PHYSICAL SCIENCES

GRADE 11

SUBJECT: PHYSICAL SCIENCES

(Page 1 of 57 )

© Gauteng Department of Education

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WAVES, LIGHT & SOUND MATTER AND MATERIALS

CHEMICAL CHANGE

Reflection

Refraction (concepts/)

Refractive index

Optical density

Ray diagrams

Snell’s law

Calculations

Critical angles

Total internal reflection

Prescribed exp

Write-up

Diffraction

Huygens’s principle

Informal test, week 11- 14

Motion of particles

Kinetic theory of gases

Boyle’s;/Charles’s/Gay-Lussac‘s law

Calculations

Ideal gas law

Calculations

Temperature& heat, pressure

Informal test, week 15 - 16

Molar gas volume

Conc. of soln

Stoichiometric calc.

Percentage yield (purity or % composition)

Empirical formula

Molecular formula

Volume relationships in gaseous reactions

Informal test, week 17 - 18 Midyear examination

Geometrical optics

(This section must be read in conjunction with the CAPS, p. 76–77.)

Refraction

Know that the speed of light is being constant when passing through a given medium and having a maximum value of c = 3 x 108 m·s-1 in a vacuum.

Define refraction of light as the change in direction of a light ray due to a change in speed when light travels from one medium into the other of different optical density.

Define optical density as a measure of the refracting power of a medium. The higher the optical density, the more the light will be refracted or slowed down as it moves through the medium.

Define the refractive index (n) of a material as the ratio of the speed of light in vacuum (c) to the speed of light in a material (v).

Solve problems using v

c = n .

Relate optical density to the refractive index of the material. Materials with a high refractive index will also have a high optical density.

Define the terms normal, angle of incidence and angle of refraction, and identify them on a ray diagram.

Normal: The line, which is perpendicular to the plane of the surface. Angle of incidence: The angle between the normal to a surface and the incident light ray.

Angle of refraction: The angle between the normal to a surface and the refracted light ray.

Draw ray diagrams to show the path of a light ray moving from one medium into another.

State Snell's Law: The ratio of the sine of the angle of incidence in one medium to the sine of the angle of refraction in the other medium is constant.

1

2

2

1

n

n

sin

sin

OR n1sinθ1 = n2sinθ2 OR

i

r

r

i

n

n

sin

sin

OR nisinθi = nrsinθr

Solve problems using the above equations for Snell's law.

Determine the refractive index of an optical medium by using 2

1

sin

sin n

for a ray of light

traveling from vacuum (or air) to the medium. Critical angles and total internal reflection

State the law of reflection: When light is reflected the angle of incidence is always equal to the angle of reflection.

Angle of incidence: The angle between the normal to a reflecting surface and the incident light ray. Angle of reflection: The angle between the normal to a reflecting surface and the reflected light ray.

Define the critical angle as the angle of incidence in the optically denser medium for which the angle of refraction in the optically less dense medium is 90°.

List conditions required for total internal reflection, i.e. when the refracted ray does not emerge from the medium, but is reflected back into the medium: o Light must travel from an optically denser medium (higher refractive index) to an

optically less dense medium (lower refractive index). o The angle of incidence in the optically denser medium must be greater than the

critical angle.

Use Snell's law to calculate the critical angle at the surface between two optically different media.

Explain the use of optical fibres in endoscopes and telecommunications.

2D and 3D wave fronts

(This section must be read in conjunction with the CAPS, p. 78.)

Diffraction

Define a wave front as an imaginary line joining points on a wave that are in phase.

State Huygens' principle: Every point of a wave front serves as a point source of spherical, secondary waves that move forward with the same speed as the wave.

Define diffraction as the ability of a wave to spread out in wave fronts as the wave passes through a small aperture or around a sharp edge.

Sketch the diffraction pattern for a single slit.

Relate the degree of diffraction to the wavelength (λ) and width of slit (w):

w ndiffractio of reedeg

State that diffraction demonstrates the wave nature of light.

Electrostatics (This section must be read in conjunction with the CAPS, p. 84–85.)

Coulomb's law

State Coulomb's law: The magnitude of the electrostatic force exerted by two point charges (Q1 and Q2) on each other is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them.

Solve problems using the equation F = 2

21

r

QkQ for charges in one dimension (1D) – restrict

to three charges.

Solve problems using the equation F = 2

21

r

QkQ for charges in two dimensions (2D) – for

three charges in a right-angled formation (limit to charges at the 'vertices of a right- angled triangle').

Electric field

Describe an electric field as a region in space in which an electric charge experiences a force. The direction of the electric field at a point is the direction that a positive test charge would move if placed at that point.

Draw electric field patterns for the following configurations: o A single point charge o Two point charges (one negative, one positive OR both positive OR both negative) o A charged sphere

NOTE: Restrict to situations in which the charges are identical in magnitude.

Define the electric field strength at a point: The electric field strength at a point is the electrostatic force experienced per unit positive charge placed at that point.

In symbols: Q

FE

Solve problems using the equation Q

FE .

Calculate the electric field strength at a point due to a number of point charges, using the

equation 2r

kQE to determine the contribution to the field due to each charge. Restrict to

three charges in a straight line.

Electric circuits (This section must be read in conjunction with the CAPS, p. 88–89.)

Ohm's law

State Ohm's law in words: The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature.

Interpret data/graphs on the relationship between current, potential difference and resistance at constant temperature.

State the difference between ohmic and non-ohmic conductors and give an example of each.

Solve problems using I

VR for circuits containing resistors that are connected in series

and/or in parallel (maximum four resistors).

Ideal gases and thermal properties (This section must be read in conjunction with the CAPS, p. 79–81.)

Motion of particles; Kinetic theory of gases

Describe the motion of individual gas molecules: o Molecules are in constant motion and collide with each other and the walls of

the container. o There are forces of attraction between molecules. o Molecules in a gas move at different speeds.

Describe an ideal gas as a gas: o That has identical particles of zero volume o With no intermolecular forces between particles o In which all collisions of the molecules with themselves or the walls of the

container, are perfectly elastic

Explain that real gases deviate from ideal gas behaviour at high pressures and low temperatures.

State the conditions under which a real gas approaches ideal gas behaviour. Ideal gas law

State Boyle's law: The pressure of an enclosed gas is inversely proportional to the volume it occupies at constant temperature.

In symbols: V

1p , therefore p1V1 = p2V2, T = constant

State Charles' law: The volume of an enclosed gas is directly proportional to its kelvin temperature provided the pressure is kept constant.

In symbols: TV , therefore 2

2

1

1

T

V

T

V , p = constant

State that the pressure of a gas is directly proportional to its temperature in kelvin at

constant volume (Gay Lussac). In symbols: Tp , therefore 2

2

1

1

T

p

T

p , V = constant

For each of the above three relationships: o Interpret a table of results or a graph o Draw a graph from given results o Solve problems using a relevant equation o Use kinetic theory to explain the gas laws

Use the general gas equation, 2

22

1

11

T

Vp

T

Vp , to solve problems.

Use the ideal gas equation, pV = nRT, to solve problems.

