Gr11 PhysicalScience RevisionDocument 2019

GRADE 11 PHYSICAL SCIENCES Guidelines Document and Revision Questions

TABLE OF CONTENTS NO CONTENT PAGE 1 Examinations Guidelines 4 2 Revision Questions – Paper 1 – Physics 37 2.1 Data Sheets – Paper 1 38 2.2 Vectors 40 2.3 Newton’s Laws 45 2.4 Waves, Sound and Light 56 2.5 Electrostatics 66 2.6 Electromagnetism 72 2.7 Electric Circuits 77 3 Revision Questions – Paper 2 – Chemistry 82 3.1 Data Sheets – Paper 2 83 3.2 Bonding and Electronegativity 88 3.3 Gas Laws 95 3.4 Energy and Chemical Change 103 3.5 Intermolecular Forces 107 3.6 Quantitative Analysis 110 3.7 Acids and Bases 114 3.8 Electrochemical Reactions 119 3.9 Chemical Systems 122

PHYSICAL SCIENCES EXAMINATION GUIDELINES GRADE 11 2015 These guidelines consist of 33 pages. COPY RIGHT RESERVED PAGE 2 PLEASE TURN OVER

Physical Sciences/Grade 11 2 DBE/2015 Examination Guidelines TABLE OF CONTENTS Page 1. Introduction 3 2. Assessment in Grade 11 4 2.1 Format of question papers 4 2.2 Numbering and sequence of questions 4 2.3 Information sheets 4 2.4 Weighting of cognitive levels 5 2.5 Weighting of prescribed content 5 2.6 Skills in Physical Sciences 5 2.7 Prior knowledge from Grade 10 6 3. Elaboration of the content for Grade 11 (CAPS) 7 3.1 Paper 1: Physics 7 3.2 Paper 2: Chemistry 13 4. General information 21 4.1 Quantities, symbols and units 21 4.2 Information sheets – Paper 1 (Physics) 23 4.3 Information sheets – Paper 2 (Chemistry) 25 5. Marking guidelines: Paper 1 29 6. Marking guidelines: Paper 2 30 7. Conclusion 33 COPY RIGHT RESERVED PAGE 3 PLEASE TURN OVER

Physical Sciences/Grade 11 3 DBE/2015 Examination Guidelines 1. INTRODUCTION The Curriculum and Assessment Policy Statement (CAPS) for Physical Sciences outlines the nature and purpose of the subject Physical Sciences. This guides the philosophy underlying the teaching and assessment of the subject in Grade 11. The purpose of these Examination Guidelines is to:  Provide clarity on the depth and scope of the content to be assessed in the common/national Grade 11 examination in Physical Sciences.  Assist teachers to adequately prepare learners for the examinations. This document deals with the final Grade 11 examinations. It does not deal in any depth with the School-Based Assessment (SBA). These Examination Guidelines should be read in conjunction with:  The National Curriculum Statement (NCS) Curriculum and Assessment Policy Statement (CAPS): Physical Sciences  The National Protocol of Assessment: An addendum to the policy document, the National Senior Certificate: A qualification at Level 4 on the National Qualifications Framework (NQF), regarding the National Protocol for Assessment (Grades R–12)  The national policy pertaining to the programme and promotion requirements of the National Curriculum Statement, Grades R–12 COPY RIGHT RESERVED PAGE 4 PLEASE TURN OVER

Physical Sciences/Grade 11 4 DBE/2015 Examination Guidelines 2. ASSESSMENT IN GRADE 11 2.1 Format of question papers Paper Types of questions Duration Total Date Marking Physics 10 multiple-choice 1 questions – 20 marks 3 hours 150 October/November Internal Structured questions – 130 marks Chemistry 10 multiple-choice 2 questions – 20 marks 3 hours 150 October/November Internal Structured questions – 130 marks 2.2 Numbering and sequence of questions QUESTION 1: Multiple-choice questions Subquestions numbered 1.1 to 1.10 (2 marks each) Questions will be set across all cognitive levels and arranged from lower to higher cognitive levels. QUESTION 2 onwards: Longer questions that will assess skills and knowledge across cognitive levels. Numbering starts with QUESTION 2 and will be continuous. Subquestions will be numbered by two digits, e.g. 2.1, 2.2. Numbering is restricted to a maximum of three digits, e.g. 2.1.1, 2.1.2. 2.3 Information sheets The separate information sheets for Paper 1 and Paper 2 are included in this document. COPY RIGHT RESERVED PAGE 5 PLEASE TURN OVER

Physical Sciences/Grade 11 5 DBE/2015 Examination Guidelines 2.4 Weighting of cognitive levels Papers 1 and 2 will include questions across all four cognitive levels. The distribution of cognitive levels in Physics and Chemistry papers is given below. Cognitive Paper 1 Paper 2 Description level (Physics) (Chemistry) Remembering/ 1 15% 15% Recall Understanding/ 2 35% 40% Comprehension Applying and 3 40% 35% analysing Evaluating and 4 10% 10% creating (synthesis) 2.5 Weighting of prescribed content Paper 1: Physics Focus Weighting of Content Marks Total Duration cognitive levels Mechanics 68 150 Waves, sound and light 32 3 hours 15 35 40 10 marks Electricity and magnetism 50 Paper 2: Chemistry Focus Weighting of Content Marks Total Duration cognitive levels Chemical change 80 150 Chemical systems 10 3 hours 15 40 35 10 marks Matter and materials 60 2.6 Skills in Physical Sciences  Identify and question phenomena: o Formulate an investigative question. o List all possible variables. o Formulate a testable hypothesis.  Design/Plan of an investigation: o Identify variables (dependent, independent and controlled variables). o List appropriate apparatus. o Plan the sequence of steps which should include, amongst others: - The need for more than one trial to minimise experimental errors. - Identify safety precautions that need to be taken. - Identify conditions that ensure a fair test. - Set an appropriate control. COPY RIGHT RESERVED PAGE 6 PLEASE TURN OVER

Physical Sciences/Grade 11 6 DBE/2015 Examination Guidelines  Graphs: o Draw accurate graphs from given data/information. o Interpret graphs. o Draw sketch graphs from given information.  Results: o Identify patterns/relationships in data. o Interpret results.  Conclusions: o Draw conclusions from given information, e.g. tables, graphs. o Evaluate the validity of conclusions.  Calculations: o Solve problems using two or more different calculations (multistep calculations).  Descriptions: o Explain/Describe/Argue the validity of a statement/event using scientific principles. 2.7 Prior knowledge from Grade 10 All skills and application of knowledge learnt in Grade 10 are applicable to assessment in Grade 11. Skills and knowledge from Grades 10 that may be assessed in Grade 11 include the following:  The scientific method  The periodic table  Writing of formulae and balanced equations  The kinetic molecular theory  Chemical bonding  The mole concept and stoichiometric calculations  Equations of motion  Vectors and scalars COPY RIGHT RESERVED PAGE 7 PLEASE TURN OVER

Physical Sciences/Grade 11 7 DBE/2015 Examination Guidelines 3. ELABORATION OF THE CONTENT FOR GRADE 11 (CAPS) The final examination in Physical Sciences will cover the topics outlined below. 3.1 Paper 1: Physics Vectors in two dimensions (This section must be read in conjunction with the CAPS, p. 61.) Resultant of vectors  Define a resultant as the vector sum of two or more vectors, i.e. a single vector having the same effect as two or more vectors together.  Determine the resultant of vectors (maximum four) on a Cartesian plane, using the component method, i.e. find the vertical and horizontal components of each vector and then add co-linear vertical components and co-linear horizontal components to obtain the resultant vertical vector (Ry) and resultant horizontal vector (Rx).  Sketch the resultant vertical vector (Ry) and the resultant horizontal vector (Rx) on a Cartesian plane.  Calculate the magnitude of the resultant using the theorem of Pythagoras.  Determine the direction of the resultant using simple trigonometric ratios.  Determine the resultant (R) of two vectors graphically using either the tail-to-head or tail- to-tail method (parallelogram method) as well as by calculation (component method) for a maximum of four vectors in both 1-dimension and 2-dimensions.  Explain the meaning of a closed vector diagram. Resolution of a vector into its horizontal and vertical components  Use Rx = Rcosθ for the resultant x component if θ is the angle between R and the x-axis.  Use Ry = Rsinθ for the resultant y component if θ is the angle between R and the x-axis. Newton's laws and application of Newton's laws (This section must be read in conjunction with the CAPS, p. 62–66.) Different kinds of forces: weight, normal force, frictional force, applied force (push, pull), tension (strings or cables)  Define normal force, N, as the force or the component of a force which a surface exerts on an object in contact with it, and which is perpendicular to the surface. NOTE: The normal force acts perpendicular to the surface irrespective of whether the plane is horizontal or inclined. For horizontal planes the only forces perpendicular to the plane should be the weight, w, and the normal force, N. All other forces should be parallel to the plane. For inclined planes the only forces perpendicular to the plane is the component of weight, wcosθ, and the normal, N. All other forces should be parallel to the plane.  Define frictional force, f, as the force that opposes the motion of an object and which acts parallel to the surface. Know that a frictional force: o Is proportional to the normal force o Is independent of the area of the surfaces that are in contact with each other  Define the static frictional force, fs, as the force that opposes the tendency of motion of a stationary object relative to a surface. The static frictional force can have a range of values from zero up to a maximum value, μsN. If a force, F, applied to an object parallel to the surface, does not cause the object to move, F is equal in magnitude to the static frictional force. COPY RIGHT RESERVED PAGE 8 PLEASE TURN OVER

Physical Sciences/Grade 11 8 DBE/2015 Examination Guidelines  State that the static frictional force is a maximum, f max s , just before the object starts to move across the surface. The maximum static frictional force, f max s , is equal to the magnitude of the maximum horizontal force that can be applied to the object without it starting to move across the surface.  Solve problems using f max s = μsN where f max s is the maximum static frictional force and μs is the coefficient of static friction. If the applied force exceeds f max s , a net force accelerates the object.  Define the kinetic frictional force, fk, as the force that opposes the motion of a moving object relative to a surface. The kinetic frictional force on an object is constant for a given surface and equals μkN.  Solve problems using fk = μkN, where fk is the kinetic frictional force and μk the coefficient of kinetic friction. Force diagrams, free-body diagrams  Draw force diagrams. In a force diagram the force is represented by an arrow. The direction of the arrow indicates the direction of the force and the length of the arrow indicates the magnitude of the force.  Draw free-body diagrams. Such a diagram shows the relative magnitudes and directions of forces acting on an object that has been isolated from its surroundings. The object is drawn as a dot and all the forces acting on it are drawn as arrows pointing away from the dot. The length of the arrows is proportional to the magnitude of the respective forces.  Resolve a two-dimensional force, e.g. the weight of an object on an inclined plane, into its parallel (F//) and perpendicular ( F ) components.  Determine the resultant/net force of two or more forces. Newton's first, second and third laws of motion  State Newton's first law of motion: A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.  Define inertia as the resistance of an object to any change in its state of motion. The mass of an object is a quantitative measure of its inertia.  Discuss why it is important to wear seatbelts using Newton's first law of motion.  State Newton's second law of motion: When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the force and inversely proportional to the mass of the object. 1 In symbols: a  Fnet, constant m and a  , constant Fnet, and therefore Fnet = ma m  Draw force diagrams and free-body diagrams for objects that are in equilibrium or accelerating.  Apply Newton's second law of motion, and therefore, to a variety of equilibrium and non- equilibrium problems including: o A single object: - Moving in a horizontal plane with or without friction - Moving on an inclined plane with or without friction - Moving in the vertical plane (lifts, rockets, etc.) o Two-body systems (joined by a light inextensible string): - Both on a flat horizontal plane with or without friction - One in a horizontal plane with or without friction, and a second hanging vertically from a string over a frictionless pulley - Both on an inclined plane with or without friction - Both hanging vertically from a string over a frictionless pulley COPY RIGHT RESERVED PAGE 9 PLEASE TURN OVER

Physical Sciences/Grade 11 9 DBE/2015 Examination Guidelines NOTE: When an object accelerates, the equation Fnet = ma must be applied separately in the x and y directions. If there is more than one object, a free-body diagram must be drawn for each object and Newton's second law must be applied to each object separately.  State Newton's third law of motion: When object A exerts a force on object B, object B SIMULTANEOUSLY exerts an oppositely directed force of equal magnitude on object A. (The forces are therefore an interaction between two bodies.)  Identify Newton III force pairs (action-reaction pairs) and list the properties of the force pairs (action-reaction pairs). When identifying the forces it must be clearly stated which body exerts a force on which body, and what kind of force it is, e.g. the earth exerts a downward gravitational force on the object, and the object exerts an upward gravitational force of equal magnitude on the earth. Newton's Law of Universal Gravitation  State Newton's law of universal gravitation: Each particle in the universe attracts every other particle with a gravitational force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Gm 1m 2  Solve problems using F  . r2 GM  Calculate acceleration due to gravity on the Earth using g  , and on another planet rE2 GM p using g  where Mp is the mass of the planet and rp is the radius of the planet. rp2  Describe weight as the gravitational force, in newton (N), exerted by the Earth on an object. Describe mass as the amount of matter in a body measured in kilogram (kg).  Calculate weight using the expression w = mg.  Calculate the weight of an object on other planets with different values of gravitational acceleration.  Explain weightlessness as the sensation experienced when all contact forces are removed, i.e. no external objects touch one's body. For example, when in free fall, the only force acting on your body is the force of gravity that is a non-contact force. Since the force of gravity cannot be felt without any other opposing forces, you would have no sensation of it and you would feel weightless when in free fall. 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). c  Solve problems using n = . v COPY RIGHT RESERVED PAGE 10 PLEASE TURN OVER

Physical Sciences/Grade 11 10 DBE/2015 Examination Guidelines  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. sin 1 n2 sin i nr  OR n1sinθ1 = n2sinθ2 OR  OR nisinθi = nrsinθr sin 2 n1 sin r ni  Solve problems using the above equations for Snell's law. sin 1  Determine the refractive index of an optical medium by using n  for a ray of light sin 2 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):  deg ree of diffraction  w  State that diffraction demonstrates the wave nature of light. COPY RIGHT RESERVED PAGE 11 PLEASE TURN OVER

Physical Sciences/Grade 11 11 DBE/2015 Examination Guidelines 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. kQ 1Q2  Solve problems using the equation F = for charges in one dimension (1D) – r2 restrict to three charges. kQ 1Q2  Solve problems using the equation F = for charges in two dimensions (2D) – for r2 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. F In symbols: E  Q F  Solve problems using the equation E  . Q  Calculate the electric field strength at a point due to a number of point charges, using the kQ equation E  2 to determine the contribution to the field due to each charge. Restrict to r 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. V  Solve problems using R  for circuits containing resistors that are connected in series I and/or in parallel (maximum four resistors). COPY RIGHT RESERVED PAGE 12 PLEASE TURN OVER

Physical Sciences/Grade 11 12 DBE/2015 Examination Guidelines Power, energy  Define power as the rate at which work is done or energy is transferred. W  Solve problems using P = . t  Recall that W = VQ, and therefore, by substituting V = IR in the equation, W = VIΔt, V2 W = I2RΔt and W = t are obtained. R W  Deduce, by substituting P = into above equations, the following equations: P = VI, t V2 P = I2R and P = . R V2  Solve problems using P = VI, P = I2R and P = . R  Solve circuit problems involving the concepts of power and electrical energy.  Deduce that the kilowatt-hour (kWh) refers to the use of 1 kilowatt of electricity for 1 hour. Know that 1 kWh is an amount of electrical energy, also known as one unit of electricity.  Calculate the cost of electricity usage given the power specifications of the appliances used, the duration and the cost of 1 kWh. Electromagnetism (This section must be read in conjunction with the CAPS, p. 86–88.) Magnetic field associated with current-carrying conductors  Use the right hand rule to determine the magnetic field (B) associated with a: o Straight current-carrying conductor o Current-carrying loop (single turn) o Solenoid  Draw the magnetic field pattern around a: o Straight current-carrying wire o Current-carrying loop (single turn) o Solenoid Faraday's law  State Faraday's law of electromagnetic induction: The magnitude of the induced emf across the ends of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor.  Use the right-hand rule to determine the direction of the current induced in a solenoid when a pole of a bar magnet moves into and out of the solenoid.  Solve problems using  = BAcosθ.  Predict the direction of the induced current in a coil.   Solve problems using   - N . t COPY RIGHT RESERVED PAGE 13 PLEASE TURN OVER

Physical Sciences/Grade 11 13 DBE/2015 Examination Guidelines 3.2 Paper 2: Chemistry Atomic combinations: molecular structure (This section must be read in conjunction with the CAPS, p. 67–71.) A chemical bond  Define a chemical bond as a mutual attraction between two atoms resulting from the simultaneous attraction between their nuclei and the outer electrons. (NOTE: The energy of the combined atoms is lower than that of the individual atoms resulting in higher stability.)  Draw Lewis dot diagrams of elements. A Lewis dot diagram is a structural formula in which valence electrons are represented by dots or crosses. It is also known as an electron dot formula, a Lewis formula, or an electron diagram.  Determine the number of valence electrons in an atom of an element. Valence electrons or outer electrons are the electrons in the highest energy level of an atom in which there are electrons.  Explain, in terms of electrostatic forces between protons and electrons, and in terms of energy considerations, why: o Two H atoms form an H2 molecule o He does not form He2 Interpret the graph of potential energy versus the distance between nuclei for two approaching hydrogen atoms.  Define a covalent bond as the sharing of electrons between two atoms to form a molecule. Molecule: A group of two or more atoms covalently bonded and that function as a unit.  Draw Lewis diagrams, given the formula and using electron configurations, for simple molecules, e.g. H2, F2, H2O, NH3, HF, OF2, HOCℓ and molecules with multiple bonds, e.g. N2, O2 and HCN.  Describe rules for bond formation: o Different atoms, each with an unpaired valence electron, can share these electrons to form a chemical bond, e.g. two H atoms form a H2 molecule by sharing an electron pair. o Different atoms with paired valence electrons, called lone pairs, cannot share these four electrons and cannot form a chemical bond, e.g. no bond forms between two He atoms. o Different atoms, with unpaired valence electrons, can share these electrons and form a chemical bond for each electron pair shared. The two atoms can form multiple bonds between them. If two pairs of electrons are shared, a double bond is formed, e.g. between two O atoms to form O2. o Atoms with an empty valence shell can share a lone pair of electrons from another atom to form a coordinate covalent or dative covalent bond, e.g. in NH 4 the lone pair of nitrogen is shared with H+ and in H3O+ the lone pair of oxygen is shared with H+.  Define a bonding pair as a pair of electrons that is shared between two atoms in a covalent bond. Define a lone pair as a pair of electrons in the valence shell of an atom that is not shared with another atom.  Describe the formation of the dative covalent (or coordinate covalent) bond by means of electron diagrams using NH 4 and H3O+ as examples. COPY RIGHT RESERVED PAGE 14 PLEASE TURN OVER

Physical Sciences/Grade 11 14 DBE/2015 Examination Guidelines Molecular shape: Valence shell electron pair repulsion (VSEPR) theory  State the major principles used in the VSEPR: o Molecular shape is determined by the repulsions between electron pairs present in the valence shell of the central atom. o The number of electron pairs around the central atom can be determined by writing the Lewis structure for the molecule. o The shape of the molecule depends on the number of bonding electron groups (or atoms bonded to the central atom) and the number of lone pairs on the central atom. o A is used to represent the central atom and X is used to represent terminal atoms. o There are five ideal shapes found when there are NO lone pairs on the central atom, ONLY bond pairs.  Use the VSEPR theory to classify given molecules as one of the five ideal molecular shapes by finding the number of atoms bonded to the central atom in molecules where there are NO lone pairs on the central atom. If this number equals: o Two, the molecular shape is linear – AX2 Examples: CO2, BeCℓ2, C2H2 o Three, the molecular shape is trigonal planar – AX3 Example: BF3 o Four, the molecular shape is tetrahedral – AX4 Example: CH4, CCℓ4 o Five, the molecular shape is trigonal bipyramidal – AX5 Example: PCℓ5 o Six, the molecular shape is octahedral – AX6 Example: SF6  Use the VSEPR theory to determine the shapes of molecules with lone pairs on the central atom (H2O, NH3, SO2) and that CANNOT have one of the ideal shapes. Electronegativity of atoms to explain the polarity of bonds  Define electronegativity as a measure of the tendency of an atom in a molecule to attract bonding electrons.  Describe a non-polar covalent bond as a bond in which the electron density is shared equally between the two atoms. An example is the bond between two H atoms.  Describe a polar covalent bond as a bond in which the electron density is shared unequally between the two atoms. An example is the bond between an H atom and a Cℓ atom.  Show polarity of bonds using partial charges, e.g. δ+ H - Cℓ δ-  Compare the polarity of chemical bonds using a table of electronegativities: o With an electronegativity difference ∆EN > 2,1 electron transfer will take place and the bond will be ionic. o With an electronegativity difference ∆EN > 1 the bond will be polar covalent. o With an electronegativity difference ∆EN < 1 the bond will be very weakly polar covalent. o With an electronegativity difference ∆EN = 0 the bond will be nonpolar. covalent  Explain that the character of a bond varies from purely covalent (when ∆EN = 0) to almost purely ionic (when ∆EN > 3). The above differences in electronegativity is only a guideline, many bonds have both ionic and covalent character.  Use difference in electronegativity and molecular shape to explain that polar bonds do not always lead to polar molecules. COPY RIGHT RESERVED PAGE 15 PLEASE TURN OVER

