1
SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2016
GRADE 12
SUBJECT:
PHYSICAL SCIENCE
LEARNER NOTES
(Page 1 of 36)
©GautengDepartmentofEducation
2
TABLE OF CONTENTS
SESSION
9
10
11
12
13
14
TOPIC
PAGE
Acids and Bases
3 – 13
Consolidation: Vertical projectile motion, Momentum
13 - 17
Consolidation: Organic Chemistry
18 - 23
Consolidation: WEP
24 - 27
Consolidation: Doppler, Rates of Reactions
27 - 32
Consolidation: Equilibrium and Acids and Bases
©GautengDepartmentofEducation
33 - 36
3
SESSION NO:
TOPIC:
9
ACIDS AND BASES
Note to Learner: You need to know the different acid-base theories. You will need
to revise all your Grade 11 acid-base theory and definitions including calculations.
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 20 minutes
1.1
Sulfuric acid is a diprotic acid.
1.1.1
1.1.2
1.2
Define an acid in terms of the Lowry-Brønsted theory.
Give a reason why sulfuric acid is referred to as a diprotic acid.
(2)
(1)
The hydrogen carbonate ion can act as both an acid and a base. It
reacts with water according to the following balanced equation:
HCO3-(aq) + H2O(ℓ) ⇌ H2CO3(aq) + OH (aq)
1.2.1
1.2.2
1.3
(Taken from NCS Feb/Mar 2015 Paper 2)
Write down ONE word for the underlined phrase.
HCO3- acts as base in the above reaction. Write down the
formula of the conjugate acid of HCO3- (aq).
(1)
(1)
A learner accidentally spills some sulfuric acid of concentration
-3
6 mol·dm from a flask on the laboratory bench. Her teacher tells her to
neutralise the spilled acid by sprinkling sodium hydrogen carbonate
powder onto it. The reaction that takes place is: (Assume that the
H2SO4 ionises completely.)
H2SO4(aq) + 2NaHCO3(s) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g)
The fizzing, due to the formation of carbon dioxide, stops after the
learner has added 27 g sodium hydrogen carbonate to the spilled acid.
1.3.1
Calculate the volume of sulfuric acid that spilled. Assume that
all the sodium hydrogen carbonate reacts with all the acid.
(6)
-3
The learner now dilutes some of the 6 mol·dm sulfuric acid solution in
-3
the flask to 0,1 mol·dm . [V = 0,03 dm3]
1.3.2
-3
Calculate the volume of the 6 mol·dm sulfuric acid solution
3
needed to prepare 1 dm of the dilute acid. [va = 0,02 dm2]
©GautengDepartmentofEducation
(2)
4
3
-3
During a titration 25 cm of the 0,1 mol·dm sulfuric acid solution is
-3
added to an Erlenmeyer flask and titrated with a 0,1 mol·dm sodium
hydroxide solution.
1.3.3
The learner uses bromothymol blue as indicator. What is the
purpose of this indicator?
(2)
1.3.4
Calculate the pH of the solution in the flask after the addition of
3
30 cm of sodium hydroxide. The endpoint of the titration is not
(8)
yet reached at this point. [1,44]
[22]
QUESTION 2: 20 minutes
(Taken from NCS Nov 2015 Paper 2)
2.1
Ammonium chloride crystals, NH4Cℓ(s), dissolve in water to form
ammonium and chloride ions. The ammonium ions react with water
according to the balanced equation below:
+
NH4+ (aq) + H2O(ℓ) ⇌ NH3(aq) + H3O (aq)
2.1.1
2.1.2
2.2
Write down the name of the process described by the
underlined sentence.
(1)
Is ammonium chloride ACIDIC or BASIC in aqueous solution?
Give a reason for the answer.
(2)
A certain fertiliser consists of 92% ammonium chloride. A sample of
3
-3
mass x g of this fertiliser is dissolved in 100 cm of a 0,10 mol·dm
sodium hydroxide solution, NaOH(aq). The NaOH is in excess.
The balanced equation for the reaction is:
NH4Cℓ(s) + NaOH(aq) → NH3(g) + H2O(ℓ) + NaCℓ(aq)
2.2.1
Calculate the number of moles of sodium hydroxide in which
the sample is dissolved. [0,01 mol]
(3)
3
During a titration, 25 cm of the excess sodium hydroxide solution is
-3
titrated with a 0,11 mol·dm hydrochloric acid solution, HCℓ(aq). At the
3
endpoint it is found that 14,55 cm of the hydrochloric acid was used to
neutralise the sodium hydroxide solution according to the following
balanced equation:
HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ)
2.2.2
2.3
Calculate the mass x (in grams) of the fertiliser sample used.
[ = 0,21 g]
-3
Calculate the pH of a 0,5 mol·dm sodium hydroxide solution at 25 °C.
[ = 13.7]
©GautengDepartmentofEducation
(8)
(4)
[18]
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SECTION B: NOTES ON CONTENT
Acids can undergo the following reactions:
1.
Acid + Metal -------------> Metal Salt + Hydrogen
This indicates that all acids contain hydrogen in their formulae.
2.
Acid + Metal Oxide ----------------> Metal Salt + water
3.
Acid + Carbonate -----------------> Metal Salt + Water + Carbon dioxide
4.
Acid + Metal hydroxide -----------------> Metal Salt + Water
5.
A base is a substance which opposes the action of an acid or neutralizes it.
Reactions 2, 3 and 4 are neutralization reactions (water is formed).
6.
Soluble bases are called alkalis e.g. NaOH, KOH.
7.
All acids form their own particular kind of salt during neutralization e.g.
HCl(aq) forms chlorides, HNO3 forms nitrates and H2SO4 forms sulfates.
8.
During neutralization, energy is liberated.
9.
When a neutralization reaction has taken place completely, we say the
reaction has reached its end point or equivalent point. The end point is
determined by the colour change of a suitable indicator.
10.
When non-metals react with oxygen, acidic oxides are formed whilst
certain metals react with oxygen to form basic oxides.
