NOTA RINGKAS CHAPTER 1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6	JAWAPAN UJIAN TEST 1 TEST 2	PEPERIKSAAN AKHIR NOV 2006 APR 2006	MARKAH UJIAN

Tahniah diucapkan kepada pelajar berikut kerana telah memperolehi keputusan yang cemerlang di dalam peperiksaan April 2007 :

Dg Noraiana Ag Kadir [A+]

Muhamad Hafiz Muhamad Sepian [A+]

Noor Hasimah Ab.llah [A]

Muhammad Fakhrurrazi Mohd Noor [A-]

Nursheha Mohd Hadzri [A-]

Siti Nadirah Abdollah [A-]

Syllabus content

Chapter 1: Measurement

-SI base units

-Significant figures and scientific notation

-Dimensional analysis

Chapter 2: Elements, compounds, chemical equations and calculation

-Atoms, molecules, elements and compounds.

-Symbols, formulas and naming of elements, molecules and compounds.

-Atomic structure and mole concept

-The empirical formula and the molecular formula

-Stoichiometry

Chapter 3: Atomic structure and periodicity

-Introduction to subatomic particles-electrons, protons and neutrons

-Planck’s quantum theory and Bohr’s theory of the hydrogen atom

Chapter 4: Introduction to the periodic table

-Classifications of elements in the periodic table

-Variations of physical properties across a period and down a group

-Variations of chemical properties

Chapter 5: Chemical bonding

-Valence electrons and the Lewis dot symbol of atoms and ions

-Formation of ionic bonds

-Formation of Covalent bonds

-Metallic bonds and the relationship between the crystal structures of metals and their properties

Chapter 6: Oxidation-reduction (redox) reactions

-The concept of oxidation states

-Definition of redox process, oxidizing and reducing agents

-Balancing oxidation-reduction reactions in acidic and basic medium by ion-electron method.

Assessment

Tests 30%

Laboratory work 10%

Final exam 60%

CHAPTER 1

Measurement

The International System of Units

SI (Le systeme International) is a metric system built on a foundation of base units for fundamental phenomena such as mass, length (distance) and time. A list of the base SI units used frequently given in table below.

Quantity	Name of unit	Symbol
Mass	Kilogram	kg
Length	Meter	m
Time	Second	s
Temperature	Kelvin	K
Amount	Mole	Mol
Electric current	Ampere	A

SI prefixes and decimal multipliers

Prefix

Symbol

Multiplication factor

Volume (v): The space occupied by a body is known as its volume.

Units of volume: Volume is expressed in meter³, dm³, cm³, mm³.

1 meter = 10 dm

1 m³ = 10³ dm³

1 meter = 100 cm

1 m³ = 100³ cm³ = 10⁶ cm³

1 meter = 1000 mm

1 m³ = 1000³ mm³ = 10⁹ mm³

Density: Mass of unit volume is called density which is mathematically expressed as the ratio of mass to volume.

Density =

CHAPTER 2

Elements, compounds, chemical equations and calculation

The matter can be classified into three main classes

Element: An element is a substance which cannot be split up into two or more simpler substances by any of the physical or chemical methods. Ex; hydrogen, oxygen, sodium, iron, silver.

Mixture: A mixture is a substance which consists of two or more elements or compounds not chemically combined together in indefinite proportions. Ex; air ( a mixture of oxygen, nitrogen inert gas, hydrogen)

Compound: A compound is a substance consisting of two or more elements in a fixed ratio which are chemically combined together. Ex; water (H₂O), sodium chloride (NaCl)

CHAPTER 3

Atomic structure and periodicity

Structure of atom: basic constituents

Proton: It is a positively charged particle present into nucleus of an atom. The charge of proton is 1.6 x 10^-19 coulombs and the mass of a proton is 1.672 x 10^-24 g.

Electron: Electron is a positively charged particle that evolves around the nucleus in fixed orbits (energy levels).The charge of an electron is 1.6 x 10^-19 coulombs and the mass of the electron is 9.1 x 10^-28g.

Neutron: It is an electrically neutral particle present in the nucleus of an atom. The mass of neutron is 1.675 x 10^-24g.

