah baldi berisi air. Mangan ialah logam
Kemudian, paku yang berkarat berwarna.
II A brown solid is dicelupkan ke dalam baldi
tersebut dan dibiarkan selama D Manganese is a
formed. sehari. Paku berkarat bertukar transition metal that
menjadi kelabu berkilat pada can form a coloured
Pepejal perang terbentuk. keesokan harinya. Terangkan compound.
keadaan ini.
III Soda lime is used to Mangan ialah logam
A Chlorine water can act peralihan yang boleh
absorb moisture in the as a cleaning agent. membentuk sebatian
berwarna.
conical flask. Air klorin boleh bertindak
sebagai sebagai bahan 1 7. Which of the following
Soda kapur digunakan untuk pencuci. statement is true about
menyerap lembapan di the hydroxonium ion,
dalam kelalang kon. B Chlorine water is AHn3tOar+a? berikut, yang
acidic.
IV The experiment is manakah benar tentang ion
Air klorin bersifat asid. hidroksonium ion, H3O+?
conducted in a fume
C Chlorine water has a A Hydroxonium ion is
chamber because bleaching effect. formed by a hydrogen
bond between
chlorine gas is Air klorin mempunyai kesan hydrogen gas and
pelunturan. oxide ion.
explosive.
D Chlorine water is Ion hidroksonium terbentuk
Eksperimen dijalankan di a strong acid that melalui ikatan hidrogen
dalam kebuk wasap kerana antara gas hidrogen dengan
gas klorin boleh meletup. dissolves iron(III) oxide ion oksida.

A I and II on the nail.

I dan II Air klorin ialah asid kuat yang
melarutkan ferum(III) oksida
B I and III pada paku.

I dan III

C II and IV

II dan IV

D III and IV

IIII dan IV



99 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

B Hydroxonium ion is 19. Idell is requested to prepare 21. Given that the
a standard solution of concentration of sulphuric
formed by a metallic acid is 0.05 mol dm–3.
1 dm3 of 1 mol dm–3 Calculate its pH value.
bond. copper(II) sulphate
solution. He is given Diberi bahawa asid sulfurik
Ion hidroksonium terbentuk a bottle of copper(II) mempunyai kepekatan 0.05
melalui ikatan logam. sulphate crystals, 1 dm3 of mol dm–3. Hitung nilai pHnya.
distilled water, a volumetric
C Hydroxonium ion is flask and other apparatus. A 0.1
Calculate the mass of B 1.0
the formation of a copper(II) sulphate crystals C 0.13
needed to prepare the D 1.3
covalent compound. standard solution.
[Relative atomic mass: 2 2. Oxalic acid is found in
Ion hidroksonium merupakan O = 16, S = 32, Cu = 64] spinach. Which of the
pembentukan sebatian following statement is
kovalen. Idell diminta untuk true about oxalic acid?
menyediakan 1 dm3 larutan
D Hydroxonium ion is piawai kuprum(II) sulfat 1 mol dm–3. Asid oksalik boleh dijumpai di
Dia dibekalkan dengan sebotol dalam sayur bayam. Antara
formed by a dative hablur kuprum(II) sulfat, 1 dm3 berikut, pernyataan yang
air suling, kelalang piawai dan manakah benar mengenai asid
bond between water radas yang lain. Hitung jisim oksalik?
hablur kuprum(II) sulfat yang
molecule with diperlukan untuk menyediakan A pH value of oxalic acid
larutan piawai tersebut. is more than 8.
hydrogen ion through [Jisim atom relatif: O = 16,
S = 32, Cu = 64] Nilai pH asid oksalik lebih
sharing of lone pair of daripada 8.
A 20 g C 80 g
electrons. B 40 g D 160 g B Oxalic acid cannot
react with magnesium
Ion hidroksonium terbentuk 20. 500 cm3 of 0.2 mol dm–3 and calcium.
daripada ikatan datif antara hydrochloric acid is added
molekul air dengan ion into excess magnesium Asid oksalik tidak dapat
hidrogen melalui perkongsian powder to produce bertindak balas dengan
sepasang elektron tunggal. magnesium chloride and magnesium dan kalsium.
hydrogen gas. What is
the maximum volume of C Oxalic acid turns moist
hydrogen gas liberated at red litmus paper to
18. Tungsten filament is used room condition? blue.
[Molar volume of gas =
in a light bulb because it 24 dm3 mol–1 at room Asid oksalik menukarkan
condition] kertas litmus merah lembap
conducts electricity very kepada biru.
500 cm3 asid hidroklorik 0.2
well. What causes the mol dm–3 ditambahkan ke D Oxalic acid ionises
dalam serbuk magnesium yang partially in water to
conductivity of electricity berlebihan untuk membentuk produce hydrogen
magnesium klorida dan gas ions.
in tungsten? hidrogen. Berapakah isi padu
maksimum gas hidrogen yang Asid oksalik mengion separa
Filamen tungsten digunakan di terhasil pada keadaan bilik?
dalam mentol kerana tungsten [Isi padu molar gas = 24 dm3 di dalam air membentuk ion
mengkonduksikan elektrik mol–1 pada keadaan bilik] hidrogen.
dengan baik. Apakah yang
menyebabkan kekonduksian A 2.4 dm3 C 1.2 dm3 2 3. Which of the following is
elektrik dalam tungsten? B 24 dm3 D 12 dm3 not the observation when
pH indicator is added
A Free-moving ions in into potassium hydroxide
tungsten solution?

Ion-ion yang bergerak bebas Antara berikut, yang manakah
dalam tungsten bukan pemerhatian apabila
penunjuk pH dimasukkan
B Free-moving electrons ke dalam larutan kalium
in tungsten hidroksida?

Elektron-elektron yang
bergerak bebas dalam
tungsten

C Protons and electrons
in tungsten

Proton dan elektron dalam
tungsten

D Vibrating atoms in
tungsten

Atom-atom yang bergetar
dalam tungsten

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Chemistry  Form 5  SPM Model Paper 

A Red litmus paper – blue A Silver nitrate – white A Hydrogen gas is formed
at the end of the
Kertas litmus merah – biru precipitate experiment.

B Methyl orange – Argentum nitrat – mendakan Gas hidrogen terhasil pada
orange putih akhir eksperimen.

Metil jingga – jingga B Lead(II) iodide – B Manganese(IV) oxide
dissolves completely
C Universal indicator – yellow precipitate at the end of the
purple Plumbum(II) iodida – experiment.

Penunjuk universal – ungu mendakan kuning Mangan(IV) oksida larut
sepenuhnya pada akhir
D Phenolphthalein – pink C Barium sulphate – eksperimen.

Fenolftalein – merah jambu white precipitate C Hydrogen peroxide
turns purple at the end
24. Which of the following Barium sulfat – mendakan of experiment.
chemical substance and putih
its colour is not correctly Hidrogen peroksida menjadi
matched? D Lead(II) chromate – warna ungu pada akhir
eksperimen.
Antara berikut, yang manakah yellow precipitate
bahan kimia dan warnanya D Time taken to obtain
tidak dipadankan dengan Plumbum(II) kromat – the maximum volume
betul? mendakan kuning of oxygen gas is
reduced.
25. Table 1 shows the steps in an anion verification.
Masa yang diambil untuk
Jadual 1 menunjukkan langkah-langkah dalam pengesahan suatu anion. mendapatkan isi padu
maksimum gas oksigen
• Add 1 cm3 dilute HNO3 solution into a test tube with berkurang.
1 cm3 solution Y.
2 7. Composite material Q is
Tambahkan 1 cm3 larutan HNO3 cair ke dalam tabung uji berisi 1 cm3 used in the camera lens as
larutan Y. shown in Diagram 3. The
properties of material Q
• Add 1 cm3 dilute sulphric acid followed by 1 cm3 of are shown in Table 2.

iron(II) sulphate solution Bahan komposit Q digunakan
pada kanta kamera seperti
Tambahkan 1 cm3 asid sulfurik cair diikuti dengan 1 cm3 larutan ferum yang ditunjukkan dalam
Rajah 3. Sifat-sifat bahan Q
(II) sulfat. ditunjukkan dalam Jadual 2.

• Tilt the test tube slightly, add few drops of concentrated

sulphuric acid slowly on the side of the test tube.

Condongkan tabung uji sedikit, tambahkan beberapa titis asid sulfurik

pekat secara perlahan-lahan pada sisi tabung uji.

Table 1 / Jadual 1

What is the anion present

in solution Y? Hydrogen Burette
Apakah anion yang hadir di peroxide Buret
dalam larutan Y? solution
Larutan Water
A SO42– C NO3– hidrogen Air
peroksida
B CO32– D Cl– Diagram 3 / Rajah 3

26. Diagram 2 shows the Manganese(IV) oxide • Transparent
apparatus set-up of an Mangan(IV) oksida
experiment to determine Lut sinar
the rate of reaction. Diagram 2 / Rajah 2
• Absorbs UV rays
Rajah 2 menunjukkan susunan Which of the following
radas bagi satu eksperimen is true about this Menyerap sinaran UV
untuk menentukan kadar tindak experiment?
balas. • The absorption of UV
Antara berikut, yang manakah rays depends on light
benar tentang eksperimen ini? intensity

Penyerapan sinar UV
bergantung pada keamatan
cahaya

Table 2 / Jadual 2

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  Chemistry  Form 5  SPM Model Paper

What is Q? C Chlorine water 3 1. Diagram 4 shows the
apparatus set-up of a
Apakah Q? Air klorin voltaic cell. The needle in
the voltmeter deflects.
A Fiber optic D Acidified potassium
manganate(VII) Rajah 4 menunjukkan susunan
Gentian optik solution radas suatu sel kimia. Jarum di
dalam voltmeter terpesong.
B Photochromic glass Larutan kalium
Kaca fotokromik manganate(VII) berasid V Copper
C Fiber glass electrode
3 0. Which of the following are Iron Elektrod
Gentian kaca not true about standard electrode kuprum
electrode potential (Eo)? Elektrod
ferum Copper(II)
2 8. Which of the following is Antara berikut, yang manakah nitrate
tidak benar tentang keupayaan Iron(II) solution
true about the balanced elektrod piawai (Eo)? nitrate solution Larutan
chemical equation below? Larutan ferum(II) kuprum(II)
I The more positive the nitrat sulfat
Antara berikut, yang manakah Eo value, the weaker
benar mengenai persamaan the oxidising power of Diagram 4 / Rajah 4
kimia seimbang di bawah? an oxidising agent.
Which of the following
CCuuOO(p(s))++HH2(2g()g→) →CuC(pu)(+s)H+2OH(c2Oe)(l) Semakin positif nilai Eo, statements are true about
semakin lemah kuasa Diagram 4?
A Copper(II) oxide pengoksidaan suatu agen
undergoes oxidation by pengoksidaan. Antara berikut, pernyataan
releasing oxygen gas. yang manakah benar mengenai
II The more positive the Rajah 4?
Kuprum(II) oksida mengalami Eo value, the higher
the ability of atom to I Iron acts as the anode
pengoksidaan dengan receive electrons. and turns thinner.
membebaskan gas oksigen.
Semakin positif nilai Eo, Ferum bertindak sebagai
B Hydrogen gas semakin tinggi keupayaan anod dan menipis.
undergoes reduction atom untuk menerima
by gaining oxygen gas elektron. II Blue copper(II) nitrate
to form water. solution turns dark
III The more negative the blue.
Gas hidrogen mengalami Eo value, the weaker
penurunan dengan the reducing power of Larutan biru kuprum(II) nitrat
menerima gas oksigen untuk a reducing agent. menjadi biru tua.
membentuk air.
Semakin negatif nilai Eo, III Electrons flow from
C Black solid turns semakin lemah kuasa iron to copper through
brown is observed. penurunan suatu agen the external circuit.
penurunan.
Pepejal hitam bertukar Elektron mengalir dari ferum
menjadi perang diperhatikan. IV The more negative the ke kuprum melalui litar
Eo value, the higher luar.
D Copper(II) oxide is a the ability of atom to
reducing agent. release electrons. IV Green iron(II) nitrate
solution turns light
Kuprum(II) oksida ialah agen Semakin negatif nilai Eo, green.
penurunan.
semakin tinggi keupayaan Larutan hijau ferum(II) nitrat
29. Which of the following atom untuk menderma menjadi hijau muda.
substance cannot convert elektron.
iron(II) ion, Fe2+ to A I and II
iron(III) ion, Fe3+? A I and II
I dan II
Antara berikut, bahan I dan II
yang manakah tidak dapat B I and III
menukarkan ion ferum(II), Fe2+ B I and III
menjadi ion ferum(III), Fe3+? I dan III
I dan III
A Bromine water C II and IV
Air bromin C II and IV
B Potassium iodide II dan IV
II dan IV
Kalium iodida D III and IV
D III and IV
IIII dan IV
IIII dan IV

© Penerbitan Pelangi Sdn. Bhd. 102

Chemistry  Form 5  SPM Model Paper 

3 2. Diagram 5 shows the 33. Compound P has the Given that the heat of
structural formulae of two following properties. displacement for this
hydrocarbons. reaction is –226.8 kJ mol–1.
Sebatian P memiliki sifat-sifat
Rajah 5 menunjukkan formula berikut: Calculate the temperature
struktur bagi dua hidrokarbon. change of this experiment.
• Reacts with aluminium
HH HHH strips to form [Specific heat capacity
&& &&& colourless bubbles of water = 4.2 J g–1 °C–1;
C"C!C!H H!C!C!C!H Density of water =
&&& &&& Bertindak balas dengan 1.0 g cm–3]
HHH   HHH kepingan aluminium untuk
membentuk gelembung tak Diberi bahawa haba
Diagram 5 / Rajah 5 berwarna penyesaran untuk tindak balas
ini ialah –226.8 kJ mol–1.
What is the similarity • Turns a moist blue Hitung perubahan suhu
litmus paper to red eksperimen ini.
between these two [Muatan haba tentu air =
Menukarkan kertas litmus 4.2 J g–1 °C–1; Ketumpatan air
hydrocarbons? biru lembap menjadi merah = 1.0 g cm–3]

Apakah persamaan antara • Reacts with egg shells A 17 °C C 34 °C
kedua-dua hidrokarbon ini? and form colourless B 27 °C D 54 °C
bubbles
A Both react with 3 5. Diagram 7 shows the
Bertindak balas dengan apparatus set-up to
bromine water. kulit telur dan membentuk investigate the heat of
gelembung yang tak combustion for propanol.
Kedua-duanya bertindak berwarna
balas dengan air bromin. Rajah 7 menunjukkan susunan
What is the homologous radas untuk mengkaji haba
B Both are soluble in series of compound P? pembakaran bagi propanol.

water. Apakah siri homolog bagi Thermometer Windshield
sebatian P? Termometer Penghadang
Kedua-duanya larut di dalam angin
air. A Alkene Copper can
Tin kuprum Tripod stand
C Both cannot conduct Alkena Tungku kaki tiga
Water Spirit lamp
electricity in any state. B Carboxylic acid Air Lampu spirit
Wooden block
Kedua-duanya tidak Asid karboksilik Propanol Bongkah kayu
mengkonduksikan elektrik Propanol
dalam semua keadaan. C Alcohol

D Both cannot oxidise to Alkohol

alcohol.

Kedua-duanya tidak boleh
teroksida kepada alkohol.

