modul kimia tingkatan 4

Nilam Publication Sdn. Bhd. (919810-T) Tingkat 1, No. 35, Jalan 5/10B, Spring Crest Industrial Park 68100 Batu Caves, Selangor, Malaysia. Tel/Fax: 03 - 6185 2402

All right reserved. No part of this publication may be reproduced, stored in a retrival system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission from Nilam Publication Sdn. Bhd. © Nilam Publication Sdn. Bhd. (919810-T), 2012

Printed by Pramaju Sdn. Bhd. No. 35, Jalan 5/10B Spring Crest Industrial Park 68100 Batu Caves Selangor Darul Ehsan

CONTENTS KANDUNGAN

1

THE STRUCTURE OF ATOMS STRUKTUR ATOM

2

CHEMICAL FORMULA AND EQUATIONS FORMULA DAN PERSAMAAN KIMIA

22

3

PERIODIC TABLE JADUAL BERKALA

49

4

CHEMICAL BOND IKATAN KIMIA

72

5

ELECTROCHEMISTRY ELEKTROKIMIA

88

6

ACID AND BASES ASID DAN BES

114

7

SALT GARAM

139

8

MANUFACTURED SUBSTANCES IN INDUSTRY BAHAN KIMIA DALAM INDUSTRI

168

Con-Chem F4 (B).indd 3

1

12/9/2011 6:00:06 PM

Chemistry Form 4 • MODULE

1

THE STRUCTURE OF ATOMS STRUKTUR ATOM MATTER / JIRIM

• PARTICLE THEORY OF MATTER / TEORI ZARAH JIRIM – To state the particle theory of matter Menyatakan teori zarah jirim

– To differentiate and draw the three types of particles i.e. atom, ion and molecule Membezakan dan melukis tiga jenis zarah jirim iaitu atom, ion dan molekul

– To describe the laboratory activity to investigate the diffusion of particles in gas, a liquid and a solid. (To prove that matter is made up of tiny and discrete particles) Menghuraikan aktiviti makmal untuk mengkaji resapan zarah dalam gas, cecair dan pepejal (Untuk membuktikan bahawa jirim terdiri daripada zarah-zarah yang halus dan diskrit)

• KINETIC THEORY OF MATTER / TEORI KINETIK JIRIM – To state the kinetic theory of matter Menyatakan teori kinetik jirim

– To relate the change of physical states of matters with energy change Menghubungkaitkan perubahan keadaan jirim dengan perubahan tenaga

– To relate the change of energy in the particles with kinetic particle theory of matter Menghubungkaitkan perubahan tenaga dalam zarah dengan perubahan tenaga kinetik zarah

THE STRUCTURE OF ATOMS / STRUKTUR ATOM

• HISTORY OF ATOMIC MODELS DEVELOPMENT / SEJARAH PERKEMBANGAN MODEL ATOM – To state the contribution of scientists in the atomic structure model such as the scientists who discovered electron, proton, nucleus, neutron and shell Menyatakan sumbangan ahli sains kepada perkembangan model struktur atom dan ahli sains yang menemui elektron, proton, nukleus, neutron dan petala

• SUBATOMIC PARTICLES / ZARAH-ZARAH SUBATOM – To compare and differentiate subatomic particles i.e. proton, neutron and electron from the aspect of charge, relative mass and location Membanding dan membezakan zarah-zarah subatom iaitu proton, neutron dan elektron dari segi cas, jisim relatif dan kedudukan

– To state the meaning of proton number and nucleon number based on the subatomic particle Menyatakan maksud nombor proton dan nombor nukleon berdasarkan zarah subatom

– To write the symbol of elements with proton number and nucleon number Menulis simbol unsur yang mengandungi nombor proton dan nombor nukleon

• ISOTOPE / ISOTOP – To state the meaning, examples and the use of isotopes Menyatakan maksud isotop, contoh-contoh isotop dan kegunaan isotop

• ELECTRON ARRANGEMENT / SUSUNAN ELEKTRON – To know the number of electron shells and number of electrons in the 1st, 2nd and 3rd shell Mengetahui bilangan petala elektron serta bilangan elektron yang diisi dalam petala 1, 2 dan 3

– To write the electron arrangement of atoms based on proton number or number of electrons and state the number of valence electron

n io

Sdn. B

m

1

. hd

Publicat

Menulis susunan elektron bagi suatu atom berdasarkan nombor proton atau bilangan elektron dan seterusnya menyatakan bilangan elektron valens

Nila

01-Chem F4 (3p).indd 1

12/9/2011 5:59:26 PM

MODULE • Chemistry Form 4

MATTER / JIRIM

Matter is any substance that has mass and occupies space. Jirim adalah sebarang bahan yang mempunyai jisim dan memenuhi ruang. The Particle Theory of Matter / Teori Zarah Jirim

atoms ,

Matter is made up of tiny and discrete particles. Three types of tiny particles are

1

atom

Jisim terdiri daripada zarah yang halus dan diskrit. Tiga jenis zarah tersebut ialah

,

ions

ion

and molecules . molekul .

dan

Matter can be classified as element or compound. / Jirim boleh dikelaskan sebagai unsur atau sebatian. Complete the following: / Lengkapkan yang berikut:

2 3

MATTER / JIRIM ELEMENT / UNSUR satu A substance made from only Bahan yang terdiri daripada

satu

type of atom. jenis atom sahaja.

COMPOUND / SEBATIAN two or more A substance made from elements which are bonded together. dua Bahan yang terdiri daripada

atau

different lebih

unsur berbeza yang terikat secara kimia.

Types of particles / Jenis zarah

Atom / Atom The smallest neutral particle of an element (Normally pure metals, noble gases and a few non-metal elements such as carbon and silicon). Zarah neutral yang paling kecil bagi suatu unsur (Biasanya logam tulen, gas adi dan beberapa unsur bukan logam seperti karbon dan silikon).

Example:

Types of particles / Jenis zarah

Molecule / Molekul A neutral particle consists of similar non-metal atoms which are covalently-bonded.

Molecule / Molekul A neutral particle consists of different non-metal atoms which are covalently-bonded.

Zarah neutral terdiri daripada atom-atom bukan logam serupa terikat secara ikatan kovalen.

Zarah neutral terdiri daripada atom-atom bukan logam berlainan terikat secara ikatan kovalen.

Example:

Example:

Contoh:

Contoh:

Oxygen gas, O2

Carbon dioxide gas, CO2

Gas oksigen, O2

Gas karbon dioksida, CO2

Contoh:

Sodium metal, Na

O O

Logam natrium, Na

O O

O

C

O

O

C

O

O

O O

Na Na Na Na Na

C

O

Natrium klorida, NaCl

Water, H2O

Na+ Cl – Na+ Cl – Na+

Air, H2O

Neon gas, Ne

H H

H H

H

O

Cl – Na+ Cl – Na + Cl – H H

Ne

Ne

Example: Sodium chloride, NaCl

Hydrogen gas, H2 Gas hidrogen, H2

Gas Neon, Ne

Zarah bercas positif atau negatif terbentuk dari logam dan bukan logam terikat secara ikatan ion. Daya tarikan antara dua ion yang berlawanan cas membentuk ikatan ion.

Contoh:

Na Na Na Na Na Na Na Na Na Na Na

Ion / Ion Positively or negatively charged particles, which are formed from metal atom and non-metal atom respectively. The force of attraction between the two oppositely charged ions forms an ionic bond.

H H

H

O

O

H

H

Ne

Na+ Cl – Na+ Cl – Na+

Calcium oxide, CaO Kalsium oksida, CaO

Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+

–

Elements can be identified as metal or non-metal by referring to the Periodic Table.

–

Formation of molecule and ion will be studied in Chapter 4 (Chemical Bond).

Unsur boleh dikenal pasti sebagai logam atau bukan logam dengan merujuk kepada Jadual Berkala Unsur. Pembentukan molekul atau ion akan dipelajari dalam Tajuk 4 (Ikatan Kimia). Publica

n Sdn.

2

tio

Nil a

m

d. Bh

01-Chem F4 (3p).indd 2

12/9/2011 5:59:27 PM

Chemistry Form 4 • MODULE

4

Determine the type of particles in the following substances: Tentukan jenis zarah bagi setiap bahan berikut: Substances

Type of particle

Substances

Type of particle

Substances

Type of particle

Bahan

Jenis zarah

Bahan

Jenis zarah

Bahan

Jenis zarah

Molecule

Sulphur dioxide (SO2) Sulfur dioksida (SO2)

Molecule

Tetrachloromethane (CCl4) Tetraklorometana (CCl4)

Molecule

Copper(II) sulphate (CuSO4) Kuprum(II) sulfat (CuSO4 )

Ion

Iron (Fe) Ferum (Fe)

Atom

Zink chloride (ZnCl2) Zink klorida (ZnCl2 )

Ion

Argon (Ar) Argon (Ar)

Atom

Carbon (C) Karbon (C)

Atom

Hydrogen peroxide (H2O2) Hidrogen peroksida (H2O2)

Molecule

Hydrogen gas (H2) Gas hidrogen (H2)

5

Diffusion Resapan

(a) The tiny and discrete particles that made up matter are constantly moving. In gases, these particles are very far apart from each other, in liquids, the particles are closer together and in solids, they are arranged closely packed. Jirim terdiri daripada zarah-zarah halus dan diskrit yang sentiasa bergerak. Dalam gas, susunan zarah-zarahnya adalah berjauhan antara satu sama lain, dalam cecair, zarah-zarahnya disusun lebih rapat dan dalam pepejal, zarah-zarahnya disusun dengan sangat padat dan teratur.

(b) Diffusion occurs when particles of a substance move between the particles of another substance. Resapan berlaku apabila zarah-zarah suatu bahan bergerak di antara zarah-zarah bahan lain.

(c) Diffusion occurs in a solid, liquid and gas. Complete the following table: Resapan berlaku dalam pepejal, cecair dan gas. Lengkapkan jadual berikut:

Experiment Eksperimen

Diffusion in a gas

Diffusion in a liquid

Diffusion in a solid

Resapan dalam gas

Resapan dalam cecair

Resapan dalam pepejal

A few drops of bromine liquid Beberapa titis cecair bromin

After few minutes

Water Air

Selepas beberapa minit

After a few hours Selepas beberapa jam

Potassium manganate(VII) Kalium manganat(VII)

Observation Pemerhatian

Explanation Penerangan

Gel Agar-agar

Copper(II) sulphate

After a day Selepas sehari

Kuprum(II) sulfat

The brown colour of bromine vapour, far Br2 spreads throughout the two jars.

The purple colour of solid potassium manganate(VII), KMnO4 spreads slowly throughout the water.

The blue colour of copper(II) sulphate, CuSO4 spreads very slowly throughout the gel.

Warna perang wap bromin, Br2 merebak cepat memenuhi kedua-dua dengan balang gas.

Warna ungu pepejal kalium manganat(VII), perlahan KMnO merebak dengan di dalam air.

Warna biru kuprum(II) sulfat, sangat perlahan CuSO4 merebak di dalam agar-agar.

Bromine vapour, Br2 and air are made molecules . up of Wap bromin, Br2 dan udara terdiri molekul daripada . Bromine molecules diffuse quickly between large

Potassium manganate(VII) is made up of potassium ions and ions manganate(VII) ions. The slowly diffuse between close space of water particles which is in liquid form.

Copper(II) sulphate, CuSO4 is made ions and up of copper(II) ions . The ions sulphate diffuse very slow between closely packed space of gel particles which is in solid form.

space of air particles which is in gas form. Molekul

pantas bromin meresap besar antara zarahmelalui ruang zarah udara yang berbentuk gas.

4

Kalium manganat(VII) terdiri daripada ion kalium dan ion manganat(VII). Ion-ion perlahan ini meresap rapat zarah air antara ruang yang berbentuk cecair.

Kuprum(II) sulfat, CuSO4 terdiri daripada ion ion kuprum(II) dan Ion-ion ini meresap dengan sulfat. perlahan antara ruang sangat padat zarah agar-agar yang

n io

Sdn. B

m

3

. hd

Publicat

berbentuk pepejal.

Nila

01-Chem F4 (3p).indd 3

12/9/2011 5:59:28 PM

MODULE • Chemistry Form 4

(d) Conclusions: Kesimpulan:

(i)

gas than in liquid. There is Diffusion occurs faster in gas gas than a liquid. Particles in a are a closer are together.

larger

space in between the particles of

further

apart. The particles in the liquid

gas Resapan berlaku lebih cepat di dalam berbanding di dalam cecair. Terdapat ruang yang gas gas berbanding dengan cecair. Zarah-zarah adalah antara zarah-zarah lebih rapat antara satu sama lain. antara satu sama lain. Zarah-zarah cecair adalah

(ii)

liquid than in solid. There is Diffusion occurs faster in a liquid of a than a solid. The particles in the solid are very cecair

Resapan berlaku lebih cepat di dalam cecair antara zarah-zarah dan

padat

larger

lebih besar berjauhan

space in between the particles

close

together. lebih besar

berbanding di dalam pepejal. Terdapat ruang yang

rapat

berbanding dengan pepejal. Zarah-zarah pepejal tersusun sangat

antara satu sama lain.

tiny

(iii) Bromine gas, potassium manganate(VII) and copper(II) sulphate are made up of particles that are constantly moving/constant motion .

and

halus

Gas bromin, kalium manganat(VII) dan kuprum(II) sulfat terdiri daripada zarah-zarah sentiasa bergerak . yang

discrete diskrit

dan

The Kinetic Theory of Matter / Teori Kinetik Jirim

solid

Matter exists in three different states which are

1

pepejal

Jirim wujud dalam tiga keadaan iaitu

Matter that made up of

2

tiny

As the temperature increases, the Apabila suhu meningkat, tenaga

discrete

and halus

Jirim terdiri daripada zarah-zarah 3

,

and

gas

gas

dan

.

.

moving

particles which are always in constantly

dan

kinetic

kinetik

liquid

,

cecair

diskrit

yang sentiasa

bergerak

.

.

energy of particles increases and the particles move

zarah-zarah akan bertambah dan zarah-zarah akan bergerak dengan

faster

.

lebih cepat

.

Particles in different states of matter have different arrangement, strength of forces between them, movement and energy content.

4

Zarah-zarah

dalam keadaan jirim yang berbeza mempunyai susunan, daya tarikan antara zarah, pergerakan dan kandungan

tenaga yang berbeza.

Complete the following table: / Lengkapkan jadual di bawah:

5

State of matter

Solid

Liquid

Gas

Keadaan jirim

Pepejal

Cecair

Gas

Draw the particles arrangement. Each particle (atom/ ion/ molecule) is represented by Lukis susunan zarah. Setiap zarah (atom / ion / molekul) diwakili dengan ‘ ’

Particles arrangement Susunan zarah

The particles are arranged closely packed in orderly

Particles movement Pergerakan zarah

padat

Particles can only vibrate rotate about their and

.

Zarah-zarah tersusun tidak teratur tetapi

The particles are very widely separated

from

each other. padat .

Particles can vibrate , rotate move and

terpisah jauh Zarah-zarah antara satu sama lain.

Particles can vibrate , rotate move and

fixed position.

throughout the liquid.

freely.

Zarah bergetar dan berputar pada kedudukan tetap.

Zarah bergetar , berputar dan bergerak dalam cecair.

Zarah bergetar , berputar dan bergerak bebas.

Publica

n Sdn.

4

tio

Nil a

m

orderly manner

manner.

Zarah-zarah tersusun teratur . dan

The particles are arranged closely packed but not in

d. Bh

01-Chem F4 (3p).indd 4

12/9/2011 5:59:28 PM

Chemistry Form 4 • MODULE

Strong

forces between the particles but weaker than the forces in the solid.

strong

Attractive forces between the particles Daya tarikan antara zarah

forces Very between the particles. Daya tarikan yang sangat kuat antara zarah-zarah.

kuat Daya tarikan yang antara zarah-zarah tetapi lebih lemah berbanding di

Weak

forces between

the perticles lemah

Daya tarikan yang antara zarah-zarah.

dalam pepejal.

Energy content of the particles Kandungan tenaga zarah

6

low .

Energy content is very Kandungan tenaga sangat rendah .

Energy content is higher than solid but less than in a gas.

very

Energy content is high.

Kandungan tenaga lebih tinggi daripada pepejal tetapi lebih rendah daripada gas.

Kandungan tenaga tinggi.

sangat

Changes in the state of matter Perubahan keadaan jirim

(a) Matter undergoes change of state when

heat

haba

Jirim mengalami perubahan keadaan apabila tenaga

(i)

(ii)

energy is

absorbed

or

serap

di

released/lose

When heat energy is absorbed by the matter (it is heated), the increases and they vibrate faster.

kinetic

diserap Apabila tenaga haba oleh jirim (semasa dipanaskan), tenaga dan zarah tersebut bergerak dengan lebih cepat.

kinetik

When matter releases heat energy (it is cooled), the they vibrate less vigorously. dibebaskan Apabila tenaga haba zarah tersebut bergerak kurang cergas.

kinetic

:

bebaskan

atau di

:

energy of the particles zarah

bertambah

energy of the particles decreases and

oleh jirim (semasa disejukkan), tenaga kinetik zarah

berkurang

dan

(b) Inter - conversion of the states of matter: Perubahan keadaan jirim: Solid Pepejal 7

Melting / Peleburan Freezing / Pembekuan

Liquid Cecair

Boiling/Evoporation / Pendidihan/Penyejatan Condensation / Kondensasi

Gas Gas

Determination of melting and freezing points of naphthalene Penentuan takat lebur dan takat beku naftalena

Materials / Bahan: Naphthalene powder, water Apparatus / Radas: Boiling tube, conical flask, beaker, retort stand, thermometer 0 – 100°C, stopwatch, Bunsen burner and wire gauze Procedure / Prosedur: I. Heating of naphthalene / Pemanasan naftalena Set-up of apparatus: / Susunan radas: Thermometer / Termometer Boiling tube / Tabung didih Water / Air Naphthalene / Naftalena

Heat n io

Sdn. B

m

5

. hd

Publicat

Haba

Nila

01-Chem F4 (3p).indd 5

12/9/2011 5:59:28 PM

MODULE • Chemistry Form 4

(a)

boiling tube A placed into it. Tabung didih

3 - 5 cm

is filled

height with naphthalene powder and a

diisi dengan serbuk naftalena setinggi

3 – 5 cm

dan

thermometer

termometer

is

diletakkan

di dalamnya.

