M1 Chemistry: Properties and Structure of Matter Practice Exam

This is an original tutorgum practice paper for mixed revision of the chemistry-2017 legacy M1 unit, Properties and Structure of Matter. It assesses only the supplied syllabus scope. It is not a full-course HSC exam and is not an official, endorsed, past, or predicted NESA paper.

01Exam instructions

Suggested working time: 120 minutes
Suggested reading time: 5 minutes
Total marks: 80

Permitted equipment

• Scientific calculator
• Periodic table
• Ruler, pens, and pencils

These are the conditions for this practice paper, not claims about current official NESA examination rules.

Assessed topics

The paper covers atomic structure, isotopes, nuclear reactions, electronic configurations, periodic trends, atomic spectra, flame tests, bonding, molecular structures, nomenclature, mixtures, separation techniques, and percentage composition by mass.

The assessment is restricted to the supplied chemistry-2017 legacy M1 scope. It does not assess material from other modules or combine content from other syllabus editions.

Working requirements

1. Show your working for calculations and explain your reasoning in written responses.
2. Include appropriate units, nuclear symbols, charges, and chemical formulae.
3. Give calculated percentages to three significant figures. Use a precision justified by the supplied data for other numerical answers.
4. In nuclear equations, conserve both nucleon number and charge number.
5. Use the approximate electronegativity classifications and all numerical data supplied in this paper.

Reference data

Approximate electronegativities:

For this paper, use the following approximate electronegativity-difference model:

These cut-offs are a simplified classification model, not universal boundaries between bonding types.

Relative atomic masses for Question 12:

02Questions

Question 1 (7 marks)

(a) Atomic structure (3 marks)

Consider the chloride ion represented by \({}^{37}_{17}\mathrm{Cl}^{-}\).

Determine the number of protons, neutrons, and electrons in this ion. (3 marks)

(b) Electronic configuration and Lewis symbols (2 marks)

Write the ground-state spdf electronic configuration of a neutral sulfur atom, and draw its Lewis dot symbol. (2 marks)

(c) Ion formation (2 marks)

Write the ground-state spdf electronic configuration of the sulfide ion, \(\mathrm{S}^{2-}\), and draw its Lewis dot symbol. (2 marks)

Question 2 (6 marks)

An element, copper, has atomic number 29. A sample contains two naturally occurring isotopes:

(a) Write the nuclear symbol for copper-65, including its mass number and atomic number. (1 mark)

(b) Calculate the relative atomic mass of this copper sample. Show your working. (3 marks)

(c) Explain why copper isotopes occupy the same position in the periodic table but can differ in mass and nuclear stability. (2 marks)

Question 3 (11 marks)

(a) Alpha decay (2 marks)

Complete the following nuclear equation by identifying the missing nuclide.

\[ {}^{238}_{92}\mathrm{U} \rightarrow {}^{4}_{2}\mathrm{He} + {}^{A}_{Z}X \]

(2 marks)

(b) Beta-minus decay (2 marks)

Complete the following nuclear equation.

\[ {}^{14}_{6}\mathrm{C} \rightarrow {}^{A}_{Z}X + {}^{0}_{-1}e \]

Identify the missing nuclide and write the completed equation. (2 marks)

(c) Another nuclear reaction (2 marks)

Complete the following nuclear reaction by identifying the missing particle.

\[ {}^{27}_{13}\mathrm{Al} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{30}_{15}\mathrm{P} + {}^{A}_{Z}X \]

Show how conservation of nucleon number and charge number determines the missing particle. (2 marks)

(d) Types of radioactive radiation (3 marks)

For each type of radiation below, state its physical nature and electrical charge.

• Alpha radiation
• Beta-minus radiation
• Gamma radiation

(3 marks)

(e) Radiation properties (2 marks)

Rank alpha, beta, and gamma radiation from least to most penetrating. Identify which type has the greatest ionising ability. (2 marks)

Question 4 (6 marks)

(a) Atomic models and spectra (2 marks)

A hydrogen discharge lamp produces several narrow emission lines rather than a continuous spectrum.

