Download Student Study Guide 1999

Student Study Guide
Physical Sciences 416/436
Student Study Guide
Physical Sciences 416/436
Dear Teachers,
This Study Guide was prepared to help English students in the province prepare for their Physical Sciences
416/436 MEQ written examination. The guide was prepared in 1996 and has undergone a first revision in
1997, a second revision in 1998 and a third revision in 1999. The 1996 and 1997 committees were funded by
P.E.O.P.T. grants and the 1998 and 1999 committees were funded by the Science Action Plan Committee
(SAPCO).
The 1999 version of the Study Guide includes references to both Discovering Matter and Energy (DME) and
Science Quest (SQ). Material on Objective 2.3 of Module II is presented in two formats — the Electron Flow/
Left Hand Rule version is found on page Module II-8a and the Conventional Flow/Right Hand Rule version
is found on page Module II-8b.
1996 TEACHERS
John Allen
Don Clark
Ken Cooke
Jim Daskalakis
Anthony Slonosky
Riverdale High School
Chambly County High School
Westmount High School
F.A.C.E. School
Chambly County High School
1996 CONSULTANTS
Michael Brennan
Carolyn Gould
Elizabeth Speyer
Protestant School Board of Greater Montreal
South Shore School Board
Lakeshore School Board
1997 TEACHERS
Ken Cooke
Marie-Thérèse Dukaczewski
Doreen Riga
Phil Ritchie
Westmount High School
St. Thomas High School
Lester B. Pearson High School
John Rennie High School
1997 CONSULTANT
Elizabeth Speyer
Lakeshore School Board
1998 TEACHERS
Ken Cooke
Jim Daskalakis
Marie-Thérèse Dukaczewski
Doreen Riga
Phil Ritchie
Anthony Slonosky
Westmount High School
F.A.C.E. School
St. Thomas High School
Lester B. Pearson High School
John Rennie High School
Chambly County High School
1998 CONSULTANT
Chris Tromp
Com. des Écoles Catholiques de Montréal
1999 TEACHERS
Phil Ritchie
Anthony Slonosky
John Rennie High School
Chambly County High School
SAPCO PROJECT
COORDINATOR
Jan Farrell
Lester B. Pearson School Board
The authors still consider this document to be a working document. They would,
therefore, ask that, in exchange for using this Guide, teachers provide feedback.
Please use the attached evaluation form to make corrections of suggestions.
Special thanks to Jill Butler of the Lester B. Pearson School Board for desktop publishing.
MODULE I
– PROPERTIES & STRUCTURES –
Key Concepts:
REVIEW OF PHYSICAL SCIENCE 214
DENSITY
The amount of mass in a given volume. The concentration of mass.
Density =
mass
volume
D = M
V
Example:
Density of water = 1 g/mL.
An object has a volume of 20 mL and a mass of 5 g. What is its density?
D = M
V
=
5g
= 0.25 g/mL
20 mL
IRREGULAR VOLUMES
The volume of a solid with an odd shape can be determined by the displacement of water.
Here’s how:
Fill a container with water to the brim, add the
object, and measure how much water overflows.
MIXTURES
•
Mixtures are formed when two or more substances are combined physically.
•
Mixtures are NOT “pure” substances. They can be separated by physical means.
For example: filtering, evaporation, etc.
•
The substances that make up mixtures are not changed, but in some cases, they are
hidden. For example: when salt is dissolved in water, the salt is only hidden and can be
recovered.
Module I – 1
TYPES OF MIXTURES
•
Physical Mixtures
For example: Sand and salt can be separated by hand picking as you can still see both
substances.
•
Solutions
For example:
•
Salt and water, a clear homogeneous mixture, cannot be separated by
filtering but by boiling.
Suspensions
For example: Chalk and water. A solid that does not dissolve when mixed in a liquid
or a gas, forms a suspension. It forms a cloudy heterogenous mixture.
The two substances will separate by themselves if left standing for long
enough. They can also be separated by decantation and filtration.
STATES OR PHASES OF MATTER
There are three phases of matter: solid, liquid and gas.
Many substances can exist in all three phases. For example:
solid water (ice) – liquid water – gaseous water (water vapor)
solid iron – liquid iron – iron vapor
•
Melting (Freezing) Point
The temperature at which a substance changes from a solid to a liquid and vice versa.
The melting point (M.P.) of water is 0°C.
•
Boiling (Condensation) Point
The temperature at which a substance changes from a liquid to a gas and vice versa.
The boiling point (B.P.) of water is 100°C.
PHASE CHANGES
•
Melting:
Solid → Liquid
•
Freezing:
Liquid → Solid
•
Boiling:
Liquid → Gas
•
Condensation:
Gas → Liquid or Gas → Solid
•
Sublimation:
Solid → Gas
The solid does not melt but changes directly to a gas.
Solid Iodine → Iodine Vapor or
Dry Ice (solid carbon dioxide) → CO2 gas
Module I – 2
SAMPLE QUESTIONS
1.
While at a flea market, you purchased a ring that was supposedly pure gold (Au). You
want to determine experimentally whether it is real gold or not. You already know the
density of pure gold.
Explain in detail the experimental procedure you will carry out to determine whether the
ring is pure gold.
2.
Copper, whose density is 8.92 g/cm3, is a metallic substance used to make pennies.
You wish to find the density of a penny to see if this value equals the density of copper.
Explain the procedure you would use to determine whether the density of the penny
equals the density of copper.
In your explanation, indicate the materials to be used and the steps involved in your
procedure and then describe everything you must do with the resulting measurements
(work involved in analyzing the measurements).
Module I – 3
2.1
To identify some of the properties of objects
and substances in the environment.
416 and 436
Key Concepts:
DME: Sections 1.8 and 1.9
SQ: pp. 9-10
PROPERTIES
A property is any characteristic of a substance. Some properties can be determined by simple
observation; others require a test or a measurement.
PHYSICAL PROPERTIES
Physical properties can be determined without changing the nature of the substance.
For example:
colour, odor, melting point, density, etc.
CHEMICAL PROPERTIES
Chemical properties require that the substance be changed in order to determine the property.
For example:
sulfur will burn in air.
USING PROPERTIES TO CLASSIFY SUBSTANCES
There are millions of chemicals and scientists like to place them into groups having properties
in common.
For example:
Some solutions will turn blue litmus paper red. Chemists call this group
of solutions acids.
USING PROPERTIES TO IDENTIFY SUBSTANCES
Chemists need to know what substances they are dealing with. Certain properties are special or
unique to a given substance and will allow you to identify the chemical. These properties are
called characteristic properties.
For example:
If a gas is mixed with limewater and turns it cloudy, we know the gas is
carbon dioxide. Turning limewater cloudy is a characteristic property of
carbon dioxide.
UNIVERSAL PROPERTIES:
There are two properties that are shared by all substances. These properties are mass and
volume. Obviously, these would be of no use in either classifying a substance or identifying it.
Module I – 4
SAMPLE QUESTIONS
1.
Louis found five unmarked bottles in a workroom. Each of the bottles contained a pure
substance. He noted the following properties for each of these colourless liquids.
1. boiling point
2. mass
3. volume
4. density
Which properties does Louis need to know to identify these liquids?
2.
a)
1 and 2
c)
1 and 4
b)
1 and 3
d)
2 and 4
Matthew carried out several tests to identify an unknown liquid. Here are the results he
obtained.
1.
colour
colourless
2.
odor
odorless
3.
temperature
22°C
4.
density
0.999 g/mL
5.
boiling point
99°C
Which of the following properties would be considered characteristic properties?
3.
a)
1, 4, and 5
c)
3, 4, and 5
b)
1, 2, and 3
d)
4 and 5
Which of the following is a characteristic property of tin (Sn)?
a)
temperature of 27°C
c)
cylindrical shape
b)
melting point of 232°C
d)
volume of 25 cm3
Module I – 5
4.
In the laboratory, Julie analyzed the properties of four solids and then completed the table
below.
Solid
Mass
Volume
Shape
Density
1
223 g
82 cm3
cylindrical
2.7 g/cm3
2
223 g
25 cm3
cubic
8.9 g/cm3
3
113 g
25 cm3
spherical
4.5 g/cm3
4
38 g
14 cm3
cubic
2.7 g/cm3
Julie found that two of these solids are probably made of the same substance. Which
solids are they?
5.
a)
1 and 2, because they have the same mass
b)
1 and 4, because they have the same density
c)
2 and 3, because they have the same volume
d)
2 and 4, because they have the same shape.
While identifying an unknown substance in the laboratory, you note that it has the following
properties:
1. Its melting point is 0°C.
2. It is colourless.
3. It does not change the colour of neutral litmus paper.
4. It does not conduct electricity.
Which of these properties most clearly indicates that the unknown substance is pure
water?
a)
1
c)
3
b)
2
d)
4
Additional Questions:
SQ: Follow-up p. 10
Module I – 6
2.2
To determine the characteristic properties of
given substances.
416 and 436
Key Concepts:
DME: Section 1.10
SQ: pp. 11-18
CHARACTERISTIC PROPERTIES
In the laboratory, you discovered a number of characteristic properties of chemicals. Some you
discovered by observation only; others you discovered by tests.
GASES
Hydrogen, oxygen and carbon dioxide gas can be distinguished by three simple tests.
They are:
1.
2.
3.
burning splint test
glowing splint test
limewater test.
•
Oxygen:
If the gas causes the glowing splint to relight, it is oxygen.
•
Hydrogen:
If the gas “pops” when you insert a burning splint, the gas is hydrogen.
•
Carbon dioxide:
If the gas causes limewater to turn cloudy when it is mixed, the
gas is carbon dioxide.
LIQUIDS
•
Sodium Hydroxide
–
turns red litmus paper blue
•
Hydrochloric Acid
–
turns blue litmus paper red
•
Water
–
turns cobalt chloride pink
SOLIDS
•
sulfur is yellow
•
Marble bubbles in acid
Module I – 7
2.3
To identify a substance on the basis of its
characteristic properties.
416 and 436
DME: Section 1.11
SQ: pp. 19-20
Key Concepts:
In the laboratory, you used the characteristic properties (observed in previous labs) to identify
an unknown substance. Below are a few possibilities:
•
If you were given a gas, you would conduct the burning splint test, the glowing splint
test and the limewater test. Suppose the gas "popped" when you did the burning
splint test, then you would identify it as hydrogen as this is the characteristic property
of hydrogen.
•
If you were given a liquid, you would do the conductivity test, the indicator paper test
and the litmus paper test. Suppose the blue litmus turned red, then you would
identify it as hydrochloric acid.
•
If you were given a solid, you would do the conductivity test, examine its appearance,
try to dissolve it in water, try to dissolve it in acid and do the ammonia solution test.
Suppose the solid bubbled in acid, you would identify it as marble.
SAMPLE QUESTIONS
1.
In the laboratory fume hood, Sylvia and Tom heated a white solid, potassium chlorate.
They observed that a gas was given off. They collected this gas by the displacement of
water. After plunging a glowing wood splint into it, they saw that the splint burst into
flames.
What gas was produced?
a)
oxygen
c)
hydrogen
b)
nitrogen
d)
carbon dioxide
Module I – 8
2.
Three well-known gases have been identified during laboratory experiments:
a.
b.
c.
oxygen (O2)
carbon dioxide (CO2)
hydrogen (H2)
The following are three characteristic properties.
1.
2.
3.
The gas turns limewater cloudy.
The gas explodes when exposed to an open flame.
The gas rekindles a glowing splint.
Match each gas with the characteristic property that identifies it.
3.
a)
a and 1
b and 2
c and 3
c)
a and 3
b and 1
c and 2
b)
a and 2
b and 1
c and 3
d)
a and 3
b and 2
c and 1
The following are the properties of an unknown gas:
Colour
Colourless
Mass
8.9 g
Reaction to a flame
None
Reaction to a glowing splint
None
Volume
5.0 L
Given the properties above and the information below, you are to identify the unknown
gas.
GAS
PROPERTIES
Ammonia
Colourless, density of 0.76 g/L
Argon
Colourless, inert
Hydrogen
Colourless, combustible
Oxygen
Colourless, brings about combustion
Which of the four above gases is the unknown gas?
a)
Ammonia
c)
Hydrogen
b)
Argon
d)
Oxygen
Module I – 9
4.
The properties of an unknown liquid are given in the following table.
Property
Observation
Reaction with litmus paper
Reaction with cobalt chloride paper
Conducts electricity
Mass
Volume
No change
No change
No
1g
1.25 mL
The following information is also provided.
Substance
Conducts electricity
Density (g/mL)
Water
Glycerine
Propanol
Salt solution
No
No
No
Quite well
1.00
1.25
0.80
1.05
Using all the above information, identify the unknown liquid.
5.
a)
Water
c)
Propanol
b)
Glycerine
d)
Salt solution
You observed and noted the properties of a solid substance.
Property
Observation
Mass
Volume
Magnetic
Conducts electricity
8.90 g
1.13 cm3
Yes
Yes
Given the observations above and the information in the table below, you are to identify
the substance you have observed.
Substance
Cobalt, Co
Copper, Cu
Iron, Fe
Sulfur, S
Density
Conducts Electricity
3
8.90 g/cm
8.95 g/cm3
7.87 g/cm3
2.07 g/cm3
Yes
Yes
Yes
No
Yes
No
Yes
No
Which substance did you observe?
a)
Cobalt
c)
Iron
b)
Copper
d)
Sulfur
Module I – 10
Magnetic
2.4
To justify the use of a substance of certain
consumer goods on the basis of the
properties of that substance.
416 and 436
DME: Section 1.12
SQ: pp. 20-23
Key Concepts:
When engineers design something, they choose materials that best fit the job at hand.
Some examples:
•
If you are building something outside, the materials should not rot or rust.
•
If you are building an electrical device, it is important whether or not a material
conducts electricity.
SAMPLE QUESTIONS
1.
2.
Which of the following properties of aluminum explain why aluminum pots may be used to
cook food?
1.
high malleability
2.
a good conductor of electricity
3.
a good conductor of heat
4.
silvery colour
5.
low density
6.
high melting point
a)
1 and 4
c)
3 and 6
b)
2 and 3
d)
4 and 5
Someone you know is talking about replacing the old windows in her house. She has a
choice of window frames made from wood, plastic or aluminum.
What choice would you recommend to her?
Justify your answer by stating at least three properties of these materials.
Module I – 11
3.
Alligator clips are used to connect the components of an electric circuit.
Alligator Clips
Jaws
Which substance can be used to make the jaws of these alligator clips?
a)
Aluminium
c)
Plastic
b)
Rubber
d)
Porcelain
Module I – 12
3.2
Classify the changes that various substances
have undergone as either physical or
chemical.
416 and 436
Key Concepts:
DME: Section 2.2
SQ: pp. 26-30
PHYSICAL AND CHEMICAL CHANGES
PHYSICAL CHANGE
This type of change does not change the basic nature of the substance.
Example:
When you tear paper, you still have paper; when you melt iron, you still have
iron, etc.
You usually can reverse physical changes.
Example:
Ice is melted. You can refreeze the water to form ice, thus reversing the
change.
Examples:
- iodine subliming
- glass breaking
- sulfur melting
- sugar dissolving in water
CHEMICAL CHANGE
This type of change does change the basic nature of the substance.
Example:
When the metal magnesium burns, it is changed into the chemical magnesium
oxide (a white powder).
You usually can't reverse a chemical change.
Example:
You can't "unburn" paper.
Examples:
- iron rusting
- paper burning
- silver tarnishing
Module I – 13
It is important that you can tell when a chemical change has taken place. If you notice any of
these things happening during a change, you know that it is a chemical change.
•
•
•
•
•
you see bubbles of gas - it fizzes
a solution turns cloudy and a solid falls to the bottom (precipitation)
a change in colour (not a change in shade as light red to dark red, but a total change, as
red to blue)
heat or light is given off or taken in (i.e. something gets hot or cold)
electricity is produced (i.e. in a battery)
SAMPLE QUESTIONS
1.
The following are some of the things we do with the food we eat:
- cooking
- cutting
- crunching
- digesting
Which of the following actions bring about chemical changes?
2.
3.
a)
cooking and crunching
c)
cutting and crunching
b)
cooking and digesting
d)
cutting and digesting
Which of the following phenomena is associated with a chemical change?
a)
The water in a lake freezes.
b)
An iron fence rusts.
c)
A heavy rain causes flooding.
d)
Dew forms.
You are making french fries for dinner.
Step 1:
Step 2:
Step 3:
Step 4:
You peel the potatoes and wash them with tap water.
You cut the potatoes into strips and pat them dry.
You fry the strips in oil for five minutes.
You put the fried strips in a bowl and sprinkle them with salt.
Module I – 14
During which step did a chemical change occur?
4.
a)
1
c)
3
b)
2
d)
4
During an experiment, a glass rod, a piece of tin, some copper powder and some rubbing
alcohol were each heated separately. The following changes were observed:
1.
the glass rod became flexible
2.
the piece of tin turned to liquid
3.
the copper powder turned black
4.
the rubbing alcohol caught on fire.
Which of these changes are chemical?
5.
a)
1 and 2
c)
2 and 4
b)
1 and 3
d)
3 and 4
A group of students identified the following as changes which occur in their environment:
1.
An automobile rusts in Montreal.
2.
The deodorant used to freshen the air in the bathroom undergoes sublimation.
3.
The limewater becomes cloudy when you blow bubbles into it using a straw.
4.
A water pipe breaks when the water in it freezes.
5.
An antacid tablet produces a gas when it is dropped into a glass of water.
Which of the above changes can be described as chemical changes?
a)
1 and 3 only
c)
1, 3, and 5 only
b)
2 and 4 only
d)
1, 2, 3, and 5 only
Module I – 15
6.
A student carried out a four-step experiment and made the following observations:
Step
Test
Observation
1
Dissolved a salt in distilled water.
The resulting solution was
colourless.
2
Poured the resulting solution into
another salt solution.
The resulting mixture contained
a precipitate.
3
Filtered the resulting mixture.
The precipitate obtained was white.
4
Weighed the precipitate.
Its mass was 5.4 g.
At which step did a chemical change occur?
a)
1
c)
3
b)
2
d)
4
Additional Questions:
SQ: Exercises p. 29
Module I – 16
3.3
Describe a pure substance as a compound or
an element, after conducting an experiment.
416 and 436
DME: Section 2.3
SQ: pp. 31-34
Key Concepts:
ELEMENTS are pure substances that cannot be broken down into simpler substances. There
are about 105 elements and they are to be found listed on the periodic table at the back of the
text.
COMPOUNDS are pure substances that can be broken down into simpler substances. They are
made of two or more elements, chemically combined. There are millions of compounds.
MIXTURES are impure substances that are combined physically and can be separated
physically.
One can tell if a pure substance is an element or a compound by conducting an experiment to
see if you can break it down or not.
Example:
– You have a red powder. Is it a compound or an element?
– You heat the powder and notice a gas is given off and a shiny silver liquid is
left behind.
– Your conclusion is that it is a compound because it has been broken down.
Chemists represent chemical change with a chemical equation.
Example:
Copper + oxygen → copper oxide
– In English, this means: copper combines with oxygen to produce copper
oxide.
– Copper and oxygen are called reactants as they are the chemicals you start
with. Copper oxide is a product because it is what you end up with.
Module I – 17
SAMPLE QUESTION
1.
A burning splint is used to test for hydrogen gas. Hydrogen gas, H2, reacts with oxygen
gas, O2, to form water, H2O.
Which of the following statements is TRUE?
a)
This water is a pure substance consisting of elements that are chemically
bonded.
b)
This water is a pure substance consisting of compounds that are chemically
bonded.
c)
This water is a pure substance consisting of a mixture of two compounds that
are chemically bonded.
d)
This water is a pure substance consisting of a mixture of elements.
Module I – 18
3.4
To analyze a compound experimentally.
416 and 436
DME: Section 2.4
SQ: pp. 31-34
Key Concepts:
In the laboratory, you took apart the compound copper oxide that you made in the previous
experiment.
You mixed the copper oxide with charcoal (carbon) and heated it. Carbon dioxide left the test
tube, leaving the copper behind.
The Word Equation:
copper oxide + carbon → carbon dioxide + copper
The carbon "stole" the oxygen from the copper, leaving the copper behind.
TYPES OF CHEMICAL REACTIONS
Synthesis:
element + element → compound
copper + oxygen → copper oxide
Decomposition:
compound → element + element
water → hydrogen + oxygen
Single Displacement:
compound 1 + element 1 → element 2 + compound 2
copper oxide + carbon → copper + carbon dioxide
SAMPLE QUESTIONS
1.
In the laboratory, you are to determine if a substance is a compound. After heating the
substance in an open container, you observe that a chemical reaction is occurring and
that the substance is undergoing certain changes.
Which of the following changes would definitely indicate that the substance was a
compound before it was heated?
a)
Its colour changes.
c)
Its mass decreases.
b)
Its physical state changes.
d)
Its texture changes.
Module I – 19
2.
The state of four substances before and after having been heated in the laboratory is
presented in the following table.
Substance
Before Heating
After Heating
1
Dark gray solid
Purple gas
2
White solid
Colourless liquid
3
Red solid
Gray liquid and colourless gas
4
Brown liquid
Orange-brown gas
According to this information, which substance was a compound before being heated?
3.
a)
1
c)
3
b)
2
d)
4
Given the following reaction:
copper + oxygen → copper oxide
Which one of the following statements about this reaction is FALSE?
4.
a)
Both copper and oxygen are elements.
b)
Copper is a solid and oxygen is a gas at room temperature.
c)
Copper retains its original properties after it has reacted with oxygen to
form copper oxide.
d)
Copper oxide is a compound.
Which of the following illustrates the formation of a compound from its elements?
a)
+
→
b)
+
→
+ Energy
+
c)
+ Energy →
d)
+ Energy →
+ Energy
+
Module I – 20
5.
Three different substances were heated separately in crucibles. The following
observations were noted during the experiments:
Experiment 1:
Experiment 2:
Experiment 3:
• Before Heating
The substance is a yellow solid with a
mass of 100 g.
• After Heating
The substance is a silvery liquid with a
mass of 92.6 g; oxygen was released during the
heating.
• Before Heating
The substance is a silvery solid with a
mass of 150 g.
• After Heating
The substance is a silvery solid with a
mass of 150 g.
• Before Heating
The substance is a silvery solid with a
mass of 50 g.
• After Heating
The substance is a white solid with a
mass of 83.3 g.
From which experiment can one conclude that the original substance was a compound?
Justify your answer by giving two reasons.
6.
A yellow solid, insoluble in water, is placed in test tube 1 and heated with oxygen gas.
This produces a gas that travels from test tube 1 to test tube 2, which contains distilled
water.
The following observations are made:
• The mass of the yellow solid decreases.
• The solid left in test tube 1 is still yellow.
• A suffocating odour is produced.
• After the gas from test tube 1 enters test tube 2, the solution formed in test tube 2
turns litmus paper red.
Module I – 21
Which of the following correctly describes the gas travelling from test tube 1 and the
solution formed in test tube 2?
7.
a)
The gas travelling from test tube 1 is an element and the solution formed in test
tube 2 is a mixture.
b)
The gas travelling from test tube 1 is an element and the solution formed in test
tube 2 is an element.
c)
The gas travelling from test tube 1 is a compound and the solution formed in
test tube 2 is a mixture.
d)
The gas travelling from test tube 1 is a compound and the solution formed in
test tube 2 is an element.
You heated four different solid substances in the laboratory and made the following
observations:
Solid Substance
1
2
3
4
Observation
The mass of the solid decreased and a gas was released.
The shape of the solid changed and a liquid was formed.
A liquid was formed and then a gas was released.
The colour of the solid changed and a gas was released.
Given these observations, which substance was definitely a compound before being
heated?
8.
a)
1
c)
3
b)
2
d)
4
A student conducted experiments involving four different liquids and made the following
observations.
Liquid
1
2
3
4
Observation
The liquid evaporated when it was heated.
The liquid became a solid when it cooled.
Neutralization occurred when this liquid was combined with another liquid.
The liquid changed colour when it was heated.
Which liquids underwent a chemical change?
a)
1 and 2
c)
2 and 4
b)
1 and 3
d)
3 and 4
Module I – 22
3.7
Describe the impact of physical and
chemical changes on the environment,
health, the economy, and society.
416 and 436
DME: Section 2.7
SQ: pp. 38-40
Sections to Read:
Questions:
DME: Section 2.7
SQ: Intermediate Objective 3.7
SQ: Exercises p. 39
Module I – 23
4.1
To compare the discontinuous (Democritus)
hypothesis and the continuous (Aristotle)
hypothesis regarding the structure of matter.
416 and 436
DME: Section 3.1
SQ: pp. 44-46
Key Concepts:
CONTINUOUS HYPOTHESIS
Matter does not contain atoms. A "magic" knife could cut forever and would still produce
smaller and smaller pieces of the substance.
DISCONTINUOUS HYPOTHESIS (ATOMIC THEORY)
Matter is made of indivisible particles called atoms.
ATOM
An atom is the smallest particle of a substance that can take part in a chemical reaction.
•
Each element contains only one type of atom.
•
Atoms have mass called the atomic mass.
•
Chemists represent atoms with symbols.
Example:
Hydrogen atoms
Oxygen atoms
- H
- O
Module I – 24
4.2
Represent a chemical change using a model.
416 and 436
DME: Section 3.2
SQ: pp. 47-49
Key Concepts:
An atom being a particle can be represented by a sphere. Different atoms could be represented
by either differently coloured or sized spheres.
Here is how we could represent a chemical reaction:
Iron atom
sulfur atom
Iron + sulfur → Iron Sulfide
+
→
Module I – 25
4.3
Describe Dalton's atomic model.
416 and 436
DME: Section 3.3
SQ: pp. 49-51
Key Concepts:
DALTON'S ATOMIC MODEL
All matter is made of tiny particles called atoms.
•
These atoms are indestructible. That is, no smaller particles exist and atoms have no
internal structure.
•
Atoms of the same element have the same size and mass while atoms of different
elements have different size and mass
Hydrogen atoms
•
Carbon atoms
Atoms combine in chemical reactions, in (ratios) simple whole numbers, to form
compounds.
SAMPLE QUESTIONS
1.
Several models have been developed to represent matter. The following information
relates to two of these models.
Model 1:
Matter is continuous. Everything is made from four elements: water, fire,
air and earth.
Model 2:
Matter is composed of atoms. An atom consists of a nucleus in which protons
and neutrons are found. Electrons spin in shells (energy levels) around the
nucleus.