3 4. Diagram 6 shows the thermometer readings when excess Diagram 7 / Rajah 7
magnesium powder is added into a polystyrene cup that
contains 200 cm3 of 1.0 mol dm–3 copper(II) sulphate solution. Which of the following
are used to increase the
Rajah 6 menunjukkan bacaan termometer apabila serbuk magnesium accuracy of heat released
berlebihan ditambahkan ke dalam cawan polistirena yang mengandungi by propanol?
100 cm3 larutan kuprum(II) sulfat 1.0 mol dm–3.
Antara berikut, yang manakah
Weighing bottle digunakan untuk meningkatkan
Botol penimbang kejituan haba yang dibebaskan
oleh propanol?
Magnesium Thermometer Stir
powder Termometer Kacau I Stir the water and
Serbuk observe the maximum
magnesium Plastic Lid level of mercury
cup Penutup achieved.
Copper(III) Cawan
sulphate plastik Plastic Kacau air dan perhatikan
solution cup tahap maksimum merkuri
Larutan Cawan yang dicapai.
kuprum(II) sulfat plastik

Diagram 6 / Rajah 6

103 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

II Add a windshield I Is an example of 38. Which of the following
around the apparatus is the correct uses of
set-up. natural polymer
thermoset?
Tambahkan pengadang angin Merupakan contoh polimer
semula jadi Antara berikut, yang manakah
di sekeliling susunan radas. benar mengenai kegunaan
II Environmentally termoset?
III Add more volume of
propanol. friendly product I Plastic chair

Tambahkan isi padu propanol. Produk mesra alam Kerusi plastik
IV Add more volume of III Produced
II Disc brake piston
water. through addition
Omboh cakera brek
Tambahkan isi padu air. polymerisation
III Compact disc (CD)
A I and II Dihasilkan melalui
and digital video disc
I dan II pempolimeran penambahan
(DVD)
B I and III IV Commonly applied in
Cakera padat (CD) dan
I dan III building, construction cakera video digital (DVD)

C II and IV and piping IV Microwavable

II dan IV Biasanya diaplikasikan dalam container
bangunan, pembinaan dan
D III and IV perpaipan Bekas gelombang mikro

IIII dan IV A I and II A I and II

36. Given that the heat of I dan II I dan II
combustion of ethanol is
–1368 kJ mol–1. What is B I and III B I and III
the fuel value of ethanol?
I dan III I dan III
[Relative atomic mass:
H = 1, C = 12, O = 16] C II and IV C II and IV

Diberi bahawa haba II dan IV II dan IV
pembakaran etanol ialah
–1368 kJ mol–1. Berapakah D III and IV D III and IV
nilai bahan api etanol?
[Jisim atom relatif: H = 1, IIII dan IV IIII dan IV
C = 12, O = 1]
3 9. Which of the following is true about saturated fats and
A 29.74 kJ g–1 unsaturated fats?
B 30.40 kJ g–1
C 45.60 kJ g–1 Antara berikut, yang manakah benar mengenai lemak tepu dan lemak
D 42.75 kJ g–1 tak tepu?

37. Diagram 8 shows the Saturated fat Unsaturated fat
structural formula for a
part of polyvinyl chloride. Lemak tepu Lemak tak tepu
Which of the following
statement is true about A Usually found in fresh fish Usually found in burgers
polyvinyl chloride?
Biasanya dijumpai dalam ikan Biasanya dijumpai dalam burger
Rajah 8 menunjukkan formula segar
struktur bagi bahagian polivinil
klorida. Antara berikut, B Can be found in cheesy Can be found in a glass of
pernyataan yang manakah pizzas fresh avocado juice
benar mengenai polivinil
klorida? Boleh didapati dalam pizza Boleh didapati di dalam segelas
berkeju jus avocado
HHHH H
&&&& & C Very helpful in protecting Easier to cause high blood
!C!C!C!C!C!
&&&& & the heart pressure
Cl H Cl H Cl Amat berguna untuk melindungi Mudah menyebabkan tekanan
jantung darah tinggi
Diagram 8 / Rajah 8
D As a seasoning in salad Use to deep-fry chicken

Sebagai perasa dalam salad Digunakan untuk menggoreng
ayam


© Penerbitan Pelangi Sdn. Bhd. 104

Chemistry  Form 5  SPM Model Paper 

4 0. Diagram 9 shows the lady Which of the following is C Cosmetic P must be
is applying cosmetic P on true about cosmetic P? removed thoroughly
her lips. before going to bed.
Antara berikut, yang manakah
Rajah 9 menunjukkan benar mengenai kosmetik P? Kosmetik P mesti
seorang wanita memakai dibersihkan dengan
kosmetik P pada bibirnya. A Cosmetic P can be sepenuhnya sebelum tidur.
applied on the lips
Cosmetic P D Cosmetic P can be
Kosmetik P overnight. used by anyone
without causing
Diagram 9 / Rajah 9 Kosmetik P boleh dipakai any side effects or
pada bibir semalaman. allergies.

B Cosmetic P can be Kosmetik P boleh
used even over 2 digunakan oleh sesiapa
sahaja tanpa menyebabkan
years. sebarang kesan sampingan
atau alahan.
Kosmetik P boleh
digunakan lagi walaupun
sudah melebihi 2 tahun.

Paper 2 / Kertas 2
[2 hours 30 minutes / 2 jam 30 minit]

Section A / Bahagian A
[60 marks / 60 markah]

Answer all questions in this section.

Jawab semua soalan dalam bahagian ini.

1. (a) Diagram 1.1 shows the arrangement of atoms in substance J and K.

Rajah 1.1 menunjukkan susunan atom dalam bahan J dan K.

Substance J Substance K
J K

Diagram 1.1 / Rajah 1.1

(i) Identify which of the substance in Diagram 1.1 is pure metal and pure alloy.

Kenal pasti bahan yang manakah dalam Rajah 1.1 ialah logam tulen dan aloi tulen.

Pure metal / Logam tulen:

Alloy / Aloi:

[2 marks / 2 markah]

105 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

(ii) State the property of substance J based on the atomic arrangement in Diagram 1.1.

Nyatakan sifat bahan J berdasarkan susunan atom dalam Rajah 1.1.

[1 mark / 1 markah]
(b) Diagram 1.2 shows the National Monument, a bronze sculpture which was built in

remembrance of the brave soldiers who died fighting for the independence of our country.

Rajah 1.2 menunjukkan Tugu Negara, arca gangsa yang dibina untuk memperingati askar-askar berani yang
terkorban demi memerdekakan negara kita.

Diagram 1.2 / Rajah 1.2

State the reason why bronze is used instead of pure copper.

Nyatakan sebab mengapa gangsa digunakan dan bukannya kuprum tulen.

[2 marks / 2 markah]

2. Diagram 2 shows the apparatus set-up used in the experiment to determine the heat of displacement
of copper by magnesium. The initial temperature of copper(II) sulphate is 28.0 °C and the highest
temperature achieved is 45.0 °C.

Rajah 2 menunjukkan susunan radas yang digunakan dalam eksperimen untuk menentukan haba penyesaran
kuprum oleh magnesium. Suhu awal kuprum(II) sulfat ialah 28.0 °C dan suhu tertinggi yang dicapai ialah 45.0 °C.

Thermometer Excess zink
Termometer powder
Serbuk zink
Polystryrene berlebihan
cup
Cawan
polistirena

© Penerbitan Pelangi Sdn. Bhd. 50.0 cm3 0.2 mol dm–3
copper(II) sulphate solution

50.0 cm3 kuprum(II) klorida
0.5 mol dm–3

Diagram 2 / Rajah 2

106

Chemistry  Form 5  SPM Model Paper 

(a) State the type of reaction that occurred based on the temperature change in the experiment.

Nyatakan jenis tindak balas yang berlaku berdasarkan perubahan suhu dalam eksperimen tersebut.

[1 mark / 1 markah]

(b) Write a balanced chemical equation for the reaction.

Tuliskan persamaan kimia seimbang bagi tindak balas tersebut.

[1 mark / 1 markah]

(c) (i) Calculate the number of moles of copper(II) sulphate solution.

Hitung bilangan mol larutan kuprum(II) sulfat.

[1 mark / 1 markah]

(ii) Calculate the heat released in this experiment.

Hitung haba yang dibebaskan dalam eksperimen ini.

[1 mark / 1 markah]
(iii) Calculate the heat of displacement of copper by magnesium in this experiment.

Hitung haba penyesaran kuprum oleh magnesium dalam eksperimen ini.

[1 mark / 1 markah]

107 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

(d) Predict the value of heat of displacement if magnesium powder is used in this experiment to
replace zinc powder. Give a reason.

Ramalkan nilai haba penyesaran sekiranya serbuk magnesium digunakan dalam eksperimen ini untuk
menggantikan serbuk zink. Berikan alasan.

[2 marks / 2 markah]
3. Diagram 3 shows the information of atoms represented by unknown M, N, Q, R, X and Y in the

Periodic Table of Elements.

Rajah 3 menunjukkan maklumat atom yang ditunjukkan oleh M, N, Q, R, X dan Y yang tidak diketahui dalam
Jadual Berkala Unsur.

XN

Y MQ
R

Diagram 3 / Rajah 3

Based on Diagram 3,

Berdasarkan Rajah 3,

(a) Suggest a name for element M.

Cadangkan nama bagi unsur M.

[1 mark / 1 markah]
(b) (i) Identify the transition metal.

Kenal pasti logam peralihan.

[1 mark / 1 markah]

(ii) State one special feature for answer in 3(b)(i).
Nyatakan satu ciri khas bagi jawapan di 3(b)(i).

[1 mark / 1 markah]

© Penerbitan Pelangi Sdn. Bhd. 108

Chemistry  Form 5  SPM Model Paper 

(c) Write the electron arrangement for atom Y.

Tuliskan susunan elektron bagi atom Y.

[1 mark / 1 markah]
(d) Write the formula of ion formed from atom Q.

Tuliskan formula ion yang terbentuk daripada atom Q.

[1 mark / 1 markah]

(e) When a small piece of element Y is burnt in gas Q, a reaction occurred to produce an ionic
compound.

Write a chemical equation for the reaction.

Apabila sekeping kecil unsur Y dibakar dalam gas Q, tindak balas berlaku untuk menghasilkan sebatian ion.
Tuliskan persamaan kimia bagi tindak balas tersebut.

[2 marsk / 2 markah]

4. Table 1 shows the number of protons and neutrons for atoms P and Q.

Jadual 1 menunjukkan bilangan proton dan neutron bagi atom P dan Q.

Atom Number of protons Number of neutrons

Atom Bilangan proton Bilangan neutron

P 12 12

Q 11 12

Table 1 / Jadual 1

(a) (i) Write the electron arrangement of atom Q.

Tuliskan susunan elektron atom Q.

[1 mark / 1 markah]
(ii) State the position of atom Q in the Periodic Table of Elements.

Nyatakan kedudukan atom Q dalam Jadual Berkala Unsur.

[1 mark / 1 markah]

109 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

(iii) Give a reason for your answer in 4(a)(i).

Berikan alasan untuk jawapan anda di 4(a)(i).

[2 marsk / 2 markah]

(b) (i) What is the nucleon number of atom P?

Apakah nombor nukleon bagi atom P?

[1 mark / 1 markah]
(ii) Write the electron arrangement of ion P.

Tuliskan susunan elektron ion P.

[1 mark / 1 markah]
(iii) Draw the electron arrangement of ion P.

Lukiskan susunan elektron ion P.

[2 marks / 2 markah]

5. Diagram 4 shows the mass spectrum of element X.

Rajah 4 menunjukkan jisim spektrum untuk unsur X.

82.8 9.1

Intensity 8.1
(% abundance)

Keamatan
(% kelimpahan)

© Penerbitan Pelangi Sdn. Bhd. 24 25 26
Isotopic mass
Jisim isotop

Diagram 4 / Rajah 4

110

Chemistry  Form 5  SPM Model Paper 

(a) How many naturally occurring isotopes does the element contain?

Berapakah isotop semula jadi yang terdapat dalam unsur ini?

[1 mark / 1 markah]

(b) Define natural abundance.

Takrifkan kelimpahan semula jadi.

[1 mark / 1 markah]

(c) State the natural abundance for the isotopes of element X.

Nyatakan kelimpahan semula jadi bagi isotop unsur X.

(d) Write the symbols for the isotopes of element X. [2 marks / 2 markah]
[2 marks / 2 markah]
Tuliskan simbol bagi semua isotop unsur X.

(e) Calculate the relative atomic mass of element X.

Hitungkan jisim atom relatif unsur X.

[2 marks / 2 markah]

6. Diagram 5 shows two sets of experiments carried out to study the effect of catalyst on the rate of
reaction between zinc and sulphuric acid.

Rajah 5 menunjukkan dua set eksperimen yang dijalankan untuk mengkaji kesan mangkin terhadap kadar tindak
balas antara zink dengan asid sulfurik.

Set I Set II

Burette Burette
Buret Buret

Sulphuric Water Sulphuric Water
acid solution Air acid + catalyst Air
Larutan asid Asid sulfurik +
sulfurik mangkin

Excess zinc granule Excess zinc granule
Ketulan zink berlebihan Ketulan zink berlebihan

Diagram 5 / Rajah 5

111 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

Table 2 shows the results obtained from the experiment.

Jadual 2 menunjukkan hasil yang diperoleh daripada eksperimen tersebut.

Set Total volume of gas collected in 3 minutes (cm3) Temperature (°C)

Set Isipadu gas terkumpul dalam 3 minit (cm3) Suhu (°C)

I 40.00 30.0

II 50.00 30.0

Table 2 / Jadual 2

(a) Sketch a graph of volume of gas (cm3) against time (s) for both sets on the same axis.

Lakarkan graf isi padu gas (cm3) melawan masa (s) untuk kedua-dua set pada paksi yang sama.



[2 marks / 2 markah]
(b) Calculate the average rate of reaction for the first 3 minutes for Set I and Set II, in cm3 min–1

(in two decimal places).

Hitung kadar tindak balas purata bagi 3 minit pertama untuk Set I dan Set II, dalam cm3 min–1 (dalam dua titik
perpuluhan).

(i) Set I / Set I

(ii) Set II / Set II

[2 marks / 2 markah]

© Penerbitan Pelangi Sdn. Bhd. 112

Chemistry  Form 5  SPM Model Paper 

(iii) Compare the average rate of reaction for the first 3 minutes for Set I and Set II.

Bandingkan kadar tindak balas purata selama 3 minit pertama untuk Set I dan Set II.



[1 mark / 1 markah]
(iv) Based on the Collision Theory, explain how catalyst affects the rate of reaction.

Berdasarkan Teori Perlanggaran, terangkan bagaimana mangkin mempengaruhi kadar tindak balas.

[3 marks / 3 markah]

7. Diagram 6 shows the U-tube consisting iron(II) sulphate solution and bromine water.

Rajah 6 menunjukkan tiub-U berisi larutan ferum(II) sulfat dan air bromin.

G

Bromine Carbon
water(aq) electrodes
Air bromin(ak) Elektrod karbon

HH22SSOO4(4a(ak)q) FFeeSSOO4(4a(ak)q)

Diagram 6 / Rajah 6

(a) (i) Identify the anode and the cathode.

Kenal pasti anod dan katod.

Anode:

Anod:

Cathode:

Katod:

[2 marks / 2 markah]

113 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

(ii) Write a half equation to support your answer in 7(a)(i).

Tuliskan persamaan setengah untuk menyokong jawapan anda di 7(a)(i).

Anode / Anod:



Cathode / Katod:

[2 marks / 2 markah]
(b) Identify the oxidising agent and the reducing agent.

Kenal pasti agen pengoksidaan dan agen penurunan.

Oxidising agent / Agen pengoksidaan:

Reducing agent / Agen penurunan:

[2 marks / 2 markah]
(c) Write the overall ionic equation.

Tuliskan persamaan ion keseluruhan.

[2 marks / 2 markah]
8. Diagram 7 shows the apparatus set-up for two types of cells, Cell A and Cell B.

Rajah 7 menunjukkan susunan radas bagi dua jenis sel, Sel A dan Sel B.

V

Zinc electrode Copper electrodes

Elektrod zink Elektrod kuprum

Porous pot

Pasu berliang

Zinc sulphate Cell A Copper(II) Cell B
solution Sel A sulphate solution Sel B

Larutan zink Larutan kuprum(II) sulfat
sulfat
Diagram 7 / Rajah 7
© Penerbitan Pelangi Sdn. Bhd.
114

Chemistry  Form 5  SPM Model Paper 

(a) (i) Name Cell A and Cell B.

Namakan Sel A dan Sel B.



[2 marks / 2 markah]

(ii) State the energy conversion in Cell B.

Nyatakan penukaran tenaga dalam Sel B.



[1 mark / 1 markah]

(b) (i) Compare the observations on copper(II) sulphate solution in Cell A and Cell B after 30
minutes.

Bandingkan pemerhatian terhadap larutan kuprum(II) sulfat dalam Sel A dengan Sel B selepas 30
minit.



[2 marks / 2 markah]

(ii) Explain your answer in 8(b)(i).

