Physical Sciences
Junior Tukkie Gr.12 Winter School 2019
Facilitator: Dr. Sarita Swanepoel
Junior Tukkie Winter School 1 Dr. S. Swanepoel (2019)
Physical Sciences
Contents
1 Chemical equilibrium 3
2 Projectile motion 10
3 Work, energy and power 16
4 Acids and bases 24
5 Mole and stoichiometric calculations 32
Junior Tukkie Winter School 2 Dr. S. Swanepoel (2019)
1 Chemical equilibrium
Chemical equilibrium is a dynamic equilibrium that exists when the rate of theforward reaction equals the rate of the reverse reaction for a reversible reaction
︸ ︷︷ ︸in
a closed system.︸ ︷︷ ︸
isolated from the surroundingsno reagents can escape
reverse reactioncan take place
N2 + 3H2 2NH3
N2 and H2 is sealed in a container:
Initially: Over time: Eventually:
High [N2] and [H2]→ Forward reaction fast[NH3] = 0→ No reverse reaction
[N2] & [H2] decrease (used)→ Forward reaction slower
[NH3] increases (produced)→ Reverse reaction faster
Chemical equilibriumreached.Forward and reverse reactionat the same rate.[N2], [H2] & [NH3] constant.Both rates stay constant.
Draw reaction rate vs time graphs for both reactions:
Reaction rate(mol.dm−3.s−1)
time(s)
Equilibrium constant (K c)
The equilibrium constant is the ratio of concentration of the products to theconcentration of the reactants in the equilibrium mixture and the value holds onlyfor a specific temperature .
aA + bB cC + dD Kc = [C]c[D]d
[A]a[B]b
Square brackets [ ] represent concentrations and Kc has no unit.Pure solids and liquids/solvents are left out.
Kc > 1: More products than reactants in the equilibrium mixture.Kc < 1: More reactants than products in the equilibrium mixture.
Junior Tukkie Winter School 3 Dr. S. Swanepoel (2019)
Le Chatelier’s Principle:
When the equilibrium in a closed system is disturbed, the system will re-instate a newequilibrium by favouring the reaction that will oppose the disturbance.
The following dynamic equilibrium is reached in a closed container:
2A(g) + B(s) ⇆ 2C(g) + D(g) ∆H is negative
Kc =[C]2[D]
[A]2c =
nV
If A is added to the mixture in the container:Disturbance: [A] increases.According to Le Chatelier’s principle the system reacts to decrease the [A] .The forward reaction is favoured. (Use A)More C and D are produced.Kc remains the same.
If C is added to the mixture in the container:Disturbance: [C] increases.According to Le Chatelier’s principle the system reacts to decrease the [C] .The reverse reaction is favoured. (Use C)D decreases.Kc remains the same.
If A is removed from the mixture in the container:Disturbance: [A] decreases.According to Le Chatelier’s principle the system reacts to increase [A] .The reverse reaction is favoured (make A).C and D decreases.Kc remains the same.
B(s) is added:Solid has no effect on equilibrium.
If the volume of the container is decreased:Disturbance: pressure increases (Boyle’s Law).According to Le Chatelier’s principle the system reacts to lower the pressure.Reacts to produce less moles of gas.Reverse reaction is favoured.C and D decreases.Kc remains the same.
If the temperature is increased:Disturbance: Temperature increases.According to Le Chatelier’s principle the system reacts to decrease the temperature.Endothermic reaction is favoured (uses energy).(∆H is negative ∴ forward reaction exothermic.)The reverse reaction is favoured.C and D decreases.Kc decreases.
Catalyst is added:No effect on equilibrium.The rate of both reactions increase equally .
Favoured does not indicate that the reaction happens faster than before, but that it happensfaster than the reaction in the opposite direction.
Action Disturbance Explanation
Adding acid [H+] increases Acid donates H+
Adding base [H+] decreases Base reacts with H+ (proton acceptor)
Adding AgNO3 [Cl−] decreases Ag+ + Cl− AgCl(s) precipitate
Adding BaCl2 [SO4−2] decreases Ba+2 + SO4
−2 BaSO4(s) precipitateor Ba(NO3)2
The colours of compounds with transition elements are well-known. For example,consider the following equation of a reversible reaction:
yellow︷ ︸︸ ︷
2CrO4−(aq) + 2H+(aq)
orange︷ ︸︸ ︷
Cr2O72−(aq) + H2O(ℓ)
1. Write down an expression for the equilibrium constant.
2. How will the equilibrium constant (Kc) be influenced in the following situations:
a) sodium chromate (Na2CrO4) is added.
b) Hydrochloric acid is added.
c) Sodium hydroxide is added
3. Explain in detail how the colour of the solution will be influenced whensodium hydroxide is added.
Junior Tukkie Winter School 5 Dr. S. Swanepoel (2019)
2A(g) + B(g) 3C(g) + D(g)
A and B intocontainer
A added B removed D added
Disturbance
Le Chatelier: Systemreacts to . . .
The . . . reaction isfavoured
[C] . . .
Kc
Reaction rate
Concentrations
Junior Tukkie Winter School 6 Dr. S. Swanepoel (2019)
∆ H < 0
Volumedoubled
Temperatuurincreased
Temperaraturedecreased
Catalystadded
Junior Tukkie Winter School 7 Dr. S. Swanepoel (2019)
Le Chatelier, Temperature and Equilibrium constants
Choose the correct word in the brackets:
1.1 The following reaction is at equilibrium at 400K:A2(g) + 3B2(g) 2AB3(g) ∆H < 0
If the temperature is increased to 600 Kthe Kc-value will (increase/decrease/stay the same).
1.2 A mixture of NO and Br2 is placed in a container. The following traction occurs:2NO(g) + Br2(g) 2NOBr(g)
After 15 minutes chemical equilibrium is reached. The following graph shows how therate of the forward reaction changes with time:
15
rate of
forward
reaction
time
(minutes)
After 15 s the temperature was (increased/decreased) andthe Kc=value (increased/fecreased/remained constant).
1.3 P and Q are sealed in a container and the following chemical equilibrium is reached after5 minutes at 500 K: P(g) + 2Q(g) R(g)
After 10 minutes the temperature is increased to 600 K and after 5 minutes a new chem-ical equilibrium is established. The following graph shows the change in concentrationwith time.
time
(minutes)
[P]
5 10 15
The Kc-value (increased/decreased/remained constant).∆H for the reaction is (positive/negative).
Junior Tukkie Winter School 8 Dr. S. Swanepoel (2019)
1.4 The following reaction mixture turns more red when the temperature is increased:3H(g)︸ ︷︷ ︸
pink
2F(g)︸ ︷︷ ︸
red
+ 2D(g)
When the temperature is increased the equilibrium constant (increases/decreases).
∆H is (positive/negative).
1.5 The following reaction reached chemical equilibrium in a closed container:2H(g) + I(g) 2 K(g) ∆H < 0
time
(minutes)
[H]
X
At time X the temperature was (increased/decreased) andthe Kc-value (increases/decreases/remains constant).
1.6 The following reaction is in chemical equilibrium:3M(g)︸ ︷︷ ︸
green
2N(g)︸ ︷︷ ︸
blue
+ 2P(g)
When the temperature increases the mixture turns more blue.
∆H is (positive/negative).
1.7 The following reaction reached chemical equilibrium in a closed container:2X(g) + Y(g) Z(g) + 2A(g)
time
(minutes)
rate of
forward
reaction
X
After time X the Kc value is (larger/smaller) than before.The heat of the reaction is (positive/negative).
Junior Tukkie Winter School 9 Dr. S. Swanepoel (2019)
2 Projectile motion
A projectile is an object upon which the only force acting is the force of gravity.
Free fall is the movement of an object upon which the only force acting is the force of gravity.The object moves at a constant downwards gravitational acceleration.
Time symmetry : The time for the upwards movement is equal to the time for the downwardsmovement .
Vectors Positive or negative
∆ x, ∆y Displacement: Directly from starting point to finishing point
v Direction of movement
a Direction of Fnet
g Direction of Fnet that is Fg
∴ always downwards
Example: Calculate the velocity at which an arrow must be shot from a15 m high building to reach the earth after 10 s.
