Learner Support Grade 11 for 2021 1

Downloaded from hlayiso.com CURRICULUM GRADE 10 -12 DIRECTORATE NCS (CAPS) LEARNER SUPPORT DOCUMENT GRADE 11 PHYSICAL SCIENCES STEP AHEAD PROGRAMME 2021

Downloaded from hlayiso.com PREFACE This support document serves to assist Physical Sciences learners on how to deal with curriculum gaps and learning losses as a result of the impact of COVID-19 in 2020. It also captures the challenging topics in the Grade 10 -12 work. Activities should serve as a guide on how various topics are assessed at different cognitive levels and also preparing learners for informal and formal tasks in Physical Sciences. It will cover the following topics: Topic Page No. 1. Vectors and Scalars 3 – 27 2. Intermolecular Forces 28 – 42 3. Quantitative Aspects 43 – 56 4. Electrostatics 57 – 73

Downloaded from hlayiso.com VECTORS IN 2D NOTES GR 11 Vector- A physical quantity with Scalars- A physical quantity with magnitude magnitude and direction. (Example: only. (Example: time, charge, work. Energy, force, velocity, acceleration, etc.) distance, speed etc.)  A vector has a magnitude and Resources direction.  Vectors can be used to 1. Relative Velocity Vectors represent many physical http://cnx.org/content/m14030/latest / quantities that have a A resource on relative velocity in two magnitude and direction, like dimensions. forces. http://www.examsolutions.net/math s  Vectors may be represented revision/mechanics/vectors/relative as arrows where the length of velocity/tutorial-1.php the arrow indicates the A video introducing relative velocity. magnitude and the arrowhead indicates the direction of the 2. Vectors of Relative Motion vector. https://www.youtube.com/watch?v=  Vectors in two dimensions can nxF47gTE2yU be drawn on the Cartesian This video discusses relative motion. plane. http://www.physicstutorials.org/hom  Vectors can be added e/mechanics/1d-kinematics/relative motion graphically using the head-to- Relative motion with examples. tail method or the tail-to-tail method. 3. Forces in Equilibrium  A closed vector diagram is a http://fiziknota.blogspot.com/2009/0 set of vectors drawn on the 5/analysing-forces-inequilibrium.html Cartesian using the tail-to- Analysing forces in equilibrium. head method and that has a https://www.uwstout.edu/physics/up resultant with a magnitude of load/Force-Equilibrium.pdf zero.  Vectors can be added A laboratory activity demonstrating the algebraically using Pythagoras' conditions under which an object is held in theorem or using static equilibrium due to concurrent forces. components. http://www.physics.arizona.edu/physics/  The direction of a vector can resources/documents/03_Ve ctors.pdf be found using simple A study of vectors in the context of force trigonometric calculations.  The components of a vector 4. Calculating Vectors using Pythagoras are a series of vectors that, http://www.mathsisfun.com/algebra/ when combined, give the vectors.html original vector as their All about calculating vectors. resultant.  Components are usually 5. Resolution of Vectors into Components created that align with the http://www.wikihow.com/Resolve-aVector- Cartesian coordinate axes. Into-Components How to resolve components can be resolved as horizontal and vertical components.

Downloaded from hlayiso.com A resultant is one vector, which has the same effect on a body as the two or more vectors that are actually acting on that body. It starts at the beginning of the first vector and ends at the end of the last one. An equilibrant is one vector, which cancels out the effect that the two or more vectors actually have on a body. It is equal in size to the resultant but opposite in direction. Positive (+) and negative (-) signs are used to indicate direction in vectors. (Not their magnitude.) Arrows are used in vectors to give magnitude (length of arrow) and direction (head of arrow). IMPORTANT NOTE IN THIS SECTION  Usage of the Protractor.  Use of the correct scale  Understanding the difference between The Tail to Tail and The Tail to Head Method.  Measuring the Angles from the reference line to the Vector. When the Learners are using the Protractor, they must ensure that Tail of the vector as at the dot centre of the protractor, and that the line is correctly placed. PRIOR KNOWLEDGE CURRENT LOOKING FORWARD GRADE 10 GRADE 11 GRADE 11 – 12 Equality of vectors, Add co-linear vertical Use these methods to find negative vectors, addition vectors and co-linear the resultant of any vector and subtraction of vectors horizontal vectors to quantity learnt in grades 10, (in one dimension only). obtain the net vertical 11 and 12. • Defined a resultant vector (Ry) and net • Resolve any vector learnt vector. horizontal vector (Rx). in grades 10, 11 and 12 • Found the resultant • Sketch Rx and Ry on a into its perpendicular vector Cartesian plane. components. graphically using the tail-to • Determine the head magnitude method as well as by of the resultant using the calculation for a maximum theorem of Pythagoras. of four force vectors in • Find resultant vector one dimension only.

Downloaded from hlayiso.com • A frame of reference has graphically using the tail- an origin and a set of to head method as well as directions e.g. East and by West or up and down. calculation (by component • Defined one dimensional method) for a maximum motion. of four force vectors in • Defined position relative both 1-dimension and to a reference point and 2-dimensions. understood that position • Understand what is a can be positive or closed negative. vector diagram. • Defined displacement as • Determine the direction a of change in position. the resultant using simple • Displacement is a vector trigonometric ratios. quantity that points from • Resolve a 2-dimensional initial to final position vector into its perpendicular components. Vectors in One Dimension – Revision • Vector – physical quantity having magnitude and direction. • Scalar – physical quantity having magnitude only. Examples: Vector Scalar Force Time Energy Weight Mass Velocity Speed Displacement Distance Acceleration Graphical Representation of a vector  Vector is represented by an arrow  The length of an arrow represents the size (magnitude) of the vector  The arrow-head represents the direction of the vector. Direction of a horizontal or vertical vector

Downloaded from hlayiso.com  A positive sign (+) or a negative sign (–) is used to indicate the direction of a vector that are either horizontal or vertical. For each example you must select the sign. Examples a) right is + +5N b) left is – -3N Three methods to describe the direction of a vector that is not horizontal or vertical 1. X and Y axes 2. Bearing 3. Compass Reading On a graph FA : 10 N at 30o above the positive x- axis (horizontal axis) FB : 8 N at 12o left of the negative y- axis (vertical axis) FC : 5 N at 65o above the negative x- axis (horizontal axis) Bearing  Only for vectors in the horizontal plane i.e. parallel to the surface of the Earth.  Use North as 0o and always measure clockwise FA : 10 N on a bearing of 60o FB : 8 N on a bearing of 192o FC : 5 N on a bearing of 335o FA : 10 N on a bearing of 60o Compass (Cardinal points or directions)

Downloaded from hlayiso.com NB: The 300 N of E means you start from east and move 300 towards the North FA : 10 N at 30o North of East FB : 8 N at 12o West of South FC : 5 N at 65o North of West RESULTANT OF VECTORS Define a resultant as the vector sum of two or more vectors, i.e. a single vector having the same effect as two or more vectors together. Resultant vector is greatest when vectors are in the same directions Resultant vector is smallest when vectors are in the opposite directions A. Two vectors acting in the same direction :(one dimension) A girl walks 120m due East and then 230m in the same direction. What is her resultant displacement? 1. By calculation: Sign of direction: Take to East to be + R = 120m + 230m = 350m East 2. By construction: B. Two vectors acting in opposite direction (one dimension) A boy walks 210m due East. He then turns and walk 60m due West. Determine his resultant displacement. By calculation: (taking East as positive) R = 210m + (- 60m) = 150m East C. Multiple vectors acting in different directions (one dimension)

Downloaded from hlayiso.com Determine the resultant(net) force when 8 N force acts to the right, a 10 N force acts to the right, a 25 N force acts to the left and a 12 N force acts to the left Let to the right be positive By Calculation: Fnet = F1 + F2 + F3 + F4 = 8 + 10 + (-25) +(-12) = -19 N Fnet = 19 N left Vectors in Two Dimension` Resultant of perpendicular vectors  Perpendicular vectors are at right angles to each other  A horizontal force of 30 N and a vertical force of 40 N that act on an object are an example of two forces that are perpendicular to each other. Adding co-linear vectors  Co-linear vectors are simply vectors that lie in the same straight line.  The net x-component (Rx ) is the sum of the vectors parallel with the x- direction: 𝑅𝑥 = 𝑅𝑥1 + 𝑅𝑥2  The net y-component (Ry) is the sum of the vectors perpendicular to the x- direction: 𝑅𝑦 = 𝑅𝑦1 + 𝑅𝑦2 Worked Example Two forces of 3N and 2N apply an upward force to an object. At the same time, two forces each of 2N act horizontally to the right. Find the resultant force acting on the object.

Downloaded from hlayiso.com Step 1: Draw a diagram and calculate the net vertical and net horizontal forces Step 2: Graphical representation of Rx and Ry Step 3: To find resultant (R) of the above vectors, one can be using tail-to-tail drawing of vectors Tail to tail method or Parallelogram: Note: When vectors are drawn tail-to-tail, a parallelogram must be completed in order to determine their resultant.  Pythagoras theorem is used to calculate the magnitude of the resultant.

