chapter 1 – atomic structure...

Chapter 1 – Atomic structure Answers

© Nora Henry, Alyn G McFarland 2017

Test yourself (page 3) 1 The plum pudding model has a positive sphere with electrons dotted through it, while today’s

model has a nucleus containing protons and neutrons with electrons orbiting in shells.

2 (James) Chadwick

3 It has no charge and so was difficult to detect.

4 An electron has relative mass of 1/1840, a proton has relative mass of 1.

An electron has relative charge of −1, a proton has relative charge of +1.

Electrons are found in shells, protons are found in the nucleus.

5 Both have relative mass of 1 and both are found in the nucleus.

6 7

7 a) Beryllium

b) Silicon

8 They have an equal number of protons and electrons.

9 30 protons, 30 electrons, 35 neutrons

Show you can (page 3)

Element Atomic number

Mass number

Number of protons

Number of electrons

Number of neutrons

73Li 3 7 3 3 4

115B 5 11 5 5 6

2412Mg 12 24 12 12 12

3919K 19 39 19 19 20

10747Ag 47 107 47 47 60

12753I 53 127 53 53 74

Test yourself (page 5) 10 2, 6

2, 7

2, 8, 3

2, 8, 6



11 chlorine

argon

12 The first shell holds a maximum of 2 electrons.

13 a)

b)

Show you can (page 6) 1 Boron

2 2, 5

3 11

4 Nucleus



Test yourself (page 8)

14

15 Both have the same number of protons (82), the same number of electrons (82) and atomic

number (82), and react in the same way.

They have different mass numbers (207 and 208) and different numbers of neutrons (125 and

126).

16 79.9 (to 1 decimal place)

(79 53) (81 47) 4187 3897 799479.94

100 1000 100rA× + × +

= = = =

17 28.1 (to 1 decimal place)

(92.2 28) (4.7 29) (3.1 30) 2581.6 136.6 93 2811.228.1

100 1000 100rA× + × + × + +

= = = =

Test yourself (page 9) 18 a) 1.1 × 10−10 m

b) 1.15 × 10−10 m

c) 1.25 × 10 −10 m

d) 7.0 × 10−11 m

19 0.025 nm

20 1.83 × 10−5 nm (0.000 0183 nm), 1.83 × 10−14 m

21 0.1 nm (1 × 10−10 m)

Protons Neutrons Electrons 1H (protium) 1 0 1 2H (deuterium) 1 1 1 3H (tritium) 1 2 1



Test yourself (page 11) 22 Answers in bold and underline

23 1−

24 3+

25 18, 1+

26 3−

27 9 protons, 10 neutrons, 10 electrons

Show you can (page 11) 1 Answers in bold and underline

2 It has an equal number of protons and electrons.

3 2−

4 D and F

Symbol Mass number

Atomic number

Number of protons

Number of neutrons

Number of electrons

Electronic structure

Calcium atom

Ca 40 20 20 20 20 2,8,8,2

Calcium ion

Ca2+ 40 20 20 20 18 2,8,8

Oxygen atom

O 16 8 8 8 8 2,6

Oxide ion

O2− 16 8 8 8 10 2,8

Nitride ion

N3− 14 7 7 7 10 2,8

Sodium atom

Na 23 11 11 12 11 2,8,1

Sodium ion

Na+ 23 11 11 12 10 2,8

Particle Atomic number

Mass number

Number of protons

Number of neutrons

Number of electrons


A 18 40 18 22 18 2,8,8 B 13 27 13 14 10 2,8 C 20 40 20 20 20 2,8,8,2 D 17 35 17 18 17 2,8,7 E 16 32 16 16 18 2,8,8 F 17 37 17 20 17 2,8,7



Practice questions (page 12) 1 a) Answers in bold and underline (4 marks)

Particle Relative charge Relative mass Proton +1 1 Electron −1 1/1840 Neutron 0 1

b) i) Nucleus (1 mark)

ii) Atomic number (1 mark)

iii) Mass number (1 mark)

iv) 13 (1 mark)

c) i) 12C and 14C (1 mark)

ii) 19F− (1 mark)

iii) 16O2−, 19F−, 20Ne (1 mark)

iv) 19F−, 20Ne (1 mark)

v) None (1 mark)

2 a) Plum pudding model: a positive sphere (1 mark) with electrons embedded in it. (1 mark)

Today’s model: a nucleus (1 mark) containing protons and neutrons (1 mark), with electrons

in shells (1 mark). (5 marks)

b) i) 11 (1 mark)

ii) 23 (1 mark)

iii) Nucleus (1 mark)

iv) 2,8,1, shown on diagram (1 mark)

v) 1.0 × 10−10 m (1 mark)

vi) (James) Chadwick (1 mark)

c) Answers in bold and underline (6 marks)

Atom/ion Number of protons Electronic configuration N 7 2,5 S2− 16 2,8,8 Ca2+ 20 2,8,8 Mg2+ 12 2,8



3 a) The total number of protons and neutrons in an atom (1 mark)

b) The number of protons in an atom (1 mark)

c) Electronic configuration 2,8,8,1 (1 mark), 19 protons (1 mark), 20 neutrons (1 mark),

protons and neutrons in the nucleus (1 mark) (4 marks)

d) 0.1 nm (1 mark)

e) i) 17 protons, 18 electrons (1 mark)

ii) Charge 1−, electronic configuration 2,8,8 (1 mark)

iii) Two or more elements chemically combined (1 mark)

4 a) 108.1 to one decimal place (2 marks)

(109 53) (107 47)

108.06100

× + ×=

(109 53) (107 47)108.06

100× + ×

=

b) Isotopes are atoms of an element with the same atomic number but a different mass

number, and thus a different number of neutrons. (1 mark) 109Ag has 62 neutrons and 107Ag has 60 neutrons. (1 mark)

Both 109Ag and 107Ag have the atomic number 47. (1 mark)

c) 1.6/10 000 nm = 1.6 × 10−5 nm (1 mark), 1.6 × 10-14 m (1 mark) (2 marks)

d) 47 protons, 62 neutrons, 46 electrons (2 marks)

e) O2− (1 mark)

5 Answers in bold and underline (5 marks)

Atom/ion Al Sn2+ Ba H+ Se2− Cl Atomic number 13 50 56 1 34 17 Mass number 27 119 137 1 79 37 Number of protons 13 50 56 1 34 17 Number of neutrons 14 69 81 0 45 20 Number of electrons 13 48 56 0 36 17

Chapter 2 – Bonding, structure and nanoparticles Answers


Test yourself (page 17) 1 They already have a full outer shell and are stable.

2

3



4



5

Show you can (page 17) Oxygen has electronic configuration 2,6 and needs to gain two electrons to become stable. Two

lithium atoms, each with electronic configuration 2,1, each transfer one electron to the oxygen.

This forms an oxide ion with a charge 2− and an electronic configuration 2,8 and two lithium ions

with a charge 1+ and an electronic configuration 2. The attraction between the oppositely

charged ions is the ionic bond.

Test yourself (page 19) 6 a)

b)

c)



d)

e)

f)

g)

7 a)

b)

c) 10



Show you can (page 19) 1

2 PCl3

3

4 Any (x) is a covalent bond and any () or (xx) is a lone pair.

Test yourself (page 19) 8 The attraction between the positive ions in a regular lattice and the delocalised electrons

9 Delocalised electrons are electrons that are free to move throughout the whole structure.

10 a) Ionic

b) Metallic

c) Covalent

d) Covalent

e) Ionic

f) Covalent

g) Metallic

h) Ionic

i) Covalent

j) Metallic

k) Covalent

l) Ionic

Test yourself (page 22) 11 a) Metallic

b) Metallic

c) Delocalised electrons can move and carry charge.

d) The layers can slide but the metallic bonding is not disrupted.

e) The substance can be hammered into shape without breaking.



12 An alloy is a mixture of two or more elements, at least one of which is a metal, and the resulting

mixture has metallic properties.

13 The different-sized zinc and copper ions distort the lattice and so the layers do not slide over

each other as easily.

14 % gold = 20

10024

× = 83.3%

15 80

24100

× = 19.2 carat

16 Gold is too soft and would not keep its shape.

Test yourself (page 23) 17 It takes a substantial amount of energy to break the strong ionic bonds.

18 The ions can move and carry charge.

19 The ions cannot move and so there are no charge carriers.

20 A giant ionic lattice is a three-dimensional structure of oppositely charged ions held by ionic

bonds.

21 MgBr2, NaF and K2S

Practical activity (page 25) 1 Place the sample to be tested in the beaker, ensure the electrodes do not touch. Use the same

volume of sample for each test.

Switch on the power pack. Using the same current for each test, and record whether or not the

bulb lights up.

Wash out the beaker thoroughly and repeat with the next solution.

2

3 Ionic solutions conduct electricity because the ions can move and carry charge.

Covalent substances do not conduct electricity as there are no free charged particles.

4 Solid copper(II) sulfate would not conduct electricity and the bulb would not light up as the ions

are held tightly by ionic bonds and cannot move.

5 Calcium nitrate solution conducts electricity and the bulb lights up as the ions can move.

Test solution Does the bulb light?

Does the substance conduct electricity?

Does the substance contain ionic or covalent bonding?

copper(II) sulfate yes yes ionic ethanol (C2H5OH) no no covalent magnesium sulfate yes yes ionic potassium iodide yes yes ionic glucose (C6H12O6) no no covalent sodium chloride yes yes ionic



6 Bromine solution cannot conduct electricity and the bulb does not light up as there are no charge

carriers.

Test yourself (page 25) 22 a) A molecule is two or more atoms covalent bonded.

b) Atoms: F, O

Molecules of elements: H2, Cl2, O2

Molecules of compounds: HF, H2O, CH4, SiCl4

23 It takes little energy to break the weak van der Waals forces between the molecules.

24 a) Do not break

b) Break

25 It is a neutral molecule and there are no charged particles to move and carry the charge.

26 H2S, CO, N2H4

Show you can (page 25) The weak van der Waals forces between the molecules of sulfur dioxide, which need to be

broken for sulfur dioxide to melt, do not take much energy to break. In calcium oxide, the strong

ionic bonds between ions of opposite charge require a substantial amount of energy to break

and it is these bonds that need to be broken for calcium oxide to melt.

Test yourself (page 28) 27 They have many strong covalent bonds and substantial energy is needed to break them.

28 In diamond there are strong covalent bonds between the atoms, which are arranged in a three-

dimensional tetrahedral structure, but in graphite there are weak forces between the layers,

which are easily broken, allowing the layers to slide.

29 Each carbon atom bonds covalently to three others in graphite, leaving one electron per carbon

atom unbonded and delocalised so that it can move and carry charge. In diamond all four outer

shell electrons of carbon are used in bonding and none are free to move.

30 A single-atom thick layer of graphite

31 Similarities: made of carbon atoms, covalently bonded and transparent

Differences: tetrahedral structure in diamond, hexagonal layer in grapheme; diamond does not

conduct electricity, graphene does; graphene is much lighter than diamond; grapheme is two-

dimensional but diamond is three-dimensional



32 graphene

graphite

Graphene is a single layer; graphite has many layers.

33 Allotropes are different forms of the same element, in the same state.

Diamond, graphite and graphene

Show you can (page 28) Answers in bold and underline

Graphite Diamond Graphene Description of structure

Layers of carbon atoms covalently bonded in hexagons, with weak forces between the layers. One electron per carbon atom is unbonded and free to move between the layers.

Each carbon atom is covalently bonded to four others in a tetrahedral structure.

A single-atom thick layer of graphite with strong covalent bonds between each carbon atom, arranged in hexagons.

Uses Pencils, lubricants. Cutting tools. Solar cells, batteries.



Test yourself (page 30) 34 Answers in bold and underline

Substance Type of bonding Type of structure Copper oxide (CuO) Ionic Giant ionic lattice Diamond (C) Covalent Giant covalent Lead carbonate (PbCO3) Ionic Giant ionic lattice Phosphorus oxide (P4O10) Covalent Covalent molecular Copper (Cu) Metallic Metallic Graphene (C) Covalent Giant covalent Ammonia (NH3) Covalent Covalent molecular

35 a) Covalent, covalent, covalent, (given covalent), ionic, metallic

b) Cu

c) CS2

d) NH3

e) Br2

f) NH3

Show you can (page 30) 1 E: high melting point and boiling point, conducts electricity when molten but not when solid

2 B: low melting point and boiling point, does not conduct electricity

3 It conducts electricity, and diamond does not conduct electricity.

4 C

Test yourself (page 32) 36 A nanoparticle is a structure that is 1–100 nm in size and contains a few hundred atoms.

37 a) Surface area = 6 × 20 × 20 =2400 nm2

Surface area = 6 × 2 × 2 = 24 nm2

b) Volume = 20 × 20 × 20 = 8000 nm3

Volume = 2 × 2 × 2 = 8 nm3

c) 2400 : 8000 = 0.3 : 1

24 : 8 = 3 : 1

d) The surface area to volume ratio increases by a factor of 10.

38 They have a larger surface area to volume ratio.

39 a) 21 nm = 21 × 10−9 = 2.1 × 10−8 m = diameter of the nanoparticle.

b) In Chapter 1 you learnt that most atoms have a radius of 0.1 nm. The diameter is twice the

radius, so the diameter of a gold atom is 0.2 nm = 0.2 × 10−9 = 2.0 × 10−10 m.



40 They could cause potential cell damage in the body – the nanoparticles are so small they may be

able to penetrate cell membranes, or be breathed in. In the body they may be more reactive or

more toxic than the bulk material.

They could cause harm to the environment.

Show you can (page 32) Not all the effects of nanoparticles are fully understood.

Nanoparticles may be harmful.

Practice questions (page 33) 1 a) Ammonia (1 mark)

NH3 (1 mark) b) A lone pair is any x x; (1 mark)

a covalent bond is any x ●. (1 mark)

c)

(1 mark)

d) Shared (1 mark)

pair of electrons (1 mark)

2 a) A, diamond (1 mark)

B, graphite (1 mark)

C, graphene (1 mark)

D, carbon dioxide (1 mark)

b) Covalent (1 mark)

c) A, giant covalent (1 mark)

B, giant covalent (1 mark)

C, giant covalent (1 mark)

D, molecular covalent (1 mark)

d) Weak van der Waals forces (1 mark)

between the molecules (1 mark)

do not take much energy to break. (1 mark)

e) B and C (1 mark)



3 a) Carbon atoms are in layers. (1 mark)

They are held by weak intermolecular forces. (1 mark)

Layers can slip off. (1 mark)

b) One electron per carbon atom (1 mark)

is free to move and carry charge. (1 mark)

c)

Electronic configuration of sodium atom (1 mark)

Electronic configuration of chlorine atom (1 mark)

Electronic configuration of sodium ion (1 mark)

Electronic configuration of chloride (one electron pair in chloride must be x ●) (1 mark)

Na+ (1 mark)

Cl– (1 mark)

d) The ions can move and carry charge in solution (1 mark). In the solid they are

held tightly by attraction between opposite ions and cannot move (1 mark). (2 marks)

e)

Type of bonding Type of structure Graphite Covalent Giant covalent Sodium chloride Ionic Giant ionic lattice



4

Indicative content:

Both have high melting points (1 mark)

due to strong bonds between the ions in magnesium chloride (1 mark)

and between the metal atoms and delocalised electrons in magnesium. (1 mark)

Magnesium is always a good conductor due to the delocalised electrons, which can

move and carry charge (1 mark)

Magnesium chloride can only conduct when molten (1 mark)

because only when it is molten can the ions move and carry charge. (1 mark)

Magnesium conducts due to the movement of electrons, magnesium chloride conducts

due to the movement of ions. (1 mark)

5 a) i) Two atoms covalently bonded in a molecule (1 mark)

ii)

Response Mark Candidates must use appropriate specialist terms to fully describe similarities and differences [5–6 indicative content points]. Relevant material is organised with a high degree of clarity and coherence. They must use excellent spelling, punctuation and grammar and the form and style are of a very high standard.

5–6 (Band A)

Candidates must use appropriate specialist terms to describe the some similarities and differences [3–4 indicative content points]. Relevant material is organised with some clarity and coherence. They use good spelling, punctuation and grammar and the form and style are of a satisfactory standard.

3–4 (Band B)

Candidates describe the bonding and structure [at least 2 indicative content points]. The organisation of material may lack clarity and coherence. They use limited spelling, punctuation and grammar and they have limited use of specialist terms. The form and style are of limited standard.

1–2 (Band C)



b) i)

Electronic configuration of calcium atom (1 mark)

Electronic configuration of chlorine atom (1 mark)

Two chlorine atoms and one calcium atom (1 mark)

Electronic configuration of calcium ion (1 mark)

Electronic configuration of chloride (one electron pair in chloride must be x ●) (1 mark)

Ca2+ (1 mark)

Cl− (1 mark)

ii) Ionic (1 mark)

iii) Giant ionic lattice (1 mark)

iv) The ions (1 mark)

are free to move (1 mark)

and carry the charge. (1 mark)

v)

(3 marks)

vi) Covalent (1 mark)

vii) Simple covalent molecular (1 mark)



c)

(2 marks) d) i) Metallic (1 mark)

strong attraction (1 mark)

between the metal ions and delocalised electrons (1 mark)

ii)

(3 marks: 1 for mark each label with correctly drawn diagram)

iii) Each carbon atom covalently bonds to three other carbon atoms (1 mark); layer

of hexagons (1 mark); weak forces between layers (1 mark). (1 mark)

e) i) Delocalised electrons (1 mark) move and carry charge (1 mark)

ii) Malleable (1 mark), ductile (1 mark)

6 a) 1 mark for each answer in bold

Carbon dioxide Nitrogen Oxygen Formula CO2 N2 O2

Dot and cross diagram

Structural formula

(7 marks)

b) An unbonded pair of electrons (1 mark)

c) There are no ions or delocalised electrons to move and carry charge. (1 mark)

7 a) A (1 mark)

b) C (1 mark)

c) A (1 mark)

d) D (1 mark)

8 a) Strong (1 mark)

metallic bond/attraction between positive ions and delocalised electrons (1 mark)

takes substantial energy to break the bond/attraction. (1 mark)

b) Delocalised electrons (1 mark)

move and carry charge. (1 mark)

Compound Solubility in water Relative melting point Calcium chloride Soluble High Tetrachloromethane Insoluble Low



c) A mixture (1 mark)

of two or more elements, at least one of which is a metal. (1 mark)

Alloys have metallic properties. (1 mark)

d) The different-sized carbon atoms (1 mark)

disrupt the structure so the layers do not slide so easily. (1 mark)

9 a) Metallic (1 mark)

b) Covalent molecular (1 mark)

c) Giant covalent (1 mark)

d) Giant ionic lattice (1 mark)

e) Covalent molecular (1 mark)

f) Giant ionic lattice (1 mark)

10 A is giant ionic lattice. (1 mark)

B is covalent molecular. (1 mark)

C is metallic. (1 mark)

D is giant ionic lattice. (1 mark)

11 a) 1–100 nm (1 mark)

b) Accept any two points of: (2 marks)

• they give better skin coverage to the sun cream

• they give more effective protection from the sun’s ultraviolet rays

• they are clear and colourless, which makes the sun cream invisible on the skin

• they do not degrade on exposure to the sun.

c) 1.9 × 10−10 m = 0.000 000 001 9 m (1 mark)

d) They have a larger surface area to volume ratio. (1 mark)

e) i) 6 × (30 × 30) = 5400 nm2 (1 mark)

30 × 30 × 30 = 27 000 nm3 (1 mark)

5400 : 27 000 = 1 : 5 (1 mark)

ii) Increases by a factor of 10 (1 mark)

Chapter 3 – The Periodic Table Answers


Test yourself (page 37) 1 Order of increasing atomic weight

2 To ensure that elements were in groups with other elements with similar properties

3 He predicted that some elements had yet to be discovered.

4 Accept any three of:

• Elements arranged in order of atomic number instead of atomic weight

• Noble gases present – none in Mendeleev’s table

• No gaps and more elements in today’s periodic table

• A transition metal block is present but was not in Mendeleev’s table

• Actinides and lanthanides are present – but were not in Mendeleev’s table.

Test yourself (page 38) 5 a) Metal

b) Metal

c) Non-metal

d) Non-metal

e) Non-metal

f) Metal

g) Non-metal

h) Non-metal

i) Metal

j) Metal

6 a) Metal

b) Non-metal

c) Non-metal

d) Metal

7 No – graphite and graphene are non-metals which are good conductors of electricity.

Show you can (page 38) Carbon as graphite conducts electricity as there is one free electron per carbon atom that can

move and carry charge. Aluminium conducts electricity as the delocalised electrons can move

and carry charge. In carbon as diamond there are no free electrons or ions to move and carry

charge, so it does not conduct electricity.



Show you can (page 39) 1 It has one electron in the outer shell. If it had five electrons in the outer shell it would be in

Group 5.

2 2 + 8 + 1 = 11

Test yourself (page 40) 8 a) Group 4

b) Group 1

c) Group 0

9 a) Beryllium

b) Bromine

c) Aluminium

d) Potassium

10 a) Mercury

b) Fluorine and chlorine

c) Any two Group 0 elements

d) Sodium

e) Bromine

f) Helium

g) Carbon

11

Test yourself (page 43) 12 The outer shell of their atoms contains one electron which is easy to lose when they react.

13 potassium + water ⟶ potassium hydroxide + hydrogen

2K + 2H2O ⟶ 2KOH + H2

14 Potassium atoms lose one electron to get a full outer shell and chlorine atoms gain one electron

to get a full outer shell.

15 Potassium atoms are bigger, so the outer shell electron is further from the nucleus. Thus the

attraction between the nucleus and the outer electron is weaker, and the outer electron is lost

more easily.

16 Soft grey metal, low melting point, low density, shiny when freshly cut.

Element Group Number of electrons in outer shell of atom

Element Group Number of electrons in outer shell of atom

Sodium 1 1 Bromine 7 7 Krypton 0 8 Barium 2 2 Magnesium 2 2 Germanium 4 4




• Group 1 products are metal hydroxide and hydrogen.