Use up as positive .
∆y = vi∆t+1
2g∆t2
− 15 = Vi(10) +1
2(−9, 8)(102)
vi = 47, 5 m · s−1 upwards
start
finish
�y = - 15 m
+
Analysis of movement graphs
Gradient of the graph
gradient = ∆y-axis∆x-axis
Gradient of x:t-graph = ∆x∆t
= vGradient of v:t-graph = ∆v
∆t= a
Area under the graph
Area: multiplying the variables of the axis.
Area under v:t-graph = v.∆t = ∆x
Area under a:t-graph = a.∆t = ∆v
gradient−−−−−−→x v a←−−−−−−
area(∆ in variable)
Junior Tukkie Winter School 10 Dr. S. Swanepoel (2019)
Gerhard shoots an arrow at a speed of X m·s−1 upwardsfrom the top of a building that is 40 m high. The arrow travels6 m up before turning around. Give the values (or just thesigns) of each of the variables for the following parts of themovement:
Take up as positive:
Movement from the . . . vi vf ∆ y(m ·s−1) (m ·s−1) (m)
starting point to highest point
starting point to same height again
starting point to the ground
highest point to the top of the building
highest point to the ground
An eagle is flying at 2 m·s−1 straight upwards. When then the eagle is65 m above the ground a flea loses his grip on theeagle’s feather and falls. Give the values (or just signs) for each of thevariables for the following parts of the flea’s movement:
Take up as positive:
Movement from the . . . vi vf ∆ y(m ·s−1) (m ·s−1) (m)
starting point to the same height
starting point to the ground
Junior Tukkie Winter School 11 Dr. S. Swanepoel (2019)
Bouncing Ball
A ball is dropped from a height of 2 m. Draw graphs for the movement.
Take upwards as positive and the earth as reference point.
Picture
of ball
Position x
(m)
time (s)
Velocity v
(m.s
-1)
time (s)
Acceleration
a (m
.s
-2)
time (s)
Junior Tukkie Winter School 12 Dr. S. Swanepoel (2019)
Redraw the following graphs according to the direction given as positive the point given asreference point.
Up positive Up positive Down positive Down positiveEarth = 0 Release point = 0 Earth = 0 Release point = 0
y(m)
t (s)
Up positive Down positive
t (��
v(��� �-1
Junior Tukkie Winter School 13 Dr. S. Swanepoel (2019)
QUESTION 1
While a hot air balloon is rising at 1,96 m·s−1 a passengerloses his grip on his binoculars. The binoculars hit the ground6 s later,
TAKE UP AS POSITVE.
1.1 Calculate the time from when the passenger releases the binoculars until it reaches itsmaximum height.
1.2 Calculate the height of the balloon at the moment that the binoculars are dropped.
1.3 Draw a sketch graph (no scale necessary) of the binoculars’ velocity against time fromthe moment that they were dropped until they reached the ground. Indicate ALL knownvalues.
1.4 Use the graph to calculate the velocity of the binoculars when they hit the earth.
Junior Tukkie Winter School 14 Dr. S. Swanepoel (2019)
QUESTION 2
John is lying on the roof of a building and throws a ball downwards. The following graph repre-sents the movement from the moment that the ball leaves John’s hand.
0,0
−2,5
−5,0
−7,5
−10,0
2,5
5,0
7,5
10,0
t(s)
v(m/s)
t A
B
C
D
2.1 Calculate time t.
2.2 During which interval was the ball in free fall? Write only AB, BC or CD.
2.3 Is the ball hard or soft? Explain.
2.4 Use the graph to calculate the height of the building.
2.5 Draw a graph of the ball’s position against time from the moment that it leaves the roofuntil it hits the ground the second time. (Only known values have to be indicated.)
Junior Tukkie Winter School 15 Dr. S. Swanepoel (2019)
3 Work, energy and power
The work done on an object by a constant force F is F ∆x cosΘ , where F is the magnitude ofthe force, ∆x the magnitude of the displacement and Θ the angle between the force and thedisplacement. W = F∆x cos θ
W = F ∆x cos θ
Workin Joulescalar
MAGNITUDEof the forcein Newton
MAGNITUDEof the displacementin meter
angle between theforce and displacement
W scalar (no direction): A negative W is energy removed from object.
Net work : Wnet = Wg + WT + Wfric + WN (use the whole force)Wnet = Fnet ∆ x cos θ (use components and calculate Fnet)
Work-energy theorem:The net/total work done on an object is equal to the change in the object’s kinetic energy.In symbols: Wnet = ∆ EK
Wnet = 12m(vf 2 - vi2)
Conservative force : The work done by the force in moving an object between 2 pointsis independent of the path taken ex. gravitational, electrostatic and elastic forces.Non-conservative force : The work done by the force in moving an object between 2 pointsdepends the path taken ex. frictional force, air resistance, tension in a chord.
Work done by non-conservative forces: Wnc = ∆ EK + ∆ EP since Wg = −∆ EP
All the W useful when ‘no corner‘except Wg is given for an inclined plane
Mechanical energy: Emech = Ek + Ep
Kinetic energy energy due to movement: Ek = 12mv2
Gravitational potential energy: energy due to position: Ep = mgh
The principle of conservation of mechanical energy:The total mechanical energy (sum of gravitational potential energy and kinetic energy) in anisolated system remains constant. Emech(i) = Emech(f) (Only Fg)
Epi + Eki = Epf + Ekf
(g and v no + or −) mgh + 12mv2 = mgh + 1
2mv2
Power : rate at which work is done/energy is expended. P = W∆t or Pave = Fvave
(A power of 200 W means 200 J energy is used/work is done per second.)
Work
The work done on an object by a constant force F is F ∆x cosΘ , where F is the magnitude ofthe force, ∆x the magnitude of the displacement and Θ the angle between the force and thedisplacement. W = F∆x cos θ
W = F ∆x cos θ
Workin Joulescalar
MAGNITUDEof the forcein Newton
MAGNITUDEof the displacementin meter
angle between theforce and displacement
Example
A crate, with mass 10 kg, is pulled 4 m up an inclinedplane that makes an angle of 30◦ with the ground. Thecrate is pulled with a force of 180 N and experiences africtional force of 10N. Calculate the work done by eachof the forces working on the crate.
30o
F
Applied force:WF = F∆x cos θ
= 180(4) cos 0◦
= 720 J
Friction:Wf = f∆x cos θ
= 10(4) cos 180◦
= −40 J
Gravity:Wg = Fg∆x cos θ
= (10× 9, 8)(4) cos(90 + 30)◦
= (10× 9, 8)(4) cos 120◦
= −196 J
Normal force:WN = N∆x cos θ
= N∆x cos 90◦
= 0 J
Junior Tukkie Winter School 17 Dr. S. Swanepoel (2019)
Horizontal Pull Push Inclined plane
-
F
Fx
Fy
FFy
Fx
Fg⊥
Fg‖
Fgθ
θ
Fx = F cos θFy = F sin θ up
Fx = F cos θFy = F sin θ down
Fg⊥ = Fg cos θFg ‖= Fg sin θ
•fk F
N
Fg
•fk
Fy
Fx
N
Fg
•fk
Fy
Fx
N
Fg
◦
◦
Fg⊥
Fg‖
Fg
N
θ
θ
N = Fg N = Fg − Fy N = Fg + Fy N = Fg⊥
Different methods to calculate W g
A 10 kg toy car is pulled 3 m up an inclined plane. The plane is at a30◦ angle to the ground and the height is1,5 m. Calculate the work done by gravity.
Method 1 Method 2 Method 3
Accoding to definition Fg components Fg conservative force
Wg = Fg∆x cos θ
= 98(3) cos(90◦ + 30◦)
= 98(3) cos(120◦)
= −147, 00J
Wg = Wg|| +Wg⊥
= Fg||∆x cos θ + 0
= (98 sin 30◦)(3) cos 180◦
= 49(3) cos 180◦
= −147, 00J
Wg routeA = Wg routeB
Wg = Wg(BI) +Wg(BII)
= 0 + Fg(h) cos 180◦
= 98(1, 5) cos 180◦
= −147, 00J
A man pulls a 50 kg-washing machine 3 m up an inclined plane byexerting a force of 2000 N parallel to the plane. The plane makes anangle of 40◦ with the horizon. The washing machine experiences20 N frictional force.