Downloaded from hlayiso.com  Considering the vector diagram above we can use Pythagoras theorem as follows: 𝑅 = 𝑅𝑥 + 𝑅𝑦2 2 2 𝑅2=42+52 𝑅=√42+52 𝑅=6.40 𝑁  Use trigonometry to find the direction of the resultant as follows: 𝑅𝑦 5 tanθ= 𝑅 = 4 𝑥 Ө = 51.240 Worked Example: A force of F1 = 5N is applied to a block in a horizontal direction. A second force F2 = 4N is applied to the object at an angle of 30⁰ above the horizontal. Determine the resultant of the two forces, by accurate scale drawing. Step 1: Draw rough sketches of the vector diagrams: Note: Forces are NOT perpendicular Step 2: Choose the suitable scale. e.g. 1cm: 1N Step 3: Draw the first vector (F1) on the horizontal, according to the scale. Step 4: Draw the second scaled vector (F2) 30⁰ above the horizontal. Step 5: Complete the parallelogram and draw the diagonal (which is the resultant) Step 6: Use the protractor to measure the angle between the horizontal and the resultant Step 7: Apply scale and convert the measured length to the actual magnitude.

Downloaded from hlayiso.com The resultant is 8,7N, 13,3⁰ above the horizontal. GRAPHICAL DETERMINATION OF THE RESULTANT VECTOR Tail-to-head method is used to find the resultant of two or more consecutive vectors (vectors that are successive) Steps to be followed:  Choose the suitable scale e.g. 10mm: 10N  Accurately draw the first vector as an arrow according to chosen scale and in the correct direction  Draw the second vector by placing the tail of the second vector at the tip of the first vector {tail – to – head method}  Complete the diagram by drawing the resultant from the tail of the first vector to the head of the last vector.  Make sure that you measure the angles correctly with a protractor.  Always add arrow heads to vectors to indicate the direction.  Measure the length and direction of the resultant vector. Use the scale to determine the real magnitude of the resultant. Worked Example 1: A ship leaves a harbour H and sails 6km north to port A. From here the ship travels 12 km east to port B, before sailing 5,5 km at 450 south-west to port C. Determine the ship’s restaurant displacement using the tail-to-head technique. Rough sketch:

Downloaded from hlayiso.com Using a scale 1cm: 2km, the accurate drawing of vectors is: Measure the angle between the North line and the resultant with a protractor to find that the direction of the resultant displacement: Resultant displacement of the ship is 9,2 km on a bearing of 72, 3⁰. Example 2: A man walks 40m East, then 30m North. Use a scale of 1 cm: 10 m and answer the following questions: 1. What was the total distance he walked? 2. Determine by construction his resultant displacement? 3. Calculate determine the direction of the resultant. 4. Calculate the magnitude of resultant displacement Solutions: 1. Rough sketch

Downloaded from hlayiso.com Total distance = 40m + 30m = 70 m 2. Scale: 1cm: 10m The resultant is 50m, 37⁰ from the horizontal 30 Tan Ф= 40 Ф=36,87⁰ R2 = x2 + y2 = 402 + 302 = 2500 R = 50m CO-LINEAR VECTORS Co-linear vectors are simply vectors that lie in the same straight line. They either act in the same direction or in the opposite direction, e.g. vectors pointing east or west are co-linear; vectors pointing north and west are not co-linear because they act at an angle (90°) to each other. Draw the diagram alongside, on the board, of four forces acting on an object.

Downloaded from hlayiso.com The 60 N and 275 N force vectors act along the horizontal axis (x-axis). It is essential to use signs (+ and -) to indicate the direction of vectors which act in the same line. In the Cartesian plane, right is positive (+) and left is negative (‒). The 275 N force acts to the right and is therefore represented as +275 N. The 60 N force acts to the left and is therefore represented as ‒60 N. The resultant horizontal force (Rx) is the vector sum of the forces acting along the horizontal (or x-) axis: Rx = (+275) + (- 60) = +215 N. The positive sign (+) in the answer indicates that the resultant horizontal force acts to the right: Rx = 215 N to the right. The 135 N and 45 N forces act in the vertical direction (y-axis). In the Cartesian plane, up is positive (+) and down is negative (‒). The 135 N force acts upwards and is therefore represented as +135 N. The 45 N force acts downwards and is therefore represented as ‒45 N. The resultant vertical force (Ry) is the vector sum of the forces acting in the vertical direction. Rx = (+135) +(- 45) = + 90 N Ry = 90N Upwards

Downloaded from hlayiso.com CONCEPT We have now simplified the given force diagram to just two perpendicular force vectors. The resultant force can be found by drawing a neat vector diagram using the tail-to-tail method or the tail-to-head method. Tail-to-tail method Place the tails of the two vectors together (tail to-tail) in the Cartesian plane and complete the parallelogram by drawing in the opposite sides. The opposite sides are PARALLEL and EQUAL IN LENGTH to each other. Tail to Tail Tail to Head The resultant force (R) is the diagonal of the parallelogram. Draw in the diagonal of the parallelogram. The diagonal MUST start from the TWO TAILS of the given forces. We now have two right-angled triangles to work with. Use Pythagoras to calculate the magnitude of the resultant force (R). R2 = 902 + 2152 R2 = 54325 R =√𝟓𝟒𝟑𝟐𝟓 =233.07N This is the magnitude of the resultant force. We now need to find the direction of the resultant force. Place a θ into your vector diagram to represent the angle between the resultant force and the horizontal plane.

Downloaded from hlayiso.com 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝟗𝟎 Use a trig. ratio to find the angle θ: tan Ө = 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 = 𝟐𝟏𝟓 = 22.71N Tail to Head Method Draw the horizontal vector Rx on your page, then draw the vertical vector Ry so that its TAIL is touching the HEAD of Rx. The resultant force (R) MUST be drawn from the TAIL of the first vector to the HEAD of the second vector. As shown in the previous method, use Pythagoras to calculate the magnitude of the resultant force (R) and a trig. ratio to find the direction of the resultant force. R = 233,08 N at 22,71° above the horizontal axis (positive x-axis) STEPS TO FOLLOW A. Draw a neat-labelled vector diagram. B. Simplify the problem by first finding the resultant of vectors in the same and opposite directions (co-linear vectors). C. When left with perpendicular vectors, decide on the method to use (tail-to-tail method, etc.) D. Draw a neat vector diagram of the perpendicular vectors. E. Use the theorem of Pythagoras and trig. ratios to find the magnitude and direction of resultant. ACTIVITY 2 1. Two forces of 120N each are exerted on a crate simultaneously as shown in the figure 1. 1 Plot a sketch of force vectors on a Cartesian plane 1.2 Use a scale of 10mm: 20N and the tail-to-tail method to determine magnitude and direction of the resultant.

Downloaded from hlayiso.com 2 force vectors P and Q are plotted on the Cartesian plane shown below 2.1 The magnitude of vector P in force units ________________________________________________________ _______________________________________________________________ 2.2 The direction of vector Q, measured clockwise from positive y-axis __________________________________________________________________ _________________________________________________________________ 3. Two Newton spring are used to exert forces on a wooden block as shown in the figure 3.1 Take the readings of the individual forces exerted on the spring balances and plot them on a Cartesian plan.

Downloaded from hlayiso.com 3.2 Plot a vector sketch diagram (not according to scale) 3.3 Calculate the magnitude and direction of the two forces and their resultant. of the resultant of the two forces. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ RESOLUTION OF A VECTOR INTO ITS PARALLEL AND PERPENDICULAR COMPONENTS  The process of breaking down the vector quantity into its components that are at right angles to each other is known as resolving a vector into its components. Worked Example A force of 400N acts at an angle 600 to the horizontal. Horizontal component Vertical component CosӨ= Adjacent /Hypotenuse SinӨ= opposite/hypotenuse Cos600= Rx/400 Sin600= Ry/400 Rx = 200N Ry = 346.41N Finding resultant of vectors acting at angles (using component method)

Downloaded from hlayiso.com Step 1: Find horizontal and vertical components of each force Components of F1: Components of F2: Components of F3: Horizontal (Fx) Horizontal Horizontal (Fx) (Fx) = F1cosФ (Fx) = F2cosФ (Fx) = F3 cosФ = 100 cos(90⁰- 400) = 80 cos(900-300) = 90 cos20⁰ = 64,28N (right = 40N (right) = 84,57N (left) (Fy) = F1 sinФ (Fy) = F2 sinФ (Fy) = F3 sinФ = 100 sin(90⁰- 400) = 80 sin(900-300) = 90 sin20⁰ = 76,60N (up = 69,28N (down) = 30,78N (down) NB : The angle α is between the Horizontal axes and the Force. Step 2: Hence, the new situation is: Step 3: Thus, the sum of horizontal components = 64,28+40+ (-84,57) = 19,71N to the right And the sum of vertical components = 69,28+30,78+ (76,20) = 23,46N downwards Step 4:

Downloaded from hlayiso.com The resultant of the two vectors (at right angles to each other) is: R2 = Rx2 + Ry2 = 19,712+23,862 = 30,64N tan Ф = 23,86/19,71 Ф = 50,47⁰ Thus, the resultant of the three forces is 30,95N in a bearing of 50,47⁰ from the horizontal. Exam MCQ questions QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A–D) next to the question number (1.1–1.10) in the ANSWER BOOK, for example 1.11 D. 1.1 Consider the following vector diagrams. Which ONE of these vector diagrams represents a zero resultant A B C D (2) 1.2 Two forces of magnitudes 3 N and 4 N respectively act on a body. The maximum magnitude of the resultant of

Downloaded from hlayiso.com these forces is … A 12 N. B 7N C 5N (2) D 1N 1.3 Three forces of magnitude 20 N each act on object P as shown below. 20 N 20 N P 20 N The resultant force on object P is … A zero. B 20 N to the left. C 20 N upwards. D 20 N downwards. (2) 1.4 Two forces of magnitudes 15 N and 20 N act at a point on an object. Which one of the following magnitudes CANNOT be the resultant of these forces? A. 35 N B. 10 N C. 4N D. 18 N (2) 1.5 Three forces, each of magnitude 7 N, act on object P as shown. 7N 7N 7N The resultant force on object P is … A zero B 7 N to the left. C 7 N upwards. D 7 N downwards. (2)