Physical Sciences/Grade 11 15 DBE/2015 Examination Guidelines Bond energy and bond length  Define bond energy of a compound as the energy needed to break one mole of its molecules into separate atoms.  Define bond length as the average distance between the nuclei of two bonded atoms.  Explain the relationship between bond energy and bond length, i.e. bonds with a shorter bond length require more energy to break than bonds with a longer bond length.  Explain the relationship between the strength of a chemical bond between two atoms and the: o Length of the bond between them If the force of attraction between two atoms is strong, the nuclei come very close together resulting in a short bond length. o Size of the bonded atoms The bond length between larger atoms is longer than the bond length between smaller atoms. o Number of bonds (single, double, triple) between the atoms Bond strength increases as the number of bonds between atoms increases, i.e. triple bonds are stronger than double bonds, which are stronger than single bonds. Intermolecular forces (This section must be read in conjunction with the CAPS, p. 71–75.) Intermolecular forces and interatomic forces (chemical bonds)  Name and explain the different intermolecular forces (Van der Waals forces): (i) Mutually induced dipole forces or London forces: Forces between non-polar molecules (ii) Dipole-dipole forces: Forces between two polar molecules (iii) Dipole-induced dipole forces: Forces between polar and non-polar molecules (iv) Hydrogen bonding: Forces between molecules in which hydrogen is covalently bonded to nitrogen, oxygen or fluorine – a special case of dipole-dipole forces (v) Ion-dipole forces: Forces between ions and polar molecules  Describe the difference between intermolecular forces and interatomic forces (intramolecular forces) using a diagram of a group of small molecules; and in words. Example: Intermolecular force H H Interatomic force/chemical bond H C H H C H H H  State the relationship between intermolecular forces and molecular mass. For non-polar molecules the strength of induced dipole forces increases with molecular size.  Explain the effect of intermolecular forces on boiling point, melting point, vapour pressure and solubility. Boiling point: The temperature at which the vapour pressure of a substance equals atmospheric pressure. The stronger the intermolecular forces, the higher the boiling point. Melting point: The temperature at which the solid and liquid phases of a substance are at equilibrium. The stronger the intermolecular forces, the higher the melting point. Vapour pressure: The pressure exerted by a vapour at equilibrium with its liquid in a closed system. The stronger the intermolecular forces, the lower the vapour pressure. Solubility: The property of a solid, liquid, or gaseous chemical substance (solute) to dissolve in a solid, liquid, or gaseous solvent to form a homogeneous solution. COPY RIGHT RESERVED PAGE 16 PLEASE TURN OVER

Physical Sciences/Grade 11 16 DBE/2015 Examination Guidelines 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. 1 In symbols: p  , therefore p1V1 = p2V2, T = constant V  State Charles' law: The volume of an enclosed gas is directly proportional to its kelvin temperature provided the pressure is kept constant. V V In symbols: VT , therefore 1  2 , p = constant T1 T2  State that the pressure of a gas is directly proportional to its temperature in kelvin at p p constant volume (Gay Lussac). In symbols: pT , therefore 1  2 , V = constant T1 T2  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 pV p V  Use the general gas equation, 1 1  2 2 , to solve problems. T1 T2  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. COPY RIGHT RESERVED PAGE 17 PLEASE TURN OVER

Physical Sciences/Grade 11 17 DBE/2015 Examination Guidelines 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. n  Calculate concentration in mol·ℓ-1 (or mol·dm-3) using c  . V 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. COPY RIGHT RESERVED PAGE 18 PLEASE TURN OVER

Physical Sciences/Grade 11 18 DBE/2015 Examination Guidelines Energy and chemical change (This section must be read in conjunction with the CAPS, p. 90–91.) Energy changes in reactions related to bond energy changes  Define heat of reaction (ΔH) as the energy absorbed or released per mole in a chemical reaction. ΔH = Hproducts – Hreactants, where Hproducts and Hreactants are the heat (energy) of the products and reactants respectively.  Define exothermic reactions as reactions that release energy.  Define endothermic reactions as reactions that absorb energy.  Classify, with reason, reactions as exothermic or endothermic. Exothermic and endothermic reactions  State that ΔH > 0 for endothermic reactions, i.e. reactions in which energy is absorbed.  State that ΔH < 0 for exothermic reactions, i.e. reactions in which energy is released. Activation energy  Define activation energy as the minimum energy needed for a reaction to take place.  Define an activated complex as the unstable transition state from reactants to products.  Draw or interpret fully labelled sketch graphs (potential energy versus course of reaction graphs) of catalysed and uncatalysed endothermic and exothermic reactions. Types of reaction (This section must be read in conjunction with the CAPS, p. 91–94.) Acid-base reactions  Define acids and bases according to Arrhenius and Lowry-Brønsted: Arrhenius theory: An acid is a substance that produces hydrogen ions (H+)/hydronium ions (H3O+) when it dissolves in water. A base is a substance that produces hydroxide ions (OH-) when it dissolves in water. Lowry-Brønsted theory: An acid is a proton/H+ ion donor. A base is a proton/H+ ion acceptor.  Identify conjugate acid-base pairs for given compounds. When the acid, HA, loses a proton, its conjugate base, A-, is formed. When the base, A-, accepts a proton, its conjugate acid, HA, is formed. These two are a conjugate acid-base pair.  Describe a substance that can act as either acid or base, as amphiprotic or as an ampholyte. Water is a good example of an ampholyte. Write equations to show how an amphiprotic substance can act as acid or base.  Write names and formulae of common acids: hydrochloric acid, nitric acid, sulphuric acid and ethanoic acid (acetic acid).  Write names and formulae of common bases: ammonia, sodium carbonate (washing soda), sodium hydrogen carbonate, sodium hydroxide (caustic soda) and potassium hydroxide.  Write reaction equations for the dissolution of acids and bases in water. Examples: HCℓ(g) + H2O(ℓ) → H3O+(aq) + Cℓ-(aq) NH3(g) + H2O(ℓ) → NH 4 (aq) + OH-(aq) H2SO4(aq) + 2H2O(ℓ) → 2H3O+(aq) + SO 24  (aq) COPY RIGHT RESERVED PAGE 19 PLEASE TURN OVER

Physical Sciences/Grade 11 19 DBE/2015 Examination Guidelines  Write the overall equations for reactions of acids with metal hydroxides, metal oxides and metal carbonates: HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ) HCℓ(aq) + KOH(aq) → KCℓ(aq) + H2O(ℓ) HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(ℓ) HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(ℓ) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(ℓ) H2SO4(aq) + KOH(aq) → K2SO4(aq) + 2H2O(ℓ) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(ℓ) 2HCℓ(aq) + CaO(aq) → CaCℓ2(aq) + H2O(ℓ) 2HCℓ(aq) + MgO(aq) → MgCℓ2(aq) + H2O(ℓ) 2HCℓ(aq) + MgO(aq) → MgCℓ2(aq) + H2O(ℓ) 2HCℓ(aq) + CuO(aq) → CuCℓ2(aq) + H2O(ℓ) 2HCℓ(aq) + Na2CO3(aq) → 2NaCℓ(aq) + H2O(ℓ) + CO2(g) 2HCℓ(aq) + CaCO3(aq) → CaCℓ2(aq) + H2O(ℓ) + CO2(g)  Describe an acid-base indicator as a weak acid, or a weak base, which colour changes as the H+ ion concentration or the OH- ion concentration in a solution changes. Know the colours of litmus, methyl orange, phenolphthalein and bromothymol blue in acids and in bases.  Identify the acid and the base needed to prepare a given salt and write an equation for the reaction.  Write down neutralisation reactions of common laboratory acids and bases. Oxidation numbers of atoms in molecules  Explain the meaning of oxidation number.  Assign oxidation numbers to atoms in various ions and molecules, e.g. H2O, CH4, CO2, H2O2 and HOCℓ, by using oxidation number guidelines or rules. Redox reactions  Describe a redox (oxidation-reduction) reaction as involving an electron transfer.  Describe a redox (oxidation-reduction) reaction as always involving changes in oxidation numbers.  Identify a redox reaction and apply the correct terminology to describe all the processes: Oxidation: A loss of electrons./An increase in oxidation number. Reduction: A gain of electrons./A decrease in oxidation number. Reducing agent: A substance that is oxidised/that loses electrons/whose oxidation number increases. Oxidising agent: A substance that is reduced/that gains electrons/whose oxidation number decreases.  Balance redox reactions by using half-reactions from the Table of Standard Reduction Potentials (Tables 4A and 4B). COPY RIGHT RESERVED PAGE 20 PLEASE TURN OVER

Physical Sciences/Grade 11 20 DBE/2015 Examination Guidelines Exploiting the lithosphere or Earth's crust: (This section must be read in conjunction with the CAPS, p. 138–140.) Mining and processing of minerals  Apply knowledge of chemistry, learnt in previous sections, to given reactions of the extraction processes of gold and iron. Gold mining and processing  State that South African gold is mined from deep level underground mines.  State that the ore undergoes a series of crushing and milling steps to reduce its size for further processing.  Describe the leaching of gold from the slurry by a dilute sodium cyanide (NaCN) or potassium cyanide (KCN) solution: 4Au + 8NaCN + O2 + 2H2O → 4NaAu(CN)2 + 4NaOH The solution containing the gold complex ion can now be filtered from the unwanted rock.  Describe the addition of zinc powder to precipitate the gold: Zn + 2NaAu(CN)2 → 2Au + Zn(CN)2 + 2NaCN Solid gold can now be filtered from the solution.  Classify the above reactions as redox reactions, giving reasons.  Identify the oxidation, reduction, oxidising agent and reducing agent in the above reactions. Iron mining and processing  Open pit mining is used to excavate iron ore that is close to the surface.  The ore undergoes a series of crushing steps to reduce the particle size.  After washing, screening and separation, the iron ore, containing 65% iron, goes into a blast furnace with coke (C), limestone (CaCO3) and hot air.  The following reactions takes place in the blast furnace: o Coke (carbon) is oxidised to carbon monoxide: 2C + O2 → 2CO o CO reacts with iron ore to form iron: 3CO + Fe2O3 → 2Fe + 3CO2 o Limestone is added to remove impurities from the molten iron to form slag: CaCO3 → CaO + CO2 CaO + SiO2 → CaSiO3 (slag) o Molten slag floats on top of the molten iron. The blast furnace produces pig iron that is sent to steel manufacturers.  Classify the above reactions as redox reactions where applicable, giving reasons.  Identify the oxidation, reduction, oxidising agent and reducing agent in the above reactions where applicable. COPY RIGHT RESERVED PAGE 21 PLEASE TURN OVER

Physical Sciences/Grade 11 21 DBE/2015 Examination Guidelines 4. GENERAL INFORMATION 4.1 Quantities, symbols and units The most common quantities, symbols and SI units used in introductory Physics are listed below. A quantity should not be confused with the units in which it is measured. Preferred Alternative Unit Quantity Unit name symbol symbol symbol mass m kilogram kg position x, y metre m displacement  x,  y s metre m -1 velocity vx, vy u, v metre per second m∙s -1 initial velocity vi u metre per second m∙s -1 final velocity vf v metre per second m∙s -2 acceleration a metre per second per second m∙s acceleration due to -2 g metre per second per second m∙s gravity time (instant) t second s time interval t second s energy E joule J kinetic energy K Ek joule J potential energy U Ep joule J work W joule J power P watt W force F newton N weight w Fg newton N normal force N FN newton N tension T FT newton N friction force f Ff newton N coefficient of friction μ, μ s , μ k (none) wavelength λ metre m frequency f ν hertz or per second Hz or s -1 period T second s -1 speed of light c metre per second m∙s refractive index n (none) charge Q, q coulomb C -1 newton per coulomb or N∙C or electric field E -1 volt per metre V∙m electric potential at VP volt V point P potential difference  V, V volt V emf E ε volt V current I, i ampere A resistance R ohm Ω magnetic field B tesla T 2 2 T∙m or magnetic flux Φ tesla∙metre or weber Wb COPY RIGHT RESERVED PAGE 22 PLEASE TURN OVER

Physical Sciences/Grade 11 22 DBE/2015 Examination Guidelines Conventions (e.g. signs, symbols, terminology and nomenclature) The syllabus and question papers will conform to generally accepted international practices. NOTE: 1. For marking purposes, alternative symbols will also be accepted. 2. Separate compound units with a multiplication dot, not a full stop, for example m·s-1. For marking purposes, m.s-1 will also be accepted. 3. Use the equal sign only when it is mathematically correct, for example: Incorrect: 1 cm = 1 m (on a scale drawing) -2 Correct: 1 cm = 10 m 1 cm represents 1 m (on a scale drawing) COPY RIGHT RESERVED PAGE 23 PLEASE TURN OVER

Physical Sciences/Grade 11 23 DBE/2015 Examination Guidelines 4.2 Information sheets – Paper 1 (Physics) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity g 9,8 m·s-2 Swaartekragversnelling Gravitational constant G 6,67 x 10-11 Nm2kg-2 Swaartekragkonstante Radius of Earth RE 6,38 x 106 m Straal van Aarde Coulomb's constant k 9,0 x 109 Nm2·C-2 Coulomb se konstante Speed of light in a vacuum c 3,0 x 108 m·s-1 Spoed van lig in 'n vakuum Charge on electron e -1,6 x 10-19 C Lading op elektron Electron mass me 9,11 x 10-31 kg Elektronmassa Mass of the Earth M 5,98 x 1024 kg Massa van die Aarde TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING v f  v i  a t x  v i t  21 at 2  v  vi  2 2 v f  v i  2ax x   f  t  2  FORCE/KRAG Fnet  ma w = mg Gm1m2 fs F  s  (max) r2 N f k  k N COPY RIGHT RESERVED PAGE 24 PLEASE TURN OVER

Physical Sciences/Grade 11 24 DBE/2015 Examination Guidelines WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG 1 vf T f ni sini  nr sinr c n v ELECTROSTATICS/ELEKTROSTATIKA kQ1Q 2 F F= (k = 9,0 x 109 Nm2·C-2) E r2 Q kQ W E 2 (k = 9,0 x 109 Nm2·C-2) V r Q ELECTROMAGNETISM/ELEKTROMAGNETISME  N   BA cos  t CURRENT ELECTRICITY/STROOMELEKTRISITEIT Q V I R t I 1 1 1 1     ... R  r1  r2  r3  ... R r1 r2 r3 W = Vq W P Δt W = VI  t P = VI W= I2R  t P = I2R 2 V Δt V2 W= P R R COPY RIGHT RESERVED PAGE 25 PLEASE TURN OVER

Physical Sciences/Grade 11 25 DBE/2015 Examination Guidelines 4.3 Information sheets – Paper 2 (Chemistry) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Avogadro's constant NA 6,02 x 1023 mol-1 Avogadro-konstante Molar gas constant R 8,31 J∙K-1∙mol-1 Molêre gaskonstante Standard pressure p 1,013 x 105 Pa Standaarddruk Molar gas volume at STP Vm 22,4 dm3∙mol-1 Molêre gasvolume by STD Standard temperature T 273 K Standaardtemperatuur TABLE 2: FORMULAE/TABEL 2: FORMULES p1V1 p2 V2  pV  nRT T1 T2 m N n n M NA V n m n c OR/OF c Vm V MV COPY RIGHT RESERVED PAGE 26 PLEASE TURN OVER

Physical Sciences/Grade 11 26 DBE/2015 Examination Guidelines TABLE 3: THE PERIODIC TABLE OF ELEMENTS 1 2 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 (I) (II) (III) (IV) (V) (VI) (VII) (VIII) 1 Atomic number 2 KEY/SLEUTEL 2,1 H Atoomgetal He 1 4 3 4 29 5 6 7 8 9 10 Symbol 1,0 Li 1,5 Be Electronegativity 1,9 Cu Simbool 2,0 B 2,5 C 3,0 N 3,5 O 4,0 F Ne 7 9 Elektronegatiwiteit 63,5 11 12 14 16 19 20 11 12 13 14 15 16 17 18 0,9 Na 1,2 Mg Approximate relative atomic mass 1,5 Aℓ 1,8 Si 2,1 P 2,5 S 3,0 Cℓ Ar 23 24 Benaderde relatiewe atoommassa 27 28 31 32 35,5 40 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 0,8 K 1,0 Ca 1,3 Sc 1,5 Ti 1,6 V 1,6 Cr 1,5 Mn 1,8 Fe 1,8 Co 1,8 Ni 1,9 Cu 1,6 Zn 1,6 Ga 1,8 Ge 2,0 As 2,4 Se 2,8 Br Kr 39 40 45 48 51 52 55 56 59 59 63,5 65 70 73 75 79 80 84 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 0,8 Rb 1,0 Sr 1,2 Y 1,4 Zr Nb 1,8 Mo 1,9 Tc 2,2 Ru 2,2 Rh 2,2 Pd 1,9 Ag 1,7 Cd 1,7 In 1,8 Sn 1,9 Sb 2,1 Te 2,5 I Xe 86 88 89 91 92 96 101 103 106 108 112 115 119 122 128 127 131 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 0,7 Cs 0,9 Ba La 1,6 Hf Ta W Re Os Ir Pt Au Hg 1,8 Tℓ 1,8 Pb 1,9 Bi 2,0 Po 2,5 At Rn 133 137 139 179 181 184 186 190 192 195 197 201 204 207 209 87 88 89 0,7 Fr 0,9 Ra Ac 58 59 60 61 62 63 64 65 66 67 68 69 70 71 226 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140 141 144 150 152 157 159 163 165 167 169 173 175 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232 238 COPY RIGHT RESERVED PAGE 27 PLEASE TURN OVER