Properties of Acids and Bases
Acids - have a sour taste
- aqueous solutions of acids conduct electricity
Well known acids:
Hydrochloric acid
Sulfuric acid
Nitric acid
Phosphoric acid
Carbonic acid
Hydrogen sulphide
Ethanoic (acetic) acid
Oxalic acid
HCl
H2SO4
HNO3
H3PO4
H2CO3
H2 S
CH3COOH
(COOH)2
©GautengDepartmentofEducation
Bases -
6
have a bitter taste
soluble bases are called alkalis
aqueous solutions of bases conduct electricity
feel soapy
Well known bases:
Sodium hydroxide
Potassium hydroxide
Calcium hydroxide
Magnesium hydroxide
Ammonium hydroxide
Sodium bicarbonate
Ammonia
NaOH
KOH
Ca(OH)2
Mg(OH)2
NH4OH
NaHCO3
NH3
Acid-Base Models – Arrhenius (1887)
An acid is a substance that releases hydrogen ions (H+) in an aqueous solution.
A base is a substance which releases hydroxyl ions (OH-) in an aqueous
solution.
Arrhenius equations: for an acid
for a base
HCl(l) ⇄ H+(aq) + Cl-(aq)
NaOH(s) ⇄ Na+(aq) + OH-(aq)
A hydrogen ion H+ (single proton) cannot exist by itself because of the high charge
density of such a particle. One or more water molecules will be attracted by the
proton to form a hydrated ion – hydronium/oxonium ion (H3O+).
Because HCl is a molecule, when it reacts with water, ionization takes place:
HCl(g) + H2O(l) ⇄ H3O+(aq) + Cl-(aq)
Compounds such as NaOH are composed of ions. These ions dissociate when
dissolved in water:
NaOH(s) ⇄Na+(aq) + OH-(aq)
Arrhenius shortcomings:
Not all acids are soluble in water.
1.
Limited acid-base reactions to taking place in aqueous solution.
2.
The OH- ion is irrelevant to acid-base process.
Acid-Base Models – Lowry-Bronsted
An acid is a substance that can donate a proton to another substance – proton
donor (giver).
A base is a substance that can accept a proton from another substance – proton
acceptor (taker).
©GautengDepartmentofEducation
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Lowry-Bronsted equations:
acid
conjugate base
HCl(g) + H2O(l) <-------------------> H3O+(aq) + Cl-(aq)
base
conjugate acid
All Lowry-Bronsted acids must contain hydrogen and the structure of these acids
must include an available, unshared electron pair.
Conjugate acid-base pairs:
Equation of an acid liberating a proton:
HA →
acid
A-
+
H+
proton
Equation of a base taking up a proton:
A- + H+ → HA
base proton
In a closed system both reactions occur and an acid-base equilibrium is
established:
HA ⇄
A- +
H+
acid
base proton
HA (the acid) and A- (the base) differ from one another by one proton – one is
obtained from the other by the removal or addition of a single proton. Such an acid
and a base are called an acid-base pair. A- is called the conjugate base of the acid
HA; HA is the conjugate acid of the base A-.
NOTE:
If the acid is strong, its conjugate base is weak, and vice-versa.
If a base is strong, its conjugate acid is weak.
Protolytic reactions:
Protolysis is a chemical reaction in which a proton transfer takes place – protolytic
reaction.
Acid1 <--------------> H+ + conjugate base1
Base 2 + H+ <--------------> conjugate acid2
Acid1 + Base2 <--------------> Conjugate base1 + Conjugate acid2
i.e.
a1 + b2 <--------------> b1 + a2
Pairs: a1 ; b1 and a2 ; b2
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Hydrolysis:
Hydrolysis is the acid-base reaction between water and the ions of a salt in solution.
Polyprotic acids:
Polyprotic acids are acids which are able to donate more than one proton per
molecule.
e.g.
monoprotic acids:
HCl
<----------------> H+
+
Cl-
diprotic acids:
H2SO4 <----------------> H+
HSO4- <----------------> H+
+
+
HSO4SO42-
triprotic acids:
H3PO4 <----------------> H+ +
H2PO4- <----------------> H+ +
HPO42- <----------------> H+ +
H2PO4HPO42PO43-
Ampholytes:
Ampholyte is a substance that can act as an acid in one reaction and then as a
base in another reaction – an amphiprotic substance.
As an acid:
As a base:
H2O <----------------> H+ + OHH2O + H+ <----------------> H3O+
HSO4- <----------------> H+ + SO42HSO42- + H+ <----------------> H2SO4
Autoprotolysis:
Since an acid reacts with a base, it follows that an ampholyte (which is an acid and a
base) can react with itself. Autoprotolysis results when a proton transfer takes place
between molecules of the same kind – self-protolysis.
As an acid:
As a base:
H2 O
H2 O + H+
Nett reaction: H2O + H2O
<---------------->
H+ + OH-
<---------------->
H3O+
<---------------->
H3O+ + OH-
H+
i.e.
H2O(l) + H2O(l) <------------> H3O+(aq) + OH-(aq)
Strong and weak acids:
The acids which we use in the lab are usually solutions of acids in water. The
original pure dissolved substance can be a gas (e.g. HCl), a liquid (e.g. H2SO4) or a
solid (e.g. oxalic acid).
A dilute acid is one which contains a lot of water in solution.
A concentrated acid contains little or no water.
©GautengDepartmentofEducation
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An acid is defined as being strong or weak depending on its ionization
equilibrium, NOT ITS CONCENTRATION i.e. a saturated solution of tartaric acid in
water is concentrated but still a weak acid; whilst a dilute solution of sulphuric acid is
still a strong acid.
Strong acids ionize almost completely in solution to form a high concentration
of hydrogen ions. They have good electrical conductivity as there are many ions in
solution.
Weak acids ionize only partially in solution and form a low concentration of
hydrogen ions. They have poor electrical conductivity as they have fewer ions in
solution.
Strong bases dissociate almost completely in water to form a high concentration of
OH- ions. They have good electrical conductivity as there are many ions in solution.
Weak bases dissociate only slightly in water to form a low concentration of OH- ions.
They have poor electrical conductivity as they have fewer ions in solution.
Ionization constant for water:
Water self-ionizes by transferring a proton from one water molecule to another,
producing a hydronium ion and a hydroxide ion.
H2O(l) + H2O(l)
The equilibrium constant is:
H3O+(aq) +OH-(aq)
Kc = [H3O+] [OH-]
[H2O]2
The [H2O] remains constant so Kc = [H3O+] [OH-]
When water ionizes, the number of moles of hydronium ions equals the
number of moles of hydroxide ions, i.e.
but
[H3O+] = [OH-]
[H3O+] [OH-] = 1.0 x 10-14 = Kw
Therefore
and
[H3O+] = 1.0 x 10-7 mol.dm-3
[OH-] = 1.0 x 10-7 mol.dm-3
Acidic:
Neutral:
Basic:
[H3O+] > 1.0 x 10-7 mol.dm-3
[H3O+] = 1.0 x 10-7 mol.dm-3
[H3O+] < 1.0 x 10-7 mol.dm-3
pH Concept and pH scale:
pH of a solution is a number that indicates the degree of acidity whilst the pOH
indicates the degree of alkalinity of a solution.
pH is defined as: the negative logarithm of the hydronium ion concentration
pH = log
1
[H3O+]
= -log [H3O+]
©GautengDepartmentofEducation
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The pH can be measured either by a pH meter or by using universal indicator.
