Gr11 PhysicalScience SupportDocument2017

Downloaded from hlayiso.com Physical Sciences Grade 11 Document 2017 KWAZULU-NATAL DEPARTMENT OF EDUCATION PHYSICAL SCIENCES GRADE 11 SUPPORT MANUAL Page 1 of 78

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 Coulomb’s Law The magnitude of the electrostatic force exerted by the Point charges on each other is directly proportional to the product of the magnitudes of the charges, and inversely proportional to the square of the distance between them. $% $& Mathematically: 𝐹 = 𝑘 '& Where: • F is the electrostatic force in newtons (N). • Q1 and Q2 are charges in coulombs (C). • r is the distance between the two charges in metres (m). • k is the proportionality constant (Coulombs Law constant) with the value of 9x109 N∙m2.C-2. Worked example 1: [in one dimension (1D)] Two point-like charges carrying charges of +3 nC and −5 nC are 2m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive. Answer Step 1: Determine what is required. We are required to find the force between two point charges given the charges and the distance between them. Step 2: Determine how to approach the problem We can use Coulomb’s Law to find the force. F = kQ1Q2 r2 Step 3: Determine what is given We are given: • Q1 = +3 nC = +3 × 10−9C • Q2 = -5 nC = −5 × 10−9C • r = 2m Page 2 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 We know that k = 9 × 109 N · m2 · C−2. Draw the diagram. Q1 =+3 X 10-9 Q2 = -5 X 10-9 Step 4: Check units and convert them. • Q1 = +3 nC = +3 × 10−9C • Q2 = -5 nC = −5 × 10−9C Step 5: Determine the magnitude of the force. Using Coulomb’s Law we have. F = kQ1Q2 r2 = (9, 0 × 109)(3 × 10−9)(5 × 10−9) (2)2 = 3, 38 × 10−8 N Thus the magnitude of the force is 3, 38 × 10−8N. However, since both point charges have opposite signs, the force will be attractive. Worked example 2 [in one dimension (1D)] The charges on two small objects, A and B, are -8µC and -2µC respectively. A and B are 340 cm apart. 1.1 Calculate the force exerted by the charges on each other. Solution $ $ 𝐹 = 𝑘 % & & ' = 9x109 x (-8 x 10-9) x (-2 x 10-9) (340 x 10-2)2 = -1, 25 x 10-8N = 1, 25 x 10-8N repulsion Page 3 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 Worked example 3 [in one dimension (1D)] Calculate the distance between two point charges of + 6 nC and – 3 nC respectively if they exert a force of 2 x 10 – 7 N on each other. +6nC • • -3nC F= 2 x 10-7 N Solution $ $ 𝐹 = 𝑘 % & & ' 2 x 10-7 = (9 x 109) (6 x 10-9)3 x 10-9 r2 = 0, 81 r = 0, 9 m Worked example 4 [in two dimensions (2D)] - Nov 2014(Question 9) Two point charges of +2 μC and +3 μC are placed a distance of 0, 2 m apart. P is a point on the line joining the two charges, a distance of x m from the 3 μC charge such that the net electric field at point P is zero. 4.1 Define the term electric field at a point in words. (2) 4.2 Calculate the distance x. (7) A -4 μC charge is now placed a distance of 0,1 m from the +3 μC charge as shown in the sketch below. 4.3 Calculate the magnitude of the electrostatic force experienced by the +3 μC (5) charge due to the presence of the other two charges. [14] Page 4 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 SOLUTION Electric field at a point is defined as the force acting per unit charge. OR It is the force experienced by a unit positive charge placed at that point. Page 5 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 ELECTRIC FIELD An electric field is a region in space in which an electric charge experiences a force. The direction of the electric field at a point is the direction that a positive test charge would move if placed at that point. Some important points to remember about electric fields: • They originate and end perpendicularly to the surface of the charged objects. • Field lines never cross. • They are most dense (closer to each other) where the field is the strongest and is least dense (further from each other) where the field is the weakest. • They surround the charged object in three dimensions. We only draw a few lines in one plane Electric field lines around a positive point charge: Electric field lines around a negative point charge: Page 6 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 Electric field lines of two equal but opposite charges: Electric field lines of two equal positive charges: Electric field lines of two equal negative charges: Electric field strength at a point The electric field strength at a point is the electrostatic force experienced per unit positive charge placed at that point. The test charge placed at a point in electric field will experience a force;; the magnitude of the force experienced will depend on the distance of the test charge(q) away from the charge(Q) setting the field. 𝑓𝑜𝑟𝑐𝑒 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ = 𝑐ℎ𝑎𝑟𝑔𝑒 Page 7 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 8 In symbols: E = $ 8 Solve problems using the equation : E = $ Unit : • The unit for electric field strength (E) is newton per coulomb( N∙C-) if the electrostatic force F acts on the charge in Newton (N), the charge Q is in coulomb (C). Direction: • Electric field strength is a vector quantity • A direction of electric field (by conversion) is the direction of the electrostatic force that a positively charged particle will experience at that point. • The positively charged particle will thus move in the direction of the field and a negatively charged particle will move against the field. The force experienced by a test charge when placed in an electric field is given by: 𝐹 = 𝑞𝐸 Worked example 1 If the magnitude of the electric field strength (intensity) is 3 x 10 6 N • C–1 at a point, calculate the magnitude of the force acting on a charge of – 7 nC placed at that point. Solution 𝐹 = 𝑞𝐸 = 7 × 10-9 × 3 × 106 = 0.021 N Electric field strength at a point due to a number of point charges ?????? From Coloumb’s Law: Page 8 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 ;<= F = (If we make the one charge Q and the other q.) >& From the definition of electric field strength: F = Eq ?$ Therefore, the electric field can be calculated as 𝐸 = & (where Q is the charge ' setting the field). Worked Example 2 Calculate the electric field strength 30cm from a 5nC charge. Solution ?$ 𝐸 = & ' = (9 x 10-9) (5 x 10-9 (0, 3)2 = =5 × 10-16 N∙C-1 Worked example 3 R and S are two points in the electric field of a small negatively charged sphere Q, Two charges of Q1 = +3nC and Q2 = −4nC are separated by a distance of 40cm. What is the electric field strength at a point that is10cm from Q1 and 30cm from Q2? The point lies between Q1 and Q2. Page 9 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 Solution ?$ 𝐸 = & ' = (9 × 109) (3 × 10−9) (0, 1)2 = 2, 70 × 103 N∙C−1 Then for Q2: ?$ 𝐸 = & ' = (9 × 109) (4 × 10−9) (0.3)2 = 4, 00 × 102 N∙C−1 Add the two electric fields because both are in the same direction. The field is away from Q1 and towards Q2. Therefore, Etotal = 2, 70 × 103 + 4, 00× 102 = 3, 10 × 103N∙C−1 ACTIVITIES ACTIVITY 1 Two pith balls shown in the diagram below have a mass of 1.0 g each and have equal charges. One pith ball is suspended by an insulating thread. The other charge is brought to within 3.0 cm. of the suspended ball (r = 0.03 m). The suspended pith ball is deflected from its rest position until the thread forms an angle of 30° with the vertical. At this angle, the ball is in equilibrium. 1.1 Draw a force diagram depicting the forces acting on the suspended ball (3) Page 10 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 1.2 Calculate Fg 1.3 Calculate Fe 1.4 Calculate the charge on the pith balls. ACTIVITY 2 Two charges P and Q with charges +4 nC and -2 nC respectively, are placed 8 cm apart. A point charge X is placed 2 cm from Q as shown in the sketch below. 2.1 Define the term electric field at a point. (2) 2.2 Draw the electric field pattern between two oppositely charged particles. (3) 2.3 Calculate the net electric field strength at point X due to P and Q (7) 2.4 Calculate the magnitude and direction of the electrostatic force on an (3) electron if it is placed at point X. ACTIVITY 3 (NOV 2016 EC) Two metal spheres charges, Q1 and Q2, on insulated stands, carry charges of +6 x 10-6 C and -4 x 10-6 C respectively are separated by a distance of 3 m from their centres. A point charge at P is 1, 2 m from Q2 in the same plane as indicated in the diagram below. 3.1 Define Coulomb’s law in words (2) 3.2 Draw the electric field pattern between the two charges. (3) 3.3 Calculate: 3.3.1. The electrostatic force exerted by Q1 on Q2 (5) 3.3.2 Calculate the electric field strength at point P as a result of (7) Page 11 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 Q1 and Q2 3.3.3. How many electrons are in excess in Q2 (2) OHM’S LAW States that: The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature This Law describes the relationship between the current flowing through a conductor and the potential difference (p.d.) across its ends. A V By increasing the number of cells in the circuit we are able to increase the p.d. across the ends of the conductor. This increased p.d. is measured using a voltmeter connected across the ends of the conductor. This changes the current flowing through the conductor, which is measured using an ammeter. If an experiment was conducted using 5 cells, each of emf 1,5V, the following is a possible set of results that could be obtained. p.d. (V) I (A) 1,5 0,075 3,0 0,15 4,5 0,225 6,0 0,3 7,5 0,375 A graph of p.d. (x-axis) vs current (y-axis) gives the following Page 12 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 0,375 current(A) 0, 3 0,225 5 0,15 0,075 0 1, 3,0 4, 6, 7, 5 5 0 5 Potential difference (V) From the above graph, the potential difference across the ends of the conductor is directly prportional to the current flowing through the conductor. NB: Potential difference is an independent variable The folowing graph is drawn: AB The gradient of this graph is a constant and yields: 𝑅 = AC Page 13 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 • R is the resistance of the conductor. • Resistance is measured in Ohms (Ω) Resistance Resistance in a wire is the opposition of a wire to the flow of electricity. It is caused by collisions between the electrons and the atoms in the wire. The hotter the wire, the more chance there is of a collision. Therefore hot wires have more resistance. One Ohm: A conductor has a resistance of one ohm if a current of one ampere passes through it when a p.d. of one volt is maintained across its ends. Ohmic and non-Ohmic conductors: Conductors that obey Ohm’s Law are said to be Ohmic conductors, for example a piece of nichrome wire. Conductors that do not obey Ohm’s Law are said to be non-Ohmic conductors, example diodes. A graph of potential difference vs current flowing for a diode may look like this: Potential difference Current Factors affecting the resistance of a conductor: • Length of the conductor. The longer the conductor the greater the resistance. • Cross sectional area. The greater the cross sectional area of a conductor the lesser the resistance. • Temperature. An increase in temperature of a conductor increases resistance. • Kind of material used Page 14 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 WORKED EXAMPLE 1(NOV 2013 DBE) The two graphs below represent the relationship between potential difference and current in a metal wire at two different constant temperatures, T and T . 1 2 1.1 Calculate the resistance of the metal wire at temperature T . (3) 1 1.2 Which graph was obtained at the higher temperature? Give a reason for the answer. (3) 1.3 The metal wire is an ohmic conductor. Justify this statement by 1.4 Calculate the power dissipated in the metal wire when the current in it is 25 mA at temperature T . (3) 2 H3) referring to the graphs. (1) Page 15 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 Series and Parallel Circuits Series Circuits In a series circuit, the electrons in the current have to pass through all the components, which are arranged in a line. Consider a typical series circuit in which there are three resistors of value R1, R2, and R3. The values may be the same, or different. There are two key points about a series circuit: • The current throughout the circuit is the same • The voltages add up to the battery voltage. Therefore: VT = V1 + V2 + V3 Page 16 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za Physical Sciences Grade 11 Document 2017 From Ohm’s Law we know: • VT = IRT;; • V1 = IR1;; • V2 = IR2;; • V3 = IR3 Þ IRT = IR1 + IR2 + IR3 Therefore: RT = R1 + R2 + R3 Parallel Resistors Parallel circuits have their components in parallel branches so that an individual electron can go through one of the branches, but not the others. The current splits into each of the branches. Look at this circuit: Page 17 of 78 https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 18 of 78 In this case, the current will split into three pathways. For a parallel circuit we know two things: • The voltage across each branch is the same • The currents in each branch add up to the total current. From this we can write: Itot = I1 + I2 + I3 From Ohm’s Law, I = V/R, we can write: I T = V ;; I1 = V;; I2 = V;; I3 = V RT R1 R2 R3 Þ V = V + V + V RT R1 R2 R3 Þ This is true for any number of parallel resistors. Relationship between Energy Transferred and the Current From the experimental investigation it was found that the energy transferred is directly proportional to the current squared. energy transferred 2 a I (Q) Energy Transferred (Dt) Energy Transferred (Dt) R constant R constant t constant t constant (oC) (oC) 2 2 2 Current (I) (A) Current (I ) (A ) Relationship between Energy Transferred and the Resistance From the experimental investigation it was found that the energy transferred is directly proportional to the resistance. Energy a R transferred(Q) Energy Transferred (Dt) I constant t constant (oC) Resistance (R) (Ω Relationship between Energy Transferred and the Time https://tvetpapers.co.za Page 18 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 19 of 78 From the experimental investigation it was found that the energy transferred is directly proportional to the time. energy a t transferred (Q) Energy Transferred (Dt) I constant R constant (oC) Time (mins) From these experiments we can conclude the following: W a I2 W a R W a t Combining these relationships we have I2 W a or W = k I2 Rt Rt In SI units, the value of k is 1 Hence W = I2 Rt W Electric work done in J I Current in A R Resistance in Ω t Time in s Using previous equations, V = IR, W = VQ and Q = It, we can write the above equation in many forms together with equations for power ( P = W ) Work Done Power I2 W = P = I2 R Rt W = VIt P = VI V2t P = V2 W = R R ACTIVITIES MULTIPLE CHOICE QUESTIONS 1.1 In the circuit shown, the resistance of the ammeter, the battery and the connecting wires are negligible. The voltmeter has a very high resistance. The ammeter reads 1,5 A. What is the reading on the voltmeter? https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 20 of 78 A 4 V B 6 V C 8 V D 12 V 1.2 What will happen to the voltmeter and ammeter readings if the switch S is closed? voltmeter reading ammeter reading A becomes zero remains the same becomes zero increases B remains the same decreases C decreases increases D 1.4 Which one of the circuits shown in the diagram below can be used to determine the resistance of R? A I B II C III D IV 1.5 A battery is connected to a small lamp. If a second identical lamp is connected in series with the first, what will happen to the current provided by the battery? A it will decrease B it will remain the same C it will increase D it will become zero 1.6 Two light bulbs are marked 230 V;; 75 W and 230 V;; 150 W. If the first bulb has a resistance R, then the resistance of the second is … A ¼ R B ½ R C 2 R https://tvetpapers.co.za Page 20 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 21 of 78 D 4 R 1.7 Consider the electric circuit shown in the diagram at the right. The reading on the voltmeter V is ... A 0 V B 2 V C 4 V D 6 V 1.7 Gugu connected four identical light bulbs in a circuit as shown to the right. She observes that the brightness of bulb A varies if some of the switches are closed. In which case will bulb A burn brightest? A S1 closed with S2 and S3 open B S1 and S2 closed with S3 open C S1, S2 and S3 closed D S1, S2 and S3 open The information given in the box refers to questions 8 and 9: The battery in the circuit has an internal resistance r and an emf E. The resistance of the ammeter and the conductors is negligible and the voltmeter has a high resistance https://tvetpapers.co.za Page 21 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 22 of 78 1.8 When the switch S is closed, the … voltmeter reading ammeter reading A decreases decreases B decreases increases C increases decreases D increases increases 1.8 When the switch S is closed the power in the resistors R1 and R2 … power in power in R1 R2 A decreases decreases B decreases increases C increases decreases D increases increases 1.10 Two resistors of resistance 2R and 3R are connected in series with a battery, which has an emf of E and an internal resistance R. What is the potential difference across the resistor of resistance 2R? A E/3 B 2E/5 C E/2 D 2E https://tvetpapers.co.za Page 22 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 23 of 78 Section B ACTIVITY 1 The ammeter and the battery have negligible internal resistance as indicated in the diagram below. The potential difference across the battery is 12 V. 12V A 9Ω 2Ω 18Ω V Calculate: 1.1 the equivalent resistance of the circuit 1.2 the voltmeter reading 1.3 the heat produced in 1 minute in the 2 Ω resistor. ACTIVITY 2 A circuit is connected as shown in the diagram below. The battery has an emf of 12 V and a significant internal resistance. The reading on the ammeter is 3 A. Calculate the: 2.1 reading on voltmeter V1 2.2 reading on the voltmeter V2 2.3 energy transferred in the 3 Ω resistor in 2 minutes https://tvetpapers.co.za Page 23 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 24 of 78 ACTIVITY 3 In the given circuit, the battery has an emf of 18 V. The resistor Z is of unknown value. The ammeter reads 0,5 A Emf = 18V A 0,5A 60 Ω 40 Ω Z 3.1 What is meant by the emf of a battery? 3.2 Calculate the equivalent resistance of the parallel branch. 3.3 Calculate the potential difference across the parallel branch. 3.4 If the potential difference across Z is 3V, calculate the value of Z. 3.5 Calculate the power developed in the 60 Ω resistor. https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 25 of 78 ACTIVITY 4 The circuit shown below consists of a number of resistors connected to a 24,0V battery of negligible internal resistance. The resistor marked R is a variable resistor with values of resistance from zero to infinity. 7,0 Ω 24,0 V 4,0 Ω 12,0 Ω R The variable resistor R is set so that the battery supplies power to the resistors at a rate of 60,0W. 4.1 Define the resistance of a conductor. 4.2 Calculate the potential difference across the 7,0 Ω resistor. 4.3 Find the value of R. ACTIVITY 5 (EC- N0V 2015) In the circuit below, the resistance of the battery, ammeter and connecting wires can be ignored . The power of the 8 Ω resistor is 0, 5 W. 5.1 Calculate the reading on the: 5.1.1 voltmeter (V) (8) https://tvetpapers.co.za Page 25 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 26 of 78 5.1.2 ammeter (A) (4) 5.2 A television is labeled: 240 V;; 750 W. 5.2.1 Calculate the resistance of the television’s resistor. (4) 5.2.2 Calculate the costs of using the television for 6 hours if electricity costs R1,04 per kWh. (3) [19] ELECTROMAGNETISM • Magnetic field associated with current-carrying conductors 1. Magnetic field around a straight conductor Magnetic field lines produced by the current form circles with, with the wire at the centre (illustrated below). Right Hand rule is used to determine the direction of the field. The current going into the paper (into the page) is represented by and the current leaving the page (out of page) is represented by . https://tvetpapers.co.za

