science clinic gr10 eng dbe smartprep v3.2 (1)

DBE ESSENTIALS
GRADE
10
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Content Acknowledgement
Many thanks to those involved in the production, translation and moderation of this book:
R Bartholomew, T Britz, S Chowles, W Cloete, L Couperwthwaite, S Dippenaar, I Govender, J Hayes,
R Lodge, J MacPhee, Q Meades, J Mitchell, G Moore, A Mouton, K Munnik, C Orchison, M Potgieter,
R Ramsugit, X Sithenjwa
www
.scienceclinic.co.za
facebook.com/scienceclinicsa
c Science Clinic (Pty) Ltd 2018
Grade 10 Science Essentials
WHY YOU SHOULD STUDY SCIENCE
SCIENCE CLINIC 2018 ©
Science is amazing! It is also one of the toughest subjects at school. Science-y careers are diverse and exciting, but require years of vigorous academic commitment.
If it’s so hard to get somewhere with Science, why should you study it? Here’s our top reasons for getting your nerd on:
4. Diversity and flexibility: From dentistry to plasma physics, Science-y careers offer vast opportunities for professional career development and diversification. Engineers are welcomed into the
financial sector, due to their problem-solving ability and analytical way of thinking. Many academic
physicists teach, perform ground-breaking research and consult private clients in the same work
week. Medical professionals diversify into the legal field to become patent attorneys or medical
lawyers. However on the flip side, it’s rare for a professional with a ‘non-Science-y’ background to
bridge into the Science-based career fields.
1. Be a modern-day hero: The single greatest reason why we should study Science, is to ensure
Humanity’s sustainable survival on Earth! Ecosystems are in crisis mode, the planetary weather
system is changing rapidly, and humanity is failing to coexist in harmony with other species.
World food production has to double in the next thirty years, in order to sustain the growing global
population. We are running out of fossil fuels which are critical to the efficiency of our industry,
farming and supply chains. Fresh water is becoming increasingly scarce, with many of the World’s
greatest rivers no longer running into the sea. Diseases are becoming increasingly resistant to antibiotics. The air in many Indian and Chinese cities are verging on unbreathable. The Great Pacific
Garbage Patch has become an unfathomable mass of floating junk that is destroying our oceans.
The use of fossil fuels is polluting our air and adding to the Greenhouse Effect.
5. Inventions: Science-y careers create an intellectual and business environment that is conducive
to problem solving and invention. Look at all the exciting inventions of the last twenty years, that
have completely transformed our lifestyles. The Internet, the everyday use of GPS, mobile phone
technology, PC’ and touch-screen displays are but a few. This technological progress was made
possible due to Science.
Before you despair, there is a silver lining: every one of these problems can be improved, and even
solved, through Science! If you are passionately concerned about this Planet and about a healthy
future for Humanity, get stuck into your Science studies and aim for a Science-y career that will
equip you to make a difference!
Visit online crowdfunding platforms such as Kickstarter and Indiegogo, and appraise the exciting
Science-y inventions that are being funded. The tech scene is mushrooming with skunkworks and
hackathons that are creating radical innovations. It is an exciting time to be part of Science and
technology, and if you want to be at the cusp of making cool things that make a big difference,
study Science!
2. Be smart: The study of Science encourages problem-solving tenacity that helps you to understand the world around you. I have always explained to my students that Science illuminates one’s
path, and that going through life without Science is similar to driving your car along dark roads your headlights might light your way forward, but they don’t illuminate the world around you. You
travel onwards without ever understanding the context of your journey.
Studying Science makes you comfortable with the unknown, and gives you the confidence to say:
“I don’t know the answers, but I will find out!”. Science is gracious to naivety but does not condone the apathy of indifference: it allows you to say “I don’t know, but I want to find out”, but
does not tolerate the attitude of “I don’t know and I don’t care”.
6. Be a modern-day hero (#2): South Africa has a growing deficit of expert Science teachers. If
you are passionate about Science, and passionate about making a difference, teaching is a massively rewarding career path that is becoming increasingly lucrative. Remember, supply and demand dictate going rates - if there are fewer expert Science teachers around, the demand for expertise leads to increased fees. Become a Science teacher, a thought leader and a role model!
7. Wealth: More than a fifth of the planet’s wealthiest people on the Forbes 2015 list studied an engineering degree, according to a recent survey by the Approved Index platform. A quarter of the
Forbes top-hundred have Science as a foundation for their work.
Science is highly structured, but welcomes change - it constantly adjusts its views based on what
is observed. This approach teaches you to evolve your thinking by constantly testing and investigating information, which makes you a well-rounded human being and empowers you with an ethical
approach to others: it enables you to discern the difference between your opinions and facts, and
to acknowledge the opinions and beliefs of others without immediately accepting or rejecting
them.
8. Discovery: Science research is a field that allows you to discover the unknown. The deep oceans
are unexplored, nanotechnology and photonic crystals have so many secrets, and we’re still not
sure whether there is any form of life outside near-Earth space. Imagine being the person that
publishes a peer-reviewed article to tell the world about a brand new discovery, or a new revelation in our understanding, or a life-altering breakthrough in technology.
3. Be adventurous: Science gets you places! I can only speak from my experience - my engineering
background, which is firmly rooted in Science, has opened a door to great adventure and exploration. I have worked on four continents and have been exposed to a diversity of incredible experiences that a ‘normal’ office job would never allow. Would you like to work in jungles? Study Natural Sciences. A life of studying volcanoes or auroras, perhaps? Study geosciences. Would you like
to ply you mind to solving massive problems and driving innovation? Study engineering! Would you
like to work with killer whales? Study zoology!
This is a call to action for young history-makers, and for a new wave of heroes to save this
world and make a difference. I encourage you to become part of it!
James Hayes
Founder – Science Clinic
Science-y careers and research allow you visit places that would not be accessible through other
fields of study. Whether you want to go to Antarctica or to outer space, Science is the way to get
there.
1
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1
TABLE OF CONTENTS
Foreword
4
6
10
Physics
Physics data
Pulses and Waves
Sound & Electromagnetic radiation
14
15
16
19
20
27
56
29
31
33
35
36
40
42
45
47
50
54
Magnetism
Electrostatics
Electricity
Vectors and Scalars
Motion in 1D
Energy
Chemistry
Chemistry data
Classification of matter
Names and formulae
Kinetic molecular theory
The atom
The periodic table
Chemical bonding
Physical and chemical change
Reactions in aqueous solutions
Quantitative aspects of chemical change
Hydrosphere
Mathematics essentials
www
GENERAL INFORMATION
Quantities, symbols and units
Unit name
Unit
symbol
kg
m
m
-1
m·s
-1
m·s
-1
m·s
-2
m·s
Alternative
symbol
kilogram
metre
metre
metre per second
metre per second
metre per second
metre per second per second
m·s
s
u, v
u
v
Preferred
symbol
m
x, y
Δ x, Δ y
v x, v y
vi
vf
a
metre per second per second
Ek
Ep
Fg
ν
ε
-2
g
λ
t
Δt
E
K
U
W
F
w
s
s
J
J
J
J
N
N
m
-1
Hz or s
s
-1
m·s
C
V
V
A
f
T
c
Q, q
Δ V, V
E
I, i
R
Ω
second
second
joule
joule
joule
joule
newton
newton
metre
hertz or per second
second
metre per second
coulomb
volt
volt
ampere
ohm
The most common quantities, symbols and SI units used in introductory Physics are listed
below. A quantity should not be confused with the units in which it is measured.
Quantity
mass
position
displacement
velocity
initial velocity
final velocity
acceleration
acceleration due to
gravity
time (instant)
time interval
energy
kinetic energy
potential energy
work
force
weight
wavelength
frequency
period
speed of light
charge
potential difference
emf
current
resistance
Conventions (e.g. signs, symbols, terminology and nomenclature)
The syllabus and question papers will conform to generally accepted international
practices.
NOTE:
1.
For marking purposes, alternative symbols will also be accepted.
2.
Separate compound units with a multiplication dot, not a full stop, for example
m·s-1.
For marking purposes, m.s-1 will also be accepted.
3.
Use the equal sign only when it is mathematically correct, for example:
Incorrect:
1 cm = 1 m
(on a scale drawing)
Correct:
1 cm = 10-2 m
1 cm represents 1 m (on a scale drawing)
Information sheets – Paper 1 (Physics)
me
e
h
c
g
SYMBOL/SIMBOOL
9,11 x 10-31 kg
-1,6 x 10-19 C
6,63 x 10-34 J·s
3,0 x 108 m·s-1
9,8 m·s-2
VALUE/WAARDE
TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES
NAME/NAAM
Acceleration due to gravity
Swaartekragversnelling
Speed of light in a vacuum
Spoed van lig in 'n vakuum
Planck's constant
Planck se konstante
Charge on electron
Lading op elektron
Electron mass
Elektronmassa
TABLE 2: FORMULAE/TABEL 2: FORMULES
Δx = v i Δt + 21 aΔt 2
MOTION/BEWEGING
v f = v i + a Δt
2
⎛ v + vi ⎞
Δx = ⎜ f
⎟ Δt
⎝ 2 ⎠
2
v f = v i + 2aΔx
K=
1
1
mv 2 or/of E k = mv 2
2
2
WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING
U = mgh or/of EP = mgh
c
λ
T=
1
f
WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG
v=fλ
E = hf or/of E = h
1
1
1
=
+
+ ...
R p R1 R 2
ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE
Q = I Δt
V=
W
Q
R s = R1 + R 2 + ...
Transverse pulses
Grade 10 Science Essentials
Pulse: A single disturbance in a medium.
Transverse pulse: A pulse in which the particles of the medium move at
right angles to the direction of motion of the pulse.
SCIENCE CLINIC 2018 ©
INTERFERENCE OF PULSES
Interference: The overlapping of two pulses when they coincide. They meet at the same point at the same time.
Superposition: The algebraic sum of the amplitudes of two pulses that occupy the same space at the same time.
Constructive interference: The phenomenon where the crest of one pulse overlaps with the crest of another to produce a pulse
of increased amplitude.
Destructive interference: The phenomenon where the crest of one pulse overlaps with the trough of another, resulting in a pulse
of reduced amplitude.
When two pulses meet:
CONSTRUCTIVE INTERFERENCE
Before
After
2a
The particles in the rope move vertically up and down as the pulse moves to
the right (particles are moving perpendicular to the direction of movement of
the pulse).
Amplitude
(a)
Rest posi)on
During
a
a
X
Y
X
Y
a
a
a
a
X+Y
Y
X
X+Y
Y
X
a
a
Pulse length
2a
Rest position: The position from which all particles start from and return to
after a pulse or wave has passed.
Amplitude: The maximum displacement of particles from the rest position.
Pulse length: The distance between the start and end of a pulse.
A pulse moves through a medium at a certain speed (v):
distance (m)
speed of pulse (m ⋅ s−1)
v=
D
Δt
DESTRUCTIVE INTERFERENCE
Before
time (s)
During
a
Y
X
a
a
Y
X-Y
After
Y
a
X
a
EXAMPLE:
A pulse travels a distance of 300m in 2 minutes. Determine the speed
of the pulse.
v=
D
Δt
v=
300
120
X
X-Y
Y
X
b-a
b
v = 2,5 m ⋅ s−1
6
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a
b
Transverse waves
Grade 10 Science Essentials
FORMATION OF TRANSVERSE WAVES
Each disturbance in a medium creates a transverse pulse. Many disturbances of
transverse pulses repeated at regular intervals creates a transverse wave.
Hand moves up → back to rest position, and then down → back to rest position.
Frequency (f)
Frequency: The number of waves per
second.
Unit: Hz (hertz)
SCIENCE CLINIC 2018 ©
CALCULATIONS
Period (T)
Period: The time taken for one complete wave.
Unit: s (seconds)
Wave speed (v)
Wave Speed: The distance travelled by
a point on a wave per unit time.
speed (m ⋅ s−1)
distance (m)
D
v=
Δt
Rest posi)on
frequency =
This motion continues at regular intervals to form a transverse wave.
Transverse wave: A wave in which the particles of the medium vibrate at right
angles to the direction of motion of the wave. A transverse wave is a succession of
transverse pulses.
number of waves
total time
period =
Example:
5 waves pass a point in 4 s,
calculate the frequency of the
number of waves
waves.
frequency =
total time
5
T = = 1,25 Hz
4
total time
number of waves
Example:
5 waves pass a point in 4 s,
calculate the period of the waves.
total time
period =
number of waves
4
T = = 0,8 s
5
T=
Rest posi9on
ONE TRANSVERSE WAVE
OR
f=
Wave speed (v)
Wave Speed: The distance travelled by a
point on a wave per unit time.
frequency (Hz)
T- period (s)
Distance (m)
or Time (s)
Wavelength (λ)
IN phase:
A–E
B–F
A–I
D–H
B
F
C
A
λ
λ
D
I
G
E
H
Distance (m)
or Time (s)
λ
-
Trough
Amplitude (a): the maximum displacement of particles from the rest position.
Wavelength (λ): the distance between two consecutive points in phase.
Crest: the highest point (peak) on a wave.
Trough: the lowest point on a wave.
wavelength (m)
MULTIPLE TRANSVERSE WAVES
λ
Amplitude
(a)
v = fλ
Example: Calculate the speed of a
wave with a frequency of 28 Hz and a
wavelength of 5 mm.
v = fλ
v = (28)(5 × 10−3)
v = 0,14 m ⋅ s−1
f- frequency (Hz)
1
T
+
Crest
(a)
-
1
f
Distance (m)
Distance (m)
+
Example: Calculate the speed of a
wave that travels 50m in 7s.
D
v=
Δt
50
v=
= 7,14 m ⋅ s−1
7
speed (m ⋅ s−1)
RELATIONSHIP BETWEEN FREQUENCY AND PERIOD
time (s)
Points in phase: Two points in phase are separated by a whole number (1; 2; 3; …) multiple
of completed wavelengths. Points in phase follow the exact same path (have identical motion).
Points totally out of phase: two points following exactly the opposite path.
Points partially out of phase: Points that are not separated by a whole number multiple of
completed wavelengths. Two points which follow different paths.
7
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TOTALLY OUT of
phase:
A–C
B–D
A-G
D–F
PARTIALY OUT of
phase:
A–B
B–E
A-F
D–E
Longitudinal waves
ion
Co
C
m
re
fa
B
Ra
Co
A
IN phase:
A–C
C–E
B–D
A–E
pr
es
s
c3
o
n
ion
pr
es
s
Ra
Longitudinal wave: a wave in which the
particles of the medium vibrate parallel to the
direction of motion of wave.
Compression: A region of high pressure in a
longitudinal wave.
Rarefaction: A region of low pressure in a longitudinal wave.
Rarefaction: The maximum displacement of a
particle from its rest position.
Co
m
re
fa
pr
es
s
c3
o
n
ion
When a series of forwards and backwards
movements are made to a spring in consecutive
intervals it creates a longitudinal wave.
When particles are pushed closer together a
compression is formed and when particles are
drawn apart a rarefaction is formed.
Wavelength is represented by the distance
between consecutive compressions or consecutive
rarefactions.
SCIENCE CLINIC 2018 ©
m
Grade 10 Science Essentials
D
TOTALLY OUT
of phase:
A–B
B–C
C–D
A–D
E
λ
λ
λ
CALCULATIONS
Frequency (Hz)
Period (T)
Frequency: the number of waves that pass a point in one second.
Unit: Hz (hertz)
frequency =
Period: the time taken for one wave to pass a point.
Unit: s (seconds).
number of waves
total time
frequency =
=
f
=
f
=
Wave Speed: the distance a wave travels in one second.
speed (m ⋅ s−1)
D
v=
Δt
number of waves
total time
Example:
8 waves pass a point in 3 s, calculate the period of the waves.
number of waves
total time
8
3
period
=
f
=
f
=
2,67 Hz
total time
number of waves
3
8
v
=
v
=
v
=
OR
f- Frequency (Hz)
T- period (s)
1
f=
T
time (s)
Example:
Determine the period of a wave
that has a frequency of 75 Hz
T
=
f
=
f
=
0,013 Hz
8
f
=
f
=
f
=
20m ⋅ s−1
frequency (Hz)
Example:
Determine the frequency of a
wave that has a period of 0,3 s
1
f
1
75
D
Δt
6000
300
Wave equation
0,38 s
Relationship between frequency and period
1
T=
f
distance (m)
Example:
Calculate the speed of a wave that travels 60 km in 5 min.
Example:
8 waves pass a point in 3 s, calculate the frequency of the
waves.
frequency
Wave speed (v)
1
T
1
0,3
speed
(m ⋅ s−1)
v = fλ
wavelength (m)
Example:
Determine the speed of a wave with a 20 mm wavelength and a
frequency of 600 Hz.
v =
fλ
v = (600)(20 × 10−3)
3,33 Hz
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v
=
12 m ⋅ s−1
Pulses and Waves- THE ESSENTIALS
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
A pulse is a single disturbance that is propagated from a source through a medium.
A wave is a repetition of pulses at regular intervals.
LONGITUDINAL
Transverse pulse: a pulse in which the particles of the medium move perpendicular to the direction of
propagation of the pulse.
Longitudinal pulse: a pulse in which the particles of the medium move parallel to the direction of
movement of the pulse.
Co
m
pr
es
s
ion
TRANSVERSE
Amplitude
(a)
Rest posi)on
Pulse length
A transverse wave is a succession of transverse pulses, eg water waves.
Crest
ion
n
m
pr
es
s
c3
o
Co
C
re
fa
pr
es
s
m
Co
B
Ra
n
c3
o
re
fa
m
Co
Ra
Rest posi9on
pr
es
s
Amplitude
(a)
ion
A longitudinal wave is a succession of longitudinal pulses. e.g. sound waves
ion
Distance (m)
+
Distance (m)
or Time (s)
Wavelength (λ)
A
D
λ
(a)
λ
λ
Trough
-
Frequency(f): the number of waves per second.
f =
1
T
Superposition: The algebraic sum of the amplitudes of two pulses that occupy the same space at the same time.
Constructive interference: The phenomenon where the crest of one pulse
overlaps with the crest of another to produce a pulse of increased amplitude.
Destructive interference: The phenomenon where the crest of one pulse
overlaps with the trough of another, resulting in a pulse of reduced amplitude
Wavelength(λ): the distance between two consecutive points in phase.
Amplitude: the maximum displacement of a particle from the rest position.
Period(T): the time taken for one complete wave. Unit: s (seconds).
T =
1
f
Wave speed (v): The speed of a wave through the medium.
v = fλ
9
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E
Sound
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
WAVE PROPERTIES & SOUND
SOUND WAVES
Sound is created by vibrations in a medium in the direction of
propagation (longitudinal wave). The vibrations cause regular variation in pressure in the medium.
Sound waves are 3D longitudinal waves that travel through solids,
liquids and gases but not through a vacuum → a medium is
necessary for the propagation of sound. The denser the medium,
the greater the speed of sound through that medium.
FREQUENCY- What the human ear perceives as pitch.
Measured in hertz (Hz)
Pitch is directly proportional to frequency
• A high pitch = high frequency.
(i.e. brakes of a car squealing)
• A low pitch = low frequency.
(i.e. a bass guitar)
A loudspeaker has a paper cone which is able to move back and
forth, producing a series of compressions and rarefactions through
the air. The small changes in pressure are detected by the ear and
we perceive this as sound. The waves move away from the
speaker in 3 dimensions getting weaker the further they travel.
The amplitude diminishes and the loudness decreases.
AMPLITUDE- what the human ear perceives as volume (loudness).
Measured in decibels (dB).
Loudness is directly proportional to amplitude
• A high volume = large amplitude.
• A low volume = small amplitude.
ULTRASOUND
ECHOES
• Humans can hear sounds between 20 Hz and 20 kHz.
• Ultrasound Refers to sound waves of a higher frequency than we
can hear. (20 kHz to 100 kHz)
Sound waves are reflected by large hard flat surfaces such as buildings walls and cliffs.
EXAMPLE:
The time taken for an echo to be heard by a listener on a ship is
measured and is found to be 0,8 s. The speed of sound in water is
1480 m·s−1 . Determine the depth of the water at this point.
Uses
•Treatment of sports injuries: waves are fed through the skin to
increase blood flow to the injured area.
•Imaging in pregnancy: When
an ultrasound wave encounters a
boundary in a medium, part of
the wave is reflected, part absorbed and part is transmitted
through the boundary. The reflected waves are detected in the
same way as echoes and can be
used to form an image (sonar
imaging).
The reflection of the sound wave is known as an echo. An echo
can be used to calculate unknown distances.
REMEMBER:
When a wave reflects/echoes, the wave had to travel to the object
and back. It has therefore travelled double the distance between
the source and the reflection surface.
SOURCE
Source
Using sonar can produce a safe method of early diagnosis of fetal
conditions and early treatment can be implemented.
Echo
LISTENER
d
Boats: use sonar to
locate fish and measure
the depth of the
ocean.