F
a. Draw a free body-diagram of all the forcesacting on the machine. (No components)
b. Calculate the work done by every force.
c. Use the previous answers to calculate thenet work.
a. Draw a free body-diagram of all the forceson the machine. Use components of Fg.
b. Calculate the net force on the machine.
c. Use the Fnet to calculate the net work.
The washing machine starts from rest. Use the work-energy principle to prove that after 3 m themagnitude of the velocity is 14,14 m·s−1.
Calculate the average power of the man withp = W
∆t
Calculate the average power of the man withpave = Fvave
Closed system Any systemNo friction or applied force With or without friction
Conservation of mechanicalenergy
Work-energy principle
∆x givenEmech(i) = Emech(f)
Epi + Eki = Epf + Ekf
mghi +1
2mv2
i = mghf +1
2mv2
f
ghi +1
2v2i = ghf +
1
2v2f
Wnet = ∆EK
WT + Wf + WN + Wg︸ ︷︷ ︸
Every W=F∆x cos θ
=1
2m(vf
2 − vi2)
No components
Pendulums & free fallInclined planes & curved planes
or Wnet = ∆EK
Fnet∆x cos θ︸ ︷︷ ︸
Use components
=1
2m(vf
2 − vi2)
v and g only magnitude (no sign) v only magnitude (no sign)
Conservation of momentum Impulse-momentum principle
Collisions and explosions ∆t givenNB: Directions!!! NB: Directions!!!
Σpi = Σpf
p1i + p2i = p1f + p2f
m1vi1 +2 mvi2 = m1vf + m2vf
Fnet∆t = ∆p
Fnet∆t = pf − pi
Fnet∆t = m(vf − vi)
Sometimes Elastic collisions Work-energy principle(Conservation of kinetic energy) for non-conservative forces
ΣEk(i) = ΣEk(f)
Ek1i + Ek2i = Ek1f + Ek2f
1
2m1v
21i +
1
2m2v
22i =
1
2m1v
21f +
1
2m2v
22f
Wnet = ∆EK
Wnc = ∆EK +∆EP
WT + Wf + WN︸ ︷︷ ︸
All W except Wg
=1
2m(vf
2 − vi2) + mg(hf − hi)
v only magnitude (no sign) v and g only magnitude (no sign)
If collision is elastic: ΣEk(i) = ΣEk(f) Inclined planes with no angleIs the collision elastic? Calculate
ΣEk(i) and ΣEk(f) and compare
Junior Tukkie Winter School 20 Dr. S. Swanepoel (2019)
The script of a new James Bond movie includes the following scenario:
James Bond (80 kg) starts from rest and skis down a 25 m slope with a villain athis heels. The slope makes an 38◦ angle with the ground and James experiences africtional force of10 N. At the bottom of the slope he covers a horizontal plane for 15 s and experiencesa 15 N frictional force. It brings him to a parcel (1 kg) fixed to an inelastic rope. Hegrabs the parcel and swings up to the window on the second floor 5,2 m above theground. He releases the parcel, breaks the window and escapes through the building.5,4 × 105 J is required to break the window.
You are the technical advisor to the producer and must determine if thescenario is possible.
Junior Tukkie Winter School 21 Dr. S. Swanepoel (2019)
Physical Sciences/P1 11 DBE/Feb.–Mar. 2015
NSC
QUESTION 5 (Start on a new page.) A 5 kg block is released from rest from a height of 5 m and slides down a frictionless incline to point P as shown in the diagram below. It then moves along a frictionless horizontal portion PQ and finally moves up a second rough inclined plane. It comes to a stop at point R which is 3 m above the horizontal.
The frictional force, which is a non-conservative force, between the surface and the block is 18 N.
5.1 Using ENERGY PRINCIPLES only, calculate the speed of the block at
point P.
(4) 5.2 Explain why the kinetic energy at point P is the same as that at point Q. (2) 5.3 Explain the term non-conservative force. (2) 5.4 Calculate the angle (θ) of the slope QR. (7)
[15]
5 kg
P Q
R
3 m
θ
5 m
Junior Tukkie Winter School 22 Dr. S. Swanepoel (2019)
Physical Sciences/P1 10 DBE/Feb.–Mar. 2015
NSC
20 g 7 kg
2 m
QUESTION 4 (Start on a new page.) The diagram below shows a bullet of mass 20 g that is travelling horizontally. The bullet strikes a stationary 7 kg block and becomes embedded in it. The bullet and block together travel on a rough horizontal surface a distance of 2 m before coming to a stop.
4.1 Use the work-energy theorem to calculate the magnitude of the velocity of the
bullet-block system immediately after the bullet strikes the block, given that the frictional force between the block and surface is 10 N.
(5) 4.2 State the principle of conservation of linear momentum in words. (2) 4.3 Calculate the magnitude of the velocity with which the bullet hits the block. (4)
[11]
Junior Tukkie Winter School 23 Dr. S. Swanepoel (2019)
4 Acids and bases
Arrhenius theory:An acid is a substance that produces hydrogen ions (H+)/hydronium ions (H3O+) when itdissolves in water. A base is a substance that produces hydroxide ions (OH−) when it dissolvesin water.
Lowry-Brønsted theory:An acid is a proton (H+ ion) donor.A base is a proton (H+ ion) acceptor.
Strong acids ionise completely in water to form a high concentration of H3O+ ions.Weak acids ionise incompletely in water to form a low concentration of H3O+ ions.
Strong bases dissociate completely in water to form a high concentration of OH− ions.Weak bases dissociate/ionise incompletely in water to form a low concentration of OH− ions.
Conjugate acid-base pairsWhen the acid, HA, loses a proton, its conjugate base, A-, is formed.When the base, A-, accepts a proton, its conjugate acid, HA, is formed.
Ampholyte or amphiprotic substance a substance that can act as either as an acid or as abase.
Concentrated acids/bases contain a large amount (number of moles) of acid/base inproportion to the volume of water.Dilute acids/bases contain a small amount (number of moles) of acid/base in proportion to thevolume of water.
Hydrolysis the reaction of a salt with water.
pH Scale is a scale with numbers 0 to 14 that is used to express the acidity of alkalinity of asolution.
Kw is the equilibrium constant for the ionisation of water or the ionic product of water or theionisation constant of water, i.e. Kw = [H3O+][OH−] = 1 x 10−14 at 298 K.
Auto-ionisation of water the reaction of water with itself to form H3O+ ions and OH− ions.
Neutralisation is the reaction between an acid an a base to produce a salt and water.
Titration is an experiment that is used to determine the concentration of an acid or a base byusing a neutralisation reaction with a standard solution.
A Standard solution is a solution of which the concentration is known and remains constantfor a time.
Equivalence point of a titration is the point at which the acid /base has completely reactedwith the base/acid.
Endpoint of a titration is the point where the indicator changes colour.
