SCIENCECLINIC SMARTPREP GR12 DBE ENG 2023 V4.1 (1)

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VIDEO LESSONS! We have free video lessons for every topic in this book! These videos can only be accessed via Science Clinic's website - follow these steps to access this amazing resource! DOWNLOAD A BOOK Visit our website to download a SmartPrep book. We'll send you an email with download links. VIDEOS WEBPAGE Go to our Videos webpage. Log in with the email address that you used to download a book. SELECT YOUR TOPIC From your videos page, select a topic. You'll be redirected to a playlist with video lessons. ABOUT OUR SMARTPREP BOOKS Our SmartPrep books provide crystal-clear summaries of the essential theory for Grade 8-12 Core Mathematics and Grade 10-12 Physical Science. Our books are available in English and Afrikaans. COPYRIGHT NOTICE AND LICENSE The content in this SmartPrep Book is the original work of Science Clinic (Pty) Ltd. You may distribute this material, provided that you adhere to the conditions in our copyright notice and license: ¢ If you copy and distribute this book electronically or in paper form, you must keep this copyright notice intact. ¢ If you use questions or syllabus summaries from this book, you must acknowledge the author. ¢ This Smart Prep Book is meant for the benefit of the community and you may not use or distribute it for commercial purposes. * You may not broadcast, publicly perform, adapt, transform, remix, add on to this book and distribute it on your own terms. ° By exercising any of the rights to this Smart Prep Book, you accept and agree to the terms and conditions of the license provided on this link: CONTENT ACKNOWLEDGEMENT Thank you to all the amazing people involved in the production, translation and moderation of this book: R_ Bartholomew, L Couperthwaite, N Cullinan, W Cloete, C de Beer, S Dippenaar, T Fairless, | Govender, C Hare, L Kroukamp, R Lodge, K Munnik, C Orchison, M Peyper, S Pienaar, Y Patterson, R Ramsugit, S Roberts, C Steyn, K Storm, S Sapsford, X Sithenjwa, A Theron, C Visser, B Ward QD ww sciencectinie.coza © exciencectinic EB ew acodook.comisciencectiniesa

TABLE OF CONTENTS CORE THEORY SUMMARIES Physics Physics Data Sheets 1 Physics Definitions 4 2D Vectors 5 Newton’s Laws of Motion 7 Newton’s Law of Universal Gravitation 11 Momentum and Impulse 12 Work, Energy and Power 16 Motion in 1D 19 Vertical Projectile Motion 20 Electricity 23 Electrostatics 25 Electromagnetism and Electrodynamics 27 Photoelectric Effect 30 Doppler Effect 32 Chemistry Chemistry Data Sheets 33 Chemistry Definitions 37 Organic Chemistry 39 Quantitative Aspects of Chemical Change 48 Energy and Chemical Change 51 Rates of Reactions 52 Chemical Equilibrium 53 Acids and Bases 56 Electrochemistry 59 Fertilisers 63 www

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DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 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 Universal gravitational constant G 6,67 x 10-11 N·m2·kg-2 Universele gravitasiekonstant Speed of light in a vacuum c 3,0 x 108 m·s-1 Spoed van lig in 'n vakuum Planck's constant h 6,63 x 10-34 J·s Planck se konstante Coulomb's constant k 9,0 x 109 N·m2·C-2 Coulomb se konstante 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 Earth M 5,98 x 1024 kg Massa van Aarde Radius of Earth RE 6,38 x 106 m Radius van Aarde

TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING vf vi  a 't Δx v i Δt  21 a't 2 or/of Δy v i Δt  21 a't 2 § vi  v f · § vi  v f · vf 2 2 v i  2a'x or/of v f 2 2 v i  2a'y Δx ¨ ¸ Δt or/of Δy ¨ ¸ Δt © 2 ¹ © 2 ¹ FORCE/KRAG Fnet ma p mv fs max = μsN fk = μk N Fnet Δt = Δp w mg Δp = mv f - mvi m1m 2 m1m 2 M M F =G or/of F =G g =G or/of g =G 2 d2 r2 d2 r WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W F'x cos T U mgh or/of EP mgh Wnet 'K or/of Wnet 'Ek 1 1 K mv2 or/of Ek mv2 2 2 'K K f  Ki or/of 'Ek Ekf  Eki W Wnc 'K  'U or/of Wnc 'Ek  'Ep P 't Pav = Fvav / Pgemid = Fvgemid WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG 1 v fO T f v r vL v r vL c fL fs fL fb E hf or /of E h v r vs v r vb O E = Wo + Ek(max) or/of E = Wo + K max where/waar 1 2 1 2 E hf and/en W0 hf0 and/en E k (max) = mvmax or/of K max = mvmax 2 2

ELECTROSTATICS/ELEKTROSTATIKA kQ1Q 2 kQ F E r2 r2 W F V E q q Q Q n= or/of n= e qe ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE emf ( ε )= I(R + r) V R I emk ( ε )= I(R + r) Rs R1  R 2  ... 1 1 1 q I 't   ... Rp R1 R 2 W = Vq W P Δt W = VI ' t P = VI W = I2R ' t 2 P = I 2R V Δt W= V2 R P R ALTERNATING CURRENT/WISSELSTROOM Pave Vrms Irms / Pgemiddeld Vwgk I wgk I max Imaks I rms / I wgk 2 2 2 Pave Irms R / Pgemiddeld I 2wgkR Vmax Vmaks Vrms / Vwgk 2 Vrms 2 Vwgk 2 2 Pave / Pgemiddeld R R

Grade 12 Science Essentials SCIENCE CLINIC 2022 © Grade 12 Physics Definitions Normal force (FN): the force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface Frictional force (Ff): the force that opposes the motion of an object and which acts parallel to the surface Static frictional force (fs): the force that opposes the tendency of motion of a stationary object relative to a surface Kinetic frictional force (fk): the force that opposes the motion of a moving object relative to a surface 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/ Newton’s Laws net force acts on it Revision 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 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 Newton's Law of Universal Gravitation: each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres Weight: the gravitational force the Earth exerts on any object on or near its surface Momentum: the product of an object's mass and its velocity Linear momentum: a vector quantity with the same direction as the velocity of the object Newton’s second law in terms of momentum: the net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force Impulse: the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object. It is a Momentum and vector quantity Impulse Isolated system (in Physics): a system on which the resultant/net external force is zero. An isolated system excludes external forces that originate outside the colliding bodies, e.g. friction. Only internal forces, e.g. contact forces between the colliding objects, are considered Principle of conservation of linear momentum: the total linear momentum of an isolated system remains constant (is conserved) Vertical Projectile: an object which has been given an initial velocity and then moves under the influence of the gravitational force only Projectile Motion Free fall: motion during which the only force acting on an object is the gravitational force Work done on an object by a constant force F: is FΔxcosθ , where F is the magnitude of the force, Δx the magnitude of the displacement and θ the angle between the force and the displacement Work-energy theorem: The work done on an object by a net force is equal to the change in the object's kinetic energy OR the work done on an object by a net force is equal to the change in the object's kinetic energy. (Wnet = ΔK = Kf - Ki) Work, Energy Conservative force: is a force for which the work done in moving an object between two points is independent of the path taken and Power Non-conservative force: is a force for which the work done in moving an object between two points depends on the path taken Principle of conservation of mechanical energy: the total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant Power: is the rate at which work is done or energy is expended Doppler Effect: the change in frequency (or pitch) of the sound detected by a listener, because the sound source and the listener Doppler Effect have different velocities relative to the medium of sound propagation Coulomb's law: the magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them Electrostatics Electric field: a region of 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 Electric field at a point: The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point Ohm's law: the potential difference across a conductor is directly proportional to the current in the conductor at constant temperature Electric Circuits Power: the rate at which work is done Emf: the maximum energy provided by a battery per unit charge passing through it rms value: the DC potential difference/current which dissipates the same amount of energy as AC rms potential difference: the AC potential difference which dissipates/produces the same amount of energy as an equivalent DC Electrodynamics potential difference. rms current: the alternating current which dissipates/produces the same amount of energy as an equivalent direct current (DC). Optical Photoelectric effect: the process whereby electrons are ejected from a metal surface when light of suitable frequency is incident Phenomena and on that surface Properties of Threshold frequency (fo): the minimum frequency of light needed to emit electrons from a certain metal surface Materials Work function (Wo): the minimum energy that an electron in the metal needs to be emitted from the metal surface Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 4

Grade 12 Science Essentials Vectors in 2D SCIENCE CLINIC 2022 © RESOLVING INTO Fx COMPONENTS ON A COMPONENTS x SLOPE θ Fg // = Fg sin θ Diagonal vectors can be broken into When forces act on objects on a components. When vectors are bro- Fx = F cos θ slope, it is useful to resolve vec- Fg ⊥ = Fg cos θ F g// ken into the x- and y-components, F Fy tors into components that are Fg θ F g⟂ we are determining the horizontal Fy = F sin θ parallel (//) or perpendicular (⟂) (x-axis) and vertical (y-axis) effect components. The most common Fg θ F g⟂ of the vector. We can use sin or cos, force resolved into components θ provided that the angle is between on a slope is weight (Fg). the force F and the horizontal. y F g// CONSTRUCTING FORCE TRIANGLE When forces are not co-linear, force triangles can be used to determine resultant forces or the equilibrant. When force triangles are formed, basic geometric rules can be used to determine vectors or resultants. Tail-to-head Parallelogram Manipulation Used for consecutive vectors (vectors that occur in sequence). Used for vectors that act concurrently on the same object. The vector arrows can be manipulated to form a force triangle to determine the The resultant is the diagonal of a parallelogram that originates resultant forces or an equilibrant. The vectors/arrows may only be moved if the Eg. A boat travels 90 m east, and then moves 50 m north. y from the tail of the vectors (common origin). magnitude and direction are both kept constant. y When manipulating the vector arrows, the following has to remain the same: • Length of arrow (magnitude) • Angle of the arrow (direction) • The direction of the arrow head t Vector 2 tan su l Re Eg. An object is suspended from a ceiling by 2 cables. Below is a free body dia- Vector 2 nt l ta gram as well as a force triangle that can be used to calculate the values of T1 and Resu T2. Free body diagram Force triangle x Vector 1 x This principle can also be applied to more than 2 vectors Vector 1 taken in order. The resultant is from the tail of the first vector to the head of the last. Eg. Two tugboats apply a force of 6 000N and 5 000N at bear- y ings of 60° and 120° respectively on a cargo ship. T1 Vect y T2 T1 or 4 nt Fg Resulta 0N or 3 6 00 Resultant Vect x x 50 Fg T2 00 N Vector 1 o r2 ct Ve Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 5

Grade 12 Science Essentials 2D Vectors- Resultant and Equilibrant SCIENCE CLINIC 2022 © RESULTANT: The single vector which has the same effect as the EQUILIBRANT: The force that keeps a system in equilibrium. original vectors acting simultaneously on an object. The equilibrant is equal in magnitude but opposite in direction to the resultant force. PYTHAGORAS (90° ONLY) COMPONENT ADDITION Pythagoras can only be applied to vector triangles that are right angle triangles. The resultant of diagonal forces can be determined using Pythagoras by determining the x-resultant and y-resultant first. This is especially useful for determining resultants when more than 2 forces act on an FOR FINDING SIDES: FOR FINDING ANGLES: object and a force triangle can not be used. o a o R2 = x2 + y2 sin θ = cos θ = tan θ = EXAMPLE: h h a Three forces act on an object as shown in the diagram below. Determine the resultant force on the ob- ject. EXAMPLE: 1.Determine the x- and y-components of each force. A boat travels 90 m due east, and then moves 50 m due north. Determine the displacement 11N force: of the boat. y Fy = F sin θ Fx = F cos θ = 11 cos 70 = 11 sin 70 11 N = 3,76 N right (90o ) = 10,34 N up (0o ) 70° 40° 35° 30N force: Fx = F cos θ Fy = F sin θ 30 N 20 N tant = 30 cos 40 = 30 sin 40 su l 22,98 N left (270o ) Re 50 m = = 19,28 N down (180o ) 20N force: Fx = F cos θ Fy = F sin θ = 20 cos 35 = 20 sin 35 θ x = 16,38 N right (90o ) = 11,47 N down (180o ) 90 m R2 = x2 + y2 2. Determine the x- and y-resultants of components. R = 2 90 + 50 2 Take left (270o) as positive Take down (180o) as positive 2,84 N R = 102,96 m Fx = −3,76 + 22,98 − 16,38 Fy = −10,34 + 19,28 + 11,47 θ = 2,84 N left (270o ) = 20,41 N down (180o ) o tan θ = 20,41 N a tan−1( 50 R 90 ) 3. Find resultant-Pythagoras. 4. Find angle- trigonometry θ = o R2 = x2 + y2 tan θ = a θ = 29,05∘ 20,41 R = 2,842 + 20,412 θ = tan−1 2,84 ∘ Remember that θ calculated is relative to the x-axis, R = 20,61 N θ = 82,08 ∘ ∘ ∘ ∴ bearing = 90 − 29,05 = 60, 95 ∴Resultant = 20,61 N at a bearing of 187,92° ∴ Displacement = 102,96 m at a bearing of 60,95∘ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 6

Grade 12 Science Essentials Newton’s Laws of Motion SCIENCE CLINIC 2022 © FORCES Non-contact force: A force exerted between Contact force: A force exerted between A force is a push or a pull action exerted on an object by another object. This action can be exerted while objects are in contact (contact force) or over a distance (non-contact force). objects over a distance without physical contact. objects that are in contact with each other. Because forces have magnitude and direction, they are vectors. Force is measured in newton (N). 1 N is Electrostatic force (FE) Applied force (FA) the force required to accelerate a 1 kg object at 1 m·s-2 in the direction of the force. We can therefore say that 1 N = 1 kg·m·s-2. Gravitational force (w/Fg) Tension (T or FT) NOTE: Magnetic force Friction (Ff or fs/fk) Free-Body Diagram: forces drawn away from a dot which represents the object Normal force (N/FN) Normal force (FN) Friction (Ff or fs/fk) The perpendicular force exerted by a surface on an object in contact with it. Frictional force due to a surface is the force that opposes the motion of an object in contact with it, parallel to the surface. FN Friction is the parallel component of the contact force on an object by the surface on which The normal force is equal to the perpendicular component of gravity if there are no other forces acting on the object. it rests. The friction between the contact surfaces is determined by the properties of both the contact surfaces of the object and surface. The coefficient of friction (µs/µk) is a description of the roughness of the surface. The rougher the surface, the greater the coefficient of friction. FN = Fg Static friction (fs) Kinetic friction (fk) Fg Static friction is the frictional force on a sta- Kinetic friction is the frictional force on a mov- If alternative forces act on the object, the normal force will change depending on the direction and mag- tionary object that opposes the tendency of ing object that opposes the motion of the nitude of the applied force. All vertical forces must be balanced if there is no acceleration in the vertical motion of the object. The magnitude of the object. The magnitude of the kinetic friction is plane. static friction will increase from 0N as the parallel constant for the specific system at all velocities FA FA component of the applied force is increased, until greater than zero, and irrespective of the applied FN FN maximum static friction is reached. fs(max) is the force. θ θ magnitude of friction when the object just starts to move. fs(max) = μs FN fk = μk FN Fg Fg fs(max) = maximum static friction (N) fk = kinetic friction (N) μs = coefficient of friction (no unit) μk = coefficient of friction (no unit) FN + FAy − Fg = 0 FN − Fg − FAy = 0 FN = normal force (N) FN = normal force (N) FN + FA sin θ = Fg FN = Fg + FA sin θ If the applied force is greater than the maximum static friction, the object will start to move. Objects suspended from a rope/string/cable have no fsmax FT normal force, as there is no surface on which the object rests. Fric%on (N) The tension is equal to the perpendicular component of ) (f s gravity if there are no other forces acting on the object n c $o OR the full magnitude of Fg for vertically suspended c fri Kine%c fric%on (fk) objects that are stationary/moving at constant velocity. a$ St FT + (−Fg ) = 0 Fg Applied force (N) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 7

Grade 12 Science Essentials Newton’s Laws of Motion SCIENCE CLINIC 2022 © Newton’s First Law of Motion Newton’s Second Law of Motion Newton’s Third Law of Motion An object continues in a state of rest or uniform (moving with When a net force is applied to an object of mass, it acceler- When object A exerts a force on object B, object B simulta- constant) velocity unless it is acted upon by an unbalanced (net ates in the direction of the net force. The acceleration is neously exerts an oppositely directed force of equal mag- or resultant) force. directly proportional to the net force and inversely propor- nitude on object A. Newton’s First Law is due to inertia- the resistance of an object to tional to the mass. change its state of rest or constant/uniform motion. NB! Newton’s Third Law describes action-reaction force pairs. These −2 Newton’s Second Law is dependent on the resultant force- Fnet = 0 N a = 0 m⋅s The vector sum of all forces acting on the same object. are forces on different objects and can not be added or sub- tracted. A 3kg object moves up an incline surface at an angle of 15º with a constant velocity. The coe"cient of friction is 0,35. Determine Fnet = m a a ≠ 0 m ⋅ s −2 FA on B = − FB on A the magnitude of the applied force. FA FN A 20 N force is applied to a 5 kg object. The object accelerates Force pairs properties: up a frictionless incline surface at an angle of 15º. Determine • Equal in magnitude the acceleration of the object. • Opposite in direction Fg// FA • Acts on different objects (and therefore DO NOT CANCEL each FN other out) fk 15° Fg T NOTE: The force pairs shown Take upwards as positive: Fg// here are gravitational Fnet⊥ = 0 Fman on earth forces. FN + (−Fg ⊥ ) = 0 FN = Fg ⊥ 15° Fg Fearth on man Gravity and Normal force are NOT force pairs. T FN = m g cos θ FN = (3)(9,8)cos 15∘ Take upwards as positive: FN = 28,40 N Fnet// = ma ∴ FN = 28,40 N ⊥ up from slope FA + (−Fg // ) = ma Fman on wall 20 − (5)(9,8)sin 15∘ = 5a Fnet// = 0 20 − 12,68 = 5a FA + (−Fg // ) + (−fk ) = 0 7,32 a = FA = Fg // + fk 5 FA = m g sin θ + μk FN ∴a = 1,46 m ⋅ s−2 // up the slope FA = (3)(9,8)sin 15∘ + (0,35)(28,40) ∴ FA = 17,55 N Effect of Newton’s Second Law on overloading: According to Newton’s Second Law, the acceleration of an object Fwall on man Importance of wearing safety belts: is directly proportional to the net force and inversely proportional According to Newton’s First Law, an object will remain in motion at a to the mass of the object. If a vehicle is overloaded, the stopping constant velocity unless a non-zero resultant force acts upon it. When a distance will increase which can lead to serious accidents. When Newton’s Third Law during an accident car is in an accident and comes to a sudden stop, the person inside the brakes are applied, the force (friction) remains the same, but the According to Newton’s Third Law, the force that two objects exert car will continue with a constant forward velocity. Without a safety increase in mass causes a decrease in negative acceleration, in- on each other is equal in magnitude but opposite in direction. If belt, the person will make contact with the windscreen of the car, caus- creasing the time (and distance) it takes for the vehicle to stop. two cars are in an accident, they will both exert the same amount ing severe head trauma. The safety belt acts as an applied force (new of force on each other irrespective of their masses. Fnet), preventing the forward motion of the person. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 8

ALL EXAMPLES: Grade 12 Science Essentials Newton’s Laws of Motion D IRECTION OF MOTION POSITIVE SCIENCE CLINIC 2022 © Horizontal Slopes The vertical resultant = 0 N. The perpendicular (⟂) resultant = 0 N. Fg // = Fg sin θ The horizontal resultant determines acceleration. The parallel (//) resultant determines acceleration. Fg ⊥ = Fg cos θ Pulled at an angle REMEMBER: Use components of weight. FA FN Force applied down the slope Force applied up the slope FAy FN θ FN Ff FA FN FA FN Ff FN Ff Ff FAx Fg// Fg// Fg// Fg// Fg Fg FA Fg⟂ Ff FA fk Fg⟂ Horizontal: Vertical: θ Fg T θ Fg T Fnet = m a Fnet = 0 FAx + (−Ff ) = m a (−Fg ) + FN + FAy = 0 Parallel: Perpendicular: Parallel: Perpendicular: Fnet = m a Fnet = 0 Fnet = m a Fnet = 0 Pushed at an angle Fg ∥ + FA + (−Ff ) = m a Fg ⊥ + (−FN ) = 0 (−Fg ∥) + (−Ff ) + FA = m a Fg ⊥ + (−FN ) = 0 FN FA FN No force applied θ Parallel: Ff FAx Ff FN Ff FN Ff Fnet = m a Fg ∥ + (−Ff ) = m a FAy Fg Fg Fg// Fg// Perpendicular: Fnet = 0 Horizontal: Fnet = m a Vertical: Fnet = 0 Fg Fg⟂ Fg ⊥ + (−FN ) = 0 θ T FAx + (−Ff ) = m a Fg + (−FN ) + FAy = 0 Lift stationary/constant velocity Lift accelerating Lift in freefall (cable snap) FT FT FT Suspended FT Vertical: Vertical: Vertical: Horizontal resultant = 0 N. Fnet = 0 Fnet = m a Fnet = m a Vertical resultant determines acceleration. Fg + (−FT ) = 0 Fg + (−FT ) = m a Fg = m a REMEMBER: No normal or friction forces. Fg Fg Fg Fg Acceleration will be in the direc- Fg Fg tion of the greatest force. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 9

ALL EXAMPLES: Grade 12 Science Essentials Newton’s Laws of Motion D IRECTION OF MOTION POSITIVE SCIENCE CLINIC 2022 © Connected objects (e.g Pulley Systems) Objects attached by rope/cable Do separate free body diagrams for each object. FN FN FN FN The velocity and acceleration of all objects are equal in Horizontal: Horizontal: magnitude and direction. Fnet = m a Ff FT FT FA Ff FA Fnet = m a Applied forces are applied to only one object at a time. (−Ff ) + FT = m a Ff FA + (−FT ) + (−Ff ) = m a Simultaneous equations for acceleration and tension are FT sometimes needed. Fg Ff Fg Fg Fg REMEMBER: Ropes/cables- The tension forces on the objects are the same in magnitude Objects in contact but opposite in direction. FN FN FN FN Touching objects- Newton’s Third Law FA Horizontal: A FBA FAB B Horizontal: Fnet = m a Ff FA Ff Ff FAB Fnet = m a Same axis Ff FA + (−FBA) + (−Ff ) = m a (−Ff ) + FAB = m a Can be horizontal (multiple objects on a surface) or vertical FBA (multiple suspended objects). Fg Fg The velocity and acceleration of all objects are equal in Fg Fg magnitude and direction. Multiple axes Multiple axes FN Horizontal (objects on a surface) AND vertical (suspended FT1 FN FN objects). In these examples, FT1 FT2 The velocity and acceleration of all objects are equal in Ff FT1 clockwise is positive: Ff magnitude NOT DIRECTION. Right positive FT2 Fg Ff Fg Down positive Fg Fg Vector direction on multiple axes FT2 Clockwise: Fg Right and Down positive Fg Left and Up negative FN FT2 FN FN FT2 FT1 Ff FT1 Ff FT1 Ff FT2 OR FT1 Anti-clockwise: Fg Fg Fg Fg Fg FT2 Fg Left and Up positive Right and Down negative Horizontal: Vertical: Vertical: Horizontal: Horizontal: Vertical: Fnet = m a Fnet = m a Fnet = m a Fnet = m a Fnet = m a Fnet = m a FT1 + (−Ff ) = m a Fg + (−FT1) + FT 2 = m a Fg + (−FT 2 ) = m a FT1 + (−Ff ) = m a FT 2 + (−FT1) + (−Ff ) = m a Fg + (−FT 2 ) = m a Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 10