Downloaded from hlayiso.com 1.6 Two forces of magnitude 50 N and 70 N respectively act on a body. The maximum magnitude of the resultant force on the body is ... A 20 N. B 60 N. C 120 N. D 140 N. (2) 1.7 Two forces of magnitudes 8 N and 6 N are added to each other. Which of the following values CANNOT be a resultant of these two forces? A 2N B 3N C 14 N D 16 N (2) 1.8 You can replace two forces, P and Q, with a single force of 7 N. If the magnitude of force P is 3 N, which one of the following can be the magnitude of force Q? A 2N B 3N C 8N D 13N (2) 1.9 Consider the following vector diagram: The vector which represents the resultant of the other two, is … A AB. B AC. C CB. D BA. (2) STRUCTURED QUESTIONS

Downloaded from hlayiso.com QUESTION 1 (Grade11 KZN MARCH 2015) The diagram below shows TWO forces P and Q of magnitude 250 N and 150 N respectively acting at a point R. 1.1 Calculate the horizontal and vertical components of vector P. (4) 1.2 Calculate the vector sum of horizontal components of P and Q. (3) 1.3 The vector sum of the vertical components of these forces is 129,45 N. (3) Using the vector sums of the horizontal and vertical components of P and Q, draw a labelled force vector diagram to show the resultant force acting on the point R. 1.4 Calculate the magnitude of the resultant of forces P and Q. (3) 1.5 Calculate the direction (measured clockwise from the positive Y axis) of (3) the resultant of vectors P and Q. 1.6 If vector P was fixed but the direction of vector Q could be changed, for (1) which value of ϴ will the resultant force have a maximum value? 17 QUESTION 2 (FS CONTROL TEST TERM 1 – 2014) Force vectors P and Q were drawn to scale on the Cartesian plane shown below. 2.1 Define the term resultant of a vector. (2) 2.2 From the graph, without using a scale drawing, CALCULATE the (no units are required): 2.2.1 Magnitude of the horizontal component of vector P (1) 2.2.2 Magnitude of the horizontal component of the resultant of vectors P (2) and Q 2.2.3 Magnitude of the vertical component of the resultant of vector P and (2) Q 2.2.4 Resultant of vectors P and Q. (6)

Downloaded from hlayiso.com [13] QUESTION 3 (Fs CONTROL TEST TERM 1 – 2015) Three forces, F1, F2 and w, act on point O as shown in the diagram below. 3.1 Define the term resultant of forces. (2) 3.2 By means of an accurate scale drawing, determine the vertical (5) component of F1. Use a scale where 10 N is represented by 10 mm. 3.3 The horizontal and vertical components of F2 are equal to 40 N and 42 N respectively. 3.3.1 Prove with calculations that the horizontal components of the (3) forces are in equilibrium. 3.3.2 Calculate the magnitude and direction of force w. (2) [12] ADDITIONAL ACTIVITIES JUNE 2014 LIMPOPO QUESTION 1 Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A – D) next to the question number (1.1 – 1.10) in the ANSWER BOOK. Which of the following pairs can be classified as vectors? 1.1 A. Friction and mass B. Mass and inertia C. Inertia and weight D. Weight and Friction (2) 1.2 Which one of the following represents a closed vector diagram?

Downloaded from hlayiso.com QUESTION 2 The diagram below shows three forces of 5 N, 4 N and 3 N acting on an object in the same Cartesian plane. Calculate the: 2.1 Magnitude of the resultant force acting on the object. (10) 2.2 Direction of the resultant force. (3) November 2015 National Vector P and vector –P are acting on a common point O. The angle between the two vectors is ... 1.1 A. 0o B. 90o C. 180o D. 270o QUESTION 2 Three forces, P, Q and R, of magnitudes 500 N, 200 N and 300 N respectively, act on a point O in the directions shown in the diagram below. The forces are NOT drawn to scale. 2.1 Refer to the information in the diagram above and give a reason why (2)

Downloaded from hlayiso.com forces P, Q and R are classified as vectors. 2.2 Determine the magnitude and direction of the resultant force, either by (8) CALCULATION or by ACCURATE CONSTRUCTION AND MEASUREMENT. (Use scale 10 mm = 50 N.) [10] November 2017 National QUESTION 1: MULTIPLE-CHOICE QUESTIONS 1. Which ONE of the following pairs of physical quantities is vector quantities? 1 A. Force and distance B. Velocity and speed C. Charge and electric field D. Electric field and force (2) 1. Which ONE of the following vector diagrams represents three forces acting on an object simultaneously while the object moves at 2 CONSTANT VELOCITY? (2) MARCH 2018 QUESTION 2 Three forces F1, F2 and F3 act at a point , as shown on the Cartesian plane in the diagram below.

Downloaded from hlayiso.com Scale : 1 square unit represents 1N 2.1 Define the “resultant vector” (2) 2.2 Write down the X-component and Y-component for force F2 (2) 2.3 Calculate the NET horizontal component for the forces F1, F2 and F3 (1) 2.4 Calculate the NET vertical component for the forces F1, F2 and F3 (1) 2.5 Calculate the magnitude and direction of the resultant force of these (4) three forces. [10]

Downloaded from hlayiso.com MATTER AND MATERIALS Intermolecular forces The covalent bond occurs when atoms of non-metal elements share unpaired valence electrons to achieve noble gas electron configuration. Example: The covalent bond is an interatomic force and the result product is a molecule. The covalent bonds are very strong and one requires a large amount of energy to break them i.e. to separate the atoms of the molecule. The intermolecular forces are the forces of attraction that exist between molecules of a chemical substance in a solid or liquid form.. The intermolecular forces are weaker than interatomic forces. Examples 1. 2. Interatomic force intermolecular forces The strength of intermolecular forces determines the physical properties (boiling points, melting point, evaporation, etc.) of a molecular substance. Molecules can be polar or non-polar. The polarity of a molecule is determined by two factors :  The difference in electronegativity between the bonding atoms.

Downloaded from hlayiso.com  The geometry (shape) of a molecule. Non-polar molecules have no dipoles (positive and negative ends) e.g. CO2 + - + Difference in electronegativity (ΔEN) = 3,5 – 2,5 = 1,0 → bond between C and O is polar but both ends of CO2 are ∂+ and the molecule is linear. The molecule is therefore non-polar. Polar molecule In polar molecule there are ∂+ and ∂- ends e.g. H2O and the geometry of the molecule is angular (bent) - + ΔEN = 3,5 – 2,1 = 1,4. The two bonds (between H and O atoms) are polar. H2O molecule has H-end being ∂+ and O-end being ∂-, they are called dipoles. The molecule is polar Non-polar bonds: • H2 is non-polar because the two atoms are identical, there is even or symmetrical distribution of charge, this makes H2 a non-polar molecule. • The electro-negativity between the two atoms is zero, this explains why the interatomic bond in Hydrogen molecule is non-polar TYPES OF INTERMOLECULAR FORCES 1. Van der Waals forces: van der Waals force is a weak attraction between molecules The force of attraction is between polar and non-polar molecules (a) Dipole-dipole force is an attraction between polar molecules. Examples: 1: between molecules of HCℓ

Downloaded from hlayiso.com Example: 2. When HCℓ dissolves in water Dipole-dipole forces are the strongest intermolecular forces (b) Dipole-induced dipole force is a weak force of attraction that exist when a polar causing a distortion in arrangement of electrons in a non-polar molecule. Temporary dipoles are induced in a non-polar molecule e.g. mixture of HCℓ and argon. (c) Induced dipole forces (London forces) occurs between atoms or molecules of non-polar substances e.g. Between atoms of noble gases, molecules of diatomic molecules (H2; N2; O2; F2; Cℓ2), non- polar compounds (CH4; CCℓ4; BF3), between molecules of CO2(s) – dry ice Induced dipole forces are the weakest van der Waals forces 2. Hydrogen bond is a special type of dipole-dipole force. Hydrogen bond is a force attraction between a hydrogen atom bonded to oxygen, nitrogen or fluorine in a molecule to a lone pair of an adjacent molecules e.g. between H2O molecules, between HF molecules. Hydrogen bonds are the strongest intermolecular forces. 3. Ion-dipole force is the attraction of an ion with a positive or negative charge and the oppositely charged end of the dipole of polar molecule e.g. NaCℓ dissolved in water 4. Ion-induced dipole force is an attraction between an ion and a non-polar molecule. As an ion approaches a non-polar molecule, it causes a distortion in the arrangement of the arrangement of the electrons of non-polar molecules. Momentary dipoles are induced in non-polar molecules e.g. dissolving NaCℓ in hexane (C6H14) PHYSICAL STATE AND DENSITY EXPLAINED IN TERMS OF THESE INTERMOLECULAR FORCES Boiling Point The hydrogen bonds between highly polar water molecules are stronger than normal dipole-dipole forces and more energy is needed to break these bonds. Energy needed for molecules to evaporate, is called