Convert temperatures in Celsius to kelvin for use in ideal and general gas equations. Temperature and heating, pressure

Explain the temperature of a gas in terms of the average kinetic energy of the molecules of the gas.

Explain the pressure exerted by a gas in terms of the collision of the molecules with the walls of the container.

Representing Chemical Change (This section must be read in conjunction with the CAPS, p. 37.)

Balanced chemical equations

Write and balance chemical equations. Use formulae for reactants and products and indicate the phases in brackets, i.e. (s), (ℓ), (g) and (aq).

Interpret balanced reaction equations in terms of: o Conservation of atoms o Conservation of mass (use relative atomic masses)

Quantitative aspects of chemical change (This section must be read in conjunction with the CAPS, p. 50–52 and 82–83.)

The mole concept

Describe the mole as the SI unit for amount of substance.

Define one mole as the amount of substance having the same number of particles as there are atoms in 12 g carbon-12.

Describe Avogadro's number, NA, as the number of particles (atoms, molecules, formula-units) present in one mole (NA = 6,023 x 1023 particles∙mol-1).

Define molar mass as the mass of one mole of a substance measured in g·mol-1.

Calculate the molar mass of a substance given its formula. Molar volume of gases

State Avogadro's Law, i.e. one mole of any gas occupies the same volume at the same temperature and pressure.

At STP: 1 mole of any gas occupies 22,4 dm3 at 0 °C (273 K) and 1 atmosphere (101,3 kPa). Thus the molar gas volume, VM, at STP = 22,4 dm3∙mol-1.

Volume relationships in gaseous reactions

Interpret balanced equations in terms of volume relationships for gases, i.e. under the same conditions of temperature and pressure, equal amounts (in mole) of all gases occupy the same volume.

Concentration of solutions

Define concentration as the amount of solute per litre of solution.

Calculate concentration in mol·ℓ-1 (or mol·dm-3) using V

nc .

More complex stoichiometric calculations

Determine the empirical formula and molecular formula of compounds.

Determine the percentage yield of a chemical reaction.

Determine percentage purity or percentage composition, e.g. the percentage CaCO3 in an impure sample of seashells.

Perform stoichiometric calculations based on balanced equations that may include limiting reagents.

DIFFRATION

The following aspects should be taken care of: Define:

A wavefront (an imaginary line that connects waves that are in phase (e.g. all at the crest of their cycle))

State Huygen’s principle

Diffraction (the ability of a wave to spread out in wave fronts as they pass through a small aperture or around a sharp edge)

Apply Huygen’s principle to explain diffraction qualitatively:

Describe light and dark areas in terms of constructive and destructive interference of secondary wavelets

Sketch the diffraction pattern for a single slit Understand that:

Degree of diffraction ∝ w

, where w = slit width

Diffraction of light demonstrates the wave nature of light

MULTIPLE CHOICE QUESTIONS

Four options shall always be given as possible answers.

TECHNIQUE IN THE ANSWERING OF MULTIPLE CHOICE QUESTIONS

Step 1: Read the question carefully.

Step 2: Underline the KEY words in the question.

Step 3: Pay attention to words that are CAPITALIZED, or words in ITALICS.

Step 4. Decide whether you are required to recall or use a fact, phenomenon, definition,

unit, formula.

Step 5. First, delete the answers that are obviously incorrect (Called Distractors)

Step 6. Finally select the correct answer from the others that remain. This is called

ELIMINATION and is particularly helpful when the answers or options are very close to

each other.

MULTIPLE CHOICE QUESTIONS

1.1 Diffraction through a narrow slit is less for the blue light than for red light because A. blue light travels faster than red light. B. the refractive index of air is bigger for red light than for blue light. C. red light has a shorter wavelength than blue light. D. blue light has a higher frequency than red light.

1.2 An example of diffraction is, light …

A. Bending as it enters a new medium. B. Entering a prism and emerging as a rainbow. C. Coming through a keyhole and spreading out in a dark

room. D. Scattering in the atmosphere.

1.3 Diffraction takes place when a light wave travels through a single slit. Which one of the following combinations of the slit width and the wavelength of the light will produce the maximum diffraction?

Slit width Wavelength of light

A wide long

B narrow long

C wide short

D narrow short

QUESTION 1

The wavelength of sound waves are much longer than light waves. Hence sound waves

are able to diffract around obstacles such as walls. The wavelength of light waves are

very short and do not diffract as much.

1.1 What is diffraction? What does diffraction proves? (3)

1.2 Which situation will show greater diffraction and why?

1.2.1 A wave of wavelength 3 cm or a wave of wavelength 5 cm going through the same

narrow gap. (3)

1.2.2 The same wave passing through a gap of 5 cm or a gap of 7 cm. (3)

1.3 Name and state the principle that predicts the behaviour of a wave behind the space

of a barrier or slit. (4)

[13]

QUESTION 2 A helium-neon laser emits red light that passes through a single slit. A diffraction pattern is observed on a screen some distance away from the slit.

2.1 2.2 2.3

Define the term diffraction. If the wavelength of red light is 644,4 nm and the slit width is 3 437 nm, calculate the angle at which the third minimum occurs. Briefly describe the diffraction pattern that will be observed on the screen.

(2)

(3)

(2)

The single slit is replaced with a double slit.

2.4 Name ONE similarity and ONE difference in the pattern observed when the single slit is replaced with a double slit.

(2)

2.5 Will this pattern be observed if the laser is replaced with a light bulb? Give a reason for your answer.

(2)

QUESTION 3

The diagram below shows wavefronts of red light incident on, and emerging from, a double slit arrangement. Each of the slits acts as a coherent light source.

3.1 In which region of the electromagnetic spectrum would you find red light? (1) 3.2 If a screen were placed at X perpendicular to the line OX, describe what would be seen

on the screen. (2) 3.3 If blue light were used instead of red light, how would the pattern differ? (1)

O

The wavefronts represent

successive crests ( )

and troughs ( ) of the

wave. The line OX joins points

of constructive. interference.

QUESTION 4 Learners use monochromatic blue light to investigate the difference between an interference pattern and a diffraction pattern.

4.1 Apart from the blue light and a screen, write down the name of ONE item that

the learners will need to obtain an interference pattern.

(1) 4.2 Briefly describe the interference pattern that will be observed on the screen. (2)

4.3 In one of their experiments they place the screen at a distance of 1,4 m from a

single slit and observe a pattern on the screen. The width of the central bright band is measured as 22 cm.

Calculate the:

4.3.1 Angle θ at which the first minimum will be observed on the screen (3)

4.3.2 The width of the slit used if the wavelength of the blue light is 470 nm

(5) 4.4 The width of the central band INCREASES when the blue light is replaced with

monochromatic red light. Explain this observation.

(2) [13]

θ

1,4 m

Monochromatic

blue light

22 c

m

Screen

QUESTION 5 Red light from two stationary narrow slits, S1 and S2, reaches a large white screen PON, indicated in the diagram below.

A dark band is observed at point P on the screen. The brightest band is observed at point O on the screen. Bands are arranged such that the band at point N on the screen is dark.