Grade 12 Science Essentials Newton’s Law of Universal Gravitation SCIENCE CLINIC 2022 © Every particle in the universe attracts every other particle with RATIOS CALCULATIONS a force which is directly proportional to the product of their 1. Write out the original formula. Gm1m2 masses and inversely proportional to the square of the distance 2. Manipulate unknown as subject. The gravitational force can be calculated using F = between their centres. 3. Substitute changes into formula (Keep symbols!). r2 4. Simplify ratio number. REMEMBER: Gm1m2 5. Replace original formula with unknown symbol. Mass in kg F= EXAMPLE: r2 Two objects, m1 and m2, are a distance r apart and experience a Radius in m force F. How would this force be affected if: Radius: centre of mass to centre of mass. F = force of attraction between objects (N) G = universal gravitational constant (6,67 ×10−11 N·m2·kg−2 ) a) One mass is doubled and the distance between the masses is Direction is ALWAYS attractive. m = object mass (kg) halved? Both objects experience the same force. r = distance between object centers (m) Gm1m2 (Newton’s Third Law of Motion) F= Write out the formula r2 A uniform sphere of matter attracts a body that is outside the shell as if G(2m1)m2 EXAMPLE: all the sphere’s mass was concentrated at its center. Fnew = Substitute changes into formula ( 1 r)2 The earth with a radius of 6,38 x 103 km is 149,6 x 106 km 2 away from the sun with a radius of 696 342 km. If the earth 2 Gm1m2 has a mass of 5,97 x 1024 kg and the sun has a mass of = Simplify ratio number 1 r2 1,99 x 1030 kg, determine the force between the two bodies. 4 Gm1m2 = 8( ) Thus, the distance is determined between the centers of the two bodies. r2 r = 6,38 × 10 3 km + 149,6 × 106 km + 696 342 km ∴ Fnew = 8 F Replace original formula = 6,38 × 106 m + 149,6 × 109 m + 696 342 × 10 3 m b) Both the two masses as well as the distance are doubled? = 1,5 × 1011 m Gm1m2 F= Write out the formula Gm1m2 r2 F= rmoon rman G(2m1)(2m2 ) r2 Fnew = Substitute changes into formula 6,67 × 10−11(5,97 × 10 24 )(1,99 × 10 30 ) NOTE: NOTE: (2r)2 F= The radius of the earth is added The radius of object 4 Gm1m2 (1,50 × 1011)2 = Simplify ratio number to the distance between the (man) on the earth is 4 r2 F = 3,52 × 10 22 N attraction earth and the moon. negligibly small. Gm1m2 = 1( r2 ) The force of gravitational attraction is a vector, therefore all vec- tor rules can be applied: KNOW THE DIFFERENCE! ∴ Fnew = 1 F Replace original formula • Direction specific g vs G DETERMINING GRAVITATIONAL ACCELERATION (g) • Can be added or subtracted g: Gravitational acceleration (9,8 m·s−2 on earth) Gm object mPlanet g is the acceleration due to gravity on a specific planet. F = m object g and F= r 2Planet G: Universal gravitational constant (6,67×10−11 N·m2·kg−2) Proportionality constant which applies everywhere in the universe. Gm o mP m og = r 2P Mass vs Weight Take right as positive: Gm o mP Mass (kg) g= Fnet on satallite = Fm on s + Fe on s m o r 2P = −( ) + ( res 2 ) A scalar quantity of matter which remains constant everywhere in the Gm m m s Gm e m s GmP universe. ∴g= rms 2 r 2P ∴ ( m 2 s ) = ( e2 s ) Weight (N) [gravitation force] Gm m Gm m Weight is the gravitational force the Earth exerts on any object. Weight Therefore the gravitational acceleration of an object only depends rms res differs from planet to planet. Fg = mg. Weight is a vector quantity. on the mass and radius of the planet. Object mass is irrelevant! Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 11

Grade 12 Science Essentials Momentum and Impulse SCIENCE CLINIC 2022 © MOMENTUM VECTOR NATURE OF MOMENTUM NEWTON’S SECOND LAW OF MOTION Momentum: the product of the mass of an Momentum is a vector quantity and has both mag- Newton’s second law in terms of momentum: The resultant/net Derivation from Newton’s object and its velocity. nitude and direction. It is therefore important to force acting on an object is equal to the rate of change of mo- Second Law always include direction in all momentum calcu- mentum of the object in the direction of the resultant/net force. Fnet = m a Momentum can be thought of as quantifying the lations. According to Newton’s Second Law, a resultant force applied to an object motion of an object. The following equation is used Δv to calculate momentum: will cause the object to accelerate. When the net force on an object Fnet = m EXAMPLE: changes, so does its velocity and hence the momentum. Δt A golf ball of mass 0,05 kg leaves a golf club mvf − mvi at a velocity of 90 m·s҆1 in an easterly direc- Fnet = Fnet = resultant force (N) Δt tion. Calculate the momentum of the golf Δp ball. Fnet Δt = Δp Δp = change in momentum (kg ⋅ m ⋅ s−1) Fnet = p = momentum (kg ⋅ m ⋅ s−1) Δt p = mv m = mass (kg) Δt = time (s) Fnet Δt = Δp Choosing east as positive: v = velocity (m ⋅ s ) −1 p = mv IMPULSE = (0,05)(90) Impulse: the product of the net force acting on an object and the time the over which the = 4,5 kg ⋅ m ⋅ s−1 east net force acts on the object. CHANGE IN MOMENTUM By rearranging Newton’s second law in terms of momentum, we find that impulse is equal to the change in momentum of an object according to the impulse-momentum theorem: When a net force acts on an object, it results in a change in velocity for the object and therefore a change in momentum (p) for the object. The change in momentum can be calculated by using: Impulse = FΔt Impulse, FΔt , is measured in N·s. Impulse = Δp Δp = change in momentum (kg ⋅ m ⋅ s−1) mΔv = Δp Δp is measured in kg·m·s−1 Δp = pf − pi pf = final momentum (kg ⋅ m ⋅ s−1) The change in momentum is directly dependent on the magnitude of the resultant force and the dura- −1 tion for which the force is applied. Impulse is a vector, ∴ direction specific. pi = initial momentum (kg ⋅ m ⋅ s ) EXAMPLE: EXAMPLE: Due to the vector nature of momentum, it is very important to choose a positive direction. A golf ball with a mass of 0,1 kg is driven from the The following graph shows the force ex- EXAMPLE: EXAMPLE: tee. The golf ball experiences a force of 1000 N while erted on a hockey ball over time. The A 1000 kg car initially moving at a constant A cricket ball with a mass of 0,2 kg approaches a in contact with the golf club and moves away from the hockey ball is initially stationary and has a velocity of 16 m·s҆1 in an easterly direction cricket bat at a velocity of 40 m·s҆1 east and golf club at 30 m·s҆1. For how long was the golf club mass of 150 g. approaches a stop street, starts breaking and leaves the cricket bat at a velocity of 50 m·s҆1 in contact with the ball? comes to a complete standstill. Calculate the west. Calculate the change in the ball’s momen- Fnet Δt = m Δv change in the car’s momentum. tum during its contact with the cricket bat. 1000t = (0,1)(30 − 0) t = 3 × 10−3 s Choosing east as positive: Choosing east as positive: EXAMPLE: Δp = pf − pi Δp = pf − pi Why can airbags be useful during a collision? State Calculate the magnitude of the impulse Δp = mvf − mvi Δp = mvf − mvi your answer by using the relevant scientific principle. (change in momentum) of the hockey ball. Δp = (1 000)(0) − (1 000)(16) Δp = (0,2)(−50) − (0,2)(40) The change in momentum remains constant, but the Fnet Δt = area under graph Δp = −16 000 Δp = −18 use of an airbag prolongs the time (t) of impact dur- 1 ing the accident. The resultant force experienced is impulse = b⊥h 2 ∴ Δp = 16 000 kg ⋅ m ⋅ s−1 west ∴ Δp = 18 kg ⋅ m ⋅ s−1 west inversely proportional to the contact time (F ∝ 1/t), 1 impulse = (0,5)(150) therefore resulting in a smaller resultant force (Fnet) 2 (Δp is constant). impulse = 37,5 N ⋅ s Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 12

Grade 12 Science Essentials Conservation of Momentum SCIENCE CLINIC 2022 © CONSERVATION OF MOMENTUM NEWTON’S THIRD LAW AND MOMENTUM Conservation of linear momentum: The total linear momentum of an isolated system re- During a collision, the objects involved will exert forces on each other. Therefore, according to Newton’s mains constant . third law, if object A exerts a force on object B, object B will exert a force on object A where the two forces are equal in magnitude, but opposite in direction. Σpbefore = Σpafter pA(before) + pB(before) = pA(after) + pB(after) The magnitude of the force, the contact time and therefore the impulse on both objects are equal in magnitude. mA viA + mBviB + . . . = mA vfA + mBvfB + . . . Forces are applied between objects during: System: A set number of objects and their interactions with each other. External forces: Forces outside of the system. Collisions: Move off together, collide and deflect, object dropped vertically on moving object. Isolated system: A system on which the net external force is zero. Explosions: Explosions, springs, firearms Collisions Explosions Move off together Explosions vA vB = 0 vA+B Objects that experience the same vA+B = 0 vA vB When objects collide and move off explosion will experience the same together, their masses can be force. added as one object mA mB Collision mA+B The acceleration, velocity and mo- mA+B Explosion mA mB mentum of the object is dependent Objects that are stationary (B) on the mass. have an initial velocity of zero. Σp before = Σpafter Objects that are stationary (A+B) Σp before = Σpafter mA viA + mBviB = (mA + mB)vf have an initial velocity of zero. (mA + mB)vi = mA vfA + mBvfB Collide and rebounds Springs Objects can collide and move off vA vB vA vB The spring will exert the same vB = 0 vB = 0 vA vB separately force on both objects (Newton’s Third Law). mA mB Collision mA mB The acceleration, velocity and mo- mA mB Push mA mB REMEMBER: The velocity and mentum of the object is dependent momentum are vectors (i.e. direc- on the mass. tion specific). Velocity substitution Σp before = Σpafter Σp before = Σpafter must take direction into account. Objects that are stationary (A+B) mA viA + mBviB = mA vfA + mBvfB have a velocity of zero. (mA + mB)vi = mA vfA + mBvfB Object dropped vertically on a moving object Firearms/ cannons Example: A stuntman jumps off a vA+B bridge and lands on a truck. vA MB The gun and bullet will experience vG+B = 0 vG vB the same force. Linear momentum= momentum vB = 0 Shoot The acceleration of the weapon is along one axis. mA Collision mA+B significantly less than the bullet mG+B mG mB A dropped object has a horizontal due to mass difference velocity of zero, Σp before = Σpafter Recoil can be reduced by increas- Σp before = Σpafter ing the mass of the weapon. ∴viB= 0m·s ҆1 mA viA + mBviB = (mA + mB)vf (m G + mB)vi = m G vfG + mBvfB Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 13

Grade 12 Science Essentials Momentum and Energy SCIENCE CLINIC 2022 © ELASTIC VS INELASTIC COLLISIONS PENDULUMS Elastic collision: a collision in which both momentum and kinetic energy are conserved. Inelastic collision: a collision in which only momentum is conserved. DOWNWARD SWING: Conservation of mechanical en- In an isolated system, momentum will always be conserved. To prove ergy (EM) to determine velocity at that a collision is elastic, we only have to prove that kinetic energy is the bottom of the swing: conserved. Kinetic energy can be calculated using the mass and velocity of an EM(top) = EM(bottom) object: 1 1 mg h + 2 mv 2 = mg h + 2 mv 2 h EK = kinetic energy (J) 1 COLLISION: EK = mv 2 m = mass (kg) 2 Conservation of linear momentum to deter- v = velocity (m ⋅ s−1) mine the velocity of the block after impact. Σp before = Σpafter Elastic collision: ΣEk(before) = ΣEk(after) pA(before) + pA(before) = pA(after) + pA(after) Inelastic collision: ΣEk(before) ≠ ΣEk(after) mA viA + mBviB + . . . = mA vfA + mBvfB + . . . (some energy is lost as sound or heat) EXAMPLE: The velocity of a moving trolley of mass 1 kg is 3 m·s҆1. A block of mass 0,5 kg is dropped vertically on to the trolley. Immediately UPWARD SWING: after the collision the speed of the trolley and block is 2 m·s҆1 in the original direction. Is the collision elastic or inelastic? Prove your Conservation of mechanical en- answer with a suitable calculation. ergy (EM) to determine height that the pendulum will reach: 1 1 ΣEk(before) = m v 2 + 2 mb vb2 2 t t EM(bottom) = EM(top) 1 = (1)(3)2 + 12 (0,5)(0)2 2 mg h + 12 mv 2 = mg h + 12 mv 2 = 4,5 J 1 ΣEk(after) = m v2 2 t+ b t+ b COLLISION: 1 Conservation of linear momentum to deter- = 2 (1 + 0,5)(2)2 mine the velocity of the pendulum after impact. = 3J Σp before = Σpafter pA(before) + pA(before) = pA(after) + pA(after) ΣEk(before) ≠ ΣEk(after) mA viA + mBviB + . . . = mA vfA + mBvfB + . . . ∴ Kinetic energy is not conserved and the collision is inelastic Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 14

Grade 12 Science Essentials Energy SCIENCE CLINIC 2022 © ENERGY PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY The ability to do work Principle of conservation of mechanical energy: The total mechanical Unit: joules (J) energy in an isolated system remains constant. EMECHA = EMECHB Scalar quantity The law of conservation of mechanical energy applies when there is no fric- (EP + EK )A = (EP + EK )B Gravitational Potential Energy (EP) Kinetic Energy (EK) tion or air resistance acting on the object. In the absence of air resistance, or 1 1 other forces, the mechanical energy of an object moving in the earth’s gravita- (m g h + 2 m v 2 )A = (m g h + 2 m v 2 )B The energy an object possesses The energy an object has due to its position relative to a as a result of the object’s tional field in free fall, is conserved. reference point. motion General law of conservation of energy: Energy cannot be created or destroyed, merely transferred. In the following instances the gravitational potential energy of an object is converted to kinetic energy (and vice versa), while the Amount of energy transferred when an Amount of energy transferred mechanical energy remains constant. object changes position relative to the to an object as it changes earth’s surface. speed. EXAMPLE 1: Object moving vertically EXAMPLE 2: Object moving on an inclined plane A 2 kg ball is dropped from rest at A, determine the maximum velocity A 2 kg ball rolls at 3 m·s−1 on the ground at A, determine the maximum 1 of the ball at B just before impact. height the ball will reach at B. EP = mg h EK = mv 2 2 g = 9,8 m·s–2, m is mass in kg, m is mass in kg, h is height in m above the ground v is velocity in m·s–1 Example: Example: Determine the gravitational potential Determine the kinetic energy (EP + EK )A = (EP + EK )B (EP + EK )A = (EP + EK )B energy of a 500 g ball when it is placed of a 500 g ball when it travels (m g h + 1 m v 2 )A = (m g h + 1 m v 2 )B 1 (m g h + m v 2 )A = (m g h + 1 m v 2 )B 2 2 2 2 on a table with a height of 3 m. with a velocity of 3 m.s–1. 1 1 (2)(9,8)(4) + (2)(0 2 ) 2 = (2)(9,8)(0) + (2)vB2 2 (2)(9,8)(0) + 1 (2)(32 ) = (2)(9,8)(h B ) + 1 (2)(0 2 ) 2 2 EP = mg h EK = 1 m v2 2 78,4 + 0 = 0 + 1vB2 0+ 9 = 19,6h B + 0 = (0,5)(9,8)(3) 1 9 = (0,5)(32 ) vB = 78,4 = hB = 14,7 J 2 19,6 = 2,25 J vB = 8,85 m ⋅ s−1 downwards hB = 0,46 m EXAMPLE 3: Pendulum EXAMPLE 4: Rollercoaster Mechanical Energy (EM) The 2 kg pendulum swings from A at 5 m·s−1 to B, on the ground, The 2 kg ball rolls on a toy rollercoaster from A, at 20 m above the The sum of gravitational potential and kinetic energy of an where its velocity is 8 m·s−1. Determine the height at A. ground, to B where its height is 8 m and velocity is 14 m·s−1. Calculate its starting velocity at A. object at a point EM = EP + EK 1 EM = mg h + 2 mv 2 EXAMPLE: A ball, mass 500 g, is thrown horizontally through the air. The ball travels at a velocity of 1,8m·s −1 and is 2,5 m from the ground. Determine the (EP + EK )A = (EP + EK )B (EP + EK )A = (EP + EK )B mechanical energy of the ball. (m g h + 1 m v 2 )A = (m g h + 1 m v 2 )B 1 (m g h + m v 2 )A = (m g h + 1 m v 2 )B 2 2 2 2 EM = EP + EK 1 (2)(9,8)(h A ) + (2)(52 ) = 1 (2)(9,8)(0) + (2)(82 ) 1 (2)(9,8)(20) + (2)(vA2 ) = 1 (2)(9,8)(16) + (2)(142 ) 2 2 2 2 1 EM = m g h + 2 m v2 19,6h A + 25 = 0 + 64 392 + vA2 = 313,6 + 196 1 64 − 25 EM = (0,5)(9,8)(2,5) + 2 (0,5)(1,82 ) 19,6 = hA vA = 313,6 + 196 − 392 EM = 13,06 J hA = 1,99 m vA = 10,84 m ⋅ s−1 to the right Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 15

Grade 12 Science Essentials Work, Energy and Power SCIENCE CLINIC 2022 © WORK No Work done on an object (moving at a constant velocity) if the force and NET WORK ON AN OBJECT Work done is the transfer of energy. Work done on an displacement are perpendicular to each other. A number of forces can act on an object at the same time. Each force object by a force is the product of the displacement Consider a man carrying a suitcase with a weight of 20 N on a ‘travelator’ can do work on the object to change the energy of the object. The net and the component of the force parallel to the dis- moving at a constant velocity. work done on the object is the sum of the work done by each force act- placement. W = FΔx cos θ FA = 20 N W = Fx Δx cos θ ing on the object. W = work (J) Fx = F cos θ If Wnet is positive, energy is added to the system. NOTE: F = force applied (N) If Wnet is negative, energy is removed from the system. Work is a scalar quantity, Direc)on Δx = displacement (m) i.e. NO DIRECTION for of mo)on θ = Angle between F and Δx F or x! Fg = 20 N Work and Energy are SCALARS, and NOT direction specific. The joule is the amount of work done when a force of one EXAMPLE: newton moves its point of application one meter in the di- Calculate the net work done on a trolley where a force of 30 N is ap- rection of the force. Δx plied to the trolley. The trolley moves 3 m to the left. The force of FA is perpendicular to the displacement: θ = 90° ; cos 90° = 0. friction is 5 N to the right. Work always involves two things: Work done by applied force: Work done by frictional force: 1. A force which acts on a certain object. (F) No force in the plane of the displacement, hence, NO WORK IS DONE by 2. The displacement of that object. (Δx / Δy) FA and Fg and no energy is transferred. We can also say that FA / Fg does WA = FΔx cos θ Wf = Ff Δx cos θ not change the potential energy (height) or kinetic energy (vertical velo- = (30)(3)cos 0 = (5)(3)cos 180 F city) of the object. = 90 J gained = − 15 J "lost" A force/force component in the direction of the displacement does positive work on the object. The force increases the energy of the object. Work done by gravity Work done by normal force: Δx Wg = FgΔx cos θ WN = FN Δx cos θ Positive work means that energy is added to the system. When a resultant force is applied to an object, the resultant = (Fg )(3)cos 90 = (FN )(3)cos 90 force accelerates the block across distance Δx. Work has been done to increase the kinetic energy of the block. F Direc+on = 0J = 0J of mo+on W = Fx Δx cos θ Wnet = WA + Wf + WN + Wg θ Fx = F cos θ = 90 − 15 + 0 + 0 Δx If a resultant force is applied to an = 75 J nett energy gained object vertically, the resultant force Alternative method for determining net work: lifts the block through distance Δy. 0° ≤ θ < 90° ; +1 ≥cos θ > 0 1. Draw a free body showing only the forces acting on the object. Work has been done to increase the F potential energy of the block. Δy A force/force component in the opposite direction of the displacement does 2. Calculate the resultant (net) force acting on the object. “Lifting” usually implies at a constant negative work on the object. The force decreases the energy of the 3. Calculate the net work using Wnet = FnetΔx cos θ velocity. object. Step 1: Freebody diagram Negative work means that energy is being removed from the FN system. W = Fx Δx cos θ F Fx = F cos θ Direc+on of mo+on FA = 30 N Ff = -5 N θ F θ Fg Δx Take left as positive: Δx 90° < θ ≤ 180° ; 0 > cos θ ≥ −1 Step 2: Calculate Fnet Step 3: Net work Work is only done in the direction of the displacement. Work is done by the component of the force that is parallel Fnet = FA + Ff Wnet = Fnet Δx cos θ to the displacement. The angle between the force and the = 30 − 5 = (25)(3)cos 0 displacement is θ. If no displacement takes place due to NB: Never use a – for F in the opposite direction. The cos θ = 25 N left = 75 J gained the applied force, no work is done. makes provision for that. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 16

Grade 12 Science Essentials Work, Energy and Power SCIENCE CLINIC 2022 © WORK ENERGY THEOREM NON-CONSERVATIVE FORCES POWER (FT, FA, Ff, Fair resistance) According to Newton’s Second Law of Motion, when a resultant force Power is the rate at which work is done OR the rate at acts on an object, the object accelerates. This means there is a which energy is transferred. change in velocity of the object, and therefore a change in kinetic A force is a non-conservative force if: W E energy of the object, since Ek = ½ mv2 1. The work done by the force in moving an object from point A to P= = WORK-ENERGY THEOREM: The work done by a net force on point B is dependent of the path taken. Δt Δt an object is equal to the change in the kinetic energy of the P = power (watt) object 2. The net work done in moving an object in a closed path which starts W = work (J) and ends at the same point is not zero. Δt = time (s) Wnet = ΔE K 1 EXAMPLE: Fnet Δx cos θ = 2 m(v 2f − v 2i ) A non-conservative force does not conserve mechanical energy. A certain amount of energy is converted into other forms such as internal Calculate the power expended by an engine when an object CONSERVATIVE FORCES energy of the particles which the objects is made of. An example of a of mass 100 kg is lifted to a height of 2,2 m in a time of 3 s, non-conservative force is the frictional force. at a constant velocity. A force is a conservative force if: Consider the crate on a rough surface being pushed with a constant W NOTE: 1. The work done by the force in mov- EP = EP force FA from position 1 to position 2 along two different paths. P= Δt ing an object from point A to point B is independent of the path taken. FA FA = FΔx cos θ Fnet = 0 N Ff Δt ∴ Wnet = 0 J 2. The net work done in moving an Ff = (100)(9,8)(2,2)cos 0 3 ∴ Pnet = 0 W object in a closed path which starts and ends at the same point is zero. = 718,67 W FA AVERAGE POWER (CONSTANT VELOCITY) 1 Ff 2 Conservative force conserve mechanical energy. Example of We can calculate the average power needed to keep an object conservative forces are gravitational force and spring force. The work done by FA is more when the longer path is taken. The work moving at constant speed using the equation: done to overcome the friction will result in the surface of the crate be- coming hotter. This energy is dissipated (as sound and/or heat) and is Paverage = Fvaverage very difficult to retrieve, i.e. not conserved. Δx Paverage = F Δt Note: the total energy of the system is conserved in all cases, whether the forces are conservative or non-conservative. EXAMPLE: Wnc = ΔE K + ΔEP A man lifts a 50 kg bag of cement from ground level up to a The work done by gravity on each ball is independent of the path 1 height of 4 m above ground level in such a way that the bag taken. Only the h⟂ is considered. = 2 m(v 2f − v 2i ) + mgΔh of cement moves at constant velocity (i.e. no work is done to change kinetic energy). Determine his average power if he does this in 10 s. EXAMPLE: ΔEP = mgΔh ΔE K = E Kf − E Ki = mg(h f − h i ) = 1 m(v 2f − v 2i ) An 800 kg car traveling at = (800)(9,8)(0 − 100 sin 30∘ ) 2 Paverage = Fvaverage 15 m·s−1 down a 30° hill needs 1 100 = (800)(0 2 − 152 ) to stop within 100 m to avoid m = − 392 000 J 2 Δy = F Δt an accident. Using energy calcu- = − 90 000 J lations only, determine the mag- Wnc = ΔE K + ΔE P 4m = (50)(9,8)( 10 4 ) nitude of the average force that FA Δx cos θ = ΔE K + ΔEP must be applied to the brakes = 196 W FA (100)cos 180 = − 90 000 − 392 000 over the 100 m. Assume that − 482 000 the surface is frictionless. FA = −100 30° This cannot be used for object in freefall, unless the object has FA = 4 820 N reached terminal velocity. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 17