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 27 of 78 2. Magnetic field around a circular wire (loop) Magnetic field lines closest to the wire are circular. Magnetic field lines at the centre of the loop are straight and parallel to each other. 3. Magnetic around a solenoid (cylindrical coil of wire) • Magnetic field lines form closed loops. • On the outside of the solenoid, the direction of the field is from north to south ( like in a bar magnet). • Inside the solenoid the direction of field is south to north. • The polarity of the current carrying solenoid is determined by Right Hand Rule. • Thumb is pointing at the direction of the field. • Curved fingers are pointing in the direction of conventional current. FARADAYS LAW The magnitude of the induced emf across the ends of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor. 𝜟𝜱 ε = -N 𝜟𝒕 • ε is the induced emf • ΔΦ is the change magnetic flux. • N is the number of loops https://tvetpapers.co.za Page 27 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 28 of 78 • Emf has a negative sign because it opposes the change in flux Current induced by changing magnetic field • When there is a change in magnetic field around the conductor, current is induced. • Current is induced when there is relative motion between the coil and the magnet • No current is induced if magnet is held stationary in the coil. • The direction of induced current is determined by Right Hand Rule. • If thumb points in the direction of field, curved fingers point in the direction of current and vice versa. • The induced current flows in a direction so as to set up the magnetic field to oppose the change in magnetic flux. Figure 1: Magnet moving into the solenoid • The distance between the magnetic and the solenoid decreases and so magnetic flux inside the solenoid increases. • The magnetic field of the magnet is towards left. • To oppose this increase towards left, the field produced by induced current is towards right. Figure 2 Magnet pulled out from the solenoid • The distance between the magnet and the solenoid increases and so the magnetic flux inside the solenoid decreases. • To oppose this decrease, the field produced by induced current flowing in the direction indicated is towards left. https://tvetpapers.co.za Page 28 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 29 of 78 Factors influencing the magnitude of the induced current • Strength of magnetic field- the stronger the magnet, the greater the induced current. • The number of turns in the coil- the greater the number of turns the greater the induced current. • The speed at which magnet and coil are moved relative to each other. The higher the speed, the greater the induced current. Magnetic flux Magnetic flux (Φ )is the product of the component of magnetic field (B cos θ) perpendicular to a surface and the area(A) of the surface the field passes through . Φ =BA cos θ Θ is the angle between the magnetic field and the normal to the cross sectional area. Unit of magnetic flux is weber (Wb). 1 Wb = 1 T Worked example 1 A 200-turn circular coil is placed in a magnetic field such that the field is perpendicular to the surface of each loop of the coil at all times. As the coil rotates, the magnetic field changes at a constant rate -2 from 0, 22 T to 0, 42 T in 3, 2 x 10 s. The emf induced in the coil during this time is -15, 2 V. 1.1 State Faraday's law of electromagnetic induction in words. (2) https://tvetpapers.co.za Page 29 of 78

Downloaded from hlayiso.com https://ultradeep.co.za GRADE 11 Page 30 of 78 1.2 Calculate the: 1.2.1 Change in magnetic flux through the circular loop. (4) 1.2.2 Radius of the coil. (4) 1.3 The coil now rotates in the opposite direction and the magnetic field changes from 0,42 T to 0,22 T in the same time interval. Write down the induced emf. (1) Solution 1.1 The magnitude of the induced emf across the ends of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor. AG 1.2 .1 ε = -N AH AG -15, 2 = - (200) I,K L MNOK 𝛥𝛷= 2.43 X 10-3 Wb or 2, 43 x 10-3 v.s-1 1.2.2 𝛥𝛷 = 𝐵𝑓 − 𝐵𝑖 𝐴 𝑐𝑜𝑠 𝜃 2.432 X 10-2 = (0, 42 – 0, 22) A cos 0° A = 0,012 m2 Area of circle = pr2 0.012 = pr2 r = 6, 22 x 10-2 m 1.3 15, 2 V ACTIVTIES ACTIVITY 1 A single circular loop of wire, 12 cm in diameter, is placed in a 0, 6 T magnetic field. It is removed from the magnetic field in 0, 04 s. 1.1 Calculate: 1.1.1 The flux which is linked to this coil. (4) 1.1.2 The average induced emf. (4) 1.2 How does the emf change if … (Write only INCREASES, DECREASES or REMAINS THE SAME) 1.2.1 the magnetic field strength changes to 0,5 T? (1) 1.2.2 the coil is removed from the field in 0,02 s? (1) https://tvetpapers.co.za Page 30 of 78

Downloaded from hlayiso.com GRADE 11 Page 31 of 78 ACTIVITY 2 2.1 Consider a flat square coil with 5 turns. The coil is 0,50 m on each side and has a magnetic field of 0,5 T passing through it. The plane of the coil is perpendicular to the magnetic field: the field points out of the page. Use Faraday's Law to calculate the induced emf, if the magnetic field is increases uniformly from 0,5 T to 1 T in 10 s. Determine the direction of the induced current. 2.2 Consider a solenoid of 9 turns with unknown radius, r r . The solenoid is subjected to a magnetic field of 0,12 T. The axis of the solenoid is parallel to the magnetic field. When the field is uniformly switched to 12 T over a period of 2 minutes an emf with a magnitude of − 0, 3 V is induced. Determine the radius of the solenoid. 2.3 Consider a circular coil of 4 turns with radius 3 × 10−2 m. The solenoid is subjected to a varying magnetic field that changes uniformly from 0, 4 T to 3, 4 T in an interval of 27 s. The axis of the solenoid makes an angle of 35° to the magnetic field. Find the induced emf. Page 31 of 78