With whom is each of these models associated?
a)
The first with Democritus and the second with Dalton.
b)
The first with Democritus and the second with Rutherford and Bohr.
c)
The first with Aristotle and the second with Rutherford and Bohr.
d)
The first with Aristotle and the second with Dalton.
Module I – 26
2.
Which of the following best describes Dalton’s atomic theory?
a)
Matter consists of atoms containing positive and negative charges.
b)
Matter consists of atoms that contain a positive nucleus and electrons that
move freely about the nucleus.
c)
Matter consists of atoms that contain a positive nucleus and electrons that
move within different energy levels.
d)
Matter consists of atoms and all atoms of the same element are identical.
Module I – 27
4.4
Demonstrate that there are two types of
electric charges – after doing experiments
and documentary research.
416 and 436
DME: Sections 3.4 and 3.5
SQ: pp. 51-56
Key Concepts:
•
Chemists found that there is a property of matter called charge.
•
This property comes in two "flavors": one is positive (+); the other is negative (–).
•
There is a force between two charged objects.
When the charges are the same, the force is repulsion (like repels like):
positive – positive
negative – negative
When the charges are different, the force is attraction (opposite charges attract):
positive – negative
The particle that is involved in static electricity is the electron. This is a small negative particle.
This is the particle that carries the charge from one object to another when the objects are
rubbed or touched.
Static charge:
Negative charge
Positive charge
No charge (neutral)
-
surplus of electrons
deficit of electrons
electrons are in balance
When chemists found the electron and realized it was part of the atom, they knew that Dalton's
Model must change.
SAMPLE QUESTIONS
1.
When any two of the substances in the list below are rubbed together, the one listed
higher becomes negatively charged.
- rubber
- silk
- wool
- glass
Module I – 28
Conduct the following experiment:
• Rub a rubber ball with a piece of wool
• Rub a glass rod with a piece of silk.
• Finally, bring the rubber ball near the piece of silk.
In light of this experiment, which of the following statements is true?
2.
a)
The positively charged rubber ball and the positively charged piece of
silk attract each other.
b)
The positively charged rubber ball and the negatively charged piece of
silk repel each other.
c)
The negatively charged rubber ball and the positively charged piece of
silk attract each other.
d)
The negatively charged rubber ball and the negatively charged piece of
silk repel each other.
Two charged spheres, A and B, are suspended.
When they are near one another, they repel each other. A third charged sphere, C, is
brought close to B and an attraction is observed.
A
B
B
C
A
C
From this experiment, what can be concluded about the charges of spheres A, B, and C?
Module I – 29
3.
Lavinia works in a bake shop during the summer. Her work consists of sprinkling icing
sugar on doughnuts using a plastic sifter.
While continuing to sift the icing sugar in the same manner, she notes the following:
•
at first, the sugar particles fall vertically
•
as time passes, the particles start to deviate from the vertical and they have a greater
tendency to stick to the sides of the sifter.
Why do the sugar particles move away from each other at the same time as they are
attracted to the sifter?
4.
A student was given the following materials:
• a wool cloth
• a vinyl ruler
• two styrofoam balls (A and B)
suspended from ring stands
A
B
Using these materials, she performed a laboratory experiment consisting of five steps.
The table below lists the five steps and the results of the first three steps.
Step
Result
1. Rub the ruler with the wool cloth.
1. The wool cloth and the ruler acquire
opposite charges.
2. Touch ball A with the ruler.
2. Ball A and the ruler have the same
charge.
3. Touch ball B with the wool cloth.
3. Ball B and the wool cloth have the same
charge.
4. Bring the ruler close to ball A, but
without touching it.
4. ?
5. Bring the ruler close to ball B, but
without touching it.
5. ?
Module I – 30
What were the results of steps 4 and 5?
a)
Result of Step 4: the ruler and ball A repelled each other.
Result of Step 5: the ruler and ball B attracted each other.
b)
Result of Step 4: the ruler and ball A repelled each other.
Result of Step 5: the ruler and ball B repelled each other.
c)
Result of Step 4: the ruler and ball A attracted each other.
Result of Step 5: the ruler and ball B repelled each other.
d)
Result of Step 4: the ruler and ball A attracted each other.
Result of Step 5: the ruler and ball B attracted each other.
Additional Questions:
SQ: Exercises p. 56
Module I – 31
4.5
Analyze the behaviour of cathode rays.
416 and 436
DME: Section 3.6
SQ: pp. 56-60
Key Concepts:
Cathode rays are electrons.
A negatively charged rod repels cathode rays.
A positively charged rod attracts cathode rays.
Thus, cathode rays are negatively charged and are electrons.
SAMPLE QUESTION
1.
Cathode ray tubes are used to observe the behaviour of cathode rays. Some of these
behaviours are described and illustrated below.
Which behaviour would suggest that cathode rays consist of negatively charged particles?
a) The cathode rays travel in a straight line.
fluorescent screen
small propeller
b) The cathode rays drive a small
propeller located in their path.
fluorescent screen
c) The cathode rays cause the shadow
of the object in their path to be
projected on the fluorescent screen.
positive charged plate
fluorescent
screen
d) The cathode rays are deflected
towards the positively charged plate.
negatively charged plate
Additional Questions:
SQ: Exercises p. 60
Module I – 32
4.6
Describe what radioactivity is.
436 only
DME: Sections 3.7, 3.8, and 3.9
SQ: pp. 61-64
Key Concepts:
RADIOACTIVITY
Some atoms are not stable and will undergo decay. That is, particles in the atom's nucleus will
fly off into the surroundings. These flying particles are called nuclear radiation.
There are three types of nuclear radiation.
Radiation Type
alpha particle
beta particle
gamma ray
Symbol
Charge
α
β
γ
2+
1–
0
If a mixture of all three types is exposed to an electric charge, the following will result:
(+)
β
γ
Source
(–)
α
PENETRATING POWER OF RADIATION
This is how hard it is to stop radiation.
•
•
alpha
beta
•
gamma
Gamma rays are hardest to stop.
Module I – 33
SAMPLE QUESTIONS
1.
The results of an experiment dealing with radioactivity are illustrated below.
negative plate
fluorescent screen
radioactive source
positive plate
What can you conclude from this experiment?
2.
a)
The atom consists of a nucleus and electrons.
b)
The alpha and beta radiation is electrically charged.
c)
The alpha and beta particles and the gamma rays have different masses.
d)
The alpha, beta, and gamma radiation can penetrate matter.
The illustration below shows radiation from a radioactive point source passing through an
electric field. Which of the following correctly describes the rays formed after the radiation
has passed through the electric field?
a)
Ray 1:
Ray 2:
Ray 3:
negatively charged gamma rays
neutral alpha particles
positively charged beta particles
b)
Ray 1:
Ray 2:
Ray 3:
negatively charged beta particles
neutral gamma rays
positively charged alpha particles
c)
Ray 1:
Ray 2:
Ray 3:
positively charged beta particles
neutral gamma rays
negatively charged alpha particles
d)
Ray 1:
Ray 2:
Ray 3:
negatively charged alpha particles
neutral beta particles
positively charged gamma rays
Module I – 34
3.
The apparatus illustrated below is used to study the behaviour of alpha, beta and gamma
radiation.
Wooden screen
+
Electric field
–
Radioactive source
The radiation that passed through the wooden screen was not deflected as it passed
through the electric field.
Which of the following types of radiation passed through the wooden screen?
a)
Alpha radiation
c)
Gamma radiation
b)
Beta radiation
d)
Alpha, beta and gamma radiation
Additional Questions:
SQ: Exercises p. 64
Module I – 35
4.7
To analyze Thomson's atomic model.
436 only
DME: Section 3.10
SQ: p. 65
While using the cathode ray tube, J. J. Thomson discovered the proton.
Protons are positively charged.
Electrons are negatively charged.
Atoms are electrically neutral since they have the same number of protons and electrons.
Using the above facts, he pictured the atom as being a sphere of positive charge with electrons
embedded in it like seeds in a watermelon.
positive sphere
negative electron
SAMPLE QUESTIONS
1.
Thomson's atomic model could account for only some of the facts about the behaviour of
matter.
Which of the following statements describe the behaviour of matter according to
Thomson’s model?
1.
Matter is electrically neutral.
2.
Matter emits radiation.
3.
Each element that makes up matter emits a unique spectrum of light.
4.
There are different isotopes of each element.
5.
An atom contains positive and negative charges.
6.
Static electricity is found in nature.
a)
1, 2, and 4
c)
2, 3, and 4
b)
1, 5, and 6
d)
3, 5, and 6
Module I – 36
2.
Which of the following are characteristics of Thomson's atomic model?
1. Electrons revolve around the nucleus.
2. The mass of an atom is concentrated in the nucleus.
3. Atoms can lose or gain electrons.
4. An atom is almost completely empty.
5. An atom is a positively charged sphere containing negative particles.
6. Electrons move within energy levels.
a)
1 and 3
c)
3 and 5
b)
2 and 4
d)
4 and 6
Module I – 37
4.8
To analyze Rutherford's atomic model.
436 only
DME: Sections 3.11 and 3.12
SQ: pp. 66-70
Key Concepts:
RUTHERFORD'S SCATTERING EXPERIMENT
The Experiment
Rutherford fired a beam of alpha particles at a piece of thin gold foil and detected the results on
a screen.
The Results
•
•
•
•
Most alpha particles went through undeflected (they missed).
Some alpha particles were slightly deflected (a glancing hit).
A very few alpha particles bounced back (a direct hit).
The nucleus was discovered.
Nucleus
•
Since most alpha particles passed through the foil, the nucleus must be a very small
part of the atom.
•
Since the nucleus repelled positive alpha particles, it must be positive. Therefore,
protons are in the nucleus.
•
Since heavy alpha particles bounced, the nucleus must be very dense.
•
When different elements other than gold were used, Rutherford found that atoms of
different elements contain different amounts of positive charge. Therefore, the number
of protons must be different for each element.
RUTHERFORD'S ATOM
A small dense positive nucleus surrounded by a cloud of electrons.
negative electron cloud
+
positive nucleus
Module I – 38
SAMPLE QUESTIONS
1.
2.
Which of the following is a characteristic common to both the Thomson and the Rutherford
models of the atom?
a)
The atom is made up of positive and negative charges.
b)
The negative charges are evenly distributed throughout the atom.
c)
The electrons revolve around the nucleus.
d)
The nucleus of atoms is made up of protons and neutrons.
Rutherford modified the atomic model after doing experiments where alpha particles were
dispersed by a sheet of gold foil.
Consider the following statements:
1. The number of protons equals the number of electrons.
2. Protons are concentrated in a small positive space at the center of the atom.
3. Atoms consist mostly of empty space.
4. Electrons are contained in a positive sphere made up of protons.
5. Electrons move about in specific energy levels (shells).
Which of these statements are based on Rutherford's experiments?
3.
a)
1 and 2
c)
2 and 3
b)
1 and 4
d)
3 and 5
The following statements refer to atomic models.
1. There is a nucleus at the center of the atom.
2. The nucleus is very small compared with the size of the entire atom.
3. The electrons are located in energy levels around the atom.
Which of the statements above apply to Rutherford's atomic model?
4.
a)
1 and 2
c)
1 and 3
b)
1, 2, and 3
d)
2 and 3
Several scientists have proposed a model to describe the structure of the atom. Which of
the following is the description of Rutherford's model?
a)
An atom is composed of evenly distributed negative and positive charges.
b)
An atom is composed of positive particles concentrated in a nucleus and
negative particles moving within different energy levels.
c)
An atom is composed of very dense positive particles concentrated in a
nucleus and negative charges moving freely around that nucleus.
d)
An atom is indivisible and the atoms of the same element are all identical.
Additional Questions: SQ: Follow-Up p. 69
Module I – 39
4.9
To describe the simplified atomic model
currently in use (Bohr-Rutherford).
416 and 436
DME: Section 3.13
SQ: pp. 70-72
Key Concepts:
SUB-ATOMIC PARTICLES
Scientists now know that the atom is made of three kinds of sub-atomic particles and that atoms
differ in the number and arrangement of these particles.
Particle
Mass
electron
1/1837 amu
–
outside the nucleus
proton
1 amu
+
inside the nucleus
neutron
1 amu
0
inside the nucleus
Definition:
amu:
Charge
a tomic m ass
Location
u nit
NUCLEUS
•
•
•
•
•
•
•
contains protons and neutrons
it is positively charged (because of protons)
contains nearly all the mass of the atom (because protons and neutrons are much
heavier than electrons)
the number of protons (atomic number) determines the type of element the atom
makes up
the nucleus is much smaller than the atom and very dense
the number of protons in the nucleus is different for each element
the number of neutrons can differ for atoms of the same element (isotopes).
ELECTRON ORBITS (ENERGY LEVELS, SHELLS)
The fact that electrons are to be found in specific energy levels (electron shells) rather than a
diffuse cloud was discovered by Neils Bohr.
Energy levels differ in size and capacity.
The energy levels are numbered from 1 to 7 starting with the one closest to the nucleus.
Module I – 40
ENERGY LEVELS
•
Electrons in different orbitals have different energy levels.
•
Orbitals far from the nucleus have higher energy than those close to the nucleus.
•
The energy does not increase evenly but in discrete jumps.
•
When an electron absorbs a photon of light, it becomes excited and jumps to a higher
energy level.
•
An excited electron will emit a photon of light and fall back to a lower level.
CAPACITY OF ENERGY LEVELS
Energy levels differ in the number of electrons that can be present at any one time. Each level
has a maximum number of electrons that can be present.
Energy Level
Capacity
1
2 electrons
2
8 electrons
3
18 electrons
4
32 electrons
The above is a maximum. However, in a given atom, some levels will be full, some empty, and
some partially full.
FILLING OF ENERGY LEVELS (first 20 elements)
Electrons "prefer" to be in the lowest energy level. However, since the lowest level can only
hold 2 electrons, the majority will be in higher levels.
Calcium (atomic number 20)
2 electrons
8 electrons
8 electrons
2 electrons
-
Atoms put 2 electrons in level 4 before filling level 3 completely.
Module I – 41
level 1
level 2
level 3
level 4
CALCULATION OF THE NUMBER OF ATOMIC PARTICLES
Protons:
The number of protons is equal to the atomic number.
Electrons: In a neutral atom, the number of electrons is equal to the number of
protons and thus the atomic number.
Neutrons: The number of neutrons is equal to the atomic mass minus the atomic
number.
Examples:
Carbon atom:
Atomic number
Atomic mass
Protons
Electrons
Neutrons
-
6
12
6
6
12 – 6 = 6
2 electrons, level 1
4 electrons, level 4
6 total
6 p+
6n
2 e-
Sodium atom:
Atomic number
Atomic mass
Protons
Electrons
Neutrons
Rest of levels empty and not shown.
4 e-
-
11
23
11
11
23 – 11 = 12
2
8
1
11
11 p+
12 n
electrons, level 1
electrons, level 2
electron, level 3
total
2 e- 8 e- 1 e-
The following notations are used to show the atomic number and atomic mass number of an
element:
12
6
C
C126
12
C6
C12
6
No matter how it is written, the larger number is the mass number, and the smaller number is
the atomic number.
Module I – 42
SAMPLE QUESTIONS
1.
Which of the following characteristics describe an atom in terms of the simplified (BohrRutherford) model?
1. The number of electrons is equal to the number of protons.
2. The number of protons is equal to the number of neutrons.
3. The nucleus is made up of neutrons, protons and electrons.
4. The nucleus is made up of neutrons and electrons.
5. The nucleus is made up of protons and neutrons.
6. Protons revolve around the nucleus.
7. Electrons revolve around the nucleus.
2.
a)
2, 5, and 7
c)
1, 2, and 3
b)
1, 4, and 6
d)
1, 5, and 7
Which of the following best represents the Bohr-Rutherford model of a potassium atom,
K39 ?
19
a)
c)
19 p+
19 p+
39 n0
20 n0
b)
2 e- 8 e- 8 e- 1 e-
d)
2 e- 8 e- 8 e- 1 e-
19 p+
19 p+
20 n0
39 n0
2 e- 8 e
9 e-
2 e- 8 e
Module I – 43
9 e-
3.
The study of the behaviour of matter has made it possible to develop simple models such
as the Bohr-Rutherford model of the atom.
If the atomic number of oxygen is 8 and its mass number is 16, which diagram represents
the oxygen atom according to the Bohr-Rutherford model?
a)
c)
2 e-
6 e-
6 e-
b)
d)
8 e-
4.
2 e-
16 e-
The atomic number of fluorine (F) is 9 and its mass number is 19.
Which of these diagrams represents the simplified model (Bohr-Rutherford) of a fluorine
atom?
a)
c)
9 p+
10 p+
10 n
19 n
2 e-
7 e-
2 e-
b)
8 e-
d)
9 p+
19 p+
10 n
10 n
2 e-
2 e- 8 e- 8 e- 1 e-
8 e-
Module I – 44
5.
Which of the following diagrams best represents the Bohr-Rutherford model of the
phosphorus atom (P)?
a)
c)
15 p+
31 n
31 p+
15 n
b)
d)
16 p+
15 n
6.
15 p+
16 n
Which of the following diagrams best represents the aluminum (Al) atom according to the
Bohr-Rutherford model?
a)
c)
12 p+
27 n
27 p+
13 n
3 e- 8 e- 3 e-
2 e- 8 e- 17 e-
b)
d)
13 p+
14 n
16 p+
13 n
2 e- 8 e- 3 e-
2 e- 3 e- 8 e-
Module I – 45
5.1
To describe the progression of the atomic
masses of the elements in the periodic table
416 and 436
DME: Section 4.1
SQ: pp. 76-78
Key Concepts:
1. The mass of a single atom is the sum of the protons and neutrons in its nucleus
expressed in atomic mass units (u or amu).
2. As the atomic number of an element increases, so does its atomic mass.
3. The atomic masses increase from left to right in a period because the number of
protons increases from left to right.
4. The atomic mass increases as one goes down a family or group because the number of
protons and neutrons increases from top to bottom.
SAMPLE QUESTION
1.
2.
In any one period of the periodic table, from left to right:
a)
the mass of the atoms increases
b)
the mass of the atoms decreases
c)
the mass of atoms does not change as all of the elements are in the same
period
d)
the mass of the atoms first decreases then increases.
In general, in the periodic table, the atomic masses of elements increase as their atomic
numbers increase. There are, however, exceptions to this rule.
Which of the following accounts for these exceptions?
a)
The number of electrons increases irregularly.
b)
The number of particles in the nucleus increases regularly.
c)
The number of neutrons increases irregularly.
d)
The number of protons increases regularly.
Additional Questions:
SQ: Follow-Up p. 77
Module I – 46
5.2
To identify the advantages and
disadvantages of using isotopes in industry,
medical science , basic research and in the
environment.
416 and 436
DME: Sections 4.2 to 4.5
SQ: pp. 79-83
Key Concepts:
1.
Isotopes are atoms of the same element. They have the same atomic number (that is, the
same number of protons ) but they have a different atomic mass (that is, a different
number of neutrons).
2.
Deuterium (2 u ) and tritium (3 u ) are isotopes of the element hydrogen and have 1 and 2
neutrons respectively.
3.
There are both natural and artificial isotopes.
4.
Natural isotopes are found in the environment. Most are stable but some are unstable.
These unstable isotopes disintegrate radioactively and are called radioisotopes. Some
radioisotopes were created during the formation of the earth and some by the
bombardment of cosmic rays in the atmosphere.
Example: uranium, thorium, radon gas
5.
Artificial isotopes are created inside nuclear reactors by bombarding atoms with other
atoms or parts of atoms. Many of these isotopes are unstable and disintegrate releasing a
great deal of radiation. These isotopes are also called radioisotopes.
6.
Radioisotopes release alpha and beta particles as well as gamma radiation and a lot of
energy.
7.
Radioactive isotopes have many beneficial uses in:
• medicine — cancer therapy
• industry — detection of welding flaws, nuclear power plants, irradiation of
gemstones, irradiation of food
• research — radioactive tracers
• the environment — insect control, carbon -14 dating
8.
The radiation from nuclear accidents or nuclear waste material can cause a variety of
illnesses depending on the amount and duration of the exposure.
Example: death, diminished immunity, cancer, mutations, sterility etc.
Module I – 47
SAMPLE QUESTIONS
1.
Which of the 5 hypothetical atoms are isotopes of the same element?
40
V20
2.
3.
40
W22
42
X20
80
Y42
40
Z22
a)
V and W
c)
X and Z
b)
W and Z
d)
V and X
What characterizes artificial isotopes ?
a)
They are stable.
b)
They are not dangerous.
c)
They are radioactive.
d)
They are formed by chemical reaction.
Which element supplies the radioactive isotope that is used to treat or study the thyroid
gland ?
a)
Iodine
c)
Calcium
b)
Thorium
d)
Strontium
Module I – 48
5.3
To analyze the irregularities in the
progression of the atomic masses of the
elements in the periodic table.
436 only
DME: Section 4.6
SQ: pp. 83-84
Key Concepts:
1.
The atomic mass number of an element is not a whole number since it is the weighted
average of the masses of the different isotopes of that element.
2.
All atoms of the same element have the same number of protons in their nuclei but the
number of neutrons may vary. Since each neutron has an atomic mass of 1 u this will result
in atoms with different atomic masses.
3.
The atomic masses of atoms do not increase regularly as their atomic numbers increase
because their neutrons increase irregularly.
4
The relative abundance of each isotope varies and is expressed as a % .
5.
The atomic mass of the element lithium can be calculated as follows:
Li- 6
Li- 7
mass =
6u
7u
7.42%
92.58%
(6 u X 7.42 ) + ( 7 u X 92.58 )
100
= 6.93 u
Module I – 49
SAMPLE QUESTIONS
1.
Calculate the atomic mass of element X from the following table.
isotope
1
2
2.
mass
number
relative
abundance
63 u
65 u
69.1%
30.9%
a)
63.54 u
c)
64.00 u
b)
63.62 u
d)
64.38 u
In nature the % abundance of isotopes of an element is very unequal. For example, the
following table represents the isotopic composition of oxygen.
mass number
% abundance
16
17
18
99.772
0.038
0.200
The atomic mass of oxygen is practically the same as one of its isotopes. Which is it and
why?
3.
The following table lists the characteristics of element A.
isotope
atomic
number
mass
number
relative
abundance
1
2
3
22
22
22
45 u
46 u
47 u
10%
75%
15%
Calculate the atomic mass of element A.
a)
22.00 u
c)
46.05 u
b)
46.00 u
d)
47.90 u
Module I – 50
4.
The atomic mass varies irregularly from one element to the next in the periodic table.
Which three statements explain why this is so?
1.
2.
3.
4.
The number of neutrons may vary irregularly from one element to the next.
The number of isotopes may vary from one element to the next.
The relative abundance of isotopes may vary from one element to the next.
Some elements have radioactive isotopes, while others do not have any at all.
a)
1, 2 and 3
c)
1, 3 and 4
b)
1, 2 and 4
d)
2, 3 and 4
Module I – 51
5.4
To locate metals , nonmetals and metalloids
in the periodic table
416 and 436
DME: Section 4.7
SQ: pp. 85-88
Key Concepts:
1.
Matter can be identified on the basis of characteristic properties.
Example: colour, density, solubility, melting point, boiling point, electrical or
thermal conductivity, etc.
2.
The elements can be classified as metals , nonmetals and metalloids on the basis of their
characteristic properties.
METALS
NONMETALS
-
lustre
conductors of heat and electricity
ductile (drawn into wire)
malleable (hammered into
different shapes)
- most metals are solids at room
temperature (except mercury)
- dense
- release gas in presence of an acid
-
dull
poor conductors
nonductile
nonmalleable
- gases, liquids, solids
- low densities
- no such reaction
3.
Metalloids (or semimetals) have some properties of metals and some of nonmetals.
4.
Metals are located to the left of the heavy steplike line in the periodic table.
Example: Na, Ba, Fe, Ag, Al
5.
Nonmetals are located to the right of the heavy steplike line in the periodic table.
Example: N, S, Cl, Ar
6.
Metalloids are located on either side of the heavy steplike line in the periodic table.
Example: B, C, Si, Ge, As, Sb, Te, Po, At
Module I – 52
SAMPLE QUESTIONS
1.
2.
3.
4.
5.
Which of the following is a characteristic property of metals?
a)
They are poor conductors of heat.
b)
They are all solids at room temperature.
c)
They generally have a high fusion temperature.
d)
They are ductile and malleable.
Which of the following is a nonmetal?
a)
Fe
c)
Cu
b)
Cl
d)
Ca
What is being separated by the solid jagged line in the periodic table?
a)
solids and gases
c)
metals and nonmetals
b)
liquids and gases
d)
stable and radioactive elements
What are the characteristics of metalloids?
a)
They are gases with metallic properties at room temperature.
b)
They have some properties of metals and some properties of nonmetals.
c)
They are chemical compounds.
d)
They are found at the bottom of the periodic table.
A student is given an unknown solid substance belonging to family II A (family 2). To
determine the nature of this substance, he conducts the tests listed in the following table.
TEST
EXPECTED RESULT
- Test to determine how it reacts
with a strong acid
- Electrical conductivity test
- Thermal conductivity test
- Malleability test
Give the expected result for each test.
Additional Questions: SQ: Follow-Up p. 86 and Exercises p. 88
Module I – 53
5.5
To locate the family of alkali metals ,
alkaline earth metals, halogens and inert
gases on the periodic table.
416 and 436
DME: Section 4.8 and 4.9
SQ: pp. 89-96
Key Concepts:
1.
Certain elements have similar chemical properties and are grouped into chemical families
or groups.
2.
A chemical family or group is represented on the periodic table in a vertical column.
3.
The 8 main groups are designated by Roman numerals such as II, IV and are often
referred to as the A group.
4.
The alkali metals are in the first vertical column (group IA ) of the periodic table. They
include:
Li, Na, K, Rb, Cs, Fr.
5.
The alkaline earth metals are in the second vertical column ( group IIA ). They include:
Be, Mg, Ca, Sr, Ba, Ra.
6.
The halogens are located in the seventh vertical column ( Group VIIA ). They include:
F, Cl, Br, I , At.
7.
The inert or noble gases are in the last vertical column of the periodic table (Group VIIIA ).
They include:
He, Ne, Ar, Kr, Xe, Rn.
SAMPLE QUESTIONS
1.
To which chemical family do the following elements belong?
Be, Sr, Ra
a)
inert gases
c)
alkaline earth
b)
halogens
d)
alkali metals
Module I – 54
2.
3.
4.
5.