Downloaded from hlayiso.com heat of vaporization. 40,7 kJ∙mol-1 energy is needed for water molecules to evaporate – hence the higher than expected boiling point. Surface tension  The surface of a liquid may be considered as an elastic film or skin. As a result of this surface phenomenon, it is possible for insects to walk on the surface of a liquid.  The curved surfaces (menisci) of liquids, most noticeable in narrow tubes, and the fact that a water drop is always spherical, are properties which are linked to surface tension.  Surface tension can be defined as the resistance that a liquid offers to a force which tries to increase its surface area. Density  The common statements that mercury is “heavier” than water or that iron is “heavier” than aluminium are actually not correct. It is not the mass of the substances that is compared, but the mass per unit volume, which is known as the density.  The density of a substance is independent of the amount and size of the sample and can therefore be used as an aid to distinguish one pure substance from another. Stronger intermolecular forces will result in substances having greater densities.  According to the kinetic molecular theory, particles in a gas are far apart with no regular motion. Particles in a liquid are close together with no regular arrangement. Particles in a solid are close together, usually in a regular pattern.  It implies that the number of particles per unit volume will decrease from the solid to the liquid state. (Refer to diagram).  The density of a substance increases as it changes from the gas to the liquid and solid state, because the intermolecular forces increase as substances changes from the gas to the solid state.  Note: The density of most substances known, decreases as the temperature rises. However, the density of water increases as the temperature is raised from 0 0C to 4 0C. (Refer to the lesson: Macroscopic properties of the three phases of water related to their microscopic structure)

Downloaded from hlayiso.com KINETIC ENERGY AND TEMPERATURE  Temperature is directly proportional to average kinetic energy of the particles. An increase in T increases K. K = ½mv3  The mass of the particles remains constant – motion increases.  T (Temperature is directly proportional to average kinetic energy of the particles).  The molar masses of the particles increase in the following order: F2 Cℓ2 Br2. The London forces between the non-polar molecule increase with an increase in molar mass. At the same temperature, the forces between F2 molecules are weaker than between Cℓ 2 molecules etc.  An increase in temperature decreases the bond strength between the molecules. The particles are further apart and the substance becomes less dense.  Thermometers work by using thermal expansion of a liquid to measure temperature; construction (bridges); it can also be used in loosening nuts and bolts; a hot-air balloon uses the thermal expansion of air to generate lift. • Thermal expansion is the expansion of a liquid on heating. • Thermal conductivity is a measure of how much a material conducts heat. Kinetic energy of particles and temperature change  Heat is a form of energy. During the heating process, the particles of the substance absorb energy that allows them to vibrate in all directions and collide with each other. Energy that causes motion is called kinetic energy. All the particles do not gain the same amount of energy and they also transfer energy to other particles when they collide with each other.  An increase in the temperature of gas particles increases the average kinetic energy of the particles and they will move faster (K = ½mv2). The average kinetic energy is directly proportional to the temperature of the particles. Thermal expansion  The most easily observed examples of thermal expansion are size changes of materials as they are heated or cooled. Almost all materials (solids, liquids, and gases) expand when they are heated, and contract when they are cooled. Increased temperature increases the frequency and magnitude of the molecular motion of the material and produces more energetic collisions. Increasing the energy of the collisions forces the molecules further apart and causes the material to expand.  Different materials expand or contract at different rates. In general, gases expand more than liquids, and liquids expand more than solids. Example:

Downloaded from hlayiso.com  Because the liquid expands at a faster rate than the tube, it rises as the temperature increases and drops as the temperature decreases.  The first step in producing a thermometer scale is to record the height of the liquid at two known temperatures (i.e. the boiling point and freezing point of water). The difference in fluid height between these points is divided into equal increments to indicate the temperature at heights between these extremes. CHEMISTRY OF WATER  The atoms in a water molecule are held together strongly by covalent bonds to for angular molecule. The electrons are not evenly around the molecule. Oxygen has a higher electronegativity than hydrogen and the oxygen side of the water molecule is partially negative, while the hydrogen is partially positive. So water molecules are polar and the form dipoles.  The water molecules are strongly attracted to one another through hydrogen bonds. At the Temperature below 0◦C at the atmospheric pressure, water is a solid (ICE), at that stage water molecules are arranged in a regular pattern to form ice crystals.  When you heat the ice, the water molecules start to vibrate faster and the kinetic energy of water molecules increases. At 0◦C, the particles are moving vigorously to break the lattice pattern. (More notes on Study and master)  At solid state – water molecules are closely parked together where hydrogen bond is very strong where they do not move (no kinetic energy)  Liquid state – water molecules can now vibrate and move (there is kinetic energy)  Gaseous state – water molecules are scattered free to move ( more kinetic energy) resulting in a very week hydrogen bonds (others they even say at this stage there no hydrogen bonds  So water has a very high specific heat capacity of 4 200 kJ.kg-1.K-1.to rise the temperature of water we need first to break many intermolecular forces (Hydrogen bonds). Therefore water molecules absorb lots of heat while its temperature rises slightly. The opposite also true – water molecule can give off much heat with only slight decrease in its temperature.  The polar water molecules vibrate, which enable them to absorb heat from the sun. The water in lakes and oceans can absorb large amount heat in summer and give off heat in winter, with only small changes in the temperature of the water. This effect moderates the climate of

Downloaded from hlayiso.com adjacent land masses. In short, oceans act as reservoirs of heat and ensure that the Earth has a moderate climate. Water vapour (H2O(g)) and carbon dioxide (CO2(g)) in the atmosphere provide a temperature buffer (greenhouse effect) which helps maintain a relatively steady surface temperature. The polarity of the covalent bonds in H2O-molecules and the strong hydrogen bond between the molecules that vibrates at the same frequency as infrared light are responsible for the fact that water absorbs sunlight. INVESTIGATE PHYSICAL PROPERTIES OF WATER (H2O) 1. INVESTIGATE SPECIFIC HEAT CAPACITY Specific heat capacity is the amount of energy needed to raise the temperature of 1kg of a substance by 1⁰C. Water has high specific heat capacity of 4200J.kg-1. ⁰C-1 A high specific heat capacity requires more energy for the temperature to change. This means water absorbs a lot of heat energy to increase the temperature of 1kg by 1⁰C. Water has higher specific heat capacity than oil because te rate at which temperature of water increases is less than that of oil with the same amount of heat added. Melting point and boiling point of water Melting point is the temperature at which the solid and liquid phases of a substance are in equilibrium. The stronger the intermolecular forces, the higher the melting point. The boiling point is the temperature at which the vapour pressure of a substance equal atmospheric pressure. Melting point of water is 𝑂0 𝐶 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑏𝑜𝑖𝑙𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 100⁰𝐶 When the boiling point of water is measured at sea level(e.g. towns like Cape town and Durban),it is often very close to 100⁰𝐶 since atmospheric pressure is almost the same as standard pressure. If you measure boiling point of water in a town at a higher altitudes (e.g. Johannesburg or Polokwane) it will have a slightly lower boiling point. 2 .WATER AS A SOLVENT A solvent is the substance that dissolves a solute to produce a homogeneous mixture.

Downloaded from hlayiso.com A solute is a substance that dissolves in a solvent to produce a homogeneous mixture. A solution is a homogeneous mixture of two or more substances. What makes water a good solvent? The polarity and the ability to form hydrogen bonds makes water a good solvent.  The polarity of water allows it to dissolve both ionic bonds and other polar molecules.  A water molecule is polar because the electric charge is distributed unevenly because the oxygen atom is on one end ,causing it to have slight negative charge and hydrogen atom is on one end causing it to have slight positive charge. Water molecules are held together by hydrogen bonds. Hydrogen bonds are a much stronger type of intermolecular force than those found in many other substances and affects the properties of water. 3. Investigating Surface tension o Intermolecular forces are forces between neighbouring molecules. . Surface tension is the energy required to increase the surface area of a liquid. o Water has strong hydrogen bonds between the molecules. These intermolecular forces hold the particles more strongly together increasing surface tension.  The molecules of water attract each other strongly.  The drop of water does not spread out easily on the paper and keeps its shape on the wax paper,  This shows that there are strong intermolecular forces on between water molecules. The oil particles spread out easily as a result of weak London forces. 4. Density of water Density is the the mass per unit volume. 𝑚𝑎𝑠𝑠 Density=𝑣𝑜𝑙𝑢𝑚𝑒 Another unusual property of water is that its solid phase (ice) is less dense than its liquid phase. This can observed by put ice into the a glass of water .The ice does not sink to the bottom of the glass but floats on top of the liquid. This phenomenon is related to hydrogen bonds between particles of water .While other material contract when they solidify, water expands. The ability of the ice to float is very important

Downloaded from hlayiso.com in the environment. When a deep body of water cools, floating ice insulates the liquid water below, preventing it from freezing and allowing life under the frozen surface to exist.