Grade 12 Science Essentials Work, Energy and Power SCIENCE CLINIC 2022 © When to use which formulae CONSERVATION OF MECHANICAL ENERGY ALTERNATIVE METHODS TO DETERMINE WORK DONE BY Fg ON A SLOPE WITHOUT RESOLVING Fg INTO Fg⟂ AND Fg// E MA = E MB 1. When an object moves up a slope 2. When an object moves down a slope (E P + EK )A = (E P + EK )B with a known angle with a known angle • When there are no external force (FA/Ff) present. • Any track/pendulum with a curve that is not a straight line. 4m 4m 120° 120° 60° 60° WORK-ENERGY PRINCIPLE 30° Fg = 100 N 30° Fg = 100 N Wnet = ΔEK 1 1 = 2 mv 2f − 2 mv 2i Wg = Fg Δx cos θ Wg = Fg Δx cos θ = (100)(4)cos 120∘ = (100)(4)cos 60∘ = −200 J = 200 J • If an object is accelerating on a horizontal/incline plane. Any of the following methods may be used: FN FA 3. When an object moves up a slope without a 4. When an object moves down a slope without a given angle, but with specified height given angle, but with specified height Fg 3m 3m 30° A. Fnet → Wnet A. Wnet = ∑W Fg = 100 N Fg = 100 N 1. Determine Fnet separately 1. Determine W of each force seperately Fnet = FA − F f − Fg // WA = FAΔx cos θ Wg = Fg Δx cos θ Wg = Fg Δx cos θ Fnet = FA − Ff − Fg sin θ WN = FN Δx cos θ = (100)(3)cos 180∘ = (100)(3)cos 0∘ Fnet = FA − Ff − (100 sin 30∘ ) Wg = Fg Δx cos θ = −300 J = 300 J Wf = Ff Δx cos θ 2. Apply Fnet to Wnet 2. Apply ∑W to Wnet 1 1 1 1 Wnet = Fnet Δx cos θ = m v 2 − m vi2 WA + Wf + Wg + WA = Wnet = m v 2 − m vi2 2 f 2 2 f 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 18

Grade 12 Science Essentials Motion in 1D SCIENCE CLINIC 2022 © VARIABLES Calculations: A racing car starting from rest on the grid, travels straight Position vs Time Velocity vs Time Acceleration vs Time vi initial velocity along the track and reaches the 400 m mark after 8,6 s. a) What was its average acceleration? Stationary (velocity = 0 m·s−1) vf final velocity a acceleration v (m·s−1) a (m·s−2) ∆x / ∆y displacement vi 0 Let forward be positive. 1 x (m) Δx = vi Δt + 2 a t 2 ∆t change in time vf / 1 EQUATIONS OF MOTION a ? 400 = (0)(8,6) + 2 a8,62 Δt (s) Δt (s) Δt (s) vf = vi + a Δt ∆x 400 m a = 10,82 m ⋅ s−2 forward Constant velocity (acceleration = 0 m·s−2) Δx = vi Δt + 1 a Δt 2 ∆t 8,6 s 2 vf2 = vi2 + 2a Δx v (m·s−1) a (m·s−2) 1 b) What was its velocity at the 400 m mark? Δx = (v + vf )Δt 2 i vf = vi + aΔt vf = 0 + (10,82)(8,6) x (m) Steps to using the equations: Δt (s) Δt (s) Δt (s) vf = 93,05 m ⋅ s−1 forward a) Draw a diagram of the motion of the object. b) Identify each stage of the motion (where the accel- Increasing velocity (constant positive acceleration) eration has changed). c) At the 400 m mark, the brakes are applied and the car c) Choose a positive direction and use the same conven- slowed down at 2 m·s−2 to come to rest. Calculate the time it tion throughout. v (m·s−1) a (m·s−2) took for the car to stop. d) Record the information given and value required by writing next to each variable. Check the unit and direc- NB! New stage of motion. x (m) tion. Find the new value of each variable. e) Select correct equation and solve for unknown. Δt (s) Δt (s) Δt (s) f) Include units and direction in your answer. Remember: vi 93,05 m·s-1 Decreasing velocity (constant negative acceleration) ‘starting from rest’ means: u or vi = 0 Let forward be positive. vf 0 ‘comes to a stop’ means: v or vf = 0 vf = vi + aΔt v (m·s−1) a (m·s−2) Slowing down means: acceleration is negative (a < 0), a -2 m·s−2 0 = 93,05 − 2t while still moving in a positive direction. Δt (s) x (m) Constant velocity means: a =0, u = v or vi = vf ∆x / t = 46,53 s Use a new set of variables for each stage of the motion. ∆t ? Conversion of units: 1 m.s-1 = 3,6 km.h-1. Δt (s) Δt (s) Displacement-Time Velocity-Time Acceleration-Time • 1 is positive acceleration • 1 is positive acceleration • 1 is positive acceleration • 2 is constant velocity • 2 is constant velocity • 2 is constant velocity (a = 0) • 3 is negative acceleration • 3 is negative acceleration • 3 is negative acceleration a (m·s–2) • 4 is at rest • 4 is at rest, v = 0 m.s–1 • 4 is at rest or constant v x (m) GRADIENT: v (m·s–1) GRADIENT: GRADIENT: Velocity Acceleration n/a AREA BELOW GRAPH: AREA BELOW GRAPH: AREA BELOW GRAPH: n/a ΔDisplacement ΔVelocity t (s) t (s) t (s) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 19

Grade 12 Science Essentials Vertical Projectile Motion SCIENCE CLINIC 2022 © PROJECTILE MOTION PARTS OF PROJECTILE PATH Only vertical movement (up and down) is considered, no hori- EXAMPLE: A projectile is an object that moves freely under the influ- zontal movement is taken into account. The path of projectile An object is projected vertically upwards. 4 seconds later, it is caught at the same ence of gravity only. It is not controlled by any mechanism motion can be analysed using the 4 sections as shown below. height (point of release) on its way downwards. Determine how long it took the (pulley or motor). The object is in free fall, but may move The combination of these 4 parts will depend on the actual ball to pass a height of 8 m in the upward direction. Choose downward as posi- upwards (thrown up) or downwards. path travelled by the projectile. Example: Dropped projectile is tive direction. Forces on a projectile sections C and D only. Object thrown upwards and falls on roof 1 is sections A to C. Δy = viΔt + 2 at 2 In the absence of friction, the gravitational force of the earth total time = 4 s is the only force acting on a free falling body. This force al- ∴ t up = 2 s −8 = − 19,6t + 12 (9,8)t 2 ways acts downwards. Time: tA = tD 0 = 4,9t 2 − 19,6t + 8 Because the gravitational force is always downward, a projec- vf = vi + aΔt tile that is moving upward, must slow down. When a projec- tB = tC tile is moving downward, it moves in the direction of the B C 0 = vi + (9,8)(2) SAME Displacement: −b ± b 2 − 4ac gravitational force, therefore it will speed up. vi = − 19,6 t = HEIGHT ΔyA = ΔyD 2a Acceleration due to gravity ΔyB = ΔyC ∴ vi = 19,6 m ⋅ s −1 u p −(−19,6) ± (−19,6) 2 − 4(4,9)(8) All free falling bodies near the surface of the earth have the A D Velocity: t = 2(4,9) same acceleration due to gravity. This acceleration is 9,8 m·s−2 downward. viA = vfD −b ± 2 b − 4a c t = 0,46 s OR 3,54* vfA = viD x= Ignoring air resistance/friction; If a marble and a rock are SAME ∴ t = 0,46 s *time for downward viB = vfC 2a released from the same height at the same time, they will VfB = ViC = 0 m·s−1 direction strike the ground simultaneously, and their final velocity will HEIGHT be the same. GRAPHS OF PROJECTILE MOTION Their momentum (mv) and kinetic energy (½mv2) are (Eg. A ball is thrown into the air and is caught at the same height.) not the same, due to a difference in mass. Graph manipulation: Area If two objects are released from different heights, they have Change in positive direction: the same acceleration, but they strike the ground at different Flip graph along x-axis Δy vs Δt v vs Δt a vs Δt times and have different velocities. Change in reference position: Shift x-axis (Δy-Δt only) Gradient vf = vi + aΔt Nega%ve a Ne ga UP POSITIVE Δy = displacement (m) % vf2 = vi2 + 2aΔy Posi%ve ve a v (m·s−1) a (m·s−2) Δt = time (s) Δy Posi% %ve Δy (m) Δt (s) Nega Ne Nega%ve Δt (s) 1 vi = initial velocity (m·s−1) ve v ga Δy Nega%ve Δv Δy = viΔt + aΔt 2 %v ea vf = final velocity (m·s−1) v 2 a = acceleration (m·s−2) vi + vf (9,8 m·s−2 downwards) Δt (s) Δy = ( )Δt Posi%ve 2 a DOWN POSITIVE a ve REMEMBER: si% Posi%ve a (m·s−2) o %ve Posi% v (m·s−1) P Posi%ve Δv Δy (m) Nega 1. Draw a sketch diagram Δy ve v Δt (s) Δt (s) Nega%ve 2. Write down given variables v Δy 3. Choose positive direction 4. Solve Δt (s) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 20

Path of a Projectile U ALL EXAMPLES: Grade 12 Science Essentials SCIENCE CLINIC 2022 © P POSITIVE, POINT OF RELEASE IS REFERENCE OBJECT DROPPED FROM HEIGHT (C+D) OBJECT THROWN DOWN FROM HEIGHT (D) OBJECT THROWN UP FROM HEIGHT (B+C+D) vi = 0 m·s−1 vf(up) = 0 m·s−1 vi(down) = 0 m·s−1 vi ≠ 0 m·s−1 Δy (m) Δy (m) Δy (m) vi(up) ≠ 0m·s−1 Δt (s) Δt (s) Δt (s) Δt (s) Δt (s) v (m·s−1) v (m·s−1) Δt (s) v (m·s−1) vf vf vf(down) a (m·s−2) Δt (s) a (m·s−2) a (m·s−2) Also applies to objects Δt (s) Also applies to objects Δt (s) dropped from a downward dropped from an upward moving reference. moving reference. OBJECT THROWN UP AND CAUGHT (A+B+C+D) OBJECT THROWN UP, LANDS AT HEIGHT (A+B+C) OBJECT THROWN UP FROM HEIGHT, BOUNCES (B+C+D) vf(up) = 0 m·s−1 = vi(down) vf(up) = 0 m·s−1 = vi(down) Δy (m) Δy (m) B A C Δt (s) Δy (m) B D v (m·s−1) A vf(down) D C C E Δt (s) Δt (s) E Δt (s) Accera&on due to force v (m·s−1) by surface during bounce v (m·s−1) Δt (s) vi(up) ≠ 0m·s−1 Vi(up) = Vf(down) a (m·s−2) Δt (s) Δt (s) A B D E C a (m·s−2) a (m·s−2) Δt (s) Δt (s) If the collision is perfectly elastic, the downward velocity before the bounce and the upward velocity after the bounce is equal in magnitude. Treat the 2 projectile paths (before and after bounce) as separate paths. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 21

Grade 12 Science Essentials Special Projectile Paths SCIENCE CLINIC 2022 © HOT AIR BALLOON LIFT BOUNCING BALL – Ball falls from rest and bounces Contact time When an object is dropped Lift moving up Lift moving down A C from a moving reference B D (hot air balloon), the initial Δy (m) Δyli% Δy (m) velocity will be equal to that of the reference. The accel- eration of the object will be li% height D downwards at B A C Δt (s) 9,8 m·s−2, regardless of the li% height Δyball Δyball Δt (s) Contact time acceleration of the refer- ence. Bounce Apex v (m·s−1) v (m·s−1) A B C D A B C D Δt (s) Δyli% EXAMPLE: Apex A hot air balloon ascends with a constant Δt (s) Bounce Δyball = lift height + Δylift velocity of 5 m·s−1. A ball is dropped from the hot air balloon at a height of 50 m and falls gradient = g = +9,8 m.s–2 gradient = g = –9,8 m.s–2 EXAMPLE: vertically towards the ground. Determine (a) the distance between the hot air balloon and A lift accelerates upwards at a rate of 1,4 m·s−2. As the lift ball after 2 seconds and (b) the velocity of the starts to move, a lightbulb falls from the ceiling of the lift. 10 ball when it reaches the ground. Determine how long it takes the lightbulb to reach the EXAMPLE: lift’s floor. The height from the ceiling of the lift to its floor The velocity-time graph below represents the 5 (a) Take downwards as positive: Δt (s) v (m·s−1) Distance travelled by balloon : is 3m. bouncing movement of a 0,1 kg ball. Use the graph to answer the questions that follow: 1 Take downwards as positive: Δy = vi Δt + 2 aΔt 2 movement of lift : 1 = (−5)(2) + 2 (0)(22 ) 1 Δylif t = vi Δt + 2 aΔt 2 a) Which direction of movement is positive? −8 = − 10 Downwards 1 ∴ Δy = 10 m up ylif t = (0)t + 2 (−1,4)t 2 b) How many times did the ball bounce? ∴ ylif t = − 0,7t 2 3 times Distance travelled by ball : 1 c) What does the gradient of the graph represent? Δy = vi Δt + 2 aΔt 2 Acceleration of the ball movement of bulb : 1 = (−5)(2) + 2 (9,8)(22 ) Δybu lb = vi Δt + 2 aΔt 2 1 d) Are the collisions between the ball and ground elastic or inelastic? After each bounce there is a decrease in magnitude of the velocity of the ball, = − 10 + 19,6 1 and therefore a change in kinetic energy. The collisions are inelastic as kinetic 3 + ylif t = (0)t + 2 (9,8)t 2 ∴ Δy = 9,6 m down energy is not conserved. 3 − 0,7t 2 = 4,9t 2 ∴ total distance = 10 + 9,6 e) If the ball is in contact with the ground for a duration of 0,08 s, determine the impulse on = 19,6 m apart 3 = 5,6t 2 the ball ∴ t = 0,73s Impulse = Δp (b) Take downwards as positive: = m (vf − vi ) vf2 = vi2 + 2aΔy Simultaneous equation is needed be- cause there are 2 unknown variables: = (0,1)(−8 − 10) vf2 = 2 (−5 ) + 2(9,8)(50) = − 1,8 •Distance that lift moved ∴ Impulse = 1,8 N ⋅ s upwards vf = 25 + 980 •Time to reach floor f) Predict why the ball stopped moving. vf = 31,70 m ⋅ s−1 downwards it was most likely caught Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 22

Grade 12 Science Essentials Electricity SCIENCE CLINIC 2022 © SERIES PARALLEL CALCULATIONS (NO INTERNAL RESISTANCE) If resistors are added in series, the If resistors are added in parallel, the total resistance will decrease Series circuit Parallel circuit total resistance will increase and the and the total current will increase, provided the emf remains total current will decrease provided constant. 10 V V1 V1 the emf remains constant. CURRENT is the rate of flow of charge. Q I= t I is the current strength, Q the charge in coulombs and t the time in seconds. SI unit is ampere (A). 3A A3 A1 A2 V2 A1 R1 A1 R1 R2 A1 R1 R2 A2 R1 2Ω 5Ω 4Ω A2 R2 A2 R3 R1 R2 12 Ω R2 IT = I1 = I2 IT = I1 + I2 IT = I1 + I2 a) Determine the total resistance. I1 = IR1 = IR2 a) Determine the total resistance. 1 1 1 = + Rp R1 R2 RESISTANCE is defined as the material’s opposition to the flow of electric current. Rtot = R1 + R2 1 1 V = IR = 2+ 5 = + 4 12 R is the electrical resistance of the conducting material, resisting the flow of charge through it. = 7Ω = 1 Resistance (R) is the quotient of the potential difference (V) across a conductor and the current (I) in 3 it. The unit of resistance is called the ohm (Ω). ∴ Rp = 3Ω b) Determine the reading on A1 and A2. R1 R1 R2 V = IR b) Determine the reading on V1 and V2. 10 = I (7) V1 = IR R1 R2 I = 1,43 A = (3)(3) R2 R3 ∴ A1 = A2 = 1,43 A = 9V ∴ V1 = V2 = 9V R s = R1 + R2 1 1 1 1 1 1 = + = + Combination circuits Rp R1 R2 Rp R1 + R2 R3 15 V Consider the circuit given. (Internal resistance is negligible) POTENTIAL DIFFERENCE (p.d.) is the work done per V2 NOTE: Calculate: unit positive charge to move the charge from one point to an- other. It is often referred to as voltage. 1. Emf (ε): voltage across cells when W no current is flowing (open circuit). a) the effective resistance of the circuit. V= Q b) the reading on ammeter A1. 2. V term or VT: voltage across cells V is Potential difference in V (volts), W is Work done or energy 3Ω R1 A1 when current is flowing. c) the reading on voltmeter V1. transferred in J (joules) and Q is Charge in C (coulombs). d) the reading on ammeter A2. 2Ω V1 V1 V2 A2 R2 a) 1 1 1 4Ω = + R1 R1 R2 RP R2 R3 R3 1 1 V1 V2 = + 2 4 V1 R2 R3 3 = R1 R2 4 V2 V3 4 b) c) d) ∴ Rp = = 1,33 Ω Vtot V1 V1 3 Rtot = RP = R2Ω = I1 I1 I2 V4 Rtot = R3Ω + RP 15 V1 4,60 V3 4,33 = 1,33 = 2 = = 3 + 1,33 I1 3,46 I2 Vs = V1 + V2 Vp = V1 = V2 = V3 Vp = V3 = V4 = (V1 + V2 ) = 4,33 Ω I1 = 3,46 A V1 = 4,60 V I2 = 2,30 A Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 23

Grade 12 Science Essentials Electricity SCIENCE CLINIC 2022 © OHM’S LAW INTERNAL RESISTANCE CALCULATIONS (WITH INTERNAL RESISTANCE) VT of external The potential difference across the conductor is directly propor- The potential difference across a bat- ε = Vint + Vext circuit tional to the current through the conductor provided that the tery not connected in a circuit is the 6V emf temperature remains constant. emf of the battery. ΔV of internal resistance V Emf is the total energy supplied R= per coulomb of charge by the cell. I Vint = Ir Vext = IRext 5V Proof of Ohm’s law: When connected in a circuit the poten- tial difference drops due to the inter- The current in the circuit is changed nal resistance of the cells. using the rheostat, thus current is the ε = I(Rext + r) independent variable and potential In reality all cells have internal resis- A difference is the dependent variable. tance (r). Internal resistors are al- EXAMPLE: It is important that the temperature of ways considered to be connected in V the resistor is kept constant. series with the rest of the circuit. Determining the emf and internal resistance of a cell r The resultant graph of potential differ- ence vs current indicates if the con- V V V ε ductor is ohmic or non-ohmic. Ohmic conductor Non-ohmic conductor r y-intecept = Emf Lost volts 5Ω 3Ω • An Ohmic conductor is a conduc- • A Non-ohmic conductor is a ε Gra Int dien = Ir tor that obeys Ohm’s law at all conductor that does not obey A ern t = al r Neg temperatures. Ohm’s law at all temperatures. esi sta ative 4Ω V V nce • Constant ratio for I . • Ratio for I change with change I When the switch is open the reading on the voltmeter is 12 V. in temperature. When the switch is closed, the current through the 3 Ω resistor is 1 A. • E.g. Nichrome wire • E.g. Light bulb a) Calculate the total current of the circuit. Independent variable Current (I) b) Calculate the internal resistance of the battery. V V Dependent variable Potential difference (V) c) How will the voltmeter reading change when the 4 Ω resistor is e removed? nc a) Rtop = R5Ω + R3Ω b) 1 = 1 + 1 ta Controlled variable Temperature si s Rp Rtop R 4Ω Re = 5+ 3 1 t = = 8Ω = + 1 en COST OF ELECTRICITY 8 4 i Rtop : R 4Ω ad 3 = Gr Electricity is paid for in terms of the amount of energy used 8 : 4 8 by the consumer. Rp = 8 = 2,67 Ω I I 2 : 1 3 Note: ∴ Itop : I4Ω POWER ε = I (R + r) Cost of electricity = power × time × cost per unit P in kW t in hours 1 : 2 12 = 3(2,67 + r) Power is the rate at which work is done. r = 1,33 Ω I4Ω = 2Itop P = power (W) c) When the 4 Ω resistor is re- W Example = 2(1) P = W = work (J) moved, the external resistance Δt = 2A increases and the current de- A geyser produces 1200 W of power. Calculate the Δt = time (s) cost of having the geyser switched on for 24 hours, if Itot = Itop + I4Ω creases. (I ∝ 1/R). Emf and P = VI I = current (A) the price of electricity is R1,55 per unit. = 1+ 2 internal resistance are constant, P = I2R Cost = power × time × cost per unit = 3A therefore a decrease in current V = potential will result in an increase in exter- difference (V) = (1,2)(24)(1,55) V2 nal VT (Voltmeter reading) P = = R44,64 R R = resistance (Ω) V = ε – Ir Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 24