• Group 2 products are metal hydroxide and hydrogen.

• Group 1 elements are more reactive than Group 2 elements – all Group 1 elements

react with cold water.

• Group 1 and Group 2 elements – reactivity increases down the group for both groups.

Test yourself (page 46) 17 a) Melting point increases down the group.

b) Solubility decreases down the group.

18 Green-yellow/yellow-green

19 a) 7

b)

c) When a solid changes directly to a gas on heating.

d) Grey-black solid changes to purple vapour.

20 In fluorine the outer shell is closer to the positive nucleus than in bromine, so an electron is

attracted more strongly.

21 a) I + e– ⟶ I–

b) Br2 + 2e– ⟶ 2Br–

c) F + e– ⟶ F–

Test yourself (page 48) 22 a) 2NaBr + Cl2 ⟶ 2NaCl + Br2 (colourless to orange)

b) Chlorine is more reactive than bromine and pushes it out of solution. The bromine is orange

in aqueous solution.

23 a) No reaction

b) 2KI + Br2 ⟶ 2KBr + I2 (colourless to brown)

c) No reaction

d) Cl2 + CaBr2 ⟶ Br2 + CaCl2 (colourless to orange)

e) Br2 + MgI2 ⟶ I2 + MgBr2 (colourless to brown)

f) No reaction



Practical activity (pages 48–49) 1 a) Fume cupboard – chlorine is toxic.

b) Chlorine is more reactive than iodine, so chlorine displaces iodine.

c) Potassium chloride and iodine

d) Colourless to brown

e) Cl2 + 2KI ⟶ 2KCl + I2

f) Colourless to brown, so observations would be the same.

2 a) Liquid in which a substance dissolves.

b) Hydrocarbon turns purple.


Group 1 Group 7

Number of electrons in outer shell of atom 1 7 Reactive or non-reactive element Reactive Reactive Trend in reactivity down the group Increases Decreases Metal or non-metal? Metal Non-metal In reactions, do atoms of the element gain electrons or lose electrons?

Lose electrons Gain elections

Test yourself (page 50) 24 They have a full outer shell and are stable so do not need to gain or lose electrons.

25 Although most elements have 8 electrons in their outer shell, helium has only 2.

26 Colourless gas, unreactive


Element Reactive or unreactive?

Metal or non-metal?

Solid, liquid or gas at room temperature?


He Unreactive Non-metal Gas 2 Ar Unreactive Non-metal Gas 2,8,8

Test yourself (page 51) 27 An alkali metal, as it is very reactive and is less dense than water.

28 Copper

29 Copper does not react with water. Potassium reacts vigorously with water.



Test yourself (page 52) 30

31

32 Place water in a trough.

Remove oil from sodium.

Using forceps, add a small piece of sodium (about 2 mm each side) to the water.

Wear safety glasses.

Use a safety screen.

33 Keep power pack away from water supply; ensure that there are no bare leads and that the

power pack has been checked for electrical safety.

Hazard Risk Control measure Hydrated copper(II) sulfate

Irritating to eyes (Hazcard 27c) Wear safety glasses

Heating in a test tube Solid may spit out of test tube and may cause a burn or hurt eyes

Wear safety glasses; point test tube away from face and others; do not cool test tube in water

Cracked glass May cause cuts Check test tube carefully before heating; do not touch broken glass

Long hair May catch fire Keep long hair tied back

Hazard Risk Control measure Concentrated sulfuric acid

Corrosive (Hazcard 22) Hazcard numbers are not required

Wear safety glasses; wear gloves; use the lowest concentration possible

Ethanol Flammable (Hazcard 60) Hazcard numbers are not required

Heat in a water bath; keep ethanol away from naked flames

Vinegar (ethanoic acid)

Corrosive (Hazcard 38A) Hazcard numbers are not required

Wear safety glasses; wear gloves; use the lowest concentration possible

Cracked glass May cause cuts Check test tube carefully before heating; do not touch broken glass

Long hair May catch fire Keep long hair tied back Bags and stools Could be a tripping hazard Tuck stools under benches; leave

bags in bag store



Practice questions (page 53) Note for all balanced symbol equations worth 3 marks:

• for left-hand side correct formula (1 mark)

• for right-hand side correct formula (1 mark)

• balancing. (1 mark)

Note for 2 mark equations:

• for left hand side correct formula (1 mark)

• for right hand side correct formula (1 mark)

• balancing either side of the equation loses the formula mark.

1 a) A and L, or G and R, or N and M (1 mark)

b) Period (1 mark)

c) i) Q (1 mark)

ii) T (1 mark)

iii) A or L (1 mark)

d) L (1 mark)

e) G (1 mark)

f) T (1 mark)

g) i) AG (1 mark)

ii) NR2 (1 mark)

iii) JG3 (1 mark)

h) N or M (also Q) (1 mark)

2 a) F (1 mark)

b) Li (1 mark)

c) F (1 mark)

d) Li (1 mark)

e) Cu (1 mark)

f) Br (1 mark)

3 a) Halogens (1 mark)

b) 7 (1 mark)

c) 1 (1 mark)

d) 7 (1 mark)

4 a) i) Potassium atom: 2,8,8,1 (1 mark)

Potassium ion: 2,8,8 (1 mark)

The ion is more stable as it has a full outer shell/noble gas electronic structure. (1 mark)

ii) To stop it reacting with water and air (1 mark)



b) Any two of the following:

• safety screen: use a large volume of water in a trough

• use a small piece of the metal

• use forceps or tweezers to lift the metal. (2 marks)

c) i) Increases (1 mark)

ii) Green, due to neutral water (1 mark)

purple, due to hydroxide/alkali formed (1 mark) (2 marks)

iii) Fizzing/bubbles (1 mark)

iv) sodium + water ⟶ sodium hydroxide + hydrogen (1 mark)

v) 2Na + 2H2O ⟶ 2NaOH + H2 (3 marks)

5 a) Number (1 mark); weight/mass (1 mark) (2 marks)

b) Accept any two of the following:

• no gaps

• more elements

• noble gases present

• lanthanides and actinides present

• transition metal block. (2 marks)

c)

(6 marks)

d) i) It decreases. (1 mark)

ii) Lithium (1 mark)

iii) Black solid (1 mark)

iv) Br + e– ⟶ Br– (2 marks)

6 a) A substance that consists of only one type of atom. (1 mark)

b)

Group number Name of group Number of electrons in the outer shell of an atom

Reactive or unreactive?

1 Alkali metals 1 Reactive 7 Halogens 7 Reactive

Response Mark Candidates must use appropriate specialist terms to fully describe key features [5–6 indicative content points]. Relevant material is organised with a high degree of clarity and coherence. They must use excellent spelling, punctuation and grammar and the form and style are of a very high standard.

5–6 (Band A)

Candidates must use appropriate specialist termsto describe some key features [3–4 indicative content points]. Relevant material is organised with some clarity and coherence. They use good spelling, punctuation and grammar and the form and style are of a satisfactory standard.

3–4 (Band B)

Candidates describe a few key features [at least 2 indicative content points].The organisation of material may lack clarity and coherence. They use limited spelling, punctuation and grammar and they have limited use of specialist terms. The form and style are of limited standard.

1–2 (Band C)



Indicative content:

• Elements in order of atomic weight/mass.

• Was prepared to go out of order if properties fitted better.

• Properties repeated at (regular) intervals.

• Elements in the same group with similar (chemical) properties.

• Left gaps for undiscovered elements.

• Predicted the properties of undiscovered elements.

• New elements were discovered that matched his predictions.

7 a) Li, Be, B, C (1 mark)

b) Carbon (1 mark)

c) Li (1 mark)

d) Ne (1 mark)

e) B (1 mark)

f) N2, O2, F2 (1 mark)

g) Li (1 mark)

h) C (as diamond) (1 mark)

8 a) i) B (1 mark)

ii) A (1 mark)

iii) D (1 mark)

b) i) Cr (1 mark)

ii) Cr (1 mark)

iii) K (1 mark)

iv) K (1 mark)

v) K (1 mark)

vi) Cr (1 mark)

vii) K (1 mark)

9 a) Accept any three of the following:

• melts into a tiny ball

• bubbles

• heat

• K disappears

• colourless solution forms

• crackles at end. (3 marks)

b) K ⟶ K+ + e– (2 marks)

c) 2K + 2H2O ⟶ 2KOH + H2 (3 marks)

d) Less vigorously (1 mark); the electron in the outer shell is closer to the nucleus and thus held

tighter than the potassium electron. (1 mark)



e) 2K + Cl2 ⟶ 2KCl (3 marks)

f) 2Cl– ⟶ Cl2 + 2e– (3 marks)

10

Indicative content:

Physical properties

• Transition metals have high melting points.

• Alkali metals have low melting points.

• Transition metals have high densities.

• Alkali metals have low densities/less dense than water.

Chemical properties

• Transition metals have low reactivity/react slowly (with water).

• Alkali metals very reactive/react quickly (with water).

• Transition metal ions with different charges are formed, e.g. iron(II) and iron(III).

• Alkali metals form 1+ ions.

Response Mark Candidates must use appropriate specialist terms to fully compare the alkali metals and transition metals [5–6 indicative content points]. Relevant material is organised with a high degree of clarity and coherence. They must use excellent spelling, punctuation and grammar and the form and style are of a very high standard.

5–6 (Band A)

Candidates must use appropriate specialist termsto compare the alkali metals and transition metals [3–4 indicative content points]. Relevant material is organised with some clarity and coherence. They use good spelling, punctuation and grammar and the form and style are of a satisfactory standard.

3–4 (Band B)

Candidates partially compare the alkali metals and transition metals [at least 2 indicative content points].The organisation of material may lack clarity and coherence. They use limited spelling, punctuation and grammar and they have limited use of specialist terms. The form and style are of limited standard.

1–2 (Band C)

Chapter 4 – Quantitative chemistry 1 Answers



1 Carbon-12

2 a) 23

b) 24

c) 19

d) 32

e) 27

3 a) 44

b) 58.5

c) 17

d) 28

e) 213

f) 331

g) 342

h) 74

i) 164

j) 400

k) 98

l) 234

m) 106

4 (2 16)

100 55.258

%

5 (2 56)

100 28.0400

%

6 (2 19)

100 48.778

%

7 (3 16)

100 61.578

%

8 (3 12)

100 60.060

%C %

(8 1)

100 13.360

%H %

(1 16)

100 26.760

%O %


439 − (4 × 80) = 119. M is tin (Sn)



Test yourself (pages 59–60)

9 a) 58.5 g

b) 40 g

c) 98 g

d) 148 g

e) 132 g

10 a) 12

0.1298

b) 16

0.2858

c) 50

0.50100

d) 54

0.16342

e) 4

0.1332

f) 5.6

0.07674

11 a) 0.5 × 152 = 76 g

b) 0.1 × 40 = 4 g

c) 0.125 × 80 = 10 g

d) 2 × 74 = 148 g

e) 4 × 106 = 424 g

12 a) 20 000 g

b) 5 000 000 g

c) 300 000 g

d) 20 g

e) 4 000 000 g

13 a) 17000

100017

b) 2 100 000

13 125160

c) 2420

16.35148

d) 3200

32100

e) 73 000 000

2 000 00036.5




0.30060

0.0050


14 a) 2

b) 1.2

15 a) 0.99

b) 1.1

16 a) 0.5

b) 0.2

c) 0.8

d) 5

17 a) 0.44

b) 1.1

c) 0.11

18 a) 10

b) 6

c) 0.5


1 In the equation C + O2 ⟶ CO2 one mole of C atoms reacts with one mole of O2 molecules to form

one mole of CO2 molecules.

2 a) In the equation C2H4 + 2O2 ⟶ CO2 + 2H2O one mole of C2H4 molecules reacts with two moles

of O2 molecules to form one mole of CO2 molecules and two moles of H2O molecules.

b) In the equation H2 + Cl2 ⟶ 2HCl one mole of H2 molecules reacts with one mole of Cl2

molecules to form two moles of HCl molecules.

c) In the equation CxHy + 3O2 ⟶ 2CO2 + 2H2O one mole of CxHy molecules reacts with three

moles of O2 molecules to form two moles of CO2 molecules and two moles of H2O molecules.

d) x = 2, y = 4


19 2

10Moles H 5mol

2

Ratio: 2H2 : 1O2

5 : 2.5

Mass =2.5 × 32 = 80 g



20 3

25Moles CaCO 0.25mol

100

Ratio: 1 : 1

0.25 : 0.25

Mass = 0.25 × 56 = 14 g

21 3

Moles 0.125mol24

Ratio: 2 : 2

0.125 : 0.125

Mass = 0.125 × 40 = 5 g

22 1.4

Moles Li 0.2 mol7

Ratio: 2Li : 1H2

0.2 : 0.1

Mass = 0.1 × 2 = 0.2 g

23 10

Moles CuO 0.125mol80

Ratio: 3CuO : 1Al2O3

0.125

0.125:3

0.125 : 0.042

Mass= 0.042 × 102 = 4.28 g

24 3.65

Moles HCl 0.1mol36.5

Ratio: 2HCl : 1CaCO3

0.1

0.1:2

0.1 : 0.05

Mass = 0.05 × 100 = 5 g

25 3 2

33.1Moles Pb(NO ) 0.1mol

331

Ratio: 2 : 4

1 : 2

0.1: 0.2

Mass = 0.2 × 46= 9.2 g



26 2 4

490Moles H SO 5mol

98

Ratio: 2NaOH : 1H2SO4

2 × 5 : 5

2 × 5 = 10 moles NaOH

Mass = 10 × 40 = 400 g

27 2 3

1020Moles Al O 10 mol

102

Ratio: 2Al2O3 : 4Al

2 × 10 : 4 × 10

20 : 40

Mass = 20 × 27 = 540 g


1 84

2 4.2

0.05mol84

3 2NaHCO3 : 1Na2CO3

0.05 : 0.025

4 Mass = 0.025 × 106 = 2.65 g

Practical activity (pages 63–64)

1

2 3

8.01Moles NaHCO 0.0954 mol

84

3 a) Moles NaOH = 0.0954 mol

b) Moles Na2CO3 = 0.0477 mol

c) Moles Na2O = 0.0477 mol



4 a) 0.0954 × 40 = 3.82 g

b) 0.0477 × 106 = 5.05 g

c) 0.0477 × 62 = 2.96 g

5 Equation 2

6 Carbon dioxide and water vapour have been lost.


28Moles of CaO 0.5mol

56

48Moles of C 4 mol

12

The reagent in excess is coke.

Moles of CaC2 formed = 0.5 mol

Mass of CaC2 = 0.5 × 64 = 32 g


28 a) 10 mol

b) 3 mol

c) 0.3 mol

29 a) 4 mol

b) 3 mol

c) 6 mol

30 a) 2 mol

b) 0.2 mol

c) 0.25 mol

31 6 8

Moles Ca 0.15,moles S = 0.2540 32

S is in excess, Ca is the limiting reactant.

Mass = 0.15 × 72 = 10.8 g

32 4

1.9 6Moles TiCl 0.01,moles Mg = 0.25

190 24

TiCl4 is the limiting reactant.

Mass of Ti produced = 0.01 × 48 = 0.48 g

33 3 2

23.2 20Moles WO 0.1,moles H = 10

232 2

WO3 is the limiting reactant.

Moles of W produced = 0.1 mol

Mass of W produced = 0.1 × 184 = 18.4 g



34 4

20 50Moles NH Cl 0.374,moles CaO = 0.893

53.5 56

NH4Cl is the limiting reactant.

2

0.374Moles of CaCl produced 0.187

2

Mass of CaCl2 produced = 0.187 × 111 = 20.76g


35 a) 150

100 75%200

b) The reaction may be incomplete; some of the product may be lost on separation; other

reactions may take place.

36 8

100 23.5%34

37 800

100 40%2000

38 0.85

100 89.5%0.95

39 a) 7.3

Moles 0.2 mol36.5

0.2 : 0.2

0.2 × 53.5 = 10.7 g

b) 4

8.3% NH Cl 100 77.6%

10.7

40 a) 6

0.25mol24

0.25 : 0.25

0.25 × 40 = 10 g

b) 4

% MgO 100 40%10

41 a) 47

0.594

0.5 : 0.5

0.5 × 197.5 = 98.75 g

b) 90.6

% TCP 100 91.7%98.75




1 actual yieldpercentage yield 100

theoretical yield

2 5.4840 100

theoretical yield

5.48 100theoretical yield 13.7

40

4 9

13.7Moles C H Br 0.1 mol

137

Mass C4H9OH = 0.1 × 74 = 7.4 g


42 a) CH3

b) C2H5OH

c) C3H8

d) CH2

e) CH4

f) P2O5

g) HO

h) C4H5N2O

i) NaS2O3

j) Na2S2O3

43 C3H6O3

44 P4O10

45 C6H3N3O6

Practical activity (page 71)

1 a) 17.36 − 16.34 = 1.02 g

b) 18.06 − 16.34 = 1.72 g

c) 1.72 – 1.02 = 0.70 g

d) 1.02 0.70

Moles Ti 0.021,moles O = 0.04448 16

, so the empirical formula is TiO2.

2 To allow oxygen in to react with the titanium

3 A glowing splint is relit.

4 The mass of the container and of the container + titanium

5 The mass would increase due to the formation of titanium oxide/oxygen combining with

the titanium.

6 Allow the container to cool before weighing and so prevent burns; wear eye protection.

7 Repeat the experiment and results should be same.




46 0.96

Moles Mg 0.04 mol24

2.84

Moles Cl 0.08 mol35.5

Ratio: 1Mg : 2Cl

MgCl2

47 2.3

Moles Na 0.1mol23

0.8

Moles O 0.05mol16

Ratio: 0.1 : 0.05

2 : 1

Na2O

48 0.72

Moles Mg 0.03mol24

0.28

Moles N 0.02mol14

Ratio: 0.03 : 0.02

3 : 2

Mg3N2

49 414

Moles Pb 2 mol207

478 414 64

Moles O 4 mol16 16

Ratio: 2 : 4

1 : 2

PbO2

50 1.44

Moles C 0.12mol12

0.36

Moles H 0.36 mol1

Ratio: 0.12 : 0.36

1 : 3

CH3



51 5.6

Moles Fe 0.1mol56

16.25 5.6 10.65

Moles Cl 0.3mol35.5 35.5

Ratio: 0.1: 0.3

1: 3

FeCl3

52 87.5

Moles Si 3.125mol28

12.5

Moles H 12.5mol1

3.125 12.5Ratio: :

3.125 3.125

1 : 4

SiH4

53 14

Moles C 1.17mol12

44.4

Moles F 2.34 mol19

100 (14.0 44.4) 41.6

Moles Cl 1.17mol35.5 35.5

Ratio: 1.17 : 2.34 : 1.17

CF2Cl


54 MgSO4·7H2O

55 Na2CO3·10H2O

56 a) 177

b) 250

c) 244

d) 147

e) 236

57 a) 4 18

100 40.7 %177

b) 8 18

100 57.6 %250

c) 2 18

100 14.8 %244

d) 2 18

100 24.5 %147

e) 4 18

100 30.5 %236




a) 14 16

% O = 100 69.6 %322

b) 20 1

% H = 100 6.2 %322

c) 2

10 18% H O = 100 55.9 %

322

Prescribed practical activity (page 76)

1 a) Water of crystallisation is water that is chemically bonded into the crystal structure.

b) Steam/beads of colourless liquid; crystals change to powder.

c) Weigh, heat, reweigh and repeat until the mass is constant.