Downloaded from hlayiso.com LEARNERS WORKSHEET 1. Which intermolecular forces are found in: 1.1 hydrogen fluoride (HF) (1) _________________________________________ 1.2 methane (CH4) (1) _________________________________________ 1.3 potassium chloride in ammonia (KCl in NH3) (1) _________________________________________ 1.4 krypton (Kr) (1) _________________________________________ 2 Given the following diagram: 2.1 Name the molecule and circle it on the diagram (2) _________________________________________ 2.2 Label the interatomic forces (covalent bonds) (1) 2.3 Label the intermolecular forces (1) 3 Given the following molecules and solutions: HCl, CO2, I2, H2O, KI(aq), NH3, NaCl(aq), HF, MgCl2 in CCl4, NO, Ar, SiO2 3.1 Complete the table below by placing each molecule next to the correct (11) type of intermolecular force. Type of force Molecules Type of force Molecules Ion-dipole Ion-induced-dipole Dipole-dipole (no hydrogen bonding) Dipole-dipole (hydrogen bonding) Induced dipole Dipole-induced-dipole 3.2 In which one of the substances listed above are the intermolecular forces: 3.2.1 strongest (2) ______________________________________________________

Downloaded from hlayiso.com 3.2.2 weakest (3) ______________________________________________________ 4 Use your knowledge of different types of intermolecular forces to explain the following statements: 4.1 The boiling point of F2 is much lower than the boiling point of NH3 (2) ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 4.2 Water evaporates slower than carbon tetrachloride (CCl4). (2) ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 4.3 Sodium chloride is likely to dissolve in methanol (CH3OH) (2) ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 5 Calvin and Jason are helping their dad tile the bathroom floor. Their (2) dad tells them to leave small gaps between the tiles. Why do they need to leave these small gaps? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 6 Hope returns home from school on a hot day and pours herself a glass of water. She adds ice cubes to the water and notices that they float on the water. ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 6.1 What property of ice cubes allows them to float in the water? (2)

Downloaded from hlayiso.com ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 6.2 Briefly describe how this property affects the survival of aquatic life (2) during winter. ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 7 Which properties of water allow it to remain in its liquid phase over a (3) large temperature range? ---------------------------------------- Explain why this is important for life on earth. ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- 8 Give one word or term for each of the following descriptions: 8.1 The attractive force that exists between molecules. (1) -------------------------------------------------------- 8.2 A molecule that has an unequal distribution of charge. (1) --------------------------------------------------------- 8.3 The amount of heat energy that is needed to increase the temperature (1) of a unit mass of a substance by one degree. ---------------------------------------------------------- 9 Refer to the list of substances below: HCl, Cl2, H2O, NH3 N2, HF Select the true statement from the list below: a) NH3 is a non-polar molecule b) The melting point of NH3 will be higher than for Cl2

Downloaded from hlayiso.com c) Ion-dipole forces exist between molecules of HF d) At room temperature N2 is usually a liquid 10 The following table gives the melting points of various hydrides: Hydride Melting point (℃) HI -34 NH3 -33 H2S -60 CH4 -164 10.1 In which of these hydrides does hydrogen bonding occur? A. HI only B. NH3 only C. HI and NH3 only D. HI, NH3 and H2S --------------------------------------------------------- 10.2 Draw a graph to show the melting points of the hydrides. 10.3 Explain the shape of the graph ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- 10.4 The respective boiling points for four chemical substances are given below: Hydrogen sulphide −60 ℃ Ammonia −33 ℃ Hydrogen fluoride 20 ℃ Water 100 ℃

Downloaded from hlayiso.com 10.4.1 Which one of the substances exhibits the strongest forces of attraction between its molecules in the liquid state? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 10.4.2 Give the name of the force responsible for the relatively high boiling points of hydrogen fluoride and water and explain how this force originates. ----------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------- 10.4.3 The shapes of the molecules of hydrogen sulfide and water are similar, yet their boiling points differ. Explain ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 10.4.4 Susan states that van der Waals forces include ion-dipole forces, dipole-dipole forces and induced dipole forces. Simphiwe states that van der Waals forces include ion-dipole forces, ion-induced dipole forces and induced dipole forces. Thembile states that van der Waals forces include dipole-induced dipole forces, dipole-dipole forces and induced dipole forces. Who is correct and why? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 11 Jason and Bongani are arguing about which molecules have which intermolecular forces. They have drawn up the following table: Compound Compound Potassium iodide in water (KI(aq)) Potassium iodide in water (KI(aq)) Hydrogen sulfide (H2S) Hydrogen sulfide (H2S) Helium (He) Helium (He) Methane (CH4) Methane (CH4)

Downloaded from hlayiso.com 11.1 Jason says that hydrogen sulfide (H2S) is non-polar and so has induced dipole forces. Bongani says hydrogen sulfide is polar and has dipole-dipole forces. Who is correct and why? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 11.2 Bongani says that helium (He) is an ion and so has ion-induced dipole forces. Jason says helium is non-polar and has induced dipole forces. Who is correct and why? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 11.3 They both agree on the rest of the table. However, they have not got the correct force for potassium iodide in water (KI(aq)). What type of force actually exists in this compound? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 11.4 Khetang is looking at power lines around him for a school project. He notices that they sag slightly between the pylons. Why do power lines need to sag slightly? ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- 11.5 Briefly describe how the properties of water make it a good liquid for life on Earth. ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------

Downloaded from hlayiso.com QUANTITATIVE ASPECT OF CHEMICAL CHANGE  The mole is the SI unit for the amount of substance  The Mole is a name for a specific number. The Mole is the SI unit for quantity of substance Abbreviation of units – the official SI abbreviation of the unit mole is mol. The mole - mass relationship is summarised if the formula: 𝑚 𝑛= 𝑀 Where -: n – number of moles of substance in mol. m – mass of sample of substance in g. M – molar mass of substance in g.mol-1. Example 1 1. Calculate the number of moles of water in 100 g of wate 𝑚 100 𝑛 = 𝑀 = 16+(2×1) = 5,56 𝑚𝑜𝑙 Example 2 2. What is the molar mass of a substance if 5 moles of the substance have a mass of 295,5g 𝑚 𝑛= 𝑀 295,5 ∴5= 𝑀 (Cross multiply and let M be the subject of the formula) 295 M= 5 M = 58, 5 g.mol-1  The mole and Avogadro constant  The mole is defined as the number of particles or atoms in 12,0 g of Carbon -12  A mole of particle is an amount of 6, 02 × 1023 particle.  6, 02 × 1023 is known as Avogadro’s number NA. WORKED EXAMPLES Calculate the molar mass for each of the following: (i) Ca (ii) MgCl2

Downloaded from hlayiso.com Solutions (i) 40 g  mol-1 (ii)  24  (35,5 2)  24  71  95 g  mol-1 Calculate the number of moles of: 1. 213g of Cl2 2. 39.5g of KMnO4 Solutions 𝑚 213 1. 𝑛 = 𝑀 = 71 = 3 mol 𝑚 39.5 2. 𝑛 = 𝑀 = 158 = 0.25 mol  The Mole and the Gases  Avogadro also determine that : 1 mole of ANY gas at STP is occupies a volume of 22,4 dm3  The Molar volume of ANY gas at STP is given the symbol Vm (Vm = 22,4 dm-3.mol-1) NOTE: STP stands for standard Temperature (0°𝑪 ) and pressure (100kPa). 𝐯  For any gas at STP n = 𝐯 𝒎  Where : n - number of moles of gas V – Volume of gas sample Vm – molar Volume of gas (22, 4 dm-1.mol-1)  The Volume of the gas sample (V) must always be measured in dm3 ( 1 dm3 = 0,001 m3 = 1000 cm3 = dm3 100ml = 1 litre ) WORKED EXAMPLES 1. Determine the volume of 2 moles of H2 at STP. 2. Determine the mass of 60 cm3 of NH3 at STP. Solutions 𝑉 1. 𝑛 = 𝑉𝑚 𝑉 = 2𝑋22.4= 44.8 dm3 2. [convert units first] V=60/1000 =0.06dm3 𝑉 0.06 𝑛= = = 0.0027 𝑚𝑜𝑙 𝑉𝑚 22.4

Downloaded from hlayiso.com 𝑚 𝑛= 𝑀 𝑚 = 0.0027𝑋17.03 = 0.046𝑔  The mole and Concentrations of solution.  Solutions are homogeneous (uniform) mixture of solute and Solvent  Solute and Solvent can be a Gas, liquid or solid.  The most common solvent is liquid water, this is called aqueous solution.  Concentration – the concentration of solution is the number of mole per unit volume of solution. 𝒏 𝒎  c =𝒗 concentration can also be calculated with c = 𝑴𝑽 Where: C – concentration ( mol.dm-3) n – number of moles (mol) V – Volume (dm3) m – mass in (g) M – Molar mass (g.mol-1) WORKED EXAMPLES 1. Calculate the of a solution of calcium chloride made by dissolving 5.55g of dry CaCl2 crystals in enough water to make 750 cm3 of solution. 2. What mass of copper (II) sulphate must be dissolved in 200ml water to yield a 0.4 mol.dm -3 solution? Solutions 𝑚 5.55 1. 𝑐 = 𝑀𝑉 = (111)(0.75) =0.067 mol.dm−3 𝑚 2. 𝑐 = 𝑀𝑉 𝑚 0.4 = (159.5)(0.2) 𝑚 = 12.76𝑔  The mole and Percentage Composition of Substances  The subscripts in a chemical formula give the mole ratio in which the elements combine.  The mole ratio enables one to calculate the percentage composition, of the elements in the compound.  The Mole and Empirical formula of compounds.  The empirical formula of a compound gives the simplest mole ratio in which the element of the compound combine.  Empirical formula simply tells us the ratio of the different elements in a compound, not number of atoms of each element in molecule.