Jelaskan jawapan anda di 8(b)(i).



[2 marks / 2 markah]

(c) (i) Suggest a method to increase the potential difference of Cell A.

Cadangkan kaedah untuk meningkatkan perbezaan keupayaan Sel A.

[1 mark / 1 markah]
(ii) Explain your answer.

Terangkan jawapan anda.

[1 mark / 1 markah]

115 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

Section B / Bahagian B
[20 marks / 20 markah]

Answer any one question from this section.

Jawab mana-mana satu soalan dalam bahagian ini.

9. In an investigaton, Bryan and Derrick carried out three experiments as shown in Table 3.

Dalam suatu penyiasatan, Bryan dan Derrick menjalankan tiga eksperimen seperti yang ditunjukkan dalam
Jadual 3.

Experiment Condition

Eksperimen Keadaan

1 100 cm3 of 0.1 mol dm–3 nitric acid acid + iron powder at room condition

100 cm3 0.1 mol dm–3 asid nitrik + serbuk besi pada keadaan bilik

2 100 cm3 of 0.1 mol dm–3 nitric acid + iron powder at 50 oC

100 cm3 0.1 mol dm–3 asid nitrik + serbuk besi pada 50 oC

3 100 cm3 of 0.1 mol dm–3 nitric acid + iron powder in 50 oC + copper(II) sulphate
solution

100 cm3 0.1 mol dm–3 asid nitrik + serbuk besi pada 50 oC + larutan kuprum(II) sulfat

Table 3 / Jadual 3

(a) Write an ionic equation to represent all the three experiments in Table 3.

Tuliskan persamaan ion untuk mewakili ketiga-tiga eksperimen dalam Jadual 3.

[2 marks / 2 markah]

(b) Explain each factor that you have identified in Experiment 1, Experiment 2 and
Experiment 3.

Terangkan setiap faktor yang telah anda kenal pasti dalam Eksperimen 1, Eksperimen 2 dan Eksperimen 3.

[3 marks / 3 markah]

(c) Based on the Collision Theory, compare and explain:

Berdasarkan Teori Perlanggaran, bandingkan dan terangkan:

(i) Experiment 1 and Experiment 2

Eksperimen 1 dan Eksperimen 2.

[5 marks / 5 markah]

(ii) Experiment 2 and Experiment 3

Eksperimen 2 dan Eksperimen 3.

[6 marks / 6 markah]

(d) Then, sketch a graph to show the comparison between experiments in 9(c)(i) and 9(c)(ii) on
the same axes.

Kemudian, lakarkan graf untuk menunjukkan perbandingan antara eksperimen di 9(c)(i) dengan 9(c)(ii) pada
paksi yang sama.

[4 marks / 4 markah]

© Penerbitan Pelangi Sdn. Bhd. 116

Chemistry  Form 5  SPM Model Paper 

1 0. (a) Diagram 8.1 shows the flow chart involving the formation of ethene gas from a cup of
pineapple juice.

Rajah 8.1 menunjukkan carta alir yang melibatkan pembentukan gas etena daripada secawan jus nanas.

I II
Ethanol Ethene
Etanol Etena

Diagram 8.1 / Rajah 8.1

(i) Name process I and process II. Then, write a balanced chemical equation for the conversion
of process I and process II.

Namakan proses I dan proses II. Kemudian, tulis persamaan kimia yang seimbang bagi penukaran
proses I dan proses II.

[6 marks / 6 markah]

(ii) Briefly describe the formation of ethene gas from ethanol. Explain how you verify the
formation of ethene gas using a suitable reagent solution.

Huraikan secara ringkas pembentukan gas etena daripada etanol. Terangkan bagaimana anda
mengesahkan pembentukan gas etena menggunakan larutan reagen yang sesuai.

[8 marks / 8 markah]

(b) Diagram 8.2 shows the coagulated latex in a cup left for more than 4 hours.

Rajah 8.2 menunjukkan getah beku di dalam cawan yang dibiarkan selama lebih daripada 4 jam.

Diagram 8.2 / Rajah 8.2

(i) Explain this situation.

Terangkan keadaan ini.

[4 marks / 4 markah]

(ii) Suggest a solution that can slow down the formation of the white solid as shown in
Diagram 8.2. Give your reason.

Cadangkan larutan yang dapat melambatkan pembentukan pepejal putih yang ditunjukkan dalam Rajah
8.2. Berikan alasan anda.

[2 marks / 2 markah]

117 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  SPM Model Paper

Section C / Bahagian C
[20 marks / 20 markah]

Answer all questions in this section.

Jawab semua soalan dalam bahagian ini.

1 1. Diagram 9 shows the conversation between Janelle and Allen.

Rajah 9 menunjukkan perbualan antara Janelle dan Allen.

Janelle, are you able to differentiate between potassium
carbonate solution and potassium sulphate solution using
the barium chloride solution given by Madam Chin?
Janelle, adakah anda dapat membezakan larutan kalium
karbonat dengan larutan kalium sulfat menggunakan larutan
barium klorida yang diberikan oleh Puan Chin?

I added 1 cm3 of barium chloride solution but cannot differentiate
between potassium carbonate solution and potassium sulphate
solution. Why?
Saya menambahkan 1 cm3 larutan barium klorida tetapi tidak dapat
membezakan antara larutan kalium karbonat dengan larutan kalium
sulfat. Mengapa?

Diagram 9 / Rajah 9

(a) Give an explanation to their problem. Suggest a change that should be done to the procedure

in order for the test to be successful. In your explanation, include ionic equations.

Berikan penjelasan tentang masalah mereka. Cadangkan perubahan yang harus dilakukan terhadap prosedur
supaya ujian tersebut berjaya. Dalam penjelasan anda, sertakan persamaan ion.

[10 marks / 10 markah]

(b) Suzannah and Danish are given different solutions. They are assigned to describe chemical
tests that can be carried out in the laboratory to differentiate the solutions.

Suzannah dan Danish diberikan larutan yang berbeza. Mereka ditugaskan untuk menerangkan ujian kimia
yang boleh dilakukan di makmal untuk membezakan larutan-larutan tersebut.

(i) Sodium chloride solution and sodium iodide solution.

Larutan natrium klorida dan larutan natrium iodida.

(ii) Ammonium sulphate solution and sodium sulphate solution.

Larutan ammonium sulfat dan larutan natrium sulfat.

[10 marks / 10 markah]

© Penerbitan Pelangi Sdn. Bhd. 118

Answers

CHA PTER Redox Equilibrium 2. free-moving ions, chemical, passes
ion bergerak bebas, kimia
1 Keseimbangan Redoks
3. molecules / molekul-molekul
4. electrical, chemical / elektrik, kimia

Oxidation and Reduction 5. Electrolyte Non-
1.1 Pengoksidaan dan Penurunan Elektrolit electrolyte
1. Definition Bukan elektrolit

Definisi Oxidation Reduction – Potassium – Sucrose
Pengoksidaan Penurunan chloride solution
(a) Transfer of oxygen solution
Pemindahan oksigen Gain Loss Larutan
Penerimaan Kehilangan Larutan kalium sukrosa
(b) Transfer of hydrogen klorida
Pemindahan hidrogen Loss Gain – Lead(II)
Kehilangan Penerimaan – Molten silver bromide
(c) Transfer of electron chloride
Pemindahan elektron Release Receive Plumbum(II)
Kehilangan Penerimaan Leburan bromida
(d) Change in oxidation number argentum
Perubahan nombor pengoksidaan Increase Decrease klorida – Molten
Bertambah Berkurang acetamide
– Nitric acid
Asid nitrik Leburan
asetamida

2. (a) NH4+ 2. (a) (i) Zinc / Zink Extraction of Metal from Its Ore
NH4 = +1 (ii) Zinc, copper / Zink, kuprum Pengesktrakan Logam daripada
N + 4(+1) = +1
N = –3 (b) (i) Copper / Kuprum 1.5 Bijihnya
(b) CO32–
CO3 = –2 (ii) Copper, zinc 1. (a) sodium, magnesium, aluminium
C + 3(–2) = –2 Kuprum, zink natrium, magnesium, aluminium
C = +4
(c) (i) Zn → Zn2+ + 2e– (b) iron, tin, zinc
(c) K2Cr2O7 ferum, stanum, zink
K2Cr2O7 = 0 (ii) Cu2+ + 2e– → Cu
2(+1) + 2Cr + 7(–2) = 0
Cr = +6 (d) (i) Zinc / Zink 2. – Cheap / Murah
– Easily available / Mudah
(d) Na2S2O3 (ii) Copper(II) ion / Cu2+
Na2S2O3 = 0 Ion kuprum(II) / Cu2+ diperoleh
2(+1) + 2S + 3(–2) = 0
S = +2 (e) (i) Zinc / Zink – Solid in room condition /
Pepejal dalam keadaan bilik
Standard Electrode Potential (ii) Copper / Kuprum
1.2 Keupayaan Elektrod Piawai
(iii) Blue, lighter blue – An effective reducing agent /
1. half cells / setengah sel biru, biru muda Agen penurunan yang berkesan
2. (a) Eº cell / Eº sel
= (–0.45) – (–0.76) (f) (i) Zinc, zinc ion / Zink, ion zink 3. (i) Consuming extremely high

= 0.31 V (ii) Copper, copper amount of electricity during
(b) Eº cell / Eº sel Kuprum, kuprum
= (+0.80) – (+0.34) extraction causing pollution.
(iii) decreases / berkurang Penggunaan jumlah elektrik
= 0.46 V yang sangat tinggi semasa
(c) Eº cell / Eº sel (g) zinc, copper / zink, kuprum pengekstrakan menyebabkan
= (–0.76) – (–2.37) pencemaran.
(h) (i) Salt bridge / Titian garam
= 1.61 V
(d) Eº cell / Eº sel (ii) – Dilute hydrochloric acid
= (+0.34) – (–0.13)
/ Asid hidroklorik cair (ii) Removal of native vegetation
= 0.47 V
– Dilute sodium chloride in mining area resulting in soil
Voltaic Cell
1.3 Sel Kimia solution / Larutan erosion.
Pembuangan tumbuhan asal
1. chemical, electrical natrium klorida cair di kawasan perlombongan
kimia, elektrik mengakibatkan hakisan tanah.
– Dilute sodium sulphate

solution / Larutan (iii) Greenhouse gas emission

natrium sulfat cair causing global warming.
Pembebasan gas rumah hijau
(iii) Zn(s) + Cu2+(aq) → menyebabkan pemanasan global.

Zn2+(aq) + Cu(s)
Zn(p) + Cu2+(ak) →
Rusting
Zn2+(ak) + Cu(p) 1.6 Pengaratan

Electrolytic Cell 1. oxidation, loss, positively-charged
1.4 Sel Elektrolisis pengoksidaan, kehilangan, bercas
positif
1. aqueous solution, molten state,
2. iron, steel / besi, keluli
electric current 3. oxygen, water / oksigen, air
larutan akueus, keadaan leburan, arus

elektrik

A1 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

4. slow down, zinc / magnesium / iron (iii) – Bring near glowing (b) – The middle region of water
/ aluminium
splinter into test tube droplet that covers the iron
melambatkan, zink / magnesium / ferum with gas X / oxygen
/ aluminium block is the anode.
gas. Kawasan tengah titisan air yang
5. speed up, tin/ lead/ copper/ silver Dekatkan kayu uji berbara menutupi blok besi ialah anod.
mempercepatkan, stanum / plumbum / ke dalam tabung uji berisi
kuprum / argentum gas X / gas oksigen. – In this area, the

SPM Practice 1 – Splinter relights concentration of oxygen gas

Paper 1 indicates the presence is lower.
of gas X / oxygen gas. Di kawasan ini, kepekatan gas
1. C 2. D 3. A 4. C 5. B Kayu uji menyala
6. D 7. C 8. C 9. D 10. C menunjukkan kehadiran oksigen lebih rendah.
11. D 12. D 13. D 14. D 15. D gas X / gas oksigen.
16. C 17. B 18. B 19. A 20. C – Iron undergoes oxidation
2 1. C 22. C 23. D 24. C 25. D
26. D 27. C 28. A (c) (i) Copper(II) ion / Ion reaction
kuprum(II) Ferum menjalani tindak balas

pengoksidaan

– by releasing electrons to

(ii) Brown solid is formed. / form iron(II) ions.
Pepejal perang terbentuk. dengan membebaskan elektron

(iii) Copper is deposited on membentuk ion ferum(II).

Paper 2 the cathode. – The edge region of water
Kuprum terenap pada katod.
Section A / Bahagian A droplet that covers the iron

1. (a – Concentration of ions 1.0 (d) 4OH– + 2Cu2+ → 2H2O + O2 + block is the cathode.
mol dm–3 2Cu Kawasan pinggir air yang

Kepekatan ion 1.0 mol dm–3 4. (a) (i) Anode: Carbon immersed menutupi blok besi ialah katod.

in iron(II) sulphate solution. – In this area, the
Anod: Karbon yang dicelup di
– Temperature 25 °C or 298 K concentration of oxygen gas
Suhu 25 °C atau 298 K dalam larutan ferum(II) sulfat.
is higher.
– Pressure of 1 atm or 101 kPa Cathode: Carbon Di kawasan ini, kepekatan gas
Tekanan pada 1 atm atau 101 kPa
immersed in acidified oksigen lebih tinggi.

potassium dichromate(VI) – Oxygen gas and water

– Platinum is used as inert solution. undergo reduction reaction
electrode Katod: Karbon yang dicelup Gas oksigen dan air menjalani

Platinum digunakan sebagai di dalam larutan dikromat(VI) tindak balas penurunan
elektrod lengai berasid.
– by receiving electrons to

[any three answers / mana- (ii) Anode / Anod: form hydroxide ions.
mana tiga jawapan] dengan menerima elektron
Fe2+ → Fe3+ + e–
(b) (i) Cl2  Ag+  Cu2+  Zn2+ Cathode / Katod: membentuk ion hidroksida.
(ii) Cl– , Ag , Cu , Zn
Cr2O72– + 14H+ + 6e– → (c) – Half-equation in anode /
(c) Eº cell = Eº cathode – Eº anode Persamaan setengah di anod:
Eº sel = Eº katod – Eº anod 2Cr3+ + 7H2O
(b) Oxidising agent / Agen Fe → Fe2+ + 2e–

= (+0.34) – (–0.76) pengoksidaan: – Iron serves as a reducing

= 1.10 V Acidified dichromate(VI) ion / agent.
Ion dikromat(VI) berasid Ferum bertindak sebagai agen
2. (a) Anode / Anod: 2O2– → O2 + 4e–
Cathode / Katod: Al3+ + 3e– → Al Reducing agent / Agen penurunan.