Physical Sciences/Grade 11 27 DBE/2015 Examination Guidelines TABLE 4A: STANDARD REDUCTION POTENTIALS TABEL 4A: STANDAARDREDUKSIEPOTENSIALE Half-reactions Eθ (V) F2(g) + 2e ⇌ 2F + 2,87 3+ 2+ Co + e ⇌ Co + 1,81 + H2O2 + 2H +2e ⇌ 2H2O +1,77  MnO 4 + 8H + 5e + ⇌ Mn 2+ + 4H2O + 1,51 Cℓ2(g) + 2e  ⇌ 2Cℓ  + 1,36 2 + Cr2O 7 + 14H + 6e ⇌ 3+ 2Cr + 7H2O + 1,33 + O2(g) + 4H + 4e  ⇌ 2H2O + 1,23 + 2+ MnO2+ 4H + 2e ⇌ Mn + 2H2O + 1,23 2+ Pt + 2e  ⇌ Pt + 1,20 Br2(ℓ) + 2e ⇌ 2Br + 1,07  + NO 3 + 4H + 3e ⇌ NO(g) + 2H2O + 0,96 2+ Hg + 2e  ⇌ Hg(ℓ) + 0,85 Increasing reducing ability/Toenemende reduserende vermoë + Ag + e ⇌ Ag + 0,80 Increasing oxidising ability/Toenemende oksiderende vermoë  + NO 3 + 2H + e ⇌ NO2(g) + H2O + 0,80 3+ 2+ Fe + e  ⇌ Fe + 0,77 + O2(g) + 2H + 2e ⇌ H2O2 + 0,68 I2 + 2e ⇌ 2I + 0,54 + Cu + e ⇌ Cu + 0,52 + SO2 + 4H + 4e ⇌ S + 2H2O + 0,45 2H2O + O2 + 4e ⇌ 4OH + 0,40 2+ Cu + 2e ⇌ Cu + 0,34 2 SO 4 + 4H + 2e + ⇌ SO2(g) + 2H2O + 0,17 2+ + Cu + e  ⇌ Cu + 0,16 4+ 2+ Sn + 2e ⇌ Sn + 0,15 + S + 2H + 2e ⇌ H2S(g) + 0,14 + 2H + 2e ⇌ H2(g) 0,00 3+ Fe + 3e ⇌ Fe  0,06 2+ Pb + 2e ⇌ Pb  0,13 2+ Sn + 2e ⇌ Sn  0,14 2+ Ni + 2e ⇌ Ni  0,27 2+ Co + 2e ⇌ Co  0,28 2+ Cd + 2e ⇌ Cd  0,40 3+ 2+ Cr + e ⇌ Cr  0,41 2+ Fe + 2e ⇌ Fe  0,44 3+ Cr + 3e ⇌ Cr  0,74 2+ Zn + 2e ⇌ Zn  0,76 2H2O + 2e ⇌ H2(g) + 2OH  0,83 2+ Cr + 2e ⇌ Cr  0,91 2+ Mn + 2e ⇌ Mn  1,18 3+ Aℓ + 3e ⇌ Aℓ  1,66 2+ Mg + 2e ⇌ Mg  2,36 + Na + e ⇌ Na  2,71 2+ Ca + 2e ⇌ Ca  2,87 2+ Sr + 2e ⇌ Sr  2,89 2+ Ba + 2e ⇌ Ba  2,90 + - Cs + e ⇌ Cs - 2,92 +  K +e ⇌ K  2,93 + Li + e ⇌ Li  3,05 COPY RIGHT RESERVED PAGE 28 PLEASE TURN OVER

Physical Sciences/Grade 11 28 DBE/2015 Examination Guidelines TABLE 4B: STANDARD REDUCTION POTENTIALS TABEL 4B: STANDAARDREDUKSIEPOTENSIALE Half-reactions Eθ (V) + Li + e ⇌ Li  3,05 + K + e ⇌ K  2,93 + Cs + e ⇌ Cs  2,92 2+ Ba + 2e ⇌ Ba  2,90 2+ Sr + 2e ⇌ Sr  2,89 2+ Ca + 2e ⇌ Ca  2,87 + Na + e ⇌ Na  2,71 2+ Mg + 2e ⇌ Mg  2,36 Increasing oxidising ability/Toenemende oksiderende vermoë 3+ Aℓ + 3e ⇌ Aℓ  1,66 2+ Mn + 2e ⇌ Mn  1,18 Increasing reducing ability/Toenemende reduserende vermoë 2+ Cr + 2e ⇌ Cr  0,91 2H2O + 2e ⇌ H2(g) + 2OH  0,83 2+ Zn + 2e ⇌ Zn  0,76 3+ Cr + 3e ⇌ Cr  0,74 2+ Fe + 2e ⇌ Fe  0,44 3+ 2+ Cr + e ⇌ Cr  0,41 2+ Cd + 2e ⇌ Cd  0,40 2+ Co + 2e ⇌ Co  0,28 2+ Ni + 2e ⇌ Ni  0,27 2+ Sn + 2e ⇌ Sn  0,14 2+ Pb + 2e ⇌ Pb  0,13 3+ Fe + 3e ⇌ Fe  0,06 + 2H + 2e ⇌ H2(g) 0,00 + S + 2H + 2e ⇌ H2S(g) + 0,14 4+ 2+ Sn + 2e ⇌ Sn + 0,15 2+ + Cu + e ⇌ Cu + 0,16 2 + SO 4 + 4H + 2e ⇌ SO2(g) + 2H2O + 0,17 2+  Cu + 2e ⇌ Cu + 0,34 2H2O + O2 + 4e ⇌ 4OH + 0,40 + SO2 + 4H + 4e ⇌ S + 2H2O + 0,45 +  Cu + e ⇌ Cu + 0,52 I2 + 2e ⇌ 2I + 0,54 + O2(g) + 2H + 2e ⇌ H2O2 + 0,68 3+  2+ Fe + e ⇌ Fe + 0,77  + NO 3 + 2H + e ⇌ NO2(g) + H2O + 0,80 +  Ag + e ⇌ Ag + 0,80 2+ Hg + 2e ⇌ Hg(ℓ) + 0,85  + NO 3 + 4H + 3e ⇌ NO(g) + 2H2O + 0,96   Br2(ℓ) + 2e ⇌ 2Br + 1,07 2+ Pt + 2 e ⇌ Pt + 1,20 + 2+ MnO2+ 4H + 2e ⇌ Mn + 2H2O + 1,23 + O2(g) + 4H + 4e ⇌ 2H2O + 1,23 2 + 3+ Cr2O 7 + 14H + 6e ⇌ 2Cr + 7H2O + 1,33   Cℓ2(g) + 2e ⇌ 2Cℓ + 1,36  +  2+ MnO 4 + 8H + 5e ⇌ Mn + 4H2O + 1,51 +  H2O2 + 2H +2 e ⇌ 2H2O +1,77 3+ 2+ Co + e ⇌ Co + 1,81 F2(g) + 2e ⇌ 2F + 2,87 COPY RIGHT RESERVED PAGE 29 PLEASE TURN OVER

Physical Sciences/Grade 11 29 DBE/2015 Examination Guidelines 5. MARKING GUIDELINES: PAPER 1 5.1 Calculations 5.1.1 Marks will be awarded for: correct formula, correct substitution, correct answer with unit. 5.1.2 No marks will be awarded if an incorrect or inappropriate formula is used, even though there may be relevant symbols and applicable substitutions. 5.1.3 When an error is made during substitution into a correct formula, a mark will be awarded for the correct formula and for the correct substitutions, but no further marks will be given. 5.1.4 If no formula is given, but all substitutions are correct, the candidate will forfeit one mark. 5.1.5 No penalisation if zero substitutions are omitted in calculations where correct formula/principle is given correctly. 5.1.6 Mathematical manipulations and change of subject of appropriate formulae carry no marks, but if a candidate starts off with the correct formula and then changes the subject of the formula incorrectly, marks will be awarded for the formula and the correct substitutions. The mark for the incorrect numerical answer is forfeited. 5.1.7 Marks are only awarded for a formula if a calculation has been attempted, i.e. substitutions have been made or a numerical answer given. 5.1.8 Marks can only be allocated for substitutions when values are substituted into formulae and not when listed before a calculation starts. 5.1.9 Final answers to all calculations, when not specified in the question, must be rounded off to a minimum of TWO decimal places. 5.1.10 If a final answer to a calculation is correct, full marks will not automatically be awarded. Markers will always ensure that the correct/appropriate formula is used and that workings, including substitutions, are correct. 5.1.11 Questions in which a series of calculations have to be made (e.g. a circuit-diagram question) do not necessarily always have to follow the same order. FULL MARKS will be awarded, provided it is a valid solution to the problem. However, any calculation that will not bring the candidate closer to the answer than the original data, will not count any marks. 5.2 Units 5.2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question. 5.2.2 Units are only required in the final answer to a calculation. 5.2.3 Marks are only awarded for an answer, and not for a unit per se. Candidates will therefore forfeit the mark allocated for the answer in each of the following situations: - Correct answer + wrong unit - Wrong answer + correct unit - Correct answer + no unit 5.2.4 SI units must be used, except in certain cases, e.g. V∙m-1 instead of N∙C-1, and cm∙s-1 or km∙h-1 instead of m∙s-1 where the question warrants this. COPY RIGHT RESERVED PAGE 30 PLEASE TURN OVER

Physical Sciences/Grade 11 30 DBE/2015 Examination Guidelines 5.3 General 5.3.1 If one answer or calculation is required, but two are given by the candidate, only the first one will be marked, irrespective of which one is correct. If two answers are required, only the first two will be marked, etc. 5.3.2 For marking purposes, alternative symbols (s, u, t, etc.) will also be accepted. 5.3.3 Separate compound units with a multiplication dot, not a full stop, e.g. m·s-1. For marking purposes, m.s-1 and m/s will also be accepted. 5.4 Positive marking Positive marking regarding calculations will be followed in the following cases: 5.4.1 Subquestion to subquestion: When a certain variable is incorrectly calculated in one subquestion (e.g. 3.1) and needs to be substituted into another subquestion (3.2 or 3.3), full marks are to be awarded for the subsequent subquestions. 5.4.2 A multistep question in a subquestion: If the candidate has to calculate, for example, current in the first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. 5.5 Negative marking Normally an incorrect answer cannot be correctly motivated if based on a conceptual mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the answer given to QUESTION 3.1, and QUESTION 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for, for example, QUESTION 3.1 is based on a calculation, the motivation for the incorrect answer in QUESTION 3.2 should be considered. 6. MARKING GUIDELINES: PAPER 2 6.1 Calculations 6.1.1 Marks will be awarded for: correct formula, correct substitution, correct answer with unit. 6.1.2 No marks will be awarded if an incorrect or inappropriate formula is used, even though there may be relevant symbols and applicable substitutions. 6.1.3 When an error is made during substitution into a correct formula, a mark will be awarded for the correct formula and for the correct substitutions, but no further marks will be given. 6.1.4 If no formula is given, but all substitutions are correct, the candidate will forfeit one mark. Example: No Kc expression, correct substitution: ( 2) 2 2 Kc = =2 ( ) (2)(1) 3 3 COPY RIGHT RESERVED PAGE 31 PLEASE TURN OVER

Physical Sciences/Grade 11 31 DBE/2015 Examination Guidelines 6.1.5 Marks are only awarded for a formula if a calculation has been attempted, i.e. substitutions have been made or a numerical answer has been given. 6.1.6 Marks can only be allocated for substitutions when values are substituted into formulae and not when listed before a calculation starts. 6.1.7 The final answer to all calculations, when not specified in the question, must be rounded off to a minimum of TWO decimal places. 6.1.8 If a final answer to a calculation is correct, full marks will not automatically be awarded. Markers will always ensure that the correct/appropriate formula is used and that workings, including substitutions, are correct. 6.1.9 Mathematical manipulations and change of subject of appropriate formulae carry no marks, but if a candidate starts off with the correct formula and then changes the subject of the formula incorrectly, marks will be awarded for the formula and the correct substitutions. The mark for the incorrect numerical answer is forfeited. Example: CORRECT ANSWER (1) POSSIBLE ANSWER (2) POSSIBLE m m n m m n  n  m  n  n  M M M M M m 52 0,01 M = 0,52 g  0,01 =  0,01 =  = m  52 m 52 n m = 0,52 g  m = 5 200 g  = 0,002 g 52  = 0,01   = 5 200  g (4) (2) (0) (3) (2) 6.2 Units 6.2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question. 6.2.2 Units are only required in the final answer to a calculation. 6.2.3 Marks are only awarded for an answer and not for a unit per se. Candidates will therefore forfeit the mark allocated for the answer in each of the following situations: - Correct answer + wrong unit - Wrong answer + correct unit - Correct answer + no unit 6.2.4 Separate compound units with a multiplication dot, not a full stop, for example mol·dm-3. Accept mol.dm-3 (or mol/dm3) for marking purposes. COPY RIGHT RESERVED PAGE 32 PLEASE TURN OVER

Physical Sciences/Grade 11 32 DBE/2015 Examination Guidelines 6.3 General 6.3.1 If one answer or calculation is required, but two are given by the candidate, only the first one will be marked, irrespective of which one is correct. If two answers are required, only the first two will be marked, etc. 6.3.2 When a chemical FORMULA is asked, and the NAME is given as answer, the candidate forfeits the marks. The same rule applies when the NAME is asked and the FORMULA is given. 6.3.3 When redox half-reactions are to be written, the correct arrow should be used. If the equation H2S → S + 2H+ + 2e- (22 ) is the correct answer, the marks must be given as follows: H2S ⇌ S + 2H+ + 2e- ( 12 ) H2S ← S + 2 H+ + 2e- (02) S + 2H+ + 2e- ← H2S ( 22 ) S + 2H+ + 2e- ⇌ H2S (02) 6.3.4 When candidates are required to give an explanation involving the relative strength of oxidising and reducing agents, the following is not accepted:  Stating the position of a substance on Table 4 only (e.g. Cu is above Mg).  Using relative reactivity only (e.g. Mg is more reactive than Cu).  The correct answer would be for instance: Mg is a stronger reducing agent than Cu, and therefore Mg will be able to reduce Cu 2+ ions to Cu. The answer can also be given in terms of the relative strength as electron acceptors and donors. 6.3.5 One mark is forfeited when the charge of an ion is omitted per equation (not for the charge on an electron). 6.3.6 The error-carrying principle does not apply to chemical equations or half-reactions. For example, if a learner writes the wrong oxidation/reduction half-reaction in the subquestion and carries the answer to another subquestion (balancing of equations or calculation of E θcell ), then the learner will not be credited for this substitution. 6.3.7 In the structural formula of an organic molecule all hydrogen atoms must be shown. Marks will be deducted if hydrogen atoms are omitted. 6.3.8 When a structural formula is required, marks will be deducted if the learner writes the condensed formula. 6.3.9 When a IUPAC name is asked and the candidate omits the hyphen(s) (e.g. instead of pent-1-ene or 1-pentene the candidate writes pent 1 ene or 1 pentene), marks will be forfeited. COPY RIGHT RESERVED PAGE 33 PLEASE TURN OVER

Physical Sciences/Grade 11 33 DBE/2015 Examination Guidelines 6.3.10 When a chemical reaction is asked, marks are awarded for correct reactants, correct products and correct balancing. If only a reactant(s) followed by an arrow, or only a product(s) preceded by an arrow, is/are written, marks may be awarded for the reactant(s) or product(s). If only a reactant(s) or only a product(s) is/are written, without an arrow, no marks are awarded for the reactant(s) or product(s). Examples: N2 + 3H2 → 2NH3  bal.  3 3 N2 + H2 → 1 3 → NH3  1 3 N2 + H2 0 3 NH3 0 3 6.4 Positive marking Positive marking regarding calculations will be followed in the following cases: 6.4.1 Subquestion to subquestion: When a certain variable is calculated in one subquestion (e.g. QUESTION 3.1) and needs to be substituted in another (QUESTION 3.2 or QUESTION 3.3), e.g. if the answer for QUESTION 3.1 is incorrect and is substituted correctly in QUESTION 3.2 or QUESTION 3.3, full marks are to be awarded for the subsequent subquestions. 6.4.2 A multistep question in a subquestion: If the candidate has to calculate, for example, current in the first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. 6.5 Negative marking Normally an incorrect answer cannot be correctly motivated if based on a conceptual mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the answer given to QUESTION 3.1, and QUESTION 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for, for example, QUESTION 3.1 is based on a calculation, the motivation for the incorrect answer in QUESTION 3.2 could be considered. 7. CONCLUSION This Examination Guidelines document is meant to articulate the assessment aspirations espoused in the CAPS document. It is therefore not a substitute for the CAPS document, which educators should teach to. Qualitative curriculum coverage as enunciated in the CAPS cannot be over-emphasised. COPY RIGHT RESERVED PAGE 34 PLEASE TURN OVER

REVISION QUESTIONS PHYSICAL SCIENCES PAPER 1 PHYSICS COPY RIGHT RESERVED PAGE 35 PLEASE TURN OVER

DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity g 9,8 m·s-2 Swaartekragversnelling Gravitational constant G 6,67 x 10-11 N⋅m2⋅kg-2 Swaartekragkonstante Radius of Earth RE 6,38 x 106 m Straal van Aarde Coulomb's constant k 9,0 x 109 N⋅m2·C-2 Coulomb se konstante Speed of light in a vacuum c 3,0 x 108 m·s-1 Spoed van lig in 'n vakuum Charge on electron e -1,6 x 10-19 C Lading op elektron Electron mass me 9,11 x 10-31 kg Elektronmassa Mass of the earth M 5,98 x 1024 kg Massa van die Aarde TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING v f = v i + a ∆t ∆x = v i ∆t + 21 a∆t 2  v + vi  2 2 v f = v i + 2a∆x ∆x =  f  ∆t  2  FORCE/KRAG Fnet = ma w = mg Gm1m 2 f s (max) F= µs = r2 N f µk = k N COPY RIGHT RESERVED PAGE 36 PLEASE TURN OVER

WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG 1 v=fλ T= f ni sin θi = nr sin θr c n= v ELECTROSTATICS/ELEKTROSTATIKA kQ 1Q 2 F F= (k = 9,0 x 109 N⋅m2·C-2) E= r2 q kQ Q E= 2 (k = 9,0 x 109 N⋅m2·C-2) n= r e ELECTROMAGNETISM/ELEKTROMAGNETISME ∆Φ ε=−N Φ = BA cos θ ∆t ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE Q V I= R= ∆t I 1 1 1 1 = + + + ... R = r1 + r2 + r3 + ... R r1 r2 r3 W = Vq W P= Δt W = VI ∆ t P = VI W= I2R ∆ t P = I2R V 2 Δt W= V2 R P= R COPY RIGHT RESERVED PAGE 37 PLEASE TURN OVER

VECTORS MULTIPLE CHOICE QUESTIONS QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A–D) next to the question number (1.1–1.10) in the ANSWER BOOK. 1.1 NOV 2014 Two vectors, P and Q, act simultaneously at point O as shown in the diagram below. The magnitude of Q is greater than the magnitude of P. y (2) Q O x P Which ONE of the following could represent the resultant R of the two vectors? y y R A O x B R O x y y R C O D x O x R (2) COPY RIGHT RESERVED PAGE 38 PLEASE TURN OVER

1.2 NOV 2014 Forces P, Q, R and S all have the same magnitude. The forces act at the same point in the directions shown in the diagram. y Q 45o P o 30 60 o x 45o R S Which ONE of the following combinations CORRECTLY shows the vectors having the greatest magnitude for the x-component and for the y-component? x-component y-component A Vector P Vector R B Vector P Vector Q C Vector R Vector Q D Vector R Vector S (2) 1.1 NOV 2015 Vector P and vector –P are acting on a common point O. The angle between the two vectors is ... A 0o B 90o C 180o D 270o (2) 1.2 NOV 2015 The statements below refer to scalars and vectors: (i) Vectors can be added together but scalars cannot. (ii) A scalar quantity can be associated with direction. (iii) A vector quantity is always associated with direction. Which of the statements above are TRUE? A (i) and (ii) only B (ii) and (iii) only C (i) only D (iii) only (2) COPY RIGHT RESERVED PAGE 39 PLEASE TURN OVER

1.1 NOV 2016 Four forces act on a point, as indicated in the diagram. 4N 1N 3N 2N The magnitudes of the components of the resultant (net) force in the horizontal (F X ) and vertical (F Y ) directions are … A FX = 3 N and FY = 6 N. B FX = 1 N and FY = 4 N. C FX = 2 N and FY = 2 N. D FX = 4 N and FY = 2 N. (2) 1.1 NOV 2017 Which ONE of the following pairs of physical quantities is vector quantities? A Force and distance B Velocity and speed C Charge and electric field D Electric field and force (2) COPY RIGHT RESERVED PAGE 40 PLEASE TURN OVER

VECTORS LONG QUESTIONS QUESTION 2 NOV 2014 The diagram below shows a rope and pulley arrangement of a device being used to lift an 800 N object. Assume that the ropes are light and inextensible and also that the pulley is light and frictionless. T2 P T1 140o 120o 800 kg Determine the: 2.1 Magnitudes of the tensions T 1 and T 2 (7) 2.2 Magnitude and direction of the reaction force at pulley P (4) [11] QUESTION 2 NOV 2015 Three forces, P, Q and R, of magnitudes 500 N, 200 N and 300 N respectively, act on a point O in the directions shown in the diagram below. The forces are NOT drawn to scale. P = 500 N y 10ο Q = 200 N 30ο O x 60ο R = 300 N 2.1 Refer to the information in the diagram above and give a reason why forces P, Q and R are classified as vectors. (2) 2.2 Determine the magnitude and direction of the resultant force, either by CALCULATION or by ACCURATE CONSTRUCTION AND MEASUREMENT. (Use scale 10 mm = 50 N.) (8) [10] COPY RIGHT RESERVED PAGE 41 PLEASE TURN OVER