Universal indicator is a mixture of indicators, which gives a colour change over a
wide range of pH values. The colour of the indicator provides the pH of the
substance by comparing the colour of the indicator to the scale on the bottle.
Neutralization:
In neutralization, water is formed as one of the products. Neutralization is the
‘reverse’ of hydrolysis (where water is one of the reactants).
Example:
Strong base, NaOH and strong acid HCl, are mixed:
NaOH(s) <-----------> Na+(aq) + OH-(aq)
HCl + H2O <-----------> H3O(aq) + Cl-(aq)
The H3O+(aq) ions and the OH-(aq) ions will immediately react with one
another to form water – neutralization.
Neutralization nett reaction:
H3O+(aq) + OH-(aq) --------------> 2 H2O
This is also a protolytic reaction – transfer of a proton
neutralization
Strong acid + strong base ---------------> salt + water
no hydrolysis
TAKE NOTE: A neutralization reaction does not always result in a neutral
solution with a pH of 7 – a pH of 7 only results from the neutralization of a strong
acid and a strong base. When a weak acid is neutralized by a strong base, the pH of
the solution > 7. Conversely, when a strong acid is neutralized by a strong base, the
pH < 7.
Indicators:
Indicators are organic compounds whose colour varies when the concentration of
hydronium ions changes.
Indicators change colours at characteristic pH values, so they can be used to
measure the approximate pH of a solution.
The following table gives some common indicators and their colour changes:
INDICATOR
ACID
BASE
pH RANGE
Litmus
Red
Blue
Bromothymol
Yellow
Blue
6,0-7,8
Methyl orange
Red
Orange
3,1-4,4
WHEN TO USE
For strong
acids
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Phenolphthalein
Colourless
Pink
8,3-10
For strong
bases
When an indicator is chosen for a reaction, the pH range must be considered.
Acid-base concentrations:
Volumetric analysis is when an acidic solution of known concentration (standard
solution) is added to a basic solution of unknown concentration until the resulting
solution is exactly neutralized. We can then calculate the unknown concentration
of the basic solution.
Titration is the process of adding an acid to a base.
Heat energy is liberated during a titration.
An indicator is used to determine the end point (exact point of neutralization).
The end point is the point where the no. of moles of acid and base are the same –
they neutralize one another.
Calculations:
c =
m =
MV
n
V
where n
m
M
c
V
=
=
=
=
=
no. moles
mass substance (g)
molar mass (g·mol-1)
concentration (mol·dm-3)
volume water (dm3)
At the end point/neutralization of a titration we use the balanced equation of
the reaction to determine the ratio na/nb in which the acid reacts with the base:
na = ca x Va and nb = cb x Vb
na = ca x Va
nb cb x Vb
where:
ca and cb are concentrations of the acid and base (mol.dm-3)
Va and Vb are volumes of acid and base – in same unit cm3 or dm3
na and nb are number of moles of acid and base (mol) in balanced
equation.
Never omit the na/nb , as not all acid – base reactions occur on a
1:1 basis.
©GautengDepartmentofEducation
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SECTION C:
HOMEWORK QUESTIONS
QUESTION 1: 25 minutes
(Taken from NCS Feb/Mar 2016 Paper 2)
1.1
Define an acid in terms of the Lowry-Bronsted theory.
(2)
1.2
Carbonated water is an aqueous solution of carbonic acid H2CO3.
H2CO3(aq) ionises in two steps when it dissolves in water.
1.2.1
1.2.2
1.2.3
1.3
Write down the FORMULA of the conjugate base of
H2CO3(aq).
Write down a balanced equation for the first step in the
ionisation of carbonic acid.
The pH of a carbonic acid solution at 25°C is 3,4. Calculate
the hydroxide ion concentration in the solution.
[= 2,51 x 10-11 mol-dm-3]
(1)
(3)
(5)
X is a monoprotic acid.
1.3.1 State the meaning of the term monoprotic.
1.3.2 A sample of acid X is titrated with a standard sodium hydroxide
solution using a suitable indicator.
(1)
At the end point it is found that 25cm3 of acid X is neutralised
by 27,5cm3 of the sodium hydroxide solution of concentration
0,1 mol·dm-3.
Calculate the concentration of acid X. [ 0,11 mol-dm-3]
1.3.3
(5)
The concentration of H3O+ ions in the sample of acid X is
2,4 x 10-4 mol·dm-3.
Is acid X a WEAK or a STRONG acid? Explain the answer by
referring to the answer in question 1.3.2.
(3)
[20]
QUESTION 2:
20 minutes
(Taken from Exemplar 2014)
A Grade 12 class wants to determine the percentage of ethanoic acid in a
certain bottle of vinegar. They titrate a sample taken from the bottle of vinegar
with a standard solution of sodium hydroxide. The equation for the reaction is:
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(ℓ)
2.1
Define an acid in terms of the Arrhenius theory.
(2)
2.2
Give a reason why ethanoic acid is classified as a weak acid.
(1)
2.3
Explain the meaning of standard solution.
(1)
©GautengDepartmentofEducation
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2.4
Write down the names of TWO items of apparatus needed to measure
accurate volumes of the acid and the base in this titration.
(2)
2.5
It is found that 40 mℓ of a 0,5 mol·dm sodium hydroxide solution is
needed to neutralise 20 mℓ of the vinegar.
Calculate the:
-3
2.5.1
2.5.2
2.6
pH of the sodium hydroxide solution. [ 13,7]
Percentage of ethanoic acid by mass present in the vinegar.
(Assume that 1 mℓ of vinegar has a mass of 1 g.) [ = 6%]
(4)
(7)
The sodium ethanoate (CH3COONa) formed during the above
neutralisation reaction undergoes hydrolysis to form an alkaline
(3)
solution. Write down an equation for this hydrolysis reaction.