(b)

The boiling tube is immersed in a water bath as shown in the diagram so that the water level in the water bath is higher than naphtalene powder in the boiling tube. Tabung didih dimasukkan ke dalam kukus air seperti di dalam gambar rajah dan pastikan aras air dalam kukus air lebih tinggi daripada aras naftalena dalam tabung didih.

(c)

heated

The water is

Air dipanaskan dan naftalena

(d)

and the naphthalene is dikacau

perlahan-lahan dengan

slowly with termometer

thermometer

.

.

60°C , the stopwatch is started. The temperature of When the temperature of naphthalene reaches 30 seconds 90°C naphthalene is recorded at intervals until the temperature of naphthalene reaches . 60°C

Apabila suhu naftalena mencapai sehingga suhunya mencapai

II.

stirred

90°C

, mulakan jam randik. Suhu naftalena dicatat setiap

30 saat

.

Cooling of naphthalene / Penyejukan naftalena

Naphthalene Naftalena

(a)

The boiling tube and its content is removed from the water bath and put into a in the diagram.

conical flask kelalang kon

Tabung didih dan kandungannya dikeluarkan daripada kukus air dan dipindahkan ke dalam dalam gambar rajah.

(b)

as shown seperti

stirred constantly with thermometer throughout cooling The content in the boiling tube is supercooling process to avoid (the temperature of cooling liquid drops below freezing point, without the appearance of a solid). dikacau Kandungan dalam tabung didih perlahan-lahan dengan termometer sepanjang proses penyejukan untuk penyejukan lampau (Suhu cecair yang disejukkan turun melepasi takat beku tanpa pembentukan mengelakkan pepejal).

(c)

The temperature of naphthalene is recorded every 60°C . to Suhu naftalena dicatat setiap

(d)

60°C

.

suhu

melawan

masa

dilukis untuk proses pemanasan dan penyejukan.

Publica

n Sdn.

6

tio

Nil a

sehingga suhunya mencapai

interval until the temperature drops

A graph of temperature against time is plotted for the heating and cooling process respectively. Graf

m

30 saat

30 seconds

d. Bh

01-Chem F4 (3p).indd 6

12/9/2011 5:59:29 PM

Chemistry Form 4 • MODULE

The Explanation of the Heating Process of Matter / Penerangan Proses Pemanasan 1

The heating curve of naphthalene: Lengkung pemanasan naftalena: Temperature/°C Suhu/°C

F D B

E

C

A Time/s Masa/s

2

faster

When a solid is heated, the particles absorb heat and move absorbed energy is , the state of matter will change.

as its energy content increases. As the heat

lebih cepat Apabila pepejal dipanaskan, zarah-zarah menyerap haba dan bergerak diserap menyebabkan perubahan keadaan jirim. Tenaga haba

Point

State of Matter

Explanation

Titik

Keadaan jirim

Penerangan

absorbed

Heat energy is kinetic A to B A ke B

Solid

energy to diserap

pepejal

oleh zarah-zarah

lebih cepat

dan zarah bergetar dengan

absorbed

Heat energy overcome

Solid and Liquid

solid

by the particles in the increase and vibrate

naphthalene causing their

faster

. The temperature

increases. Tenaga haba bertambah

B to C B ke C

disebabkan kandungan tenaga bertambah.

naftalena menyebabkan tenaga meningkat . Suhu semakin

by the particles in the

forces between particles so that the remains constant

temperature

liquid solid

turn to

C to D C ke D

Liquid

increases Tenaga haba bertambah

Heat energy overcome D to E D ke E

Liquid and Gas

to form a Tenaga haba

absorbed

liquid by the particles in the increase energy to and move

Gas

. The

digunakan

untuk mengatasi tetap . Suhu adalah .

naphthalene causing their faster . The temperature

oleh zarah-zarah

cecair

dan zarah-zarah bergerak dengan

absorbed

kinetik naftalena menyebabkan tenaga lebih cepat meningkat . Suhu semakin .

by the particles in the

liquid

naphthalene is

the forces of attraction between particles. The particles begin to move gas diserap

. The temperature oleh zarah-zarah dalam

remains constant cecair

absorbed

energy to incerease and move

akan

used

to

freely

.

naftalena untuk mengatasi bebas gas untuk membentuk . Suhu

gas by the particles in the faster . The temperature

digunakan

causing their increases .

diserap oleh zarah-zarah gas naftalena menyebabkan tenaga Tenaga haba lebih cepat meningkat . Suhu semakin dan zarah-zarah bergerak dengan

kinetik

kinetic

akan bertambah

. n io

Sdn. B

m

7

. hd

Publicat

E to F E ke F

to

. diserap

daya tarikan antara zarah-zarah. Zarah-zarah mula bergerak tetap . adalah

Heat energy is

liquid

.

pepejal oleh zarah-zarah dalam naftalena pepejal cecair berubah menjadi daya tarikan antara zarah-zarah supaya

Heat energy kinetic

akan

used

naphthalene is

diserap

Tenaga haba yang

kinetik .

Nila

01-Chem F4 (3p).indd 7

12/9/2011 5:59:29 PM

MODULE • Chemistry Form 4

completely changes to become a liquid is called the melting point . absorbed by the particles During the melting process, the temperature remains unchanged because heat energy used liquid . to overcome the forces between particles so that the solid change to turn into a is The constant temperature at which a

3

solid

takat lebur berubah kepada keadaan cecair dipanggil diserap oleh zarah-zarah Semasa proses peleburan, suhu tidak berubah kerana haba yang mengatasi cecair daya tarikan antara zarah supaya pepejal berubah menjadi .

Suhu tetap di mana suatu

pepejal

. digunakan

untuk

completely changes to become a gas is called the boiling point . absorbed by the particles During the boiling process, the temperature remains unchanged because heat energy used is to overcome the forces between particles so that the liquid change to turn into a gas.

The constant temperature at which a

4

liquid

cecair takat didih berubah kepada keadaan gas dipanggil . diserap digunakan oleh zarah-zarah untuk Semasa proses pendidihan, suhu tidak berubah kerana haba yang mengatasi daya tarikan antara zarah supaya cecair berubah menjadi gas.

Suhu tetap di mana suatu bahan dalam keadaan

The Explanation for the Cooling Process of Matter: / Penerangan Proses Penyejukan Bahan:

The cooling curve of naphthalene:

1

Lengkung penyejukan naftalena: Temperature/°C Suhu/°C

P Q

R

S Time/s Masa/s

slower When the liquid is cooled, the particles in the liquid release energy and move released decreases. As the energy is to the surrounding, the state of matter will change.

2

cecair Apabila cecair disejukkan, zarah membebaskan tenaga dan dibebaskan ke persekitaran. berubah semasa tenaga

bergerak

Point

State of matter

Explanation

Titik

Keadaan jirim

Penerangan

Heat is

released/given out liquid

P to Q P ke Q

Liquid

The particles in the temperature decreases

Q to R Q ke R

R to S R ke S

Solid

by the

The temperature Haba tenaga

naphthalene. slower. The

. cecair naftalena. Zarah-zarah dalam semakin perlahan. Suhu semakin menurun

.

liquid to the surrounding by the particles in naphthalene is balanced released solid energy as the particles attract one another to form a . remains constant .

dibebaskan cecair diimbangi ke persekitaran oleh zarah-zarah dalam naftalena oleh haba terbebas yang apabila zarah-zarah tertarik antara satu sama lain untuk membentuk pepejal tetap . Suhu adalah .

The particles in the solid naphthalene decreases . Zarah-zarah dalam pepejal naftalena menurun Suhu semakin .

releases

membebaskan

heat and vibrate

slower

tenaga dan bergetar dengan

. The temperature lebih perlahan

.

Publica

n Sdn.

8

tio

Nil a

m

Liquid and Solid

released heat

semakin perlahan. Keadaan jirim

liquid to the surrounding by the particles in the kinetic move lose their energy and

dibebaskan ke persekitaran oleh zarah-zarah dalam Haba cecair kinetik kehilangan tenaga dan bergerak

The heat

as its energy content

d. Bh

01-Chem F4 (3p).indd 8

12/9/2011 5:59:29 PM

Chemistry Form 4 • MODULE

3

freezing point changes to a solid is called . During the freezing released to the surrounding is balanced by the process, the temperature remains unchanged because the heat

The constant temperature at which a

liquid

heat released when the liquid particles rearrange themselves to become a

solid

.

takat beku berubah kepada keadaan pepejal dipanggil . Semasa proses dibebaskan diimbangi ke persekitaran oleh haba yang terbebas pembekuan, suhu tidak berubah kerana haba yang pepejal . apabila zarah-zarah cecair menyusun semula untuk membentuk

Suhu tetap di mana suatu

cecair

Keadaan Fizik Bahan pada Sebarang Suhu: / Physical State Of A Substance At Any Given Temperature: 1

A substance is in

solid

state if the temperature of the substance is below melting point pepejal

Suatu bahan berada dalam keadaan 2

A substance is in

liquid

state if the temperature of the substance is between melting and boiling points. cecair

Suatu bahan berada dalam keadaan 3

A substance is in

gas

jika suhu bahan tersebut lebih rendah daripada takat leburnya.

jika suhu bahan tersebut berada antara takat lebur dan takat didihnya.

state if the temperature of the substance is above boiling point.

Suatu bahan berada dalam keadaan

gas

jika suhu bahan tersebut lebih tinggi daripada takat didihnya.

EXERCISE / LATIHAN 1

The table below shows substances and their chemical formula. Jadual di bawah menunjukkan bahan dan formula kimia masing-masing. Substance / Bahan

Chemical formula / Formula kimia

Type of particle / Jenis zarah

Silver / Argentum

Ag

Atom

Potassium oxide / Kalium oksida

K2O

Ion

Ammonia / Ammonia

NH3

Molecule

Chlorine / Klorin

Cl2

Molecule

(a) State the type of particles that made up each substance in the table. Nyatakan jenis zarah yang membentuk bahan dalam jadual di atas.

(b) Which of the substances are element? Explain your answer. Yang manakah antara bahan tersebut merupakan suatu unsur? Jelaskan jawapan anda.

Silver and chlorine. Silver and chlorine are made up of one type of atom (c) Which of the substance are compound? Explain your answer. Yang manakah antara bahan tersebut merupakan suatu sebatian? Jelaskan jawapan anda.

Potassium oxide and ammonia. Potassium oxide and ammonia are made up of two different elements 2

The table below shows the melting and boiling points of substance P, Q and R. Jadual di bawah menunjukkan takat lebur dan takat didih bagi bahan P, Q dan R. Melting point / Takat lebur / °C

Boiling point / Takat didih / °C

P

–36

6

Q

–18

70

R

98

230

n io

Sdn. B

m

9

. hd

Publicat

Substance / Bahan

Nila

01-Chem F4 (3p).indd 9

12/9/2011 5:59:29 PM

MODULE • Chemistry Form 4

(a) (i)

What is meant by ‘melting point’? Apakah yang dimaksudkan dengan ‘takat lebur’?

The constant temperature at which a solid charges to a liquid at particular pressure (ii)

What is meant by ‘boiling point’? Apakah yang dimaksudkan dengan ‘takat didih’?

The constant temperature at which a liquid changes to a gas at particular pressure (b) Draw the particles arrangement of substances P, Q and R at room condition. Lukis susunan zarah P, Q dan R pada keadaan bilik.

Substance P / Bahan P

(c)

(i)

Substance Q / Bahan Q

Substance R / Bahan R

What is the substance that exist in the form of liquid at 0°C. Nyatakan bahan yang wujud dalam keadaan cecair pada suhu 0°C.

P, Q (ii)

Give reason to your answer. Jelaskan jawapan anda.

The temperature 0°C is above the melting point of Q and below the boiling point of Q (d) (i)

Substance Q is heated from room temperature to 100°C. Sketch a graph of temperature against time for the heating of substance Q. Bahan Q dipanaskan dari suhu bilik hingga 100°C. Lakarkan graf suhu melawan masa bagi pemanasan bahan Q terhadap masa untuk pemanasan bahan Q.

Temperature/°C

70

Time/s (ii)

What is the state of matter of substance Q at 70°C? Apakah keadaan fizik bahan Q pada 70°C?

Liquid and gas (e) Compare the melting point of substances Q and R. Explain your answer. Bandingkan takat lebur bahan Q dan R. Terangkan jawapan anda.

The melting point of substance R is higher than subtance Q. The attraction force between particles in substance R is stronger than Q. More heat is needed to overcome the force between particles in substance R.

Publica

n Sdn.

10

tio

Nil a

m

d. Bh

01-Chem F4 (3p).indd 10

12/9/2011 5:59:30 PM

Chemistry Form 4 • MODULE

3

The melting point of acetamide can be determined by heating solid acetamide until it melts as shown in the diagram below. The temperature of acetemide is recorded every three minutes when it is left to cool down at room temperature. Takat lebur asetamida boleh ditentukan dengan memanaskan pepejal asetamida sehingga lebur seperti dalam rajah di bawah. Suhu asetamida dicatatkan setiap tiga minit semasa disejukkan pada suhu bilik. Thermometer / Termometer Boiling tube / Tabung didih Water / Air Acetamide / Asetamida

(a) What is the purpose of using water bath in the experiment? Apakah tujuan menggunakan kukus air dalam eksperimen ini?

To ensure even heating of acetemide. Acetamide is easily combustible. (b) State the name of another substance which its melting point can also be determined by using water bath as shown in the above diagram. Namakan satu bahan lain yang mana takat leburnya boleh ditentukan dengan menggunakan kukus air seperti rajah di atas.

Naphthalene (c) Sodium nitrate has a melting point of 310°C. Can the melting point of sodium nitrate be determined by using the water bath as shown in the diagram? Explain your answer. Natrium nitrat mempunyai takat lebur 310°C. Bolehkah takat lebur natrium nitrat ditentukan dengan menggunakan kukus air seperti yang ditunjukkan dalam rajah di atas? Jelaskan jawapan anda.

No, because the melting point of water is 100°C which is less than the melting point of sodium nitrate. (d) Why do we need to stir the acetemide in the boiling tube in above experiment? Mengapakah asetamida dalam tabung didih itu perlu dikacau sepanjang eksperimen?

To make sure the heat is distributed evenly (e) The graph of temperature against time for the cooling of liquid acetamide is shown below. Rajah di bawah menunjukkan graf suhu melawan masa untuk penyejukan cecair asetamida. Temperature / Suhu/ °C

T3 T2

Q

R

T1

(i)

Time / Masa/s

What is the freezing point of acetamide? Apakah takat beku asetamida?

T2°C (ii)

The temperature between Q and R is constant. Explain. Suhu antara titik Q dan R adalah tetap. Jelaskan.

The heat lost to the surrounding is balanced by the heat released when the liquid particles rearrange themselves to become solid. (f)

Acetemide exists as molecules. State the name of another compound that is made up of molecules. Asetamida wujud sebagai molekul. Namakan sebatian lain yang terdiri daripada molekul.

Water/naphthalene (g) What is the melting point of acetamide? Apakah takat lebur asetamida?

n io

Sdn. B

m

11

. hd

Publicat

T2°C

Nila

01-Chem F4 (3p).indd 11

12/9/2011 5:59:30 PM

MODULE • Chemistry Form 4

The Atomic Structure / Struktur Atom History of the development of atomic models:

1

Sejarah perkembangan model atom: Scientist

Atomic Model

Discovery

Saintis

Model atom

Penemuan

(i)

Matter is made up of particles called

atoms

.

atom

.

Jirim terdiri daripada zarah-zarah dipanggil

(ii)

Dalton

created

Atoms cannot be

dicipta

Atom tidak boleh

,

destroyed

dimusnah

,

Positively charged sphere Sfera bercas

(i) (ii)

Thomson Electron charges negative Elektron

(i) Electron moves outside the nucleus Elektron nukleus

Rutherford mengandungi Nucleus that contain proton Nukleus mengandungi proton

elektron

, zarah subatom yang pertama.

positive

charge which embedded with negatively charged particles called electrons .

Atom is sphere of

yang mengandungi zarah

Discovered the nucleus as the centre of an atom and positively charged . Menjumpai

bergerak di luar

.

.

positif Atom adalah sfera yang bercas elektron . bercas negatif dipanggil

bercas negatif

.

.

electrons , the first subatomic particle.

Discovered the Menjumpai

positif

atau

sama

Atom daripada unsur sama adalah

dibahagi

identical

(iii) Atoms from the same element are

divided

or

nukleus

bercas positif

(ii)

Proton

(iii)

Electron

yang merupakan pusat bagi atom dan

.

is a part of the nucleus.

Proton

adalah sebahagian daripada nukleus.

move outside the nucleus.

Elektron

bergerak di sekeliling nukleus.

(iv) Most of the mass of the atom found in the Nukleus

nucleus

.

mempunyai hampir semua jisim atom.

Shell

Neils Bohr

Nucleus that contain proton

(i)

Nukleus mengandungi proton

(ii)

shells

Discovered the existence of electron petala

Menjumpai kewujudan

Electrons move in the Elektron

.

elektron.

shells

around the nucleus. nukleus

bergerak di dalam petala mengelilingi

.

Electron

Shell

James Chadwick

Nucleus that contain proton and neutron Nukleus mengandungi proton dan neutron

Electron

(i)

Discovered the existence of

(ii)

proton

neutron

dan

(iii) The mass of a neutron and proton is almost the same. neutron

dan

proton

adalah hampir sama.

a

n Sdn.

12

.

Nukleus mengandungi zarah-zarah neutral dipanggil proton . zarah-zarah bercas positif dipanggil

tio

Nil a

m

.

.

Nucleus of an atom contains neutral particles called neutron and positively charged particles called

Jisim Public

neutron

neutron

Menjumpai kewujudan

d. Bh

01-Chem F4 (3p).indd 12

12/9/2011 5:59:31 PM

Chemistry Form 4 • MODULE

2

The structure of an atom: / Struktur Atom: Shell / Petala Nucleus that contain proton and neutron Nukleus yang mengandungi proton dan neutron Electron / Elektron

nucleus

(a) An atom has a central Atom mempunyai

nucleus

(b) The

Nukleus

nukleus

shells

and electrons that move in the

around the nucleus.

petala

di tengahnya dan elektron bergerak di dalam

mengelilingi nukleus tersebut.

contains protons and neutrons. mengandungi proton dan neutron.