Atomic number: Atomic number of an element is the number of protons present in the nucleus of the atom of that element.

Atomic number = number of protons

Mass number: It is equal to the total number of protons and neutrons (ex; nucleus) present in the atom of an element.

Mass number = number pf protons + number of neutrons

Isotopes: Atoms of the same element having same atomic number but different atomic weights (mass numbers) are known as isotopes. Ex; ³⁷Cl, ³⁶Cl and ³⁵Cl are the three isotopes of chlorine.

Bohr’s Model of atom

Niels Bohr proposed this model in 1913. He slightly modified the rutherford’s model to overcome its limitations. According to this model:

1. An atom consists of a small positively charged nucleus situated at its centre.
2. The negatively charged electrons revolve around the nucleus in certain definite circular paths called energy levels.
3. Each energy level is associated with a definite amount of energy.
4. The energy levels are either numbered n = 1,2,3,4….. or designated as K,L,M,N,.. outwards from the nucleus.
5. The change in energy of an electron takes place only when it jumps from lower energy level to higher level (gain of energy) or when it jumps from a higher energy level to a lower energy level (loss of energy). This means that as long as the electrons remain in a given energy and atom remains stable.

Dual Nature of electrons: According to De Broglie, all materials particles like electrons, protons, neutrons, atoms and molecules exhibit dual character.

De Broglie’s relationship: He gave the following relationship between wave length (λ) and momentum (p) of a particle.

λ = h or h where λ = wavelength of the particle

mv p m = mass of the particle

v = velocity of the particle

p = momentum of the particle

h = Planck’s constant (6.626 x 10³⁴ JS) or (6.626 x 10³⁴ kgm²s^-1)

Velocity of radiations: Mathematically it can be expressed as v = f x λ

where v = velocity of all types of electromagnetic radiations

f = frequency of radiations

λ = wavelength of radiations

velocity of all types of electromagnetic radiations is equal to 3 x 10^-8 ms^-1

Wave number: The reciprocal of wavelength (λ) is known as wave number. Mathematically, wave number v = 1 = v

λ c

or v = R [1/n²₁ -1/n²₂] where n1 and n2 are intergers such that n2 > n1, R is known as Rydberg constant

for hydrogen

Energy of proton: E = hv = h c = hcv where E = energy of proton

λ h = Planck’s constant

Bohr model for the hydrogen atom: The electron moves in circular orbits (stationary states) without irradiation energy. Permitted orbits have an angular momentum.

The emission and absorption of electromagnetic waves occurs only when transitions take up between two stationary states and the frequency of radiation emitted or absorbed is expressed by : hf = E_n-E_n’ here n > n’

E_n, E_n’ are the energies of the stationary states characterized by the principal quantum numbers n and n’.

f = CR_H [1/n² – 1/n’²] n > n’

f = frequencies emitted or absorbed by the hydrogen atom

C = speed of light

R_H = Rydberg’s constant for hydrogen = 1.097 x 10⁷ m^-1

Emission spectrum of hydrogen:

Lyman’s series n = 1; n’ = 2,3,….

Balmer’s series n = 2; n’ = 3,4,….

Paschen’s series n = 3; n’ = 4,5,….

Brackett’s series n = 4; n’ = 5,6,….

Pfund’s series n = 5; n’ = 6,7,….

The Balmer formula: Wavelength of the visible lines in the atomic spectrum of hydrogen follows the following formula

v = 1 = R_H [1/2²₁ -1/n²₂] , n = 3,4,5…..

λ where v = is the wave number

Rydberg formula: Balmer formula was generalized to account for new sets of spectral lines of hydrogen atom.