[20]
SESSION NO:
10
TOPIC:
CONSOLIDATION OF MOMENTUM, IMPULSE AND
VERTICAL PROJECTILE MOTION
Note to Learner:
MOMENTUM: ALWAYS SELECT A DIRECTION AS POSITIVE OR NEGATIVE!!!
Try and place all the given information onto the diagram given or draw a sketch of
the collision situation. Then when you are writing down the final answer always
convert the positive or negative sign into a direction in your final answer!!!!
VERTICAL PROJECTILE MOTION: Always draw a diagram of the situation and
enter all the numerical values onto your diagram. Remember to SELECT A
DIRECTION AS POSITIVE OR NEGATIVE. A general rule is to take the initial
direction of motion of the object as the positive direction.
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
20 minutes
(Taken from NCS Feb/Mar 2016)
A man throws ball A downwards with a speed of 2 m·s-1 from the edge of a window, 45m
above a dam of water. One second later he throws a second ball, ball B, downwards and
observes that both balls strike the surface of the water in the dam at the same time. Ignore
air friction.
1.1
1.2
1.3
Calculate the speed with which ball A hits the surface of the water.
[vr = 29,80 ms-1]
Calculate the time it takes for ball B to hit the surface of the water.
[ = 1,83 s]
Calculate the initial velocity of ball B.
[vi = -15,62 m-s-1]
©GautengDepartmentofEducation
(3)
(3)
(5)
1.4
14
On the same set of axes, sketch a velocity versus time graph for the motion
of balls A and B. Clearly indicate the following on your graph:
• Initial velocities of both balls A and B
• The time of release of ball B
• The time taken by both balls to hit the surface of the water
(5)
[16]
DO NOT WRITE THE EQUATIONS OF MOTION OUT OF YOUR HEAD ….USE
YOUR INFORMATION SHEET OTHERWISE YOU WILL MAKE MISTAKES!!!
QUESTION 2: 15 minutes
(Taken from GDE Prep Exam 2014 P1)
Two identical objects P and Q with a mass of 12 kg each, are moving side by side
with an initial velocity of 5,5 m⋅s-1 east on a horizontal surface.
The following graphs show the net force experienced by each object respectively
during the same time interval.
2.1
2.2
2.3
(3)
Calculate the total impulse experienced by object Q in 10 s. [ = 60 N ⋅ s]
Compare without using any calculations the total impulse for object P with
that of object Q. Write down only GREATER THAN, LESS THAN or EQUAL (1)
TO.
(4)
Calculate the final velocity of object Q. [vf = 0,5 m ⋅ s-1]
[8]
ALWAYS REMEMBER THAT ONCE YOU HAVE SELECTED A DIRECTION AS
POSITIVE OR NEGATIVE, YOU STAY WITH THAT CONVENTION
THROUGHOUT THE QUESTION AND REMEMBER TO APPLY IT TO YOUR
GRAPH AS WELL!!!
QUESTION 3:
20 minutes
(Taken from Feb/Mar 2014 P1)
The momentum versus time graph of object A, originally moving horizontally EAST,
is shown below.
©GautengDepartmentofEducation
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3.1
Write down the definition of momentum in words.
(2)
3.2
The net force acting on object A is zero between t = 10 s and t = 20 s.
Use the graph and a relevant equation to explain why this statement is (2)
TRUE.
3.3
3.4
Calculate the magnitude of the impulse that object A experiences between
t = 20 s and t = 50 s. [ = 170 N ⋅ s / kg ⋅ m s-1]
(3)
At t = 50 s, object A collides with another object, B, which has a momentum
of 70 kg·m·s-1 EAST. [ = 100 kg ⋅ m s-1]
Use the information from the graph and the relevant principle to calculate the
momentum of object B after the collision.
REMEMBER THAT IN ALL COLLISIONS, MOMENTUM IS CONSERVED. IN
ELASTIC COLLISIONS KINETIC ENERGY IS CONSERVED. IN AN INELASTIC
COLLISION, KINETIC ENERGY IS NOT CONSERVED!!!
SECTION C:
HOMEWORK QUESTIONS
QUESTION 1: 25 minutes
(Taken from NSC Exemplar 2014)
A ball of mass 0,5 kg is projected vertically downwards towards the ground from a
-1
height of 1,8 m at a velocity of 2 m·s . The position-time graph for the motion of the
ball is shown below.
©GautengDepartmentofEducation
(5)
[12]
1.1
16
What is the maximum vertical height reached by the ball after the second
bounce?
(1)
Calculate the:
1.2
Magnitude of the time t1 indicated on the graph. [= 0,44 s]
(5)
1.3
Velocity with which the ball rebounds from the ground during the first bounce
[ vi = 4,2 m-s upwards]
(4)
The ball is in contact with the ground for 0,2 s during the first bounce.
1.4
Calculate the magnitude of the force exerted by the ground on the ball during
-1
(4)
the first bounce if the ball strikes the ground at 6,27 m·s .
[Fnet = 26,175 N]
1.5
Draw a velocity-time graph for the motion of the ball from the time that it is
projected to the time when it rebounds to a height of 0,9 m.
Clearly show the following on your graph:
• The time when the ball hits the ground
• The velocity of the ball when it hits the ground
• The velocity of the ball when it rebounds from the ground
(3)
[17]
REMEMBER THAT THE GRADIENT OF A v-t GRAPH GIVES YOU
ACCELERATION AND THE GRADIENT OF A x-t GRAPH GIVES YOU
VELOCITY!!!!
QUESTION 2: 10 minutes
(Taken from NSC Exemplar 2014)
Two boys, each of mass m, are standing at the back of a flatbed trolley of mass 4 m.
The trolley is at rest on a frictionless horizontal surface.
2.1
Write down the principle of conservation of linear momentum in words.
©GautengDepartmentofEducation
(2)
17
2.2
Calculate the final velocity of the trolley. [ vf = 1 m-s-1]
2.3
The two boys jump off the trolley one at a time. How will the velocity of the
trolley compare to that calculated in QUESTION 4.2? Write down only
GREATER THAN, SMALLER THAN or EQUAL TO.
(4)
ALWAYS START YOUR CALCULATION WITH “Σp before = Σp after” AND
THEN EXPAND IT INTO THE CONSERVATION OF MOMENTUM EQUATION!!!
©GautengDepartmentofEducation
(1)
[8]
18
SESSION NO:
TOPIC:
11
CONSOLIDATION OF ORGANIC MOLECULES AND THEIR
REACTIONS
Note to Learner: The functional groups, homologous series and IUPAC naming system
needs to be learnt well and practiced regularly in order to do well in this section. A good
knowledge of intermolecular forces is required in order to explain the relationship between
physical properties and the intermolecular forces, the type of functional groups, chain
length and branched chains.
