Gr 12 Physical Sciences Presentation General Notes

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CONTENT

:

1. Work, energy and power 2. Doppler Effect 3. Organic Chemistry 4. Exam guidelines

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1. Work, Energy and Power 1.1 Work

When a force exerted on an object causes it to move, WORK is done on the object (except if the force and displacement are at right angles to each other).

W

net

= F

net

Δ

x

cosθ WHERE

: W net = ΣW (sum of each individual force that is exerted on the system )

F

net Δ

x

θ

= the size of the net force = size of the displacement = angle between the force displacement Δ

x

F

net and the

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(1) (2) W = F Δ

x

cosθ Fcosθ Δ

x

θ F The work done by the force F on the lawnmower is FΔ

x

cosθ.

When F is exerted parallel in the same direction as Δ

x

, then cos θ = cos 0º = 1

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Δ

x = 0

A person suitcase is work on because holding the doing there is motion Δ

x

= 0 m thus W = 0 J a no suitcase no

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(3) θ = 90° cos θ = 0 Δ

x

(4) θ

Fcos

θ Δ

x

When perpendicular to cos F is Δ exerted

x

, then θ = cos 90º = 0, and then the force is not doing work on the suitcase.

Work will be done if a person carry a suitcase up a staircase because W = F Δ

x

cos θ and cos θ will be between 0 and 1.

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(5) (6)

Electrical motor

cos θ = cos 0° = 1 Δ

x

Work will be done by (f) on the suitcase that is displaced over a floor, Δ

x

. But f is parallel and in the opposite direction as Δ

x

, so: cos θ = cos 180º = -1 Force F, that the electric motor exerts on the suitcase, is doing work. But F is parallel and in the opposite direction to cos θ = cos 180º = -1 Δy, so:

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An object is moving DOWN a frictionless surface

A free body diagram to indicate ALL the forces exerted on the object as it moves

DOWN

the surface.

Δ

x

REMEMBER w = F g = force due to gravity (weight)

h W net = W w// + W N = w // Δ

x

cos θ + NΔ

x

= (mgsin30 º)Δ

x

= (mgsin30 + W w┴ º)Δ

x

cos θ + w ┴ Δ

x

cos0 º + NΔ

x

cos0 º + 0 + 0 cos θ cos90 º + w ┴ Δ

x

= (mgsin30 º)Δ

x

cos0 º cos90 º

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An object is moving UP a frictionless surface

A free body diagram to indicate ALL the forces exerted on the object as it moves

UP

the surface.

Δ

x

h W net = W = w // w// Δ

x

+ W cos = (mgsin30 = (mgsin30 N + W w┴ θ + NΔ º)Δ º)Δ

x x x

cos θ + w ┴ cos180 º + NΔ

x

Δ

x

cos180 º + 0 + 0 cos cos90 θ º + w ┴ = (mgsin30 º)Δ

x

cos180 º Δ

x

cos90 º

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An object is moving DOWN a surface with friction

A free body diagram to indicate ALL the forces exerted on the object as it moves DOWN the surface.

Δ

x

h W net = W w// = w // Δ

x

+ W f + W N = (mgsin30 º)Δ

x

+ W w┴ cos θ + fΔ

x

cos cos0 θ + 0 + 0 º + fΔ

x

cos180 º

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An object is moving UP a surface with friction

A free body diagram to indicate ALL the forces exerted on the object as it moves UP the surface.

Δ

x

h W net = W w// = w // Δ

x

+ W f + W N cos θ + fΔ

x

= (mgsin30 º) Δ

x

+ W w┴ cos θ + 0 + 0 cos180 º + fΔ

x

cos180 º

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1.2 WORK ENERGY THEOREM

The netto work done on an object is equal to the change of the kinetic energy of the objects .

OR: The work done on an object by a net force is equal to the change in the kinetic energy of the object.

W

net

= ΔK = E

kf

– E

ki

but

W net = F net Δ

x

cosθ

therefore F

net Δ

x

cosθ = ΔK = E kf – E ki = ½mv f 2 ½mv i 2 Also for W net = ΣW (of each individual force exerted on the system)

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LAW OF CONSERVATION OF ENERGY Energy cannot be destroyed or created but can only be transferred from one FORM to another .