2 a) Ni(NO3)2·xH2O ⟶ Ni(NO3)2 + xH2O

b) 2.91 – 1.83 = 1.08 g

c) 1.08

0.0618

d) 1.83

0.01183

e) 6

f) Not all of the water has been removed.

g) It may cause it to decompose.

h)


58 a) Water of crystallisation is water that is chemically bonded into the crystal structure.

b) Heat to constant mass.

c) 23.11 – 21.50 = 1.61 g

d) 161

e) 1.61

0.01161

f) 24.37 – 23.11 = 1.26 g

g) 1.26

0.0718

h) 7



59

Formula: Li2SO4·H2O

Practice questions (page 77)

1 a) PbS = 239 (1 mark)

Fe2O3 =160 (1 mark)

CaMg(CO3)2 = 184 (1 mark)

b) 102 – (3 × 16) = 54 = 2x (1 mark)

x = 27 (1 mark)

c) i) 2.39

0.01239

(1 mark)

ii) 12

0.37532

(2 marks)

2 a) Carbon (1 mark); 12 (1 mark) (2 marks)

b) 152 (1 mark)

c) 152 g (1 mark)

d) 4

0.2Moles FeSO 0.001 32mol

152 (2 marks)

3 a) CH2O (1 mark)

b) 16 6

% O 100 53.3%180

(2 marks)

c) 2

1080Moles H O 60 mol

18 (1 mark)

Ratio: 1 glucose : 6 water (1 mark)

? : 60 (1 mark)

60

10 mol6 (1 mark)

Mass = 10 × 180 = 1800 g (1 mark)

4 a) 4780

Moles PbS 20 mol239

(1 mark)

Ratio 1 : 1 (1 mark)

Mass PbO = 20 × 223 = 4460 g (1 mark)

b) Moles Pb = 20 (1 mark)

Mass Pb = 20 × 207 = 4140 g (1 mark)

Anhydrous Li2SO4 Water (H2O)

Mass in g 3.23 3.76 – 3.23 = 0.53

Moles = mass/Mr 3.230.029

110

0.530.029

18

Ratio (divide by the smallest number of moles (0.029))

0.0291

0.029

0.0291

0.029



5 a) 4.54 kg = 4540 g (1 mark)

4 10

4540Moles C H 78.3mol

58 (1 mark)

b) 2

2000Moles O 62.5mol

32 (2 marks)

6 a) Mr = 84 (1 mark)

3.360.04 mol

84 (1 mark)

b) 0.02 (1 mark)

c) Mr = 106 (1 mark)

0.02 × 106 = 2.12 g (1 mark)

d) 3 16% O 100 45.3%

106

(2 marks)

7 a) 3

2.5Moles ZnCO 0.02mol

125 (1 mark)

Moles ZnSO4 = 0.02 mol (1 mark)

Mass ZnSO4 = 0.02 × 161 = 3.22 g (1 mark)

b) 2.8

100 87.0%3.22

(2 marks)

c) Accept any two of the following:

Some product may be lost in filtration, transfer between apparatus, etc.

some of the reactants may react in different ways from the expected reaction

the reaction may be incomplete

the reaction may be reversible

the zinc sulfate may be impure. (1 mark)

8 a) Mr = 138 (1 mark)

40.0290

128 (1 mark)

b) Mr = 102 (1 mark)

6.5

0.0637102

(1 mark)

c) 0.0290 (1 mark)

d) Mr = 180 (1 mark)

0.0290 × 180 = 5.22 g (1 mark)

e) 2.90

100 55.6%5.22

(1 mark)



9 a) 2

7000Moles N 250 mol

28 (1 mark)

2

60 000Moles H 30 000 mol

2 (1 mark)

Nitrogen is the limiting reactant. (1 mark)

Moles NH3 = 250 × 2 = 500 (1 mark)

Mass = 500 × 17 = 8500 = 8.5 tonnes (1 mark)

b) 2 14

% N 100 35%80

(2 marks)

c) 20.00

Moles C 1.67mol12

(1 mark)

6.66

Moles H 6.66 mol1

(1 mark)

46.67

Moles N 3.33mol14

(1 mark)

26.67

Moles O 1.67mol16

(1 mark)

Empirical formula: CH4N2O (1 mark)

10 a) HO (1 mark)

b) 2 2

5.1Moles H O 0.15mol

34 (1 mark)

2

0.15Moles O 0.075mol

2 (1 mark)

Mass of O2 = 0.075 × 32 = 2.4 g (1 mark)

11 a) Contains water of crystallisation. (1 mark)

b) 2 3

2.04Moles Al O 0.02mol

102 (1 mark)

Mass of H2O = 3.12 – 2.04 = 1.08 g (1 mark)

2

1.08Moles H O 0.06 mol

18 (1 mark)

n = 3 (1 mark)

c) Marks are for recognisable apparatus with labels:

Evaporating basin (1 mark)

Tripod gauze (1 mark)

Heat (1 mark)

12 a) Contains no water of crystallisation. (1 mark)

b) 2 3

5.3Moles Na CO 0.05mol

106 (2 marks)

c) Mass of water = 14.3 – 5.3 = 9 g (1 mark)

d) 2

9Moles H O 0.5mol

18 (1 mark)

e) x = 10 (1 mark)

Chapter 5 – Acids, bases and salts Answers



1

2 a) An acid contains hydrogen ions when dissolved in water, an alkali contains hydroxide ions.

b) Hydrogen ions and sulfate ions

c) Calcium ions and hydroxide ions

d) HNO3

3 a) Colourless

b) Colourless

c) Pink


4 a) Weak alkali

b) Weak acid

c) Weak alkali

d) Strong alkali

5 a) pH 14

b) pH 1

c) Dip in pH meter and record to 1 decimal place.

6

7 a) Red

b) Red

c) Orange

Indicator Colour of the indicator in a solution of:

Hydrochloric acid

Sodium hydroxide

Citric acid Calcium hydroxide

Ethanoic acid

Red litmus paper Red Blue Red Blue Red

Blue litmus paper Red Blue Red Blue Red

Phenolphthalein Colourless Pink Colourless Pink Colourless

Methyl orange Red Yellow Red Yellow Red

Substance Acid or alkali? Strong or weak? pH range Colour of pH paper

Hydrochloric acid Acid Strong 0–2 Red

Potassium hydroxide

Alkali Strong 12–14 Dark blue/purple

Ammonia Alkali Weak 8–11 Green-blue

Nitric acid Acid Strong 0–2 Red

Ethanoic acid Acid Weak 3–6 Orange/yellow

Citric acid Acid Weak 3–6 Orange/yellow

Sodium hydroxide Alkali Strong 12–14 Dark blue/purple




1 Using a pH meter; values were recorded to 1 decimal place.

2 4 minutes

3 8 minutes

4 45 minutes


8 There are more alkali particles per unit volume in the concentrated alkali.

9 a) An acid which is fully ionised in water.

b) An acid which is partially ionised in water.

c) They are the same concentration.

d) Nitric acid, because it is a strong acid.

e) i) Corrosive

ii) Flammable

10 a) C

b) Dip in pH paper/add a few drops of universal indicator solution, and compare the colour

against the colour chart. Note: the pH is not recorded to a decimal place, hence a pH meter

has not been used.

c) 0.5 mol/dm3

d) Equal concentration


1 By comparing the colour obtained against the universal indicator colour chart.

2 A = acidic, B = acidic.

3 There would be more H+ ions in solution so the pH as recorded by the pH meter would be lower.

The colours would be much the same.

4 Neutral solutions also give a red result with litmus. Universal indicator results can be used to give

a pH value.


11 Solid white magnesium carbonate disappears, heat released, colourless solution forms, bubbles.

12 Colourless solution remains, heat released.

13 Solid grey calcium disappears, heat released, colourless solution formed, bubbles.

14 Solid white magnesium oxide disappears, colourless solution produced, heat released.

15 Black copper(II) oxide disappears, blue solution formed.




1 a) Independent – volume of acid added

b) Dependent – temperature

c) Controlled – volume of sodium hydroxide solution, concentration of hydrochloric acid and

concentration of sodium hydroxide

2 To ensure the results are reliable

3 To minimise heat loss

4 To support it and prevent it toppling over and spilling contents

5 To ensure the solutions mix and react

6 Heat loss from surface – use a lid. Heat loss from polystyrene cup – insulate with cotton wool.

7 The x axis is the volume of acid added (cm3), the y axis is temperature (°C).

The best-fit line should be drawn, or the two lines could be extrapolated to find the highest

temperature.

8 As the volume of acid increases, the temperature increases to a maximum and then decreases.

9 Add a few drops of indicator to the sodium hydroxide solution and repeat the experiment, noting

the temperature at which the indicator changes colour. If it is the same as the highest

temperature then neutralisation has occurred at this point.

10 a) Independent – type of acid

b) Dependent – temperature

c) Controlled – volume of sodium hydroxide solution, concentration of acid and concentration

of sodium hydroxide solution

11 Strong acids (hydrochloric and sulfuric acids) have a similar highest temperature.

Strong acids (sulfuric and hydrochloric acids) have a higher temperature (at neutralisation) than

weaker acids (ethanoic acid).

Neutralisation gives out heat/is accompanied by a temperature increase/is exothermic.


16 a) zinc + nitric acid ⟶ zinc nitrate + hydrogen

b) sodium hydroxide + sulfuric acid ⟶ sodium sulfate + water

c) magnesium + sulfuric acid ⟶ magnesium sulfate + hydrogen

d) hydrochloric acid + sodium carbonate ⟶ sodium chloride + water + carbon dioxide

e) copper(II) oxide + sulfuric acid ⟶ copper(II) sulfate + water

f) potassium hydrogencarbonate + sulfuric acid ⟶ potassium sulfate + water + carbon dioxide

g) calcium carbonate + hydrochloric acid ⟶ calcium chloride + water + carbon dioxide

h) sodium hydroxide + nitric acid ⟶ sodium nitrate + water



17 a) Heat released, bubbles, solid grey zinc disappears, colourless solution produced.

b) Solution remains colourless, heat released.

c) Heat released, bubbles, solid grey magnesium disappears, colourless solution produced.

d) Heat released, bubbles, solid white sodium carbonate disappears, colourless solution

produced.

e) Black solid disappears, blue solution produced.

f) Bubbles, solid white potassium hydrogencarbonate disappears, colourless solution produced.

g) Heat released, bubbles, solid white calcium carbonate disappears, colourless solution

produced.

h) Solution remains colourless, heat released.

18 a) Zn + 2HNO3 ⟶ Zn(NO3)2 + H2

b) 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O

c) Mg + H2SO4 ⟶ MgSO4 + H2

d) 2HCl + Na2CO3 ⟶ 2NaCl + H2O + CO2

e) CuO + H2SO4 ⟶ CuSO4 + H2O

f) 2KHCO3 + H2SO4 ⟶ K2SO4 + 2H2O + CO2

g) CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

h) NaOH + HNO3 ⟶ NaNO3 + H2O

19 a) Base

b) Base

c) Base

d) Alkali

e) Base

f) Base

g) Alkali


1 sodium hydroxide + sulfuric acid ⟶ sodium sulfate + water

2NaOH + H2SO4 ⟶ Na2SO4 + H2O

2 A; contains sodium carbonate.

3 Carbon dioxide

4 Bubble into limewater; limewater turns from colourless to milky.

5 Na2SiO3



Prescribed practical activity (pages 94–95)

1 Measuring cylinder

2 To ensure all the acid has reacted

3 Solid remaining/no more bubbles

4 Residue

5 To ensure the water of crystallisation is not removed

6 Solubility decreases on cooling.

7 Between two pieces of filter paper, in a low-temperature oven or in a desiccator

8 magnesium carbonate + sulfuric acid ⟶ magnesium sulfate + carbon dioxide + water

MgCO3 + H2SO4 ⟶ MgSO4 + H2O + CO2

9


20 a) Hydrochloric acid, water

b) Potassium carbonate/potassium hydrogencarbonate, sulfuric acid

c) Hydrochloric acid, carbon dioxide, water

d) Nitric acid, aluminium

e) Sulfuric acid, water

21 a) To make sure all the acid is used up

b) Filtering

c) iron + sulfuric acid ⟶ iron(II) sulfate + hydrogen

Fe + H2SO4 ⟶ FeSO4 + H2



22 a) Copper does not react with hydrochloric acid / copper has low reactivity.

b) copper(II) oxide + hydrochloric acid ⟶ copper(II) chloride + water

CuO + 2HCl ⟶ CuCl2 + H2O

copper(II) carbonate + hydrochloric acid ⟶ copper(II) chloride + water + carbon dioxide

CuCO3 + 2HCl ⟶ CuCl2 + H2O + CO2

copper(II) hydroxide + hydrochloric acid ⟶ copper(II) chloride + water

Cu(OH)2 + 2HCl ⟶ CuCl2 + 2H2O

23 a) nitric acid + calcium/calcium oxide/calcium hydroxide/calcium carbonate

b) hydrochloric acid + copper(II) oxide/copper(II) hydroxide/copper(II) carbonate

c) sulfuric acid + zinc/zinc oxide/zinc hydroxide/zinc carbonate.


1 A: magnesium + hydrochloric acid ⟶ magnesium chloride + hydrogen

B: magnesium oxide + hydrochloric acid ⟶ magnesium chloride + water

C: magnesium hydroxide + hydrochloric acid ⟶ magnesium chloride + water

D: magnesium carbonate + hydrochloric acid ⟶ magnesium chloride + water + carbon dioxide

2 Accept any two of the following:

heat released

bubbles

white solid magnesium carbonate disappears

colourless solution produced.

Practice questions (pages 97–99)

1 a)

(7 marks)

b) i) Changes to blue (1 mark)

ii) Stays blue (1 mark)

c) A substance that dissolves in water to produce hydrogen ions (1 mark)

d) 0.5 mol/dm3 (1 mark) as it has a lower concentration of hydrogen ions (1 mark). (2 marks)

Substance pH Classification

Baking soda 8–10 Weak alkali

Ethanoic acid 3–6 Weak acid

Caustic soda solution 12–14 Strong alkali

Hydrochloric acid 0–2 Strong acid

Lemon juice 3–6 Weak acid



2 a) Burette (1 mark)

b) Conical flask (1 mark)

c) Pipette (1 mark)

d) Place a pH meter in the solution (1 mark); record the result to one decimal place. (1 mark)

e) To ensure the solution is mixed and X and Y have reacted together (1 mark)

f) Acidic – low pH (1 mark)

g) Alkaline – on adding to X the pH increases. (1 mark)

h) The pH starts low (1 mark) and the solution is acidic. When Y is added the pH rises and the

solution becomes neutral (1 mark). When the acid X and alkali Y have cancelled out and when

more Y (alkali) is added the pH continues to rise. (1 mark)

i) The pH is high initially as Y is an alkali, and the pH falls as acid is added; so the curve

on the graph would be the opposite way around to Figure 5.24. (1 mark)

j) 10 cm3 (1 mark)

3 a) i)

(4 marks)

ii) H+(aq) + OH–(aq) ⟶ H2O(l) (3 marks)

b) i) 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O (3 marks)

ii) CuO + 2HCl ⟶ CuCl2 + H2O (2 marks)

iii) Reaction in (i): colourless solution remains (1 mark); heat released (1 mark)

Reaction in (ii): black solid disappears (1 mark); blue solution forms (1 mark)

c) i) Accept any two of the following:

heat

bubbles

grey solid Mg disappears

colourless solution forms. (2 marks)

ii) Lighted splint (1 mark); pop (1 mark) (2 marks)

d) Evaporate off some of the solution. (1 mark)

Cool and crystallise. (1 mark)

Dry between two pieces of filter paper/in a desiccator/in a low-temperature oven. (1 mark)

Ion present in all acids Ion present in all alkalis

Name Hydrogen (1 mark) Hydroxide (1 mark)

Formula H+ (1 mark) OH– (1 mark)



4 a) i) 0–2 (1 mark)

ii) 3–6 (1 mark)

iii) 0–2 (1 mark)

iv) 12–14 (1 mark)

b) Corrosive (1 mark)

c) One which only partially ionises in water (1 mark)

d) One which fully ionises in water (1 mark)

e) i) Colourless (1 mark)

ii) Pink (1 mark)

f) A salt is formed when the hydrogen ion of an acid (1 mark) is replaced by metal ions or

ammonium ions. (1 mark)

g) Contains water of crystallisation. (1 mark)

h) i) pH meter (1 mark); record pH to one decimal place. (1 mark)

ii) pH 3.3 – weak acid (1 mark); pH 11.4 – strong alkali (1 mark)

i) i) White (1 mark)

ii) Blue (1 mark)

iii) Colourless (1 mark)

5 a)

(18 marks)

b) i) Base (1 mark)

ii) Base (1 mark)

iii) Salt (1 mark)

iv) Salt (1 mark)

v) Salt (1 mark)

vi) Base (1 mark)

vii) Salt (1 mark)

viii) Base (1 mark)

ix) Base (1 mark)

Reaction Balanced symbol equation Observations

Magnesium + hydrochloric acid

Mg + 2HCl ⟶ MgCl2 + H2 (3 marks)

Any three of: bubbles, heat released, grey solid magnesium disappears, colourless solution produced

(3 marks)

Calcium carbonate + hydrochloric acid

CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

(3 marks)

Any three of : bubbles, heat released, white solid calcium carbonate disappears, colourless solution produced

(3 marks)

Copper(II) oxide + sulfuric acid

CuO + H2SO4 ⟶ CuSO4 + H2O (3 marks)

Black solid disappears, blue solution produced

(3 marks)



c) Soluble base (1 mark)

d) Sodium hydroxide (1 mark); potassium hydroxide (1 mark) (2 marks)

6

Indicative content:

Same salt in both

Products for first reaction: calcium chloride and water

Products for second reaction: calcium chloride and hydrogen

Observations for reaction 1: test tube gets warm/heat released. (Any extra incorrect

observation loses this mark.)

Observations for reaction 2 – any two of: bubbles; effervescence/gas produced; test

tube gets warm/heat increased; solid disappears; colourless solution formed.

7 a) A salt is the compound formed when some or all of the hydrogen ions of an acid are

replaced by metal or ammonium ions. (1 mark)

b) Accept any three of the following:

green solid disappears

heat released

blue solution produced

bubbles. (3 marks)

c) CuCO3 + 2HCl ⟶ CuCl2 + H2O + CO2 (3 marks)

d) Bubble into colourless (1 mark) limewater (1 mark), changes to milky (1 mark) (3 marks)

e) CuCl2·2H2O (1 mark)

8 a) Pipette (burette would also work, but not usually used) (1 mark)

b) Pink to colourless (2 marks; if put colourless to pink, 1 mark)

c) To remove the indicator/colour (1 mark)

d) They are less soluble at lower temperature. (1 mark)

e) Between two pieces of filter paper/in a desiccator/in a low-temperature oven (1 mark)

f) KOH + HCl ⟶ KCl + H2O (2 marks)

Response Mark

Candidates use appropriate specialist terms to fully compare and contrast the products and observations of both reactions [5–6 indicative content points]. They use good spelling, punctuation and grammar, and the form and style are of a high standard.

5–6 marks (Band A)

Candidates use appropriate specialist terms to compare a number of products and observations [3–4 indicative content points]. They include at least one comparison of properties. They use good spelling, punctuation and grammar, and the form and style are of a high standard.

3–4 marks (Band B)

Candidates describe briefly some products or observations [at least 2 indicative content points]. They use limited spelling, punctuation and grammar, and have limited use of specialist terms. The form and style are of limited standard.

1–2 marks (Band C)

Response not worthy of credit. (0 marks)



9 a) Measuring cylinder (1 mark)

b) To ensure that all the acid is used up (1 mark)

c) No more bubbles/excess solid on bottom of beaker (1 mark)

d) Residue (1 mark)

e) Filtrate (1 mark)

f) Solubility decreases as temperature falls. (1 mark)

g) Any two of the following:

between two pieces of filtered paper

in a desiccator

in a low-temperature oven (2 marks)

h) Any two of the following:

zinc oxide

zinc hydroxide

zinc metal (2 marks)

10 Measure out 25.0 cm3 (1 mark)

of hydrochloric acid using a pipette (1 mark)

and place in a conical flask. (1 mark)

Add three drops of phenolphthalein. (1 mark)

Slowly add sodium hydroxide from a burette (1 mark)

until the phenolphthalein turns from colourless (1 mark)

to pink. (1 mark)

Either

Record the volume of sodium hydroxide. (1 mark)

Repeat the procedure using the same volumes of reactants but no indicator. (1 mark)

Or

Add a spatula of charcoal heat to remove the indicator. (1 mark)

Filter. (1 mark)

Heat in an evaporating basin to evaporate and make the solution more concentrated. (1 mark)

Cool and crystallise. (1 mark)

Dry the crystals: between sheets of filter paper/in a low temperature oven

/in a desiccator. (1 mark)

(12 marks)

Chapter 6 – Chemical analysis, assessing purity and separating mixtures Answers


Test yourself (page 102) 1 a) They might say that it is a natural substance that has had nothing added to it.

b) It is a mixture of several different substances and not a single element or compound.

c) Copper, sodium chloride, magnesium

2 a) The water is not pure. It would freeze at 0 °C if it was pure.

b) The boiling point is increased.

3 C only – it melts at a sharp specific temperature.

4 The aspirin is impure as the melting point has a range and is lower than the actual melting point.

5 It is a mixture that has been designed as a useful product and is formed by mixing together

several different substances in carefully measured quantities to ensure the product has the

required properties.

Show you can (page 103) a) Liquid

b) Solid

c) Solid

d) An alloy is a mixture of two or more elements, at least one of which is a metal.

e) A = element, B = element, C = alloy. Elements are pure substances with exact boiling and

melting points; alloys are mixtures and change state over a range.

Test yourself (page 106) 6 a) Water

b) Copper carbonate

7 a) Copper(II) sulfate

b) Evaporation

c) Water

d) Anti-bumping granules




Practical activity (pages 107–108) 1 a) Sodium chloride is soluble (in water).

b) Heat is needed to produce hot water.

c) The land may collapse.

d) Evaporation

2 Photo A = b, photo B = e, photo C = f

3 a) The salt, clay and sand are not chemically combined.

b) To increase the surface area to speed up dissolving (see Chapter 10 on Rates of reaction)

c) To speed up dissolving

d) Salt dissolved in water

e) Clay and sand

f) Some finer sand may have passed through filter paper; paper may have ripped; refilter (use

finer filter paper)

Test yourself (page 109) 8 a) 1, 3, 4 and 6; only one spot

b) 3, 4

c) Spot 3 = 11/32 = 0.34; spot 4 = 24/32 = 0.75

d) Spot 6

e) Solvent is the liquid that a solute dissolves in. Solvent front is the furthest distance travelled

by the solvent.

f) The stationary phase is the paper.

g) Draw line in pencil, 1–2 cm from bottom of paper.

Filtration Distillation Fractional distillation Type of mixture separated

Insoluble solid and liquid

Soluble solid dissolved in liquid

Miscible liquids

Important word and definition

Filtrate – liquid passing through filter paper

Condenser – apparatus to cool the vapour and form liquid

Miscible – liquids that mix

Important word and definition

Residue – solid left in filter funnel

Distillate – the product collected at end of condenser

Fractionating column – allows good separation as it keeps different liquids apart as they boil/condense



9 a) Solvent 1 = 10/26 = 0.38; solvent 2 = 18/26 = 0.69

b) Substance had stronger attraction to solvent 2/more soluble in solvent 2 so moved faster.

Show you can (page 110) a) Chromatography

b) Distillation

c) Crystallisation/evaporation

d) Fractional distillation

e) Distillation

f) Filtration

g) Fractional distillation

Test yourself (page 112) 10 a) Filtration removes solids from the water; the water is passed through filter beds made of

sand.

b) Aluminium sulfate, allows small particles to clump together so they are easily removed.

c) Kills microbes

11 The removal of dissolved substances from sea water

12 a) There is little fresh water in Saudi Arabia but lots of sea water. They also have cheap

energy supplies.

b) Sea water is heated so that it boils; the water molecules are turned to steam leaving behind

the dissolved substances; the water vapour is then cooled and condensed.

13 White anhydrous copper(II) sulfate turns blue in the presence of water.

Test yourself (page 113) 14 Make a loop on the end of a piece of nichrome wire. Dip the loop into concentrated hydrochloric

acid and then into A. Place the loop into a blue Bunsen burner flame and if a lilac flame is

observed, potassium ions are present in the substance.