QUESTION 2 NOV 2016 heavy object A heavy objectis islifted lifted using using two two ropesropes and two and two pulleys, pulleys, as as in shown shown the in the diagram diagrambelow. The two pulleys are a distance x apart. The force FA, in rope A, is 730 N and the force FB, in rope B, is 1 440 N. Rope A makes an angle of 70° with the horizontal and rope B makes an angle of 10° with the vertical. x FA = 730 N FB = 1 440 N 10° 70° Object 2.1 Define the term resultant vector. (2) 2.2 Explain why the vector diagram of force F A , force F B and the weight will NOT be a closed vector diagram. (2) 2.3 Calculate the: 2.3.1 Vertical component of F A (2) 2.3.2 Horizontal component of F A (2) 2.4 Calculate the maximum weight that force F A and force F B will be able to lift from the ground. Show ALL calculations. (4) 2.5 Explain why the rope and pulley system will be less effective if the distance x between the pulleys is increased. (2) [14] QUESTION 2 NOV 2017 Block A, which is at rest on a horizontal rough surface, is used as an anchor to hold block B, with a mass of 56 kg, in the air at a certain height above the ground. The two blocks are connected with rope R, which makes an angle of 35° with the vertical. Block B is suspended from the ceiling with cable C. Refer to the diagram below. θ C A 35° R 56 kg B Block A experiences a frictional force of magnitude 200 N. The system is stationary. 2.1 Define the term resultant vector. (2) 2.2 What is the magnitude of the resultant force acting on block B? (1) 2.3 What is the magnitude of the resultant force acting on block B? (3) 2.4 Determine the horizontal component of the force in rope R. (1) 2.5 Calculate the vertical component of the force in cable C. (4) 2.6 Calculate the angle θ between the cable and the ceiling. (2) [13] COPY RIGHT RESERVED PAGE 42 PLEASE TURN OVER

NEWTON'S LAWS MULTIPLE CHOICE QUESTIONS 1.3 NOV 2014 If the resultant (net) force acting on an object is zero, the object … A slows down. B accelerates uniformly. C changes its direction of motion. D continues moving with constant velocity. (2) 1.4 NOV 2014 A graph of the gravitational force versus the mass of an object is shown below. Gravitational force 0 Mass of object Which ONE of the following CORRECTLY represents the slope of the graph? A Velocity of the object B Weight of the object C Acceleration due to gravity (g) D Universal gravitation constant (G) (2) 1.3 NOV 2015 A net force, F, is applied on an object of mass m kg and causes an acceleration of a m∙s-2. When the net force, F, on the same object is doubled, the resulting acceleration, in m∙s-2, will be ... A a 3 B a 2 C 2a D 3a (2) COPY RIGHT RESERVED PAGE 43 PLEASE TURN OVER

1.4 NOV 2015 A person stands on a bathroom scale in a stationary elevator. The reading on the scale is 490 N. When the elevator is in motion, the reading on the scale changes to 470 N. Which ONE of the following combinations best describes the DIRECTION OF THE MOTION and the DIRECTION OF THE ACCELERATION of the elevator during the motion? DIRECTION OF THE DIRECTION OF THE MOTION ACCELERATION A Upwards Upwards B Downwards Downwards C Upwards Downwards and then upwards D Downwards Upwards and then downwards (2) 1.2 NOV 2016 A lawnmower is pushed across the ground with a force of F at an angle of θ with the horizontal. The weight of the lawnmower is w. N F θ w The normal force, in N, on the lawnmower is … A w B w + FY C w - FY D w + FX (2) COPY RIGHT RESERVED PAGE 44 PLEASE TURN OVER

1.3 NOV 2016 A car is travelling along a road. The driver has his seat belt on. The driver sees an obstruction in the road ahead and suddenly applies the brakes. Obstruction An action-reaction pair is the force of the seat belt on the driver and the force of the … A driver on the seat. B wheels on the road. C driver on the seat belt. D seat belt on the seat. (2) 1.4 NOV 2016 Planet X has a mass twice the mass of the Earth and a radius one and a half times that of the Earth. If the acceleration due to gravity on the Earth is g, then the gravitational acceleration on planet X will be … 8 A g 9 9 B g 8 4 C g 3 3 D g 4 (2) 1.3 NOV 2017 A block with a mass of 4 kg is pulled upwards along a frictionless slope, inclined at an angle θ, with a force F, as shown in the sketch below. 4 kg F θ Which ONE of the following equations can be used to calculate the magnitude of the normal force (N)? A N = (4)(9,8)sinθ B N = F - (4)(9,8)cosθ C N = F + (4)(9,8)cosθ D N = (4)(9,8)cosθ (2) COPY RIGHT RESERVED PAGE 45 PLEASE TURN OVER

1.4 NOV 2107 A satellite orbits Earth at a height where the gravitational force is a quarter (¼) of the force it experiences on the surface of the Earth. If the radius of Earth is R, the height of the satellite ABOVE THE SURFACE of Earth is … A 4R B 2R C R D ½R (2) COPY RIGHT RESERVED PAGE 46 PLEASE TURN OVER

NEWTON'S LAWS LONG QUESTIONS QUESTION 3 NOV 2014 A block Q of mass 70 kg is at rest on a table. It is connected to block P by means of two light inextensible strings knotted at S. A third string is arranged in such a way that the string connecting block Q is horizontal as shown in the diagram below. The coefficient of static friction between block Q and the surface of the table is 0,25. The knot S is in equilibrium. T1 T2 S 35o Q ● P The tension in the string connecting block Q is T 2 and that for the string that pulls at 35o is T 1 as shown in the diagram. 3.1 Define the term static frictional force in words. (2) 3.2 Explain what is meant by the knot S is in equilibrium. (2) 3.3 Draw a labelled free-body diagram to show all the forces acting on: 3.3.1 The knot at S (3) 3.3.2 Block Q (4) 3.4 Calculate the maximum weight of block P for which block Q will just begin to slip. (7) [18] QUESTION 4 NOV 2014 A block of mass 8 kg resting on a rough horizontal table is connected by a light inextensible string which passes over a light frictionless pulley to another block of mass 5 kg. The 5 kg block hangs vertically as shown in the diagram below. A 15 N force is applied to the 8 kg block at an angle of 30o to the horizontal, causing the block to slide to the left. 15 N 30o 8 kg 5 kg COPY RIGHT RESERVED PAGE 47 PLEASE TURN OVER

The coefficient of kinetic friction between the 8 kg block and the surface of the table is 0,25. Ignore the effects of air friction. 4.1 Draw a free-body diagram showing ALL the forces acting on the 8 kg block. (5) 4.2 Write down Newton's second law of motion in words. (2) 4.3 Calculate the magnitude of the: Normal force acting on the 8 kg block (3) 4.4 Tension in the string connecting the two blocks (6) [16] QUESTION 3 NOV 2015 A constant force, F, pulls a 50 kg block at a constant speed over a rough horizontal surface, AB, as shown in the diagram below. The coefficient of kinetic friction (μ k ) between the block and the surface is 0,4. F 50 kg 20o A B 3.1 Draw a labelled free-body diagram showing ALL the forces acting on the block. (4) 3.2 State Newton's first law of motion in words. (2) Calculate the magnitude of the: 3.3 Force F (6) 3.4 Normal force (2) 3.5 Frictional force (2) The same constant force, F, is now applied to the block over the same rough horizontal surface as before, but in the direction shown below. F 20o 50 kg A B 3.6 Give a reason why the block will move slower than before. (2) [18] COPY RIGHT RESERVED PAGE 48 PLEASE TURN OVER

QUESTION 4 NOV 2105 4.1 In the diagram below, a 2 kg block is connected to a 1 kg block by means of a light inextensible string. The blocks are pulled up an inclined plane, which forms an angle of 28o to the horizontal. The sides of the blocks which touch the inclined plane have the same area. 30 N 1 kg 2 kg 28o The kinetic frictional force between the 1 kg block and the inclined plane is 4 N while the kinetic frictional force between the 2 kg block and the inclined plane is 8 N. 4.1.1 Draw a labelled free-body diagram showing ALL the forces acting on the 1 kg block. (5) 4.1.2 State Newton's second law of motion in words. (2) 4.1.3 Calculate the magnitude of the tension in the string connecting the blocks. (5) The two blocks are interchanged so that the SAME 30 N force is now acting on the 2 kg block along the inclined plane, as shown in the diagram below. 30 N 2 kg 1 kg 28o 4.1.4 How will the acceleration of the system change? Write down only INCREASE, DECREASE or REMAIN THE SAME. (1) 4.1.5 How will the tension in the string connecting the two blocks change? Write down only INCREASE, DECREASE or REMAIN THE SAME. (1) 4.2 In the diagram below, a 1 kg mass and a 2 kg mass are connected by an inextensible string of negligible mass. The string is passed over a light frictionless pulley so that the masses hang down as shown. Initially the system is held stationary. 4.2.1 Draw a labelled free-body diagram showing ALL the forces acting on the 2 kg mass. (2) 1 kg 4.2.2 Calculate the time it will take the 1 kg mass to move 2 kg a distance of 1 m when the system is released. (7) [23] COPY RIGHT RESERVED PAGE 49 PLEASE TURN OVER

QUESTION 3 NOV 2016 A hovering rescue helicopter has a container of supplies, with a weight of 1 960 N, hanging from a cable. The tension in the cable is 2 100 N. container 3.1 State Newton's First Law of Motion in words. (2) 3.2 Draw a labelled force-diagram of ALL the forces acting on the container. (3) 3.3 Why does the container remain stationary despite the tension being greater than the weight? (2) Now the winch inside the helicopter starts to pull the container upwards with an acceleration of 0,13 m.s-2, while the helicopter remains in its position. 3.4 Calculate the mass of the container. (2) 3.5 Calculate the magnitude of the tension in the cable while the container is being pulled upwards. (4) After an acceleration of a few metres the container is pulled up at a constant velocity of 0,8 m.s-1. 3.6 What will be the magnitude of the tension in the cable while the container moves upwards at a constant velocity? (1) [14] QUESTION 4 NOV 2016 Two blocks, of mass 8 kg and 4 kg respectively, are joined with an inelastic string of negligible mass. The string runs over a frictionless pulley. The 8 kg block is on a horizontal surface while the 4 kg block is on an inclined plane of 40° with the horizontal. The coefficient of kinetic friction for both blocks is 0,2. The 4 kg block accelerates down the slope. 8 kg 4 kg 40° 4.1 State Newton's Second Law of Motion in words. (2) 4.2 Draw a labelled free-body diagram of ALL the forces acting on the 4 kg block. (4) 4.3 Calculate the frictional force between the surface and the 4 kg block. (4) 4.4 Calculate the magnitude of the acceleration of the system. (6) 4.5 How will the acceleration compare if the positions of the 8 kg block and 4 kg block are switched? Choose from GREATER THAN, LESS THAN or THE SAME. Explain the answer. (4) [20] COPY RIGHT RESERVED PAGE 50 PLEASE TURN OVER

QUESTION 3 NOV 2017 Learners investigate the relationship between net force and acceleration by pulling a trolley across a surface which is slightly inclined to compensate for friction. The trolley is connected to different masses by a string of negligible mass. The string passes over a frictionless pulley. Refer to the diagram below. trolley tape ticker-timer mass Ticker-tape attached to the trolley passes through the ticker-timer. The acceleration of the trolley is determined by analysing the ticker-tape. The results of the net force produced by the different masses and the acceleration of the trolley were recorded in the table below. NET FORCE (N) a (m∙s-2) 0,3 0,36 0,6 0,73 0,9 1,09 1,2 1,45 3.1 Write down a hypothesis for this experiment. (2) 3.2.1 Identify the independent variable. (1) 3.2.2 Identify the controlled variable. (1) 3.3 Use the graph paper on the ANSWER SHEET and draw a graph of the acceleration versus net force. (4) 3.4 Calculate the gradient of the graph. (3) 3.5 Use the gradient of the graph calculated in QUESTION 3.4 to determine the mass of the trolley. (2) [13] QUESTION 4 NOV 2017 A tow truck pulls a car along a gravel road. The force applied by the engine of the tow truck is 9 000 N. The mass of the tow truck is 1 300 kg and the mass of the car is 950 kg. The vehicles are connected to each other by an inelastic tow bar of negligible mass. See the diagram below. Tow truck 1 300 kg Car 950 kg 9 000 N T The tow truck and car move at a CONSTANT VELOCITY. 4.1 Define the term frictional force. (2) COPY RIGHT RESERVED PAGE 51 PLEASE TURN OVER

4.2 NAME AND STATE the law that explains why the force exerted by the tow truck on the car is the same as the force exerted by the car on the tow truck. (3) Draw a labelled free-body diagram indicating all the forces acting on the tow truck. (5) 4.4 If the coefficient of kinetic friction between the tow-truck tyres and the road surface is 0,45, calculate the: 4.3 4.4.1 Magnitude of the tension in the tow bar (5) 4.4.2 Coefficient of kinetic friction between the CAR tyres and the road surface (5) Suddenly the tow bar between the car and the tow truck disconnects and the car comes loose. 4.5 Using a relevant law of motion, explain why the car continues moving forward for a short distance. (3) 4.6 Calculate the acceleration of the car as it comes to a stop after a short distance. (3) [26] NEWTON'S LAW OF UNIVERSAL GRAVITATION QUESTION 5 NOV 2014 5.1 Write down Newton's law of universal gravitation in words. (2) An object weighing 140 N on the surface of the earth is moved to a position which is 6,7 x 106 m above the surface of the earth. 5.2 Calculate the percentage by which its weight will change. (8) [10] QUESTION 5 NOV 2015 Gravitational force exists between the sun and the Earth. 5.1 State Newton's law of Universal Gravitation in words. (2) 5.2 The mass of the sun is 330 000 times greater than that of the Earth. The distance between the centres of the sun and the Earth is 1,38 x 109 m. Calculate the gravitational force that the sun exerts on the Earth. (4) 5.3 How will the gravitational force that the Earth exerts on the Sun compare to the answer to QUESTION 5.2? Write only GREATER THAN, LESS THAN OR EQUAL TO. Give a reason for the answer. (2) [8] COPY RIGHT RESERVED PAGE 52 PLEASE TURN OVER

QUESTION 5 NOV 2016 Two satellites orbiting the Earth are situated on opposite sides of the Earth. Satellite A has a mass of 3 800 kg and Satellite B has a mass of 4 500 kg. Satellite A is at a height of 25 000 km above the surface of the Earth. Earth Satellite A Satellite B 5.1 State Newton's Universal Gravitational Law in words. (2) 5.2 Explain the term weightlessness. (2) 5.3 Calculate the force between the Earth and Satellite A. (4) 5.4 What distance above the surface of the Earth should Satellite B be to experience the same force towards the Earth as Satellite A? Choose from: GREATER THAN, LESS THAN or EQUAL TO the distance above the Earth. Explain how you arrived at the answer. (4) [12] QUESTION 5 NOV 2017 The acceleration due to gravity on planet X is 2,7 m∙s-2. The radius of this planet is a third (⅓) of the radius of Earth. 5.1 Explain the difference between weight and mass. (2) 5.2 Calculate the mass of planet X. (4) 5.3 Determine the factor by which the weight of an object on planet X will differ from the weight of the same object on Earth. (2) [8] COPY RIGHT RESERVED PAGE 53 PLEASE TURN OVER

WAVES, SOUND AND LIGHT MULTIPLE CHOICE QUESTIONS 1.5 NOV 2014 A light wave travels obliquely from air into a glass block and its speed changes. Which ONE of the combinations below CORRECTLY describes the changes in the FREQUENCY of the wave and REFRACTIVE INDEX of the block compared to that of air? FREQUENCY REFRACTIVE INDEX A Remains the same Increases B Remains the same Decreases C Increases Decreases D Decreases Increases (2) 1.6 NOV 2014 Sound waves bend readily around buildings whereas light waves only bend very slightly around buildings. Which ONE of the following statements BEST explains this observation? A Sound waves have much longer wavelengths than light waves. B Sound waves have much shorter wavelengths than light waves. C Sound waves have higher frequencies compared to light waves. D Sound waves have greater amplitudes compared to light waves. (2) 1.5 NOV 2015 A light ray is refracted when it travels from air into water. Which ONE of the following combinations relating to the DIRECTION OF THE REFRACTED RAY and the SPEED OF THE LIGHT RAY is CORRECT? DIRECTION OF THE SPEED OF THE LIGHT RAY REFRACTED RAY A Towards the normal Decreases B Away from the normal Decreases C Towards the normal Increases D Away from the normal Increases (2) COPY RIGHT RESERVED PAGE 54 PLEASE TURN OVER

1.6 NOV 2015 A learner hiding behind an open door of a room can hear people talking outside the room although he cannot see them. The reason for this is that ... A sound has a shorter wavelength than a light wave. B sound has a longer wavelength than a light wave. C light travels in straight lines while sound travels only around corners. D the energy in a sound wave is less than the energy in a light wave. (2) 1.5 NOV 2016 A light ray travels from medium A to medium B. Medium B has a lower refractive index than medium A. The critical angle for medium A in terms of medium B is 42°. Which ONE of the sketches below represents the CORRECT path of the light ray? A B Medium B Medium B 40º Medium A 40º Medium A C D Medium B Medium B Medium A 40º Medium A 40º (2) 1.6 NOV 2016 The ability of a wave to spread out in wave fronts as the wave passes around a sharp edge, is known as … A Snell's law. B total internal C reflection. diffraction. D Huygens' principle. (2) COPY RIGHT RESERVED PAGE 55 PLEASE TURN OVER

1.5 NOV 2017 A light ray passes from glass to air. The angle of incidence is 35°. The critical angle of glass is 38°. The light ray will undergo … A diffraction. B refraction and bend away from the C normal. total internal reflection. D refraction and bend towards the normal. (2) 1.6 NOV 2017 The path of a light ray passing from air through a rectangular Perspex block is shown below. 39° Perspex θ The magnitude of angle θ will be … A less than 39°. B equal to 39°. C equal to 51°. D equal to 90°. (2) 1.7 NOV 2017 A monochromatic red light ray passes through a single slit of width d. The diffraction pattern is projected on a screen. The red light is then replaced with monochromatic blue light and passed through the same single slit. The degree of diffraction will … A increase because the degree of diffraction is directly proportional to wavelength. B decrease because blue light has a shorter wavelength than red light. C increase because blue light has a longer wavelength than red light. D decrease because the degree of diffraction is inversely proportional to wavelength. (2) COPY RIGHT RESERVED PAGE 56 PLEASE TURN OVER

WAVES, SOUND AND LIGHT LONG QUESTIONS : REFRACTION QUESTION 6 NOV 2014 Learners investigate how the path of a light ray incident on an air-glass boundary changes as it enters the glass medium. Their results are shown in the table below. o o angle i angle r sin i sin r 15 10 0,259 0,174 25 16 0,423 0,276 45 28 0,707 0,469 55 33 0,819 0,545 60 35 0,866 0,574 70 39 0,940 0,629 6.1 For this investigation, write down the: 6.1.1 Dependent variable (1) 6.1.2 Independent variable (1) 6.1.3 Constant (control) variable (1) 6.2 Draw an appropriate graph of the data in the table and use it to obtain the refractive index of the glass material. USE THE GRAPH PAPER ATTACHED TO YOUR QUESTION PAPER TO ANSWER THIS QUESTION. (8) 6.3 Use the result in QUESTION 6.2 to calculate the speed of light through the glass material. (3) [14] QUESTION 7 NOV 2014 In the diagram below (not to scale), a ray of light, PO, is travelling from flint glass towards the boundary with crown glass. The angle of incidence of ray PO at the boundary between the two surfaces (∠ PON/) is 21o. Crown glass (n = 1,52) N O 21o Flint glass Q (n = 1,66) P N/ 7.1 Write down Snell's law in words. (2) COPY RIGHT RESERVED PAGE 57 PLEASE TURN OVER