LAW OF CONSERVATION OF MECHANICAL ENERGY

The sum of the potential energy and the kinetic energy (the mechanical energy) in a closed system remains constant

Mechanical energy (i) = Mechanical energy (f) (U + K) i = (U + K) f CLOSED SYSTEM No external forces (like friction) act on the system or have an effect on the system.

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IF THERE ARE NO FRICTIONAL FORCES

Use the law of conservation of mechanical energy:

Mechanical energy (i) = Mechanical energy (f) (U + K) i = (U + K) f …OR…

Use the work energy principle

W net = ΔK = E kf = – E ki ½mv f 2 ½mv i 2 IF THERE ARE FRICTIONAL FORCES

Use the work energy principle:

W net = = E kf = ΔK – E ki ½mv f 2 ½mv i 2 REMEMBER

: W net = ΣW (sum of each individual force exerted on the system)

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EXAMPLES

EXAMPLE 1

A 5 kg crate moves at a speed of 10 m

.

s -1 at point A. The crate moves further along the slide as in the sketch until it has moved passed E. All parts of the slide is frictionless except part DE. Make use of energy methods when doing calculations.

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ANSWERS

1.1 Calculate the speed of the block at point B.

There is no friction: Use the law of conservation of mechanical energy.

Mechanical energy (A)= Mechanical energy (B) (U + K) A = (U + K) B

mgh A + + ½mv A 2 = mgh B + ½mv B 2 (5)(9,8)(12) + ½(5)(10) 2 = (5)(9,8)(3) + ½(5)

v B 2

588 + 250 = 147 + 2,5

v B 2 v B = 16,63 m .

s -1

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ANSWERS

(Continued)

1.2 Calculate the speed of the block at point C.

There is no friction: Use the conservation of mechanical energy.

law

Mechanical energy (A)= Mechanical energy (C) (U + K) A = (U + K) C

mgh A + + ½m

v A 2

= mgh C + ½m

v C 2

(5)(9,8)(12) + ½(5)(10) 2 = (5)(9,8)(6) + ½(5)

v C 2

588 + 250 = 294 + 2,5

v C 2 v C = 14,75 m .

s -1

of

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ANSWERS

(Continued)

1.3 Calculate the speed of the block at point D.

There is no friction: Use the law of conservation of mechanical energy.

Mechanical energy (A)= Mechanical energy (D) (U + K) A = (U + K) D

mgh A + ½m

v A 2

= mgh C + ½m

v D 2

(5)(9,8)(12) + ½(5)(10) 2 = 0 + ½(5)

v D 2

588 + 250 = 0 + 2,5

v D 2 v D = 18,31 m .

s -1

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1.4 Draw a free body diagram of ALL the forces on the block as it moves up the plane.

N (Force of surface on crate) f (Frictional force)

OR

w (Force of earth on crate)

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1.5 The crate reaches the top of the slope (point E) with a

speed of 15,38 m .

s -1 where-after it moves further at this speed over the horizontal surface.

1.5.1 Calculate the work done on the crate by the gravitational force.

W w = w Δ

x

cos θ

= mg

Δ

x

cos120 º

…OR… W w = w Δ

y

cos θ

= mghcos180 º = (5)(9,8)(6)(-0,5) = -147 J

…OR…

= (5)(9,8)(3)(-1) = -147 J

W w = w // Δ

x

cos θ

= (mgsin35 º)(6)(cos180º) = (5)(9,8)sin30 º(6)(-1) = -147 J

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ANSWERS (Continued)

1.3.2 Calculate the work done on the crate by the frictional force.

There is friction now: Use the work energy theorem

W net = ΔK W w -147 + N Δ

x

+ W N cos90

º

+ W f + W f = = ΔK ½mv f 2 ½mv i 2

-147 + 0+

W f

= ½(5)(15,38) 2 ½(5)(18,31) 2

W f

= -99,78 J

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ANSWERS (Continued)

1.3.3 The size of the friction that is exerted on the crate.

W f = f Δxcosθ

-99,78 = f(6)cos180

f = 16,36 N

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1.3 POWER

Power is the rate at which work is done or energy is used.

P = W Δt

If a force that is exerted on an object moves the object at a constant velocity, we can calculate the instantaneous power or average power by using:

P = Fv

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Example 2

A lift with a mass 1000 kg carry a maximum load of 800 kg. A constant frictional force of 4000 N works against the motion.

The force in the cable is called tension (T). Weight is w = F g .