Downloaded from hlayiso.com E.g In a combustion reaction 0.48 g of Mg ribbons is burnt. The amount of MgO produced is 0.8g. Calculate the empirical formula for MgO Steps Magnesium Oxygen Step 1 0.48 g 0.80- 0.48= 0.32 g Mass of element Step 2 𝑚 0.48 𝑚 0.32 𝑚 𝑛= = = 0.02 mol 𝑛= = = 0.02 mol 𝑀 24 𝑀 16 Mol (𝑛 = 𝑀 ) Step 3 0.02 0.02 =1 =1 Atom ratio 0.02 0.02 (divide by smallest number in ratio) Therefore the empirical formula is MgO Worked example 2 The action of bacteria on meat and fish produces a stinking compound called CADAVERINE. The compound has a composition of 58, 77% C; 13,81% H and 27, 42% N by mass. Determine the empirical formula of CADAVERINE. In 100 g of compound we have 58,77 g C; 13,81 g H; and 27,40 g N m n= M 58,77 n(C) =  = 4,8975 mol C 12 13,81 n(H) =  = 13,81 mol H 1 27,40 n(N) =  = 1,9571 mol N 14 Mole ratios = C: H : N = 2,50 : 7,06 : 1,00 x2 Nearest whole number ratios = 5 : 14 : 2 ∴ empirical formula is C5H14N2  Empirical formula to Molecular formula  Molecular formula is the actual ratio of an atom in a molecular mass.  The molecular formula can be calculated from the empirical formula and the relative molecular mass. STEPS TO DETERMINE MOLECULAR FORMULAR  Determine the empirical formula ( if not given).  Determine the molar mass of the empirical formula.  Determine the ratio between molecular formula and empirical formula

Downloaded from hlayiso.com WORKED EXAMPLE Butene has the empirical formula 𝐶𝐻2 .The molecular mass of butene is 56 g.mol-1 Determine the molecular formula of butene. Step 1 Empirical formula given 𝐶𝐻2 . Step 2 M (𝐶𝐻2 )= 12+2(1)= 14 g.mol-1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 56 Step 3 ratio number = = =4 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 14 Step 4 𝐶𝐻2 × 4 = 𝐶4 𝐻8  Limiting Reaction  In a reaction between two substances, one reaction is likely to be used up completely before the other and this limit the amount of product formed.  The amount of limiting reactant will determine :  The amount of product formed.  The amount of other (excess) reactant used. Determining limiting reactants  Calculate the number of moles of each element.  Determine the ratio between reactants.  Determine limiting reactant using the ratio. NOTE: If one reactant is in excess, it means that there is more enough of it. If there are only two reactants and one is in excess, it means that the other is the limiting reactant. WORKED EXAMPLES. 1. A 8,4g sample of nitrogen reacts with 1,5g of hydrogen. The reaction is represented with the unbalanced equation below. 𝑁2 (𝑔) + 𝐻2 (𝑔) → 𝑁𝐻3 (𝑔) 1.1. Balance the equation. 1.2. Determine: 1.2.1. Which reactant is a limiting reactant? 1.2.2. The mass of ammonia that can be produced. SOLUTIONS 1. 1. 𝑁2 + 3𝐻2 → 2𝑁𝐻3 𝑚 8,4 1.2.1. n(𝑁2 ) = = = 0,3 𝑚𝑜𝑙 𝑀 28 𝑚 1,5 n(𝐻2 ) = = = 0,75 𝑚𝑜𝑙 𝑀 2

Downloaded from hlayiso.com 𝑁2 ∶ 𝐻2 1 ∶ 3 0,3 : x (cross multiply) ∴ 𝑥 = 0,9 𝑚𝑜𝑙 If all nitrogen is used, 0,9 mol of hydrogen is needed, however, only 0,75 mol of hydrogen is available. The hydrogen will run out first therefore hydrogen is the limiting reactant. 1.2.2. Because the hydrogen is the limiting reactant, it will determine the mass of ammonia produced: 𝐻2 ∶ 𝑁𝐻3 3 ∶ 2 0.75 : x ( cross multiply) ∴ x = 0,5 mol 𝑚 n(𝑁𝐻3 ) = 𝑀 m= (0,5)(17)= 8,5g  Percentage purity  Sometimes chemicals are not pure and one needs to calculate the percentage purity.  Only the pure component of the substance will react.  For impure sample of a substance : 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐩𝐮𝐫𝐞 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 Percentage purity = × 𝟏𝟎𝟎 % 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐢𝐦𝐩𝐮𝐫𝐞 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 Steps to determine the percentage purity  Determine moles of a product.  Balance the equation.  Determine the ratio between reactants and products.  Using the ratio, determine the number of moles of reactants.  Determine the mass of pure substance.  Calculate the percentage purity of the sample.  Percentage Yield  The percentage yield shows how much product is obtained compare to the maximum possible mass.  Some of the product may be lost due to evaporation into the surrounding air, or to a little being left in solution. This results in the amount of produced being less than maximum theoretical amount you would expect.  We can express this by the percentage yield : 𝐀𝐜𝐭𝐮𝐚𝐥 𝐲𝐢𝐞𝐥𝐝 Percentage yield = × 𝟏𝟎𝟎% 𝐓𝐡𝐞𝐨𝐫𝐢𝐭𝐢𝐜𝐚𝐥 𝐲𝐢𝐞𝐥𝐝  Percentage yield is usually determined using mass, but can also be determined with mol and volume.

Downloaded from hlayiso.com STEPS TO DETERMINE THE PERCENTAGE YIELD  Determine the moles of reactant  Balance the equation.  Using the ratio from the balance equation, determine the numbers of moles of product.  Determine the theoretical mass of product.  Calculate the percentage yield.

Downloaded from hlayiso.com WORKED EXAMPLE Emphases that for percentage yield the focus on actual yield and theoretical yield An excess of Pb(NO3)2 reacts with 0.75g of KI according to the reaction: Pb(NO3) + KI  PbI2 + KNO3 After titration and drying, a mass of 0.583g of PbI2 is measured. Determine the percentage yield of PbI2 SOLUTIONS Step 1: (balance chemical equation) Pb(NO3) + 2KI  PbI2 + 2KNO3 Step 2: (convert all given information to mole) 𝑚 0.75 𝑛(𝐾𝐼) = = = 4.52𝑋10−3 𝑚𝑜𝑙 𝑀 166 Step 3: (use stoichiometric ratio) From the balance equation (n)KI : (n)PbI2 2 :1 4.52𝑋10−3 : ? 1 (n)PbI2= 2 (4.52𝑋10−3 ) = 2.26𝑋10−3 𝑚𝑜𝑙 Step 4: (convert the number of moles to mass) 𝑚 𝑛= 𝑀 𝑚 2.26𝑋10−3 = 461 𝑚 = 1.04𝑔 Step 5: (percentage yield) 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑚𝑎𝑠𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑦𝑖𝑒𝑙𝑑 = 𝑋100 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑒𝑖𝑙𝑑 𝑚𝑎𝑠𝑠 0.583 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑦𝑖𝑒𝑙𝑑 = 𝑋100 1.04 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑦𝑖𝑒𝑙𝑑 = 56.1%

Downloaded from hlayiso.com LESSON 1- REVISION OF GRADE 10 STOICHIOMETRY 1.1 Calculate the molar mass for each of the following: 1.1.1 Ca (1) 1.1.2 MgCl2 (2) 1.1.3 H2SO4 (2) 1.1.4 Ca(NO3)2 (2) 1.2 Determine the percentage composition for each of the following substances: 1.2.1 MgCl2 (2) 1.2.2 CuSO4 (2) 1.3 Given 80 g of NaOH 1.3.1 Calculate the number of moles of 80 g of NaOH. (3) 1.3.2 Calculate the number of NaOH particles in 80 g of NaOH. (3) 1.4 Given 0,05 kg of CaCO3.Calculate 1.4.1 the number of moles of 0,05 kg of CaCO3. (2) 1.4.2 the number CaCO3 particles in 0,05 kg of CaCO3. (2) 1.5 Given 11,2 dm3 of nitrogen gas at STP. Calculate 1.5.1 The number of moles of 11,2 dm3 of nitrogen gas at STP. (2) 1.5.2 The number of N2 (g) molecules at STP. (2) 1.5.3 The number of nitrogen(N) atoms at STP (2) LESSON 2- MOLAR VOLUME OF GASES 2.1. Choose the correct answer 2.1.1 ….of a gas is the volume of one mole of a substance at STP. (2) A. Molar mass B. Molar volume C. Atomic weight D. Molar weight

Downloaded from hlayiso.com 2.1.2 Equal volumes of all gases at the same temperature and pressure (2) contain the same number of.. A. Protons B. Neutrons C. Electrons D. Molecules 2.2 2.2.1 Calculate the number of moles of water in 12 dm3 of water vapour at (3) STP. 2.2.2 Calculate the volume of hydrogen gas that combines with 12 cm3 of (3) chloride. LESSON 3- CONCENTRATION 3.1 3.1.1 Calculate the concentration of a solution of calcium chloride made by (3) dissolving 5.55g of dry CaCl2 crystals in enough water to make 750 cm 3 of solution. 3.1.2 What mass of copper (II) sulphate must be dissolved in 200ml water to (3) yield a 0.4 mol.dm-3 solution? 3.2 3.2.1 How many moles of chloride ions are present in 111 g of calcium (2) chloride? A 0,5 B2 C1 D 1,47 3.2.3 What amount of oxygen gas (in moles) contains 1,8 x 1022 molecules? (2) A 0,03 B 33,34 C 1,2 X 1024 D 1,08 X046 3.3 A solution of Mg(OH)2 is made up so that it will have a volume of 0,25 dm 3 and a (2) concentration of 0,5 mol.dm-3. The solution is made up using distilled water. 3.3.1 Define the term concentration. (1) 3.3.2 Name the solute used to make this solution. (5)

Downloaded from hlayiso.com 3.3.3 Calculate the mass of solid Mg(OH)2 required to make up the solution. (3) 3.4. 20 cm3 of a 0,1 mol·dm-3 nitric acid solution neutralises 25 cm3 sodium carbonate solution according to the following balanced equation: 2HNO3(aq) + Na2CO3(aq) 2NaNO3(aq) + H2O(ℓ) + CO2(g) 3.4.1 Write down the NAME of the salt formed in the reaction. (5) 3.4.2 Calculate the mass of sodium carbonate used to prepare 250 cm3 of (3) sodium carbonate solution of this concentration. LESSON 4-EMPIRICAL FORMULA 4.1 Methyl benzoate is a compound used in the manufacture of perfumes. It is found that a 5,325 g sample of methyl benzoate contains 3,758 g of carbon, 0,316 g of hydrogen and 1,251 g of oxygen. 4.1.1 Define the term empirical formula (2) 4.1.2 Determine the empirical formula of methyl benzoate (5) -1 4.1.3 If the molar mass of methyl benzoate is 136 g mol , what is its molecular (3) formula? 4.2 The action of bacteria on meat and fish produces a stinking compound called CADAVERINE. The compound has a composition of 58, 77% C; 13,81% H and 27, 42% N by mass. 4.2.1. Determine the empirical formula of CADAVERINE. 5 LESSON 5- STOICHIOMETRIC CALCULATIONS 5.1. 8 g of zinc react completely with an excess hydrochloric acid at STP to form hydrogen gas. 5.1.1 Calculate the mass of hydrochloric acid that reacts. (4) 5.1.2 Calculate the volume of hydrogen gas that is produced. (3) 5.2 Sodium thiosulphate, Na2S2O3(s), reacts with 200 cm3 of a hydrochloric acid solution, HCℓ(aq), of concentration 0,2 mol·dm-3 according to the following balanced equation: Na2S2O3(s) + HCℓ(aq) → NaCℓ(aq) + S(s) + SO2(g) + H2O(ℓ) 5.2.1 Define the term concentration of a solution (2) 5.2.2 Write down the balance chemical equation. (3) 5.2.3 Calculate the number of moles of HCℓ(aq) added to the sodium (3) thiosulphate. 5.2.4 Calculate the volume of SO2(g) that will be formed if the reaction takes (3) place at STP.