Grade 12 Science Essentials Electrostatics SCIENCE CLINIC 2022 © FLASHBACK!! Unit of charges: CALCULATIONS- Electrostatic force Prefix Conversion Electrostatic force is a vector, therefore all vec- The force can be calculated using • Substitute charge magnitude only. tor rules can be applied: • Direction determined by nature of Principle of conservation of charge. centi− (cC) ×10−2 kQ1Q2 charge (like repel, unlike attract). • Direction specific F= Q1 + Q2 milli− (mC) ×10 −3 r2 • Both objects experience the same Q new = • Can be added or subtracted force (Newton’s Third Law of Motion). 2 micro− (µC) ×10−6 Principle of charge quantization. 1 Dimensional nano− (nC) ×10−9 Q Determine the resultant electrostatic force on QB. Ne− = pico− (pC) ×10−12 qe + 3 cm 5 cm COULOMB’S LAW A B C 3 FAB FCB The magnitude of the electrostatic force exerted by one point charge (Q1) −2 nC +3 nC −1 nC on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the kQ A Q B kQ C Q B Fnet = FAB + FCB distance (r) between them. 1 FAB = 2 FCB = 4 r2 r2 = (−6 × 10−5) + (1,08 × 10−5) 9 × 109 (2 × 10−9 )(3 × 10−9 ) 9 × 109 (1 × 10−9 )(3 × 10−9 ) kQ1Q2 = (3 × 10−2 ) 2 = (5 × 10−2 ) 2 = − 4,92 × 10−5 F= ∴ Fnet = 4,92 × 10−5 N left r2 = 6 × 10−5 N left (A attracts B) = 1,08 × 10−5 N right (C attracts B) 2 Dimensional F = force of attraction between charges Q1 and Q2 (N) k = Coulomb’s constant (9×109 N·m2·C−2) Determine the resultant electrostatic force on QB. Q = magnitudes of charge (C) r = distance between charges (m) -10 μC +7 μC 15 mm 2 FCB FCB RATIOS B C θ θ In ratio questions, the same process is used as with Newton’s Law of Uni- 10 mm versal Gravitation. OR EXAMPLE: Fnet FAB FAB Fnet Two charges experience a force F when held a distance r apart. How would this force be affected if one charge is doubled, the other charge is tripled and the distance is halved. A +5 μC PYTHAGORAS : 3 kQ1Q 2 F 2net = F 2AB + F 2BC F= r2 kQ A Q B 1 FAB = Fnet = 4 500 2 + 2 800 2 k(2Q1)(3Q 2 ) r2 = Fnet = 5 300 N ( 1 r)2 = 9 × 109 (5 × 10−6 )(10 × 10−6 ) 2 (10 × 10−3) 2 6 kQ1Q 2 = = 4 500 N down (A attracts B) tanθ = o 1 r2 4 a 4 FAB kQ1Q 2 θ = tan−1 = 24( ) kQ C Q B FCB FCB = r2 r2 −1 4 500 θ = tan = 24F 9 × 109 (7 × 10−6 )(10 × 10−6 ) 2 800 = θ = 58,11∘ (15 × 10−3) 2 = 2 800 N right (C attracts B) ∴ Fnet = 5 300 N at 58,11∘ clockwise from the positive x − axis Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 25

Grade 12 Science Essentials Electrostatics SCIENCE CLINIC 2022 © ELECTRIC FIELDS ELECTRIC FIELD STRENGTH An electric field is the region around an electric charge where another charge (or Electric field strength at any point in space is the force per unit charge experienced by a charged object) will experience an electrostatic force. The direction of the electric positive test charge at that point. field at a point is the direction that a positive test charge (+1C) would move if placed at that point. F kQ E= E= REMEMBER: Field lines leave a conducting surface at 90 o q r2 Single point charges E = electric field strength (N·C−1) E = electric field strength (N·C−1) F = force on charge q (N) k = Coulomb’s constant (9×109 N·m2·C−2) q = charge (C) Q = charge (C) r = distance from charge Q (m) + − q is the charge that experiences the force. Q is the charge that creates the electric field. Force due to charge Q Certain point Q in space r X Unlike charges (of equal magnitude) F Charge experiencing Q the electric field due Distance between q to charge Q charge Q and point X + − EXAMPLE: EXAMPLE: Charge A experiences a force of 2 N due to charge B. Determine the electric field strength at point P due to Determine the electric field strength at point B. charge Q. A B Q 5mm P +2μC −5μC +3μC Like charges F kQ E = q E = r2 − − 2 = 9 × 109(3 × 10−6 ) –2C –2C –C 2 × 10−6 = (5 × 10−3) 2 = 4 × 10 N ⋅ C−1 to the right 6 = 1,08 × 109 N ⋅ C−1 to the right Different magnitude charges Equal magnitude charges DIRECTION OF E: DIRECTION OF E: Direction that charge q would move Direction that point in space ( X ) IF it was positive. would move IF it was positive. +C +2C + + NOTE: Electric field strength is a VECTOR. All vector rules and calculations apply. (linear addition, 2D arrangement, resultant vectors, etc.) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 26

Grade 12 Science Essentials Electromagnetism GRADE 11 RECAP SCIENCE CLINIC 2022 © INDUCTION OF A MAGNETIC FIELD INDUCTION OF AN ELECTRIC CURRENT INCREASING THE INDUCED EMF •Increase the rate of change of magnetic flux, ie. decrease When current passes through a conductor, a magnetic field is induced When a magnet is brought close to a metal wire, and relative motion the time it takes to change the flux, ie. increase speed of around the wire. occurs between the conductor and magnet, it causes movement of movement. charge in the wire. As a result, an EMF is induced in the wire, which The direction of the magnetic field can be determined by the right hand •Increase the number of loops in the coil. causes an induced current. Only a change in magnetic flux will in- thumb rule. For a straight, single wire, point the thumb of your right hand •Increase the strength of the magnet. duce a current. in the direction of the conventional current and your curled fingers will point •Increase the surface area of the loops in the coil. in the direction of the magnetic field around the wire. Faraday’s law states that the emf induced is directly propor- •Increase the change of flux by changing the angle, θ, tional to the rate of change of magnetic flux (flux linkage). from a minimum of 0° to a maximum of 90°. Magnetic flux linkage is the product of the number of turns on ENVIRONMENTAL IMPACT OF I I Out of page Into page the coil and the flux through the coil. OVERHEAD CABLES •Birds fly into power lines as they cannot see them from a NOTE THAT NO CALCULATIONS ARE distance. REQUIRED, ONLY RELATIONSHIP •This places birds at risk of becoming extinct because of B BETWEEN VARIABLES the increase in unnatural mortality. •Trees fall onto power lines and can cause fires to erupt. •Trees have to be cut down to make space for the power lines. For a wire loop, the magnetic field is the sum of the individual magnetic ε = emf (V) •There is no evidence that the electromagnetic effect of fields around the single wires at each end. Use the right hand rule for a sin- −NΔϕ N= number of turns/windings in coil the power lines has any negative impact on people and the gle wire at each end of the loop. ε= Δɸ= change in magnetic flux (Wb) Δt surroundings as the strength of the field is low. F F Δt= change in time (s) •The electromagnetic effect of the power line can disrupt radio signals and for emergency services, this can be a major problem. The magnetic flux (ɸ) is the result of the product of the perpendicular component of the magnetic field (B) and cross-sectional area (A) the DIRECTION OF INDUCED CURRENT field lines pass through. As a bar magnet moves into a solenoid the needle of the I galvanometer is deflected away from the 0 mark. As the ɸ= magnetic flux (Wb) bar magnet is removed, the needle deflects in the oppo- B= magnetic flux density (T) site direction. The magnetic energy is converted to electri- F F ϕ = BA cos θ A= area (m2) cal energy. The direction of the induced current can be determined using Lenz’s law. θ= angle between magnetic field line and normal N B S Normal Normal S I θ B θ θ N N For a solenoid, curl your fingers around the solenoid in the direction of the θ S N S conventional current and your thumb will point in the direction of the North A=l ×b pole. This is know as the right hand solenoid rule. A=πr2 Lenz’s Law states that the induced current flows in I a direction so as to set up a magnetic field to op- B pose the change in magnetic flux. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 27

Grade 12 Science Essentials Electrodynamics SCIENCE CLINIC 2022 © GENERATORS ALTERNATING CURRENT Generators convert mechanical energy into electrical energy. A generator works on the principle of mechanically rotating The potential difference and current for an alternating current fluctuates between a posi- a conductor in a magnetic field. This creates a changing flux which induces an emf in the conductor. tive and negative maxima. The average potential difference and current is zero and so the average power is used to define the potential difference and current for an alternat- Direct Alternating ing current. A direct current generator uses a split ring commutator to The alternating current generator is connected to the exter- Power versus time graph for a AC circuit. connect the conductor to the external circuit instead of a nal circuit by 2 slip rings which is connected to the conduc- slip ring. tor. The slip rings make contact with brushes which are connected to the external circuit. Pmax = VmaxImax B C B C Pave = ½ VmaxImax N S N S A D A D Δt (s) 1 V V Pave = P 2 max The current in the external circuit does not change direc- The direction of the current changes with every half turn of 1 tion and is known as direct current (DC). the coil. The current that is produced is known as alternat- = (Vmax ⋅ I max ) 2 ing current (AC). 1 1 θ = 0° θ = 90° θ = 180° θ = 270° θ = 360° θ = 0° θ = 90° θ = 180° θ = 270° θ = 360° = Vmax ⋅ I max B C B B B 2 2 C C B B C C B B C = Vrms ⋅ I rms C B C C B C + – – + The potential difference for an alternating current is therefore defined by the root mean square of the maximum potential difference (Vrms) and the current for an alternating current is defined as the root mean square of the maximum current (Irms). These values can be calculated from the maximum potential difference and maximum current as fol- Induced Inducedemf EMF(V) lows: Induced EMF Rotations Rotations Vmax Vrms = 0 1/4 1/2 3/4 1 0 1/4 1/2 3/4 1 2 I max I rms = ɸ ɸ 2 FLEMING’S RIGHT HAND-DYNAMO RULE The right hand rule is used to predict the direction of the induced emf in the coil. Using your right hand, F B C B C hold your first finger, second finger and thumb at right angles to each other. Point your first finger in B the direction of the magnetic field (N to S), your thumb in the direction of the motion (or force) of the N S N S A D A D conductor. The middle finger will point in the direc- I tion of the induced current (from + to -). V V Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 28

Grade 12 Science Essentials Electrodynamics SCIENCE CLINIC 2022 © MOTORS Electric motors convert electrical energy to mechanical energy. It consists of a current carrying armature, connected to a source by a commutator and brushes and placed in a magnetic field. Direct Alternating A direct current motor uses a split ring commutator to con- The alternating current motor is connected to the external nect the conductor to the external circuit instead of a slip circuit by a slip ring. The slip ring makes contact with ring. brushes which are connected to the external circuit at- tached to an alternating current source. The split ring commutator allows the current in the coil to The direction of the current in the coil is constantly chang- alternate with every half turn, which allows the coil to con- ing, which allows the coil to continue to rotate in the same tinue to rotate in the same direction. direction. When a charge moves in a magnetic field it experiences a force. The force experienced on both sides of the armature creates torque which makes it turn. The direction of the force can be explained using the left hand rule. FLEMING’S LEFT HAND MOTOR RULE The Left hand Rule is used to predict the direction of the movement of the coil in the motor. Using your left hand, hold your first finger, second finger and thumb right angles to each other. Point your first finger in the direction of the magnetic field, your second finger in the direction of the conventional current and your thumb will then point in the direction of the force. F B C B I N S A D Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 29

Grade 12 Science Essentials Photoelectric Effect SCIENCE CLINIC 2022 © PARTICLE NATURE OF LIGHT THRESHOLD FREQUENCY (f0), WORK FUNCTION (W0) AND FREQUENCY AND WAVELENGTH The photoelectric effect occurs when light is shone on a metal’s ELECTRON ENERGY An increase in frequency will increase the kinetic energy of surface and this causes the metal to emit electrons. The frequency required to produce enough energy to emit an electron is called the electrons. On a graph of Ek(max) vs frequency, the the threshold frequency (f0). The threshold frequency (f0) is the minimum x-intercept indicates the threshold frequency. Metals are bonded in such a way that they share their valence electrons in a sea of frequency of incident radiation at which electrons will be emitted from a particular metal. The minimum amount of energy needed to emit an No emission Emission delocalized electrons. In order to get an electron to be removed from the surface of electron from the surface of a metal is known as the work function f < f0 f > f0 EK(max) (J) a metal, you would have to provide it with (W0). The work function is material specific. enough energy in order to escape the If the energy of the photons exceed the work function (i.e. the frequency of light bond. exceeds the threshold frequency), the excess energy is transferred to the liber- ated electron in the form of kinetic energy. The energy that light provides enables the electron to escape and f0 f (Hz) this phenomenon is called photoelectric effect. The kinetic energy of the electron can be determined by: NOTE: Similarly, on the graph of Ek(max) vs wavelength, the PHOTON ENERGY E = W0 + E K(max) Sometimes work function is given x-intercept indicates the maximum wavelength of light in eV. Convert from eV to J: that can emit an electron (wavelength is inversely propor- Photons are “little packets” of energy called quanta, which act as tional to frequency). particles. The energy of the photon(light) can be calculated in one f < f0 f = f0 f > f0 J = eV × 1,6×10−19 of two ways: Emission No emission hc c/! > f0 c/! < f0 EK(max) (J) E = hf E= λ E = energy of the photon measured in joules (J) e- EK(max) = ½ mv(max)2 h = Planck’s constant, 6,63 × 10−34 (J·s) e- f = frequency measured in hertz (Hz) W0 = hf0 𝜆 = wavelength measured in meters (m) e- Maximum ! (m) wavelength c = speed of light, 3 × 108 (m·s−1) INTENSITY AND FREQUENCY INTENSITY, FREQUENCY AND CURRENT Increasing the intensity (brightness) of the light (radiation) means that there are more photons with the same frequency. This The higher the intensity of the radiation, the more electrons are emitted per sec- will result in more electrons with the same amount of energy being emitted. Thus the number of electrons emitted per second ond. Therefore an increased intensity will increase the current produced. increases with the intensity of the incident radiation The higher the frequency, the more kinetic energy is provided to the same number Light LOW INTENSITY HIGH INTENSITY of electrons. The rate of electron flow REMAINS THE SAME. source Metal . . .. . .. . Number of electrons . . . . . . . . . . . . . . . . . . . f0 . . . High-intensity light G V ejected per second (current) Metal Electron Photon Note that the energy of the electrons remain the same. If the frequency of the incident radiation is below the cut-off frequency, Low-intensity light then increasing the intensity of the radiation has no effect i.e. it does not cause electrons to be ejected. To increase the energy, the frequency of the radiant light needs to be increased. INCREASE INTENSITY AT CONSTANT FREQUENCY = INCREASE AMOUNT OF EMITTED ELECTRONS INCREASE FREQUENCY AT CONSTANT INTENSITY = INCREASE KINETIC ENERGY OF EMITTED ELECTRONS Frequency of light on cathode (Hz) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 30

Grade 12 Science Essentials Emission and Absorption Spectra SCIENCE CLINIC 2022 © CONTINUOUS SPECTRUM ENERGY LEVELS ATOMIC ABSORPTION SPECTRA When white light shines through a prism, the light is dispersed into a The lines on the atomic spectrum (line emission spectrum) are as a result spectrum of light. It is called the continuous spectrum. White light is of electron transitions between energy levels. As electrons gain energy As white light is passed through a cold gaseous element, made up of all the colours of the spectrum, and when dispersed we are (heat) they transition to a higher state of energy, hence a higher energy the electrons inside the atoms of the element absorb the able to see the components of white light. This is the same range as the level. As the element cools, the electrons drop back down to the lower en- energy of the light (photons). The energy is absorbed in visible spectrum. ergy state. As they do this they emit the absorbed energy in the form of specific bands which corresponds to the energy required light (photons). Therefore, each electron gives off a frequency of light by electrons to transition between energy levels. which corresponds with the line emission spectrum. R O Y G B I V n=4 The absorbed energy furthermore corresponds to a spe- cific frequency of light. As the electrons absorb energy, red orange yellow green blue indigo violet n=3 the specific frequency of light disappears, resulting in an absorption spectrum. The black lines below represent the The white light is emitted from a bulb, therefore the continuous spectrum is an energy absorbed by the element. emission spectrum. Red has the lowest frequency, and violet the highest. n=2 ATOMIC EMISSION SPECTRA When an element in the gaseous phase is heated, it emits light. If the n=1 light produced is passed through a prism, a spectrum is produced. How- Electrons in the n = 1 energy level are closest to the nucleus and have the The absorption spectra for a specific element will be the ever, this spectrum is not continuous, but consists of only some lines of lowest potential energy. Electrons in the energy level furthest from the nu- inverse of its emission spectra. colour. This is known as line emission spectrum. cleus (eg. n = 4) has the highest potential energy. If an electron is free Each element has their own light signature as no two elements have the from the atom then it has zero potential energy. The potential energy is same spectrum. given as a negative value due to a decrease in potential as the electrons get close to the nucleus. As the electrons transition from their energized state back to their normal state they give off energy that is equal to the difference in potential energy between the energy levels. ΔE = E 2 − E1 ΔE = difference in potential energy between two energy levels E2 = the highest energy state E1 = the lowest energy state CONTINUOUS EMISSION SPECTRA The amount of energy that is released relates directly to a specific fre- quency or wavelength (thus colour) of light. hc E = hf E= λ ATOMIC ABSORPTION SPECTRA EXAMPLE: A sample of hydrogen gas is placed in a discharge tube. The electron from the hydrogen atom emits energy as it transitions from energy level E6 (−0,61×10−19 J) to E2 (−5,46×10−19). Determine the wavelength of light emitted. E = hc ATOMIC EMISSION SPECTRA λ ΔE = E 6 − E 2 −19 (6,63 × 10−34 )(3 × 108 ) = − 0,61 × 10−19 − (−5,46 × 10−19 ) 4,85 × 10 = λ = 4,85 × 10−19 J λ = 4,1 × 10−7 = 410 nm Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 31

Grade 12 Science Essentials Doppler Effect SCIENCE CLINIC 2022 © As a sound source moves through space, it moves relative to the waves that it has already produced. This causes an apparent change in wavelength, and therefore also a change in the perceived frequency. Listener away from source (−) Source away from listener (+) Doppler effect: The change in the frequency of the sound observed by a listener, because the sound source and the listener have different Decrease in observed frequency. Decrease in observed frequency. speeds relative to the medium in which the sound is propagated. v ± vL v ± vL fL = f fL = f v ± vs s v ± vs s v − vL v fL = fs fL = fs v v + vs x Listener to source (+) Source to listener (−) When a source moves towards an ob- When a source moves away from an server OR an observer moves towards observer OR an observer moves away Increase in observed frequency. Increase in observed frequency. a source, the distance between the wave- from a source, the distance between the fronts (wavelength) is decreased and wavefronts (wavelength) is increased and v ± vL v ± vL the perceived frequency increases. the perceived frequency decreases. fL = f fL = fs v ± vs s v ± vs A change in wavelength/frequency can be observed as: • Sound- A change in pitch (higher frequency, higher pitch) v + vL v fL = fs fL = f • Light- A change in light colour (red-shift OR blue shift) v v − vs s EQUATIONS DOPPLER EFFECT AS PROOF OF EXPANDING UNIVERSE The visible range of the electromagnetic spectrum The shift can be observed by comparing the ab- REMEMBER: v ± vL ALWAYS start with the is as follow: sorption spectra of moving stars with a stationary fL = f f reference sample. v ± vs s FULL original formula (on data sheet)! Red shi3 R O Y G B I V fL = frequency of listener (Hz) v = speed of sound (m·s−1) Red light has a longer wavelength than blue light, Sta$onary OR thus a shift towards the red spectrum (longer wave- reference speed of light (3 ×108 m·s−1) length, lower frequency) indicates that the source vL = velocity of listener (m·s−1) of the light is moving away from the observer. vS = velocity of source (m·s−1) Blue shi3 fS = frequency of source (Hz) Red shift: object moving away from observer. Blue shift: object moving towards observer. Because most stars exhibit a red shift, we can con- To convert between frequency and wavelength: clude that they are moving away from the earth and thus that the universe is expanding. Material waves (i.e. sound) Electromagnetic radiation APPLICATIONS OF DOPPLER EFFECT v = fλ c = fλ Medicine Maritime Law enforcement • Measures blood flow rate • SONAR positioning underwater • Radar monitoring of speed- v = speed of wave (m·s−1) c = speed of light (3 ×108 m·s−1) • Observation of foetus heart rate ing cars f = frequency of sound (Hz) f = frequency of sound (Hz) 𝜆 = wavelength (m) 𝜆 = wavelength (m) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 32

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DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 2 (CHEMISTRY) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 VRAESTEL 2 (CHEMIE) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Standard pressure pT 1,013 x 105 Pa Standaarddruk Molar gas volume at STP Vm 22,4 dm3∙mol-1 Molêre gasvolume by STD Standard temperature Standaardtemperatuur TT 273 K Charge on electron e -1,6 x 10-19 C Lading op elektron Avogadro's constant NA 6,02 x 1023 mol-1 Avogadro-konstante TABLE 2: FORMULAE/TABEL 2: FORMULES m N n n M NA n m V c or/of c n V MV Vm cava na pH = -log[H3O+] cbvb nb Kw = [H3O+][OH-] = 1 x 10-14 at/by 298 K Eθcell Eθcathode  Eθanode / Eθsel θ Ekatode  Eθanode or/of Eθcell Ereduction θ  Eθoxidation / Eθsel θ Ereduksie  Eθoksidasie or/of Eθcell agent  Ereducingagent / E sel Eθoksideermiddel  Ereduseermi θ θ θ θ Eoxidising ddel

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) Atomic number 1 KEY/SLEUTEL Atoomgetal 2 2,1 H He 1 29 4 3 4 Electronegativity Symbol 5 6 7 8 9 10 1,0 1,5 1,9 Cu 2,0 2,5 3,0 3,5 4,0 Li Be Elektronegatiwiteit 63,5 Simbool B C N O F Ne 7 9 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

TABLE 4A: STANDARD REDUCTION POTENTIALS TABEL 4A: STANDAARDREDUKSIEPOTENSIALE Half-reactions/Halfreaksies E θ (V) F2(g) + 2e ⇌ 2F + 2,87 3+  2+ Co +e ⇌ Co + 1,81 H2O2 + 2H +2e + ⇌ 2H2O +1,77  +  2+ MnO 4 + 8H + 5e ⇌ Mn + 4H2O + 1,51   Cℓ2(g) + 2e ⇌ 2Cℓ + 1,36 2 3+ Cr2O 7 + 14H + 6e + ⇌ 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  + 4H + 3e + NO 3 ⇌ NO(g) + 2H2O + 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 ⇌ 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

TABLE 4B: STANDARD REDUCTION POTENTIALS TABEL 4B: STANDAARDREDUKSIEPOTENSIALE Half-reactions/Halfreaksies E θ (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 + 2e 2+ Cr ⇌ Cr  0,91 Increasing oxidising ability/Toenemende oksiderende vermoë   2H2O + 2e ⇌ H2(g) + 2OH  0,83 + 2e 2+ Zn ⇌ Zn  0,76 Increasing reducing ability/Toenemende reduserende vermoë + 3e 3+ Cr ⇌ Cr  0,74 2+  Fe + 2e ⇌ Fe  0,44 2+ + e 3+ Cr ⇌ 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 + 3e 3+ Fe ⇌ 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 +2e ⇌ 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

Grade 12 Science Essentials SCIENCE CLINIC 2022 © Grade 12 Chemistry Definitions Intermolecular Boiling point: The temperature at which the vapour pressure equals atmospheric pressure Forces Revision Melting point: The temperature at which the solid and liquid phases of a substance are at equilibrium Vapour pressure: The pressure exerted by a vapour at equilibrium with its liquid in a closed system Organic molecules: molecules containing carbon atoms Functional group: a bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds Hydrocarbon: organic compounds that consist of hydrogen and carbon only Homologous series: a series of organic compounds that can be described by the same general formula OR in which one member differs from the next with a CH2 group Saturated compounds: compounds in which there are no multiple bonds between C atoms in their hydrocarbon chains Organic chemistry Unsaturated compounds: compounds with one or more multiple bonds between C atoms in their hydrocarbon chains Functional group: A bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds Molecular formula: a chemical formula that indicates the element and numbers of each of the atoms in a molecule Structural formula: a structural formula of a compound shows which atoms are attached to which within the molecule. Atoms are represented by their chemical symbols and lines are used to represent ALL the bonds that hold the atoms together Condensed structural formula: this notation shows the way in which atoms are bonded together in the molecule, but DOES NOT SHOW ALL bond lines Structural isomer: Organic molecules with the same molecular formula, but different structural formulae Macromolecule: A molecule that consists of a large number of atoms Polymer: A large molecule composed of smaller monomer units covalently bonded to each other in a repeating pattern Monomer: Small organic molecules that can be covalently bonded to each other in a repeating pattern Polymerisation: A chemical reaction in which monomer molecules join to form a polymer Addition polymerisation: A reaction in which small molecules join to form very large molecules by adding on double bonds Polymers Addition polymer: A polymer formed when monomers (usually containing a double bond) combine through an addition reaction. Condensation polymerisation: Molecules of two monomers with different functional groups undergo condensation reactions with the loss of small molecules, usually water Condensation polymer: A polymer formed by two monomers with different functional groups that are linked together in a condensation reaction in which a small molecule, usually water, is lost Heat of reaction (ΔH): the energy absorbed or released in a chemical reaction Exothermic reactions: reactions that release energy Energy change Endothermic reactions: reactions that absorb energy Revision Activation energy: the minimum energy needed for a reaction to take place Activated complex: the unstable transition state from reactants to products Rate and Extent Reaction rate: as the change in concentration of reactants or products per unit time of reaction Positive catalyst: a substance that increases the rate of a chemical reaction without itself undergoing a permanent change. Open system: one which continuously interacts with its environment Closed system: one which is isolated from its surroundings Chemical Chemical equilibrium: it is a dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction Equilibrium Le Chatelier’s principle: when the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 37