The refractive indices of crown glass and flint glass are 1,52 and 1,66 respectively as shown in the diagram above. 7.2 Calculate the critical angle for the boundary between the two glass materials. (3) Copy the diagram into your ANSWER BOOK. 7.3 On your diagram draw a ray to show what happens to light ray PO at the boundary between the two glass surfaces. Label the ray OX. (1) 7.4 Ray QO is incident at the boundary at 40o. Draw a ray to show what happens to light ray QO at the boundary between the two glass surfaces. Label the ray OY. Include the angle ∠ N/OY on your drawing. (2) 7.5 How does the speed of light in the crown glass compare to that in the flint glass? Write down only GREATER THAN, LESS THAN or EQUAL TO. (1) [9] QUESTION 6 NOV 2015 The table below shows accurate values for the refractive indices (relative to air) of some optical media obtained during refraction experiments. The optical densities of the media increase from water to diamond. MEDIUM REFRACTIVE INDEX Water 1,33 Crown glass 1,52 Cubic zirconium 2,20 Diamond 2,42 6.1 What important deduction about the relationship between a medium and its refractive index can be made from the information above? (2) 6.2 Define the term refraction. (2) The diagram below shows a light ray moving from medium A into water. 35° water medium A 20,28° 6.3 Provide a suitable calculation to identify medium A. (5) Light travels from water into crown glass. 6.4 Will the refractive index of the crown glass relative to water be GREATER THAN, LESS THAN or EQUAL TO the refractive index of crown glass in the table above? (1) [10] COPY RIGHT RESERVED PAGE 58 PLEASE TURN OVER

QUESTION 7 NOV 2015 The diagram below shows a light ray travelling through an optical fibre. The fibre is made from two different glass materials known as the core glass and the outer cladding glass. These two types of glass materials have different refractive indices. Core: n1 i Cladding: n2 7.1 State the TWO conditions necessary for total internal reflection to occur. (4) 7.2 Which part of the optical fibre will have a higher refractive index? Write down only CORE or CLADDING. (1) The refractive index of the core glass is 1,5. 7.3 Calculate the critical angle for the boundary between the two glass materials. (3) 7.4 State TWO advantages of optical fibre cables over copper cables when used in telecommunication. (2) [10] QUESTION 6 NOV 2016 Olive oil floats on water because its density is less than the density of water. However, the nair = 1 Air refractive index of olive oil is 1,47 while the refractive index of water is 1,33. Olive oil nolive oil = 1,47 Water nwater = 1,33 37° 6.1 Define the term refractive index. (2) 6.2 Calculate the speed of light through olive oil. (3) 6.3 If light travels through water and hits the water-olive oil boundary at an angle of 37°, calculate the angle of refraction of the light in olive oil. (4) 6.4 What does the refractive index of oil tell us about the optical density of olive oil compared to the optical density of water? (2) 6.5 6.5.1 If the light ray exits the oil into the air, will it refract TOWARDS or AWAY from the normal? (1) 6.5.2 Give a reason for the answer to QUESTION 6.5.1. (2) 6.6 6.6.1 At which surface is total internal reflection most likely to occur? Choose from oil and air or water and oil. (1) 6.6.2 Give a reason for the answer to QUESTION 6.6.1. (2) 6.7 Which other wave characteristic, WAVELENGTH or FREQUENCY, will change together with the speed, when a wave moves from one medium to another? (1) [18] COPY RIGHT RESERVED PAGE 59 PLEASE TURN OVER

QUESTION 6 NOV 2017 Experiments are performed to compare the refractive indices of different materials. In one experiment a light ray passes from air to material A and the angles of incidence and refraction are measured. The refractive index for air is 1. The graph below was drawn using the results of material A. sin θr A 0,37 0 sin θi 0,56 6.1 Define the term angle of incidence. (2) 6.2 Calculate the refractive index of material A using the data in the graph. C (3) 6.3 Calculate the refractive index of material A using the data in the graph. (3) 6.4 If material A is replaced by material B, the angle of refraction is 31° when the angle of incidence is 40°. 6.4.1 Calculate the refractive index of material B. (4) 6.4.2 Redraw the graph of material A, and on the same set of axis, draw the graph you expect for material B. Label the graphs of material A and material B clearly. (2) 6.5 Total internal reflection occurs when a light ray passes from material A to material B. The critical angle of material A is 49°. 6.5.1 Which range of angles will make it possible for total internal reflection to occur? (2) 6.5.2 What OTHER condition is necessary for total internal reflection to take place? (2) [18] COPY RIGHT RESERVED PAGE 60 PLEASE TURN OVER

LONG QUESTIONS : DIFFRACTION QUESTION 8 NOV 2014 Diffraction provides evidence that light can behave as a wave. 8.1 Define the term diffraction in words. (2) In the diagram below a plain wave front of light of wavelength 6 x 10-7 m approaches a narrow opening. Diffraction effects are observed on a screen placed some distance from the slit as shown in the diagram below. slit screen 8.2 Describe the pattern observed on the screen. (2) 8.3 Two important principles explain the diffraction pattern. (2) 8.4 Write down the NAME of each of these principles. The width of the slit (opening) is increased slightly. Describe how this change will affect the: 8.4.1 Diffraction pattern observed (1) 8.4.2 Brightness of the diffraction pattern observed 8.5 The width of the slit is kept constant but light of wavelength 4 x 10-7 m is now (1) used. Describe how this change will affect the diffraction pattern obtained. (1) [9] QUESTION 8 NOV 2015 8.1 All types of waves undergo diffraction. 8.1.1 Define the term diffraction. (2) 8.1.2 State the condition necessary for diffraction to occur in a single slit. (2) 8.2 Light of wavelength 460 nm screen passes through a single slit of width 7 x 10-6 m. A diffraction pattern is observed on a screen, as shown in the diagram below. slit X COPY RIGHT RESERVED PAGE 61 PLEASE TURN OVER

8.2.1 Describe the pattern observed on the screen. (2) 8.2.2 Describe how each of the following changes will affect the width (broadness) of portion X in the diagram above: (a) Light of wavelength 470 nm is used while the slit width is kept constant (1) -7 (b) The slit width is changed to 7 x 10 m while the original wavelength is left unchanged (1) [8] QUESTION 7 NOV 2016 Two learners investigate the effect of the slit width on the degree of diffraction. They use a green light with a wavelength 520 nm. They set up an experiment, as shown below, and measure the position of the first dark line from the centre of the bright, central band when changing the slit width. Screen Barrier Light source Position of first dark line measured from the centre of the bright, central band The following results are obtained: SLIT WIDTH (mm) POSITION OF FIRST DARK LINE FROM THE CENTRE (mm) 5,3 x 10-5 19,6 4,9 x 10-5 21,2 4,2 x 10-5 24,8 7.1 State Huygens' principle in words. (2) 7.2 For this experiment, write down: 7.2.1 The independent variable (1) 7.2.2 The dependant variable (1) 7.2.3 ONE controlled variable (1) 7.3 Give a conclusion for the results obtained in this experiment. (2) 7.4 7.4.1 How will the pattern on the screen change if red light, with a wavelength of 660 nm, is used instead of green light? (2) 7.4.2 Explain the answer to QUESTION 7.4.1. (1) [10] COPY RIGHT RESERVED PAGE 62 PLEASE TURN OVER

QUESTION 7 NOV 2017 An experiment is set up, as shown below, to investigate the effect of slit width on the degree of diffraction. Distance y on the diagram represents the distance between the screen and the single slit. Distance x on the diagram represents the width of the central bright band. x screen y single slit Light source 7.1 Write down an investigative question for this experiment. (2) 7.2 State Huygens' principle in words. (2) 7.3 How will distance x be affected if the slit width is increased? Choose from INCREASE, DECREASE or REMAIN THE SAME. (1) 7.4 Explain the answer to QUESTION 7.3. (2) 7.5 How will distance x be affected if distance y is increased? Choose from INCREASE, DECREASE or REMAIN THE SAME. (1) [8] COPY RIGHT RESERVED PAGE 63 PLEASE TURN OVER

ELECTROSTATICS MULTIPLE CHOICE QUESTIONS 1.7 NOV 2014 The electrostatic force between two charged spheres, a distance r apart, is F. When the charge on each sphere is doubled and the distance between the spheres is also doubled, the force between the spheres will now be … A ½F B F C 2F D 4F (2) 1.8 NOV 2014 The electrostatic force F between two charged particles is positive. Which ONE of the following is CORRECT? A The magnitudes of the two charges are equal. B One charge is positive while the other is negative. C The electrostatic force between the charges is attractive. D The electrostatic force between the charges is repulsive. (2) 1.7 NOV 2015 The leaves of the electroscope in the diagram below are positively charged. When an object is brought close to the plate, the leaves diverge more. We can therefore conclude that the object … A is positively charged. B is negatively charged. C is not charged at all. D releases positive charges. (2) COPY RIGHT RESERVED PAGE 64 PLEASE TURN OVER

1.8 NOV 2015 Two charged objects repel each other with a force F when they are separated 1 by a distance d. The distance between the charges is reduced to d. 3 The new force, in terms of F, will now be ... A F B 3F C 6F D 9F (2) 1.7 NOV 2016 Which ONE of the graphs below represents the correct relationship between force F on a charge and the electric field E? A B F (N) F (N) E (N.C-1) E (N.C-1) F (N) F (N) C D E (N.C-1) E (N.C-1) (2) 1.8 NOV 2016 A negative charge Q is placed at a distance of 5d from another charge R. Q P 2d R 5d If the net electric field at point P, at a distance of 2d from R, is ZERO, which ONE of the following combinations concerning the ratio of the charges Q and R and the charge on R, is CORRECT? RATIO OF THE CHARGES CHARGE ON R Q:R A 4:9 Positive B 3:2 Negative C 5:2 Positive D 9:4 Negative (2) COPY RIGHT RESERVED PAGE 65 PLEASE TURN OVER

1.8 NOV 2017 A negative charge of 1 µC, which is free to move, is placed at a distance 2r from a positive charge of 4 µC. +4 µC -1 µC + - l l l r 2r Which ONE of the following statements regarding the -1 µC charge, when it is at distance r, is CORRECT? The electrostatic force experienced by the -1 µC charge will … A remain the same. B be halved. C be doubled. D increase four times. (2) COPY RIGHT RESERVED PAGE 66 PLEASE TURN OVER

ELECTROSTATICS LONG QUESTIONS QUESTION 9 NOV 2014 Two point charges of +2 μC and +3 μC are placed a distance of 0,2 m apart. P is a point on the line joining the two charges, a distance of x m from the 3 μC charge such that the net electric field at point P is zero. x +2 μC • P +3 μC 0,2 m 9.1 Define the term electric field at a point in words. (2) 9.2 Calculate the distance x. (7) A -4 μC charge is now placed a distance of 0,1 m from the +3 μC charge as shown in the sketch below. 0,2 m +2 μC +3 μC 0,1 m -4 μC 9.3 Calculate the magnitude of the electrostatic force experienced by the +3 μC charge due to the presence of the other two charges. (5) [14] QUESTION 9 NOV 2015 9.1 Two identical metal spheres, P and Q, separated by a distance d, are suspended from an insulated rigid wooden bar, as shown in the diagram below. The charge on the spheres are –4 μC and +8 μC respectively. d –4 μC +8 μC P Q 9.1.1 Sphere Q experiences an electrostatic force. In which direction will sphere Q move? Write down only TO THE LEFT or TO THE RIGHT. (1) COPY RIGHT RESERVED PAGE 67 PLEASE TURN OVER

The spheres are now made to touch and are then separated. 9.1.2 Were electrons REMOVED FROM or TRANSFERRED TO P? Give a reason for the answer. (2) 9.1.3 Calculate the distance d between the two spheres if the magnitude of the force experienced by P is 0,8 N. (3) 9.2 Charges of +4 μC and –3 μC are placed a distance 0,2 m apart on a straight line, as shown below. Point P is located 0,1 m to the right of the –3 μC charge. +4 μC –3 μC P x 0,2 m 0,1 m Calculate the: 9.2.1 Net electric field at point P (5) 9.2.2 Mass of a +6 μC charge placed at P that will experience an acceleration of magnitude 5 x 102 m∙s-2 (3) [14] QUESTION 8 NOV 2016 Two charged spheres, X and Y, are placed in a vacuum at a distance of 0,04 m apart. X +5 µC Y -5 µC 0,04 m 8.1 Draw the resulting electric field pattern between the two charges. (3) 8.2 Calculate the electrostatic force sphere X experiences due to the charge on sphere Y. (4) 8.3 8.3.1 If sphere Y is at a fixed position and sphere X is free to move, will the acceleration experienced by sphere X towards sphere Y be constant? Write down YES or NO. (1) Explain the answer to QUESTION 8.3.1 by referring to the electric field and the force. (2) A third sphere, Z, with a charge of -4 µC, is placed at right angles to sphere X and at a distance of 0,03 m from sphere X. 8.3.2 -4 µC Z 0,03 m +5 µC X 0,04 m Y -5 µC 8.4 Calculate the magnitude of the net force on sphere X due to sphere Y and sphere Z. (4) [14] COPY RIGHT RESERVED PAGE 68 PLEASE TURN OVER

QUESTION 8 NOV 2017 Two IDENTICAL point charges, X and Y, are placed 2 mm apart. Point P is 3 mm to the right of charge Y. The net electric field at point P is 5,44 × 106 N∙C-1 to the left. X Y P 2 mm 3 mm 8.1 Define the term electric field at a point. (2) 8.2 Are the charges NEGATIVE or POSITIVE? (1) 8.3 Draw the resultant electric field pattern for charges X and Y. (3) 8.4 Calculate the magnitude of the charge X. (5) 8.5 Charge Y is now replaced by an identical oppositely charged point charge. How will the magnitude of the net electric field at point P be affected? Choose from INCREASE, DECREASE or REMAIN THE SAME. Give a reason for the answer. (2) [13] COPY RIGHT RESERVED PAGE 69 PLEASE TURN OVER

ELECTROMMAGNETISM MULTIPLE CHOICE QUESTIONS 1.9 NOV 2014 A conducting wire, XY, moves between two magnets as shown below. Which ONE of the following actions can lead to an increased induced current in wire XY? Move the wire… A quickly and parallel to the magnetic field. B slowly and parallel to the magnetic field. C quickly and perpendicular to the magnetic field. D slowly and perpendicular to the magnetic field. (2) 1.9 NOV 2015 In the diagram below, the north pole of a bar magnet approaches end A of a solenoid. S N Which ONE of the following statements about the polarity of A and the direction of the magnetic field INSIDE the solenoid is CORRECT as the NORTH POLE approaches A? POLARITY OF A DIRECTION OF FIELD IN SOLENOID A South pole A to B B North pole B to A C North pole A to B D South pole B to A (2) COPY RIGHT RESERVED PAGE 70 PLEASE TURN OVER

1.9 NOV 2016 In which ONE of the sketches below is the induced polarity of the coil CORRECTLY indicated? A B N N N S Magnet is Magnet is pushed into pushed into the coil the coil C D N S N N Magnet is Magnet is pulled out of pulled out of the coil the coil (2) 1.9 NOV 2017 A circular coil is placed inside a magnetic field and rotated clockwise to induce an emf. Which ONE of the following changes will increase the induced emf? A Rotating the coil slower B Decreasing the number of turns/windings of the C coil Increasing the speed of rotation of the coil D Changing the polarity of the magnets (2) COPY RIGHT RESERVED PAGE 71 PLEASE TURN OVER

ELECTROMAGNETISM LONG QUESTIONS QUESTION 10 NOV 2014 In the diagram below a bar magnet is being pushed into a coil. The current induced in the coil is in the direction indicated. 10.1 Write down the polarity (north pole or south pole) of the end of the coil facing the bar magnet, as the bar magnet approaches the coil. (2) 10.2 Which end of the bar magnet is approaching the coil? Write down only NORTH POLE or SOUTH POLE (1) 10.3 Write down what will be observed on the galvanometer if the bar magnet is held stationary inside the coil. Give a reason for the answer. (2) Faraday's law of electromagnetic induction plays a very important role in the generation of electricity. 10.4 Write down Faraday's law of electromagnetic induction in words. (2) -4 2 A coil of 100 turns, each of area 4,8 x 10 m , is made from insulated copper wire. The coil is placed in a uniform magnetic field of 4 x 10-4 T in such a way that the angle between the magnetic field and the normal to the plane of the coil is 30°. The coil is then rotated so that the angle changes to 70° in a time interval of 0,2 s. Calculate the: 10.5 Magnitude of the emf induced in the coil (5) 10.6 Current induced in the coil if it has an effective resistance of 2 Ω (3) [15] QUESTION 10 NOV 2015 10.1 The arrangement of apparatus to demonstrate Faraday's law of electromagnetic induction is shown below. magnet 10.1.1 State Faraday's law of electromagnetic induction in words. (2) 10.1.2 State TWO ways in which the deflection on the galvanometer can be increased. (2) COPY RIGHT RESERVED PAGE 72 PLEASE TURN OVER

10.2 A coil with area 0,6 m2 is held with its axis coinciding with the direction of a magnetic field of strength 0,4 T. 10.2.1 Calculate the magnetic flux linkage. (3) In order to produce an emf of 9 V, the area of the coil, with its axis coinciding with the direction of a magnetic field, is halved from 0,6 m2 to 0,3 m2 in 2 minutes. 10.2.2 Calculate the number of turns in the coil. (4) [11] QUESTION 9 NOV 2016 A circular coil with 250 windings (turns) and a radius of 0,04 m, is rotated clockwise inside a magnetic field with a field strength of 3,2 T. rotation N S 9.1 Calculate the magnetic flux through the coil at the position indicated on the diagram, where the coil is perpendicular to the field. (3) 9.2 If the coil rotates clockwise through 25°, and the potential difference induced is 2,8 V, calculate the time in which this rotation took place. (4) 9.3 Which law can be used to explain the phenomenon described in QUESTION 9.2? Name and state the law. (2) 9.4 9.4.1 If the circular coil is replaced with a square coil with a side length of 0,04 m, and the same movement is made in the same amount of time, will the induced emf be the same as, larger than or smaller than the circular coil? Write down only THE SAME AS, LARGER THAN or SMALLER THAN. (1) 9.4.2 Explain the answer to QUESTION 9.4.1. (2) [12] COPY RIGHT RESERVED PAGE 73 PLEASE TURN OVER

QUESTION 9 NOV 2017 An induction coil of area 48,6 cm2 and 200 windings is rotated clockwise in a constant magnetic field of magnitude 2,4 T. Refer to the diagram below. θ S N The graph below shows how the induced emf varies with the inverse of time. ε (V) 3 2 1 0 1 1 2 3 ∆t 9.1 State Faraday's law in words. (2) 9.2 Use the information in the graph to calculate the change in magnetic flux. (5) 9.3 The coil rotates through an angle θ to a position where the magnetic flux becomes zero. Calculate angle θ. (4) [11] COPY RIGHT RESERVED PAGE 74 PLEASE TURN OVER

ELECTRIC CIRCUITS MULTIPLE CHOICE QUESTIONS 1.10 NOV 2014 A learner wants to measure the current in and the potential difference across a resistor R in a circuit. In which ONE of the following circuits will the learner be able to take these readings? V A V A A B R R A R V C R V D A (2) 1.10 NOV 2015 Which ONE of the following graphs best represents the relationship between current and potential difference for a conductor which obeys Ohm's Law? A B I I V V C D I I V V (2) COPY RIGHT RESERVED PAGE 75 PLEASE TURN OVER

1.10 NOV 2016 The potential difference of the battery in the circuit below is 12 V. The internal resistance of the battery is negligible. Two voltmeters, V 1 and V 2 , are connected to the circuit, as shown in the diagram. When the switch is open, the correct readings on V 1 and V 2 will be as follows: V1 V2 A 12 V 12 V B 0V 12 V C 12 V 0V D 0V 0V (2) 1.10 NOV 2017 In the circuit diagram below, the battery has negligible internal resistance. The resistance of the ammeter and wires may also be ignored. A V1 4Ω V2 8Ω 8Ω V3 The reading on voltmeter V 3 will be equal to … A V1 B ½ V1 C V1 + V2 D V2 - V1 (2) COPY RIGHT RESERVED PAGE 76 PLEASE TURN OVER

ELECTRIC CIRCUITS LONG QUESTIONS QUESTION 11 NOV 2014 In the circuit below the internal resistance of the 6 V battery is negligible. The resistance of the connecting wires is negligible. When switch S is closed, the current in the 6 Ω resistor is 0,6 A. 6V A ● S 4Ω X 6Ω V1 V2 11.1 State Ohm's law in words. (2) Calculate the: 11.2 Current passing through the 4 Ω resistor (4) 11.3 Total current in the circuit (2) 11.4 Resistance X (3) The 4 Ω resistor gets hotter than the 6 Ω resistor after a while. 11.5 Explain this observation. (3) [14] QUESTION 11 NOV 2015 11.1 The circuit below is used to determine the resistance of resistor X. A 8Ω X ●S ● 12 V The 12 V battery has negligible internal resistance. When switch S is closed, the reading on the ammeter is 0,5 A. 11.1.1 State Ohm's law in words. (2) 11.1.2 Calculate the resistance of resistor X. (5) COPY RIGHT RESERVED PAGE 77 PLEASE TURN OVER