What is the minimum power that the lift motor must deliver to move the lift upwards at a constant speed of 3 m·s -1 ?

ANSWER

T F net = ma = 0 (since constant speed; a = 0 – f – w = 0 T = f + mw

= f + mg = 4 000 + (1 800)(9,8)

= 21 640 N P = Fv = Tv

=(21 640)(3)

= 64 920 W

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2.2 What is the power that is delivered by the lift motor if the lift is designed to accelerate upwards at 1 m·s

-2

?

T F net = ma – f – w = ma T = w + ma + f

= mg + ma + f = (1 800)(9,8) + (1 800)(1) + 4 000

= 23 440 N P = Fv = Tv

= (23 340)(3)

= 64 920 W

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Problem 1

A 2 kg block, which is initially at rest, is moved by a constant force of 20 N over a distance of 2 m. Assume all the surfaces is frictionless unless indicated differently.

Use energy principles to answer the following questions: 1.1 Calculate the speed of the block of each of the points A to E.

When the block reaches the bottom of the slope at F the speed is 9,5 Calculate the following (when the block moves down the slope): m·s -1 .

1.2 The work done on the block by the gravitational force.

1.3 The work done on the block by the frictional force.

1.4 The size of the frictional force which is exerted on the block.

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Problem 2

Calculate the power needed for the 1 400 kg car under the following conditions: 2.1 The car moves up a hill with a constant speed of 80 km·h -1 ?

2.2 The car accelerates on a smooth parallel surface from 90 110 km·h in 6 s to pass another car.

km·h -1 to

Assume that there are in both cases a frictional force of 700 N exerted on the car.

TIPS:

2.1 To move a car at a constant speed up a hill, the motor of the car must exert a force equal to the sum of the frictional force and the component of the gravitational force parallel to the hill.

2.2 Because the car is accelerating the motor of the car must exert a force that must overcome the force of friction of 700 N plus the force necessary to accelerate the motor.

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2. DOPPLER EFFECT

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GENERAL DOPPLER EQUATION

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GENERAL DOPPLER EQUATION

We use the equation in the following manner:

•

+v L when the listener is approaching the source (because f L needs to be increased)

•

-v L when the listener is moving away from the source (because f L needs to be decreased)

•

-v s when the source is moving towards a listener (because f L needs to be increased)

•

+v s when the source is moving away from a listener (because f L needs to be decreased)

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November 2009 Exams Question 7

A fire truck, with its siren on, is moving at 20 m ·s -1 towards a burning building. A person standing next to the road with a detector, measures the frequency of the sound emitted by the siren to be 450 Hz. The measured frequency is HIGHER than the frequency of the sound emitted by the siren.

7.1 Is the fire truck moving towards or away from the person? 7.2 Explain why the registered frequency is higher.

7.3 Calculate the frequency of the siren if the speed of sound in air is 340 m∙s -1 .

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3. ORGANIC CHEMISTRY

3.1 IUPAC name, structural formulae, terminology 3.2 Physical properties 3.3 Reactions

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3.1.1 Structural formulae

The structural formula of a compound shows which atoms are attached to which within the molecule.

Atoms are represented by chemical symbols and lines are used to represent the bonds that hold the atoms together. A structural formula does not represent the geometry of the molecule.

Example:

Structural formula of ethanol (the O-H can be straight)

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3.1.2 Condensed structural formula

This notation represents the way that the atoms are bonded in a molecule but does not indicate all the bonds.

Example: Condensed structural formula for ethanol CH 3 CH 2 OH or CH 3 – CH 2 –OH or

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3.1.3 Molecular formula

A chemical formula that indicates the types of atoms and the correct number of atoms in a molecule.

Example:

Molecular formula for ethanol C 2 H 6 O

IUPAC names

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3.1.4 November 2009 Question 4

Both esters and amides are considered derivatives of carboxylic acids and can be prepared by using carboxylic acids as one of the reactants. Esters are known for their pleasant smells. Amides are the building blocks of proteins.

4.1

4.2

Write down the structural formula for the functional group of a primary amide.

An ester with six carbon atoms is prepared using propanoic acid as one of the reactants.

4.2.1

Use structural formulae to write a balanced equation for the preparation of this ester.

4.2.2 Write down the IUPAC name of this ester.

4.2.3 Write down the name of the catalyst needed for this preparation.

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3.2 PHYSICAL PROPERTIES

What is required?