5.1 State Huygens' principle in words. (2)

5.2 Write down the type of interference that occurs at point O. Write down only

DESTRUCTIVE or CONSTRUCTIVE. Briefly explain your answer.

(3) 5.3 Describe the change in brightness, if any, of the light bands formed on the

screen as you walk closer to the screen from point M to point O. Briefly explain your answer.

(3)

The red light is now replaced with a green light.

5.4

How will the new pattern differ from the previous one? (2) [10]

P

O

N

M

S1

S2

S

C

R

E

E

CRITICAL ANGLE Learners should be able to: Explain the concept of critical angle List the conditions required for total internal reflection Calculate the critical angle at the surface between a given pair of media -use

Snell’s Law Explain the use of optical fibers in endoscopes and telecommunications

MULTIPLE CHOICE QUESTIONS

Four options shall always be given as possible answers.

TECHNIQUE IN THE ANSWERING OF MULTIPLE CHOICE QUESTIONS

Step 1: Read the question carefully.

Step 2: Underline the KEY words in the question.

Step 3: Pay attention to words that are CAPITALIZED, or words in ITALICS.

Step 4. Decide whether you are required to recall or use a fact, phenomenon, definition,

unit, formula.

Step 5. First, delete the answers that are obviously incorrect (Called Distractors)

Step 6. Finally select the correct answer from the others that remain. This is called

ELIMINATION and is particularly helpful when the answers or options are very close to

each other.

MULTIPLE CHOICE QUESTIONS

1.1. A monochromatic ray of light moves from medium P to Q as shown in the diagram.

Medium Q is less dense than medium P.

Which relationship is the correct relationship between Ø𝑖 and Ø𝑟 ( Ø𝑖 is the angle of incidence

and Ø𝑟is the angle of refraction).

Angle of refraction 𝒔𝒊𝒏 Ø𝒊

𝒔𝒊𝒏Ø𝒓

A B C D

Ø𝑟=∅𝑖

Ø𝑟=∅𝑖

Ø𝑟> ∅𝑖

Ø𝑟<∅𝑖

Constant Not constant Constant Not constant

1.2 A light ray from air strikes the boundary between air and water

perpendicularly. The ray

A is bent towards the normal.

B is bent away from the normal.

C reflected.

D does not change direction.

Ø𝑖

Q

P

1.3 The critical angle between air and water is 490. Total internal reflection

only occurs when a light ray moves from water to air and the angle of

incidence is:

A 900

B 550

C 490

D 350

1.4 Light enters a block of glass as shown. Which one of the following diagrams is the correct one for the ray of

refraction of the light in and through the glass block? A B C D

1.5 Perspex and glass has a refractive index of 1,49 and 1,52 respectively. Which one of the following will be the correct combination for the speed of the light and the deviation relative to the normal of the incident ray if light passes from glass to perspex?

Speed of light ray Deviation of incident light ray relative to

normal

A increase away from normal

B decrease away from normal

C increase towards normal

D decrease towards normal

QUESTION 1

A light ray travelling through air goes through a glass window as indicated in the sketch below. The refractive index of air is 1 and of the glass window is 1,5.

Air Glass Air

X

Y

Z

K

1.1 Use Snell’s Law to calculate the size of angle X if angle K is 300.

(7)

1.2 Write down the relationship between the incident ray and the

emergent ray?

(2)

1.3 Is it possible that total reflection be attained by changing the size of

angle X? Explain your answer.

(3)

[12]

Question 2 2.1.

2.1.1. What conditions must be met before total internal reflection can occur? (2)

2.1.2. Determine the magnitude of the angle of incidence at the first boundary (2)

2.1.3. Determine the angle of refraction at the first boundary (2)

2.1.4. Determine the critical angle for the substance used to make the prism. (4)

2.2. Two learners designed a practical investigation to determine the relationship between

the angle of refraction and the angle of incidence. They set up the investigation using

a semi-circular glass block. The learners measure the angle of incidence and the

angle of refraction. They draw up the table with their results.

Angle of Incidence Angle of Refraction

10 20 30 40 ?

15.10 30.56 48.59 74.62 90.00

2.2.1 Draw a graph of the angle of refraction Vs angle of incidence. (4)

2.2.2 Identify the dependent and independent variables in this investigation. (2)

2.2.3 From the graph give the relationship between angle of refraction and angle of

incidence. (2)

1320

260

Glass block

QUESTION 3

A ray of light strikes a glass-air interface, as shown in the diagram below. The refractive

index of the glass is 1.44. The ray then travels from the glass block into air.

Air

30°

3.1 Calculate the angle of refraction at which the light ray emerges from the

glass block. Take the refractive index of air as 1,00. (3)

3.2 Give a reason why the answer to QUESTION 7.1 differs from the angle of

incidence on the glass-air interface. (1)

3.3 How is the speed of the light ray affected as it moves from the glass block

into air? Write down only INCREASES, DECREASES or NO EFFECT. (1)

QUESTION 4

A glass block is covered with a layer of oil. The refractive index of oil is 1,47 and that of

glass 1,52. A light ray travels obliquely from the oil to the glass.

4.1 Does the refracted ray bend towards or away from the normal? Explain your

answer. (3)

4.2 Draw the ray diagram to show the refraction from oil to glass and from the glass

to the air. (3)

4.3 The angle of incidence from the oil to the glass is 20. Calculate the angle of

refraction. (5)

4.4 The angle of incidence is increased. What will happen to the angle of refraction?

Explain your answer. (3)

Glass

Oil

Air

QUESTION 5 Light travelling in air falls on a diamond at an unknown angle of incidence. The refractive index for diamond is 2,42. 5.1 Define “refractive index”. (2) 5.2 State Snell’s Law in your own words. (3) 5.3 Complete the refraction for a light ray incident on a diamond until it

passes through into the air again. Label all important angles. (4)

5.4 Calculate the angle of incidence if the angle of refraction is 11. (4)

CHEMICAL BONDING

LEARNING OBJECTIVES

The following definitions and explanations must be known.

Draw Lewis diagrams for a given chemical formulae and using electron configuration

to represent atoms

to deduce number of valence electrons in an atom of an element

for the hydrogen molecule

Simple molecules eg ( F2;H2O;NH3;HF OF2 HOCl)

Molecules with multiple bonds. Eg ( O2;N2& HCN)

Molecular shape as predicted using the Valence Shell Electron Pair Repulsion (VSEPR) theory.

Electronegativity of atoms- Tendency of atoms to attract electrons.

Chemical Bond---The net electrostatic force that exists between two or more atoms sharing elecrons:

Describe a covalent chemical bond (a shared pair of electrons) Recall the role of models in Science and describe the explanation of

chemical bonding in the course as an application of a model. Deduce the number of valence electrons in an atom of an element. Represent atoms using Lewis diagram. Explain, referring to diagrams showing electrostatic forces between

protons and electrons, and in terms of energy considerations, why:

two Hydrogen (H) atoms form and H2 molecule, but,

Helium (He) does not form He2. 1.Covalent bonding: happens when two atoms form the same element or from different element share one or more pairs of electrons. The result can be simple molecule structure or giant covalent structures.