Which element is in the same family as Ar and Kr ?
a)
I
c)
N
b)
Ne
d)
Co
Which of the following series of elements represents the halogen family?
a)
Fe, Co, Ni, Cu, Zn
c)
Li, Na, K, Rb, Cs
b)
F, Cl, Br, I, At
d)
Li, Be, C, N, O
From left to right, what names are given to the shaded columns in the periodic table ?
a)
alkali metals, alkaline earth metals, halogens, inert gases
b)
inert gases, alkaline earth metals, alkali metals, halogens
c)
halogens, alkaline earth metals, alkali metals, inert gases
d)
halogens, alkali metals, inert gases, alkaline earth metals
From left to right in the periodic table, the elements are found in the following order:
a)
inert, nonmetal, metalloid, metal
b)
nonmetal, inert, metalloid, metal
c)
metal, nonmetal, metalloid, inert
d)
metal, metalloid, nonmetal, inert
Additional Questions:
SQ: Exercises p. 95
Module I – 55
5.6
To describe the progression of certain
properties of elements in a given period on
the basis of their atomic number.
436 only
DME: Section 4.10
SQ: pp. 96-99
Key Concepts:
1.
Each element of the periodic table is numbered in sequence using its atomic number.
2.
Each horizontal row in the periodic table is called a period.
3.
The relationship between atomic number and the value of a particular property
(e.g. atomic radius, electronegativity) can be graphed.
4.
When the unit values of a particular property are plotted along the y-axis of a graph and
the atomic numbers of the elements along the x-axis, periodic trends are observed.
DME: pp. 105-106
SQ: Figure 1 – 5.24 on p. 97
5.
Within a period , from left to right, the following are observed:
• atomic radius
• ionization energy
• density
–
–
–
•
•
•
•
–
–
–
–
conductivity
melting point
boiling point
electronegativity -
decreases
increases irregularly
increases and then decreases with a peak in
the middle
highest for the metals
peaks in the middle
peaks in the middle
increases
Module I – 56
SUMMARY OF TRENDS
The arrows in the following graphs indicate the direction of increase for each property by
period:
A)
Electronegativity
B)
Ionization energy
C)
Boiling point
D)
Melting point
E)
Atomic radius
SAMPLE QUESTIONS
1.
What can be said of the atomic radius within a period ?
a)
It increases in the elements from left to right.
b)
It decreases and then increases in the elements from left to right.
c)
It is a constant in the elements.
d)
It decreases in the elements from left to right.
Module I – 57
2.
What can be said of the melting point within a cycle of the periodic table?
a)
It is higher for the alkali metals than for the inert gases
b)
It increases from left to right
c)
It is a constant for all the elements
d)
It is lower for the alkali metals than for the inert gases.
Atomic Number vs Melting Point
H
3.
He Li Be
B C
N O
F Ne Na Mg Al Si
P S
Cl
Ar K
Ca Sc Ti
V Cr
Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
The properties of the elements in the periodic table vary from one element to another. Four
of these variations are:
1.
increase in electrical conductivity
2.
increase in chemical activity
3.
increase in atomic radius
4.
increase in metallic lustre
Which of these variations occur(s) as one goes from one element with a lower atomic
number to one with a higher atomic number within the same period?
4.
a)
1
c)
2 and 4
b)
1 and 3
d)
1, 2, and 3
The properties of the elements in the periodic table vary from one element to another. Four
of these variations are:
1.
increase in electrical conductivity
2.
increase in chemical activity
3.
increase in atomic radius
4.
increase in metallic lustre
Which of these variations occur(s) as one goes from one element with a lower atomic
number to one with a higher atomic number within the same period?
a)
1 and 2
c)
2 and 4
b)
1 and 3
d)
3 and 4
Module I – 58
5.
The following graph shows the ionization energies of certain elements as a function of
their atomic numbers:
According to this graph, which of the following statements is TRUE?
a)
Within a period, the ionization energy usually increases as the atomic number
increases.
b)
Within a period, the ionization energy usually decreases as the atomic number
increases.
c)
In general, the ionization energy of the elements in Period 3 is greater than the
ionization energy of the elements in Period 2.
d)
The ionization energy of the elements in Period 4 varies regularly when the
atomic number increases regularly.
Module I – 59
5.8
To justify the classification of alkali metals,
alkaline earth metals, halogens, and inert
gases in the periodic table on the basis of
the simplified atomic model currently in
use.
416 and 436
DME: Section 4.13
SQ: pp. 103-105
Key Concepts:
1.
The chemical properties of an atom are determined by the atom's outermost electrons
which are called valence electrons.
2.
The elements of the same chemical group or family all have the same number of valence
electrons.
3.
The group number of each chemical family also corresponds to the number of valence
electrons each atom in the group has.
4.
Chemical reactivity of elements is determined by their valence electrons.
5.
For metals, reactivity increases from right to left in a period and increases from top to
bottom in a family.
For nonmetals, reactivity increases from left to right in a period and increases from bottom
to top in a family .
Example:
6.
most reactive metal
most reactive nonmetal
– Francium
– Fluorine
In chemical reactions , atoms tend to become as stable as possible. They try to acquire the
electron configuration of the closest inert gas in the periodic table by either losing or
gaining electrons.
Example:
Na loses 1 electron to have the configuration of Ne
Cl gains 1 electron to have the configuration of Ar
7.
Metals lose electrons.
8.
Nonmetals tend to gain electrons.
9.
The inert gases are nonreactive since they have an octet number of electrons (8).
Module I – 60
SAMPLE QUESTIONS
1.
2.
3.
4.
In the third period, which metal is the most reactive ?
a)
Al
c)
Cl
b)
Mg
d)
Na
What is a characteristic of an element possessing 8 valence electrons
a)
It easily captures electrons
b)
It is very reactive
c)
It easily gives up an electron
d)
It is chemically inert.
Which of the following elements is most likely to gain an electron ?
a)
Li
c)
Ne
b)
F
d)
B
Five elements are identified in the following periodic table:
IA
1
VIII A
18
II A
2
III A IV A V A VI A VII A
13
14 15 16 17
5
B
10.81
8
O
16.00
18
Ar
39.95
11
Na
22.99
20
Ca
40.08
Each of the following characteristics describes one of these elements.
1.
Its outermost energy level contains 6 electrons.
2.
It is an inert gas that does not react with metals or nonmetals.
3.
It contains 1 more proton than an alkaline earth metal.
4.
It is a metal that reacts vigorously with water.
5.
It has electrons located in 4 energy levels.
Match each of these five elements with the appropriate characteristic. Write the chemical
symbol of the element beside the corresponding characteristic.
Module I – 61
5.
Using the information below as well as the periodic table, give the chemical symbol of
each of the four unknown elements.
Element
Number
of valence
electrons
1
1
2
1
3
7
Number
of protons
Other
characteristics
1
11
3
4
6.
Number
of energy
levels
2
Reacts vigorously
with water
Potassium, K, and calcium, Ca, are located next to each other in the periodic table of
elements.
These two elements belong to the same period, but not to the same family.
Explain why potassium and calcium belong to the same period, but not to the same family.
Additional Questions: SQ: Follow-Up p. 104
Module I – 62
5.9
To justify the structure of the periodic table
on the basis of the simplified atomic model
currently in use.
416 and 436
DME: Section 4.13
SQ: pp. 106-108
Key Concepts:
1.
The atom has a nucleus, core electrons and the valence electrons.
2.
Using the 2n2 rule (2 x level number2), we can determine the maximum number of
electrons in each energy level.
• first level
• second level
• third level
3.
– 2
– 8
– 18
The following elements have the following atomic structure:
He
4.
N2
Ar
Each energy level in an atom corresponds to one of the periods in the periodic table.
energy level 1
energy level 2
energy level 3
=
=
=
period 1
period 2
period 3
11p
12n
2e-
8e- 1e-
Module I – 63
SAMPLE QUESTIONS
1.
2.
Which element corresponds to each of the definitions below?
a)
The element has electrons in 2 energy levels and the outer level is full.
b)
The element has electrons in 3 energy levels and it has 2 valence electrons.
c)
The element has an atomic mass of 28 and its nucleus contains 14 neutrons.
d)
The element reacts vigorously with water and the electric charge of its nucleus
is +19 .
Which element of the second period has 5 valence shell electrons?
a)
N
b)
B
c)
O
d)
3.
How many layers of electrons does an element in the third period have?
a)
3
4.
Which of the following is the electron configuration of an alkali metal?
5.
b)
4
c)
C
1 d) 2
a)
) ) )
2e- 8e- 7e-
c)
) )
2e- 4e-
b)
) ) )
2e- 8e- 8e-
d)
) ) )
2e- 8e- 1e-
Which of the following represents the electron configuration of a halogen?
a)
b)
) ) )
2e- 8e- 7e-
)
) )
2e- 8e- 8e-
Module I – 64
c)
) )
2e- 4e-
d)
) ) )
2e- 8e- 1e-
6.1
To illustrate the structure of a water
molecule after conducting a laboratory
experiment in accordance with the
suggested procedure.
436 and 436
DME: Section 5.1
SQ: pp. 114-116
Key Concepts:
1.
The electrolysis of water results in the formation of 2 gases.
2.
The gas that accumulates at the cathode (negative electrode) is hydrogen because it pops
in the presence of a lighted splint.
3.
The gas that accumulates at the anode (positive electrode) is oxygen because it rekindles a
glowing splint.
4.
The ratio of the volume of hydrogen to oxygen is 2 : 1
5.
The compound water decomposes into its elements hydrogen and oxygen.
6.
A molecule of water is made up of 2 hydrogen atoms and 1 oxygen atom.
7.
The molecular formula for water is H2O.
SAMPLE QUESTIONS
1.
Which of the following best represents the electrolysis of water experimental results?
-
+
a)
-
+
-
b)
+
c)
Module I – 65
-
+
d)
2.
Pick out the correct statements
1. A molecule is the smallest part of a chemical compound.
2. The written depiction of a molecule is called a symbol.
3. A molecule is formed by the bonding of atoms.
4. A molecule contains two or more atoms.
3.
a)
1, 2
c)
1, 2, 3, 4
b)
3, 4
d)
1, 3, 4
What is the role of the electric current in the electrolysis of water ?
a)
It charges up the water.
b)
It decomposes the water.
c)
It evaporates the water.
d)
It combines with the sulfuric acid.
Module I – 66
6.2
To verify the molecular formula for water by
referring to the periodic table.
436 only
DME: Sections 5.2 and 5.3
SQ: pp. 117-120
Key Concepts:
1.
An atom with 8 electrons in its outer energy level is chemically stable and unreactive. This
is known as the Octet Rule.
2.
The number of bonds that an atom of an element (metal or nonmetal) can form is called
the valence of the element.
3.
The valence of an element in Family 1, 2 or 3 is the same as the Family or Group number.
(Hydrogen = 1)
The valence of an element in Family 4, 5, 6 or 7 is equal to 8 minus the Family number.
(Oxygen = 2)
4.
Oxygen can form 2 bonds and would therefore combine with 2 hydrogen atoms to form a
molecule of water.
SAMPLE QUESTIONS
1.
2.
3.
What is the most probable formula for water ?
a)
HO
c)
H2O
b)
HO2
d)
O2H
In what ratio do hydrogen and oxygen combine ?
a)
1 to 1
c)
1 to 2
b)
2 to 1
d)
2 to 2
How many bonds do group 6A atoms make when forming compounds?
a)
6
c)
4
b)
2
d)
8
Module I – 67
6.3
To write the molecular formula of
substances composed of two types of
elements based on the group numbers of the
representative elements of the periodic table
436 only
DME: Section 5.4
SQ: pp. 117-120
Key Concepts:
1.
Binary compounds contain only 2 elements.
2.
Use the Cross-over Rule to write the formulas of binary compounds.
Example 1
a)
write the symbols of the two elements
placing the metal first
Mg O
Al S
write the valence of each element
as a superscript
Mg2 O2
Al3 S2
c)
cross over the valences
Mg2 O2
Al2 S3
d)
divide the valences by the
greatest common factor
Mg1 O1
Al2 S3
e)
drop any subscript that is a 1
MgO
Al2 S3
b)
3.
Example 2
Domestic, industrial and pharmaceutical products with common names have a chemical
name and a molecular formula.
Example:
peroxide is the common name for hydrogen peroxide (H2O2)
Module I – 68
SAMPLE QUESTIONS
1.
2.
3.
4.
What is the most probable formula for boron and fluorine ?
a)
BF
c)
BF3
b)
B3 F
d)
B3 F7
If we had a compound with formula X2O to which group would element X belong?
a)
halogen
c)
noble gas
b)
alkali
d)
alkaline earth
Which formulas are probably correct?
1.
HCl
2.
NaI
3.
CO2
4.
H2 F
a)
1, 4
c)
1, 2, 3
b)
1, 2
d)
2, 3, 4
An element X from Group III A (13) reacts with an element Z from Group VI A (16).
What is the molecular formula of the resulting compound?
Explain your answer by taking into account bonding or valence electrons.
Additional Questions: SQ: Follow-Up p. 120
Module I – 69
6.4
To illustrate the structure of the molecule of
a pure substance, given its chemical name
436 and 436
DME: Section 5.4
SQ: pp. 121-124
Key Concepts:
1.
Binary compounds composed of a metal and a nonmetal are named as follows:
name of metal + name of nonmetal with suffix -ide
Example:
sodium chloride
NaCl
DO NOT USE PREFIXES!
2.
Binary compounds composed of 2 nonmetals are named as follows:
(prefix) nonmetal + (prefix) nonmetal with suffix -ide
The following prefixes are used:
1.
mono4.
2.
di5.
3.
tri6.
tetrapentahexa-
Note 1: the most metallic nonmetal is written first
Note 2: the prefix “mono-” is not used for the first nonmetal
Example:
CO
N2O3
carbon monoxide
dinitrogen trioxide
SAMPLE QUESTIONS
1.
What is the name of the compound with the following formula?
Na2S
a)
disodium sulfate
c)
sodium sulfide
b)
disodium sulfide
d)
sulfur disodium
Module I – 70
2.
3.
4.
What is the formula for aluminum oxide?
a)
Al2O3
c)
Al3O2
b)
O3Al2
d)
O2Al3
What is the name for the compound having the formula CF4 ?
a)
tetrafluoro carbide
c)
tetrafluoride carbon
b)
carbon fluoride
d)
carbon tetrafluoride
What is the formula for dihydrogen monoxide ?
a)
H2O
c)
HO2
b)
H2O2
d)
HO
Additional Questions: SQ: Follow-Up p. 122
Module I – 71
MODULE II
– ELECTRICAL PHENOMENA –
2.1
To distinguish magnetic substances from
ferromagnetic, and non-magnetic
substances.
416 and 436
DME: Sections 6.2 and 6.3
SQ: pp. 144-146
Key Concepts:
1.
There are 3 categories of substances:
• magnetic
• ferromagnetic
• non-magnetic
2.
Every magnet has two poles:
North (N) and South (S)
3.
Like poles repel.
N
(Repulsion)
N
S
(Repulsion)
S
4.
5.
N
S
Unlike poles attract.
S
(Attraction)
N
N
(Attraction)
S
As the distance between poles increases, the magnetic force decreases.
Strong
Attraction
N
N
S
Weak
Attraction
Module II – 1
S
Magnetic Substance
• any material which acts like a magnet,
(can be attracted and repelled by another magnet)
Examples: loadstone, bar magnet.
Ferromagnetic Substance
• any material which is strongly attracted by a magnet.
It can be magnetized. It must contain at least one of the
following: nickel, cobalt, or iron.
Examples: steel, nichrome, iron
Non-magnetic Substance
• any material which is not attracted or repelled by a magnet.
A magnet will not affect it.
Examples: plastic, wood, glass, aluminum, paper, silver...
SAMPLE QUESTIONS
1.
In the laboratory, you are given three different substances:
1.
a magnetic substance
2.
a ferromagnetic substance
3.
a non-magnetic substance
You bring these substances close to one another and note your observations.
Which of the following observations is correct?
a)
Substances 1 and 2 repel each other.
b)
Substances 1 and 2 attract each other.
c)
Substances 1 and 3 attract each other.
d)
Substances 2 and 3 attract each other.
Module II – 2
2.
You have two ten-cent coins, one from 1965 and the other from 1994.
To determine whether these coins are magnetic, ferromagnetic or non-magnetic, you
conduct tests and note your observations.
Step
Test
Observation
1
Bring a magnet near the 1965 coin.
No reaction
2
Bring a magnet near the 1994 coin.
Attraction
3
Bring each coin near an iron nail.
No reaction
Given these observations, what can you say about these coins?
a) The 1965 coin is non-magnetic and the 1994 coin is magnetic.
b) The 1965 coin is non-magnetic and the 1994 coin is ferromagnetic.
c) The 1965 coin is magnetic and the 1994 coin is ferromagnetic.
d) The 1965 coin is ferromagnetic and the 1994 coin is magnetic.
3.
Four circular pieces of metal were brought close to one another during a laboratory
experiment. Only one of these pieces of metal is a magnet.
The following table shows whether or not these pieces of metal attracted on another.
Attraction
Yes No
Combination
Piece 1 with piece 2
Piece 1 with piece 3
Piece 1 with piece 4
Piece 2 with piece 3
Piece 2 with piece 4
Piece 3 with piece 4
√
√
√
√
√
√
Which piece of metal is a magnet?
a)
Piece 1
c)
Piece 3
b)
Piece 2
d)
Piece 4
Module II – 3
4.
Four objects W, X, Y, and Z were brought close together two at a time. One of these
objects is magnetic, one is nonmagnetic and two are ferromagnetic. The results were as
follows:
OBJECTS
RESULT
W and X
X and Y
W and Y
X and Z
The objects attracted each other.
The objects attracted each other.
Nothing happened.
Nothing happened.
Which of these objects is nonmagnetic?
a)
W
c)
Y
b)
X
d)
Z
Additional Questions:
DME: p. 165
SQ: Exercises p. 150 and p. 174, #1-4
Module II – 4
2.2
To map the magnetic fields of magnetized
objects.
416 and 436
DME: Section 6.6 and p. 154 “Extensions”
SQ: pp. 146-150
Key Concepts:
1.
All magnets have a magnetic field. A magnetic field is the space around a magnet where
magnetic forces are felt (both attraction and repulsion).
2.
Lines of Force show you the shape, direction, and strength of the magnetic field around a
magnet.
Magnetic Field
around a Bar Magnet
SHAPE is shown by lines of force which can be
straight, curved, circular, etc.
DIRECTION is shown by arrowheads. The
direction is always from North to South.
N
S
STRENGTH is shown by how close the lines
are to each other. The closer the lines of force
are, the stronger the magnetic field.
Remember: Lines of force never cross!
line of force
Magnetic field is weak here.
(Lines are far apart.)
Magnetic field is strong here.
(Lines are close together.)
Magnetic Field
around the Earth
S m = South magnetic pole
N m = North magnetic pole
Module II – 5
Nm
The Earth
Sm
3.
A compass can be used to determine the direction of the magnetic field lines around a
magnetized object.
→N
N
N
N
S
N
N
→N
4.
Iron filings can also be used to determine the shape of a magnetic field
SAMPLE QUESTIONS
1.
Two magnets are placed end to end.
Which diagram correctly illustrates the magnetic fields around these magnets?
a)
c)
N
S
N
S
N
S
N
S
b)
N
S
N
S
N
S
N
S
d)
Module II – 6
2.
A straight magnet always produces an external magnetic field.
In which of the following diagrams is this magnetic field correctly represented?
a)
c)
N
3.
S
N
S
S
b)
d)
N
N
S
Two magnets are placed side by side.
Which of the following diagrams correctly represents the magnetic fields produced by
these magnets?
a)
c)
N
S
N
S
b)
S
N
S
N
N
S
S
N
d)
S
N
N
S
Additional Questions:
DME: p. 165, #10, 17
SQ: Exercises p. 150, #2-5 and p. 174, #5, 6
Module II – 7
Electron Flow (DME Version)
2.3
To map the magnetic fields around a current
bearing:
a) straight line conductor
b) solenoid
416 and 436
DME: Lab. 6.7, 6.8, pp. 153-156
Key Concepts:
1.
Straight line conductors (wires which have a current flowing through them), also have a
magnetic field around them. (Diagram A, below.)
2.
Straight line conductors have circular lines of force. We determine the direction using the
Left Hand Rule. Your thumb points in the direction of the electron flow and your fingers
wrap in the direction of the magnetic field lines.
See DME: p. 154.
Diagram A
Magnetic Field around a
Straight Line Conductor
–
Diagram B
Magnetic Field around a Solenoid
+
e-
–
Magnetic field lines are
circular around a straight line
+
e-
3.
Solenoids (coiled wires which have a current flowing through them), also have a magnetic
field around them. (Diagram B, above.)
4.
Solenoids have magnetic fields which look like the magnetic field around a bar magnet.
Again, we determine the direction using the Left Hand Rule but there is an important
difference from above. Your fingers wrap in the direction of the electron flow and your
thumb points in the direction of the magnetic field lines (points N).
Module II – 8 a
Conventional Current (SQ Version)
2.3
To map the magnetic fields around a current
bearing:
a) straight line conductor
b) solenoid
416 and 436
SQ: pp. 151-156
Key Concepts:
1.
Straight line conductors (wires which have a current flowing through them), also have a
magnetic field around them. (Diagram A, below.)
2.
Straight line conductors have circular lines of force. We determine the direction using the
Right Hand Rule. If the thumb of your right hand points in the direction of the
conventional (positive charge) current flow, then the fingers of your right hand will wrap
around the wire in the direction of the magnetic field lines.
See SQ: p. 154.
Diagram A
Diagram B
Magnetic Field around a
Straight Line Conductor
Magnetic Field around a Solenoid
I (current direction)
–
+
e-
–
Magnetic field lines are
circular around a straight line
I (current direction)
e-
+
I (current direction)
3.
Solenoids (coiled wires which have a current flowing through them), also have a magnetic
field around them. (Diagram B, above.)
4.
Solenoids have magnetic fields which look like the magnetic field around a bar magnet.
Again, we determine the direction using the Right Hand Rule but there is an important
difference from above. Your fingers wrap around the coil in the direction of the current
flow and your thumb points in the direction of the magnetic field lines (points N).
Module
Module II
II -–8b
9
SAMPLE QUESTIONS
1.
2.
Nathalie wants to draw a sketch representing the magnetic field she observed around a
current-bearing solenoid she used in the laboratory.
Which sketch should she use?
a)
c)
b)
d)
Wire
A copper wire, with a current flowing through it,
passes through a piece of cardboard as shown in
the diagram to the right.
−–
++
A magnetic compass is placed on the piece
of cardboard near the wire. Which of the
following diagrams shows the direction in
which the compass needle will point?
Cardboard
−
−
a)
c)
N
W
W
E
S
S
+
b)
N
E
−
+
d)
S
E
−
E
W
N
N
S
W
+
+
Module II – 9
3.
An electric current flows through a solenoid.
Which diagram correctly illustrates the magnetic field produced by this solenoid?
4.
a)
c)
b)
d)
A compass is placed at one end of a solenoid.
In which illustration is the compass needle pointing in the right direction?
a)
b)
c)
d)
Additional Questions:
DME: p. 165, #33
SQ: Exercises pp. 155-156 and p. 174, #7, 8
Module II – 10
2.4
To demonstrate the effect of a core on the
behavior of an electromagnet.
416 and 436
DME: Section 6.9
SQ: pp. 156-158
Key Concepts:
1.
An electromagnet is created when an object is inserted into the centre of a current-bearing
solenoid. This object is called a core.
Core
2.
A ferromagnetic core will strengthen the magnetic field of an electromagnet. The
ferromagnetic core becomes magnetized when the current is turned on.
3.
The ferromagnetic cores you could use would be: iron, steel, nickel, cobalt. However,
only iron will demagnetize quickly once the current is turned off. Therefore, iron is the
most commonly used core in electromagnets.
4.
The strength of the magnetic field around an electromagnet can be measured by seeing
how many ferromagnetic objects it can pick up. The more it picks up, the stronger its
magnetic field.
Additional Questions:
DME: p. 157 (top)
SQ: Exercises p. 158 and p. 174, #9, 10
Module II – 11
2.6
To identify the factors that affect the
magnetic field of an electromagnet.
416 and 436
DME: Section 6.12
SQ: pp. 163-165
Key Concepts:
1.
The strength of an electromagnet is affected by these factors:
• the current intensity (I) in the coil of the solenoid
• the number of turns (loops) in the solenoid
• the core material.
2.
As the current intensity increases → strength of field increases.
As the number of turns increases → strength of field increases.
3.
If both current intensity and number of turns change simultaneously,
STRENGTH OF FIELD = (current intensity) x (number of turns)
Module II – 12
SAMPLE QUESTIONS
1.
Julie performed several experiments in the laboratory investigating the magnetic field
produced by a solenoid. She plotted the following four graphs:
Force
Force
Force
Force
Al
Current Intensity
Time
Number of loops
Fe
Cu
Nature of core
According to the graphs, which variables affect the strength of the magnetic field?
2.
a)
the current intensity, the time, the number of loops, and the nature of the core
b)
the time, and the number of loops only
c)
the current intensity, and the number of loops only
d)
the current intensity, the number of loops, and the nature of the core only
The diagrams below, illustrate electromagnets all consisting of the same core. One of
these electromagnets produces a magnetic field that is more intense than that of all the
others.
Which electromagnet is it?
10 turns
a)
10 turns
c)
I=5A
I=2A
5 turns
b)
d)
I=2A
5 turns
I=5A
Module II – 13
3.
Which of the following electromagnets produces the strongest field?
Wooden core
Wooden core
a)
c)
I=5A
I = 10 A
Iron core
b)
I = 10 A
I=5A
4.
Iron core
d)
The electromagnets illustrated below produce magnetic fields of different intensities.
Which electromagnet produces the strongest field?
a)
Soft Iron
c)
Aluminum
I = 10 A
b)
I = 10 A
Soft Iron
d)
Aluminum
I=8A
I = 10 A
5.
If you are making an electromagnet, which combination will produce the strongest
magnetic field:
1.
A solenoid with 100 turns
2.
A solenoid with 200 turns
3.
A current of 5 amperes
4.
A current of 10 amperes
a)
1 and 3
c)
2 and 3
b)
1 and 4
d)
2 and 4
Additional Question:
SQ: p. 174, #11
Module II – 14
2.8
To justify applying the principles of
magnetism and electromagnetism in the
manufacture of consumer goods.
416 and 436
DME: Section 6.12 + “Tidbits” pp. 146, 150, 162
SQ: pp. 168-171
Key Concepts:
1.
