0620 Chemistry November 2025 Question Paper 41

1 Protons, neutrons and electrons are particles found in atoms.

(a) Complete Table 1.1 to show the relative mass and relative charge of a proton, a neutron and an electron. Table 1.1 particle relative mass relative charge proton +1 neutron electron 1 1840

[2]

(b) Some elements have many isotopes.

(i) Define the term isotopes [2]

(ii) Explain why all isotopes of the same element have the same chemical properties [1]

(c) Complete Table 1.2. Table 1.2 atom or ion number of protons number of neutrons number of electrons 4 1 0 8Ar 18 18 3 1 2 6S2– 16 22 28 20

[5] , ,

(d) The term mass number is defined as the total number of protons and neutrons in the nucleus of an atom.

State the name of one other term which is defined as the total number of protons and neutrons in the nucleus of an atom [1]

(e) Calculate the number of atoms in 2.00 g of argon.

Give your answer in standard form.

number of atoms = [2] [Total: 13] , ,

2 Calcium is an element in Group II of the Periodic Table.

(a) Identify the element in Group II which has only five occupied electron shells [1]

(b) Name and describe the type of bonding found in elements in Group II. name description [4]

(c) When a piece of calcium is added to some cold water containing universal indicator a reaction takes place.

(i) Give three observations when this reaction takes place. 1 2 3 [3]

(ii) Name the two products of this reaction and [2]

(d) Calcium burns in oxygen.

(i) State the colour of the flame [1]

(ii) Write the symbol equation for this reaction [2] , ,

(e) Crystals of hydrated calcium nitrate contain water molecules.

The formula of hydrated calcium nitrate is Ca(NO3)2•xH2O.

x is a whole number.

(i) State the term given to the water molecules present in hydrated crystals [1]

(ii) When hydrated calcium nitrate is heated gently, the following reaction occurs. Ca(NO3)2•xH2O Ca(NO3)2 + xH2O

A sample of hydrated calcium nitrate is heated gently. 3.28 g of Ca(NO3)2 forms and the mass of the crystals decreases by 1.44 g.

[M r: Ca(NO3)2, 164; H2O, 18]

Determine the value of x in Ca(NO3)2•xH2O using the following steps. • Calculate the number of moles of Ca(NO3)2 that remain mol • Calculate the number of moles of H2O given off mol • Determine the value of x.

x = [3] [Total: 17] , ,

3 A student makes crystals of the salt sodium sulfate, Na2SO4 . The student reacts 0.200 mol / dm3 dilute sulfuric acid, H2SO4(aq), with aqueous sodium hydroxide, NaOH(aq).

The student uses the following steps.

step 1 The student places 40.0 cm3 of NaOH(aq) into a conical flask. This volume contains 0.0100 moles of NaOH.

step 2 The student adds a few drops of methyl orange indicator to the NaOH(aq) in the conical flask.

step 3 The student adds 0.200 mol / dm3 H2SO4(aq) to the flask until the end-point is reached.

step 4 The student transfers the mixture from the conical flask to an evaporating basin and obtains dry crystals.

(a) Complete the symbol equation for the reaction. Include state symbols. H2SO4(aq) + 2NaOH(aq) Na2SO4( ) + ( )

[2]

(b) State the type of exothermic reaction taking place [1]

(c) Calculate the concentration of NaOH(aq) used in step 1.

concentration of NaOH(aq) = mol / dm3 [1]

(d) Name the item of apparatus the student uses to add H2SO4(aq) in step 3 [1]

(e) Calculate the volume of H2SO4(aq), in cm3, added in step 3.

volume of H2SO4(aq) = cm3 [2]

(f) State the colour change observed in step 3.

from to [2] , ,

(g) The dry crystals formed in step 4 are coloured and not white. This is because the student should do an additional step between step 3 and step 4.

Suggest what the student should do in this additional step to produce white crystals [1]

(h) In step 4, the student gently heats the solution in the evaporating basin until the solution is saturated. The student then stops heating and leaves the hot solution to cool. Crystals start to appear.

(i) Explain the term saturated solution [2]

(ii) Explain why crystals start to appear as the hot solution cools [1]

(iii) Suggest the effect, if any, on the mass of crystals collected in step 4 if the solution in the evaporating basin is allowed to dry without gentle heating [1] [Total: 14] , ,

4 Gaseous sulfur tetrafluoride, SF4 , reacts with steam in a reversible reaction. SF4(g) + 2H2O(g) SO2(g) + 4HF(g) ΔH = –54 kJ / mol

(a) Complete the reaction pathway diagram in Fig. 4.1 for this reaction.

Include in your diagram: • the position and the formulae of the products • an arrow, labelled Ea , to show the activation energy • an arrow, labelled ΔH, to show the enthalpy change of the reaction. SF4(g) + 2H2O(g) energy progress of reaction Fig. 4.1

[4] , ,

(b) The equation for the reaction can be represented as shown in Fig. 4.2. S F H O H F F F H O H H F H F H F H F O S O + + ΔH = –54 kJ / mol Fig. 4.2

Table 4.1 shows some bond energies. Table 4.1 bond S – F O – H H – F bond energy in kJ / mol 330 460 570

Use the bond energies in Table 4.1 and the value of ΔH of the reaction to calculate the S=O bond energy in kJ / mol.

Use the following steps. • Calculate the energy needed to break the bonds in the reactants kJ • Calculate the energy released when the bonds in the products form kJ • Calculate the S=O bond energy kJ / mol

[4] , ,

(c) The equation for the reaction is shown. SF4(g) + 2H2O(g) SO2(g) + 4HF(g) ΔH = –54 kJ / mol

State the effect, if any, on the position of equilibrium when the following changes are made.