Acid Formula Strength
Hydrochloric acid HCl strong
Sulphuric acid H2SO4 strong
Nitric acid HNO3 strong
Phosphoric acid H3PO4 strong
Sulphurous acid H2SO3 weak
Carbonic acid H2CO3 weak
Oxalic acid (COOH)2 weakH2C2O4
Acetic acid CH3COOH weak(etanoic acid)
Base Formula Strength
Sodium hydroxide NaOH strong
Potassium hydroxide KOH strong
Lithium hydroxide LiOH strong
Calcium hydroxide Ca(OH)2 weak
Magnesium hydoxide Mg(OH)2 weak
Ammonia NH3 weak
Potassium carbonate K2CO3 weak
Sodium bicarbonate NaHCO3 weak
Sodium carbonate Na2CO3 weak
Acids BasesBrønsted-Lowry: Proton donor Brønsted-Lowry: Proton acceptor
Strong acids Strong bases(ionise completely) (dissociate completely)
covalent → no ions → ionises ionic → has ions → dissociate
HCl(g) + H2O(l) H3O+(aq) + Cl−(aq) KOH K+(aq) + OH−(aq)hydrochloric acid H2O
HNO3(g) + H2O(l) H3O+(aq) + NO3−(aq) NaOH Na+(aq) + OH−(aq)
nitric acid H2O
H2SO4(l) + 2H2O(l) 2H3O+(aq) + SO4−2(aq) LiOH Li+(aq) + OH−(aq)
sulfuric acid H2O
Ka = [Cℓ−][H3O+]
[HCl] = VERY BIG Kb = VERY BIG
Weak acids Weak bases(ionise incompletely) (dissociate/ionise incompletely)
covalent → no ions → ionises Covalent NH3 ionises and forms NH4OH that weakly dissociates
CH3COOH + H2O(l) H3O+(aq) + CH3COO−(aq) NH3 + H2O NH4+(aq) + OH−(aq)
acetic acid acetate ionethanoic acid ethanoate ion
H2CO3 + H2O(l) H3O+(aq) + HCO3−(aq) Na2CO3 dissociates completely and give CO3
−2
carbonic acid that hydrolyses incompletely
(COOH)2 + 2H2O 2H3O+(aq) + (COO)2−2 Na2CO3 2Na+(aq) + CO3−2(aq)
oxalic acid oxylate ion
Ka = [HCO−
3 ][H3O+]
[H2CO3]= small Kb = small
Junior Tukkie Winter School 25 Dr. S. Swanepoel (2019)
Reactions of acids
Acid and reactive metal → salt + hydrogen gas
2HCl(aq) + Zn(s)→ ZnCl2(aq) + H2(g)H2SO4(aq) + Mg(s)→ MgSO4(aq) + H2(g)
Acid and metal hydroxide → salt + water
2HCl(aq) + Zn(OH)2(s)→ ZnCl2(aq) + 2H2O(l)H2SO4(aq) + 2NaOH(aq)→ Na2SO4(aq) + 2H2O(l)
If the base is NH3:2HCl(aq) + NH3(g)→ NH4Cl(aq) or2HCl(aq) + NH3(g)→ NH4
+(aq) + Cl−(aq)
Acid and metal oxide → salt + water
2HCl(aq) + ZnO(s)→ ZnCl2(aq) + H2O(l)H2SO4(aq) + Na2O(aq)→ Na2SO4(aq) + H2O(l)
Acid and metal carbonate → salt + water + carbon dioxide
2HCl(aq) + ZnCO3(s)→ ZnCl2(aq) + H2O(l) + CO2(g)H2SO4(aq) + Na2CO3(aq)→ Na2SO4(aq) + 2H2O(l) + CO2(g)
[Strong acid]
c = m
M V Dilute
c1v1 = c2v2
- dissociates completely
- equation gives ratio
- ionises completely
- equation gives ratio
pH
Kw = [ OH- ][ H3O+ ]
pH = - log[ H3O + ]
Calculations with acids and bases
Indicator Colour in acid Colour in base coulour chage pH
Phenolphthalein Colourless Pink 8,3 - 10,0
Bromothymol blue Yellow Blue 6,0 - 7,7
Methyl orange Orange Yellow 3,1 - 4,4
Junior Tukkie Winter School 26 Dr. S. Swanepoel (2019)
Concentration c = nV en n = m
V of c = mM V (V in dm3)
Dilutions C1 V1︸ ︷︷ ︸
old
= C2 V2︸ ︷︷ ︸
newNB. New volume = original volume + water added !!
Conjugatedacid base pairs
base conj acidNH3(g) + H2O(ℓ) NH4
+(aq) + OH−(aq)acid conj base
conjugated pair 1
conjugated pairs 2
The conjugated base of a strong acid is weak and the conjugatedacid of a strong base is weak.
Outo-protolysisof water
Water is an ampholite and can undergo protolysis:H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
Kw = [H3O+][OH−] = 1× 10−14 by 25◦ C
Bigger [H3O+]→ smaller [OH−]
Neutral solution [H3O+] = [OH−]Acidic solution [H3O+] > [OH−]Basic solution [H3O+] < [OH−]
pH scale pH = − log [H3O+] (IEB no calculations)
Bigger pH → lower [H3O+]
[H3O+] 10−2 10−3 10−4 mol.dm−3
pH 2 3 4
If [H3O+] decreases by factor 10 the pH increases by 1.
Titration nanb
=Ca.VaCb.Vb
Ion Hydrolysis Reason
Cl−, SO4−2, NO3
− None Conjugated base of strong acid
Na+, K+, Li+ None Ion from a strong base
CO3−2 CO3
−2 + H2O HCO3− + 2OH− Conjugated base of weak acid
carbonate
CH3COO− CH3COO− + H2O CH3COOH + OH−
acetate / ethanoate
(COO)2−2 (COO)2−2 + H2O H(COO)2−1 + OH−
C2O4−2 C2O4
−2 + 2H2O HC2O4−1 + OH−
oxalate ion
NH4+ NH4
+ + H2O NH3 + H3O+ Conjugated acid of weak base
Junior Tukkie Winter School 27 Dr. S. Swanepoel (2019)
Write balanced equations for the following:
a. Nitric acid and water
b. Sodium hydroxide dissolves un water
c. Reaction of oxalic acid and lithium oxide
d. Reaction of ammonia with water
e. Reaction of sulphuric acid and water
f. Neutralisation that produces potassium sulphate
g. Ionisation of oxalic acid in water
h. Reaction between lithium oxide and water
i. Reaction between ammonia and hydrochloric acid
Junior Tukkie Winter School 28 Dr. S. Swanepoel (2019)
QUESTION 3
3.1 Calculate how much water must be added to 30 cm3 of a 0,2 mol.dm−3 HCl solution tochange the concentration to 0,03 mol.dm−3.
3.2 A solution of acid HX with concentration 0,15 mol.dm−3 is prepared. The concentrationof hydronium ions in the solution is 3,2 × 10−6 mol·dm−3.
a. Write an equation for the reaction of HX with water.
b. Is HX a strong acid? Explain.
c. Name the conjugated acid base pairs in the reaction.
3.3 X cm3 of water is added to 300 cm3 of a 0,15 mol.dm−3 sulphuric acid solution. Theconcentration decreases to 0,045 mol.dm−3.
a. Calculated the volume X of water that is added.
b. Calculate the concentration of the hydronium ions in the solution.
c. Calculate the pH of the diluted solution.(Not IEB)
Junior Tukkie Winter School 29 Dr. S. Swanepoel (2019)
3.4 0,5 g of lithium hydroxide is used to prepare 200 cm3 of solution.
a. Calculate the concentration of the lithium hydroxide
b. What is the concentration of the hydroxide ions in the solution?
c. What is the concentration of the lithium ions in the solution?
d. Calculate the hydronium ion concentration in the solution.
e. Calculate the pH of the solution. (Not IEB)
3.5 The pH of household ammonia is 11,61. Calculate the [H+] and the [OH−] of this dilutesolution. (Not IEB.)
Junior Tukkie Winter School 30 Dr. S. Swanepoel (2019)
Hydrolysis
Hydrolysis is the reaction of a salt with water.
Example: ammonium chloride
What acid and base can be used to prepare the salt?NH4Cl can be prepared by the reaction of ammonia with hydrochloric acid.It is a reaction between a strong acid and a weak base and the solution is acidic.
Consider the reactions of the ions with water to explain why the solution is acidic.Cl− does not hydrolyse. It is the weak conjugated base of a strong acid.
NH4+ is the conjugated acid of a weak base.
It hydrolyses to form H3O+, that makes the solution acidic:NH4
+ + H2O NH3 + H3O+ (Table p.26)
a. Give the formulas of the acid and the base that can be used to prepare the salt and indicateif the solution is acidic, basic or neutral.
Compound Acid Base Solution
ammonium nitrate HNO3 NH3
Strong Weak Acidic
potassium chloride
sodium oxalate
ammonium carbonate
lithium acetate
b. Sodium oxalate is dissolved in water. Is the solution acidic, basic or neutral? Use equationsto explain the answer.