Magnets are used in many consumer goods.
Examples:
2.
magnetic compasses, magnetic door latches, magnetic screwdrivers,
refrigerator door magnets, paper clip dispensers, etc.
Electromagnets are used in many consumer goods, specifically those which move
something (or move) using electricity.
Examples:
electric lawnmowers, blenders, electric doorbells, electric drills,
electric toothbrushes, VCR’s, cassette recorders, CD players,
hydroelectric generators, electric fans, vacuum cleaners, scrap yard
cranes (for lifting and moving scrap iron), NMR imaging devices, etc.
These devices are important because when they are “turned on”, they generate strong
magnetic fields. When they are “turned off”, the magnetic field disappears.
SAMPLE QUESTION
1.
In a steel mill, a mechanical crane with a powerful electromagnet suspended from the end
of a cable is used to load and unload pieces of iron of all shapes and sizes.
Why is an electromagnet used rather than a natural magnet?
Additional Question:
SQ: p. 174, #12
Module II – 15
3.1
To justify using certain substances, to
assemble electric circuits, based on the
properties of those substances.
416 and 436
DME: Section 7.2
SQ: pp. 176-178
Key Concepts:
1.
Insulators are substances which do not easily allow electricity to flow through them.
Examples:
wood, paper, plastics, rubber, ceramics, glass, etc.
Conductors are substances which easily allow electricity to flow through them.
Examples:
metal wires, graphite rods, etc.
2.
Electric circuits contain both insulators (plastic coating on wires) and conductors (copper
wires). This is so that electricity will flow only through selected parts of the circuit.
3.
Current electricity is the movement of negative electric charges (electrons) through wires.
Static electricity is a collection of stationary electric charges.
SAMPLE QUESTIONS
1.
Alligator clips are used to connect the components of an electric circuit.
Which substance can be used to make the jaws of these alligator clips?
a) Aluminum
c) Plastic
b) Rubber
d) Porcelain
Module II – 16
2.
Porcelain is used to support electrical wires on poles.
Which properties of porcelain make it desirable for this use?
1.
2.
3.
4.
3.
Is a good insulator.
Is non-ductile.
It rusts.
Breaks easily.
a) 1 and 2
c) 2 and 4
b) 1 and 3
d) 3 and 4
A manufacturer wants to wrap an electric wire with material that does not conduct
electricity.
Which one of the following materials CANNOT be used for this purpose?
a)
Ceramic
c)
Plastic
b)
Graphite
d)
Glass
Additional Questions:
DME: p. 169, #2-7
SQ: p. 224, #1-5
Module II – 17
3.2
To determine how various factors affect the
conductivity of given materials.
416 and 436
DME: Section 7.3
SQ: pp. 178-183
Key Concepts:
1.
Conductance is a characteristic property of a substance.
2.
Conductance is a number which tells you how well electricity flows through a substance.
3.
Conductance of a wire depends on:
- the type of material used (copper has a higher conductance than iron)
- the length of the wire
- the diameter of the wire (thickness or cross-sectional area)
- the temperature of the wire.
4.
As the length increases → conductance DECREASES
As the diameter increases → conductance INCREASES
As the temperature increases → conductance DECREASES
5.
Conductors have a high conductance.
Insulators have a low conductance.
6.
The unit of conductance is the “siemens”.
Module II – 18
SAMPLE QUESTIONS
1.
The four conductors shown below are made out of copper.
Which one has the greatest conductance?
a)
b)
c)
d)
2.
Several factors can influence the electrical conductivity of a wire.
Which of the following diagrams shows the wire with the best electrical conductivity?
10 cm
a)
copper
copper
3.
25˚C
10 cm
b)
10 cm
copper
c)
10 cm
d)
100˚C
nichrome
Which of the following would increase the conductivity of a circuit ?
1234-
A thicker wire
A longer wire
A decrease in the temperature of the wire
The use of porcelain wire
a)
1 and 2
c)
2 and 4
b)
1 and 3
d)
3 and 4
Additional Questions:
DME: p. 208, #1-3
SQ: Exercises p. 183 and p. 224, #6-8
Module II – 19
25˚C
100˚C
3.3
To measure current intensities in a circuit
element.
416 and 436
DME: Section 7.4
SQ: pp. 183-186
Key Concepts:
1.
Current intensity measures the amount of electrons which flow through an electric circuit
in one second. It is often called “current”.
2.
The symbol for current intensity is “I”.
3.
The unit of current intensity is the ampere (often called “amps”).
The symbol for ampere is “A”.
Example:
If the current is 0.5 amperes, we would write
I = 0.5 A
4.
The instrument used to measure current intensity is the ammeter.
An ammeter is always put inside a circuit to measure all the flowing electrons.
It must be connected in series. (See the diagram at the bottom of this page.)
5.
The symbol used for an ammeter is A
6.
The direction of flow of electrons must be considered when connecting the ammeter into a
circuit (positive to positive, negative to negative).
.
+
+
−
Module II – 20
A
−
SAMPLE QUESTIONS
1.
The diagram of a parallel electric circuit is shown below.
R1
R2
You have to connect ammeter A into this circuit so that you will be able to read the total
current flowing through resistors R1 and R2.
Which diagram shows the right way to connect the ammeter?
a)
A
c)
R1
R2
R1
R2
A
b)
d)
R1
R2
A
R1
R2
A
2.
You are to assemble a series circuit consisting of resistors R1 and R2. Using an ammeter,
A, you are to verify the hypothesis that the current intensity, I, is the same in all parts of the
circuit.
Draw a circuit diagram showing all the places you would connect the ammeter.
Additional Questions:
DME: p. 175, #1-4
SQ: p. 224, #9-10
Module II – 21
3.4
To measure potential differences in simple
circuits.
416 and 436
DME: Section 7.5
SQ: pp. 187-189
Key Concepts:
1.
Potential difference causes electrons to flow through a circuit.
Potential difference is provided by either a battery or a power supply.
Potential difference is often referred to as “voltage”.
2.
The symbol for potential difference is “V”.
3.
The unit of potential difference is the volt.
The symbol for volt is “V”.
Example:
If the potential difference across a battery is 12 volts, we would
write, V = 12 V
4.
The instrument used to measure potential difference is the voltmeter.
A voltmeter is always put outside a circuit to measure potential difference between two
points in the circuit. It must be connected in parallel. (See the diagram at the bottom of this
page.)
5.
The symbol used for a voltmeter is V .
6.
The direction of flow of electrons must be considered when connecting the voltmeter
across a circuit (positive to positive, negative to negative).
+
V
−
+
−
Module II – 22
SAMPLE QUESTION
1.
You have to connect a voltmeter to determine the potential difference across the terminals
of a resistor in a simple circuit.
In which diagram below is the voltmeter properly connected?
a)
b)
V
c)
V
d)
Additional Questions:
DME: p. 178, #1-5
SQ: p. 224, #11-15
Module II – 23
V
V
3.5
To determine the conductance of a circuit
element, using a graph.
416 and 436
DME: Section 7.6
SQ: pp. 190-192
Key Concepts:
1.
The symbol for conductance is “G”.
2.
The conductance of an object can be determined by plotting Current Intensity (on Y-axis)
versus x (on X-axis) and then calculating the slope of the resulting line.
Y axis
I (A)
Y2
Y1
slope = G
X1
X axis
X2
V (V)
G=
Y2 – Y1
X2 – X1
OR
Conductance =
Notice that the units of conductance must be amperes
volts.
3.
One ampere = 1 siemens
volt
4.
The symbol for siemens is “S”.
5.
The formula for calculating conductance is
G =
Module II – 24
∆I
∆V
difference in current intensity
difference in voltage
SAMPLE QUESTIONS
1.
The following graph illustrates the change in electric current intensity, I, as a function of
potential difference (voltage), V, for a given resistor.
I (A)
1.8
0.9
6
12
V (V)
According to the graph, what is the conductance of the resistor?
2.
a) 0.15 S
c)
6.7 S
b) 5.4 S
d) 21.6 S
The following graph describes the behavior of three conductors subjected to different
voltages.
I (A)
1
4.0
3.5
2
3.0
2.5
3
2.0
1.5
1.0
0.5
0
1
2
3
4
5
6
7
8
V
Which is the conductor with the best electrical conductance? Justify your answer using
calculations.
Additional Questions:
DME: p. 208, #15a
SQ: Exercises p. 192 and p. 224, #16-18
Module II – 25
3.6
To distinguish between the conductance and
the resistance of a circuit element.
416 and 436
DME: Sections 7.7, 7.8 and 7.9
SQ: pp. 192-194
Key Concepts:
1.
Recall that conductance tells how easily current flows through an object.
Resistance tells how difficult it is for current to flow through an object.
2.
The symbol for resistance is “R”.
3.
Resistance is the reciprocal of conductance.
so
4.
R= V
I
In general:
This formula is called Ohm’s Law.
Material
Conductance
Resistance
Conductor
high
low
Insulator
low
high
1 ohm = 1
volt
ampere
5.
The unit for resistance is the “ohm”.
6.
The symbol for ohms is “Ω” (pronounced “omega”).
7.
Some resistors have a ceramic coating with colour coded bands to indicate the resistance.
(See DME: p. 184; SQ: p. 194)
1Ω = 1 V
A
You must be able to determine the resistance of this type of resistor.
• The 1st and
• 2nd bands give you the first two digits of the resistance.
• The 3rd band is the exponent on ten by which the first two digits must be
multiplied.
• The 4th band indicates the tolerance of the resistor. This tells you how close the
actual resistance will be to the value indicated by the bands.
Example:
A resistor having bands of red, brown, orange,
gold has a resistance of 21 x 103 ± 5%.
Module II – 26
red
gold
brown
orange
SAMPLE QUESTION
1.
The resistance of a resistor can be determined using the three coloured bands on the
resistor as well as a colour code.
How to Read the Coloured Bands
Colour
• The first band indicates the first digit of the
resistance value.
Black
0
Brown
1
Red
2
Orange
3
Yellow
4
Green
5
• The second band indicates the second digit of
the resistance value.
• The third band indicates the number of zeros
after the first two digits.
Codes
Which of these resistors illustrated below has a resistance of 4200 Ω?
a)
c)
yellow
red
black
b)
yellow
red
yellow
d)
yellow
red
red
red
yellow
black
Additional Questions:
DME: p. 183, #1-3; p. 185, #2-3; p. 208, #4, 10-13, 15b, 16-17
SQ: Exercises p. 194 and p. 225, #19-20
Module II – 27
3.7
To determine the equivalent resistance of
series and parallel circuits kept at constant
temperature.
416 and 436
DME: Sections 7.10 and 7.11
SQ: pp. 195-197
Key Concepts:
1.
In a series circuit, all the current travels through each resistor, one after the other. The
current is the same at every point in the circuit.
R1
R2
R3
2.
Equivalent Resistance (Req) is the resistance of a single resistor that could replace all the
resistors in a circuit without changing the total current in the circuit.
R1
Req
becomes
R2
R3
3.
The equivalent resistance of a series circuit is given by the following:
Req = R1 + R2 + R3 + ...
4.
In a parallel circuit, the current splits up so that part of it travels through each resistor at
the same time. The current is not the same at every point in the circuit.
R1
R2
Module II – 28
R3
5.
The equivalent resistance of a parallel circuit is given by the following:
1 = 1 + 1+
Req
R1
R2
1+ ...
R3
SAMPLE QUESTIONS
1.
A parallel circuit is illustrated below.
R1 = 5 Ω
R2 = 5 Ω
What is the equivalent resistance of this circuit?
2.
a)
0.4 Ω
c)
5Ω
b)
2.5 Ω
d)
10 Ω
The electric circuit illustrated below consists of a power supply and resistors
R1, R2, R3 and R4.
R1 = 10 Ω
R2 = 10 Ω
R3 = 10 Ω
R4 = 10 Ω
What is the equivalent resistance of the circuit?
a)
0.4 Ω
c)
10 Ω
b)
2.5 Ω
d)
40 Ω
Module II – 29
3.
The following circuit consists of three resistors (R1 , R2 , and R3).
R1
5Ω
R2
10Ω
R3
30Ω
What is the equivalent resistance Req of this circuit ?
4.
a) 0.33 Ω
c) 15 Ω
b) 3 Ω
d) 45 Ω
The following parallel circuit consists of two resistors (R1 and R2 ) and two ammeters A
and A2. The potential difference (voltage) across the terminals of the power supply is 24 V.
R2
R2
Vt = 24 V
R1
R
1
A2
I2 = 4 A
A
It = 12A
What is the resistance of resistor R1? Show all your work.
Additional Questions:
DME: p. 187, #5; p. 189, #1 to 6; p. 208, #6, 7, 8, 9, 14
SQ: Exercises p. 197, pp. 200-201 and p. 225, #21, 22
Module II – 30
3.8
To determine the equivalent resistance of a
series-parallel circuit kept at constant
temperature.
436 Only
DME: Section 7.12
SQ: pp. 198-199
Key Concepts:
1.
In a series-parallel circuit, both series and parallel connections are present in the same
circuit.
R1
R2
R3
2.
Recall that the equivalent resistance is the resistance of a single resistor that could replace
all the resistors in a circuit without changing the total current in the circuit.
3.
You must first identify a section of the circuit whose equivalent resistance can be
determined using only a series calculation or a parallel calculation. The choice you make
will be determined by the example that you are given.
STEP 1:
In the following example, you must determine the equivalent
resistance of the parallel section first using
1
1
=
R1, 2
R1
R1
R3
+
1
R2
R2
becomes
You have now simplified the circuit so that it is now a series circuit.
Module II – 31
R1,2
R3
STEP 2:
Now, the equivalent resistance of the entire circuit may be
determined, using
becomes
R1,2
Req
R3
so: Req = R3 + R1,2
SAMPLE QUESTIONS
1.
A series-parallel electric circuit is illustrated below.
R1 = 10 Ω
R2 = 15 Ω
R5 = 10 Ω
R3 = 20 Ω
R4 = 10 Ω
What is the equivalent resistance of this circuit?
2.
a)
2.4 Ω
c)
30.0 Ω
b)
24.6 Ω
d)
65.0 Ω
The electric circuit illustrated below consists of a power supply and resistors R1, R2, R3,
and R4.
R1 = 20 Ω
60 V
R4 = 20 Ω
R2 = 20 Ω
R3 = 20 Ω
What is the equivalent resistance, R, of this circuit?
a)
5.0 Ω
c)
50 Ω
b)
40.1 Ω
d)
80 Ω
Additional Questions:
DME: p. 190, #4, 5, 6; p. 206, #1
SQ: Exercises pp. 204-206
Module II – 32
3.9
To evaluate the significance of the error in
the measurement of electrical resistance.
416 and 436
DME: Section 7.13
SQ: pp. 206-209
Key Concepts:
1.
Absolute error is the difference between the measured and accepted values for a quantity.
Absolute error = measured value – accepted value
The absolute error can be either positive or negative.
2.
Relative error (percentage error) is the absolute error divided by the accepted value,
expressed as a percentage.
Relative error = absolute error
accepted value
x 100%
Note: The relative error can be either positive or negative.
Example:
According to its bands, a resistor has a resistance of 12 x 101 Ω.
During an experiment, you determined that the resistor had a
measured resistance of 110 Ω. Find the absolute and relative errors.
Absolute error = 110 Ω – 120 Ω = –10 Ω
Relative error = –10 Ω x 100% =
120 Ω
8.3% (approx.)
Additional Questions: SQ: p. 225, #23, 24
Module II – 33
3.11
To analyze the distribution of current in
various combinations of circuit elements.
416 and 436
DME: Section 7.15
SQ: pp. 210-213
Key Concepts:
1.
Kirchoff’s Laws for current intensity.
In a series circuit: the current is the same at every point in the circuit
IS = I1 = I2 = I3 = I4 . . .
In a parallel circuit: the current from the battery equals the sum of
the current intensities through each of the resistors.
IS = I1 + I2 + I3 + I4 . . .
Note: These laws are NOT given on the formula sheet on your exam.
You must remember them!!
DME: See examples 1 and 2 on p. 197, 198.
Module II – 34
SAMPLE QUESTIONS
1.
The following electric circuit consists of two resistors, R1 and R2, and a power source.
R1
R2
Using an ammeter, you measured the current intensity, I, through each resistor.
Here are the results:
Resistor
Intensity (A)
R1
0.75
R2
0.75
Given this information, what is the current intensity provided by the power source, IS?
2.
The electric circuit in the diagram below consists of 3 resistors and 5 ammeters numbered
1 to 5.
A2
I2 = 3 A
What are the readings
of ammeter 3 and
ammeter 5?
A3
I3 = ?
A1
A4
I1 = 6 A
I4 = 1 A
A5
I5 = ?
Module II – 35
3.
An electric circuit consists of a power source, two switches (S1 and S2) and two light bulbs
(L1 and L2).
The following table shows what happens to both light bulbs:
Switch
S1
open
Switch
S2
open
Light Bulb
L1
out
Light Bulb
L2
out
closed
open
bright
out
Which of the following circuit diagrams illustrates the results shown in the table above ?
A)a)
c)
S2
S1
C)
L1
S1
L1
L2
L2
S2
B)
D)
b)
d)
S1
S1
L1
L2
L1
S2
S2
Additional Questions:
DME: p. 200, #1
SQ: Exercises p. 213 and p. 226, #25, 26
Module II – 36
L2
3.12 To explain the distribution of electric
potential in various combinations of circuit
elements.
416 and 436
DME: Section 7.16
SQ: pp. 213-215
Key Concepts:
1.
Kirchoff’s Laws for potential difference.
In a series circuit: the potential difference across the battery is the sum
of the potential differences across the individual resistors.
VS = V1 + V2 + V3 + V4 . . .
potential difference from the battery (source)
In a parallel circuit: the potential difference across the battery is the
same as the potential difference across each resistor.
VS = V1 = V2 = V3 = V4 . . .
Note: These laws are NOT given on the formula sheet on your exam.
You must remember them!!
DME: See examples 3 and 4 on p. 198, 199
SQ: See “For Your Information”, p. 214
Module II – 37
SAMPLE QUESTIONS
1.
A student assembled the two circuits illustrated below. Each of these circuits consists of
two identical light bulbs, a 1.5 V battery and a voltmeter.
V1
V2
1.5 V
1.5 V
What is the reading displayed on voltmeters V1 and V2?
2.
a)
V1 = 0.75 V and V2 = 0.75 V
c)
V1 = 1.5 V and V2 = 0.75 V
b)
V1 = 0.75 V and V2 = 1.5 V
d)
V1 = 1.5 V and V2 = 1.5 V
Two electric circuits each consist of a power supply and resistors R1 and R2.
For each circuit, the following table gives the potential difference (voltage), V, across the
terminals of the power supply; the potential difference, V1, across resistor, R1; and the
potential difference, V2, across resistor R2.
Circuit
V (V)
V1 (V)
V2 (V)
1
10
8
2
2
10
10
10
Draw a circuit diagram to illustrate each of these circuits.
Give one reason to justify each of your diagrams.
Additional Questions:
SQ: Exercises p. 215 and p. 226, #27
Module II – 38
3.14 To explain the behaviour of unknown
circuits, using the rules of electric circuits.
416 and 436
DME: Section 7.20
SQ: pp. 218-219
Key Concepts:
1.
Unscrewing a light bulb in a series circuit will stop any current from flowing.
All bulbs in this circuit will go out.
2.
Opening a switch in a series circuit will stop the current flow to each bulb.
All the bulbs will go out.
Closing a switch in a series circuit will allow the current flow to each bulb.
All the bulbs will light up.
3.
Opening a switch in a parallel circuit will stop the current flow only to the branch that
the switch controls. The bulb(s) on that branch will go out.
Closing a switch in a parallel circuit will allow the current to flow to the bulb that the
switch controls. The bulb(s) on that branch will light up.
4.
Always draw a circuit diagram when solving your unknown circuit.
Module II – 39
3.15 To solve numerical problems that deal with
the mathematical relationships between the
different physical quantities associated with
series and parallel circuits.
416 and 436
DME: Section 7.17
SQ: pp. 200-202
Key Concepts:
1.
Kirchoff’s Laws for current intensity and potential difference are usually used together
with Ohm’s Law to solve problems.
2.
Always draw a circuit diagram which includes all the resistances, currents and potential
differences that are given to you. As you calculate resistances, currents or potential
differences, write them on your diagram
SAMPLE QUESTIONS
1.
In the electric circuit illustrated below, the current intensity, I, is 0.25 A.
R1 = 10 Ω
Vs
R2 = 20 Ω
R3 = 40 Ω
What is the potential difference across the terminals of the power source, VS?
Show all your work.
2.
The electric circuit shown below consists of an ammeter, A, a power supply, and resistors
R1 and R2 connected in parallel.
A
20 V
R1 = 40 Ω
R2 = 40 Ω
What is the current intensity, I, flowing through the ammeter?
Show all your work.
Module II – 40
3.
The following circuit is connected to a source that can provide a current of 18 A when the
potential difference (voltage) is 36 V.
R
1
R2
R3
R2 = 4 Ω
VS = 36 V
IS = 18 A
R3 = 6 Ω
I3 = 6 A
What is the current intensity I1 flowing through resistor R1?
Show all your work.
4.
The following circuit is connected to a source that can provide a current of 2 A when the
potential difference (voltage) is 12 V.
R1
VS = 12 V
IS = 2 A
R2
R1 = 1 Ω
R2 = 2 Ω
R3
What is the resistance of resistor R3?
Show all your work.
Additional Questions:
DME: p. 201, #2 to 7; p. 208, #18; p. 209, #26
SQ: Exercises pp. 200-202
Module II – 41
3.16 To solve numerical problems that deal with
the mathematical relationships between the
different physical quantities associated with
series-parallel circuits.
436 only
DME: Section 7.18
SQ: pp. 203-206
Key Concepts:
1.
Kirchoff’s Laws for current intensity and potential difference are usually used together
with Ohm’s Law to solve problems.
2.
Always draw a circuit diagram which includes all the resistances, currents and potential
differences that are given to you. As you calculate resistances, currents or potential
differences, write them on your diagram
3.
Always draw a new circuit diagram after you have calculated the equivalent resistance of
one section of the circuit. (You may need several circuit diagrams.)
4.
Always check that your answers are reasonable when you have finished.
DME: See examples 1 and 2 on pp. 201, 202.
SQ: See example on p. 203.
SAMPLE QUESTIONS
1.
A series-parallel electric circuit is illustrated below.
R2 = 5 Ω
VS = 12 V
R1 = 30 Ω
R3 = 10 Ω
R4 = 20 Ω
What is the potential difference across the terminals of resistor R1?
a)
4V
c)
8V
b)
6V
d)
12 V
Module II – 42
2.
A series-parallel electric circuit is illustrated below.
R1 = 75 Ω
A 0.5 A
VS
R2 = 75 Ω
R3 = 100 Ω
R4 = 50 Ω
What is the intensity of the current flowing from the power source, IS?
Show all your work.
3.
The electric circuit illustrated below, consists of a power supply and resistors R1, R2 and
R3.
R2 = 200 Ω
R1 = 100 Ω
R3 = 300 Ω
Using an ammeter, you measured the current intensity, I, flowing through each resistor.
The results are shown in the following table.
Resistor
R1
R2
R3
Current (A)
2.0
1.2
0.8
You forgot to measure the total current intensity, I, flowing through the circuit and the
potential difference (voltage), V, across the terminals of the power supply. However, you
can calculate them using the above results.
What values will you obtain?
a)
I = 2 A and V = 440 V
c)
I = 4 A and V = 440 V
b)
I = 2 A and V = 680 V
d)
I = 4 A and V = 680 V
Additional Questions:
DME: p. 203, #1 to 4; p. 208, #19; p. 209, #20, 21, 24, 25
SQ: Exercises pp. 204-206
Module II – 43
4.1
To determine the cost of using an electrical
appliance.
416 and 436
DME: Sections 8.2 and 8.3
SQ: pp. 228-232
Key Concepts:
1.
Electrical appliances transform electrical energy into many different forms of energy (heat,
sound, light, mechanical energy (movement)).
Every appliance has a rating plate, indicating its power rating, the maximum voltage that
can be applied, the current that it draws, and the line frequency.
voltage
frequency
60 Hz
1.7 A
120 V
200 W (0.2 kW)
Model No. A-1-150
power rating
current
2.
The symbol we use for energy is “E”.
The “joule” is the SI unit of energy. (symbol: J)
One joule is a tiny amount of energy.
The “kilowatt•hour” is a much larger unit of energy. (symbol: kW•h)
1 kW•h = 3 600 000 J
3.
Power is the rate at which energy is produced or used.
The symbol for power is “P”.
The “watt” is the SI unit of power. (symbol: W)
1 W = 1 joule
second
The “kilowatt” is a larger unit of power. (symbol: kW)
1 kW = 1000 W
Module II – 44
4.
Energy, power and time are related by the following formula:
E = Pxt
5.
therefore
1 J = 1 W x 1s
In general:
The cost of using an appliance = cost of energy x power rating x time used.
Cost of energy is usually given in
¢
kW•h
so the power must be expressed in kilowatts, and the time in hours.
Example:
How much does it cost to operate an applicance with the
following rating plate for 2 hours?
(Cost of electricity energy = 5¢ per kW•h)
60 Hz
1.7 A
120 V
200 W (0.2 kW)
Model No. A-1-150
Solution:
Cost = cost of energy x P x t
=
5¢ x 0.2 kW x 2h
kW•h
units of time must be hours
units of power must be kilowatts
= 2¢
Module II – 45
SAMPLE QUESTIONS
1.
The rating plate on a water heater gives the following information:
Model CR-RG
Serial #LC 10 U
240 V
4500 W (4.5 kW)
This water heater is used an average of two hours a day. Electricity costs $0.0454 / kW•h.
How much does it cost per day to use this water heater?
2.
a)
$0.20
c)
10.90
b)
$0.41
d)
$21.79
The rating plate on a particular computer gives the following information:
60 Hz
1.7 A
120 V
200 W (0.2 kW)
Model No. A-1-150
The computer is used for 5 hours. The cost of electrical energy is $0.05 per kW•h.
How much does it cost to use the computer?
a)
$0.01
c)
$0.25
b)
$0.05
d)
$1.02
Module II – 46
3.
The rating plate on an electrical appliance gives the following information:
Model:
Series:
240 V
SM65
CRJB3
600 W (0.6 kW)
This appliance is used 100 hours a month, and electricity costs $0.05/kW•h.
How much does it cost to use this appliance over a 12-month period?
