IGCSE Chemistry 0620 Paper 42, Oct/Nov 2023

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0620/42 · Oct/Nov 2023

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1

Table 1.1 gives the electronic configurations of some atoms and ions, A to G. Table 1.1 electronic configuration A 2,5 B 2,8 C 2,8,2 D 2,8,4 E 2,8,5 F 2,8,6 G 2,8,18,7 Answer the following questions about A to G. Each letter may be used once, more than once or not at all. State which of the atoms or ions, A to G, could be:

a

a noble gas atom

[1]
b

an atom of an element in Group VI

[1]
c

an atom with an atomic number of 14

[1]
d

atoms from the same group .................................................................... and ..................................................................

[1]
e

a halogen atom

[1]
f

an atom of an element which is a good conductor of electricity

[1]
g

a stable ion of a Group V element

[1]
h

an atom that forms an ion with a 2– charge.

[1]
2

Cobalt and copper are transition elements.

a(i)

Copper has two naturally occurring isotopes, 63Cu and 65Cu. Cobalt has only one naturally occurring isotope, 59Co. Complete Table 2.1 to show the number of protons, neutrons and electrons in the 59Co atom and the 65Cu2+ ion. Table 2.1

59Co65Cu2+
protons
neutrons
electrons
[3]
a(ii)

Table 2.2 shows the relative abundance of the two naturally occurring isotopes of copper. Table 2.2

isotope63Cu65Cu
relative abundance70%30%

Calculate the relative atomic mass of copper to one decimal place. relative atomic mass = ..............................

[2]
b

One physical property of transition elements such as copper and cobalt is that they are hard. Other metals such as lithium are softer. State two other physical properties of copper and cobalt which are significantly different from lithium. 1 ................................................................................................................................................. 2 .................................................................................................................................................

[2]
c(i)

Both copper and cobalt can form coloured compounds. Some of these compounds contain water of crystallisation. Define the term water of crystallisation.

[2]
c(ii)

State the colour and formula of hydrated cobalt(II) chloride crystals. colour .................................................................................................................................. formula ................................................................................................................................

[2]
c(iii)

State the colour change seen when a few drops of water are added to anhydrous copper(II) sulfate. from ............................................................. to .............................................................

[2]
c(iv)

State how this colour change can be reversed.

[1]
3

Iron is manufactured in a blast furnace.

a

Three of the starting materials added to the blast furnace are coke, iron ore and limestone. Name the other starting material added to the blast furnace.

[1]
b(i)

The source of iron in the blast furnace is Fe2O3. Fe2O3 is found in iron ore. Name the main ore of iron which contains Fe2O3.

[1]
b(ii)

The iron in Fe2O3 is reduced by reaction with carbon monoxide. The unbalanced symbol equation is shown. Complete the equation. Fe2O3+.....CO→.....CO2+.....Fe

[1]
b(iii)

State the change in oxidation number of iron in the reaction in (ii). from ............................................................. to .............................................................

[2]
b(iv)

Explain how the change of oxidation number shows that iron has been reduced.

[1]
c

The major impurity in iron ore is silicon(IV) oxide. Limestone is added to the blast furnace to remove this impurity. Write two symbol equations to show how silicon(IV) oxide is removed. For each equation, state the type of chemical reaction that takes place. equation 1 .................................................................................................................................. type of chemical reaction ........................................................................................................... equation 2 .................................................................................................................................. type of chemical reaction ...........................................................................................................

[4]
d(i)

Iron is converted to steel by mixing it with carbon and other elements. State the term given to a substance which is a mixture of a metal and other elements.

[1]
d(ii)

Name one element, other than carbon, mixed with iron in the making of stainless steel.

[1]
e

Preventing the rusting of steel is important. State the chemical name of rust.

[1]
f(i)

Steel can be coated with zinc to prevent rusting. This provides both a barrier method and sacrificial protection. State the term used for coating steel with zinc.

[1]
f(ii)

Describe another barrier method for preventing rusting.

[1]
f(iii)

Explain how zinc provides sacrificial protection.

[2]
4

This question is about lead(II) chloride, PbCl2.

Figure 1 of 1 for question 4
a(i)

A student prepares a sample of insoluble lead(II) chloride, PbCl2, by mixing aqueous solutions of two salts in a beaker. Identify two soluble salts suitable for making lead(II) chloride when mixed together.

[2]
a(ii)

Write the ionic equation for the formation of lead(II) chloride by mixing aqueous solutions. Include state symbols.