2.Simple molecule structures are molecules that contain a small number of atoms that function together as a unit. They can consist of atoms of the same element (H2, O2, Cl2) or be a combination of different elements (H2O, H2SO4, NH3). Simple molecules generally have low melting and boiling points, and they do not conduct electricity and heat. Giant covalent substances exist as large,

three-dimensional structures held together by covalent bonds. The atoms in the structures are joined to each other in an extensive lattice. These structures are very strong, because the atoms are interlinked by strong covalent bonds. Diamond and graphite are allotropes of carbon, and quartz consists of silicon dioxide molecules bonded in a network structure. Giant covalent structures are strong, but brittle, with a very high melting and boiling point. Generally covalent substances do not conduct heat or electricity. The delocalised electrons in graphite make it an electrical conductor, and the structural of diamond allows it to be an excellent thermal conductor.

Ionic bonding: one or more electrons are

transferred from the less electronegative

atom (metal atom) to the more

electronegative atom (non-metal atom). Ions

(cations or anion) are formed, and pack

together in a crystal lattice. The electrostatic

forces between the ions are strong, and for this reason, ionic

compounds have high melting and boiling points. Both molten ionic

compounds and solutions of ionic compounds can conduct electricity.

Metallic bonding: comprises the sharing of a sea of delocalised

electrons by metal atoms that are packed in an orderly fashion in a

metallic lattice. The sea of delocalised electrons allows metals to be

good conductors of both heat and electricity.

The type of chemical bond in a compound determines the physical

and chemical prosperities of that compound.

In Grade 10 we look at three types of bond: covalent, ionic and

metallic.

A model is a representation containing the essential structure of

some object or event in the real world.

The valance electrons are the electrons in the highest occupied

energy levels of an atom.

Core electrons are the electrons in an atom that are not valence

electrons and therefore do not participate in bonding.

The number of electrons that an atom of an element gains, loses or

share to reach noble gas configuration is called its valency.

A Lewis dot symbol consists of the symbol of an element and one dot

for each valence electron in an atom of the element

- THE TEACHER MUST EXPLAIN HOW TO REPRESENT ATOMS

USING LEWIS DIAGRAM.

Atoms combine in order to achieve a more stable electron

configuration. And atoms have maximum stability when its electron

configuration is similar to that of a noble gas. When atoms react to

form a chemical bond, only the valance electrons are involved.

Gilbert Lewis invented a way to represent the valance electrons in an

atom. He suggested using a system of dots to show valence

electrons, called Lewis dot symbols. A Lewis dot symbol consists of

the symbol of an element and one dot for each valence electron in an

atom of the element.

CLASS ACTIVITY

Use the given Periodic table given below to draw Lewis diagram for

the following Compounds

CO2;NH3; SO2 H2O and NaCl

F2;H2O;NH3;HF OF2 HOCl

O2;N2& HCN

Learning Objectives

Learners must be able to:

Describe a covalent chemical bond as a shared pair of electrons.

Draw Lewis diagram for the hydrogen molecule.

Describe and apply simple rules to deduce bond formation, viz:

o different atoms, each with an unpaired valence electron can

share these electrons to form a chemical bond.

o different atoms with pared valence electrons called lone pairs

of electrons, cannot share these four electrons and cannot

form a chemical bond.

o different atoms, with unpaired valence electrons can share

these electrons and form a chemical bond for each electron

pair shared (multiple bond formation).

o atoms with an incomplete complement of electrons in their

valence shell can share a lone pair of electrons form another

atoms to form a co-ordinate covalent or dative covalent bond.

(e.g. NH4+, H3O+).

Baseline assessment

Define the term valance electrons.

The valance electrons are the electrons in the highest occupied

energy levels of an atom.

Define the term core electrons

Core electrons are the electrons in an atom that are not valence

electrons and therefore do not participate in bonding

Define the term Covalent Bond

A covalent bond is a form of chemical bond where pairs of

electrons are shared between atoms.

RULES FOR THE FORMATION OF COVALENT BONDS

Rule 1: If two atoms have an unpaired valance electron, they can

share this pair of electrons to form a chemical bond. For example,

two hydrogen atoms form an H2 molecule by sharing an electron pair.

Rule 2: If two atoms each have a pair of electrons, or lone pair, they

cannot share these four electrons and do not forms a chemical bond.

Example not bond form between to helium atoms.

Rule 3: If two atoms each have two or three unpaired electrons, they

can share these electrons and form a chemical boned for each

electron pair shared. The two atoms form multiple bonds between

them. If they share two pair of electrons, it is called a double bond,

for example in O2. If they three pairs of electrons, it is called a triple

bond, for example in N2.

Rule 4: If atoms have an empty valance shell, they can share a lone

pair of electrons from another atoms to form a coordinate of dative

covalent bond, for example in the ion NH4+, the lone pair of nitrogen

is shared with H+.

APPLICATION OF THE ABOVE RULES

Represent hydrogen chloride (HCl) using a Lewis diagram.

Step 1 For each atom, determine

the number of valence

electrons in the atom, and

represent these using

dots and crosses.

Step 2 Arrange the electrons so

that the outermost energy

level of each atom is full.

Hydrogen chloride

is represented below

Conclusion The dot and cross in

between the two atoms,

represent the pair of

electrons that are shared

in the covalent bond. We

can also show this bond

using a single line:

Rule 2: Atoms with lone pairs. (A lone pair is an unshared electron

pair. A lone pair stays on the atom that it belongs to.)

Represent water (H2O) using a Lewis diagram

Step 1: The electron configuration

of hydrogen is (1s1) and the

electron configuration for oxygen

is ([He]2s22p4). Each hydrogen

atom has 1 valence electron and

the oxygen atom has 6 valence

electrons.

Notice how in this example we wrote a 2 in

front of the hydrogen. Instead of writing the

Lewis diagram for hydrogen twice,

we simply write it once and use the 2 in

front of it to indicate that two hydrogen

are needed for each oxygen.

Step 2: Arrange the electrons so

that the outermost energy level of

each atom is full.

Conclusion: The water molecule is

represented below.

Rule 3: Atoms with multiple bonds:

Represent oxygen (O2) and hydrogen cyanide (HCN) using a Lewis

diagram:

Step 1: For each atom, determine the

number of valence electrons that the atom

has from its electron configuration.

The electron configuration of

oxygen is ([He]2s22p4). Oxygen has

6 valence electrons.

Step 2: Arrange the electrons in the O2

molecule so that the outermost energy

level in each atom is full.

The O2 molecule is represented below.

Notice the two electron pairs between the

two oxygen atoms (highlighted in blue).

Because these two covalent bonds are

between the same two atoms, this is a

double bond.

Conclusion: Each oxygen atom uses its

two unpaired electrons to form two bonds.

This forms a double covalent bond (which

is shown by a double line between the two

oxygen atoms).

Step 1 For each atom, determine

the number of valence electrons

that the atom has from its electron

configuration.