Grade 12 Science Essentials SCIENCE CLINIC 2022 © Grade 12 Chemistry Definitions Arrhenius theory Acid: a substance that produces hydrogen ions (H+) when it dissolves in water Base: a substance that produces hydroxide ions (OH–) when it dissolves in water Lowry-Brønsted theory: Acid: a proton (H+ ion) donor Base: a proton (H+ ion) acceptor Strong acids: ionise completely in water to form a high concentration of H3O+ ions. Weak acids: ionise incompletely in water to form a low concentration of H3O+ ions. Strong bases: dissociate completely in water to form a high concentration of OH– ions. Acids and Bases Weak bases: dissociate/ionise incompletely in water to form a low concentration of OH– ions. Concentrated acids/bases: contain a large amount (number of moles) of acid/base in proportion to the volume of water Dilute acids/bases: contain a small amount (number of moles) of acid/base in proportion to the volume of water Ampholyte (amphiprotic substance): a substance that can act as either an acid or a base. Eg. Water Equivalence point (of a titration): the point at which the acid/base has completely reacted with the base/acid. Endpoint (of a titration): the point where the indicator changes colour Hydrolysis: the reaction of a salt with water Auto-ionisation of water: the reaction of water with itself to form H3O+ ions and OH– ions Kw: the equilibrium constant for the ionisation of water or the ionic product of water or the ionisation constant of water, i.e. Kw = [H3O+][OH–] = 1 x 10-14 by 298 K Oxidation: a loss of electrons, an increase in oxidation number Reduction: a gain of electrons, a decrease in oxidation number Oxidising agent: a substance that is reduced/gains electrons. Reducing agent: a substance that is oxidised/loses electrons. Anode: the electrode where oxidation takes place Electrochemical Cathode: the electrode where reduction takes place Reactions Electrolyte: a substance of which the aqueous solution contains ions OR a substance that dissolves in water to give a solution that conducts electricity Electrolysis: the chemical process in which electrical energy is converted to chemical energy OR the use of electrical energy to produce a chemical change Galvanic cell: a cell in which chemical energy is converted to electrical energy Electrolytic cell: a cell in which electrical energy is converted into chemical energy Eutrophication: the process by which an ecosystem, e.g. a river or dam, becomes enriched with inorganic plant nutrients, Fertilizers especially phosphorus and nitrogen, resulting in excessive plant growth. As plant growth becomes excessive, the amount of dead and decaying plant material increases rapidly. Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 38

Grade 12 Science Essentials Organic molecules SCIENCE CLINIC 2022 © Organic compounds contain carbon and hydrogen atoms. These compounds can be in the REPRESENTING ORGANIC COMPOUNDS gaseous, liquid, or solid phase. All living matter contains organic compounds. We use a variety of ways to draw or write organic compounds. We either make use of the molecular formula, con- UNIQUENESS OF CARBON densed formula or we use full structural formulae. Carbon is very unique and is the basic building block of all organic compounds. It’s atoms have a valency of four in a tetrahedral arrangement. A chemical formula that indicates the type of Molecular This means it is able to make four bonds. NOTE: formula atoms and the correct number of each in the molecule. C3H8 Carbon atoms can form single, double or triple bonds Carbon atoms have to form 4 bonds, but not necessarily with C C C C C C 4 other atoms Condensed formula The notation shows the way in which atoms are bonded to each other in a molecule, but CH3CH2CH3 HYDROCARBONS does not show all bond lines. A hydrocarbon is a compound that contains only carbon and hydrogen atoms. These H H H compounds can be saturated (single bonds) and unsaturated (double or triple bonds). Structural This formula shows all the bonds and atoms H C C C H Hydrocarbon: A compound containing only carbon and hydrogen atoms. formula in the molecule. Saturated compound: Unsaturated compound: H H H A compound in which all of the bonds be- A compound in which there is at least one tween carbon atoms are single bonds. double and/or triple bond between carbon General Describes the homologous series formula for atoms. formula similar compounds. Alkanes: CnH2n+2 H H H H H H H C C C H (Alkanes) C C C H (Alkenes and alkynes) ISOMERS H H H H H Isomers: Compounds having the same molecular formula but different structural formulae. Test to differentiate between saturated and unsaturated solutions butane 2-methylpropane 1) Unknown solution H H H Brown colour vanishes Does not discolour Chain isomers have the same immediately H H H H Add brown Br2 immediately Chain molecular formula but H C C C H isomers different arrangements of H C C C C H chains in the molecule. H H Unsaturated ∴ alkene/alkyne Saturated ∴ alkane H H H H H C H 2) H Purple colour vanishes Unknown solution immediately. Brown Purple colour does not propan-1-ol propan-2-ol precipitate forms vanish immediately H Add purple KMnO4 These have the same H Positional molecular formula but the H H O H O H Unsaturated ∴ alkene/alkyne Saturated ∴ alkane isomers functional group is in a different position H C C C H H C C C H CLASSIFICATION OF ALCOHOL H H H H H H Primary Secondary Tertiary One C bonded to the C Two C’s bonded to the C Three C’s bonded to the C These have the same propanoic acid methyl ethanoate bonded to the OH bonded to the OH bonded to the OH molecular formula but a H H R’’ different functional group. H H O H O H Functional Aldehydes and ketone are isomers R C O H R C O H R C O H functional isomers as well as H C C C O H H C C O C H carboxylic acids and esters H H H H H R’ R’ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 39

Grade 12 Science Essentials Organic molecules- Naming SCIENCE CLINIC 2022 © All organic compounds belong to a specific group which allows us to identify or name the compound. The group that compounds belong Number of to, known as the homologous series, depends on the the func- carbon atoms Root name Substituent Formula Structural formula Name tional group of the compound. in main chain Homologous series: A series of organic compounds that can be H 1 meth− described by the same general formula OR in which one member differs from the next with a CH2 group CH3− C C H methyl− Functional group: A bond or an atom or a group of atoms that 2 eth− H determine(s) the physical and chemical properties of a group of or- ganic compounds 3 prop− H H 4 but− Alkyl CH3CH2− C C C H ethyl− NAMING ORGANIC COMPOUNDS H H Every distinct compound has a unique name, and there is only one 5 pent− possible structure for any IUPAC (International Union of Pure and H H H Applied Chemistry) name. The IUPAC method for naming is a set 6 hex− CH3CH2CH2− propyl− C C C C H pattern. It indicates the longest chain (the longest continuous chain), the functional group and names of substituent groups 7 hept− H H H (side chains) or atoms attached to the longest chain. 8 oct− X repesents a halogen: Three parts of an IUPAC name: Fluorine: fluoro− Substituent Functional 9 non− Halogen X− C X Chlorine: chloro− Root name Bromine: bromo− Prefix group suffix 10 dec− Iodine: iodo− The root name indicates the number of carbon atoms in the long- est chain. This chain must contain the functional group. The EXAMPLE: prefix indicates the number and location of atoms or groups Number of Substituent substituents prefix Write down the name of the molecule below: (substituents) attached to the longest chain. The suffix identifies the functional group. Steps to naming organic compounds: di 2 1. Identify the longest continuous carbon chain which must contain eg. dimethyl the functional group. 2. Number the longest carbon chain beginning at the carbon (car- tri 3 bon 1) nearest to the functional group with the alkyl substitu- eg. triethyl ents on the lowest numbered carbon atoms of the longest chain. 3. Name the longest chain according to the number of carbons in tetra the chain. (the root name) 4 eg. tetramethyl 4. The suffix of the compound name is dependent on the func- tional group. NOTE: Substituents Main chain Functional group A maximum of THREE substituent 5. Identify and name substituents (alkyl and halogen substituents), 1−chloro 7 = hept 1,4−diene chains (alkyl substituents) are al- indicating the position of the substituent lowed on the main chain 2,4−diethyl 6. For several identical side chains use the prefix di-, tri-, tetra- NOTE: 3−methyl 7. Arrange substituents in alphabetical order in the name of the comma between numbers compound, ignore the prefix di-, tri-, tetra- (substituent pre- 5,6−dibromo number , number fix) dash between letter and number letter − number − letter 5,6−dibromo−1−chloro−2,4−diethyl−3−methylhept−1,4−diene 8. Indicate position using numbers. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 40

Grade 12 Science Essentials Organic Functional Groups SCIENCE CLINIC 2022 © Homologous Examples series and Functional Suffix Properties General group Condensed formula Structural formula Name formula Single bonds H H H Polarity: Non-Polar Alkanes IMF: Weak London CnH2n+2 R C C R’ -ane H C C C H CH3CH2CH3 propane Reactions: Substitution, Elimination, H H H Combustion Double bonds H H H Polarity: Non-polar Alkenes CnH2n R C C R’ -ene H C C C H CH3CH=CH2 propene IMF: London Reactions: Addition, combustion H Triple bonds H Polarity: Non-polar Alkynes CnH2n−2 R C C R’ -yne H C C C H CH≣CCH3 propyne IMF: London Reactions: Addition H Br Cl H Haloalkane/ X (Group Halogens 17) fluoro− chloro− Polarity: Polar Haloalkene H C C C H CH2BrCHClCH3 1−bromo−2−chloropropane IMF: Dipole−Dipole bromo− (Alkyl halide) Reactions: Elimination, Substitution R C R’ iodo− H H H Hydroxyl H H H Polarity: Polar Alcohols IMF: Strong Hydrogen bonds CnH2n+2O R O H -ol H C C C O H CH3CH2CH2OH propan−1−ol Reactions: Substitution, Elimination, Esterification, Combustion H H H Carboxyl H H O O Polarity: Polar Carboxylic acids -oic acid H C C C O H CH3CH2COOH propanoic acid IMF: Strong Hydrogen bonds CnH2nO2 R C O H Reactions: Esterification H H Carbonyl H H O H H O Polarity: Polar Esters -yl -oate CH3CH2COOCH2CH3 H C C C O C C H ethyl propanoate IMF: Dipole−Dipole R−COO−R’ (alch.) (carbox.) (carbox.) (alch.) R C O R’ Reactions: Formed by esterification H H H H Formyl H H O O Aldehydes Polarity: Polar -al H C C C H CH3CH2CHO pronanal CnH2n+1CHO IMF: Dipole−Dipole R C H H H H O H O Carbonyl Ketone Polarity: Polar -one H C C C H CH3COCH3 propan−2−one R−COC−R’ IMF: Dipole−Dipole R C R’ H H Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 41

Grade 12 Science Essentials Organic Intermolecular Forces SCIENCE CLINIC 2022 © Intermolecular forces are forces that exist between molecules in the solid, liquid and gaseous phases. They are electrostatic COMPARING IMF attractive forces. The strength of the IMF will determine the freedom of the particles, determining the phase of the sub- 1. Identify the type of intermolecular force. stance (solid, liquid, gas). 2. Discuss the difference between the two compounds (ɠ → ɣ ). Intermolecular force are a weak force of attraction between molecules or between atoms of noble gases 3. Discuss how this difference either ↑ or ↓ the strength of the intermolecular force. 4. Discuss how the physical property is affected (↑ or ↓ ). The types of intermolecular forces that exists between different types of organic molecules and the strength of the intermo- 5. Discuss energy required to overcome forces. lecular forces will affect the physical properties of a molecule. ●Strongest of all the intermolecular forces Alcohols ɠ TYPE OF FUNCTIONAL GROUP ●Act over shorter distances. (1 bonding site) The more polar the molecule, the stronger the IMF Hydrogen Bonds ●Between molecules that are strongly polar that contain hydrogen Carboxylic Acids TYPES OF IMF bonded to a small highly electronegative atom such as N, O or F. (2 bonding sites) Haloalkane ●Stronger than Dispersion forces/Induced Dipole forces. Aldehydes Carboxylic Alcohol Hydro- Dipole-Dipole ●Between slightly polar molecules. Ketones carbon Ester acid Forces ●Force of attraction between the δ+ end of the one molecule and the Esters δ– end of another. Alkyl Halides INCREASING IMF, MORE ENERGY REQUIRED TO OVERCOME London Dipole-dipole Hydrogen ●Very weak Van der Waals forces. Alkanes (dispersion) bonding Induced Dipole ●Between non-polar molecules that form induced (temporary) dipoles Alkenes Forces (London) and these temporary dipoles attract each other Alkynes ɡ NUMBER OF FUNCTIONAL GROUPS An increase in functional groups increase the IMF RELATIONSHIP BETWEEN PHYSICAL PROPERTIES AND IMF OH H H OH OH H OH OH OH RELATIONSHIP PHYSICAL PROPERTY TO IMF H C C C H H C C C H H C C C H H H H H H H H H H Melting Point: The temperature at which the solid and liquid phases of a substance are Directly at equilibrium. It is the temperature where solid particles will undergo a phase change (melt) INCREASING IMF, MORE ENERGY REQUIRED TO OVERCOME proportional and become a liquid. Boiling Point: The temperature at which vapour pressure of the substance equals Directly ɢ CHAIN LENGTH: MOLECULAR MASS atmospheric pressure. It is the temperature where liquid boils and turns into a vapour (gas). proportional The greater the number of carbon atoms in the chain, the greater the molecular mass. An increase in molecular mass increases the IMF OPTIONAL H H H H H H Vapour Pressure: This is the pressure that an enclosed vapour at equilibrium exerts on Inversely the surface of its liquid. proportional H C H H C C H H C C C H H H H H H H Viscosity: this is the measure of a liquid’s resistance to flow. A liquid with high viscosity Directly resists motion e.g. syrup. A liquid with low viscosity is runny e.g water. proportional INCREASING IMF, MORE ENERGY REQUIRED TO OVERCOME Solubility: Substances will only dissolve in substances that are like bonded. A non-polar Inversely ɣ CHAIN LENGTH: BRANCHES substance will dissolve in a non-polar substance. A polar substance will dissolve only in polar proportional substances. More branching results in a smaller contact surface area and lower the strength of the IMF Density: Density is a measure of the mass per unit volume. The solid phase of the Directly substance is generally more dense than the gaseous and liquid phase. proportional CH3 CH3 Flammability: The ability to burn in air or ignite causing combustion. Most organic Inversely CH3 CH2 CH2 CH2 CH3 CH3 CH CH2 CH3 CH3 C CH3 compounds are flammable and burn in oxygen to form carbon dioxide and water. proportional CH3 Odour: Different functional groups attach differently to different receptors in our nose. Different Inversely DECREASING IMF, LESS ENERGY REQUIRED TO OVERCOME organic substances give off odour quicker based on their intermolecular forces and distinct odours. proportional Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 42

Grade 12 Science Essentials Organic Reactions SCIENCE CLINIC 2022 © ADDITION REACTIONS (UNSATURATED → SATURATED) ELIMINATION REACTIONS (SATURATED → UNSATURATED) Addition reactions are reactions where atoms are added to an organic molecule. The double or triple An elimination reaction is a reaction where atoms or groups of atoms are removed from an organic mole- bonds break open and the new atoms are added to the carbon atoms on either side of the double or cule to form either a double or triple bonded compound. triple bond. This reaction is the opposite reaction to the addition reactions and is the removal of hydrogen (H2), a We can add hydrogen (H2), a halogen (Group 7 – e.g. Cl2), a Hydrogen halide or water H2O. In addition halogen (Group 7 – e.g. Cl2), a hydrogen halide or water H2O. In elimination reactions, alkanes form ei- reactions alkenes or alkynes form an alkane chain as a product. ther an alkene or alkyne chain as a product. 1) Hydrogenation – add H2 1) Dehydrogenation – remove H2 Reaction conditions: Reaction conditions: H H H H H H H H H H H H The alkene needs to be dis- The alkane needs to be in the H2 + H C C C H H C C C H solved in a non-polar solvent H C C C H H C C C H + H2 presence of a catalyst eg. Pt, Ni and needs to have a catalyst or Pd. H H H H present eg. Pt, Ni or Pd. H H H H hydrogen + alkene/alkyne → alkane alkane → alkene + hydrogen 2) Halogenation – add X (X = Halogen: F2, Cl2, Br2, I2) 2) Dehalogenation – remove X (X = Halogen: F2, Cl2, Br2, I2) H H H H H H H H H H H H Reaction conditions: Reaction conditions: X2 + H C C C H H C C C H No water to be present if it is to H C C C H H C C C H + X2 Takes place in an unreactive H X X take place solvent H H X X H halogen + alkene/alkyne → haloalkane haloalkane → alkene + halogen 3) Hydrohalogenation – add HX (X = Halogen: F2, Cl2, Br2, I2) 3) Dehydrohalogenation – remove HX (X = Halogen: F2, Cl2, Br2, I2) (Major product) (Major product) Reaction conditions: H H H H H H Reaction conditions: H H H H H H H H Takes place in the presence of No water to be present if it is to H C C C C H HX + concentrated NaOH/KOH in HX + H C C C H H C C C H H C C C C H take place. ethanol as the solvent. Strong H H X H H H X H H H heating under reflux. (Minor product) (Minor product) Markovnikov’s rule: The H H H H H Zaitzev’s rule: H atom is re- H H H atom will bond to the carbon moved from the carbon atom H C C C H atom which has the greater H C C C C H with the least number of H at- number of H atoms bonded to oms. (Keeps biggest H groups) H H X H H it. (Form biggest H groups) hydrogen halide + alkene/alkyne → haloalkane haloalkane → hydrogen halide + alkene 4) Hydration – add of H2O 4) Dehydration – remove H2O Reaction conditions: (Major product) (Major product) Requires the heating of an alco- Reaction conditions: H H H H H H H H H H H H H H hol with concentrated acid cata- Strong but dilute acid catalyst lyst eg. H2SO4 or H3PO4. The H2O + H C C C H H C C C H e.g. H2SO4 or H3PO4 H C C C C H H2O + H C C C C H acid should be in excess Heat in the form of steam H H O H (H2O) reactant H H O H H H H Sulfuric acid is known (Minor product) Markovnikov’s rule: The H H as a dehydrating agent. H H H atom will bond to the carbon (Minor product) atom which has the greater H H H H H C C C H number of H atoms bonded to Zaitzev’s rule: H atom is re- H H O it. (Form biggest H groups) H C C C C H moved from the carbon atom with the least number of H at- H H H oms. (Keeps biggest H groups) water (H2O) + alkene/alkyne → alcohol alcohol → water (H2O) + alkene Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 43

Grade 12 Science Essentials Organic Reactions SCIENCE CLINIC 2022 © SUBSTITUTION REACTIONS (SATURATED → SATURATED) COMBUSTION/OXIDATION REACTIONS Substitution reactions is when an atom or group of atoms in an organic molecule are replaced or Hydrocarbons are the main source of fuel in the world at the moment. They are used in the production swopped/exchanged for another atom or group of atoms. Substitution reactions take place between com- of electrical energy and as fuel for various engines. When hydrocarbons and alcohols react with oxygen pounds that are saturated alkanes, haloalkanes and alcohols. they form water and carbon dioxide. These reactions are exothermic and produce large quantities of heat. 1) Alkanes: Halogenation- Substitute X2 (X = Halogen: F2, Cl2, Br2, I2) Complete combustion (excess oxygen): C3H8 + 5O2 → 3CO2 + 4H2O + energy H H H H H H Reaction conditions: Incomplete combustion (insufficient oxygen): 2C3H8 + 7O2 → 6CO + 8H2O + energy Reaction takes place in the H C C C H + X2 H C C C H + HX Balancing: C → H → O presence of sunlight/heat H H H H X H CRACKING alkane + halogen → haloalkane + hydrogen halide Hydrocarbons can be made up of very long chains of hundreds of carbons. Crude oil is a mixture of many large hydrocarbons and each source of crude oil is different resulting in different types and 2) Haloalkanes: Hydrolysis - Substitute H2O amounts of hydrocarbons. Shorter chain hydrocarbons are more useful to use as fuels as they burn more readily and are more flam- H H H H H H H2O / HX / Reaction conditions: mable. Cracking is the breaking up of long hydrocarbon chains into smaller more useful hydrocarbons. H C C C H + NaOH / H C C C H + NaX / The haloalkane is dissolved An alkene and alkane will be the products as a result of cracking. Cracking is a type of elimination reac- KOH KX in an ethanol solution and tion. H X H H O H treated with hot aqueous H H H H H H H H H H H H H NaOH/KOH solution H C C C C C C H H C C C H + H C C C H haloalkane + H2O/NaOH/KOH → alcohol + HX/NaX/KX H H H H H H H H H H 3) Alcohols: Hydrohalogenation- Substitute HX (X = Halogen: F2, Cl2, Br2, I2) alkane → alkane + alkene Tertiary alcohols Thermal cracking This method makes use of high pressures (up to 70 atmospheres) and high temperatures to crack the H H long hydrocarbon chains with no catalyst. Products: alkanes and alkenes. H C H H C H Catalytic cracking H H H H Reaction conditions: This method uses a catalyst to crack long carbon chains at low pressure (up to 1 atmosphere) and low Require HX present at room temperature. The heated crude oil is passed into a fractional distillation column and passed over a cata- H C C C H + HX H C C C H + H2O temperature. lyst. The column is hottest at the bottom and coolest at the top. The crude oil separates according to H O H H X H boiling points and condenses as the gas rises up the column. The substances/chains with the longest chains will have higher boiling points and condense at the bottom and vice versa. Products: alkanes H and alkenes OR alkenes and H2. alcohol + hydrogen halide → haloalkane + water ESTERIFICATION Primary and secondary alcohols This is a reaction between an alcohol and a carboxylic acid in the presence of a concentrated acid catalyst, H2SO4. This reaction is a type of an elimination reaction and is also known as a condensation reaction as two organic molecules form one organic molecule and water is removed from the reactants Reaction conditions: and forms as a product in the reaction. Esters are responsible for the various smells which occur in na- H H H H H H High temperatures and need ture and they are generally pleasant smells like banana and apple etc. H C C C O H + HX H C C C X + H2O to be treated with NaBr and H H O H H O H H H H H H H H H H concentrated H2SO4. H C C O H + H O C C C H H C C O C C C H + H2O H H H H H H H H alcohol + hydrogen halide → haloalkane + water alcohol + carboxylic acid → ester + water Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 44