Junior Tukkie Winter School 31 Dr. S. Swanepoel (2019)
5 Mole and stoichiometric calculations
Mass Particles Gas at STP Solutions
n = mM
n =N
NA
n =V
Vm
c = nV or c = m
M V
m mass g NA = 6,02 × 1023 VM = 22,4 dm3mol−1 c concentration mol.dm−3
M molar mass g.mol−1 N number of particles v volume dm3 v volume dm3
a) 300 cm3 solution contains 100 gNaCl. Calculate the solution con-centration.
b) How many moles is 9,03 x 1024
NH3 molecules?c) How many mol CO2(g) are therein 4,48 dm3 at STP?
d) What is the volume of 2,7 molN2(g) at STP?
e) What is the mass of3,6 mol potassium sulphate?
f) How many molecules are therein 4,2 mol ammonia?
Junior Tukkie Winter School 32 Dr. S. Swanepoel (2019)
Stoichiometry
When the substance given is not the substance asked, the rati o must be used.This step must be shown even if the ratio is 1:1
The amount of substance can be given/asked as volume/mass or concentration.It is converted to moles before/after working with the ratio .
Calculate the mass Na2O that can be produced when 4,93 dm3 oxygen gas (at STP)reacts with an excess of sodium. Na + O2 Na2O
4Na + O2 2Na2On n
v = 4, 93 dm3 m =?
n =v
Vm
=4, 93
22, 4
= 0, 22 molO2
O2 : Na2O1 : 2
0,22 : x1 × x = 2 × 0,22
x = 0,44 mol Na2O
n =m
M
0, 44 =m
62
m = 27, 28 gNa2O
1. What mass of C4H10 is required to react completely with 4,48 dm3 oxygen gas at STP?2C4H10 + 13O2 CO2 + 10H2O
2. What mass of glucose will be produced when 100 g carbon dioxidereacts completely?
6CO2(g) + 6H2O (ℓ)→ C6H12O6(s) + 6O2(g)
3. Calculate the volume nitogen dioxide (at STP) that will be produced when1, 5× 1024 N2O5 molecule decompose.
2N2O5(g)→ 4NO2(g) + O2(g)
4. Calculate the volume of a 0,2 mol.dm−3 HCℓ -solution that is needed to produce3,36 dm3 Cℓ2(g) at STP.
2KMnO4 + 16HCℓ→ 2KCℓ+ 2MnCℓ2 + 8H2O + 5Cℓ2(g)
Junior Tukkie Winter School 34 Dr. S. Swanepoel (2019)
Exception: Gas volume to gas volume.
Avogadro’s law: Equal volumes of all gasses, at the same temperature and pressure, havethe same number of molecules (and same mole).
Reagents are in the same container and therefore are at the same temperature and pressure.The volume ratio is the same as the mole ratio .What volume ammonia gas will be produced when 2,24 dm3 nitrogen gas reacts completelywith an excess of hydrogen gas?
N2(g) + 3H2 (g) → 2NH3(g)
n n
V = 2,24 dm3 V = ?
N3 : NH3
mole ratio 1 : 2
volume ratio 1 : 2
2,24 dm3 : x
4,48 dm3 NH3 is produced
4 The following reaction take place in a container where CONDITIONS ARE NOT STP!Calculate the volume nitogen dioxide that will be produced when 4,86 dm3
N2O5 decompose.
2N2O5(g)→ 4NO2(g) + O2(g)
5. The following reaction occurs under non-standard conditions:
N2(g) + 3H2(g) 2NH3(g)
What volume ammoniak gas is produced when 4,48 dm3 hydogen gas reacts completelywith an excess of nitrogen gas?
Junior Tukkie Winter School 35 Dr. S. Swanepoel (2019)
Neutralisation and stochiometry
Any question or part of a question involving a titration or where an acid neutralises a baseimplies that the reactants react in the ratio according to the balanced reaction. (No reactant inexcess or limiting agent.)
In a titration experiment 35 cm3 of a 0,5 mol.dm−3 NaOH solution neutralises 25 cm3 of asulphuric acid solution. Calculate the concentration of the acid solution.
H2SO4 + 2NaOH Na2SO4 + 2H2O
n n
v =35 cm3 v = 25 cm3
c= ? c= 0,5 mol.dm−3
Three separate steps:
c =nV
0, 5 =n
25 × 10−3
n = 1, 25× 10−2 mol
NaOH
H2SO4 : NaOH
1 : 2
x : 1,25 × 10−2
6,25 ×10−3 mol H2SO4
c =nV
=6, 25 × 10−3
35× 10−3
= 0, 179 mol.dm−3
H2SO4
Or titration equation:
na
nb=
Ca.Va
Cb.Vb
1
2=
Ca.35
0, 5.25
Ca = 0, 179 mol.dm−3 H2SO4
a. Sulphuric acid (concentration 0,01 mol.dm−3 is neutralised by a ammonia solution. If20 cm3 of the base is used to neutralise 30 cm3 of the acid, determine the concentration ofthe base.
Junior Tukkie Winter School 36 Dr. S. Swanepoel (2019)
Percentage yield and percentage purity
% purity = mass of compound(theoretical)mass of sample × 100%
% yield = actual yieldtheoretical yield(possible) × 100%
1) N2(g) + 3H2(g)→ 2NH3(g)
21 g H2 reacts and produces 90 g ammonia. Calculate the percentage yield.
2) 12 g unpure copper reacts with HNO3 according to the following equation and 2,24 dm3
NO(g) is produced at STP.
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO(g) + 4H2OProve that the copper is 79,38 % pure.
Junior Tukkie Winter School 37 Dr. S. Swanepoel (2019)
Limiting agents
Sodium burns in oxygen according to the following equation:
4Na + O2 → 2Na2O
483 g sodium is placed in a container with 129,92 dm3 oxygen gas.
a. Determine the limiting agent. Show all calculations.
b. Calculate the mass of sodium oxide that can be form.
c. Calculate the mass that remain of the reagent that is in excess.
4Na + O2 (g) → 2Na2O
n n n
m = 483 g V = 129,92 dm3 m = ?
a.
n =mM
=483
23
= 21 mol Na
available
n =V
Vm
=129, 92
22, 4
= 5, 8 mol O2
available
Na : O2
4 : 1
21 : x
5,25 mol O2 requiredO2 available > O2 required∴ O2 in excess∴ Na limiting agent
b. Work with limiting agent
Na : Na2O
4 : 2
21 : x
x = 10,5 mol Na2O
n =mM
10, 5 =m62
m = 651 g Na2O
c. O2 remain = 5,8 - 5,25= 0,55 mol
n =mM
0, 55 =m32
m = 17, 60 g O2
Junior Tukkie Winter School 38 Dr. S. Swanepoel (2019)
Limiting agents
1) 100 g nitrogen and 20 g hydrogen are available. N2(g) + 3H2(g)→ 2NH3(g)
a. Which reactant is the limiting agent?
b. What volume of ammonia can be produced at STP?
c. What mass of the reactant in excess will remain?
Junior Tukkie Winter School 39 Dr. S. Swanepoel (2019)
2) Calsium oxide reacts with hydrochloric acid: CaO + 2HCl→ CaCl2 + H2O
19,6 g CaO and 400 cm3 of a 2 mol.dm−3 HCl solution are mixed.
(a) Calculate the mass of CaCl2 that can be produced.
(b) Calculate how many mol of the reactant in excess will remain unreacted.
Junior Tukkie Winter School 40 Dr. S. Swanepoel (2019)
3) During a titration 25 cm3 of the 0,1 mol.dm−3 sulphuric acid solution is added to anErlenmeyer flask and titrated with a 0,1 mol.dm−3 sodium hydroxide solution.
Calculate the pH of the solution in the flask after the addition of 30 cm3 of sodiumhydroxide. The endpoint of the titration is not yet reached at this point.