+1 . Each electron has an electrical charge of –1 . The neutron has no (c) Each proton has charge of charge neutral (it is ). An atom has the same number of protons and electrons, so the overall charge zero neutral of atom is . Atom is . (If an atom loses or gains electrons it is called an ion – formation of ion will be studied in Chapter 4) (ianya adalah neutral ). sifar . Atom Setiap atom mempunyai bilangan proton dan elektron yang sama, oleh itu cas keseluruhan bagi atom adalah neutral . (Suatu atom akan membentuk ion apabila ia kehilangan atau menerima elektron – pembentukan ion akan adalah

Setiap proton bercas

+1

–1

. Setiap elektron bercas

. Neutron tidak mempunyai

cas

dipelajari dalam Tajuk 4.)

(d) The relative mass of a neutron and a proton which are in the nucleus is 1. The mass of an atom is obtained mainly proton and neutron . from the number of proton

Jisim relatif proton dan neutron di dalam nukleus ialah 1. Jisim suatu atom diperoleh daripada jumlah bilangan neutron dan bilangan .

1 (e) The mass of an electron can be ignored as the mass of an electron is about times the size of a proton or 1 840 neutron. Jisim elektron boleh diabaikan kerana ia terlalu kecil iaitu 3

1 daripada jisim proton dan neutron. 1 840

Complete the following table: Lengkapkan jadual di bawah:

4

Subatomic particles

Symbol

Charge

Relative atomic mass

Position

Zarah subatom

Simbol

Cas

Jisim atom relatif

Kedudukan

Electron/Elektron

e

– (negative)

1 =0 1 840

In the shells

Proton/Proton

p

+ (positive)

1

In the nucleus

Neutron/Neutron

n

neutral

1

In the nucleus

Atom is the smallest neutral particle of an element. Atom adalah zarah neutral paling kecil dalam suatu unsur.

Complete the following diagram: / Lengkapkan yang berikut: Na

Na

Na

Na

Sodium element natrium

Sodium element Unsur

natrium

Na

Na

Na

Na

Na

Na

Sodium element Unsur

natrium

Na

Na

Na

Sodium Atom

atom natrium n io

Sdn. B

m

13

. hd

Publicat

Unsur

Na

Nila

01-Chem F4 (3p).indd 13

12/9/2011 5:59:32 PM

MODULE • Chemistry Form 4

Proton number of an element (Refer to Periodic table of an element)

5

Nombor proton sesuatu unsur (Rujuk Jadual Berkala Unsur)

(a) Proton number of an

element

atom

is the number of proton in its

.

atom

Nombor proton sesuatu unsur adalah bilangan proton yang terdapat dalam

.

neutral .

(b) The number of proton of an atom is also equal to the number of electrons in the atom because atom is Bilangan proton sesuatu atom adalah sama dengan bilangan elektron dalam atom kerana atom adalah

neutral

.

(c) Every element has its own proton number: Setiap unsur mempunyai nombor protonnya tersendiri:

atom

– Proton number of potassium, K is 19. Potasium in the shells. Atom

Nombor proton untuk kalium, K ialah 19. 19 elektron di dalam petala.

19 proton

kalium mempunyai

atom

– Proton number of oxygen, O is 8. Oxygen in the shells.

Atom

Nombor proton untuk oksigen, O ialah 8. 8 elektron di dalam petala.

has 19 protons in the nucleus and 19 electrons

8 protons

has

di dalam nukleus dan

in the nucleus and 8 proton

oksigen mempunyai

8 electrons

di dalam nukleus dan

Nucleon number of an element (Refer to Periodic table of an element)

6

Nombor nukleon sesuatu unsur (Rujuk Jadual Berkala Unsur)

(a) Nucleon number of an element is the total number of protons and neutrons in the nucleus of its Nombor nukleon sesuatu unsur adalah jumlah bilangan proton dan neutron di dalam nukleus sesuatu

atom

atom

.

.

(b) Nucleon number is also known as a mass number. Nombor nukleon juga dikenali sebagai nombor jisim.

(c) Nucleon number = number of proton + number of neutron. Nombor nukleon = bilangan proton + bilangan neutron. Symbol of Element And Standard Representation For An Atom of Element Simbol Unsur dan Perwakilan Piawai bagi Atom Sesuatu Unsur

The symbol of an element is a short way of representing an element. If the symbol has only one letter, it must be a capital letter. If it has two letters, the first is always a capital letter, while the second is always a small letter.

1

Simbol unsur adalah cara mudah untuk mewakilkan unsur. Jika simbol hanya terdiri daripada satu huruf, maka ia mesti ditulis dengan huruf besar. Tetapi jika simbol terdiri daripada dua huruf, maka huruf pertama merupakan huruf besar dan huruf kedua merupakan huruf kecil.

Example: / Contoh: Element

Symbol

Element

Symbol

Element

Symbol

Unsur

Simbol

Unsur

Simbol

Unsur

Simbol

O

Nitrogen/Nitrogen

N

Calcium/Kalsium

Ca

Mg

Sodium/Natrium

Na

Copper/Kuprum

Cu

Potassium/Kalium

K

Chlorine/Klorin

Cl

Oxygen/Oksigen Magnesium/Magnesium Hydrogen/Hidrogen

H

The first letter of each element is capitalised to show that it is a new element. This is helpful when writing a chemical formula. For example KCl. There are two elements chemically bonded in KCl because there are two capital letters represent potassium and chlorine. Huruf yang pertama bagi setiap unsur ditulis dengan huruf besar untuk menunjukkan ia adalah unsur yang baru. Ini sangat berguna semasa menulis formula kimia. Contohnya KCl. Terdapat dua unsur yang terikat secara kimia dalam KCl kerana adanya dua huruf besar yang mewakili kalium dan klorin.

Standard representation symbol represents

2

Simbol perwakilan piawai mewakili

one atom

of an element. It can be written as:

sesuatu unsur. Ianya boleh ditulis sebagai:

Nucleon number/Nombor nukleon

A

Proton number/Nombor proton

Z

X

Symbol of an element/Simbol unsur

Publica

n Sdn.

14

tio

Nil a

m

satu atom

d. Bh

01-Chem F4 (3p).indd 14

12/9/2011 5:59:32 PM

Chemistry Form 4 • MODULE

Example: / Contoh: 27 A1 13 – The element is Aluminium. Unsur itu adalah Aluminium.

27

– The nucleon number of Aluminium is 27

Nombor nukleon Aluminium adalah

.

13

– The proton number of Aluminium is 13

Nombor proton Aluminium adalah

– Aluminium has

13 protons

. .

14 neutrons

13 proton

Atom Aluminium mempunyai 3

,

.

13

and

14 neutron

,

electrons. 13

dan

elektron.

Isotope / Isotop (a) Isotopes are atoms of the same element with same number of protons but different number of neutrons. Isotop ialah atom-atom unsur yang mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza.

Or / Atau Isotopes are atoms of the same element with same

proton

Isotop ialah atom-atom unsur yang mempunyai nombor berbeza.

proton

number but different

nucleon

number.

nukleon

yang sama tetapi nombor

yang

Example: / Contoh: 1 1 H

2 1 H

Nucleon number/Nombor nukleon = 1

Nucleon number/Nombor nukleon = 2

Proton number/Nombor proton = 1

Proton number/Nombor proton = 1

Number of neutron/Bilangan neutron = 0

Number of neutron/Bilangan neutron = 1

– Hydrogen-1 and Hydrogen-2 are isotopes. Hydrogen-1 and Hydrogen-2 atoms have the same proton number or the same number of protons

but

different

in nucleon number because of the difference in the number of

Atom Hidrogen-1 dan Hidrogen-2 mempunyai nombor proton atau bilangan neutron . kerana perbezaan

– Isotopes have the same arrangements. Isotop mempunyai sifat

chemical kimia

bilangan proton

properties but different

yang sama tetapi nombor nukleon yang

physical

neutron

.

berbeza

properties because they have the same electron

yang sama kerana mempunyai susunan elektron yang sama tetapi sifat

fizik

yang berbeza.

(b) Examples of the usage of isotopes: Contoh kegunaan isotop:

i.

Medical field Bidang perubatan

–

To detect brain cancer.

–

To detect thrombosis (blockage in blood vessel).

–

Sodium-24 is used to measure the rate of iodine absorption by thyroid gland.

–

Cobalt-60 is used to destroy cancer cells.

–

To kill microorganism in the sterilising process.

Untuk mengesan barah otak. Untuk mengesan trombosis (saluran darah tersumbat). Untuk mengukur kadar penyerapan iodin oleh kelenjar tiroid. Contoh: Natrium-24 Untuk memusnahkan sel barah. Contoh: Kobalt-60 Untuk membunuh mikroorganisma semasa proses pensterilan.

ii.

In the industrial field Bidang industri

–

To detect wearing out in machines.

–

To detect any blockage in water, gas or oil pipes.

Untuk mengesan kehausan enjin.

n io

Sdn. B

m

15

. hd

Publicat

Untuk mengesan saluran paip air, gas atau minyak yang tersumbat.

Nila

01-Chem F4 (3p).indd 15

12/9/2011 5:59:32 PM

MODULE • Chemistry Form 4

–

To detect leakage of pipes underground.

–

To detect defects/cracks in the body of an aeroplane.

Untuk mengesan kebocoran paip bawah tanah. Untuk mengesan keretakan atau kecacatan pada badan kapal terbang.

iii.

In the agriculture field Bidang pertanian

–

To detect the rate of absorption of phosphate fertilizer in plants.

–

To sterile insect pests for plants.

Untuk mengesan kadar penyerapan baja fosfat oleh tumbuhan. Untuk memandulkan serangga perosak tumbuhan.

iv.

In the archeology field Bidang arkeologi

–

Carbon-14 can be used to estimate the age of artifacts. Karbon-14 untuk menentukan usia sesuatu artifak.

Electron Arrangement

4

Susunan elektron

(a) The electrons are filled in specific shells. Every shell can be filled only with a certain number of electrons. For the elements with atomic numbers 1-20: Elektron diisi dalam petala tertentu. Setiap petala hanya boleh diisi dengan bilangan elektron tertentu. Bagi unsur-unsur yang mempunyai nombor proton 1–20:

2

– First shell can be filled with a maximum of

electrons. 2

Petala pertama boleh diisi dengan bilangan maksimum

8

– Second shell can be filled with a maximum of Petala kedua boleh diisi dengan bilangan maksimum

Petala ketiga boleh diisi dengan bilangan maksimum

electrons.

8

8

– Third shell can be filled with a maximum of

elektron.

elektron.

electrons.

8

elektron.

First shell is filled with 2 electrons (duplet) Petala pertama diisi 2 elektron (duplet)

Second shell is filled with 8 electrons (octet) Petala kedua diisi 8 elektron (oktet)

Third shell is filled with 8 electrons (octet) Petala ketiga disi 8 elektron (oktet)

(b) Valence electrons are the electrons in the outermost shell of an atom. Elektron valens: Elektron yang diisi dalam petala paling luar suatu atom.

Complete the following table:

5

Lengkapkan jadual berikut:

(a) Draw the electron arrangement and complete the description for each element: Lukis susunan elektron bagi atom dan penerangan bagi setiap unsur berikut: Standard representation of an element Perwakilan piawai unsur

Electron arrangement of an atom Lukiskan susunan elektron bagi atom

Hydrogen Atom Atom Hidrogen

1 H 1

1

Number of eletrons/Bilangan elektron

1

Number of neutrons/Bilangan neutron

0

Proton number/Nombor proton

1

Nucleon number/Nombor nukleon

1

Electron Arrangement/Susunan elektron

1

H

a

n Sdn.

16

Number of protons/Bilangan proton

tio

Nil a

m

Public

Description Penerangan

d. Bh

01-Chem F4 (3p).indd 16

12/9/2011 5:59:32 PM

Chemistry Form 4 • MODULE

Sodium Atom

Number of protons/Bilangan proton

11

Number of electrons/Bilangan elektron

11

Number of neutrons/Bilangan neutron

12

Proton number/Nombor proton

11

Nucleon number/Nombor nukleon

23

Atom Natrium

23 Na 11

Na

Electron Arrangement/Susunan elektron

2.8.1

(b) Choose the correct statement for the symbol of element X. Pilih pernyataan yang betul bagi simbol unsur X. 23 Na 11 Statement

Tick ( ✓ / ✗ )

Pernyataan

Tanda ( ✓ / ✗ )

Element X has 11 proton number. Unsur X mempunyai 11 nombor proton.

The proton number of element X is 11. Nombor proton unsur X ialah 11.

The proton number of atom X is 11. Nombor proton atom X ialah 11.

The number of proton of element X is 11. Bilangan proton unsur X ialah 11.

The number of proton of atom X is 11. Bilangan proton atom X ialah 11.

Nucleon number of element X is 23. Nombor nukleon unsur X ialah 23.

Nucleon number of atom X is 23. Nombor nukleon atom X ialah 23.

Number of nucleon of element X is 23. Bilangan nukleon unsur X ialah 23.

Atom X has 23 nucleon number. Atom X mempunyai 23 nombor nukleon.

Neutron number of atom X is 12. Nombor neutron atom X ialah 12.

Number of neutron of atom X is 12. Bilangan neutron atom X ialah 12.

Number of neutron of element X is 12.

✓ ✓ ✗ ✓ ✓ ✓ ✗ ✗ ✗ ✓ ✗

n io

Sdn. B

m

17

. hd

Publicat

Bilangan neutron unsur X ialah 12.

✗

Nila

01-Chem F4 (3p).indd 17

12/9/2011 5:59:32 PM

MODULE • Chemistry Form 4

EXERCISE / LATIHAN

Complete the following table:

1

Lengkapkan jadual berikut:

Element Unsur

Hydrogen Hidrogen

Helium Helium

Boron Boron

Carbon Karbon

Nitrogen Nitrogen

Neon Neon

Sodium Natrium

Magnesium Magnesium

Calcium Kalsium

Symbol of element Simbol unsur

Number of proton Bilangan proton

Number of electron Bilangan elektron

Number of neutron Bilangan neutron

Proton number Nombor proton

Nucleon number Nombor nukleon

Electron arrangement Susunan elektron atom

Number of valence electron Bilangan elektron valens

1 1 H

1

1

0

1

1

1

1

4 He 2

2

2

2

2

4

2

2

11 5 B

5

5

6

5

11

2.3

3

12 6 C

6

6

6

6

12

2.4

4

14 7 N

7

7

7

7

14

2.5

5

20 Ne 10

10

10

10

10

20

2.8

8

23 Na 11

11

11

12

11

23

2.8.1

1

24 Mg 12

12

12

12

12

24

2.8.2

2

40 Ca 20

20

20

20

20

40

2.8.8.2

2

The diagram below shows the symbol of atoms P, R and S.

2

Rajah di bawah menunjukkan simbol atom P, R dan S.

35 P 17

12 R 6

37 S 17

(a) What is meant by nucleon number / Apakah maksud nombor nukleon? Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom (b) What is the nucleon number of P / Apakah nombor nukleon atom P? 35 (c) State the number of neutron in atom P / Nyatakan bilangan neutron atom P. 18 (d) State number of proton in atom P / Nyatakan bilangan proton atom P. 17 (e)

(i)

What is meant by isotope / Apakah maksud isotop? Isotopes are atoms of the same element with same number of proton but different number of neutrons

Publica

n Sdn.

18

tio

Nil a

m

d. Bh

01-Chem F4 (3p).indd 18

12/9/2011 5:59:33 PM

Chemistry Form 4 • MODULE

(ii)

State a pair of isotope in the diagram shown / Nyatakan sepasang isotop dalam rajah yang ditunjukkan. P and S

(iii) Give reason for your answer in (e)(ii) / Berikan sebab bagi jawapan di (e)(ii). Atom P and S have same proton number but different nucleon number//number of neutron (f)

An isotope of R has 8 neutron. Write the symbol for the isotope R. Isotop bagi atom R mempunyai 8 neutron. Tuliskan simbol bagi isotop R.

14 R 6 3

The table below shows the number of proton and neutron of atoms of elements P, Q and R. Jadual di bawah menunjukkan bilangan proton dan neutron bagi atom unsur P, Q dan R. Element Unsur

Number of proton Bilangan proton

Number of neutron Bilangan neutron

P

1

0

Q

1

1

R

6

6

(a) Which of the atoms in the above table are isotope? Explain your answer. Berdasarkan jadual di atas, atom yang manakah merupakan isotop? Terangkan jawapan anda.

P and Q. Atom P and Q have same number of proton but different number of neutron // nucleon number. (b) (i)

Write the standard representation of element Q. Tuliskan perwakilan piawai untuk unsur Q.

2 Q 1 (ii)

State three information that can be deduced from your answer in (b)(i). Nyatakan tiga maklumat yang boleh didapati daripada jawapan anda di (b)(i).

The proton number of element Q is 1 // Number of proton of atom Q is 1 Nucleon number of element Q is 2 // Atomic mass of atom Q is 2 Number of neutron of atom Q is 1 Nucleus of atom Q contains 1p and 1n (c)

(i)

Draw atomic structure for atom of element R. Lukiskan struktur atom bagi atom unsur R.

6 protons + 6 neutrons

(ii)

Describe the atomic structure in (c)(i). Huraikan struktur atom di (c)(i).

– The atom consists of 2 parts: the centre part called nucleus and the outer part called electron shell. – The nucleus consists of 6 protons which are positively charged and 6 neutrons which are neutral. – The electrons are in two shells, the first shell consists of two electrons and the second shell consists of four electrons.

n io

Sdn. B

m

19

. hd

Publicat

– Electrons move around nucleus in the shells.

Nila

01-Chem F4 (3p).indd 19

12/9/2011 5:59:33 PM

MODULE • Chemistry Form 4

(d) Element R react with oxygen and to produce liquid Z at room temperature. The graph below shows the sketch of the graph when liquid Z at room temperature, 27°C is cooled to –5°C. Unsur R bertindak balas dengan oksigen dan menghasilkan cecair Z pada suhu bilik. Rajah di bawah menunjukkan lakaran graf apabila cecair Z pada suhu bilik, 27°C disejukkan sehingga –5°C. Temperature /°C Suhu /°C

Time /s

0

t1

Masa /s

t2

−5

(i)

What is the state of matter of liquid Z from t1 to t2? Explain why is the temperature remain unchanged from t1 to t2. Apakah keadaan jirim Z daripada t1 hingga t2? Terangkan mengapa suhu tidak berubah daripada t1 hingga t2.