Downloaded from hlayiso.com GRADE 11 Page 32 of 78 ACTVITY 3 3.1 State Faraday's Law of electromagnetic induction in words and write down a mathematical relationship. 3.2 Describe what happens when a bar magnet is pushed into or pulled out of a solenoid connected to an ammeter. Draw pictures to support your description. 3.3 Use the Right Hand Rule to determine the direction of the induced current in the solenoid below ACTIVITY 4 (Nov 2014 DBE) In the diagram below a bar magnet is being pushed into a coil. The current induced in the coil is in the direction indicated. 4.1. Write down the polarity (north pole or south pole) of the end of the coil facing the bar magnet, as the bar magnet approaches the coil. (2) 4.2 Which end of the bar magnet is approaching the coil? Write down only NORTH POLE or SOUTH POLE (1) 4.3. Write down what will be observed on the galvanometer if the bar magnet is held stationary inside the coil. Give a reason for the answer. (2) Faraday's law of electromagnetic induction plays a very important role in the generation of electricity. 4.4 Write down Faraday's law of electromagnetic induction in words. (2) A coil of 100 turns, each of area 4,8 x 10-4 m2, is made from insulated copper wire. The coil is placed in a uniform magnetic field of 4 x 10-4 T in such a way that the angle between the magnetic field and the normal to the plane of the coil is 30°. The coil is then rotated so that the angle changes to 70° in a time interval of 0,2 s. Calculate the: 4.5 Magnitude of the emf induced in the coil (5) 4.6 Current induced in the coil if it has an effective resistance of 2 Ω (3) [15] Page 32 of 78

Downloaded from hlayiso.com GRADE 11 Page 33 of 78 MATTER AND MATERIALS IDEAL GASES AND THERMAL PROPERTIES MATTER: Is anything that occupies space and has mass - Matter is made up of very small particles in constant motion - KINETIC THEORY determines the movement of particles in the three phases of matter: 1. Solids- particles vibrate in fixed position 2. Liquids- particles move around freely in a confined space 3. Gases- particles move rapidly and are far apart Gas particles have enough kinetic energy enabling them to move randomly  A true gas such as hydrogen, oxygen, nitrogen, carbon dioxide, sulphur dioxide etc, will only behave ideally at: 1. High temperature 2. Low pressure Reason: at high temperature • kinetic energy is very high • molecules move extremely fast resulting in intermolecular forces decreasing as the temperature increases • Particles moving faster result in more collisions with each other and with the walls of the container, causing volume to increases  Real gases deviate from ideal gas behavior under these conditions: 1. Low temperature 2. High pressure Reason: at low temperature • Kinetic energy of particles is lower • Intermolecular forces exist between particles • Particles are attracted to each other and gas liquifies At high pressure • gas molecules are close to each other • intermolecular forces increase as there are now repulsion forces between molecules. There scientists who studied gas laws include the following: 1. Boyle: studied pressure and volume relationship 2. Charles: studied temperature and volume relationship 3. Gay- Lussac: studied pressure and temperature relationship Avogadro: studied volume and amount relationship Page 33 of 78

Downloaded from hlayiso.com GRADE 11 Page 34 of 78 Boyle’s law: States that the volume of an enclosed mass of a gas is inversely proportional to the pressure provided that the temperature remains constant Worked example: 1. An amount of gas has a volume 23dm3 at a pressure of 150kPa. The pressure on the gas is increased to 200kPa while the temperature remains constant. Calculate the new volume. p1V1=p2V2 150x23 = 200 V2 V2= 17,25 dm3 QUESTIONS PART A 1. A fixed amount of gas is in a cylinder with an airtight piston. The piston is pushed down, to halve the volume of the enclosed gas. (Temperature remains constant) Gas a) Would the pressure of the enclosed gas decrease, increase or would it remain the same? Use the Kinetic Theory of Matter to explain your answer. b) If the original pressure and volume of the gas was 100 kPa and 100 dm3 respectively, calculate the pressure of the gas. c) Name and state the law applied to answer question 1 b). d) Draw a graph of pressure versus volume to illustrate this relationship as stated in this Law (question c). 2. A fixed number of moles of helium gas are in a closed container at STP. The container has a capacity of 50 cm3. If the pressure remains constant, what will be the volume of the helium gas if the temperature is doubled? 3. A diver dives down into the sea, where the temperature is 5oC and the pressure is 150 kPa. She releases a bubble of air from her lungs. The volume of the bubble is10 cm3. Page 34 of 78

Downloaded from hlayiso.com GRADE 11 Page 35 of 78 Calculate the volume of the bubble when it reaches the surface of the sea, where the temperature is 30oC. 4. A fluorescent advertising sign outside a shop has a capacity of 20 dm3. During the night, when the temperature is 12oC, the pressure in the sign is 120 kPa. The sign can withstand a pressure of 140 kPa before breaking. If the temperature rises to 32oC on a hot day, will the sign be able to withstand the pressure? 5. Name two differences between ideal gases and real gases. 6. Which real gas behaves most like an ideal gas? Explain your answer. 7. Copy the following pV versus p graph pV for an ideal gas and complete it by using a dotted line to show the deviation of a real gas. Briefly explain the deviation. p 8 Calculate the number of moles of an ideal gas present in a 100dm3 closed container at a pressure of 150kPa and a temperature of 100oC. 9 A mass of 64 g of oxygen gas is in a closed gas cylinder, at a pressure of 120kPa and a temperature of 25oC. Calculate the volume of the cylinder. 10. A certain gas with a volume of 5,6 dm3 is at STP. a) Calculate the number of moles of this gas present in the container. b) To what temperature must the gas be heated to expand to a volume of 22,4 dm3? (Assume the pressure remains constant) 11 A steel cylinder contains 1000cm3 gas at a pressure of 250kPa. What volume of gas will escape if the cylinder is opened to let out the gas? The local atmospheric pressure is 80kPa. Assume the temperature remains constant. 12. An enclosed gas occupies volume of 360dm3 at a pressure of 100kPa calculate the volume of the gas at 120kPa and constant temperature. Page 35 of 78

Downloaded from hlayiso.com GRADE 11 Page 36 of 78 13 Calculate what pressure is needed to compress 6cm3 of hydrogen gas at 100kPa to a volume of 4cm3 without changing the temperature of the gas. 14 A sealed gas syringe lying on a work bench contains 75cm3 air .the plunger is airtight, but moves freely. The atmospheric pressure is 100kPa. 14.1 What is the pressure of gas in the syringe? 14.2 what additional pressure has to be applied on the plunger so that the air is compressed to a volume of 50cm3 while the temperature remains constant. 15 A hot air balloon with a volume of 20m3 has to be filled with hydrogen at pressure of 100kPa. The hydrogen is kept in cylinders that have a volume of 0,02m3 at a pressure of 2000kPa. How many cylinders are needed to fill the balloon if the temperature is kept constant? PART B QUESTION 1 There are four possible options for each answer in the following questions. Each question has only ONE correct answer. Choose the correct answer and write only A, B, C or D next to the question number. 1.1 The temperature of a gas is defined as … A the heat of the gas B the product of the pressure and volume of the gas C a measure of the average speed of the molecules of the gas D a measure of the average kinetic energy of the molecules of the gas (2) 1.2 Which one of the following is NOT a property of an ideal gas? A The volume occupied by the ideal gas is equal to the total volume of all the individual molecules. B The collisions between the molecules are perfectly elastic. C There are no forces of attraction between the molecules. D The product of the pressure and the volume of the ideal gas is constant at constant temperature. (2) Page 36 of 78

Downloaded from hlayiso.com GRADE 11 Page 37 of 78 1.3 Which one of the following graphs represents the relationship between the volume (V) and the absolute temperature (T) of a real gas? A B C D (2) 1.4 Under which conditions do real gases behave like ideal gases? A low temperature and low pressure B high temperature and high pressure C low temperature and high pressure D high temperature and low pressure (2) 1.5 You store 1 mol of an ideal gas in a rigid tank at a temperature T in Kelvin. If you halve the average speed of the gas particles in the tank, what is the new temperature of the ideal gas? A 4T B T C 0,5T D 0,25T (2) 1.6 Tumi performed an experiment with separate samples of oxygen gas and neon 1 gas. Her results produced the following graphs of pressure versus for volume each gas. If Tumi used the same number of moles of gas in each experiment, which of the following statements about the temperature of the gases is true? Page 37 of 78

Downloaded from hlayiso.com GRADE 11 Page 38 of 78 A The temperatures of both gases were the same. B The temperature of neon gas was lower than the temperature of oxygen gas. C The temperature of neon gas was higher than the temperature of oxygen gas. D Tumi cannot make a deduction about the temperature from the graph. (2) 1.7 A container of volume 2 dm3 contains a fixed mass of gas at a temperature of 20 °C and a pressure of 100 kPa. Nombulelo increases the temperature of the container and the gas to 40 °C. The pressure of the gas in the container … 100 ´ 40 A increases by 20 B doubles 100 ´ 313 C increases to 293 D halves (2) 1.8 The following graphs represent possible relationships between pressure and volume for an ideal gas (solid line) and oxygen (dashed line). Which graph correctly illustrates the deviation of the oxygen gas from that of an ideal gas? (Hint: Assume that the real gas did NOT liquefy and that the temperature remained constant when the data was obtained.) A B C D (2) QUESTION 2 23 At a temperature of 27 °C, 3,01 ´ 10 molecules of nitrogen gas occupy a container with a volume of 10 dm3. Calculate the pressure in the container. [6] Page 38 of 78

Downloaded from hlayiso.com GRADE 11 Page 39 of 78 QUESTION 3 Calculate the mass of helium gas that you need to occupy a volume of 2,5 dm3 at a pressure of 190 kPa and a temperature of 21 °C. Give your answer in grams and round your answer to three decimal places. [6] QUESTION 4 1 4.1 Draw a labelled graph of pressure versus in an ideal gas. Assume that all volume other factors remain constant. (3) 4.2 On your graph in question 4.1, show how the graph for a real gas differs from the graph of the ideal gas. Label both graphs clearly. (2) 4.3 Show that the unit of pV is equivalent to a joule. (4) [9] QUESTION 5 Sizakele fills a weather balloon to a volume of 300 dm3 with helium gas at sea level where the temperature is 18 °C. She releases the weather balloon and it reaches a height of 15 000 m above sea level. At this height, the pressure in the weather balloon decreases to 12,3 kPa and the temperature decreases to –48 °C. (Hint: At sea level, atmospheric pressure is 101,3 kPa.) 5.1 Describe the conditions of temperature and pressure under which real gases behave like ideal gases. (2) 5.2 Calculate the volume of the balloon at 15 000 m above sea level. (5) 5.3 Assume that the helium gas behaves like an ideal gas. Calculate the mass of helium gas that Sizakele needed to fill the balloon at sea level. (6) [13] QUESTION 6 The following diagram represents the apparatus Thabang used to investigate a fixed mass of air. Thabang pumped air from a bicycle pump into the pressure gauge on the right. When he did this, the air from the pump pushed the level of oil in the oil reservoir downwards. The oil in the glass tube then moved upwards. This reduced the volume of air trapped at the top of the glass tube. Thabang used the volume scale to measure the change in the volume of trapped air. Page 39 of 78

Downloaded from hlayiso.com GRADE 11 Page 40 of 78 Thabang obtained the following results. Pressure (kPa) Volume (cm3) 105,0 30 126,0 25 157,5 20 210 15 6.1 Write a hypothesis for Thabang’s experiment. (2) 6.2 Draw sketch graphs of the following relationships. a) pressure versus volume (3) 1 b) pressure versus (3) volume 6.3 Name the factor that Thabang keeps constant in his experiment. (1) 6.4 State ONE experimental procedure that Thabang must remember so that he obtains reliable readings. (1) 6.5 State a conclusion for Thabang’s experiment. (2) 6.6 Name and state the law that Thabang verified in his experiment. (2) [14] QUESTION 7 Mrs Kgopa demonstrates an experiment about gases to her Grade 11 class. The following diagram represents the apparatus she used. She attaches a bicycle pump to the apparatus. She then increases the pressure on the liquid when she applies pressure to the bicycle pump. Page 40 of 78

Downloaded from hlayiso.com GRADE 11 Page 41 of 78 7.1 Write an investigative question for Mrs Kgopa’s experiment. (2) 7.2 Identify the independent variable in this experiment. (1) 7.3 After each increase in pressure, explain why Mrs Kgopa should leave the apparatus alone for a few seconds before she measures the volume of the gas.(2) 1 7.4 Copy the following table. Calculate the values of to complete the table. V 1 p (kPa) V (cm3) (cm–3) V 105,0 30 126,0 25 157,5 20 210,0 15 (4) 1 7.5 Draw a graph of versus p. (4) V 7.6 State a conclusion to the experiment. (2) [15] QUESTION 8 8.1 A cylinder contains 10 mol of gas at STP. a) Calculate the temperature and pressure of the gas. (2) b) Calculate the volume of the gas. (5) c) Determine how many molecules of gas there are in the cylinder. (3) 8.2 An air bubble on the ocean bed has a volume of 1,5 cm3. The temperature of the water on the ocean bed is 5 °C. The air bubble rises towards the surface of the Page 41 of 78