(IEB Calculate the hydronium ion concentration ...)DOE Mrt 2015 (8)
Junior Tukkie Winter School 41 Dr. S. Swanepoel (2019)
Reactant in excess and a second reaction
Some reagent is left over from the first reaction and then used in the second reaction. Remem-ber that the ratio given by the balanced equation is the ratio in which the reagents react .
X g magnesium reacts with 200 cm3 of a 0,15 mol·dm−3 HCl solution. Theacid is in excess .
Mg + 2HCℓ MgCℓ2 + H2(g)
The acid that is in excess is titrated with a 0,8 mol·dm−3 NaOH solution. 25cm3 of the NaOH is needed to neutralise the acid:
HCℓ+ NaOH NaCℓ+ H2O
Calculate mass X.
Mg + 2HCℓ MgCℓ2 + H2(g)
n
m = X g
n start 1 c v
n react 5
n excess
HCℓ + NaOH NaCℓ + H2On n
v = 0,025 dm3
c = 0,8 mol·dm−3
1 HCℓadded:
c =nV
0, 15 =m0, 2
n = 0, 03 mol
2 NaOH react:
c =nV
0, 8 =m
0, 025
n = 0, 02 mol
3 HCℓ : NaOH
1 : 10,02 mol HCℓ reacts in 2nd reaction
4 0,02 mol HCℓ excess in 1st reaction
5 nHCℓ react = nHCℓ added - nHCℓ excess= 0,03 - 0,02= 0,01 mol
6 Mg : HCℓ
1 : 2
x : 0,010,005 mol Mg
7 n =mM
0, 005 =m24
m = 0, 12 g
7
6
4
3 2
Junior Tukkie Winter School 42 Dr. S. Swanepoel (2019)
1. A learner adds a sample of calcium carbonate to 50,0 cm3 of hydrochloric acid of concen-tration 1,0 mol·dm−3. The hydrochloric acid is in excess.
CaCO3 + 2HCl CaCl2 + CO2 + H2O
The excess hydrochloric acid is now neutralised by 28,0 cm3 of a 0,5 mol·dm−3 sodiumhydroxide solution. The balanced equation for this reaction is:
HCl + NaOH NaCl + H2O
Calculate the mass of calcium carbonate in the sample.
Junior Tukkie Winter School 43 Dr. S. Swanepoel (2019)
Memorandum
p.3
Forward
Reverse
p.5
1 Kc = [Cr2O72−]
[CrO4−][H+]2
2a Stays constant2b Stays constant2c Stays constant
3 Disturbance: [H+] decreases. According toLe Chatelier’s Principle the system reactsto increase the [H+] by favouring the reversreaction. More yellow CrO4
− and less or-ange Cr2O7
2− are produced and the mix-ture looks more yellow.
p.8&91.1 decrease1.2 increased, increased1.3 decreased, negative1.4 increases, positive1.5 decreased, increased1.6 positive1.7 larger, negative
p.11
Movement from the . . . vi vf ∆ y(m ·s−1) (m ·s−1) (m)
starting point to highest point +X 0 +6
starting point to same height again +X -X 0
starting point to the ground +X -big -40
highest point to the top of the building 0 -X -6
highest point to the ground 0 -big -46
Movement from the . . . vi vf ∆ y(m ·s−1) (m ·s−1) (m)
starting point to the same height +2 -2 0
starting point to the ground +2 -big -65
p.6&7
2A(g) + B(g) ⇄ 3C(g) + D(g) �H < 0
A and B in acontainer
A added B removed D added Volume doubled
Temperatureincreased
Temperature decreased
Catalyst added
Disturbance
Le Chatelier’s principle: the system reacts to ...
The ... reaction is favoured
[C] will ...
Kc ...
Rate of
reaction
time
Concentration
time
[A]�
[A]� [B]�
[B]�
[D]�
[D]�
forward
� � �
constant
P�
P�
�
T�
T�
�
�
T�
T�
�
�constant constant
forward forwardreverse reverse
reverse
constant
forward
reverse
A
B
C
D
forwards exothermic
more mol gas endothermic exothermicB
oth
rea
ctions
equally
faste
r
constant
Junior Tukkie Winter School 44 Dr. S. Swanepoel (2019)
p.12
Position x
(m)
time (s)
Velocity v
(m.s
-1)
time (s)
Acceleration
a (m
.s
-2)
time (s)
-9,8
p.13
y(m)
t (s)
30
y(m)
t (s)-30
y(m)
t (s)-30
y(m)
t (s)
30
t (s)
v(m.s )-1
t (s)
v(m.s )-1
p.141.1 vf = vi + g∆t
0 = 1.96 + (−9, 8)∆t
∆t = 0, 2s
1.2 ∆y = vi∆t+1
2g∆t
2
= 1.96(6) +1
2(−9, 8)(62)
= −164, 64
height = 164, 64 m
1.3
t (s)
v(m.s )-1
_
_
_
1,96
0,2 6
1.4 Gradient =∆v
∆t
−9, 8 =v − 0
6− 0, 2
(or − 9, 8 =v − 1, 96
6− 0)
v = −56, 84
= 56, 84 m·s−1, down
p.152.1 vf = vi + g∆t
−10 = −2, 5 + (−9, 8) ∆t
∆t = 0, 77 s
or g = gradient
−9, 8altX
=∆v
∆t
−9, 8 =−10− (−2, 5))
t− 0t = 0, 77s
2.2 AB
2.3 Hard acceleration is instantly.
2.4 ∆y = area under graph
= (0, 77× 2, 5) + (1
2× 0, 77× 7, 5)
= 4, 81m
2.5
Position
(m)
4,81
0,77 Time (s)
Junior Tukkie Winter School 45 Dr. S. Swanepoel (2019)
A man pulls a 50 kg-washing machine 3 m up an inclined plane byexerting a force of 2000 N parallel to the plane. The plane makes anangle of 40◦ with the horizon. The washing machine experiences20 N frictional force.
F
a. Draw a free body-diagram of all the forces actingon the machine. (No components)
b. Calculate the work done by every force.
c. Use the previous answers to calculate the network.
a. Draw a free body-diagram of all the forceson the machine. Use components of Fg.
b. Calculate the net force on the machine.
c. Use the Fnet to calculate the net work.
a. a.
Fg
f
NF
f
NF
Fg ll
Fg �
b. WF = F∆xcosΘ
= 2000(3)cos0◦
= 6000 J
Wf = fk∆xcosΘ
= 20(3)cos180◦
= −60 J
Wg = Fg∆xcosΘ
= (50)(9, 8)(3)cos(90 + 40)◦
= −994, 90 J
WN = fk∆xcos90◦
= 0 J
b. Fnet = F + (−f) + (−Fg||)
= 2000 − 20− 50(98)sin40◦
= 1665, 03 N,up the slope
c. Wnet = WF +Wf +Wg +WN
= 6000 − 60− 944, 90 + 0
= 4995, 10 J
c. Wnet = Fnet∆xcosΘ
= (1665, 10)(3)cos0◦
= 4995, 10 J
The washing machine starts from rest. Use the work-energy principle to prove that after 3 m themagnitude of the velocity is 14,14 m·s−1.