11.2 Study the circuit below. The battery has an emf of 12 V with negligible internal resistance. 12 Ω 8Ω A 16 Ω 4Ω ● ● S 12 V Switch S is closed. 11.2.1 Write down the potential difference across the 4 Ω resistor. (1) 11.2.2 Calculate the reading on the ammeter. (5) 11.2.3 Calculate the energy dissipated in the 12 Ω resistor in 2 minutes. (5) [18] QUESTION 10 NOV 2016 An experiment is done to verify that the potential difference across a conductor is directly proportional to the current in the conductor if the temperature stays constant. Four cells, each with an emf of 1,5 V, are connected in series with an ammeter, switch S and a combination of a resistor R and resistors of 4 Ω and 6 Ω, as shown in the diagram. Voltmeters V1 and V2 are connected across the battery and the parallel resistors respectively. The internal resistance of the battery and wires are negligible. V1 S R A V2 4Ω 6Ω 10.1 Which law is represented by the underlined phrase above? (1) COPY RIGHT RESERVED PAGE 78 PLEASE TURN OVER

10.1 Which law is represented by the underlined phrase above? (1) The switch is now closed and six resistors (R 1 –R 6 ), each with a different resistance, are placed in the place of R, one at a time. The voltmeter and ammeter readings are recorded. The results are as follows: RESISTORS READING ON READING ON AT R VOLTMETER V 2 (V) AMMETER (A) R1 1,2 0,5 R2 1,4 0,6 R3 1,9 0,8 R4 2,4 1 R5 2,9 1,2 R6 3,6 1,5 10.2 Use the attached graph paper and draw a graph of potential difference versus current using the data in the table. (4) 10.3 What does the gradient of the graph represent? (1) 10.4 If voltmeter V 2 is only connected across the 4 Ω resistor, how will the gradient of the graph change? Write down only INCREASES, DECREASES or STAYS THE SAME. (1) 10.5 If the 4 Ω resistor is removed, how will the gradient of the graph change? Write down only INCREASES, DECREASES or STAYS THE SAME. (1) 10.6 Calculate the resistance of resistor R 3 using the values in the table. (5) 10.7 Calculate the energy dissipated in resistor R 4 in 10 seconds. (3) [16] QUESTION 10 NOV 2017 10.1 The circuit below consists of a 6 Ω and 15 Ω resistor connected in parallel and an unknown resistor R, in series. An ammeter, a high-resistance voltmeter, a closed switch and battery are connected, as shown. The resistance of the battery and wires can be ignored. R V A 6Ω 15 Ω The total power dissipated in the parallel part of the circuit is 50 W. COPY RIGHT RESERVED PAGE 79 PLEASE TURN OVER

10.1.1 Define the term power. (2) 10.1.2 Calculate the effective resistance of the parallel combination. 10.1.3 Calculate the potential difference across the resistors in parallel. (3) 10.1.4 Calculate the current through resistor R. (3) (2) The switch in the circuit is now OPENED. 10.1.5 How will the reading on the voltmeter (V) be influenced? Choose from INCREASE, DECREASE or REMAIN THE SAME. (1) 10.1.6 Explain the answer to QUESTION 10.1.5. (3) The cost of electricity is 80 cents per kWh. 10.2.1 Calculate the energy used by the geyser for 5 hours per day. (4) 10.2.2 Calculate the cost of electricity to operate the geyser for a month with 30 days. (2) [20] COPY RIGHT RESERVED PAGE 80 PLEASE TURN OVER

REVISION QUESTIONS PHYSICAL SCIENCES PAPER 2 CHEMISTRY COPY RIGHT RESERVED PAGE 81 PLEASE TURN OVER

DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 2 (CHEMISTRY) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 2 (CHEMIE) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Avogadro's constant NA 6,02 x 1023 mol-1 Avogadro-konstante Molar gas constant R 8,31 J∙K-1∙mol-1 Molêre gaskonstante Standard pressure pθ 1,013 x 105 Pa Standaarddruk Molar gas volume at STP Vm 22,4 dm3∙mol-1 Molêre gasvolume by STD Standard temperature Tθ 273 K Standaardtemperatuur TABLE 2: FORMULAE/TABEL 2: FORMULES p1V1 p 2 V2 = pV = nRT T1 T2 m N n= n= M NA V n m n= c= OR/OF c= Vm V MV COPY RIGHT RESERVED PAGE 82 PLEASE TURN OVER

TABLE 3: THE PERIODIC TABLE OF ELEMENTS/TABEL 3: DIE PERIODIEKE TABEL VAN ELEMENTE 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 (I) (II) (III) (IV) (V) (VI) (VII) (VIII) 1 Atomic number 2 KEY/SLEUTEL Atoomgetal 2,1 H He 1 4 3 4 29 5 6 7 8 9 10 Electronegativity Symbol 1,0 Li 1,5 Be Elektronegatiwiteit 1,9 Cu Simbool 2,0 B 2,5 C 3,0 N 3,5 O 4,0 F Ne 7 9 63,5 11 12 14 16 19 20 11 12 13 14 15 16 17 18 0,9 Na 1,2 Mg Approximate relative atomic mass 1,5 Aℓ 1,8 Si 2,1 P 2,5 S 3,0 Cℓ Ar 23 24 Benaderde relatiewe atoommassa 27 28 31 32 35,5 40 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 0,8 K 1,0 Ca 1,3 Sc 1,5 Ti 1,6 V 1,6 Cr 1,5 Mn 1,8 Fe 1,8 Co 1,8 Ni 1,9 Cu 1,6 Zn 1,6 Ga 1,8 Ge 2,0 As 2,4 Se 2,8 Br Kr 39 40 45 48 51 52 55 56 59 59 63,5 65 70 73 75 79 80 84 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 0,8 Rb 1,0 Sr 1,2 Y 1,4 Zr Nb 1,8 Mo 1,9 Tc 2,2 Ru 2,2 Rh 2,2 Pd 1,9 Ag 1,7 Cd 1,7 In 1,8 Sn 1,9 Sb 2,1 Te 2,5 I Xe 86 88 89 91 92 96 101 103 106 108 112 115 119 122 128 127 131 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 0,7 Cs 0,9 Ba La 1,6 Hf Ta W Re Os Ir Pt Au Hg 1,8 Tℓ 1,8 Pb 1,9 Bi 2,0 Po 2,5 At Rn 133 137 139 179 181 184 186 190 192 195 197 201 204 207 209 87 88 89 0,7 Fr 0,9 Ra Ac 58 59 60 61 62 63 64 65 66 67 68 69 70 71 226 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140 141 144 150 152 157 159 163 165 167 169 173 175 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232 238 COPY RIGHT RESERVED PAGE 83 PLEASE TURN OVER

TABLE 4A: STANDARD REDUCTION POTENTIALS TABEL 4A: STANDAARD-REDUKSIEPOTENSIALE θ Half-reactions/Halfreaksies E (V) F 2 (g) + 2e− ⇌ 2F− + 2,87 3+ 2+ Co + e− ⇌ Co + 1,81 + H 2 O 2 + 2H +2e− ⇌ 2H 2 O +1,77 − + 2+ MnO 4 + 8H + 5e− ⇌ Mn + 4H 2 O + 1,51 Cℓ 2 (g) + 2e− ⇌ 2Cℓ− + 1,36 2− + 3+ Cr 2 O 7 + 14H + 6e− ⇌ 2Cr + 7H 2 O + 1,33 + − O 2 (g) + 4H + 4e ⇌ 2H 2 O + 1,23 + 2+ MnO 2 + 4H + 2e− ⇌ Mn + 2H 2 O + 1,23 2+ Pt + 2e− ⇌ Pt + 1,20 Br 2 (ℓ) + 2e− ⇌ 2Br− + 1,07 − + NO 3 + 4H + 3e− ⇌ NO(g) + 2H 2 O + 0,96 Increasing reducing ability/Toenemende reduserende vermoë 2+ − Hg + 2e ⇌ Hg(ℓ) + 0,85 + Ag + e− ⇌ Ag + 0,80 Increasing oxidising ability/Toenemende oksiderende vermoë − + NO 3 + 2H + e− ⇌ NO 2 (g) + H 2 O + 0,80 3+ − 2+ Fe + e ⇌ Fe + 0,77 + O 2 (g) + 2H + 2e− ⇌ H2O2 + 0,68 I 2 + 2e− ⇌ 2I− + 0,54 + Cu + e− ⇌ Cu + 0,52 + SO 2 + 4H + 4e− ⇌ S + 2H 2 O + 0,45 2H 2 O + O 2 + 4e− ⇌ 4OH− + 0,40 2+ Cu + 2e− ⇌ Cu + 0,34 2− + SO 4 + 4H + 2e− ⇌ SO 2 (g) + 2H 2 O + 0,17 2+ + Cu + e− ⇌ Cu + 0,16 4+ 2+ Sn + 2e− ⇌ Sn + 0,15 + S + 2H + 2e− ⇌ H 2 S(g) + 0,14 + 2H + 2e− ⇌ H 2 (g) 0,00 3+ Fe + 3e− ⇌ Fe − 0,06 2+ Pb + 2e− ⇌ Pb − 0,13 2+ Sn + 2e− ⇌ Sn − 0,14 2+ Ni + 2e− ⇌ Ni − 0,27 2+ Co + 2e− ⇌ Co − 0,28 2+ Cd + 2e− ⇌ Cd − 0,40 3+ 2+ Cr + e− ⇌ Cr − 0,41 2+ Fe + 2e− ⇌ Fe − 0,44 3+ Cr + 3e− ⇌ Cr − 0,74 2+ Zn + 2e− ⇌ Zn − 0,76 2H 2 O + 2e− ⇌ H 2 (g) + 2OH− − 0,83 2+ Cr + 2e− ⇌ Cr − 0,91 2+ Mn + 2e− ⇌ Mn − 1,18 3+ Aℓ + 3e− ⇌ Aℓ − 1,66 2+ Mg + 2e− ⇌ Mg − 2,36 + Na + e− ⇌ Na − 2,71 2+ Ca + 2e− ⇌ Ca − 2,87 2+ Sr + 2e− ⇌ Sr − 2,89 2+ Ba + 2e− ⇌ Ba − 2,90 + - Cs + e ⇌ Cs - 2,92 + − K +e ⇌ K − 2,93 + Li + e− ⇌ Li − 3,05 COPY RIGHT RESERVED PAGE 84 PLEASE TURN OVER

TABLE 4B: STANDARD REDUCTION POTENTIALS TABEL 4B: STANDAARD-REDUKSIEPOTENSIALE θ Half-reactions/Halfreaksies E (V) + Li + e− ⇌ Li − 3,05 + K + e− ⇌ K − 2,93 + Cs + e− ⇌ Cs − 2,92 2+ Ba + 2e− ⇌ Ba − 2,90 2+ Sr + 2e− ⇌ Sr − 2,89 2+ Ca + 2e− ⇌ Ca − 2,87 + Na + e− ⇌ Na − 2,71 2+ Mg + 2e− ⇌ Mg − 2,36 3+ Aℓ + 3e− ⇌ Aℓ − 1,66 2+ Mn + 2e− ⇌ Mn − 1,18 2+ Cr + 2e− ⇌ Cr − 0,91 2H 2 O + 2e− ⇌ H 2 (g) + 2OH− − 0,83 2+ Zn + 2e− ⇌ Zn − 0,76 3+ Cr + 3e− Cr Increasing oxidising ability/Toenemende oksiderende vermoë ⇌ − 0,74 2+ Fe + 2e− ⇌ Fe − 0,44 3+ 2+ Cr + e− ⇌ Cr − 0,41 Increasing reducing ability/Toenemende reduserende vermoë 2+ Cd + 2e− ⇌ Cd − 0,40 2+ Co + 2e− ⇌ Co − 0,28 2+ Ni + 2e− ⇌ Ni − 0,27 2+ Sn + 2e− ⇌ Sn − 0,14 2+ Pb + 2e− ⇌ Pb − 0,13 3+ Fe + 3e− ⇌ Fe − 0,06 + 2H + 2e− ⇌ H 2 (g) 0,00 + S + 2H + 2e− ⇌ H 2 S(g) + 0,14 4+ 2+ Sn + 2e− ⇌ Sn + 0,15 2+ + Cu + e− ⇌ Cu + 0,16 2− + SO 4 + 4H + 2e− ⇌ SO 2 (g) + 2H 2 O + 0,17 2+ − Cu + 2e ⇌ Cu + 0,34 2H 2 O + O 2 + 4e− ⇌ 4OH− + 0,40 + SO 2 + 4H + 4e− ⇌ S + 2H 2 O + 0,45 + Cu + e− ⇌ Cu + 0,52 I 2 + 2e− ⇌ 2I− + 0,54 + O 2 (g) + 2H + 2e− ⇌ H2O2 + 0,68 3+ 2+ Fe + e− ⇌ Fe + 0,77 − + NO 3 + 2H + e− ⇌ NO 2 (g) + H 2 O + 0,80 + − Ag + e ⇌ Ag + 0,80 2+ Hg + 2e− ⇌ Hg(ℓ) + 0,85 − + NO 3 + 4H + 3e− ⇌ NO(g) + 2H 2 O + 0,96 − − Br 2 (ℓ) + 2e ⇌ 2Br + 1,07 2+ Pt + 2 e− ⇌ Pt + 1,20 + 2+ MnO 2 + 4H + 2e− ⇌ Mn + 2H 2 O + 1,23 + O 2 (g) + 4H + 4e− ⇌ 2H 2 O + 1,23 2− + 3+ Cr 2 O 7 + 14H + 6e− ⇌ 2Cr + 7H 2 O + 1,33 − − Cℓ 2 (g) + 2e ⇌ 2Cℓ + 1,36 − + 2+ MnO 4 + 8H + 5e− ⇌ Mn + 4H 2 O + 1,51 + H 2 O 2 + 2H +2 e− ⇌ 2H 2 O +1,77 3+ 2+ Co + e− ⇌ Co + 1,81 F 2 (g) + 2e− ⇌ 2F − + 2,87 COPY RIGHT RESERVED PAGE 85 PLEASE TURN OVER

MATTER AND MATERIALS MULTIPLE CHOICE QUESTIONS BONDING AND ELECTRONEGATIVITY 1.1 NOV 2014 A substance that donates electrons during a chemical reaction is a/an … A Arrhenius acid. B Arrhenius base. C reducing agent. D oxidising agent. (2) 1.2 NOV 2014 Which ONE of the following species contains a dative covalent bond? A NH 3 B CH 4 C H 3 O+ D NF 3 (2) 1.6 NOV 2014 Consider the Lewis structure of a compound below: Which ONE of the following is CORRECT? Name of Name of Molecular shape element X element Y of compound A Chlorine Oxygen Angular B Oxygen Chlorine Linear C Chlorine Sulphur Linear D Sulphur Chlorine Angular (2) 1.2 NOV 2015 The number of neutrons in an atom of 157 N is ... A 5 B 7 C 8 D 15 (2) COPY RIGHT RESERVED PAGE 86 PLEASE TURN OVER

1.4 NOV 2015 Which ONE of the following has a tetrahedral shape? A H3 O + B NH +4 C CO 2 D AℓCℓ 3 (2) 1.1 NOV 2016 The type of bond formed between a H+ ion and H 2 O is called a/an … A hydrogen bond. B dative covalent C bond. ionic bond. D covalent bond. (2) 1.2 NOV 2016 The shape of the molecule in which the central atom is surrounded by two lone pairs and two bonding pairs is … A linear. trigonal B planar. C tetrahedral. D bent. (2) 1.3 NOV 2017 The graph below shows how the potential energy varies with distance between the nuclei of two nitrogen atoms when a double bond between the nitrogen atoms (N = N) is formed. Potential energy (kJ∙mol-1) 0 L l l l l Distance between 120 125 130 135 140 nuclei (pm) -100 -418 Choose from the table the bond length and bond energy for N = N. BOND LENGTH BOND ENERGY (pm) (kJ∙mol-1) A 120 0 B 125 518 C 125 418 D 130 -100 (2) COPY RIGHT RESERVED PAGE 87 PLEASE TURN OVER

1.1 NOV 2017 Which ONE of the bonds between the atoms below has the highest polarity? A H-C B H - Cℓ C H-O D H-N (2) COPY RIGHT RESERVED PAGE 88 PLEASE TURN OVER

MATTER AND MATERIALS LONG QUESTIONS BONDING AND ELECTRONEGATIVITY QUESTION 2 NOV 2014 The graph below shows the change in energy that takes place when a hydrogen (H) atom approaches a bromine (Br) atom. Potential energy kJ·mol-1 100 120 240 360 480 600 0 Distance between nuclei -100 (pm) -200 -300 E -400 2.1 Define the term bond length. (2) 2.2 From the graph, write down the: 2.2.1 Bond length, in pm, of the H-Br bond (2) -1 2.2.2 Energy, in kJ·mol , needed to break the H-Br bond (2) 2.2.3 Name of the potential energy represented by E (1) 2.3 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. (2) [9] QUESTION 3 NOV 2014 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. 3.1 Explain the difference between a covalent bond and an ionic bond. (2) 3.2 Name the type of chemical bond between particles in: 3.2.1 AℓF 3 (1) 3.2.2 PF 3 (1) 3.3 Draw the Lewis structures for: 3.3.1 AℓF 3 (3) 3.3.2 PF 3 (2) 3.4 Write down the molecular shape of PF 3 . (1) 3.5 The melting point of AℓF3 is 1 291 °C and that of PF3 is -151 °C. Fully explain this difference in melting point. (4) [14] COPY RIGHT RESERVED PAGE 89 PLEASE TURN OVER

QUESTION 2 NOV 2015 Molecules such as CO2 and H2O are formed through covalent bonding. 2.1 Define the term covalent bonding. (2) 2.2 ONE of the above molecules has lone pairs of electrons on the central atom. Draw the Lewis diagram for this molecule. (2) 2.3 H 3 O+ is formed when H 2 O forms a dative covalent bond with an H+ ion. 2.3.1 Draw the Lewis diagram for H3O+. (1) 2.3.2 State TWO conditions for the formation of such a bond. (2) 2.4 The polarity of molecules depends on the DIFFERENCE IN ELECTRONEGATIVITY and the MOLECULAR SHAPE. 2.4.1 Define the term electronegativity. (2) 2.4.2 Calculate the difference in electronegativity between: (a) C and O in CO2 (1) (b) H and O in H2O (1) 2.4.3 Explain the difference in polarity between CO 2 and H 2 O by referring to the polarity of the bonds and the shape of the molecules. (6) [17] QUESTION 2 NOV 2017 Consider the following two reactions of methane (CH4): Reaction 1: CH 4 (g) + HCℓ (g) → CH 3 Cℓ (g) + H 2 (g) Reaction 2: CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) 2.1 Define the term covalent bond. (2) 2.2 Draw Lewis structures for: 2.2.1 CH 3 Cℓ (2) 2.2.2 CO 2 (2) 2.3 How many lone-pair electrons are on the central atom in the CO2 molecule? (1) 2.4 Identify ONE of the substances in Reaction 2 that can form a dative covalent bond when reacting with an acid. (1) 2.5 Write down the shape of the: 2.5.1 H 2 O molecule (1) 2.5.2 CO 2 molecule (1) 2.6 Although the molecules of CH 4 and CH 3 Cℓ have the same shape, CH 4 is non-polar, while CH 3 Cℓ is polar. Give a reason for the difference in molecular polarity. (1) [11] COPY RIGHT RESERVED PAGE 90 PLEASE TURN OVER

QUESTION 4 NOV 2014 The boiling points of four compounds of hydrogen, represented by the letters P, Q, R and S, are given in the table below. Formula Boiling point (°C) P CH 4 -164 Q NH 3 -33 R H2O 100 S SiH 4 -112 4.1 Define the term boiling point. (2) 4.2 Fully explain the difference in boiling points between compound P and: 4.2.1 Compound Q (3) 4.2.2 Compound S (3) 4.3 Explain why the boiling points of compounds Q and R differ by referring to ELECTRONEGATIVITY and DEGREE OF POLARITY. (2) 4.4 Write down the letter from the table that represents the following: 4.4.1 ONE polar compound (1) 4.4.2 ONE non-polar compound (1) [12] QUESTION 3 NOV 2015 3.1 The boiling point of compounds A to E are given in the table below. COMPOUND FORMULA BOILING POINT (°C) A CH 4 –164 B C2H6 –89 C C 5 H 12 36 D C 6 H 14 69 E C 20 H 42 343 3.1.1 Define the term boiling point. (2) 3.1.2 Calculate the molecular mass of compound D. (1) 3.1.3 In what phase is compound B at 25 °C? (1) 3.1.4 Name the type of intermolecular force present in compound A. (1) 3.1.5 Explain why the boiling point increases from compound A to E. (3) 3.1.6 How does the vapour pressure of compound B compare to the vapour pressure of compound C? Write down only HIGHER THAN, LOWER THAN or EQUAL TO. (1) COPY RIGHT RESERVED PAGE 91 PLEASE TURN OVER