15 a) Na+

b) Cu2+

16 A positive ion

17 a) Brick-red flame

b) Crimson flame



18 Answers in bold and underline

Test yourself (pages 115–116) 19 Dissolve in water; add sodium hydroxide solution/ammonia solution; green ppt for Fe2+; brown

ppt for Fe3+.

20 Dissolve in water; add sodium hydroxide solution and a white ppt forms; add sodium hydroxide

until in excess; white ppt dissolves to give a colourless solution.Add ammonia solution and a

white ppt forms; add ammonia solution until in excess; white ppt remains on excess.

21 a) i) Blue ppt which dissolves to give a deep blue solution

ii) White ppt

b) Cu2+(aq) + 2OH–(aq) ⟶ Cu(OH)2(s)

Mg2+(aq) + 2OH–(aq) ⟶ Mg(OH)2(s)

c) For copper(II) ions the blue ppt remains; for magnesium, same observations.

22 a) Blue ppt: Cu2+(aq) + 2OH–(aq) ⟶ Cu(OH)2(s)

Cu(NO3)2(aq) + 2NaOH(aq) ⟶ Cu(OH)2(s) + 2NaNO3(aq)

b) White ppt: Mg2+(aq) + 2OH–(aq) ⟶ Mg(OH)2(s)

MgSO4(aq) + 2NaOH(aq) ⟶ Mg(OH)2(s) + Na2SO4(aq)

c) Green ppt: Fe2+(aq) + 2OH–(aq) ⟶ Fe(OH)2(s)

FeCl2(aq) + 2NaOH(aq) ⟶ Fe(OH)2(s) + 2NaCl(aq)

d) White ppt (which dissolves in excess to give a colourless solution):

Al3+(aq) + 3OH–(aq) ⟶ Al(OH)3(s)

Al(NO3)3(aq) + 3NaOH(aq) ⟶ Al(OH)3(s) + 3NaNO3(aq)

23 No ppt as potassium hydroxide is soluble.

Ion name Ion formula Cation or anion? Sodium Na+ Cation Hydrogencarbonate HCO3

– Anion

Iron(II) Fe2+ Cation Aluminium Al3+ Cation Sulfate SO4

2– Anion Chloride Cl– Anion Carbonate CO3

2– Anion Iodide I– Anion Copper(II) Cu2+ Cation



Test yourself (page 118) 24 a) Cl–

b) SO42–

c) Br–

d) CO32– (or HCO3

−)

25 a) Ba2+(aq) + SO42–(aq) ⟶ BaSO4(s)

b) Ag+(aq) + Cl–(aq) ⟶ AgCl(s)

c) Ag+(aq) + I–(aq) ⟶ AgI(s)

d) Ba2+(aq) + SO42–(aq) ⟶ BaSO4(s)

e) Ba2+(aq) + SO42–(aq) ⟶ BaSO4(s)

f) Ag+(aq) + Cl–(aq) ⟶ AgCl(s)

26 a) BaCl2(aq) + CuSO4(aq) ⟶ BaSO4(s) + CuCl2(aq)

b) AgNO3(aq) + KCl(aq) ⟶ AgCl(s) + KNO3(aq)

c) AgNO3(aq) + NaI(aq) ⟶ AgI(s) + NaNO3(aq)

d) Na2SO4(aq) + BaCl2(aq) ⟶ BaSO4(s) + 2NaCl(aq)

e) H2SO4(aq) + BaCl2(aq) ⟶ BaSO4(s) + 2HCl(aq)

f) HCl(aq) + AgNO3(aq) ⟶ AgCl(s) + HNO3(aq)

27 a) White

b) White

c) Yellow

d) White

e) White

f) White

28 C

Show you can (page 119) Chloride, sulfate, carbonate



Prescribed practical activity (page 120) 1

2 Magnesium chloride and magnesium sulfate

Test yourself (page 121) 29 a) Iron(II) ions, Fe2+

b) Iodide

c) Iron(II) iodide, FeI2

d) Fe2+(aq) + 2OH–(aq) ⟶ Fe(OH)2(s)

e) Ag+(aq) + I–(aq) ⟶ AgI(s)

30 a) Lithium

b) Carbonate (or hydrogencarbonate)

c) Lithium carbonate, Li2CO3 (or lithium hydrogencarbonate, LiHCO3)

d) CO32–(s) + 2H+(aq) ⟶ CO2(g) + H2O(l) (or HCO3

−(s) + H+(aq) ⟶ CO2(g) + H2O(l))

31 a) Test 3: no carbonate (or hydrogencarbonate) ion present, chloride ion present; Test 4: no

sulfate ion present; Test 5: copper(II) ion present; Test 6: copper(II) ion present

b) Copper(II) chloride, CuCl2

c) AgCl

d) Cu(OH)2

Test Observation Deduction 1 Make a solution of Z by

dissolving a spatula of Z in a test tube half full of deionised water. Warm gently.

White solid dissolves and colourless solution formed.

Could contain a group 1, 2 or ammonium compound; not a transition metal compound

2 Add a few drops of sodium hydroxide to the solution prepared in Test 1, then add a further 3 cm3 of the sodium hydroxide solution.

A white ppt forms which is insoluble in excess sodium hydroxide.

Magnesium ions are present

3 Make a solution of Z by dissolving half a spatula of Z in a test tube half full of nitric acid solution.

No effervescence Z is not a carbonate or hydrogencarbonate

4 Transfer 1 cm3 of the solution from Test 3 into each of two separate test tubes. Add a few drops of silver nitrate solution to the first test tube. Add a few drops of barium chloride solution to the second test tube.

A white ppt forms A white ppt forms

Chloride present Sulfate present



Practice questions (pages 122–124) 1 a) i) Water that is safe to drink. (1 mark)

ii) A solution is a solute dissolved in a solvent. (1 mark)

iii) Filtration to remove insoluble solids (1 mark)

sedimentation to clump tiny particles together into large particles

which settle out (1 mark)

chlorination to kill microbes (1 mark)

b) i) It is a mixture of several substances in carefully measured quantities to ensure

the tablet has the required properties. (1 mark)

ii) Test it to see if it had a sharp melting point. (1 mark)

iii) Alloys, fertilisers (2 marks)

2 a) 1 = evaporation/crystallization (1 mark)

2 = filtration (1 mark)

3 = (simple) distillation (1 mark)

b) 3 (1 mark)

c) 2 (1 mark)

d) A = filtrate (1 mark); B = residue (1 mark) (2 marks)

e) Copper(II) sulfate is soluble and would pass through the filter paper. (1 mark)

3 a) Chromatography (1 mark)

b) There is a spot at the same height as the one for E102. (1 mark)

c) E160 (1 mark)

d) A spot drawn at the same height as E160 and directly above orange drink Y, labelled. (1 mark)

e) The ink would separate in the solvent and ruin the chromatogram. (1 mark)

4 a) Sodium chloride (1 mark)

b) Water (1 mark)

c) Anti-bumping granules (1 mark) which aid smoother boiling (1 mark) (2 marks)

d) B ‘water out’ (1 mark); C ‘water in’ (1 mark) (2 marks)

e) D is a condenser. (1 mark)

f) 100 °C (1 mark); it is pure water that is distilling over (1 mark). (2 marks)

g) It remains in the flask. (1 mark)

h) Evaporation is a change from liquid to gas on heating. (1 mark)

It occurs in the flask. (1 mark)

i) Condensation is a change from gas to liquid on cooling. (1 mark)

It occurs in D. (1 mark)



j) In Diagram 1, temperature of distillate can be recorded,

condensation is more efficient,

larger quantities can be distilled.

In Diagram 2, there is the disadvantage that some salt solution may splash into the

delivery tube and some steam may escape before condensation. (1 mark)

k) To condense the steam (1 mark)

l) X = conical flask, Y = delivery tube, Z = test tube (3 marks)

m) Boiling point at 100 °C (1 mark)

n) Removal of dissolved substances from sea water (1 mark)

o) Anhydrous copper(II) sulfate (1 mark); turns from white to blue (1 mark) (2 marks)

5 a) The pencil line is insoluble and will not move with the solvent or interfere with

the results. (1 mark)

b) If the solvent is too deep, the spots will be under it and will dissolve in it. (1 mark)

c) Use a capillary tube to add a spot of the solution to the base line. (1 mark)

Allow it to dry and reapply the solution to make a concentrated spot. (1 mark)

d) Copper(II) (1 mark)

Iron(III) (1 mark)

e) Fe(OH)3 (1 mark)

f) fdistance moved by spot 8.5

R 0.25distance moved by solvent 34

= = = (2 marks)

g) Rf values vary depending on the solvent used. (1 mark)

6 a) Sulfate ion (1 mark)

b) When sodium hydroxide solution is added, a white ppt is formed with magnesium,

aluminium and zinc ions. (1 mark)

To make the test more valid, excess sodium hydroxide could be added: the ppt will

dissolve and form a colourless solution if aluminium and zinc ions are present; the

white ppt will remain if magnesium ions are present. This is a more valid test, but

two ions still cannot be distinguished. (1 mark)

For a valid experiment, the solutions should also be tested with ammonia solution,

followed by excess; the white ppt will dissolve if aluminium ions are present. (1 mark)

c) Ba2+(aq) + SO42–(aq) ⟶ BaSO4(s) (3 marks)

7 a) i) HCl = white precipitate (1 mark)

HBr = cream precipitate (1 mark)

HI = yellow precipitate (1 mark)

ii) Ag+ + Cl– ⟶ AgCl (2 marks)



b) i) Carry out a flame test. (1 mark)

Yellow/orange flame for sodium (1 mark)

ii) Add a few drops of sodium hydroxide solution. (1 mark)

White precipitate forms. (1 mark)

Add excess sodium hydroxide solution. (1 mark)

White precipitate remains. (1 mark)

Valid test: it confirms magnesium ions as no other ion gives same result with

sodium hydroxide solution. (1 mark)

Note that if ammonia is used, it is not a valid test as aluminium ions give the

same result.

8 a) Potassium ions (1 mark)

b) Zinc ions (1 mark)

c) SO42– (1 mark)

d) Zinc sulfate (1 mark); potassium sulfate (1 mark) (2 marks)

e) Use nichrome wire. (1 mark)

Dip in conc. hydrochloric acid and then into the salt; (1 mark)

place wire into blue Bunsen burner flame. (1 mark)

9 a) i) A negative ion (1 mark)

ii) A solid formed when two solutions are mixed (1 mark)

iii) Barium chloride (1 mark)

iv) Yellow (1 mark)

b)

c)

i) Green ppt

ii) Fe2+(aq) + 2OH–(aq) ⟶ Fe(OH)2(s) (4 marks)

iii) Brown ppt (1 mark)

iv) Ammonia solution (1 mark)

Metal ion Copper ion Aluminium ion Flame test result Blue-green/green-blue

flame (1 mark)

Result on adding a few drops of sodium hydroxide solution, followed by excess, to the metal ion solution

Blue precipitate (1 mark) White precipitate which dissolves to form colourless solution (1 mark)

Result on adding a few drops of ammonia solution, followed by excess, to the metal ion solution

Blue precipitate, which dissolves to form blue solution (1 mark)

White precipitate (1 mark)

Chapter 7 – Solubility Answers



1 a) Soluble

b) Insoluble

c) Soluble

d) Insoluble

e) Insoluble

f) Soluble

g) Insoluble

h) Soluble

i) Soluble

j) Insoluble

k) Soluble

2 a) Dissolves to give a colourless solution

b) Dissolves to give a blue solution

c) Dissolves to give a colourless solution

d) Does not dissolve, cloudy green

e) Dissolves to give a colourless solution

3 a) No precipitate

b) Barium sulfate

c) Silver chloride

d) Calcium carbonate

e) Aluminium hydroxide

4 a) NaCl(aq) + KNO3(aq) ⟶ KCl(aq) + NaNO3(aq)

b) BaCl2(aq) + Na2SO4(aq) ⟶ 2NaCl(aq) + BaSO4(s)

c) AgNO3(aq) + NaCl(aq) ⟶ AgCl(s) + NaNO3(aq)

d) Ca(OH)2(aq) + CO2(g) ⟶ CaCO3(s) + H2O(l)

e) 3KOH(aq) + AlCl3(aq) ⟶ 3KCl(aq) + Al(OH)3(s)


5 a) i) 50 g

ii) 12.5 g

b) i) 20 g

ii) 400 g

6 a) It is not saturated: 4.525 g saturates 25 g water so 1.525 g more solid is needed.

b) It is saturated: 54.3 g saturates 300 g of water, and 59 g is greater than this.

c) It is not saturated: 9.05 g saturates 50 g water so 2.05 g more solid is needed.




1 B

2 C


7 The mass of a solid that saturates 100 g of water at a particular temperature

8 a) NaNO3

b) 28 °C

c) The solubility of sodium chloride increases very gradually with temperature; it is almost

constant.

The solubility of KNO3 increases much more with temperature.

d) 80 °C

e) 50 °C

f) No. At 80 °C 52 g of KNO3 saturates 100 g of water. This solution contains 40 g in 100 g.


1

2

Formula of potassium nitrate

Colour of potassium nitrate solution

Colour of potassium nitrate crystals

KNO3 Colourless White



3 Balance

4 a)

b)

5 a) 70 g/100 g

b) 33 °C

c) 60 – 15 = 45 g/100 g

d) 100 g is needed to saturate at 60 °C, so it is not a saturated solution; 25 g more solute is

needed.

e) Original solution would have 60 g in 100 g of water. This is not saturated at 55 °C. At 20 °C,

solubility is 30 g in 100 g water, so would be 15g in 50 g of water. Solution originally had 30 g

in 50 g of water, so 30 – 15 = 5g deposited on cooling.

Mass of potassium nitrate dissolved in 25 g of water (g)

Temperature at which crystals first appear (°C)

Solubility (g/100 g water)

7.5 20 30

15.0 40 60

25.0 60 100

37.5 80 150

55.0 100 220




9 a) 40 g/100 g water

b) 24°C

c) 18 g in 50 g water is equivalent to 36 g in 100 g.

At 40 °C the solubility is 26 g/100 g water, so the solution is saturated (10 g extra solid).

d) 58 – 18 = 40 g

e) 49 – 22 = 27 g in 100 g water. So in 50 g water there is 13.5 g.

10 a) 48.5 – 37 =11.5 g

b) Solubility at 20 °C is 34.0 g/100 g water. This gives 17.0 g in 50 g of water. Solution originally

had 18 g, so

18.0 – 17.0 = 1.0 g

c) 10 – 31

4

= 2.25 g

d) At 40 °C 40 g saturates 100 g water. This solution contains 25 g/100 g of water, which is less

than 40 g, so the solution is not saturated.


1

(1 mark for each correct column)

2 a) i) The mass of solute (1 mark) which saturates (1 mark) 100 g of water (1 mark)

at a particular temperature (1 mark)

ii) (NH4)2Cr2O7 (1 mark)

b) i) 67 °C (1 mark)

ii) 24 °C (1 mark)

iii) Water is only liquid between 0 °C and 100 °C. (1 mark)

c) i) Solubility at 60 °C = 85 g/100 g water (1 mark)

In 40 g of water, mass = 34 g (1 mark)

Substance Soluble Insoluble

Copper(II) oxide

Silver nitrate

Sodium carbonate

Zinc hydroxide

Lead nitrate

Magnesium chloride

Barium sulphate



ii) At 20 °C solubility = 36 g/100 g water (1 mark)

In 40 g of water, mass = 14.4 g (1 mark)

Mass deposited = 30 – 14.4 = 15.6 g (1 mark)

3 a) 40 g/100 g water (1 mark)

b) Increases (1 mark)

c) 50 – 36.4 = 13.6 g (1 mark)

13.6

4 (1 mark) = 3.4 g (1 mark) (3 marks)

4 a) Lead(II) chloride or silver chloride (1 mark)

b) 162 × 20 (1 mark) = 3240 g (1 mark) (2 marks)

c) i) 144

8 = 18 g (1 mark)

20 g > 18 g (1 mark)

ii) 20 – 18 = 2 g (1 mark)

5 a) i) D (1 mark)

ii) 28 °C (1 mark)

iii) 72 °C (1 mark)

b) i) A = 49 g/100 g water

B = 40 g/100 g water

C = 5 g/100 g water

D = 36 g/100 g water (2 marks if all correct)

ii) A and C (1 mark)

iii) Solubility of B at 60 °C = 46 g/100 g water (1 mark)

46 – (6 × 2) = 34 g (1 mark)

Temperature at which B has a solubility of 34 g/100 g water = 24 °C (1 mark)

Chapter 8 – Metals and reactivity series Answers



1 Calcium reacts vigorously, magnesium reacts very slowly; both produce hydrogen and

a hydroxide.

2 a) 2K(s) + 2H2O(l) ⟶ 2KOH(aq) + H2(g)

b) Mg(s) + H2O(g) ⟶ MgO(s) + H2(g)

c) 2Al(s) + 3H2O(g) ⟶ Al2O3(s) + 3H2(g)

3 a) Heat released, bubbles, Ca sinks (and rises), cloudy solution.

b) calcium + water ⟶ calcium hydroxide + hydrogen

c) Ca + 2H2O ⟶ Ca(OH)2 + H2

d) The calcium atoms lose electrons.

e) The copper atoms do not lose electrons.

4 a) Sodium: not as vigorous, no flame, no crackle at end.

b) Na ⟶ Na+ + e–

K ⟶ K+ + e–

c) Loses electrons more readily (the outer electron is further from the nucleus).

5 Magnesium burns with a white light. It burns completely in air producing a white solid. Copper

glows red and burns with a blue-green flame producing a black layer, only reacts on surface, not

completely.


1 a) Magnesium

b) Greater tendency to form ions

2 Mg + H2SO4 ⟶ MgSO4 + H2

Mg ⟶ Mg2+ + 2e–


6 a) calcium + copper(II) sulfate ⟶ calcium sulfate + copper

Ca + CuSO4 ⟶ CaSO4 + Cu

b) The calcium is more reactive than copper and so it pushes the copper out of solution.

c) Ionic equation: Ca + Cu2+ ⟶ Ca2+ + Cu

d) Blue solution fades, heat.



7 a) Mg + Zn(NO3)2 ⟶ Zn + Mg(NO3)2

Mg + Zn2+ ⟶ Mg2+ + Zn

b) FeCl3 + Al ⟶ AlCl3 + Fe

Fe3+ + Al ⟶ Al3+ + Fe

c) No reaction

d) 3ZnSO4 + 2Al ⟶ Al2(SO4)3 + 3Zn

3Zn2+ + 2Al ⟶ 2Al3+ + 3Zn

e) No reaction

f) Cu(NO3)2 + Zn ⟶ Zn(NO3)2 + Cu

Zn + Cu2+ ⟶ Zn2+ + Cu

8 a) Aluminium is more reactive than chromium.

b) 2Al + Cr2O3 ⟶ Al2O3 + 2Cr

c) Al + Cr3+ ⟶ Al3+ + Cr

9


1 Mg, Zn, Ni, Cu

2 NiO + Mg ⟶ MgO + Ni

3 nickel + hydrochloric acid ⟶ nickel chloride + hydrogen

Ni + 2HCl ⟶ NiCl2 + H2

[zinc + cold water ⟶ no reaction]

[nickel + cold water ⟶ no reaction]

zinc + sulfuric acid ⟶ zinc sulfate + hydrogen

Zn + H2SO4 ⟶ ZnSO4 + H2

magnesium + zinc oxide ⟶ magnesium oxide + zinc

Mg + ZnO ⟶ MgO + Zn

Prescribed practical (page 143)

1 Same volume of copper(II) sulfate, same concentration of copper(II) sulfate, same mass of metal,

same surface area of metal

2 Magnesium – there was a big difference in the results.

3 Silver is less reactive than copper, and so no reaction occurred.

4 Magnesium, biggest temperature rise

Metal solution

Metal

Mg Al Zn Cu

Magnesium sulfate ✖ ✖ ✖

Aluminium chloride

✔ ✖ ✖

Zinc sulfate ✔ ✔ ✖

Copper(II) sulfate ✔ ✔ ✔

Iron (II) sulfate ✔ ✔ ✔ ✖



5 Two metals (silver and gold) gave the same result.

6 CuSO4 + Zn ⟶ ZnSO4 + Cu

7 Blue colour of solution fades, red-brown solid produced.

8 To reduce heat loss to the surroundings

9 Put a lid on the cup.


10 a) Low reactivity

b) A rock that contains a metal compound from which the metal can be extracted

11 a) Heat with carbon

b) Electrolysis

c) Electrolysis

d) Heat with carbon.

12 Tin is extracted from the ore tin oxide by heating with carbon.

a) tin(II) oxide + carbon ⟶ tin + carbon dioxide

2SnO + C ⟶ 2Sn + CO2

b) Reduction is loss of oxygen. SnO loses oxygen.


1 Gold


sodium

zinc

copper.

3 Carbon

4 Sodium


13 a) An ore containing a low percentage of metal compounds.

b) Because we have run out of high-grade ores of copper.