Recognize and apply specific examples on, the relationship between physical properties (meting point, boiling points, vapour pressure, viscosities) and: 3.2.1 Intermolecular forces (hydrogen bonds and van der Waal’s forces) 3.2.2 Type and number of functional groups 3.2.3 Chain length 3.2.4 Branched chains (as on page 16 of the 2009 exam guidelines) 36

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3.3 Intermolecular forces

1. Melting and boiling points

• The stronger the intermolecular forces, the higher the boiling and melting points.

• More energy is needed to break the bonds.

Example: pentane Melting point = -130 °C; boiling point = 36°C pentan-1-ol melting point = -90 °C; boiling point = 118 °C The hydrogen bonds between pentan-1-ol molecules are stronger than the intermolecular forces between pentane molecules.

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2. Vapour pressure

The weaker the intermolecular forces the higher the vapour pressure.

Example:

Methylmethanoate has a higher vapour pressure than ethanoic acid .

• The intermolecular forces (hydrogen bonds ) between ethanoic acid molecules are stronger than the intermolecular forces (van der Waal’s forces) between methylmethanoate.

• More energy is needed to break the bonds between the molecules of ethanoic acid 38

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3. Viscosity

•

Low viscosity

There are weak intermolecular forces between molecules •

High viscosity

There are strong intermolecular forces between molecules

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4. Number and type of functional groups 4.1 Melting and boiling points

• The more of the same functional groups present, the stronger the intermolecular forces.

• The stronger the intermolecular forces the higher the melting and boiling points.

• Different functional groups have different strengths of intermolecular forces.

• The boiling and melting points depend on the intermolecular forces.

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2. Vapour pressure

Type of functional groups:

• Hexane has a higher vapour pressure than hexan-1-ol because the intermolecular forces (van der Waal’s forces) between the hexane molecules are weaker than the intermolecular forces (hydrogen bonds) between the hexan-1-ol molecules.

Number of functional groups:

• Vapour pressure between propan-1-ol is higher than that of 1,2,3-propantriol because there are 3 –OH groups that form stronger intermolecular forces than one -OH group.

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3. Viscosity

• The more functional groups present the stronger the intermolecular forces become and the higher the viscosity.

• Functional groups with strong intermolecular forces between molecules have a higher viscosity than functional groups with weak intermolecular forces between the molecules.

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Chain length 1. Melting and boiling point

• The longer the chain of a molecule the bigger the contact surface will be and then the strength of the intermolecular forces increases.

• The stronger the intermolecular forces , the higher the boiling and melting points.

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2. Vapour pressure

• The longer the chain length of a molecule the bigger the contact surface and therefore the stronger the intermolecular forces .

• The stronger the intermolecular forces the lower the vapour pressure.

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3. Viscosity

• Viscosity increases with an increase in chain length.

• Longer C-chain has a bigger contact surface and then the strength of the intermolecular forces increase.

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Branched chains 1. Melting and boiling points

• Branched chains are more compact (more spherical) with smaller contact area and thus weaker intermolecular forces.

• More branches mean weaker intermolecular forces and the boiling and melting points decrease.

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2. Vapour pressure

• More branches mean weaker intermolecular forces and higher vapour pressure.

3. Viscosity

• Viscosity decreases with more branches.

• The more branches present the weaker the intermolecular forces become.

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Intermolecular forces

• Carboxylic acids One polar side

= O

One hydrogen bond

-OH

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Intermolecular forces

• Alcohol One hydrogen bond

–OH

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Intermolecular forces

• Aldehyde

Weak intermolecular forces (van der Waal’s forces – polar molecule that can bond to other molecules that can form hydrogen bonds) 50

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Intermolecular forces

The strength of the intermolecular forces for different homologous series can be arranged as follows:

amide > carboxylic acid > alcohol> aldehide / ketene > amine > ester > hydrocarbons.

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November 2009 Question 3.2

Which ONE of these compounds has the highest vapour pressure at room temperature?

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16 31 32 Langer kettings – Van der Waalskragte werk oor groter oppervlak

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16 31 32

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Give IUPAC name and explain the difference in boiling points

b.p. 78 °C b.p. 51 °C 1-chlorobutane 2-chloro-2-methylpropane The BRANCHED COMPOUND has a smaller contact surface and therefore weaker intermolecular forces than a straight chain.