3.2 Consider the boiling points of the compounds in the table below. SUBSTANCE BOILING POINT (°C) H2S –60 NH 3 –33 H2O 100 3.2.1 Which ONE of the substances in the table above has the weakest forces between its molecules? (1) 3.2.2 Name the type of intermolecular force found between NH 3 molecules. (1) 3.2.3 Explain the following statement: Although the shape of the molecules of H 2 S and H 2 O is similar, there is a remarkable difference in their boiling points. (4) [15] QUESTION 2 NOV 2016 Electronegativity of atoms may be used to explain the polarity of bonds. 2.1 Define the term electronegativity. (2) 2.2 Draw the Lewis diagram of an oxygen difluoride molecule. (2) 2.3 Calculate the electronegativity difference between O and F in oxygen difluoride and predict the polarity of the bond. (2) 2.4 A polar bond does not always lead to a polar molecule. Explain the statement by referring to OF 2 and CO 2 molecules. In your explanation, include the polarity of the bonds and the shape of the molecules. (4) 2.5 The diagram below shows the energy change that takes place when two atoms move towards each other. 2.5.1 What does X and Y represent? (2) 2.5.2 Define the concept represented by X. (2) 2.5.3 Explain the relationship between bond order, bond length and bond energy. (3) [17] COPY RIGHT RESERVED PAGE 92 PLEASE TURN OVER

MATTER AND MATERIALS MULTIPLE CHOICE QUESTIONS GAS LAWS 1.8 NOV 2014 In which ONE of the following graphs does the dotted line CORRECTLY represent the deviation of a real gas from ideal gas behaviour? A p B p 1 1 V V C p D p 1 1 V V (2) 1.10 NOV 2014 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. X Y 10 cm3 20 cm3 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. (2) COPY RIGHT RESERVED PAGE 93 PLEASE TURN OVER

1.5 NOV 2015 Two moles of H 2 gas at STP occupy a volume of ... A 2 dm3 B 11,2 dm3 C 22,4 dm3 D 44,8 dm3 (2) 1.9 NOV 2015 The volume of a gas at a certain temperature and pressure is V. If the temperature is doubled and the pressure is halved, the volume of the gas will be … A 4V B 2V C V D ½V (2) 1.10 NOV 2015 Two identical containers, M and N, are shown below. Container M contains H 2 (g) and container N contains CO 2 (g). Both gases are at a temperature of 25 °C and a pressure of 101,3 kPa. M N H2(g) CO2(g) 25 °C 25 °C 101,3 kPa 101,3 kPa Consider the following statements: (i) The average kinetic energy of the molecules is the same in both containers. (ii) Container M contains more gas molecules than container N. (iii) The mass of the gas in container N is greater than the mass of the gas in container M. Which of the above statements is/are CORRECT? A (i) only B (iii) only C (i) and (ii) only D (i) and (iii) only (2) COPY RIGHT RESERVED PAGE 94 PLEASE TURN OVER

1.5 NOV 2016 A gas of volume V is at a temperature T 1 and pressure P 1 in a gas syringe. If the pressure on the gas is doubled and the temperature halved, then the volume that the gas will occupy is … A ¼V B ½V C V D 2V (2) 1.6 NOV 2016 Which ONE of the graphs below CORRECTLY represents the deviation of a real gas from ideal gas behaviour at very high pressures? The dotted line represents the graph of the real gas. A V B V 1 1 p p C D V V 1 1 p p (2) 1.4 NOV 2017 According to Boyle's law, … 1 A pα if T is constant. V B V α T if p is constant. 1 C Vα if p is constant. T D p α V if n is constant. (2) COPY RIGHT RESERVED PAGE 95 PLEASE TURN OVER

1.5 NOV 2017 One mole of any gas occupies the same volume at the same temperature and pressure. This statement is known as … A Charles's law. B Gay Lussac's law. C Avogadro's law. D the ideal gas LAW. (2) 1.6 NOV 2017 One mole of a gas, SEALED in a container, has volume V at temperature T and pressure p. If the pressure is increased to 3p, the ratio between the volume and temperature (V : T) is … A 1:⅓ B 3:1 C ⅓:3 D 1:3 (2) COPY RIGHT RESERVED PAGE 96 PLEASE TURN OVER

MATTER AND MATERIALS LONG QUESTIONS GAS LAWS QUESTION 5 NOV 2014 5.1 Explain what is meant by the term temperature of a gas. (1) 5.2 Two learners investigate the relationship between the temperature and the pressure of an enclosed gas. The learners use different samples of the same gas in two identical containers of fixed volumes. Graph P and Graph Q below represent the results obtained by the learners. P 150 Pressure (kPa) Q 100 298 X Temperature (K) 5.2.1 State Guy Lussac's law in words. (2) 5.2.2 Use the law in QUESTION 5.2.1 to determine the value of temperature X, shown on the graph, in °C. (4) 5.2.3 Explain, using the relevant formulae, why graph Q has a smaller gradient than graph P. (4) 5.3 A certain gas with a mass of 2,2 g occupies a volume of 0,831 dm3 at 27 °C and pressure 150 kPa. 5.3.1 Calculate the molar mass of the gas. Assume that the gas behaves like an ideal gas. (6) 5.3.2 Write down the MOLECULAR FORMULA or NAME of the gas in QUESTION 5.3.1. (1) [18] QUESTION 4 NOV 2015 A learner investigates the relationship between the pressure and volume of an enclosed DIATOMIC gas at 25 °C. He records the volume of the gas for different pressures in the table below. PRESSURE VOLUME 1 (kPa) (cm3) (cm-3) V 40 43 0,02 80 27 0,04 100 22 (a) 120 18 (b) COPY RIGHT RESERVED PAGE 97 PLEASE TURN OVER

4.1 Write down the name of the gas law being investigated. (1) Answer QUESTION 4.2 and QUESTION 4.3 on the attached ANSWER SHEET. 1 Two values in the table, (a) and (b), have not been calculated. 4.2 V Calculate these values. (1) 1 4.3 Draw a graph of pressure versus on the attached ANSWER SHEET. (4) V 4.4 Use the graph to determine the volume of the gas at 68 kPa. (2) 4.5 The mass of the enclosed DIATOMIC gas is 2,49 x 10-2 g. 4.5.1 Use the conditions at a pressure of 100 kPa and calculate the molar mass of the enclosed gas. (6) 4.5.2 Write down the molecular formula of the enclosed gas. (1) 4.6 The sketch graph below shows the relationship between volume and temperature for an ideal gas. Volume (cm3) Temperature (K) 4.6.1 Redraw the above graph in the ANSWER BOOK. On the same set of axes, use a BROKEN LINE to sketch the graph that will be obtained for the diatomic gas above. (1) 4.6.2 Fully explain why this diatomic gas deviates from ideal behaviour. (3) [19] QUESTION 4 NOV 2016 4.1 Charles' Law describes the relationship between the volume and temperature of an enclosed mass of gas at constant pressure. The volumes of a gas at different temperatures at constant pressure are given in the table below. VOLUME TEMPERATURE (cm3) (ºC) 114 0 124 25 134 50 145 75 155 100 COPY RIGHT RESERVED PAGE 98 PLEASE TURN OVER

4.1.1 Explain Charles' Law in terms of the kinetic molecular theory of gases. (2) 4.1.2 Draw a graph of volume versus temperature at constant pressure on the attached graph paper. (4) 4.1.3 Extrapolate the graph (extend the graph) and state at what temperature the graph intersects the x-axis. (1) 4.1.4 What is significant about this temperature? (2) 4.1.5 Calculate the volume that this gas will occupy at 120 °C at constant pressure. (3) 4.1.6 The experiment is now conducted at a lower constant pressure. On the graph drawn for QUESTION 4.1.2, sketch the graph that will be obtained at a lower pressure. Label this graph P. (2) 3 4.2 An unknown gas with a mass of 0,77 g occupies a volume of 0,32 dm at a temperature of 27 °C and a pressure of 96 kPa. Assume that the gas behaves as an ideal gas. 4.2.1 Calculate the molar mass of the gas. (5) 4.2.2 Write down the MOLECULAR FORMULA or NAME of the gas named in QUESTION 4.2.1. (2) [21] QUESTION 4 NOV 2017 In an experiment to investigate the relationship between pressure and temperature of an enclosed gas, 48 g of oxygen gas was sealed in a container. The results obtained are recorded in the table below. PRESSURE (kPa) TEMPERATURE (K) 155,8 250 187,0 300 218,1 350 249,3 400 280,5 450 4.1 Draw a graph of pressure versus temperature on the attached ANSWER SHEET. Extrapolate the graph so that it intersects the y-axis. (4) 4.2 What conclusion can be made from the final graph? (2) 4.3 Explain why it will not be possible to obtain accurate values at very low temperatures. (2) 4.4 Use the kinetic molecular theory to explain the effect of an increase in temperature on the pressure of a gas. (4) 4.5 Under which conditions of temperature and pressure will a real gas act as an ideal gas? (2) 4.6 Calculate the gradient of the graph. (3) 4.7 Use the answer to QUESTION 4.6 to determine the volume of the container. (5) [22] COPY RIGHT RESERVED PAGE 99 PLEASE TURN OVER

QUESTION 5 NOV 2017 Weather balloons are sent into space to gather data. The balloons usually burst at a pressure of 27 640 Pa and a volume of 36,3 m3. The data collector then falls back to Earth. balloon data collector The gas in a certain weather balloon has an initial volume of 12,6 m3 and pressure of 105 000 Pa at a temperature of 25 °C when it is released into space. Calculate the: 5.1 Temperature of the gas, in °C, in the balloon when it bursts (4) 5.2 Initial amount of gas (in moles) in the balloon (4) [8] COPY RIGHT RESERVED PAGE 100 PLEASE TURN OVER

CHEMICAL CHANGE MULTIPLE CHOICE QUESTIONS ENERGY AND CHEMICAL CHANGE 1.7 NOV 2014 Consider the reaction represented by the equation below: 2Fe(s) + 3CO 2 (g) → Fe 2 O 3 (s) + 2CO(g) ΔH = + 53,2 kJ Which ONE of the following statements is TRUE? For each mole of Fe that reacts, … A 26,6 kJ of energy are released. B 26,6 kJ of energy are absorbed. C 53,2 kJ of energy are released. D 53,2 kJ of energy are absorbed. (2) 1.6 NOV 2015 Which ONE of the following statements CORRECTLY describes the characteristics of an endothermic reaction? A ΔH is positive and the products have less potential energy than the reactants. B ΔH is positive and the products have more potential energy than the reactants. C ΔH is negative and the products have less potential energy than the reactants. D ΔH is negative and the products have more potential energy than the reactants. (2) 1.4 NOV 2016 The bond energy of a C–Cℓ bond is 338 kJ.mol-1 whereas the bond energy of a C–I bond is 238 kJ.mol-1. The difference in bond energy exists because … A the bond length of the C–Cℓ bond is greater than that of the C–I bond. B chlorine is more electronegative than iodine. C the bond length of the C–I bond is greater than that of the C–Cℓ D bond. the chlorine atom is bigger than the iodine atom. (2) COPY RIGHT RESERVED PAGE 101 PLEASE TURN OVER

1.8 NOV 2016 Cellular respiration occurs inside the cells of all living organisms. Oxygen reacts with glucose in cellular respiration to produce the following compounds according to the balanced equation below: C 6 H 12 O 6 (aq) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(ℓ) ΔH = -2 830 kJ The potential energy versus progress of reaction diagram for this reaction is … A B Ep Ep Progress of reaction Progress of reaction C Ep D Ep Progress of reaction Progress of reaction (2) 1.7 NOV 2017 The chemical equation that represents an endothermic reaction: A NH4NO3(s) + H2O(ℓ) → NH4+(aq) + NO3-(aq) ∆H > 0 B 2Mg(s) + O2(g) → 2MgO(s) ∆H < 0 C Zn(s) + 2HCℓ(aq) → ZnCℓ2(aq) + H2(g) + heat D H 2 (g) + Cℓ 2 (g) → 2HCℓ(g) ∆H = -131 kJ∙mol-1 (2) COPY RIGHT RESERVED PAGE 102 PLEASE TURN OVER

CHEMICAL CHANGE LONG QUESTIONS ENERGY AND CHEMICAL CHANGE QUESTION 8 NOV 2014 Hydrogen gas and oxygen gas react to form water according to the following balanced equation: 2H 2 (g) + O 2 (g) → 2H 2 O(g) + 241,8 kJ·mol-1 The activation energy (E A ) for this reaction is 1 370 kJ·mol-1. 8.1 Define the term activation energy. (2) 8.2 Sketch a potential energy versus reaction coordinate graph for the above reaction. Clearly label the axes and indicate the following on the graph: • ΔH • E A for the forward reaction • Reactants (R) and products (P) • Activated complex (X) (5) 8.3 Write down the value of the: 8.3.1 Heat of reaction 8.3.2 Activation energy for the following reaction: (1) 2H 2 O(g) → 2H 2 (g) + O 2 (g) (2) [10] QUESTION 7 NOV 2015 A barium hydroxide solution, Ba(OH) 2 (aq), reacts with a nitric acid solution, HNO 3 (aq), according to the following balanced equation: Ba(OH) 2 (aq) + 2HNO 3 (aq) → Ba(NO 3 ) 2 (aq) + 2H 2 O(ℓ) The potential energy graph below shows the change in potential energy for this reaction. Potential energy A (kJ⋅mol-1) B 116 kJ⋅mol-1 C Course of reaction COPY RIGHT RESERVED PAGE 103 PLEASE TURN OVER

7.1 Is this reaction ENDOTHERMIC or EXOTHERMIC? Give a reason for the answer. (2) 7.2 Use energy values A, B and C indicated on the graph and write down an expression for each of the following: 7.2.1 The energy of the activated complex (1) 7.2.2 The activation energy for the forward reaction (1) 7.2.3 ∆H for the reverse reaction (1) 7.3 Calculate the amount of energy released during the reaction if 0,18 moles of Ba(OH)2(aq) reacts completely with the acid. (3) [8] QUESTION 7 NOV 2016 Methane is used as an alternative fuel. The combustion of methane releases carbon dioxide and water. The balanced equation for this reaction is: CH 4 (g) + O 2 (g) → CO 2 (g) + 2H 2 O (g) + 891 kJ.mol-1 The activation energy for this reaction is 172 kJ.mol-1. 7.1 Is this reaction ENDOTHERMIC or EXOTHERMIC? Give a reason for the answer. (2) 7.2 Explain why all chemical reactions need activation energy. (2) 7.3 Why is this reaction not considered to be environmentally friendly? (2) [6] QUESTION 7 NOV 2017 The equation for the combustion of butane gas is given below. butane(g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g) ∆H < 0 7.1 Define the term activation energy. (2) 7.2 Is the combustion reaction of butane exothermic or endothermic? Give a reason for the answer. (2) 7.3 Draw a sketch graph of potential energy versus course of reaction for the reaction above. Clearly indicate the following on the graph: • Activation energy • Heat of reaction (∆H) • Reactants and products (3) 7.4 Determine the empirical formula of butane gas if it consists of 82,76% carbon and 17,24% hydrogen. (4) [11] COPY RIGHT RESERVED PAGE 104 PLEASE TURN OVER

CHEMICAL CHANGE MULTIPLE CHOICE QUESTIONS INTERMOLECULAR FORCE 1.3 NOV 2014 Which ONE of the following compounds has dipole-dipole forces between their molecules? A CO 2 B HCℓ C Cℓ 2 D CCℓ 4 (2) 1.3 NOV 2015 Which ONE of the following has the strongest forces between its molecules? A F2 B Cℓ 2 C Br 2 D I2 (2) 1.3 NOV 2016 The intermolecular forces in dry ice (CO 2 ) are … A ion-induced dipole forces. B hydrogen bonding. C ion-dipole forces. D London forces. (2) 1.2 NOV 2017 Solid iodine sublimes easily. The intermolecular forces present in iodine are … A London forces. B hydrogen bonding. C ion-dipole forces. D dipole-dipole forces. (2) COPY RIGHT RESERVED PAGE 105 PLEASE TURN OVER

CHEMICAL CHANGE LONG QUESTIONS INTERMOLECULAR FORCE QUESTION 3 NOV 2016 Learners conduct an experiment to investigate the effects of intermolecular forces on boiling points. They use 20 mℓ of each of the following compounds in their investigation: water, sunflower oil, nail polish remover, glycerine and methylated spirits. The results are shown in the table below: NAME OF BOILING COMPOUND POINT (°C) water 93 sunflower oil 230 nail polish remover 56 glycerine 290 methylated spirits 62 3.1 Define the term boiling point. (2) 3.2 Formulate an investigative question for this experiment. (2) 3.3 In which compound in the table above will the strongest intermolecular forces occur? Give a reason for the answer. (2) 3.4 The learners now use 40 mℓ of each of the compounds above in the experiment. Will it affect the boiling points? Choose YES or NO. Give a reason for the answer. (2) 3.5 Methylated spirits is highly flammable. State TWO safety precautions that should be taken when using methylated spirits in the laboratory. (2) 3.6 Which compound in the table above will have the highest rate of evaporation? Give a reason for the answer. (2) 3.7 Sunflower oil is a non-polar compound with induced dipole forces between the molecules, while water is a polar molecule with hydrogen bonds between its molecules. Explain why the boiling point of sunflower oil is higher than the boiling point of water. (2) [14] QUESTION 3 NOV 2017 Consider the list of six substances with their formulae and boiling points in the table below. NAME OF SUBSTANCE FORMULA BOILING POINT (°C) Water H2O 100 Ethanol CH 3 CH 2 OH 78 Bromine Br 2 58,8 Iodine I2 184,3 Ammonia NH 3 -33,3 Phosphine PH 3 -87,7 COPY RIGHT RESERVED PAGE 106 PLEASE TURN OVER

3.1 Explain why ethanol is soluble in water. Refer to the relative strength of the intermolecular forces in ethanol and water. (3) 3.2 Explain why the boiling point of iodine is higher than that of bromine. Refer to the intermolecular forces present in EACH substance in the explanation. (3) 3.3 Explain why phosphine will evaporate faster than ammonia by referring to the types of intermolecular forces present in EACH substance. (4) 3.4 Water, ethanol and bromine are all liquids at room temperature. Which ONE will have the highest vapour pressure? (1) 3.5 Give a reason for the answer to QUESTION 3.4 by referring to the relative strength of the intermolecular forces and boiling points. (2) [13] COPY RIGHT RESERVED PAGE 107 PLEASE TURN OVER

CHEMICAL CHANGE MULTIPLE CHOICE QUESTIONS QUANTITATIVE ASPECTS OF CHEMICAL CHANGE 1.5 NOV 2014 Which ONE of the following represents the greatest mass? A One chlorine atom B One chlorine molecule C One mole of chlorine D One gram of chlorine (2) 1.9 NOV 2014 10 moles of hydrogen gas (H 2 ) and 2,5 moles of nitrogen gas (N 2 ) are mixed and allowed to react to form ammonia (NH 3 ) according to the following balanced equation: 3H 2 (g) + N 2 (g) → 2NH 3 (g) If 4 moles of NH 3 (g) is formed during the reaction, the number of moles of H 2 (g) and N 2 (g) that remain in the container are respectively: Moles of H 2 (g) Moles of N 2 (g) A 0 0 B 7 1,5 C 4 0,5 D 4 2 (2) COPY RIGHT RESERVED PAGE 108 PLEASE TURN OVER