Liquid and solid. Heat lost to the surrounding is balanced by the heat released when the particles at 0 °C (ii)

Draw the arrangement of particles of Z at 20°C. Lukiskan susunan zarah-zarah Z pada suhu 20°C.

(iii) Describe the change in the particles movement when Z is cooled from room temperature to –5°C. Nyatakan perubahan dalam pergerakan zarah-zarah apabila cecair Z disejukkan daripada suhu bilik ke –5°C.

The particles move slower Objective Questions / Soalan Objektif The diagram shows the arrangement of particles for a type of matter that undergoes a change in physical state through process X.

1

3

The diagram below shows the heating curve for substance X. Rajah di bawah menunjukkan lengkung pemanasan bahan X.

Temperature / Suhu °C

Rajah di bawah menunjukkan susunan zarah sejenis bahan yang mengalami perubahan keadaan fizik melalui proses X.

U S T

Q

X

R P

Time (m) Masa (m)

Which region of the graph does boiling process occur?

What is process X?

Bahagian manakah pada graf berlaku proses pendidihan?

Apakah proses X ?

A

Melting

C

Peleburan

B

Boiling

A B

Freezing Pembekuan

D

Pendidihan

Sublimation Pemejalwapan

4

PQ QR

C D

ST TU

Which of the following information is true? Antara pernyataan berikut, yang manakah adalah betul?

Which of the following substances can undergo sublimation when heated?

2

Antara bahan berikut, yang manakah mengalami pemejalwapan apabila dipanaskan?

A

Sulphur

B

Ammonium chloride

C

Sulfur Ammonium klorida

A

Glucose Glukosa

D

Change of state Perubahan keadaan

Sodium chloride Natrium klorida

B C D

Publica

Heat energy Tenaga haba

Solid → Liquid

Melting

Released

Pepejal → Cecair

Peleburan

Dibebaskan

Liquid → Gas

Evaporation

Released

Cecair → Gas

Penyejatan

Dibebaskan

Gas → Solid

Sublimation

Released

Gas → Pepejal

Pemejalwapan

Dibebaskan

Gas → Liquid

Condensation

Absorbed

Gas → Cecair

Kondensasi

Diserap

n Sdn.

20

tio

Nil a

m

Process Proses

d. Bh

01-Chem F4 (3p).indd 20

12/9/2011 5:59:33 PM

Chemistry Form 4 • MODULE

5

The diagram below shows the graph of temperature against time when a liquid Y is cooled. Rajah di bawah menunjukkan graf suhu melawan masa apabila cecair Y disejukkan.

Substance Bahan

Melting point/°C Takat lebur/°C

Boiling point/°C Takat didih/°C

S

–182

–162

T

–23

77

U

–97

65

V

41

182

W

132

290

Temperature / Suhu °C t3

P Q

t2

R

Which substance exists as liquid at room temperature? t1

Bahan yang manakah wujud sebagai cecair pada suhu bilik?

S

A

Time (m)

B

Which of the following statements are true about the curve? Antara pernyataan berikut, yang manakah adalah betul tentang lengkung itu?

I

At Q, liquid Y begins to freeze.

II

At PQ, particles in Y absorb heat from the surroundings.

C

S only

T dan U sahaja

D

S and T only S dan T sahaja

8

Pada Q, cecair Y mula membeku.

Liquid Y freezes completely at S.

IV

The freezing point of Y is t2°C.

The diagram below shows standard representation of an atom copper. Rajah di bawah menunjukkan perwakilan piawai atom kuprum.

64 Cu 29

Cecair Y membeku dengan lengkap pada S.

A

C

I dan III sahaja

B

II and III only

Antara berikut, yang manakah betul berdasarkan rajah di atas?

II dan III sahaja

D

I and IV only I dan IV sahaja

6

Which of the following is correct based on the symbol the diagram?

Takat beku bagi Y adalah t2°C.

I and III only

V and W only V dan W sahaja

Pada PQ, zarah dalam Y menyerap haba dari persekitaran.

III

T and U only

S sahaja

Masa (m)

Proton number Nombor proton

Nucleon number Nombor nukleon

Number of electron Bilangan elektron

A

29

64

29

B

35

29

64

C

64

35

29

D

29

64

35

II and IV only II dan IV sahaja

The diagram below shows the graph of temperature against time when solid Z is heated. Rajah di bawah menunjukkan graf suhu melawan masa apabila pepejal Z dipanaskan.

Temperature / Suhu °C 9

The diagram below shows the standard representation of beryllium atom. Rajah di bawah menunjukkan perwakilan piawai atom berillium.

80

9 Be 4 What is the number of valence electrons of beryllium atom?

Time (m) 0

1

2

3

4

5

6

7

8

9

Apakah bilangan elektron valens bagi atom berillium?

Masa (m)

A B

Which of the following is true during the fourth minute? Antara berikut, yang manakah adalah benar pada minit keempat?

A B

C D

7

All the molecules are in random motion. Semua molekul bergerak secara rawak. All the molecules are closely packed and in random motion. Semua molekul sangat rapat dan bergerak secara rawak. All the molecules are vibrating at fixed positions. Semua molekul bergetar pada kedudukan tetap. Some of the molecules are vibrating at fixed positions but some are in random motion. Sebahagian molekul bergetar pada kedudukan tetap dan sebahagian bergerak secara rawak.

The table shows the melting points and boiling points of substances S, T, U, V and W.

C D

4 7

The table below shows the proton number and the number of neutrons for atoms of elements W, X, Y and Z. Jadual di bawah menunjukkan nombor proton dan bilangan neutron bagi atom unsur W, X, Y dan Z. Element Atom

Proton number Nombor proton

Number of neutrons Bilangan neutron

W 7 7 X 8 8 Y 8 9 Z 9 10 Which of the following pair of elements is isotope? Antara pasangan berikut, yang manakah adalah isotop?

A

W and X

C

W dan X

B

W and Y W dan Y

X and Y X dan Y

D

Y and Z Y dan Z

n io

Sdn. B

m

21

. hd

Publicat

Jadual di bawah menunjukkan takat lebur dan takat didih bahan S, T, U, V dan W.

10

2 3

Nila

01-Chem F4 (3p).indd 21

12/9/2011 5:59:34 PM

MODULE • Chemistry Form 4

2

CHEMICAL FORMULA AND EQUATIONS FORMULA DAN PERSAMAAN KIMIA

RELATIF MASS / JISIM RELATIF

• RELATIVE ATOMIC MASS / JISIM ATOM RELATIF (JAR) – To state the meaning of relative mass and solve numerical problems Menyatakan maksud jisim atom relatif dan menyelesaikan masalah pengiraan

• RELATIVE FORMULA MASS / JISIM FORMULA RELATIF (JFR) – To state the meaning of RAM, RMM and RFM based on carbon-12 scale Menyatakan maksud JAR, JMR dan JFR berdasarkan skala karbon-12

• RELATIVE MOLECULAR MASS / JISIM MOLEKUL RELATIF (JMR) – To calculate RAM, RMM and RFM using the chemical formulae of various substances Menghitung JAR, JMR dan JFR menggunakan formula kimia beberapa bahan

MOLE CONCEPT / KONSEP MOL

• MOLE AND THE NUMBER OF PARTICLES / MOL DAN BILANGAN ZARAH – To solve numerical problems involving mole and the number of atoms/ ions/ molecules Menyelesaikan masalah pengiraan melibatkan mol dan bilangan atom, ion dan molekul

• MOLE AND THE MASS OF SUBSTANCES / MOL DAN JISIM BAHAN – To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole concept Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol

• MOLE AND THE VOLUME OF GAS / MOL DAN ISIPADU GAS – To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole concept Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol

CHEMICAL FORMULA AND EQUATIONS / FORMULA DAN PERSAMAAN KIMIA

• EMPIRICAL FORMULA / FORMULA EMPIRIK – Stating the purpose and describe the empirical formula laboratory activities to determine the formula empirical Menyatakan maksud formula empirik dan menghuraikan aktiviti makmal untuk menentukan formula empirik

• MOLECULAR FORMULA / FORMULA MOLEKUL – Solve calculation problems involving empirical formula Menyelesaikan masalah pengiraan melibatkan formula empirik

• CHEMICAL FORMULAE / FORMULA KIMIA – To write formula of anion and cation and to write chemical formula for ionic compounds Menulis formula kation dan anion dan menulis formula kimia untuk sebatian ion

• CHEMICAL EQUATIONS / PERSAMAAN KIMIA – Write a balanced chemical equation and solve problems arrangements involving the mole concept Menulis persamaan kimia seimbang dan menyelesaikan masalah pengiraan yang melibatkan konsep mol

Publica

n Sdn.

22

tio

Nil a

m

d. Bh

02-Chem F4 (3P).indd 22

12/9/2011 5:59:05 PM

Chemistry Form 4 • MODULE

RELATIVE ATOMIC MASS (RAM) / JISIM ATOM RELATIF (JAR) 1

A single atom is too small and light and cannot be weighed directly. Satu atom adalah terlalu ringan, kecil dan tidak dapat ditimbang secara langsung.

2

The best way to determine the mass of a single atom is to compare its mass to the mass of another atom of an element that is used as a standard. Cara yang paling sesuai untuk menentukan jisim satu atom ialah dengan membandingkan jisimnya dengan jisim suatu atom unsur lain yang dianggap sebagai piawai.

3

Hydrogen was the first element to be chosen as the standard for comparing mass because the hydrogen atom is the lightest atom with a mass of 1.0 a.m.u (atomic mass unit). Hidrogen adalah unsur pertama dipilih sebagai piawai untuk membandingkan jisim kerana atom hidrogen adalah unsur yang paling ringan dengan jisim 1.0 u.j.a (unit jisim atom).

Example: Contoh:

• The mass of one helium atom is four times larger than one hydrogen atom. Jisim satu atom Helium adalah 4 kali lebih besar daripada satu atom hidrogen.

• RAM for He is 4. JAR untuk He ialah 4. 4

On the hydrogen scale, the relative atomic mass of an element means the mass of one atom of the element compared to the mass of a single hydrogen atom: Pada skala hidrogen, jisim atom relatif suatu unsur ditakrifkan sebagai jisim satu atom unsur berbanding jisim satu atom hidrogen:

Relative atomic mass of an element (RAM) / Jism atom relatif suatu unsur (JAR) The average mass of one atom of the element / Jisim purata satu atom unsur Mass of one hydrogen atom / Jisim satu atom hidrogen

=

• RAM has no unit. JAR tiada unit.

• The new standard used today is the carbon-12 atom. Piawai yang digunakan sekarang adalah berdasarkan atom karbon-12.

1 • RAM based on the carbon-12 scale is the mass of one atom of the element compared with of the mass of an 12 atom of carbon-12: JAR berdasarkan skala atom karbon-12 adalah jisim satu atom unsur berbanding dengan

n io

Sdn. B

m

23

. hd

Publicat

• Relative atomic mass of an element (RAM) / Jisim atom relatif suatu unsur (JAR) The average mass on one atom of the element / Jisim purata satu atom unsur = 1 × The mass of an atom of carbon-12 / Jisim satu atom karbon-12 12

1 jisim satu atom karbon-12: 12

Nila

02-Chem F4 (3P).indd 23

12/9/2011 5:59:05 PM

MODULE • Chemistry Form 4

RELATIVE MOLECULAR MASS (RMM) / RELATIVE FORMULA MASS (RFM) JISIM MOLEKUL RELATIF (JMR) / JISIM FORMULA RELATIF (JFR)

RMM / JMR =

1

The average mass on one atom of the element / Jisim purata satu molekul

1 × The mass of an atom of carbon-12 / Jisim satu atom karbon-12 12 RMM is obtained by adding up the RAM of all the atoms that are present in the molecule.

2

JMR diperoleh dengan menambahkan JAR semua atom yang terdapat dalam satu molekul. Molecular substance

Molecular formula

Relative molecular mass

Bahan molekul

Formula molekul

Jisim molekul relatif

O2

2 × 16 = 32

Water / Air

H2O

2 × 1 + 16 = 18

Carbon dioxide / Karbon dioksida

CO2

12 + 2 × 16 = 44

Ammonia / Ammonia

NH3

14 + 3 × 1 = 17

Oxygen / Oksigen

[Relative atomic mass / Jisim atom relatif : O = 16, H = 1, C = 12, N = 14]

For ionic substances, RMM is replaced with Relative Formula Mass (RFM).

3

Untuk sebatian ion, JMR digantikan dengan Jisim Formula Relatif (JFR). Substance

Chemical formula

Relative molecular mass

Bahan

Formula kimia

Jisim formula relatif

Sodium chloride / Natrium klorida

NaCl

23 + 35.5 = 58.5

Potassium oxide / Kalium oksida

K 2O

2 × 39 + 16 = 94

CuSO4

64 + 32 + 4 × 16 = 160

(NH4)2CO3

2 [14 + 4 × 1] + 12 + 3 × 16 = 96

Aluminium nitrate / Aluminium nitrat

Al(NO3)3

27 + 3 [14 + 3 × 16] = 213

Calcium hydroxide / Kalsium hidroksida

Ca(OH)2

40 + 2 [16 + 1] = 74

Lead(II) hydroxide / Plumbum(II) hidroksida

Pb(OH)2

207 + 2 [16 + 1] = 241

CuSO45H2O

64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250

Copper(II) sulphate / Kuprum(II) sulfat Ammonium carbonate / Ammonium karbonat

Hydrated copper(II) sulphate / Kuprum(II) sulfat terhidrat

[Relative atomic mass / Jisim atom relatif : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12, Al = 27, Ca = 40, Pb = 207]

(i)

The formula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of metal M? Oksida logam M mempunyai formula M2O3. Jisim formula relatif ialah 152. Apakah jisim atom relatif logam M?

M = RAM for M 2M + 3 × 16 = 152 M = 52 (ii) Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x. Fosforus membentuk sebatian klorida dengan formula PClx. Jisim molekul relatifnya adalah 208.5. Hitungkan nilai x. [Relative atomic mass / Jisim atom relatif : P = 31, Cl = 35.5]

31 + x × 35.5 35.5x 35.5x x Publica

n Sdn.

24

tio

Nil a

m

= 208.5 = 208.5 – 31 = 177.5 =5

d. Bh

02-Chem F4 (3P).indd 24

12/9/2011 5:59:05 PM

Chemistry Form 4 • MODULE

MOLE CONCEPT / KONSEP MOL Mole and the Number of Particles / Bilangan Mol dan Bilangan Zarah 1 2

To describe the amount of atoms, ions or molecules, mole is used. Untuk menyatakan jumlah atom, ion atau molekul, unit mol digunakan. A mole is an amount of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12. Satu mol ialah jumlah bahan yang mengandungi bilangan zarah seperti mana yang terdapat dalam 12 g atom karbon-12.

3

A mole of a substance is the amount of substance which contains a constant number of particles (atoms, ions, molecules), which is 6.02 × 1023. Satu mol bahan adalah jumlah bahan yang mengandungi bilangan zarah yang tetap (atom, molekul, ion) iaitu 6.02 × 1023.

4

The number 6.02 × 1023 is called the Avogadro Constant or Avogadro Number (NA). Nombor 6.02 × 1023 dikenali sebagai Pemalar Avogadro atau Nombor Avogadro (NA ).

5

For compounds that exist as molecules/ions, the number of atoms/ions in that compound must be known. Bagi sebatian yang wujud dalam bentuk molekul/ion, bilangan atom/ion dalam sebatian itu mestilah diketahui.

6 7

The symbol of mole is mol. Simbol untuk mol ialah mol. Complete the following table: Lengkapkan jadual berikut: Substance

Formula

Bahan

Formula

Type of particles

Model / Figure

Number of atom per molecule/ Number of positive and negative ion

Model / Rajah

Jenis zarah

Bilangan atom per molekul/ Bilangan ion positif dan negatif

8

Chlorine / Klorin

Cl2

Molecule

Cl Cl

Water / Air

H2O

Molecule

H O H

Ammonia / Ammonia

NH3

Molecule

H H N H

Sulphur dioxide / Sulfur dioksida

SO2

Molecule

O S O

Magnesium chloride / Magnesium klorida

MgCl2

Ion

[Cl]– [Mg]2+ [Cl]–

Aluminium oxide / Aluminium oksida

Al2O3

Ion

[O]2– [A1]3+ [O]2– [A1]3+ [O]2–

Cl : 2 H: 2 O:1 N:1 H: 3 S:1 O:2 Mg2+ : 1 Cl–

:2

Al3+ : 2 O2–

:3

Relationship between number of moles and number of particles (atoms/ions/molecules): Hubungan bilangan mol dan bilangan zarah (atom/ion/molekul):

Number of moles Bilangan mol 9

× Avogadro Constant / Pemalar Avogadro ÷ Avogadro Constant / Pemalar Avogadro

Number of particles Bilangan zarah

Complete the following: [Differentiate between “mole” dan “molecule”] Lengkapkan yang berikut: [Bezakan antara “mol” dan “molekul”]

(a) 1 mol of Cl2 [Chlorine gas] 1 mol Cl2 [Gas klorin]

(b) 1 mol of NH3 [Ammonia gas]

molecules of chlorine, Cl2 / molekul klorin, Cl2

2 × 6.02 × 1023 atoms of chlorine, Cl / atom klorin, Cl 6.02 × 1023 4

molecules of ammonia, NH3 / molekul ammonia, NH3 1 mol of nitrogen atom, N / mol atom nitrogen, N

mol atoms / mol atom

3

mol of hydrogen atoms, H / mol atom hidrogen, H n io

Sdn. B

m

25

. hd

Publicat

1 mol NH3 [Gas ammonia]

6.02 × 1023

Nila

02-Chem F4 (3P).indd 25

12/9/2011 5:59:05 PM

MODULE • Chemistry Form 4

0.25 × 6.02 × 1023

1 mol of NH3 4 [Ammonia gas]

(c)

1 mol NH3 4 [Gas ammonia]

1

mol of atoms

1

mol atom

molecules of ammonia, NH3 / molekul ammonia, NH3 0.25 mol of N atoms / mol atom N, 23 number of N atoms / bilangan atom N = 0.25 × 6.02 × 10 0.75 mol of H atoms / mol atom H, number of H atoms / bilangan atom H =

2 mol of Mg2+ ions / mol ion Mg2+, number of Mg2+ ions / bilangan ion Mg2+ =

(d) 2 mol of MgCl2 [Magnesium chloride] 2 mol MgCl2 [Magnesium klorida]

4 mol of Cl– ions / mol ion Cl–, number of Cl- ions / bilangan ion Cl– =

0.75 × 6.02 × 1023

2 × 6.02 × 1023

4 × 6.02 × 1023

2 × 6.02 × 1023

(e) 2 mol of SO2 [Sulphur dioxide]

molecules of SO2 / molekul SO2 2 mol of S atoms / mol atom S, number of S atoms / bilangan atom S = 3 × 2 = 6 mol of atoms

2 mol SO2 [Sulfur dioksida]

3×2=6

mol atom

2 × 6.02 × 1023

4 mol of O atoms / mol atom O, number of O atoms / bilangan atom O =

4 × 6.02 × 1023

10 Complete the table below: Lengkapkan jadual berikut: Number of moles

Number of particles

Bilangan mol

Bilangan zarah

0.5

mole of carbon, C

3.01 × 1023 atoms of carbon

0.5

mol atom karbon, C

3.01 × 1023 atom karbon

0.2 moles of hydrogen gas, H2

(i)

0.2 mol gas hidrogen, H2

(ii) 1

molecules of hydrogen / molekul hidrogen

2 × 0.2 × 6.02 × 1023 atoms of hydrogen / atom hidrogen

6.02 × 1023 molecules of carbon dioxide contains:

mole of carbon dioxide molecules, CO2

1

0.2 × 6.02 × 1023

6.02 × 1023 molekul karbon dioksida mengandungi:

mol molekul karbon dioksida, CO2

6.02 × 1023 23

6.02 × 10

atoms of C and atom C dan

2 × 6.02 × 1023

2 × 6.02 × 1023

atoms of O.

atom O.