1 = R_H [1/n²₁ -1/n²₂] , where n₂ > n₁

λ

This is known as Rydberg’s equation

CHAPTER 4

Periodic Table

Modern Periodic Law: According to the law ,”The properties of elements are the periodic function of their atomic numbers”

Periodicty:

CHAPTER 5

CHAPTER 6

Gambarajah yang terdapat di dalam nota ini diambil dari Chemistry matter and its changes, 4^th ed.Brady and Senese.John Wiley and sons

ANSWER SCHEME TEST 1

1. a) Kilogram,kg b) meter,m c) second,s d) mole e) Kelvin,K

2. a) centi b) nano c) milli

3. a) 6.3 b) 23000

4. Y ⁰C = (23^oF-32^oF) x 5^oC

9^oC

= -5 ^oC

5. a) 15 b) 0.40 c) 36.4

6. 8.30 x 2.70 g = 22.4g

7. 25 atom Ni x 1 mol Ni x 58.7 g Ni = 2.44 x 10^-21 g Ni

6.02 x 1023 atom Ni 1mol Ni

8. 0.3683 g Cr x 100 % = 39.70 % Cr, 17.55 % Na, 42.76 % O

0.9278 g sample

9. 0.250 mol C₃H₈ x 4 mol H₂O = 1 mol H₂O

1 mol C₃H₈

10. 1 mol HCl = 36.46 g 1 mol Mg(OH)₂ = 58.32 g

2.88 g HCl = 0.079 mol HCl 2.48 g Mg(OH)₂ = 0.0425 mol Mg(OH)₂

36.46 g 58.32 g

0.079 mol HCl x 1 mol Mg(OH)₂ = 0.0395 mol Mg(OH)₂ amount needed

2 mol HCl

0.0425 mol Mg(OH)₂ available ↔ 0.0395 mol Mg(OH)₂ needed

therefore HCl limiting reactant.

0.0790 mol HCl x 1mol MgCl₂ x 95.2 g MgCl₂ = 3.76 MgCl₂ produced

2 mol HCl 1 mol MgCl₂

ANSWER SCHEME TEST 2

ANSWER : FINAL NOV 06

Part A

1.D 2.D 3.A 4.C 5.C 6.C 7.A 8.A 9.B 10.B 11.C 12.B 13.B 14.A 15.B 16.B 17.D 18.A 19.B 20.B

Part B (hanya jawapan akhir sahaja diberikan selebihnya perlu cuba dan akan dibincangkan dalam kelas)

QUESTION 1

d) 1.54 x 10² pm

e) 6.25 x 10^-4 gcm^-3

QUESTION 2

bii) 40g

QUESTION 3

bi) C₃H₃O₂

bii) C₉H₉O₆

c) % C = 79.9 %

% H = 20.1 %

QUESTION 4

aii) 33.15g

aiii) 12.32g

bi) 0.26 M

bii) 0.26 M

QUESTION 5

b) 656.33 nm

ANSWER : FINAL APR 06

PART A

1.A 2.D 3.B 4.C 5.B 6.C 7.C 8.D 9.A 10.C 11.D 12.D 13.C 14.C 15.D 16.A 17.B 18.C 19.A 20.D

PART B

QUESTION 1

a) i)0.73 M ii)710 g iii) 2700mL

b) 7 L

c) i)83.1^oF ii) 301.6K

e) 3.6 x 10¹ cm² ii) 7128.1 cm

QUESTION 2

a) Sulfur hexafluoride, SF₆

b) i) HI ii) H₃PO₄ iii) CH₃COOH

c) i) 8.82 mol NH₃ ii) 26.5 mol H iii) 1.60 x 10²⁵ atoms

d) 754 g

QUESTION 3

a) 36 %

b) Fe₃O₄

c) 140.4 g

d) 0.5 M

QUESTION 4

a) i) AgNO₃ ii) 10.6 g

b) valence electron = 5

c) i) 4.86 x 10^-7 m ii) 9.30 x 10^-8 m

QUESTION 5

a) e1 : 6 1 -1 +1/2

e 2: 6 1 0 +1/2

e3: 6 1 1 +1/2

e4: 6 1 -1 +1/2

e5: 6 1 0 +1/2

QUESTION 6

b) ¹⁷ Y

⁸

ii) group 16 @ 6A ; Period 2

iii) e1 : 2 1 0 +1/2

e2 : 2 1 1 +1/2

iv) H₂Y bent shape

QUESTION 7

b) i) +1 ii) +4 iii) +5

c) 3Fe⁺² + 4H₂O + CrO₄^-2 → Cr⁺³ + 3Fe⁺³ + 8OH^-

RESULT : TEST 1 and 2

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