14 a) Plants are grown in soil rich in copper compounds; the plants absorb the copper compounds;

the plants are burned and acid added to the ash; a solution is produced.

b) i) Iron is more reactive than copper.

ii) Cu2+(aq) + Fe(s) ⟶ Cu(s) + Fe2+(aq)

iii) Cu2+(aq) + 2e– ⟶ Cu(s)




1 a) Cut off a small piece (1 mark), remove the oil (1 mark). (2 marks)

b) Accept any three of the following:

potassium disappears

moves on the surface

crackles at the end

heat released

bubbles. (3 marks)

c) 2K + 2H2O ⟶ 2KOH + H2 (3 marks)

d) K ⟶ K+ + e− (2 marks)

e) Ca + 2H2O ⟶ Ca(OH)2 + H2 (3 marks)

f) Any three of: bubbles, heat released, calcium disappears, cloudy solution (3 marks)

g)

(3 marks)

h) Potassium loses an electron more easily and is more reactive. (1 mark)

i) Zn(s) + H2O(g) ⟶ ZnO(s) + H2(g) (3 marks)

j)

(5 marks)



2 a) Red–brown (1 mark)

b) The copper is more reactive than silver and replaces the silver from solution, forming

silver metal and copper nitrate solution, which is blue. (1 mark)

c) Zinc, iron, copper, silver (1 mark)

d) Zinc (1 mark)

e) Cu(s) + 2AgNO3(aq) ⟶ Cu(NO3)2(aq) + 2Ag(s) (3 marks)

f) Cu + 2Ag+ ⟶ Cu2+ + 2Ag (3 marks)

g) Zinc is more reactive than copper. (2 marks)

3 a) Mg, Mn, Cr, Ni (1 mark)

b) Mg + Ni(NO3)2 ⟶ Mg(NO3)2 + Ni (2 marks)

c) They are soluble. (1 mark)

4 a) Q, R, P, S (1 mark)

b) i) Zn + CuSO4 ⟶ Cu + ZnSO4 (2 marks)

ii) Accept any two of the following:

blue colour fades

heat released

red-brown solid formed. (2 marks)

iii) Zinc is more reactive than copper. (1 mark)

c) i) Hydrogen (1 mark)

ii) Lighted splint (1 mark), pop (1 mark)

iii) Zn ⟶ Zn2+ + 2e– (2 marks)

d) Reduction with carbon. (1 mark)

5 a) i) More traffic/noise pollution/more solid waste/dust pollution (1 mark)

ii) CuS (1 mark)

b) i) Plants absorb copper compounds through their roots. (1 mark)

The plants are then burned. (1 mark)

The ash is reacted with acid to form a metal compound solution. (1 mark)


energy is released when plants are burned

not an eyesore

no dust or noise pollution. (2 marks)

iii) Any one of: it takes a long time for plants to grow; supply is not continuous. (1 mark)

c) i) Iron is more reactive than copper and displaces it. (1 mark)

ii) CuSO4 + Fe ⟶ Cu + FeSO4 (2 marks)

d) i) 2Cu + O2 ⟶ 2CuO (3 marks)

ii) Black layer (2 marks)

Chapter 9 – Redox, rusting and iron Answers



1 a) Oxidation is the gain of oxygen. Calcium has gained oxygen, and so is oxidised.

b) Oxidation is the loss of hydrogen. HI has lost hydrogen, and so is oxidised.

c) Reduction is the gain of hydrogen. Na has gained hydrogen, and so is reduced.

d) Reduction is the gain of hydrogen. N2 has gained hydrogen, and so is reduced.

e) Oxidation is the gain of oxygen. Carbon has gained oxygen, and so is oxidised.

2 a) Reduction is the loss of oxygen. The copper(II) oxide loses oxygen and forms copper, so the

copper(II) oxide is reduced.

Oxidation is the gain of oxygen. The Mg gains oxygen and forms MgO, so the magnesium is

oxidised.

This reaction is a redox reaction because both oxidation and reduction occur at the same

time.

b) Reduction is the loss of oxygen. The zinc oxide loses oxygen and forms zinc, so the zinc oxide

is reduced.

Oxidation is gain of oxygen. The hydrogen gains oxygen and forms water, so the hydrogen is

oxidised.

This reaction is a redox reaction because oxidation and reduction occur at the same time.

c) Oxidation is the gain of oxygen. The hydrogen gains oxygen and is oxidised.

Reduction is the gain of hydrogen. The oxygen gains hydrogen and is reduced.


3 a) In this reaction, the Fe atoms become Fe2+ ions in FeSO4 while the Cu2+ ions in CuSO4 become

Cu atoms.

The Fe atoms lose electrons to form Fe2+ ions: Fe ⟶ Fe2+ + 2e–

This loss of electrons is oxidation.

The Cu2+ ions in CuSO4 gain electrons to form Cu atoms: Cu2+ + 2e– ⟶ Cu

This gain of electrons is reduction.

Both reduction and oxidation occur, so this is a redox reaction.

b) 2Al + 3Zn2+ ⟶ 2Al3+ + 3Zn

The Al atoms lose electrons to form Al3+ ions: Al ⟶ Al3+ + 3e–

This loss of electrons is oxidation.

The Zn2+ ions in ZnSO4 gain electrons to form Zn atoms: Zn2+ + 2e– ⟶ Zn

This gain of electrons is reduction.




c) Zn + Cu2+ ⟶ Zn2+ + Cu

The Zn atoms lose electrons to form Zn2+ ions: Zn ⟶ Zn2+ + 2e–

This is oxidation.

The Cu2+ ions in CuO gain electrons to form Cu atoms: Cu2+ + 2e– ⟶ Cu

This is reduction.


4 a) Oxidation is gain of oxygen. Mg gains oxygen

b) Mg ⟶ Mg2+ + 2e–

c) Oxidation is loss of electrons. Mg loses electrons.

d) Reduction is loss of oxygen. CuO loses oxygen.

e) Cu2+ + 2e– ⟶ Cu

f) Cu2+ gains electrons. Gain of electrons is reduction.

5 a) Zn + Cu2+ ⟶ Zn2+ + Cu

Cu2+ + 2e– ⟶ Cu

Zn ⟶ Zn2+ + 2e–

b) Zn + 2Ag+ ⟶ Zn2+ + 2Ag

Ag+ + e– ⟶ Ag

Zn ⟶ Zn2+ + 2e–

c) 2Al + 3Cu2+ ⟶ 2Al3+ + 3Cu

Cu2+ + 2e– ⟶ Cu

Al ⟶ Al3+ + 3e–

6 a) zinc + iron(II) sulfate ⟶ zinc sulfate + iron

b) Zn + FeSO4 ⟶ ZnSO4 + Fe

c) Zn + Fe2+ ⟶ Zn2+ + Fe

d) Fe2+ + 2e– ⟶ Fe; Zn ⟶ Zn2+ + 2e–

e) Fe2+ + 2e– ⟶ Fe

f) Zn ⟶ Zn2+ + 2e–

g) Both reduction and oxidation take place.

7 a) magnesium + silver nitrate ⟶ magnesium nitrate + silver

b) Mg + 2AgNO3 ⟶ Mg(NO3)2 + 2Ag

c) Mg + 2Ag+ ⟶ Mg2+ + 2Ag

d) Ag+ + e– ⟶ Ag; Mg ⟶ Mg2+ + 2e–

e) Ag+ + e– ⟶ Ag

f) Mg ⟶ Mg2+ + 2e–

g) Both reduction and oxidation take place.




1 CH4

2 CuO is reduced as it has lost oxygen.

3 H2 is oxidised as it has gained oxygen.

4 Cu2+ + Mg ⟶ Cu + Mg2+

Mg ⟶ Mg2+ + 2e–

Magnesium is oxidised as it has lost electrons to form magnesium ions.


8 a) Hydrated iron(III) oxide

b) Oxygen/air and water

9 a) iron + water + oxygen ⟶ rust

b) Set up three stoppered boiling tubes with a nail and boiled water in each tube. One tube

should contain oxygen, one argon and one nitrogen. The nail will only rust in the tube

containing oxygen.

10 a) Magnesium is more reactive than steel/iron and so reacts instead.

b) Paint forms a protective layer/barrier that prevents water/oxygen coming into contact with

iron/steel.

c) Thin layer of chromium forms a protective layer/barrier that prevents water/oxygen coming

into contact with iron/steel.

d) Galvanising forms a protective layer/barrier that prevents water/oxygen coming into contact

with iron/steel, and would react instead of the iron/steel if the surface was broken as zinc is

more reactive than iron/steel.


1 The same volume of water, the same size and mass of nail, the same temperature, the same

mass and surface area of metal

2 Tube P – magnesium and zinc are more reactive than iron so they react instead of the iron; in

tube P the copper is less reactive than iron, so the iron corrodes.



Practice questions (page 157–158)

1 a) i) Oxidation is gain of oxygen. (1 mark)

Calcium gains oxygen, gain of oxygen is oxidation. (1 mark)

ii) 2Ca + O2 ⟶ 2CaO (4 marks)

iii) Ca ⟶ Ca2+ + 2e– (3 marks)

b) Oxidation is gain of oxygen. (1 mark)

SO2 gains oxygen and forms SO3. (1 mark)

c) Reduction is loss of oxygen. (1 mark)

The Fe2O3 loses oxygen and forms iron, so the Fe2O3 is reduced. (1 mark)

Oxidation is gain of oxygen. (1 mark)

The CO gains oxygen and forms CO2, so the CO is oxidised. (1 mark)

The reaction is a redox reaction because oxidation and reduction occur at the

same time. (1 mark)

2 a) Mg + Cu(NO3) ⟶ Mg(NO3)2 + Cu (2 marks)

b) Mg + Cu2+ ⟶ Mg2+ + Cu (2 marks)

c) Cu2+ + 2e– ⟶ Cu (3 marks)

Mg ⟶ Mg2+ + 2e– (3 marks)

d) The copper half equation is reduction. (1 mark)

It is a gain of electrons. (1 mark)

e) Both reduction and oxidation take place. (1 mark)

3 a) Zn(s) + CuCl2(aq) ⟶ Cu(s)+ ZnCl2(aq) (3 marks)

b) Zn + Cu2+ ⟶ Cu + Zn2+ (2 marks)

c) Oxidation is loss of electrons (1 mark); zinc loses electrons (1 mark).

Reduction is gain of electrons (1 mark); copper ions gain electrons (1 mark).

Redox, as both oxidation and reduction occur (1 mark). (5 marks)

4 a) Stops oxygen/water reaching the metal (1 mark)

b) Magnesium is more reactive than iron, so it reacts in place of the iron. (2 marks)

c) i) Oiling (1 mark)

ii) Plating (1 mark)

d) Zinc is more reactive than iron. (1 mark)

It corrodes instead/sacrificial protection. (1 mark)



5 a) i) Rust (1 mark)

ii) Oxidation is gain of oxygen. (1 mark)

Iron gains oxygen and is oxidized. (1 mark)

iii) Oxidation is loss of electrons (1 mark); iron loses electrons (1 mark).

Reduction is gain of electrons (1 mark); copper ions gain electrons (1 mark).

Redox, as both oxidation and reduction occur (1 mark). (5 marks)

iv) Fe + 2HCl ⟶ FeCl2 + H2 (3 marks)

v) Fe ⟶ Fe2+ + 2e– (3 marks)

Oxidation is loss of electrons (1 mark). (4 marks)

vi) Contains water of crystallisation (1 mark)

b) i) Haematite (1 mark)

ii) Aluminium is more reactive. (1 mark)

iii) Carbon reacts with oxygen to form carbon dioxide; (1 mark)

carbon reacts with carbon dioxide (1 mark)

to form carbon monoxide, which is the reducing agent. (1 mark)

iv) CaCO3 ⟶ CaO + CO2 (2 marks)

CaO + SiO2 ⟶ CaSiO3 (2 marks)

6 a) i) Iron (1 mark)

ii) Oxygen (1 mark)

iii) Water (1 mark)

iv) Oxidation is gain of oxygen; (1 mark)

iron gains oxygen and forms iron oxide. (1 mark)

v) Red-brown (1 mark); solid/flakes (1 mark). (2 marks)

b) i) Form a barrier and prevent air and water reaching the iron (1 mark)

ii) Galvanised gate is covered with a layer of zinc. (1 mark)

c) i) Iron(II) oxide (FeO) (1 mark); reduction is loss of oxygen (1 mark). (2 marks)

ii) Carbon (1 mark); oxidation is gaining oxygen (1 mark). (2 marks)

d) A more reactive metal is attached to iron; (1 mark)

it reacts instead of iron. (1 mark)

7 a) i) Sulfate (1 mark)

ii) Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq) (3 marks)

iii) Magnesium (1 mark)

iv) Mg ⟶ Mg2+ + 2e– (3 marks)

v) Accept any two of the following:

blue solution fades

heat released

red-brown solid forms. (2 marks)



b) i) Mg + 2HCl ⟶ MgCl2 + H2 (3 marks)

ii) Chloride/Cl– (1 mark)

iii) 2H+ + 2e– ⟶ H2 (3 marks)

8 a) Haematite (1 mark)

b) Carbon/coke (1 mark); calcium carbonate/limestone (1 mark) (2 marks)

c) To allow coke to burn (1 mark)

d) Carbon monoxide (1 mark)

C + O2 ⟶ CO2 (2 marks)

CO2 + C ⟶ 2CO (2 marks)

e) Fe2O3 + 3CO ⟶ 2Fe + 3CO2 (3 marks)

f) The calcium oxide (1 mark)

reacts with silicon dioxide impurities (1 mark)

to form calcium silicate. (1 mark)

g) Molten/liquid, so is run off (1 mark)

h) Carbon dioxide (1 mark)

Chapter 10 – Rates of reaction Answers


Test yourself (page 160) 1 0.067 s–1

2 0.017 s–1

3 a) 0.0313 s–1

b) Increasing the concentration increases the rate.

Show you can (page 160) 1 2 minutes

2 1120

= 0.008

3 Zn + 2HCl ⟶ ZnCl2 + H2

4 Bubbles, heat, Zn disappears, colourless solution produced.

Test yourself (page 162) 4, 5

Show you can (page 163) 1 The result at 4.5 minutes

2 Mass of flask and contents/g

3 Time/minutes

4 Time

5 100.3 g

Test yourself (page 164) 6 a) A, the slope is steepest at this point.

b) F, this is the first place where the slope is zero/the line starts to be horizontal.

c) B, the slope is steeper than at E.

d) Na2CO3 + 2HCl ⟶ 2NaCl + H2O + CO2

Time /s 10 20 30 40 50

Volume 1/cm3 30 49 59 63 63

Volume 2/cm3 32 51 59 63 65

Average volume/cm3 31 50 59 63 64



7 a) P

b) R

8 a)

b) 55 seconds

c) Red line on graph in a)

d) magnesium + sulfuric acid ⟶ magnesium sulfate + hydrogen

e) Mg + H2SO4 ⟶ MgSO4 + H2

Practical activity (page 166) 1 2H2O2 ⟶ 2H2O + O2

2 Allows gas to escape but prevents reaction mixture spraying out.

3 Unlabelled line on graph

4 0.41 g

5 Gas syringe to collect the oxygen, measure the volume

6 The graph rises to the same height but is less steep, e.g. graph B.



7

8 Dry the manganese(IV) oxide and find its mass at the end of the experiment. The mass should

remain the same.

9 Catalyst

Show you can (page 167) 1 Carbon dioxide gas escapes from flask.

2 To prevent loss of mass due to spray

3, 4

Prescribed practical activity (page 171) 1 Sulfur

2 0.03125

3 Accept any three of the following:

• same concentration of HCl

• same volume of HCl

• same volume of sodium thiosulfate solution

• same temperature

• same cross.



4 Use a thermostatically controlled water bath as it is difficult to control the temperature of the

room and increased temperature increases the rate of reaction.

5 The rate increases as the concentration increases.

6 There are more particles per unit volume, so there are more successful collisions in a given time.

Prescribed practical activity (page 172) 1 Gas syringe

2 So no gas escapes

3 Stopwatch/measuring cylinder/balance

4 Mg + 2HCl ⟶ MgCl2 + H2

5 Bubbles

6 Use the same volume of acid, the same temperature, the same mass of magnesium, the same

surface area of magnesium.

7 Temperature control/loss of gas on inserting bung

8 8.2 minutes, 12.2 minutes

9 At 6 minutes for 1.0 mol dm–3 concentration, it does not fit the trend.

10 0.5 mol dm–3

11 Increasing the concentration of acid increases the rate of reaction.

12 There are more particles present per unit volume/particles are closer together and so there is a

greater frequency of successful collisions/ more successful collisions per unit time.

Practice questions (pages 173–174) 1 a) D (1 mark)

b) Increasing concentration increases the rate. (1 mark)

A and B, or any combination of A, B and C (1 mark)

c) Increasing temperature increases rate. (1 mark)

B and D (1 mark)

d) 175

= 0.013 s–1 (1 mark)

2 a) calcium carbonate + hydrochloric acid ⟶ calcium chloride + water + carbon dioxide (1 mark)

b) CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2 (3 marks)

c) 0.2 s–1 (1 mark)

d) Increasing surface area increases rate. (1 mark)

3 a) A substance that speeds up the rate of a reaction (1 mark)

and is not used up during the reaction. (1 mark)

b) Manganese(IV) oxide (1 mark)

c) Iron (1 mark)



d) It provides a pathway of lower activation energy (1 mark)

so there are more successful collisions (1 mark)

in a given time. (1 mark)

4 a) A substance that speeds up the rate of a reaction (1 mark)

and is not used up during the reaction. (1 mark) (2 marks)

b) i) A (1 mark)

ii) D (1 mark)

iii) The curve stays above the original (1 mark)

but finishes at an earlier time (1 mark)

at half the volume of oxygen produced. (1 mark)

c) C (1 mark)

5 a) There are more particles in the same volume (1 mark)

and so there are more successful collisions (with the activation energy) (1 mark)

between particles in a given time/the successful collisions are more frequent. (1 mark)

b) Accept any two of the following:

• temperature

• presence of a catalyst

• surface area of magnesium. (2 marks)

6 a) Experiment 1: volume of rhubarb = 5 cm × 1 cm × 1 cm =5 cm3

Experiment 2: volume of rhubarb = 5 × (1 cm × 1 cm × 1 cm) = 5 cm3 (3 marks)

b) Experiment 1: surface area of rhubarb = 22 cm2 (1 mark)

Experiment 2: surface area of rhubarb = 30 cm3 (1 mark)

c) Experiment 1: ratio = 22 : 5 = 4.4 : 1 (1 mark)

Experiment 2: ratio = 30 : 5 = 6 : 1 (1 mark)

d) The greater the surface area the greater the rate of reaction. (1 mark)

e) The greater the surface area the more particles available for collisions (1 mark)

and so the more successful collisions (1 mark)

in a given time. (1 mark)

f) Temperature, mass of rhubarb, volume of solution, concentration of solution (3 marks)

7 a) 70 s (1 mark)

b) 30 cm3 (1 mark)

c) B, faster reaction but same volume of gas produced (2 marks)

d) D, slower reaction but same volume of gas produced (2 marks)

e) E, half the mass of magnesium used, so half the volume of gas produced (2 marks)

f) More particles in the same volume, (1 mark)

more successful collisions (1 mark)

in unit time (1 mark)

Chapter 11 – Equilibrium Answers



1 a) A reversible reaction is one in which the products, once made, can react to reform the

reactants.

b) i) Irreversible

ii) Reversible

iii) Irreversible

iv) Reversible

2 a) A + B ⟶ C + D

b) C + D ⟶ A + B

3 a) NH3+ HCl ⇌ NH4Cl

b) NH4Cl ⟶ NH3+ HCl

c) Endothermic


Answer is C.


4 a) +43 kJ

b) Ethene, steam and ethanol

c) Equal

d) They are constant.

e) No substances enter or leave.

5 a) The rates of the forward and reverse reactions are equal.

b) Closed system; the amounts of reactants and products are constant.


Answer is D.


6 a) Reaction 2

b) Reactions 1 and 2

c) Reaction 4

7 Goes darker brown; equilibrium moves to the right, in the endothermic direction, to oppose the

increase in temperature.



8 a) If a change is made to the conditions of a system at equilibrium the position of the

equilibrium moves to oppose the change in conditions.

b) It goes blue; the equilibrium moves to the right to remove the added Cl– ions.

c) Endothermic; when cooled the equilibrium moves to the left, in the exothermic direction, to

oppose the decrease in temperature.


1 a) Reversible arrow (1 mark)

b) +42 kJ (1 mark)

c) A substance that speeds up a chemical reaction (1 mark)

but is not used up in the reaction. (1 mark)

d) CO + H2O ⇌ CO2 + H2 (2 marks)

e) CO2(g) + H2(g) ⟶ CO(g) + H2O(g) (1 mark for equation with arrow, 1 mark for state symbols)

2 a) A reversible reaction is one in which the products, once made, can react

to reform the reactants/can occur in both directions. (1 mark)

b) Heat is given out. (1 mark)

c) Closed system; (1 mark)

the amounts of reactants and products are unchanged. (1 mark)

d) When the rates of the forward and reverse reactions are equal (1 mark)

e) Increasing the pressure causes the equilibrium position to move to the

right as this is the side where there are fewer molecules (1 mark)

to reduce the effect of the increase in pressure. (1 mark)

f) The equilibrium position moves left, the endothermic direction (1 mark)

to reduce the effect of the increase in temperature. (1 mark)

g) A catalyst speeds up the rate of a chemical reaction (1 mark)

but is not used up during the reaction. (1 mark)

h) No effect (1 mark)

3 a) 3 molecules ⇌ 1 molecule, so the equilibrium shifts right, (1 mark)

to the side with fewest gas molecules, to oppose the increase in pressure; (1 mark)

this gives more methanol. (1 mark)

b) The use of high pressure is very expensive due to the expense of the thick

pipes/valves (1 mark) and the energy required to compress the gases (1 mark). (2 marks)

c) The reaction rate is too slow at low temperatures. (1 mark)

d) The position of the equilibrium moves right (1 mark)

to reduce the concentration of hydrogen. (1 mark)



4 a) Any two of the following:

closed system

rates of forward and backward reactions are equal

amounts of reactant and product are constant. (2 marks)

b) If a change is made to the conditions of a system at equilibrium (1 mark)

the position of the equilibrium moves to oppose that change in conditions. (1 mark)

c) If the temperature is decreased, the equilibrium position moves right, (1 mark)

in the exothermic direction, to produce heat (1 mark)

and hence more NO2 is formed. (1 mark)

d) 1 molecule ⇌ 2 molecules. Decreasing the pressure causes the position of

equilibrium to move right, (1 mark)

to the side with more molecules, to increase the pressure. (1 mark)

5 a) 3 molecules ⇌ 2 molecules. The position of equilibrium moves right (1 mark)

to the side where there are fewer molecules, to reduce the pressure

and so the amount of NO2 increases. (1 mark)

b) Decreasing the temperature means the equilibrium position moves right (1 mark)

in the exothermic direction, to increase the temperature (1 mark)

and so more NO2, which is brown, is formed. (1 mark)

6 a) Goes orange; (1 mark)

equilibrium position shifts right (1 mark)

to remove added H+. (1 mark)

b) Goes yellow; (1 mark)

equilibrium position shifts left (1 mark)

to replace H+ ions removed by OH–. (1 mark)

7 a) Low pressure (1 mark)

gives more tetrafluoroethene as there are more gas molecules on the

right-hand side, so equilibrium moves right (1 mark)

at low pressure to increase the pressure; however, the reaction rate

would be slower. (1 mark)

b) High temperature (1 mark)

gives more tetrafluoroethene as the reaction is endothermic and so

the equilibrium position moves right, (1 mark)

in the endothermic direction, to lower the temperature; the reaction

rate is increased. (1 mark)



8 a) N2 + 3H2 ⇌ 2NH3 (3 marks)

b) Iron (1 mark)

c) Haber process (1 mark)

d) Increase pressure, increase yield (1 mark)

e) Increase temperature, decrease yield (1 mark)

f) 200 atm (1 mark)

450 °C (1 mark)

32% (1 mark)

Chapter 12 – Organic chemistry Answers



1 a) 2

b) 3

c) 2

d) 3

e) 2

f) 4

g) 1

h) 4

i) 3

2 a) Propane

b) Methanol

c) Ethanoic acid

d) Propene

3 a) C3H8, propane

b) C8H18

c) C4H10, butane

d) C10H22

4 A homologous series is a family of organic molecules that have the same general formula, show

similar chemical properties, show a gradation in their physical properties and differ by a ‘CH2’

unit.