Grade 12 Science Essentials Summary of Organic Reactions SCIENCE CLINIC 2022 © Haloalkane ( Subs ( ) Alkane + Alkene [d H2 O with 1 halogen H H H H H H ilu /N te aO Add ( + ) ) ds H C H + C C ct C C C tro /K Hyd od u Ha ng OH ro Elim ( − ) H H pr H X H lo , m bas HX ysis l X by ge na od e /N te er dis aX , H V] Dehydrohalogenation m [te tio pe rt n at so /K X2 –U Hydrohalogenation ra ia ( e h lv X ( ght t u ry ea ed by Thermal cracking n li ( − HX) re al HX t] in pro ati o S u n alc . Pr coh , H oh im ol 2 O et d h a uc gen [ [Concentrated strong ols ar s a by no t ) ( + HX) y l, alo t - h an ro pro H base, strongly heated, ea d om du te sec ct) d] dissolved in ethanol] on da ry Alcohol Hydrogenation ( + H2) Alkene Alkane Hydration ( + H2O) H H H H H H [Pt] H H H [H2SO4] C C C C C C C C C Catalytic cracking ( − H2) Dehydration ( − H2O) H O H H H H H [Pt] [concentrated H2SO4] H Dehalogenation Halogenation Esterification ( + X2 ) Combustion ( − X2) [Sunlight – UV] ( + Carboxylic acid ) ( reacts with O2 ) [concentrated H2SO 4 ] Haloalkane with 2 halogens H H H C C C CO2 + H2O Ester X X H Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 45

Grade 12 Science Essentials Polymers SCIENCE CLINIC 2022 © POLYMERS ADDITION POLYMERIZATION Polymers are very large organic molecules. They are made up of hundreds or thousands of atoms. Addition polymers are made by joining two or more monomers which have double bonds- unsaturated Polymers are structured from repeated smaller units called monomers. molecules. (These are alkene’s and alkyne’s) and form long single bonded chains. Monomer: Small organic molecules that can be covalently bonded to each other in a repeating pattern The double bond breaks open and the electrons join the two monomers together to create the long chain. No atoms are added or lost. Monomers may have different groups that are attached to the car- A B bon atoms. Polymer: A large molecule composed of smaller monomer units covalently bonded to each other in a H H H H H H repeating pattern C C C C C C A B A B A B A B H H H H H H n CONDENSATION POLYMERIZATION ethene monomer ethene polymer (polyethene) Condensation polymerization occurs when two different monomers which have functional groups (alcohols and carboxylic acids) on both ends of their molecules react. In this reaction a water molecule is Steps of addition polymerization of polyethene (polyethylene) eliminated. Polyethene is used to make squeeze bottles, plastic bags, films, toys and molded objects, and electric insulation. Polythene has the recycling number 4. Formation of polyesters 1) Initiation A polyester forms in a similar way as to a normal ester i.e. an alcohol and carboxylic acid reacting. The The presence of a free radical initiates the double bond of the monomer to break open. A free radical is only difference is that the alcohol and carboxylic acids have a functional group on both their ends and a molecule that has an unpaired electron. When the monomer reacts with the free radical it creates an- are diols and dicarboxylic acids. other free radical. Monomer (ester) formation H H H H * : Free radical H H Esterifica1on O O H H O O R* + C C R C C* (unpaired electron) H O C C O H +H O C R C O H H O C C O C R C O H + H2O H H H H 2) Propagation H H H H This new free radical will add to the double bond of another monomer and create another free radical. Polymer formation The chain will continue to grow and get longer and longer. These additions give rise to a macromole- H H O O Esterifica0on H H O O cule. This reaction takes place very quickly. H O C C O C R C O H + H O C C O C R C O H H H H H H H H H H H * : Free radical H H R C C* + C C R C C C C* (unpaired electron) H H H H H H H H 3) Termination O H H O O H There are two ways in which the polymerization can terminate. C R C O H H O C •Two radicals join together and terminate. C O H H O C C O H H H H H H H H H H H H H H H H H H H R C C C C* + *C C C C R R C C C C C C C C R H H H H H H H H H H H H H H H H • One radical removes a hydrogen from another radical and forms a saturated molecule (alkane) and O H H O O H forces the other radical to form an unsaturated molecule(alkene). C O C C O C R C O C H H H H H H H H H H H H H H H H H H H R C C C C* + *C C C C R R C C C C H + C C C C R H H H H H H H H H H H H H H H Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 46

Grade 12 Science Essentials Polymers SCIENCE CLINIC 2022 © MONOMER FREE RADICAL MONOMER STRUCTURE POLYMER STRUCTURE POLYMER NAME NAME BUILDING BLOCK Addition polymerization H H H H H H H H H H Propene C C C C C C C C C C Polypropylene H CH3 H CH3 H CH3 H CH3 H CH3 H H H H H H H H H H Styrene C C C C C C C C C C Polystyrene H H H H H H H H H H H C H H C H H C H H C H H C H O C O C O C O C O C Vinyl acetate PVA (polyvinyl acetate) H O H O H O H O H O C C C C C C C C C C H H H H H H H H H H Cl H Cl H Cl H Cl H Cl H Chloroethene C C C C C C C C C C PVC (polyvinyl chloride) H H H H H H H H H H Condensation polymerization H H H H PLA (polylactic acid) PLA monomer formed from Polymer formed by H H C H C H H C H H C H biological fermentation of esterification. Lactic acid plant materials is REMEMBER: Water is O C C O C C O C C biodegradable. Used for H O C C O H formed as byproduct. packaging material, has the H O H O H O potential to alleviate land- H O fill disposal problems. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 47

Grade 12 Science Essentials Quantitative aspects of chemical change SCIENCE CLINIC 2022 © The Mole Concentrations of solutions Molar Volumes of Gases Atoms, molecules and ions are too small to count, and there are so many Solutions are homogeneous (uniform) mixtures of two or more sub- If different gases have the same volume under the particles in even the smallest sample of a substance. stances. A solution is formed when a solute dissolves in a solvent. same conditions of temperature and pressure, they The solvent and solute can be a gas, liquid, or solid. The most common sol- will have the same number of molecules. The mole is the SI unit for the amount of substance. A mole of particles is vent is liquid water. This is called an aqueous solution. The molar volume of a gas, VM, is the volume an amount of 6,02 x 1023 particles. 6,02 x 1023 is known as Avo- occupied by one mole of the gas. gadro’s number, NA. number of particles Solution Solute Solvent N salt water salt water VM for all gases at STP is 22.4 dm3·mol−1. n= number of mole (mol) NA Avogadro’s number soda water carbon dioxide water Standard Temperature and Pressure (STP) is (6,02 x 1023) Concentration 273 K (0°C) and 1,01×105 Pa (1 atm). The concentration of a solution is the number of moles of solute EXAMPLE: This also means that for reactions at constant tem- per unit volume of solution. Determine the amount of H+ ions in 3 mol of H2SO4. perature and pressure, gas volumes will react in the N(H2 SO4 ) = nNA H2 SO4 : H + number of moles (mol) same ratio as the molar ratio. 23 = 3(6,02 × 10 ) 1:2 n of solute 24 = 1,81 × 10 H SO molecules 1,81 × 10 24 : 3,62 × 10 24 concentration (mol ⋅ dm−3) c= volume (dm3) V 2 4 ∴ N(H + ) = 3,62 × 10 24 ions of solution Molar Mass can also be calculated with Particles are too small to weigh individually. Molar mass (M) is defined as the mass in grams of one mole of that m c= substance (atoms, molecules or formula units) and is measured in the MV unit g.mol-1. m mass of substance (g) Concentration is the number moles of solute per 1 dm3 of solution number of mole (mol) n= i.e. mol·dm-3. If a solution of potassium permanganate KMnO4 has a M molar mass (g ⋅ mol−1) concentration of 2 mol.dm-3 it means that for every 1 dm3 of solution, there N2 + 2O2 → 2NO2 EXAMPLE: are 2 moles of KMnO4 dissolved in the solvent. Determine the number of moles in 13 g of CuSO4. 1 mol + 2 mol → 2 mol M(CuSO4 ) = 63,5 + 32 + 4(16) m n= EXAMPLE: EXAMPLE: −1 M 1 dm3 + 2 dm3 → 2 dm3 = 159,5 g ⋅ mol = 13 A solution contains 10 g of sodium Calculate the mass of solute in 159,5 = 0,082 mol hydroxide, NaOH, in 200 cm3 of 600 cm3 of 1,5 mol·dm-3 sodium volume of gaseous solution. Calculate the concentra- chloride solution. V substance (dm3) Percentage Composition tion of the solution. n= n(NaOH) = m V = 600 cm3 = 0,6 dm3 VM molar mass of element M Percentage composition of element= × 100 M(NaCl) = 23 + 35,5 MR of compound 10 number of moles (mol) molar gas volume at STP = 23 + 16 + 1 = 58,5 g ⋅ mol−1 (22,4 dm3 ⋅ mol−1) Consider these iron ores: haematite and magnetite – which con- = 0,25 mol tains more iron by mass? EXAMPLE: 3 3 n = cV A gas jar with a volume of 224 cm3 is full of chlo- Ore Haematite Magnetite V = 200 cm = 0,2 dm rine gas, at STP. How many moles of chlorine gas = 1,5 × 0,6 Formula Fe2O3 Fe3O4 are there in the gas jar? n = 0,9 mol Relative (2 x 56) + (3 x 16) (3 x 56) + (4 x 16) c(NaOH) = V n = V molecular mass =160 =232 VM 0,25 = m = nM 0,224 % iron by mass [(2 x 56) /160] x 100 [(3 x 56) / 232] x 100 0,2 = 0,9 × 58,5 = 22,4 = 70% = 72% = 1,25 mol ⋅ dm−3 = 52,65 g = 0,01 m ol ∴ magnetite contains more iron Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 48

Grade 12 Science Essentials Quantitative aspects of chemical change SCIENCE CLINIC 2022 © Calculating Empirical Formula Empirical formula to Molecular Formula Limiting Reactants from Mass The empirical formula is the simplest whole number ratio of atoms in a molecule. In a reaction between two substances, one reac- The molecular formula is actual ratio of the atoms in a molecule. tant is likely to be used up completely before the Empirical formula is the chemical formula of a compound The molecular formula can be calculated from the empirical formula and the relative other. This limits the amount of product formed. that shows the smallest whole number ratio of the atoms. molecular mass. Consider the reaction between magnesium and STEPS TO DETERMINE MOLECULAR FORMULA: dilute sulphuric acid. The balanced chemical equa- 1.Determine the mass of the elements. tion is 2.Determine mol of each substance. 1. Determine the empirical formula (if not given). 3.Simplify the atomic ratio. 2. Determine the molar mass of the empirical formula. Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) 3. Determine the ratio between molecular formula and empirical formula. This means that 1 mole of magnesium reacts with EXAMPLE: 4. Multiply the ratio into the empirical formula 1 mole of sulphuric acid. Both reactants will be A sample of an oxide of copper contains 8 g of copper combined EXAMPLE: completely used up by the time the reaction stops. with 1 g of oxygen. Find the empirical formula of the compound. An unknown organic compound has the empirical formula CH2. What happens if 1 mole of magnesium and 2 mole Steps Copper Oxygen The molecular mass of butene is 56g·mol-1. Determine the mo- of sulphuric acid are available to react? There is lecular formula of the compound. now insufficient magnesium to react with all of the Step 1: 8g 1g sulphuric acid. 1 mole of sulphuric acid is left after Mass of element 1. Empirical formula given: CH2 the reaction. n=m/M n=m/M 2. Determine the molar mass of the empirical formula. Step 2: Mol M(CH2 ) = 12 + 1 + 1 All of the magnesium is used up, We say the mag- = 8 / 63,5 = 1 / 16 (divide by mass of 1 mol) nesium is the limiting reactant. Some sulphuric = 0,126 mol = 0,0625 mol −1 = 14g ⋅ mol acid is left after the reaction. We say the sulphuric Step 3: Atom ratio 0,125/0,0625 0,0625/0,0625 acid is in excess. (divide by smallest no in ratio) ≈2 =1 3. Determine the ratio between molecular and empirical. molecular formula mass Empirical formula: Cu2O ratio number = empirical formula mass The amount of limiting reactant will deter- mine: 56 Calculating Empirical Formula from = • The amount of product formed. 14 Percentage Composition = 4 • The amount of other (excess) reactants used. The empirical formula of a compound can also be found from its per- centage composition. We assume that 100 g of the compound is ana- lysed, then each percentage gives the mass of the element in grams 4. Multiply the ratio into the empirical formula in 100 g of the compound. CH2 × 4 = C4 H8 Determining limiting reactants An oxide of sulphur contains 40% sulphur and 60% oxygen by 1. Calculate the number of moles of each reactant. mass. Determine the empirical formula of this oxide of sulphur. Approach to reaction stoichiometry 2. Determine the ratio between reactants. 1. Write a balanced chemical equation. 3. Determine limiting reactant using the ratios. Steps Sulphur Oxygen 2. Convert the ‘given’ amount into mole (use limiting reactant if applicable). Step 1: 40 60 3. Determine the number of mole of the ‘asked’ substance using the mole ratio. % of element 4. Determine the ‘asked’ amount from the number of mole. NOTE: Step 2: 40 60 Mass of element (g) Concentra0on Concentra0on Concentra0on Concentra0on If one reactant is in excess, it means that there is more than enough of it. Step 3: n=m/M n=m/M If there are only 2 reactants and one is in Mol = 40 / 32 = 60 / 16 excess, it means that the other is the limit- = 1,25 mol = 3,75 mol MOL + MOL MOL + MOL ing reactant. Step 4: 1,25 / 1,25 3,75 /1,25 Divide by smallest =1 =3 mol ratio Mass Volume Mass Volume Mass Volume Mass Volume Empirical formula: SO3 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 49

Grade 12 Science Essentials Quantitative aspects of chemical change SCIENCE CLINIC 2022 © EXAMPLE: Percentage Purity Percentage Yield A 8,4 g sample of nitrogen reacts with 1,5 g hydro- Sometimes chemicals are not pure and one needs to calculate the Yield is the measure of the extent of a reaction, generally measured by comparing gen. The balanced equation is: percentage purity. Only the pure component of the substance the amount of product against the amount of product that is possible. will react. For an impure sample of a substance: N2(g) + 3H2(g) → 2NH3(g) Some of the product may be lost due to evaporation into the surrounding air, or due to a little being left in solution. Some of the reactants may not react. We say that Determine (a) which reactant is the limiting reac- the reaction has not run to completion. Mass of pure substance tant, and (b) the mass of ammonia that can be Percentage purity = × 100 produced. Mass of impure substance This results in the amount of the product produced being less than the maximum theoretical amount you would expect. We can express this by the percentage yield: (a) 1. STEPS TO DETERMINE THE PERCENTAGE PURITY n(N2 ) = m Actual yield M 1. Determine moles of products. Percentage yield = × 100 8,4 2. From the balanced formula, determine the ratio between reactants Theoretical yield = 28 and products. = 0,3 mol 3. Using the ratio, determine the number of moles of reactants. Percentage yield is usually determined using mass, but can also be determined with 4. Determine the mass of pure reactant. mol and volume. 5. Calculate the percentage purity of the sample. m n(H2 ) = M STEPS TO DETERMINE THE PERCENTAGE YIELD 1,5 = EXAMPLE: 1. Determine moles of reactants. 2 2. From the balanced formula, determine the ratio between reactants and products. = 0,75 mol An impure sample of calcium carbonate, CaCO3, contains calcium 3. Using the ratio, determine the number of moles of products. sulphate, CaSO4, as an impurity. When excess hydrochloric acid 4. Determine the theoretical mass of product. 2. was added to 6g of the sample, 1200 cm3 of gas was produced 5. Calculate the percentage yield. N2 : H2 (measured at STP). Calculate the percentage purity of the cal- 1 : 3 cium carbonate sample. The equation for the reaction is: EXAMPLE: 0,3 mol : 0,9 mol CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2 O(l) + CO2(g) 128g of sulphur dioxide, SO2, was reacted with oxygen to produce sulphur trioxide, 0,25 mol : 0,75 mol SO3. The equation for the reaction is: V 1,2 2SO2(g) + O2(g) → 2SO3(g) 3 . 1. n(CO2 ) = = = 0,054mol If all nitrogen is used, 0,9 mol hydrogen is VM 22,4 140g of SO3 was produced in the reaction. Calculate the percentage yield of the needed, However, only 0,75 mol hydrogen is avail- reaction. 2. 3. able. The hydrogen will run out first, therefore CaCO3 : CO2 hydrogen is the limiting reactant. CaCO3 : CO2 m 128 1 : 1 1. n(SO2 ) = = = 2mol 1 : 1 M 64 0,054 : 0,054 (b) 2. 3. ∴ 0,054mol CaCO3 reacted SO2 : SO3 SO2 : SO3 Because the hydrogen is the limiting reactant, it will determine the mass of ammonia produced: 4. 2 : 2 1 : 1 m(CaCO3) = nM = (0,054)(40 + 12 + 16 + 16 + 16) = 5,4g 1 : 1 2 : 2 H2 : NH3 ∴ 2mol SO3 5. 3 : 2 Mass of pure substance 4. m(SO3) = nM = (2)(32 + 16 + 16 + 16) = 160g Percentage purity = × 100 0,75mol : 0,5mol Mass of impure substance 5. 5,4 Mass of product produced Percentage purity = × 100 Percentage yield = × 100 m(NH3) = nM 6,0 Maximum theoretical mass of product = (0,5)(17) Percentage purity = 90 % Percentage yield = 140 × 100 160 = 8,5g Percentage yield = 87.5 % Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 50

Grade 12 Science Essentials Energy and Chemical Change SCIENCE CLINIC 2022 © ENERGY CHANGE ENTHALPY AND ENTHALPY CHANGE The energy change that takes place occurs because of bonds being broken and new bonds being formed. Enthalpy (H) is the total amount of stored chemical energy (potential energy) of the reactants and the products. During chemical reactions, energy can be exchanged between the chemical system and the When bonds are broken, energy is absorbed from the environment. (endothermic) environment, resulting in a change in enthalpy. This change in enthalpy, 𝚫H, represents the heat of the When bonds are formed, energy is released into the environment. (exothermic) reaction measured in kJ·mol−1. The net energy change will determine if the reaction is endothermic or exothermic. The heat of reaction (𝚫H) is the net change of chemical potential energy of the system. ACTIVATION ENERGY CHEMICAL SYSTEM AND THE ENVIRONMENT In order to start a reaction, energy first needs to be absorbed to break the bonds. This energy is known as the activation energy – the minimum energy required to start a chemical reaction OR the The chemical system is the reactant and product molecules. energy required to form the activated complex. The environment is the surroundings of the chemical system, including the container in which the reac- Once the bonds have been broken, the atoms in the chemical system form an activated complex – a tion takes place, or the water in which the molecules are dissolved. high energy, unstable temporary transition state between the reactants and the products. CATALYST ENDOTHERMIC EXOTHERMIC In order for a reaction to occur, enough en- ergy has to be provided (activation energy) A reaction that absorbs heat energy A reaction that releases heat energy for particles to collide effectively. The amount of required energy can be de- More energy absorbed than released More energy released than absorbed creased by using a catalyst. A catalyst is a chemical substances that lowers the activa- Net energy change is energy absorbed from the environment Net energy change is energy released into the environment tion energy required without undergoing chemical change. By lowering the activation The chemical system’s energy increases (𝚫H>0) The chemical system’s energy decreases (𝚫H<0) energy, the rate of the reaction can also be increased. The environment’s energy decreases The environment’s energy increases A catalyst is a substance that increases the rate of the reaction but remains Temperature of the environment decreases (test tube gets colder) Temperature of the environment increases (test tube gets hotter) unchanged at the end of the reaction POTENTIAL ENERGY PROFILE GRAPH OF AN ENDOTHERMIC REACTION POTENTIAL ENERGY PROFILE GRAPH OF AN EXOTHERMIC REACTION IMPORTANT REACTIONS Ac+vated Ac+vated ENDOTHERMIC Complex Complex Photosynthesis Effect of Effect of 6CO2 + 6H2 O light C6 H12 O6 + 6O2 ; 𝚫H>0 Poten+al Energy- EP (kJ) Poten+al Energy- EP (kJ) catalyst catalyst EA Products EXOTHERMIC Reactants Cellular respiration EA ΔH ΔH C6 H12 O6 + 6O2 → 6CO2 + 6H2 O ; 𝚫H<0 Reactants Products Combustion CH 4 + 2O2 → CO2 + 2H2 O ; 𝚫H<0 Course of reac+on Course of reac+on C 2 H5OH + 3O2 → 2CO2 + 3H2 O ; 𝚫H<0 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 51

Grade 12 Science Essentials Rates of Reactions SCIENCE CLINIC 2022 © RATES OF REACTIONS FACTORS INFLUENCING REACTION RATE Temperature Nature of reactants The change in concentration per unit time of either a reac- Increasing temperature increases rate of reaction. When the temperature increases, The physical and chemical properties of certain tant or product. the kinetic energy of the particles increases. More particles have sufficient energy to molecules make them more likely to react. overcome the activation energy, and more effective collisions can take place per Δ amount of product For example: unit time. Rate = • O2 has many effective orientations Δt EA • F’s electronegativity makes it more reactive Number of particles Δ amount of reactant Amount of product te H • Tertiary alcohols have limited effective collision Rate = m ig pe h orientations due to molecule structure Δt Hi ra w re gh T tu re Lo ratu • Simple (Ca2+) and complex (C2O4−) ions. May be given terms of change in mass per unit time (g·s−1) OR of pe Low p. te m State of division / surface area (solids only) (mol)/(mol·dm−3) change in volume per unit time (dm3·s−1) OR change in number of em mole per unit time (mol·s−1) p. Increased state of division (powder instead of Tem chunks) increases rate of reaction. Increasing the surface area exposed to collisions increases that The gradient of a concentration/mass/volume versus time graph number of particles that will undergo collisions, and gives the rate of a reaction, thus a steeper gradient means a higher Time (s) more effective collisions can take place per unit Energy rate of reaction. time. Concentration (gases and solutions only) Increasing concentration increases rate of reaction. The greater the concentration, the Amount of product COLLISION THEORY more particles occur per unit volume. More particles have sufficient energy to over- The conditions for successful collisions are: come the activation energy, and more effective collisions can take place per unit Po wd lar time. If the concentration of a limiting reactant is increased, more product can be er 1. Particles must collide with correct orientation Gran (mol)/(mol·dm−3) The structure of the molecules and their relative orientations to formed. High conc. u 1M each other is important for effective collisions. Some catalysts EA Limiting react. Number of particles Amount of product function by improving molecular orientation. onc. 2. Particles must collide with sufficient energy (kinetic High c act. s s re energy ≥ activation energy) 0,5 M Exce Time (s) The molecules have to collide with sufficient amount of energy (mol)/(mol·dm−3) Pressure (gases only) for bonds to break and the reaction to occur (activation energy). . nc Increased pressure (by decreasing volume) in- MAXWELL-BOLTZMAN DISTRIBUTION CURVE co creases the concentration of the gas thus increas- w The Maxwell-Boltzman distribution curve shows the distribution of Lo ing the rate of reaction. the kinetic energy of molecules. The area under the graph to the (See ‘Concentration’) right of the EA line represents the particles with sufficient kinetic Energy Time (s) energy Amount of product Catalyst most particles have moderate energy The presence of a catalyst decreases the activation energy (EA). More particles have H sufficient energy to overcome the lowered activation energy, and more effective colli- pre igh w ∴ average EK ssu re Lo sure sions can take place per unit time. es (mol)/(mol·dm−3) p r number of EA (catalyst) Number of particles few particles have Amount of product EA very high energy Ca ta particles ∴ high EK EA (no catalyst) lys t t Time (s) ou ith lyst W ta WAYS TO MEASURE RATE (mol)/(mol·dm−3) ca energy 1. Change in mass 2. Volume of gas produced All calculated few particles have Particles with sufficient energy 3. Change in colour per unit time very little energy for an effective collision 4. Turbidity (precipitation) ∴ low EK Energy Time (s) 5. Change in pH Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 52