4.
a)
$3.00
c)
$144.00
b)
$36.00
d)
$720.00
The rating plate on an electrical appliance gives the following information:
MODEL: LB-97CR
120 V
60 Hz
1200 W (1.2 kW)
This appliance is used 90 minutes a day, and electricity costs $0.046/kW•h.
How much does it cost to use this appliance over a 60-day period?
a)
$4.97
c)
$596.16
b)
$298.08
d)
$4968
Additional Questions:
DME: p. 318, #1 to 4 and p. 219, #5
SQ: Exercises p. 231 and p. 247, #1-4
Module II – 47
4.2
To define the unit for measuring electric
current, after observing a demonstration.
436 only
DME: Section 8.4
SQ: pp. 232-236
Key Concepts:
1.
The symbol for the amount of charge is “Q”.
The SI unit of charge is the “coulomb”. (symbol “C”)
One coulomb is a very large amount of charge. (charge on 6.25 x 1018 electrons)
2.
Current intensity = amount of charge
time
or
I = Q
t
The SI unit of current intensity is the “ampere” (symbol “A”)
where 1 ampere = 1 coulomb
second
SAMPLE QUESTIONS
1.
You are to measure the number of coulombs that flow through a component of an electric
circuit in one second (C/s).
Which instrument should you use in this case?
2.
a)
Ammeter
c)
Ohmmeter
b)
Calorimeter
d)
Voltmeter
What is the definition of the unit for current intensity?
a)
1 volt of potential difference measured for 1 second
b)
1 joule of energy consumed in 1 second
c)
1 coulomb of charge per second
d)
1 watt of power per second
Module II – 48
3.
4.
Which of the following is the definition of current intensity, I ?
a)
The amount of electric charge flowing through a specific point in a circuit per
unit of time.
b)
The energy flowing through a circuit in one second.
c)
The energy carried by a certain quantity of electric charge.
d)
The force with which electric charges flow through a circuit.
You are to measure the number of coulombs that flow through a component of an electric
circuit in one second (C/s).
Which instrument should you use in this case?
a)
Ammeter
c)
Ohmmeter
b)
Calorimeter
d)
Voltmeter
Additional Questions:
DME: p. 178, #1 to 5
SQ: Exercises p. 236 and p. 247, #8-13
Module II – 49
4.3
To define the unit for measure for electrical
potential difference.
436 only
DME: Section 8.5
SQ: pp. 236-240
Key Concepts:
1.
The SI unit of potential difference is the “volt” (symbol “V”).
The potential difference across a resistor is 1 volt if:
1 joule of energy is lost by every coulomb of charge flowing through the resistor.
so:
2.
1 volt =
1 joule
coulomb
or
1V = 1J
C
For batteries connected in series, the total potential difference equals the sum of the
individual potential differences of each battery.
Vs = V1 + V2 + V3 + ...
3.
Alessandro Volta’s invention of the battery contributed research in the areas of
electromagnetism and electrochemistry in the 1800’s. The battery made it possible to
decompose water into hydrogen gas and oxygen gas. This aided in the discovery of
elements which now make up our present day periodic table of elements.
SAMPLE QUESTIONS
1.
Which statement corresponds to the definition of electric potential?
a)
... is the quantity of electric charge that flows through an electric circuit in one
second.
b)
... is the energy associated with the movement of electrical charges flowing
from a source.
c)
... is the energy that flows through an electric circuit in one second.
d)
... is the current intensity that flows through an electric circuit.
Module II – 50
2.
Which of the following is the definition of a volt (unit of measurement of potential
difference)?
a)
The energy carried by a certain quantity of electric charge.
b)
The amount of electric charge that flows through a specific point in a circuit per
unit of time.
c)
The force with which electric charges flow through a conductor.
d)
The amount of electric charge flowing through a circuit.
Additional Questions:
DME: p. 250, #10
SQ: p. 247, #14-16
Module II – 51
4.4
To identify the units that define electric
energy, using the variables associated with
electric circuits.
416 and 436
DME: Sections 8.6 and 8.7
SQ: pp. 240-242
Key Concepts:
1.
Recall:
The symbol we use for energy is “E”.
The “joule” is the SI unit of energy (symbol “J”).
= power x time
or
E = Pxt
2.
Power = potential difference x current intensity
3.
Energy = potential difference x current intensity x time
4.
5.
or
P = VxI
or
E = V x I x t
Unit
Unit of ...
Definition
Ampere
current intensity
1 coulomb per second
Volt
potential difference
1 joule per coulomb
Watt
power
1 joule per second
1 kilowatt•hour (kW•h) = 3 600 000 joules (J).
1 joule (J) = 1 watt•second (W•s) = 1 volt•ampere•second (V•A•s).
SAMPLE QUESTION
1.
Which of the following is a unit of electrical energy?
a)
V•A
c)
W•s
b)
Ω•A2
d)
kW/h
Additional Questions:
DME: p. 218, #1 to 4 and p. 219, #5
SQ: Exercises p. 242 and p. 247, #5-7
Module II – 52
4.5
Justify the use of high tension wires to
transport electrical energy.
436 only
DME: Sections 8.8, 8.9 and 8.10
SQ: pp. 243-244
Key Concepts:
In Quebec, most electrical sources are situated long distances from major population and
industrial centers. We have seen that resistance along a wire increases with its length. There is
the potential for huge losses of electrical energy along these long electrical lines. Fortunately,
there is a way to solve this problem.
The equation P = RI2 determines the amount of energy lost. Note that current intensity (I) is
related to resistance (R). Further note that, because current intensity is squared, the ratio of
current intensity to resistance in this equation will increase as power increases.
In point of fact, the voltage produced by our electrical power plants is increased by use of
transformers. Thus, these power lines are known as “high tension wires”.
Mathematically, this can be demonstrated with the following problem:
Problem:
The transmission lines used to distribute 100 kW of power have a resistance of 1 Ω. For a
potential difference of 100 V and 10 kV, calculate:
a)
the current intensity in the lines
b)
the power loss in the lines
c)
the percent power loss.
Solution:
For 100 V
a)
P = IV
I = P/V
= 1 x 105 W ÷ 1 x 102 V
= 1000 W/V
= 1000 A
For 10 kV
P = IV
I = P/V
= 1 x 105 W ÷ 1 x 104
= 10 W/V
= 10 A
I2R
(1000 A)2 x 1 Ω
1 x 106 A2 Ω
1 x 106 W
b)
P =
=
=
=
P =
=
=
=
c)
% power loss = 1 x 106 ÷ 1 x 105 x 100%
= 1000%
I2R
(2 A)2 x 1 Ω
4 A2 Ω
4W
% power loss = 4 W ÷ 1 x 105 x 100%
= 0.004%
Note the dramatic difference in power loss between the two examples.
Module II – 53
SAMPLE QUESTIONS
1.
Hydro-Québec uses high-tension lines to reduce the loss of power caused by the Joule
effect.
Why do high-tension lines reduce the loss of power?
2.
a)
When the power generated is constant, higher voltages result in greater current
intensity.
b)
When the power generated is constant, higher voltages result in lower current
intensity.
c)
When the power generated is constant, higher voltages result in an increase in
the resistance of the lines.
d)
When the power generated is constant, higher voltages result in a decrease in
the conductance of the lines.
An electric circuit has a total resistance of 2 Ω. This circuit is connected to a source with a
maximum power of 180 W. The source can provide three different voltages, namely:
120 V, 180 V and 240 V.
Which of these voltages will result in the lowest power loss?
Justify your answer using calculations.
Additional Questions:
SQ: Exercises p. 244 and p. 247, #17, 18
Module II – 54
4.6
To solve numerical problems that involve
calculating the electric energy used in a
circuit.
436 only
DME: Sections 8.10 and 8.11
SQ: p. 245
Key Concepts:
1.
Several formulas are used to determine energy or power.
E = P x t
where
E = V x I x t
P = V x I
“E” is energy
“P” is power
“V” is potential difference (voltage)
“I” is current intensity
“R” is resistance
P = I2 x R
P = V2
R
2.
Make sure that the units of your variables are consistent before plugging them into your
formula.
Example:
A toaster uses 10 A of current on a 120 V line.
a)
What is the power rating of the toaster?
b)
How much energy does the toaster use if it takes 3 minutes
to make toast?
Solution: V = 120 V
I = 10 A
t = 180 s (3 min.)
a)
P
= VI
b)
E
= Pt
= (120 V) (10 A)
= (1200 W) (180 s)
= (120 J) (10 C )
C
s
(1200 J) (180 s )
s
= 1200 W
= 216 000 J
Module II – 55
or 216 kJ
SAMPLE QUESTIONS
1.
You connect a fan to a 12-V power source. The total resistance of the fan wires used is
10Ω. You operate the fan for 20 minutes.
How much energy is used by the fan wires during this period?
2.
a)
4.8 J
c)
2400 J
b)
288 J
d)
17280 J
The following information is found on the back of a television:
Model SFMCL
Serial #: 181920
120 V
60 Hz
1.5 A
This television is used an average of 8 hours a day.
How much electrical energy does this television use during this period?
3.
a)
1.44 kJ
c)
86.4 kJ
b)
22.5 kJ
d)
5184 kJ
The resistance of a heating element is 10 Ω and the potential difference (voltage) across
its terminals is 240 V. This element is used for 3 hours.
How much electric energy was used during this period?
a)
17.3 kJ
c)
17 280.0 kJ
b)
108.0 kJ
d)
62 208.0 kJ
Module II – 56
4.
An electrical appliance with an internal resistance of 50 Ω is connected to a 110-volt power
source. The applicance is used for 30 minutes.
How much energy was used?
5.
a)
121 J
c)
89 100 J
b)
7260 J
d)
435 600 J
An electric heater connected to a 120 V power supply has an internal resistance of 12 Ω.
After being on for a certain period of time, this heater produced 3600 kJ of heat energy.
How long (in minutes) was this electric heater in operation?
Additional Questions:
DME: p. 228, #1 to 3; p. 229, #4 to 6; p. 250, #3, 4, 5, 12, 18
SQ: Exercises p. 245
Module II – 57
5.1 To determine the amount of heat energy used by a
resistor, by conducting an experiment.
5.2 To calculate heat energy, using a mathematical
relationship.
5.3 To analyze the transformation of electrical energy,
based on measurements and calculations.
416 and 436
DME: Sections 8.10 and 8.13
SQ: pp. 249-256
Key Concepts:
1.
Recall:
The amount of electrical energy used by an appliance is found using:
E = V x I x t
or
E = P x t
2.
Objects which change temperature either gain or lose Heat Energy.
Heat energy is given the symbol “Q” and is measured in joules.
3.
Specific heat capacity indicates how much heat energy 1 gram of a material will absorb in
order to raise its temperature 1°C.
Specific heat capacity has the symbol “c”.
The units of heat capacity are J/(g • C)
Water has a specific heat capacity of 4.19 J/(g • C)
4.
Heat energy (Q) is determined using the following formula:
Q = m x c x ∆T
Where:
“m” is the mass of the material
“c” is the specific heat capacity of the material
“∆T” is the change in temperature of the material
In most problems, the material is water.
Module II – 58
SAMPLE QUESTION
1.
A student heated a certain amount of water in a calorimeter fitted with a resistor and made
the following observations during the experiment:
Mass of the water
200 g
Initial temperature of the water
20°C
Final temperature of the water
45°C
Duration of the experiment
15 min.
How much energy was absorbed by the water?
Module II – 59
5.4
To solve numerical problems that require
the student to apply the law of conservation
of energy to situations in which electric
energy is being transformed into heat
energy.
436 only
DME: Sections 8.14 and 8.15
SQ: pp. 257-258
Key Concept:
If all the electrical energy that an appliance uses is converted into heat energy, then:
E = Q
so
V x I x t = m x c x ∆T
Most problems involve using this formula.
Example:
An electric heater on a 120 V line uses 8 A of current. It heats water from
18°C to 40°C in 20 minutes. If all the electrical energy is converted to
heat, what amount of water was heated?
STEP 1
Write down all of your variables with their values.
Make sure your units are consistent. (Standard units are best.)
STEP 2
V = 120 V
m = ?
I = 8A
c = 4.19 J/(g • °C)
t = 1200 s (20 min.)
∆T = 40°C – 18°C = 22°C
Put all these values into the formula and solve for the missing variable.
Make sure to modify your units so that they will cancel leaving the
desired unit in the answer.
VIt =
m c ∆T
(120 V) (8 A) (1200 s) =
m (4.19
(120 J ) (8 C ) (1200 s)
C
s
=
(4.19 J ) (22°C)
g • °C
so m =
J ) (22°C)
g • C°
m
12 500 g or 12.5 kg of water
Module II – 60
SAMPLE QUESTIONS
1.
Using an electric calorimeter, you conducted an experiment to determine the specific heat
capacity of a liquid. You noted the following information during the experiment.
Initial temperature of the liquid
20°C
Mass of the liquid
100 g
Final temperature of the liquid
50°C
Current Intensity
0.25 A
Potential Difference across the
terminals of the power source
15 V
Duration of the experiment
15 minutes
Given this information, what is the specific heat capacity of this liquid? (Ignore the energy
absorbed by the calorimeter.)
2.
A water heater containing 160 L of water operates at a voltage of 240 V. It can heat the
water from 20°C to 75°C in 2.5 hours.
What is the internal resistance of the water heater?
3.
An electric calorimeter containing 125 mL of water is connected to a 6-V power source.
After the appliance had been in operation for 15 minutes, the temperature of the water
increased by 5°C.
What is the current intensity, I, in this calorimeter?
a)
0.12 A
c)
2.1 A
b)
0.48 A
d)
29 A
Module II – 61
4.
Using a calorimeter, a student conducts an experiment on the conservation of energy.
The results of this experiement are given in the following table:
Variable
Mass of the water in the calorimeter
Result
50 g
Voltage across terminals of power source
Current intensity
10 V
0.5 A
Initial temperature of the water
Final temperature of the water
20°C
?
Duration of the experiment
30 min.
If energy is conserved, what will be the final temperature of the water?
a)
180°C
c)
43°C
b)
63°C
d)
20.7°C
Additional Questions:
DME: p. 237, #1 and p. 250, #8, 14
SQ: Exercises pp. 257-258 and p. 262, #8
Module II – 62
6.1, 6.2 & 6.3
To identify how electricity is produced and
its environmental impact.
416 and 436
DME: Sections 8.18 & 8.19
SQ: pp. 264-273
Key Concepts:
Means of Production
Method of Energy
Transformation
Environmental Impact
Hydroelectricity
kinetic energy of water →
mechanical energy (turbine) →
electrical energy (generator)
flooding of huge areas of
land, upsetting of ecosystem,
disruption of native
peoples’ culture
Thermonuclear Plants
atomic energy → thermal
energy (steam) → mechanical
energy (turbine) → electrical
energy (generator)
danger of nuclear accident
such as the Chernobyl
disaster, disposal of
radioactive wastes
Thermal Plants
chemical energy from burning coal
or oil → thermal energy (steam)
→ electrical energy (generator)
acid rain caused by
emissions of SO2 and NO2
Solar Plants
radiant (light) energy → thermal
energy → kinetic energy (steam)
→ electrical energy (generator)
take up large expanses of
land
Wind Power
air’s kinetic energy →
mechanical energy (turbine)
Module II – 63
SAMPLE QUESTIONS
1.
Methods of producing electric energy vary throughout the world.
• Nuclear power stations are widely used in Europe.
• Solar panels have been tested in some deserts in the southern United States.
• Hydro-electric power stations are generally used in Quebec.
• Oil-fired thermal power stations are common in the West Indies.
In which part of the world does the production of electric energy cause the least amount of
damage to the environment?
2.
3.
a)
West Indies
c)
Quebec
b)
Europe
d)
southern United States
Which type of power plant DOES NOT USE steam to produce electrical energy?
a)
oil-fired power plant
c)
nuclear power plant
b)
coal-burning power plant
d)
hydroelectric power plant
To which type of electric power station do the following statement apply?
1. It uses a form of non-renewable energy.
2. It is located near major areas of electricity consumption.
3. It is partially responsible for the production of acid rain.
4.
a)
wind power station
c)
nuclear power station
b)
hydroelectric power station
d)
coal-fired power station
Nuclear-powered, hydro-electric, coal-burning and diesel-powered generating stations
produce electrical energy by different methods. All of them have negative effects on the
environment.
Explain one negative effect that each type of generating station has on the environment.
5.
Which type of generating station is partly responsible for producing acid rain?
a)
hydro-electric station
c)
nuclear power station
b)
wind power station
d)
thermal power station
Module II – 64
MODULE III
— IONIC PHENOMENA —
2.1
To identify the properties normally used to
classify substances in aqueous solutions as
either acids, bases, or neutral salts
416 and 436
DME: Sections 9.2 to 9.9
SQ: pp. 313-316
Key Concepts:
1.
Indicators are used to detect acids or bases (show a change in colour).
Examples:
blue litmus paper, red litmus paper, phenolphthalein, universal
indicator, methyl orange, etc.
• Acids turn blue litmus paper red.
• Bases turn red litmus paper blue.
• Salts have no effect on litmus paper.
• Bases turn clear phenolphthalein pink.
2.
Acids and bases can change the texture and colour of certain foods.
3.
Acids react with most metals to form hydrogen gas.
4.
Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas.
calcium carbonate (limestone) + hydrochloric acid → calcium chloride + carbon dioxide gas
5.
Acids and bases neutralize each other forming a salt and water.
acid + base → salt + water
HCl NaOH
NaCl
H2O
6.
Acid, base, and salt solutions conduct electricity (because they release ions in their
solutions): they are called electrolytes.
• Electrolytes are solutions that conduct electricity.
• Non-electrolytes are solutions that do not conduct electricity.
Module III – 1
SAMPLE QUESTIONS
Below are the observations from two tests done on four different solutions.
1.
2.
3.
4.
Solution
Reaction to litmus
Test for conductivity
a
red to blue
yes
b
blue to red
yes
c
no change
yes
d
no change
no
Which of the above solution(s) contain(s) an electrolytic substance?
a)
a, b and c
c)
a and d
b)
a and b
d)
d only
Which of these solutions is a non-electrolyte?
a)
a, b and c
c)
a and d
b)
a and b
d)
d only
Which solution contains a neutral salt?
a)
a
c)
c
b)
b
d)
d
Anna often uses a white powder when cleaning the house. She is curious and wonders if
this powder is acidic, basic or neutral. What must she do FIRST to find out?
a)
put a piece of blue litmus paper on the solid
b)
put a piece of red litmus paper on the solid
c)
check to see if the solid conducts electricity
d)
dissolve a small amount of the solid in water
Module III – 2
2.2
To determine the conditions under which
substances will show the properties of acids,
bases, or salts.
416 and 436
DME: Section 9.10
SQ: pp. 316-318
Key Concepts:
1.
Distilled water does not conduct electricity because it has no free ions.
2.
Solutions of acids, bases, and salts do conduct electricity.
Module III – 3
2.3
To differentiate among acids, bases, and
salts based on their formulas.
DME: Section 9.11
SQ: pp. 318-319
2.4
To justify the need to produce various acids,
bases, and salts based on their uses/
properties.
416 and 436
DME: Section 9.12 and 9.13
SQ: pp. 320-323
Key Concepts:
1.
Acids have a formula that starts with H.
2.
Bases have a formula that ends with OH.
3.
Salts neither start with H nor end with OH: they start with the first part of the formula of
a base and end with the last part of the formula of an acid (because they are formed by the
neutralization of an acid with a base).
4.
Tests can be used to determine if a substance is an acid, a base, or a salt.
5.
Acids, bases, and salts have numerous uses in the everyday world.
SAMPLE QUESTIONS
1.
One of the properties of bases is that they dissolve fats. Which of the following substances
would you use to clean greasy dishes?
a)
Na2SO4
c)
H3PO4
b)
MnO2
d)
LiOH
2.
Which of the following is a salt?
a)
KBr
c)
LiOH
b)
HNO3
d)
SO2
Module III – 4
3.
The lab technician stores chemicals according to their type. Classify the following
substances as acids, bases or salts.
NaCl
H2SO4
KOH
Na2SO4
HC2H3O2
KCIO3
a) acid:
base:
salt:
H2SO4, Na2SO4, KClO3
KOH, Ca(OH)2
NaCl, HC2H3O2, HCl
b) acid:
base:
salt:
H2SO4, HC2H3O2, HCl
KOH, Ca(OH)2
NaCl, Na2SO4, KClO3
c) acid:
base:
salt:
KOH, Ca(OH)2
NaCl, Na2SO4, KClO3
H2SO4, HC2H3O2, HCl
d) acid:
base:
salt:
HC2H3O2, HCl
KOH
NaCl, Na2SO4, H2SO4, Ca(OH)2, KCIO3
Additional Questions:
SQ: Exercises p. 323
Module III – 5
HCl
Ca(OH)2
2.5
To distinguish between ionic and covalent
bonds.
436 only
DME: Sections 10.1, 10.2, 10.3 and 10.5
SQ: pp. 323-327
Key Concepts:
1.
An ion is an atom or group of atoms that has gained or lost electrons.
2.
An ion is a charged particle.
3.
Positive ions are formed by the loss of electrons (these are called cations).
Negative ions are formed by the gain of electrons (these are called anions).
4.
An ionic bond is the bond formed between oppositely charged ions.
Ionic bonds are usually formed between metals and non-metals.
5.
A covalent bond is formed by the sharing of electrons between atoms.
Covalent bonds are usually formed by two non-metals or two identical atoms such as in
Cl2.
6.
A polyatomic ion (radical) is a group of atoms of different elements, usually non-metals,
bonded together to form a single ion.
SAMPLE QUESTIONS
1.
2.
What type of bond is present in a molecule of O2?
a)
polar covalent
c)
covalent
b)
ionic
d)
static
What type of bond is formed between fluorine and magnesium?
a)
polar covalent
c)
covalent
b)
ionic
d)
static
Module III – 6
3.
4.
Which substance has only covalent bonding?
a)
KCl
c)
SO2
b)
MgO
d)
Na2O
Which substance has only ionic bonding?
a)
H2O
c)
P4O10
b)
Na2O
d)
CO2
Additional Questions:
SQ: p. 325
Module III – 7
2.6
To write the molecular formula of various
compounds based on the ionic charges.
436 only
DME: Sections 10.4, 10.5 and 10.6
SQ: pp. 327-330
Key Concepts:
1.
Review the crossover rule: Section 5.4.
2.
Treat polyatomic ions (radicals) as if they are a single element and use the crossover rule to
get the formula. Remember to use brackets if more than one polyatomic ion is needed.
3.
Some examples of polyatomic ions are:
hydroxide ion
OH-1
-1
NO3
nitrate ion
SO4-2
CO3-2
sulfate ion
carbonate ion
NH4+1
PO4–3
ammonium ion
phosphate ion
SAMPLE QUESTIONS
1.
2.
Which of the following is a polyatomic ion?
a)
H+
c)
H2O
b)
OH-
d)
CO2
Which is the correct formula for the compound formed by the Al+3 and SO4-2 ions?
a)
AlSO4
c)
Al2(SO4)3
b)
Al3(SO4)2
d)
Al(SO4)3
Module III – 8
3.
4.
What would be the correct formula for ammonium phosphate?
a)
NH4PO4
c)
(NH4)3PO4
b)
NH4(PO4)3
d)
PO4(NH4)3
The formula for aluminum oxalate is Al2(C2O4)3. What is the charge of the oxalate ion,
C2O4x ?
a)
-1
c)
-6
b)
-3
d)
-2
Additional Questions:
SQ: Exercises p. 330
Module III – 9
2.7
To distinguish between electrolytic and nonelectrolytic substances.
416 and 436
DME: Section 10.7
SQ: pp. 331-332
Key Concepts:
1.
Electrolytes are classified as strong or weak. Electrolytes form positive and negative ions
when dissolved in water.
2.
Strong electrolytes conduct very well because they have a greater degree of ionization.
3.
Weak electrolytes conduct less well because they have a lower degree of ionization.
4.
The fewer the ions present, the weaker the electrolyte.
SAMPLE QUESTIONS
1.
a)
Predict which of the following compounds will be electrolytes in water.
H2
b)
CH3OH
KCl
Sr(OH)2
MgSO4
C6H12O6
Na3PO4
Which ones are non-electrolytes?
Note:
CH3OH looks like a base because of the presence of OH.
However, in this case, it is methanol, a non-conductor or electrolyte.
Methanol is an alcohol.
2.
3.
Which of the following, when dissolved in water, will be an electrolyte?
a)
CO2
c)
O2
b)
HNO3
d)
C6H12O6
Which of the following, when dissolved in water, will be a non-electrolyte?
a)
NaCl
c)
KOH
b)
HCl
d)
C2H5OH
Additional Questions:
SQ: Exercises p. 331
Module III – 10
2.8
To explain the electrolytic property of a
solute using ion dissociation.
436 only
DME: Section 10.8
SQ: pp. 333-336
Key Concepts:
1.
Acids, bases, and salts ionize in water.
2.
Free ions must be present to allow conductivity.
3.
Conductivity depends on the number of ions present.
4.
Acids, bases, and salts that ionize readily form strong electrolytes.
5.
Acids, bases, and salts that ionize to a lesser degree form weak electrolytes.
6.
Substances that do not ionize form non-electrolytes.
SAMPLE QUESTIONS
1.
Explain why acids, bases, and salts conduct electricity?
2.
Why does sulfuric acid, H2SO4, conduct electricity better than acetic acid, HC2H3O4,
(vinegar)?
Module III – 11
2.10 To analyze how non-neutral salts in solution
affect litmus paper.
416 and 436
DME: Section 10.10
SQ: pp. 340-341
Key Concepts:
1.
Not all salts are neutral.
2.
Salts can be acidic or basic.
3.
Litmus paper can be used to identify if the salt is acidic or basic.
SAMPLE QUESTION
1.
You are given a white solid. How would you go about confirming that it is a neutral salt?
Module III – 12
3.1
To prepare specific volumes of solutions at a
specific concentration.
416 and 436
DME: Sections 11.2, 11.3, 11.4
SQ: pp. 349-352
Key Concepts:
Concentration is the measurement of the amount of solute that is dissolved in a
given amount of solution. Concentration can be expressed in several ways;
among these are the two following:
Definition:
1.
Mass Percent:
Mass percent is expressed as:
Mass Percent = Mass of Solute
Mass of Solution
Example:
x 100
A solution labeled 5% acetic acid, contains 5g of acetic acid and 95 g water.
5 g of acetic acid
100 g of solution
2.