Downloaded from hlayiso.com GRADE 11 Page 42 of 78 ocean. The temperature of the water at the surface is 15 °C and the pressure is 101,3 kPa. The volume of the air bubble increases to 4,5 cm3 at the instant the bubble reaches the surface of the water. Calculate the pressure that the water exerts on the air bubble on the ocean bed.(6) [16] QUESTION 9 9.1 Give THREE statements on which the ‘kinetic molecular theory’ of gases is based. (6) 9.2 Rachel measured the mass of an empty 250 cm3 flask. She then filled the flask with a gas of unknown molar mass at an atmospheric pressure of 100 kPa and a temperature of 27 °C. She found that the mass of the flask and contents increased by 0,44 g. a) Calculate the molar mass of the gas Rachel used to fill the flask. (7) b) Assume that the gas molecules are only made up of carbon and oxygen. Determine the molecular formula of this gas. (3) 9.3 Copy the following set of axes onto your page. Draw a line to show the relationship between the pressure and the inverse of the volume of an ideal gas. (2) 9.4 On your graph, draw a line for the same ideal gas, but at a lower temperature. Use a broken line to represent this line. (2) [20] QUESTION 10 The following diagram represents the apparatus that Tyrone used to investigate a fixed mass of dry air. The round-bottomed glass flask has a fixed volume of 1,0 dm3. This flask traps the air as Tyrone increases the temperature in steps by heating the water. He then measures the pressure on the gauge to obtain his results. Page 42 of 78

Downloaded from hlayiso.com GRADE 11 Page 43 of 78 Tyrone obtained the following results during his experiment. Pressure (kPa) Temperature (°C) 175 77 225 177 275 277 325 377 10.1 Write a hypothesis for Tyrone’s experiment. (2) 10.2 Draw a graph of pressure versus temperature (in °C). (5) 10.3 State the relationship between pressure and temperature for the fixed mass of air. (2) 10.4 Explain what Tyrone needs to do to the graph so that he can find the temperature when the pressure is 0 kPa. (2) 10.5 In terms of the kinetic molecular theory, explain why the pressure increases when the temperature increases. (3) 10.6 Assume that the molar mass of air is approximately 28 g•mol–1. Calculate the mass of the air in the 1,0 dm3 round-bottomed flask. (6) [20] Page 43 of 78

Downloaded from hlayiso.com GRADE 11 Page 44 of 78 QUESTION 11 11.1 State Boyle’s law in words. (4) 11.2 Explain why the pressure of an enclosed gas increases as the volume decreases. (4) 11.3 Calculate the pressure of 0,12 g of helium if the gas is enclosed in a 10,0 dm3 container at a temperature of 100 °C. (7) 11.4 The calculated value of pressure in question 18.3 is almost exactly the same as the value measured under these conditions experimentally. Give THREE reasons why this is so. (6) 11.5 Copy the following diagram onto your page. Draw the line that shows the relationship between volume and absolute temperature for an ideal gas. (2) 11.6 On your graph, draw a line for the same ideal gas at a higher pressure. Use a broken line to represent this graph. Label the two lines clearly. (2) [25] QUANTITATIVE ASPECTS OF CHEMICAL CHANGE MOLE CONCEPT 1. The mole and Avogadro’s Number (NA) Although atoms are too small to count individually, chemists need to know how many are present in a sample. To do this they use a set number of particles as a base unit of measurement. This unit is called a mole. One MOLE of a substance contains the same number of particles as there are atoms in 12g of carbon-12, which is 6,02 x 1023. This is called the AVOGADRO NUMBER (NA). 2. Particle Formula n = no. of moles No No. = no. of particles (atoms/ molecules/ions) NA = Avogadro’s Number, 6,023 x 1023 n NA Example 1 1 mole of H2O contains 6,023 x 1023 molecules of H2O 1 mole of He contains 6,023 x 1023 atoms of He Page 44 of 78

Downloaded from hlayiso.com GRADE 11 Page 45 of 78 1 mole of NO3– contains 6,023 x 1023 NO3– ions Example 2 1. Calculate the number of moles represented by 1,2 x 1024 molecules of CO2 n = No = 1,2 x 1024 = 2 moles NA 6 x 1023 Calculator: 1,2 exp 24 ÷ 6 exp 23 = 2. Calculate the number of molecules in 0,5mol of SO2 gas. No. = n x NA = 0,5 x 6 x 1023 = 3 x 1023 molecules of SO2 Calculator: 0,5 x 6 exp 23 = 3 x 1023 (NOT 323) 3. Calculate the number of atoms in 0,5 mol of SO2 gas 1 molecule of SO2 contains 3 atoms [i.e 1(S) + 2(O)] Therefore, 3 x 10 molecules of SO2 contains 3 x 3 x 1023) atoms = 9 x 1023 atoms 23 Activities on Particle Formula 1. Calculate the number of moles represented by 1.1 1,5 x 1023 atoms of He 1.2 2,4 x 1025 molecules of H2O 1.3 3 x 1021 molecules of CO 1.4 3,6 x 1023 “molecules” of NaCℓ 2. Calculate the number of particles in 2.1 0,2 mol NH3 2.2 5 mol O2 2.3 100 mol CaSO4 2.4 4 mol Ne 3. For each of the examples given in 2 determine the number of atoms present in the sample. Answers 1.1 0,25 1.2 40 1.3 0,0005 1.4 0,6 23 24 25 2.1 1,2 x 10 2.2 3 x 10 2.3 6 x 10 2.4 2,4 x 1024 3.1 4,8 x 1023 3.2 6 x 1024 3.3 3,6 x 1026 3.4 2,4 x 1024 3. Relative Atomic Mass (R.A.M.) The R.A.M. of an element (as given on the periodic table) is the average mass of all the isotopes of atoms of the element compared to an atom of carbon-12 (i.e it is how many times the average mass of one atom of the element is heavier than 1/12 of the mass of a 12C atom.) Page 45 of 78

Downloaded from hlayiso.com GRADE 11 Page 46 of 78 4. Relative Formula Mass (R.F.M.) The R.F.M. of a substance can be obtained by adding relative atomic masses of the constituent atoms, as given on the periodic table. Example: R.F.M. of H2SO4 = 2(H) + S + 4(O) = 2(1) + 32 + 4(16) = 98u 5. Molar Mass (M) The molar mass of a substance is simply the Relative Formula Mass expressed in grams per mol (g.mol-1) and it represents that mass of a particular substance which contains Avogadro’s Number of particles. Examples Substance Ar H2O KF R.F.M. 40u 18u 58u Molar Mass (M) 40g.mol-1 18g.mol-1 58g.mol-1 Number of moles (n) 1 mole 1 mole 1 mole 23 23 Number of particles 6 x 10 6 x 10 6 x 1023 atoms molecules “molecules” 6. Mass Formula n = no. of moles m m = mass M = Molar mass (from Periodic Table) n M Examples 1. Calculate the number of moles in 7g of nitrogen gas (N2) m 7 n = = = 0,25mol M 28 2. Calculate the mass of 3 moles of oxygen molecules (O2) m = n x M = 3 x 32 = 96g Activities Mass Formula 1. Calculate the number of moles represented by 1.1 90g of H2O 1.2 3,4g of NH3 1.3 110g of CO2 1.4 4,9g of H2SO4 2. Calculate the mass of 2.1 0,2 mol of Na2CO3 2.2 10 mol of N2O 2.3 0,4 mol of H2 2.4 5 mol of H2S Page 46 of 78

Downloaded from hlayiso.com GRADE 11 Page 47 of 78 3. Calculate the number molecules in 3.1 36g of H2O 3.2 2,2g of CO2 3.3 70g of N2 3.4 8,5g of K2S Answers 1.1 5 mol 1.2 0,2 mol 1.3 2,5 mol 1.4 0,05mol 2.1 21,2g 2.2 440g 2.3 0,8g 2.4 550g 3.1 1,2 x 1024 3.2 22 3 x 10 3.3 1,5 x 1024 3.4 1,5 x 1023 7. Molar Gas Volume (Vo) Ø One mole of any gas at STP will occupy a volume of 22,4 dm3 Ø STP = Standard Temperature (O °C) and Pressure (100 kPa) Examples SUBSTANCE O2 CO2 Ne No. of moles (n) 1 1 1 -1 Molar mass (g.mol ) 32 44 20 No. of particles 6 x 1023 molecules 6 x 1023 molecules 6 x 1023 atoms Volume occupied at 22,4dm3 22,4dm3 22,4dm3 STP Equal volumes of gases measured at the same temperature and pressure, contain the same number of molecules. Example 5dm3 of oxygen at 25°C and 80kPa pressure contains exactly the same number of molecules of 5dm3 of hydrogen gas at 25°C and 80kPa 8. Volume Formula n = no. of moles V V = volume measured in dm3 Vo = molar volume at STP, 22,4dm3 n Vo NOTE: To convert cm3 to dm3 divide by 1000, eg 200cm3 = 0,2dm3 Examples 1. Calculate the number of moles of gas in 11,2dm3 of SO2 at STP. n = V = 11,2 = 0,5mol Vo 22,4 Page 47 of 78

Downloaded from hlayiso.com GRADE 11 Page 48 of 78 2. Calculate the volume occupied by 4 mols of H2 gas at STP V = n x Vo = 4 x 22,4 = 89,6dm3 3. Calculate the volume occupied by 11g of CO2 gas at STP. n = m = 11 = 0,25 mol;; V = n x Vo = 0,25 x 22,4 = 5,6dm3 M 44 Activities Volume Formula 1. Calculate the number of moles in the following volumes of gas measured at STP. 1.1 56dm3 1.2 1,12dm3 1.3 44,8dm3 1.4 672dm3 2. Calculate the volume, at STP, occupied by 2.1 2 mol of Ne 2.2 0,1 mol of CO2 2.3 34g of NH3 2.4 4,2 x 1023 atoms of He Answers 1.1 2,5 1.2 0,05 1.3 2 1.4 30 3 3 3 2.1 44,8dm 2.2 2,24dm 2.3 44,8dm 2.4 15,68dm3 Page 48 of 78

Downloaded from hlayiso.com GRADE 11 Page 49 of 78 MOLE CALCULATIONS As shown the number of moles can be calculated by three different methods: Ø particle formula;; Ø the mass formula;; Ø the volume formula. Since all three formulae contain the number of moles, it is best when solving the problems to always try and determine the number of moles first, after which the other information can easily be calculated. Activities 1. Determine the number of moles in 1.1 60g of carbon 1.2 2,8g of N2 1.3 8,8g of CO2 1.4 10g of CaCO3 1.5 3 x 1023 atoms He 1.6 1,5 x 1024 molecules H2S 1.7 1,8 x 1022 atoms of Fe 1.8 2,4 x 1025 molecules of CH4 1.9 11,2dm3 NH3 at STP 1.10 224dm3 CO2 at STP 1.11 5,6dm3 O2 at STP 1.12 448dm3 Ne at STP 1.13 5,3g of Na2CO3 1.14 3 x 1024 molecules of H2O 3 1.15 89,6dm Ar at STP 2. Determine the mass of 2.1 0,5 moles Mg(NO3)2 2.2 3 moles of NaCℓ 2.3 20 moles of CuSO4 2.4 224dm3 CO2 at STP 2.5 1,12dm3 NH3 at STP 2.6 6 x 1022 molecules of F2 23 2.7 1,5 x 10 molecules of NO2 2.8 56dm3 SO2 at STP 2.9 2,4 x 1023 molecules of PH3 2.10 44,8dm3 NO at STP 3. Determine the volume, at STP, 3.1 6 moles of NH4 3.2 0,4 moles N2O4 3.3 10 moles of O2 3.4 14g N2 3.5 6,8g H2S 3.6 3 x 1023 molecules NO 3.7 9 x 1024 molecules SO3 3.8 1,5 x 1026 atoms Ne 4. Determine the number of (A) molecules and B (atoms) in 4.1 0,1 moles of CH4 4.2 50g of H2 4.3 5 moles of N2 4.4 112dm3 N2O at STP 4.5 340g NH3 4.6 102g H2S 4.7 4,9g H2SO4 4.8 4480dm3 Cℓ2 at STP 3 4.9 2,24dm O2 at STP Answers 1.1 5 1.2 0,1 1.3 0,2 1.4 0,1 1.5 0,5 1.6 2,5 1.7 0,03 1.8 40 1.9 0,5 1.10 10 1.11 0,25 1.12 20 1.13 0,05 1.14 5 1.15 4 Page 49 of 78