Echo
Sound
1480
=
D
D
=
=
depth
=
depth
=
depth
=
v
=
v
=
v
=
=
=
2 × 250
500 m
D
Δt
500
1,5
333,33 m ⋅ s−1
∴ speed of sound = 333,33 m ⋅ s−1
10
D
Δt
D
0,8
(1480)(0,8)
1184 m
D
2
1184
2
592 m
EXAMPLE:
A man stands 250 m from a wall and hears the echo of his gunshot
after 1,5 s. Calculate the velocity of sound under the circumstances.
distance traveled by sound
•Bats: use ultrasound to locate insects and to navigate in the dark.
=
The sound has to travel to the bottom and then reflect back to the
ship at the surface.
sound
http://en.wikipedia.org/wiki/Ultrasound#/media/File:CRL_Crown_
rump_lengh_12_weeks_ecografia_Dr._Wolfgang_Moroder.jpg
v
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Electromagnetic radiation
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
Electromagnetic radiation behaves
like a
WAVE
when it travels
PARTICLE
when it interacts with other materials
WAVE NATURE OF ELECTROMAGNETIC RADIATION
PARTICLE NATURE OF ELECTROMAGNETIC RADIATION
• Accelerating charges are the source of all Electromagnetic Radiation.
• The accelerating charges create a constantly changing electric field which travels away from the source
in 3 dimensions. The continuously changing electric field then induces a changing magnetic field that is
perpendicular to the electric field.The changing magnetic field in turn produces an electric field.
Electromagnetic waves are created by oscillating magnetic and electric fields which move at right angles
to each other and to the direction of the propagation of the wave.
Accelerating
charges
produce a…
Continuously
changing electric
field
which then
induces a…
Continuously
changing magnetic
field at 90⁰
•
•
•
•
Electromagnetic radiation transfers energy to other matter in “packets” called photons.
Photons have a set amount of energy → called a quantum.
The amount of energy is directly proportional to the penetrating ability of the specific EM-radiation.
A Photon is a quantum of EM radiation (which carries a set amount of energy)
Energy of a photon can be calculated using the formula:
energy (J)
Electromagnetic wave is
created
E = hf
Electric field
Magne/c
field
frequency (Hz)
Planck's constant (6,63 × 10−34 )
Direc/on of
propaga/on
Planck's constant (6,63 × 10−34 )
hc
E=
λ
energy (J)
speed of light (3 × 108 m ⋅ s−1)
wavelength (λ)
WAVE EQUATION (EM)
The wave is propagated at a speed of 3×108 m·s−1.
This is called c (speed of light).
All electromagnetic waves travel at the speed of
light.
wave speed (m ⋅ s−1)
wavelength (m)
c = fλ
frequency (Hz)
EXAMPLE:
Calculate the energy of a microwave with a frequency of 4,5 × 1010 Hz.
EXAMPLE:
Determine the wavelength of ultraviolet waves
with a frequency of 1,6 × 1016 Hz
c
3 × 108
=
=
fλ
(1,6 × 1016 )λ
λ
= 1,88 × 10−8 m
EXAMPLE:
Determine the frequency of a radio wave with a
wavelength of 300 m
c
3 × 108
=
=
f
=
E
E
=
=
E
=
EXAMPLE:
Determine the energy of gamma ray with a wavelength of 4 × 10−14 m.
hf
(6,63 × 10−34 )(4,5 × 10−10 )
2,98 × 10−43 J
E
=
E
=
E
=
hc
λ
(6,63 × 10−34 )(3 × 108 )
(4,5 × 10−14 )
4,42 × 10−12 J
PENETRATING ABILITY
The penetrating ability of a wave refers to its ability to
move through matter.
fλ
f(300)
1 × 106 m
The higher the energy, the greater the penetrating ability.
11
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penetrating ability α Energy
penetrating ability α Frequency
Grade 10 Science Essentials
Electromagnetic radiation- Spectrum
SCIENCE CLINIC 2018 ©
INCREASING FREQUENCY
INCREASING WAVE ENERGY
INCREASING WAVELENGTH
Ultraviolet light
(UV)
X-rays
Gamma rays
4 x 1014 to 7 x 1014
1 x 1016
1 x 1019
1 x 1022
1 x 10-4
7 x 10-9 to 4 x 10-9
1 x 10-8
1 x 10-10
1 x 10-14
Remote controls;
op4cal fibers
Objects reflect, refract
or transmit light that we
are able to see;
photosynthesis
Light bulbs; steriliza4on
x-rays; CT scans;
security scans
Radia4on of cancer
Damage to eyes and
skin
Damage to skin and
underlying 4ssue
Damage to 4ssue;
nuclear radia4on
Radio waves
Frequency (Hz):
1 x 106
1 x 1010
1 x 1012
Wavelength (m):
1 x 102
1 x 10-2
Radio and TV
broadcasts; radio
telescopes
Telephone and cell
phone connec4ons;
communica4on
satellites;
microwave ovens;
radar systems
Advantages:
Microwaves
Visible light
Infrared
radiation (IR)
EM Radiation:
R O Y G
Disadvantages:
12
B
I
V
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Sound and EM Waves-THE ESSENTIALS
Grade 10 Science Essentials
Sound
EM Radiation
•Sound waves are created by vibrations in a medium.
•
•
•
•
•
•
•The speed of sound in air = 340 m·s-1
•Sound is a longitudinal wave that travels through solids, liquids and gases but not through a vacuum.
Frequency:
Accelerating charges are the source of all Electromagnetic Radiation.
Propagates as electric and magnetic fields which are perpendicular to each other.
The wave is propagated at a speed of 3x108 m.s-1 in a vacuum. This is called c (speed of light)
Does not need a medium to propagate through; can travel through a vacuum.
Electromagnetic radiation behaves like a transverse wave when it propagates.
Electromagnetic radiation behaves like a particle when it interacts with other materials
Amplitude:
What the human ear perceives as pitch.
What the human ear perceives as volume.
Measured in decibels (dB).
•A high pitch = high frequency.
(I.e. brakes of a car squealing)
•A low pitch = low frequency.
(I.e. a bass guitar)
WAVE NATURE
wave speed
•A high volume = large amplitude.
•A low volume = small amplitude.
(m ⋅ s−1)
PARTICLE NATURE
wavelength (m)
c = fλ
• Energy of a photon can be calculated using:
energy (J)
Sound waves are reflected by large hard flat surfaces such as buildings walls and cliffs. The reflection of
the sound wave is known as an echo. An echo can be used to calculate unknown distances.
Source
• Electromagnetic radiation transfers energy in
“packets” called photons
Photon: quanta (energy packets) that transfer
energy to particles of matter
frequency (Hz)
ECHOES
SOURCE
SCIENCE CLINIC 2018 ©
As the frequency increases, the wavelength
decreases.
E = hf
In any given medium, the speed remains
constant.
Planck′s constant (6,63 × 10−34 )
sound
Planck′s constant (6,63 × 10−34 )
Electromagnetic Spectrum
f
Radiowaves
Echo
LISTENER
Waves
λ
frequency (Hz)
E=
hc
λ
speed of light (3 × 108 m ⋅ s−1)
wavelength (λ)
Microwaves
Infrared
d
Visible light
In the picture:
distance to wall and back
2D
=
=
energy (J)
Ultraviolet
v ×t
v ×t
X-Rays
Gamma Rays
where v is the speed of sound in air;
v = 340 m ⋅ s−1
13
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The greater the energy of a photon, the higher
its frequency and its penetrating ability (and the
smaller/ shorter the wavelength)
Magnetism
Grade 10 Science Essentials
Magnetic field: region in space where another magnet or ferromagnetic material will experience a non-contact force
Non-contact force: A force exerted on an object without touching the object.
Magnet: an object with a pair of opposite poles, called north and south
Magnets can be made from ferromagnetic materials; materials which are strongly attracted by magnets and are easily magnetised. Common ferromagnetic materials include iron, cobalt, nickel and their alloys.
When two magnets are brought close to each other:
~ Like poles → REPEL each other
~ Unlike poles → ATTRACT each other
Even if a magnet is cut into little pieces it will form smaller magnets, each with their own north and south poles
N
NS N
S
S N S
• Magnetic field lines are imaginary lines around a magnet indicating the direction and the strength of the magnetic field
surrounding the magnet (iron filings on paper can be used to indicate the field lines)
The field lines occur inside the magnet and 3 dimensionally outside it. They come out of the magnet near the north pole
and enter near the south pole again forming a loop.
SCIENCE CLINIC 2018 ©
Earth’s Magnetic Field
The Earth behaves as if it has a giant bar magnet running through it, with its
magnetic field looping around the earth
The arrow of the compass is the North pole of the compass, and points towards
magnetic North Pole of the Earth.
In reality, the North pole of the Earth is actually the South pole of the imaginary magnet through the Earth.
The point in the Northern Hemisphere where the Earth’s axis meets the surface of
the Earth is the geographical or true North Pole. A compass will not point to this, but
rather to the magnetic North Pole – which is actually the South pole of the imaginary
magnet through Earth.
The angle between the true North Pole and the magnetic North Pole is called the angle of declination.
Geographical north pole: Point in the northern hemisphere where the rotation
axis of the earth meets the surface.
Magnetic north pole: The Point where the magnetic field lines of the earth enters
the earth.It is the direction in which
Magne)c
North Pole
Bar magnet
N
S
Geographical
North Pole
Other “fields” in physics
Properties of magnetic field
lines
• Imaginary lines
• Continuous
• 3 dimensional
• Never cross or touch
• Arrows indicate the direction
of the field (N to S)
• Are more concentrated at the
poles where field is stronger
Magnetic field
Area in which a magnetic material
experiences a magnetic force
Electric field
Area in which an electric charge
experiences an electric force
Gravitational field
Area in which a mass experiences
a gravitational force
Magne&c
North Pole
S
Geographical
South Pole
Geographical
North Pole
Angle of
declina&on
S
N
N
Magne)c
South Pole
Attraction between unlike poles
Repulsion between like poles
Magnetic Phenomena
The sun releases particles (mainly electrons and protons) which travel outwards
in all directions at speeds of up to 900 m.s⁻¹.
N
S
N
S
N
S
S
N
More particles are released during a solar storm called a coronal mass ejection.
When these particles come close to the Earth, the Magnetosphere protects the
Earth by deflecting most of the particles, however some penetrate the Magnetosphere and are pulled towards our North and South Poles. They enter the upper
atmosphere at high speed and collide with nitrogen and oxygen molecules
causing them to glow with different colours.
These are known as the Aurora Borealis (Northern lights) and Aurora Australis
(Southern lights).
14
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Electrostatics
Grade 10 Science Essentials
• All materials contain positive charges
(protons) and negative charges (electrons)
• A neutral object has an equal number of
protons and electrons and has no net
charge.
• A positively charged object has lost
electrons (electron deficient)
• A negatively charged object has gained
electrons (excess of electrons)
• A charged object can also be attracted to
a neutral conductor because of the
movement of charges in the conductor
• A charged object can also be attracted to
an insulator because of the movement of
polarised molecules present in the insulator
• Charged objects exert forces on each
other when brought close together
• Like charges exert REPULSIVE forces on
each other
• Opposite charges exert ATTRACTIVE
forces on each other
QUANTISATION OF CHARGE
The charge on a single electron is
Methods by which a neutral object can become charged
Friction
• Also known as tribo-electric charging
• Rub two objects together such as a glass rod and
a woolen cloth. Electrons are transferred from one
object to the other (this depends on which materials
are used)
• The one object loses electrons and the other gains
them
Glass rod and woolen cloth
3e¯
3p⁺
charge of e− ( − 1,6 × 10−19 )
Q = nqe
charge (C)
number of electrons
Examples:
Calculate the number of electrons gained or
lost by (a) +4,5 nC and (b)-9 µC
Q
=
+4,5 × 10−9
=
n
=
∴
=
neutral
Q
4e¯
4p⁺
−9 × 10−6
=
n
=
electrons gained
0C
neutral
+1,4C
+1,4C
Neutral
7e¯
4p⁺
2e¯
5p⁺
+
Transfers nega5ve
charge
-2,3C
0C
neutral
neutral
Nega-vely
charged
(e¯ excess)
neutral
-2,3C
-2,3C
Temporary charge
(polarised)
+
---- - -
Neutral
+
+
+
+
+ +
+ +
+
+
e⁻ are attracted to the positive rod. Positively charged
leaves repel each other
Nega%ve
rod
-+ -+ -+ + +
- -+
+
-
--
Semi-permanent
charge
--
+ + + ++
+ +
+ +
+
+
e¯
A negative rod is brought
close to a neutral electroscope. The e-s are repelled.
touch
Posi-vely
charged
(e¯ deficient)
+
-+ -+ -+ + +
- -+
+
-
+
+
A positive rod is
brought close to a
neutral electroscope
Negative conductor touches a neutral
conductor → negative charges are transferred
from negative to neutral charge
-4,6C
Electroscope is earthed allowing
extra e-s to come towards the rod. A
semi-permanent charge is created.
PRINCIPLE OF CONSERVATION OF CHARGE
Principle of conservation of charge: the net charge of an isolated system remains constant during any physical process. Eg. When identical objects with different charges touch, charges will be transferred between conductors. The two conductors will then have equal charge.
EXAMPLE: Two identical charges on insulated stands, +5 mC and -9 mC respectively, touch and move apart again. Determine (a) the new charge on each, (b) the
amount of charge transferred and (c) the amount of electrons transferred.
Amount of charge transferred:
a) New charge on each:
=
5,63 × 1013
Nega-vely
charged
(e¯ excess)
Fric-on
e¯
Q new
n(−1,6 × 10−19 )
Posi-vely
charged
(e¯ deficient)
5e¯
5p⁺
−2,81 × 1010
nq e
+
Transfers nega5ve
charge
+2,8C
PVC rod and woolen cloth:
=
=
Posi%ve rod
-1,4C
neutral
Q new
2,81 × 1010 electrons lost
Induction
A method of charging whereby the objects do not touch
one another. Neutral electroscope is charged by induction.
touch
nq e
n(−1,6 × 10−19 )
Positive conductor touches a neutral
conductor → negative charges are transferred
from neutral to positive charge
7e¯
5p⁺
1e¯
3p⁺
5e¯
5p⁺
Contact
When a charged conductor touches an identic a l n e u t ra l c o n d u c t o r, e l e c t r o n s a r e
transferred to make the charge on the
conductors equal.
Fric-on
e¯
q e = − 1,6 × 10−19 C
Principle of charge quantisation: all
charges in the universe consist of an integer
multiple of the charge of an electron:
SCIENCE CLINIC 2018 ©
Q new
=
Q1 + Q 2
2
(+5 × 10−3) + (−9 × 10−3)
2
−2 × 10−3 C each
ΔQ = Qf − Qi
Charge
transferred
(C)
new
charge
on
each
original
charge
before
contact
(you can use either of the original charges in this step)
15
c) Number of electrons transferred:
ΔQ = Q f − Q i
Q
=
7 × 10−3
=
ΔQ = − 7 × 10−3 C
OR
ΔQ = Q f − Q i
n
=
4,38 × 1016
∴n
=
n(−1,6 × 10−19 )
ΔQ = (−2 × 10−3) − (+5 × 10−3)
ΔQ =
(−2 × 10−3) − (−9 × 10−3)
ΔQ =
− 7 × 10−3
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C
Q
−7 × 10−3
n
nq e
n(−1,6 × 10−19 )
= 4,38 × 1016
OR
=
nq e
=
4,38 × 1016
e−
e−
Electricity
Grade 10 Science Essentials
Electricity is the process of giving energy to a charge and then using that energy
to do work. The most basic circuit is made of a POWER SOURCE, a LOAD and the
conducting wires carrying the charge around the circuit.
The POWER SOURCE
gives electrical energy
to the charge.
The POWER SOURCE gives
electrical energy to the charge.
Some examples are generators,
batteries, photovoltaic cells.
The LOAD is a device that uses the
energy of a charge.
The LOAD is a device which
dissipates/uses the energy.
Some examples are resistors, light
bulbs, motors, speakers, chargers.
A battery goes “flat” when all of the stored chemical energy in the battery has
been converted to electrical energy.
The electrical energy is converted into other forms: heat, light, kinetic, chemical
CURRENT is the rate of flow of charge.
I=
Q
t
Current
Lightning
10 000 A
Starter motor in car
200 A
Fan heater
10 A
Toaster
3A
Light bulb
0,4 A
Pocket calculator
0,005 A
Nerve fibre in the body
0,000 001 A
DEFINITION: The coulomb(C) is that quantity of charge
which passes a fixed point in a conductor in one second when
the conductor carries a current of one ampere.
EXAMPLE:
An electric current of 7.5 A flows through an electric
circuit for 3 minutes. Calculate the amount of charge
flowing through the circuit.
Q
Q
Q
=
=
=
It
(7,5)(3 × 60)
1350 C
A torch is an example of a simple circuit. The arrows indicate the flow of conventional current.
Conductor takes energized
charge to load
Cell
(Chemical Energy into
Electrical Energy)
Light Bulb
(Electrical Energy
into Light Energy)
Conductor returns charge to power
source to get more energy
POTENTIAL DIFFERENCE (p.d.) across the ends of a conductor is
the energy transferred per unit electric charge that flows
through it.
V=
I is the current strength, Q is the charge in coulombs and t is
the time in seconds. The SI unit of current is the ampere (A).
Typical values for electric current:
Situation
SCIENCE CLINIC 2018 ©
W
Q
V is Potential difference in V (volts), W is Work done or energy
transferred in J (joules) and Q is Charge in C (coulombs).
Note that 1V = means 1 joule per coulomb of charge.
RESISTANCE & OHM’S LAW: Opposition to the flow of electric current.
The greater the potential difference across
the ends of a particular conductor, the
greater the current in it. In a metal
conductor, the charge carriers are electrons.
The valence electrons of the metal atoms
swarm around randomly in between the positive metal ions. When a voltage is applied
across two points of the metal, the electrons
migrate towards the positive point and collide with the particles of the metal. The
kinetic energy of the electrons is transferred
to the positive metal ions causing them to
vibrate faster and get hot.
Chemical energy →
In cell
Note:
1. Emf is the voltage measured across the terminals of a battery when no
current is flowing through the battery (the switch is open). The emf is
the work done per unit charge by the source
2. V term or potential difference (pd) is the voltage measured across the
terminals of a battery when current is flowing through the battery. This
is also known as the operating voltage.
EXAMPLE:
A battery transfers 60 kJ of energy to 5 kC of electric charge.
Calculate the voltage of the battery.
V
=
V
=
V
=
→
Heat energy
metal particles
DEFINITION: Ohm's Law states that the current in a conductor is
directly proportional to the potential difference across it provided its
temperature remains constant.
V = IR
R is referred to as the electrical resistance of the conducting material, resisting the
flow of charge through it.
Resistance (R) is the ratio of the potential difference (V) across a resistor and
the current (I) in it. The unit of resistance is called the ohm (Ω).
EXAMPLE:
A battery has a voltage of 12 V. The current that flows through the resistor in the circuit is
6 A. Calculate the resistance of the resistor.
W
Q
60000
5000
12 V
16
Electrical energy
charge carriers
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R
=
R
=
R
=
V
I
12
6
2Ω
Electricity- Series and Parallel
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
Series Circuits: A 1,5 V cell is connected to 3 identical bulbs in series. This creates 3 times more
resistance than one bulb, so the current is 3 times less
Parallel circuits: A 1,5 V cell is connected to 3 identical bulbs in parallel. The cell has to supply 1,5 V to
each bulb at the same time, so the current is 3 times more
1.
2.
3.
4.
1. The current splits into three equal parts for each of the branches.
2. If one branch breaks, the other two will still conduct current.
3. Each bulb receives the full 1,5 V provided by the cell across it
There is only one path for the current to follow.
The current is the same everywhere.
If the circuit is broken at any point, the current will stop flowing.
The 1,5V supplied by the cell is divided up into 3 equal parts (0,5 V) across each bulb.
RESISTORS IN SERIES
R1
• Current is split up through resistors.
• The current is inversely proportional to the
resistance. The smaller resistance, the more
current going through it.
• Voltage is the same across each resistor in the
parallel combination.
• “Current divider”
• Cars, houses
• Current is the same through each resistor
• Voltage is split up across resistors.
•The voltage is directly proportional to the resistance in
the series combination. Most voltage will be across the
largest resistance.
•“Voltage divider”
•Dimmer switches, Christmas lights
RTOTAL = R1 + R2 . . .
R2
RESISTORS IN PARALLEL
COMBINATION CIRCUITS.
TO SOVLE: Identify the separate series and parallel combinations in the circuit.
Apply Ohms Law to each resistor, or each combination of resistors or the whole circuit consistently.
STEP 1:
Find the total resistance
1
RP
1
RP
=
=
1
RP
=
∴ RP
=
RTOT
RTOT
RTOT
=
=
=
1
R2
1
8
+
+
1
R3
1
8
2
8
4Ω
RP + R1 + R 4
5+4+6
15 Ω
EXAMPLE:
Calculate the current through each resistor and the voltage across each resistor in the combination circuit.