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November 2009 Question 5

Two experiments were performed to determine the

boiling points of compounds from three different homologous series

under the same conditions.

Experiment I II Organic compound

A CH B CH C CH D CH E CH F CH 3 3 3 3 3 3 COOH CH CH (CH (CH (CH 2 2 2 2 2 CH CHO ) ) ) 2 3 3 2 OH COOH CH 2 CHO OH

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Molar mass (g·mol -1 )

60,5 60,1 58,1 88,1 88,1 86,1

Boiling point ( ° C)

118 97 48 163 137 103 60

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November 2009 Question 5

5.1 Name the homologous series to which each of the following pairs of compounds belong: 5.2

Write down the IUPAC name for: 5.2.1 C propanal 5.2.2 E pentan-1-ol

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November 2009 Question 5

5.3 Formulate an investigative question for this practical investigation.

5.4 Which other variable, apart from the conditions for determining boiling points, was kept constant?

5.5 What conclusion can be drawn from the results in Experiment II?

5.6

Refer to intermolecular forces to explain the trend in boiling points, as shown in the table.

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ORGANIC REACTIONS

Addition:

Adding of atoms to an alkene (double bond breaks)

Substitution:

Substitution of one atom in a molecule with another atom (alkane, halogen, alcohol)

Elimination :

One or more atoms are removed (haloalkane or alcohol) from a molecule (Double bond forms)

Oxidation:

Reaction of alkanes with oxygen.

Esterification:

Reaction of an alcohol and carboxylic acid to form an ester.

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ORGANIC REACTIONS

Addition:

During addition of HX to unsaturated hydrocarbons, the H atom attaches to the C atom already having the greater number of H atoms. The X atom attaches to the substituted C atom. (Exam guideline 2009 page 16) more

Elimination:

The major product is the one where the H atom is removed from the C atom with the least number of H atoms. (Exam guideline 2009 page 17) 64

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November 2009 Question 6

Propene HBr Compound X H 2 O Substitution Secondary alcohol

6.1

6.2

Give a reason why propene unsaturated organic compound.

is classified as an Use structural formulae to write a balanced equation for the formation of compound X.

6.3

Name the type of reaction that takes place when propene is converted to compound X.

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November 2009 Eksamen Question 6

Propene Compound X H 2 O Substitution Secondary alcohol

6.4

Write down the structural formula and IUPAC name for the secondary alcohol that is formed.

6.5

Name the type of substitution reaction that takes place when compound X is converted to the secondary alcohol.

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November 2009 Question 6

Propene Compound X H 2 O Substitution Secondary alcohol

6.6

With the aid of a catalyst, propene can be converted directly to the secondary alcohol, without the formation of the intermediate compound X.

6.6.1 Besides propene, write down the NAME of the reactant needed for this direct conversion.

6.6.2 Write down the FORMULA of a catalyst that can be used.

6.6.3 Name the type of reaction that will take place during this direct conversion.

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November 2009 Question 6

Propene

6.7

Compound X H 2 O Substitution Secondary alcohol

Instead of adding water to compound X, concentrated sodium hydroxide is added and the mixture is heated.

6.7.1 Write down the IUPAC name of the organic product that is formed.

6.7.2 Name the type of reaction that takes place.

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4. HINTS regarding marking guidelines 1. Practical investigations

1.1

Identify variables

Q 7.3 & 5.4 Nov 2009: learners could not identify or explain the concept of controlled variable (that which has to be kept constant in order for the test to be fair)

.

Also Q 7.3.4 Febr / March 2010.

Independent variable:

The one that changes under your control.

Dependent variable:

The variable that changes as a result of the change of the independent change 69

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4. HINTS regarding marking guidelines

(Continued)

1. Practical investigations 1.2 Investigative question

The following marking rubric is used in the NSC exam:

Criteria for investigative question

Refers to

relationship

between

dependent

and

independent variables

Is stated as a

question

, not an aim

Mark √√

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4. HINTS regarding marking guidelines

(Continued)

1.3 Hypothesis

The following marking rubric is used in the NSC exam.

Criteria for hypothesis

Refers to

relationship

between

dependent

and

independent variables

.

Statement

that can be proved correct or incorrect –

prediction based on (prior) knowledge.