(b) (i) Oxide ion undergoes – Half-equation / Persamaan
setengah di katod: O2 + 2H2O
oxidation reaction by penurunan:
+ 4e– → 4OH–
releasing electrons to Iron(II) ion / Ion ferum(II)

form oxygen gas. (c) Electron flows from carbon – Oxygen gas serves as an

Ion oksida menjalani tindak immersed in iron(II) sulphate oxidising agent.
balas pengoksidaan dengan Gas oksigen bertindak sebagai
membebaskan elektron solution to carbon immersed
membentuk gas oksigen. agen pengoksidaan.
in acidified potassium

dichromate(VI) solution (d) – Coil magnesium on the iron

(ii) Aluminium ion undergoes through external circuit. block as a sacrificial metal.
Lilitkan magnesium pada blok
reduction reaction by Elektron mengalir dari karbon yang
dicelup di dalam larutan ferum(II) besi sebagai logam korban.
receiving electrons to sulfat ke karbon yang dicelup di
dalam larutan kalium dikromat(VI) – Magnesium is more
form aluminium. berasid melalui litar luar.
electropositive than iron,
Ion aluminium menjalani
tindak balas penurunan thus it will be oxidised to
dengan menerima elektron
(d) 6Fe2+ + Cr2O72– + 14H+ → 6Fe3+ protect iron.
membentuk aluminium. + 2Cr3+ + 7H2O Magnesium lebih elektropositif
daripada besi, maka magnesium
(c) (i) Aluminium ion / Ion Section B / Bahagian B akan teroksida untuk melindungi
aluminium besi.
5. (a) Presence of water and oxygen
(ii) Oxide ion / Ion oksida gas. – Apply paint on the surface

3. (a) Oxygen / Oksigen Kehadiran air dan gas oksigen. of iron.
(b) (i) 4OH– → 2H2O + O2 + 4e– Sapukan cat pada permukaan
(ii) Hydroxide ion / Ion
besi.
hidroksida

© Penerbitan Pelangi Sdn. Bhd. A2

Chemistry  Form 5  Answers 

– Paint prevents the (b) Materials / Bahan: 3. Each iron nail is
placed into a test tube
penetration of water and Iron nails, zinc strip, respectively. Then, three
drops of phenolphthalein
oxygen gas into the iron magnesium strip, copper strip, indicator and three
drops of potassium
block. sandpaper, hot agar solution, hexacyanoferrate(III)
Cat menghalang penembusan solution are added into
phenolphthalein indicator the hot agar solution and
air dan gas oksigen pada blok stirred well.
besi. and 0.1 mol dm–3 potassium
Setiap paku besi diletakkan
hexacyanoferrate(III) solution ke dalam tabung uji masing-
masing. Kemudian, tiga
Section C / Bahagian C Paku besi, jalur zink jalur titis petunjuk fenolftalein
6. (a) Electrolyte: Silver nitrate solution magnesium, jalur kuprum, kertas dan tiga titis larutan
Elektrolit: Larutan argentum nitrat pasir, larutan agar panas, petunjuk kalium heksasianoferat(III)
fenolftalein, dan larutan kalium ditambahkan ke dalam larutan
Anode: Pure silver rod heksasianoferat(III) 0.1 mol dm–3 agar panas dan dikacau hingga
Anod: Rod argentum tulen sebati.
Apparatus / Radas:
Procedure / Prosedur: Test tubes, test tube rack, and 4. The hot agar solution
1. The iron key is cleaned dropper is poured into each test
Tabung uji, rak tabung uji dan tube until the iron nails
using sandpaper. penitis are covered with hot agar
Kunci besi dibersihkan dengan solution.
Procedure / Prosedur:
kertas pasir. 1. All the metals and iron Larutan agar panas itu
2. 150 cm3 1.0 mol dm–3 silver dituangkan ke dalam setiap
nail are cleaned with tabung uji sehingga semua
nitrate solution is measured sandpaper. paku besi ditutupi dengan
and poured into a beaker. Semua logam dan paku besi larutan agar panas.
150 cm3 larutan argentum dibersihkan menggunakan
nitrat 1.0 mol dm–3 disukat dan kertas pasir. 5. The test tubes are left
dituang ke dalam sebuah bikar. on the rack for a day. All
2. Thin strip of magnesium is observations are recorded.
3. Pure silver is connected coiled on iron nail B, zinc
to the positive terminal of on iron nail C, and copper Semua tabung uji tersebut
a battery and cleaned iron on iron nail D. dibiarkan selama sehari.
key is connected to the Semua pemerhatian dicatatkan.
negative terminal of the Jalur magnesium yang nipis
battery using a connecting dililit pada paku besi B, jalur
wire. zink pada paku besi C dan
jalur kuprum pada paku besi D.
Argentum tulen disambung ke
terminal positif bateri dan kunci Result / Keputusan:
yang bersih ke terminal negatif
bateri dengan menggunakan Test tube Observation Inference
wayar penyambung. Tabung uji Pemerhatian Inferens

4. Both electrodes are A • Dark blue spots are observed • Rusting occurs
immersed into silver nitrate Fe on the agar gel. Pengaratan berlaku
solution.
B Bintik-bintik biru tua kelihatan di • Rusting does not occur
Kedua-dua elektrod direndam Fe + Mg atas agar gel Pengaratan tidak berlaku
ke dalam larutan argentum
nitrat sulfat. C • Pink colouration is found on • Rusting does not occur
Fe + Zn agar gel Pengaratan tidak berlaku
5. The switch is turned on and
the electricity is allowed to D Warna merah jambu terbentuk di • Rusting occurs the
flow for 30 minutes. Fe + Cu atas agar gel fastest

Suis dihidupkan dan elektrik • Pink colouration is found on Pengaratan berlaku paling
dibiarkan mengalir selama agar gel cepat
30 minit.
Warna merah jambu terbentuk di
6. All changes that occur at atas agar gel
the anode, cathode and
electrolyte are recorded. • The whole agar gel turns
dark blue
Semua perubahan yang
berlaku pada anod, katod dan Keseluruhan agar gel menjadi
elektrolit dicatatkan. biru tua

Observations / Pemerhatian: HOTS Challenge – Polish the rusted grill with sandpaper
– Pure silver anode becomes
– Conditions to cause rusting: to remove all the brown solid on the
thinner. Presence of oxygen gas and water.
Anod argentum tulen semakin grill.
Keadaan yang menyebabkan pengaratan: Gosokkan jeriji yang berkarat dengan
nipis. Kehadiran gas oksigen dan air.
kertas pasir untuk menyingkirkan semua
– A layer of shiny grey metal pepejal perang pada jeriji tersebut.
is coated on the antique key.

Lapisan logam kelabu berkilat
menyaluti kunci antik.

A3 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers (d) C3H7OH(l) + 3O2(g) → 3CO(g)

– Paint the cleaned iron grill. + 4H2O(g) //
Catkan jeriji besi yang telah dibersihkan.
– Paint can prevent the contact of water and oxygen gas with the iron grill. C3H7OH(ce) + 3O2(g) → 3CO(g) +
Cat dapat mengelakkan sentuhan air dan gas oksigen dengan jeriji besi. 4H2O(g) //

CHA PTER Carbon Compound C3H7OH(l) + 3232OO22(g(g) )→→343C4HCH(p2(2O)Os+()(gg+))
C3H7OH(ce) +
2 Sebatian Karbon
(e) C2H4(g) + H2O(g) → C2H5OH(l)
Types of Carbon Compound C2H4(g) + H2O(g) → C2H5OH(ce)

2.1 Jenis-jenis Sebatian Karbon (f) C2H5OH(l) + 2[O] →

1. carbon / karbon CH3COOH(l) + H2O(l)
C2H5OH(ce) + 2[O] →
2. (a) carbon, living / karbon, hidup
CH3COOH(ce) + H2O(ce)
(b) saturated, unsaturated / tepu, tak tepu
(g) C2H5OH(l) + C3H7COOH(l) →
(c) natural gas, petroleum / gas asli, petroleum
C3H7COOC2H5(l) + H2O(l)
3. Organic compounds Characteristics Inorganic compounds C2H5OH(ce) + C3H7COOH(ce) →
Sebatian organik Ciri-ciri Sebatian tak organik
C3H7COOC2H5(ce) + H2O(ce)
Living things Source Non-living things
Benda hidup Sumber Benda bukan hidup 2. (a) Ethanol / Etanol
Lower Higher (b) 1,2-dibromopropane /
Lebih rendah Melting point and Lebih tinggi
boiling point 1,2-dibromopropana
Takat lebur dan takat didih (c) Ethene / Etena

Homologous Series 3. (a) Mg(s) + 2H+(aq) → Mg2+(aq) +
2.2 Siri Homolog H2(g)

1. General formula CnH2n+2 CnH2n CnH2n–2 Mg(p) + 2H+(ak) → Mg2+(ak) +
Formula am H2(g)
Carbon-carbon Carbon-carbon
Functional Carbon-carbon double bond triple bond Colorless bubbles are
group Ikatan karbon Ikatan karbon observed. / Gelembung tak
Kumpulan single bond ganda dua ganda tiga berwarna kelihatan.
berfungsi Ikatan karbon
tunggal (b) CaCO3(s) + 2H+(aq) →

2. (a) Pentane / Pentana Chemical Properties and Ca2+(aq) + CO2(g) + H2O(l)
(b) Propanol / Propanol Interconversion of Compounds CaCO3(p) + 2H+(ak) → Ca2+(ak) +
between Homologous Series
(c) Hexanoic acid / Asid heksanoik Sifat Kimia dan Saling Pertukaran CO2(g) + H2O(ce)
2.3 Sebatian antara Siri Homolog
(d) Butyne / Butuna Colorless bubbles are
observed. / Gelembung tak
3. Octane Odourless 1. (a) C4H10(l) + 123O2(g) → 4CO2(g) berwarna kelihatan.
Oktana Tidak berbau C4H10(ce) + → + 5H2O(g)
4CO2(g) + (c) H+(aq) + OH–(aq) → H2O(l)
Heptene Flammable 13 O2(g) H+(ak) + OH–(ak) → H2O(ce)
Heptena Mudah terbakar 2 5H2O(g)
Blue cobalt(II) chloride paper

turns pink.
Kertas kobalt(II) klorida yang

berwarna biru bertukar menjadi
merah jambu.

Pentyne Soury smell (b) C3H6(l) + 3O2(g) → 3CO(g) + 4. (a) Esterification / Pengesteran
Pentuna Berbau masam
3H2O(g) // (b) Concentrated sulphuric acid /
Asid sulfurik pekat
Ethanol Fragrant smell C3H6(ce) + 3O2(g) → 3CO(g)
Etanol Berbau wangi + 3H2O(g) // (c) Methyl ethanoate / Metil etanoat
3 (d) – sweet / wangi
C3H6(l) + 2 O2(g) → 3C(s) + – low / rendah
C3H6(ce) + → 3H2O(g) – Insoluble / Tidak larut
Methanoic Soluble in 3 O2(g)
acid water 2 3C(p) +
Asid metanoik Larut dalam air 3H2O(g) Isomers and Naming Based on
IUPAC Nomenclature
Ethyl Very volatile (c) C5H8(l) + 7O2(g) → 5CO2(g) + Isomer dan Penamaan mengikut
butanoate Mudah meruap 2.4 IUPAC
Etil butanoate 4H2O(g)
1. chemical formula, structural
C5H8(ce) + 7O2(g) → 5CO2(g) + formula
4H2O(g)
formula kimia, formula struktur

© Penerbitan Pelangi Sdn. Bhd. A4

Chemistry  Form 5  Answers 

2. Butyne / Butuna (ii) Compound P has carbon- Ion hidrogen bertindak
balas dengan kalsium
HH HH carbon double bond. karbonat daripada kulit
&& '& &' Sebatian P mempunyai ikatan telur untuk menghasilkan
H!C#C!C!C!H C!C gas karbon dioksida.
&& karbon ganda dua.
HH
H!C C !H (d) (i) Orange to green / Jingga Section B / Bahagian B
& & kepada hijau
H H
(ii) C2H5OH(l) + 2[O] →
1–butyne 1,3–butadiene CH3COOH(aq) + H2O(l) 4. (a) (i) – Structural formula of
1–butuna 1,3–butadiena butene / Formula struktur
C2H5OH(ce) + 2[O] → butena:
3. (a) (i) Ethyl butanoate / Etil CH3COOH(ak) + H2O(ce)
butanoat HH
3. (a) (i) Carbon, hydrogen and &&
(ii) Butanoic acid / Asid
butanoik oxygen H!C"C!C!C!H
Karbon, hydrogen dan oksigen &&
(iii) Ethanol / Etanol
(b) (i) Methyl ethanoate / Metil (ii) Molecular formula / HH

etanoat Formula molekul: C9H8O4 – C4H8(l) + 6O2(g) →
Empirical formula / Formula
(ii) Ethanoic acid / Asid etanoik 4CO2(g) + 4H2O(g)
(iii) Methanol / Metanol empirik: C9H8O4 C4H8(ce) + 6O2(g) →
(iii) Molar mass / Jisim molar
4CO2(g) + 4H2O(g)
SPM Practice 2 = 9(12) + 8(1) + 4(16) (ii) Catalyst X / Mangkin X:

Paper 1 = 180 g mol–1 Nikel powder / Serbuk nikel
Temperature Y / Suhu Y:
1. A 2. D 3. B 4. B 5. D (b) (i) Colorless bubbles are
6. A 7. C 8. D 9. B 10. B
1 1. D 12. C 13. D 14. B 15. A formed. 180 oC
1 6. C 17. A 18. B 19. C 20. D Gelembung gas tak berwarna Compound Z / Sebatian Z:
21. A 22. C 23. B 24. C 25. C terbentuk.
26. C 27. C 28. B Butane / Butana
(ii) – Aspirin dissolved in

warm water releases (iii)

Paper 2 hydrogen ion. HHH H H
Aspirin yang larut di dalam &&& & &
Section A / Bahagian A H!C!C!C!C!H
air suam membebaskan &&& & H!C!H
ion hidrogen. HHH H &

– Hydrogen ion reacts H& H
&& &
with calcium carbonate
Butane / Butana H!C!C!C!H
1. (a) X: CnH2n+2 from the eggshells to && &
Y: CnH2n
Z: CnH2n–2 produce carbon dioxide HH H
(b) X: Propane / Propana
Y: Propene / Propena gas. 2-methylpropane / 2-metilpropana
Z: Propyne / Propuna
(c) (i) C3H4(g) + 4O2(g) → (b) Butyne Butene Butanol Butanoic acid
Butuna Butena Butanol Asid butanoik
3CO2(g) + 2H2O(g) Carboxylic acid
C3H4(g) + 4O2(g) → 3CO2(g) + Homologous Alkyne Alkene Alcohol Asid karbosilik
Alkuna Alkena Alkohol CnH2n+1COOH
2H2O(g ) series
Siri homolog Carboxyl group
(ii) – Number of moles of Kumpulan
propyne / Bilangan mol General CnH2n–2 CnH2n CnH2n+1OH karboksil
propuna formula
Formula am

= 4 Functional Carbon- Carbon- Hydroxyl
[3(12) + group group
4(1)] Kumpulan carbon carbon Kumpulan
berfungsi hidroksil
= 0.1 mol triple bond double bond
– Mole ratio / Nisbah mol Ikatan karbon Ikatan karbon
ganda tiga ganda dua
= 1 : 3

= 0.1 : 0.3 Section C / Bahagian C
– Volume of gas / Isi padu
5. (a) (i) Element
gas Unsur
Composition by mass
= 0.3 × 24 Komposisi dengan jisim C H O

= 7.2 dm3 Number of moles 52.3 13.3 34.4
(iii) Carbon dioxide / Karbon Bilangan mol
52.3 13.3 34.4
dioksida Mole ratio 12 1 16
2. (a) P: Ethene / Etena Nisbah mol = 4.36 = 2.15
Simplest ratio 4.36 = 13.3 2.15
Q: Ethanol / Etanol Nisbah teringkas 2.15 13.3 2.15
R: Ethanoic acid / Asid etanoik = 2.03 2.15 =1
(b) I: Hydration / Penghidratan = 6.19
II: Oxidation / Pengoksidaan 2 1
6
(c) (i) Brown bromine water

turns colourless.
Warna perang air bromin

menjadi tak berwarna.