The electron configuration of hydrogen is

(1s1) , the electron configuration of

nitrogen is ([He]2s22p3) and for carbon is

([He]2s22p2). Hydrogen has 1 valence

electron, carbon has 4 valence

electrons and nitrogen has 5 valence

electrons.

Step 2: Arrange the electrons in the HCN

molecule so that the outermost energy

level in each atom is full.

The HCN molecule is represented below.

Notice the three electron pairs (highlighted

in red) between the nitrogen and carbon

atom. Because these three covalent bonds

are between the same two atoms, this

is a triple bond.

Conclusion: As we have just seen carbon

shares one electron with hydrogen and

three with nitrogen. Nitrogen keeps its

electron pair and shares its three unpaired

electrons with carbon.

Molecular shape as predicted using the Valence Shell Electron

Pair Repulsion (VSEPR) theory.

Valence Shell Electron Pair Repulsion (VSEPR) theory: It is a

model that is used to predicts the shape of molecules, based on the

tendency of electron – pairs to minimize the repulsion between them.

The following steps should be used when using the theory:

o Draw the Lewis structures of bonding electron of the

molecules

o Determine the number of bonding electrons pairs.

o Determine if there are lone pairs.

o Work out the shape of the molecule using the given tables

below.

Table 1: Central atom with no lone pairs on the central atom

Number of

bonding

electron

pairs

Geometry

Shape

General

formula

Example

1 linear

AX HCl

2 linear

AX2 CO2 &

BeF2

3 trigonal

AX3 BF3

4 tetrahydral

AX4 CH4

5 Trigonal

bipyramid

al

AX5 PCl5

6 Octahedra

l

AX6 SF6

o In this table A stands for central atom, X for terminal

atom(outer atoms that are bonded to the central atom)

o Note: Double bonds are treated the same way as single

bonds when determining the molecular shape

Table 2: Central atom with lone pairs on the central atom

Number of bonding

Electron pairs

Number of

lone pairs

Geometr

y

Shape Gener

al

formul

a

Exampl

e

2 2 Angular

AX2E2 SO2 &

H2O

3 1 Trigonal

planar

AX3E NH3

NOTES:

o In this table A stands for central atom,

o X for terminal atom(outer atoms that are bonded to the central atom)

o E stand for lone pairs

o Molecules with lone pairs do not have one of the ideal shapes.

o Lone pairs take up more space than bonding pairs.

o They also exert a repulsive force on the bonding pairs

Electronegativity (∆EN): (Refer to Periodic Table)

Gives the strength of the atom to attract electrons from other atoms react

with it.

If the electronegativity values of two bonded atoms are:

Similar i.e

∆EN< 𝟐. 𝟏 𝒕𝒉𝒆𝒏 𝒄𝒐𝒗𝒂𝒍𝒆𝒏𝒕 𝒃𝒐𝒏𝒅𝒔 𝒂𝒓𝒆 𝒇𝒐𝒓𝒎𝒆𝒅.

∆EN= 𝟎 𝒕𝒉𝒆𝒏 𝒏𝒐𝒏𝒑𝒐𝒍𝒂𝒓 𝒄𝒐𝒗𝒂𝒍𝒆𝒏𝒕 𝒃𝒐𝒏𝒅𝒔 𝒂𝒓𝒆 𝒇𝒐𝒓𝒎𝒆𝒅.

∆𝑬𝑵 < 𝟏 then weakly covalent bonds are formed.

𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕:

∆EN > 𝟐. 𝟏 THEN IONIC BONDS ARE FORMED

The strength of the intermolecular forces is summarised in the given table:

Type of intermolecular

force

Kinds of particles

involved

Example

Ion -dipole Ions and polar molecules NaCl + H2o

hydrogen bond polar molecules (WITH

HYDROGEN ATOM) and

polar molecules

H2o + CO

dipole-dipole polar molecules and polar

molecules

H2o + HCl

ion-induced dipole NON polar molecules

and ions

dipole-induced -dipole polar molecules and non-

polar

London forces Non-polar molecules

and Non- polar molecules

MULTIPLE CHOICE QUESTIONS

Dry ice (solid carbon dioxide), when heated, changes directly from the solid

phase to the gas phase. Which type of intermolecular forces would one

expect to be present between the carbon dioxide molecules to allow for this

to happen?

A.Covalent forces

B.Dipole-dipole forces

C. Hydrogen bonding forces

D. London forces

1.2 The molecular shape of a molecule with the formular AB2 is _______.

A. Linear or bent.

B. Linear or trigonal planar.

Linear or tetrahedral.

D. Linear or trigonal bipyramidal.

1.3 Which ONE of the following species contains a dative covalent bond?

A. NH3

B. CH4

C H3O+

D. NF3

1.4 Which ONE of the following compounds has dipole-dipole forces

between their molecules?

CO2

HCℓ

Cℓ2

CCℓ4

1.5 Consider the Lewis structure of a compound below:

Which ONE of the following is CORRECT?

Name of element

X

Name of element

Y

Molecular Shape

of the Compound

A Chlorine Oxygen Angular

B Oxygen Chlorine Linear

C Chlorine Sulphur Linear

D Sulphur Chlorine Angular

1.6 In which ONE of the following graphs does the dotted line CORRECTLY

represent the deviation of a real gas from ideal gas behaviour?

1.7 10 moles of hydrogen gas (H2) and 2,5 moles of nitrogen gas (N2) are

mixed and allowed to react to form ammonia (NH3) according to the

following balanced equation: 3H2(g) + N2(g) → 2NH3(g)

Moles of H2 (g) Moles of N2 (g)

A 0 0

B 7 1.5

C 4 0.5

D 4 2

1.8 Two gas syringes, X and Y, each contains the same gas at STP. The

volume

of syringe X is 10 cm3 and that of syringe Y is 20 cm3 as shown below.

Assume ideal gas behaviour.

Which ONE of the following statements is CORRECT?

A .The average kinetic energy of the molecules in X is less than that of the

molecules in Y.

B .The total kinetic energy of the molecules in X is less than that of the

molecules in Y.

C .The number of gas molecules in X is equal to the number of gas

molecules in Y.

D The product pV in X is equal to the product pV in Y.

QUESTION 2

Consider the substances listed below:

H2O C KCℓ CH4 CO2 HCℓ

Which substance or substances matches each of the statements below?

(Each substance may be used more than once in your answer or not at all).

2.1 Made up of polar molecules.

6.2 Is made up of non-polar molecules.

2.3 Has mainly hydrogen bonding between the particles in the solid and

liquid phases.

2.4 Will form a network solid.

2.5 Made up of an ionic crystal lattice.

2.6 Is a poor conductor of electricity in the solid phase, but conducts well in

aqueous solution.

QUESTION 3

The table below gives the boiling points and the molar masses of the

hydrides of some of the group 16 elements.

Formula of

Compound

Molar mass (g∙mol-1)

Boiling Point (0C)

H2O

18

100

H2S

34

-61

H2Se

80

-41

H2Te

130

-2

A learner in a science class performs an experiment to investigate how

the boiling point varies with the molar mass for the group 16 hydrides. Use

the information in the table to answer the following quesTIONS

3.1 Name the type of intermolecular force that occurs between the

molecules in the compounds H2S to H2Te.