Give a reason for each of your answers. • The temperature is increased • The pressure is increased • A catalyst is added [5]

(d) Explain, in terms of collision theory, why reducing the temperature decreases the rate of the reverse reaction [3] [Total: 16] , , Question 5 starts on the next page. , ,

5 This question is about the homologous series of alcohols.

(a) A homologous series is a family of organic compounds whose members have the same general formula.

(i) State the general formula for alcohols [1]

(ii) Give one other characteristic that is the same for all members of a homologous series [1]

(b) Ethanol can be manufactured by two methods: • method 1 uses glucose as the starting material • method 2 uses ethene as the starting material.

(i) Complete Table 5.1. Table 5.1 method 1 glucose as starting material method 2 ethene as starting material typical temperature used / °C two other essential conditions 1 1 2 2

[6]

(ii) Write the symbol equation for the reaction in method 1 [2]

(iii) Write the symbol equation for the reaction in method 2 [2]

(c) Butane-1,4-diol has the structural formula HO — CH2 — CH2 — CH2 — CH2 — OH.

(i) Deduce the molecular formula of butane-1,4-diol [1]

(ii) Butane-1,4-diol reacts with ethanoic acid.

Determine the number of moles of ethanoic acid which react fully with one mole of butane-1,4-diol [1] , ,

(d) Butanedioic acid has the structural formula HOOC — CH2 — CH2 — COOH.

(i) Deduce the empirical formula of butanedioic acid [1]

(ii) Name the gas formed when butanedioic acid reacts with sodium [1]

(e) Butane-1,4-diol can be represented as shown. HO OH

Butanedioic acid can be represented as shown. HOOC COOH

Butane-1,4-diol reacts with butanedioic acid to form a polymer.

(i) Draw two repeat units of the polymer formed from the reaction of butane-1,4-diol with butanedioic acid.

Show all the atoms and all the bonds in the ester linkages.

[3]

(ii) State the type of polymerisation when butane-1,4-diol reacts with butanedioic acid [1] [Total: 20] , , Group The Periodic Table of Elements 1 H hydrogen 1 2 He helium 4 I II III IV V VI VII VIII 3 Li lithium 7 4 Be beryllium 9 atomic number atomic symbol Key name relative atomic mass 11 Na sodium 23 12 Mg magnesium 24 19 K potassium 39 20 Ca calcium 40 37 Rb rubidium 85 38 Sr strontium 88 55 Cs caesium 133 56 Ba barium 137 87 Fr francium – 88 Ra radium – 5 B boron 11 13 Al aluminium 27 31 Ga gallium 70 49 In indium 115 81 Tl thallium 204 113 Nh nihonium – 6 C carbon 12 14 Si silicon 28 32 Ge germanium 73 50 Sn tin 119 82 Pb lead 207 22 Ti titanium 48 40 Zr zirconium 91 72 Hf hafnium 178 104 Rf rutherfordium – 23 V vanadium 51 41 Nb niobium 93 73 Ta tantalum 181 105 Db dubnium – 24 Cr chromium 52 42 Mo molybdenum 96 74 W tungsten 184 106 Sg seaborgium – 25 Mn manganese 55 43 Tc technetium – 75 Re rhenium 186 107 Bh bohrium – 26 Fe iron 56 44 Ru ruthenium 101 76 Os osmium 190 108 Hs hassium – 27 Co cobalt 59 45 Rh rhodium 103 77 Ir iridium 192 109 Mt meitnerium – 28 Ni nickel 59 46 Pd palladium 106 78 Pt platinum 195 110 Ds darmstadtium – 29 Cu copper 64 47 Ag silver 108 79 Au gold 197 111 Rg roentgenium – 30 Zn zinc 65 48 Cd cadmium 112 80 Hg mercury 201 112 Cn copernicium – 114 Fl flerovium – 116 Lv livermorium – 7 N nitrogen 14 15 P phosphorus 31 33 As arsenic 75 51 Sb antimony 122 83 Bi bismuth 209 115 Mc moscovium – 8 O oxygen 16 16 S sulfur 32 34 Se selenium 79 52 Te tellurium 128 84 Po polonium – 9 F fluorine 19 17 Cl chlorine 35.5 35 Br bromine 80 53 I iodine 127 85 At astatine – 117 Ts tennessine – 10 Ne neon 20 18 Ar argon 40 36 Kr krypton 84 54 Xe xenon 131 86 Rn radon – 118 Og oganesson – 21 Sc scandium 45 39 Y yttrium 89 57–71 lanthanoids 89–103 actinoids 57 La lanthanum 139 89 Ac lanthanoids actinoids The volume of one mole of any gas is 24 dm3 at room temperature and pressure (r.t.p.). actinium – 58 Ce cerium 140 90 Th thorium 232 59 Pr praseodymium 141 91 Pa protactinium 231 60 Nd neodymium 144 92 U uranium 238 61 Pm promethium – 93 Np neptunium – 62 Sm samarium 150 94 Pu plutonium – 63 Eu europium 152 95 Am americium – 64 Gd gadolinium 157 96 Cm curium – 65 Tb terbium 159 97 Bk berkelium – 66 Dy dysprosium 163 98 Cf californium – 67 Ho holmium 165 99 Es einsteinium – 68 Er erbium 167 100 Fm fermium – 69 Tm thulium 169 101 Md mendelevium – 70 Yb ytterbium 173 102 No nobelium – 71 Lu lutetium 175 103 Lr lawrencium – , ,