Wnet = ∆Ek
4995, 10 =1
2m(v2f − v2i )
4995, 10 =1
2(50)(v2f − 0)
vf = 14, 14 m · s−1
Calculate the average power of the man with p = W∆t
Calculate the average power of the manwith pave = Fvave
∆x = (vi + vf
2)∆t
3 = (0 + 14, 41
2)∆t
∆t = 0, 424 s
P =W
∆t
=6000
0, 424= 14140 W
Pave = F vave
= 2000(0 + 14, 41
2)
= 14140 W
Junior Tukkie Winter School 46 Dr. S. Swanepoel (2019)
p.21
3o
f = 10 N
x = 25 m
Wnet = Ek
f = 15 N
t = 15 s
Fnet t =p Σpi = Σpf
Emech i = E mech f
Wnet = ∆EK
(WN ) +Wg +Wf =1
2m(v2f − v2i )
(0)− (50)(9, 8)(25)cos(90 − 38)◦ + 10(25)cos180◦ =1
2m(v2f − 0)
12066, 965 − 250 = 40v2f
vf = 17, 19 m · s−1
+Fnet∆t = ∆p
Fnet∆t = (vf − vi)
(−15)(15) = (vf − 17, 19)
vf = 14, 38 m · s−1
Σpi = ΣpfpJi + ppi = pJf + ppf
mvJi +mvpi = (mJ +mp)vf80(14, 38) + 0 = (80 + 1)vf
vf = 14, 20 m · s−1
Emechi = Emechf
mghi +1
2mv2i = mghf +
1
2mv2f
ghi +1
2v2i = ghf +
1
2v2f
0 +1
2(14, 20)2 = 9, 8(5, 2) + v2f
vf = 9, 99 m · s−1
Ek =1
2mv2
=1
2(50)(9, 99)2
= 3992, 00 J (not enough)
Junior Tukkie Winter School 47 Dr. S. Swanepoel (2019)
p.225.1 Emechi = Emechf
mghi +1
2mv2i = mghf +
1
2mv2f
ghi +1
2v2i = ghf +
1
2v2f
(9, 8)h + 0 = 0 + 12v
2f
vf = 9, 90 m · s−1
5.2 Fnet=0 ∴ a = 0 (Newton I) ∴ v constant ∴ Ek constant orFnet=0 ∴ Wnet = 0 ∴ Ek constant (Wnet =∆ Ek
5.4 Wnc = ∆EK +∆EP
(WN ) +Wf =1
2m(v2f − v2i ) +mg(hf − hi)
(0) + 18∆xcos180◦ =1
2(5)(0 − (9, 90)2) + 9, 8(5)(3 − 0)
∆x = 5, 45 m
sinΘ =3
5, 45
Θ = 33, 40◦
p.23
4.1
Wnet = ∆EK
(WN ) +Wg +Wf =1
2m(v2f − v2i )
0 + 0 + 10(2)cos180◦ =1
2(7, 02)(0 − v2i )
vi = 2, 39 m · s−1
4.2
Σpi = Σpfp1i + p2i = p1&2
mv1i +mv2i = (m1 +m2)vf0, 02v1i + 0 = (7 + 0, 02)(2, 39)
v1i = 838, 89 m · s−1
p.28a HNO3 + H2O NO3
−(aq) + H3O+(aq)
b NaOH Na+ (aq) + OH−(aq)H2O
c (COOH)2 + Li2OH (COOLi)2 + H2OH2C2O4 + Li2OH Li2C2O4 H2O
d NH3 + H2O NH4+(aq) + OH−(aq)
e H2SO4 + 2H2O SO4−2(aq) + 2H3O+(aq)
f H2SO4 + 2KOH K2SO4 + 2H2O
g (COOH)2 + 2H2O (COO)2−2(aq) + 2H3O+(aq)H2C2O4 + 2H2O C2O4
−2(aq) + 2H3O+(aq)(or donate only 1 proton)
h LiOH Li+ (aq) + OH−(aq)H2O
i NH3 + HCl NH4+(aq) + Cl−(aq)
Junior Tukkie Winter School 48 Dr. S. Swanepoel (2019)
p.29 p.30
3.1c1V1 = c2V2
(0, 2)(30) = 0, 03V2
V2 = 200 cm3
Vadded = 200− 30
= 170 cm3
3.2a) HX + H2O X−(aq) + H3O+(aq)
3.2b) No
[H3O+] < HXionise incompletely
2.2c) HX and X−
H2O and H3O+
3.3a)c1V1 = c2V2
(0, 15)(300) = 0, 045V2
V2 = 1000 cm3
Vadded = 1000 − 300
= 700 cm3
3.3b) H2SO4 is a strong acid and ionises completelyH2SO4 + 2H2O SO4
−2(aq) + 2H3O+(aq)
H2SO4 : H3O+
1 : 2
[H3O+] = 2(0, 045) = 0, 09mol.dm−3
3.3c)pH = −log[H3O
+]
= −log(0, 09)
= 1, 05
3.4ac =
m
M × V
=0, 5
24× 0, 2
= 0, 104 mol.dm−3
3.4b) LiOH is a strong base and dissociates completelyLiOH Li+ (aq) + OH−(aq)
H2O
LiOH : OH−
1 : 1
[OH−] = 0, 104mol.dm−3
3.4c) [Li+] = 0, 104mol.dm−3
3.4d)
Kw = [OH−][H3O+]
1× 10−14 = (0, 104)[H3O+]
[H3O+] = 9, 62× 10−14 mol.dm−3
3.4e)pH = −log[H3O
+]
= −log(9, 62× 10−14)
= 13, 023.5
pH = −log[H3O+]
11, 61 = −log[H3O+]
[H3O+] = 2, 45× 10−12 mol.dm−3
Kw = [OH−][H3O+]
1× 10−14 = [OH−](2, 45 × 10−12)
[OH−] = 4, 08× 10−13 mol.dm−3
BasicNa+ does not hydrolise(COO)2−2 + 2H2O (COOH)2 + 2OH− or(COO)2−2 + H2O H(COO)2− +OH−
p.31 Compound Acid Base Solution
ammonium nitrate HNO3 NH3
Strong Weak Acidic
potassium chloride HCl KOHStrong Strong Neutral
sodium oxalate (COOH)2 NaOHWeak Strong Basic
ammonium carbonate H2CO3 NH3
Weak Weak Neutral
lithium acetate CH3COOH LiOHWeak Strong Basic
p.32
a) c =m
M × V
=100
58, 5× 0, 3
= 5, 70 mol.dm−3
b) n =N
NA
=9, 03× 1024
6, 02× 2023
= 15 mol
c) n =V
Vm
=4, 48
22, 4
= 0, 20 mol
d) n =V
Vm
2, 7 =V
22, 4
V = 60, 48 dm3
e) M(K2SO4) = 174g.mol−1
n =m
M
3, 6 =m
174m = 626, 40 g
f) n =N
NA
4, 2 =N
6, 02× 2023
N = 2, 53× 1024
Junior Tukkie Winter School 49 Dr. S. Swanepoel (2019)
p.33
2C4H10(g) + O2(g) → 4NO2(g)
n n
m - ? V = 4,48 dm3
n =V
VM
=4, 48
22, 4 dm3.mol−1
= 0, 2 molO2
C4H10 : O2
2 : 1
x : 0, 2
x = 0, 4 mol C4H10
M(C4H10 = 4(12) + 10(1)
= 58 g,mol−1
n =m
M
0, 4 =m
58 g.mol−1
m = 23, 20 g C4H10
6CO2(g) + 6H2O (ℓ) → C6H12O6(s) + 6O2(g)
n n
m = 100 g m = ?
M(CO)2 = 12 + 2(16)
= 44 g,mol−1
n =m
M
=100 g
44 g.mol−1
= 2, 27 mol CO2
CO2 : C6H12O6
6 : 1
2, 27 : x
x = 0, 378mol C6H12O6
M(C6H12O6) = 6(12) + 12(1) + 6(16)
= 180 g,mol−1
=m
M
0, 378 =m
180 g.mol−1
m = 68, 04g C6H12O6
2N2O5(g) → 4NO2(g) + O2(g)
n n
N = 1, 5× 1024 V = ?
n =N
NA
=1, 5× 1024
6, 02× 1023
= 2, 492mol N2O5
N2O5 : NO2
2 : 4
2, 492 : x
x = 4, 984 mol Cl2
n =V
VM
4, 984 =V
22, 4 dm3.mol−1
V = 111, 64 dm3
2KMnO4(s) + 16HCl(aq) → 2KCl + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
n n
m = ? V = 33,6 dm3
n =V
Vm
=3, 36 dm3
22, 4 dm3.mol−1
= 0, 15 mol Cl2
KMnO4 : Cl2
16 : 5
x : 0, 15
x = 0, 480 mol Cl2
c =n
V
0, 2 =0, 48
V
V = 2, 40 dm3
Junior Tukkie Winter School 50 Dr. S. Swanepoel (2019)
p.35
N2O5 : NO2
2 : 4
4, 86 : x
x = 9, 72 dm3 NO2
H2 : NH3
3 : 2
4, 68 : x
x = 2, 99 dm3 NH3
p.36H2SO4 + 2NH3 → (NH4)2SO4
na
nb
=CaVa
cbVb
1
2=
0, 1(30)
Cb(20)
cb = 0, 03 mol.dm−3
p.37 Percentage Yield and percentage purity
N2(g) + 3H2 (g) → NH3
n n
m = 21 g m = ?