3.2 Explain why the boiling points of the compounds from H2S to H2Te

increase. (3)

3.3 Water does not fit in with the trend shown by the other compounds in the

table. Use intermolecular forces to explain why water has a much higher

boiling point than the other group 16 hydrides. (3)

QUESTION 5

The graph below shows the change in energy that takes place when a

hydrogen (H) atom approaches a bromine (Br) atom.

5.1 Define the term bond length. (2)

5.2 From the graph, write down the:

5.2.1 Bond length, in pm, of the H-Br bond (2)

5.2.2 Energy, in kJ·mol-1, needed to break the H-Br bond (2)

5.2.3 Name of the potential energy represented by E (1)

5.3 How will the bond length of an H-F bond compare to that of the H-Br

bond? Write down EQUAL TO, SHORTER THAN or LONGER THAN. Give

a reason for the answer.

QUESTION 6

Both aluminium fluoride (AℓF3) and phosphorous trifluoride (PF3) contain

fluorine. Aluminium fluoride is a colourless solid used in the production of

aluminium, whilst phosphorous trifluoride is a poisonous, colourless gas.

6.1 Explain the difference between a covalent bond and an ionic bond. (2)

6.2 Name the type of chemical bond between particles in:

6.2.1 AℓF3 (1)

6.2.2 PF3 (1)

6.3 Draw the Lewis structures for:

6.3.1 AℓF3 (3)

6.3.2 PF3 (2)

6.4 Write down the molecular shape of PF3. (1)

6.5 The melting point of AℓF3 is 1 291 °C and that of PF3 is -151 °C. Fully

explain this difference in melting point.

Ideal Gases and Thermal Properties

KINETIC MODEL OF MATTER

movement

anything that has got a mass and occupies

space

It describes the movement of the particles in the 3 states of matter, solids

liquids and gases.

This movement of particles determines the properties of matter in the 3

phases.

Particles in all the 3 phases of matter move and the movement is determined

by the amount of kinetic energy possessed by the particles.

The amount of kinetic energy in the particles can be increased by adding

heat energy and be reduced by removing heat energy.

Solid matter

Particles closely packed

regularly arranged

No free movement the

particles about their fixed

positions

Particles held together by

strong forces of attraction

Particles have little kinetic

energy

Liquid matter

Gaseous matter

Ideal gas It is a gas that has got the following properties:

No intermolecular forces between the gas particles and the particles move in

straight lines between collisions.

Not all the particles have the same energy and not all of them move at the

same speed. The average kinetic energy of the particles is determined by the

temperature of the gas.

The particles of a certain type of gas are identical and all in a constant

motion.

The gas particles fill up the whole space that is available to them and takes

the shape and volume of a container.

Particles exert pressure when they collide with the walls of the container and

the collisions are inelastic (after the collisions the particles continue to move

with the same speed)

The volume of a gas increases when the gas is heated. The particles will

move far away from each other but their size remaining the same.

Particles loosely packed and

irregularly arranged.

Particles move about freely

within a confined space.

Particles held together by

forces of attraction that are

weaker than those in solids.

Particles have more kinetic

energy than the particles in a

solid.

Particles irregularly arranged

and far away from each

other.

Particles move randomly.

No forces of attraction

between the particles.

Particles have a high amount

of kinetic energy.

The pressure of a gas in a closed container increases when heated due to an

increase in the amount of kinetic energy.

Gases can be compressed due to the presence of large spaces between the

particles.

The gases can diffuse.

Pressure of a gas Is due to the collision of the gas particles with the walls of the container.

The pressure of the gas is determined by the number of collisions as well as

the intensity of the collisions.

An increase in the number and intensity of the collisions increases the

pressure.

The pressure of gas is measured in Pascals (Pa) or in N.m-2 using an

instrument called the manometer.

NB At times the pressure will be given in kilopascals so the following

relationship must be known.

1kPa = 1000 Pa

Atmospheric Pressure

It is the pressure that the Earth’s atmosphere exerts. (Remember the

atmosphere consists of gases and these gases exert a pressure resulting in the

atmospheric pressure)

The atmospheric pressure is determined by the location, temperature and

weather conditions.

It is measured by instrument called the barometer.

The Gas Laws

These are the laws that describe the relationships between the variables,

temperature, pressure, volume and the number of moles of a gas.

Two variables are compared at a time. To do this one variable is made the

dependent variable and the other variable the independent one. The two

other variables will be kept constant.

The Boyle’s Law It is a law that gives the relationship between pressure and volume.

The law was named Boyle’s Law after the name of person who carried out

the experiment to investigate the relationship, Robert Boyle. In his experiment, Robert Boyle increased the pressure of a gas and looked

at the effect of the increase of the pressure on the volume of the gas.

The pressure is measured using the Bourdon pressure gauge and the gas is

enclosed in a tube with a volume scale.

Pressure

Results of the Boyle’s experiment

Pressure / Pa Volume / cm3 pV

5 20 100

10 10 100

15 6.67 100.05

20 5 100

25 4 100

30 3.33 99.9

From the results, an increase in pressure is resulting to a decrease in the

pressure.

The product of pressure and volume (pressure × volume) is a constant for all

the sets of values.

Boyle concluded that, Volume of a fixed amount of a gas is inversely

proportional to the pressure on a gas if the temperature is kept

constant. This is the Boyle’s law.

Mathematically:

V = k where k is a constant.

p

If k is made subject of the formula it becomes

k = pV

Since different sets of pV are giving a constant value it means that

P1V1 = p2V2

Calculations using the Boyle’s Law

1. 0

1

2

3

4

5

6

7

8

9

Question

Before a journey the volume of a car tyre was 50dm3 at a pressure of 120 kPa.

After the journey, the tyre pressure decreased to 100kPa. Calculate the new

volume of the tyre.

Solution

Step 1: identify the given information

-initial volume, V1 = 50 dm3

-initial pressure, p1 = 120 kPa

-final volume, V2 =?

-final pressure, P2 = 100 kPa

Step 2: write the formula for the Boyle’s law

p1V1 = p2V2

Step 3: substitute the values in step 1 into the formula

120 kPa × 50dm3 = 100 kPa × V2

6000 kPa. dm3 = 100 V2 kPa

100 kPa 100 kPa

Answer V2 = 60 dm3

The Charles’ Law

It is a law that gives the relationship between temperature and volume at a

constant pressure and mass of the gas.

To investigate the relationship, the following apparatus set up is used:

thermometer capillary tube

mercury bead

gas

water bath

If the water is heated the heat energy is transmitted to the gas. The will

expand when heated up and its volume increases.

Example of results that can be produced,

Temp /oC 20 25 30 35 40 45

Volume /cm3 5 10 15 20 25 30

From the graph, the values for volume are increasing as the values of

temperature increase. Therefore, volume is being directly proportional to

temperature.