You need to know the different types of reactions, the conditions under which they occur
as well as the type of products that are formed. Only basic polymerization as an
application of organic chemistry is required.
SECTION A: TYPICAL EXAM QUESTIONS
QQ
QUESTION 1: 25 minutes
(Taken from NSC Exemplar 2014 Paper 2)
The letters A to G in the table below represent seven organic compounds.
©GautengDepartmentofEducation
1.1
19
Write down the:
1.1.1
Name of the homologous series to which compound F belongs
(1)
1.1.2
Name of the functional group of compound D
(1)
1.1.3
Letter that represents a primary alcohol
(1)
1.1.4
IUPAC name of compound A
(2)
1.1.5
Structural formula of the monomer of compound B
(2)
1.1.6
Balanced equation, using molecular formulae, for the combustion of
compound E in excess oxygen
(3)
1.2
Briefly explain why compounds C and D are classified as POSITIONAL
ISOMERS.
(2)
1.3
Compound G is prepared using an alcohol as one of the reactants. Write
down the balanced equation for the reaction using structural formulae for all
the organic reagents.
(7)
[19]
ALWAYS IDENTIFY THE MOLECULES AND THEIR FUNCTIONAL GROUPS
GIVEN IN THE TABLE BEFORE TRYING TO ANSWER THE QUESTIONS!!!
QUESTION 2: 15 minutes
(Taken from NSC Feb/Mar 2015 Paper 2)
Learners use compounds A to C, shown in the table below, to investigate a factor
which influences the boiling point of organic compounds.
2.1
Which ONE of the compounds (A, B or C) has the highest boiling point?
(1)
2.2
For this investigation, write down the:
2.2.1
Independent variable
2.2.2
Dependent variable
(1)
(1)
2.3
Write down the name of the type of Van der Waals force that occurs between
the molecules of compound B.
(1)
2.4
How will the vapour pressure of 2-methylpentane compare to that of
compound C? Write down only HIGHER THAN, LOWER THAN or EQUAL (1)
TO.
©GautengDepartmentofEducation
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The learners now compare the boiling points of compounds D and E, shown in the
table below.
2.5
How does the boiling point of compound D compare to that of compound E?
Write down HIGHER THAN, LOWER THAN or EQUAL TO. Fully explain the
answer.
(4)
[9]
ALWAYS REMEMBER THAT ALCOHOLS HAVE ONE HYDROGEN BOND
BETWEEN THEIR MOLECULES AND CARBOXYLIC ACIDS HAVE TWO
HYDROGEN SITES. THE ALKANES, ALKENES AND ALKYNES HAVE WEAK
VAN DER WAALS FORCES. THE KETONES AND ALDEHYDES HAVE DIPOLEDIPOLE FORCES!!!
QUESTION 3: 15 minutes
(Taken from NSC Nov 2015 Paper 2)
3.1
The flow diagram below shows two organic reactions. The letter P
represents an organic compound.
Use the information in the flow diagram to answer the questions that follow.
Write down the:
3.1.1
Type of reaction of which Reaction 1 is an example
(1)
3.1.2
STRUCTURAL FORMULA of the functional group of ethyl
propanoate
(1)
3.1.3
IUPAC name of compound P
(1)
Reaction 2 takes place in the presence of an acid catalyst and heat.
Writedownthe:
3.1.4
Type of reaction of which Reaction 2 is an example
(1)
3.1.5
NAME or FORMULA of the acid catalyst
(1)
©GautengDepartmentofEducation
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3.1.6
3.2
STRUCTURAL FORMULA of the alkene
(2)
The condensed formula of a polymer is shown below.
Write down the:
3.2.1
3.2.2
STRUCTURAL FORMULA of the monomer that is used to prepare
the above polymer.
Type of polymerisation reaction (ADDITION or CONDENSATION)
that is used to prepare this polymer.
ADDITION REACTIONS: during the addition of HX and H2O to unsaturated
hydrocarbons, the H atom attaches to the C atom which has the greater
number of H atoms. The X or OH groups attaches to the C atom, which has the
more substitutions.
ELIMINATION REACTIONS: if more than one elimination product is possible,
the major product is the one where the H atom is removed from the C atom
with the least number of H atoms.
©GautengDepartmentofEducation
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SECTION C:
HOMEWORK QUESTIONS
QUESTION 1: 20 minutes
(Taken from NSC Nov 2015 Paper 2)
Four compounds of comparable molecular mass are used to investigate the effect of
functional groups on vapour pressure. The results obtained are shown in the table below.
1.1
Define the term functional group of an organic compound.
1.2
Which ONE of the compounds (A, B, C or D) in the table has the:
1.2.1
1.2.2
(2)
Highest boiling point
(Refer to the vapour pressures in the table to give a reason for the
answer.)
Weakest intermolecular forces
(2)
(1)
1.3
Refer to the type of intermolecular forces to explain the difference between
the vapour pressure of compound A and compound B.
(3)
1.4
The vapour pressures of compounds C and D are much lower than those of
compounds A and B. Name the type of intermolecular force in A and B that
is responsible for this difference.
(1)
1.5
Briefly explain the difference in vapour pressure between compound C and
compound D.
1.6
During a combustion reaction in a closed container of adjustable volume,
3
8 cm of compound A (butane) reacts in excess oxygen according to the
following balanced equation:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
©GautengDepartmentofEducation
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3
If the initial volume of the oxygen in the container was 60 cm , calculate the
TOTAL volume of the gases that are present in the container at the end of
the reaction. All the gases in the container are at the same temperature and
(5)
pressure. [vtot = 80 cm3]
[16]
QUESTION 2: 20 minutes
(Taken from NSC Nov 2014 Paper 2)
The flow diagram below shows the preparation of an ester using prop-1-ene as a
starting reagent. P, Q, R and S represent different organic reactions.
2.1
2.2
Write down the type of reaction represented by:
2.1.1
Q
2.1.2
R
(1)
(1)
For reaction P write down the:
2.2.1
Type of addition reaction.
(1)
2.2.2
(3)
Balanced equation using structural formulae.
2.3
Write down the structural formula of the haloalkane formed in reaction Q.
(2)
2.4
In reaction S propan-1-ol reacts with ethanoic acid to form the ester.
For this reaction write down the:
2.4.1
Name of the reaction that takes place.