Downloaded from hlayiso.com LESSON 6- EXPERIMENTAL YIELD 6.1. An excess of Pb(NO3)2 reacts with 0.75g of KI according to the reaction: Pb(NO3) + KI  PbI2 + KNO3 After titration and drying, a mass of 0.583g of PbI2 is measured. 6.1.1 Determine the percentage yield of PbI2 (6) 6.2. It is found that 40 cm3 of a 0,5 mol.dm-3 sodium hydroxide solution is needed to neutralise 20 cm3 of the vinegar with a mass of 20,8 g. Vinegar is a solution of ethanoic acid in water. The balanced chemical equation for this reaction is: NaOH + CH3COOH  CH3COONa + H2O 6.2.1 Calculate the number of moles of sodium hydroxide that reacted. (3) 6.2.2 Calculate the mass of ethanoic acid present in the vinegar. (4) 6.2.3 Calculate the percentage (by mass) of ethanoic acid present in the (2) 6.3 37 g C react with an excess of oxygen and produce 65 dm3 CO2 gas at STP. The balanced equation for the reaction is as follows: C(s) + O2(g)  CO2(g) 6.3.1 Calculate the percentage purity of the carbon (5) 6.4 22,5 dm3 H2 and 30 dm3 N2 is placed in a container and produces 12 dm3 NH3. The balanced equation for the reaction is N2(g) +3 H2(g)  2NH3(g) 6.4.1 Calculate the percentage yield for this reaction. (5) LESSON 7- REACTING SOLUTIONS- STANDARD SOLUTION 7.1 25 cm3 of HCl of concentration 0.12 mol.dm-3 reacts with 28.4cm3 NaOH (3) to form water and NaCl. Calculate the concentration of sodium hydroxide. 7.2 30 cm3 of HCl of concentration 0.5 mol.dm-3 is diluted with 100ml of (3) water. Calculate the concentration of the diluted solution. 7.3 25 cm3 of BaCl2 reacts with 20 cm3 of a standard solution of 0.05 mol.dm-3 sulfuric acid to form barium sulphate and hydrochloric acid. 7.3.1 Define the term standard solution. (2) 7.3.2 Write down the balanced equation (3)

Downloaded from hlayiso.com 7.3.3 Calculate the concentration of barium chloride solution. (3) 7.3.4 (3) Calculate the mass of the precipitate (BaSO4) that is formed. 7.4 30 ml of an NaOH solution with concentration of 0,2 mol·dm -3 is (5) mixed thoroughly with 50 ml of an NaOH solution with a concentration of 0,3 mol·dm-3. Calculate the concentration of the final solution. 7.5 25 g NaCl is added to 100 cm of a NaCl solution with a concentration of (7) 0,5 mol·dm-3 and thoroughly mixed. Assume that the volume of the solution does not change. Calculate the concentration of the final solution. LESSON 8-LIMITING REAGENTS 8.1. 50 g of magnesium carbonate is added to 500 cm 3 of hydrochloric acid with a concentration of 0,75 mol.dm-3. The equation for the reaction is given below: MgCO3 (aq) + 2HCl (aq) → MgCl2 (aq) + CO2 (aq) +H2O The carbon dioxide gas is collected at STP. 8.1.1. What are the standard conditions used when conducting an experiment (2) at STP? 8.1.2. Determine which reactant is the limiting reactant. (6) 8.2. Consider the following balanced chemical reaction: 2HNO3 (aq) + Ca(OH)2 (aq) → Ca(NO3)2 (aq) + 2H2O (l) 25,0 ml of the nitric acid of concentration of 0,15 mol.dm -3 reacts with the calcium hydroxide solution. 8.2.1. How many moles of the acid are used? (2) 8.2.2. What mass of calcium hydroxide reacted with the nitric acid? (3) 8.2.3. 13,6 ml of calcium hydroxide solution was used. What was the (4) concentration of the calcium hydroxide solution? LESSON 9-MORE COMPLEX STOICHIOMETRIC CALCULATIONS 9.1 9.1.1 Define the term concentration. (2) 9.2 Eight (8) grams of Na2S2O3 is dissolved in water to prepare 500 cm3 of (3) solution. Calculate the concentration of the Na2S2O3 solution 9.3 A 10 g sample of a compound contains 2,66 g of potassium, 3,54 g of chromium and 3,81 g of oxygen.

Downloaded from hlayiso.com 9.3.1 Define the term empirical formula (2) 9.3.2 Determine the empirical formula of this compound. (7)

Downloaded from hlayiso.com ELECTROSTATICS Principle of conservation of charge. 𝑄1 + 𝑄2 𝑄𝑛𝑒𝑤 = 2 Principle of charge quantization. 𝑄 𝑛 = 𝑞𝑒 Prefix Conversion centi− (cC) ×10−2 milli− (mC) ×10−3 micro− (μC) ×10−6 nano− (nC) ×10−9 pico− (pC) ×10−12 COULOMB’S LAW The electrostatic force between two point charges is directly proportional to the product of the two charges and inversely proportional to the square distance between them. Mathematical representation: F α Q1Q2, Fα1/r2 Then F α Q1Q2/r2 𝑄 𝑄 F = k 𝑟1 2 2 F = force of attraction between objects (N)S k = Coulomb’s constant (9×109 N·m2·C−2) Q = object charge (C) r = distance between objects (m) EXAMPLES: 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? 𝑞 𝑞 F = k 12 2 𝑟 2𝑞 3𝑞 Fnew = k 11 22 ( 𝑟) 2 𝑞 𝑞 Fnew = 24k 𝑟1 2 2 Fnew = 24F 2. Two charges experience a force F when held a distance r apart. How would this force be affected if BOTH charges are doubled, and the distance is halved? 𝑞 𝑞 F = k 𝑟1 2 2 2𝑞 2𝑞 Fnew = k 11 22 ( 𝑟) 2 𝑞1 𝑞2 Fnew = 16k 𝑟 2

Downloaded from hlayiso.com Fnew = 16F 3. Two charges experience a force F when held a distance r apart. How would this force be affected if one charge is halved, and the distance is doubled? 𝑞 𝑞 F = k 𝑟1 2 2 1/2𝑞 𝑞 Fnew = k (2𝑟)12 2 𝑞 𝑞 Fnew = 1/4k 𝑟1 2 2 Fnew = 1/4F Graphical representation of Coulombs law: Electrostatic force and 1/d2

Downloaded from hlayiso.com Electrostatic force vs product of charges CALCULATIONS- Electrostatic force Electrostatic force is a vector quantity, therefore all vector rules can be applied: • Direction specific • Can be added or subtracted. The force can be calculated using 𝒒 𝒒 F = k 𝒓𝟏𝟐 𝟐 • Substitute charge magnitude only. • Direction determined by charge (like charges repel, unlike charges attract). • Both objects experience the EQUAL but opposite forces (Newton’s Third Law of Motion). 1-Dimensional Determine the resultant electrostatic force on QB.

Downloaded from hlayiso.com 2-Dimensional ELECTRIC FIELDS An electric field is 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 (+1C) would move if placed at that point. ELECTRIC FIELD LINE PATTERNS Field lines around the single point charge Field lines between two unlike charges

Downloaded from hlayiso.com Electric Field Lines between the two like charges ELECTRIC FIELD STRENGTH Electric field strength at any point in space is the electrostatic force per unit positive charge experienced by a positive test charge at that point. 𝐹 𝐸= 𝑞 E = electric field strength (N·C−1) F = force (N) q = charge (C) q is the charge that experiences the force. EXAMPLE: Charge B experiences a force of 2 N due to charge A. Determine the electric field strength at point B. DIRECTION: Direction that point in space ( X )would move IF it was positive. 𝑸 𝑬=𝒌 𝒓𝟐 E = electric field strength (N·C−1) k = Coulomb’s constant (9×109 N·m2·C−2) Q = object charge (C) r = distance between objects (m) Q is the charge that creates the electric field.

Downloaded from hlayiso.com EXAMPLE: Determine the electric field strength at point P due to charge Q. DIRECTION: Direction that point in space (X) would move IF it was positive. NOTE: Electric field strength is a VECTOR. All vector rules and calculations apply. ( linear addition, 2D arrangement, resultant vectors, etc.) ACTIVITIES SQUESTION 1 The diagram shows two small identical metal spheres, R and S, each placed on a wooden stand. Spheres R and S carry charges of + 8μC and - 4μC respectively. Ignore the effects of air. After R and S have been in contact and separated, a third sphere, T, of charge +1μC is now placed between them as shown in the diagram below.