Explain what this observation indicates about electron energies, and contrast the Bohr model with the Schrödinger model of the atom. (2 marks)

(b) Flame tests (2 marks)

Explain why different metal ions can produce characteristic colours in a flame test. (2 marks)

(c) Evaluating an investigation (2 marks)

A student uses flame tests to identify an unknown metal ion.

State one limitation of this method and one practical improvement that could make the identification more reliable. (2 marks)

Question 5 (7 marks)

The following table gives approximate first ionisation energies for Period 3 elements.

(a) Describe the overall trend in first ionisation energy from sodium to argon. (1 mark)

(b) Explain this trend using nuclear charge and electron shielding. (2 marks)

(c) Explain why aluminium has a lower first ionisation energy than magnesium. (1 mark)

(d) Explain why sulfur has a lower first ionisation energy than phosphorus. (1 mark)

(e) Compare the atomic radii of Na and Cl, and of Li and K. State which atom is larger in each pair and explain each comparison using atomic structure. (2 marks)

Question 6 (5 marks)

(a) Reactivity with water (2 marks)

Potassium and sodium are both Group 1 metals.

Predict which metal reacts more vigorously with cold water. Explain your prediction using electron arrangement and shielding. (2 marks)

(b) Classifying elements (1 mark)

Classify each element as a metal, non-metal, or noble gas:

• Mg
• S
• Ar

(1 mark)

(c) States of matter in a group (2 marks)

At room temperature, chlorine is a gas, bromine is a liquid, and iodine is a solid.

Describe and explain this trend down Group 17. (2 marks)

Question 7 (6 marks)

Use the electronegativity data and classification model provided in the exam instructions.

(a) Bond classification (3 marks)

Classify each bond as predominantly ionic, polar covalent, or non-polar covalent.

(i) Na-Cl (1 mark)

(ii) C-O (1 mark)

(iii) Cl-Cl (1 mark)

(b) Bond polarity (1 mark)

Show the partial charges on the H-Cl bond and identify the direction in which the bonding electron density is shifted. (1 mark)

(c) Electronegativity trends (2 marks)

Explain how electronegativity changes across a period and down a group. Refer to nuclear attraction, atomic radius, and shielding in your answer. (2 marks)

Question 8 (8 marks)

(a) Ionic structures (2 marks)

Solid sodium chloride does not conduct electricity, but molten sodium chloride does.

Explain this difference using the arrangement and movement of ions. (2 marks)

(b) Metallic structures (2 marks)

Copper is a good electrical conductor and can be hammered into different shapes without shattering.

Explain both properties using the metallic bonding model. (2 marks)

(c) Giant covalent and molecular structures (2 marks)

Explain why much more energy is needed to separate the particles in silicon dioxide than in solid carbon dioxide during sublimation or other phase changes. Refer to their structures and bonding. (2 marks)

(d) Allotropy (2 marks)

Diamond and graphite are two structural forms of carbon.

Define allotropy and explain one reason why diamond and graphite differ in electrical conductivity. (2 marks)

Question 9 (7 marks)

(a) Lewis dot diagrams (2 marks)

Draw a Lewis dot diagram for an ammonia molecule, \(\mathrm{NH_3}\). Show all bonding pairs and lone pairs of electrons. (2 marks)

(b) Molecular shapes (2 marks)

Predict the molecular shapes of \(\mathrm{NH_3}\) and \(\mathrm{CO_2}\). Use electron-pair repulsion to explain each shape. (2 marks)

(c) Intermolecular and intramolecular forces (3 marks)

Water has a much higher boiling point than hydrogen sulfide, \(\mathrm{H_2S}\).