[3]
a(iii)

List the steps the student should take in preparing a pure sample of lead(II) chloride from the mixture in the beaker.

[3]
b(i)

The student carries out an electrolysis experiment on molten lead(II) chloride using the apparatus shown in Fig. 4.1. Chlorine gas forms at the anode and escapes from the apparatus. Explain why lead(II) chloride needs to be molten before it will conduct electricity.

[1]
b(ii)

Write the ionic half-equation for the reaction occurring at the anode.

[2]
b(iii)

State the test for chlorine gas. test ...................................................................................................................................... observations .......................................................................................................................

[2]
b(iv)

Describe what is observed at the cathode.

[1]
5

Chemical reactions can involve transfer of thermal energy.

Figure 1 of 1 for question 5
a

State the term used for the transfer of thermal energy during a reaction.

[1]
b(i)

Tetrachloromethane gas, CCl4(g), reacts with steam as shown. CCl4(g)+2H2O(g)⇌CO2(g)+4HCl(g) The reaction is reversible. The forward reaction is exothermic. State what happens, if anything, to the rate of the forward reaction if the concentration of CCl4 is increased. Explain your answer in terms of collision theory.

[3]
b(ii)

State what happens to the position of equilibrium, if anything, when the pressure is increased. Explain your answer.

[2]
b(iii)

Fig. 5.1 shows an incomplete reaction pathway diagram for the forward reaction. CCl4(g)+2H2O(g)⇌CO2(g)+4HCl(g) On Fig. 5.1: ● insert the formulae of the reactants and products ● draw an arrow, labelled Ea, to show the activation energy ● draw an arrow, labelled ΔH, to show the transfer of energy in the reaction.

[3]
b(iv)

Define the term activation energy.

[2]
b(v)

State one way in which the activation energy of a reaction can be changed.

[1]
c

The equation for the reaction between tetrachloromethane gas and steam can be represented as shown in Fig. 5.2. Table 5.1 shows some bond energies. Table 5.1

bondC−ClH−OC=O
bond energy in kJ/mol340460805

Use the bond energies in Table 5.1 and the ΔH value for the reaction to calculate the H−Cl bond energy using the following steps. ● Calculate the energy needed to break the bonds in the reactants. .............................. kJ ● Calculate the energy released when the bonds in carbon dioxide form. .............................. kJ ● Calculate the H−Cl bond energy. .............................. kJ/mol

[4]
6

A homologous series is a family of organic compounds whose members have similar chemical properties.

Figure 1 of 1 for question 6
a

Give two characteristics that are the same for all members of a homologous series. 1 ................................................................................................................................................. 2 .................................................................................................................................................

[2]
b

In terms of structure, state how one member of a homologous series differs from the next member of that homologous series.

[1]
c(i)

A, B and C are organic compounds. A has the molecular formula C12H24. B has the name tetradecane. C has three carbon atoms and is in the homologous series with the general formula CnH2n+1COOH. Name the homologous series each organic compound belongs to. A ......................................................................................................................................... B ......................................................................................................................................... C .........................................................................................................................................

[3]
c(ii)

Name C and draw its displayed formula. name ................................................................................................................................... displayed formula

[2]
d(i)

Amino acids are a homologous series where each member has the general structure shown in Fig. 6.1. The R side chain contains carbon and hydrogen atoms only. An amino acid has a relative molecular mass of 103. Deduce the formula of the R side chain in this amino acid. Show your working. .......................................................................................................................................

[2]
d(ii)

State the name given to the natural polyamides formed from amino acid monomers.

[1]

Mark scheme

0620/42 · Oct/Nov 2023

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1a · 1 mark

  • 1 mark

    B

1b · 1 mark

  • 1 mark

    F

1c · 1 mark

  • 1 mark

    D

1d · 1 mark

  • 1 mark

    A AND E

1e · 1 mark

  • 1 mark

    G

1f · 1 mark

  • 1 mark

    C

1g · 1 mark

  • 1 mark

    B

1h · 1 mark

  • 1 mark

    F

2a(i) · 3 marks

  • 1 mark

    protons: 27 and 29

  • 1 mark

    neutrons: 32 and 36

  • 1 mark

    electrons: 27 and 27

2a(ii) · 2 marks

  • M1

    (63  70) + (65  30) = 6360

  • M2

    6360 / 100 = 63.6

2b · 2 marks

  • M1

    (high) density

  • M2

    (high) melting point

2c(i) · 2 marks

  • M1

    water molecules

  • M2

    (the water present) in hydrated crystals

2c(ii) · 2 marks

  • M1

    pink

  • M2

    CoCl2∙6H2O

2c(iii) · 2 marks

  • M1

    white

  • M2

    blue

2c(iv) · 1 mark

  • 1 mark

    heating (the hydrated copper(II) sulfate)