Downloaded from hlayiso.com QUESTION 2 Two identical negatively charged spheres, A and B, having charges of the same magnitude, are placed 0,5m apart in vacuum. The magnitude of the electrostatic force that one sphere exerts on the other 144 X1\

Downloaded from hlayiso.com QUESTION 3 Three point charges, Qi, Q2 and Qs, carrying charges of +6 UC, -3 C and +5 UC respectively, are arranged in space as shown in the diagram below. The distance between Qs and Q: is 30 cm and that between Qs and Q2 is 10 cm. Qa = +5 uC 30cm Qet6 uc 10cm | I I 4 Q.=-3 uC | 3.1 State Coulomb's law in words. (2) 3.2 Calculate the net force acting on charge Qs due to the presence of Q: and Qo. (7) [9] QUESTION 4 Two identical neutral spheres, M and N, are placed on insulating stands. They are brought into contact and a charged a rod is brought near sphere M. Charged When the spheres are separated it is found that 5 x 10° electrons were transferred from sphere M to sphere N. 441 What is the net charge on sphere N after separation? (3) 4.2 Write down the net charge on sphere M after separation. (2) The charged spheres, M and N, are now arranged along a straight line, in space, such that the distance between their centres is 15 cm. A point P lies 10 cm to the right of N as shown in the diagram below. M N P Hae KA ----- eH LA 15cm | 10cm + 43 Define the electric field at a point. (2) 44 Calculate the net electric field at point P due to M and N. (6) [13] QUESTION 5 A very small graphite-coated sphere P is rubbed 7 with a cloth. It is found that the sphere acquires a charge of + 0,5 uC. 5.1 Calculate the number of electrons removed from sphere P during the 20cm charging process. (3) P Now the charged sphere P is suspended from a light, inextensible string. Another sphere, R, with a charge of — 0,9 UC, on an insulated stand, is brought close to sphere P. As a result sphere P moves to a position where it is 20 cm from sphere R, as shown. The system is in equilibrium and the angle between the string and the vertical is 7°. 5.2 Draw a labelled free-body diagram showing ALL the forces acting on sphere P. (3) 5.3 State Coulomb's law in words. (2) 5.4 Calculate the magnitude of the tension in the string. (5) [13]

Downloaded from hlayiso.com QUESTION 6 Two charged particles, Qi and Q», are placed 0,4 m apart along a straight line. The charge on Qi is + 2 x 10°C, and the charge on Q2 is — 8 x 10°C. Point X is 0,25 m east of Qi, as shown in the diagram below. Calculate the: s 61 Net electric field at point X due to the two charges (6) 6.2 Electrostatic force that a — 2 x 10° C charge will experience at point X (4) The — 2 x 10° C charge is replaced with a charge of — 4 x 10° C at point X. 6.3 Without any further calculation, determine the magnitude of the force (1) that the — 4 x 10° C charge will experience at point X. [11] QUESTION 7 Ceiling Two identical spherical balls, P and Q, each of mass 100 g, are suspended at the same point from a ceiling by means of identical light, inextensible insulating strings. Each ball carries a charge of +250 nC. The balls come to rest in the positions shown in the ho\20 diagram. 7.1 In the diagram, the angles between each string and the vertical are the same. Give a reason why the angles are the P Q same. (1) 72 State Coulomb's law in words. \ (2) 7.3 The free-body diagram, not drawn to scale, of the | forces acting on ball P is shown alongside. Calculate H the: \ 73.1 Magnitude of the tension (T) in the string (3) 7.3.2 Distance between balls P and Q (5) Fe [11] WIFg QUESTION 8 A sphere Qi, with a charge of -2,5 uC, is placed 1 m away from a second sphere Qz, with a charge +6 uC. The spheres lie along a straight line, as shown in the diagram below. Point P is located a distance of 0,3 m to the left of sphere Qi, while point X is located between Q; and Q2. The diagram is not drawn to scale. -2,5 uC +6 UC P Qa x oo ------ oe —-- a a \aa o * ° «—0,3m ie 1m > 8.1 Show, with the aid ofa VECTOR DIAGRAM, why the net electric field at point X cannot be zero. (4) 8.2 Calculate the net electric field at point P, due to the two charged spheres Q: and Qo. (6) [10]

Downloaded from hlayiso.com QUESTION 9 Asmall sphere, Q:, with a charge of + 32 x 10° C, is suspended from a light string secured toa support. A second, identical sphere, Qo, with a charge of — 55 x 10°C, is placed in a narrow, cylindrical glass tube vertically below Q:. Each sphere has a mass of 7 g. Both spheres come to equilibrium when Q;> is 2,5 cm from Q;, as shown in the diagram. Ignore the effects of air friction. support Qi =+32x109C + Glass tube — > 2,5cem Qo =-55x109C IC) a 9.1 Calculate the number of electrons that were removed from Q; to give it a charge of + 32 x 10° C. Assume that the sphere was neutral before being charged. (3) 9.2 Draw a labelled free-body diagram showing all the forces acting on sphere Q:. (3) 93 Calculate the magnitude of the tension in the string. (5) [11] QUESTION 10 10.1 Define electric field at a point in words. (2) 10.2 Draw the electric field pattern for two identical positively charged spheres placed close to each other. (3) 10.3 A-30 uC point charge, Qi, is placed at a distance of 0,15 m from a + 45 UC point charge, Qz, in space, as shown in the diagram below. The net electric field at point P, which is on the same line as the two charges, is zero. Qe =+ 45 uC Qi =- 30 uC P nn et e 0,15m x >! Calculate x, the distance of point P from charge Q:. (5) [10]

Downloaded from hlayiso.com WORKSHEET 1 (20 Min) 1.1 Two identical small metal spheres on insulated stands carry equal charges and (2) are a distance d apart. Each sphere experiences an electrostatic force of magnitude F. The spheres are now placed a distance d apart. The magnitude of the electrostatic force each sphere now experiences is: A. 1/2F B. F C. 2F D. 4F 1.2 Two charged objects on insulated stands have charges of 3Q and 5Q (2) respectively. The Objects are a distance R apart and exert a force F on each other. They are moved so that they are now 1/3 R apart. What is the new force that they exert on each other? A. F B. 1/3 F C. 3F D. 9F 2 (3) Calculate the new charge on each of the spheres in the following diagram after they have touched. _______________________________________________________

Downloaded from hlayiso.com 3. Two metal spheres Q1 and Q2 on insulated stands, carry charges of +4nC and −12nC respectively. 3.1 Using free body diagrams, show the electrostatic force on each sphere. (2) 3.2 State the Principle of conservation of charge. (2) 4 4.1 A neutral object obtains a charge by gaining 1 800 electrons. What is the (2) charge that the object obtains? 4.2 Two metal spheres Q1 and Q2 on insulated stands, carry charges of +8nC and −10n respectively. The two spheres are now brought together to touch each other. They are then placed back in their original positions. (3) 4.2.1 Calculate the new charge on each sphere. 4.2.2 What quantity of charge is transferred between the two spheres during (2) contact?

Downloaded from hlayiso.com WORKSHEET 2 - COULOMB’S LAW (30min) 1. Two identical metal spheres, A and B have charges of +8μC and -16μC respectively. They are suspended from a horizontal wooden pole and placed a distance d apart from each other, as shown below. 1.1 Sphere B experiences an electrostatic force. In which direction does (1) the force act? 2. The spheres are touched together and then separated and placed the same distance d apart from each other. 2.2.1 When they touched each other, were electrons removed from or (2) added to A? Give a reason for your answer. (3) 2.2.2 Sphere B now experiences a force of 1,6 N. Calculate the distance d. 3. (3

Downloaded from hlayiso.com 4. 4.1 Two small metal spheres have charges of +2pC and +5pC respectively. If the force between the two charges is 9 x 10 -12 N, calculate the distance between them. (4) 5. Two charged spheres X and Y are placed 0,04m apart, as shown in the diagram below. . 5.1 Calculate the electrostatic force that Y experiences due to charge X 5.2 Thus state the force that X experiences due to Y 6 Three point charges are in a straight line. The charges and distances between them are shown.

Downloaded from hlayiso.com 1.6. What is the resultant electrostatic force on Q2 as a result of the other two charges? 7. Three point charges are placed in close proximity to one another. The charges and distances between them are shown. What is the resultant electrostatic force on Q2 as result of the other two charges?

Downloaded from hlayiso.com WORKSHEET 3 - THE ELECTRIC FIELD ( 25min) 1. Define an electric field at a point. 2. 2.1 Draw the electric field pattern for a single positive point charge. 2.2 draw the resultant electric field pattern of the following charges 3. A charge of 5nC experiences a force of 4 x10-6 N at a point in an electric (3) field. Calculate the strength of the electric field. 4. Two charges, Q1 of +6nC and Q2 of -7,5nC are separated by a distance of 150 mm. What is the electric field strength at point X, which is 70 mm from Q1 and (6) 80 mm from Q2?

Downloaded from hlayiso.com 5 A and B are two small spheres separated by a distance of 0,70 m. Sphere A carries a charge of +1,5 x 10-6 C and sphere B carries a charge of -2,0 x 10-6 C. P is a point between spheres A and B and is 0,40 m from sphere A, as shown in the diagram below. 5.1 Calculate the magnitude of the net electric field at point P. (4) (3) 5.2 A point charge of magnitude 3,0 x 10-9C is now placed at point P. Calculate the magnitude of the electrostatic force experienced by this charge.