STEP 2:
STEP 3:
STEP 4:
Find the total current
Apply known variables
Apply Ohm’s law to unknown resistors
to circuit components
V
I =
I
=
I
=
R
60
15
R1 and R2 are in series
∴ ITOT = IR = IR = 4 A
1
4A
4
IR + IR = 4A
2
3
and IR = IR
2
3
∴ IR = IR = 2 A
2
3
VR1
=
IR
VR1
=
(4)(5)
VR1
=
20 V
VR
=
VR
=
IR
=
VR
=
(2)(8)
=
VR
=
16 V
VR
VR
17
2
2
2
3
3
3
VR 4
=
IR
=
(4)(6)
VR 4
=
24 V
VR 4
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R1
R2
1
RTOTAL
=
1 1
+
R1 R2 . . .
Electricity - THE ESSENTIALS
Grade 10 Science Essentials
Quantity
Current (I)
SCIENCE CLINIC 2018 ©
Potential Difference (V)
Resistance (R)
Definition
The rate of flow of charge.
The energy transferred per unit electric charge.
The ratio of potential difference across a resistor
to the current through it.
Unit
A (amperes)
V (volts)
Ω (ohms)
Equation
I = Q /t
V=W/Q
R = V /I
Meaning
How much charge passes every second
How much energy each coulomb of charge
transfers between 2 points
How many volts are required to sustain
a current of 1 ampere.
The coulomb (C) is that quantity of charge
which passes a fixed point in a conductor
in one second when the conductor carries
a current of one ampere
Emf is the voltage measured across the
terminals of a battery when no current is flowing
through the battery
V term or potential difference (pd) is the
voltage measured across the terminals of a
battery when current is flowing through the
battery.
Ohm's Law states that the current in a
conductor is directly proportional to the
potential difference across it provided its
temperature remains constant.
Resistance of a material is dependent on the
type of material, as well as the length, thickness
and temperature of the conductor
RESISTORS IN SERIES
RESISTORS IN PARALLEL
R1
R2
RTOTAL = R1 + R2 . . .
R1
•Current is the same through each resistor
•Voltage is split up across resistors. The voltage
is directly proportional to the resistance in the
series combination.
•‘Voltage divider’
•Eg. Dimmer switches
R2
1
RTOTAL
COMBINATION CIRCUITS.
=
1 1
+
R1 R2 . . .
•Current is split up through resistors. The current
is inversely proportional to the resistance. The
smaller resistance, the more current going
through it.
•Voltage is the same across each resistor in the
parallel combination.
•‘Current divider’
• Eg Cars, houses.
Current is measured with an ammeter,
which is connected in series.
Potential difference is measured in parallel using
a voltmeter.
18
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• Identify the separate series and parallel
combinations in the circuit.
• Apply Ohms Law consistently to:
o each resistor,
o or each combination of resistors
o or the whole circuit.
Vectors and Scalars
Grade 10 Science Essentials
NET OR RESULTANT VECTOR
Scalar
Vector
A physical quantity with
magnitude only
A physical quantity with both
magnitude and direction
E.g.
mass (kg); distance (m);
speed (m.s⁻¹); time (s);
energy (J); temperature (K)
E.g.
Force (N); weight (N);
displacement (m); velocity
(m.s⁻¹); acceleration (m.s⁻²)
Graphical representation of a vector
Vector is represented by an arrow.
Length of arrow = magnitude of vector.
Direction of arrow = direction of vector.
ud
nit
ag
e
M
Tail = Origin
+ and – is used to indicate direction of a vector.
+5 N
left is –
• Net or resultant vector: the single vector that has the same effect as two or more vectors together.
• Net vector is greatest when vectors are in the same direction.
• Net vector is smallest when vectors are in the opposite direction.
Vectors in same direction
Vectors in opposite directions
Determine the net force when a 5 N force
acts to the right and a 10 N force also acts
to the right.
Determine the net force when a 12 N force
acts to the right and a 7 N force acts to
the left.
Let to the right be positive:
Let to the right be positive:
Multiple vectors in different
directions
Determine the net force when a 8 N force
acts to the right, a 10 N force acts to the
right, a 25 N force acts to the left and a 12
N acts to the left.
Head =
Direc3on
θ
e.g. right is +
SCIENCE CLINIC 2018 ©
Fnet
=
=
=
F1 + F2
5 + 10
15 N right
Fnet
=
=
=
F1 + F2
12 + (−7)
5 N right
Let to the right be positive:
Fnet
=
=
=
=
F1 + F2 + F3 + F4
8 + 10 + (−25) + (−12)
−19 N
19 N left
–3N
3 METHODS TO DESCRIBE THE DIRECTION OF A VECTOR
(the same vectors are used and described in each example)
On a graph
Bearing
Use North as 0⁰ and always measure clockwise
Compass (Cardinal Points or directions)
FA: 10 N at 30⁰ above the positive x-axis
FA: 10 N on a bearing of 60⁰
FA: 10 N at 30⁰ North of East
FB: 8 N at 12⁰ left of the negative y-axis
FB: 8 N on a bearing of 192⁰
FB: 8 N at 12⁰ West of South
FC: 5 N at 65⁰ above the negative x-axis
FC: 5 N on a bearing of 335⁰
FC: 5 N at 65⁰ North of West
19
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NOTE:
South-West,
North-West,
North-East and
South east can
only be used if
the angle is 45°
Motion in 1 dimension
Grade 10 Science Essentials
FRAME OF REFERENCE AND POSITION
SCIENCE CLINIC 2018 ©
DISTANCE AND DISPLACEMENT
• Position (x): the place where an object is relative to a specific reference point.
Position is a vector. Unit is meter(m)
• Reference point: a point from which the position of the object is measured (zero point).
• One dimensional motion: motion that occurs either vertically or horizontally along a straight line.
EXAMPLE:
Let to the right be positive:
Distance
Displacement
Total path length travelled
Difference in position in space. Displacement
is measured from start to final position
Scalar
Vector
Symbol: D
Symbol: ∆x (or ∆y)
Unit: meter (m)
Unit: meter (m)
EXAMPLE:
A boy runs from position A to E in a clockwise motion as shown in the diagram. Describe his distance
and displacement at position B,C and E using A as the starting (reference) point.
(a) Describe the position of objects A and C by using object B as a reference point as in the diagram.
Object A is -3 m from B (3m to the left)
Object C is +5 m from B (5m to the right)
B: distance is 100 m, displacement is 80 m at 25⁰ East of South
C: distance is 200 m, displacement is 120 m South
E: distance is 400 m, displacement is 0 m
(b) Describe the position of objects B and C by using object A as a reference point as in the diagram.
Object B is 3 m right of A
Object C is 8 m right of A
SPEED
• Speed ( v ): rate of change of distance with time.
Scalar, unit is m.s⁻¹
In other words → how much distance is travelled every second.
• Constant speed: covers equals distances in equal time intervals.
• Average speed: total distance covered over the total time taken.
speed (m ⋅ s−1)
distance (m)
v=
D
t
VELOCITY
ACCELERATION
• Acceleration ( a ⃗ ): rate of change of velocity.
Vector, unit is m.s⁻2
In other words → how much the velocity changes every second.
• Constant acceleration: a constant increase or decrease in velocity
• Constant velocity: covers equals displacement in equal time in equal time intervals.
intervals.
change in velocity (m ⋅ s−1)
acceleration (m ⋅ s−2 )
• Average velocity: total displacement covered over the total time
taken.
vf − vi
Δv
⃗
a
=
=
−1
displacement (m)
velocity (m ⋅ s )
Δt
Δt
time (s)
• Velocity ( v ⃗ ): rate of change of position.
Vector, unit is m.s⁻¹
In other words → how much the position changes every second.
v⃗=
Δx
Δt
Remember + and – refers to direction of the vectors
Let us say right is + and left is – in this case:
1. Positive acceleration:
Increasing velocity in + direction (speeding up right) OR decreasing
velocity in the – direction (slowing down left)
2. Negative acceleration:
time (s)
time (s)
Decreasing velocity in + direction (slowing down right) OR increasing
• Instantaneous velocity: the velocity of an object at a specific velocity in the – direction (speeding up left)
[if change in velocity and acceleration is in the same direction it is
• Instantaneous speed: the speed of an object at a specific moment in time
moment in time.
e.g. 1. Let the specific time be 2 seconds – then the instantaneous speeding up and if in opposite directions it is slowing down]
3. Deceleration:
e.g. 1. Let the specific time be 2 seconds – then the instantaneous
velocity will be annotated by V2.
Non-scientific term that means an object is slowing down. In science
speed will be annotated by V2.
e.g.2. the initial instantaneous velocity will be Vi.
we rather refer to acceleration → direction and context describe
e.g.2. the initial instantaneous speed will be Vi.
whether it is speeding up or slowing down.
20
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Motion in 1D- Equations of motion
Grade 10 Science Essentials
VARIABLES
u
vi
Initial velocity
v
vf
Final velocity
a
a
acceleration
s
∆x
Displacement
t
∆t
Time taken
CALCULATIONS:
A racing car starting from rest on the grid, travels straight along the track and reaches the 400 m mark
after 8,6 s.
a) What was its average acceleration?
Let forward be positive.
Equations for constant acceleration in one dimension:
Old symbols
v = u + at
New symbols
vf = vi + aΔt
Leaves out
s or Δx
s = 12 (u + v)t
Δx = 12 (vi + vf )Δt
a
s = ut + 12 at 2
v 2 = u 2 + 2as
Δx = 12 aΔt 2
vf2 = vi2 + 2aΔx
SCIENCE CLINIC 2018 ©
u
0
v
/
a
?
s
400 m
t
8,6 s
Δx
=
400
=
a
=
viΔt + at 2
1
a8,62
2
−2
10,82 m ⋅ s
forward
b) What was its velocity at the 400 m mark?
v or vf
t or Δt
vf
vf
=
=
vi + aΔt
0 + (10,82)(8,6)
vf
=
93,05 m ⋅ s−1 forward
c) At the 400 m mark, the brakes are applied and the car slowed down at 2 m.s-2 to come to rest.
Calculate the time it took for the car to stop.
Steps to using the equations:
a) Identify each stage of the motion, where the acceleration has changed.
b) Choose a positive direction and stick to your convention.
c) Record the information given and value required by writing next to each variable. Check the unit and
direction.
d) Select correct equation and solve for unknown.
e) Include units and direction in your answer.
NB! New stage of motion. Find the new value of each variable.
Let forward be positive.
Remember:
‘starting from rest’ means: u or vi = 0
‘comes to a stop’ means: v or vf = 0
‘Slowing down’ means: acceleration is negative (a < 0), while still moving in a positive
direction.
Constant velocity means: a =0, u = v or vi = vf
Use a new set a variables for each stage of the motion.
Conversion of units: 1 m.s-1 = 3,6 km.h-1.
21
u
93.05 m.s-1
v
0
a
-2 m.s-2
s
/
t
?
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vf
0
t
=
=
=
vi + aΔt
93,05 − 2t
46,53 s
Grade 10 Science Essentials
Motion in 1D- Graphs of Motion
1.
2.
3.
4.
5.
6.
SCIENCE CLINIC 2018 ©
Tips on reading graphs:
Check which variable is on the vertical axis
Check which direction is positive.
The velocity at any specific time, can be calculated from gradient of the x-t graph at that time.
The acceleration at any specific time, can be calculated from gradient of the v-t graph at that time.
The displacement up to a specific time can be calculated from the area under the v-t graph up to that time.
The change in velocity up to a specific time can be calculated from the area under the a-t graph up to that time.
gr a d i e n t
=
=
area of triangle =
Δy
Δx
y2 − y1
1
l ×b
2
area of rectangle = l × b
x 2 − x1
Identify the separate stages of the motion (Forward direction positive):
1. Positive acceleration (accelerating forward) from rest.
2. Constant positive velocity (forward).
3. Negative acceleration (decelerating forward) to rest.
4. At rest
5. Negative acceleration (accelerating backwards) from rest.
6. Constant negative velocity (backwards).
7. Positive acceleration (decelerating backwards) to rest. ( i.e. Accelerating forwards)
EXAMPLE:
The graph shows two cars A and B are traveling at 40 m.s⁻¹ when the brakes are applied and they decelerate to rest.
a) What is the initial speed of the cars in kmh-1?
40 × 3,6
= 144 km ⋅ h−1
b) Which car had the greatest deceleration?
Car A, because the gradient is more steep.
c) How far did car B travel after the brakes were applied?
Area under triangle B
=
=
=
22
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1
b×h
2
1
(6)(40)
2
120 m
Grade 10 Science Essentials
Examples of graphs of motion
Consider the racing car we used to illustrate the equations of motion:
A racing car starting from rest on the grid, travels straight along the track and reaches the 400m mark
after 8,6 s. At the 400 m mark, the brakes are applied and the car slowed down at 2 m.s-2 to come to
rest.
The two stages of the motion can be
represented by x –t, v – t and a – t
graphs:
x-t:
The displacement is always increasing, the
car always moves forward.
a. Describe the motion of the object from 0 to 8 seconds.
0-3 s: Negative acceleration from 12 to 3 m.s-1 in forward direction
3-8 s: Constant velocity of 3 m.s-1 forward.
Gradient of s-t at point = velocity at that
point.
b. What does the gradient of the line AB represent?
The acceleration from A to B.
The car starts from rest and ends at rest
where gradient = 0.
c. Calculate this value.
a =
gradient
The max gradient is at 8,6 s.
a
=
a
=
a
=
v-t:
The velocity is always positive, the car
always moves forward.
The gradient from 0 to 8,6 s is positive,
the car is accelerating forward.
The gradient from 8,6 to 55,1 s is
negative, the car is decelerating while traveling forward.
vf − vi
t
3 − 12
3
−3 m ⋅ s−2
∴ a = 3 m ⋅ s−2 in the opposite direction to motion
d.What is the acceleration between B and C?
a(BC) = 0 m.s-2 because gradient = 0
e. Calculate the total displacement of the object in 8 seconds.
1
Δx = area =
bh + lb + lb
=
=
=
2
1
(3)(9) + (3)(3) + (5)(3)
2
13,5 + 9 + 15
37,5 m
f. Draw the corresponding a – t graph for the motion:
a-t:
Constant positive acceleration of +10,82
m.s-2 from 0 to 8,6 s.
Constant negative acceleration of -2 m.s-2
from 8,6 s to 55,1 s.
23
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SCIENCE CLINIC 2018 ©
Grade 10 Science Essentials
Measuring Instantaneous Velocity
Ticker timers
A ticker timer produces a series of dots on a tape which can be analyzed to determine the
displacement between two points and the time taken between the two points.
SCIENCE CLINIC 2018 ©
Measuring Acceleration
If we know the instantaneous velocity at two times, we can calculate the acceleration between
the points using
Δx
vaverage =
t
a=
∆x is measured in meters with a ruler.
1
t can be calculated from the period of the timer. T =
and is measured in seconds.
f
vf − vi
Δt
Here is a section of tape from a ticker timer which vibrates at 50 Hz.
When t is small we are calculating the instantaneous velocity over the time taken. When t is large
we are calculating the average velocity over the time taken.
To calculate the instantaneous velocity at a certain point (Z) on a tape:
1. Mark off one or more spaces before the point Z.
2. Mark off the same number of spaces after the point Z.
3. Z will occur at half the time of the interval.
4. Measure the length of the interval in meters. This is the displacement s.
5. Count the spaces in the time interval to calculate t. NOTE: t = number of spaces x T.
6. Velocity at Z is called VZ.
The time interval between the dots is
Divide the tape into two intervals, C to E and A to C, to find vf and vi respectively.
vCE
Δx
VZ =
t
REMEMBER:
The footnotes used are simply a descriptor of the
time point to which the velocity refers to.
1
= 0,02 s
50
=
0,042
0,04
=
1,05 m ⋅ s−1
vAC
=
0,018
0,04
=
0,45 m ⋅ s−1
At which dot was the instantaneous velocity = 1,05 m.s-1 ? D
At which dot was the instantaneous velocity = 0,45 m.s-1? B
a
vi means velocity at the initial time point
v4 means velocity at the 4th time point (4th second)
vb means velocity at time point B
=
vD − vB
t B to D
a
=
1,05 − 0,45
0,04
a
=
15 m ⋅ s−2
Use the pattern of the dots to describe the motion of each object:
REMEMBER:
The average velocity across the a time section (from A to C) is equal to the instantaneous
velocity in the middle of that time interval (at B)
a) acceleration from rest
The instantaneous velocity at B = average velocity from A to C
The instantaneous velocity at C = average velocity from B to D OR from A to E
The instantaneous velocity at D = average velocity from C to E
b) constant speed, and then
suddenly accelerates
This only applies to constant acceleration or zero acceleration. Because acceleration is rarely
constant in real life, this could lead to inaccuracy.
In real life, the variability in acceleration is overcome by making the time across which the
average velocity is calculated shorter.
c) accelerating, then moving
at constant velocity and
then decelerating
Shorter average velocity time = more accurate instantaneous velocity
d) slowly decelerating
24
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Motion in 1D - THE ESSENTIALS
Grade 10 Science Essentials
Quan&ty
Defini&on
Posi&on (x)
SCIENCE CLINIC 2018 ©
Vector/scalar
Unit
The place where an object is rela4ve to a
specific reference point.
Vector
m
Distance(D)
Length of path taken by a moving object
Scalar
m
Displacement (∆x)
Change in posi4on with respect to a start
point
Vector
m
Speed (v)
Rate of change of distance.
Scalar
m.s⁻¹
Velocity (v)
Rate of change of posi4on.
Vector
m.s⁻¹
Accelera&on (a)
Rate of change of velocity.
Vector
m.s⁻2
Average speed
!!"#$!%# =
Average velocity
!!"#$!%# =
Instantaneous
velocity
Posi&ve
accelera&on
∆!
∆!
the velocity of an object at a specific moment in 4me.
Eg. Vi is ini4al velocity. Vf is final velocity.
Accelera&on
Nega&ve
accelera&on
!
∆!
!=
!! − !!
∆!
Decreasing velocity in + direc4on OR increasing velocity in the – direc4on
Increasing velocity in + direc4on OR decreasing velocity in the – direc4on
FORMULAE
Old symbols
v = u + at
New symbols
vf = vi + aΔt
Leaves out
s or Δx
s = 12 (u + v)t
Δx = 12 (vi + vf )Δt
a
s = ut + 12 at 2
v 2 = u 2 + 2as
Δx = 12 aΔt 2
vf2 = vi2 + 2aΔx
25
v or vf
t or Δt
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Grade 10 Science Essentials
Graphs of motion- THE ESSENTIALS
SCIENCE CLINIC 2018 ©
Displacement-Time
•
1 is positive acceleration
•
2 is constant velocity
•
3 is negative acceleration
•
4 is at rest
GRADIENT:
Velocity
AREA:
n/a
Velocity-Time
•
1 is positive acceleration
•
2 is constant velocity
•
3 is negative acceleration
•
4 is at rest
GRADIENT:
Acceleration
AREA:
Displacement
Acceleration-Time
• 1 is positive acceleration
• 2 is constant velocity (a = 0)
GRADIENT:
n/a
• 3 is negative acceleration
• 4 is at rest, but negative
acceleration still applies
26
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AREA:
Velocity
Energy
Grade 10 Science Essentials
PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY
ENERGY
The ability to do work
Unit: joules (J)
Scalar quantity
Gravitational Potential Energy (EP)
The energy an object possesses
due to its position in the gravitational field relative to a reference
point.
Amount of energy transferred when an
object changes position relative to the
earth’s surface.
Kinetic Energy (EK)
The energy an object possesses as a result of its motion
Amount of energy transferred
to an object as it changes
speed.
EP = mgh
EK =
-2
1 2
mv
2
g = 9,8m.s ,
m is mass in kg,
h is height in m
m is mass in kg,
v is velocity in m.s-1
Example:
Determine the gravitational potential
energy of a 500 g ball when it is placed
on a table with a height of 3 m.
Example:
Determine the kinetic energy
of a 500 g ball when it travels
with a velocity of 3 m.s -1.
EP
=
=
=
mgh
(0,5)(9,8)(3)
14,7 J
EK
=
=
=
1
mv 2
2
1
(0,5)(32 )
2
2,25 J
Mechanical Energy (EM)
The sum of gravitational potential and kinetic energy at a point
EM
=
EP + EK
EM
=
mgh + 12 mv 2
EM
=
=
EM
=
EM
=
EP + E K
1
2
mgh + mv
(0,5)(9,8)(2,5) +
2
1
(0,5)(1,82 )
2
13,06 J
Principle of conservation of mechanical energy: The total
mechanical energy in an isolated system remains constant. The law
of conservation of mechanical energy applies when there is no friction or air
resistance acting on the object. In the absence of air resistance, or other
forces, the mechanical energy of an object moving in the earth’s gravitational field in free fall, is conserved.
EMECHA
(EP + EK )A
1
(m gh + 2 m v 2 )A
=
EMECHB
=
(EP + EK )B
=
(m gh + 2 m v 2 )B
1
Law of conservation of energy: The total energy of an isolated system remains constant.