CHEMICAL CHANGE MULTIPLE CHOICE QUESTIONS QUANTITATIVE ASPECTS OF CHEMICAL CHANGE QUESTION 6 NOV 2014 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) 6.3 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) -1 6.3.3 If the molar mass of methyl benzoate is 136 g⋅mol , what is its molecular formula? (2) [15] QUESTION 6 NOV 2015 6.1 Sodium thiosulphate, Na 2 S 2 O 3 (s), reacts with 200 cm3 of a hydrochloric acid solution, HCℓ(aq), of concentration 0,2 mol·dm-3 according to the following balanced equation: Na 2 S 2 O 3 (s) + 2HCℓ(aq) → 2NaCℓ(aq) + S(s) + SO 2 (g) + H 2 O(ℓ) 6.1.1 Define the term concentration of a solution. (2) 6.1.2 Calculate the number of moles of HCℓ(aq) added to the sodium thiosulphate. (3) 6.1.3 Calculate the volume of SO 2 (g) that will be formed if the reaction takes place at STP. (3) 6.2 Menthol, the substance we can smell in mentholated cough drops, is composed of carbon (C), hydrogen (H) and oxygen (O). During combustion of a 9,984 g sample of menthol, it is found that 28,160 g of CO 2 (g) and 11,520 g of H 2 O(g) is produced. 6.2.1 Calculate the mass of carbon (C) in the CO 2 . (4) 6.2.2 Use relevant calculations to determine the empirical formula of menthol. (7) 6.2.3 The molar mass of menthol is 156 g·mol-1. Determine the molecular formula of menthol. (2) [21] COPY RIGHT RESERVED PAGE 109 PLEASE TURN OVER

QUESTION 5 NOV 2016 5.1 Define the term concentration. (2) 3 5.2 Eight (8) grams of Na 2 S 2 O 3 is dissolved in water to prepare 500 cm of solution. Calculate the concentration of the Na 2 S 2 O 3 solution. (3) 5.3 A 10 g sample of a compound contains 2,66 g of potassium, 3,54 g of chromium and 3,81 g of oxygen. 5.3.1 Define the term empirical formula. (2) 5.3.2 Determine the empirical formula of this compound. (7) [14] QUESTION 6 NOV 2016 Learners made a mini volcano in a science laboratory by adding sodium bicarbonate to ethanoic acid. They added 100 mℓ of a 0,2 mol.dm-3 ethanoic acid solution to 10 g of NaHCO 3 to start the reaction of the volcano. The balanced equation for this reaction is: CH 3 COOH(aq) + NaHCO 3 (s) → CH 3 COONa(aq) + H 2 O(ℓ) + CO 2 (g) 6.1 Define the term limiting reagent. (2) 6.2 Determine the limiting reagent in this reaction. (6) 6.3 Calculate the mass of the other substance in excess. (3) 6.4 Calculate the volume of CO 2 produced at STP. (4) [15] QUESTION 6 NOV 2017 6.1 The decomposition of hydrogen peroxide in the presence of a catalyst at standard pressure and room temperature is given by the unbalanced chemical equation below. H 2 O 2 (aq) → H 2 O(ℓ) + O 2 (g) The oxygen gas is collected and the volume is recorded over a period of time. The reaction is completed at time t. The results are plotted on a graph of volume O 2 versus time, as shown below. 600 Volume O2 (cm ) 3 Time (s) t COPY RIGHT RESERVED PAGE 110 PLEASE TURN OVER

Take the molar gas volume (V m ) as 24,45 dm3 at room temperature and standard pressure. 6.1.1 Balance the equation. (2) 6.1.2 How would a catalyst affect the reaction? (2) 6.1.3 Use the information on the graph to calculate the mass of hydrogen peroxide that decomposed. (6) 6.2 In an experiment, a learner adds 500 cm3 hydrochloric acid (HCℓ), with a concentration of 0,36 mol·dm-3, to 1,2 g of magnesium in a test tube. She records the change in the mass of magnesium as the reaction proceeds at regular intervals. The balanced chemical equation for the reaction is: Mg(s) + 2HCℓ(aq) → MgCℓ 2 (aq) + H 2 (g) The change in the mass of magnesium during the reaction is shown on the graph below. 1,2 Magnesium (g) Time (min.) 1 2 3 6.2.1 Identify the limiting agent in this reaction. Give a reason for the answer. (2) 6.2.2 Calculate the number of moles of unreacted hydrochloric acid in the test tube after 3 minutes. (7) [19] COPY RIGHT RESERVED PAGE 111 PLEASE TURN OVER

CHEMICAL CHANGE MULTIPLE CHOICE QUESTIONS ACIDS AND BASES 1.4 NOV 2014 Consider the following chemical reaction: HCO 3− + HC 2 O −4 ⇌ H 2 CO 3 + C 2 O 24− Which ONE of the following CORRECTLY identifies the order of Lowry-Brönsted acids and bases in the above reaction? A Base, acid, acid, base B Acid, base, base, acid C Acid, base, acid, base D Base, acid, base, acid (2) 1.1 NOV 2015 According to the Arrhenius theory, all bases … A are proton donors. B are proton acceptors. C form H3O+ ions in solution. D form OH‾ ions in solution. (2) 1.7 NOV 2016 The flowers of hydrangeas are natural indicators of soil pH. A natural indicator is made in a laboratory by using hydrangea flowers. NaOH and HCℓ are added to the indicator and the colour change is recorded in the table below. INDICATOR NATURAL COLOUR NaOH HCℓ Hydrangea flowers Blue Purple Pink If orange juice is added to the indicator above, the observed colour may be … A brown. B pink. C purple. D blue. (2) 1.8 NOV 2017 The CORRECT formula for nitric acid: A H 2 SO 4 B CH 3 COOH C NH 3 D HNO 3 (2) COPY RIGHT RESERVED PAGE 112 PLEASE TURN OVER

1.10 NOV 2017 Which ONE of the reactions below will produce the salt sodium ethanoate (sodium acetate)? A HCℓ(s) + CH 3 COOH(aq) → B CH3COOH(aq) + H2O(ℓ) → C CH3COOH(aq) + NaOH(aq) → D H2CO3(aq) + NaOH(aq) → (2) COPY RIGHT RESERVED PAGE 113 PLEASE TURN OVER

CHEMICAL CHANGE LONG QUESTIONS ACIDS AND BASES QUESTION 9 NOV 2014 9.1 Limestone, or sometimes ash, is used in pit latrines (long drops) to neutralise acidic waste. Limestone reacts with hydrochloric acid according to the following UNBALANCED equation: CaCO 3 (s) + HCℓ(aq) → CaCℓ 2 (aq) + H 2 O(ℓ) + CO 2 (g) 9.1.1 Define an acid in terms of the Arrhenius theory. (2) 9.1.2 Is ash acidic or basic? (1) 9.1.3 Rewrite the above equation into your ANSWER BOOK and then balance the equation. (1) 9.2 Sulphuric acid reacts with water in two steps as represented by the equations below. Equation I: H 2 SO 4 (aq) + H 2 O(ℓ) ⇌ H 3 O+(aq) + HSO −4 (aq) Equation II: HSO −4 (aq) + H 2 O(ℓ) ⇌ H 3 O+(aq) + SO 24 − (aq) 9.2.1 Define the term ampholyte. (2) 9.2.2 Write down the FORMULA of a species that acts as ampholyte in the above reactions. (1) 9.2.3 Write down the NAME of the conjugate base of the hydrogen sulphate ion. (1) 9.3 A standard sodium carbonate solution is prepared in a 250 cm3 volumetric flask. During a titration, 20 cm3 of a 0,1 mol·dm-3 nitric acid solution neutralises 25 cm3 of the above standard solution according to the following balanced equation: 2HNO 3 (aq) + Na 2 CO 3 (aq) → 2NaNO 3 (aq) + H 2 O(ℓ) + CO 2 (g) 9.3.1 Write down the NAME of the salt formed in the above reaction. (1) 9.3.2 Calculate the mass of sodium carbonate used to prepare the standard solution in the volumetric flask. (5) [14] COPY RIGHT RESERVED PAGE 114 PLEASE TURN OVER

QUESTION 8 NOV 2015 When sulphuric acid reacts with water, it ionises in two steps, as shown in the two balanced equations below. I: H 2 SO 4 (aq) + H 2 O(ℓ) → H 3 O+(aq) + HSO −4 (aq) + II: HSO −4 (aq) + H 2 O(ℓ) → H 3 O (aq) + SO 2− 4 (aq) 8.1 Define an acid in terms of the Lowry-Brønsted theory. (2) 8.2 Write down the FORMULA of: 8.2.1 The conjugate base of HSO −4 (aq) (1) 8.2.2 The conjugate acid of HSO −4 (aq) (1) 8.2.3 A substance that acts as ampholyte in these reactions (1) 8.3 A few drops of bromothymol blue indicator are added to a potassium hydroxide solution in a beaker. A dilute sulphuric acid solution is now gradually added to this solution until the colour of the indicator changes. Write down the: 8.3.1 Type of reaction that takes place (Write down only REDOX, PRECIPITATION or NEUTRALISATION.) (1) 8.3.2 Balanced equation for the reaction that takes place (3) 8.3.3 Colour change of the indicator (2) 8.3.4 NAME of the salt formed in this reaction (1) [12] QUESTION 8 NOV 2016 8.1 Define an acid in terms of the Lowry-Brønsted theory. (2) 8.2 Consider the following acid-base reaction: H 2 PO 4 - + NH 3 → HPO 4 2- + NH 4 + 8.2.1 Identify the conjugate acid-base pairs in the above reaction. (4) 8.2.2 Define the term ampholyte. (1) 8.2.3 Choose an ampholyte in the above reaction. (2) 8.3 Ten grams (10 g) of an impure sample of sodium carbonate is added to 100 cm3 of a 0,2 mol.dm-3 solution of hydrochloric acid. The acid is in excess. The equation for the reaction is: Na 2 CO 3 + HCℓ → NaCℓ + H 2 O + CO 2 8.3.1 Balance the equation above. (1) 8.3.2 Calculate the number of moles of hydrochloric acid. (3) The excess acid neutralises 20 cm3 of a solution of 0,1 mol.dm-3 of magnesium hydroxide. 2HCℓ(aq) + Mg(OH) 2 (aq) → MgCℓ 2 (aq) + H 2 O(ℓ) 8.3.3 Calculate the percentage purity of the sodium carbonate solution. (8) [21] COPY RIGHT RESERVED PAGE 115 PLEASE TURN OVER

QUESTION 8 NOV 2017 8.1 Two reactions of sulphuric acid are shown in the diagram below. Reaction 1 H2SO4 Salt X + Product Y NaOH Reaction 2 H2O Ampholyte A + H3O+ 8.1.1 Define a Lowry-Brønsted base. (2) 8.1.2 Write down a balanced equation for Reaction 1. (3) 8.1.3 Write down the NAME of the salt represented by X. (2) 8.1.4 Write down the FORMULA of ampholyte A. (2) 8.1.5 Write down the formulae of the TWO conjugate acid-base pairs in Reaction 2. (4) 8.2 A solution of sodium hydroxide (NaOH) is prepared by dissolving 6 g solid NaOH in 500 cm3 water. This solution reacts completely with 10 g impure ammonium chloride (NH 4 Cℓ) according to the equation below. NaOH(aq) + NH 4 Cℓ(s) → NaCℓ(aq) + H 2 O(ℓ) + NH 3 (aq) 8.2.1 Calculate the concentration of the NaOH solution. (4) 8.2.2 Calculate the percentage impurities in the NH 4 Cℓ. (6) [23] COPY RIGHT RESERVED PAGE 116 PLEASE TURN OVER

CHEMICAL CHANGE MULTIPLE CHOICE QUESTIONS ELECTROCHEMICAL REACTIONS 1.7 NOV 2015 Which ONE of the following balanced equations represents a redox reaction? A H+(aq) + OH-(aq) → H2O(ℓ) B Mg(s) + CuSO4(aq) → Cu(s) + MgSO4(aq) C 2NaCℓ(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbCℓ2(s) D H2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2HNO3(aq) (2) 1.9 NOV 2016 The oxidation number of phosphorus in H 3 PO 4 is … A +3 B -2 C +2 D +5 (2) 1.9 NOV 2017 Consider the reaction below. Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s) Which substance is the oxidising agent? A Zn B Cu 2+ C Zn 2+ D Cu (2) COPY RIGHT RESERVED PAGE 117 PLEASE TURN OVER

CHEMICAL CHANGE LONG QUESTIONS ELECTROCHEMICAL REACTIONS QUESTION 10 NOV 2014 Nitric acid and copper reacts according to the following unbalanced equation: HNO 3 (aq) + Cu(s) → Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(ℓ) 10.1 Define reduction in terms of oxidation numbers. (2) 10.2 For this reaction, write down the FORMULA of the: 10.2.1 Substance that is oxidised (1) 10.2.2 Reducing agent Allocate oxidation numbers to the relevant species and then explain the answer. (3) 10.2.3 Oxidising agent Explain the answer in terms of electron transfer. (2) 10.3 Balance the equation using the ion-electron method. Show the oxidation and reduction half-reactions during the balancing. (5) [13] QUESTION 9 NOV 2015 9.1 Oxidation numbers make it easier to determine whether an element or a substance is oxidised or reduced during a chemical reaction. 9.1.1 D efine the term oxidation with reference to oxidation numbers. (2) 9.1.2 C alculate the oxidation number of chromium in Cr2O2− 7 . (1) 9.1.3 Calculate the oxidation number of oxygen in H2O2. (1) 9.2 Consider the UNBALANCED equation below: - Fe2+(aq) + Cℓ2(g) → Fe3+(aq) + Cℓ 9.2.1 Define the term reducing agent with reference to electron transfer. (2) From the above equation, write down the: 9.2.2 FORMULA of the reducing agent (1) 9.2.3 FORMULA of the oxidising agent (1) 9.2.4 Reduction half-reaction (2) 9.2.5 Oxidation half-reaction (2) 9.2.6 Balanced net redox reaction (2) [14] COPY RIGHT RESERVED PAGE 118 PLEASE TURN OVER

QUESTION 9 NOV 2016 A silver Christmas tree can be made by placing copper wire, shaped in the form of a tree, into a silver nitrate solution. The unbalanced equation for the reaction is: Cu(s) + AgNO 3 (aq) → Cu(NO 3 ) 2 (aq) + Ag(s) 9.1 Define the term oxidation in terms of oxidation number. (2) 9.2 Write down the following for the reaction above: 9.2.1 Formula of the reducing agent (2) 9.2.2 Name of the oxidising agent (2) 9.2.3 Oxidation half-reaction (2) 9.2.4 Balanced net ionic equation using the ion-electron method (4) 9.3 Use oxidation numbers to explain your choice of oxidising agent in QUESTION 9.2.2. (2) [14] QUESTION 9 NOV 2017 The reaction between dichromate ions (Cr 2 O 7 -2) and iron(II) ions (Fe2+) in an acidic medium is given below. Cr 2 O 7 -2(aq) + Fe2+(aq) + H+(aq) → Cr3+(aq) + Fe3+(aq) + H 2 O(ℓ) 9.1 Determine the oxidation number of CHROMIUM in Cr2O7-2(aq). (2) 9.2 Define reduction in terms of electron transfer. (2) 9.3 Write down the FORMULA of the substance that undergoes oxidation. Explain the answer in terms of oxidation numbers. (2) 9.4 Write down the FORMULA of the oxidising (2) 9.5 agent. Write down the reduction half-reaction. (2) 9.6 Write down the net balanced ionic equation for the reaction, using the ion-electron method. (3) [13] COPY RIGHT RESERVED PAGE 119 PLEASE TURN OVER

CHEMICAL SYSTEMS MULTIPLE CHOICE QUESTIONS 1.8 NOV 2015 During the extraction of gold, zinc powder is added to a solution of gold cyanide to produce gold according to the following balanced equation: Zn(s) + 2NaAu(CN) 2 (aq) → 2Au(s) + Zn(CN) 2 (aq) + 2NaCN(aq) The oxidising agent in this reaction is ... A Au+ B Zn C Na+ D CN − (2) 1.10 NOV 2016 During the processing of gold ore, zinc is added to the gold cyanide solution to produce gold according to the balanced equation below: Zn(s) + 2NaAu(CN) 2 (aq) → 2Au(s)+ Zn(CN) 2 (aq) + 2NaCN(aq) The reducing agent in this reaction is … A Na+ B Au+ C Zn D CN- (2) COPY RIGHT RESERVED PAGE 120 PLEASE TURN OVER

CHEMICAL SYSTEMS LONG QUESTIONS QUESTION 11 NOV 2014 Mining has been the main driving force behind the history and development of South Africa. Diamond and gold production may now be well down from their peaks, though South Africa is still number 5 in gold. It is the world's largest producer of chrome, manganese, platinum and vanadium, as well as the world's third largest coal exporter. In 2012 South Africa overtook India to become the world third biggest iron ore supplier to China, the world's largest consumer of iron ore. 11.1 Choose from the list below the mining activity that you have studied and then answer the questions that follow. gold; iron; phosphate; coal; diamonds; copper; platinum; zinc; chromium; asbestos; manganese 11.1.1 What is the location of the major mining activity in South Africa? (1) 11.1.2 What type of mining is used to recover the selected mineral? (1) 11.2 Mining has advantages and disadvantages. 11.2.1 Give TWO reasons why the mining industry is so important to the South African economy. (2) 11.2.2 Write down TWO negative impacts that mining has on the environment. (2) 11.3 A large deposit of a precious metal is discovered in South Africa. Write down TWO factors which have to be considered before developing the site for mining. (4) [10] QUESTION 10 NOV 2015 The simplified diagram below shows a blast furnace used for the extraction of iron from iron ore. P represents a reactant added to the blast furnace. Q and R represent products that leave the blast furnace. Raw materials Iron ore Limestone Coke Waste Waste gases gases out out P R Q COPY RIGHT RESERVED PAGE 121 PLEASE TURN OVER

Limestone and coke are added to the blast furnace, as shown in the diagram above. Write down the function of: 10.1.1 Limestone (1) 10.1.2 Coke (1) 10.2 Write down the NAME of the substance represented by: 10.2.1 P (1) 10.2.2 Q (1) 10.2.3 R (1) 10.3 Write down the NAME or FORMULA of ONE waste gas formed during the extraction of iron from iron ore. (1) 10.4 The balanced equation for the extraction of iron from iron ore is: Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) Write down the: 10.4.1 Function of carbon monoxide in this reaction (1) 10.4.2 FORMULA of the substance that is reduced (1) [8] QUESTION 10 (Start on a new page.) The discovery of gold played a significant role in the economic development of South Africa. In 1970 gold mining in South Africa contributed 68 per cent to global production. 10.1 Which TWO mining methods are used in South Africa? (2) 10.2 Give TWO reasons why mining is important for the South African economy. (2) 10.3 State TWO negative effects of mining, with respect to the environment. (2) 10.4 Most alloys of gold are used in jewellery and dentistry. The gold content in jewellery is expressed in carat. The term indicates the number of parts of gold present in each 24 parts of alloy. 10.4.1 Pure gold, which is 24 carat, is not used to make jewellery. Give ONE reason why, referring to the properties of gold. (1) 10.4.2 Apart from gold being used in jewellery and dentistry, name ONE other use of gold. (1) >@ COPY RIGHT RESERVED PAGE 122 PLEASE TURN OVER

QUESTION 10 NOV 2017 Gold and iron are two of many minerals mined in South Africa. Iron is mined in open-cast mines, while gold is usually found in deep-shaft (underground) mines. During the process of refining, the following chemical reactions take place to extract the metal from the ore: Gold is dissolved in a solution containing cyanide ions (CN-) to extract it from the ore. The balanced chemical equation for the reaction is: 4Au(s) + 8NaCN(aq) + 2H 2 O(ℓ) + O 2 (g) → 4NaAu(CN) 2 (aq) + NaOH(aq) Iron(Vl) oxide and carbon are heated in a furnace to extract iron from the ore. The balanced chemical equation for the reaction is: 2Fe 2 O 3 (s) +3C(s) → 4Fe(ℓ) + 3CO 2 (g) 10.1 State TWO advantages of open-cast mining when compared to deep-shaft (underground) mining. (2) Consider the iron extraction reaction. 10.2 Is iron oxidised or reduced during the reaction? Give a reason for the answer. (2) 10.3 State TWO disadvantages of using carbon in this reaction. (2) Consider the gold extraction reaction. 10.4 Give ONE reason why gold is present as an element in the ore. (2) 10.5 What role does oxygen gas (O 2 ) play in the reaction? (2) [10] COPY RIGHT RESERVED PAGE 123 PLEASE TURN OVER