Downloaded from hlayiso.com GRADE 11 Page 50 of 78 2.1 74g 2.2 175,5g 2.3 3190g 2.4 440g 2.5 0,85g 2.6 3,8g 2.7 11,5g 2.8 160g 2.9 13,6g 2.10 60g 3.1 134,4dm3 3.2 8,96dm3 3.3 224dm3 3.4 11,2dm3 3.5 4,48dm3 3.6 11,2dm3 3.7 336dm3 3.8 5600dm3 3.9 44,8dm3 3.10 4,48dm3 A (molecules) 4.1 6 x 1022 4.2 1,5 x 1025 4.3 3 x 1024 4.4 3 x 1024 4.5 1,2 x 1025 4.6 1,8 x 1024 4.7 3 x 1022 4.8 1,2 x 1026 4.9 6 x 1022 4.10 3 x 1023 B (molecules) 4.1 3 x 1023 4.2 3 x 1025 4.3 6 x 1024 4.4 9 x 1024 4.5 4,8 x 1025 4.6 5,4 x 1024 4.7 2,1 x 1023 4.8 2,4 x 1026 4.9 1,2 x 1023 4.10 1,2 x 1024 Stoichiometric Calculations • Balanced equations contain the same number of each kind of atom on each side of the equation. • The total calculated mass of the reactants is equal to the total calculated mass of the products in a balanced equation. • The co-efficients in a balanced equation give mole relationships between reacting chemicals. • The quantity of product formed in the reaction is determined by the reactant which is not in excess (i.e. limiting reactant). Example N2(g) + 3H2(g) ® 2NH3(g) Mole ratio 1 mol 3 mol 2 mol Mass ratio 28g 6g 34g Volume ratio 22,4dm3 67,2dm3 44,8dm3 (at STP) From the above equation, it can be said that one mole of N2 will react with 3 moles of H2 to form 2 moles of NH3. Since the ratio in which the masses react is directly related to the mole ratio it is possible to relate these variables in a balanced equation by converting the given information to moles using one of the 3 mole formulae. 1.1 How many moles of NH3 are formed when 6 moles of H2 react with excess nitrogen? Rewrite the equation and fill in the mole ratio. N2(g) + 3H2(g) ® 2NH3(g) Mol ratio 1 3 2 When H2 = 6 mol 2 6 4 (x2) 4 mols of NH3 are formed from 6 mols of H2 Page 50 of 78

Downloaded from hlayiso.com GRADE 11 Page 51 of 78 1.2 What volume of NH3 would be formed when 24g of H2, react with excess N2 at STP? Step 1 Convert given information to MOLES, if not already in moles. 24g of H2 n = m = 24 = 12mol of H2 M 2 Step 2 Find MOLE RATIO between what is given and what is required. Mole ratio H2 : NH3 3 : 2 (from balanced equation) 12 : 8 (x4) Step 3 Calculate the required quantity using the appropriate mole formula Volume of 8 mols of NH3? V = n x 22,4 = 8 x 22,4 = 179,2dm3 of NH3 1.3 What mass of N2 is required to produce 4,48dm3 of NH3 at STP? Step 1 Convert given information to MOLES 4,48dm3 of NH3 n = V = 4,48 = 0,2 mols of NH3 22,4 22,4 Step 2 Find MOLE RATIO between what is given and what is required. Mol ratio : N2 : NH3 1 : 2 (from balanced equation) 0,1 : 0,2 (x0,1) Step 3 Calculate the required quantity using the appropriate mole formula Mass of N2 m = n x M = 0,1 x 28 = 2,8g of N2 Activities on mass-mass, volume-volume & mass-volume calculations 1. Consider the balanced equation: 2H2(g) + O2(g) ® 2H2O(g) 1.1 How many moles of H2O are formed when 5 moles of O2 react with excess H2? 1.2 What mass of H2O is formed when 5 moles of O2 react with excess H2? 1.3 What mass of H2 is needed to react with 64g of O2? 1.4 What volume of O2 at STP is required to react with 11,2dm3 of H2? 1.5 What volume of H2O is formed when 44,8dm3 of O2 react with excess H2 at STP? 2. Consider the balanced equation: 4NH3(g) + 5O2(g) ® 4NO(g) + 6H2O (g) 2.1 How many moles of NO are formed when 1 mole of O2 reacts with excess NH3? 2.2 What volume of NO is formed when 1 mole of O2 reacts with excess NH3 at STP? 2.3 What mass of H2O is formed when 34g of NH3 reacts with excess O2? 2.4 What volume of NO is formed when 16g of O2 react with excess NH3 at STP? 2.5 What mass of NH3 is needed to react with 224dm3 of O2 at STP? 2.6 425g of NH3 and 80g of O2 are allowed to react together. Which reactant is used up first and what ass of the other reactant is in excess? 3. Consider the balanced equation: Aℓ4C3(s) + 12H2O(ℓ) ® 3CH4(g) + 4Aℓ(OH)3(s) 3.1 How many moles of CH4 are formed when 4 moles of H2O react with excess Aℓ4C3? 3.2 What volume of CH4 is formed when 4 moles of H2O react with excess Aℓ4C3 at STP? 3.3 What mass of Aℓ4C3 is needed to form 4,8g of CH4? Page 51 of 78

Downloaded from hlayiso.com GRADE 11 Page 52 of 78 3.4 What volume of CH4 is formed when 288g of Aℓ4C3 react with excess H2O? 3.5 What mass of Aℓ(OH)3 is formed from 108g of H2O? 3.6 28,8g of Aℓ4C3 is allowed to react with 45g of H2O. Which reactant is used up first and what mass of the other reactant is in excess? 4. Consider the balanced equation: SO2(g) + 2H2S ® 3S(s) + 2H2O(ℓ) 4.1 How many moles of H2S are needed to form 9 moles of S? 4.2 What mass of H2S is needed to form 9 moles of S? 4.3 What volume of SO2 is needed to form 3,6g of H2O at STP? 4.4 What mass of S will be formed when 16g of SO2 react with excess H2S? 4.5 What volume of H2S is needed to form 9,6g of S? 4.6 6dm3 of SO2 is allowed to react with 11,2dm3 of H2S at STP. Which reactant is used up first and what volume of the other reactant is in excess? 5. Consider the balanced equation: N2(g) + 3H2 ® 2NH3(g) 5.1 How many moles of NH3 are formed from 12 moles of H2? 5.2 What mass of N2 is needed to form 4 moles of NH3? 5.3 What volume of N2 is needed to completely react with 1,5 moles of H2? 5.4 What mass of NH3 is formed when 2,8g of N2 reacts with excess H2? 5.5 What volume of H2 is needed to form 5,6dm3 of NH3 at STP? 5.6 What mass of N2 is needed to form 5,6dm3 of NH3 at STP? 5.7 44,8dm3 of N2 is allowed to react with 14g of H2 at STP. Which reactant is used up first and what mass of the other is in excess? Answers 1.1 10 mol 1.2 180g 1.3 8g 1.4 5,6dm3 1.5 72g 2.1 0,8 mol 2.2 17,92dm3 2.3 54g 2.4 8,96dm3 2.5 136g 2.6 8,5g NH3 ix XS 3.1 1 mol 3.2 22,4dm3 3.3 14,4g 3.4 134,4dm3 3.5 156g 3.6 1,8 x H2O in XS 4.1 6 mol 4.2 204g 4.3 2,24dm3 4.4 24g 4.5 4,48dm3 4.6 0,4dm3 SO2 in XS 5.1 8 mol 5.2 56g 5.3 11,2dm3 5.4 3,4g 5.5 6,72dm3 5.6 3,5g 5.7 2g H2 in XS Page 52 of 78

Downloaded from hlayiso.com GRADE 11 Page 53 of 78 Concentration of Solutions Concentration of a solution refers to moles of a solute per unit volume of a solvent. The concentration of a solution is determined by use of the following formulae: c = n n = m m V M c M V and which combine to give Symbol Quantity Unit M mass g M molar mass g.mol-1 To convert cm3 to dm3 divide by N no. of moles mol 1000 C concentration mol.dm-3 V volume dm3 NOTE: 1cm3 = 1ml ;; 1dm3 = 1litre Example 1: Calculating Concentration Calculate the concentration of the solution formed when 10,6g of sodium carbonate (Na2CO3) is dissolved in water and the solution is made up to 200cm3. Method A Method B -1 M of Na2CO3 = 2(23) + 12 + 3(16) = 106g.mol c = m = 10,6 = 0,5mol.dm-3 V = 200 ÷ 1000 = 0,2dm3 MV 106 x 0,2 -3 n = m = 10,6 = 0,1mol ;; c = n = 0,1 = 0,5mol.dm M 106 V 0,2 Example 2: Calculating Mass Calculate the mass of sodium hydroxide (NaOH) required to make a solution of concentration 0,1mol.dm-3 in a 250cm3 flask. Method A Method B -1 M of NaOH = 23 + 10 + 1 = 40g.mol m = c x M x V V = 250 ÷ 1000 = 0,25dm3 = 0,1 x 40 x 0 n = c x V = 0,1 x 0,25 = 0,025mol = 1g m = n x M = 0,025 x 40 = 1g Example 3 :Diluting a solution 20cm3 of hydrochloric acid (HCl) of concentration 0,4mol.dm-3 is diluted by adding it to 80cm3 of distilled water. Calculate the concentration of the diluted acid. Method The number of moles of HCℓ stays the same regardless of the amount of water added. n = c x V = 0,4 x 0,02 = 0,008mol of HCl in original solution Page 53 of 78

Downloaded from hlayiso.com GRADE 11 Page 54 of 78 The volume has increased, Vnew = 0,02 + 0,08 = 0,1 dm3 c = n = 0,008 = 0,08mol.dm-3 V 0,1 Example 4 : Mixing solutions (HG) 200cm3 of NaOH of concentration 0,2mol.dm-3 is added to 300cm3 of Na2CO3 of concentration 0,15mol.dm-3. Calculate the concentration of Na+ ions in the solution. Method NaOH(s) ® Na+(aq) + OH – 1 mol 1mol 1mol n = c x V = 0,2 x 0,2 = 0,04 mol of NaOH, therefore, 0,04mol of Na+ ions (refer to dissociation equation) Na2CO3(S) ® 2Na+(aq) + CO32– (aq) 1mol 2mol 1mol n = c x V = 0,15 x 0,3 = 0,045mol of Na2CO3, therefore, (2 x 0,045) = 0,09mol of Na+ ions (refer to dissociation equation) c = n total = 0,04 + 0,09 = 0,13 = 0,26 mol.dm -3 Vtotal 0,2 + 0,3 0,5 Activities : Concentration Calculations 1. Calculate the mass of NaOH in 100cm3 of a solution of concentration 0,15mol.dm-3. 2. Calculate the concentration of a solution containing 5,3g of Na2CO3 in 500cm3 of water. 3. Calculate the concentration of 50cm3 of a solution containing 0,49g of H2SO4 . 4. If 3,15g of HNO3 was dissolved in 250cm3 of solution what would the concentration be? 5. Calculate the mass of CaSO4 needed to make 100cm3 of a solution of concentration 0,2mol.dm-3. 6. 87g of K2SO4 is dissolved to make a solution of concentration 0,8mol.dm-3. What is the volume of this solution? 7. 2,84g of Na2SO4 is dissolved in water to form a solution of volume 250cm3. 7.1 Write an equation showing how the Na2SO4 dissociates in water. Calculate: 7.2 the concentration of the Na2SO4 solution;; and + 7.3 the concentration of Na ions in the solution. 8. Calculate the concentration of NO3 - ions in a solution containing 4,44g of Mg(NO3)2 in 200cm3. 9. 36g of MgSO4 and 19g of MgCl2 are mixed together to form a solution of total volume 250cm3. What is the concentration of Mg2+ ions in this solution? Page 54 of 78