Distinguish between intermolecular and intramolecular forces, then explain this boiling-point difference. (3 marks)

Question 10 (6 marks)

(a) Naming compounds (2 marks)

Give the correct names for:

(i) \(\mathrm{FeCl_3}\) (1 mark)

(ii) \(\mathrm{N_2O_5}\) (1 mark)

(b) Writing formulae (2 marks)

Write the correct chemical formulae for:

(i) Calcium nitrate (1 mark)

(ii) Sulfur dioxide (1 mark)

(c) Oxidation states and nomenclature (2 marks)

Name \(\mathrm{CuO}\) using the appropriate oxidation-state naming convention. Determine the oxidation state of copper and show how you obtained it. (2 marks)

Question 11 (8 marks)

(a) Classifying substances and mixtures (3 marks)

Classify each sample as a pure substance, a homogeneous mixture, or a heterogeneous mixture.

(i) Distilled water (1 mark)

(ii) A solution of sodium chloride in water (1 mark)

(iii) Oil and water in a beaker after the mixture has settled (1 mark)

(b) Separation of a mixture (3 marks)

A mixture contains sand, dissolved sodium chloride, and water.

Describe a sequence of steps that would allow you to obtain dry sand and pure water. (3 marks)

(c) Crystallisation (1 mark)

If the aim were instead to recover solid sodium chloride from the salt solution, name a suitable separation technique. (1 mark)

(d) Chromatography (1 mark)

Name a technique that can separate the coloured dyes in ink, and state why the dyes separate during the process. (1 mark)

Question 12 (3 marks)

(a) Percentage composition of a compound (2 marks)

Calculate the percentage by mass of oxygen in pure sodium sulfate, \(\mathrm{Na_2SO_4}\).

Use the relative atomic masses supplied in the exam instructions. Show your working. (2 marks)

(b) Percentage composition of a mixture (1 mark)

A solution is prepared by mixing 18.0 g of sodium chloride with 42.0 g of water.

Calculate the percentage by mass of sodium chloride in the solution. (1 mark)

03Worked solutions and marking guide

Solution 1

(a) Atomic structure

The nuclear symbol is \({}^{37}_{17}\mathrm{Cl}^{-}\).

• Protons: the atomic number is 17, so there are 17 protons.
• Neutrons: mass number minus atomic number gives \(37-17=20\) neutrons.
• Electrons: the negative charge means the ion has one more electron than the neutral atom, giving \(17+1=18\) electrons.

..

..

Solution 2

(a) Nuclear symbol

Copper has atomic number 29. Copper-65 has mass number 65.

\[ \boxed{{}^{65}_{29}\mathrm{Cu}} \]

Marking guide (1 mark):

• 1 mark for the correct nuclear symbol.

(b) Relative atomic mass

Relative atomic mass is the weighted mean of the isotopic masses, using the fractional abundance of each isotope.

Convert the percentages to fractions:

\[ 69.2\%=0.692,\qquad 30.8\%=0.308 \]

Then calculate:

\[ \begin{aligned} A_r &= (62.93\times0.692)\\ &\quad +(64.93\times0.308)\\ &=43.54756+19.99844\\ &=63.546 \end{aligned} \]

Therefore, to four significant figures:

\[ \boxed{A_r=63.55} \]

The relative atomic mass has no unit.

Marking guide (3 marks):

• 1 mark for using the fractional abundances correctly.
• 1 mark for the correct weighted-average calculation.
• 1 mark for the final value, approximately 63.55.

(c) Isotope position and stability

Isotopes of the same element have the same number of protons. They therefore have the same atomic number and occupy the same position in the periodic table.

Their neutron numbers differ, which changes their mass numbers. Differences in neutron-to-proton ratio can also affect nuclear stability, so some isotopes are stable while others are radioactive.

Marking guide (2 marks):

• 1 mark for linking the same proton number to the same element and periodic-table position.
• 1 mark for explaining that different neutron numbers change mass and can affect nuclear stability.

Solution 3

(a) Alpha decay

Alpha decay releases a helium nucleus, \({}^{4}_{2}\mathrm{He}\).

The missing nuclide must have mass number:

\[ A=238-4=234 \]

Its atomic number must be:

\[ Z=92-2=90 \]

Element 90 is thorium, so the completed equation is:

\[ \boxed{ {}^{238}_{92}\mathrm{U} \rightarrow {}^{4}_{2}\mathrm{He} + {}^{234}_{90}\mathrm{Th} } \]

Check both conserved quantities:

\[ 238=4+234 \]

\[ 92=2+90 \]

Marking guide (2 marks):

• 1 mark for conserving nucleon number.
• 1 mark for identifying the correct thorium nuclide and conserving charge number.