Concentration in grams per liter (g/L).
C(g/L) =
mass of solute (g)
volume of solution (L)
A solution of 5 g/L concentration will contain 5 g of solute for every 1 litre or 1000 mL of
solution.
Example:
In order to prepare 600 mL of a salt water solution at 10 g/L concentration, we
need to do the following: We calculate the amount of salt needed for this
concentration:
10 g
= xg
1000 mL
600 mL
x =
6000
1000
=
6g
We place the 6 g of salt in a container and fill it up to 600 mL of solution. We
now have 600 mL of solution with a concentration of 10 g/L.
Module III – 13
SAMPLE QUESTIONS
1.
What mass of solute is needed to prepare 200 mL of a 50 g/L solution?
2.
You are to prepare 300 mL of a 30 g/L aqueous solution of potassium chloride, KCl.
What mass of solute will you need?
3.
a)
9g
c)
100 g
b)
10 g
d)
9000 g
You prepared three solutions in the laboratory. The following table gives information about
each of the solution.
Solution
Mass of Solute
Volume of Solution
1
2.0 g
0.1 L
2
0.6 kg
3.0 L
3
0.2 g
2.0 mL
Arrange these solutions in increasing order of their concentration (from least concentrated
to most concentrated).
4.
What volume of solution would you prepare if you dissolved 20 g of solute and made a
5 g/L solution?
a)
4L
c)
100 L
b)
0.4 L
d)
0.1 L
5.
How much water is there present in 200 g of a 45% by mass solution?
6.
You prepared an aqueous solution of sodium hydroxide, NaOH, that has a concentration
of 15 g/L. To do this, you used 60 g of NaOH.
What is the volume of the solution you prepared?
a)
0.25 L
c)
1.5 L
b)
0.90 L
d)
4.0 L
Additional Questions:
SQ: Exercises p. 351
Module III – 14
3.2
To dilute a solution of known concentration
to a given concentration.
416 and 436
DME: Section 11.5, 11.6
SQ: pp. 352-354
Key Concepts:
A dilute solution contains less solute than a more concentrated solution. To prepare a dilute
solution from a more concentrated solution we use the formula:
C1 x V1 (Concentrated) = C2 x V2 (Diluted)
Note: C is the concentration and V is the volume.
Using this formula you can prepare a given volume of a solution of specific concentration by
diluting a solution of known concentration.
Example: Given 450 mL of a calcium chloride water solution concentrated to 10 g/L,
what will be its concentration if we add 50 mL of water to the solution?
STEP 1
Write down all your variables and their values.
C1 = 10 g/L
C2 = ? g/L
V1 = 0.450 L
V2 = 0.450 L + 0.050 L = 0.5 L
Notice that mL were converted to liters.
STEP 2
Put all these values into the formula and solve for the missing variable.
C1 x V 1
= C2 x V2
10gL X 0.450 L = ? g/L x 0.500 L
? g/L = 10 g/L x 0.450 L
0.500 L
= 9 g/L
Module III – 15
SAMPLE QUESTIONS
1.
You have 500 mL of a 45 g/L solution. You add 1000 mL of distilled water to this solution.
What is the concentration of the diluted solution?
2.
3.
You dilute 50 mL of a 2.8 g/L solution to 200 mL. What is the new concentration of the
solution?
a)
0.7 g/L
c)
7 g/L
b)
11.2 g/L
d)
1.12 g/L
How much of a 5 g/L solution do you need to make 1.5 L of a 3 g/L solution?
Additional Questions:
SQ: Exercises p. 354
Module III – 16
3.3
To associate the concept of the mole with a
measurement of a quantity of matter.
DME: Section 11.7
SQ: pp. 354-356
3.5
To associate the concept of the mole with a
measure of the molar mass of a substance.
436 only
DME: Section 11.10
SQ: pp. 359-360
Key Concepts:
The mole is the amount of substance that contains 6.02 x 1023 (Avogadro’s number) particles
of that substance. It can be abbreviated as mol.
Molar mass of a substance is the mass of a mole of any element or compound.
The molar mass of an uncombined element is derived directly from the periodic table.
To calculate the molar mass of a compound, we add up the molar masses of the elements shown
in the formula.
The number of moles of a given substance can be calculated with the formula:
n
=
mass of substance (g)
molar mass of substance (g/mol)
Examples of moles of different substances:
12 g of carbon is 1 mol
4 g of helium is 1 mol
18 g of water is 1 mol
44 g of carbon dioxide is 1 mol
40 g of argon is 1 mol
40 g of calcium is 1 mol
Examples:
1.
The molar mass of water is:
The molecular formula of water is H2O.
The molar mass
= 2 (1.0 g/mol) + 16.0 g/mol
= 18.0 g/mol
2.
To calculate the molar mass of KNO3:
1 x K = 39.1 g/mol
1 x N = 14.0 g/mol
3 x O = 3 x 16.0 g/mol = 48 g/mol
Therefore, 1 mol KNO3 = 39.1 g + 14.0 g + 48 g
= 101.1 g
Module III – 17
3.
How many moles of hydrochloric acid are in 100 g of hydrochloric acid?
The molecular formula of hydrochloric acid is HCl.
The molar mass
= 1.0 g/mol + 35.5 g/mol
= 36.5 g/mol
Number of moles (n) =
100 g
36.5 g/mol
=
2.74 mol of HCl
There are 2.74 moles.
4.
How many moles are there in 200 g of water?
Water has a molar mass of 18 g.
18 g
200 g
x
= 1 mol
x
=
11.11 mol
SAMPLE QUESTIONS
1.
The molecular formula of alum is Al2(SO4)3.
What is the molar mass of this substance?
2.
3.
a)
170 g
c)
342 g
b)
278 g
d)
450 g
What is the mass of 0.25 mol of potassium nitrate, KNO3?
a)
17.3 g
c)
69.0 g
b)
25.3 g
d)
101.0 g
How many moles are present in 100 g of (NH4)2SO4?
a)
1.32 mol
c)
1.14 mol
b)
0.76 mol
d)
0.88 mol
Additional Questions:
SQ: Exercises pp. 358 and 360
Module III – 18
3.6
To learn to express concentration as moles/liter.
3.7
State the law regarding the concentration of a
solution that contains a constant quantity of
solute.
3.8
Solve numerical problems related to the physical
quantities associated with a solution.
436 only
DME: Sections 11.11, 11.13, 11.14
SQ: pp. 361-366
Key Concepts:
Concentration, using moles as the basic unit of measure, is expressed as mole/liter of solution.
(mol/L).
We can use the same law to dilute a solution in moles per liter (mol/L) as in grams per liter
(g/L). Thus, we can use the formula:
C 1 x V1
= C2 x V2
Example 1:
If you need 4.00 L of 0.200 mol/L of solium chloride solution, what mass of sodium chloride
must you use?
Solution:
0.200 mol/L = amount of solute
4.00 L
amount of solute = 0.200 mL x 4.00 L
1L
= 0.800 mol
The molar mass of NaCl is: 23 g + 35.5 g = 58.5 g
1 mol = mass of NaCl
58.5 g/mol
0.800 mol x 58.5 g/mol
46.8 g
= mass of NaCl
= mass of NaCl
Therefore, you must use 46.8 g of NaCl
Module III – 19
Example 2:
In an experiment, 80 mL of 0.02 mol/L are needed. A 0.1 mol/L solution of HCl is available.
How much of this solution must be added to distilled water to obtain the needed concentration?
Solution:
C1 x V1 = C2 x V2
C1 = 0.1 mol/L
C2 = 0.02 mol/L
V2 = 80 mL
0.1 mol/L x V1 = 0.02 mol/L x 80 mL
Therefore V1 = 16 mL
Example 3:
Suppose you want to make as large a volume of 6.00 mol/L sodium nitrate solution as possible.
You have a bottle with 1.5 kg of this substance. What volume of solution can you make?
Solution:
The formula of sodium nitrate is NaNO3.
The molar mass of NaNO3.
23.0 g + 14.0 g + (3 x 16.0) = 85 g/mol
Amount of available NaNO3 (mol) =
mass of NaNO3
molar mass of NaNO3
= 1.5 x 103 g
85 g/mol
= 17.65 mol
Molar concentration =
6.00 mol/L =
amount of solute
volume of solution
17.65 mol
volume of solution
Volume of solution = 17.65 mol
6.00 mol/L
= 2.94 L
So you can make 2.94 L of solution.
Module III – 20
SAMPLE QUESTIONS
1.
You are to prepare 300 mL of an aqueous solution of sodium hydroxide, NaOH, that will
have a concentration of 0.8 mol/L.
What mass of NaOH do you need?
2.
a)
0.24 g
c)
9.6 g
b)
6.0 g
d)
32 g
What is the concentration of the solution when 125 g of MgSO4 are dissolved to form 2.5 L
of solution?
a)
0.38 mol/L
c)
2.40 mol/L
b)
0.42 mol/L
d)
2.60 mol/L
3.
500 mL of a 4 mol/L solution are diluted to form a 1 mol/L solution. What volume of water
is added to the original solution?
4.
The lab technician is asked to prepare 2 L of 0.3 mol/L HNO3 solution. He has a stock
supply of 12 mol/L HNO3 solution. How much of the stock solution does he need?
Additional Questions:
SQ: Exercises pp. 362 and 366
Module III – 21
4.1
Describe how acids and bases in solution
affect certain colour indicators.
416 and 436
DME: Section 12.1
SQ: pp. 371-372
Key Concepts:
An indicator is a substance capable of changing the colour of a basic or acidic solution.
Common examples are:
• Red litmus turns blue in the presence of a base.
• Blue litmus turns red in the presence of an acid.
• Phenolphthalein turns pink in the presence of a base.
Other indicators such as methyl orange, methyl red, and bromothymol blue are further
examples of indicators.
SAMPLE QUESTIONS
1.
In a laboratory, a student has prepared three solutions: HCl, NaCl and NaOH. He has
forgotten to label his three solutions. To identify each, he has used litmus paper. These
are his results:
Solution A:
Solution B:
Solution C:
turns red litmus to blue
no reaction
turns blue litmus to red
Identify the three substances.
2.
The following are some characteristics of a certain liquid:
• produces a gas when in contact with a piece of metal
• conducts electricity
• turns litmus paper red
How would you describe this liquid?
a)
This liquid is a neutral solution.
c)
This liquid is a basic solution.
b)
This liquid is an acidic solution.
d)
This liquid is a neutral salt solution.
Module III – 22
3.
To determine the pH of fruit juice, one would normally use universal indicator paper.
However, none of this paper is available. You use litmus paper instead and observe that
the juice turns blue litmus paper red.
What could the pH of the juice be?
Module III – 23
4.2
Determine the pH of a solution using a
universal indicator.
416 and 436
DME: Section 12.2
SQ: pp. 372-374
Key Concepts:
There exists a special type of indicator called “universal indicator paper” which turns a different
colour depending on how acidic or basic a solution is.
It is possible with this indicator to determine how acidic or basic a solution is by matching
different colours of the indicator paper with specific levels of acidity or basicity.
The level of acidity or basicity is determined by its pH. The pH scale looks like the following:
pH Values
0
1
2
3
4
more acidic
5
6
7
8
9
10
11
12
neutral
14
more basic
Some examples of common substances and their pH:
Substance
pH
Acid, Base or Neutral
Distilled Water
Tomato Juice
7
4.2
neutral
acid
Salt Water
Human Blood
8.4
7.3 - 7.5
base
base
Floor Cleaner
Vinegar
11
2.8
base
acid
Egg
Gastric Juice
7.8
2.0
base
acid
Lemon
Orange
2.8
3.5
acid
acid
Bread
Milk
5.5
6.5
acid
acid
Soft Drink
3.0
acid
Module III – 24
13
SAMPLE QUESTIONS
1.
The following table gives the colours of a universal indicator in solutions that have pH
values ranging from 1 to 13.
Colour
Red
Yellow
Turquoise Blue
Violet
pH
1
5
9
13
A given solution turns orange when this indicator is added.
Which of the following statements is definitely TRUE?
2.
a)
This solution is acidic.
c)
This solution is neutral.
b)
This solution is basic.
d)
This solution is saline.
A universal indicator changes colour as indicated by the table below.
pH
1
3
5
7
9
11
13
Colour
green
blue
violet
colourless
red
orange
yellow
What are the indicators’ colours in the presence of an acid ?
3.
a)
Green, blue, violet
c)
Colourless
b)
Green, blue, violet, colourless
d)
Red, orange, yellow
Blood in the body has a pH between 7.2 and 7.4. This means that blood is:
a)
a strong acid
c)
a strong base
b)
slightly acidic
d)
slightly basic
Module III – 25
4.
Kim and Sebastien measured the pH of different solutions using a universal
indicator. The following results were recorded:
SOLUTIONS
Salt water
pH
8
Soft drink
Cleaning liquid
3
11
Window cleaner
Antacid
9
10
Grape juice
Potato juice
3
6
Vinegar
3
Which solutions are acidic ?
a)
Cleaning liquid, grape juice, antacid and salt water.
b)
Soft drink, grape juice, potato juice and vinegar.
c)
Cleaning liquid, window cleaner, soft drink and vinegar.
d)
Potato juice, window cleaner, antacid and salt water.
5.
Given NaOH, CaCl2, H2SO4, NH4OH, H2O, circle which have a pH over 8 ?
6.
The table below indicates the colour of the indicator phenol red in solution
with a pH varying from 1 to 12.
Color
pH
Yellow
2
4
6
Orange
Red
8
10
12
A drop of this indicator is added to some lemon juice.
What colour is the indicator after being added to the lemon juice? Explain your answer.
Additional Questions:
SQ: Exercises p. 374
Module III – 26
4.3 and 4.4 Determine the turning point of one or
more indicator(s).
416 and 436
DME: Section 12.3, 12.4
SQ: pp. 374-377
Key Concepts:
The pH at which an indicator changes colour is called its turning point.
This point may not be a sudden colour change but a gradual change from one colour to another.
Example:
Methyl orange remains red up to and including pH 3, starts to turn orange up to
pH 4, then changes to yellow at pH 5 and above.
TABLE OF INDICATORS
INDICATOR
COLOUR
pH OF TURNING POINT
Methyl violet
Methyl orange
Bromophenol blue
yellow .... violet
red .... yellow
yellow .... blue
0.2 - 2.0
3.0 - 4.4
3.0 - 4.6
Bromocresol green
Methyl red
yellow .... blue
red .... yellow
3.8 - 5.4
4.4 - 6.2
p-nitrophenol
Bromocresol violet
colourless .... yellow
yellow .... violet
5.0 - 7.0
5.2 - 6.8
Bromothymol blue
Phenol red
yellow .... blue
yellow .... red
6.0 - 7.6
6.4 - 8.2
Litmus paper
Violet m-cresol
red .... blue
yellow .... violet
7.6 - 9.2
Phenolphthalein
Alizarin R yellow
colourless .... fuchsia
yellow .... red
8.2 - 10.0
10.1 - 11.1
Carmine indigo
blue .... yellow
12.0 - 14.0
One indicator may not give a precise value for the pH of a solution, therefore, it is advantageous
to mix indicators to obtain more than one turning point which will help identify the pH.
Module III – 27
SAMPLE QUESTIONS
1.
The following table gives the colours of the indicator bromothymol blue in solutions whose
pH values vary from 0 to 14. What is the turning point of the indicator?
pH
1
3
Colour
2.
5
7
YELLOW
9
11
13
BLUE
GREEN
a)
from pH 0 to pH 7.6
c)
from pH 6.0 to pH 14.0
b)
from pH 6.0 to pH 7.6
d)
from pH 7.6 to pH 14.0
The following table gives the colours of the indicators methyl orange and bromothymol
blue in solutions whose pH values vary from 0 to 14.
Colour
pH
RED
1
Colour
YELLOW (Methyl Orange)
ORANGE
3
5
7
YELLOW
9
11
13
BLUE
GREEN
(Bromothymol Blue)
A solution turns yellow when methyl orange is added; it also turns yellow when
bromothymol blue is added.
What could the pH of this solution be?
a)
4
c)
7
b)
5
d)
9
Module III – 28
3.
The following table gives the colours of the indicators phenolphthalein and methyl red in
solutions that have pH values ranging from 1 to 14.
Colourless
Phenolphthalein
pH
1
Methyl Red
2
3
4
Red
5
Pink
6
7
8
Red
9
10
11
12
13
14
Yellow
Orange
A mixture of these two indicators is added to a neutral solution.
What colour will the mixture of the two indicators be in this neutral solution?
4.
a)
colourless
c)
orange
b)
yellow
d)
pink
The following table gives the colours of the indicator phenol red in solutions whose pH
values vary from 0 to 14. A few drops of this indicator are added to a basic solution. What
colour does the phenol red become ?
pH
1
33
Color
55
7 7
9 9
11 11 13
Red
Orange
Yellow
13
5.
A solution in the lab has pH 9.5. Using the Table of Indicators, which would be the best
indicator to identify this pH?
6.
Below is the colour scale of three indicators.
pH
A
B
C
11
33
55
Red
7 7
9
11 13
13
Yellow
Orange
Yellow
9 11
Blue
Green
Yellow
Red
Orange
a) What is the colour of each indicator in the presence of an acid?
b) What is the colour of each indicator in the presence of a base?
c) What is the turning point of each indicator?
Module III – 29
Refer to the Table of Indicators to answer questions 7, 8 and 9.
7.
8.
9.
A certain salt is very soluble in an acidic solution. Which indicator should be used to
demonstrate that the solution is of the highest acidity ?
a)
Methyl orange
c)
Bromothymol blue
b)
Methyl red
d)
Carmine indigo
Indicate which of the above indicators would be the best to use to determine that the pH of
water in an aquarium is neutral.
a)
Methyl orange
c)
Bromothymol blue
b)
Methyl red
d)
Carmine indigo
Which of the following mixtures would be the best to determine the pH of an unknown
solution ?
a) Methyl orange and methyl red
c) Methyl orange and violet m-cresol
b) Methyl red and bromothymol blue
d) Bromothymol blue and violet m-cresol
This table gives the behavior of certain indicators. Use it to answer questions 10 and 11.
pH
2
3
A
6
7
8
9
Red
D
Red
10
Green
Orange
11
12
Blue
Colourless
C
11.
5
Yellow
B
10.
4
Pink
Fuchsia
Yellow
Orange
Yellow
Green
You are given a solution and told it is neutral. You would like to check if this is true. Which
indicators in the table will you use ?
a)
Indicator A
c)
Indicator C
b)
Indicator B
d)
Indicator D
The pH of a given solution is unknown. Indicators A and C turn yellow in this solution.
What colour will indicator D become in the solution ?
a)
Red
c)
Yellow
b)
Orange
d)
Green
Additional Questions:
SQ: Exercises pp. 376 and 377
Module III – 30
4.6
Associate the pH scale with the scale of the
molar concentrations of H1+ and OH1-
4.7
Explain the molar concentration of H+ and
OH- ions in pure water, based on its pH
value and molecular formula.
436 only
DME: Sections 12.6 and 12.7
SQ: pp. 380-382
Key Concepts:
ACIDS contain a greater number of H+ ions than OH-.
NEUTRAL SOLUTIONS (water) contain an equal number of H+ and OH- ions.
BASES contain less H+ ions than OH- ions.
[H ] mol/L
RANGE
10 .... 10-7 .... 10-14
pH
[OH ] mol/L
0 .... 7 .... 14
10 .... 10-7 .... 100
Nature
Acid .... Neutral .... Base
+
0
-
-14
The level of acidity or basicity is based on the nature of the acid or base and its concentration.
The pH can be expressed as the positive value of an exponent in H+ ion concentration.
Example:
In HCl 0.1 mol/L, the concentration of H+ will be [H+] = 1.0 x 10-1 = pH 1.
In HCl 0.001 mol/L, the concentration H+ will be [H+] = 1 x 10-3 = pH 3.
The product of [H+] and [OH-] is always 1 x 10-14.
Therefore, 0.1 mol/L contains H+ at 1 x 10-1 and OH- at 1 x 10-13 because
(1 x 10-1) x (1 x 10-13) = 1 x 10-14.
For NaOH, 0.1 mol/L:
[OH] = 0.1 mol/L = 1 x 10-1 mol/L
[OH-] x [H+] = 1 x 10-14
l x 10-1 x [H+] = 1 x 10-14.
Therefore
[H+] = 1 x 10-13
pH = 13
Module III – 31
For distilled water, the pH is 7 and so the concentration H+ is 1 x 10-7 mol/L.
To calculate the concentration of OH-, we use the equation:
[H+] x [OH-] = 1 x 10-14
1 x 10-7 x [OH-] = 1 x 10-14
therefore,
[OH-] = 1 x 10-7 mol/L.
SAMPLE QUESTIONS
1.
If solution A has a pH 3.5, solution B has pH 4.0, solution C has pH 4.6, and solution D has
pH 5.6, which solution has the greatest H+ ion concentration?
a)
Solution A
c)
Solution C
b)
Solution B
d)
Solution D
2.
What is the pH of a solution with a [H+] of 0.01 mol/L ?
3.
In solution A, the [H+] is 10-3 mol/L and in solution B, the value is 10-4 mol/L. Which solution
is more acidic ?
4.
What is the [OH-] in a solution with pH 6 ?
5.
Fill in the blank space for the following three solutions.
+
[H ] mol/L
[OH-] mol/L
pH
A
10-1
B
C
10-10
13
nature
6.
The pH of a swimming pool was found to be 7.2. It then rained for two days and when the
pH was checked again it was found to be 6.1. What do you think caused the change in pH
and how would you re-establish the original pH? Explain your answers.
Module III – 32
7.
Fill in the table below for each solution.
[H+]
A
B
C
8.
[OH-]
10-5 mol/L
pH
Nature
0.001 mol/L
6
You are given two colourless solutions, A and B. You place a piece of litmus paper in
solution A and it becomes blue. You then place a couple of drops of an unknown indicator
in solution A and it becomes green. Then you place a couple of drops of the same
indicator in solution B and solution B becomes yellow.
What is the nature (acid or base) of solution B? Justify your answer.
Module III – 33
5.1
Neutralize an acidic substance in his/her
environment.
416 and 436
DME: Sections 12.10 and 12.11
SQ: pp. 390-392
Key Concepts:
An acid is characterized by the presence of H+ (hydrogen ions) and a base characterized by the
presence of OH- (hydroxide ions). If we mix an acid and a base and the number of H+ and OHare equal, we have neutralization.
If [H+] = [OH-], the solution is neutral.
HCl
Hydrochloric
Acid
+
NaOH
Sodium
Hydroxide
→
NaCl
Sodium
Chloride
+
H2O
Water
By adding a base to an acid or vice versa, we can neutralize the base or acid and the products
are a salt and water. By using indicators such as litmus, universal indicator or phenolphltalein,
we can obtain exact neutralization.
SAMPLE QUESTIONS
1.
When hydrochloric acid, HCl, is neutralized by magnesium hydroxide, Mg(OH)2,
magnesium chloride, MgCl2, and water, H2O, are formed.
Write the balanced equation that represents this neutralization reaction.
2.
In the laboratory, you are to neutralize a basic solution before disposing of it.
Explain how you would neutralize this solution.
In your explanation, indicate the materials used, the steps involved and the
observation which shows that the solution has been neutralized.
3.
You find a bottle containing an unidentified liquid. By using universal indicator paper, you
determine that the pH of this liquid is 11. Therefore you have to neutralize it before
disposing of it by adding a solution whose pH is _____ ?
a)
pH 3
c)
pH 7
b)
pH 11
d)
pH 0
Additional Questions:
SQ: Exercises p. 392
Module III – 34
5.2
To represent a chemical change, after
conducting an experiment involving a
neutralization reaction.
416 and 436
DME: Section 12.12
SQ: pp. 392-394
Key Concepts:
Neutralization of the following acids and bases can be represented in the following way:
Sodium hydroxide + Hydrochloric acid → Sodium Chloride + Water
NaOH
base
+
HCl
acid
→
NaCl
salt
+
H2O
water
Sulfuric acid + Potassium hydroxide → Potassium sulfate + Water
H2SO4
+
2 KOH
→
K2SO4
+
2 H2O
SAMPLE QUESTIONS
1.
2.
Which equation below corresponds to the neutralization of hydroiodic acid, HI, by
potassium hydroxide, KOH?
a)
HI + KOH → KI + H2O
c)
HI + KOH → KO + H2I
b)
HI + KOH → KH + IOH
d)
HI + KOH → KIOH2
A product used as an oven cleaner contains the strong base potassium hydroxide, KOH.
After using this product to clean the oven, you are told to rinse with a solution of vinegar,
CH3COOH, so that the excess product is neutralized.
a)
Why is vinegar used to neutralize the product ?
b)
Knowing that the neutralization reaction forms the salt potassium acetate,
CH3COOK, write the balanced equation for this reaction.
Module III – 35
5.3
To explain the law of conservation of matter
in a chemical change, after conducting an
experiment involving a neutralization
reaction.
416 and 436
DME: Section 12.13
SQ: pp. 394-396
Key Concepts:
In a chemical reaction, the mass of the reactants is always equal to the mass of the products.
This statement is called “The Law of Conservation of Mass”.
Example:
Reactants
Products
CH4 + 202
→
CO2 + 2 H2O
8 g + 32 g
→
22 g + 18 g
40 g
=
40 g
SAMPLE QUESTIONS
1.
In a chemical reaction of hydrochloric acid and calcium carbonate, the mass of the
reactants was 11 g. What will be the mass of the products?
2.
Given the following equation:
2 H2 + 02 → 2 H2O
If 8 g of hydrogen reacts completely with 64 g of oxygen, what will be the mass of water
formed?
Additional Questions:
SQ: Exercises p. 396
Module III – 36
5.4
To represent chemical changes in the form
of balanced equations.
416 and 436
DME: Sections 12.14 and 12.16
SQ: pp. 397-400
Key Concepts:
In all chemical reactions, the number of atoms of each element in the reactants will be equal to
the number of atoms of each element in the products (in a balanced equation).
Thus:
NaOH
+
HCl
→
NaCl
+
H2O
In this reaction, even though the atoms have been arranged in different compounds, the number
of atoms of each element remains the same on each side of the equation.
In another example, if we count the number of atoms of each element in the reactants, we find it
is the same in the products (in a balanced equation).
Zn
+
2HCl
→ ZnCl2 +
H2
Reactants
Products
Zn
1
1
H
2
2
Cl
2
2
In the example above, the equation has been balanced by using 2 HCl in the reactants.
Module III – 37
SAMPLE QUESTIONS
1.