(1)
2.4.2
FORMULA or NAME of the catalyst needed
(1)
2.4.3
Structural formula of the ester formed
(2)
2.4.4
IUPAC name of the ester formed
(2)
2.5
The propan-1-ol formed in reaction R can be converted to prop-1-ene. Write
down the FORMULA or NAME of the inorganic reagent needed.
©GautengDepartmentofEducation
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24
SESSION NO:
TOPIC:
12
CONSOLIDATION OF WORK, ENERGY AND POWER
Note to Learner: Remember that only the component of the applied force that is
parallel to the motion does work on an object. Forces perpendicular to the object’s
displacement does no work on the object. Forces parallel to the object’s
displacement does positive work on the object while forces anti-parallel i.e. in the
opposite direction to the object’s displacement does negative work on the object.
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 20 minutes
(Taken from NSC Exemplar 2014 Paper 1)
A 3 kg trolley is at rest on a horizontal frictionless surface. A constant horizontal
force of 10 N is applied to the trolley over a distance of 2,5 m.
When the force is removed at point P, the trolley moves a distance of 10 m up the
incline until it reaches the maximum height at point Q. While the trolley moves up the
incline, there is a constant frictional force of 2 N acting on it.
1.1
Write down the name of a non-conservative force acting on the trolley as it
moves up the incline.
(1)
Draw a labelled free-body diagram showing all the forces acting on the
trolley as it moves along the horizontal surface.
(3)
1.3
State the WORK-ENERGY THEOREM in words.
(2)
1.4
Use the work-energy theorem to calculate the speed of the trolley when it
reaches point P. [vf = 4,08 m⋅s-1]
(4)
1.2
1.5
Calculate the height h, that the trolley reaches at point Q. [h = 0,17 m]
REMEMBER YOU CAN USE EQUATIONS OF MOTION HERE IF IT IS MOTION IN
ONE DIRECTION!!!!
©GautengDepartmentofEducation
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ALSO REMEMBER THAT A FREE BODY DIAGRAM MUST HAVE A KEY OR
LEGEND AND IT MUST BE A BIG DOT WITH ALL THE FORCES ACTING FROM
IT!!!
REMEMBER TO FIRST CALCULATE THE FORCE WHETHER IT IS NET FORCE,
FORCE OF FRICTION FIRST ETC AND THEN SUBSTITUTE INTO THE WORK
EQUATION!!! THE ANGLE IN THE WORK EQUATION WILL THEN ONLY BE 0°,
180° OR 90 °!!
QUESTION 2:
20 minutes
(Taken from MP SEPT 2015 Paper 1)
A toy canon, mass 1,6 kg, is at rest on a rough horizontal surface as shown in the
diagram. A steel marble, mass 0,8 kg, is fired horizontally to the east from the canon.
Immediately after firing the marble, the canon moves at 0,26 m·s-1 to the west.
2.1
Calculate the speed of the steel marble immediately after firing the marble.
[vf = 0,52 m⋅s-1]
2.2
The steel marble experiences a force F during the firing. Explain in terms of
F how the force experienced by the CANON compares with that experienced
by the steel marble.
(4)
(3)
The canon reaches point A with a speed of 0,2 m·s-1 and then moves down
a rough 0,5 m long slope AB.
2.3
Explain why this is NOT a closed system.
(1)
2.4
Calculate the kinetic frictional force experienced by the canon as it moves
from A to B if the coefficient of kinetic friction (µk) is 0,12. [= 1,63 N]
(3)
2.5
Using ENERGY PRINCIPLES only, calculate the velocity of the canon at
point B. [vf = 1,98 m⋅s-1]
IF IT IS A NON-CONSERVATIVE FORCE, LIKE AIR RESISTANCE, FRICTION
ETC ACTING ON AN OBJECT, YOU CANNOT USE THE CONSERVATION OF
MECHANICAL ENERGY TO CALCULATE ANY UNKNOWNS!!! YOU CAN ONLY
USE THE CONSERVATION OF MECHANICAL ENERGY IF THE FORCE IS A
CONSERVATIVE LIKE GRAVITATIONAL FORCE!!
©GautengDepartmentofEducation
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SECTION C:
HOMEWORK QUESTIONS
QUESTION 1:
20 minutes
(Taken from NSC Feb/Mar 2015 paper 1)
A 5 kg block is released from rest from a height of 5 m and slides down a frictionless
incline to point P as shown in the diagram below. It then moves along a frictionless
horizontal portion PQ and finally moves up a second rough inclined plane. It comes
to a stop at point R which is 3 m above the horizontal.
The frictional force, which is a non-conservative force, between the surface and the
block is 18 N.
1.1
Using ENERGY PRINCIPLES only, calculate the speed of the block at point
P. [ = 9,90 m⋅s-1]
(4)
1.2
Explain why the kinetic energy at point P is the same as that at point Q.
(2)
1.3
Explain the term non-conservative force.
(3)
1.4
Calculate the angle (θ) of the slope QR. [ = 39,430]
(7)
[15]
QUESTION 2: 15 minutes
(Taken from GAUT SEPT 2014 Paper 1)
A cyclist pushes his bicycle of mass 6,1 kg up an incline with a force of 20 N. The
bicycle is pushed from an initial velocity of 5 m⋅s-1 from point A to point B. The road is
inclined at 10° to the horizontal and the distance from A to B is 32 m as shown in the
diagram.
©GautengDepartmentofEducation
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The road surface exerts a force of friction of 11 N on the bicycle tyres.
2.1
Calculate the work done by the cyclist on the bicycle. [ = 630,28 J]
(3)
2.2
Use the work-energy theorem and calculate the magnitude of the velocity of
the bicycle at 32 m. [ = 3,24 m⋅s-1]
(5)
2.3
Explain why frictional forces are regarded as non-conservative forces.
SESSION NO:
TOPIC:
(2)
[10]
13
CONSOLIDATION OF DOPPLER EFFECT AND RATES OF
REACTION
Note to Learner: For Doppler equation, always write the equation down exactly as it
is given on the information sheet. Always include the ±. Only once you have
substituted into the equation do you make the unknown the subject of the formula. It
is suggested that you do not substitute the zeros into the equation. Rather leave
them out. This is the only time that you do not substitute in the zero values.
For rates of reactions, remember you need to revise some of your Grade 11 content
such as exothermic and endothermic reactions. You must be able to draw these
graphs and label them as well as being able to read off these graphs and answer
questions on them. You will need to know the definitions of bond energy, activation
energy, enthalpy etc.