A5 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

– Empirical formula / (ii) C2H5OH(l) → C2H4(g) + CHA PTER
Formula empirik = C2H6O H2O(g)
3 Thermochemistry
– n[C2H6O] = [2(12) 46 + 16] C2H5OH(ce) → C2H4(g) +
+ 6(1) H2O(g) Termokimia

n = 1 Compound N: Ethene Heat Change in Reactions
Sebatian N: Etena 3.1 Perubahan Haba dalam Tindak Balas
– Molecular formula /
(iii) Test 1 / Ujian 1
Formula molekul = – 2 cm3 of bromine water

C2H6O is added into a test 1. (a) energy changes / perubahan
tube containing ethene tenaga
(ii) – General formula / / compound N and
Formula am: CnH2n+1OH shaken. (b) heat changes / perubahan haba
2 cm3 air bromin (c) (i) Exothermic / eksotermik
– Homologous series: ditambahkan ke dalam
tabung uji yang berisi (ii) Endothermic / endotermik
Alcohol etena / sebatian N dan (d) released, absorbed
Siri homolog: Alkohol digoncang.
– Brown bromine water dibebaskan, diserap
(iii) – Struktural formula / turns colourless.
Formula struktur: Air bromin perang bertukar Heat of Reaction
menjadi tak berwarna. 3.2 Haba Tindak Balas
HH
&& Test 2 / Ujian 2 1. (a) precipitation, double
H!C!C!O!H – 2 cm3 of acidified
&&
HH potassium
maganate(VII) solution
– IUPAC nomenclature: is added into a test decomposition
tube containing ethene pemendakan, penguraian ganda dua
/ compound N and
Ethanol shaken. (b) 1 mole of precipitate is formed
2 cm3 larutan kalium
Penamaan IUPAC: Etanol manganate(VII) berasid from the ions in the solution.
ditambahkan ke dalam 1 mol mendakan terbentuk
(b) (i) 1. Glass wool is soaked tabung uji yang berisi
etena / sebatian N dan daripada ion-ion di dalam larutan.
in ethanol / compound digoncang. 2. (a) AgNO3(aq) + KCl(aq) →
M and inserted into the – Purple acidified
potassium AgCl(s) + KNO3(aq)
end of a combustion manganate(VII) AgNO3(ak) + KCl(ak) → AgCl(p)
solution turns
tube. colourless. + KNO3(ak)
Wul kaca direndam di Larutan ungu kalium
dalam etanol / sebatian M manganate(VII) berasid (b) Ag+(aq) + Cl–(aq) → AgCl(s)
dan dimasukkan ke dalam bertukar menjadi tak Ag+(ak) + Cl–(ak) → AgCl(p)
hujung tiub pembakaran. berwarna.
(iv) – Soluble in water (c) – Number of moles of AgNO3 /
2. Porcelain chips are Larut dalam air Bilangan mol AgNO3
– Low melting point and
arranged in the middle boiling point (0.5)(50)
Takat lebur dan takat didih 1 000
of the combustion tube yang rendah =

and the combustion = 0.025 mol

tube is sealed with a

rubber stopper connected – Number of moles of KCl /
Bilangan mol KCl
to a delivery tube.
Serpihan porselin disusun (0.5)(50)
di tengah tiub pembakaran = 1 000
dan tiub pembakaran
ditutup dengan penyumbat = 0.025 mol
getah yang disambungkan
(d) Average initial temperature /
kepada tiub penghantar. Purata suhu awal

3. The porcelain chips are = (28.0 + 29.0)
2
heated strongly until

hot red, while the glass = 28.5 oC

wool is heated gently. Temperature change
Serpihan porselin Perubahan suhu

dipanaskan dengan HOTS Challenge = (37.5 – 28.5)
kuat sehingga merah
panas manakala wul = 9.0 oC
kaca dipanaskan secara
perlahan. – Diesel / Diesel (e) Heat change, Q
Ethene / compound N Perubahan haba, Q
– Percentage of carbon in diesel, C12H24 = mcθ
is collected using water Peratusan karbon dalam diesel, C12H24
= (50 + 50)(4.2)(9.0)
displacement technique = 12(12) × 100% = 85.71%
[12(12) + 24(1)] = 3780 J
in an inverted test tube
= 3.78 kJ
placed inside the water. – Percentage of carbon in RON 95
Etena / sebatian (f) Heat of precipitation, ΔH
N dikumpulkan petrol, C8H20 Haba pemendakan, ΔH
menggunakan teknik Peratusan karbon dalam petrol RON 95,
sesaran air di dalam = 3.78
tabung uji yang C8H20 0.025
diterbalikkan di dalam air.
= 8(12) × 100% = 82.76% = –151.2 kJ mol–1
[8(12) + 20(1)]

© Penerbitan Pelangi Sdn. Bhd. A6

Chemistry  Form 5  Answers 

(g) 5. (a) alkali, base, salt, water (d) 1 mole of fuel is burnt in
Energy / Tenaga alkali, bes, garam, air
excess oxygen gas to form
Ag+(aq) + Cl–(aq)
Ag+(ak) + Cl–(ak) (b) when 1 mole of water carbon dioxide gas and water

ΔH = –151.2 kJ mol–1 is formed through the 1 mol bahan api terbakar dalam
gas oksigen yang berlebihan untuk
AgCl(s) neutralization reaction between membentuk gas karbon dioksida
AgCl(p)
acid solution and alkali dan air

solution 8. (a) CH3OH(l) + 3 O2(g) → CO2(g) +
apabila 1 mol air terbentuk melalui CH3OH(ce) + 2 → 2H2O(g)
tindak balas peneutralan antara
larutan asid dan larutan alkali 3 O2(g) CO2(g) +
2 2H2O(g)
6. (a) HCl(aq) + KOH(aq) → KCl(aq)

+ H2O(l) (b) Mass of methanol / Jisim
HCl(ak) + KOH(ak) → KCl(ak) +
(h) – Heat of precipitation / Haba metanol
pemendakan H2O(ce)
= 54.87 – 53.75
– silver chloride / argentum (b) H+(aq) + OH–(aq) → H2O(l)
klorida H+(ak) + OH–(ak) → H2O(ce) = 1.12 g

– number of moles / Bilangan (c) Number of moles of HCl / Number of moles of methanol /
mol Bilangan mol HCl Bilangan mol metanol

3. (a) more, less (1.0)(100) = 1.12
lebih, kurang 1 000 [12 + 3(1) + 16 + 1]
=
(b) 1 mole of metal is displaced = 0.035 mol
from its metal ion solution by a
more electropositive metal = 0.1 mol Heat change, Q / Perubahan
Number of moles of KOH / suhu, Q
1 mol logam disesarkan daripada Bilangan mol KOH = mcθ
larutan ion logamnya oleh logam = 500 × 4.2 × (40 – 29)
yang lebih elektropositif = (1.0)(100) = 23100 J
1 000 = 23.1 kJ
4. (a) Zn(s) + Pb(NO3)2(aq) →
Zn(NO3)2(aq) + Pb(s) = 0.1 mol

Zn(p) + Pb(NO3)2(ak) → (d) Heat change, Q / Perubahan Heat of combustion, ΔH / Haba
Zn(NO3)2(ak) + Pb(p) haba, Q pembakaran, ΔH

(b) Zn(s) + Pb2+(aq) → Zn2+(aq) + = mcθ = 23.1
Pb(s) = (100 + 100)(4.2)(6.5) 0.035
= 5460 J
Zn(p) + Pb2+(ak) → Zn2+(ak) + Pb(p) = 5.46 kJ = –660 kJ mol–1
(c)
(e) Heat of neutralisation, ΔH
(c) Number of moles of Pb(NO3)2 / Haba peneutralan, ΔH Energy / Tenaga
Bilangan mol Pb(NO3)2
5.46 CH3OH(l) + —32 O2(g)
= (1.0)(200) = 0.1 CH3OH(ce) + —32 O2(g)
1 000
= –54.6 kJ mol–1
= 0.2 mol (f)
(d) Heat change, Q / Perubahan
Energy / Tenaga ΔH = –660 kJ mol–1
haba, Q
= mcθ

= (200)(4.2)(54.7 – 28.0) HCl(aq) + KOH(aq) CCOO22((gg) )++2H22HO2(Og)(g)
= 22428 J HCl(ak) + KOH(ak)
= 22.428 kJ
(e) Heat of displacement, ΔH ΔH = –54.6 kJ mol–1
Haba penyesaran, ΔH HKCCll((aakq) )++KOHH2O(c(el))

= 22.428 Application of Endothermic and
0.2 Exothermic Reactions in Daily
Life
= –112.14 kJ mol–1 Aplikasi Tindak Balas Endotermik dan
3.3 Eksotermik dalam Kehidupan Harian
(f)
1. (a) Fuel / Bahan api
Energy / Tenaga (g) Some heat is loss to the (b) fuel value / nilai bahan api
environment. (c) the amount of heat energy
Zn(s) + Pb2+(aq) released when 1 g of fuel is
Zn(p) + Pb2+(ak) Sebahagian haba terbebas ke completely burnt in excess
persekitaran. oxygen gas
ΔH = –112.14 kJ mol–1 jumlah tenaga haba yang
Zn2+(aq) + Pb(s) 7. (a) carbon dioxide gas, water dibebaskan ketika 1 g bahan api
Zn2+(ak) + Pb(p) gas karbon dioksida, air terbakar dengan lengkap dalam
gas oksigen yang berlebihan
(b) carbon monoxide gas, carbon,
water.

gas karbon monoksida, karbon, air
(c) Alcohol, alkane, alkene
Alkohol, alkana, alkena

A7 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

2. (a) Molar mass / Jisim molar C4H10 2. (a) Mg(NO3)2(aq) + K2CO3(aq) → Energy / Tenaga

= 58 g mol–1 MgCO3(s) + 2KNO3(aq) MgCO3(s)
Fuel value / Nilai bahan api Mg(NO3)2(ak) + K2CO3(ak) → MgCO3(p)

= 2 880 MgCO3(p) + 2KNO3(ak)
58
(b) Mg2+(aq) + CO32–(aq) →
= 49.66 kJ g–1 ΔH = +50.4 kJ mol–1
MgCO3(s) Mg2+(aq) + CO32–(aq)
(b) Molar mass / Jisim molar CH4 Mg2+(ak) + CO32–(ak) → MgCO3(p) Mg2+(ak) + CO32–(ak)

= 16 g mol–1 (c) Average of initial temperature
Fuel value / Nilai bahan api
of Mg(NO3)2 and K2CO3
= 52 × 16 solutions

= –832 kJ mol–1 Purata suhu awal larutan Mg(NO3)2
dan K2CO3
(c) Molar mass / Jisim molar
C2H5OH = (28.0 + 29.0) 3. (a) Heat change when 1 mole
2
= 46 g mol–1 of water is formed from
Fuel value / Nilai bahan api
= 28.5 oC the neutralization between

1 370 Temperature change, θ / potassium hydroxide solution
46 Perubahan suhu, θ
= and hydrochloric acid.
Perubahan haba apabila 1 mol air
= 29.78 kJ g–1 = 28.5 – 16.5
terbentuk daripada peneutralan
(d) Molar mass / Jisim molar C3H8 = 12 oC antara larutan kalium hidroksida
dengan asid hidroklorik.
= 44 g mol–1 Heat change, Q / Perubahan
Fuel value / Nilai bahan api haba, Q (b) (i) Final thermometer reading
= mcθ
= 51 × 44 is higher than the initial

= –2244 kJ mol–1 = (50 + 50)(4.2)(12) thermometer reading.
Bacaan termometer akhir
= 5040 J
adalah lebih tinggi daripada
SPM Practice 3 (d) Number of moles of Mg(NO3)2 / bacaan termometer awal.
Bilangan mol Mg(NO3)2, (ii) • Heat of neutralisation
Paper 1
n = (2.0)(50) for Experiment II is
1. A 2. C 3. C 4. C 5. B 1 000
6. B 7. D 8. C 9. C 10. A n = 0.1 mol higher / doubled the
11. A 12. C 13. D 14. A 15. B
1 6. A heat of neutralisation

Number of moles of K2CO3 / for Experiment I.
Bilangan mol K2CO3,
Haba peneutralan bagi
Paper 2 n = (2.0)(50) Eksperimen II lebih
1 000 tinggi / dua kali ganda
Section A / Bahagian A n = 0.1 mol haba peneutralan bagi
Eksperimen I.

1. (a) Pb2+(aq) + SO42–(aq) → PbSO4(s) Number of moles of MgCO3 • Sulphuric acid is a
Pb2+(ak) + SO42–(ak) → PbSO4(p) formed / Bilangan mol MgCO3
(b) Heat is released. / Haba yang terbentuk strong diprotic acid

dibebaskan. = Number of moles of Mg2+, that ionises completely
(c) Temperature of the reaction CO32– / Bilangan mol Mg2+, CO32–
in water to produce
increases. / Suhu tindak balas = 0.1 mol
meningkat. a higher / double
(d) – Exothermic / Eksotermik Heat of precipitation of MgCO3
– Heat is released. / Haba / Haba pemendakan MgCO3 the concentration of

dibebaskan. = +50.4 kJ mol–1 hydrogen ions.
(e) Heat change, Q / Perubahan Asid sulfurik ialah asid
(e) Mg(NO3)2(aq) + K2CO3(aq) → diprotik kuat yang mengion
haba, Q MgCO3(s) + 2KNO3(aq) sepenuhnya di dalam
= 50 × 0.5 ΔH = +50.4 kJ mol–1 // air untuk menghasilkan
kepekatan ion hidrogen
= 25 kJ yang lebih tinggi /
berganda.
(f)
Mg(NO3)2(ak) + K2CO3(ak) → • More heat is produced
Energy / Tenaga MgCO3(p) + 2KNO3(ak)
when a higher number
ΔH = +50.4 kJ mol–1 //
of moles of water is

Pb2+(aq) +SOS4O2–(4a2–k()aq) Mg2+(aq) + CO32–(aq) → formed.
Pb2+(ak) + MgCO3(s) Lebih banyak haba
dihasilkan apabila bilangan
ΔH = +50.4 kJ mol–1 mol air yang terbentuk
lebih tinggi.
ΔH = –50 kJ mol–1 Mg2+(ak) + CO32–(ak) → MgCO3(p)
PPbbSSOO4(4p(s) ) (f) ΔH = +50.4 kJ mol–1 (c) • Nitric acid / Asid nitrik

• Nitric acid is a strong

monoprotic acid that ionises

completely in water to form

a higher concentration of

© Penerbitan Pelangi Sdn. Bhd. A8

Chemistry  Form 5  Answers 

hydrogen ions just like T1 lebih tinggi daripada Heat of combustion (kJ mol–1)
29.0 oC. Haba pembakaran (kJ mol–1)
hydrochloric acid.
Asid nitrik ialah asid monoprotik • Zinc is more
kuat yang mengion sepenuhnya
di dalam air untuk membentuk electropositive than 4000
kepekatan ion hidrogen x
yang lebih tinggi seperti asid iron.
hidroklorik. Zink lebih elektropositif 3000

daripada besi. 2000

• The same number of • More heat energy is 1000

moles of water formed as produced at the end of 0

hydrochloric acid. the reaction.
Bilangan mol air yang
sama terbentuk seperti asid Lebih banyak tenaga haba
hidroklorik. dihasilkan pada akhir
tindak balas. 123456
Section B / Bahagian B
4. (a) (i) • Fe(s) + Cu2+(aq) → • T2 remains as 29.0 oC. Number of carbon atoms
T2 masih kekal pada per alcohol molecule
Fe2+(aq) + Cu(s) Bilangan atom karbon
Fe(p) + Cu2+(ak) → Fe2+(ak) 29.0 oC. per molekul alkohol

+ Cu(p) • Silver is less

electropositive than (b) Materials and apparatus /

copper. Bahan dan radas:
Argentum kurang
• Blue copper(II) sulphate Pentane, water, copper can,
elektropositif daripada
solution turns green. kuprum. thermometers, spirit lamp,
Larutan kuprum(II) sulfat
yang berwarna biru • Silver cannot displace windshield, wooden block,
bertukar menjadi hijau.
copper from copper(II) tripod stand, electric weighing

• Concentration of Cu2+ sulphate solution. balance and 50 cm3 measuring

ions decreases. Argentum tidak dapat cylinder
Kepekatan ion Cu2+ menyesarkan kuprum
daripada larutan Pentana, air, tin kuprum,
menurun. kuprum(II) sulfat. termometer, lampu pelita,
penghadang angin, blok kayu,
• Concentration of Fe2+ (ii) • Magnesium / Magnesium tungku kaki tiga, penimbang elektrik
dan silinder penyukat 50 cm3
ions increases. • Blue copper(II)
Kepekatan ion Fe2+
sulphate solution Procedures / Prosedur:
meningkat.
turns lighter blue /
• Brown solid – copper 1. Measure and pour 500 cm3
Pepejal perang – kuprum colourless.
Larutan biru kuprum(II) of water into a copper can
(ii) • Number of moles sulfat bertukar kepada biru
muda / tak berwarna. and leave it for 5 minutes.
of copper formed / Sukat dan tuangkan 500 cm3
Bilangan mol kuprum yang air ke dalam tin kuprum dan
terbentuk • Final thermometer biarkan selama 5 minit.