Downloaded from hlayiso.com GRADE 11 Page 55 of 78 10. 200cm3 of a 0,25mol.dm-3 of NaOH is mixed with 300cm3 of a 0,1mol.dm-3 solution of Na2CO3. Calculate the total concentration of Na+ ions in this solution. 11. What volume of water must be added to 200cm3 of a 0,56mol.dm-3 NaOH solution to change the concentration to 0,4mol.dm-3 ? 12. 150cm3 of distilled water is added to 250cm3 of NaOH which has a concentration of 0,1mol.dm-3. What is the final concentration of the diluted NaOH? Answers 1. 0,6g 2. 0,1mol.dm-3 3. 0,1mol.dm-3 4. 0,2mol.dm-3 5. 2,72g 6. 0,625dm3 7.2 0,08mol.dm-3 7.3 0,16mol.dm-3 -3 8. 0,3mol.dm 9. 2mol.dm-3 10. 0,22mol.dm-3 11. 0,08dm3 12. 0,0625mol.dm-3 Percentage Composition This gives the percentage of each element which makes up a compound. Example Find the percentage composition of H2SO4 Molar mass of H2SO4 = 98 g.mol-1 2 Mass of H 2 x 1 = 2g % of H = x 100 = 2,04% 98 Mass of S 1 x 32 = 32g 32 % 0f S = x 100 = 32,65% 98 Mass of O 4 x 16 = 64g 64 % of O = x 100 = 65,31% 98 Empirical Formula This is the simplest formula of a compound. The simplest ratio of the elements found from the % composition and the atomic masses of elements. The real formula can be a multiple of the simplest formula Example (a) The percentage composition of a compound is 80% C;; 20% H. Find the empirical formula. In 100g of the compound there are 80g C and 20g H 80 20 Convert to moles : = 6,66mol C;; = 20 mol H 12 1 Page 55 of 78

Downloaded from hlayiso.com GRADE 11 Page 56 of 78 Ratio C : H = 6,66 : 20 (divide by the smallest number to get the simplified ratio) 6,66 Simplified ratio = : 20 = 1 : 3 6,66 6,66 Empirical formula = CH3 (b) If the molecular mass of the above compound is 30g.mol-1 find the true formula. Molecular mass = 30g.mol-1 (given) Empirical formula mass = 15g.mol-1 Number of formula units = 30/15 = 2 True formula = 2(CH3) = C2H6 Activities 1. Find the % composition of a) Cu(NO3)2 b) caffeine C8H10N4O2 2. Find the percentage of water of crystallization in Epsom salts MgSO4.7H2O. 3. Ammonium nitrate NH4NO3 and ammonium sulphate (NH4)2SO4 are both used as nitrogen fertilizers. Which compound contains the greater percentage of nitrogen? 4. Find the empirical formulae of the compounds which have the following percentage composition. Name these compounds. 4.1 54% Ca 43% O 2,7% H 4.2 63,4% Ag 8,25% N 28,35% O 4.3 56% Ba 29,2% Cl 14,8% H2O 5. Compounds of hydrogen and boron are called boranes. One of these has the percentage composition 78,55% B and 21,45% H. If its molar mass is 28g.mol-1 what is its molecular formula? 6. Xylene is the solvent used in marker pens. The % composition of xylene is 88,88% C and 11,11% H. If the molecular mass of xylene is 108g.mol-1 find the empirical and molecular formula. 7. Gypsum (hydrated calcium sulphate) is heated to make Plaster of Paris. When 1g of gypsum was heated to drive off the water 0,79g of calcium sulphate remained. 7.1 How many moles of calcium sulphate are there in this sample? 7.2 How many moles of water are there in this sample? Page 56 of 78

Downloaded from hlayiso.com GRADE 11 Page 57 of 78 Water of Crystallisation One needs to be able to: Ø explain the terms anhydrous, hydrated and water of crystallisation. Ø calculate the formula of a hydrated salt. Some substances can trap water molecules inside of them. The water contained is called the water of crystallisation. There are two ways in which a formula can be written if it is a hydrated compound: 1) Empirical formula, which you know and love;; 2) Dot formula, so copper sulfate pentahydrate becomes CuSO4 • 5H2O. Example A sample of hydrated calcium sulphate, CaSO4●xH2O, has a relative formula mass of 172. Calculate the value of x. 40+32 + (4×16) + x(2+16) = 172 x = 2 Percentage Yield Yield: the amount of product. Theoretical yield: the amount of product we expect, based on stoichiometric calculations. Actual yield: amount of product from a procedure or experiment (this is given in the question). WXHYWZ [\]Z^ Percent yield = ×100% H_]`']H\XWZ [\]Z^ Example What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: Write the balanced chemical equation 2H2 + O2 ® 2H2O Step 2: Determine actual and theoretical yield. Mc ×Ic theoretical amount of water to be produced = = 144 g d Step 3: Calculate % yield WXHYWZ [\]Z^ % yield = ×100% H_]`']H\XWZ [\]Z^ MIe = ×100% Mdd = 95,83% Page 57 of 78

Downloaded from hlayiso.com GRADE 11 Page 58 of 78 Percentage Purity Ø Chemicals are a mixture of different compounds Ø Chemical analysis can determine the composition of minerals Ø Then composition can be used to work out purity of products. Ø Can be calculated using: fWgg `h iY'] i'`^YXHg % purity= x100% fWgg `h H_] gWfiZ] Example 1) 3,68g of washing soda crystals(Na2CO3.5H2O) are dissolved in water and made up to a volume of 275cm3. 25 cm3 of this sample is neutralised by 23,5 cm3 of HCl of concentration of 0,11 mol.dm-3. Calculate the percentage purity of Na2CO3 in the commercial washing soda. 2) An impure sample of table salt that is contaminated by sand has a mass of 5,5 g. The sample is dissolved in distilled water and then filtered. The residue that stays behind after being filtered is dried and its mass is 1,5g. Calculate the percentage purity of the table salt in the sample. PREVIOUS EXAM QUESTIONS Question 5 (EC November 2016) 5.1 In order to determine the empirical and molecular formula of a compound, CxHy, a certain mass of the compound is burnt completely in excess oxygen to produce 47,1 g CO2 and 19,35 g H2O as the only products. 5.1.1 Define the term empirical formula. (2) 5.1.2 Use relevant calculations to determine the empirical formula of the compound. (8) 5.1.3 The molar mass of the compound is 28 g.mol-1. Determine by using calculations The values of x and y. (2) 5.2 A sample of IMPURE calcium carbonate (limestone) of unknown mass required a continuous supply of strong heat to decompose according to the following equation: CaCO3(s) → CaO(s) + CO2(g) After the completion of reaction, 11,76 g CaO was produced. The percentage purity of calcium carbonate is found to be 80%. 5.2.1 Calculate the mass of the impure calcium carbonate. (6) Page 58 of 78

Downloaded from hlayiso.com GRADE 11 Page 59 of 78 Question 5 (EC November 2012) A learner wants to determine the molecular formula of a compound found in cigarettes. The learner asks for your help with the calculation. The following is known about the compound: It is responsible for cigarettes being such an addictive substance, it has a molecular mass of 162,2 g∙mol-1 and its percentage composition is 74,07% C;; 8,65% H and 17,28% N. 5.1 Name the compound being referred to here. (2) 5.2 As much as the substance referred to in QUESTION 5.1 is responsible for the addiction of smokers to cigarettes, it is used in some products to help them overcome their addiction. Name ONE example of such a product. (2) 5.3 Cigarette packs contain various warnings on them concerning the harmful effects of smoking. Name ONE harmful effect of smoking on humans. (1) 5.4 Show, by means of calculation, how you would go about helping the learner determine the molecular formula of this compound. (7) Question 8 (GP June 2016) 8.1 A laboratory analysis of a compound shows that it has the following chemical composition: Carbon: 76 % Hydrogen: 12,5% Oxygen: 11,5% 8.1.1 Determine the empirical formula for the compound. (5) 8.1.2 Determine the molecular formula if the molar mass of the compound is 282 g.mol-1 (4) 8.2 14,5 g of iron reacts with 12 g of water during a corrosion reaction to form Fe3O4 and H2, according to the following equation. Fe + H2O → Fe3O4 + H2 8.2.1 Balance the equation. (3) 8.2.2 Find the substance which is the limiting reactant? Show your calculations. (6) 8.2.3 Using the calculations in your answer to Question 8.2.2, determine the mass of Fe3O4 that will be formed. (3) 8.3 Chalk is almost pure calcium carbonate. If 10 g of chalk was reacted with an excess of dilute hydrochloric acid, 2.128 litres of carbon dioxide gas is collected at STP. CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) Determine the purity of calcium carbonate. (5) Page 59 of 78

Downloaded from hlayiso.com GRADE 11 Page 60 of 78 Question 8 (Hilton College 2013) 8.1 When a 50 g sample of impure calcium carbonate is heated, 10 dm3 of carbon dioxide, measured at STP, are evolved. Refer to the balanced chemical equation below and answer the questions that follow. CaCO3 ® CaO + CO2. 8.1.1 Calculate the number of moles of carbon dioxide gas that forms? (2) 8.1.2 What mass of pure calcium carbonate has reacted? (3) 8.1.3 Calculate the percentage purity of calcium carbonate. (2) 8.2 40 cm3 of sulphur dioxide react with 30 cm3 of oxygen gas to form sulphur trioxide. Assume that the reaction has run to completion and all volumes are measured at the same temperature and pressure. 8.2.1 Write a balanced chemical equation for the reaction. (3) 8.2.2 Which of the reactants, sulphur dioxide or oxygen, is the limiting reagent? (2) 8.2.3 Calculate the volume of the remaining gases at the end of the reaction. (5) 8.3 Chromium, a transition element, forms a variety of compounds of various colours. Among these compounds sodium dichromate is commercially the most important compound. A chemical analysis revealed the following information about the chemical composition of sodium dichromate by mass: Na: 17,5% Cr: 39,7% O: 42,8% 8.3.1 Define the concept empirical formula. (2) 8.3.2 Determine the empirical formula of the above compound. (5) 8.4 Muriatic acid, an industrial grade of concentrated HCℓ, is used to clean bricks and to etch cement for painting. Its concentration is 11,7 mol∙dm-3. 8.4.1 How much water must be added to the concentrated muriatic acid to produce 5 litres of a diluted muriatic acid of concentration 3,5 mol∙dm-3 acid for routine use. (5) -3 8.4.2 Determine the volume of muriatic acid solution of concentration 11.7 mol.dm which contains 9,55 g of HCℓ. (3) Page 60 of 78

Downloaded from hlayiso.com GRADE 11 Page 61 of 78 Energy changes in reactions related to bond energy changes • Define heat of reaction (ΔH) as the energy absorbed or released per mole in a chemical reaction. ΔH = Hproducts – Hreactants, where Hproducts and Hreactants are the heat (energy) of the products and reactants respectively. • Define exothermic reactions as reactions that release energy. • Define endothermic reactions as reactions that absorb energy. • Classify, with reason, reactions as exothermic or endothermic. Exothermic and endothermic reactions • State that ΔH > 0 for endothermic reactions, i.e. reactions in which energy is absorbed. • State that ΔH < 0 for exothermic reactions, i.e. reactions in which energy is released. Activation energy • Define activation energy as the minimum energy needed for a reaction to take place. • Define an activated complex as the unstable transition state from reactants to products. • Draw or interpret fully labelled sketch graphs (potential energy versus course of reaction graphs) of catalysed and uncatalysed endothermic and exothermic reactions. Chemical Change/Reaction Is the change that involves the formation of a new substance e.g combustion of petrol, rusting of iron. ü can be represented using a chemical equation Reactants(R) → Products(P) • The arrow shows that the reactants are used up to form products. • Reactant is the substance that is used up during a chemical reaction • Product is the substance that is formed during a chemical reaction Energy and Chemical Change HEAT OF REACTION/ENTHALPY CHANGE • is the energy absorbed or released in a chemical reaction? • Symbol is ∆H • Unit is kJ or kJ.mol-1 BOND ENERGY Is defined as the amount of energy required to break a bond in 1 mole of a molecule. Page 61 of 78