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 10 minutes
(Taken from WC SEPT 2014 Paper 1)
1.1
Use the diagram below to answer the following questions.
©GautengDepartmentofEducation
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1.1.1
1.1.2
1.2
Identify the medical device shown in the diagram.
Explain very briefly how the device functions and what it may be
used for.
(1)
(2)
A fire truck with its siren on, moves away at constant velocity from a person
standing next to the road. The person measures a frequency which is 90% of
the frequency of the sound emitted by the siren of the fire truck.
1.2.1
Name the phenomenon observed.
(1)
1.2.2
If the speed of sound in air is 340 m·s-1, calculate the speed of the
(4)
fire truck. [ v = 37,78 m⋅s-1]
[8]
WHEN YOU USE THE DOPPLER EQUATION, ALWAYS WRITE IT DOWN EXACTLY
OFF THE INFORMATION SHEET! IF THE LISTENER OR THE SOUND IS
STATIONARY, DO NOT SUBSTITUTE THE ZERO!!
QUESTION 2: 10 minutes
(Taken from WCP SEPT 2014 Paper 2)
You are asked to do an investigation into how the concentration of reactants affects
reaction rate. The reaction used for the investigation is that between sodium
thiosulfate, Na2S2O3, and hydrochloric acid (HCℓ). When combined, a substance is
produced which causes the solution to become cloudy:
Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + H2O(l) + SO2(g)
List of chemicals: 250 ml of sodium thiosulfate solution; distilled water;
hydrochloric acid solution of concentration 2 mol·dm-3
Method: (incomplete)
1. Use a black pen to draw a large cross on a piece of paper
2. Put 50 ml of sodium thiosulfate solution into a conical flask.
3. Add 5 ml of hydrochloric acid, starting the stopwatch; swirl the mixture once or
twice.
4. Place the reaction flask on the paper with the cross and note the time when the
cross can no longer be seen.
©GautengDepartmentofEducation
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5. Repeat the experiment but ….
6. Repeat the experiment a third time but …
7. Record the results in a table.
Results:
2.1
Give the definition of reaction rate.
(1)
2.2
For the experiment described above, name the
2.2.1
independent variable
2.2.2
dependant variable
(1)
(1)
Consider the method:
2.3.1
Rewrite points 5 and 6 describing what you would change each
time so that the aim of the experiment can be achieved.
(Note how the Time taken changes- do not be concerned with
how much it changes).
2.3.2
What needs to be done in order to ensure that this is a fair test?
(3)
(1)
2.3
2.4
2.5
Write down the name or formula of the substance responsible for
the cloudiness.
Consider the table of results: which experiment (1, 2 or 3) occurred at the
highest reaction rate?
©GautengDepartmentofEducation
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(1)
[9]
30
QUESTION 3: 20 minutes
(Taken from NSC Exemplar 2014 Paper 2)
3
-3
Zinc granules are added to 100 cm of a 0,2 mol·dm hydrochloric acid solution in an
Erlenmeyer flask. The equation for the reaction that takes place is:
Zn(s) + 2HCℓ(aq) → ZnCℓ2(aq) + H2(g)
The rate of the reaction is followed by measuring the loss in mass of the flask and its
contents at regular time intervals. After completion of the reaction, it is found that
0,12 g zinc granules did not react.
3.1
Which reactant is the limiting reagent?
(1)
3.2
Give a reason for the loss in mass of the flask and its contents.
(1)
3.3
Sketch a graph of the mass of zinc versus time for the above reaction. Label
this graph P.
(2)
On the same set of axes as in QUESTION 5.3, sketch graph Q which
represents the same reaction at a HIGHER TEMPERATURE.
(1)
3.5
Use the collision theory to explain why graph Q differs from graph P.
(2)
3.6
Calculate the mass of zinc initially present in the flask. [ = 0,77 g]
(6)
[13]
3.4
GRADE 11 STOICHIOMETRY, IS TESTED HERE AND YOU MUST ALSO BE ABLE TO
DETERMINE THE RATE OF THE REACTION BY READING VALUE OFF A GRAPH!!!
THIS IS A VERY THEORETICAL SESSION AND YOU NEED TO KNOW THE FIVE
FACTORS INFLUENCING RATE VERY WELL!! YOU MUST BE ABLE TO APPLY THE
COLLISION THEORY!!!
©GautengDepartmentofEducation
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SECTION C:
HOMEWORK QUESTIONS
QUESTION 1:
20 minutes
(Taken from MP SEPT 2015 Paper 1)
A man mounts a siren, which produces a constant frequency of 800 Hz, on the roof
of his car. He drives at a constant speed up and down a straight road while a
stationary learner measures the observed sound. At a certain stage of the journey,
the learner obtains the following pressure-time graph of the sound wave:
1.1
What is the period of the detected sound wave?
(1)
1.2
Calculate the frequency of the detected sound wave. [ = 1000 Hz]
(3)
1.3
State the Doppler-effect in words.
(2)
1.4
Calculate the speed of the moving car. Take the speed of sound in air as
340 m·s-1. [Vs = 68 m⋅s-1]
(5)
1.5
While the car is stationary, the frequency of the siren is changed to 900 Hz.
Will the wavelength of the detected sound wave INCREASE, DECREASE
or REMAIN THE SAME? Explain the answer.
(3)
[14]
QUESTION 2:
25 minutes
(Taken from NSC Feb/Mar 2016 Paper2)
Methanol and hydrochloric acid react according to the following balanced equation:
CH3OH(aq) + HCℓ(aq) → CH3Cℓ(aq) + H2O(ℓ)
2.1
State TWO factors that can INCREASE the rate of this reaction.
(2)
2.2
Define the term reaction rate.
(2)
©GautengDepartmentofEducation
2.3
32
The rate of the reaction between methanol and hydrochloric acid is
investigated. The concentration of HCℓ(aq) was measured at different time
intervals. The following results were obtained:
2.3.1
2.3.2
2.3.3
2.3.4
2.3.5
TIME (MINUTES)
HCℓ CONCENTRATION (mol⋅dm-3)
0
15
55
100
215
1,90
1,45
1,10
0,85
0,60
Calculate the average reaction rate, in (mol·dm-3)·min-1 during the
first 15 minutes. [ = 0,03 (mol⋅dm-3) ⋅ min-1]
(3)
Use the data in the table to draw a graph of concentration versus
time.
(3)
From the graph, determine the concentration of HCℓ(aq) at the
40th minute.