In the laboratory, a potassium hydroxide, KOH, solution is neutralized with sulfuric acid,
H2SO4.
Which balanced equation correctly represents the neutralization reaction?
2.
3.
a)
KOH + H2SO4 → H2O + KSO4
c)
KOH + H2SO4 → OHSO4 + KH2
b)
KOH + H2SO4 → H2O + K2SO4
d)
2 KOH + H2SO4 → 2 H2O + K2SO4
Which of the following equations is balanced?
a)
2Fe2O3 + 3C → 3CO2 + 2Fe
b)
CH4 + O2 → CO2 + 2H2O
c)
Cu + 4HNO3 → Cu(NO3)2 + 4NO2 + 2H2O
d)
4NH3 + 3O2 → 2N2 + 6H2O
Identify the missing substance in the following balanced equation:
CaCO3 + 2NaCl →
4.
_____ + CaCl2
a)
NaCO
c)
Na2CO3
b)
NaCO3
d)
2NaCO3
Balance the following equations:
a)
KOH + H2SO4 → K2SO4 + H2O
b)
NH3 + Cl2 → N2 + NH4Cl
c)
Al + HCl
d)
Mg + H3PO4
→
e)
Fe2S3 + O2
→
Fe2O3 + SO2
f)
PbO2 + HCl
→
PbCl2 + H20 + Cl2
g)
Cu(NO3)2
→
→
AlCl3 + H2
Mg3(PO4)2 + H2
CuO + NO2 + O2
Additional Questions:
SQ: Exercises p. 400
Module III – 38
5.5
Verify that various chemical changes are
consistent with the law of conservation of
mass.
436 only
SQ: pp. 400-402
Key Concepts:
In the reaction:
2H2 + O2 → 2H2O
the balanced equation tells us that it takes two mol of H2 for every 1 mol of O2 to produce 2 mol
of H2O.
The molar mass of the reactants is:
2 H2 :
O2 :
2 H2 O :
2x2g
=
2 x 16 g
=
2 x 1 g + 1 x 16 g =
Reactants
2H2 + O2
→
Products
2H2O
4g
=
36 g
32 g
4g
32 g
36 g
Thus, the sum of the masses of the reactants (or reagents) is equal to the sum of the masses of
the products.
Additional Questions:
SQ: Exercises p. 402
Module III – 39
5.6
To determine, using stoichiometric
calculations, the quantity of a substance
involved in a chemical reaction.
5.8
To solve numerical problems related to
stoichiometry.
436 only
DME: Sections 12.18 and 12.19
SQ: pp. 402-406
Key Concepts:
The law of conservation of mass can be used to predict the mass of a substance in a chemical
reaction.
Example:
The oxidation of iron is:
4 Fe + 3 O2 → 2 Fe2O3
Calculate the quantity of iron used to form 100 g of ferric oxide:
1.
Starting with a balanced equation, determine the number of moles of each substance.
4 mol
4 Fe
+
3 O2
→
x
2 mol
2 Fe2O3
100 g
2.
Underline the substances you are working with and write the experimental amount under
each substance.
3.
Find the mass for the substances in grams.
224 g
4 Fe
+
3 O2 →
x
4.
100 g
Use ratios to calculate the unknown mass.
224 g of Fe
x of Fe
320 g
2 Fe2O3
→
→
320 g of Fe2O3
100 g of Fe2O3
x = 70 g of Fe.
Module III – 40
Example:
A student wants to produce 10 g of calcium from the reduction of chalk (CaO) by
aluminium, given by the following equation:
3 CaO + 2 Al → Al2O3 + 3 Ca
What mass of aluminium must be used in this reaction?
2 mol
3 mol
→
3 CaO + 2 Al
x
Al2O3 + 3 Ca
10 g
54 g
120 g
3 CaO + 2 Al
x
→
54 g of Al →
x of Al →
120 g of Ca
10 g of Ca
Al2O3 + 3 Ca
10 g
x = 4.5 g of Al
SAMPLE QUESTIONS
1.
Complete the chart:
CaCO3 + 2 HCl
→ CaCl2
+
H2 O
+
CO2
100 g
100 g
100 g
100 g
100 g
Module III – 41
2.
The reaction caused by the burning of butane in air is represented by
the following equation:
2C4H10(g) + 13O2(g)
→ 8CO2(g) + 10H2O(g) + Energy
During a laboratory experiment you reacted 29 g of butane (C4H10) in the presence of
oxygen. What mass of oxygen was needed for this reaction?
3.
What mass of carbon dioxide is needed in the neutralization of 100 g of nitric acid given
the equation:
2 HNO3 + CaCO3
→
Ca(NO3)2 + H2O + CO2
4.
By burning methane, CH4, in air containing oxygen gas, O2, you produce carbon dioxide,
CO2, and water vapour, H2O. You are to burn 192 g of methane. What mass of carbon
dioxide gas will be produced?
5.
Iron, Fe, and carbon monoxide, CO, are produced when iron oxide Fe2O3, reacts with
carbon, C. You would like to produce 50 moles of iron. What mass of iron oxide is
required?
6.
Octane, C8H18, is one of the main components of gasoline. When a car engine is running,
octane burns by reacting with oxygen gas, O2, in the air. Carbon dioxide, CO2, and water
vapour, H2O, are produced during this combustion reaction. The balanced equation for this
reaction is;
2C8H18 + 25O2 → 16CO2 + 18H2O
The car engine ran for a certain period of time and 57 g of octane were burned. What
mass of carbon dioxide was released into the atmosphere?
7.
One of the substances responsible for acid rain is nitrogen dioxide (NO2). In the
atmosphere, nitrogen dioxide reacts with water vapour in the air to produce nitric acid,
HNO3, according to the following balanced equation:
3NO2 + H2O
→
2HNO3 + NO
If 1000 g of NO2 reacts in the atmosphere during a certain period of time, what mass of
nitric acid is produced ?
Module III – 42
8.
In industry, water vapour is decomposed by passing the water vapour over hot iron.
3Fe + 4H2O
→
Fe3O4 + 4H2
The iron reacts with the water to form an oxide of iron, Fe3O4, and hydrogen gas, H2. What
mass of iron is required if you produced 2 moles of hydrogen?
9.
Calculate the mass of sodium hydroxide that neutralizes 14.7 g of sulfuric acid. The
equation of this reaction is:
2NaOH + H2SO4
→
Na2SO4 + 2H2O
Additional Questions:
SQ: Exercises p. 406
Module III – 43
6.4
Identify the effects of chemical substances
released into the environment.
7.3
Analyze the social, economic and
environmental consequences of consumer
product recycling.
416 and 436
DME: Sections 13.1 – 13.4, 13.6, 13.9
SQ: pp. 443-445
Key Concepts:
One of the principal problems facing our society today stems from chemical reactions which
occur in our environment which do not simply disappear. These environmental problems are
due to human cause.
ACID RAIN
Sulfur is emitted by:
–
burning of fuels (cars, coal, power stations to drive turbines,
incineration, forest fires, volcanoes)
mineral refineries (e.g. Noranda, Algoma Steel, INCO)
–
• sulfur combines with O2 in atmosphere to produce sulfur dioxide
S
+
O2
→
SO2
• sulfur dioxide combines with more O2 in the atmosphere to produce sulfur trioxide
2SO2
+
O2
→
2SO3
• sulfur trioxide combines with water vapor in atmosphere to produce diluted sulfuric
acid which falls as acid rain
SO3
Note:
+
H2O
→
H2SO4
The same reaction occurs with nitrogen oxides which combine with water
vapor in the atmosphere to produce nitric acid.
Normal rain has a pH of 5.6 but acid rain has a pH of 2.5.
Tall smoke stacks send pollutants high into the atmosphere and wind carries these pollutants
great distances. The pollutants combine with water to produce acid rain far away from the
original source.
Module III – 44
Effects of acid rain on environment:
1.
Acid rain corrodes roads, buildings, statues and monuments.
2.
Acid rain affects the growth of plants. Leaves, germination of seeds, growth of vegetables
are all harmed.
3.
Acid rain destroys plants and animals in lakes. The growth cycles, the respiration in fish,
and plant tissue are particularly affected.
4.
Acid rain causes leaching of heavy metals (Hg, Al, Cd, Cu…) into the drinking water.
OTHER POLLUTANTS
Principal pollutants in the atmosphere:
POLLUTANTS
PRINCIPAL SOURCE
EFFECTS
Hydrocarbons
gasoline combustion
smog, respiratory disease
Carbon monoxide
incomplete combustion of
gasoline
toxicity
Chlorofluorocarbon (CFC)
Refrigeration, aerosol sprays,
plastic manufacture
destruction of ozone layer
Polychlorophenols (PCB)
transformers
aquatic plants and animals
Organochlorides:
dioxins and furanes
incinerators
cancers
Dust
mines, industries, cities
respiratory disease
Radioactive fallout
nuclear reactors, nuclear
energy experimentation
mutations, cancer
Carbon dioxide
fossil fuels, plants
contributes to greenhouse
effect
Principal Pollutants in aquatic environment:
POLLUTANTS
PRINCIPAL SOURCE
EFFECTS
Hydrocarbons
(petroleum, benzene)
industries, transportation,
organic solvents
destruction of aquatic fauna,
polluted water
Human waste and oils
industries, municipal sewage,
gasoline
pollution of potable water,
bacterial and viral disease
Inorganic material
(acid, bases, salts, metals)
industries, mines, municipal
sewage
corrosion, toxicity
Fertilizers
(nitrates, phosphates)
agriculture, industries
toxicity
Pesticides, herbicides
(DDT)
agriculture, forestry
cancers, toxicity, disruption
in food chain
Module III – 45
Principal pollutants in soil:
POLLUTANTS
PRINCIPAL SOURCE
EFFECTS
Radioactive wastes
nuclear reactors, nuclear
mutations, cancers
energy experimentation
Domestic solids
tires, cars ...
industries, municipal sewage,
transportation
alteration of scenery, risk
of fire
Toxic metals
(Pb, Hg, Al, Cu ...)
industries, mines, municipal
sewage
disruption in food chain,
toxicity, cancer
Toxic liquid wastes
mines, industries,
municipalities
filtration, vegatation,
toxicity
Organochlorides
agriculture, industries,
electrical transformers
air pollution, water
pollution, cancers
Pesticides, herbicides
(DDT)
agriculture, forestry, domestic
use
cancers, toxicity, disruption
in food chain
Microbes
agricultural wastes
infectious diseases
REDUCTION OF POLLUTANTS:
1.
Recycle:
• use biodegradable materials,
• introduce recycling programs
• use recyclable materials
2.
Reuse:
• find alternative methods of transportation (bus, metro, cycle, car pool)
• buy reusable products
3.
Reduce:
• car emissions
• the use of non biodegradable substances
4.
Develop better filtration systems for industries.
5.
Find alternative methods for the generation of electrical power. (aeolian, wind, solar,
geothermal)
6.
Find alternative energy sources for the combustion fossil fuels for cars.
Module III – 46
SAMPLE QUESTIONS
1.
A factory was built on a lake close to a town. Before the factory was built, the lake was full
of plants and animal life, and the pH of the water was 6.0. The factory burns carbon and
sulfur to produce CO, CO2, SO2.
For the past several months, the town people have complained about an increase amount
of algae in the lake and dead fish often floating on the surface of the lake. The
municipality supported its citizens by having the water in the lake analyzed. The results of
this analysis showed the pH was 4.5.
Explain how the burning of carbon and sulfur and the release of gases into the air can
change the pH of the water in the lake.
Explain in what ways the gases released have toxic effects on humans and the
environment.
2.
Studies show that traces of pesticides are present in streams and rivers and in the human
body. How can you explain that pesticides sprayed on lawns are found first in streams
and rivers and then in the human body?
3.
Incinerators do not eliminate all substances harmful to the environment since they produce
some themselves. By referring to the information below, name two substances harmful to
the environment that are produced by incinerators. For each substance, state why it is
harmful.
carbon dioxide
water vapor
oxygen
nitrogen
carbon monoxide
sulfur dioxide
nitrogen oxides
dioxins
furanes
=
99.95 %
=
0.05 %
=
0.0002 %
Module III – 47
4.
5.
6.
Which gases are responsible for producing acid rain?
a)
Carbon monoxide and sulfur dioxide
b)
Carbon dioxide and methane
c)
Nitrogen oxide and sulfur dioxide
d)
Methane and nitrogen oxide
All of the following except one is a primary cause of acid rain. Which of the following is
NOT a primary cause of acid rain?
a)
the refinement of metals
b)
transportation (cars, boats, planes, trucks, etc.)
c)
burning of coal for heating
d)
the liberation of CFC in atmosphere
The tables below show four substances released into the environment by various human
activity. Which table matches correctly each substance with the damage it does
to the environment ?
a)
Substance
SO2
lead
CO2
freon
Damage of Environment
acid rain
soil contamination
greenhouse effect
depletion of the ozone layer
b)
Substance
SO2
lead
CO2
freon
Damage to Environment
acid rain
depletion of the ozone layer
soil contamination
greenhouse effect
c)
Substance
SO2
lead
CO2
freon
Damage to Environment
soil contamination
acid rain
depletion of the ozone layer
greenhouse effect
d)
Substance
SO2
lead
CO2
freon
Damage to Environment
greenhouse effect
soil contamination
depletion of the ozone layer
acid rain
Module III – 48
7.
8.
Which type of generating station is responsible for the production of acid rain ?
a)
Hydro-electric
c)
Nuclear power
b)
Wind power
d)
Thermal power
Which of the following is NOT a major cause of acid rain ?
a)
Garbage in open dumps
b)
The chemical transformation of certain ores in a metallurgical plant.
c)
Automobile traffic
d)
Residential oil heating systems
9.
Nuclear-powered, hydro-electric, coal-burning and diesel-powered generating stations
produce electrical energy by different methods. All of them have negative effects on the
environment. Explain one negative effect that each type of generating station has on the
environment.
10.
How could the burning of coal for the steel making industry in Detroit, be responsible for
acid rain in Quebec ?
11.
How would you test if the precipitation falling in your neighborhood is acid rain ?
12.
Explain the greenhouse effect.
13.
Give another two ways to reduce pollution in your city.
14.
Explain how CFC effect the ozone layer.
Module III – 49
MODULE I
— ANSWER KEY —
REVIEW OF PHYSICAL SCIENCE 214
1.
•
•
•
•
mass the ring
use displacement of water to measure volume
calculate density
compare to known density of gold
2.
Same procedure as above but now you must see if the density of the penny is 8.92 g/cm3.
OBJECTIVE 2.1
1.
c)
Mass and volume are universal properties.
2.
d)
Only density and boiling point are characteristic; the others are shared by many.
3.
b)
4.
b)
Density is a characteristic property. 1 and 4 have the same density.
5.
a)
Only 1 is characteristic. The others are properties shared by many substances.
OBJECTIVES 2.2 AND 2.3
1.
a)
Glowing splint relights, shows gas is oxygen.
2.
c)
3.
b)
The gas is not oxygen or hydrogen as no reaction to flame or glowing splint. It is
not ammonia as the density is not 0.76 g/L
D = M = 8.9 g = 1.78 g/L
V
5L
Thus it is argon.
4.
c)
The density of unknown liquid is D = M =
1 g = 0.8 g/mL
V
1.25 mL
5.
c)
The substance is iron because the density of unknown solid is 7.87 g/cm3 and it is
magnetic and conducts electricity.
ANSWER KEY
Module I – 1
OBJECTIVE 2.4
1.
c)
Being both a good heat conductor and having a high melting point are essential in a
cooking pot.
2.
Aluminum won't rust or rot and is lightweight and strong.
Wood will rot.
Plastic may break.
3.
a)
Aluminum is chosen because it is a good conductor of electricity.
OBJECTIVE 3.2
1.
b)
They change the basic nature of the food and thus are chemical changes.
2.
b)
A new substance forms, others are changes in state.
3.
c)
This step changes the basic nature of the potatoes.
4.
d)
Colour changes and production of heat and light are evidence of chemical change.
5.
c)
New substances are produced. See evidence for chemical change.
6.
b)
Production of a solid (precipitate) is evidence of chemical change.
OBJECTIVE 3.3
1.
a)
Water is a compound, which is a pure substance with elements chemically bonded.
OBJECTIVE 3.4
1.
c)
The substance has been broken down as a chemical substance has been lost.
2.
c)
Two new substances produced from one.
3.
c)
Yellow copper turns to black copper oxide.
4.
a)
Element + element → compound.
5.
Experiment 1: Shows it is a compound as it produces two substances and it loses mass.
6.
c)
The yellow solid combines with oxygen to form a gaseous compound. All solutions
are mixtures.
7.
d)
Only this shows evidence of chemical change and since a gas was released, it shows
the original was broken down.
8.
d)
1 and 2 are changes in state. Only 3 and 4 show evidence of chemical change.
ANSWER KEY
Module I – 2
OBJECTIVES 4.1, 4.2, 4.3
1.
c)
Aristotle
Democritus
Dalton
Bohr-Rutherford
–
–
–
–
continuous matter
discontinuous matter
no internal structure
nucleus and shells
2.
d)
Dalton’s atoms had no internal structure.
OBJECTIVE 4.4
1.
d)
Rubber – negative
Silk – negative
Wool – positive
Glass – positive
2.
Because A and B repel, they have like charges. Because C attracts both A and B, it has a
different charge.
3.
The sugar deviates from the vertical because the sugar has developed an electric charge
which, being the same, causes the sugar grains to repel each other. The sifter has
developed a charge opposite to the sugar grains and thus attracts them.
4.
a)
The ruler has the same charge as Ball A and an opposite charge to Ball B.
OBJECTIVE 4.5
1.
d)
Because positive charge attracts negative charge.
OBJECTIVE 4.6
1.
b)
Though the other statements are true, the experiment only shows particles are
charged.
2.
b)
1 attracted to positive, 2 no effect, 3 attracted to negative
3.
c)
Only gamma radiation has no electric charge and is not affected by an electric field.
OBJECTIVE 4.7
1.
b)
The other facts are found in the theories of Bohr and Rutherford.
2.
c)
The others are part of other atomic theories.
ANSWER KEY
Module I – 3
OBJECTIVE 4.8
1.
a)
2.
c)
3.
a)
4.
c)
•
•
•
•
Dalton’s atom had no internal structure.
Thomson’s atom had positive and negative charges mixed.
Bohr added the concept of energy levels.
Rutherford’s atom had a dense positive nucleus surrounded by an electron cloud.
OBJECTIVE 4.9
1.
d)
2.
c)
3.
a)
4.
a)
5.
d)
6.
d)
See calculation of the number of atomic particles in the study guide.
OBJECTIVE 5.1
1.
a)
- because number of protons increases from left to right
2.
c)
Since some elements have isotopes with more neutrons, the atomic mass may be
higher than the element next to it.
OBJECTIVE 5.2
1.
d)
same atomic number but different atomic masses
2.
c)
are unstable and disintegrate radioactively
3.
a)
I-131 is radioactive isotope of iodine.
OBJECTIVE 5.3
1.
b)
See method of calculation in study guide.
2.
Oxygen -16:
Because of its large relative abundance, it represents most of the
mass of natural oxygen.
ANSWER KEY
Module I – 4
3.
c)
See method of calculation in study guide.
4.
a)
Isotopes with more neutrons may have a higher atomic mass than the element next
to it and a higher percentage of heavy isotopes will give an element a higher atomic
mass than its neighbour.
OBJECTIVE 5.4
1.
d)
2.
b)
3.
c)
4.
b)
5.
The unknown substance is an alkaline earth metal with the following experimental
results: gas produced when it reacts with acid; conducts heat and electricity; bends but
does not break.
Cl is to the right of the step line.
OBJECTIVE 5.5
1.
c)
2.
b)
3.
b)
4.
a)
5.
d)
OBJECTIVE 5.6
1.
d)
2.
a)
Although melting point peaks in the middle of a period, the graph values for the
alkali metals are higher than those for the inert gases.
3.
a)
Chemical activity, lustre and atomic radius decrease from left to right in a period.
4.
d)
Electrical conductivity and chemical activity decrease from left to right in a period
— not increase.
5.
a)
ANSWER KEY
Module I – 5
OBJECTIVE 5.8
1.
d)
- because Na is in Family 1
2.
d)
- because it has a completed valence shell
3.
b)
It is the only non-metal.
4.
1.
2.
3.
4.
5.
O
Ar
B
Na
Ca
5.
Element 1
Element 2
Element 3
Element 4
6.
K and Cl belong to the same period because they have the same number of energy levels
but belong to different families because they have different numbers of valence electrons.
H
Na
Cl
Li
OBJECTIVE 5.9
1.
a) Ne
b) Mg
c) Si
d) K
The number of valence electrons gives the family number.
The number of energy levels gives the period number.
The number of protons gives the atomic number.
2.
a)
3.
a)
4.
d)
- because it has 1 valence electron
5.
a)
- because it has 7 valence electrons
OBJECTIVE 6.1
1.
b)
2.
d)
3.
b)
2 volumes of hydrogen gas to 1 volume of oxygen
ANSWER KEY
Module I – 6
OBJECTIVE 6.2
1.
c)
2.
b)
3.
b)
8 – 6 = valence of 2
OBJECTIVE 6.3
1.
c)
- use crossover rule
2.
b)
- using reverse crossover, x has a valence of 1
3.
c)
- use crossover rule
4.
Element X has 3 valence electrons to donate or share and element Y can receive or share 2
valence electrons. Because 6 electrons are involved, 2 atoms of X and 3 atoms of Y are
needed. The formula is X2Y3.
OBJECTIVE 6.4
1.
c)
2.
a)
3.
d)
4.
a)
ANSWER KEY
Module I – 7
MODULE II
— ANSWER KEY —
OBJECTIVE 2.1
1.
b)
A magnetic substance will always attract a ferromagnetic substance.
2.
b)
1965 coin is not attracted, therefore it is nonmagnetic. 1994 coin is attracted to a
magnet, but does not affect an iron nail, therefore it is ferromagnetic.
3.
b)
Piece 2 is the only one to attract any of the other metal pieces (pieces 3 and 4). Pieces
3 and 4 do not attract each other, therefore, they are not magnets but they are
ferromagnetic.
4
d)
W and Y are ferromagnetic because they are not attracted to each other but are
attracted to X. X is magnetic and Z is nonmagnetic.
OBJECTIVE 2.2
1.
b)
Magnetic field lines always point away from the north pole and towards the south
pole of a magnet.
2.
a)
Magnetic field lines always point away from the north pole and towards the south
pole of a magnet.
3.
a)
Magnetic field lines always point away from the north pole and towards the south
pole of a magnet.
OBJECTIVE 2.3
1.
d)
Magnetic lines of force always enter and leave at the poles.
2.
c)
Using the left-hand rule, the thumb of your left hand is pointed in the direction of
electron flow (from negative to positive). The curl of your fingers is the direction of
the magnetic field. The compass aligns itself along the magnetic field lines.
3.
d)
The fingers of your left hand would grasp the solenoid from behind to follow the
electron current. Your left thumb would then point to the left end of the solenoid.
This is the north pole of the magnetic field.
4.
d)
The fingers of your left hand would grasp the solenoid from behind in the direction
of electron current. Your left thumb would then point towards the north pole at the
left end of the solenoid. The north pole of the compass points towards the south
pole of the solenoid.
ANSWER KEY
Module II – 1
OBJECTIVE 2.6
1.
d)
There is no change in the force vs time graph.
2.
c)
The strength of the magnetic field depends on the product of the current intensity
and the number of turns.
3.
d)
The strength of the magnetic field depends on the nature of the core, the product of
the current intensity and the number of turns.
4.
b)
The strength of the magnetic field depends on the nature of the core, the product of
the current intensity and the number of turns.
5.
d)
The strength of the magnetic field depends on the product of the current intensity
and the number of turns.
OBJECTIVE 2.8
1.
The magnetic field of the electromagnet can be "turned off", so the iron and steel can be
dropped when the crane operator turns off the current. However, you would not be able
to "turn off" the field of a natural magnet. The iron and steel would stick to the crane and
someone would have to physically remove it. This means that the natural magnet cannot
be used for this task.
OBJECTIVE 3.1
1.
a)
Aluminum is the only electrical conductor.
2.
a)
An insulator is needed to prevent grounding ; it must not easily lose its
shape or bend.
3.
b)
Graphite is a conductor.
OBJECTIVE 3.2
1.
b)
This is the shortest and thickest (largest diameter) wire.
2.
a)
The thickest wire at the lowest temperature is the best conductor.
3.
b)
A larger diameter and a lower temperature increase the conductivity of a wire.
OBJECTIVE 3.3
1.
a)
The ammeter must be connected in series before the parallel section.
2.
1st position
2nd position
R1
A
3rd position
R2
A
A
ANSWER KEY
Module II – 2
OBJECTIVE 3.4
1.
a)
The voltmeter is always connected in parallel across the resistor measured.
OBJECTIVE 3.5
1.
a)
conductance = slope; ∆I/∆V = (1.8 - 0.9) A/ (12 - 6) V = 0.15 S .
2.
Conductor 1:
G1 = 2 A = 1 S
2V
Conductor 2:
G2 = 2 A = 0.5 S
4V
Conductor 3:
G3 = 2 A = 0.25 S
8V
Conductor 1 has the best electrical conductance.
OBJECTIVE 3.6
1.
b)
1st band : Yellow = 4 ;
2nd band: red = 2; 1st two digits: 42;
3rd band : red = 2, multiplier = 102 =100; resistance = 42 x 102 Ω = 4200 Ω.
OBJECTIVE 3.7
1.
b)
Parallel circuit:
1 = 1 + 1
Req
5Ω
5Ω
=
2 ;
5Ω
Req = 5Ω = 2.5 Ω
2
2.
d)
Series circuit: Req = R1 + R2 + R3 + R4 ; Req = 10Ω + 10Ω + 10Ω + 10Ω = 40Ω
3.
b)
In parallel circuits:
1
Req
Req
4.
R1
=
3Ω
=
1
R1
+
1
R2
=
1
5Ω
+
1 +
10 Ω
=
0.33 Ω
=
3Ω
In a parallel circuit:
1t
11
In a parallel circuit:
From Ohm’s Law:
Vt
R1
ANSWER KEY
+
1
R3
1
30 Ω
=
=
=
=
=
11
+
12
12 A – 4 A
8A
V1
=
V2
V1
I1
=
24 V
8A
=
3Ω
Module II – 3
OBJECTIVE 3.8
1.
c)
Req (parallel): 1/R2 + 1/(R3 + R4) = 1/15 + 1/30 =3/30; Req (parallel) = 10Ω
Req (circuit) = Req (parallel) + Req (series) = 10Ω + (10Ω + 10Ω) = 30Ω .