1 A = methanoic acid, B = butanol


5 CnH2n+2

6 A molecule that contains carbon and hydrogen only

7 a) Butane, C4H10

b) Methane, CH4

8 Seven

9 a) 4

b) 1



10 a) That all the carbon–carbon bonds are single bonds

b) C6H14

c)


1 C5H12

2 16


11 a) Carbon dioxide and water

b) Carbon monoxide, water and soot

c) Carbon dioxide and water

d) Carbon monoxide, water and soot

12 a) C2H6 + 3½O2 ⟶ 2CO2 + 3H2O

2C2H6 +7O2 ⟶ 4CO2 + 6H2O

b) C3H8 + 5O2 ⟶ 3CO2 + 4H2O

c) C4H10 + 4½O2 ⟶ 4CO + 5H2O

2C4H10 +9O2 ⟶ 8CO + 10H2O

d) CH4 + 1½O2 ⟶ CO + 2H2O

2CH4 + 3O2 ⟶ 2CO + 4H2O

e) C4H10 + 6½O2 ⟶ 4CO2 + 5H2O

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

13 a) C8H18 + 12½O2 ⟶ 8CO2 + 9H2O

2C8H18 + 25O2 ⟶ 16CO2 + 18H2O

b) C10H22 + 15½O2 ⟶ 10CO2 + 11H2O

2C10H22 + 31O2 ⟶ 20CO2 + 22H2O

c) C6H14 + 9½O2 ⟶ 6CO2 + 7H2O

2C6H14 + 19O2 ⟶ 12CO2 + 14H2O

14 Bubble the gas into colourless limewater, which turns milky in the presence of carbon dioxide.

15 White anhydrous copper(II) sulfate changes to blue in the presence of water.




1 SO2

2 It reacts with water to give sulfurous acid.

3 Remove sulfur from coal, petrol and diesel/use alternative fuels.

4 Destroys vegetation/kills fish/destroys buildings


a) The hot gas is cooled in the U tube and condenses to form a liquid.

b) Water

c) Calcium hydroxide/limewater. The carbon dioxide produced reacts with the limewater.

d) Water and carbon dioxide

e) Carbon, incomplete combustion of fuel

f) i) Sulfur dioxide

ii) Reacts with liquid A (calcium hydroxide)

iii) The sulfur dioxide produced causes acid rain.


16 a) C3H6, propene

b) C8H16

c) C2H4, ethene

d) C10H20

17 a) But-2-ene, C4H8

b) Propene, C3H6

18 a) 9 (the double bond = 2 covalent bonds)

b) It is a molecule containing carbon and hydrogen only.


1 C5H10

2 Either of:




19 a) Carbon dioxide and water

b) Carbon monoxide and water (soot)

20 a) C2H4 + 3O2 ⟶ 2CO2 + 2H2O

b) C3H6 + 4½O2 ⟶ 3CO2 + 3H2O

2C3H6 + 9O2 ⟶ 6CO2 + 6H2O

c) C4H8 + 4O2 ⟶ 4CO + 4H2O

d) C2H4 + 2O2 ⟶ 2CO + 2H2O

21 a) C8H16 + 12O2 ⟶ 8CO2 + 8H2O

b) C10H20 + 15O2 ⟶ 10CO2 + 10H2O

c) C6H12 + 9O2 ⟶ 6CO2 + 6H2O

22 a) Contains some carbon–carbon double bonds

b) A molecule containing carbon and hydrogen only.

c) They have a functional group (C=C).

d) Shake with bromine water.

If the bromine water changes from orange to colourless it is chloroethene.

If the bromine water stays orange it is chloroethane.

23 a) C6H12

b) Orange ⟶ colourless

c) Hexane

24 a) A reaction where two reactant molecules form a single product

b) i) C2H4 + H2 ⟶ C2H6

ii) C2H4 + H2O⟶ C2H5OH

iii) C2H4 + Br2 ⟶ C2H4Br2



c) Ethane

d) Ethanol

e) All three parts

f) b(ii)

g) Nickel


1 + hydrogen ⟶ C2H6

+ bromine ⟶ C2H4Br2

+ steam ⟶ C2H5OH

+ plentiful oxygen ⟶ CO2 + H2O

2 Orange solution ⟶ colourless

3 C

4 A

5

6


25 a) A polymer is a long-chain molecule made from joining small molecules together.

b) A monomer is a small molecule that combines with other monomers to make a polymer.

c) C=C

d)

e)

f) Poly(styrene)

g) Bromoethene



26

27

28


1

2

3 No, because it contains chlorine, not just carbon and hydrogen.


29 a) A reactive group in a molecule.

b) hydroxyl (–OH)

c) No, as the molecule contains oxygen and not just carbon and hydrogen.

Structure of monomer Repeating unit of polymer

Structure of polymer



d) Alcohols

e) The same general formula; show similar chemical properties; show a gradation in their

physical properties; and differ by a CH2 unit.

30 a) Ethanol

b) Propan-2-ol

c) Methanol

31 a) Colourless limewater changes to milky.

b) Carbon dioxide and water

c) C2H5OH + 3O2 ⟶ 2CO2 + 3H2O

d) C4H9OH + 6O2 ⟶ 4CO2 + 5H2O

e) Carbon, carbon monoxide and water


32 a) Breakdown of sugar by yeast to form carbon dioxide and ethanol

b) The sugars are dissolved in solution; in the presence of yeast; at a warm temperature (not

above 37 °C); in the absence of air.

c) C2H4 + H2O(g) ⟶ C2H5OH

33 a) False

b) True

c) False

d) False

e) True

f) False

g) False


Ethene (+ steam)

(Sugar +) yeast (⟶ ethanol +) carbon dioxide

Carbon dioxide + water


34 a) CH3COONa

b) (CH3COO)2Ca

c) (HCOO)2Mg

d) (HCOO)3Al

e) (C2H5COO)2Cu

f) C2H5COONa



35 a) Calcium + ethanoic acid ⟶ calcium ethanoate + hydrogen

b) Magnesium carbonate + methanoic acid ⟶ magnesium methanoate + carbon dioxide +

water

c) Sodium hydroxide + propanoic acid ⟶ sodium propanoate + water

d) Magnesium + methanoic acid ⟶ magnesium methanoate + hydrogen

e) Copper(II) carbonate + propanoic acid ⟶ copper(II) propanoate + carbon dioxide + water

f) Sodium carbonate + propanoic acid ⟶ sodium propanoate + water + carbon dioxide

36 a) Ca + 2CH3COOH ⟶ (CH3COO)2Ca + H2

b) MgCO3 + 2CH3COOH ⟶ (CH3COO)2Mg + H2O + CO2

c) NaOH + C2H5COOH ⟶ C2H5COONa + H2O

d) Mg + 2HCOOH ⟶ (HCOO)2Mg + H2

e) CuCO3 + 2C2H5COOH ⟶ (C2H5COO)2Cu + H2O + CO2

f) Na2CO3 + 2C2H5COOH ⟶ 2C2H5COONa + CO2 + H2O

37 a) Grey solid disappears; colourless solution produced; bubbles; heat released.

b) White solid disappears; colourless solution produced; bubbles; heat released.

c) Heat released; colourless solution remains.

d) Grey solid disappears; colourless solution produced; bubbles; heat released.

e) Green solid disappears; blue solution produced; bubbles; heat released.

f) White solid disappears; colourless solution produced; bubbles; heat released.

38 Lighted splint pop


39 a) Copper sulfate + magnesium ⟶ copper + magnesium sulfate

b) The observations should be that the grey magnesium turned red-brown, the blue colour of

the solution faded, and heat was released. The student simply recorded the names of the

products, not observations.

40 a) CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

b) Calcium carbonate disappears; bubbles; solution forms; heat released.

41

Reaction Observations

Ethanoic acid + sodium carbonate Bubbles, carbonate disappears

Potassium iodide solution and silver nitrate solution

Yellow precipitate

Bromine water + alkene Orange solution to colourless solution

Hydrochloric acid + magnesium Bubbles, magnesium disappears, heat released

Acidified barium chloride solution and sulfuric acid

White precipitate




1 a) Heat released; white solid sodium carbonate disappears; colourless solution produced.

b) Bubble it into colourless limewater; in a positive test the limewater turns milky.

2 a) Magnesium + ethanoic acid ⟶ magnesium ethanoate + hydrogen

b) Mg +2CH3COOH ⟶ (CH3COO)2Mg + H2

c) Faster/more vigorous; more heat given out

3 a) A substance that speeds up a reaction and is not used up.

b)

c) Concentrated sulfuric acid is corrosive, so wear gloves and eye protection.

d) It is not soluble in water; it is less dense than water.


42 Alcohol

43 Alkene

44 Warm with acidified potassium dichromate. The colour will change from orange to green in an

alcohol; there will be no change in an alkane.

45 Add a spatula of sodium carbonate. If it is a carboxylic acid bubbles will form; if it is an alkane

there is no change.

46 –COOH




1 Incomplete

2 C=C

3 COOH

4

5 COOH


47 a) A molecule containing hydrogen and carbon only

b) Different boiling points

c) Oil is vaporised; put into tower that is hot at the bottom and cool at the top; alkanes rise, cool

and condense at different heights; smaller molecules are separated nearer the top.

48 a) Less flammable, do not burn as well

b) Cracking is the breakdown of larger saturated hydrocarbons (alkanes) into smaller more

useful ones, some of which are unsaturated (alkenes).

c) Breaking down a substance using heat

49 a) C18H38 ⟶ C12H26 + 2C3H6

b) C20H42 ⟶ C12H26 + 4C2H4

c) C18H38 ⟶ C10H22 + C4H8 + 2C2H4

d) C25H52 ⟶ C11H24 + 2C4H8 + 2C3H6


C10H22 ⟶ C6H14 + 2C2H4


1 a) CnH2n+2 (1 mark)

b) Differ by a CH2 unit (1 mark)

Gradation in physical properties down the group (1 mark)

Similar chemical reactions (1 mark)

c)

(1 mark)

Experiment Observation

1 Place 2 cm3 of ethanoic acid in a test tube and add 2 cm3 of water. Shake well.

Mixes completely/dissolves

2 Place 2 cm3 of ethanoic acid in a test tube and add 1 spatula of sodium carbonate.

Bubbles



d) They are molecules which contain carbon and hydrogen only. (1 mark)

e)

(1 mark)

f) C3H8 + 5O2 ⟶ 3CO2 + 4H2O (3 marks)

2 a) Gas (1 mark)

b) C2H4 (1 mark)

c) i)

(1 mark)

ii)

(1 mark)

iii)

(1 mark)

iv) Water (1 mark)

v) It is colourless (1 mark)

Bubble through limewater (1 mark)

Limewater turns milky (1 mark)

vi) Addition (1 mark)

3 a) It contains chlorine, not carbon and hydrogen only. (1 mark)

b) A polymer is a long-chain molecule made by joining small molecules together. (1 mark)

c) Vinyl chloride/chloroethene (1 mark)

d)

chloroethene poly(chloroethene) (monomer) (polymer) (4 marks)



4 a) Addition (1 mark)

Polymerisation (1 mark)

b) Poly(ethene)/polythene (1 mark)

c) A large number (1 mark)

d)

Indicative content:

Incineration, burning

Advantage – heat energy is produced during incineration, which can be used to

generate electricity.

Disadvantage – polluting gases are released; for example CO2, which can cause global

warming, and also toxic gases such as dioxins.

Disadvantage – costs to buy incinerator

Dumping in landfill

Advantage – cheap

Disadvantage – eyesore

Disadvantage – wastes land

5 a)

(4 marks)

b) i) Limited oxygen (1 mark)

ii) Toxic/prevents blood carrying oxygen (1 mark)

iii) Colourless/odourless (1 mark)

iv) Carbon (1 mark)

v) Lung damage to humans (1 mark)

Response Mark

Candidates must use appropriate specialist terms to fully state each method and the advantages and disadvantages of each [6–8 indicative content points]. They must use good spelling, punctuation and grammar, and the form and style must be of a high standard.

5–6 (Band A)

Candidates must use appropriate specialist terms to state each method, and some advantages and disadvantages of each [4–5 indicative content points]. They must use good spelling, punctuation and grammar, and the form and style must be of a high standard.

3–4 (Band B)

Candidates describe briefly a method and some advantages and disadvantages of it [at least 3 indicative content points]. They use limited spelling, punctuation and grammar, and they have limited use of specialist terms. The form and style are of limited standard.

1–2 (Band C)

Response not worthy of credit 0

At the beginning of experiment

5 minutes after the start of the experiment

Limewater Colourless (1 mark) Milky (1 mark)

Anhydrous copper(II) sulfate

White (1 mark) Blue (1 mark)



6 a)

(1 mark)

b) i) Polyvinylchloride /poly(chloroethene) (1 mark)

ii) Vinyl chloride/chloroethene (1 mark)

(1 mark)

iii)

(1 mark)

7 a) C8H18 + 12½O2 ⟶ 8CO2 + 9H2O

OR

2C8H18 + 25O2 ⟶ 16CO2 + 18H2O (3 marks)

b) Carbon (1 mark)

Carbon monoxide (1 mark)

Water (1 mark)

c) An impurity in petrol is sulfur (1 mark)

when petrol burns the sulfur burns and forms sulfur dioxide. (1 mark)

d) Kills fish (1 mark)

Destroys buildings (1 mark)

Destroys vegetation (1 mark)

8 a) i) A mixture of hydrocarbons (1 mark)

ii) Fractional distillation (1 mark)

iii) Accept any two of the following:

petrol

diesel

kerosene

bitumen. (2 marks)

b) i) Contains only single carbon–carbon bonds (1 mark)

Contains carbon and hydrogen only (1 mark)

ii) CnH2n+2 (1 mark)



c) i) Cracking (1 mark)

ii) C8H18 ⟶ C3H6 + C5H12 (2 marks)

iii) Bromine water (1 mark)

changes from orange (1 mark)

to colourless. (1 mark)

iv) C2H4 + Br2 ⟶ CH2BrCH2Br (or could use structural formulae) (2 marks)

9 a) i) Contains carbon and hydrogen only (1 mark)

ii) C=C (1 mark)

iii) Unsaturated (1 mark)

iv) Orange (1 mark)

to colourless (1 mark)

b) i) —OH and C=C (2 marks)

ii) C10H17OH + 14O2 ⟶ 10CO2 + 9H2O (2 marks)

c) i)

(1 mark)

ii) C2H4 + H2O ⟶ C2H5OH (2 marks)

Steam/hot (1 mark)

iii) Addition (1 mark)

iv) Orange to green (2 marks)

d) i)

(1 mark)


bubbles

white solid disappears

heat released

colourless solution produced. (2 marks)

iii) Na2CO3 + 2CH3COOH ⟶ 2CH3COONa + CO2 + H2O (3 marks)

iv) One that is partially ionised in water (1 mark)



10

(21 marks)

Name Molecular formula

Structural formula State at room temperature and pressure

Ethene C2H4

Gas

Propene C3H6

Gas

Butane C4H10

Gas

But-1-ene / But-2-ene

C4H8

OR

Gas

Propan-2-ol C3H7OH

Liquid

Methanoic acid

HCOOH

Liquid

Butanoic acid C3H7COOH

Liquid




1 a) 45 g/dm3

b) 72 g/dm3

c) 70 g/dm3

2 360 g

3 1.5 g


200 cm3

Test yourself (page 225–226)

4 a) 4 mol/dm3

b) 0.8 mol/dm3

c) 1.5 mol/dm3

d) 0.8 mol/dm3

e) 0.5mol/dm3

f) 3.0 mol/dm3

g) 1.44 mol/dm3

h) 2.0 mol/dm3

5 a) 1.5 mol/dm3

b) 0.5 mol/dm3

c) 0.1 mol/dm3


1 1.06 g

2 42.4 g


6 21.9 g/dm3

7 1.48 g/dm3

8 122.5 g/dm3

9 0.1 mol/dm3

10 0.05 mol/dm3

11 1.2 mol

12 3.0 mol

13 0.15 mol



14 0.0125 mol

15 0.25 mol

16 2.5 mol

17 1.125 mol


0.4 mol/dm3; 25.2 g/dm3


18 Rinse the pipette with the sodium hydroxide solution.

Using a pipette filler, draw up the sodium hydroxide until the bottom of meniscus is on the line.

Transfer the pipette to a conical flask and release the liquid.

Touch the pipette on the surface of the sodium hydroxide to remove the last drops in

the pipette.

19 Rinse the burette with the sulfuric acid solution, and discard the rinsings.

Fill the burette with the sulfuric acid solution.

Making sure that the jet is filled and there are no air bubbles.

Record the volume at the bottom of the meniscus at eye level.

20 Pink to colourless

21 One of known concentration

22 Add the acid dropwise near the end point.

23 Add the two accurate titres and divide by 2.

24




1 A: Use a burette to measure the volume of acid.

B: Use a pipette to measure the volume of potassium hydroxide.

D: Add the acid drop by drop at the end point.

E: Swirl the flask during titration.

2 C: It is only used to give a rough idea of the volume needed.

Test yourself (age 232)

25 Moles NaOH = 25.0 ×0.100

1000= 0.0025 mol

Moles HCl = 0.0025 mol

Concentration = 0.0025 ×1000

30.0= 00833 mol/dm3

26 Moles H2SO4 = 27.5 ×0.500

1000= 0.01375 mol

Moles KOH = 0.01375 × 2 = 0.0275 mol


25.0= 1.1 mol/dm3

27 Moles HCl = 24.0 ×0.400

1000= 0.0096 mol

Moles Na2CO3 =0.00964

2= 0.0048 mol


20.0= 0.24 mol/dm3

Concentration = 0.24 × 106 = 25.44 g/dm3

28 Moles NaOH = 33.5 ×0.050

1000= 0.001675 mol

Moles citric acid =0.001675

3 = 0.000558 mol


25.0= 0.0223 mol/dm3

29 Moles = 32.0 ×0.1

1000= 0.0032 mol


25= 0.128 mol/dm3



30 a) Moles = 17.5 ×0.1

1000= 0.00175 mol

b) 0.00175

2= 0.000875

c) 0.000875 ×1000

25.0= 0.035 mol/dm3

d) Phenolphthalein, colourless to pink; or methyl orange, red to yellow



31 a) Moles = 18.0 ×0.250

1000= 0.0045 mol

b) 0.009 mol

c) 0.009 ×1000

25.0= 0.36 mol/dm3

d) Pipette

32 a) Moles = 22.8 ×0.0500

1000= 0.00114 mol

b) 0.001114

2= 0.00057

c) 0.0057 ×1000

25.0= 0.0228 mol/dm3


33 Concentration of NaOH =6

40= 0.15 mol/dm3

Moles NaOH = 25.0 ×0.15

1000= 0.00375 mol

Moles H2SO4 = 0.001875 mol

0.001875 ×1000

30= 0.0625 mol/dm3


34 Concentration of H2SO4 =4.9

98= 0.05 mol/dm3

Moles H2SO4 = 20.0 ×0.05

1000= 0.001 mol

Moles NH3 = 2 × 0.001 = 0.002 mol

Concentration of NH3 = 0.002 ×1000

25= 0.08 mol/dm3

Concentration of NH3 = 0.08 × 17 = 1.36 g/dm3

35 Concentration H2SO4 =2.45

98= 0.025 mol/dm3

Moles H2SO4 = 0.025 ×20.0

1000= 0.0005 mol

Moles KOH 0.0005 × 2 = 0.001 mol

Concentration of KOH = 0.001 ×1000

10.0= 0.1 mol/dm3

Concentration of KOH = 0.1 × 56 = 5.6 g/dm3


Moles H2SO4 = 0.05 ×26.2

1000= 2000 = 0.00131

Moles MOH = 2 × 0.00131 = 0.00262 mol

Concentration of MOH = 0.00262 ×1000

25.0= 0.1048 mol/dm3

3

3

10.7 g/dm

0.1048 mol/dm=Mr

Mr = 102 Ar(M) = 102 – 17 = 85 M is Rb




1 Average titre 326.2 26.426.3 cm

2

2 3

3

r

g/dm 80.2 mol/dm

40M

Moles = 26.3 ×0.2

1000= 0.00526

3 0.00526

2= 0.00263

4 0.00263 ×1000

25.0=0.1052 mol/dm3


36 a) Moles H2SO4 = 12.5 ×0.05

1000= 0.000625 mol

b) Moles of ammonia = 2 × 0.000625 = 0.00125 mol

c) Concentration of ammonia = 0.00125 ×1000

25= 0.0500 mol/dm3

d) Dilution factor = ×100

Concentration of original ammonia = 0.0500 × 100 = 5.00 mol/dm3

Concentration of original ammonia = 5.0 × 17 = 85 g/dm3

37 Moles KOH = 0.1 ×25

1000= 0.0025 mol

Moles H2SO4 =0.0025

2= 0.00125 mol

Concentration of diluted acid = 0.00125 ×1000

10.0= 0.125 mol/dm3

Dilution factor = ×10

Concentration of original acid = 0.125 × 10 = 1.25 mol/dm3

38 a) Moles NaOH = 30.3 ×0.1

1000= 0.00303 mol

b) 1 : 1 ratio, so 0.00303 mol ethanoic acid in 25.0 cm3 diluted wood vinegar

c) Dilution factor = ×50

50 × 0.00303 = 0.1515 mol ethanoic acid in 25.0 cm3 undiluted wood vinegar

d) 0.1515 × 60 g = 9.09 g ethanoic acid in 25.0 cm3 undiluted wood vinegar


39 a) Moles HCl =20.00×0.125

1000= 0.0025 mol

b) For 2 HCl : X2CO3

Moles of X2CO3 =0.0025

2= 0.00125 mol

c) 0.00125 mol X2CO3 in 25 cm3

Concentration of X2CO3 = 0.00125 ×1000

25= 0.05 mol/dm3



d) Moles = mass (g)

rM

Mr of X2CO3 =3.7

0.05= 74

e) Mr of CO3 = 12 + (3 × 16) = 60

Mr of X2 = 74 – 60 = 14

Ar of X =14

2= 7

X is lithium (from the Periodic Table).