The Charles’ Law therefore states that: the volume of a fixed amount of

a gas at a constant pressure is directly proportional to the absolute

temperature of the gas. Mathematically:

Volume = constant × Temperature V = kT or k = V/T

For two sets of values of temperature and volume, V1/T1 = V2/T2

Calculations using the Charles’ Law

Question

A soccer ball contains air of 200cm3 volume at a temperature of 300C before

the match. After a tough match the temperature of air inside the ball increased

to 350C, calculate the new volume of the air in the ball.

Gay-Lussac’s Law

It is a law that gives the relationship between pressure and temperature of a

fixed amount of gas at a constant volume.

This can be investigated using the following apparatus set up:

Bourdon meter

Ice water

gas

Temperature of the ice and the corresponding pressure is recorded.

The ice is replaced by tape water and the temperature and the corresponding

pressure is recorded.

The tap water is heated until it boils and the corresponding pressure is

recorded.

The following table can be used to record the results:

kPa

Ice Tap water Boiling water

Pressure (kPa) 20 28 50

Temperature (0C) 0 25 100

Temperature (K) 273 298 373

From the results, pressure is increasing as the temperature increases.

Therefore, the pressure of a fixed amount of a gas is directly

proportional to the absolute (Kelvin) temperature.

p = kT or k = p/T

For two sets of values of temperature and pressure,

p1/T1 = p2/T2

The general gas equation

It is an equation obtained by combining the Boyle, Charles and Gay-

Lussac’s Laws.

It combines pressure, volume and temperature for a fixed amount of a

gas.

Mathematically it can be written as,

p1V1/T1 = p2V2/T2

The Ideal gas equation

It is an equation that combines the 3 gas laws, the Boyle, Charles and the

Avogadro’s laws.

pV = nRT

where R = 8.31 J.K-1

NB: -The equation or law used depends on the variables given.

-Temperature given in 0C must be converted to Kelvins.

-When using the general gas equation p1V1/T1 = p2V2/T2 p

and V can be expressed in any units as long as V1 and

V2 are expressed in the same units and p1 and p2 with

the same units as well.

Questions

1. 300cm3 of oxygen is at temperature of 200C and pressure of 100kPa. During the

process of liquefaction of air, the oxygen is cooled down to a temperature of

˗2000C and the pressure decreased to 60kPa. Calculate the new volume of the

oxygen.

2. A car is travelling in tarred road and one of its tyres has got a volume of 200dm3,

temperature of 800C and a pressure of 300kPa. The driver applied emergency

brakes and the temperature increased to1000C and the volume to 250dm3.

Calculate the pressure of the air in the tyre when the emergency brakes were

applied.

Question 9

Charles' law can be demonstrated with the

apparatus shown in the figure alongside. A

syringe and a thermometer are inserted through

a rubber stopper into a flask that has been

cooled to 0 ºC. The ice bath is then removed

and the flask is immersed in a warm-water bath

at various temperatures. The gas in the flask

expands as it warms, slowly pushing the piston

out of the syringe. The total volume of the gas

in the system is equal to the volume of the flask

plus the volume of the syringe.

The results obtained are tabulated below.

Temperature (oC) Volume (cm3)

0 24,6

10 25,5

20 26,4

30 27,3

60 30,0

100 33,6

9.1 Use the data in the table above and plot a graph of volume versus

temperature on the graph paper provided.

(5)

The line does not pass through the origin; hence the volume is not proportional to the

temperature on the Celsius scale. Now produce (extrapolate) the straight line until

it intersects the temperature axis. This temperature is known as the 'absolute zero'.

Clamp Syringe

Thermometer

Erlenmeyer-flask

Ice-water bath

Clamp

Block

9.2 Explain why this temperature is known as the 'absolute zero' and what is its

value? Show ALL the calculations.

(4)

9.4

9.5

When asked to state Charles' law, the learner replies: 'The

volume of a gas is directly proportional to temperature.'

How is this statement incomplete? Give the correct

statement of Charles' law.

A gas cylinder of 10 dm3 contains 16 g oxygen gas at 77'oC.

Calculate the pressure in the cylinder.

(2)

9.3 A combination Charles' law and Boyle's law leads to the 'general gas law'.

Give the TWO forms of this gas law.

(2)

STOICHIOMETRY

GRADE 11

• 1 mole of gas occupies 22.4 dm3 at 00C (273 K) and 1 atmosphere (101.3 kPa)

• Interpret balanced reaction equations in terms of volume relationships for gases

under the same conditions of temperature and pressure (volume of gases is

directly proportional to the number of particles of the gases).

• Calculate molar concentration of a solution

• Perform stoichiometric calculations using balanced equations that may include

limiting reagents

• Do stoichiometric calculation to determine the percent yield of a chemical

reaction.

• Do calculations to determine empirical formula and molecular formula of

compounds (revise empirical formula calculations done in grade 10).

• Determine the percent CaCO3 in an impure sample of sea shells (purity or

percent composition)

• Do stoichiometric calculations with explosions as reactions during which a great

many molecules are produced in the gas phase so that there is a massive

increase in volume e.g. ammonium nitrate in mining or petrol in a car cylinder.

2NH4NO3 → 2N2(g) + 4H2O(g) + O2(g)

2C8H18 + 25O2 → 16CO2 + 18H2O

Give the reactions and use it in stoichiometric calculations.

• Do as application the functioning of airbags. Sodium azide reaction:

2NaN3(s) → 2Na(s) + 3N2(g) Reaction must be given when used in

calculations.

The Sandwich Theory

Basic Recipe

Two slices of bread + one slice of polony makes up one sandwich.

2B + 1P 1S

The MOLE

1 mole of any substance contains 6.02 x 1023

Particles / things/ atoms/ molecules/.

602 000 000 000 000 000 000 000

• The relative formula mass of a substance is the sum of the relative atomic

masses of the elements present in a formula unit.

• The symbol for relative formula mass is Mr

If the substance is made of simple molecules, this mass may also be called the

relative molecular mass.

• Limiting Reagent

• 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

• 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting

reactant and how much excess reactant remains after the reaction has stopped?

• • 4 NH3(g) • 5O2(g) • 4NO(g) • 6 H2O(g)

• Relative mass[ Mr[

• 14+3=17 • 32 • •

• Given mass [m]

• 2 • 4 • •

• n • 0.12 • 0.13 • •

ONE 0.03 x 4 = 0.12 0.03 x 5=0.15

We were given sufficient NH3 but not sufficient O2 therefore the O2 limited this reaction

% Yield

An excess of Pb(NO3)2 reacts with 0,75 of KI according to the reaction:

Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)

After filtration and drying, a mass of 0,583 grams of PbI2 is measured.

Determine the % yield of PbI2

Pb(NO3)2 2KI PBI2 2KNO3

m 0.75 1.04

n 4.49 x 10-3 2.25 x 10 -3

Mr 39 + 128 = 167 207 + (128)2 = 463

If everything was perfect we should have got 1.04 g

We only got 0.583g

Therefore 0.583/1.04 x 100 = 56.06% was our true yield.

Empirical & Molecular formula

A compound consists of 85.7% carbon and 14,3% hydrogen.

Its molar mass is 56,12 g. mol-1.

Determine the molecular formula of the compound.