In the following instances the gravitational potential energy of an object is converted to kinetic energy (and vice versa), while the
mechanical energy remains constant
EXAMPLE 1: Object moving vertically
A 2 kg ball is dropped from rest at A, determine the maximum velocity of the ball at B just before impact.
(EP + EK )A
=
(EP + EK )B
(m gh + m v )A
=
(2)(9,8)(4) + 12 (2)(0 2 )
(m gh + 12 m v 2 )B
=
78,4 + 0
=
v
=
v
=
1
2
2
(2)(9,8)(0) + 12 (2)v 2
0 + 1v 2
78,4
8,85 m ⋅ s−1 downwards
EXAMPLE 2: Object moving on an inclined plane
A 2 kg ball rolls at 3 m·s−1 on the ground at A, determine the maximum height the ball will reach at B.
(EP + EK )A
=
(EP + EK )B
(m gh + m v )A
=
(2)(9,8)(0) + 12 (2)(32 )
(m gh + 2 m v 2 )B
=
9
19,6
=
h
=
0,46 m
1
2
2
0+9
h
EXAMPLE:
A ball, mass 500 g, is thrown horizontally through the air. The ball travels
at a velocity of 1,8 m·s−1 and is 2,5 m from the ground. Determine the
mechanical energy of the ball.
EM
SCIENCE CLINIC 2018 ©
1
(2)(9,8)(h) + 12 (2)(0 2 )
=
19,6h + 0
EXAMPLE 3: Rollercoaster
The 2 kg ball rolls on a toy rollercoaster from A, at 20 m above the ground, to B where its height is 8 m and velocity is 14 m·s−1.
Calculate its starting velocity at A.
(EP + EK )A
=
(EP + EK )B
=
1
(m gh + 2 m v 2 )B
=
392 + v 2
(2)(9,8)(16) + 2 (2)(142 )
=
313,6 + 196
v
=
v
=
1
2
2
(m gh + m v )A
(2)(9,8)(20) + 2 (2)(v 2 )
27
1
1
313,6 + 196 − 392
10,84 m ⋅ s−1 to the right
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www
Information sheets – Paper 2 (Chemistry)
NA
e
Tθ
Vm
pθ
SYMBOL/SIMBOOL
6,02 x 1023 mol-1
-1,6 x 10-19 C
273 K
22,4 dm3·mol-1
1,013 x 105 Pa
VALUE/WAARDE
TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES
NAME/NAAM
Standard pressure
Standaarddruk
Molar gas volume at STP
Molêre gasvolume by STD
Standard temperature
Standaardtemperatuur
Charge on electron
Lading op elektron
Avogadro's constant
Avogadro-konstante
OR c =
m
MV
n=
n=
V
Vm
N
NA
TABLE 2: FORMULAE/TABEL 2: FORMULES
m
n=
M
n
V
c=
TABLE 3: THE PERIODIC TABLE OF ELEMENTS/TABEL 3: DIE PERIODIEKE TABEL VAN ELEMENTE
6
28
29
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
8
9
B
C
N
O
F
Ne
19
17
20
18
11
13
12
14
Aℓ
Si
14
15
16
16
Cℓ
Ar
30
27
31
28
32
31
33
32
34
S
35,5
35
40
36
Zn
Ga
Ge
As
Se
Br
Kr
3,0
P
4,0
7
3,5
6
2,8
27
1,9
26
1,8
25
1,8
24
1,8
23
1,5
22
1,6
21
1,6
24
20
5
2,5
63,5
18
(VIII)
4
10
3,0
Symbol
Simbool
Cu
Approximate relative atomic mass
Benaderde relatiewe atoommassa
1,5
17
(VII)
He
2,5
29
Electronegativity
Elektronegatiwiteit
1,3
16
(VI)
2
23
19
1,0
15
(V)
2,4
Mg
14
(IV)
2,1
Na
13
(III)
2,0
9
12
1,2
7
11
12
1,8
Be
11
1,5
Li
10
Atomic number
Atoomgetal
1,9
4
1,5
1
3
9
1,8
H
8
7
KEY/SLEUTEL
79
52
80
53
84
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
91
72
92
73
Ra
Ac
0,9
Fr
226
103
77
106
78
108
79
112
80
115
81
119
82
Tℓ
Pb
122
83
128
84
127
85
131
86
Bi
Po
At
Rn
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
179
181
184
186
190
192
195
197
201
204
207
209
58
59
60
61
62
63
64
65
66
67
68
69
70
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140
141
144
150
152
157
159
163
165
167
169
173
175
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232
238
2,5
139
89
101
76
2,0
La
137
88
75
1,9
Ba
96
74
1,8
Cs
133
87
89
57
1,8
88
56
1,6
86
55
2,5
75
51
2,1
73
50
1,9
70
49
1,8
65
48
1,7
63,5
47
1,7
59
46
1,9
59
45
2,2
56
44
2,2
55
43
2,2
52
42
1,9
51
41
1,8
48
40
1,4
45
39
1,2
40
38
1,0
39
37
0,9
0,7
0,8
0,8
0,9
1,0
2,1
1
0,7
5
4
2,0
3
1,6
2
(II)
1,6
1
(I)
71
Classification of Matter
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
CLASSIFICATION
All matter is made up of particles whose properties determine the observable
characteristics and reactivity.
Elements of the periodic table can be classified into 3 different categories:
Ions
Atoms
Non-metals, Metals and Metalloids
Molecules
Par:cles
Non-metals
All ma9er
Pure
substances
Mixtures
Elements
Metals
Compounds
Metalloids
Nonmetals
Eg. Silicon
Poor electrical conductors
(except graphite)
Good electrical conductors
Poor electrical conductors but conductivity
increases with an increase in temperature
Poor thermal conductors
(Except diamond)
Good thermal conductors
Varied thermal conductivity
Dull appearance
Lustre (shiny)
Lustre (shiny)
Brittle
Malleable and ductile
Brittle
Low melting and boiling points
High melting and boiling points
Melting and boiling points differ vastly
Can be solids, liquids or gases at
room temperature
All solids except mercury (Hg) at
room temperature
Solids at room temperature
ELECTRICAL CHARACTERISTICS
Pure substances
Compound
Metalloids
Eg. Copper
MIXTURES AND PURE SUBSTANCES
Mixture
Metals
Eg. Sulphur
Element
Consists of different
particles, not
chemically joined.
Consists of two or more
elements chemically bonded
Consists of only
one type of atom
Components can be
separated into
simpler substances
by physical means
Components can only be
separated into simpler substances by chemical means
Cannot be
separated into
simpler
substances
Smallest sample
will reflect the
composition of
mixture
Smallest particles are
molecules or ions or formula
units
Smallest particles
are atoms or
molecules
No fixed ratio
between the
different particles.
Fixed ratio between the
smallest particles
Eg. Sand in water
Eg. H2O, MgSO4
Insulator
Electrical conductor
Semi-conductor
A material that prevents the flow
of charge
A material that allows the flow of
charge
A substance that can conduct electricity under
certain conditions
Non-metals
Metals
Metalloids
Eg. Plastic and rubber
Eg. Electrical cables
Eg. Diodes and LEDs (light emitting diodes) and
integrated circuits.
MAGNETIC PROPERTIES
Ferromagnetic materials
Materials can be influenced by a magnet
Non-magnetic materials
Materials which are not influenced by a magnet
Soft magnetic materials
Hard magnetic materials
Are easily magnetised but lose
magnetism easily
Not easily magnetised but retain
magnetism for a long time
Cannot become magnetised
Eg. Iron (Fe)
Uses: Temporary magnets such as
electromagnets in metal scrap
yards and loud speakers
Eg. Cobalt (Co), Nickel (Ni),
Steel
Uses: Permanent magnets and
compasses
Eg. Other elements including metals such as
aluminium (Al), zinc (Zn),Copper(Cu)
Eg. Cu, H2
31
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Classification of Matter-THE ESSENTIALS
Grade 10 Science Essentials
Physical Properties of materials
Ions
Strength: the ability of a material to resist stress and strain
Atoms
SCIENCE CLINIC 2018 ©
Molecules
Thermal conductivity: the ability of material to conduct heat
Par:cles
Electrical conductivity: the ability of a material to conduct electricity
Brittle: hard but likely to break
Malleable: Ability to be hammered or pressed into shape without
breaking or cracking
All ma9er
Ductile: the ability to be stretched into a wire
Pure
substances
Magnetic: a material which can be attracted or repelled by a magnet
(its domains are aligned)
Non-magnetic: a material which is not attracted or repelled by a magnet (its domains are misaligned)
Elements
Density: the mass per unit volume of a substance
Mixtures
Compounds
Melting point: The temperature at which a solid becomes a liquid.
Boiling point: the temperature of a liquid at which its vapour pressure
equals the external (atmospheric) pressure
Metals
Metalloids
Pure Substance
A substance that cannot be separated into simpler
components by physical methods
Pure copper, O2
Element
Pure substance consisting of only one type of
atom
Argon, mercury, silicon,
gold
Compound
Pure substance consisting of two or more elements
chemically bonded in a fixed ratio
Water, carbon dioxide,
sodium chloride
Mixture
Consists of different particles mixed together, but
not chemically joined.
Milk, air, salt water
Homogenous
mixture
A mixture of uniform composition and in
which all components are in the same phase
Eg: air, brine, steel
Heterogeneous
mixture
A mixture of non-uniform composition and of
which the components can be easily identified
Sand and rock mixture,
Pizza toppings
32
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Nonmetals
Names and formulae
Grade 10 Science Essentials
BASIC CONCEPTS
• Atoms combine in very specific ratios to
form certain compounds.
E.g. carbon dioxide consists of 1 carbon
and 2 oxygen atoms.
•These ratios remain fixed for that
compound. The ratio is given by the
formula. e.g. CO2
SCIENCE CLINIC 2018 ©
TABLE OF COMMON ELEMENTS
CARBON
OXYGEN
OXYGEN
•Simple ions are single atoms which carry a charge. They are also called monatomic ions.
Polyatomic ions are groups of atoms bonded together which collectively carry a
charge.
H
Hydrogen
Ne
Neon
K
Potassium
Zn
Zinc
He
Helium
Na
Sodium
Ca
Calcium
Br
Bromine
Li
Lithium
Mg
Magnesium
V
Vanadium
Sn
Tin
Be
Beryllium
Al
Aluminium
Cr
Chromium
Pt
Platinum
B
Boron
Si
Silicon
Mn
Manganese
Ag
Silver
C
Carbon
P
Phosphorus
Fe
Iron
Au
Gold
N
Nitrogen
S
Sulphur
Co
Cobalt
Hg
Mercury
O
Oxygen
Cl
Chlorine
Ni
Nickel
Pb
Lead
F
Fluorine
Ar
Argon
Cu
Copper
Xe
Xenon
TABLE OF COMMON CATIONS AND ANIONS
TABLE OF COMMON COMPOUNDS
Formula
Chemical name
Common name
H2O
Hydrogen oxide
Water
ANIONS (NEGATIVE IONS)
CO2
Carbon dioxide
Carbon dioxide
All Group 7
NH3
Hydrogen nitride
Ammonia
OH¯
Hydroxide ion
SO4²¯
Sulphate ion
HCℓ
Hydrogen chloride
Hydrochloric acid
NO3¯
Nitrate ion
SO3²¯
Sulfite ion
H2SO4
Hydrogen sulphate
Sulphuric acid
NO2¯
Nitrite ion
CO3²¯
Carbonate ion
HNO3
Hydrogen nitrate
Nitric acid
MnO4¯
Permanganate ion
CrO4²¯
Chromate ion
H2CO3
Hydrogen carbonate
Carbonic acid
CℓO3¯
Chlorate ion
Cr2O4²¯
Dichromate ion
H3PO4
Hydrogen phosphate
Phosphoric acid
CℓO¯
Hypochlorite ion
O2²¯
Peroxide ion
CH3COOH
Ethanoic acid
Vinegar
HCO3¯
Hydrogen carbonate ion
NaCℓ
Sodium chloride
Table salt
HSO3¯
Hydrogen sulphate ion
NaOH
Sodium hydroxide
Caustic soda
HSO4¯
Hydrogen sulfite ion
NaHCO3
Sodium hydrogen carbonate
Baking soda (bicarbonate of soda)
H2PO4¯
Dihydrogen phosphate ion
Na2CO3
Sodium carbonate
Washing soda
CH3COO¯
Acetate ion
NaNO3
Sodium nitrate
Chile saltpetre
CATIONS (POSITIVE IONS)
KOH
Potassium hydroxide
Caustic potash
All Group 1
KNO3
Potassium nitrate
Saltpetre
NH4+
CaCO3
Calcium carbonate
Marble/ chalk/ lime stone
CaSO4
Calcium sulphate
Gypsum
MgSO4
Magnesium sulphate
Epsom salts
CuSO4
Copper sulphate
Blue vitriol
CH4
Methane
Natural gas
SINGLE CHARGE
DOUBLE CHARGE
All Group 6
All Group 2
Ammonium ion
+
H3O
Hydronium ion
33
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TRIPLE CHARGE
All Group 5
PO4³¯
Phosphate ion
All Group 3
Names and formulae
Grade 10 Science Essentials
WRITING CHEMICAL FORMULAE
SCIENCE CLINIC 2018 ©
BALANCING CHEMICAL EQUATIONS
Follow these basic steps to write chemical formulae for ionic substances:
A chemical equation represents the chemical changes that occur when a reaction takes place.
1. Write symbol for the positive ion first, then for the negative ion.
The ions could be monatomic ions or polyatomic ions.
The Law of conservation of matter means that matter cannot be created or destroyed, merely transferred from one
substance to another.
•A chemical equation must be balanced in order to conform to the above law.
•Compounds have fixed ratios, thus only the number of compounds on either side of the arrow can be changed .
•Numbers are placed in front of compounds on either side of the arrow to balance the equation.
•The total number of each atom should be the same on either side of the equation.
2. Write ionic charges at top right of symbols
• Monatomic ion → the ion of one single element only, so you can use periodic table to
determine the charge.
• Polyatomic ion → you must memorise the charges.
3. “Cross multiply” the numbers so that the total charge of the
compound is 0.
4. Write final compound:
• ends in –ide → monatomic ion
Eg. Hydrogen Sulphide = H2S
• the name ends in –ate or –ite if it includes oxygen
Steps for balancing reaction equations:
1.Identify all reactants and products of the reaction.
2.Start from the word equation where possible.
3.Write the correct formula for the compounds of the reactant and products and separate them with an arrow.
4.Indicate the phases of each compound.
5.Count the number of atoms of each element on either side of the arrow, and add numbers in front of
compounds in order to get it to balance.
6.Check the total number of atoms of a certain element in reactants must equal the total number atoms of the
same element in the product.
• Roman numerals (II) → ionic charge of the metal
Eg. Copper(II)sulphate → Cu2 SO4
EXAMPLE:
In the Synthesis of water, Hydrogen reacts with oxygen to form
water.
EXAMPLE:
Aluminium
Aℓ3+
Aℓ 2 O
Step 1. Hydrogen + oxygen → water.
oxide
O23
Step 2. H2 + O2→ H2 O
(“cross-multiply” numbers )
Step 3. H2 (g) + O2(g) → H2 O(g)
Step 4a. Add a 2 in front of H2O to balance the oxygen atoms.
H2 (g) + O2(g) → 2H2 O(g)
(the total charge must be zero)
Step 4b. Add a 2 in front of H2 to balance the hydrogen atoms.
2H2 (g) + O2(g) → 2H2 O(g)
Step 5. Reactants have 4 H atoms and 2 oxygen atoms.
Products have 4 H atoms and 2 oxygen atoms.
Equation is balanced.
PHASES OF A COMPOUND
NOTE: These steps are a systematic approach to balancing
chemical equations. You don’t have to show all the steps in your
working out. Usually only the final balanced equation is required.
Phases are represented as subscripts after the formulae as follow:
(s) solid state
(ℓ) liquid state
(aq) aqueous solution
(g) gaseous state
34
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Kinetic Theory of Matter
STATES OF MATTER
CHANGES OF STATE
HEATING AND COOLING CURVES FOR WATER
Mel/ng
•Matter occurs in one of three states, solid, liquid or gas,
depending on the temperature.
•The state of matter depends on its melting point (MP)
and boiling point (BP). If the temperature is below the
MP, it is a solid. If the temperature is above the MP, but
below the BP, it is a liquid. If the temperature is above
the BP, it is a gas.
Solid
Temperature (oC)
Sublima/on
•All matter has mass and takes up space (volume).
Boiling/
evapora/on
Liquid
Freezing
Gas
boiling
mel+ng
Condensa/on
water
molecules
equilibrium
Properties of
states of
matter
freezing
Time (min)
Cooling Curve
The Kinetic Theory of Matter states that:
EVIDENCE FOR THE KINETIC THEORY OF MATTER
dye
molecules
condensing
Time (min)
Hea+ng Curve
Deposi/on
Diffusion is the movement of atoms or molecules
from an area of high concentration to an area of
lower concentration. The higher the temperature,
the faster diffusion takes place.
SCIENCE CLINIC 2018 ©
Temperature (oC)
Grade 10 Science Essentials
Brownian motion is the random
movement of microscopic particles
suspended in a gas or liquid. When
we use a microscope to study smoke
particles in air or pollen particles in
water, these particles jiggle about in a
completely random way. The tiny
particles of air or water, which are too
small to be seen with the microscope,
are in constant motion. They collide
with the bigger particles of smoke and
pollen, causing the bigger particles to
change direction and speed.
•All matter is made of tiny particles with spaces in between them.
•The tiny particles are constantly moving, but the particles lose no kinetic
energy when they collide with other particles or with the walls of their container.
•The kinetic energy of the particles depends on how fast they are moving.
At any given time, some particles are moving slowly while others are moving
fast. The temperature is a measure of the average kinetic energy of the
particles.
•The potential energy of the particles depends on how far they are apart
and depends on their state (solid, liquid or gas).
•There are attractive forces between the particles which become stronger as
the particle move closer.
Volume
Shape
Spaces between
particles
Forces between
particles
Arrangement of
particles
Movement of
particles
Solid
Fixed volume
Fixed shape
Touching each other
Strong
Close together. Fixed
positions in a set pattern.
Vibrate about fixed
positions
Liquid
Definite
volume
Takes the shape
of the container
Touching each other
Medium
Random – no fixed pattern
Slide past each other
switching places.
Far apart
Very weak
Gas
Particle diagram
No definite shape or volume.
Expands to fill container
35
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On their own
Fast and free
The atom- Atomic models
Grade 10 Science Essentials
J.J Thomson (1897)
1.Thomson used a cathoderay tube to show that there
are small negative
particles inside atoms. He
called them “electrons”.
2. Electrons are found in all
atoms. They have the same
charge and mass regardless
of which substance they
come from.
3. He used the current bun
(A.K.A Plum pudding) model
to explain that the negative
electrons are found inside a
solid positive mass.
SCIENCE CLINIC 2018 ©
James Chadwick (1932)
Electrons
-
-
-
Uniform, posi1vely
charged sphere
1. Discovered the neutron which is a
particle in the nucleus.
2. Neutrons have no charge, and their
mass is almost the same as a proton.
Niels Bohr (1913)
•Electrons travel around the nucleus in circular, definite paths
called orbits- like planets orbit
around the sun.
•Each orbit has a ‘certain energy.’
Electrons in the same orbit have
the same energy.
•The further the orbit is from the
nucleus, the higher the energy.
•Electrons can jump from one orbit to another.
•When an electron moves to higher
energy level it gains energy.
Schrödinger (1926)
1. It is not possible to find the exact position of an
electron, but only the most probable regions where
the electrons move.
2. The regions inside an atom where electrons are
likely to be found are called orbitals.
TIMELINE OF THE HISTORY OF THE ATOM
Democritus
( +/- 440 BC)
1. Democritus
proposed that if you
kept cutting a
substance in half
repeatedly, you would
eventually end up
with an “uncuttable”
particle.
2. He called these
particles atoms.
John Dalton
(1803)
Ernest Rutherford (1909)
1. All elements are
made up of small,
indestructible,
solid spheres
called atoms.
2. A t o m s o f t h e
same element
are identical:
same size, shape
and mass.
3. An atom is the
smallest particle of an element which can take
part in a chemical reaction.
4. Compounds are formed when the atoms of two or
more elements combine with one another in
fixed whole number ratios.
Rutherford conducted an experiment in which he shot a beam of alpha particles towards a sheet of very thin gold foil .
Most of the particles continued in a straight line, but some were deflected He deduced that:
i.The positive charge is not evenly distributed as suggested by Thomson but is in the centre of the atom, in the nucleus.
ii.Mass is concentrated in the nucleus. The nucleus has a large enough mass to deflect the alpha particles.
iii.The nucleus is small compared to the atom because most of the particles passed undisturbed through the gold foil.
iv.Electrons are extremely small and far from the nucleus and make up practically the whole volume of the atom.
v.The volume of the atom is 10 000 greater than the volume of the nucleus. The atom is mostly empty space.
flourescent
screen
metal foil
α-par&cle source
undeflected
α-par&cles
solid
nucleus
Shortcomings:
1. Atoms are not solid spheres.
2. All atoms of an element are not identical. There
are different isotopes of elements.