reading is higher than 2. Place the copper can filled

= 3.2 the initial thermometer with water on a tripod
6.4
reading.
= 0.05 mol Bacaan termometer akhir stand. Then, insert a
lebih tinggi daripada
• Mole ratio between Cu bacaan termometer awal. thermometer into the copper

and CuSO4 • Brown solid is formed. can to measure the initial
Bilangan mol antara Cu Pepejal perang terbentuk.
dengan CuSO4 thermometer reading and
=1:1 Section C / Bahagian C record as T1.
Letakkan tin kuprum berisi air di
= 0.05 : 0.05 5. (a) • Correct scales, labels and atas tungku kaki tiga. Kemudian,
• Heat change, Q / unit in both axes. masukkan termometer ke dalam
tin kuprum untuk menyukat
Perubahan haba, Q Skala, label dan unit yang betul bacaan awal termometer dan
= mcθ pada kedua-dua paksi. catatkan sebagai T1.
= 100 × 4.2 × (57.5 – 28.5)
• Transferring data correctly 3. Fill up a spirit lamp with
= 12180 J
onto graph. pentane and measure its
= 12.18 kJ Pemindahan data yang betul ke initial mass as m1.
Isikan lampu pelita dengan
• Heat of displacement, atas graf. pentana dan timbang jisim
ΔH / Haba penyesaran, awalnya sebagai m1.
ΔH • Shape of the graph.
Bentuk graf.
= 12.18 4. Place the spirit lamp
0.05 • Showing dotted line on the
containing pentane on
graph to determine the heat
= –243.6 kJ mol–1 of combustion for x. a wooden block under
(iii) Concentration / Kepekatan Menunjukkan garis putus-putus
pada graf untuk menentukan the copper can. Put a
haba pembakaran bagi x.
= 0.05 × 1 000 • x = –3350 kJ mol–1 windshield around the
100
copper can and the spirit

= 0.5 mol dm–3 lamp.
(b) (i) • T1 is higher than Letakkan lampu pelita berisi

29.0 oC. pentana di atas blok kayu di

A9 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

bawah tin kuprum. Letakkan (ii) 4. additional, condensation
penghadang angin di sekeliling Energy / Tenaga penambahan, kondensasi
tin kuprum dan lampu pelita. 5. long-chained molecules,
CC55HH1122((cl)e+) +8O82O(g2()g)
5. Light up the wick in the monomers
ΔH = –3528 kJ mol–1 molekul berantai panjang, monomer
spirit lamp to heat water 55CCOO2(2g(g) )++6H62HO2(Og)(g) 6. (a) Glucose / Glukosa
(b) Natural / Semula jadi
in the copper can and stir (c) Protein / Protein
(d) Isoprene / Isoprena
the water in the copper can (e) Natural / Semula jadi
(f) Polyvinyl chloride / Polivinil
throughout the experiment
klorida
until the final thermometer (g) Styrene / Stirena
(h) Synthetic / Sintetik
reading achieves 70 oC. HOTS Challenge 7. (a) Polypropylene
Nyalakan sumbu di dalam
– Mak Minah’s selection – Palm oil Polipropilena
lampu pelita untuk memanaskan Pilihan Mak Minah – Minyak kelapa sawit (b) Food packaging
air di dalam tin kuprum dan – Reason 1 – Palm oil has a higher
kacau air di dalam tin kuprum Pembungkus makanan
sepanjang eksperimen sehingga fuel value that is able to produce (c) Epoxy glue
bacaan termometer akhir more heat when 1 g of palm oil is
mencapai 70 oC. burnt. Gam epoksi
Alasan 1 – Minyak kepala sawit memiliki (d) Waterproof jacket
6. Put out the flame in the nilai bahan api yang lebih tinggi yang
boleh menghasilkan lebih banyak tenaga Jaket kalis air
spirit lamp and measure the apabila 1 g minyak kelapa sawit terbakar.
– Reason 2 – The cost price of palm Natural Rubber
final mass of spirit lamp with oil is cheaper than olive oil. 4.2 Getah Asli
pentane again as m2. Alasan 2 – Harga minyak kelapa sawit
Matikan nyalaan lampu spirit lebih murah berbanding minyak zaitun. 1 2-methybut-1,3-diene
dan timbang jisim akhir lampu 2-metilbut-1,3-diena
spirit bersama pentana sebagai
m2. 2. &CH3
H!&C"C!&C"CH2
Calculation for heat of HH
combustion / Penghitungan haba
pembakaran: 3. Bakteria, negative charges, protein
membrane
– Number of moles of pentane
Bilangan mol petana Bakteria, cas-cas negatif, membran
m1 – m2 protein
= 72
4. (a) methanoic acid / ethanoic acid
– Heat change, Q / Perubahan / formic acid
suhu, Q
= mcθ asid metanoik / asid etanoik / asid
= 500 × c × (70 – T1) PTER formik
– Heat of combustion / Haba CHA
4 Polymer Chemistry (b) ammonia solution
pembakaran larutan ammonia
–[500 × c × (70 – T1)] Kimia Polimer
5. heating the natural rubber with
= 1000 kJ Polymer sulphur, crosslinks
mol–1 (m1 – m2) 4.1 Polimer
pemanasan getah asli dengan sulfur,
72 taut silang

(c) (i) Number of moles of 1. natural, synthetic
semula jadi, sintetik
pentane 2. plants, animals
Bilangan mol petana tumbuh-tumbuhan, haiwan
3. man-made, petroleum
= 1.8 buatan manusia, petroleum
72

= 0.025 mol

Heat change, Q / 6. Vulcanised rubber Properties Unvulcanised rubber
Perubahan suhu, Q Getah tervulkan Sifat-sifat Getah tak tervulkan
= mcθ
= 500 × 4.2 × (70 – 28) Harder Hardness Softer
Lebuh keras Kekerasan Lebih lembut
= 88200 J
More elastic
= 88.2 kJ Lebih kenyal

– Heat of combustion / More resistant to heat Elasticity Less elastic
Haba pembakaran Lebih tahan haba Kekenyalan Kurang kenyal

= 88.2 More resistant to Resistance to heat Less resistant to heat
0.025 oxidation Daya tahan haba Kurang tahan haba
Lebih tahan terhadap
= –3528 kJ mol–1 pengoksidaan Resistance to oxidation Less resistant to
Daya tahan pengoksidaan oxidation
Kurang tahan terhadap
pengoksidaan

© Penerbitan Pelangi Sdn. Bhd. A10

Synthetic Rubber yang paling merosakkan Chemistry  Form 5  Answers 
4.3 Getah Sintetik alam sekitar.
(ii) – When thermoplastic is
1. a type of artificial elastomer which – Disposal of PVC heated, polymer chains
is synthesised from petroleum slide over one another
byproducts releases toxic causing the plastics to
soften and melt.
sejenis elastomer tiruan yang disintesis chemicals and chlorine
daripada hasil sampingan petroleum Apabila termoplastik
to the environment. dipanaskan, rantai
2. thermal stability, resistance to oils Pembuangan PVC polimer menggelongsor
kestabilan terma, ketahanan terhadap antara satu sama lain
membebaskan bahan dan menyebabkan plastik
minyak kimia beracun dan klorin menjadi lembut dan cair.
3. styrene, buta-1,3-diene kepada alam sekitar.
stirena, buta-1,3-diena – When thermoset is
4. (a) Elastic / Kenyal 2. (a) Petroleum / Petroleum heated, the crosslinks
(b) (i) – Plastic A does not prevent the polymer
(b) Less elastic / Kurang kenyal chains from sliding
(c) Less / Kurang have a crosslink over each other. This
(d) More / Lebih will prevent the plastic
(e) Easily / Mudah between polymer from melting.
(f) Not easily / Tidak mudah
chains. Apabila termoplastik
SPM Practice 4 Plastik A tidak mempunyai dipanaskan, taut silang
menghalang rantai
Paper 1 taut silang di antara rantai polimer daripada saling
polimer. menggelongsor antara
1. C 2. A 3. A 4. B 5. B – Plastic B has crosslinks satu sama lain. Hal ini
6. C 7. C 8. C 9. A 10. D dapat mengelakkan plastik
11. D 12. B 13. D 14. B 15. B between polymer daripada menjadi cair.
16. C 17. D 18. D
chains.
Paper 2 Plastik B mempunyai taut
silang di antara rantai
Section A / Bahagian A polimer.

1. (a) (i) Both undergo additional (c) (i) Characteristic Thermoplastic Thermoset
polymerisation. Ciri-ciri Termoplastik Termoset

Kedua-duanya menjalani Hardness Soft Hard
pempolimeran penambahan. Kekerasan Lembut Keras
Ability to be Can only be moulded
(ii) – Monomer A: moulded Can be moulded once
2-methybuta-1,3-diene Keupayaan untuk repeatedly Hanya boleh dibentuk
dibentuk Boleh dibentuk berulang sekali sahaja
Monomer A: 2-metilbuta- kali
1,3-diena Do not melt when
Effect on heat Melt when heated and heated
– Monomer B: Kesan terhadap haba harden again when Tidak cair apabila
chloroethene / vinyl cooled dipanaskan
chloride Cair apabila dipanaskan
dan mengeras semula
Monomer B: kloroetena / apabila disejukkan
vinil klorida
(ii) – Plastic A: (ii) Polyethene / Polietena
(iii) – Natural polymer: Thermoplastic
natural rubber – Recycle the polyethene
Plastik A: Termoplastik
Polimer semula jadi: getah – Plastic B: Thermoset by heating, melting and
asli Plastik B: Termoset
remoulding it into a
– Synthetic polymer: Section B / Bahagian B
polychloroethane / 3. (a) (i) Polyethene / Polietena new product.
polyvinyl chloride – Type: thermoplastic Kitar semula polietena
Jenis: termoplastik dengan memanaskan,
Polimer sintetik: – Monomer: ethene mencairkan dan
polikloroetana / polivinil Monomer: etena membentuknya menjadi
klorida – Use: plastic bag produk baru.
Kegunaan: beg plastik
(b) (i) – As a water pipe / – Reuse the polyethene.
Sebagai paip air SBR Guna semula polietena.
– Type: synthetic rubber
– As an electrical wire Jenis: getah sintetik SBR
case / Sebagai kotak – Monomer: styrene and – Donate to Fisheries
wayar elektrik
butadiene Department as building
(ii) – Polychloroethane / Monomer: stirena dan blocks for artificial
Polyvinyl chloride reefs.
(PVC) is the most butadiena Sumbangkan kepada
environmental – Use: synthetic tyres Jabatan Perikanan
damaging polymer. Kegunaan: tayar sintetik sebagai binaan tukun
tiruan.
Polikloroetana / polivinil
klorida (PVC) ialah polimer – Sell back to tyre
manufacturers for
recycling.

A11 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

Menjual semula kepada diproses semula menjadi – Repulsive force prevents
pengeluar tayar untuk tin aluminium baru.
dikitar semula. the collision among rubber
Reuse / Guna semula
(b) (i) – Plastics substitute – Reuse the plastic bags particles.
Daya tolakan menghalang
woods and papers for shopping. perlanggaran antara zarah
Gunakan semula beg getah.
to prevent a massive
plastik untuk membeli-
deforestation. belah. – Protein membranes do not
Plastik menggantikan
Reduce / Kurangkan break, rubber polymers are
kayu dan kertas untuk • Reduce the usage of
mengelakkan penebangan not released, no coagulation
hutan secara berleluasa. seedling plastic bags
and substitute with occurs and white liquid
biodegradable seedling
– Plastics substitute bags. remains white.
Kurangkan penggunaan Membran protein tidak pecah,
metals to conserve beg plastik semaian dan molekul getah tidak terbebas,
gantikan dengan beg penggumpalan tidak berlaku dan
natural resources from semaian terbiodegradasi. cecair putih kekal putih.

the core of the Earth. (b) Material / Bahan:
Plastik menggantikan
– 5 cm rubber strip, disulphur
logam untuk memulihara
sumber semula jadi dari dichloride in methylbenzene
teras Bumi.
Jalur getah 5 cm, disulfur
– Plastics substitute Section C / Bahagian C diklorida dalam metilbenzena

some parts of the Apparatus / Radas:

vehicles to make them 4. (a) Experiment I / Eksperimen I – Bulldog clips, beakers,
– Ethanoic acid contains
lighter and improve fuel retort stand and clamp, 50 g
hydrogen ions.
efficiency. Asid etanoik mengandungi ion weight and meter ruler
Plastik menggantikan Klip bulldog, bikar, kaki retort
hidrogen. dan pengapit, pemberat 50 g
beberapa bahagian pada dan pembaris meter
kenderaan supaya lebih – Hydrogen ions neutralise
ringan dan memperbaiki negative charges on the
kecekapan bahan api. protein membrane surface Procedure / Prosedur:
(ii) – Clogged drains causing
of rubber particles. 1. Dip a 5 cm of rubber strip
flash flood. Ion hidrogen meneutralkan
Longkang yang tersumbat cas negatif pada permukaan into disulphur dichloride
membran protein zarah getah.
menyebabkan banjir kilat. in methylbenzene for 5

– Marine animals like – Neutral rubber particles are minutes to produce a

turtles mistake the formed. vulcanised rubber.
Zarah getah yang neutral Celupkan jalur getah 5 cm ke
plastics for food dalam disulfur diklorida dalam
terbentuk. metilbenzena selama 5 minit
causing them to die of untuk menghasilkan getah
tervulkan.
food poisoning. – When neutral particles
Haiwan laut seperti
penyu menganggap collide among each other, 2. Hang the vulcanised rubber
plastik sebagai makanan protein membranes break.
dan menyebabkannya Apabila zarah neutral strip using a retort stand
mati akibat keracunan berlanggaran antara satu sama
makanan. lain, membran protein pecah. and a clamp.
Gantung jalur getah tervulkan
– Clogged soil prevents – Rubber polymers are dengan menggunakan kaki
retort dan pengapit.
the free flow of water released, entangled and
coagulated to form a white 3. Measure and record the
in soil, thus depleting
solid. initial length of the rubber
the soil fertility. Polimer getah terbebas,
Tanah yang tersumbat bersimpul dan bergumpal untuk strip.
mencegah pengaliran membentuk pepejal putih. Ukur dan catatkan panjang awal
air yang lancar di dalam
tanah dan mengakibatkan jalur getah itu.
kesuburan tanah
berkurang. Experiment II / Eksperimen II 4. Add 50 g weight to the

– Ammonia solution contains rubber strip and leave it for

– Production of toxic hydroxide ions. 3 hours.
Larutan ammonia mengandungi Tambahkan pemberat 50 g pada
gases during open jalur getah dan biarkan selama
ion hidroksida. 3 jam.
burning.
– Hydroxide ions neutralise 5. Remove the 50 g weight.
Penghasilan gas toksik
semasa pembakaran hydrogen ions produced by Then, measure and record
terbuka.
bacteria. the final length of the rubber
(iii) Recycle / Kitar semula Ion hidroksida meneutralkan ion
hidrogen yang dihasilkan oleh strip.
bakteria.
– Return the aluminium Alihkan pemberat 50 g
kemudian ukur dan catatkan
cans to manufacturers – Negative charges remain panjang akhir jalur getah.

to reprocess into new on the protein membrane 6. Repeat step 1-5 with

aluminium cans. surface. unvulcanised rubber.
Kembalikan tin aluminium Cas negatif kekal pada
Ulangi langkah 1-5 menggunakan
kepada pengilang untuk permukaan membran protein. getah tak tervulkan.