40 Moles Na2CO3 = 23.0 ×0.050

1000= 0.00115 mol

Moles HX 2 × 0.00115 = 0.0023 mol

0.0023 mol in 25 cm3 so concentration = 0.0023 ×1000

25.0= 0.092 mol/dm3

3

r 3

g/dm 1.725 237.5

0.092mol/dmM

Ar(X) = 37.5 − 1 = 26.5

X is chlorine (from the Periodic Table).


Moles of KOH = 25.0 ×0.5

1000= 0.0125 mol

Ratio 1 KOH : 1 HX

0.0125 mol = 1.1 g/Mr

Mr =0.

1.1

0125= 88


1 3.92 ×1000

250= 15.68 g/dm3

2 Rinse a pipette with XOH. Using a pipette filler, fill the pipette with XOH solution, until the

bottom of the meniscus is on the mark at eye level. Drain the contents of the pipette into a

conical flask.

3 To make the colour change easier to detect

4 Blue to yellow

5 Add the acid drop-wise, swirling the contents of the flask until the colour changes.

6 To mix the solution and ensure the solutions have reacted fully

7 14.0, 14.0

8 14.0; do not use reading 1

9 0.500 ×14

1000= 0.007 mol HCl

10 0.007 mol XOH (1 : 1 ratio)



11 Concentration of XOH = 0.007 ×1000

25= 0.28 mol/dm3

12 15.68

0.28= 56 for XOH; X is 39, which is K (potassium)


41 a) Moles HCl = 18.0 ×0.175

1000= 0.00315 mol

b) Moles Na2CO3 =0.00315

2= 0.001575 mol

c) 0.01575 mol

d) 0.0

2.52

1575= 160

e) Mr(Na2CO3) = 106

160 – 106 = 56

Mr(H2O) = 18

x =56

18= 3

42 a) Moles HCl = 30.0 ×0.2

1000= 0.006 mol

b) Moles Na2CO3 =0.006

2= 0.003 mol

c) 0.03 mol

d) Mr(Na2CO3) = 106

0.03 × 106 = 3.18 g

e) 6.42 – 3.18 =3.24 g

f) Mr(H2O) = 18

3.24

18= 0.18 mol

g) 0.03 : 0.18

1 : 6

x = 6


43 a) 72 dm3

b) 9.6 dm3

c) 6 dm3

d) 240 dm3

44 a) 0.75 mol

b) 0.005 mol



45 a) Moles =72

24= 3

Mass = 3 × 32 = 96 g

b) Moles =6

24= 0.25

Mass = 0.25 × 16 = 4 g


O2 H2 NH3

Mass of 1 mol of gas/g 32 2 17

Volume of 1 mol of gas at room temperature and pressure/dm3

24 24 24

Number of moles in 12 cm3 at room temperature and pressure

= 0.0005

= 0.0005

= 0.0005

Number of moles in 12 g = 0.375

= 6

= 0.7


46 4 dm3

47 6 dm3

48 3.5 dm3

49 750 cm3


2NH3 + 2½O2 ⟶ 2NO + 3H2O

or

4NH3 + 5O2 ⟶ 4NO + 6H2O


50 Moles HCl = 25.0 ×2.00

1000= 0.05 mol

Moles H2 =0.05

2= 0.025 mol

Volume H2 = 0.025 × 24 = 0.6 dm3

51 a) Moles chloroethane =1.94

64.5= 0.03 mol

Ratio = 1 : 1

Moles ammonia = 0.03 mol

Mass ammonia = 0.03 × 17 = 0.51 g

b) 1.94 g = 0.03 mol

Volume = 0.03 × 24 = 0.72 dm3

12

24000

12

24000

12

24000

12

32

12

17

12

2



52 Moles Mg =8

24= 0.33 mol

Moles hydrogen = 0.33 mol

Volume hydrogen = 0.33 × 24 000 = 8000 cm3


1 Mr(ZnCO3) = 125

12.5

125= 0.1 mol

2 Ratio = 1: 1

0.1 mol

3 0.1 × 24 dm3 = 2.4 dm3


53 Mr(CaO) = 56, Mr(CaCO3) = 100

Atom economy = 100 ×56

100= 56.0%

54 Mr(Al2O3) = 102, Ar(Al) = 27

Ratio = 2 : 4

Atom economy = 100 × (4 × 27)/(2 × 102) = 52.9%

55 Mr(CH4 + 2Cl2) = 158, Mr(CH2Cl2) = 85


158= 53.8%

56 a) 400 g

b) 200 g

c) Less waste (to dispose of)/less resource used.


1 Method 1:

Mr(TiO2 + 2Mg) = 128, Ar(Ti) = 48


128= 37.5%

Method 2:

Mr(TiO2) = 80, Ar(Ti) = 48


80= 60%

2 The second reaction has a higher atom economy so there is less waste. However, it does use a

substantial amount of electricity. Other useful information is the toxicity or environmental

hazard of any products. You also might want to know how the magnesium is made; this could be

by electrolysis too.

3 100%




1 a) Pipette (1 mark)

b) Accept any four of the following:

rinse with NaOH

fill the burette with the NaOH solution

make sure that the jet is filled

no air bubbles

record the volume at the bottom of the meniscus at eye level. (4 marks)

c) Colourless (1 mark)

To pink (1 mark)

2 a) 1.25 mol/dm3 (1 mark)

b) 0.40 × 56 (1 mark)

= 22.4 g/dm3 (1 mark)

3 Moles of Mg =3

24= 0.125 mol (1 mark)

Ratio = 2: 1, so 2 × 0.125 = 0.25 moles of HCl (1 mark)

0.25 ×1000

2.0= 125 cm3 (1 mark)

4 Moles of KOH = 28.9 ×0.200

1000= 0.000578 mol (1 mark)

Ratio = 1 : 1, so moles HNO3 = 0.000578 mol (1 mark)

Concentration of HNO3 = 0.000578 ×1000

25.0= 0.231 mol/dm3 (1 mark)

5 Moles of NaOH = 16.4 ×0.400

1000= 0.00656 mol (1 mark)

Ratio = 1 : 2, so moles of H2SO4 =0.00656

2= 0.00328 mol (1 mark)

Concentration of H2SO4 = 0.00328 ×1000

20.0= 0.164 mol/dm3 (1 mark)

Concentration of H2SO4 = 0.164 × 98 = 16.072 g/dm3 (1 mark)

6 a) Moles HNO3 = 20.0 ×2.4

1000= 0.048 mol (1 mark)

b) Ratio = 4 : 1, so moles NO =0.048

4= 0.012 mol (1 mark)

c) Volume of NO = 0.012 × 24 000 = 288 cm3 (1 mark)

7 a) i) Moles Ca =0.40

40= 0.010 mol (1 mark)

Ratio = 1 : 1, so moles H2 = 0.010 mol (1 mark)

Volume H2 = 0.010 × 24 = 0.24 dm3 (1 mark)

ii) Moles HNO3 = 2 × 0.010 = 0.020 mol (1 mark)

Volume HNO3 = 1000 ×0.020

2.00= 10 cm3 (1 mark)



b) i) Accept any four of the following:

rinse the pipette with the barium hydroxide solution

use a pipette filler

draw up the solution into the pipette until the bottom of meniscus is on the line

transfer to a conical flask and release the liquid

touch the pipette on the surface of the liquid to remove the last drops in the

pipette. (4 marks)

ii)

Indicative content:

Add a few drops of methyl orange/phenolphthalein.

Add the HCl from the burette, with swirling.

Indicator changes from yellow to red/pink to colourless.

Record the volume, reading to the bottom of the meniscus at eye level.

Repeat the experiment and add the solution dropwise near the end point. Several

accurate titrations are carried out and the average of the accurate titrations calculated.

iii) Results 2 and 3 (1 mark)

iv) 22.4 cm3 (1 mark)

v) Moles HCl = 22.4 ×0.2

1000= 0.00448 mol (1 mark)

vi) Ratio = 1 : 2 so moles barium hydroxide =0.00448

2= 0.00224 mol (1 mark)

vii) 0.00224 ×1000

25.0= 0.0896 mol/dm3 (1 mark)

viii) 0.0896 × 171 = 15.3 g/dm3 (1 mark)

Response Mark

Candidates use appropriate specialist terms to explain fully how to

accurately carry out a titration [using 5–6 points of indicative

content]. They use good spelling, punctuation and grammar, and the

form and style are of a high standard.

5–6

(Band A)

Candidates use appropriate specialist terms to explain how to

accurately carry out a titration [using 3–4 points of indicative

content]. They use satisfactory spelling, punctuation and grammar,

and the form and style are of a satisfactory standard.

3–4

(Band B)

Candidates partially describe how to carry out a titration [using 2–4

points of indicative content]. They use limited spelling, punctuation

and grammar, and make little use of specialist terms. The form and

style are of a limited standard.

1–2

(Band C)




8 a) 12.5 cm3 (2 marks, allow 1 mark if rough is used – 12.7 cm3)

b) 12.5 ×0.100

1000= 0.00125 mol (1 mark)

c) Ratio = 2 : 1, so moles of sodium carbonate =0.00125

2= 0.000625 mol (2 marks)

d) 0.000625 ×1000

25.0= 0.025mol/dm3 (1 mark)

e) g/dm3 = 31.576.28 g/dm

0.250 (1 mark)

f) 3

3

g/dm 6.28Formula mass 251.2

mol/dm 0.025 (1 mark)

g) Mr(Na2CO3) = 106

251.2 – 106 (1 mark)

= 145.2 = 18x; x = 8 (1 mark)

9 Mr(BaCO3) = 197

Moles barium carbonate = 0.6600.003

197 (1 mark)

Moles CO2 = 0.003 = 3volume (dm )

24 (1 mark)

Volume = 0.072 dm3 (1 mark)

10 a) Ratio is 4 NO : 1 O2, so 1.6 dm3 : ?

Volume of oxygen =1.6

4= 0.4 dm3 (1 mark)

b) Mr((NH4)2CO3 = 96

Moles (NH4)2CO3 3.20.0333 mol

96 (1 mark)

Moles NH3 = 0.0333 × 2 = 0.0666 (1 mark)

Volume = 0.0666 × 24 = 1.6 dm3 (1 mark)

11 a) 90 × (2

3) = 60 dm3 (2 marks)

b) 60

24= 2.5 mol (2 marks)

12 a) Moles NaOH = 19.0 ×0.100

1000= 0.0019 mol (1 mark)

b) i) Pipette (1 mark)

ii) Burette (1 mark)

c) Ratio = 1 : 2, so moles tartaric acid =0.0019

2= 0.00095 mol (1 mark)

d) 0.00095 ×1000

25.0= 0.038 mol/dm3 (2 marks)

e) Mr(tartaric acid) = 150 (1 mark)

0.038 × 150 = 5.7 g/dm3 (1 mark)

13 Mr(NH4NO3) = 80

Moles NH4NO3 =2000

80= 25 mol (1 mark)

Ratio = 1 : 1 , so moles N2O = 25 mol (1 mark)

Volume = 25 × 24 = 600 dm3 (1 mark)



14 a) Moles NaOH = 20.0 ×0.25

1000= 0.005 mol (1 mark)

Ratio = 2 : 1, so moles H2X =0.005

2= 0.0025 mol (1 mark)

Concentration of H2X = 0.0025 ×1000

25.0= 0.1 mol/dm3 (1 mark)

b) 3

r3

g/dm

mol/dmM

9

900.1

(1 mark)

c) 90 – 2 = 88 (1 mark)

X is Sr (1 mark)

15 Moles HCl = 2.00 ×25.0

1000= 0.05 mol (1 mark)

Ratio = 1 : 2, so moles M(OH)2 =0.05

2= 0.025 mol (1 mark)

0.025 =1.45

rM

Mr =0

1.45

.025= 58 (1 mark)

Subtract the Mr of (OH)2 = 34

58 – 34 = 24 (1 mark)

Mg (1 mark)

16 Atom economy = 2 64100

(2 64) 44

(1 mark)

= = 128

100 74%172

(1 mark)

17 a) Atom economy = 4 2100

(4 2) 44

(1 mark)

8= 100 = 15.4%

52 (1 mark)

b) Find a use for carbon dioxide. (1 mark)

Chapter 14 – Electrochemistry Answers


Test yourself (page 254) 1 Electrolysis is the decomposition of electrolytes using electricity.

2 An electrolyte is an ionic liquid or solution that conducts electricity and is decomposed during

the process of electrolysis.

3 Molten lead bromide, calcium chloride solution, magnesium chloride solution

4 The negative ions are attracted to the positive electrode (anode) and the positive ions are

attracted to the negative electrode (cathode).

5 It is a good conductor of electricity and is an unreactive solid with a high melting point.

Show you can (page 254) 1 Sodium ions and chloride ions

2 Cl–

Test yourself (page 258) 6 a) The ions can move and carry charge.

b) The ions cannot move and so cannot carry charge.

c) When the potassium chloride melts it splits into potassium ions and chloride ions. The

potassium ions move to the negative cathode, where they gain electrons and form potassium

atoms. The negative chloride ions move to the positive anode, where they lose electrons and

form chlorine molecules.

7 a) They lose electrons and form atoms.

b) Conductors conduct electricity by electrons moving and are unchanged by the process.

Electrolytes conduct electricity by ions moving and are decomposed by the process.

8 a)

b) KI, purple gas; ZnBr2, red-brown gas; MgO, colourless gas; bubbles for all three

c) Grey liquid for all three

Ionic compound (molten)

Product at the negative electrode (cathode)

Product at the positive electrode (anode)

Cathode equation Anode equation

Potassium iodide (KI)

Potassium Iodine K+ + e– ⟶ K 2I– ⟶ I2 + 2e–

Zinc bromide (ZnBr2)

Zinc Bromine Zn2+ + 2e– ⟶ Zn 2Br– ⟶ Br2 + 2e–

Magnesium oxide (MgO)

Magnesium Oxygen Mg2+ + 2e– ⟶ Mg 2O2– ⟶ O2 + 4e–




Anode Cathode

Product Iodine Lithium Observation Purple gas, bubbles Molten grey liquid Half equation 2I– ⟶ I2 + 2e– Li+ + e– ⟶ Li Oxidation or reduction Oxidation Reduction

Test yourself (page 259) 9 The metal is too reactive.


• sodium

• potassium

• magnesium

• calcium.

11 Bauxite

12 Alumina

13 Aluminium oxide/alumina dissolved in cryolite

14 900 °C

15 The positive aluminium ions gain electrons and form aluminium.

16 The hot carbon anode reacts with the oxygen produced there and changes into carbon dioxide gas.

17 The decomposition of an electrolyte using electricity


Anode Cathode

Product Oxygen (then forms carbon dioxide)

Aluminium

Half equation 2O2– ⟶ O2 + 4e– Al3+ + 3e– ⟶ Al Oxidation or reduction Oxidation Reduction

Test yourself (page 260) 18 Hydrogen, hydroxide and sulfate

19 a) 4OH– ⟶ O2 + 2H2O+ 4e–

b) Relights a glowing splint

c) Insert a lighted splint and the gas explodes with a ‘pop’.



Practice questions (pages 261–262) 1 a) The ions (1 mark)

Can move and carry charge (1 mark)

b) i)

Tripod, pipe clay triangle (1 mark)

Bunsen burner (1 mark)

Lead(II) bromide in crucible (1 mark)

ii)

(10 marks)

2 a) Decomposition of electrolytes using electricity (1 mark)

b) Positive electrode (1 mark)

c) Sodium (1 mark)

d) Ions are held tightly in a lattice. (1 mark)

They cannot move to carry the charge. (1 mark)

e) An electrolyte is an ionic liquid or solution. (1 mark)

It conducts electricity and is decomposed in the process. (1 mark)

f) 2Cl– ⟶ Cl2 + 2e– (3 marks)

g) Graphite (1 mark)

3 a) A = anode (1 mark)

B = cathode (1 mark)

C = evaporating basin (1 mark)

b) Lamp / ammeter (1 mark)

c)

d) Bromine and lead are toxic. (1 mark)

Anode Cathode Name of product Bromine (1 mark) Lead (1 mark) Observations Red-brown pungent gas, bubbles

(1 mark) Molten grey liquid (1 mark)

Half-equation 2Br– ⟶ Br2 + 2e– (3 marks) Pb2+ + 2e– ⟶ Pb (3 marks)

Electrode Name of product Half equation A Bromine (1 mark) 2Br– ⟶ Br2 + 2e– (3

marks) B Lead (1 mark) Pb2+ + 2e– ⟶ Pb (3 marks)



e) Copper conducts by delocalised electrons moving. (1 mark)

It is not changed during the process. (1 mark)

Molten lead bromide conducts by ions moving. (1 mark)

It is decomposed during the process. (1 mark)

f) Anode – oxygen (1 mark)

Cathode – hydrogen (1 mark)

4 a) Decomposition of an electrolyte using electricity (1 mark)

b) Bauxite (1 mark)

c) i) X = anode (1 mark)

Z = cathode (1 mark)

ii) Alumina (1 mark)

Dissolved in cryolite (1 mark)

iii) 900 °C (1 mark)

iv) Oxygen at positive electrode oxygen (1 mark)

Aluminium at negative electrode (1 mark)

v) 2O2– ⟶ O2 + 4e– (3 marks)

vi) Al3+ + 3e– ⟶ Al (3 marks)

vii) Anode (1 mark)

C + O2 ⟶ CO2 (2 marks)

d) Aluminium gains electrons. (1 mark)

The gain of electrons is reduction. (1 mark)

5 a) i) Bromine (1 mark)

ii) Zn2+ + 2e– ⟶ Zn (3 marks)

Reduction (1 mark)

iii) 2Br– ⟶ Br2 + 2e– (3 marks)

b) i) Oxygen (1 mark)

ii) 2H+ + 2e ⟶ H2 (3 marks)

iii) Ions (1 mark)

c) i) Cathode – potassium (1 mark)

Anode – iodine (1 mark)

ii) Cathode – calcium (1 mark)

Anode – oxygen (1 mark)

iii) Cathode – sodium (1 mark)

Anode – bromine (1 mark)

6 a) i) Negative electrode (1 mark)

ii) Unreactive electrode (1 mark)



b) i) Al3+ + 3e– ⟶ Al (3 marks)

ii) They discharge by losing electrons. (1 mark)

Oxygen atoms form. (1 mark)

They combine to form O2. (1 mark)

iii) The hot carbon reacts (1 mark)

With the oxygen produced (1 mark)

This forms carbon dioxide. (1 mark)

iv) Accept any two of the following:

• recycling aluminium is much cheaper than producing aluminium

from bauxite because it uses only a fraction of the energy

• recycling saves waste

• recycling saves the natural resources of bauxite. (2 marks)

v) Accept any two of the following:

• it acts like a lid and keeps heat in

• it stops impurities from entering

• it prevents the aluminium formed from reacting with the air. (2 marks)

vi) Accept any one of the following:

• the high cost of electricity for the process

• the high cost of the heat energy to melt the metal compounds and

keep them molten. (1 mark)

Chapter 15 – Energy changes in chemistry Answers


Test yourself (page 264) 1 a) Exothermic

b) Endothermic

c) Exothermic

2 a) Exothermic

b) Endothermic

c) Endothermic

d) Exothermic

3 a) Heat energy is transferred from the chemicals to the surroundings.

b) Heat energy is transferred from the surroundings to the chemicals.

4 a) Exothermic

b) Endothermic

c) Endothermic

d) Exothermic


Combustion Decomposition Neutralisation Oxidation Respiration Exothermic Endothermic

P Q

Test yourself (page 266) 5 a) Reactions 2 and 3

b) Reaction 1

6 a), b), c)

7 a) C

b) D

c) A



Show you can (page 266) 1 Exothermic

2 A

Test yourself (page 268) 8 The energy taken in to break bonds in hydrogen and iodine is more than the energy given out

when bonds are made in hydrogen iodide.

9 The energy taken in to break bonds in hydrogen and oxygen is less than the energy given out

when bonds are made in water.

10 The energy taken in to break bonds in ethanol is greater than the energy given out when bonds

are made in C2H4 (ethene) and water.