3. Atoms consist of sub-particles.
deflected
α-par&cles
36
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gold
atom
The atom
Grade 10 Science Essentials
WHAT IS AN ATOM?
•Atoms are the very small particles of which all
elements are made- they are the basic building
block of all matter – including your own body,
your hair, your organs, the earth, the sun are all
made of different combinations of atoms.
•E.g. Your body is made of 65% Oxygen, 19%
Carbon and 10% Hydrogen by mass.
SCIENCE CLINIC 2018 ©
Consider the atom shown here:
Atomic Number (Z) and Atomic Mass (A)
An atom is identified by the contents of its nucleus.
An atom is represented by the notation:
A
Z
•All known elements are arranged on the
periodic table in order of increasing atomic
number.
E
E
Symbol of the element
A
Atomic Mass (A) is the number of nucleons
Z
Atomic Number (Z) is the number of protons
•Elements in the periodic table are arranged in order of increasing atomic number (Z).
•The number of protons (Z) gives each element its unique
properties.
•Z also indicates how many electrons are present in the atom
as the atom is electrically neutral.
•A indicates the mass of the atom how many nucleons are
present in the atom.
•Number of neutrons = Atomic mass number - Number of
protons (N = A - Z)
•Periodic refers to ‘repeating patterns.’ Elements
are arranged into groups (vertical columns) and
periods (horizontal rows).
•Elements in within a group have similar physical
and chemical properties.
•Atomic theory is the basis for understanding the
interactions and changes in matter.
ATOMS AND SUB-ATOMIC PARTICLES
•There are three subatomic particles that are
found inside the atoms, i.e. protons, neutrons
and electrons.
•The nucleus is in the centre of the atom and
contributes most of the mass. The nucleus is
made of the protons and neutrons which are
called nucleons.
•Because subatomic particles have extremely
small masses, the atomic mass unit amu is used.
1 amu = 1,677 × 10−27 kg.
•The mass of an atom is measured relative to
the mass of carbon-12 atom.
Carbon-12 is
exactly 12 amu, therefore 1 amu is one-twelfth
the mass of carbon-12 atom.
proton p+
neutron n0
electron e-
1,677 × 10−27
(1 amu)
1,677 × 10−27
( 1 amu)
9,11 × 10−31
Units of
charge
+1
0
Isotopes have different:
number of neutrons
different masses
1
1
-1
+
+ n
n
+ n
-
-
-
H
Hydrogen
2
1
3
1
H
Deuterium
The diameter of an atom varies for
each type of atom. The diameter is
about 1 x 10 -10 m.
+
+
-
Carbon Atom
7
3 Li
EXAMPLE 1: Lithium
(Z) Atomic number is 3: 3 protons, therefore 3 electrons
(A) Atomic Mass is 7: 3 protons and (7 - 3) = 4 neutrons
32
16 S
EXAMPLE 2: Sulphur
13
6 C
EXAMPLE 3: Carbon-13
(Z) Atomic number is 6: 6 protons, therefore 6 electrons
(A) Atomic Mass is 13 : 6 protons and (13 - 6) = 7 neutrons
EXAMPLE
Hydrogen exists as isotopes.
99,65% of H atoms are 1 1 H
0,30% of H atoms are 1 2 H
0,05% of H atoms are 1 3 H
RAM(H)
H
Tri2um
37
+
-
Relative Atomic Mass (RAM)
Isotopes of elements
number of protons
number of electrons
chemical properties.
It has 6 electrons so it has a total
charge of 6 + (-6) = 0. It is neutral.
-
RAM indicates how many times the average mass of an atom of a certain
element is heavier than 1/12 the mass of a carbon atom.
Isotopes in an element are responsible for the RAM of elements NOT being
whole numbers.
Consider these three atoms of hydrogen. They each have
one proton, but they have different numbers of neutrons.
They are called isotopes.
Isotopes have the same:
-
It has 6 neutrons. This means it has 12
nucleons and a mass of 12 amu.
(Z) Atomic number is 16: 16 protons, therefore 16 electrons
(A) Atomic Mass is 32 : 16 protons and (32 - 16) = 16 neutrons
Table showing the mass and charge of subatomic particles
Mass
(kg)
6 Protons
6 Neutrons
It has 6 protons. This means it is a
Carbon atom.
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=
=
99,65 × 1
100
+
0,30 × 2
100
1,004
+
0,05 × 3
100
Grade 10 Science Essentials
ELECTRON ARRANGEMENT IN THE ATOM
The atom
SCIENCE CLINIC 2018 ©
IONS
1. In neutral atoms, the number of electrons in the cloud around the nucleus equals
the number of protons in the nucleus.
2. The electrons closer to the nucleus have less energy than the electrons further
away from the nucleus.
3. The electrons occur in energy levels (n).
4. Different energy levels are able to accommodate different numbers of electrons
(Total electrons = n2).
5. The electrons are found in certain regions within an energy level, referred to as
orbitals. Each orbital can accommodate 2 electrons.
CATIONS
Cations are positive ions which are formed when the
electrons from the outermost energy level are separated
from the atom.
ANIONS
Anions are negative ions which are formed when electrons
attach to an atom to fill the outermost energy level of the
atom.
The atoms become charged atoms known as ions. These
charges need to be considered when completing Aufbau
diagrams or electron configurations.
The atoms become charged atoms known as ions. These
charges need to be considered when completing Aufbau
diagrams or electron configurations.
Electrons are always found in positions of lowest possible energy (as close to the
nucleus as possible)
A neutral atom of aluminium contains 13 protons and 13
electrons.
This aluminium ion has a charge of +3, indicating it has 3
protons more than electrons.
The number of protons does not change for a given element, thus it has lost 3 electrons and has 10 electrons.
•
•
•
energy level 1: 1 s orbital
energy level 2: 1 s and 3 p orbitals
energy level 3: 1 s, 3 p and 5 d orbitals
Shapes of the orbitals
1s orbital
Energy level 1
Spherical in shape
2px, 2py , 2pz orbitals
Energy level 2
Dumbbell shapes along
x,y and z axes
+3
13Al
Orbital box (Aufbau) diagrams
Aufbau diagrams is most detailed description of the way the
orbitals are filled with electrons. The orbitals are filled from
the lowest energy at the bottom in the following way:
4s
3p
Energy
2 s orbital
Energy level 1
Spherical in shape
Electrons spend more
time further from nucleus
EXAMPLE:
Valence electrons are
in the outer most
energy level and take
part in bonding.
•the size of their s and p orbitals also
increases.
•The number of orbitals increases.
•We encounter d and f orbitals which are more
complicated shapes.
Spectroscopic Electron
Configuration (spd) notation:
This is a concise way to indicate the electron
arrangement in an atom. This is similar to the
Aufbau diagram, but represents it in a simpler
way.
EXAMPLE:
2
2
5
9F = 1s 2s 2p
2
2
6
2
1
13Al = 1s 2s 2p 3s 3p
2
2
6
2
6
2
20Ca = 1s 2s 2p 3s 3p 4s
Negative ions have gained electrons
Positive ions have lost electrons
2p
Core electrons are in
full energy levels and
are not involved
in bonding.
1s
Flourine
As the energy level(n) increases
A neutral atom of fluorine contains 9 protons and 9
electrons.
This fluorine ion has a charge of -1, indicating it has 1
electron more than protons.
The number of protons does not change for a given element, thus it has gained electron and has 10 electrons.
We can also concisely represent the Electron
Configuration of ions in this way:
3s
2s
EXAMPLE: 9F-
Pauli’s exclusion principle: Maximum of two electrons per
orbital provided that they spin in opposite directions. (arrows
in opposite directions)
Hund’s rule: No pairing in p orbitals before at least one
electron in each of them.
38
EXAMPLE:
2
2
6
9F = 1s 2s 2p
+3
= 1s2 2s2 2p6
13Al
+2
= 1s2 2s2 2p6 3s2 3p6
20Ca
We can also write the electron configuration
according to noble gas notation.
Noble gas notation makes use of the first preceding noble gas, which has a completely filled
outer energy level.
EXAMPLE:
5
9F = [He] 2p
2
1
13Al = [Ne]3s 3p
2
20Ca = [Ar]4s
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Lewis dot diagrams
Lewis dot diagrams are used to show the
position of the valence electrons around an
element. The valence electrons are related to
the group of the element.
In all of the examples below, X represents any
element in the group.
Group 1
Group 2
X
X
Group 3
Group 4
X
X
Group 5
Group 6
X
X
Group 7
Group 8
X
X
The atom-THE ESSENTIALS
Grade 10 Science Essentials
Atomic models
John Dalton
(1803)
J.J Thomson
(1897)
Ernest
Rutherford
(1909)
Niels Bohr
(1913)
James
Chadwick
(1932)
Mass and charge of subatomic particles
1. All elements are made up of
small, indestructible, solid spheres
called atoms like billiard balls.
2. Atoms of the same element are
identical.
Mass
(kg)
Electrons are negative particles,
found inside a solid positive mass,
like raisins in a current bun.
1. The positive charge is in the
centre of the atom, in the nucleus.
2. Mass is concentrated in the
nucleus.
3. The nucleus is small compared
to the atom.
4. Electrons are extremely small
and travel far from the nucleus.
5. Most of the atom is empty
space.
Electrons travel around the nucleus
in circular, definite paths called
orbits. Each orbit has a ‘certain
energy’.
SCIENCE CLINIC 2018 ©
proton p+
neutron n0
electron e-
1,677 × 10−27
(1 amu)
1,677 × 10−27
( 1 amu)
9,11 × 10−31
Units of
charge
+1
0
Isotopes have the same number of protons (Z) but different number of neutrons (A)
Relative Atomic Mass (RAM)
RAM indicates how many times the average mass of an
atom of a certain element is heavier than 1/12 the mass
of a carbon atom.
Electron arrangement (neutral atom)
1.
Number of electrons = Z
2.
The electrons occur in energy levels (n).
3.
The closer to the nucleus, the less the energy.
4.
Total electrons in a level = n2
5.
The electrons are found in orbitals. Each orbital
can accommodate 2 electrons.
-1
Representing atoms
A
Z
E
E
Symbol of the element
A
Atomic Mass (A) is the number of nucleons
Z
Atomic Number (Z) is the number of protons
Orbitals
•
level 1: 1 s orbital
•
level 2: 1 s and 3 p orbitals
•
level 3: 1 s, 3 p and 5 d orbitals
Valence electrons are found in the outer energy level.
Notations for Electron Configuration
Aufbau: orbitals are shown as boxes. Electrons are shown
as arrows in the boxes.
spd: 1s2 2s2 2p6 3s2 3p6 4s2
In a neutral atom:
Number of protons = Z
Number of electrons = Z
Number of neutrons = A - Z
The neutron is a particle in the
nucleus with no charge, and almost
the same mass as a proton.
4s
Valence electrons are
in the outer most
energy level and take
part in bonding.
Energy
3p
IONS
CATIONS
ANIONS
Cations are positive ions which are formed when the
electrons from the outermost energy level are separated
from the atom.
Anions are negative ions which are formed when
electrons attach to an atom to fill the outermost energy
level of an atom.
The atoms become charged atoms known as ions. These
charges need to be considered when completing Aufbau
diagrams or electron configurations.
The atoms become charged atoms known as ions. These
charges need to be considered when completing Aufbau
diagrams or electron configurations.
3s
2p
2s
1s
Flourine
39
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Core electrons are in
full energy levels and
are not involved
in bonding.
The periodic table
Grade 10 Science Essentials
s-block
GROUPS (COLUMNS)
•Groups are the vertical columns in the periodic table.
•Elements in a group have similar outer electron
configurations.
•This means they have similar chemical properties.
•The group number corresponds to the number of
valence electrons.
•Eg All of Group 2 elements have two valence
electrons.
p-block
1s
2s
3s
4s
5s
6s
7s
IMPORTANT GROUPS TO REMEMBER:
1: Alkali metals
2: Alkali Earth metals
17: Halogens
18: Noble gases
Group 1:
Alkali
metals
SCIENCE CLINIC 2018 ©
PERIOD (ROWS)
1s
2p
3p
4p
5p
6p
d-block
3d
4d
5d
6d
•Eg Period 3: outermost electrons in the 3rd energy
level.
EXAMPLE: Locating an element.
Group 18:
Group 17: Noble gases
Halogens
Group 2:
Alkali Earth
metals
•Periods are the horizontal rows in the periodic table.
•There is a gradual change in the physical and
chemical properties as we move across a period.
•The Period number tells us the energy level where
the outermost electrons are found.
1. Silicon is in group 4 and period 3.
2. Calcium is in group 2 and period 4.
TRENDS OF THE PERIODIC TABLE
IONISATION ENERGY (IE)
Energy needed per mole to remove an electron from an atom
in the gaseous phase.
First ionization energy is the energy required to remove the
first electron (outermost electron in the highest energy sublevel)
Na
Loses
outer
electron
Sodium atom
ATOMIC RADIUS
The distance from the nucleus to the border of the outer orbital.
•Atomic radii decrease across a period because as electrons are being
added to the same energy level, protons are being added to the nucleus,
which pulls the electrons with a stronger force.
•Atomic radii increase down a group because
a)electrons enter a whole new energy level in the next period
b)the core electrons shield the outer electron from the pull of the nucleus.
Na+
1A
The smaller the atom, the more the force of attraction by the
nucleus on the outer electron, hence the higher the ionization
energy.
•IE increases across a period
•IE decreases down a group
ELECTRONEGATIVITY (EN)
The tendency of an atom in a molecule to attract bonding
electrons. Electronegativity range from 0,7 (Cs) to 4,0 (F).
F
Atomic radii (pm)
2A
3A 4A 5A 6A 7A 8A
Li
Be
B
C
N
O
37
F
152
112
85
77
75
73
72
71
Na
Mg
Al
Si
P
S
Cl
Ar
186
160
143
118
110
100
98
Sodium ion
ELECTRON AFFINITY (EA)
The energy released when an electron is is attached to an
atom or molecule to form a negative ion.
31
Ne
Gains
outer
electron
Fluorine atom
F−
Fluorine ion
The electron affinity is also a description of how likely an atom
is to accept an electron
The smaller the atom, the greater the ability of an atom to
receive an electron due to proximity to the nucleus.
• EA increase across a period
• EA decreases down a group
Interesting trend:
103
• EN increases across a period
• EN decreases down a group
40
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Melting point, Boiling point and Density increase from
group 1 to 4, then drop in Group 5 and start increasing again
from Group 5 to 8.
Periodic table-THE ESSENTIALS
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
TABLE 3: THE PERIODIC TABLE OF ELEMENTS
6
Y
2
26
27
28
Cr
Mn
Fe
Co
Ni
56
44
Mo Tc Ru
METALS
96
101
Zn
Ga
Rh
Pd
Ag
Cd
28
32
Ge
In
Sn
Sb
181
184
186
190
Ra
Ac
58
59
60
61
Ce
Pr
Nd
Pm
140
141
144
90
91
92
93
94
95
96
97
98
99
100
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
0,9
Se
4,0
79
52
Te
I
Xe
122
83
128
84
127
85
131
86
Bi
Po
At
Rn
Ir
Pt
Au
Hg
192
195
197
201
204
207
209
62
63
64
65
66
67
68
69
70
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
150
152
157
159
163
165
167
169
173
175
101
102
103
Md
No
Lr
1,9
112
80
1,8
108
79
238
AR DECREASES
IE INCREASES
The energy released
when an electron is
added to a neutral atom
to form a negative ion.
41
Atomic radius (AR)
The distance between
the nucleus and the
outermost orbital.
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IE INCREASES
IE DECREASES
AR INCREASES
IE DECREASES
EN DECREASES
Electron affinity (EA)
40
36
106
78
1,8
1,6
1,9
179
1,8
139
89
1,7
137
88
1,7
Os
1,9
Re
2,2
W
2,2
Ta
2,2
Hf
1,9
76
La
1,8
75
Ba
1,4
74
Cs
1,2
92
73
1,0
Nb
3,0
84
54
91
72
226
3,5
Kr
Zr
Fr
20
18
80
53
89
57
Pb
Ne
Br
75
51
Tℓ
F
19
17
As
73
50
119
82
O
16
16
Ar
70
49
115
81
4
10
NONS
Cℓ
31
32
35,5
33 METALS
34
35
88
56
103
77
9
8
2,5
3,0
P
2,1
Si
Cu
65
48
2,5
Aℓ
1,8
14
15
30
63,5
47
N
12
14
29
59
46
C
11
13
27
31
59
45
7
2,8
25
B
6
2,4
24
55
43
2,0
1,5
1,9
Approximate relative atomic mass
Benaderde relatiewe atoommassa
52
42
5
86
55
EN INCREASES
The tendency of an atom
to attract a bonding pair
of electrons. Electronegativity range from 0,7 (Cs)
to 4,0 (F).
63,5
18
(VIII)
He
2,5
Sr
51
41
17
(VII)
2,5
Rb
V
48
40
16
(VI)
2,1
Ti
45
39
15
(V)
2,0
Sc
40
38
14
(IV)
2,0
Ca
13
(III)
1,8
K
39
37
23
12
1,6
22
1,5
21
1,3
24
20
1,0
23
19
11
1,6
Mg
10
Symbol
Simbool
Cu
1,8
Na
Electronegativity
Elektronegatiwiteit
1,8
9
12
1,2
7
11
9
Atomic number
Atoomgetal
1,5
Be
1,6
Li
8
29
1,6
4
1,5
1
3
232
Electronegativity (EN)
7
KEY/SLEUTEL
133
87
0,7
5
1,9
4
1,8
3
H
0,9
2,1
1,0
0,8
0,7
IMPORTANT GROUPS TO REMEMBER:
1: Alkali metals
2: Alkali Earth metals
17: Halogens
18: Noble gases
2
(II)
1
0,8
PERIOD (ROWS)
•There is a gradual change in the physical
and chemical properties as we move across a
period.
•The Period number tells us the energy level
where the outermost electrons are found.
1
(I)
0,9
GROUPS (COLUMNS)
•Elements in a group have similar outer
electron configurations and similar chemical
properties.
•The group number corresponds to the
number of valence electrons.
Ionization Energy (IE)
The energy required to
remove an electron from
an atom in the gaseous
phase.
71
Chemical bonding
Grade 10 Science Essentials
CHEMICAL BONDS are INTRAMOLECULAR BONDS which occur between
atoms within molecules.
Atoms tend to gain, lose, or share electrons until they are surrounded by a
full electron shell. A full shell makes the atom more stable like a noble gas.
This is known as The Octet Rule: Each atom requires 8 electrons in the
outer shell, except for H and He.
A) Covalent Bonding
1. Sharing of at least one pair of electrons by two non-metal atoms.
2. Results in simple molecules with only a few atoms connected to
each other.
Eg. H2, H2O, CO2, Cl2 , CH4.
Single Covalent
E.g.
H−H
Each H atom needs one more electron to fill
the outer shell with 2 electrons.
As the hydrogen atoms come closer so their
orbitals overlap. The sharing of electrons results in a lower energy for both electrons. A
single bond is formed.
SCIENCE CLINIC 2018 ©
• If two non-metal atoms approach, they tend to share their
valence electrons and form a covalent bond.
• If a metal and a non-metal atom approach, the metal transfers
its valence electrons to the non-metal. They form an ionic bond.
• If two metal atoms approach, they release their valence
electrons to surround them and form a metallic bond.
B) Ionic Bonding
1. Involves a complete transfer of electron(s).
2. Metal atom gives e- to non-metal.
3. Metal forms a positive cation.
4. Non-metal forms a negative anion.
5. Electrostatic attraction of ions leads to formation of giant crystal
lattice.
Ionic Bonding takes place in two steps.
1. Donation of e-(s) to form ions
2. Electrostatic attraction
C) Metallic Bonding
1. Occurs between a metal and a metal atom.
2. The metal atoms release their valence electrons to surround
them. There is a strong but flexible bond between the positive
metal kernels and a sea of delocalised electrons.
Metal atoms are very closely packed, so that the outermost energy
levels overlap.
The valence electrons are able to move from their respective atoms.
They are called delocalised electrons or free electrons.
The bond is between the positive metal kernels and the sea of
electrons.
Double Covalent
E.g.
O=O
Each O atom needs two more electrons to
fill the outer shell with 8 electrons.
The diagram shows two oxygen atoms coming closer so their orbitals overlap. They
each share 2 outer electrons. A double
bond is formed.
Na+
Cl-
Triple Covalent
E.g.
N≡N
Each N atom needs three more electrons
to fill the outer shell with 8 electrons.
The diagram shows two nitrogen atoms
coming closer so their orbitals overlap.
They each share 3 outer electrons. A
triple bond is formed.
Diagrams show the crystal lattice
made of alternate ions.
IONIC BONDING
maintains the la2ce
structure between ions
Delocalised
electrons
42
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Chemical bonding- Properties of substances
Giant Covalent Network Structures:
Properties of simple covalent (molecular) substances
• Low Melting and Boiling Points because molecules are not bonded to
each other, but only loosely held together. They are easily separated. E.g.
nitrogen has a BP of -1830C.