Mark √√

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4. HINTS regarding marking guidelines

(Continued)

1.4 Relationships

• Proportional to • Directly proportional to • Inversely proportional to (NOT: indirectly!!) • Exponential relationship • Linear relationship 72

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4. HINTS regarding marking guidelines

(Continued)

1.5 Graphs

Questions regarding conclusions from graphs: always refer to the

gradient

of the graph.

In Physics the following checklist below was used for the drawing of a graph in Q 14 March 2010.

Checklist

Criteria for graph Relevant heading Axes labelled correctly with units Appropriate scale Plotting all points Line of best fit

Marks √ √ √ √ √

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4. HINTS regarding marking guidelines

(Continued)

1.6 Conclusion from results

This is the answer to the

relationship

investigative question between the variables refer to the

1.7 Explain trends

e.g. refer to intermolecular forces as in Q 5.6 Nov 2009 5.6 Refer to the intermolecular forces to explain the trend in boiling points, as shown in the table.

Be very specific! Do not give a general rule/statement – use the compounds referred to in the question and apply the facts/principles to the specific compound (e.g. carboxylic acids....) 74

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4. HINTS regarding marking guidelines

(Continued)

2. Calculations

2.1

No marks will be awarded if an incorrect or inappropriate used, even though there may substitutions formula is be relevant symbols and applicable 2.2

If

no

formula is given, but all substitutions are correct, a candidate will forfeit one mark.

2.3

No marks will be awarded if no formula is given, but correct substitutions OMITTING ZERO SUBSTITUTIONS, are given.

2.4

2.5

Marks are only awarded for a formula if a calculation had been

attempted,

i.e. substitutions have been made or a numerical answer given Start all calculations with a formula from the data sheet – NO abbreviations (E OA or E RA is not accepted!) use: E cell = E oxidising agent - E reducing agent

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4. HINTS regarding marking guidelines

2.6

(Continued) K c calculations:

always write the

K c expression

! Use the substances in the chemical equation and

NOT

general like 2 [AB] 3 [produkte] [reaktante] or something The expression must relate to the reaction, [NH ] 3 2 eg: K c = [N ][H ] 2 2 3 If learners use the table to solve K c calculations, headings of rows and columns must be clearly indicated – many learners only put values into the table without any indication whether it is mole, concentration, etc. Do not teach the table as only method for solving K c problems! Only stating initial or equilibrium without the mole on concentration will result in no marks

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4. HINTS regarding marking guidelines

(Continued)

2. Calculations

2.7

Stoichiometric calculations

Marks are only given for the calculation of molar mass (M) if

used

in the formula: n = m M Learners loose marks because they do not show / use the mol ratio. Learners struggle with stoichiometric calculations – more emphasis should be put on teaching mole calculations in the right context.

2.8 All calculations, when not specified in the question, must be done to two decimal places.

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4. HINTS regarding marking guidelines

(Continued)

3. Units

A lot of marks are forfeited because answers are written without units or for using the incorrect units. Even when the question asks for values from graphs / tables, the answers need to be written with the correct units.

4. Definitions

An unnecessary number of marks are forfeited for wrong definitions (learning work!!). Definitions obtain two marks or zero. (Electrolyte, electrolysis, reaction rate, unsaturated hydrocarbon, homogeneous, etc). Definitions are awarded two marks or nil.

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4. HINTS regarding marking guidelines

(Continued)

5. General

5.1

If one answer or calculation is required, but two given, only the first one will be marked, irrespective of which one is correct.

5.2 C andidates will be penalised when a NAME is asked and a FORMULA given or visa versa.

5.3

Many terminology mistakes – chloride vs chlorine, substance, atom, molecule, compound, etc. used in the wrong context .

5.4

Learners talk about forming ions instead of oxidised.

5.5

Learners do not know the standard conditions of galvanic cells, 273 K vs 298 K. They give 1 atm as an answer when no gases are present.

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4. HINTS regarding marking guidelines

(Continued)

5. General

Organic chemistry

IUPAC naming, structural formulae and the marking rules for organic chemistry will be addressed in the power point presentation.

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Acknowledgements

• Janetta Nel for documents with regard to Organic Chemistry • Dr. M. Blackie for IUPEC PowerPoint presentations • Department of Basic Education for the use of question papers and other materials from www.thutong.doe.gov.za

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CONTACT DETAILS

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Tel Fax Email 012 348 9641 086 687 9996 [email protected]

[email protected]

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