Test yourself (page 270) 11 a) (436 + 193) – (2 × 366) = 629 – 732 = –103 kJ

b) (944 + 3 × 436) – 2 × (388 × 3) = 2252 – 2328 = –76 kJ

c) (4 × 412 + 612 + 193) – (4 × 412 + 348 + 2 × 276) = 2453 – 2548 = –95 kJ

d) (8 × 412 + 2 × 348 + 5 × 496) − {[3 × (2 × 743)] + [4 × (2 × 463)]} = 6472 – 8162 = –1690 kJ

12 (436 + 242) − 2 × 432 = 678 – 864 = –186 kJ

Show you can (page 270) (150 + 242) − 2 × ??? = −30 kJ

2 × ??? = 30 + (150 + 242) = 422

The bond energy of I–Cl is 211 kJ

Practice questions (pages 271–272) 1 a) A Decomposition (1 mark)

B Neutralisation (1 mark)

C Oxidation (1 mark)

D Combustion/oxidation (1 mark)

E Neutralisation (1 mark)

b) An exothermic reaction is one that gives out heat (1 mark); an endothermic reaction

is one that takes in heat. (1 mark) (2 marks)

c) A Endothermic (1 mark)

B Exothermic (1 mark)

C Exothermic (1 mark)

D Exothermic (1 mark)

E Exothermic (1 mark)



d) Measure a volume of hydrochloric acid (1 mark)

Into a polystyrene cup. (1 mark)

Measure the temperature. (1 mark)

Add magnesium. (1 mark)

Measure the temperature again. (1 mark)

If the temperature has increased the reaction is exothermic. (1 mark)

2 a) An endothermic reaction is one that takes in heat. (1 mark)

b) The minimum amount of energy needed for a reaction to occur (1 mark)

c)

Axes labelled correctly (1 mark)

Products above reactants (1 mark)

Activation energy and energy change labelled (1 mark)

d) The energy taken in to break bonds in carbon dioxide and water(1 mark) is

greater than (1 mark) the energy given out when bonds are made in C6H12O6

and O2. (1 mark) (3 marks)

3 a) Endothermic (1 mark)

b)

Axes labelled correctly (1 mark)

Products above reactants (1 mark)

Activation energy and energy change labelled (1 mark)



c) Moles of N2 = 28/28 = 1 (1 mark)

Ratio is 2 : 3, so moles NaN3 = 0.67 (1 mark)

Mass = 0.67 × Mr = 0.67 × 65= 43.6 g (1 mark)

4 a) (436 + 158) – (2 × 568) = 594 − 1136 = –542 kJ (3 marks)

b) Exothermic (1 mark)

5 a) Exothermic (1 mark)

b) No, the graph shows that the overall energy change is the same. (1 mark)

c)

Correct labels (2 marks)

d) The presence of a catalyst provides a pathway of lower activation energy. (1 mark)

6 a) Reaction 1, endothermic (1 mark)

Reaction 2, exothermic (1 mark)

Reaction 3, exothermic (1 mark)

b) Increases, endothermic, decreases (3 marks)

c) Collide, activation (2 marks)

7 a) C (1 mark)

b) F (1 mark)

c) Exothermic (1 mark)

8 a) Bonds broken: 436 + 242 = 678 (1 mark)

Bonds formed: 2 × 431 = 862 (1 mark)

Overall: 678 – 862 = –184 kJ (1 mark)

b) Exothermic (1 mark)

c) The energy taken in to break bonds in hydrogen and chlorine (1 mark) is less

than (1 mark) the energy given out when bonds are made in HCl. (1 mark) (3 marks)

9 Bonds broken: 4 × 388 + 158 + 496 = 2206 (1 mark)

Bonds formed: 2 × (2 × 463) + 944 = 2796 (1 mark)

Overall: 2206 – 2796 = –590 kJ (1 mark)



10 a) Bonds broken: 4 × 412 + 2 × 496 = 2640 (1 mark)

Bonds formed: 2 × 743 + 2 × (2 × 463) = 3338 (1 mark)

Overall: 2640 – 3338 = –698 kJ (1 mark)

b) CH4 + 2O2 ⟶ CO2 + 2H2O (1 mark)

Chapter 16 – Gas chemistry Answers


Test yourself (page 275) 1 a)

b) Water vapour

2

3 a) Nitrogen

b) Ammonia

c) Nitrogen

d) Glass rod dipped in concentrated hydrochloric acid, white smoke

e) NH3 + HNO3 ⟶ NH4NO3

f) It has a strong triple covalent bond.

Test yourself (page 278) 4 a) Two atoms covalently bonded

b) Nitrogen and oxygen

5 a) Hydrogen peroxide

b) 2H2O2 ⟶ O2 + 2H2O

c) The hydrogen peroxide breaks up.

d) A substance that speeds up a chemical reaction and is not used up in the reaction

e) Manganese(IV) oxide; a black solid

f) Relights a glowing splint

Gas Symbol/formula Element or compound % Nitrogen N2 Element 78 Oxygen O2 Element 21 Argon Ar Element 1 Carbon dioxide CO2 Compound 0.03–0.04

Gas Formula Dot and cross diagram Colour Acidic, basic or neutral

Nitrogen N2

Colourless Neutral

Ammonia NH3

Colourless Basic



6 a) i) magnesium + oxygen ⟶ magnesium oxide

2Mg(s) + O2(g) ⟶ 2MgO(s)

ii) copper + oxygen ⟶ copper(II) oxide

2Cu(s) + O2(g) ⟶ 2CuO(s)

iii) iron + oxygen⟶ iron oxide

3Fe(s) + 2O2 ⟶ Fe3O4

iv) sulfur + oxygen ⟶ sulfur dioxide

S(s) + O2(g) ⟶ SO2(g)

v) carbon + oxygen ⟶ carbon dioxide

C(s)+ O2(g) ⟶ CO2(g)

b) MgO is basic, CuO is basic, Fe3O4 is basic, SO2 is acidic, CO2 is acidic.

c) i) Grey solid burns with bright white light, forms white powder

ii) Red-brown solid, blue–green flame glows red, gives black layer

iii) Grey solid, burns with orange sparks, gives black solid

Show you can (page 278) Carbon burns to produce carbon monoxide: 2C + O2 ⟶ 2CO


7

8 a) Zinc and hydrochloric acid

b) Zn + 2HCl ⟶ ZnCl2 + H2

9 a) It burns to produce water, which does not pollute.

b) It is flammable and difficult to store; fuel cells are expensive; the hydrogen is often made

using fossil fuels, and this also causes pollution.

c) It changes liquid oil to solid fat.

10 a) Calcium carbonate and hydrochloric acid

b) CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

c) Carbonic acid

Hydrogen Carbon dioxide Formula H2 CO2 Element or compound? Element Compound Acidic, basic or neutral? Neutral Acidic Lighter or denser than air? Lighter Denser Test for the gas Apply a lighted splint,

explodes with a ‘pop’ Colourless limewater changes to milky



d) Fire extinguishers, fizzy drinks




Gas Acidic, basic or neutral? Formula Solubility in water Nitrogen Neutral N2 No Ammonia Basic NH3 Yes Carbon dioxide Acidic CO2 Yes Oxygen Neutral O2 Slightly Hydrogen Neutral H2 No

Prescribed practical activity (page 281) 1

2

3 a) Red litmus stays red, blue litmus changes to red.

Carbonic acid is produced.

b) If bubbled into colourless limewater, it changes from colourless to milky.

Practice questions (pages 282–283) 1

Observations Name of gas produced Zinc is added to some hydrochloric acid.

Bubbles, heat released, zinc disappears, colourless solution produced

Hydrogen

Calcium carbonate is added to some hydrochloric acid.

Bubbles, heat released, calcium carbonate disappears, colourless solution produced

Carbon dioxide

Hydrogen peroxide is added to some manganese(IV) oxide.

Bubbles, black solid remains Oxygen

Symbol equation Colour of red litmus paper

Colour of blue litmus paper

A spatula of sulfur was burnt in a gas jar of oxygen.

S + O2 ⟶ SO2 Red Red

A piece of carbon was burnt in a gas jar of oxygen.

C + O2 ⟶ CO2 Red Red

A piece of calcium ribbon was burnt in a gas jar of oxygen.

2Ca + O2 ⟶ 2CaO Blue Blue

Hydrogen Carbon dioxide Oxygen Formula H2 (1 mark) CO2 (1 mark) O2 (1 mark) Names of substances used to prepare the gas

1. Zinc 2. Hydrochloric acid (2 marks)

1. Calcium carbonate 2. Hydrochloric acid (2 marks)

1. Hydrogen peroxide 2. Manganese(IV) oxide (2 marks)

Test for the gas Lighted splint Pop (2 marks)

Colourless limewater Turns milky (3 marks)

Glowing splint Relights (2 marks)

Equation for preparation of the gas

Zn + 2HCl ⟶ ZnCl2 + H2 (3 marks)

CaCO3 + 2HCl ⟶ CaCl2 + CO2 + H2O (3 marks)

2H2O2 ⟶ 2H2O + O2 (3 marks)



2 a) Accept any two of the following:

• neutral

• less dense than air

• insoluble in water. (2 marks)

b) Apply a lighted splint (1 mark)

Explodes with a ‘pop’ (1 mark)

c) acid + metal ⟶ salt + hydrogen (1 mark)

d) Zinc (1 mark)

Hydrochloric acid (1 mark)

e) Accept any two of the following:

• heat released

• bubbles

• zinc disappears

• colourless solution produced. (2 marks)

f)

(5 marks)

g) Burns to produce water, (1 mark) which is non-polluting. (1 mark) (2 marks)

h) Lighter than air (1 mark)

3 a) A is conical flask (1 mark)

B is thistle funnel (1 mark)

C is delivery tube (1 mark)

D is gas jar (1 mark)

E is trough (1 mark)

F is beehive shelf (1 mark)

b) Not very soluble, as collected over water (1 mark)

c) Carbon dioxide (1 mark)

Hydrogen (1 mark)

d) Calcium carbonate (1 mark)

Zinc (1 mark)



4 a) Nitrogen, 78% (1 mark)

Oxygen, 21% (1 mark)

Carbon dioxide, 0.03–0.04% (1 mark)

b) The triple covalent bond is strong. (1 mark)

It requires a lot of energy to break. (1 mark)

c) N2 + 2O2 ⟶ 2NO2 (2 marks)

d) N2 + 3H2 ⟶ 2NH3 (3 marks)

e) Dip a glass rod in concentrated hydrochloric acid. (1 mark)

White smoke is produced with ammonia. (1 mark)

f) 2NH3 + H2SO4 ⟶ (NH4)2SO4 (3 marks)

g) CO2 + H2O ⟶ H2CO3 (2 marks)

Carbonic acid (1 mark)

h) i) Calcium hydroxide solution (1 mark)

ii) Colourless (1 mark)

iii) Insoluble (1 mark)

Calcium carbonate is produced. (1 mark)

iv) Soluble (1 mark)

Calcium hydrogencarbonate is produced. (1 mark)

i) CaCO3 + 2HCl ⟶ CaCl2 + CO2 + H2O (3 marks)

j) Fire extinguishers (1 mark)

Fizzy drinks (1 mark)



5 a) Welding (1 mark)

b) i) Colourless liquid/solution (1 mark)

ii) Manganese(IV) oxide (1 mark)

Black solid (1 mark)

iii) 2H2O2 ⟶ 2H2O + O2 (3 marks)

c)

Indicative content:

Names of products:

Cu: copper(II) oxide

Mg: magnesium oxide

Observations:

Cu: Red brown solid at start

Blue–green flame

Glows red

Black layer

Mg: Grey solid at start

White light

White powder

Comparison of reactivity:

Magnesium is more reactive; copper only reacts on the surface; magnesium reacts fully.

d) i) Accept any three of the following:

• yellow solid

• burns with blue flame

• red-brown liquid

• colourless gas. (3 marks)

ii) Colourless (1 mark)

iii) C + O2 ⟶ CO2 (2 marks)

iv) Acidic (1 mark)

Response Marks

A well-structured answer comparing the reaction of copper and magnesium,

with very good spelling, punctuation and grammar [using 8–10 indicative

content points]. Written with a high degree of clarity and coherence.

5–6

(Band A)

A reasonably structured answer comparing the reaction of copper and

magnesium, with good spelling, punctuation and grammar [using 5–7

indicative content points]. Written with some clarity and coherence.

3–4

(Band B)

An attempt has been made to compare the reactions of copper and

magnesium but the form is poor [a minimum of two indicative content

marks] with little comparison of reactions and poor clarity and coherence.

1–2

(Band C)




6 Carbon dioxide:

• colourless (1 mark)

• limewater (1 mark)

• turns milky (1 mark)

Oxygen:

• relights (1 mark)

• glowing splint (1 mark)

Hydrogen:

• apply a lighted splint (1 mark)

• explodes with a ‘pop’ (1 mark)

Ammonia:

• glass rod dipped in concentrated hydrochloric acid (1 mark)

• white smoke (1 mark)

Nitrogen:

• no test, so negative result with the fours test above (1 mark)

Appendix Answers



1 a) K, b) P, c) Si, d) Na, e) S, f) F, g) Zn, h) Cu, i) Ca, j) C

2 Four of: hydrogen H2, oxygen O2, nitrogen N2, chlorine Cl2, fluorine F2

3 a) 1 atom sodium, 1 atom oxygen, 1 atom hydrogen

b) 1 atom hydrogen, 1 atom nitrogen, 3 atoms oxygen

c) 3 atoms carbon, 6 atoms hydrogen, 2 atoms oxygen

d) 1 atom magnesium, 2 atoms hydrogen, 2 atoms oxygen

e) 1 atom aluminium, 3 atoms chlorine

f) 2 atoms nitrogen, 8 atoms hydrogen, 1 atom sulfur, 4 atoms oxygen

g) 2 atoms nitrogen, 4 atoms hydrogen, 3 atoms oxygen

h) 1 atom aluminium, 3 atoms oxygen, 3 atoms hydrogen

i) 2 atoms aluminium, 3 atoms carbon, 9 atoms oxygen

4 a) 1+, b) 1+, c) 1−, d) 1−, e) 2−, f) 3+, g) 3−, h) 2−

5 a) 2+, b) 1−, c) 3−, d) 2−, e) 1+, f) 2+, g) 2−, h) 1−, i) 2−, j) 1−, k) 1+

6 a) potassium chloride b) sodium nitrate

c) calcium chloride d) calcium sulfate

e) aluminium carbonate f) magnesium nitrate

g) sodium sulfate h) calcium carbonate

i) iron(III) sulfate j) calcium hydrogencarbonate

k) potassium dichromate l) silver nitrate

m) magnesium bromide n) lead sulfate

o) zinc hydroxide p) sodium hydrogencarbonate

q) iron(II) iodide


7 KCl

8 CaI2

9 CuO

10 Al2S3

11 Ca(NO3)2

12 Cu(OH)2

13 (NH4)2CO3

14 Fe(NO3)2

15 Fe(OH)2

16 Al(HCO3)3

Appendix Answers


17 Na2SO4

18 CaS

19 CuSO4

20 MgO

21 NH4Cl

22 NaNO3

23 Cr(OH)3

24 FeO

25 Al2(Cr2O7)3

26 NaHCO3

27 Na2CO3

28 K2SO4

29 NaS

30 Li2Cr2O7

31 Fe2O3

32 Mg(Cr2O7)2

33 CuCl2

34 Zn(NO3)2

35 (NH4)2CO3

36 Na2Cr2O7

37 FeCl2

38 Al(OH)3

39 FeBr3

40 Mg(OH)2

41 AgNO3


42 ⟶ magnesium chloride + hydrogen

43 ⟶ calcium nitrate + carbon dioxide + water

44 ⟶ potassium sulfate + hydrogen

45 ⟶ sodium chloride + water

46 ⟶ potassium nitrate + water

47 ⟶ magnesium sulfate + water + carbon dioxide

48 ⟶ copper(II) chloride + water

49 ⟶ ammonium chloride

50 ⟶ sodium hydroxide + hydrogen

51 ⟶ magnesium oxide

Appendix Answers


52 ⟶ potassium hydroxide + hydrogen

53 ⟶ zinc oxide

54 ⟶ aluminium oxide

55 ⟶ copper oxide

56 ⟶ sodium sulfate + water

57 ⟶ calcium nitrate + water

58 ⟶ potassium chloride + water + carbon dioxide

59 ⟶ zinc chloride + hydrogen

60 ⟶ magnesium nitrate + water

61 ⟶ aluminium chloride + hydrogen


62 1 Ca, 2 Cl

63 1 Ca, 2 O, 2 H

64 4 H, 2 S, 8 O

65 2 H, 2 N, 6 O

66 6 H

67 2 Ca, 2 S, 6 O

68 3 Mg, 6 O, 6 H

69 2 N, 8 H, 1 C, 3 O

70 2 Al, 6 N, 18 O

71 6 K, 3 S, 12 O

72 4 Na, 4 O, 4 H

73 4 Al, 6 C, 18 O

74 4 Na, 4 Al, 16 O, 16 H

75 4 Fe, 6 C, 18 O


76 3Mg + N2 ⟶ Mg3N2

77 H2 + Cl2 ⟶ 2HCl

78 MgO + 2HCl ⟶ MgCl2 + H2O

79 N2 + 3H2 ⟶ 2NH3

80 2Na + O2 ⟶ Na2O

81 2Ca + O2 ⟶ 2CaO

82 2K + 2H2O ⟶ 2KOH + H2

83 Ca + 2HCl ⟶ CaCl2 + H2

Appendix Answers


84 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O

85 Ca(OH)2 + 2NH4Cl ⟶ CaCl2 + 2NH3 + H2O

86 C5H12 + 8O2 ⟶ 5CO2 + 6H2O

87 SO2 + O2 ⟶ SO3

88 Ca + 2HNO3 ⟶ Ca(NO3)2 + H2

89 C2H5OH + 3O2 ⟶ 2CO2 + 3H2O

90 Ca + 2H2O ⟶ Ca(OH)2 + H2

91 Ca(OH)2 + 2HNO3 ⟶ Ca(NO3)3 + H2O

92 2Fe + 1½O2 ⟶ Fe2O3 (or 4Fe + 3O2 ⟶ 2Fe2O3)

93 2Al + 6HCl ⟶ 2AlCl3 + 3H2

94 C2H6 + 3½O2 ⟶ 2CO2 + 3H2O (or 2C2H6 + 7O2 ⟶ 4CO2 + 6H2O)

95 Mg(OH)2 + 2HCl ⟶ MgCl2 + 2H2O

96 Na2CO3 + HCl ⟶ 2NaCl + H2O + CO2

97 CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

98 Na + Cl2 ⟶ 2NaCl

99 CH4 + 2O2 ⟶ CO2 + 2H2O

100 Li + 2HNO3 ⟶ 2LiNO3 + H2

101 2Al + 1½O2 ⟶ Al2O3 (or 4Al + 3O2 ⟶ 2Al2O3)

102 3Pb + 2O2 ⟶ Pb3O4

103 2Na + 2H2O ⟶ 2NaOH + H2

104 C2H4 + 3O2 ⟶ 2CO2 + 2H2O

105 2NO + O2 ⟶ 2NO2

106 Zn + 2HCl ⟶ ZnCl2 + H2

107 2KHCO3 + H2SO4 ⟶ K2SO4 + 2CO2 + 2H2O


108 Na2CO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2

109 CuCO3 + 2HNO3 ⟶ Cu(NO3)2 + H2O + CO2

110 2Al + 6HCl ⟶ 2AlCl3 + 3H2

111 2Na + 2H2O ⟶ 2NaOH + H2

112 Ca + 2H2O ⟶ Ca(OH)2 + H2

113 2HNO3 + Ca(OH)2 ⟶ Ca(NO3)2 + H2O

114 NH3 + HNO3 ⟶ NH4NO3

115 2NH3 + H2SO4 ⟶ (NH4)2SO4

116 CuCO3 ⟶ CuO + CO2

117 Na2O + H2SO4 ⟶ Na2SO4 + H2O

118 Ca(OH)2 + 2HCl ⟶ CaCl2 + 2H2O

Appendix Answers


119 K2CO3 + 2HCl ⟶ 2KCl + H2O + CO2

120 Mg + 2HCl ⟶ MgCl2 + H2

121 2HNO3 + Mg(OH)2 ⟶ Mg(NO3)2 + 2H2O

122 2Al + 3H2SO4 ⟶ Al2(SO4)3 + 3H2


123 H+(aq) + OH−(aq) ⟶ H2O(l) neutralisation


125 Ca2+(aq) + 2OH−(aq) ⟶ Ca(OH)2(s) precipitation

126 Ag+(aq) + Br−(aq) ⟶ AgBr(s) precipitation

127 Cu2+(aq) + Mg(s) ⟶ Cu(s) + Mg2+(aq) displacement

128 2Ag+(aq) + Mg(s) ⟶ 2Ag(s) + Mg2+(aq) displacement

129 Fe3+(aq) + 3OH−(aq) ⟶ Fe(OH)3(s) precipitation

130 Ba2+(aq) + SO42−(aq) ⟶ BaSO4(s) precipitation

131 Pb2+(aq) + 2I−(aq) ⟶ PbI2(s) precipitation



134 3Zn2+(aq) + 2Al(s) ⟶ 3Zn(s) + 2Al3+(aq) displacement


135 a) Mg2+ + 2e− ⟶ Mg b) S2− ⟶ S + 2e−

c) K+ + e− ⟶ K d) 2Br− ⟶ Br2 + 2e−

e) 2O2− ⟶ O2 + 4e− f) 2H+ + 2e− ⟶ H2

g) Li+ + e− ⟶ Li h) Fe2+ + 2e− ⟶ Fe

I) Ca ⟶ Ca2+ + 2e−

136 a) 2I− ⟶ I2 + 2e− b) Al3+ + 3e− ⟶ Al

c) 2N3− ⟶ N2 + 6e− d) Na+ + e− ⟶ Na

e) Fe ⟶ Fe3+ + 3e− f) Cu2+ + 2e− ⟶ Cu

chapter 1 – atomic structure...

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