n =m
M
=21 g
2 g.mol−1
= 10, 5 mol H2
H2 : NH3
3 : 2
10, 5 : x
x = 7 mol NH3
n =m
M
7 =m
17 g.mol−1
m = 119 g NH3
% yield =actual yield
possible yield)× 100
=90
119× 100
= 75, 63%
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO2(g) + 4H2O
n n
m = ? V = 2,24 dm3
n =V
Vm
=2, 24
22, 4
= 0, 10 mol NO2
Cu : NO2
3 : 2
x : 0, 10
x = 0, 15 mol Cu
n =m
M
0, 15 =m
63, 5
m = 9, 53 g NH3
% purity =mass Cu)
mass sample× 100
=9, 53
12× 100
= 79, 38%
Junior Tukkie Winter School 51 Dr. S. Swanepoel (2019)
p.39 Limiting agent
N2 + 3H2 (g) → 2NH3
n n n
m=10 g m = 20 g m = ?
a) n =m
M
=100
28
= 3, 57 molN2
available
n =m
M
=20
2
= 10 molH2
available
N2 : H2
1 : 310 : x
30 mol H2 required
H2 available < H2 required
∴ H2 is limiting agent
b) Work with limiting agent
H2 : NH3
3 : 210 : x
x = 6, 67 mol NH3
n =V
Vm
6, 67 =V
22, 4
V = 149, 41 dm3 NH3
c) N2 : H2
1 : 3x : 10x = 3, 33 mol N2 used
N2 remain = 3, 57− 3, 33= 0, 24 mol
n =m
M
0, 24 =m
28
m = 6, 72 gN2
p.40CaO + 2HCl → CaCl2 + H2O
n n n
m=19,6 g c = 2 mol.dm−3 m = ?V = 0, 2 dm3
a) n =m
M
=19, 6
56
= 0, 35 mol N2
available
c =n
V
2 =n
0, 4
= 0, 8 mol H2
available
For all the CaO:
CaO : HCl
1 : 20, 35 : x
0, 7 mol HCl required
HCl available > HCl required
∴ H2 is in excess
CaO is limiting agent
Work with limiting agent
CaO : CaCl21 : 1
0, 35 : x
x = 0, 35 mol CaCl2
n =m
M
0, 35 =m
111
m = 38, 85 g CaCl2
b) HCl remain = 0, 8− 0, 7
= 0, 1 mol
Junior Tukkie Winter School 52 Dr. S. Swanepoel (2019)
p.41H2SO4 + 2 NaOH Na2SO4 + H2O
n n
v = 25 cm3 v = 30 cm3
c = 0,1 mol.dm−3 c = 0,1 mol.dm−3
c =n
V
0, 1 =n
0, 025
= 2, 5× 10−3 mol
H2SO4
c =n
V
0, 1 =n
0, 03
= 3× 10−3 mol
NaOH
H2SO4 : NaOH
1 : 22, 5× 10−3 : x
5× 10−3 mol NaOH required
NaOH available < NaOH required
∴ NaOH is limiting agent
H2SO4 : NaOH
1 : 2x : 3× 10−3
1, 5× 10−3 mol H2SO4 reacted
H2SO4 remain = 3× 10−3 − 1, 5× 10−3
= 1× 10−3 mol
c =n
V
=1× 10−3
0, 025 + 0, 03
=1× 10−3
0, 055
= 0, 018 mol.dm−3
H2SO4 is a strong acid
H2SO4 ionises completely
H2SO4 + 2H2O SO−2
4 (aq) + 2H3O+(aq)
H2SO4 : H3O+
1 : 2
[H3O+] = 2(0, 018) = 0, 036mol.dm−3
pH = −log[H3O+]
= −log(0, 036)= 1, 44
p.43c =
n
V
1 =n
0, 05
= 0, 05 mol
HCl begin
c =n
V
0, 5 =n
0, 028
= 0, 014 mol
NaOHin 2
HCl : NaOH
1 : 1x : 0, 014
0, 014mol HCl reacts in 2
0, 014mol HCl remain after 1
HCl in 1 reacts
= 0, 05− 0, 014= 0, 036 mol
CaCO3 : HCl
1 : 2
x : 0, 036
0, 018 mol CaCO3
n =m
M
0, 018 =m
100
m = 1, 8 g CaCl2
Junior Tukkie Winter School 53 Dr. S. Swanepoel (2019)
Fis
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3 K
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x 1
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: FO
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V mc
mV V
n
b a
bb
aa
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vc
vc
pH
= -lo
g[H
3 O+]
Kw =
[H3 O
+][OH
-] = 1
x 1
0-1
4 at/b
y 2
98 K
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Fisiese Wetenskappe/V2 DBE/November 2015 NSS
Kopiereg voorbehou
TABLE 3: THE PERIODIC TABLE OF ELEMENTS TABEL 3: DIE PERIODIEKE TABEL VAN ELEMENTE
1(I)
2(II)
3 4 5 6 7 8 9 10 11 12 13 (III)
14 (IV)
15 (V)
16 (VI)
17 (VII)
18 (VIII)
2,1
1
H1
2
He4
1,0
3
Li7
1,5
4
Be9
2,0
5
B11
2,5
6
C12
3,0
7
N14
3,5
8
O16
4,0
9
F19
10
Ne20
0,9
11
Na23
1,2
12
Mg24
1,5
13
27
1,8
14
Si28
2,1
15
P31
2,5
16
S32
3,0
17
35,5
18
Ar 40
0,8
19
K39
1,0
20
Ca40
1,3
21
Sc45
1,5
22
Ti48
1,6
23
V51
1,6
24
Cr52
1,5
25
Mn55
1,8
26
Fe 56
1,8
27
Co 59
1,8
28
Ni59
1,9
29
Cu 63,5
1,6
30
Zn 65
1,6
31
Ga 70
1,8
32
Ge 73
2,0
33
As 75
2,4
34
Se79
2,8
35
Br80
36
Kr84
0,8
37
Rb 86
1,0
38
Sr88
1,2
39
Y89
1,4
40
Zr91
41
Nb 92
1,8
42
Mo96
1,9
43
Tc 2,2
44
Ru 101
2,2
45
Rh 103
2,2
46
Pd106
1,9
47
Ag 108
1,7
48
Cd 112
1,7
49
In115
1,8
50
Sn119
1,9
51
Sb122
2,1
52
Te 128
2,5
53
I127
54
Xe131
0,7
55
Cs133
0,9
56
Ba137
57
La 139
1,6
72
Hf 179
73
Ta 181
74
W184
75
Re186
76
Os 190
77
Ir192
78
Pt195
79
Au 197
80
Hg 201
1,8
81
204
1,8
82
Pb207
1,9
83
Bi209
2,0
84
Po 2,5
85
At 86
Rn
0,7
87
Fr 0,9
88
Ra226
89
Ac
58
Ce140
59
Pr141
60
Nd 144
61
Pm62
Sm150
63
Eu152
64
Gd 157
65
Tb 159
66
Dy 163
67
Ho 165
68
Er167
69
Tm 169
70
Yb173
71
Lu 175
90
Th 232
91
Pa92
U238
93
Np 94
Pu95
Am 96
Cm 97
Bk98
Cf 99
Es100
Fm 101
Md102
No 103
Lr
Electronegativity Elektronegatiwiteit
24 25 26 27 28 29
Approximate relative atomic mass Benaderde relatiewe atoommassa
Atomic number Atoomgetal
29
Cu 63,5
2
C6
1,9
Symbol Simbool
KEY/SLEUTEL
Junior Tukkie Winter School 54 Dr. S. Swanepoel (2019)
Junior Tukkie Winter School 55 Dr. S. Swanepoel (2019)
Junior Tukkie Winter School 56 Dr. S. Swanepoel (2019)