Facilitator: Dr. Sarita Swanepoel · 2019-10-19 · Facilitator: Dr. Sarita Swanepoel Junior Tukkie Winter School 1 Dr. S. Swanepoel (2019) Physical Sciences Contents 1 Chemical equilibrium

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


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.


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.


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


∆ H < 0

Volumedoubled

Temperatuurincreased

Temperaraturedecreased

Catalystadded


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).


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).


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)


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:


starting point to the same height

starting point to the ground


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)


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


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.


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.)


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


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


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.


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


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]


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


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


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


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


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)


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.)


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.


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?


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)


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?


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.


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.


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


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?


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.


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)


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


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.


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


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


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


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)


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


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)


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)


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


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


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%


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


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


Fis

iese W

ete

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2

DB

E/N

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(CH

EM

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GE

GE

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FIS

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SK

AP

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AA

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VR

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bb

aa

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vc

vc

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g[H

3 O+]

Kw =

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+][OH

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x 1

0-1

4 at/b

y 2

98 K

an

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tho

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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




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Facilitator: Dr. Sarita Swanepoel · 2019-10-19 · Facilitator: Dr. Sarita Swanepoel Junior Tukkie Winter School 1 Dr. S. Swanepoel (2019) Physical Sciences Contents 1 Chemical equilibrium