• Highly volatile. (High tendency of a substance to vaporize.)
• Insulators – there are no free charge carriers.
Properties of ionic substances
• Hard, crystalline solids because the Ions held together by strong ionic
bonds.
• flat sides and regular shapes because the ions are arranged in
straight rows in a pattern.
• Melting points & Boiling points are very high because there are
strong ionic bonds between cations and anions. A large amount of
energy is required to separate the ions e.g. NaCl melts at 808ºC and
boils at 1465ºC.
• Low volatility because there are strong ionic bonds between the
ions.
• Solid ionic compounds don’t conduct electricity, because the ions
cannot move, there are no free charge carriers.
• Molten or dissolved ionic compounds conduct electricity because
in these situations the ions can move and carry charge, which allows
current to flow.
Properties of metallic substances
The sea of delocalised electrons and tightly packed metal kernels are responsible for the properties of metals:
• High density
• Good conductors of electricity
• The delocalised electrons are free to move randomly throughout the
metal, but if a cell is connected across the terminals, the electrons will
drift towards the positive end.
• Good conductors of heat
• Delocalised electrons can also transfer heat energy from a region of high
temperature to a region of low temperature, because they move so freely.
• Metals have high melting and boiling points because of the strength
of the metallic bond.
• Metals are malleable and ductile- they can be easily flattened, shaped
or drawn into threads without breaking. This is because the metals kernels
are not attracted to one specific electron. This makes them different from
an ionic compound which is brittle.
• Metallic lustre: The sea of electrons would reflect all frequencies of
light to give a shiny surface.
SCIENCE CLINIC 2018 ©
These occur when a non-metal atom forms covalent bonds with more than
one other non-metal atom. This allows a giant covalent lattices to form.
EXAMPLE 1: Diamond.
Each C is bonded to four other carbon atoms.
Properties:
1. Hard: Many strong covalent bonds holding the structure
together-making it a very strong crystal.
2. Very high melting point: Many strong covalent bonds hold
the structure together. It requires massive amounts of energy to
pull it apart- about 35000C
3. Insulator: All of the valence electrons are used in bonding.
None of the electrons are free to move.
EXAMPLE 2: Graphite.
Each C is bonded to three other atoms to form a layer.
Properties:
1. The bonds between the layers are weak and thus the layers
can slide over each other- making graphite slippery and can be
used as a lubricant.
2. Because only three electrons out four valence electrons are
used for bonding, the fourth electron is free to move around.
This is why graphite, unlike other non-metals, can conduct electricity.
EXAMPLE 3: Silicon Dioxide.
Silicon dioxide (SiO2), also called silica. Many precious gems
contain silica.
There are no discrete SiO2 units. Every Si is bonded to two O
atoms.
43
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Diamonds and Graphite are both allotropes of carbon.
Allotropes are different forms of the same element but has different physical
and chemical properties due to the difference in particle arrangement.
Grade 10 Science Essentials
Grade 10 Science Essentials
Chemical bonding-THE ESSENTIALS
SCIENCE CLINIC 2018 ©
Inter-atomic bonding/
Intramolecular bonding
COVALENT
BONDING
IONIC
BONDING
METALLIC
BONDING
Between 2 or more non-metals
Between 2 or more metal and
non-metal atoms
Between 2 or more metal atoms
Molecules of an element
Covalent network structures
Example:
Diatomic gases
P4
S8
Example:
diamond
graphite
SiO2
C60 (buckyballs)
Ionic network structures
Metallic network structures
Crystals made from a
network of cations and
anions.
Positive kernels and a sea of
electrons
Example:
NaCl
KMnO4
Molecules of compounds
Example:
water (H2O)
Ammonia (NH3)
44
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Physical and Chemical change
Grade 10 Science Essentials
PHYSICAL CHANGE
SCIENCE CLINIC 2018 ©
CHEMICAL CHANGE
1. Usually easy to reverse.
2. No new substances formed.
3. Small energy changes involved.
4. Eg. Melting, evaporation, separation of mixtures.
1. Usually hard to reverse.
2. New substances formed.
3. Large energy changes involved.
4. Eg. Decomposition, synthesis, burning, rotting, rusting.
The particles stay the same, but their arrangement changes.
The particles change, because the bonding between their atoms changes.
Example of Physical change: Water
EXAMPLE: Decomposition of Hydrogen peroxide
Consider liquid water evaporating to form water
vapour. Each water molecule consists of one
oxygen atom bonded to two hydrogen atoms.
O
H
Reactants
H
Products
Word Equation
Hydrogen peroxide
forms
water and oxygen
Balanced chemical
equation
2 H2 O2
→
2 H2 O + O2
Particle diagram
Evapora'on
2H2O2
Hydrogen Peroxide
Water
Spaces between
water molecules
2H2O
+
Water
O2
Oxygen
water vapor
EXAMPLE: Synthesis of water
Touching each other
Far apart
Reactants
Forces between
water molecules
Medium forces
No forces, except when
they come close to each
other there are weak forces
Movement of
water molecules
Slide past each other
switching places slowly.
Flying freely at high speeds
Hydrogen + oxygen
forms
Water
Balanced chemical
equation
2 H2 + O2
→
2 H2 O
Particle diagram
Chemical
Reac+on
During evaporation, the actual water molecules do not change. The mass of water and the number
of molecules does not change. Energy is absorbed to overcome the forces between the molecules.
This changes their arrangement and movement. This change could be represented as follows:
Reactants
Products
Word Equation
Products
Word Equation
water
forms
water vapor
Balanced chemical equation
H2O (l)
→
H2O (g)
These two reactions are chemical changes because new substances are formed. Huge energy
changes are involved. The bonds between the atoms of the reactants are broken and new bonds are
formed to create the molecules of the products. Every atom in the reactants becomes an atom in
the products. The mass is conserved, but the atoms have been rearranged.
45
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Chemical change- Chemical reactions
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
Law of constant composition
Law of conservation of matter
In any particular chemical compound, all samples of that
compound will be made up of the same elements in the
same proportion or ratio as per its chemical formula.
In a chemical reaction, the sum of the mass of the reactants equals the sum of the mass of the
products.
This is also know as the Law of conservation of matter, which indicates that matter particles can
not be created or destroyed, but can be rearranged in chemical reactions.
For example, any water molecule is always made up of two hydrogen atoms and one oxygen atom in a ratio according to the formula H2O.
+
2 Parts Hydrogen
+
OR
1 Part Oxygen
1 Part Water
4 Parts Hydrogen
2 Parts Oxygen
2 Parts Water
Consider the following reaction to form ammonia: any ammonia molecule is always made up of three hydrogen atoms and one nitrogen atom in a ratio according to the formula NH3.
The different parts of the reaction are summarised in the table below:
Reactants before
Word equation
Chemical equation
Hydrogen + nitrogen
3 H2
+
N2
Products after
React to
form
ammonia
→
2 NH3
Atoms
6H
2N
→
6H
2N
Total mass of all
atoms
3 x ( 1+1) + 1 x (14+14)= 34
→
2 x (14 + 1+1+1) =
34
Energy
Energy used to break:
3 x H-H bonds
And 1 N-N bond
→
Energy released when forming 6 x
N-H bonds
46
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Reactions in aqueous solutions
Grade 10 Science Essentials
Ionic network structure
Because cations and anions are oppositely
charged, they attract each other. In solid
form, giant ionic crystals are created, such as
sodium chloride and magnesium nitrate.
Ions in aqueous solution
Ions are atoms or groups of atoms which carry charge. Cations are positively charged,
and anions are negatively charged. The table below has examples of common ions:
Ion
Formula
Charge
Nitrate
NO3 -1
-1
Chloride
-1
Cl
-1
Ammonium
NH4+1
+1
Sodium
Na+1
+1
Magnesium
Mg+2
+2
+3
+3
Aluminium
Al
Polar nature of water
SCIENCE CLINIC 2018 ©
Reactions in aqueous solutions
Reactions in aqueous solutions are reactions of chemicals that have
been dissolved.
A solution is formed when a solute dissolves in a solvent. Many
reactions in chemistry and in living systems are carried out in aqueous
solution. This means that water is the solvent.
In order for a reaction to occur the particles of the reactant need to collide with each other.
For the collision to be successful, there needs to be a transfer of either :
A. ions – these are known as ion exchange reactions
B. electrons – these are known as redox reactions.
Hydration of ions
The ions of the solute become totally surrounded by water molecules. This is called hydration.
When the water molecules surround the ions, they keep the ions apart.
Water is the solvent in all aqueous solutions. The
charge in a water molecule is not evenly distributed. The oxygen atom is slightly negative (δ-),
and the hydrogen atoms are slightly positive δ+).
We say that water is polar.
The negative dipole of one water molecule is attracted to the positive dipole of another water
molecule.
Dissolving of ionic substances
Water molecules are attracted to each other, but also to ionic compounds with the oxygen (δ-) closest to
the cation(+) and the hydrogen (δ+) closest to the anions(-). If the ions are pulled free from the crystal,
the crystal dissolves.
Electrolytes
Electrolytes are aqueous solutions which contain
ions, such as NaCl (aq) or KNO3 (aq). Because the
ions are free to move, electrolytes conduct electricAmmeter
ity. A simple circuit can be used to test the conductivity of the electrolyte, by measuring the current
through the solution with an ammeter. The more
ions in solution the better the electrolyte conducts
the current.
Dissolving is a physical process because no new
substances are formed. When all the ions are separated,
the ionic crystal has dissolved.
EXAMPLE:
Certain substances, like sugar, are also soluble in
water, but do not form ions in solution, and do not
affect conductivity.
NaCl (s) → Na+(aq) + Cl-(aq)
47
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Ba0ery
Crocodile
clip
A
Test substance
Circuit diagram for testing the
conductivity of a solution.
Reactions in aqueous solutions- Reaction types
Grade 10 Science Essentials
SCIENCE CLINIC 2018 ©
A. Ion exchange reactions
An ion exchange reaction is where the cations exchange their respective anions.
There are 3 types of ion-exchange reaction, classified according to the reactants and products involved.
1. Precipitation reactions
In precipitation reactions, anions and cations bond with new partners to form
insoluble salts. Certain cations and anions attract each other so well that they will
not be pulled from the ionic crystal. These salts are insoluble and form a precipitate.
EXAMPLE:
NaI (aq) is added to AgNO3 (aq). There are Na+, l-, Ag+ and NO3- ions present in
solution. When the I- and Ag+ ions attract, they from an insoluble salt AgI(s)
which a yellow cloudy precipitate.
NaI (aq) + Ag NO3 (aq) → AgI(s) + NaNO3 (aq)
Test
Positive result
Chloride
1. Add a few drops of AgNO3
2. Acidify with Dilute HNO3
White precipitate forms
after step 1, and
remains after step 2.
AgCl(s)
Bromide
Add a few drops of AgNO3
Cream precipitate
AgBr(s)
Iodide
Add a few drops of AgNO3
Yellow precipitate
AgI(s)
Sulphate
1. Add a few drops of BaNO3
2. Acidify in Dilute HNO3
White precipitate forms
after step 1, and
remains after step 2.
BaSO4
1. Add a few drops of BaNO3
2. Acidify in Dilute HNO3
White precipitate forms
after step 1, and
dissolves after step 2.
BaCO3
Carbonate
There are 4 types of acid-base reactions:
1. Acid + metal → salt + hydrogen
2. Acid + metal oxide → salt + water
3. Acid + metal hydroxide → salt + water
4. Acid + metal carbonate → salt + water + carbon dioxide
EXAMPLE:
Acid+ Metal carbonate → salt + water + carbon dioxide
Na2CO3 (s) + 2HCl(aq)→ 2NaCl(aq) + H2O(ℓ) + CO2 (g)
The sodium ions from the sodium hydroxide swap
places with the hydrogen in the hydrogen chloride
forming sodium chloride. At the same time the
hydroxide and the hydrogen combine to form water.
Precipitate
3. Gas forming reactions
In gas forming reactions, anions and cations have bonded
with new partners and through the exchange of ions a gas is
formed.
EXAMPLE:
Metal hydroxide + acid → salt + water
NaOH(aq) + HCl(aq)→NaCl(aq) + H2O(ℓ)
Precipitation reactions can be used to test for the presence of certain anions. An
environmental scientist analyses water for pollutants. The following table shows
how to test for some of these anions.
Anion
2. Acid-base reactions
Acid-base reactions take place between acids and bases.
Through the exchange of ions, the products formed will be
water and an ionic salt.
EXAMPLE:
Metal oxide + acid → salt + water
CuO(aq) + 2HCl(aq) → CuCl2(aq) + H2O(l)
The copper ions from the copper oxide swap places
with the hydrogen in the hydrogen chloride forming
copper(II) chloride. At the same time the oxide ion(-)
and the hydrogen combine to form water.
EXAMPLE:
acid + metal → salt + hydrogen
H2SO4+Mg → MgSO4 + H2
Gas
Test
Result
Hydrogen (H2)
Plunge a lighted
wooden splint into a
test tube of
hydrogen.
makes a popping
sound
Oxygen (O2)
Plunge a glowing
wooden splint into a
test tube of oxygen.
splint relights
Carbon dioxide
(CO2)
Bubble the test gas
through limewater calcium hydroxide
solution.
Carbon dioxide
turns the limewater
cloudy white
B. Redox reactions
Redox reactions occur when one reactant gains electron(s) called reduction, and another reactant loses
electron(s) called oxidation.
The atom or group of atoms which loses electron(s) becomes more positive, while the atom or group of
atoms which gains electron(s) becomes more negative.
EXAMPLE:
Hydrochloric acid reacts with zinc to form zinc
chloride and hydrogen gas.
2HCl(aq) + Zn(s) → ZnCl2 + H2
We show the transfer of electrons as follow:
Zn(s) → Zn 2+ + 2e- (Oxidation)
2 H+ + 2 e- → H2 (g) ( Reduction)
O
I
L
R
I
G
OIL RIG
oxidation
is a
loss of electrons
reduction
is a
gain of electrons
48
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EXAMPLE:
Copper Sulphate reacts with zinc to form zinc
sulphate and Copper.
CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)
We show the transfer of electrons as follows:
Zn(s) → Zn 2+ + 2e- (Oxidation)
Cu2+ + 2 e- → Cu(s) ( Reduction)
Reactions in aqueous solutions-THE ESSENTIALS
Grade 10 Science Essentials
AQUEOUS SOLUTIONS
SCIENCE CLINIC 2018 ©
REACTIONS
Water
Water is the solvent in all aqueous solutions.
Water is polar. Polar molecules are molecules that have
oppositely charged poles and are known as dipoles.
A. Ion exchange reactions
1.Precipitation reactions
In precipitation reactions, anions and cations bond with new partners to form insoluble salts.
NaI (aq) + Ag NO3 (aq) → AgI(s) + NaNO3 (aq)
Water dissolves ionic substances which are made up of
Cations (positively charged), and anions (negatively
charged)
Precipitation reactions can be used to test for the presence of certain anions
2. Acid-base reactions
There are 4 types of acid-base reactions:
1. Acid + metal → salt + hydrogen
2. Acid + metal oxide → salt + water
3. Acid + metal hydroxide → salt + water
4. Acid + metal carbonate → salt + water + carbon dioxide
Water molecules pull the ions free from the ionic crystal. This is called dissolution.
3. Gas forming reactions
In gas forming reactions, anions and cations have bonded with new partners and through the exchange of ions a gas is formed
The ions become totally surrounded by water molecules. This is called hydration.
Electrolytes
Electrolytes are aqueous solutions which contain ions, such as NaCl (aq) or KNO3 (aq). Because the
ions are free to move electrolytes conduct electricity. The more ions in solution the better the electrolyte conducts the current. This can be tested with the use of a simple circuit as shown below.
Gas
Test
Result
Hydrogen (H2)
Plunge a lighted wooden splint
into a test tube of hydrogen.
makes a popping sound
Oxygen (O2)
Plunge a glowing wooden splint
into a test tube of oxygen.
splint relights
Carbon dioxide (CO2)
Bubble the test gas through
limewater - calcium hydroxide
solution.
Carbon dioxide turns
the limewater cloudy
white
Ba0ery
Ammeter
Crocodile
clip
A
B. Redox reactions
One reactant gains electron(s) → reduction; another reactant loses electron(s) → oxidation.
Iron + copper(II)sulphate → Copper + iron (II) sulphate
Eg. Fe(s) + CuSO4 (aq) → Cu (s) + FeSO4 (aq)
Cu 2+ ions gained 2 electrons → Cu atoms
Fe atoms lost 2 electrons → Fe 2+ ions
Test substance
49
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Quantitative aspects of chemical change
Grade 10 Science Essentials
The Mole
Atoms, molecules and ions are too small to count, and there are so many particles in even the smallest sample of a
substance.
Percentage Composition
Percentage composition is the mass of each atom present in the compound expressed as a
percentage of the total mass of the compound.
There are more particles of water in a teaspoon then there are teaspoons of water in all the oceans.
Percentage composition of element =
Rather than dealing with the particles individually, we deal with a special number of particles.
The mole is a name for a special number. Many numbers have names, such as:
2 = pair
3 = hat-trick
12 = dozen
Ore
This many grains of sand, piled on the surface of the earth would almost reach the moon.
The mole is defined as the amount of substance having the same of particles or atoms in 12,0 g of Carbon -12.
Molar Mass
Particles are too small to weigh individually.
Molar mass (M) is defined as the mass of one mole of particles (atoms, molecules or formula units) and is
measured in the unit g.mol-1.
mass of substance (g)
m
M
Example
Formula
Molar mass
(g.mol-1)
Atoms
Neon
Ne
20
Covalent
compound
Usually
Molecules
Carbon
dioxide
CO2
12+32
= 44
Ionic
compound
Ions (formula
units)
Salt
NaCl
23+35,5
= 58,5
Metallic
compound
Positive kernels
and delocalized
electrons
Gold
Au
197
Element
Particles
EXAMPLE:
What is the relative formula mass of Calcium sulphate (CaSO4)?
MR (CaSO4) = AR (Ca) + AR (S)+ (4 x AR (O))
= 40 + 32 + (4 x 16)
= 136 (no unit)
Magnetite
Fe2O3
Fe3O4
Relative
molecular mass
(2 x 56) + (3 x 16)
=160
(3 x 56) + (4 x 16)
=232
% iron by mass
[(2 x 56) /160] x 100
= 70%
[(3 x 56) / 232] x 100
= 72%
∴ magnetite contains more iron
Different types of Chemical Formulae
Consider the substance ethane.
molar mass (g·mol−1)
Ball and stick
model of ethane
Relative atomic mass (Ar) is the average mass of an atom compared to the mass of a Carbon 12 atom. It is
measured in atomic mass units (amu).
Molar mass (M) of an element is equal to the magnitude of relative atomic mass (Ar) in amu. This is found
on the periodic table. See the table below for other substances:
Type of
substance
Haematite
Formula
602 000 000 000 000 000 000 000 .
n=
molar mass of element
× 100
MR of compound
Consider these iron ores: haematite and magnetite – which contains more iron by
mass?
A mole of particles is an amount of 6,02 x 1023 particles. 6,02 x 1023 is known as Avogadro’s number, NA.
Avogadro’s number (NA) is too big to imagine.
number of mole (mol)
SCIENCE CLINIC 2018 ©
It also can be represented using a formula. There are three types of formulas we use:
Molecular formula
Actual number of each
atom.
Eg. C2H6
Empirical formula
Simplest whole number ratio
of the atoms.
Eg.C1H3 → CH3
Shows how the atoms are
joined.
Structural formula
EXAMPLE:
What is the molar mass of
Calcium sulphate (CaSO4)?
136 g.mol-1
EXAMPLE:
What is the relative molecular mass of sucrose (C12H22O11)?
MR (C12H22O11) = 12 x AR (C) + 22 x AR (H)+ 11 x AR (O)
= (12x 12) + (22x1) + (11 x 16)
= 342 (no unit)
50
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EXAMPLE:
What is the molar mass of
sucrose(C12H22O11) ?
342 g.mol-1
Quantitative aspects of chemical change
Grade 10 Science Essentials
Concentrations of solutions
Solutions are homogeneous (uniform) mixtures of two or more substances. A solution
is formed when a solute dissolves in a solvent.
The solvent and solute can be a gas, liquid or solid. The most common solvent is
liquid water. This is called an aqueous solution.
Solution
Solute
Solvent
salt water
Salt
water
Soda water
Carbon dioxide
water
SCIENCE CLINIC 2018 ©
Molar Volumes of Gases
If different gases have the same volume under the same conditions of temperature and pressure, they will have the
same number of molecules.
This is known as Avogadro’s Law: One mole of any gas occupies the same volume at the same
temperature and pressure.
Concentration
The concentration of a solution is the number of mole of solute per unit
volume of solution.
number of moles (mol)
n
c=
V
concentration (mol ⋅ dm-3)
volume (dm3)
VM for all gases at STP is 22.4 dm3·mol−1.
can also be calculated with
m
c=
MV
i.e. The number moles of solute per 1 dm3 of
solution i.e. mol. dm-3. If a solution of potassium
permanganate KMnO4 has a concentration of 2
mol.dm-3 it means that for every 1 dm3 of solution,
there are 2 moles of KMnO4 dissolved in the solvent.