NUMBER OF MOLES AND MASS OF SUBSTANCE / BILANGAN MOL DAN JISIM BAHAN Molar mass / Jisim molar (a) Molar mass is the mass of one mole of any substance / Jisim molar adalah jisim satu mol sebarang bahan. (b) Molar Mass is the relative atomic mass, relative molecular mass and relative formula mass of a substance in g mol–1.

1

Jisim molar adalah jisim atom relatif, jisim molekul relatif dan jisim formula relatif suatu bahan dalam g mol–1.

(c) Molar mass of any substance is numerically equal to its relative mass (Relative atomic mass/ relative formula mass/relative molecular mass). Jisim molar sebarang bahan mempunyai nilai yang sama dengan jisim relatif (Jisim atom relatif/ jisim formula relatif/ jisim molekul relatif).

Example / Contoh: Molar mass of H2O = 18 g mol–1

2

Jisim molar H2O = 18 g mol–1

× RAM/ /RFM/RMM

Mass of 1 mol of H2O = 18 g Jisim 1 mol H2O = 18 g

Mass of 2 mol of H2O = 2 mol × 18 g mol–1 = 36 g

Jisim 2 mol H2O = 2 mol ×

Mass of Jisim Publica

36

g

Bilangan mol

÷ RAM/ /RFM/RMM

Mass in gram Jisim dalam gram

÷ JAR/JFR/JMR

mol of H2O = 45 g mol H2O = 45 g

n Sdn.

26

2.5

g mol–1 =

× JAR/JFR/JMR

tio

Nil a

m

2.5

18

Number of moles

d. Bh

02-Chem F4 (3P).indd 26

12/9/2011 5:59:06 PM

Chemistry Form 4 • MODULE

3

Complete the following table: Lengkapkan jadual berikut: Element/ Compound

Chemical formula

Unsur/Sebatian

Formula kimia

Copper

RAM/RMM/RFM

Calculate

JAR/JMR/JFR

Penghitungan

Cu

RAM/JAR = 64

NaOH

RFM/JFR = 40

–1 (a) Mass of 1 mol of Cu / Jisim 1 mol Cu : 1 mol × 64 g mol = 64 g 2 mol × 64 g mol–1 = 128 g (b) Jisim 2 mol / Jisim 1 mol : 1 mol × 64 g mol–1 = 32 g 1 1 2 (c) Jisim mol / Jisim mol: 2 2 32 g (d) Mass of 3.01 × 1023 Cu atoms / Jisim 3.01 × 1023 atom Cu:

Kuprum

Sodium hydroxide

(a) Mass of 3 mol of sodium hydroxide:

Natrium hidroksida

Jisim 3 mol natrium hidroksida:

120 g

120 g

(b) Number of moles in 20 g sodium hydroxide: Bilangan mol natrium hidroksida dalam 20 g:

Oxygen gas Gas oksigen

O2

RMM/JMR =

32

(a) Mass of 2.5 mol of oxygen gas: Jisim 2.5 mol gas oksigen:

0.5 mol

0.5 mol

2.5 mol × 32 g mol–1 = 80 g

2.5 mol × 32 g mol–1 = 80 g

(b) Number of moles is 1.5 mol oxygen gas: Bilangan molekul dalam 1.5 mol gas oksigen:

1.5 mol × 6.02 × 1023 1 (c) Number of molecules in mol of oxygen gas: 2 1 Bilangan molekul dalam mol gas oksigen: 2 0.5 mol × 6.02 × 1023 (d) Number of atoms in 2 mol of oxygen gas: Bilangan atom dalam 2 mol gas oksigen:

2 × 2 × 6.02 × 1023 Sodium chloride

NaCl

Natrium klorida

Zinc nitrate Zink nitrat

Zn(NO3)2

RFM/JFR = 58.5

RFM/JFR =

189

Mass of 0.5 mol of NaCl / Jisim bagi 0.5 mol NaCl: 0.5 mol × 58.5 g mol–1 = 29.25 g Number of moles in 37.8 g of zinc nitrate: Bilangan mol dalam 37.8 g zink nitrat:

37.8 g/189 g mol–1 = 0.2 mol [Relative atomic mass / Jisim atom relatif: Cu = 64, Na = 23, O = 16, H = 1, Cl = 35.5, Zn = 65, N = 14]

NUMBER OF MOLES AND VOLUME OF GAS / BILANGAN MOL DAN ISI PADU GAS 1

Molar volume of a gas: Volume occupied by one mole of any gas is 24 dm3 at room conditions and 22.4 dm3 at

standard temperature and pressure (STP). Isi padu molar gas: Isipadu yang dipenuhi oleh satu mol sebarang gas iaitu 24 dm3 pada keadaan bilik dan 22.4 dm3 pada suhu dan tekanan piawai (STP). 2

The molar volume of any gas is 24 dm3 at room conditions and 22.4 dm3 at STP. Isi padu molar sebarang gas adalah 24 dm3 pada keadaan bilik dan 22.4 dm3 pada STP.

3

Generalisation: One mole of any gas always occupies the same volume under the same temperature and pressure: Umumnya: satu mol sebarang jenis gas menempati isi padu yang sama pada suhu dan tekanan yang sama.

Example / Contoh: (i) 1 mol of oxygen gas, 1 mol ammonia gas, 1 mol helium gas dan 1 mol sulphur dioxide gas occupy the same volume of 24 dm3 at room conditions. 1 mol gas oksigen, 1 mol gas ammonia, 1 mol gas helium dan 1 mol gas sulfur dioksida menempati isi padu yang sama iaitu 24 dm3 pada keadaan bilik.

44.8

(ii) 2 mol of carbon dioxide gas occupies 44.8

dm3 pada STP.

n io

Sdn. B

m

27

. hd

Publicat

2 mol gas karbon dioksida menempati

dm3 pada STP.

Nila

02-Chem F4 (3P).indd 27

12/9/2011 5:59:06 PM

MODULE • Chemistry Form 4

(iii) 16 g of oxygen gas = 0.5 mol of oxygen gas. Therefore, 16 g of oxygen gas occupies a volume of at room conditions [Relative atomic mass: O =16] 0.5 mol gas oksigen. Oleh itu, 16 g gas oksigen menempati isi padu 16 g gas oksigen = [Jisim atom relatif; O = 16]

Number of moles of gas Bilangan mol gas

× 24 dm3 mol–1/ 22.4 dm3 mol–1

12

12

dm3

dm3 pada keadaan bilik.

Volume of gas in dm2 Isi padu gas dalam dm3

÷ 24 dm3 mol–1/ 22.4 dm3 mol–1

Formula for conversion of unit: Formula untuk penukaran unit:

Volume of gas in dm3 Isi padu gas dalam dm3

÷ 24 dm3 mol–1/ 22.4 dm3 mol–1 ÷ (RAM/ /RFM/RMM) g mol–1 Mass in gram (g) Jisim dalam gram (g)

× 24 dm3 mol–1/ 22.4 dm3 mol–1

÷ (JAR/JFR/JMR) g mol–1

Number of moles

÷ (6.02 × 1023)

× (RAM/ /RFM/RMM) g mol–1

Bilangan mol

× (6.02 × 1023)

–1

× (JAR/JFR/JMR) g mol

Number of particles Bilangan zarah

EXERCISE / LATIHAN

Relative atomic mass of calcium is 40 based on the carbon-12 scale.

1

Jisim atom relatif kalsium berdasarkan skala karbon-12 ialah 40.

(a) State the meaning of the statement above. Nyatakan maksud penyataan di atas.

Mass of calcium atom is 4 times greater than

1 mass of carbon-12 atom. 12

(b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16] Berapa kalikah satu atom kalsium lebih berat daripada satu atom oksigen? [JAR: O = 16]

Relative atomic mass of calcium 40 = = 2.5 times Relative atomic mass of oxygen 16 (c) How many calcium atoms have the same mass as two atoms of bromine? [RAM Br = 80] Berapakah bilangan atom kalsium yang mempunyai jisim yang sama dengan dua atom bromin? [Jisim atom relatif: Br = 80]

Number of calcium atom × 40 = 2 × 80 2 × 80 Number of calcium atom = =4 40 A sampel of chlorine gas weighs 14.2 g. Calculate / Suatu sampel gas klorin berjisim 14.2 g. Hitungkan: [Relative atomic mass / Jisim atom relatif : Cl = 35.5] (a) Number of moles of chlorine atoms / Bilangan mol atom klorin. 14.2 Number of mol of chlorine atoms, Cl = = 0.4 mol 35.5

2

(b) Number of moles of chlorine molecules (Cl2) / Bilangan mol molekul klorin (Cl2 ). 14.2 Number of mol of chlorine molecule, Cl2 = = 0.2 mol 71 (c) Volume of chlorine gas at room conditions / Isi padu gas klorin pada keadaan bilik. [Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure] [Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan piawai]

Volume of chlorine gas = 0.2 mol × 24 dm3 mol–1 = 4.8 dm3 Publica

n Sdn.

28

tio

Nil a

m

d. Bh

02-Chem F4 (3P).indd 28

12/9/2011 5:59:06 PM

Chemistry Form 4 • MODULE

3

(a) Calculate the number of atoms in the following substances / Hitungkan bilangan atom yang terdapat dalam bahan berikut: [Relative atomic mass: N = 14; Zn = 65; Avogadro Constant = 6.02 × 1023] [Jisim atom relatif: N = 14; Zn = 65; Pemalar Avogadro = 6.02 × 1023] (i) 13 g of zinc / 13 g zink

13 = 0.2 mol 65 Number of zinc atom = 0.2 × 6.02 × 1023 = 1.204 × 1023 Number of mol of zinc atom =

(ii)

5.6 g of nitrogen gas / 5.6 g gas nitrogen 5.6 = 0.4 mol 14 Number of N atom = 0.4 × 6.02 × 1023 = 2.408 × 1023 Number of mol of N atom =

(b) Calculate the number of molecules in the following substances / Hitungkan bilangan molekul dalam bahan berikut: [Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant = 6.02 × 1023] [Jisim atom relatif: N = 14, H = 1, Cl = 35.5, Pemalar Avogadro = 6.02 × 1023] (i) 8.5 g of ammonia gas, NH3 / 8.5 g gas ammonia, NH3

8.5 × 6.02 × 1023 17 = 2.408 × 1023 (ii)

4

14.2 g of chlorine gas, Cl2 / 14.2 g gas klorin, Cl2 14.2 × 6.02 × 1023 71 = 1.2 × 1023

A gas jar contains 240 cm3 of carbon dioxide gas. Calculate: Suatu balang gas berisi 240 cm3 gas karbon dioksida. Hitungkan:

[Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions] [Jisim atom relatif: C = 12, O = 16; Isi padu molar gas: 24 dm3 mol–1 pada keadaan bilik] (a) Number of moles of carbon dioxide gas / Bilangan mol gas karbon dioksida:

Number of moles of CO2 =

240 = 0.01 mol 24 000

(b) Number of molecules of carbon dioxide gas / Bilangan molekul gas karbon dioksida: Number of molecules of CO2 = 0.01 × 6.02 × 1023 = 6.02 × 1021 (c) Mass of carbon dioxide gas / Jisim gas karbon dioksida: Mass of CO2 = 0.01 mol × [12 + 2 × 16] g mol–1 = 0.44 g 5

What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as that found in 3.6 g of water? Berapakah jisim molekul klorin (Cl2 ) yang mengandungi dua kali ganda bilangan molekul yang terdapat dalam 3.6 g air? [Relative atomic mass / Jisim atom relatif : H = 1, O = 16, Cl = 35.5]

Number of moles of chlorine molecule = 2 × no of mol in H2O 3.6 =2× = 0.4 mol 18 Mass of Cl2 = 0.4 × 71= 28.4 g 6

Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium. Hitungkan jisim karbon yang mempunyai bilangan atom yang sama seperti yang terdapat dalam 4 g magnesium. [Relative atomic mass / Jisim atom relatif : C = 12, Mg = 24]

n io

Sdn. B

m

29

. hd

Publicat

2g

Nila

02-Chem F4 (3P).indd 29

12/9/2011 5:59:06 PM

MODULE • Chemistry Form 4

Compare the number of molecule in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your answer.

7

Bandingkan bilangan molekul dalam 32 g sulfur dioksida (SO2 ) dengan 7 g gas nitrogen (N2 ). Terangkan jawapan anda. [Relative atomic mass / Jisim atom relatif : S = 32, O = 16, N = 14]

Number of moles of molecules in 32 g SO2 =

32 = 0.5 mol 64

7 = 0.25 mol 28 Number of molecule in 32 g SO2 is two times more than 7 g N2. Number of mole in sulphur dioxide molecule is two times more than number of mole of nitrogen molecule. Number of moles of molecules in 7 g N2 =

Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer.

8

Bandingkan bilangan atom dalam 1.28 g oksigen dengan bilangan atom dalam 1.3 g zink. Terangkan jawapan anda. [Relative atomic mass / Jisim atom relatif : O = 16, Zn = 65]

1.28 = 0.08 mol 16 1.30 Number of mol of Zn atoms in 1.3 g Zn = = 0.04 mol 65 Number of oxygen atoms in 1.28 g oxygen is 2 times more than number of zinc atoms in 1.3 g zinc. Number of mol of oxygen atom is 2 times more than zinc atom.

Number of mol of O atoms in 1.28 g SO2 =

CHEMICAL FORMULAE AND CHEMICAL EQUATIONS / FORMULA KIMIA DAN PERSAMAAN KIMIA

Symbol of elements – use capital letters for the first alphabet and use small letters if there is a second alphabet.

1

Simbol unsur – gunakan huruf besar untuk huruf pertama dan huruf kecil jika ada huruf kedua.

Example / Contoh: Potassium / Kalium – K, Calcium / Kalsium – Ca, – Fe, Iron / Ferum

Sodium / Natrium – Na Nitrogen / Nitrogen – N Fluorine / Fluorin – F

Chemical Formula – A set of chemical symbols for atoms of elements in whole numbers representing chemical substances. Formula kimia – Satu set simbol kimia bagi atom-atom unsur dengan gandaan nombor bulat yang mewakili bahan kimia. Chemical substance

Chemical formula

Notes

Bahan kimia

Formula kimia

Catatan

Water Air

Ammonia Ammonia

Propane Propana

NH3 C3H8

2 atoms of H combines with 1 atom of O. 2 atom H bergabung dengan 1 atom O.

3 atoms of H combines with 1 atom of N. 3 atom H bergabung dengan 1 atom N.

3 atoms of C combines with 8 atoms of H. 3 atom C bergabung dengan 8 atom H.

2

Information that can be obtained from the chemical formula / Maklumat yang diperoleh daripada formula kimia: (i) All the elements present in the compound / Jenis unsur yang terdapat dalam sebatian, (ii) Number of atoms of each element in the compound / Bilangan atom setiap unsur yang terdapat dalam sebatian, (iii) Calculation of RMM/RFM of the compound / Pengiraan JMR/JFR bagi sebatian.

3

Two types of chemical formula / Dua jenis formula kimia: (i) Empirical formula / Formula empirik, (ii) Molecular formula / Formula molekul.

Publica

n Sdn.

30

tio

Nil a

m

H2O

d. Bh

02-Chem F4 (3P).indd 30

12/9/2011 5:59:06 PM

Chemistry Form 4 • MODULE

EMPIRICAL FORMULA / FORMULA EMPIRIK 1 2

A formula that shows the simplest whole number ratio of atoms of each element in a compound. Formula yang menunjukkan nisbah nombor bulat teringkas bagi bilangan atom setiap unsur yang terdapat dalam sebatian. The formula can be determined by calculating the simplest ratio of moles of atoms of each element in the compound. Formula itu boleh ditentukan dengan menghitung nisbah bilangan mol atom bagi setiap unsur yang terdapat dalam sebatian.

3

Experiments to determine empirical formula of metal oxide / Formula empirik bagi oksida logam diperoleh dengan cara: Empirical formula of magnesium oxide

Empirical formula of copper(II) oxide

Formula empirik magnesium oksida

Formula empirik kuprum(II) oksida

Set-up of apparatus / Susunan radas:

Set-up of apparatus / Susunan radas: Copper(II) oxide Kuprum(II) oksida

Magnesium Magnesium

Hydrogen gas Gas hidrogen

Heat

Heat

Panaskan

4

Panaskan

Reaction occurs / Tindak balas yang berlaku:

Reaction occurs / Tindak balas yang berlaku:

Magnesium is burnt in a crucble to react with oxygen to form magnesium oxide.

Hydrogen gas is passed through heated copper(II) oxide. Hydrogen reduces copper(II) oxide to form copper and water.

Magnesium dipanaskan dengan kuat di dalam mangkuk pijar untuk bertindak balas dengan oksigen membentuk magnesium oksida.

Gas hidrogen dilalukan melalui kuprum(II) oksida yang dipanaskan. Hidrogen menurunkan kuprum(II) oksida kepada kuprum dan air.