2.
c)
Req (parallel): 1/R3 + 1/R4 = 1/20Ω + 1/20Ω = 2/20Ω; Req (parallel) = 10Ω
Req (circuit) = Req (parallel) + Req (series) = 10Ω +(20Ω + 20Ω) = 50Ω
OBJECTIVE 3.11
1.
1.5 A
2.
Is = I1 + I2
= 0.75 A + 0.75 A
= 1.5 A
I3 = 2 A, I5 = 6 A; Parallel circuit: Iparallel
Series circuit: I1
3.
a)
= I2 + I3 + I4 = I1
= 3A + I3 +1A = 6A; I3 = 2A
= I5 = 6A.
In circuit (A) only L1 is on when S1 is closed.
In circuit (C) L1 will always be on whether S1 and S2 are open or closed.
In circuit (B) only L2 will be on when S1 is closed.
In circuit (D) L1 and L2 will always be out even if S1 is closed.
OBJECTIVE 3.12
1.
b)
1st circuit is series; Veq = V1 + V2 ; 1.5 V = 0.75 V + 0.75 V; V1 = 0.75V
2nd circuit is parallel; Veq = V1 = V2 ; V2 = 1.5 V
2.
Circuit 1
8V
Circuit 2
V1
10 V
R1
10 V
R1
V1
10 V
R2
V2 10 V
2V
V2
R2
Circuit 1 must be series because V1 + V2 = 8V + 2V = 10V= V(power supply)
Circuit 2 must be parallel because V1 = V2 = V( power supply) = 10V.
ANSWER KEY
Module II – 4
OBJECTIVE 3.15
1.
17.5 V
Req
Vs
2.
0.5 A
=
=
=
=
=
=
R1 + R 2 + R 3
10 Ω + 20 Ω + 40 Ω
70 Ω
Is x Req
0.25 A x 70 Ω
17.5 A
V1 = Vs = 20 V
I1 = V1 = 20 V
R1
40 Ω
= 0.5 A
3.
3A
I2 = V2 = 36 V = 9 A
R2
4Ω
I1 = Is – (I2 + I3)
= 18 A – (9 A + 6 A)
= 3A
4.
3Ω
Req
=
Vs = 12 V = 6 Ω
Is
2A
R3
= Req – (R1 + R2)
= 6 Ω – (1 Ω + 2 Ω)
= 3Ω
OBJECTIVE 3.16
1.
a)
Parallel portion: R2 and R3 in series with each other and parallel to R1
R2 + R3 = 5Ω +10Ω = 15Ω
Req (parallel) : 1/R1 + 1/(R2 + R3) = 1/30Ω + 1/15Ω
= 3/30Ω
Req (parallel) = 30Ω/3 = 10Ω
Req (circuit) = Req (parallel) + Req (series) = 10Ω + 20Ω = 30Ω
Ieq (circuit) = Vs / Req = 12 V / 30Ω = 0.4A
V4 = I4 x R4 = 0.4A x 20Ω = 8V
Vs = Vseries + V eq (parallel)
12 V = 8V + V eq (parallel)
V eq (parallel) = V1 = 4V
ANSWER KEY
Module II – 5
2.
0.875 A
V2
V2
I3
=
=
=
=
I2 x R2
0.5 A x 75 Ω
37.5 V
V3 = 37.5 V
= V3
R3
= 37.5 V
100 Ω
= 0.375 A
Is
3.
a)
= I2 + I3
= 0.5 A + 0.375 A
= 0.875 A
I1
= I circuit ( I1 is in series with the power supply)
I circuit = 2 A
V1
= I1 x R1 = 2A x 100Ω = 200V
V2
V2
= V3 ( parallel resistors)
= I2 x R2 = 1.2 A x 200Ω = 240 V
Vs
= 240 V + 200 V = 440 V
OBJECTIVE 4.1
1.
b)
5.5 kW x 2 hrs x $0.0454/kw•h = $0.41
2.
b)
0.2 kW x 5 hrs x $0.05/kw•h = $0.05
3.
a)
0.6 kW x 100 hrs x $0.05 = $3.00
4.
a)
90 minutes ÷ 60 minutes = 1.5 hours
60 days x 1.5 hours x $0.046/kW•h = $4.97
OBJECTIVE 4.2
1.
a)
Coulomb per second is an ampere. An ammeter measures current intensity.
2.
c)
A coulomb per second is equal to an ampere.
3.
a)
The unit of current intensity is the ampere. One ampere is equivalent to one
coulomb of charge per second. (charge per second)
4.
a)
The ammeter is used to measure current intensity which is equal to coulombs per
second.
ANSWER KEY
Module II – 6
OBJECTIVE 4.3
1.
b)
The volt is equal to 1 joule of energy lost by every coulomb of charge flowing
through the resistor.
2.
a)
Similar to the previous question, the amount of charge carried by a certain quantity
of electric charge is a definition of the volt.
OBJECTIVE 4.4
1.
c)
E = P x t, kW/h is the amount of energy used in one hour.
OBJECTIVE 4.5
1.
a)
If the power is kept constant, current intensity will increase as voltage is increased
and the effect of resistance will decrease.
2.
240 V
I = P ÷ V
For 120 V, the current is 180 W ÷ 120 V = 1.5 A
For 180 V, the current is 180 W ÷ 180 V = 1 A
For 240 V, the current is 180 W ÷ 240 V = 0.75 A
P = RI2
2 Ω x (1.5)2 = 4.5 W, % power loss = 4 W ÷ 180 W x 100% = 2.22%
2 Ω x (1 A)2 = 3 W, % power loss = 3 W ÷ 180 W x 100% = 1.67%
2 Ω x (0.75 A)2 = 1.125 W, % power loss = 1.125 W ÷ 180 W x 100% = 0.625%
Therefore, 240 V provides the smallest power loss (0.625%).
OBJECTIVE 4.6
1.
d)
I = V = 12 V = 1.2 A
t
10 Ω
P = V x I
= 12 V x 1.2 A
= 14.4 W
E = P x t
= 14.4 W x 20 minutes x 60 seconds
= 17 280 J
ANSWER KEY
Module II – 7
2.
d)
P = V x I
= 120 V x 1.5 A
= 180 W
E = P x t
= 180 W x (60 minutes x 60 seconds x 8 hours) = 5 184 000 joules
= 5 184 000 J ÷ 1000 = 5184 kJ
3.
d)
E = P x T
P = V x I
First find I,
I = V/R
P = V x I,
E = P x t
240 x 24 = 5760 watts
5760 W x (3 hours x 60 minutes x 60 seconds) = 62 208 000 joules
240 volts ÷ 10 Ω = 24 A
62 208 000 J ÷ 1000 = 62 208 kJ
4.
d)
I = V/R
P = V x I
E = P x t
5.
I = V ÷ R,
P = V x I,
t = E ÷ P,
110 V ÷ 50 Ω = 2.2 A
110 V x 2.2 A = 242 Watts
240 x (30 minutes x 60 seconds) = 435 600 J
120 V ÷ 12 Ω = 10 A
120 V x 10 A = 1200 watts
(3600 kJ x 1000) ÷ 1200 W = 3000 seconds, 3000s ÷ 60 = 50 minutes
OBJECTIVES 5.1 – 5.3
1.
Q
= mc∆t
= 200 g x 4.19 J/g•°C x 25°C
= 20 950 J
OBJECTIVE 5.4
1.
1.125 J/(g • ˚C)
E
Q
c
=
=
=
=
V x I x t
15 V x 0.25 A x 900 s
3375 J
E = 3375 J
= Q
m∆T
=
3375 J
100 g x 30° C
= 1.125 J/(g • °C)
ANSWER KEY
Module II – 8
2.
Q = m x c x ∆t
= 160 000 g x 4.19 x 55
= 36 872 000 J
E
36 872 000 J
I
=
=
=
=
V x I x t
240 V x I x 1500 s
36 872 000 J ÷ (240 V x 1500 s)
102.4 A
R = V ÷ I
= 240 V ÷ 102.4 A
= 2.34 Ω
3.
b)
Q
2618.75 J
I
4.
c)
= m x c x ∆t
= 125 g x 4.19 J/g•°C x 5°C
= 2618.75 J
= 6 V x I x (15 min x 60 s)
= 2618.75 J ÷ (6 V x 900 s)
= 0.48 A
Electrical energy =
=
=
9000 W
∆t
=
=
=
V x I x t
10 V x 0.5 A x (30 min x 60 s)
9000 W
50 g x 4.19 J/g•°C x ∆t
9000 W ÷ (50 g x 4.19 J/g•°C)
42.95 °C
OBJECTIVES 6.2 & 6.3
1.
d)
Solar panels use solar energy, a renewable resource which produces no pollution.
2.
d)
Hydroelectricity uses no heat to produce electricity.
3.
d)
Coal fired power stations are responsible for all three statements.
4.
Nuclear power stations – disposal of radioactive wates is difficult, accidents can be very
harmful.
Hydroelectricity floods huge amounts of land and upsets the ecology.
Coal-burning and diesel-powered power plants cause acid rain by producing SO2 and
NO2.
5.
d)
Thermal-power uses fossil fuels such as coal and oil which produce SO2 and NO2
which produce acid rain.
ANSWER KEY
Module II – 9
MODULE III
— ANSWER KEY —
OBJECTIVE 2.1
1.
2.
3.
4.
a)
d)
c)
d)
based on electrical conductivity
based on electrical conductivity
does conduct electricity but does not change litmus paper
A solid cannot be tested with litmus paper and electrical conductivity is present for
an acid, a base, and a salt
OBJECTIVE 2.3
1.
2.
3.
d)
a)
b)
the formula for a base ends in OH
A salt usually involves a metal; HNO3 is an acid, LiOH is a base, and SO2 is a gas.
Acid formulas start with H, base formulas end with OH.
OBJECTIVE 2.5
1.
2.
3.
4.
c)
b)
c)
b)
a bond between two atoms of the same element is covalent
a bond between a metal and a non-metal is ionic
Covalent bonding usually involves non-metals.
Ionic bonding usually occurs between metals and non-metals.
OBJECTIVE 2.6
1.
2.
3.
b)
c)
c)
4.
d)
a polyatomic ion involves two or more symbols and has a charge
use the crossover rule
The ammonium ion has a charge of +1 and the phosphate ion has a charge of -3 and
the crossover rule is applied.
There are two aluminum ions each with a charge of +3 for a +6 total; the three
oxalate ions total -6, making each a -2.
OBJECTIVE 2.7
1.
2.
3.
Electrolytes: KCl, Sr(OH)2, MgSO4, Na3PO4 - these release ions in solution.
Non-electrolytes: H2, CH3OH, C6H12O6 - these do not release ions in solution.
CH3OH is the alcohol: methanol.
b)
Acids, bases, and salts form electrolytes.
d)
The other substances are acid, base or salt (see note in #1).
ANSWER KEY
Module III – 1
OBJECTIVE 2.8
1.
2.
Acids, bases, and salts, in solution, conduct electricity because ions are released in the
solutions.
Sulfuric acid is a very strong electrolyte and acetic acid (vinegar) is a very weak
electrolyte. Sulfuric acid produces very many ions but acetic acid only produces a small
number of ions.
OBJECTIVE 2.10
1.
Dissolve a small amount of the solid in water.
Test for electrical conductivity: If the solution conducts, the solid is an acid, a base or a
salt.
Test with blue and red litmus paper: If the blue paper turns red, the solid is an acid salt.
If the red paper turns blue, the solid is a base salt. If neither litmus paper turned colour,
the solid is a neutral salt.
OBJECTIVE 3.1
1.
2.
3.
50 g
= xg
1000 mL 200 mL
a)
30 g
=
1000 mL
x = 10 g of solute
xg
300 mL
x = 9 g of solute
Write all solutions in g/L.
Solution 1:
2g =
0.1 L
20 g =
1L
Solution 2:
0.6 kg = 600 g
600 g =
3L
Solution 3:
0.2 g =
2 mL
100 g
1L
20 g/L
200 g
1L
= 200 g/L
= 100 g/L
Solution 1, Solution 3, and Solution 2 are in order of increasing concentrations.
ANSWER KEY
Module III – 2
4.
a)
C (g/L)
=
mass of solute
volume (L)
5 g/L
=
20 g
V
=
20 g
5 g/L
=
4L
Mass percent
=
mass of solute x 100
mass of solution
45
=
mass of solute
200 g
=
45 x 200 g
100
=
90 g
mass of water
=
=
=
mass of solution - mass of solute
200 g - 90 g
110 g
15 g
1000 mL
60 g
x mL
V
5.
mass of solute
6.
=
x 100
x = 4000 mL = 4 L
OBJECTIVE 3.2
1.
C1 V1 = C2 V2
45 g/L x 500 mL = C2 x 1500 mL
45 g/L x 500 mL
1500 mL
=
→
? g/L
15 g/L
C2 = 15 g/L
2.
a)
C1 x V 1
=
50 mL x 2.8 g/L =
C2
=
=
C2 x V 2
C2 x 200 mL
50 mL x 2.8 g/L
200 mL
0.7 g/L
ANSWER KEY
Module III – 3
Answer - d)
3.
C1 x V 1
5 g/L x V1
V1
=
=
=
=
C2 x V2
3 g/L x 1.5 L
3 g/L x 1.5 L
5 g/L
0.9 L or 900 mL
OBJECTIVE 3.3 and 3.5
1.
c)
2 Al = 2 x 27 = 54
3 S = 3 x 32 = 96
12 O = 12 x 16 = 192
– the sum of the above is 342
2.
b)
1 mol
0.25 mol
3.
b)
molar mass:
n
=
=
=
=
101 g
x
2N = 2 x 14
8H = 8x1
1 S = 1 x 32
4 O = 4 x 16
28
8
32
64
132 g/mol
mass / molar mass
100 g / 132 g/mol
0.76 mol
OBJECTIVE 3.6 and 3.7
1.
2.
c)
b)
40 g
1 mol/L
=
xg
0.8 mol/L
=
32 g
32 g
1000 mL
=
x
300 mL
=
9.6 g
molar mass of MgSO4 is 120 g/mol
n
=
mass / molar mass
=
125 g / 120 g/mol
=
1.04 mol
Concentration (mol/L) = Molarity = mol of solute
V (L)
ANSWER KEY
=
1.04 mol
2.5 L
=
0.416 mol/L
Module III – 4
3.
C1 x V 1
4 mol/L x 500 mL
V2
=
=
=
=
C2 x V 2
1 mol/L x V2
4 mol/L x 500 mL
1 mol/L
2000 mL
The volume of water needed is V2-V1 = 2000 mL - 500 mL = 1500 mL
4.
C1 x V 1
12 mol/L x V1
V1
=
=
=
=
C2 x V 2
0.3 mol/L x 2 L
0.3 mol/L x 2 L
12 mol/L
0.05 L or 50 mL
OBJECTIVE 4.1
1.
2.
3.
Solution A:
base – NaOH
Solution B:
salt – NaCl
Solution C:
acid – HCl
b)
Solution is acidic: pH 0 – 6.9
OBJECTIVE 4.2
1.
2.
3.
4.
5.
6.
a)
a)
d)
b)
NH4OH and NaOH
Yellow, lemon juice is acidic
OBJECTIVE 4.3 AND 4.4
1.
2.
3.
4.
5.
6.
7.
8.
b)
b)
b)
Orange or red
Phenolphthalein
a)
A: Red or orange
B: Yellow
b)
A. Yellow
B: Green or blue
c)
A: pH 4 - 7
B: pH 7.2 - 10
a)
c)
ANSWER KEY
C: Yellow
C: Yellow or orange or red
C: pH 10 - 11
Module III – 5
9.
10.
11.
c)
a)
b)
OBJECTIVES 4.6 and 4.7
1.
2.
3.
4.
5.
a)
pH = 2
Solution A has pH = 3; Solution B has pH = 4
pH = 6
[H+] = 1 x 10-6 mol/L
[OH-] = 1 x 10-8 mol/L
[H+] mol/L
[OH-] mol/L
pH
Nature
6.
A
10-1
10-13
1
acid
B
10-4
10-10
4
acid
C
10-13
10-1
13
base
The rain was acidic. Add a basic solution.
OR
Some of the water evaporated so the pool water left became more concentrated, that is,
more acidic. Add more water.
OR ....
7.
A
B
C
8.
[H+]
10-9 mol/L
0.001 mol/L
10-6 mol/L
[OH-]
10-5 mol/L
10-11 mol/L
10-8 mol/L
pH
9
3
6
Nature
base
acid
acid
Solution A is basic because it turns litmus paper blue. Solution B is acidic because if it
was basic as Solution A, then it would turn green in the presence of the unknown
indicator, but it did not, so, it is acidic.
OBJECTIVE 5.1
1.
2HCl + Mg (OH)2 → MgCl2 + 2H2O
2.
1) Add phenolphthalein to the base. The solution will turn pink.
2) Add acid drop by drop to the base and stir.
3) When the solution turns colourless, it is neutral.
3.
a)
ANSWER KEY
Module III – 6
OBJECTIVE 5.2
1.
2.
a)
a)
b)
Vinegar is an acid and is added to neutralize the base (KOH).
KOH + CH3COOH → CH3COOK + H2O
OBJECTIVE 5.3
1.
The mass of the products will be the same as the mass of the reactants, that is, 11 g.
2.
2 H2 + O2 → 2 H2O
8 g + 64 g → 72 g
OBJECTIVE 5.4
1.
2.
3.
4.
d)
d)
c)
a)
2 KOH + H2SO4 → K2SO4 + 2 H2O
b)
8 NH3 + 3 Cl2 → N2 + 6 NH4Cl
c)
2 Al + 6 HCl → 2 AlCl3 + 3 H2
d)
3 Mg + 2 H3PO4 → Mg3(PO4)2 + 3 H2
e)
2 Fe2S3 + 9 O2 → 2 Fe2O3 + 6 SO2
f)
PbO2 + 4 HCl → PbCl2 + 2 H2O + Cl2
g)
2 Cu(NO3)2 → 2 CuO + 4 NO2 + O2
OBJECTIVE 5.6 & 5.8
1.
CaCO3
+
2 HCl
→
CaCl2
+
H2O
+
CO2
100 g
73 g
111 g
18 g
44 g
140 g
100 g
152 g
25 g
60 g
90 g
66 g
100 g
16 g
40 g
556 g
406 g
617 g
100 g
244 g
227 g
166 g
252 g
41 g
100 g
ANSWER KEY
Module III – 7
2.
2 mol
2 C4H10
29 g
13 mol
+
13 O2
x
116 g
2 C4H10
29 g
→
8 CO2
+
10 H2O
→
8 CO2
+
10 H2O
416 g
+
116 g of C4H10
29 g of C4H10
13 O2
x
reacts with 416 g of O2
reacts with x g of O2
x = 104 g of O2
3.
2 mol
2 HNO3
100 g
1 mol
+
CaCO3
→
Ca(NO3)2
+
H2O +
126 g
2 HNO3
100 g
44 g
+
126 g of HNO3
100 g of HNO3
CaCO3
→
→
→
Ca(NO3)2
+
H2O +
44 g of CO2
x g of CO2
x = 34.9 g of CO2
4.
1 mol
CH4 +
192 g
1 mol
2 O2 →
16 g
CH4 +
192 g
CO2
x
CO2 +
x
2 H2O
44 g
2 O2 →
16 g of CH4
192 g of CH4
→
→
CO2 +
x
2 H2O
44 g of CO2
x g of CO2
x = 528 g of CO2
ANSWER KEY
Module III – 8
CO2
x
5.
1 mol
2 mol
Fe2O3
x
+
1 mol of Fe2O3
x mol of Fe2O3
3C
→
→
→
2 Fe
50 mol
+
3 CO
2 mol of Fe
50 mol of Fe
x = 25 mol of Fe2O3
6.
2 mol
16 mol
2 C8H18
57 g
+
25 O2 →
228 g
16 CO2
x
+
18 H2O
+
18 H2O
+
NO
+
NO
704 g
2 C8H18
57 g
+
228 g of C8H18
57 g of C8H18
25 O2 →
→
→
16 CO2
x
704 g of CO2
x g of CO2
x = 176 g of CO2
7.
3 mol
2 mol
3 NO2
1000 g
+
H2O →
138 g
2 HNO3
x
126 g
3 NO2
1000 g
+
138 g of NO2
1000 g of NO2
H2O →
→
→
2 HNO3
x
126 g of HNO3
x g of HNO3
x = 913.04 g of HNO3
8.
3 mol
4 mol
3 Fe
x
+
3 mol of Fe
x mol of Fe
x
4 H2 O
→
→
→
Fe3O4
+
4 H2
2 mol
4 mol of H2
2 mol of H2
= 1.5 mol of Fe
= 56 g/mol x 1.5 mol
= 84 g
ANSWER KEY
Module III – 9
9.
2 mol
2 NaOH
x
1 mol
+
80 g
2 NaOH
x
H2SO4
14.7 g
→
Na2SO4
+
2 H2O
→
Na2SO4
+
2 H2O
98 g
+
80 g of NaOH
x g of NaOH
H2SO4
14.7 g
98 g of H2SO4
14.7 g of H2SO4
reacts with
reacts with
x = 12 g of NaOH
OBJECTIVES 6.4 & 7.3
1.
See explanation on pages 44 and 46 of Module III.
2.
Pesticides are washed into streams where they get into fish through the food chain.
Humans ingest the pesticides by eating the fish.
3.
carbon dioxide → greenhouse effect
carbon monoxide → poisonous gas
sulfur dioxide → acid rain
nitrogen oxides → acid rain
dioxins and furones → cancers
4.
c)
5.
d)
6.
a)
7.
8.
d)
a)
9.
Nuclear power: radiation leaks, etc.
ANSWER KEY
Module III – 10
Coal-burning:
Diesel power:
carbon dioxide emission, SO2 emission, etc.
carbon dioxide emission, etc.
10.
Weather systems from Detroit bring acid rain to Quebec.
11.
Collect the rain and test its pH with pH paper.
12.
Polluting gases allow more of the sun’s heat to remain in our atmosphere and increase the
temperature of the Earth.
13.
Recycle more garbage rather than burn it.
Use public transportation.
Buy reusable products.
Reduce car emissions.
etc.
14.
CFC’s react with ozone in the ozone layer breaking down this protective ozone layer and
exposing the Earth to ultraviolet radiation.
ANSWER KEY
Module III – 11
Appendix I
FORMULAS AND QUANTITIES
FORMULAS
ρ = m
V
Q = m c ∆t
ρ : density
m : mass
V : volume
Q
m
c
∆t
I=
: quantity of heat
: mass
: specific heat capacity
: change in temperature
R e = R 1 + R 2 + ... R : resistance
1 = 1 + 1 +
... R e :equivalent resistance
Re R 1 R 2
R= V
I
R : resistance
V : potential difference
I •: current intensity
G= I
V
G : conductance
I : current intensity
V : potential difference
Q
∆t
E = VI ∆ t
I •: current intensity
Q : electric charge
∆t : time
E : energy
V : potential difference
I •: current intensity
∆t : time
P = E
∆t
P : power
E : energy
∆t : time
P = VI
P : power
V : potential difference
I •:current intensity
P = R I2
P : power
R : resistance
I •: current intensity
Note : Potential difference, V , mais be represented by U .
QUANTITIES
NAME
Specific heat capacity
Density
SYMBOL
VALUE (for water)
c
4190 J / (kg • °C)
or
4,19 J / (g • °C)
ρ
1,0 g / mL
or
1,0 kg / L
or
1000 kg / m 3
PERIODIC TABLE OF THE ELEMENTS
IA
VIII A
1
Atomic number
18
Element symbol
1
2
3
4
5
6
7
2
1
1
II A
III A
IV A
VA
VI A
VII A
Atomic mass
He
H
H
2
13
14
15
17
16
4,00
1,01
1,01
5
10
3
6
8
9
4
7
B
Ne
Li
C
O
F
Be
N
10,81 12,01 14,01 16,00 19,00 20,18
6,94
9,01
VIII
14
11
15
13
17
12
18
16
III B
IV B
VB
VI B
VII B
IB
II B
Si
Na
P
Al
Cl
Mg
Ar
S
10
7
12
8
9
3
4
6
11
5
22,99 24,31
26,98 28,09 30,97 32,07 35,45 39,95
31
34
29
24
27
19
33
35
30
23
28
20
36
32
26
21
22
25
Ga
Se
Cu
Cr
Co
K
Zn
As
Br
Ni
V
Ca
Kr
Ge
Ti
Fe
Sc
Mn
39,10 40,08 44,96 47,90 50,94 52,00 54,94 55,85 58,93 58,71 63,55 65,39 69,72 72,59 74,92 78,96 79,90 83,80
40
45
50
52
38
39
41
43
44
46
48
49
51
53
54
37
42
47
Zr
Rh
Sn
Te
Sr
Y
Nb
Tc
Ru
Pd
Cd
In
Sb
I
Xe
Rb
Mo
Ag
85,47 87,62 88,91 91,22 92,91 95,94 98,91 101,07 102,91 106,40 107,87 112,41 114,82 118,71 121,75 127,60 126,90 131,30
57-71
73
76
81
85
55
56
72
74
75
77
79
80
82
84
86
78
83
La-Lu
Ta
Os
Tl
At
Cs
Ba
Hf
W
Re
Ir
Au
Hg
Pb
Po
Rn
Pt
Bi
132,91 137,33
178,49 180,95 183,85 186,21 190,20 192,22 195,09 196,97 200,59 204,37 207,20 208,98 (209) (210) (222)
88
87
105
104
89-103
Ac-Lr
Ra
Fr
Ha
Rf
(223) (226)
(261) (262)
6
Note. –
The term “metalloid” will continue to be used, but will be followed by the term “semimetal”.
The term “inert gases” will continue to be used for the elements in Group VIII A (18).
Appendix II
7
59
60
57
61
64
68
69
58
62
63
65
66
67
70
71
Pr
Nd
La
Pm
Gd
Er
Tm
Ce
Sm
Eu
Tb
Dy
Ho
Yb
Lu
138,91 140,12 140,91 144,24 145 150,35 151,96 157,25 158,92 162,50 164,93 167,26 168,93 173,04 174,97
93
96
97
98
99
100
101
102
103
89
91
92
94
95
90
Np
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Ac
Pa
U
Pu
Am
Th
(247) (251) (254) (257) (258) (259) (260)
227,03 232,04 231,04 238,03 237,05 (244) (243) (247)