Standard Temperature and Pressure (STP) is 273 K (0°C) and 1,01x105 Pa.
Note:
1000 cm3 = 1 dm3 (= 1 litre)
1 cm3 = 0,001 dm3
1 cm3 = 1 ml
EXAMPLE:
EXAMPLE:
A solution contains 10 g of sodium
hydroxide, NaOH, in 200 cm 3 of
solution. Calculate the concentration of
the solution.
Calculate the mass of solute in 600 cm3
of 1,5 mol·dm -3 sodium chloride
solution.
n(NaOH)
=
=
=
m
M
10
23 + 16 + 1
0,25 mol
V = 200 cm3 = 0,2 dm3
c(NaOH)
=
n
V
0,25
0,2
=
1,25 mol ⋅ dm−3
=
V = 600 cm3 = 0,6 dm3
M(NaCl)
=
=
n
m
The molar volume of a gas, VM, is the volume occupied by one mole of the gas.
This also means that for reactions at constant temperature and pressure, gas volumes will react in the same ratio as
the molar ratio.
N2 + 2O2
1 mol + 2 mol
1 dm3 + 2 dm3
V
n=
VM
number of moles (mol)
23 + 35,5
58,5 g ⋅ mol−1
=
=
=
cV
1,5 × 0,6
0,9 mol
=
=
=
nM
0,9 × 58,5
52,65 g
→
→
→
2NO2
2 mol
2 dm3
volume of gaseous substance (dm3)
molar gas volume at STP (22,4 dm3·mol−1)
EXAMPLE:
A gas jar with a volume of 224 cm3 is full of chlorine gas, at STP. How
many moles of chlorine gas are there in the gas jar?
n
=
V
VM
=
0,224
22,4
=
51
0,01 m ol
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Grade 10 Science Essentials
Water of crystallization
Quantitative aspects of chemical change
Some ionic crystals trap a certain number of water
molecules between the ions as they are forming.
These water molecules are known as “Water of
crystallization”.
Eg. Hydrated copper sulphate:
CuSO4 · 5 H2O has 5 water molecules per formula
unit. When the hydrated salt crystals are heated, the
water molecules evaporate off, leaving the
anhydrous salt behind.
EXAMPLE:
13,2 g of a sample of zinc sulphate, ZnSO4.xH2O, was
heated in a crucible. Calculate the number of moles of
water of crystallisation if 7.4 g of solid remained.
1 . m(H2 O) = 13,2 g − 7,4 g
= 5,8 g
=
CuSO4 · 5 H2O(s) → CuSO4 (s) + 5 H2O(g)
=
To calculate the number of moles of water of
crystallization:
1.
2.
3.
4.
5.
Note:
The dot in the formula ( · ) between the
salt and the water means that a light bond
is formed. It is NOT a multiplication dot.
1.Determine the mass of the elements.
2.Determine mol of each substance.
3.Simplify the atomic ratio.
=
EXAMPLE:
In a combustion reaction 0,48 g of magnesium ribbon is burnt. The amount of
magnesium oxide produced is 0,80 g.
Calculate the empirical formula for magnesium oxide.
m
M
7,4
161
=
0,046 mol
Steps
mol water
mol anhydrous salt
4 . ratio =
0,32
0,046
=
=
1:7
5 . ∴ formula = Z n SO4 ⋅ 7H2O
Calculating the Empirical Formula from Percentage Composition
The empirical formula of a compound can also be found from its percentage composition. We assume that 100 g of the
compound is analysed, then each percentage gives the mass of the element in grams in 100 g of the compound.
An oxide of sulphur contains 40% sulphur and 60% oxygen by mass. Determine the empirical
formula of this oxide of sulphur.
Steps
Sulphur
Empirical formula is the chemical formula of a compound that shows
the smallest whole number ratio of the atoms.
To calculate the empirical formula of a compound from mass:
0,32 mol
3 . n(ZnSO4 ) =
Calculate the mass of water that evaporated off.
Calculate the moles of water.
Calculate the moles of anhydrous salt.
Determine the ratio of water to anhydrous salt.
Write the formula for the hydrated salt.
Calculating the Empirical Formula of a Compound from Mass
m
M
5,8
18
2 . n(H2 O) =
SCIENCE CLINIC 2018 ©
Magnesium
Oxygen
Step 1:
Mass of element
0.48 g
0.80 – 0.48 = 0.32 g
Step 2:
Mol (divide by mass of 1 mol)
n=m/M
= 0.48 / 24
= 0,02 mol
n=m/M
= 0.32 / 16
= 0,02 mol
Step 3:
Atom ratio
(divide by smallest no in ratio)
1
1
Empirical formula: MgO
EXAMPLE:
A sample of an oxide of copper contains 8 g of copper combined with 1 g of
oxygen.
Find the empirical formula of the compound.
Oxygen
Step 1:
% of element
40
60
Step 2:
Mass of element (g)
40
60
Step 3:
Mol
n=m/M
= 40 / 32
= 1,25 mol
n=m/M
= 60 / 16
= 3,75 mol
Step 4:
Smallest mol ratio
1,25 / 1,25
=1
3,75 /1,25
=3
Steps
Oxygen
Step 1:
Mass of element
8g
1g
Step 2:
Mol
(divide by mass of 1 mol)
n=m/M
= 8 / 63,5
= 0,126 mol
n=m/M
= 1 / 16
= 0,0625 mol
Step 3:
Atom ratio
(divide by smallest no in ratio)
0,125/0,0625
≈2
0,0625/0,0625
=1
Empirical formula: Cu2O
Empirical formula: SO3
52
Copper
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Quantitative aspects of chemical change
-THE ESSENTIALS
Grade 10 Science Essentials
Calculating the Empirical Formula of a Compound from
SCIENCE CLINIC 2018 ©
To calculate the number of moles of water of crystallization:
Empirical formula is the chemical formula of a compound that shows the
smallest whole number ratio of the atoms.
1.
2.
3.
4.
5.
Mass
Calculate the mass of water that evaporated off
Calculate the moles of water
Calculate the moles of anhydrous salt
Determine the ratio of water to anhydrous salt.
Write the formula for the hydrated salt.
To calculate the empirical formula of a compound from mass:
1.Determine the mass of the elements
2.Determine mol of each substance
3.Simplify the atomic ratio
Molar mass (M) is defined as the mass
of one mole of particles
Number of
par,cles
Mass of pure
substance (in grams)
!=
The concentration of a solution
is the amount of solute per unit
volume of solution.
Solu,on
!
!
! = !"
Volume of gas
at STP (in dm3)
!
22,4
M v for all gases at STP is 22.4
dm3.mol-1.
Standard Temperature and Pressure
(STP) is 0°C and 101.3 kPa
53
#
6,02×10!"
Number of
moles
!=
The molar volume of a gas, Mv, is
the volume occupied by one mole of
the gas.
!=
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The mole is the same
number of particles or
atoms in 12,0 g of Carbon
Hydrosphere
Grade 10 Science Essentials
Water
Water exists in 3 forms:
•Solid: snow caps at the poles, icebergs and permafrost
•Liquid: surface water (oceans, lakes, dams, rivers) and ground water
•Vapour: water vapour in the atmosphere
Water is important for:
•All living organisms depend on water → cells consist mainly of water, substances in cells are suspended or
dissolved in water, substances are transported to and from cells in water
•Drinking water, habitat, photosynthesis.
•Cooking, cleaning, manufacturing processes, agriculture etc.
SCIENCE CLINIC 2018 ©
SYSTEMS OF THE EARTH
Systems influenced
by the
HYDROSPHERE
ATMOSPHERE
LITHOSPHERE
BIOSPHERE
Components of
system
All gaseous
components of the
earth
Hard crust of the earth,
semi-solid rocks, hot
liquid rocks and molten
iron core
All living organisms on
earth
Processes
influenced by
water
Evaporation,
condensation and
precipitation
Surface run-off and
percolation
Photosynthesis,
metabolism etc.
Description
Energy from the sun
causes water to
evaporate and form
water vapour which
condenses to form
clouds and becomes
too heavy and falls to
the earth as
precipitation.
Precipitation filters
through the soil. A
certain portion flows
across the surface as
run-off and the other
portion seeps deeper
into underwater rivers.
Water collects in lakes,
dams and oceans.
Water is essential for all
living organisms. Plants
require water for
photosynthesis, animals
and humans require
water for metabolism
and various other
intracellular processes.
HYDROSPHERE
•The water cycle (also known as hydrosphere) describes the continuous movement of water through the ground
(lithosphere), living organisms (biosphere) and the atmosphere.
•Importance of water cycle
-Water is recycled
-Water is purified
CLOUDS
Condensa8on
Water vapour collects and
droplets form which s.ck
together and form clouds
Precipita8on
Water droplets
become too heavy
and fall to the earth
Water vapour
Precipita.on:
Rain, hail, snow
Transpira8on
Water is released
from plants and
evaporates
PLANTS
SOIL
Surface run-off
Water runs over the surface of
the ground
Evapora8on
Liquid water is
converted to a vapour
OCEANS, LAKES,
DAMS, RIVERS,
STREAMS
Surface and ground water
Percola8on
Water filters through the
ground and collects in
underground streams and dams
INFLUENCE OF DAMS
Environmental Influence
•Flooding in surrounding areas → affects the plant and animal life
•A dam has a large surface area resulting in increased evaporation which may affect the local
climate in the area
•Erosion and deepening of rivers due to initial emptying of rivers and periodical release of
water from the dam
•Silting up of dams
•Altering of temperature of water in dam due to large surface area
•Fish migration is prevented
•Dams provide water for agricultural use
•Dams produce water used in households and various industrial processes
•Production of hydroelectricity
Economical Influence
•Cost of buying private land to build the dam on
•Cost of building the dam
•Cost of buying electricity (energy) to run the dam
54
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www
Scientific Notation
N is a number
between 1 and 9,99
3)
N × 10n
n is the integer
power of 10.
1000 = 1 × 103
10 000 = 1 × 104
54 500 = 5,45 × 104
4)
Very large numbers such as 352 000 000 or very small numbers such as 0,00000000000000085 are difficult to read and write. Scientific
notation is used to write these very large or very small numbers. In scientific notation, a number is written with only 1 digit to the left of
the comma. The number is expressed as the product of 2 numbers as follows:
Positive Exponents
100 = 1 × 102
2500 = 2,5 × 103
2)
125 = 1,25 × 102
4)
1
SYSTEMMATHS | ©
0,00000018 = 1,8 × 10-7
0,0000018 = 1,8 × 10-6
0,000018 = 1,8 × 10-5
0,00018 = 1,8 × 10-4
300 000 000 m.s-1
= 7,856157236 billion
7 856 157 236 = 7,856157236 × 109
1 000 000 000 = 1 × 109 [109 = a billion]
17 = 1,7 × 101
1 = 0,001 = 10-3
1000
7)
31572 = 3,1572 × 104
8 157 236 = 8,157236 × 106
1 000 000 = 1 × 106 [106 = a million]
8971 = 8,971 × 103
6)
389 = 3,89 × 102
100 000 = 1 × 105
85 = 8,5 × 101
10 = 1 × 101
Examples:
1)
5)
715 723 = 7,15723 × 105
= 8,157236 million
Real Life Example: The speed of light is written in scientific notation as 3 × 108 m.s-1
Negative Exponents
Numbers less than 1 (fractional numbers) have negative exponents.
1 = 0,01 = 10-2
100
0,0027 = 2,7 × 10-3
3)
1 = 0,1 = 10-1
10
0,051 = 5,1 × 10-2
0,0091 = 9,1 × 10-3
2)
0,12 = 1,2 × 10-1
0,037 = 3,7 × 10-2
Examples:
1)
0,85 = 8,5 × 10-1
Type 9 . 1
Example 2: When typing 9,1 × 10-31
1.
Do not type: 9 . 1 × 1 0 Exp +/– 3 1
3. Type +/– 3 1
2. Press the Exp / ×10
x
0,000 000 000 000 000 000 000 000 000 009 1 kg
Real Life Example: The mass of an electron is written in scientific notation is 9,1 × 10-31 kg.
Calculator Skills:
x
button
How to use the “ Exp ” or “ ×10 ” button on a calculator:
Type 6 . 0 2
Example 1: When typing 6,02 × 1023
1.
x
2. Press the Exp / ×10
3. Type 2 3
Do not type: 6 . 0 2 × 1 0 Exp 2 3
Grade 10 Maths Essentials
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Not for commercial use. Accreditation required.
For Mathematics textbooks, please visit www.systemmaths.co.za.
The Metric System
Time – second (s)
Temperature – Kelvin (K)
Mass – kilogram (kg)
Light intensity – candela (cd)
2
The System International d’Unites, abbreviated as SI units is the system of units used by scientists. There are seven base units and all other
units are derived from these.
SI Base Units
Length – metre (m)
Electric Current – Ampere (A)
Amount of matter – mole (mol)
nano
μ
micro
× 10-3
pico
n
10-6
× 10-3
p
10-9
× 103
10-12
× 103
× 10-3
× 10-3
deci
k
× 10-3
centi
d
kilo
milli
c
unit
m
103 or 1000
100
× 103
10-1 or 1
1
1000
× 103
10-2 or 1
10-3 or
× 103
10
The metric system is a decimal system of units. Units increase or decrease in size by factors of 10. These factors are indicated by prefixes
e.g. kilogram is 1000 grams.
Metrix Prefixes
Prefix
abbreviation
value
Conversions to SI units
2)
875 × 10-9 J
= 8,75 × 10-7 J
45 × 10-3 s
= 4,5 × 10-2 s
45 ms = 45 ÷ 1000 s
= 0,045 s
or
2)
or
p p p
p-2
p p p p p
p3 × p-5 = p-2
because
3)
4)
p3 × p-2 = p1
875 nJ = 875 ÷ 1 000 000 000 J
= 0,000000875 J
To convert to the SI unit, replace the prefix by its numerical value. All final answers below are written in Scientific Notation.
25 × 103 m
= 2,5 × 104 m
Examples:
25 km = 25 × 1000 m
= 25 000 m
or
or
1)
43 μC = 43 ÷ 1 000 000 C
p × p2 = p3
43 × 10-6 C
= 4,3 × 10-5 C
3)
= 0,000043 C
Multiplying powers
p × p = p2
p × p × p = p3
p3 × p2 = p5 (not p6)
Rule: “When bases are the same, add the exponents.”
1)
p p
because p p p p
[All final answers below are written in Scientific Notation.]
because p3 × p2 = p × p × p × p × p = p5
Conversions between any unit
To convert between any units, count the number of “jumps” left or right on the Metrix prefixes table above.
kilo 2 jumps right
kilo 2 jumps right
× 10-6
× 10-6
5)
2)
micro 2 jumps right
kilo 4 jumps right
× 10-6
× 10-12
2,7 × 1015 nm = 2,7 × 1015 × 10-12 km
= 2,7 × 103 km
nano
2071 pA = 2071 × 10-6 μA
= 2,071 × 10-3 μA
pico
6)
3)
milli 3 jumps right
micro 2 jumps right
× 10-9
× 10-6
SYSTEMMATHS | ©
= 1,5 × 10-11 μC
1,5 × 10-5 pC = 1,5 × 10-5 × 10-6 μC
pico
0,17 pJ = 0,17 × 10-9 mJ
= 1,7 × 10-10 mJ
pico
One jump right means × 10-3. Two jumps right means 10-3 × 10-3 = 10-6. Four jumps right means 10-3 × 10-3 × 10-3 × 10-3 = 10-12 etc.
(Note: this is not applicable for centi- and deci-)
milli
785 mm = 785 × 10-6 km
= 7,85 × 10-4 km
milli
Examples:
1)
4)
7,8 × 104 mm = 7,8 × 104 × 10-6 km
= 7,8 × 10-2 km
Grade 10 Maths Essentials
This material may be copied and distributed freely according to the copyright notice.
Not for commercial use. Accreditation required.
For Mathematics textbooks, please visit www.systemmaths.co.za.
milli 2 jumps left
milli 2 jumps left
106
× 106
5)
2)
nano 1 jump left
pico 3 jumps left
= 8,3 × 102 ps
(p-2) 3 = p-6
× 103
× 109
8,3 × 10-7 ms = 8,3 × 10-7 × 109 ps
milli
12,7 μC = 12,7 × 103 nC
= 1,27 × 104 nC
micro
(p3) 3 = p9
(p-3) 3 = p-9
1 km = 1000 m
6)
3)
pico 2 jumps left
nano 1 jump left
= 1,8 nC
×
× 103
25 cm2 = 25 × (10-2)2 m2
= 25 × 10-4 m2
= 2,5 × 10-3 m2
1 km = 1000 m
3)
1,8 × 10-3 μC = 1,8 × 10-3 × 103 nC
micro
0,85 μC = 0,85 × 106 pC
= 8,5 × 105 nC
106
micro
One jump left means × 103. Therefore two jumps left means 103 × 103 = 106. Three jumps left means 103 × 103 × 103 = 109 etc.
(Note: this is not applicable for centi- and deci-)
kilo
37 km = 37 × 106 mm
= 3,7 × 107 mm
kilo
Examples:
1)
4)
4,5 × 104 km = 4,5 × 104 × 106 mm
= 4,5 × 1010 mm
Raising a power to a power
(p2) 2 = p2 × p2 = p4
(p3) 2 = p3 × p3 = p6
(p2) 3 = p6
Rule: “When raising a power to a power, multiply the exponents.”
Examples:
Conversions of square units
If the unit is squared, as for area, the conversion factors must also be squared.
1 m = 100 cm
Reasoning:
1 cm = 10 mm
1 km2 = 1 km × 1 km
= 1000 m × 1000 m
= 10002 m2
or = 1 000 000 m2
2)
13 dm2 = 13 × (10-1)2 m2
= 13 × 10-2 m2
= 1,3 × 10-1 m2
1 m2 = 1 m × 1 m
= 100 cm × 100 cm
= 1002 cm2
or = 10 000 cm2
or
5 km2 = 5 × (103)2 m2
= 5 × 106 m2
1 cm2 = 1 cm × 1 cm
= 10 mm × 10 mm
= 102 mm2
or = 100 mm2
Examples:
1)
5 km2 = 5 × 10002 m2
= 5 000 000 m2
Conversions of cubic units
If the unit is cubed, as for volume, the conversion factors must also be cubed.
1 m = 100 cm
Reasoning:
1 cm = 10 mm
× 10-3
× 103
m3 ( kl )
1 km3 = 1 km × 1 km × 1 km
= 1000 m × 1000 m × 1000 m
= 10003 m3
or = 1 000 000 000 m3
× 10-3
dm3 ( l )
1 m3 = 1 m × 1 m × 1 m
= 100 cm × 100 cm × 100 cm
= 1003 cm3
or = 1 000 000 cm3
cm3 ( ml )
× 103
2)
13 dm3 = 13 × (10-1)3 m3
= 13 × 10-3 m3
= 1,3 × 10-2 m3
3)
25 cm3 = 25 × (10-2)3 m3
= 25 × 10-6 m3
= 2,5 × 10-5 m3
3
SYSTEMMATHS | ©
Note:
◦ An object with a volume of 1 cm3 will displace exactly 1 ml of water.
◦ An object with a volume of 1 dm3 will displace exactly 1 l of water.
◦ An object with a volume of 1 m3 will displace exactly 1 kl of water.
1 cm3 = 1 cm × 1 cm × 1 cm
= 10 mm × 10 mm × 10 mm
= 103 mm3
or = 1000 mm3
Study:
Examples:
1)
7 km3 = 7 × (103)3 m3
= 7 × 109 m3
Grade 10 Maths Essentials
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Solving an equation for an unknown variable
then
x–3=y–3
x+3=y+3
x=y
4
3
2
1
1
4
3x
x
=
=
3y
y
2
x+3
5
=
=
y+3
3
x-3
=
y-3
4
An equation is like a pair of scales balancing on the equals sign. To maintain the balance means that whatever you do to the left–hand side
you must also do to the right–hand side. If the same operation is carried out on both sides of the equal sign, the equality is maintained.
and
3x = 3y
If
and
5
x
3
y
3
and
Examples
In all examples below the unknown variable is “p”. Each equation is solved for p by making it the subject of the formula (writing “p” on its
own on one side of the equal sign.)
Remember: Only if the same operation is carried out on both sides of the equal sign will equality be maintained.
4
=4 p 1
Example 2
p-2=8
p-2+2=8+2 1
p = 10
18
2
p = 12
3
2
3
p = 12
3
2
Example 6
1 Add 2 to both sides.
3
2
p
1 Multiply both sides by
3
2
1
Example 4
3
Example 3
1
p
because
2
3
5
7
3
2
7
5
1
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1
5 is the reciprocal of 7
7
5
because
3 is the reciprocal of 2
2
3
Note: Example 6 and 7
When the coefficient of the
variable is a fraction, multiply
both sides of the equation by
the fraction’s reciprocal.