Balanced equation / Persamaan kimia seimbang:

Balanced equation / Persamaan kimia seimbang:

2Mg + O2 → 2MgO

CuO + H2 → Cu + H2O

This method can also be used to determine the empirical formulae of reactive metals such as aluminium oxide and zinc oxide.

This method can also be used to determine the empirical formulae of less reactive metals such as lead(II) oxide and tin(II) oxide.

Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida logam reaktif seperti aluminium oksida dan zink oksida.

Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida logam kurang reaktif seperti plumbum(II) oksida and stanum(II) oksida.

Experiment to Determine Empirical Formula of Magnesium Oxide Eksperimen untuk Menentukan Formula Empirik Magnesium Oksida

In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide: Semasa eksperimen ini, magnesium bertindak balas dengan oksigen dalam udara untuk membentuk asap putih, magnesium oksida:

Magnesium + Oxygen → Magnesium oxide Magnesium + Oksigen → Magnesium oksida Material / Bahan: Magnesium ribbon, sand paper Apparatus / Radas:

Crucible with lid, tongs, Bunsen burner, tripod stand and balance

Set-up of apparatus / Susunan radas:

Magnesium ribbon

n io

Sdn. B

m

31

. hd

Publicat

Heat

Nila

02-Chem F4 (3P).indd 31

12/9/2011 5:59:07 PM

MODULE • Chemistry Form 4

Procedure / Langkah:

crucible

(a) A

(c) The

dengan

ditimbang.

sand paper

is cleaned with

pita magnesium

dibersihkan dengan menggunakan

magnesium ribbon

coiled

Pita magnesium

is gulung

di

crucible

(d) The

are weighed.

penutup

magnesium ribbon

(b) 10 cm of 10 cm

lid

and its

Mangkuk pijar

..

loosely and placed in the crucible.

dan diletakkan dalam mangkuk pijar.

magnesium ribbon

together with the lid and

Mangkuk pijar

.

kertas pasir

pita magnesium

bersama dengan penutup dan

are weighed again. ditimbang.

(e) The apparatus is set up as shown in the diagram. Radas disusun seperti dalam gambar rajah.

(f)

strongly

The crucible is heated burn

Mangkuk pijar dipanaskan dengan terbakar , mangkuk pijar ditutup dengan

lid

pita magnesium Apabila kuat selama 2 minit lagi.

penutup

weighed again

, lid and its content are heating

constant

is removed and the crucible is

dibuka dan mangkuk pijar dipanaskan dengan

cooling

,

suhu bilik

.

.

ditimbang sekali lagi

, penutup dan kandungannya

(k) The process of

.

repeated

and weighing are

until a

mass is obtained.

pemanasan

Proses

,

, penutup dan kandungannya dibiarkan sejuk ke

Mangkuk pijar

lid

, the

lid and its content are allowed to cool down to room temperature .

crucible

The

burning terbakar

berhenti

crucible Mangkuk pijar

(j)

. Apabila pita magnesium mula

.

dibuka sekali sekala dengan menggunakan penyepit.

(h) When the magnesium ribbon stops heated strongly for another 2 minutes.

The

tanpa penutup

.

penutup

of the crucible is lifted from time to time using a pair of tongs.

Penutup

(i)

. When the magnesium starts to

lid

, the crucible is covered with its kuat

(g) The

lid

without its

tetap

penyejukan

,

dan penimbangan

diulang

beberapa kali sehingga jisim

diperoleh.

Observation / Pemerhatian:

Magnesium burns

brightly

Magnesium terbakar dengan

white fumes

to release

terang

membebaskan

and

wasap putih

white solid

is formed. pepejal putih

dan kemudiannya membentuk

.

Inference / Inferens:

Magnesium is a

reactive

metal. reaktif

Magnesium adalah logam yang

Magnesium reacts with

oxygen

Magnesium bertindak balas dengan

in the air to form oksigen

magnesium oxide

dalam udara membentuk

.

magnesium oksida

.

Publica

n Sdn.

32

tio

Nil a

m

.

d. Bh

02-Chem F4 (3P).indd 32

12/9/2011 5:59:07 PM

Chemistry Form 4 • MODULE

Precaution steps / Langkah berjaga-jaga: Step taken / Langkah yang diambil Magnesium ribbon is cleaned with Pita magnesium perlu digosok dengan

Purpose / Tujuan

sand paper . kertas pasir

To remove the ribbon.

.

oxide layer

on the surface of the magnesium

Untuk membuang lapisan oksida pada permukaan magnesium oksida.

The

crucible lid

is lifted from time to time.

crucible lid

replaced

then

from the air to react with magnesium .

Untuk membenarkan oksigen masuk dan bertindak balas dengan magnesium .

Penutup mangkuk pijar dibuka sekali sekala.

The

oxygen

To allow

To prevent fumes of

quickly.

magnesium oxide

from escaping.

Untuk mengelakkan wasap magnesium oksida dari terbebas.

Penutup mangkuk pijar kemudian ditutup semula dengan cepat.

The process of

heating , cooling and weighing are repeated until a constant mass is obtained.

To ensure magnesium react for magnesium oxide .

completely

oxygen

pemanasan , penyejukan penimbang dan Proses jisim tetap diperoleh. diulang beberapa kali sehingga

lengkap Untuk memastikan semua magnesium telah bertindak balas oksigen untuk membentuk magnesium oksida . dengan

with

to

Result / Keputusan: Description / Penerangan

Mass (g) / Jisim (g)

Mass of crucible + lid

x

Jisim mangkuk pijar + penutup

Mass of crucible + lid + magnesium

y

Jisim mangkuk pijar + penutup + magnesium

Mass of crucible + lid + magnesium oxide

z

Jisim mangkuk pijar + penutup + magnesium oksida

Calculation / Pengiraan: Element / Unsur

Mg

O

Mass (g) / Jisim (g)

y–x

z–y

Number of mole of atoms / Bilangan mol atom

y–x 24

z–y 16

Simplest ratio of moles / Nisbah mol teringkas

p

q

Empirical formula of magnesium oxide is Formula empirik magnesium oksida ialah 5

MgpOq MgpOq

. .

Experiment to Determine Empirical Formula of Copper(II) Oxide Eksperimen untuk Menentukan Formula Empirik Kuprum(II) Oksida

Copper(II) Oxide + Hidrogen → Copper + Water Kuprum(II) oksida + Hidrogen → Kuprum + Air Set-up of apparatus / Susunan radas:

Copper(II) oxide Burning of hydrogen gas Hydrogen gas Combustion tube Heat

n io

Sdn. B

m

33

. hd

Publicat

Anhydrous calcium chloride, CaCl2

Nila

02-Chem F4 (3P).indd 33

12/9/2011 5:59:07 PM

MODULE • Chemistry Form 4

Observation / Pemerhatian:

The

black

Warna

colour of copper(II) oxide turns

hitam

brown

perang

kuprum(II) oksida menjadi

.

.

Inference / Inferens:

copper metal

Copper(II) oxide reacts with hydrogen to produce the brown

.

logam kuprum

Kuprum(II) oksida bertindak balas dengan hidrogen untuk menghasilkan

yang berwarna perang.

Precaution steps / Langkah berjaga-jaga: Step taken / Langkah yang ambil

Purpose / Tujuan

Hydrogen gas is passed through anhydrous calcium chloride.

hydrogen gas.

Gas hidrogen dialirkan melalui kalsium klorida kontang.

To

remove

all the

air

gas hidrogen.

in the combustion tube. air explodes when lighted).

(The mixture of hydrogen gas and

kering dialirkan melalui tabung pembakaran Gas hidrogen selama 5 hingga 10 minit.

udara dalam tabung pembakaran. Untuk mengeluarkan semua udara (Campuran hidrogen dan menghasilkan letupan apabila dinyalakan)

If the gas burns quietly without ‘pop’ sound , all the has been removed from the combustion tube.

The gas that comes out from the small hole is collected in the test tube. Then, a lighted wooden splinter is

bunyi ‘pop’ Jika gas terbakar tanpa daripada tabung pembakaran.

at mouth of the test tube.

Gas yang keluar daripada lubang kecil dikumpul dalam sebuah tabung uji. Kayu uji menyala di letakkan di mulut tabung uji.

The flow of hydrogen gas must be throughout the experiment.

mengering

Kalsium klorida kontang menyerap wap air untuk

Dry hydrogen is passed through the combustion tube for 5 to 10 minutes.

placed

dry

Anhydrous calcium chloride absorb water vapour to

continuous

dikeluarkan

, semua gas telah

To prevent hot copper from reacting with copper(II) oxide again.

air

oxygen

to form

Gas hidrogen dialirkan secara berterusan sepanjang eksperimen.

Untuk mengelakkan kuprum panas daripada bertindak balas dengan oksigen dan membentuk kuprum(II) oksida .

heating , cooling and weighing are constant mass is obtained. repeated until a

To ensure all copper(II) oxide has changed to

The process of

pemanasan , penyejukan Proses diulang beberapa kali sehingga jisim

dan tetap

copper .

Untuk memastikan semua kuprum(II) oksida telah bertukar kepada

kuprum .

penimbang diperoleh.

Result / Keputusan: Description / Penerangan

Mass (g) / Jisim (g)

Mass of combustion tube + porcelain dish

x

Jisim tabung pembakaran + piring tanah liat

Mass of combustion tube + porcelain dish + copper(II) oxide

y

Jisim tabung pembakaran + piring tanah liat + kuprum(II) oksida

Mass of combustion tube + porcelain dish + copper

z

Jisim tabung pembakaran + piring tanah liat + kuprum

Calculation / Pengiraan: Element / Unsur

Cu

O

Mass (g) / Jisim (g)

z–x

y–z

Number of mole of atoms / Bilangan mol atom

z–x 64

y–z 16

Simplest ratio of moles / Nisbah mol teringkas

p

q

Empirical formula of copper(II) oxide is Formula empirik kuprum(II) oksida ialah

. .

Publica

n Sdn.

34

tio

Nil a

m

CupOq CupOq

d. Bh

02-Chem F4 (3P).indd 34

12/9/2011 5:59:07 PM

Chemistry Form 4 • MODULE

6

Explain why the set-up of apparatus to determine the empirical formula in both the experiments is different. Terangkan mengapa susunan radas untuk menentukan formula empirik dalam kedua-dua eksperimen itu berbeza.

reactive

(a) Magnesium is

magnesium oxide

metal (above hidrogen in reactivity series). Magnesium

by

hydrogen

hydrogen gas

Kuprum di bawah gas hidrogen 7

easily to form

.

reaktif Magnesium adalah logam membentuk magnesium oksida .

(b) Copper is below

reacts

teroksida

(terletak di atas hidrogen dalam siri kereaktifan. Magnesium mudah

in the metal reactivity series. Oxygen in copper(II) oxide can be

reduced/removed

to form copper and water. hidrogen

dalam siri kereaktifan. Kuprum(II) okida boleh

diturunkan/disingkirkan

oleh

untuk membentuk kuprum dan air.

To calculate the empirical formula of a compound, use the following table: Untuk menghitung formula empirik suatu sebatian, jadual di bawah boleh digunakan sebagai panduan:

Calculation steps / Langkah pengiraan:

Element / Unsur

(a) Calculate the mass of each element in the compound. Hitungkan jisim setiap unsur dalam sebatian.

Mass of element (g) / Jisim unsur (g)

(b) Convert the mass of each element to number of mole of atom.

Number of mole of atom / Bilangan mol atom

Tukar jisim setiap unsur kepada bilangan mol atom.

(c) Calculate the simplest ratio of moles of atom of the elements.

Simplest ratio of moles / Nisbah mol teringkas

Hitungkan nisbah bilangan mol atom teringkas unsur-unsur tersebut.

EXERCISE / LATIHAN

1

When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical formula of metal X oxide. Apabila 11.95 g oksida logam X diturunkan oleh hidrogen, 10.35 g logam terhasil. Hitungkan formula empirik bagi oksida logam X. [RAM / JAR: X = 207, O = 16] X

O

Mass of element (g) / Jisim unsur (g)

10.35

1.6

Number of mole of atoms / Bilangan mol atom

0.05

0.1

Ratio of moles / Nisbah mol

1

2

Simplest ratio of moles / Nisbah mol teringkas

1

2

Element / Unsur

Empirical formula / Formula empirik: 2

XO2

.

A certain compound contains the following composition / Satu sebatian mengandungi komposisi unsur seperti berikut: Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass / Jisim atom relatif: O = 16, Na = 23, Br = 80] (Assume that 100 g of substance is used / Anggap 100 g bahan digunakan) Element / Unsur

Na

Br

O

Mass of element (g) / Jisim unsur (g)

15.23

52.98

31.79

Number of mole of atoms / Bilangan mol atom

0.66

0.66

1.99

Ratio of moles / Nisbah mol

1

1

3.01

Simplest ratio of moles / Nisbah mol teringkas

1

1

3

NaBrO3

. n io

Sdn. B

m

35

. hd

Publicat

Empirical formula / Formula empirik:

Nila

02-Chem F4 (3P).indd 35

12/9/2011 5:59:07 PM

MODULE • Chemistry Form 4

2.08 g of element X combines with 4.26 g of element Y to form a compound with formula XY3. Calculate the relative atomic mass of element X. [RAM: Y = 35.5]

3

2.08 g unsur X bergabung dengan 4.26 g unsur Y untuk membentuk sebatian dengan formula XY3 . Hitung jisim atom relatif unsur X. [JAR: Y = 35.5] Element / Unsur Mass of element (g) Jisim unsur (g)

Number of mole of atoms Bilangan mol atom

Simplest ratio of moles Nisbah mol teringkas

X

Y

2.08

4.26

2.08 x

4.26 = 0.12 35.5

1

3

x = relative atomic mass of X Mol X = 1 Mol Y 3 2.08 x 1 = 0.12 3 x = 52

2.07 g of element Z reacts with bromine to form 3.67 g of a compound with empirical formula ZBr2. Calculate the relative atomic mass of element Z. [RAM: Br = 80]

4

2.07 g unsur Z bertindak balas dengan bromin membentuk 3.67 g sebatian dengan formula empirik ZBr2. Hitung jisim atom relatif bagi unsur Z. [JAR: Br = 80] Element / Unsur Mass of element (g) Jisim unsur (g)

Number of mole of atoms Bilangan mol atom

Simplest ratio of moles Nisbah mol teringkas

Z

Br

2.07

1.6

2.07 z

1.6 = 0.02 80

1

2

z = relative atomic mass of Z Mol Z Mol Br 2.08 z 0.02

= 1

2

1 2 z = 207 =

The statement below is about compound J / Pernyataan berikut adalah mengenai sebatian J.

5

• It is black solid / Merupakan pepejal hitam. • Contains 12.8 g copper and 0.2 mol of oxygen / Mengandungi 12.8 g kuprum dan 0.2 mol oksigen. [Relative atomic mass / Jisim atom relatif : Cu = 64] (a) What is meant by empirical formula / Apakah maksud formula empirik? A formula that shows the simplest whole number ratio of atoms of each element in a compound. (b) (i)

Calculate the number of mol of copper atom / Hitung bilangan mol atom kuprum. 12.8 = 0.2 mol 64

(ii)

What is the empirical formula of compound J / Apakah formula empirik sebatian J ? 0.2 mol Cu : 0.2 mol O. 1 mol Cu : 1 mol O. Empirical formula of Compound J is CuO.

(c) Compound J reacts completely with hydrogen to produce copper and compound Q. Sebatian J bertindak balas lengkap dengan hidrogen menghasilkan kuprum dan sebatian Q.

(i)

State one observation for the reaction / Nyatakan satu pemerhatian daripada tindak balas tersebut. Black solid change to brown

(ii)

Name two the substances that can be used to prepare hydrogen gas. Namakan dua bahan yang digunakan untuk menyediakan gas hidrogen.

Zinc/magnesium and hydrochloric acid/nitric acid/sulphuric acid. (iii) Name compound Q / Nama sebatian Q. Water (iv) Write a balanced equation for the reaction. Tuliskan persamaan kimia yang seimbang bagi tindak balas tersebut.

CuO + H2 → Cu + H2O Publica

n Sdn.

36

tio

Nil a

m

d. Bh

02-Chem F4 (3P).indd 36

12/9/2011 5:59:08 PM

Chemistry Form 4 • MODULE

(d) Draw a labelled diagram of the set-up of apparatus for the experiment. Lukiskan gambar rajah berlabel susunan radas bagi tindak balas tersebut.

Gas hidrogen

Compound J

Heat (e) (i)

Why is hydrogen gas passed through the combustion tube after heating has stpopped? Mengapakah gas hidrogen dilalukan melalui tiub pembakaran selepas pemanasan dihentikan?

To avoid copper produced react with oxygen to form copper(II) oxide. (ii)

State how to determine that the reaction between compound J and hydrogen has completed. Nyatakan bagaimana menentukan tindak balas antara sebatian J dengan hidrogen telah lengkap.

By repeating the process of heating, cooling and weighing until constant mass is obtained. (f)

(i)

Can the empirical formula of magnesium oxide be determined by the same method? Explain your answer. Bolehkah formula empirik bagi magnesium oksida ditentukan dengan cara yang sama? Jelaskan jawapan anda.

Cannot. Magnesium is more reactive than hydrogen. Hydrogen cannot reduce magnesium oxide to form magnesium. (ii)

Magnesium can reduce copper oxide to copper. Explain why the empirical formula of the copper oxide cannot be determined by heating the mixture of copper oxide and magnesium powder. Magnesium boleh menurunkan kuprum oksida kepada kuprum. Terangkan mengapa formula empirik kuprum oksida tidak boleh ditentukan dengan pemanasan campuran kuprum oksida dengan serbuk magnesium.

Magnesium oxide and copper produced are in solid form, copper cannot be separated from magnesium oxide. The mass of copper cannot be weighed.

MOLECULAR FORMULA / FORMULA MOLEKUL

1

Molecular formula of a compound shows the actual number of atoms of each element that are present in a molecule of the compound. Formula molekul suatu sebatian menunjukkan bilangan sebenar atom bagi setiap unsur yang terdapat dalam satu molekul sebatian.

