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-]
Chapter 1: Measurement
-SI base units
-Significant figures and scientific notation
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
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.
Laboratory work 10%
Final exam 60%
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.
Name of unit
SI prefixes and decimal multipliers
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.
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)
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:
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
or v = R [1/n21 -1/n22] where n1 and n2 are intergers such that n2 > n1, R is known as Rydberg constant
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 = En-En’ here n > n’
En, En’ are the energies of the stationary states characterized by the principal quantum numbers n and n’.
f = CRH [1/n2 – 1/n’2] n > n’
f = frequencies emitted or absorbed by the hydrogen atom
C = speed of light
RH = 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 = RH [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 = RH [1/n21 -1/n22] , where n2 > n1
This is known as Rydberg’s equation
Modern Periodic Law: According to the law ,”The properties of elements are the periodic function of their atomic numbers”
Gambarajah yang terdapat di dalam nota ini diambil dari Chemistry matter and its changes, 4th 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 = (23oF-32oF) x 5oC
= -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
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
d) 1.54 x 102 pm
e) 6.25 x 10-4 gcm-3
c) % C = 79.9 %
% H = 20.1 %
bi) 0.26 M
bii) 0.26 M
b) 656.33 nm
ANSWER : FINAL APR 06
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
a) i)0.73 M ii)710 g iii) 2700mL
b) 7 L
c) i)83.1oF ii) 301.6K
e) 3.6 x 101 cm2 ii) 7128.1 cm
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
a) 36 %
c) 140.4 g
d) 0.5 M
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
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
b) 17 Y
ii) group 16 @ 6A ; Period 2
iii) e1 : 2 1 0 +1/2
e2 : 2 1 1 +1/2
iv) H2Y bent shape
b) i) +1 ii) +4 iii) +5
c) 3Fe+2 + 4H2O + CrO4-2 → Cr+3 + 3Fe+3 + 8OH-
RESULT : TEST 1 and 2