1 Multiply both sides by 3.
p = 12
3p = 15
15
3
7
5
p
=4
3
3
4 3 1
3p
3
p=5
22, 4
5
p = 16
7
5
7
p = 16
7
5
Example 7
1 Divide both sides by 3.
7
5
p
1 Multiply both sides by
1
Please note: All the steps are written out in the examples, however the numbered steps 1 , 2 and 3 in grey are only shown for
completeness. These steps can and should be done mentally or on a calculator.
1) Basic examples:
Example 1
p + 8 = 15
p + 8 - 8 = 15 - 8 1
p=7
p
3
p
Example 5
1 Subtract 8 from both sides.
3
p
3 = 4p
3
4p
=
2
4
4
3
p=
or 0,75
4
1 Multiply both sides by p. (p 0)
2 Divide both sides by 4.
Grade 10 Maths Essentials
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Example 8
p
7
=
2
9
p 2 7 2
1
2
9
14
p=
or 1,56
9
1 Multiply both sides by 2.
1
LCD is 9p
Example 9
2
5
=
p
9
2×9×p
5×9×p
=
p
9
1
LCD is 8p
Example 10
5
9
=
8
p
2
5×8×p
9×8×p
=
8
p
72
5p
5p
18
5p
72
=
5
5
14, 4
1 Multiply both sides by 8p
2 Divide both sides by 5.
p
18
5p
=
2
5
5
18
or 3,6
5
p
1 Multiply both sides by 9p
2 Divide both sides by 5.
2p = 16
2p
16
=
2
2
p=8
p
3
Example 14
p
3
3
3 3 2
3
p
+1=4
3
p
+1-1 =4-1 1
3
p
3
p=9
1 Subtract 1 from both sides.
2 Multiply both sides by 3.
) alone on one side first.
1 Add 1 to both sides.
2 Divide both sides by 2.
2
2p - 1 + 1 = 15 + 1 1
2p - 1 = 15
Example 13
1
LCD is 10p
Example 11
3
5
=
10
2p
25
2
3 × 10 × p
5 × 10 × p
=
10
2p
3p
8, 3
3p
25
=
3
3
p
1 Multiply both sides by 10p
2 Divide both sides by 3.
p
3
p
p
7
7 p 2
3 = 7p
3
7p
=
3
7
7
3
or 0,43 (2 dec. places)
7
p=
5
SYSTEMMATHS | ©
1 Subtract 1 from both sides.
2 Multiply both sides by p.
3 Divide both sides by 7.
3
3
+1-1 =8-1 1
p
3
+1=8
p
Example 15
Example 9, 10 and 11: When the variable is in the denominator, always multiply both sides of the equation by the Lowest Common
Denominator (LCD).
Example 12
2p + 7 = 15
2
2p + 7 - 7 = 15 - 7 1
2p = 8
2p
8
=
2
2
p=4
1 Subtract 7 from both sides.
2 Divide by 2 on both sides.
Examples 12 - 15: Always get the variable (e.g. 2p or
Grade 10 Maths Essentials
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Example 2
-2p + 12 = 4
2
-2p + 12 - 12 = 4 - 12 1
-2p = -8
-2p
-8
=
-2
-2
p=4
1 Subtract 12 from both sides.
2 Divide both sides by -2.
Example 6
-0,2p - 1,8 = 1,5
2
-0,2p - 1,8 + 1,8 = 1,5 + 1,8 1
-0,2p
3,3
=
-0,2
-0,2
p = -16,5
1 Add 1,8 to both sides.
2 Divide both sides by -0,2.
Example 3
p
3
3
-3 3 2
-3
p
+4=1
3
p
+4-4 =1-4 1
3
p
3
p = -9
p
0,3
0,3
0,3
-46,7
-46,7 0,3 2
p = -14, 01
1 Subtract 1,7 from both sides.
2 Multiply both sides by 0,3.
p
p
+ 1,7 - 1,7 = -45 - 1,7 1
0,3
p
+ 1,7 = -45
0,3
Example 7
1 Subtract 4 from both sides.
2 Multiply both sides by 3.
2) Examples including negative numbers and decimal fractions:
Example 1
2p + 15 = 8
2
2p + 15 - 15 = 8 - 15 1
2p = -7
2p
-7
=
2
2
p = -3,5
1 Subtract 15 from both sides.
2 Divide both sides by 2.
Example 5
2,5p + 0,7 = 5,25
2
2,5p + 0,7 - 0,7 = 5,25 - 0,7 1
2,5p = 4,55
2,5p
4,55
=
2,5
2,5
p = 1,82
1 Subtract 0,7 from both sides.
2 Divide both sides by 2,5.
Grade 10 Maths Essentials
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Example 4
3
+9=3
p
p
p
3
p
-6 p 2
-6
3
+9-9 =3-9 1
p
3
3 = -6p
3
-6p
=
3
-6
-6
1
p = - or -0,5
2
1 Subtract 9 from both sides.
2 Multiply both sides by p.
3 Divide both sides by -6.
Example 8
5
- 2,3 = 4,5
p
p
5
p
p
47,3
47,3 p
47,3p
3
2
5
- 2,3 + 2,3 = 45 + 2,3 1
p
5
5
5
47,3p
=
47,3
47 ,3
6
p = 0,11 (2 dec. places)
1 Add 2,3 to both sides.
2 Multiply both sides by p.
3 Divide both sides by 47,3.
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= 4
2
1
1
2
2
Example 2
2
2
7 - 13
p
4
Example 2
25 =
-6
p
2
p
25 =
-6
25p =
p
25p = - 6
25p
-6
=
25
25
p = - 0,24
1 Multiply both sides by p.
2 Divide both sides by 25
1
1 Subtract 4 from both sides.
2 Find the square root of both
sides.
p=
p = 16 2
p = 16
p + 4 - 4 = 20 - 4 1
p + 4 = 20
2
3) Examples including squares and square roots:
Example 1
2
2
p =4
p
2
=3
p =3
2
2
and
p=
p
p=9
1 Find the square root of both
sides.
2 Square both sides.
p+3
2
Example 1
19 = 5
19
5 p+3
=
5
5
2
19
p+3
=
5
2
19
p+3
=
2
5
2
3
4) Examples including more complex fractions:
2
7,6 = p + 3
7,6 - 3 = p + 3 - 3
p = 4,6
1 Divide both sides by 5
2 Multiply both sides by 2.
3 Subtract 3 from both sides.
Grade 10 Maths Essentials
Example 3
1 2
18 = 5 + p
2
1 2
18 - 5 = 5 - 5 + p
2
1 2
13 = p
2
1 2
p
2
2
3
2
1
5,1 (1 dec. place)
p
2
13 2 = 2
26 = p
26 =
p=
Example 4
2
9 = 3 + 0,7p
2
3
2
9 - 3 = 3 - 3 + 0,7p
2
6 = 0,7p
2
0,7p
6
=
0,7
0,7
2
1
2, 93 (2 dec. place)
p
6
2
=p
0,7
6
=
0 ,7
p=
p
=
2
4 × 5 × 10
p × 10
p
1
1
4×5
=
LCD = 10p
2
p × 10
1 Subtract 3 from both sides.
2 Divide both sides by 0.7.
3 Find the square root of both
sides.
2
5p = 20
5p
20
=
5
5
p=4
7
SYSTEMMATHS | ©
1 Multiply both sides by 10p.
2 Divide both sides by 5.
10
1 Subtract 5 from both sides.
2 Multiply both sides by 2.
3 Find the square root of both
sides.
= 45,5
3
2
1
Example 4
13
p2
2
p = 45,5p
2
Example 3
13
p2
2
2
0 ,53 (2 dec. place)
p
45,5p2
45,5
13 = 45,5p
13
45,5
13
2
=p
45,5
13
45,5
p=
1 Multiply both sides by p2.
2 Divide both sides by 45,5.
3 Find the square root of both
sides.
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Ratios
The Basics
If a certain substance consists of 3 kg of X and 5 kg of Y we say, “the ratio of the mass of X to the mass of Y is 3 to 5”.
X 3kg 3
=
= . NB: No units in the answer.
Y 5kg 5
This can be written as X : Y = 3 : 5 or in fraction form
If another substance consists of 9 kg of X and 12 kg of Y we say, “the ratio of the mass of X to the mass of Y is 9 to 12”.
X 9kg 3
=
= in simplest fraction form.
Y 12kg 4
This can be written as X : Y = 9 : 12 = 3 : 4 in simplest form or
Ratios can only be determined for quantities of the same kind i.e. the quantities must have the same units.
60c
60c
6 1
For example, the ratio of 60c to R5,40 = 60c : 540c = 6 : 54 = 1 : 9 in simplest form or R5,40 = 540c = 54 = 9 .
1 : 2 = 5 : 10
and
or
and
or
1
5
=
2
10
sec ond sec ond
=
first
first
first
first
=
sec ond sec ond
3
6
=
2
4
2
4
=
3
6
sec ond sec ond
=
first
first
first
first
=
sec ond sec ond
“The ratio of 2 parts to 3 parts is equal to the ratio of 4 parts to 6 parts” can be written in the following ways.
2
10
=
1
5
“The ratio of 1 part to 2 parts is equal to the ratio of 5 parts to 10 parts” can be written in the following ways.
Writing ratios in different ways
1)
2)
2:3 = 4:6
Solving problems involving ratios
Example 2
Example 3
0,6
0,6
3
4
Example 4
12 : b 12 =
6 : 12b = 8 : 9
6
3 Divide both sides by 2.
8
SYSTEMMATHS | ©
12b 9
2
6
8
9
2b
3
8
9
b=
or 0,5625
16
1 Multiply each fraction by the LCD.
2 Simplify 12
1
2
1
2 3
:b=
:
LCD = 12
2
3 4
2
3
12 :
12 1
3
4
of 2 to 3 .
The ratio of 1 to b is equal to the ratio
2
NB: When solving a ratio for an unknown variable, always write the variable in the numerator of the fraction as it makes solving the
equation much easier.
Example 1
0,5 : 0,3 = 0,6 : b
The ratio of 0,5 parts to 0,3
parts is equal to the ratio of
0,6 parts to b parts.
4 : 3 = b : 13
b
0,3
b
=
0,5
0,6
0,3 0,6
0,5
1
The ratio of 4 parts to 3
parts is equal to the ratio of
b parts to 13 parts.
4
=
4
b
=
3
13
13 b 13
13
3
b = 0,36
0,36 = b
b = 17,3
1 Multiply both sides by 0,6.
17,3 = b
1 Multiply both sides by 13.
1
The ratio of 2 parts to b
parts is equal to the ratio of
5 parts to 3 parts.
2:b=5:3
b
3
=
2
5
b 2 3 2
1
2
5
6
b=
or 1,2
5
1 Multiply both sides by 2.
Grade 10 Maths Essentials
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Add:
2 + 3 = 5 [we have 5 equal parts]
Divide: 15 ÷ 5 = 3
Multiply both sides of the ratio by 3:
2×3:3×3
6 : 9
o
3 : 6 = 1 : 2 NB: Simplify the ratio first
1 × 0,5 : 2 × 0,5
0,5A : 1A
Add:
1 + 2 = 3 [we have 5 equal parts]
Divide: 1,5 ÷ 3 = 0,5
Multiply both sides of the ratio by 0,5:
Answer is:
5
Divide a current of 33 A in a ratio of 1 : 3 .
1 3
:
(LCD = 10)
2 5
1
3
10 :
10
2
5
5 : 6
5×3:6×3
15A : 18A
Add:
5 + 6 = 11
Divide: 33 ÷11 = 3
Multiply both sides of the ratio by 3:
2
o
o
o
Divide a current of 1,5 A in a ratio of 3 : 6.
Answer is:
o
o
o
Divide 15 in a ratio of 2 : 3. [think “2 parts to 3 parts”]
Dividing a number into a ratio:
1)
3)
5)
o
o
o
Answer is:
Grade 10 Maths Essentials
2)
4)
4)
Add:
1 + 2 = 3 [we have 3 equal parts]
Divide: 0,75 ÷ 3 = 0,25
Multiply both sides of the ratio by 0,25:
4
4
9
SYSTEMMATHS | ©
Divide 0,75 in a ratio of 2 : 1. [think “2 parts to 1 part”]
o
o
o
2 × 0,25 : 1 × 0,25
Answer is: 0,5 : 0,25
2
Divide 10m in a ratio of 1 : 3
2×2:3×2
4m : 6m
Add:
2 + 3 = 5 [we have 5 equal parts]
Divide: 10 ÷ 5 = 2
Multiply both sides of the ratio by 2:
1 3
:
(LCD = 4)
2 4
1
3
4:
4
2
4
2 : 3
NB: Simplify the ratio first:
o
o
o
Answer is:
3
Divide 55m in a ratio of 2 : 1
8×5:3×5
40m : 15m
Add:
8 + 3 = 11
Divide: 55 ÷ 11 = 5
Multiply both sides of the ratio by 5:
2 1
:
(LCD = 12)
3 4
2
1
12 :
12
3
4
8 : 3
o
o
o
Answer is:
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Straight line graphs and the Cartesian plane
1
3
2
6
3
9
4
12
The rule is
y = 3x
or
x:y=1:3
The table below shows a relationship between input values (x values) and output values (y values).
x
y
(1 ; 3)
2
(2 ; 6)
3
(3 ; 9)
4
(4 ; 12)
5
X
or
y
x
3
1
Quadrant 2
0 Quadrant 4
Quadrant 1
Y
Quadrant 3
y
x
k
10
If the points are joined with a straight line, it means that each point on
the line represents the rule y = 3x.
The point (0 ; 0), the point of intersection of the X-axis and the Y-axis, is
called the origin.
The point (2 ; 6) is 2 units from the Y-axis and 6 units from the X-axis.
The y co-ordinate of a point indicates how far the point is from the X-axis.
The x co-ordinate of a point indicates how far the point is from the Y-axis.
The x- and y-values are called the co-ordinates of the points.
The input (x) values are represented on the horizontal axis and the
output (y) values on the vertical axis.
X
Another way to express the relationship between the x-values and the y-values is as ordered number pairs, namely (1 ; 3) , (2 ; 6) , (3 ; 9)
and (4 ; 12). The input value (x-value) is always written 1st and the output value (y-value) 2nd inside the round brackets.
1
Graphs are visual representations of relationships where the shape of the graph indicates
the properties of the relationship. Graphs are drawn on the Cartesian plane. The vertical
number line, called the Y-axis, and the horizontal number line, called the X-axis, divide
the Cartesian plane into 4 quadrants.
Y
12
10
8
6
4
2
0
or
Graph is a straight line passing
through the origin.
y = kx
In the Cartesian plane:
Horizontal change to the right is in a positive direction and to the left is in a negative direction.
Vertical change upwards is in a positive direction and downwards is in a negative direction.
NB: DIRECTLY PROPORTIONAL RELATIONSHIPS
k.
Consider the equation y = 3x above. This is a directly proportional relationship.
y
This means that y = kx where k is a constant. This can also be written as x
In the above example, the value of this constant is 3.
Increasing linear function.
As the x -values increase the y -values
decrease.
Decreasing linear function.
Types of straight line graphs:
Graphs are used to communicate information and relationships. To read a graph, one must understand how the quantities in the graph
relate to each other e.g. look to see if the quantities make the graph go up or down or remain constant.
Constant linear function
As the x -values increase the
y -values increase.
SYSTEMMATHS | ©
As the x-values increase the y-values
remain constant.
Grade 10 Maths Essentials
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The gradient of a straight line graph
y2
x2
y1
.
x1
(2;4)
A
2
Y
(2;2)
B
(3;3)
3
(2;-1)
2
y1
x1
y1
x1
( 1)
1 1
y2
x2
2
1 1
11
) means “change in”.
3
3 1
1
Gradient of graph B
y y
y2
2
1
x2 x1
x2
3 1
1
2
4
2
1
2
2 1
2
3
2
1
y1
x1
Gradient of graph C
y2
x2
1 2
1 ( 1)
1 1
2 1
2
or
2
0, 6
3
1 2
or
1
X
0, 6
3
Note: (-1;1) can be used as the
starting point or (2;-1) can also be
used to calculate the gradient of
line C.
2
2
1
SYSTEMMATHS | ©
1 0
1 1
Gradient of graph D
y2 y1
x2 x1
Note: Any two co-ordinates on a graph
can be used to work out its gradient.
2
Gradient of graph A
y2 y1
x2 x1
Note:
Delta (
The vertical change divided by the horizontal change between any 2 points on a straight line indicates the steepness of its slope and
whether it slopes to the left or the right. This value is known as the gradient of the straight line.
y
change in y
The gradient is written as follow The vertical change or
or most commonly
x
change in x
The horizontal change
Example 1
(1;2)
(1;1)
1
X
The two straight line graphs below, A and B, both slope to the right and therefore have positive gradients.
The gradients can easily be worked out using the formula y y2 y1 , as shown below.
x x2 x1
Graph A is visibly “steeper” than graph B and therefore has a larger gradient.
Y
4
3
2
1
0
Example 2
D
(0;1)
0
(1;-1)
The two straight line graphs below, C and D, both slope to the left and therefore have negative gradients.
The gradients can easily be worked out using the formula y y2 y1 , as shown below.
x x2 x1
Graph D is visibly “steeper” than graph C and therefore has a larger gradient.
C
(-1;1)
Grade 10 Maths Essentials
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Example 3
12
The horizontal change
The two straight line graphs below, E and F, are both constant graphs. Remember, the gradient of the graphs means The vertical change .
0
3
0
2
1 2
2
Gradient of graph E
3
2
3
4
4
1,5
5
6
6
1
or
The rule is
k
x
y
6
x
xy = k.
or
2
0
0
1 ( 1)
1 1
2
2
0
undefined
k
x
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Graph is a curved line and does
not pass through the origin.
xy = k or y
xy = 6
Gradient of graph F
. Therefore a vertical line’s gradient is undefined.
0
. Therefore a horizontal line’s gradient is 0.
The horizontal change
E
Graph E is horizontal and therefore has no vertical change i.e.
Y
(0;2)
0
(1;2)
The vertical change
0
(-1;2)
(-1;1)
(-1;0)
(-1;-1)
X
Graph F is vertical and therefore has no horizontal change i.e.
F
Inversely proportional relationships
1
6
2
X
The table below shows a relationship between input values (x values) and output values (y values).
x
y
1
This relationship is represented graphically as follows:
Y
6
5
4
3
2
1
0
6 above.
x
NB: INVERSELY PROPORTIONAL RELATIONSHIPS
Consider the equation y
This is an inversely proportional relationship. This is written as y
In the above example k = 6.
Grade 10 Maths Essentials
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B
B
Trigonometry
Trigonometry is the branch of Mathematics which deals with the relationship between the sides and angles of a triangle.
Understanding the terms “Opposite”, “Adjacent” and “Hypotenuse”.
4cm
8cm
A
AB
LEARN:
sin A =
opposite
Example 4:
adjacent
Example 2:
B
C
AB
AD
10
AD
opposite
hypotenuse
5
B
C
5
With respect to  :
13
o BC is the opposite side
o AC is the adjacent side
o AB is the hypotenuse
o AC is the opposite side
o BC is the adjacent side
o AB is the hypotenuse
With respect to B̂ :
A
A
AB
adjacent
AD
AB
cos A =
adjacent
hypotenuse
This constant ratio is called the cosine of  or cos A.
LEARN:
adjacent
A
For any angle  , the ratio of the length of the side adjacent to  to the length of the
hypotenuse is a constant.
AD
In the adjacent triangles, AC = 3 and AE = 6 = 3 . Therefore AC = AE .
10
hypotenuse is a constant. This constant ratio is called the sine of  or sin A.
For any angle  , the ratio of the length of the side opposite  to the length of the
5
In the adjacent triangles, BC = 4 and DE = 8 = 4 . Therefore BC = DE .
o BC is the opposite side
o AC is the adjacent side
o AB is the hypotenuse
D
E
D
E
5
Definition: The Hypotenuse is the longest side of a right-angled triangle. It is always the side opposite the right angle.
With respect to B̂ :
o AC is the opposite side
o BC is the adjacent* side
o AB is the hypotenuse
C
With respect to  :
*adjacent means next to or adjoining
A
C
A
Consider the right-angled triangles below:
Example 1:
adjacent
Example 3:
opposite
B
C
B
C
opposite
Trigonometric ratios: Sine, Cosine and Tangent
1) Sine
A
3cm
6cm
opposite
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opposite
2) Cosine
A
Grade 10 Maths Essentials
adjacent
opposite
adjacent
3cm
B
C
6cm
4cm
D
E
8cm
AC
3
AE
6
3
AC
AE
In the adjacent triangles, BC = 4 and DE = 8 = 4 . Therefore BC = DE .
adjacent
A
14
For any angle  , the ratio of the length of the side opposite  to the length of the side
adjacent to  is a constant.
tan A =
This constant ratio is called the tangent of  or tan A.
LEARN:
opposite
adjacent
Trigonometric ratios are independent of the lengths of the sides of a triangle and depend only on the angle size.
Hence trigonometric ratios are considered to be functions of the angles.
3) Tangent
A
NB:
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opposite