Molecular Formula = (empirical formula)n, where n is a integer. Formula molekul = (Formula empirik)n, di mana n adalah integer. 2

Example / Contoh: Compound

Molecular formula

Empirical formula

Value of n

Sebatian

Formula molekul

Formula empirik

Nilai n

Water / Air

H2O

H2O

1

Carbon dioxide / Karbon dioksida

CO2

CO2

1

H2SO4

H2SO4

1

Ethene / Etena

C2H4

CH2

2

Benzene / Benzena

C6H6

CH

6

Glucose / Glukosa

C6H12O6

CH2O

6

Sulphuric acid / Asid sulfurik

n io

Sdn. B

m

37

. hd

Publicat

The molecular formula and the empirical formula of a compound may be the same if the value of n = 1 but different if the value is n > 1. Formula molekul dan formula empirik suatu sebatian akan sama sekiranya nilai n = 1 tetapi akan berbeza sekiranya nilai n > 1.

Nila

02-Chem F4 (3P).indd 37

12/9/2011 5:59:08 PM

MODULE • Chemistry Form 4

EXERCISE / LATIHAN

The empirical formula of compound X is CH2 and relative molecular mass is 56. Determine the molecular formula of compound X. [Relative atomic mass: H = 1; C = 12]

1

Formula empirik sebatian X adalah CH2 dan JMR adalah 56. Tentukan formula molekul sebatian X. [Jisim atom relatif: H = 1; C = 12]

(12 + 2)n = 56 56 n= =4 14 Molecular formula = (CH2)4 = C4H8 2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86.

2

2.58 g suatu hidrokarbon mengandungi 2.16 g karbon. Jisim molekul relatif bagi hidrokarbon ini ialah 86. [Relative atomic mass / Jisim atom relatif : H = 1; C = 12] (i) Calculate the empirical formula of the hydrocarbon / Hitungkan formula empirik bagi hidrokarbon ini. Element

C

H

Mass of element (g)

2.16

0.42

Number of mole of atoms

0.18

0.42

Ratio of moles

1

21 = 7 3 3

Simplest ratio of moles

3

7

Empirical formula = C3H7 (ii) Determine the molecular formula of the hydrocarbon / Tentukan formula molekul hidrokarbon tersebut. (12 × 3 + 7 × 1)n = 86 86 n= =2 43 Molecular formula = (C3H7)2 = C6H14 The diagram below shows the structural formula for benzene molecule.

3

Rajah di bawah menujukkan formula struktur bagi benzena. H H

C

H

C

C C

C

H

C

H

H

(a) Name the element that make up benzene / Namakan unsur yang membentuk benzena. Carbon and hydrogen (b) What are the molecular formula and empirical formula for benzene? Apakah formula molekul dan formula empirik bagi benzena?

Molecular formula / Formula molekul: C6H6 Empirical formula / Formula empirik: CH (c) Compare and contrast the molecular formula and empirical formula for benzene. Banding dan bezakan formula molekul dan formula empirik bagi benzena.

• Both empirical formula and molecular formula shows benzene is made up of elements. Kedua-dua fomula molekul dan formula empirik menunjukkan benzena terdiri dari unsur

actual

• Molecular formula shows the molecule . Each benzene molecule

carbon

number of consists of

6 carbon

sebenar Formula molekul menunjukkan bilangan bagi atom molekul 6 benzena terdiri daripada benzena. Setiap a

atom

dan

hydrogen hidrogen

.

hydrogen

atoms in benzene 6 hydrogen atoms and atoms. dan atom hidrogen dalam molekul karbon 6 dan atom

.

n Sdn.

38

hidrogen

karbon

and

tio

Nil a

m

Public

karbon

atoms and

carbon

d. Bh

02-Chem F4 (3P).indd 38

12/9/2011 5:59:08 PM

Chemistry Form 4 • MODULE

• Empirical formula shows the simplest ratio of number carbon atoms to hydrogen atoms, the simplest carbon hydrogen 1:1 ratio of number of atoms to atoms in benzene is . Formula empirik benzena menunjukkan

nisbah paling ringkas

Nisbah paling ringkas bilangan atom

karbon

kepada

hidrogen

karbon bilangan atoms kepada atom hidrogen 1:1 adalah .

.

PERCENTAGE COMPOSITION BY MASS OF AN ELEMENT IN A COMPOUND PERATUS KOMPOSISI UNSUR MENGIKUT JISIM DALAM SEBATIAN

Total RAM of the element in the compound × 100% 1

% composition by mass of an element = % komposisi unsur mengikut jisim

2

Jumlah JAR unsur dalam suatu sebatian × 100%

RMM/RFM of compound/JMR/JFR sebatian

Example / Contoh: Calculate the percentage composition by mass of nitrogen in the following compounds: Hitungkan peratusan nitrogen mengikut jisim dalam sebatian berikut: [Relative atomic mass / Jisim atom relatif : N = 14, H = 1, O = 16, S = 32, K = 39]

(i)

(NH4)2SO4 2 × 14 × 100% 132 = 21.2%

%N =

(ii) KNO3 14 × 100% 101 = 13.9%

%N =

CHEMICAL FORMULA FOR IONIC COMPOUNDS / FORMULA KIMIA BAGI SEBATIAN ION

1

Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is by exchanging the charges on each ion. The formula obtained will be XmYn.

Formula kimia sebatian ion yang mengandungi ion X m + dan Y n– boleh diperoleh melalui pertukaran bilangan cas setiap ion. Formula yang diperoleh ialah XnYm. 2

Example / Contoh: (i) Sodium oxide / Natrium oksida Ion / Ion

Na+

O2–

+1

–2

Exchange of charges / Pertukaran bilangan cas

2

1

Smallest ratio / Nisbah teringkas

2

Charges / Bilangan cas

1 +

O2–

2 Na

Number of combining ions / Bilangan ion yang bergabung Formula / Formula

Na2O

(ii) Copper(II) nitrate / Kuprum(II) nitrat 2+

Cu +2

1

NO –1

2 (Ratio / Nisbah)

(iii) Zinc oxide / Zink oksida Zn2+ +2

O2– –2

2

2

1 ⇒ ZnO

1 (Ratio / Nisbah) n io

Sdn. B

m

39

. hd

Publicat

⇒ Cu(NO3)2

– 3

Nila

02-Chem F4 (3P).indd 39

12/9/2011 5:59:08 PM

ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS Aktiviti 1: TULIS FORMULA KIMIA DAN NAMA BAGI BAHAN KIMIA BERIKUT

tio

n Sdn.

K+ Potassium ion Ion kalium

Na+ Sodium ion Ion natrium

O2–, Oxide ion

CO32–, Carbonat ion

SO42–, Sulphate ion

Cl–, Chloride ion

Br–, Bromide ion

I–, Iodide ion

OH–, Hydroxide ion

NO3–, Nitrate ion

Ion oksida

Ion karbonat

Ion sulfat

Ion klorida

Ion bromida

Ion iodida

Ion hidroksida

Ion nitrat

K2O Potassium oxide

K2CO3 Potassium carbonate

K2SO4 KCl KBr Potassium sulphate Potassium chloride Potassium bromide

KI Potassium iodide

KOH KNO3 Potassium hydroxide Potassium nitrate

Na2O Sodium oxide

Na2CO3 Sodium carbonate

Na2SO4 Sodium sulphate

NaCl Sodium chloride

NaBr Sodium bromide

NaI Sodium iodide

NaOH Sodium hydroxide

H2CO3 Carbonic acid

H2SO4 Sulphuric acid

HCl Hydrocloric acid

HBr Hydrobromic acid

HI Hydroiodic acid

Ag2CO3 Silver carbonate

Ag2SO4 Silver sulphate

AgCl Silver chloride

AgBr Silver bromide

AgI Silver iodide

(NH4)2CO3 Ammonium carbonate

(NH4)2SO4 Ammonium sulphate

NH4Cl Ammonium chloride

NH4Br Ammonium bromide

NH4I Ammonium iodide

CaO Calcium oxide

CaCO3 Calcium carbonate

CaSO4 Calcium sulphate

CaCl2 Calcium chloride

CaBr2 Calcium bromide

CaI2 Calcium iodide

Ca(OH)2 Calcium hydroxide

CuO Copper(II) oxide

CuCO3 Copper(II) carbonate

CuSO4 Copper(II) sulphate

CuCl2 CuBr2 Copper(II) chloride Copper(II) bromide

CuI2 Copper(II) iodide

Cu(OH)2 Cu(NO3 )2 Copper(II) hydroxide Copper(II) nitrate

MgCO3 Magnesium carbonate

MgSO4 Magnesium sulphate

MgCl2 Magnesium chloride

MgBr2 Magnesium bromide

Mg(OH)2 MgI2 Magnesium Magnesium iodide hydroxide

Mg(NO3 )2 Magnesium nitrate

ZnO Zinc oxide

ZnCO3 Zinc carbonate

ZnSO4 Zinc sulphate

ZnCl2 Zinc chloride

ZnBr2 Zinc bromide

ZnI2 Zinc iodide

Zn(OH)2 Zinc hydroxide

Zn(NO3 )2 Zinc nitrate

PbO Lead(II) oxide

PbCO3 Lead(II) carbonate

PbSO4 Lead(II) sulphate

PbCl2 Lead(II) chloride

PbBr2 Lead(II) bromide

PbI2 Lead(II) iodide

Pb(OH)2 Lead(II) hydroxide

Pb(NO3 )2 Lead(II) nitrate

AlCl3 Aluminium chloride

Al(OH)3 AlBr3 AlI3 Aluminium Aluminium bromide Aluminium iodide hydroxide

H+ Hydrogen ion Ion hidrogen

Ag+ Silver ion Ion argentum

Ag2O Silver oxide

NH4 + Ammonium ion Ion ammonium

Ca2+ Calcium ion Ion kalsium

Cu2+ Copper(II) ion Ion kuprum(II)

MgO Mg2+ Magnesium ion Magnesium Ion magnesium oxide Zn2+ Zinc ion Ion zink

Pb2+ Lead(II) ion Ion plumbum(II)

Al 3+ Aluminium ion 12/9/2011 5:59:08 PM

Ion aluminium

Al2(SO4 )3 Al2O3 Al2(CO3 )3 Aluminium Aluminium oxide Aluminium carbonate sulphate

NaNO3 Sodium nitrate HNO3 Nitric acid

AgOH Silver hydroxide

AgNO3 Silver nitrate NH4NO3 Ammonium nitrate Ca(NO3 )2 Calcium nitrate

Al(NO3)3 Aluminium nirate

MODULE • Chemistry Form 4

Publica

40

d. Bh

m

02-Chem F4 (3P).indd 40

Nil a

02-Chem F4 (3P).indd 41

ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULAE OF THE FOLLOWING COMPOUNDS AKTIVITI 2: TANPA MERUJUK KEPADA JADUAL AKTIVITI 1, TULISKAN FORMULA KIMIA BAGI SEBATIAN BERIKUT

Potassium ion Ion kalium

Sodium ion Ion natrium

Oxide ion

Carbonat ion

Sulphate ion

Chloride ion

Bromide ion

Iodide ion

Hydroxide ion

Nitrate ion

Ion oksida

Ion karbonat

Ion sulfat

Ion klorida

Ion bromida

Ion iodida

Ion hidroksida

Ion nitrat

K2O

K2CO3

K2SO4

KCl

KBr

KI

KOH

KNO3

Na2O

Na2CO3

Na2SO4

NaCl

NaBr

NaI

NaOH

NaNO3

H2CO3

H2SO4

HCl

HBr

HI

Ag2CO3

Ag2SO4

AgCl

AgBr

AgI

(NH4 )2CO3

(NH4 )2SO4

NH4Cl

NH4 Br

NH4 I

CaO

CaCO3

CaSO4

CaCl2

CaBr2

CaI2

Ca(OH)2

Ca(NO3 )2

CuO

CuCO3

CuSO4

CuCl2

CuBr2

CuI2

Cu(OH)2

Cu(NO3 )2

MgO

MgCO3

MgSO4

MgCl2

MgBr2

MgI2

Mg(OH)2

Mg(NO3 )2

ZnO

ZnCO3

ZnSO4

ZnCl2

ZnBr2

ZnI2

Zn(OH)2

Zn(NO3 )2

PbO

PbCO3

PbSO4

PbCl2

PbBr2

PbI2

Pb(OH)2

Pb(NO3 )2

Al2O3

Al2(CO3)3

Al2(SO4 )3

AlCl3

AlBr3

AlI3

Al(OH)3

Al(NO3 )3

Hydrogen ion Ion hidrogen

Silver ion Ion argentum

Ag2O

Ammonium ion Ion ammonium

Calcium ion Ion kalsium

Copper(II) ion Ion kuprum(II)

Magnesium ion Ion magnesium

Zinc ion

Lead(II) ion Ion plumbum(II)

Nila

Ion aluminium Sdn. B

41

. hd

m

n io

12/9/2011 5:59:09 PM

Aluminium ion Publicat

AgOH

AgNO3

NH4 NO3

Chemistry Form 4 • MODULE

Ion zink

HNO3

MODULE • Chemistry Form 4

ACTIVITY 3: WRITE THE CHEMICAL FORMULAE AND TYPE OF PARTICLES FOR THE FOLLOWING ELEMENT/COMPOUND AKTIVITI 3: TULIS FORMULA KIMIA DAN JENIS ZARAH UNTUK UNSUR/SEBATIAN BERIKUT

Compound / Element

Formula

Type of particles

Compound / Element

Formula

Type of particles

Sebatian/Unsur

Formula

Jenis zarah

Sebatian/Unsur

Formula

Jenis zarah

Na2SO4

Ion

ZnCO3

Ion

(NH4 )2CO3

Ion

(NH4 )2CO3

Ion

Mg(NO3 )2

Ion

AgCl

Ion

HCl

Ion

H2SO4

Ion

K2O

Ion

Cu(NO3 )2

Ion

MgO

Ion

H2

Molecule

PbCO3

Ion

CO2

Molecule

Fe2(SO4)3

Ion

O2

Molecule

MgCl2

Ion

Al2(SO4 )3

Ion

ZnSO4

Ion

PbCl2

Ion

AgNO3

Ion

KI

Ion

(NH4 )2SO4

Ion

CuCO3

Ion

ZnO

Ion

K2CO3

Ion

HNO3

Ion

NaOH

Ion

NH3

Molecule

NH3(aq)

Ion and molecule

Mg

Atom

NH4Cl

Ion

Zn

Atom

NO2

Molecule

CuSO4

Ion

NaCl

Ion

I2

Molecule

Ag

Atom

Cl2

Molecule

Br2

Molecule

Sodium sulphate Natrium sulfat

Ammonium carbonate Ammonium karbonat

Magnesium nitrate Magnesium nitrat

Hyrochloric acid Asid hidroklorik

Potassium oxide Kalium oksida

Magnesium oxide Magnesium oksida

Lead(II) carbonate Plumbum(II) karbonat

Iron(III) sulphate Ferum(III) sulfat

Magnesium chloride Magnesium klorida

Zinc sulphate Zink sulfat

Silver nitrate Argentum nitrat

Ammonium sulphate Ammonium sulfat

Zinc oxide Zink oksida

Nitric acid Asid nitrik

Ammonia gas Gas ammonia

Magnesium Magnesium

Zinc Zink

Copper(II) sulphate Kuprum(II) sulfat

Iodine Iodin

Chlorine Klorin

Ammonium carbonate Ammonium karbonat

Silver chloride Argentum klorida

Sulphuric acid Asid sulfurik

Copper(II) nitrate Kuprum(II) nitrat

Hydrogen gas Gas hidrogen

Carbon dioxide gas Gas karbon dioksida

Oxygen gas Gas oksigen

Aluminium sulphate Aluminium sulfat

Lead(II) chloride Plumbun(II) klorida

Potassium iodide Kalium iodida

Copper(II) carbonate Kuprum(II) karbonat

Potasium carbonate Kalium karbonat

Sodium hydroxide Natrium hidroksida

Aqueous ammonia Ammonia akueus

Ammonium chloride Ammonium klorida

Nitrogen dioxide gas Gas nitrogen dioksida

Sodium chloride Natrium klorida

Silver Argentum

Bromine Bromin

Publica

n Sdn.

42

tio

Nil a

m

Zinc carbonate Zink karbonat

d. Bh

02-Chem F4 (3P).indd 42

12/9/2011 5:59:09 PM

Chemistry Form 4 • MODULE

CHEMICAL EQUATIONS / PERSAMAAN KIMIA

1

Two types of equation / Dua jenis persamaan: • Equation in words / Persamaan perkataan – using names of reactants and products / menggunakan nama bahan tindak balas dan hasil tindak balas; • Equation using symbols / Persamaan menggunakan simbol – reactants and products are represented by chemical formulae and have certain meanings menggunakan formula kimia untuk mewakili bahan tindak balas dan hasil tindak balas serta menggunakan pelbagai jenis simbol yang membawa makna tertentu. Symbol / Simbol +

⇌

2

Meaning / Maksud

Symbol / Simbol

Meaning / Maksud

Separating 2 reactants / products

(g)

Gaseous state

Mengasingkan 2 bahan / hasil

(g)

Keadaan gas

Produces

(aq)

Aqueous state

Menghasilkan

(ak)

Keadaan akueus

Reversible reaction

Gas released

Tindak balas berbalik

Gas terbebas

(s)

Solid state

Precipitation

(p)

Keadaan pepejal

Bahan termendap

(l)

Liquid state

(ce)

Keadaan cecair

Heating / Heat energy is given

∆

Pemanasan / Haba dibekalkan

Information obtained from chemical equation using symbols / Maklumat yang diperoleh daripada persamaan kimia bersimbol: (a) Qualitative aspect / Aspek kualitatif : type of reactants and products involved in the chemical reaction and the state of each reactant and product. jenis bahan / hasil tindak balas yang terlibat dalam tindak balas dan keadaan fizikal bagi setiap bahan / hasil tindak balas. (b) Quantitative aspect / Aspek kuantitatif : number of moles of reactants and products involved in the chemical reaction

that is the coeffficients involved in a balanced equation of the formulae of reactants and products. bilangan mol yang bertindak balas dan hasil tindak balas yang terbentuk iaitu pekali bagi setiap formula bahan dan hasil tindak balas.

Example / Contoh:

Zn (s) + 2HCl (aq)

ZnCl2 (aq) + H2 (g)

Zn (p) + 2HCl (ak)

ZnCl2 (ak) + H2 (g)

1 mol

2 mol

1 mol

1 mol

Interpretation / Tafsiran: 1 mol of zinc reacts with 2 mol of hydrochloric acid to produce 1 mol of zinc chloride and 1 mol of hydrogen. 1 mol zink bertindak balas dengan 2 mol asid hidroklorik menghasilkan 1 mol zink klorida dan 1 mol hidrogen. 3

Writing balanced chemical equations / Menulis persamaan kimia seimbang:

Step 1 / Langkah 1 : Write the correct chemical formulae for each reactant and product. Tulis formula kimia bagi setiap bahan dan hasil tindak balas.