3a · 1 mark

  • 1 mark

    (hot) air

3b(i) · 1 mark

  • 1 mark

    hematite

3b(ii) · 1 mark

  • 1 mark

    Fe2O3 + 3CO → 3CO2 + 2Fe

3b(iii) · 2 marks

  • M1

    from +3

  • M2

    to 0

3b(iv) · 1 mark

  • 1 mark

    decrease (in oxidation number)

3c · 4 marks

  • M1

    CaCO3 → CaO + CO2

  • M2

    thermal decomposition

  • M3

    SiO2 + CaO → CaSiO3

  • M4

    acid-base reaction

3d(i) · 1 mark

  • 1 mark

    alloy

3d(ii) · 1 mark

  • 1 mark

    nickel / chromium

3e · 1 mark

  • 1 mark

    hydrated iron(III) oxide

3f(i) · 1 mark

  • 1 mark

    galvanising

3f(ii) · 1 mark

  • 1 mark

    painting / greasing / coating with plastic

3f(iii) · 2 marks

  • M1

    zinc more reactive (than iron)

  • M2

    zinc corrodes / oxidises / reacts in preference to / instead of iron

4a(i) · 2 marks

  • M1

    lead nitrate

  • M2

    soluble chloride e.g. sodium chloride

4a(ii) · 3 marks

  • M1

    PbCl2 as only product

  • M2

    Pb2+ + 2Cl– as only reactants

  • M3

    state symbols

4a(iii) · 3 marks

  • M1

    filter

  • M2

    wash (the residue) using water

  • M3

    dry the residue between filter papers / in a warm place

4b(i) · 1 mark

  • 1 mark

    mobile ions

4b(ii) · 2 marks

  • M1

    any negative Cl species losing electron(s)

  • M2

    correct ionic half equation

4b(iii) · 2 marks

  • M1

    (damp) litmus (paper)

  • M2

    is bleached / goes white

4b(iv) · 1 mark

  • 1 mark

    (shiny) grey AND solid

5a · 1 mark

  • 1 mark

    enthalpy change (of reaction)

5b(i) · 3 marks

  • M1

    rate of reaction increases

  • M2

    more (CCl4) particles per unit volume

  • M3

    frequency of collisions between (CCl4 and H2O) particles increases

5b(ii) · 2 marks

  • M1

    (position of) equilibrium moves to left hand side

  • M2

    more (gaseous) moles / molecules on right hand side

5b(iii) · 3 marks

  • M1

    Labels mark CCl4(g) + 2H2O(g) on reactant line AND CO2(g) + 4HCl(g) on product line

  • M2

    Activation E mark upward arrow labelled Ea from energy level of reactants to top of ‘hump’

  • M3

    Energy change mark one downward arrow labelled H AND energy change starting from E level of reactants and finishing at E level of products

5b(iv) · 2 marks

  • M1

    minimum energy

  • M2

    that colliding particles must have to react

5b(v) · 1 mark

  • 1 mark

    (decreased by adding) catalyst

5c · 4 marks

  • M1

    energy needed to break bonds in reactants [(4  340) + (4  460) = 3200 (kJ / mol)

  • M2

    energy released when bond in carbon dioxide form 2  805 = 1610 (kJ / mol)

  • M3

    calculate H–Cl bond energy 3200 – (1610 + E(4H–Cl)) = – 130 E(4H–Cl) = 3330 – 1610 = 1720 (kJ / mol)

  • M4

    1720 / 4 = 430 (kJ / mol)

6a · 2 marks

  • M1

    (same) general formula

  • M2

    (contain the same) functional group

6b · 1 mark

  • 1 mark

    -CH2- unit

6c(i) · 3 marks

  • M1

    alkenes

  • M2

    alkanes

  • M3

    carboxylic acids

6c(ii) · 2 marks

  • M1

    propanoic acid

  • M2

    displayed formula of propanoic acid

6d(i) · 2 marks

  • M1

    working to give the mass of atoms shown in Fig 6.1 103 – 14 – (2  12) – (2  16) – (4  1) (= 29)

  • M2

    ‘R’ = C2H5

6d(ii) · 1 mark

  • 1 mark

    proteins

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