Grade 12 Science Essentials Chemical Equilibrium SCIENCE CLINIC 2022 © An open system is one in which both energy and matter can LE CHATELIER’S PRINCIPLE be exchanged between the system and its surroundings - it interacts continuously with its environment. When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring A closed system is one in which energy can enter or leave the reaction that will oppose the disturbance. the system freely, but no reactant or products can leave or Factors which affect equilibrium position enter the system. 1. Concentration A reaction is a reversible reaction when products can be Equilibrium will shift to decrease any increase in Removing HI (t1): converted back to reactants. Reversible reactions are repre- concentration of either reactants or products. H2(g) + I2(g)
2HI(g) When HI is removed, the system re- sented with double arrows. • Adding reactant: forward reaction favoured HI H2 Concentra/on (mol·dm−3) H2(g) establishes equilibrium by favouring the reac- • Adding product: reverse reaction favoured removed added For example: tion that will produce more HI. Because the Hydrogen reacts with iodine to form hydrogen iodide: forward reaction is favoured, some of the reac- Equilibrium will shift to increase any decrease in H2 (g) + I2 (g) → 2HI(g) tants are used. concentration of either reactants or products. I2(g) Hydrogen iodide can decompose to form hydrogen and iodine: • Removing reactant: reverse reaction favoured Adding H2 (t2): 2HI(g) → H2 (g) + I2 (g) • Removing product: forward reaction favoured HI(g) When adding H2, the system re-establishes Therefore the reversible reaction can be written as: equilibrium by favouring the reaction that uses The concentration can be changed by adding/ H2. Because the forward reaction is favoured, Forward reac*on removing reactants/products that are in solution the reactants are used and more products (aq) or a gas (g). Changing the mass of pure sol- H2(g) + I2(g) ⇌ 2HI(g) Time (min) t1 t2 form. ids (s) or volume of liquids (l) will not disrupt the Reverse reac*on equilibrium or change the rate of the reactions. Dynamic chemical equilibrium refers to a reversible 2. Pressure (gases only) reaction in which the forward reaction and the reverse reaction A change in pressure can be identified on a Pressure decrease (t1): N2(g) + 3H2(g)
2NH3(g) are taking place at the same rate. The concentrations of the graph by an instantaneous change in concentra- When the pressure is decreased, the system reactants and products are constant. Chemical equilibrium can tion of all gasses due to a change in volume. Pressure Pressure re-establishes equilibrium by favouring the Concentra/on (mol·dm−3) N2(g) decrease increase only be achieved in a closed system. Equilibrium will shift to decrease any increase in reaction that will produce more moles of gas. Equilibrium pressure by favouring the reaction direction that The reverse reaction is favoured, reacting 2 Concentra/on (mol·dm−3) H2(g) produces less molecules. mol gas and forming 4 mol gas. reached H2(g) Equilibrium will shift to increase any decrease in Pressure increase (t2): Concentra/on pressure by favouring the reaction that produces more molecules. When the pressure is increased, the system I2(g) remains constant To determine which reaction is favoured, com- re-establishes equilibrium by favouring the a?er equilibrium pare the total number of mol of gaseous reaction that will produce less moles of gas. reached reactants to the total number of mol of gaseous NH3(g) The forward reaction of favoured, reacting 4 HI(g) products. mol of gas and forming 2 moles of gas. Time (min) t1 t2 3. Temperature If the temperature is increased, the endo- 2NO2(g) ⇌ N2O4(g); ΔH = -57 kJ Temperature increase (t1): Time (min) t1 thermic reaction will be favoured. brown colourless When the temperature is increased, the system Forward reac0on Temperature Temperature If the temperature decreases, the exother- re-establishes equilibrium by favouring the reac- Concentration (mol·dm−3) H2(g) + I2(g) → 2HI(g) Equilibrium increase decrease mic reaction is favoured. NO2(g) tion that will decrease the temperature (i.e. the reached Rate of reac0on endothermic reaction). The reverse reaction will The heat of the reaction, ΔH, is always be favoured because the forward reaction is exo- Equilibrium: used to indicate the forward reaction. thermic. The gas mixture becomes darker brown. H2(g) + I2(g)
2HI(g) NOTE: An increase in temperature in- Temperature decrease (t2): Reac0on rate creases the rate of both the forward remains constant When the temperature is decreased, the system and the reverse reaction, but shifts aAer equilibrium re-establishes equilibrium by favouring the reac- Reverse reac0on the equilibrium position. reached tion that will increase the temperature (i.e. the 2HI(g) → H2(g) + I2(g) N2O4(g) NOTE: Temperature change is the only exothermic reaction). The forward reaction will be Time (min) t1 change that affects Kc. Time (min) t1 t2 favoured. The gas mixture becomes lighter brown. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 53

Grade 12 Science Essentials Chemical Equilibrium SCIENCE CLINIC 2022 © COMMON ION EFFECT EQUILIBRIUM CONSTANT (KC) EQUILIBRIUM CONSTANT TABLE When ionic substances are in solution, they form ions: General equation: aA + bB ⇌ cC The equilibrium constant table assists in calculating the concentration of reactants and products when the reaction has reached equilibrium. Where A,B,C are chemical substances (ONLY aq and g, NOT s or l !) NaCl(s) ⇌ Na+ (aq) + Cl−(aq) EXAMPLE: and a,b,c are molar ratio numbers Note: The factors of the wh i t e c ol ou r l ess numerator and denominator 3 moles of NO2 are placed in a 1,5 dm3 container and the following equilibrium [C]c must be MULTIPLIED is established: Kc = If no volume is given, If HCl is added to this solution, the concentration of [A]a[B]b respectively (NOT added) 2NO2(g) ⇌ N2O4(g) Cl− ions will increase because Cl− is a common ion. assume volume = 1 dm3 Kc value is a ratio and therefore has no units. The system will attempt to re-establish equilibrium by At equilibrium it was found that 0,3 mol of NO2 was present in the container. favouring the reverse reaction, forming a white If Kc > 1 then equilibrium lies to the right – there are more products Calculate the value of the equilibrium constant for this reaction. sodium chloride precipitate. than reactants. If Kc < 1 then equilibrium lies to the left – there are more reactants The disturbance of a system at equilibrium that occurs NO2 N2O4 than products. when the concentration of a common ion is increased, is known as the common ion effect. Kc values are constant at specific temperatures. If the temperature of Ratio 2 1 the system changes then the Kc value will change. [N2 O4 ] CATALYST AND EQUILIBRIUM Initial (mol) 3 0 Kc = [NO 2 ]2 When a catalyst is added, the rate of the forward, as Temperature Kc of Exothermic Kc of Endothermic (0,9) well as the reverse reaction, is increased. The use of a Change (mol) -2,7 +1,35 = (0,2) 2 catalyst does not affect the equilibrium position or the Increase Decrease Increase Equilibrium (mol) 0,3 1,35 = 22,5 Kc value at all. Decrease Increase Decrease Forward reac0on Eq. concentration 0,3/1,5 1,35/1,5 H2(g) + I2(g) → 2HI(g) EXAMPLE: in (mol·dm−3) = 0,2 =0,9 Rate of reac0on Consider the following equilibrium system: Catalyst added CoCl 4 2−(aq) + H2 O(l) ⇌ Co(H2 O)6 2+ (aq) + 4Cl−(aq) ΔH < 0 EXAMPLES OF APPLICATIONS OF CHEMICAL EQUILIBRIUM bl u e pi nk The Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH<0 For each change made to the system state and explain the colour Reverse reac0on change seen. Temperature: The forward reaction is exothermic, thus a decrease in tempera- 2HI(g) → H2(g) + I2(g) ture will favour the forward exothermic reaction. However, a decrease in tempera- a) Water added ture will also decrease the rate of the reaction. Therefore, a compromise be- Time (min) t1 Adding water decreases the concentration of all the ions. According tween rate and yield is found at a temperature of around 450 °C to 550 °C. to Le Chatelier’s principle, the equilibrium will shift in such as way so as to produce more ions. Thus the forward reaction will be Pressure: As there are fewer reactant gas particles than product gas particles, DESCRIBING EQUILIBRIUM SHIFT AC- an increase in pressure will therefore favour the production of products. Thus this favoured, forming more product and the solution turns pink. CORDING TO LE CHATELIER reaction is done under a high pressure of 200 atm. 1. Identify the disturbance b) AgNO3 added Catalyst: Iron or Iron Oxide Adding/removing reactants or products, pressure Adding AgNO3 creates an insoluble precipitate of AgCl. This change, temperature change. decreases the Cl− ion concentration. According to Le Chatelier’s The Contact Process: 2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH < 0 principle the equilibrium will shift in such a way so as to produce 2. State Le Chatelier’s principle Temperature: The forward reaction is exothermic, thus a decrease in tempera- more Cl− ions. The forward reaction will be favoured, more product ture will favour the forward exothermic reaction. However, a decrease in tempera- 3. System response is made and the solution turns pink. ture will also decrease the rate of the reaction. Therefore, a compromise be- Use up/create more products or reactants, make tween rate and yield is found at a temperature of around 450 °C more/less gas molecules, increase/decrease tem- c) Temperature is increased perature. The temperature of the system is increased. According to Pressure: There are more reactant gas particles than product gas particles, there- Le Chatelier’s principle, the equilibrium will shift in such a way to fore an increase in pressure will favour the production of product. However, in 4. State favoured reaction reduce the temperature. The reverse endothermic reaction is practice a very high yield is obtained at atmospheric pressure. 5. Discuss results favoured. This decreases the temperature and produces more reac- Catalyst: Vanadium pentoxide (V2O5) Equilibrium shift, change in colour/concentration/ tant and the solution turns blue. pressure/temperature. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 54

Grade 12 Science Essentials Chemical Equilibrium - Rate and Concentration SCIENCE CLINIC 2022 © 1: Increase in pressure 2: Decrease in pressure Le Chatelier: The forward reaction is favoured, Le Chatelier: The reverse reaction is favoured, less gas particles are formed and the pressure more gas particles are formed and the pressure decreases. increases. 2 NO + O2 ⇌ 2 NO2 ∆H<0 Concentration: ɠConcentration of all gasses increases sharply. Concentration: ɠConcentration of all gasses decreases sharply. ɡThe forward reaction is favoured. ɡThe reverse reaction is favoured. 1 2 3 4 5 6 ɢ[NO 2 ] gradually increases, [NO], [O 2 ] ɢ[NO 2 ] gradually decreases, [NO], [O 2 ] gradually decreases. gradually increases. NO Rate: Concentration of all gasses increases, Rate: Concentration of all gasses decreases, therefore the rate of both the forward and the therefore the rate of both the forward and the reverse reaction increases. reverse reaction decreases. According to Le Chatelier the forward reaction According to Le Chatelier the reverse reaction is Concentration will be favoured, therefore the forward reaction favoured, therefore the reverse reaction O2 experiences a greater increase in rate. experiences a smaller decrease in rate. 3: O2 is added 4: NO is removed Le Chatelier: The forward reaction is favoured Le Chatelier: The reverse reaction is favoured to decrease the [O2]. to increase the [NO]. Concentration: Concentration: NO2 ɠConcentration of O2 decreases sharply. ɠConcentration of NO decreases sharply. ɡThe forward reaction is favoured. ɡThe reverse reaction is favoured. ɢ[NO 2 ] increases gradually, [NO], [O 2 ] ɢ[NO 2 ] decreases gradually, [NO], [O 2 ] decreases gradually. increases gradually. Rate: An increase in concentration of a reactant Rate: A decrease in concentration of a reactant Forward leads to an increase in the rate of the and leads to an decrease in the rate of the forward reaction forward reaction. and reverse reaction. Rate of reaction 5: Increase in temperature 6: Decrease in temperature Le Chatelier: The reverse (endothermic)reac- Le Chatelier: The forward (exothermic) reac- tion is favoured to decrease the temperature. tion is favoured to increase the temperature. Concentration: Concentration: ɠThe reverse reaction is favoured. ɠThe forward reaction is favoured. ɡ[NO 2 ] decreases gradually, [NO], [O 2 ] ɡ[NO 2 ] increases gradually, [NO], [O 2 ] Reverse increases gradually. decreases gradually. reaction Rate: An increase in temperature increases the Rate: A decrease in temperature decreases the rate of both the forward and the reverse rate of both the forward and the reverse reactions. reactions. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 55

Grade 12 Science Essentials Acids and Bases SCIENCE CLINIC 2022 © ACID/BASE DEFINITIONS CONCENTRATED VS DILUTE ACIDS AND BASES INFLUENCE OF ACID/BASE STRENGTH Arrhenius Concentration is the number of moles of solute per unit volume Reaction rate An acid is a substance that produces hydrogen ions (H+)/ n m of solution. ( c = of c = ) Reaction rates increase as the strength of the acid/base in- hydronium ions (H3O+) when it dissolves in water. V MV creases. A base is a substance that produces hydroxide (OH−) when A concentrated acid/base contains a large amount (mol) of acid/ Stronger acid = higher concentration of ions = greater rate of dissolved in water. base in relationship to the volume of water. reaction. Lowry-Brønsted An acid is a proton (H+ ion) donor. A dilute acid/base contains a small amount (mol) of acid/base in Conductivity A base is a proton (H+ ion) acceptor. relationship to the volume of water. Conductivity increases as the strength of the acid/base increases. ACID PROTICITY Concentrated strong acid - 1mol.dm-3 of HCl Stronger acid = higher concentration of H+ = greater Some acids are able to donate more than one proton. The number of protons conductivity. Concentrated weak acid - 1mol.dm-3 of CH3COOH that an acid can donate is referred to as the acid proticity. 1 proton- monoprotic 2 protons- diprotic 3 protons- triprotic Dilute strong acid - 0,01mol.dm-3 of HCl Dilute weak acid - 0,01mol.dm-3 of CH3COOH COMMON ACIDS COMMON BASES HCl → Cl− + H+ H2SO4 → HSO4− + H+ H3PO4 ⇌ H2PO4− + H+ (Similar for bases) Hydrochloric acid Sodium Hydroxide S HSO4− → HSO42− + H+ H2PO4− ⇌ HPO42− + H+ (HCl) (NaOH) T S HPO42− ⇌ PO43− + H+ STRONG VS WEAK ACIDS AND BASES T R CONJUGATE ACID-BASE PAIRS The strength of an acid/base refers to the ability of the sub- R Potassium hydroxide O Nitric acid (HNO3) N stance to ionise or dissociate. O (KOH) An acid forms a conjugate base when it donates a proton. N G acid ⇌ conjugate base + H+ G ACIDS Sulfuric acid HA ⇌ A– + H+ (H2SO4) A strong acid will ionise completely in water. Sodium hydrogen A base forms a conjugate acid when it accepts a proton. HCl (g) + H2O (l) → H3O+ (aq) + Cl− (aq) Oxalic acid ((COOH)2) carbonate (NaHCO3) base + H+ ⇌ conjugate acid (strong acid → weak conjugate base) Hydrofluoric acid (HF) W HA ⇌ A– + H+ Sulfurous acid (H2SO3) W Calcium carbonate E A weak acid will only partially ionise in water. Carbonic acid E (CaCO3) A Conjugate acids and bases differ only by a proton (H+). 2H2CO3 (l) + 2H2O (l) ⇌ 2H3O+ (aq) + CO32− (aq) (H2CO3) A K EXAMPLE: (weak acid → strong conjugate base) Acetic acid / ethanoic acid K Sodium carbonate Identify the conjugate acid-base pair in the following example: (CH3COOH) (Na2CO3) H+ Conjugate acid-base pair BASES Ammonia Phosphoric acid (H3PO4) (NH3) A strong base will dissociate completely in water. HNO3 (g) + H2O (l) → NO3− (aq) + H3O+ (aq) NaOH (s) → Na+ (aq) + OH− (aq) THE pH SCALE acid base conjugate base conjugate acid (strong base → weak conjugate acid) The pH of a solution is a number that represents the acidity or alkalinity of a solution. Conjugate acid-base pair A weak base will dissociate only partially in water. The greater the concentration of H+ ions in solution, the more AMPHOLYTE/ AMPHIPROTIC SUBSTANCES 2Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH− (aq) acidic the solution and the lower the pH. The lower the concen- tration of H+ in solution, the more alkali the solution and the Ampholyte- A substance that can act as either an acid or a base. (weak base → strong conjugate acid) higher the pH. Ampholytes/amphiprotic substances can therefore either donate or accept The pH scale is a range from 0 to 14, and is a measure of the protons. Common ampholytes include H2O, HCO3− and HSO4−. [H3O+] at 25 °C (in water). pH = − log[H3O+ ] NH3 is an exception, it ionises. HSO4− as an ampholyte: Acid: HSO4− + H2O → SO42− + H3O+ NH3 (g) + H2O (l) ⇌ OH− (aq) + NH4+ (aq) Neutral Base: HSO4− + H2O ⇌ H2SO4 + OH− 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 56

Grade 12 Science Essentials Acids and Bases SCIENCE CLINIC 2022 © ACID REACTIONS HYDROLYSIS OF SALTS INDICATORS Acid reactions are reactions during which Hydrolysis is the reaction of An indicator is a compound that changes colour according to the pH of the sub- ACID BASE SALT stance. During titrations, the indicator needs to be selected according to the pH protons (H+) are transferred. an ion with water. During hydrolysis, the salt will form of the salt solution that will be produced (see hydrolysis). Strong Weak Acidic acid + metal → salt + H2 an acidic, alkali or neutral HA + M → A— + M+ + H2 solution. The pH of the salt Strong Strong Neutral 2HNO3(aq) + 2Na(s) → 2NaNO3(aq) + H2(g) solution is determined by the COLOUR AT COLOUR COLOUR PH RANGE OF relative strength of the acid Weak Strong Alkali INDICATOR EQUIVALENCE IN ACID IN BASE EQUIVALENCE and base that is used to form POINT A weak acid and a weak base do not acid + metal hydroxide (base) → salt + H2O the salt. form ions readily. HA + MOH → A— + M++ H2O H2SO4(aq) + Mg(OH)2(aq) → MgSO4(aq) + 2H2O(l) How to determine the pH of a salt Litmus Red Blue 4,5 - 8,3 NH4Cl Na2CO3 NH4Cl → NH4+ + Cl– Na2CO3 → 2Na+ + CO32– NH4+ + H2O ⥂ NH3 + H3O+ Methyl orange Red Yellow Orange 3,1 - 4,4 acid + metal oxide → salt + H2O Na+ + H2O ⥃ NaOH + H+ HA + MO → A— + M+ + H2O Cl– + H2O ⥄ HCl + OH– CO32– + 2H2O ⥂ H2CO3 + 2OH– 2HCl (aq) + MgO(s) → MgCl2(aq) + H2O(l) Bromothymol Yellow Blue Green 6,0 - 7,6 ·NH3 is a weak base ·NaOH is a strong base blue ·Equilibrium lies to the right ·Equilibrium lies to the left acid + metal carbonate → salt + H2O + CO2 ·High concentration H3O+ ·Low concentratioin H+/H3O+ Phenolphthalein Colourless Pink Pale Pink 8,3 - 10,0 HA + MCO3 → A— + M+ + H2O + CO2 2HCl (aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g) ·HCl is a strong acid ·H2CO3 is a weak acid Neutral: pH 7 ·Equilibrium lies to the left ·Equilibrium lies to the right Neutralisation is when the equivalence point is reached. Equivalence point is The ionic salt is made up of A— (from the acid) and ·Low concentration OH– ·High concentration OH– NOT when the solution is at pH 7, but when the acid/base has completely re- M+ (from the base). High [H3O+], low [OH–], Low [H3O+], high [OH–], acted with the base/acid according to the molar ratio. The pH at neutralisation ∴NH4Cl(aq) is acidic (pH<7) ∴Na2CO3(aq) is alkali (pH>7) is dependent on the salt that is formed. (see hydrolysis). TITRATIONS Titration setup EXAMPLE: A titration is a practical laboratory method to de- During a titration, 25 cm3 of dilute H2SO4 neutralises 40 cm3 of NaOH solution. Va Burette termine the concentration of an acid or base. The If the concentration of the H2SO4 solution is 0,25 mol.dm−3, calculate the con- concentration of an acid or base can be deter- Acid centration of the NaOH. mined by accurate neutralisation using a standard solution- a solution of known con- Stopcock H2 SO4 + 2NaOH → Na 2 SO4 + 2H2 O centration. Neutralisation occurs at the equiva- na = 1 nb = 2 lence point, when the acid/base has completely ca = 0,25 mol ⋅ dm−3 cb = ? reacted with the base/acid according to the molar Conical flask Va = 25 cm3 = 0,025 dm3 Vb = 40 cm3 = 0,04 dm3 ratio. Vb na cV Base & By titration equations By first principles: = a a Indicator na c aVa nb cbVb nb = c bVb Acid: n = cV 1 (0,25)(0,025) = (0,25)(0,025) n- number of mole of substance (mol) / mol ratio 1. Pipette a specific amount of the standard solution into conical flask = 2 c b (0,04) from the balanced equation (usually base). = 6,25 × 10−3 mol c- concentration of acid/base (mol.dm−3) cb = 0,31 mol ⋅ dm−3 2. Add appropriate indicator to flask (according to pH of final solution). V- volume of solution (dm3) 3. Fill the buret with the solution of unknown concentration (usually Base: n 3 acid). c = 1 mL = 1 cm V 1 L = 1 dm3 4. Add solution from burette to conical flask at a dropwise rate (remem- 6,25 × 10−3 × 2 ber to swirl). = 1000 mL = 1 L 0,04 1000 cm3 = 1 dm3 5. Stop burette when indicator shows neutralisation/equivalence point = 0,31 mol ⋅ dm−3 has been reached. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 57