Step 2 / Langkah 2 : Detemine the number of atoms for each element / Tentukan bilangan atom setiap unsur. Step 3 / Langkah 3 : Balance the number of atoms for each element by adjusting the coefficients in front of the chemical formulae.

n io

Sdn. B

m

43

. hd

Publicat

Imbangkan bilangan atom setiap jenis unsur dengan menambahkan pekali di hadapan setiap formula kimia.

Nila

02-Chem F4 (3P).indd 43

12/9/2011 5:59:09 PM

MODULE • Chemistry Form 4

EXERCISE / LATIHAN Write a balanced chemical equation for each of the following reactions: Tulis persamaan kimia seimbang bagi setiap tindak balas yang berikut:

Zinc carbonate

1

ZnCO3

Zinc oxide + Carbon dioxide / Zink karbonat

Sulphuric acid + Sodium hydroxide

2

H2SO4 + 2NaOH

Sodium sulphate + Water / Asid sulfurik + Natrium hidroksida

Argentum nitrat + Natrium klorida AgNO3 + NaCl AgCl + NaNO3

Silver chloride + Sodium nitrate Argentum klorida + Natrium nitrat

Copper(II) oxide + Hydrochloric acid

4

Kuprum(II) oksida + Asid hidroklorik CuO + 2HCl CuCl2 + H2O

Magnesium + Oxygen

5

2Mg + O2

2Na + 2H2O

K 2O + H 2 O

Natrium hidroksida + Hidrogen

Potassium hydroxide / Kalium oksida + Air

Kalium hidroksida

2KOH

ZnO + 2HNO3

Lead(II) nitrate

9

Magnesium oksida

2NaOH + H2

Zinc oxide + Nitric acid

8

Magnesium oxide / Magnesium + Oksigen

Sodium hydroxide + Hydrogen / Natrium + Air

Potassium oxide + Water

7

Copper(II) chloride + Water

Kuprum(II) klorida + Air

2MgO

Sodium + Water

6

Natrium sulfat + Air

Na2SO4 + 2H2O

Silver nitrate + Sodium chloride

3

Zink oksida + Karbon dioksida

ZnO + CO2

Zinc nitrate + Water / Zink oksida + Asid nitrik

Zink nitrat + Air

Zn(NO3 )2 + H2O

Lead(II) oxide + Nitrogen dioxide + Oxygen

Plumbum(II) nitrat Plumbum (II) oksida + Nitrogen dioksida + Oksigen 2Pb(NO3 )2 2PbO + 4NO2 + O2 10 Aluminium nitrate Aluminium oxide + Nitrogen dioxide + Oxygen Aluminium nitrat Aluminium oksida + Nitrogen dioksida + Oksigen 4Al(NO3 )3 2Al2O3 + 12NO2 + 3O2

NUMERICAL PROBLEMS INVOLVING CHEMICAL EQUATIONS / PENGHITUNGAN BERKAITAN PERSAMAAN KIMIA Calculation steps / Langkah perhitungan: S1 / L1 : Write a balanced equation / Tulis persamaan kimia seimbang. S2 / L2 : Write the information from the question above the equation / Tulis maklumat daripada soalan di atas persamaan. S3 / L3 : Write the information from the chemical equation below the equation (information about the number of moles of

reactants/products). Tulis maklumat daripada persamaan kimia di bawah persamaan (Maklumat perhubungan bilangan mol bahan/hasil tindak balas terlibat).

S4 / L4 : Change the information in S2 into moles by using the method shown in the chart below. Tukarkan maklumat L2 kepada mol menggunakan carta di bawah.

S5 / L5 : Use the relationship between number of moles of substance involved in S3 to find the answer. Gunakan perhubungan bilangan mol bahan terlibat dalam L3 untuk mencari jawapan.

S6 / L6 : Change the information to the unit required using the chart below. Tukar maklumat kepada unit yang dikehendaki dengan menggunakan carta di bawah.

Mass (g) Jisim (g)

× (RAM/FRM/RMM) g mol–1

No. of moles (n) Bilangan mol (n)

× 24 dm3 mol–1 / 22.4 dm3 mol–1 ÷ 24 dm3 mol–1 / 22.4 dm3 mol–1

Volume of gas (dm3) Isipadu gas (dm3)

Publica

n Sdn.

44

tio

Nil a

m

÷ (RAM/FRM/RMM) g mol–1

d. Bh

02-Chem F4 (3P).indd 44

12/9/2011 5:59:09 PM

Chemistry Form 4 • MODULE

EXERCISE / LATIHAN

1

The equation shows the reaction between zinc and hydrochloric acid. Persamaan menunjukkan tindak balas antara zink dengan asid hidroklorik.

Zn + 2HCl

ZnCl2 + H2

Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at room conditions. [Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions] Hitungkan jisim zink yang perlu ditindakbalaskan dengan asid hidroklorik berlebihan untuk menghasilkan 6 dm3 gas hidrogen pada keadaan bilik. [Jisim atom relatif: Zn = 65, Cl = 35.5, 1 mol gas menempati 24 dm3 pada suhu bilik]

Mol of H2 =

6 dm3 = 0.25 mol 24 dm3 mol–1

From the equation, 1 mol of H2 : 1 mol of Zn 0.25 mol of H2 : 0.25 mol of Zn Mass of Zn = 0.25 × 65 = 16.2 g

2

The equation shows the reaction between potassium and oxygen. Persamaan berikut menunjukkan tindak balas antara kalium dengan oksigen.

4K + O2

2K2O

Calculate the mass of potassium required to produce 23.5 g of potassium oxide. [Relative atomic mass: K = 39, O = 16] Hitungkan jisim kalium yang diperlukan untuk menghasilkan 23.5 g kalium oksida. [Jisim atom relatif: K = 39, O = 16]

Mol of K2O =

23.5 23.5 = = 0.25 mol (2 × 39 + 16) 94

From the equation, 2 mol of K2O : 4 mol of K 0.25 mol of K2O : 0.5 mol of K Mass of K = 0.5 mol × 39 g mol–1 = 19.5 g

3

The equation shows the decomposition of hydrogen peroxide. Persamaan menunjukkan penguraian hidrogen peroksida.

H2O2

H2O + O2

Balance the equation above. Calculate the number of moles of H2O2 that decomposes if 11.2 dm3 oxygen gas is collected at STP. [Relative Atomic mass: H = 1, O = 16, molar volume of gas = 22.4 dm3 mol–1 at STP] Seimbangkan persamaan di atas. Hitung bilangan mol H2O2 yang telah terurai sekiranya 11.2 dm3 gas oksigen dikumpulkan pada STP. [Jisim atom relatif: H = 1, O = 16, isi padu molar gas = 22.4 dm3 mol–1 pada STP]

Mol of O2 =

11.2 dm3 = 0.5 mol 22.4 dm3 mol–1

n io

Sdn. B

m

45

. hd

Publicat

From the equation, 1 mol of O2 : 2 mol of H2O2 0.5 mol of O2 : 1.0 mol of H2O2

Nila

02-Chem F4 (3P).indd 45

12/9/2011 5:59:09 PM

MODULE • Chemistry Form 4

8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate produced. [Relative atomic mass: N = 14, O = 16, Cu = 64]

4

8.0 g serbuk kuprum(II) oksida dicampurkan kepada asid nitrik cair yang berlebihan dan dihangatkan. Hitungkan jisim kuprum(II) nitrat yang terhasil. [Jisim atom relatif: N = 14, O = 16, Cu = 64]

CuO + 2HNO3 Mol of CuO =

Cu(NO3 )2 + H2O 8g = 0.1 mol (64 + 16)g mol–1

From the equation, 1 mol of CuO : 1 mol of Cu(NO3)2 0.1 of CuO : 0.1 mol of Cu(NO3)2 Mass of Cu(NO3)2 = 0.1 mol × 188 g mol–1 = 18.8 g

1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1at STP]

5

1.3 g zink bertindak balas dengan asid sulfurik cair yang berlebihan. Hasil tindak balas adalah zink sulfat dan hidrogen. Hitungkan isi padu hidrogen yang terbebas pada STP. [Jisim atom relatif: Zn = 65, isipadu molar gas = 22.4 dm3 mol–1 pada STP]

Answer/Jawapan:

448 cm3

0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate the volume of chlorine gas that has reacted. [Relative atomic mass: Na = 23, Molar volume of gas = 24 dm3 mol–1 at room conditions]

6

0.46 g natrium terbakar lengkap dalam gas klorin pada keadaan bilik menghasilkan natrium klorida. Hitungkan isi padu klorin yang diperlukan untuk bertindak balas lengkap. [Jisim atom relatif: Na = 23, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

Answer/Jawapan:

0.24 dm3

The equation shows the combustion of propane gas.

7

Persamaan menunjukkan pembakaran gas propana.

C3H8 + 5O2

3CO2 + 4H2O

720 cm3 of propane gas (C3H8) at room conditions burns in excess oxygen. Calculate the mass of carbon dioxide formed. [Relative atomic mass: C = 12, O = 16, Molar volume of gas = 22.4 dm3 mol–1 at room conditions] 720 cm3 gas propana (C3H8) pada keadaan bilik terbakar dalam oksigen berlebihan. Hitungkan jisim karbon dioksida yang terbentuk. [Jisim atom relatif: C = 12, O = 16, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

Answer/Jawapan:

3.96 g

Publica

n Sdn.

46

tio

Nil a

m

d. Bh

02-Chem F4 (3P).indd 46

12/9/2011 5:59:09 PM

Chemistry Form 4 • MODULE

Objective Questions / Soalan Objektif 1

2

The mass of one atom of element Y is two times more than an atom of oxygen. What is the relative atomic mass of element Y? [Relative atomic mass: O = 16] Jisim satu atom unsur Y adalah dua kali lebih daripada satu atom oksigen. Apakah jisim atom relatif bagi unsur Y? [Jisim atom relatif: O = 16] A 12 B 24 C 32 D 36

5

Jadual berikut menunujukkan jisim atom relatif bagi neon, karbon, oksigen dan kalsium. Element/Unsur Relative atomic mass/Jisim atom relatif

The chemical formula for butane is C4H10. Which of the following statements are true about butane? [Relative atomic mass: H = 1, C =12 and O =16, Avogadro Constant = 6 × 1023 mol–1]

The empirical formula for butane is CH2.

Jumlah bilangan atom dalam 1 mol butana adalah 8.4 × 1024.

Calcium / Kalsium

40

D

6

A I and II only II and III only II dan III sahaja

II, III and IV only

A 4.515 × 1022 B 4.515 × 1023

II, III dan IV sahaja

D I, II, III and IV I, II, III dan IV

7

A bottle contains 3.01 × 1023 of gas particles. What is the number of moles of the gas in the bottle? C 3.0 mol D 6.0 mol

Which of the following gases contains 0.4 mol of atoms at room temperature and pressure? [Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure] Antara gas berikut, yang manakah mengandungi 0.4 mol atom pada suhu dan tekanan bilik? [Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan bilik]

What is the number of hydrogen atom in 0.1 mol of water? [Avogadro constant: 6.02 × 1023 mol–1] A 6.02 × 1022 B 60.2 × 1023

8

C D

6.02 × 1023 3.01 × 1023

5 g of element X reacted with 8 g of element Y to form a compound with the formula XY2. What is the relative atomic mass of element X? [Relative atomic mass: Y = 80] 5 g unsur X bertindak balas dengan 8 g unsur Y membentuk sebatian dengan formula XY2. Apakah jisim atom relatif unsur X? [Jisim atom relatif: Y = 80]

A 25 B 40

C 50 D 100

n io

Sdn. B

m

47

. hd

Publicat

C 4.8 dm3 CO2 D 4.8 dm3 NH3

C 8.03 × 1022 D 8.03 × 1021

Berapakah bilangan atom oksigen dalam 0.1 mol air? [Pemalar Avogadro = 6.02 × 1023 mol–1]

Sebuah botol mengandungi 3.01 × 1023 zarah gas. Berapakah bilangan mol zarah gas dalam botol itu?

A 4.8 dm3 Ne B 4.8 dm3 O2

A bulb is filled with 1 800 cm3 of argon gas at room conditions. What is the number of argon atom in the bulb? [Molar volume of gas = 24 dm3 mol–1 at room conditions, Avogadro constant = 6.02 × 1023 mol–1] Sebuah belon diisi dengan 1 800 cm3 gas argon pada keadaan bilik. Berapakah bilangan atom argon dalam belon itu? [Isipadu molar gas = 24 dm3 mol–1 pada keadaan bilik, Pemalar Avogadro = 6.02 × 1023 mol–1]

I dan II sahaja

A 0.5 mol B 1.0 mol

Mass of one oxygen atom is 16 times bigger than one carbon atom Jisim satu atom oksigen adalah 16 kali lebih besar daripada satu atom karbon

Satu molekul butana mempunyai jisim 84 kali lebih daripada jisim satu atom hidrogen.

4

16

molecule

IV One butane molecule has a mass of 84 times higher than the mass of 1 hydrogen atom.

3

Oxygen / Oksigen

16 g oksigen mengandungi 6.02 × 1023 molekul oksigen

III 1 mol of butane contains a total of 8.4 × 1024 atoms.

C

12

Each butane molecule is made up of 4 carbon atoms and 10 hydrogen atoms. Setiap molekul butana terdiri dari 4 atom karbon dan 10 atom hidrogen.

B

20

Carbon / Karbon

Antara pernyataan berikut, yang manakah adalah benar? [Pemalar Avogadro = 6.02 × 1023 mol–1] A Mass of one calcium atom is 40 g Jisim satu atom kalsium ialah 40 g B Mass of 1 mol of neon is 20 g Jisim 1 mol neon ialah 20 g C 16 g of oxygen contains 6.02 × 1023 oxygen

Formula empirik butana ialah CH2.

II

Neon / Neon

Which of the following statements is true? [Avogadro constant = 6.0 × 1023 mol–1]

Formula kimia bagi butana ialah C4H10. Antara pernyataan berikut, yang manakah adalah benar tentang butana? [Jisim atom relatif: H = 1, C = 12 dan O = 16, Pemalar Avogadro = 6 × 1023 mol–1]

I

The table below shows the relative atomic mass of neon, carbon, oxygen and calcium.

Nila

02-Chem F4 (3P).indd 47

12/9/2011 5:59:10 PM

MODULE • Chemistry Form 4

9

The diagram below shows the set-up of apparatus to determine the empirical formula of an oxide metal X.

11 The equation shows a decomposition of magnesium nitrate when heated.

Rajah di bawah menunjukkan susunan radas bagi menentukan formula empirik oksida logam X.

Persamaan di bawah menunjukkan penguraian nitrat apabila dipanaskan.

2Mg(NO3)2

Metal X Logam X

Heat Panaskan

Which of the following is metal X? Antara berikut, yang manakah mungkin bagi logam X?

A Zinc

C

Zink

B

Tin Stanum

D Copper

Lead Plumbum

Kuprum

10 The following equation shows the decomposition reaction of lead(II) nitrate when heated at room temperature and pressure. Persamaan tindak balas di bawah menunjukkan penguraian plumbum(II) nitrat apabila dipanaskan pada suhu dan tekanan bilik.

2Pb(NO3)2

What is the number of oxygen molecules is produced when 7.4 g magnesium nitrate decomposed when heated. [Relative formula mass of Mg(NO3)2 = 148; Avogadro constant = 6.02 × 1023 mol–1] Berapakah bilangan molekul oksigen apabila 7.4 g magnesium nitrat terurai apabila dipanaskan? [Jisim formula relatif Mg(NO3)2 = 148; Pemalar Avogadro = 6.02 × 1023 mol–1]

A B C D

Antara berikut, yang manakah adalah benar apabila 0.1 mol plumbum(II) nitrat terurai? [Jisim atom relatif: N = 14, O = 16, Pb = 207 dan 1 mol gas menempati isipadu 24 dm3 pada suhu dan tekanan bilik]

I

66.2 g of lead(II) oxide is formed

II

22.3 g of lead(II) oxide is formed

66.2 g plumbum(II) oksida terbentuk 22.3 g plumbum(II) oksida terbentuk

1.505 × 1022 3.010 × 1022 1.505 × 1023 3.010 × 1023

12 The equation below shows the chemical equation of the combustion of ethanol in excess oxygen. Persamaan di bawah menunjukkan persamaan kimia pembakaran etanol dalam oksigen berlebihan.

2PbO + 4NO2 + O2

Which of the following are true when 0.1 mol of lead(II) nitrate is decomposed? [Relative atomic mass: N = 14, O = 16, Pb = 207 and 1 mol gas occupies the volume of 24 dm3 at room temperature and pressure]

2MgO + 4NO2 + O2

2C2H5OH + 6O2

4CO2 + 6H2O

What is the volume of carbon dioxide gas released when 9.20 g ethanol burnt completely? [Relative atomic mass of H = 1, C = 12, O = 16, 1 mol of gas occupies 24 dm3 at room condition] Apakah isi padu gas karbon dioksida dibebaskan apabila 9.20 g etanol terbakar lengkap? [Jisim atom relatif: H = 1, C = 12, O = 16, 1 mol gas menempati 24 dm3 pada keadaan bilik]

A B C D

4.8 cm3 9.6 cm3 96.0 cm3 9 600 cm3

III 2.4 dm3 of oxygen gases is given off 2.4 dm3 gas oksigen dibebaskan

IV 4 800 cm3 of nitrogen dioxide given off 4 800 cm3 nitrogen dioksida dibebaskan

A I and III only I dan III sahaja

B

I and IV only

C

II and III only

I dan IV sahaja II dan III sahaja

D II and IV only

13 What is the percentage by mass of nitrogen content in urea, CO(NH2)2? [Relative atomic mass: C = 12, N = 14, H = 1 and O = 16] Apakah peratus kandungan nitrogen mengikut jisim dalam urea, CO(NH2)2? [Jisim atom relatif: C = 12, N = 14, H = 1 dan O = 16]

A B C D

23.3% 31.8% 46.7% 63.6%

II dan IV sahaja

Publica

n Sdn.

48

tio

Nil a

m

d. Bh

02-Chem F4 (3P).indd 48

12/9/2011 5:59:10 PM