Downloaded from hlayiso.com GRADE 11 Page 62 of 78 BOND ENERGIES OF SOME COVALENT BONDS Type of bond Bond Energy (kJ.mol-1) H – H 436 C-H 414 O-H 460 N-H 390 C-C 347 O=O 499 C=O 799 NΞN 946 There are two types of reaction based on energy and chemical change i.e. 1. Exothermic reaction 2. Endothermic reaction Exothermic reaction is the reaction that releases energy Endothermic reactionis the reaction that absorb energy CHEMICAL REACTIONS TAKE PLACE IN TWO STAGES: 1. The bonds in the reactant molecules are broken to form free atoms. This step requires more energy and is endothermic. 2. The free atoms bond together to form the product molecules. This step releases energy and is exothermic. The stage with the highest energy determines whether an overall energy change is endothermic or exothermic. ENERGY CHANGES IN FUELS Fuels undergo combustion reactions in the presence of oxygen to produce water and carbon dioxide. Combustion reactions are highly exothermic and release large amounts of heat energy and light energy. Example: When methane burns in the presence of oxygen: CH4 + 2O2 → CO2 + 2H2O + heat Energy needed to break the bonds: C-H BONDS =4 X 414 =1656 kJmol O=O BONDS =2 X499= 998 kJ/mol TOTAL = 1656 +998 = 2654 kJ/mol Page 62 of 78

Downloaded from hlayiso.com GRADE 11 Page 63 of 78 ENERGY RELEASED TO FORM NEW BONDS: C=O BONDS = 2 x 799 = 1598 kJ/mol O-H BONDS = 4 X 460 = 1840kJ/mol TOTAL ENERGY RELEASED = 1598+ 1840 =3438kJ/mol More energy is released therefore the reaction is exothermic. Change in energy = energy needed – energy released =2654-3438 = -784kJ/mol ENERGY CHANGES IN BIOLOGICAL SYSTEMS 1 Photosynthesis is highly endothermic as it requires huge amounts of light energy. 6CO2 + 6H2O + light energy → C6H12O6 + 6 O2 2 Respiration is the reverse of photosynthesis and releases energy and is exothermic. C6H12O6 + 6 O2 → 6CO2 + 6H2O + energy Energy profiles Reactions are classified as exothermic and endothermic • Enthalpy(H) is the amount of energy of reactants or products • Enthalpy cannot be measured • Enthalpy change (∆H) can be calculated using: ∆H= Hproducts-Hreactants • exothermic reactions have a negative value of ∆H • endothermic reactions have a positive value of ∆H ΔH < 0 for exothermic reactions, i.e. reactions in which energy is released. Reactants ΔH is negative/ Heat released Potential Energy Products kJ.mol-1 Reaction Progress Endothermic reactions Page 63 of 78

Downloaded from hlayiso.com GRADE 11 Page 64 of 78 • Heat energy is taken in, ∆H>0, ∆H is positive • Enthalpy of reactants is less than the enthalpy of products Products Potential Energy kJ.mol-1 ΔH is positive/ Heat absorbed Reactants Reaction Progress EXOTHERMIC REACTIONS: reactions in which energy is released ∆H<0, ∆H is negative Activated complex Activation Reactants energy ΔH is negative/ Heat released Products • Activation energy is the minimum energy needed for a reaction to take place • Activated complex is the unstable transition state from reactants to products or is the temporary unstable state that is formed during the course of a chemical reaction ACIDS AND BASES • Acids and bases are chemicals that exhibit their chemical behaviour when dissolved in water. Hence, when dealing with them the word aqueous will be repeatedly used, so is in the chemical equations for these substances. Acids and bases when dissolved in water they produce ions;; therefore, will produce electrolytes. • Electrolytes are ionic solution that is able to conduct electricity. • There are different types of acids and bases that are occurring naturally;; however, also have laboratory derived acids Page 64 of 78

Downloaded from hlayiso.com GRADE 11 Page 65 of 78 • Acids and bases are defined based of two theories i.e. Lowry-Bronsted and Arrhenius Table1: Illustrates acid and base definition using different theories. definition Acid Base + Lowry- An acid is a proton donor (H ) A base is a proton acceptor (H+) Bronsted HCl + H2O ↔H3O+(aq) + Cl-(aq) NH3 + H20 ↔NH4+(aq) + OH-(aq) theory Arrhenius An acid liberates hydrogen ions A base liberates hydroxyl ions (OH- theory (H+) in aqueous solutions ) in aqueous solutions + - HCl ↔H (aq) + Cl (aq) NaOH ↔ Na+(aq) + OH-(aq) Table 2: illustrates properties of acids and bases Acid Base 1. Sour taste 1. Bitter taste 2. Corrosive 2. Caustic 3. have pH lower than 7 3. Have pH greater than 7 4. As pH decreases the acidic strength 4. As pH increases the basic strength increases increases. Ampholyte An ampholyte is a substance that can act as a base and as an acid E.g. HSO4- + H2O → H2SO4 + OH- HSO4- acts as a base HSO4- + H2O → SO42- + H2O HSO4- acts as an acid Activity: 1. With the use of relevant Equations show that water is an Ampholyte. 1.1. H2PO3- 1.2. H2O Conjugate Acid base pairs . b2 a2 E.g. HCl + H2O ↔ H3O+(aq) + Cl- (aq) a1 b1 Page 65 of 78

Downloaded from hlayiso.com GRADE 11 Page 66 of 78 Activity 1. Write down the equation and identify conjugate acid-base pairs using the following: 1.1. H2SO4 dissolved in H2O 1.2. NH3 dissolved in H2O IONIC STRENGTH of Acids and Bases Table 4: Illustrates the difference between a strong acid and a weak acid Strong Acid Weak Acid 1. Ionises completely in water 1. Ionises partially in water 2. Has a lower pH value 2. Has pH value closer to 7 3. has bigger Ka 3. Has smaller Ka 4. Stronger electrolytes, high 4. Weak electrolytes, lower conductivity conductivity Table 5: Illustrates the difference between a strong base and a weak base Strong bases Weak bases 1. Ionises completely in water 1. Ionises partially in water 2. Has a pH value closer to 14 2. Has pH value closer to 7 3. has bigger Kb 3. Has smaller Kb 4. Stronger electrolytes, high 4. Weak electrolytes, lower conductivity conductivity Indicators • An indicator is a chemical substance that changes colour when the solution has a specific pH • There are three indicators that are of concern i.e. Methyl orange, Phenolphthalein and Bromothymol blue • Indicators are used in titration (neutralisation reactions) to determine an equivalence point or an end point. • An equivalence point is a stage whereby an acid/base has completely reacted with an acid/base. Therefore, in solution there is only salt, water and an indicator available Table 3: Illustrates the selection of a suitable indicator Combination Indicator pH range Expected colour Strong Acid + Weak base Methyl Orange 3,1 – 4,4 Yellow / orange Strong acid + Strong Bromothymol blue 6,0 – 7,6 Blue / dark blue base Weak acid + strong base Phenolphthalein 8,3 – 10 Pink / red Page 66 of 78

Downloaded from hlayiso.com GRADE 11 Page 67 of 78 Activity Which indicator would you use if solutions of the following acid and base were titrated against one another? 1.1. HNO3 and NaOH 1.2. CH3COOH and Ca(OH)2 1.3. H2SO4 and CaCO3 Redox reactions In Grade 11 we need to: • Determine the oxidation number from a chemical formula and electronegativities • Identify a reduction - oxidation reaction and apply the correct terminology to describe all the processes • Describe oxidation reduction reactions as involving electron transfer • Describe oxidation - reduction reactions as always involving changes in oxidation number • Balance redox reaction equations by using oxidation numbers via the ion-electron method • Explain the meaning of • Assign oxidation numbers to atoms in various molecules like H2O, CH4, CO2, H2O2, HOCℓ by using oxidation number guidelines or rules • Use rules of oxidation to assign oxidation numbers to atoms in a variety of molecules and ions Electrochemical Reactions Oxidation and Reduction: Oxidation is: • the loss of electrons. • an increase in oxidation state. Reduction is: • the gain of electrons. • a decrease in oxidation state. Oxidation and reduction reactions always occur together and are sometimes called redox reactions. The substance that is oxidized is the reducing agent, and the substance that is reduced is the oxidizing agent Oxidation States: The oxidation state is a fictitious charge assigned to each atom in a compound. It is calculated by assigning all bonding electrons in a compound to the most electronegative element. The Activity Series: The activity series is a listing of metals from those that are easiest to oxidize to those that are most difficult to oxidize. Any half-reaction in the activity series is spontaneous when paired with the reverse of a half-reaction below it on the list. Corrosion: Corrosion is the oxidation of iron and other metals by atmospheric oxygen. Corrosion can be prevented by keeping the metal dry, sealing it with a protective coating, or depositing a more active metal onto the surface of the metal to be protected. Let a Good Chemistry Story be told: Oxidation and Reduction: Definitions Page 67 of 78

Downloaded from hlayiso.com GRADE 11 Page 68 of 78 Consider the following redox reactions: 1. 4Fe(s) + 3O2(g) à2Fe2O3(s) (rusting of iron) 2. CH4(g) + 2O2(g) àCO2(g) + 2H2O(g) (combustion of methane) What do they all have in common? Each of these reactions involves one or more elements gaining oxygen. In the rusting of iron, iron gains oxygen as it turns into iron oxide, the familiar orange substance we call rust. In the combustion of methane, carbon gains oxygen to form carbon dioxide, producing the brilliant blue flame we see on gas stoves. In each case, the substance that gains oxygen is oxidized in the reaction. One definition of oxidation—though not the most fundamental one—is simply the gaining of oxygen. Now consider these same three reactions in reverse: 1. 2H2O(g) à2H2(g) + O2(g) 2. 2Fe2O3(s)à4Fe(s) + 3O2(g) Each of these reactions involves loss of oxygen. In the first reaction, hydrogen loses oxygen;; in the second reaction, iron loses oxygen. One definition of reduction is simply the loss of oxygen. Redox reactions need not involve oxygen, however. Consider, for example, the similarities between the following two reactions: 1. 4Li(s) + O2(g)à2Li2O(s) 2. 2Li(s) + Cl2(g) à 2LiCl(s) In both cases, lithium (a metal with a strong tendency to lose electrons) reacts with an electronegative non metal (which has a tendency to gain electrons). In both cases, lithium atoms lose electrons to become positive ions—lithium is oxidized. Li àLi+ + e- The electrons lost by lithium are gained by the non metals, which become negative ions the non metals are reduced. O2 + 4e-à2O2- Cl2 + 2e- à2Cl- A more fundamental definition of oxidation, then, is simply the loss of electrons, and a more fundamental definition of reduction is the gain of electrons. Notice that oxidation and reduction must occur together. If one substance loses electrons (oxidation), then another substance must gain electrons (reduction). The substance that is oxidized is called the reducing agent because it causes the reduction of the other substance. Page 68 of 78

Downloaded from hlayiso.com GRADE 11 Page 69 of 78 Similarly, the substance that is reduced is called the oxidizing agent because it causes the oxidation of the other substance. For example, consider our hydrogen–oxygen fuel-cell reaction. 2H2(g) + O2(g) à 2H2O(g) Reducing agent Oxidizing agent In this reaction, hydrogen is oxidized, making it the reducing agent. Oxygen is reduced, making it the oxidizing agent. Substances such as oxygen, which have a strong tendency to attract electrons, are good oxidizing agents—they tend to cause the oxidation of other substances. Substances such as hydrogen, which have a strong tendency to give up electrons, are good reducing agents—they tend to cause the reduction of other substances. NB: Helpful mnemonics: OIL RIG—Oxidation Is Loss (of electrons);; Reduction Is Gain (of electrons). LEO the lion says GER—Lose Electrons Oxidation;; Gain Electrons Reduction. Activity Identifying Oxidation and Reduction Identify the substance being oxidized and the substance being reduced in each reaction. (a) C(s) + O2(g) à CO2(g) (b) 2K(s) + Cl2(g) à 2KCl(s) (c) 2Al(s) + 3Sn2+(aq)à 2Al3+(aq) + 3Sn(s) Oxidation States Rules for Assigning Oxidation States Examples • The oxidation state of an atom in a free element is 0, e.g. Cu Cl2 • The oxidation state of a monoatomic ion is equal to its charge. e.g. Ca2+ +2 ox state, Cl- -1 ox state • The sum of the oxidation states of all atoms in: ü a neutral molecule or formula unit is 0. e.g. H2O (2xH ox state) + (1xO ox state) = 0 ü an ion is equal to the charge NO3- e.g. (1xN ox state) + (3xO ox state) = -1 In their compounds, • Group I metals have an oxidation state of +1. e.g. NaCl +1 ox state • Group II metals have an oxidation state of +2. e.g. CaF2 +2 ox state In their compounds, non metals are assigned oxidation states according to the hierarchical table at left. Entries at the top of the table have priority over entries at the bottom. Page 69 of 78