© Penerbitan Pelangi Sdn. Bhd. A12

Tabulation of data / Penjadualan data: Chemistry  Form 5  Answers 

Type of rubber strip Initial length of rubber Final length of rubber Length of rubber strip after
Jenis jalur getah strip (cm) strip (cm) removal of 50 g weight
Panjang jalur getah selepas
Panjang awal jalur getah Panjang akhir jalur getah pemberat 50 g dialihkan
(cm) (cm) 5.2
10.2 5.0
Unvulcanised rubber 5.0
Getah tak tervulkan 10.0

Vulcanised rubber 5.0
Getah tervulkan

HOTS Challenge (d) Hypertension / Tekanan darah sayur / zaitun, natrium hidroksida /
tinggi kalium hidroksida
– Dip the latex gloves into disulphur
dichloride in methylbenzene. 8. renewable, biodegradable, non-toxic (b) heat / panaskan
diperbaharui, boleh terbiodegradasi, (c) room temperature / suhu bilik
Celupkan sarung tangan getah ke dalam (d) sodium chloride / natrium klorida
disulfur diklorida dalam metilbenzena. tidak beracun (e) soap / sabun
3. (a) petroleum fractions / pecahan
– Vulcanisation of latex gloves will occur. Cleaning Agents
Pemvulkanan sarung tangan lateks akan 5.2 Bahan Pencuci petroleum

berlaku. 1. (a) long-chain fatty acid, ROO-K+ (b) (i) Sodium alkyl sulphate
– Texture of the latex gloves will asid lemak berantai panjang,
ROO–K+ detergent
become stronger and more elastic. Detergen natrium alkil sulfat
Tekstur sarung tangan getah akan (b) alkyl / alkil
(c) animal fats, vegetable oils (ii) Sodium alkylbenzene
menjadi lebih kuat dan lebih kenyal. lemak haiwan, minyak sayuran
(d) saponification / saponifikasi sulphonate detergent
CHA 5PTER Consumer and 2. (a) vegetable / olive, sodium Detergen natrium
Industrial Chemistry
hydroxide / potassium hydroxide alkilbenzena sulfonat
Kimia Konsumer dan 4. (a) (i) Hydrophobic / Hidrofobik
Industri (ii) Hydrophilic / Hidrofilik
(b) (i) Hydrophobic / Hidrofobik
(ii) Hydrophilic / Hidrofilik

Oils and Fats 5. Soap Comparison Detergent
5.1 Minyak dan Lemak Sabun Perbandingan Detergen
Petroleum fraction
1. carbon, hydrogen, oxygen Plants and animals Source Pecahan petroleum
karbon, hidrogen, oksigen Tumbuhan dan haiwan Sumber Effective
Not effective Effectiveness in hard water Berkesan
2. carbon-carbon double bond Tidak berkesan Keberkesanan dalam air liat Non-biodegradable
ikatan karbon ganda dua Biodegradable Biodegradability Tidak terbiodegradasi
3. unsaturated, carbon-carbon double Terbiodegradasi Kebolehan terbiodegradasi

bond Food Additives
tak tepu, ikatan karbon ganda dua 5.3 Bahan Tambah Makanan
4. solid, liquid
pepejal, cecair

5. Hydrogenation, unsaturated, 1. synthetic, spoilage, appearance / sintetik, kerosakan, rupa
saturated
2. Preservatives Prevent the oxidation that causes rancidity of oils and
penghidrogenan, tak tepu, tepu Pengawet fats and fruit browning
Menghalang pengoksidaan yang menyebabkan minyak dan
6. (a) Energy supply to the body lemak berbau tengik dan buah bertukar menjadi warna perang
Bekalan tenaga kepada badan

(b) Supports cell growth Antioxidants Prevent an emulsion from separating out
Menyokong pertumbuhan sel Pengantioksida Mengelakkan pemisahan emulsi

(c) Protects internal organs
Melindungi organ dalaman

(d) Keeps the body warm Flavours Stabilise an emulsion
Memanaskan badan Perisa Menstabilkan emulsi

(e) Absorbs of nutrients and Stabilisers Improve or restore the taste loss due to processing
Penstabil Memperbaiki atau memulihkan rasa yang hilang disebabkan
vitamins A,D,E, K oleh pemprosesan
Menyerap nutrien dan vitamin A, D,
E, K Dyes Improve or restore the colour loss due to processing
Pewarna Memperbaiki atau memulihkan warna yang hilang disebabkan
(f) Produces hormones oleh pemprosesan
Menghasilkan hormon
7. (a) Heart diseases / Penyakit jantung Thickeners Slow down the bacterial growth
(b) Stroke / Strok Pemekat Melambatkan pertumbuhan bakteria

(c) Weight gain / Obesity Emulsifiers Thicken the food texture
Pertambahan berat badan / Pengemulsi Memekatkan tekstur makanan
Kegemukan

A13 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers Application of Green Technology in Industrial Waste
Manangement
Medicines and Cosmetics 5.6 Aplikasi Teknologi Hijau dalam Pengurusan Sisa Industri
5.4 Ubat-ubatan dan Bahan Kosmetik
1. Green Technology is defined as the development and
1. (a) pain / kesakitan application of products, equipment and systems to
(b) bacterial growth / pertumbuhan bakteria preserve the environment and nature as well as to
(c) thinking, behaviours / pemikiran, tingkah laku minimise the negative effects of human activities.
(d) allergic effects / kesan alergi
(e) inflammation / keradangan Teknologi Hijau didefinisikan sebagai pembangunan dan
aplikasi produk, peralatan serta sistem untuk memelihara alam
2. sekitar dan alam semula jadi serta meminimumkan kesan
Treat skin problems or wounds negatif akibat aktiviti manusia.
Merawat masalah kulit atau luka
2. human waste, soap
Treat bloated stomach sisa manusia, sabun
Merawat perut kembung
3. Reduce, Reuse and Recycle, Restore
Reduce high blood pressure Kurangkan, Guna semula, Kitar semula dan Pulihkan
Mengurangkan tekanan darah tinggi
4. Type of wastewater Example
Jenis air sisa Contoh

Domestic wastewater
Air sisa domestik

Treat inflammation
Merawat keradangan

Build up stamina and reduce Industrial wastewater
fatigue Air sisa daripada
Membina stamina dan perindustrian
mengurangkan keletihan

3. Example Type of Function Stormwater runoff
Contoh cosmetic Fungsi Air larian ribut
Jenis kosmetik
Products to SPM Practice 5
Make-up treats body or
cosmetic face Paper 1
Kosmetik rias Produk untuk
merawat tubuh 1. B 2. A 3. B 4. C 5. C
Fragrance atau muka 6. D 7. A 8. C 9. C 10. B
Pewangi 11. D 12. C 13. C 14. B 15. C
Products to 16. C 17. A 18. B 19. A 20. B
Treatment remove body 21. A
cosmetic odour.
Kosmetik Produk untuk Paper 2
perawatan mewangikan
badan. Section A / Bahagian A

Products to 1. (a) (i) – Soft water: tap water
beautify the Air lembut: air paip
face. – Hard water: well water
Produk hiasan Air liat: air perigi
untuk wajah. (ii) A – Soap is an effective cleansing agent in

Application of Nanotechnology in Industry soft water.
5.5 Aplikasi Nanoteknologi dalam Industri Sabun ialah agen pembersih yang berkesan

1. 1-100 / 1-100 dalam air lembut.
2. very small scale B – Soap is not an effective cleansing agent

skala yang sangat kecil in hard water.
Sabun bukan agen pembersih yang berkesan
3. carbon, two-dimensional honeycomb lattice
karbon, kekisi sarang lebah dua dimensi dalam air liat.
(b) – Hydrophobic part of soap dissolves in the oil
4. (a) ✓
(b) ✓ while hydrophilic part of soap dissolves in water.
(c) ✓ Bahagian hidrofobik sabun larut di dalam minyak

manakala bahagian hidrofilik sabun larut di dalam air.

© Penerbitan Pelangi Sdn. Bhd. A14

Chemistry  Form 5  Answers 

– Agitation breaks the oil into – The hydrophilic part of (b) (i) – P: Preservative /

smaller oil droplets. soap dissolves in water Pengawet
Kocakan memecahkan minyak Q: Preservative /
menjadi titisan minyak yang / The hydrophilic part
lebih kecil.
of soap is attracted to Pengawet
R: Preservative /
– Oil droplets are suspended water molecules.
Bahagian hidrofilik Pengawet
and rinsed off easily. sabun larut di dalam air / – P is sugar. / P ialah gula.
Titisan minyak diapungkan dan Bahagian hidrofilik sabun
tertarik kepada molekul – that draws
dibilas dengan mudah. air.
2. (a) Sunburn / Selaran matahari water out of the
(b) (i) Aloe vera / Lidah buaya
– The hydrophobic part microorganisms’ cells

(ii) Leave / Daun of soap dissolves in to retard the growth of

(iii) – As a moisturiser / grease. microorganisms.
Sebagai pelembap Bahagian hidrofobik sabun yang menyingkirkan

– Promotes hair growth larut di dalam gris. air daripada sel
mikroorganisma untuk
/ Menggalakkan – Mechanical agitation merencatkan pertumbuhan
mikroorganisma.
pertumbuhan rambut helps to pull the grease – Q is vinegar. / Q ialah
cuka.
(c) – Make use of natural free and
products / Memanfaatkan Kocakan mekanikal

produk semula jadi membantu menarik keluar – which is acidic and
gris dan
Has less side effects / inhibits the growth of
Mempunyai kurang kesan – break the grease into
sampingan microorganisms.
smaller droplets. yang bersifat asid dan
3. (a) (i) Substance of the same memecahkan minyak
menjadi titisan yang lebih menghalang pertumbuhan
element that exists in kecil. mikroorganisma.
– R is sodium nitrite /
more than one form of – The grease droplets
nitrate
arrangements. do not coagulate R ialah natrium nitrik /
Bahan bagi unsur yang
and redeposit on nitrat
sama wujud dalam keadaan
lebih daripada satu bentuk the surface of the – that slows down
susunan.
handkerchief due to the growth of

(ii) Diamond and graphite / the repulsion between microorganisms.
Berlian dan grafit yang melambatkan
negative charges on pertumbuhan
mikroorganisma.
(b) (i) – To produce the surface.
Titisan minyak tidak
semiconductor bergumpal dan terkumpul (ii) May cause cancer /
Untuk menghasilkan pada permukaan sapu
tangan kerana tolakan asthma / allergies /
semikonduktor antara cas negatif pada
permukaannya.
– To produce graphene- hyperactive
Boleh menyebabkan kanser /
infused carbon fibre asma / alahan / hiperaktif

helmet – These droplets [any one answer / mana-
Untuk menghasilkan mana satu jawapan]
are suspended in
helmet gentian karbon
grafen water forming an

(ii) – Best electrical emulsion. Section C / Bahagian C
conductor / Konduktor Titisan ini terapung dalam
elektrik terbaik 5. (a) (i) Plastic pollution /
air membentuk emulsi. Pencemaran plastik

– Thin but strong / Nipis – Rinse away these (ii) – Formation of breeding
tetapi kuat sites for Aedes
droplets and the mosquitoes

[any one answer / mana- handkerchief is clean. Pembentukan tapak
Bilaskan titisan ini dan pembiakan nyamuk Aedes
mana satu jawapan]
sapu tangan menjadi – Open burning of the
Section B / Bahagian B bersih. plastic bottles produces
toxic gases
(ii) – Soap is biodegradable.
Pembakaran botol
4. (a) (i) – When the soap / Soap can be plastik secara terbuka
menghasilkan gas-gas
molecules dissolve in decomposed naturally beracun

water, by bacteria or (iii) – Reduce / Kurangkan
Apabila molekul sabun – Reduce the purchase
microorganisms.
larut dalam air Sabun terbiodegradasi. of materials / foods
/ drinks in plastic
– the surface tension / Sabun boleh terurai bottles / containers
secara semula jadi / packagings.
of water reduces / oleh bakteria atau Consumers are
mikroorganisma. encouraged to bring
the ability of water to

wet the handkerchief – Soap does not cause

increases. pollution. / Soap is
ketegangan permukaan air
berkurang / keupayaan air environmental friendly.
membasahi sapu tangan Sabun tidak menyebabkan
meningkat. pencemaran. / Sabun
bersifat mesra alam.

A15 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

their own bottles sabun, minyak dan bahan SPM Model Paper
kimia.
/ containers when (ii) – Domestic wastewater
Air sisa domestik
they are going to the – Industrial wastewater Paper 1

restaurants / shops. Air sisa daripada 1. C 2. A 3. B 4. C 5. A
Kurangkan pembelian perindustrian 6. A 7. A 8. D 9. A 10. A
– Stormwater runoff 1 1. D 12. D 13. B 14. D 15. D
bahan / makanan / Air larian ribut 1 6. D 17. D 18. B 19. D 20. C
minuman di dalam (iii) – Produces alternative 2 1. B 22. D 23. B 24. A 25. C
botol plastik / bekas / energy through 2 6. D 27. B 28. C 29. B 30. B
pembungkus. Pengguna methane gas collection 3 1. B 32. C 33. B 34. B 35. A
digalakkan membawa Menghasilkan tenaga 3 6. A 37. D 38. C 39. B 40. C
botol / bekas sendiri alternatif melalui
semasa berkunjung ke pengumpulan gas metana
restoran / kedai. – Produces clean and
reusable water
– Reuse / Guna semula Menghasilkan air Paper 2
yang bersih dan boleh
Consumers can use digunakan semula Section A / Bahagian A
– Produces natural
plastic containers to fertilisers
Menghasilkan baja semula
keep snack / plastic jadi 1. (a) (i) Pure metal / Logam tulen: J
(iv) – Increases in water- Alloy / Aloi: K
bottles for sauce / oil / borne diseases like (ii) Substance J is ductile.
cholera or typhoid Bahan J bersifat mulur.
plastic bag as dustbin fever.
Peningkatan penyakit
liners. bawaan air seperti kolera (b) – The presence of zinc atoms
Pengguna boleh atau demam kepialu.
– Water source in different sizes disrupt
menggunakan bekas contamination by
plastik untuk menyimpan E. coli bacteria that the orderly arrangement of
makanan ringan / botol causes diarrhea /
plastik untuk kuah / cramping stomach pain copper atoms.
minyak / beg plastik / vomiting. Kehadiran atom zink yang
sampah. Pencemaran sumber air
oleh bakteria E. coli yang berbeza saiz mengganggu
– Recycle / Kitar semula menyebabkan cirit-birit / susunan atom kuprum yang
sakit perut / muntah. teratur.
Consumers are
HOTS Challenge – Minimise the layers of
encouraged to return
– Mineral supplement contains nano- copper atoms from sliding.
/ resell used plastic size calcium ions and magnesium Meminimumkan lapisan
ions.
bottles / containers atom kuprum daripada
Mineral tambahan mengandungi ion menggelongsor.
to manufacturers to kalsium dan ion magnesium bersaiz nano.

reprocess into new – This allows a more efficient calcium 2. (a) Exothermic reaction / Tindak
ions and magnesium ions absorption balas eksotermik
products. into the bloodstream due to their
Pengguna digalakkan nano-size. (b) Zn(s) + CuSO4(aq) →
ZnSO4(aq) + Cu(s)
untuk mengembalikan Membenarkan penyerapan ion kalsium
/ menjual kembali botol dan ion magnesium dengan lebih Zn(p) + CuSO4(ak) → ZnSO4(ak)
/ bekas plastik terpakai berkesan ke dalam saluran darah kerana + Cu(p)
kepada pengeluar untuk bersaiz nano.
diproses semula menjadi (c) (i) Number of moles /
produk baru. – Enrichment of magnesium helps Bilangan mol
building up stronger bones for a
– Restore / Pulihkan better absorption of calcium ions. = 0.2 × 50
1 000
Plastic manufacturers Kaya dengan magnesium membantu
membina tulang yang lebih kuat untuk
restore the used plastic penyerapan ion kalsium yang lebih baik. = 0.01 mol
(ii) Heat released / Haba
bottles / containers
terbebas
by converting them

into fuels / industrial = 50 × 4.2 × (45.0 – 28.0)

products instead of = 3570 J

piling them up on = 3.57 kJ

landfills. (iii) Heat of displacement of
Pengilang plastik
copper
menukarkan botol / Haba penyesaran kuprum
bekas plastik terpakai
kepada bahan bakar / = 3.57
produk industri daripada 0.01
mengakibatkan timbunan
di tempat pembuangan = 3.57 kJ mol–1
sampah.
(d) – Heat of displacement is

(b) (i) Wastewater is water that higher than –357 kJ mol–1.
Haba sesaran lebih tinggi
has been used including
daripada –357 kJ mol–1.

human wastes, food – Magnesium is more

wastes, soaps, oils and electropositive than zinc.
Magnesium lebih elektropositif
chemical wastes.
Air sisa ialah air yang telah daripada zink.
digunakan termasuklah sisa
manusia, sisa makanan, 3. (a) Aluminium / Aluminium
(b) (i) R