C H

m 85,7 14,3

Mr 12 1

n 7.14 14.15

Ratio 7.14 / 7.14 1

14.15 / 7.14 2

We only get a molar mass of 12 + 2 = 14 with this formula. We need to get a molar

mass of 56,12.

56.12 / 14.3 = 4

That means we must multiply the formula by 4 to get the correct molar mass.

C4H8

Volume using 22,4

If 44,8l Nitrogen and 134l Hydrogen at STP are placed in a container and allowed

to react. Calculate the volume of Ammonia produced.

N2(g) + 3H2(g) 2NH3 (g)

N2(g) 3H2(g) 2NH3 (g)

Ratio Ratio 1 3 2

Volume 44,8 l 134 l 89.6 l

Moles 2 6 4

Water of crystallisation

Calculate the % water in CuSO4 ∙ 5H2O

Mr of whole unit = 64+32+4(16) +5(2+16) = 250

Mr of water only = 5(2+16) = 90

(90/250 )100 = 36%

% composition

Calculate the % composition of sulphuric acid.

H2SO4

Mr %

H2SO4 2(1) + 32 + 4(16) = 98 100

H 2(1) = 2 (2/98) 100 = 2

S 32 (32 /98)100 = 32,7

O 4(16) =64 (64/98)100 = 65,3

CONCENTRATION

25cm3 of hydrochloric acid of concentration

0,12 mol∙dm-3 reacts with 28,4cm3 of sodium hydroxide solution to form water and

sodium chloride.

Calculate the concentration of the sodium hydroxide.

NaOH + HCl NaCl + H2O

NaOH HCl NaCl H2O

V (volume) 28,4 25

C (concentration) XXXXXXXXX 0,12

n (number of moles)

NaOH HCl NaCl H2O

V (volume) 28,4 25

C (concentration) 0,12

n (number of moles) 3 3

NaOH HCl NaCl H2O

V (volume) 28,4 25

C (concentration) 0,11 0,12

n (number of moles) 3 3

QUESTION 1

1.1 Given below is an unbalanced chemical equation:

BaCl2 + Na2CO3 → BaCO3 + NaCl

Calculate the mass of sodium chloride formed if 58g of sodium carbonate reacts

with 138g barium chloride. (6)

1.2 Study the following chemical equation:

2P + 5O2 →2P2O5

Phosphorous burns in oxygen to form P2O5. If 18,6g phosphorous reacts, calculate:

a. How many mol phosphorous reacted? (3)

b. How many mol oxygen will form in this reaction? (3)

c. What volume of oxygen will form if 18,6g phosphorous reacted? (3) d. the percentage P in P2O5. (3) [18] QUESTION 2

Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic

acid has the following percentage composition: 39,9% carbon 6,7% hydrogen 53,4%

oxygen

2.1 Determine the empirical formula of acetic acid. (4)

2.2 Determine the molecular formula of acetic acid if the molar mass of acetic acid is

60 g·mol-1. (2)

[6]

QUESTION 3 Ozone (O

3) reacts with nitrogen monoxide gas (NO) to produce NO

2 gas. The NO gas

forms largely as a result of emissions from the exhausts of motor vehicles and from certain jet planes. The NO

2 gas also causes the brown smog (smoke and fog), which is

seen over most urban areas. This gas is also harmful to humans, as it causes breathing (respiratory) problems. The following equation indicates the reaction between ozone and nitrogen monoxide: O

3 (g) + NO (g) → O

2 (g) + NO

2 (g)

In one such reaction 0,74 g of O3

reacts with 0,67 g NO.

3.1 Calculate the number of moles of O

3 and of NO present at the start of the

reaction. (6)

3.2 Identify the limiting reagent in the reaction and justify your answer. (2) 3.3 Calculate the mass of NO

2 produced from the reaction. (4)

[12]

QUESTION 4 The stomach secretes gastric juice, which contains hydrochloric acid. The gastric juice helps with digestion. Sometimes there is an overproduction of acid, leading to heartburn or indigestion. Antacids, such as milk of magnesia, can be taken to neutralize the

excess acid. Milk of magnesia is only slightly soluble in water and has the chemical formula Mg(OH)

2.

4.1 Write a balanced chemical equation to show how the antacid reacts with the acid.

(3) 4.2 The directions on the bottle recommend that children under the age of 12 years

take one teaspoon of milk of magnesia, whereas adults can take two teaspoons of the antacid. Briefly explain why the dosages are different. (2)

4.3 Why is it not advisable to take an overdose of the antacid in the stomach?

Refer to the hydrochloric acid concentration in the stomach in your answer. (2)

In an experiment, 25,0 cm3

of a standard solution of sodium carbonate [Na(CO3)2] of

concentration 0,1 mol·dm-3

was used to neutralize 35,0 cm3 of a solution of hydrochloric

acid [HCl]. 4.4 Write a balanced chemical equation for the reaction. (3) 4.5 Calculate how many moles of sodium carbonate were used. (2) 4.6 Use the information from 9.4 and 9.5 to now calculate the concentration of the

acid. (3) [15]

QUESTION 5

The balanced chemical equation for the complete COMBUSTION OF PROPANE IS

GIVEN BELOW:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

108 g of C3H8 initially reacted completely with oxygen.

5.1 Determine the number of moles of C3H8 that was initially present. (2)

5.2 Calculate the volume of O2(g), at STP, that was used up in the reaction. (3)

5.3 Calculate the mass of CO2 that was formed when 108 g of C3H8 reacted completely.

(3)

5.4 When an additional amount of C3H8 was added to the reaction mixture, an extra

67,2 dm3 of oxygen, at STP, was required to react completely with it. Calculate the

additional mass of C3H8 that was added.

QUESTION 6

6.1 Define the term molar mass of a substance. (1)

6.2 Calculate the number of moles of water in 100 g of water. (3)

7.3 Methyl benzoate is a compound used in the manufacture of perfumes.

It is found that a 5,325 g sample of methyl benzoate contains 3,758 g of carbon, 0,316

g of hydrogen and 1,251 g of oxygen.

6.3.1 Define the term empirical formula. (2)

6.3.2 Determine the empirical formula of methyl benzoate. (7)

6.3.3 If the molar mass of methyl benzoate is 136 g⋅mol-1, what is its molecular formula?

QUESTION 7

7.1 Define the term limiting reactant. (2)

8.2 Iron (Fe) reacts with sulphur (S) to form iron sulphide (FeS) according to the

following balanced equation:

Fe(s) + S(s) → FeS

7.2.1 Calculate which of the two substances will be used up completely if 20 g of Fe and

10 g of S are mixed and heated.

7.2.2 How many grams of the other substance are in excess? (2)

7.3 Magnesium burns in air to form magnesium oxide according to the following

balanced equation:

2Mg(s) + O2(g) → 2MgO(s)

7.4 If the percentage yield of this reaction is only 80%, calculate the mass of

magnesium that needs to be burned to produce 30 g of magnesium oxide.

7.5 Calculate which of the two substances will be used up completely if

20 g of Fe and 10 g of S are mixed and heated.

7.6 How many grams of the other substance are in excess? (2)