Downloaded from hlayiso.com GRADE 11 Page 70 of 78 Balancing Redox Equations Using the Half-Reaction Method 1. Assign oxidation states to all atoms and identify the substances being oxidized and reduced. 2. Separate the overall reaction into two half-reactions, one for oxidation and one for reduction. 3. Balance each half-reaction with respect to mass in the following order. • Balance all elements other than H and O. • Balance O by adding H2O. • Balance H by adding H+. 4. Balance each half-reaction with respect to charge by adding electrons to the right side of the oxidation half-reaction and the left side of the reduction half-reaction. (The sum of the charges on both sides of each equation should then be equal.) 5. Make the number of electrons in both half-reactions equal by multiplying one or both half- reactions by a small whole number. 6. Add the two half-reactions together, cancelling electrons and other species as necessary. 7. Verify that the reaction is balanced with respect to both mass and charge. The above rules work with reactions in acid medium. If reactions are in basic solutions the rules are the same except that after adding H+ to balance the hydrogens, OH- must he added to neutralise the acidic H+. Balancing when reactions are occurring in basic solutions 1. Follow the half-reaction method for balancing redox reactions. Begin by assigning oxidation states. 2. Separate the overall reaction into two half-reactions . 3. Balance each half-reaction with respect to mass. • Balance all elements other than H and O. • Balance O by adding H2O. • Balance H by adding H+ • Neutralize by adding OH- (Add the same number of OH- to each side of the equation (to preserve mass balance). • Cancel any water molecules that occur on both sides of a half reaction. 4. Balance each half-reaction with respect to charge. 5. Make the number of electrons in both half - reactions equal. 6. Add the half-reactions together and cancel. 7. Verify that the reaction is balanced. The Activity Series: Predicting the spontaneous nature of reactions Page 70 of 78

Downloaded from hlayiso.com GRADE 11 Page 71 of 78 As we have seen, redox reactions depend on the gaining of electrons by one substance and the losing of electrons by another. Is there a way to predict whether a particular redox reaction will spontaneously occur? Mg(s) + Cu2+(aq)àMg2+(aq) + Cu(s) We see this as the fading of blue (the colour of the Cu2+ions in solution), the dissolving of the solid magnesium, and the appearance of solid copper on the remaining magnesium surface. This reaction is spontaneous—it occurs on its own when Mg(s) and Cu2+ come into contact. On the other hand, if we put Cu(s) in a solution containing Mg2+ions, no reaction occurs. Cu(s) + Mg2+ (aq) à no reaction No reaction occurs because, as we said previously, Mg atoms have a greater tendency to lose electrons than do Cu atoms;; Cu atoms will therefore not lose electrons to ions. Table 16.1 shows the activity series of metals. • This table lists metals in order of decreasing tendency to lose electrons. • The metals at the top of the list have the greatest tendency to lose electrons—they are most easily oxidized and therefore the most reactive. The metals at the bottom of the list have the lowest tendency to lose electrons—they are the most difficult to oxidize and Page 71 of 78

Downloaded from hlayiso.com GRADE 11 Page 72 of 78 therefore the least reactive. It is not a coincidence that the metals used for jewellery, such as silver and gold, are near the bottom of the list. • They are among the least reactive metals and therefore do not form compounds easily. Instead, they tend to remain as solid silver and solid gold rather than being oxidized to silver and gold cations by elements in the environment (such as the oxygen in the air). • Each reaction in the activity series is an oxidation half-reaction. The half reactions at the top are most likely to occur in the forward direction, and the half reactions at the bottom are most likely to occur in the reverse direction. Consequently, if we pair a half-reaction from the top of the list with the reverse of a half-reaction from the bottom of the list, we get a spontaneous reaction. More specifically, any half-reaction on the list will be spontaneous when paired with the reverse of any half-reaction below it. Activity Are the following redox reactions spontaneous? (a) Zn(s) + Ni2+(aq) à Zn2+(aq) + Ni(aq) (b) Zn(s) + Ca2+(aq) à Zn2+(aq) + Ca(s) (c) Fe(s) + Pb2+(aq) àFe2+(aq) + Pb(s) (d) Fe(s) + Mg2+ (aq) à Fe2+ (aq) + Mg(s) Page 72 of 78

Downloaded from hlayiso.com GRADE 11 Page 73 of 78 Typical Examination Question QUESTION 1 There are four possible options for each answer in the following questions. Each question has only ONE correct answer. Choose the correct answer and write only A, B, C or D next to the question number. 1.1 Which one of the pairs below is a conjugated acid-base pair for the following reaction? H2SO4(aq) + H2O(l) HSO4–(aq) + H3O+(aq) A H3O+ and H2O B H2SO4 and H3O+ C HSO4– and H3O+ D HSO4– and H2O (2) 1.2 Which ONE of the following equations represents an acid-base reaction? A H2SO4 + 3H2S ® 4S + 4H2O B 2HCl + Na2CO3 ® 2NaCl + H2O + CO2 C H2SO4 + 2NaCl ® Na2SO4 + 4HCl D 2HCl + FeS ® FeCl2 + H2S (2) 1.3 During the final process of gold extraction Zn is used. Which of the following redox reactions takes place? A Zn is reduced and 2NaAu(CN)2 is oxidized. B Zn is oxidized and Zn(CN)2 is reduced. C Zn is oxidized and 2NaAu(CN)2 is reduced. D Zn is reduced and Zn(CN)2 is oxidized. (2) 1.4 Make use of oxidation numbers to determine which one of the following reactions is a redox reaction. A NaCl + AgNO3 ® NaNO3 + AgCl B Mn + CuCl2 ® Cu + MnCl2 C H2SO4 + 2KNO3 ® K2SO4 + 2HNO3 D HCl + H2O ® Cl– + H3O+ (2) 1.5 When an acid reacts with a substance, the reaction produces carbon dioxide gas. The substance is most likely a … A base B carbonate C metal D Non-metal (2) Page 73 of 78

Downloaded from hlayiso.com GRADE 11 Page 74 of 78 1.6 The following energy profile diagram represents the reaction W + X ® Y + Z. Which one of the following statements is true? A DH = H3 – H1 B DH = H1 – H2 C The activation energy is H1 – H2. D The activation energy is H1 – H3 (2) 1.7. The following balanced equation describes the reaction for the formation of ammonia. N2 + 3H2 ® 2NH3 DH = – 47 kJ·mol–1 The reverse reaction … A is endothermic B is exothermic C needs less energy to activate than the forward reaction D has a rate that is not affected by temperature. (2) 1.8. Which of the following is the oxidation number of Cl in potassium chlorate (KClO3)? A –1 B 0 C +5 D +7 (2) [18] Page 74 of 78

Downloaded from hlayiso.com GRADE 11 Page 75 of 78 QUESTION 2 When methane gas burns in oxygen, the products are carbon dioxide and water. CH4 + 2O2 ® CO2 + 2H2O 2.1 Define bond length. (2) 2.2 Use the table of average bond energies to calculate the total energy needed to break all the bonds in the reactant molecules (2) Bond Energy (kJ·mol–1) Bond length (pm) H–H 436 - C–H 414 - O–H 460 - C–C 347 154 C=C 614 134 C≡C 9 120 C=O 799 - O=O 499 - 2.3 Calculate the total energy released when all the bonds in the product molecules form. (2) 2.4 Determine whether the reaction is exothermic or endothermic reaction? (2) 2.5 Use the information for different carbon bonds to describe the relationship between bond energy and bond length. (2) [10] QUESTION 3 In the energy profile diagram, the x-axis represents the extent of the reaction. The y-axis represents the energy of the reactants or products. Page 75 of 78

Downloaded from hlayiso.com GRADE 11 Page 76 of 78 3.1 Provide names for labels 1 to 4. (4) 3.2 Calculate the activation energy for the forward reaction shown on the graph (2) 3.3 Calculate the overall enthalpy of this reaction. (3) [10] QUESTION 4 Methane gas reacts with chlorine gas in the presence of light. The following equation describes the reaction. CH4 + Cl2 ® CH3Cl + HCl DH = –99,5 kJ 4.1 Decide whether this reaction is endothermic or exothermic. (1) 4.2 Explain how you reached your decision in question 4.1? (2) 4.3 Draw an energy profile diagram for this reaction. On your diagram, clearly indicate the reactants and products, the activation energy, the enthalpy change and the activated complex. (6) 4.4 Describe how methane gas in the atmosphere negatively affects life on Earth. (2) [11] QUESTION 5 Study the equation 2Al + 6HCl ® 2AlCl3 + 3H2. 5.1 Copy the equation and fill in the oxidation number of each atom above the symbol of the atom. (6) 5.2 Write the oxidation half reaction. (3) 5.3 Write the reduction half reaction. (3) 5.4 Identify the spectator ion. (2) 5.5 Write the formula of the substance that acts as the oxidising agent. (2) [16] QUESTION 6 Consider the equation for the reaction between zinc metal and a copper(II) sulfate solution. Zn + CuSO4 ® Cu + ZnSO4 6.1 Write the equation for the oxidation half reaction. (2) 6.2 Write the equation of the half reaction for the reducing agent (2) 6.3 Identify the oxidising agent. (2) [6] Page 76 of 78

Downloaded from hlayiso.com GRADE 11 Page 77 of 78 QUESTION 7 In the compound KMnO4, the oxidation number of Mn is +7. 7.1 Define oxidation number. (2) 7.2 Apply of the rules for oxidation numbers to prove that the oxidation number of Mn in KMnO4 is +7. (2) 7.3 Rui dissolved 20 g KMnO4 in 300 cm3 of distilled water. Calculate the concentration of the solution that he made. (4) [8] Question 8: Consider the following reactions NH 3 + HCO3- ® NH 4+ + CO32- 8.1 Identify: 8.1.1 the conjugate base of HCO3- (1) 8.1.2 a substance that is an ampholyte (1) 8.2 Give the complete balanced equation for the reaction between sulphuric acid and lithium. (3) Consider the following reaction: 4 NH3 + 5O2 ® 4 NO + 6H 2O 8.3 Rewrite the equation and above each element, write down its oxidation number. (4) 8.4 Write out the reduction half reaction for this reaction. (2) [11] QUESTION 9 Michael dissolves 1,4 g potassium hydroxide (KOH) in water. He makes up the solution to a final volume of 250 cm3. 9.1 Calculate the concentration of the KOH solution in units of mol•dm–3. (4) 9.2 An amount of 25 cm3 of Michael’s KOH solution neutralises 30 cm3 of a diluted hydrochloric acid solution. The balanced equation for the reaction is as follows. KOH + HCl ® KCl + H2O Calculate the concentration of the hydrochloric acid solution. (4) [8] Page 77 of 78

Downloaded from hlayiso.com GRADE 11 Page 78 of 78 QUESTION 10 Jakob prepares a standard basic solution of sodium hydroxide (NaOH). He dissolves 12g NaOH powder in 500 cm3of water. 10.1 Use the Lowry-Brfnsted theory to define a base. (2) 10.2 Explain what a standard solution is. (2) 10.3 Name the type of glassware that Jakob should use to prepare the standard solution. (2) 10.4 Calculate the concentration of Jakob’s NaOH solution. (6) [12] Question 11 11.1 Define the acid using Lowry-Bronsted theory (2) 11.2 HSO4-(aq) acts as a base and an acid 11.2.1 Name the underlined sentence (1) 11.2.2. Write down an ionic equation showing HSO4-(aq) acting as the base. (4) 11.2.3 Indicate conjugate acid-base pairs in question 11.2.2. (2) 11.3. Mg(OH)2 is an active ingredient in tablets used for indigestion, and is a STRONG BASE 11.3.1 What is meant by the term STRONG BASE (2) 11.3.2. 50cm3 3 of gastric juice (Hydrochloric acid) is titrated against 45cm3 of 0,01mol.dm-Magnesium chloride to neutralise fully. Calculate the concentration of a Gastric Juice. (7) [18] Page 78 of 78

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