(b) Beta-minus decay

The nucleon number remains 14 because a beta-minus particle has mass number zero.

The atomic number must satisfy:

\[ 6=Z+(-1) \]

Therefore, \(Z=7\), which identifies nitrogen.

The completed equation is:

\[ \boxed{ {}^{14}_{6}\mathrm{C} \rightarrow {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}e } \]

The total nucleon number is 14 on both sides, and the total charge number is 6 on both sides.

Marking guide (2 marks):

• 1 mark for identifying nitrogen-14 as the daughter nuclide.
• 1 mark for the completed, balanced equation.

(c) Another nuclear reaction

The total nucleon number before the reaction is:

\[ 27+4=31 \]

The phosphorus product has nucleon number 30, so the missing particle must have nucleon number:

\[ A=31-30=1 \]

The total charge number before the reaction is:

\[ 13+2=15 \]

Phosphorus already has charge number 15, so the missing particle has charge number zero.

The missing particle is a neutron, \({}^{1}_{0}\mathrm{n}\).

The completed equation is:

\[ \boxed{ {}^{27}_{13}\mathrm{Al} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{30}_{15}\mathrm{P} + {}^{1}_{0}\mathrm{n} } \]

Marking guide (2 marks):

• 1 mark for identifying the missing neutron.
• 1 mark for showing that both nucleon number and charge number are conserved.

(d) Types of radiation

Marking guide (3 marks):

• 1 mark for the nature and charge of alpha radiation.
• 1 mark for the nature and charge of beta-minus radiation.
• 1 mark for the nature and charge of gamma radiation.

(e) Radiation properties

The order of increasing penetrating ability is:

\[ \boxed{\alpha < \beta < \gamma} \]

Alpha radiation has the greatest ionising ability of the three. Its relatively large charge and mass make it effective at ionising atoms, although it loses energy over a short distance.

Marking guide (2 marks):

• 1 mark for the correct order of penetrating ability.
• 1 mark for identifying alpha radiation as the most strongly ionising.

Solution 4

(a) Atomic models and spectra

An emission line occurs when an atom emits a photon with a particular energy. The presence of distinct lines rather than a continuous spectrum indicates that the allowed electron energy changes are discrete, or quantised.

The Bohr model places electrons in specific allowed energy levels, represented by fixed orbits in the model. The Schrödinger model instead describes electrons using orbitals, which represent probability distributions for where an electron may be found. It does not describe electrons as travelling along fixed circular paths.

Marking guide (2 marks):

• 1 mark for linking discrete emission lines to quantised energy changes.
• 1 mark for contrasting Bohr’s allowed orbits or levels with Schrödinger’s probability-based orbitals.

(b) Flame tests

Energy from the flame excites electrons in the metal ions, moving them into higher-energy states. When the electrons return to lower-energy states, they emit photons.

The photon energies correspond to particular wavelengths. Some of these wavelengths fall in the visible region, producing colours characteristic of the ions involved.

Marking guide (2 marks):

• 1 mark for explaining electron excitation and the return to lower energy.
• 1 mark for linking the emitted photon energies to characteristic visible colours.

(c) Evaluating an investigation

One limitation is that contamination can produce misleading colours. For example, traces of sodium may produce a strong yellow emission that masks the colour from another ion.

An improvement is to clean the wire thoroughly between samples or use a spectroscope to examine the emitted wavelengths rather than relying only on visual colour judgement.

Other valid limitations and improvements are acceptable.

Marking guide (2 marks):

• 1 mark for a valid limitation.
• 1 mark for a practical improvement that addresses an experimental limitation.

Solution 5

(a) Overall trend

First ionisation energy generally increases from sodium to argon, although the increase is not perfectly smooth.

Marking guide (1 mark):

• 1 mark for identifying the overall increase.