Grade 12 Science Essentials Acids and Bases SCIENCE CLINIC 2022 © CALCULATING PH VALUE OF AUTO-IONISATION OF WATER AND KW KA AND KB VALUES A STRONG ACID/BASE Water reacts with water to form hydronium and hydroxide ions When weak acids or bases are dissolved in water, only partial ionisation/ in the following reaction: dissociation occurs. There is a mixture of the original reactant as well as the ionic pH = − log[H3O+ ] products that were formed. The extent of ionisation can be treated in the same H2O (ℓ) + H2O (ℓ) ⇌ H3O+ (aq) + OH− (aq) Strong acids ionise completely in water, meaning that all way as the extent to which an equilibrium reaction takes place. The acid acids donate their protons (H+). The concentration of the The concentration of H3O+ (aq) and OH− (aq) are equal, and the dissociation constant (Ka) and base dissociation constant (Kb) values are like the H+ ions can be determined from the initial concentration of equilibrium constant for the ionisation of water (Kw) is 1,00 ✕ equilibrium constant (Kc), but specifically described the extent of ionisation/ the acids, taking the proticity of the acid into account. 10−14 (at a temperature of 25°C or 298 K), therefore: dissociation, and therefore the strength of the acid/base. Monoprotic: HCl(g) → H+(aq) +. Cl−(aq) [H3O+] [OH−] = 1,00 ✕ 10−14 mol·dm−3 Calculating Ka General equation Example 1 mol : 1 mol : 1 mol [H3O+] = 1,00 ✕ 10−7 mol·dm−3 HA + H2O ⇌ A− + H3O+ CH3COOH + H2O ⇌ CH3COO− + H3O+ 0,3 mol·dm−3 : 0,3 mol·dm−3 : 0,3 mol·dm−3 [OH−] = 1,00 ✕ 10−7 mol·dm−3 [A−][H3O+ ] [CH3COO−][H3O+ ] Ka = Ka = When the concentration of H3O+ (aq) and OH− (aq) are equal, [HA] [CH3COOH] the solution is neutral and has a pH of 7. −l og [1 × 10−7] = 7 Calculating Kb Diprotic: H2SO4(g) → 2H+(aq) + SO42−(aq) General equation Example 1 mol : 2 mol : 1 mol The pH of a base can be determined by first determining the B + H2O ⇌ BH+ + OH− NH3 + H2O ⇌ NH4+ + OH− [OH–], then the [H3O+] [NH +4 ][OH−] 0,3 mol·dm−3 : 0,6 mol·dm−3 : 0,3 mol·dm−3 [BH + ][OH−] Kb = Kb = [B] [NH3] EXAMPLE: Triprotic: H3PO4(g) →. 3H+(aq) +. PO43− Calculate the pH of a Ba(OH)2 solution with a concentration DETERMINING CONCENTRATION FROM PH (aq) of 0,35 mol·dm−3. To determine the [H3O+] when the pH is known: 1 mol : 3 mol : 1 mol Step 1: Write down the dissociation pH = − log[H3O+ ] 0,3 mol·dm−3 : 0,9 mol·dm−3 : 0,3 mol·dm−3 reaction ∴ [H3O+ ] = 10−pH H2 O + − EXAMPLE: Ba(OH)2 (s) → Ba (aq) + 2OH (aq) EXAMPLE: Calculate the pH of a H2SO4 solution with a concen- 0,1 mol HF is dissolved in 1 dm3 water. Determine the Ka of HF if the pH is found tration of 0,25 mol·dm−3 Step 2: Determine the mole ratio of base to OH− ions to be 2,1. HF dissociates according to the following chemical equation: Step 1: Write down the ionisation reaction: 1 mol Ba(OH)2 : 2 mol OH− H2O (ℓ) + HF (aq) ⇌ H3O+ (aq) + F− (aq) H2 SO4 + 2H2 O → 2H3O+ + SO 2− 4 pH = − log[H3O+ ] Step 3: Determine the [OH−] [OH−] = 2[Ba(OH)2 ] [H3O+ ] = 10−2,1 Step 2: determine the ratio of acid to H3O+ ions H2 SO4 : H3O+ = 2(0,35) [H3O+ ] = 7,94 × 10−3 mol ⋅ dm−3 1:2 = 0,7 mol ⋅ dm−3 HF H3O+ F− Step 4: Determine the [H3O ] + Ratio 1 1 1 Step 3: determine the [H3O+] [H3O+ ][OH−] = 1,00 × 10−14 Initial (mol) 0,1 0 0 [H3O+ ] = 2[H2 SO4 ] [H3O+ ](0,7) = 1,00 × 10−14 Change (mol) −7,94 ×10−3 +7,94 ×10−3 +7,94 ×10−3 = 2(0,25) Equilibrium (mol) 0,092 7,94 ×10−3 7,94 ×10−3 [H3O+ ] = 1,43 × 10−14 mol ⋅ dm−3 Eq. concentration in = 0,5mol ⋅ dm−3 (mol·dm−3) 0,092 7,94 ×10−3 7,94 ×10−3 Step 5: Calculate the pH Step 4: Calculate the pH [H3O+ ][F −] pH = −log [H3O+] Ka = pH = − log[H3O+ ] [HF] = −log (1,43 x 10−14) Ka/Kb > 1; Strong acid/base (7,94 × 10−3)(7,94 × 10−3) = − log(0,5) = = 13,84 0,092 Ka/Kb < 1; Weak acid/base = 0,3 = 6,85 × 10−4 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 58

Grade 12 Science Essentials Electrochemistry-Redox SCIENCE CLINIC 2022 © REDOX REACTION TABLE OF STANDARD REDUCTION POTENTIALS (REDOX TABLE) A redox reaction is a reaction in which there is a transfer of electrons The oxidation and reduction half reactions can also be found using the Table of Standard Reduction Potentials. (We will use Table 4B). between elements/compounds. The reactions shown on the table are all written as reduction half reactions, with the reversible reaction arrow (⇌) shown. This Oxidation is the loss of electrons (oxidation number increases) means that each reaction is reversible. When a reaction is written from the table the arrow must only be one way (i.e. →). Reduction is the gain of electrons (oxidation number decreases) The reduction half reaction is written from the table from left to right and the oxidation half reaction is written from right to left. (Oxidation reaction) The oxidising agent is the substance which accepts electrons. (It is the substance which is reduced and causes oxidation.) Oxidising agent Reducing agent (weak) Li+ + e- Li (strong) The reducing agent is the substance that donates electrons. (It is the substance which is oxidised and causes reduction.) Positive gradient Spontaneous reaction The anode is the electrode where oxidation takes place. The cathode is the electrode where reduction takes place. F2 + 2e- 2F- OIL: Oxidation is loss (strong) (weak) RIG: Reduction is gain (Reduction reaction) LEO: Loss of electrons is oxidation Once the half-reactions are identified it is possible to write a balanced reaction, without the spectator ions. A spectator ion is an ion in GER: Gain of electrons is reduction a redox reaction that does not take part in electron transfer. Remember that the number of electrons lost or gained by each substance must be the same. REDCAT: Reduction at cathode ANOX: Oxidation at anode If the line drawn between the two reactants has a positive gradient, the reaction is spontaneous. If the line between the reactants are negative, the reaction is non-spontaneous. STEPS TO WRITING BALANCED REDOX EXAMPLE: EXAMPLE: REACTIONS USING REDOX TABLE + Zinc metal reacts with an acid, H (aq) to produce hydrogen gas. Using Magnesium ribbon is burnt in a gas jar containing chlorine 1. IDENTIFY THE REACTANTS the oxidation and reduction half reactions write a balanced equation for gas. Using half reactions write a balanced chemical equation this reaction. for this reaction. 2. UNDERLINE REACTANTS ON THE REDOX TABLE STEP 1: IDENTIFY THE REACTANTS 3. DRAW ARROWS IN DIRECTION OF REACTION Ox half-reaction: Mg → Mg2+ + 2e− Zn(s) and H+(aq) 4. WRITE THE OXIDATION AND REDUCTION HALF-REACTIONS Red half-reaction: Cl2 + 2e− → 2Cl− STEP 2: UNDERLINE REACTANTS ON THE REDOX TABLE 5. BALANCE ELECTRONS IF NECESSARY Overall reaction: Mg + Cl2 → Mg2+ + 2Cl− STEP 3: DRAW ARROWS IN DIRECTION OF REACTION 6. WRITE OVERALL REACTION (LEAVE OUT SPECTATORS) EXAMPLE: STEP 4: WRITE THE OXIDATION AND REDUCTION HALF- REACTIONS Using half reactions, complete and balance the following reac- tion: Pb + Ag+ Ox half-reaction: Zn → Zn2+ + 2e− Ox half reaction: Pb → Pb2+ + 2e− Red half-reaction: 2H + 2e− → H2 + Red half reaction: Ag+ + e− → Ag STEP 5: BALANCE ELECTRONS IF NECESSARY STEP 6: WRITE OVERALL REACTION (LEAVE OUT SPECTATORS) Ox half reaction: Pb → Pb2+ + 2e− Overall: Zn + 2H → Zn + 2+ + H2 (x2 red half reaction): 2Ag+ + 2e− → 2Ag Overall reaction: Pb + 2Ag+ → Pb2+ + Ag Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 59

Grade 12 Science Essentials Electrochemistry- Galvanic/Voltaic cell SCIENCE CLINIC 2022 © A galvanic cell reaction is always a spontaneous, EMF OF THE CELL ZINC-COPPER CELL exothermic reaction during which chemical energy is converted to electrical energy. A electric cell/ The emf of the cell is calculated using one of the following equations: e– Voltmeter battery is an example of a galvanic cell. V Eθcell = Eθcathode − Eθanode STRUCTURE Eθcell = Eθreduction − Eθoxidation A positive εθ value indicates Anode Cathode Two half-cells (usually in separate containers): a spontaneous reaction KCl Anode – where oxidation takes place – negative electrode Eθcell = Eθoxidising agent − Eθreducing agent − Salt bridge + Cathode – where reduction takes place – positive The emf of the half-cells are determined using the standard hydrogen electrode half-cell Zn Cu The anode and cathode connected together through an external circuit, which allows for current to flow Standard hydrogen half-cell from the anode to the cathode The hydrogen half-cell is allocated a reference potential of 0,00 V. All Zn(NO3)2 Cu(NO3)2 other half-cells will have a potential which is either higher or lower Voltmeter e– than this reference. This difference is the reading on the voltmeter V The zinc half-cell (–): The copper half-cell (+): placed in the circuit. • Zinc electrode • Consists of a copper electrode Anode Cathode • Zinc salt solution • Copper salt solution Salt bridge H2 at Temperature (e.g. zinc (II) nitrate) (e.g. copper (II) nitrate) − + 1 atm = 298 K • Zn is a stronger RA than • Cu2+ is a stronger OA than Zn2+, Cu, ∴Zn oxidises ∴Cu2+ reduces • Oxidation reaction occurs: • Reduction reaction occurs: Zn → Zn2+ + 2e− Cu2+ + 2e− → Cu • Anode • Cathode Electrolyte Electrolyte • Electrode decreases in mass • Electrode increases in mass Pla3num Dilute H2SO4 SALT BRIDGE electrode [H+] = 1 mol·dm−3 Ox: Zn (s) → Zn2+ (aq) + 2e− The salt bridge connects the two half-cells. It is filled Red: Cu2+ (aq) + 2e− → Cu (s) with a saturated ionic solution of either KCl, NaCl, H2 is bubbled through the electrolyte over the inert platinum electrode. KNO3 or Na2SO4. A concentrated solution is used to Reduction potentials are measured under standard conditions: Nett cell: Zn (s) + Cu2+(aq) → Zn2+ (aq) + Cu (s) reduce the internal resistance. The ends of the tubes temperature 25 °C; 298 K For the zinc-copper cell: are closed with a porous material such as cotton wool concentration of the solutions 1 mol·dm−3 or glass wool. The anode reaction is: Zn (s) → Zn2+ (aq) + 2e−; Eθ = −0,76 V pressure 1 atm; 101,3 kPa (only relevant for gases) The cathode reaction is: Cu2+ (aq) + 2e− → Cu (s); Eθ = +0,34 V Functions of the salt bridge: Completes the circuit (which allows current to flow) Eθ cell = Eθ cathode − Eθ anode Maintains the electrical neutrality of the electrolyte The cell-notation for the hydrogen half-cell is: solutions. Pt | H2 (g) / H+ (aq) (1 mol⋅dm−3) = 0,34 − (−0,76) EQUILIBRIUM IN A CELL = + 1,1 V (spontaneous) When the circuit is complete the current will begin to EXAMPLE flow. The current and potential difference of the cell CELL NOTATION Consider the cell notation of the following electrochemical cell: Anode Salt bridge Cathode is related to the rate of the reaction and extent to which the reaction in the cell has reached equilibrium. Pt | H2(g)/H2SO4(aq) (0,5 mol⋅dm−3)// CuSO4(aq) (1 mol⋅dm−3)/Cu(s) As the chemical reaction proceeds, the rate of the The experimentally determined cell potential is 0,34 V at 25 °C. Zn(s) / Zn2+(aq) (1 mol·dm−3) // Cu2+(aq) (1 mol·dm−3) / Cu(s) forward reaction will decrease, so the rate of transfer of electrons will also decrease which results in the If a value of 0,00 V is given to the hydrogen half-cell, it means that Eθcell value decreasing. the value of the copper half-cell must be 0,34 V. Zn(s) → Zn2+(aq) + 2e− Cu2+(aq) + 2e− → Cu(s) The cell potential will continue to decrease gradually When the half-reactions do not include conductors (metals), unreactive electrodes Eθ cell = Eθ cathode − Eθ anode until equilibrium is reached at which point the cell are used, e.g. carbon or platinum. potential will be zero and the battery is “flat”. 0,34 = Eθ (Cu) − 0,00 Pt(s)/H2(g)(1 atm)/H+(aq)(1 mol·dm–3)//Br2(g)(1 atm)/Br–(aq)(1 mol·dm–3)/Pt(s) Le Chatelier’s principle can be applied to increase the θ EMF, with conditions that favour the forward reaction. E (Cu) = + 0,34 V Ca(s)/Ca2+(aq)(1 mol·dm–3 )//Fe3+(aq)(1 mol·dm–3), Fe2+(aq)(1 mol·dm–3)/C(s) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 60

Grade 12 Science Essentials Electrochemistry- Electrolytic cells SCIENCE CLINIC 2022 © An electrolytic cell reaction is always a non-spontaneous, ELECTROREFINING OF COPPER EXTRACTION OF ALUMINIUM (HALL-HEROLT PROCESS) endothermic reaction which requires a battery. The electrical energy is converted to chemical energy. Electrolytic cells When copper is purified, the process is similar to electroplating. Aluminium is found in the mineral known as bauxite which contains primarily require a DC power source. Impure (blister) copper is used as the anode and the cathode is aluminium oxide (Al2O3) in an impure form. pure copper. STRUCTURE Bauxite is not found in South Africa so is imported from Australia for refining. At the anode the copper is Step 1: Converting impure Al2O3 to pure Al2O3 Two electrodes (in the same container): oxidised to produce Cu2+ Pure Impure Anode – where oxidation takes place – positive electrode Bauxite treated with NaOH – impure Al2O3 becomes Al(OH)3 ions in the electrolyte. The copper copper Cathode – where reduction takes place – negative electrode mass of the impure copper The anode and cathode are connected to an external circuit, anode decreases. − + Al(OH)3 is heated (T > 1000 °C) which is connected to a DC power source. Al(OH)3 becomes pure Al2O3 – alumina At the cathode the Cu2+ Cu2+ ions in the electrolyte is Step 2: Melting Al2O3 Cu2+ reduced to form a pure Alumina is dissolved in cryolite (sodium aluminium hexafluoride – Na3AlF6). copper layer on the cathode. The mass of the Ag Pt Au Melting point reduced from over 2000 °C to 1000 °C. cathode anode cathode increases. − + Reduces energy requirements, costs and less environmental impact. The less reactive elements and compounds found in the impure Step 3: Anion copper anode are precipitated to the bottom of the reaction ves- Molten Alumina (Al2O3) – cryolite mixture placed in reaction vessel (−) sel. The more reactive metals (stronger reducing agents) are also Anodes (+) are carbon rods in mixture oxidised. The ions of these metals remain in the solution. Cu2+ is a Ca#on Electrolyte stronger oxidising agent, thus it is preferentially reduced at the Cathode (-) is the carbon lining of the tank (+) cathode. Cu2+ stays constant. At cathode Al3+ ions are reduced to Al metal ELECTROPLATING Oxidation (anode): Cu (s) → Cu2+ (aq) + 2e− Carbon anodes (+) Electroplating is the process of depositing a layer of one metal Reduction (cathode): Cu2+ (aq) + 2e− → Cu (s) onto another metal. Steel tank Nett cell: Cu2+ (aq) + Cu (s) → Cu2+ (aq) + Cu (s) EXAMPLE: Silver plating of a metal spoon impure pure Solu9on of aluminium oxide in molten cryolite ELECTROLYSIS OF COPPER (II) CHLORIDE The anode is silver, it will be oxi- dised to Ag+ ions. The mass of the Molten aluminium silver electrode decreases. Carbon lining for The cathode is the object (spoon) cathode (−) to be plated. The Ag+ ions from the Ag+ − + electrolyte will be reduced to form silver metal, which plates the Cl− spoon. The mass of the cathode Ag+ Oxidation (Anode): 2O2− (aq) → O2(g) + 4e− ×3 (spoon) increases. Cu2+ Reduction (Cathode): Al3+ (aq) + 3e− → Al (l) ×4 AgNO3 (aq) CuCl2 (aq) Nett cell: 2Al2O3 (aq) → 4Al (l) + 3O2 (g) The anode and electrolyte always contains the plating metal. Oxidation (anode): 2Cl− (aq) → Cl2 (g) + 2e− Reduction (cathode): Cu2+ (aq) + 2e− → Cu (s) Due to the high temperature of the reaction, the oxygen produced reacts Oxidation (anode): Ag (s) → Ag+ (aq) + e− Reduction (cathode): Ag (s) + e− → Ag (s) + Nett cell: 2Cl− (aq) + Cu2+ (aq) → Cl2 (g) + Cu (s) with the carbon electrodes to produce carbon dioxide gas. The carbon electrodes (anodes) therefore need to be replaced regularly. Nett cell: Ag+ (aq) + Ag (s) → Ag+ (aq) + Ag (s) Aluminium extraction uses a large amount of electrical energy, therefore the ∴ [Ag ] stays constant + Chlorine gas is produced at the anode, while copper metal is pro- cost of aluminium extraction is very high. duced at the cathode. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 61

Grade 12 Science Essentials Electrochemistry- Electrolytic cells SCIENCE CLINIC 2022 © ELECTROLYSIS OF SOLUTIONS MEMBRANE CELL In the electrolysis of NaCl, the Na+ ions are not reduced as might be An ion-exchange membrane is used to separate the sodium and chloride ions of the sodium chloride. The selectively permeable ion- expected. To identify which ions are oxidised/reduced apply the exchange membrane is a fluoro-polymer which allows only Na+ ions to pass through it. The cell consists of two half cells separated by the following rules (based on relative strength of OA/RA): membrane. The electrolytic cell has the lowest environmental impact. It is also the most cost effective to run, as the internal resistance is far lower than that of the diaphragm and mercury cells. OXIDATION (ANODE): Either the anion or H2O will be oxidised. Chlorine gas Hydrogen gas − − − − Anode (+) Cathode (−) If a HALOGEN ION (Cl , Br , I , not F ) is present, the HALOGEN ION is oxidised. The cathode is filled with pure water Dilute If no halogen ion is present, or the concentration is very low, water is oxidised according to: At cathode - Reduction: NaCl (aq) NaOH (aq) 2H2O (l) → O2 (g) + 4H+ (aq) + 4e− Cl2 (g) H2 (g) TITANIUM 2H2O (l) + 2e− → H2 (g) + 2OH− STEEL Membrane REDUCTION (CATHODE): H2O Either the cation or H2O will be reduced. The anode is filled with the brine solution Conc. If a GROUP I OR GROUP II METAL CATION is present, WATER will be NaCl (aq) Cl− (aq) At anode - Oxidation: OH− (aq) H2O reduced according to: 2Cl− (aq) → Cl2 (g) + 2e− 2H2O (l) + 2e− → H2 (g) + 2OH− (aq) Na+ (aq) Na+ (aq) Water is reduced because it is a stronger oxidising agent than other group I and II elements. If any other cation is present, the cation will Nett cell: 2Cl− (aq) + 2H2O (l) → Cl2 (g) + H2 (g) + 2OH− (aq) ELECTROLYSIS OF NaCl (CHLOR-ALKALI INDUSTRY) Overall reaction: 2NaCl (aq) + 2H2O (l) → Cl2 (g) + 2NaOH (aq) + H2 (g) Brine (concentrated NaCl solution) is placed in an electrolytic cell to produce chlorine gas, hydrogen gas and sodium hydroxide solution. Overall reaction: 2NaCl(aq) + 2H2O(l) → Cl2(g) + 2NaOH(aq) + H2(g) At the anode, Cl− ions are oxidised to form Cl2 (g). Cl2 gas bubbles form on the electrode. At the cathode, water is reduced to form H2 (g) and OH− (aq). H2 (g) bubbles form on the electrode. Cl− Na+ H2O Oxidation (anode): 2Cl (aq) → Cl2 (g) + 2e− − Reduction (cathode): 2H2O (l) + 2e− → H2 (g) + 2OH− (aq) Nett cell: 2Cl− (aq) + 2H2O (l) → Cl2 (g) + H2 (g) + OH− (aq) Overall reaction: 2NaCl (aq) + 2H2O (l) → Cl2 (g) + 2NaOH (aq) + H2 (g) The electrolysis is conducted in specialised electrolytic cells to control the reaction process and allow reactions to occur under controlled con- ditions. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 62

Grade 12 Science Essentials Fertilisers SCIENCE CLINIC 2022 © PLANT NUTRIENTS NPK FERTILISERS Fractional Distillation SASOL Non-mineral nutrients Fertilisers are classified according to Some non-mineral nutrients are not derived from their NPK content ratio. The ratio is of Liquid Air soil: given as: N2 H2 Carbon (C): From CO2 in the atmosphere N:P:K (%) Oxygen (O): From CO2 and O2 in the atmosphere N- Ratio of nitrogen (N) Hydrogen (H): From rainwater in the soil P- Ratio of phosphorus (P) HARBER PROCESS ( NH3 ) OSTWALD PROCESS ( HNO3 ) K- Ratio of potassium (K) These non-mineral nutrients are needed in great N2 + 3H2 → 2NH3 Catalytic oxidation of ammonia EXAMPLE 1 quantities for the synthesis of glucose through pho- 4NH3 + 5O2 → 4NO + 6H2O 3:2:1 (25) tosynthesis. 3 of 6 parts are N Mineral nutrients 2 of 6 parts are P Ammonium nitrate Oxidation of nitrogen monoxide 1 of 6 parts are K Mineral nutrients are dissolved in water in the soil NH3 + HNO3 → NH4NO3 and are absorbed by the roots of the plants. These 25% of the total mass is pure 2NO + O2 → 2NO2 fertiliser. nutrients can be supplemented by means of fertilis- ers to ensure healthy plant development. EXAMPLE 2 Absorption of NO2 in water 2:6:3 (40) Potassium nitrate Nitrogen (N): 3NO2 + H2O → 2HNO3 + NO Nitrogen is absorbed from nitrate- and ammonium 2 of 11 parts are N KCl + HNO3 → KNO3 + HCl 6 of 11 parts are P salts in the soil. 3 of 11 parts are K • Assists protein formation 40% of the total mass is pure • Promotes leaf growth fertiliser. UREA ( CO(NH2)2 ) CONTACT PROCESS ( H2SO4 ) • Improves the rate of above-ground growth The percentages of each nutrient can NH3 + CO2 → CO(NH2)2 Combustion of sulfur Nitrogen can be found naturally in guano, or syn- be determined as follows: S + O2 → SO2 thetically in nitrate- and ammonium salts. (FROM EXAMPLE 2) 2 Phosphorus (P): % N: × 40 = 7,27% Ammonium sulphate Catalytic oxidation of SO2 11 Phosphorus is absorbed from bonemeal, superphos- 2SO2 + O2 → 2SO3 phate and dissolved fertilsers. 6 NH3 + H2SO4 → (NH4)2SO4 % P: × 40 = 21,82% • Stimulates early root growth 11 • Assists in photosynthesis Formation of oleum (H2S2O7) • Early formation of flowers and maturation of 3 % K: × 40 = 10,91% Monoammonium phosphates 2SO3 + H2SO4 → H2S2O7 fruits and seeds 11 NH4H2PO4 Phosphorus can be found naturally in bonemeal, or EUTROFICATION Diammonium phosphates Conversion of oleum to H2SO4 synthetically in superphosphate fertilisers. Due to over-irrigation dissolved minerals and phosphates are ‘leached’ from the (NH4)2HPO4 H2S2O7 + H2O → 2H2SO4 Potassium (K): soil and ends up in rivers and dams mak- Potassium is found is small quantities in organic ing them nutrient rich. plant and animal material (compost and manure). • Promotes protein and carbohydrate synthesis The nutrient rich water allows algae to Triple super phosphates ( Ca(H2PO4)2∙H2O ) Superphosphates • Assists photosynthesis grow rapidly. When algae die, it de- Ca(H2PO4)2∙H2O No CaSO4 byproduct, ∴Higher P concentration • Improves quality of flowers and fruit. pletes the dissolved O2 in the water. • Makes plants more heat, drought and disease Fish and other water-life die. resistance. Excess nitrates in drinking water is also PHOSPHORIC ACID ( H3PO4 ) Potassium can be found naturally in compost and harmful to humans. manure, but is primarily imported from Germany in Ca5(PO4)3F + 5H2SO4 + 5H2O → 5CaSO4∙H2O + 3H3PO4 + HF the form of potash. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 63