(1)
Use the collision theory to explain why the reaction rate
decreases with time. Assume the temperature remains constant.
(3)
Calculate the mass of CH3Cℓ(aq) in the flask at the 215th minute.
The volume of the reagents remains 60cm3 during the reaction.
[ = 3,54 g]
(5)
[19]
THIS IS A VERY THEORETICAL SESSION AND YOU NEED TO KNOW THE FIVE
FACTORS INFLUENCING RATE VERY WELL!! YOU MUST BE ABLE TO APPLY THE
COLLISION THEORY!!!
©GautengDepartmentofEducation
33
SESSION NO:
TOPIC:
14
CONSOLIDATION OF CHEMICAL EQUILIBRIUM AND ACIDS
AND BASES
Note to Learner: For chemical equilibrium, you need to be able to distinguish
between an open and closed system, as well as to identify a reversible reaction and
understand dynamic equilibrium. You need to list the factors that influence the
position of an equilibrium. The calculation of Kc is normally the higher order question.
You will need to be able to interpret equilibrium and rate graphs as well as apply Le
Chatelier’s Principle.
For acids and bases, you need to know the different acid-base theories. You will
need to revise all your Grade 11 acid-base theory and definitions including
calculations.
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 20 minutes
(Taken from NSC Feb/Mar 2015 Paper 2)
3
Pure hydrogen iodide, sealed in a 2 dm container at 721 K, decomposes
according to the following balanced equation:
-1
2HI(g) ⇌ H2(g) + I2(g) ΔH = + 26 kJ·mol
The graph below shows how reaction rate changes with time for this reversible
reaction.
1.1
Write down the meaning of the term reversible reaction.
©GautengDepartmentofEducation
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1.2
1.3
34
th
How does the concentration of the reactant change between the 12 and
th
the 15 minute? Write down only INCREASES, DECREASES or NO
CHANGE.
(1)
The rates of both the forward and the reverse reactions suddenly change
at t = 15 minutes.
1.3.1
Give a reason for the sudden change in reaction rate.
(1)
1.3.2
Fully explain how you arrived at the answer to QUESTION
1.3.1.
(3)
The equilibrium constant (Kc) for the forward reaction is 0,02 at 721 K.
1.4
At equilibrium it is found that 0,04 mol HI(g) is present in the container.
Calculate the concentration of H2(g) at equilibrium.
[2,83 x 10-3 mol⋅dm-3]
(6)
1.5
Calculate the equilibrium constant for the reverse reaction. [ = 50]
(1)
1.6
The temperature is now increased to 800 K. How will the value of the
equilibrium constant (Kc) for the forward reaction change? Write down only
INCREASES, DECREASES or REMAINS THE SAME.
(1)
[14]
REMEMBER ONLY TEMPERATURE INFLUENCES Kc!!!!
REMEMBER TO CALCULATE THE EQUILBRIUM NUMBER OF MOL
AND THEN TO CONVERT IT TO CONCENTRATION BY USING
c = n/V!!!!
IF YOU HAVE TO EXPLAIN LE CHATELIER’S PRINCIPLE ALWAYS
STATE THE CHANGE, STATE HOW THE REACTION GIVEN IS
AFFECTED BY THE CHANGE AND THEN ANSWER THE QUESTION
OR GIVE A CONCLUSION!!
QUESTION 2:
30 minutes
(Taken from Gauteng Sept 2014 Paper 2)
You are given the following ionisation reaction of ethanoic acid in water:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
Ka=1,8 x 10-5
2.1
What does the ionisation constant indicate about the strength of the acid?
(1)
2.2
What is meant by a concentrated acid?
(2)
2.3
H2PO4- is an ampholyte.
Write an equation to indicate its role as a base.
(2)
Write down a conjugate acid/base pair from your equation in QUESTION
2.3.
(1)
2.4
©GautengDepartmentofEducation
2.5
35
Milk of Magnesia has been used over the ages to relieve stomach ailments
caused by excess stomach acid. The active ingredient in Milk of Magnesia
is magnesium hydroxide (Mg(OH)2).
A group of learners prepare a solution of magnesium hydroxide.
2.5.1
2.5.2
2.5.3
2.6
What mass of Mg(OH)2 must be dissolved in distilled water to
prepare 500 cm3 of a solution with a concentration of 0,20
mol·dm-3?
[ = 5,8 g]
What will the concentration of the hydroxide ions in the solution
be? Assume 100% dissociation of magnesium hydroxide in
water.
(5)
(2)
The pH of any medicine safe for human consumption must lie
between pH = 4 and pH =9.
Will this solution that the learners prepare be safe for human
consumption? Show all calculations. [ = 13,3]
(5)
The learners now check the concentration of their solution that they made.
They run a titration using hydrochloric acid of concentration 0,1 mol·dm-3.
They transfer 25 cm3 of the magnesium hydroxide solution into a conical
flask.
The balanced chemical equation for this reaction is:
Mg(OH)2 (aq) + 2HCℓ (aq)→ MgCℓ2 (aq) + 2 H20(ℓ)
Determine what volume of HCℓ will be needed to fully neutralise the Milk of
Magnesia if the concentration is actually 0,2 mol·dm-3. [ V = 100 cm3]
(4)
[22]
NO SECTION B. REFER TO PREVIOUS CONTENT NOTES PROVIDED.
SECTION C:
HOMEWORK QUESTIONS
QUESTION 1:
20 minutes
(Taken from NSC Feb/Mar 2014 Paper 2)
The reaction of methane gas (CH4) with steam (H2O) produces hydrogen gas. The
equation for the reaction is shown below.
CH4(g) + 2H2O(g) ⇌ CO2(g) + 4H2(g)
1.1
Briefly explain why the CO2 gas may be harmful to the environment.
Initially, 1 mol of methane and 2 mol of steam are sealed in a 5,0 dm3
container. When equilibrium is established at temperature T1, the mixture
contains 0,3 mol of CO2(g).
©GautengDepartmentofEducation
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36
1.2
Define the term chemical equilibrium.
(2)
1.3
Calculate the equilibrium constant (KC) at T1. [Kc = 0,02 (0,018)]
(7)
1.4
A new equilibrium is now established at a higher temperature T2. The
value of the equilibrium constant (KC) at this new temperature is 0,01.
Is this reaction exothermic or endothermic? Use Le Chatelier's principle
and the value of KC at T1 and T2 to explain the answer.
©GautengDepartmentofEducation
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[15]