© Penerbitan Pelangi Sdn. Bhd. A16

Chemistry  Form 5  Answers 

(ii) – Has more than one = 24.263 Reducing agent / Agen
6. (a)
oxidation number penurunan: Iron(II) ion / Ion
Mempunyai lebih Volume of gas (cm3) ferum(II)
daripada satu nombor Isi padu gas (cm3)
pengoksidaan (c) 2Fe2+ + Br2 → 2Fe3+ + 2Br–
8. (a) (i) Cell A / Sel A: Voltaic cell /
– Can form a coloured Set II / Set II
Set I / Set I Sel kimia
ion / compound 50.00 Cell B / Sel B: Electrolytic
Boleh membentuk ion / 40.00
cell / Sel elektrolisis
sebatian berwarna
(ii) Electrical energy to
– Can form a complex
chemical energy
ion / compound Time (minute) Tenaga elektrik kepada
Boleh membentuk ion / 3 Masa (minit) (b) (i) tenaga kimia
Cell A – Blue copper(II)
sebatian kompleks (b) (i) Set I / Set I

– Can act as a catalyst = 40 sulphate solution turns
Boleh bertindak sebagai 3
mangkin light blue / colorless.
Sel A – Larutan biru
[any one answer / mana-mana = 13.33 cm3 min–1
(ii) Set II / Set II kuprum(II) sulfat bertukar
satu jawapan] menjadi biru muda / tak
berwarna.
(c) 2.8.1 = 50 Cell B – Blue copper(II)
(d) Q– 3
(e) 2Y(s) + Q2(g) → 2YQ(s)
= 16.67 cm3 min–1 sulphate solution remains
2Y(p) + Q2(g) → 2YQ(p)
(iii) Rate of reaction for Set II blue.
4. (a) (i) 2.8.1 Sel B – Larutan biru
is higher than Set I.
(ii) Group 1, Period 3 Kadar tindak balas bagi Set II kuprum(II) sulfat kekal biru.
Kumpulan 1, Kala 3 lebih tinggi daripada Set I. (ii) Cell A – Concentration of

(iii) – Group 1 because it has (iv) – Catalyst provides an Cu2+ ion decreases.
Sel A – Kepekatan ion Cu2+
one valence electron. alternative pathway
Kumpulan 1 kerana berkurang.
mempunyai satu elektron with lower activation Cell B – Concentration
valens.
energy. of Cu2+ ion remains the
Mangkin menyediakan
– Period 3 because it same.
lintasan alternatif dengan Sel B – Kepekatan ion Cu2+
has three electron-filled tenaga pengaktifan yang
lebih rendah. kekal sama.
shells.
Kala 3 kerana mempunyai – Frequency of collision (c) (i) Replace zinc with

tiga petala berisi elektron. between zinc atoms magnesium / aluminium.
Gantikan zink menjadi
(b) (i) 24 2+ and hydrogen ions is
(ii) 2.8 magnesium / aluminium.
higher.
(iii) Frekuensi perlanggaran (ii) Magnesium / aluminium is

antara atom zink dengan more electropositive than
ion hidrogen lebih tinggi.
P zinc.
Magnesium / aluminium lebih
– Frequency of effective
elektropositif daripada zink.
collision between zinc

atoms and hydrogen Section B / Bahagian B

5. (a) 3 ions is higher. 9. (a) 2H+(aq) + Fe(s) → Fe2+(aq) +
(b) Natural abundance is the Frekuensi perlanggaran H2(g)

berkesan antara atom zink 2H+(ak) + Fe(p) → Fe2+(ak) + H2(g)
dengan ion hidrogen lebih
percentage of isotopes tinggi.

presents in a natural sample of 7. (a) (i) Anode: Carbon that (b) Experiment 1 – When

an element. is immersed in iron(II) powdered iron is used, total

Kelimpahan semula jadi ialah sulphate solution. surface area exposed to
peratusan isotop yang hadir dalam Anod: Karbon yang dicelup di
suatu sampel semula jadi unsur. dalam larutan ferum(II) sulfat. hydrogen ions is larger.
Eksperimen 1 – Apabila serbuk
(c) X24 – 82.8%, ferum digunakan, jumlah luas
permukaan yang terdedah kepada
12 Cathode: Carbon that

X54 – 8.1%, is immersed in bromine ion hidrogen lebih besar.

12 water.
Katod: Karbon yang dicelup
X26 – 9.1% di dalam air bromin. Experiment 2 – When HNO3
X26 is heated to 50 °C, the kinetic
12
(d) 2142X, 2152X, 12

(e) The relative atomic mass of (ii) Anode / Anod: energy of hydrogen ion is
element X
Fe2+ → Fe3+ + e– higher.
Jisim atom relatif unsur X Eksperimen 2 – Apabila HNO3
Cathode / Katod: dipanaskan sehingga 50 °C,
tenaga kinetik ion hidrogen lebih
[(24 × 82.8) + (25 × 8.1) + Br2 + 2e– → 2Br– tinggi.
(b) Oxidising agent / Agen
= (26 × 9.1)]
1 000 pengoksidaan: Bromine water /

Air bromin

A17 © Penerbitan Pelangi Sdn. Bhd.

  Chemistry  Form 5  Answers

Experiment 3 – When CuSO4 and hydrogen ions is (ii) 1. Soak a glass wool
is added as a catalyst, an higher.
Frekuensi perlanggaran in ethanol and insert
antara atom ferum dengan
alternative pathway with lower ion hidrogen lebih tinggi. it into the end of a
– Frequency of effective
activation energy is provided. collision between iron combustion tube.
atoms and hydrogen Rendamkan wul kaca
Eksperimen 3 – Apabila CuSO4 ions is higher. dalam etanol dan
ditambahkan sebagai mangkin, Frekuensi perlanggaran masukkan ke dalam
lintasan alternatif dengan tenaga berkesan antara atom hujung tiub pembakaran.
pengaktifan yang lebih rendah ferum dengan ion hidrogen
disediakan. lebih tinggi. 2. Clamp the combustion

(c) (i) In Experiment 2 / Dalam (d) tube horizontally using
Eksperimen 2,
VoIlsuimpaeduofHH2 (2c(mc3m) 3) a retort stand and

– Rate of reaction is clamp.
Pasangkan tiub
higher.
Kadar tindak balas lebih pembakaran secara
mendatar dengan
tinggi. menggunakan kaki retort
dan pengapit.
– Temperature of nitric

acid, HNO3 is higher. 3. Put aluminium oxide
Suhu asid nitrik, HNO3
in the centre of the
lebih tinggi.
2 combustion tube and
– Kinetic energy of 1
close the combustion
hydrogen ion is higher. VoIlsuimpaeduofHH2 (2c(mc3m) 3)
Tenaga kinetik ion tube with a rubber

hidrogen lebih tinggi. Time taken (s) stopper connected to a
Masa yang
– Frequency of collision diambil (s) delivery tube.
Letakkan aluminium
between iron atoms
oksida di dalam bahagian
and hydrogen ions is tengah tiub pembakaran
dan tutup tiub pembakaran
higher. dengan penyumbat getah
Frekuensi perlanggaran yang disambungkan ke
salur penghantar.
antara atom ferum dengan
ion hidrogen lebih tinggi.

– Frequency of effective 4. Insert the other end of

collision between iron 3 the delivery tube into
2
atoms and hydrogen an inverted test tube.
Masukkan hujung salur
ions is higher. penghantar yang lain ke
Frekuensi perlanggaran dalam tabung uji yang
Time taken (s) diterbalikkan.
berkesan antara atom Masa yang
ferum dengan ion hidrogen diambil (s)
lebih tinggi.
5. Heat the aluminium
(ii) In Experiment 3 / Dalam
Eksperimen 3, 10. (a) (i) – Process I is oxide strongly until
fermentation.
– Rate of reaction is brick red and gently
Proses I ialah penapaian.
higher. – Process II is heat the glass wool
Kadar tindak balas lebih
dehydration. with ethanol.
tinggi. Proses II ialah Panaskan aluminium

– Copper(II) sulphate pendehidratan. oksida dengan kuat
– Chemical equation for sehingga merah bata
solution acts as a dan panaskan wul kaca
process I: dengan etanol secara
catalyst. Persamaan kimia untuk perlahan.
Larutan kuprum(II)
proses I: 6. Collect the colourless
sulfat bertindak sebagai
mangkin. C6H12O6(l) yeast C2H5OH(l) + 2H2O(l) bubbles via inverted

– Catalyst provides an C6H12O6(ce) yis C2H5OH(ce) + 2H2O(ce) water displacement.
Kumpulkan gelembung-
alternative pathway – Chemical equation for
process II: gelembung tak berwarna
with lower activation melalui penyesaran air
Persamaan kimia untuk berbalik.
energy. proses II:
Mangkin menyediakan 7. Add a few drops of
heated
lintasan alternatif yang C2H5OH(l) Al2O3 CH2=CH2(g) + H2O(g) bromine water into the
mempunyai tenaga
pengaktifan yang lebih Al2O3 test tube containing the
rendah. C2H5OH(ce) dipanaskan CH2=CH2(g) + H2O(g)
colorless gas.
– More iron atoms and
Tambahkan beberapa
hydrogen ions achieve titis air bromin ke
dalam tabung uji yang
the activation energy. mengandungi gas tak
Lebih banyak atom berwarna.
ferum dan ion hidrogen
mencapai tenaga 8. Brown bromine water
pengaktifan.
that turns colourless
– Frequency of collision
indicates the presence
between iron atoms
of ethene gas.

© Penerbitan Pelangi Sdn. Bhd. A18

Chemistry  Form 5  Answers 

Air bromin berwarna – Ba2+(aq) + CO32–(aq) → Tambahkan 1 cm3 larutan
perang yang bertukar BaCO3(s) argentum nitrat ke dalam
menjadi tak berwarna kedua-dua tabung uji.
menunjukkan kehadiran Ba2+(ak) + CO32–(ak) → BaCO3(p)
gas etena. 4. White precipitate is
– Barium ion reacts with
(b) (i) – Bacteria / formed in the test tube
sulphate ion to form white
microorganisms containing sodium
precipitate, barium sulphate.
present in the air. Ion barium bertindak balas chloride solution.
Bakteria / mikroorganisma Mendakan putih terbentuk
dengan ion sulfat untuk
hadir dalam udara. membentuk mendakan putih, di dalam tabung uji yang
barium sulfat. berisi larutan natirum
– Bacteria / – Ba2+(aq) + SO42–(aq) → klorida.

microorganisms in the BaSO4(s) 5. Yellow precipitate is
Ba2+(ak) + SO42–(ak) → BaSO4(p)
air produce hydrogen formed in the test tube

ions that neutralise containing sodium

the negative charges Changes that should be done, iodide solution.
Perubahan yang harus dilakukan, Mendakan kuning
around the protein
terbentuk di dalam tabung
membranes of rubber 1. Measure and pour 1 cm3 uji yang berisi larutan
natrium iodida.
molecules. of potassium carbonate
Bakteria / mikroorganisma
dalam udara menghasilkan solution and potassium (ii) 1. Measure and pour 1 cm3
ion hidrogen yang
meneutralkan cas negatif sulphate solution into two of both solutions
di sekitar membran protein
molekul getah. different test tubes. into two test tubes
Sukat dan tuangkan 1 cm3
– Neutral rubber larutan kalium karbonat dan separately.
larutan kalium sulfat ke dalam Sukat dan tuangkan 1 cm3
molecules collide dua tabung uji yang berbeza. kedua-dua larutan tersebut
dalam dua tabung uji yang
among one another 2. Add 1 cm3 of nitric acid into berasingan.

and break the protein both test tubes. 2. Add 1 cm3 of sodium
Tambahkan 1 cm3 asid nitrik ke
membranes. hydroxide solution
Molekul getah neutral dalam kedua-dua tabung uji.
berlanggar antara into both test tubes.
satu sama lain dan 3. Add 1 cm3 of barium
memecahkan membran Then, heat the mixture
protein. chloride solution into both
solution.
test tubes. Tambahkan 1 cm3 larutan
Tambahkan 1 cm3 larutan natrium hidroksida ke
– Rubber polymers are barium klorida ke dalam kedua- dalam kedua-dua tabung
dua tabung uji. uji. Kemudian, panaskan
released and entangled larutan campuran tersebut.
4. White precipitate is formed
causing the formation
in the test tube containing 3. Test tube with pungent
of the white solid.
Polimer getah terbebas potassium sulphate solution. smell gas produced
Mendakan putih terbentuk di
dan bersimpul dalam tabung uji yang berisi contains ammonium
menyebabkan larutan kalium sulfat.
pembentukan pepejal sulphate solution.
putih. 5. Colourless solution is Tabung uji dengan gas
berbau sengit yang
(ii) – Ammonia solution formed in the test tube terhasil mengandungi
Larutan ammonia larutan ammonium sulfat.
containing potassium

– Hydroxide ions from carbonate solution. 4. A moist red litmus
Larutan tak berwarna terbentuk
ammonia solution di dalam tabung uji yang berisi paper is inserted into
larutan kalium karbonat.
neutralise the hydrogen the test tube containing

ions produced 6. Barium carbonate formed is the pungent smell gas.

by the bacteria / soluble in nitric acid. Moist red litmus turns
Barium karbonat yang terbentuk
microorganisms. blue.
larut dalam asid nitrik. Kertas litmus merah
Ion hidroksida daripada lembap dimasukkan ke
larutan ammonia (b) (i) 1. Measure and pour 1 cm3 dalam tabung uji yang
meneutralkan ion hidrogen berisi gas berbau sengit
yang dihasilkan oleh of both solutions into itu. Kertas litmus merah
bakteria / mikroorganisma. lembap bertukar menjadi
two test tubes separately. biru.
Sukat dan tuangkan 1 cm3
kedua-dua larutan tersebut
Section C / Bahagian C dalam dua tabung uji yang 5. Test tube with no
berasingan.
1 1. (a) – Barium ion reacts with observable change
2. Add 1 cm3 of nitric acid
carbonate ion to form contains sodium
into both test tubes.
white precipitate, barium Tambahkan 1 cm3 asid sulphate solution.
nitrik ke dalam kedua-dua Tabung uji tanpa
carbonate. tabung uji. perubahan yang kelihatan
Ion barium bertindak balas mengandungi larutan
dengan ion karbonat untuk 3. Add 1 cm3 silver nitrate natrium sulfat.
membentuk mendakan putih,
barium karbonat. solution into both test

tubes.

A19 © Penerbitan Pelangi Sdn. Bhd.

© Penerbitan Pelangi Sdn. Bhd. Periodic Table of Elements

Group

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Alkali Alkaline Key: Halogens Inert
metals earth gases

metals

1 Proton number Metal

H Symbol of element Semi-metal

Hydrogen Name of element
1 Relative atomic mass

1 Non-metal 2

1 HHydrogen He
1
34 Helium
4

5 6 7 8 9 10

Li Be2 Lithium B C N O F Ne
Beryllium
79 Boron Carbon Nitrogen Oxygen Fluorine Neon
11 12 14 16 19 20

A20 11 12 13 14 15 16 17 18

Na Mg3 Sodium Transition metals Al Si P S Cl Ar
Magnesium
23 24 Aluminium Silicon Phosphorus Sulphur Chlorine Argon
27 28 31 32 35.5 40

Period 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

4 K Ca Sc Ti V Cr Mn FePotassium Co Ni Cu Zn Ga Ge As Se Br Kr
Calsium Scandium Titanium Vanadium Cromium Manganese Iron Zinc
39 40 45 48 51 52 55 56 Cobalt Nickel Copper 65 Gallium Germanium Arsenic Selenium Bromine Krypton
59 59 64 70 73 75 79 80 84

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd InRubidium Sn Sb Te I Xe
Strontium Ytrium Zirconium Niobium Molybdenum Technetium Ruthenium Rodium Palladium Silver Cadmium Indium Tin
85.5 88 89 91 93 96 101 103 106 108 112 115 119 Antimony Tellurium Iodine Xenon
122 128 127 131

55 56 57 – 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

6 Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At RnCesium
Barium Lanthanoids Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon

133 137 178.5 181 184 186 190 192 195 197 201 204 207 209 210 210 222

87 88 89 – 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

7 Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts OgFrancium
Radium Actinoids Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Copernicium Nohonium Flerovium Moscovium Livermorium Tennessine Oganesson

57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

Lanthanides La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Actinides
Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium
139 140 141 144 145 150 152 157 159 162.5 165 167 169 173 175

89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
232 231 238

5CLASS Form

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