(b) Explanation of the trend

Across Period 3, the number of protons increases. Added electrons enter the same principal energy level, so shielding does not increase enough to cancel the rise in nuclear charge.

The stronger attraction between the nucleus and the outer electrons makes an electron harder to remove. More energy is therefore required for first ionisation.

Marking guide (2 marks):

• 1 mark for identifying the increase in nuclear charge and the broadly similar shielding.
• 1 mark for linking stronger nuclear attraction to higher first ionisation energy.

(c) Magnesium compared with aluminium

Magnesium has a \(3s^2\) outer configuration, while aluminium has a \(3s^2\,3p^1\) outer configuration.

The electron removed from aluminium is in the higher-energy \(3p\) subshell, so it is easier to remove than a \(3s\) electron from magnesium.

Marking guide (1 mark):

• 1 mark for linking aluminium’s lower value to the higher-energy \(3p\) electron.

(d) Phosphorus compared with sulfur

Phosphorus has a \(3p^3\) arrangement with one electron in each of its three \(p\) orbitals. Sulfur has a \(3p^4\) arrangement, so one orbital contains a pair of electrons.

Repulsion between the paired electrons makes an electron in sulfur easier to remove, explaining why sulfur has a slightly lower first ionisation energy.

Marking guide (1 mark):

• 1 mark for explaining the difference using the paired-electron repulsion in sulfur.

(e) Atomic radii

• Na is larger than Cl. Across Period 3, electrons enter the same main shell while nuclear charge increases, pulling the electron cloud closer to the nucleus.
• K is larger than Li. Moving down Group 1 adds an occupied electron shell and increases shielding, so the outer electrons are farther from the nucleus.

Marking guide (2 marks):

• 1 mark for the Na-Cl comparison with an appropriate reason.
• 1 mark for the Li-K comparison with an appropriate reason.

Solution 6

(a) Reactivity with water

Potassium reacts more vigorously with cold water than sodium.

Both metals lose their outer electron during the reaction. Down Group 1, the outer electron is farther from the nucleus and more shielded by inner electrons. It is therefore easier to remove from potassium.

The reactions can be represented by:

\[ \ce{2Na + 2H2O -> 2NaOH + H2} \]

\[ \ce{2K + 2H2O -> 2KOH + H2} \]

Marking guide (2 marks):

• 1 mark for predicting that potassium reacts more vigorously.
• 1 mark for explaining the trend using distance, shielding, and ease of electron removal.

(b) Classifying elements

• Mg: metal
• S: non-metal
• Ar: noble gas

Marking guide (1 mark):

• 1 mark for all three classifications.

(c) States of matter in Group 17

Down Group 17, the halogens change from gases to liquids to solids at room temperature.

The molecules become larger and contain more electrons down the group. Their electron clouds become more polarisable, strengthening London dispersion forces between molecules. More energy is needed to separate the molecules, so their boiling points increase.

Marking guide (2 marks):

• 1 mark for describing the trend in physical state or boiling point.
• 1 mark for explaining the trend using stronger London dispersion forces.

Solution 7

(a) Bond classification

Calculate the electronegativity difference for each bond.

(i) Na-Cl:

\[ \Delta EN=3.0-0.9=2.1 \]

This exceeds 1.7, so the bond is classified as predominantly ionic.

(ii) C-O:

\[ \Delta EN=3.5-2.5=1.0 \]

This lies within the supplied polar-covalent range.

(iii) Cl-Cl:

\[ \Delta EN=3.0-3.0=0 \]

The electrons are shared equally in this bond, so it is non-polar covalent.

Marking guide (3 marks):

• 1 mark for correctly classifying Na-Cl.
• 1 mark for correctly classifying C-O.
• 1 mark for correctly classifying Cl-Cl.

(b) Bond polarity

The hydrogen atom is less electronegative than chlorine. The bond can be represented as:

\[ \mathrm{H^{\delta+}-Cl^{\delta-}} \]

The bonding electron density is shifted towards chlorine, giving it a partial negative charge and hydrogen a partial positive charge.

Marking guide (1 mark):

• 1 mark for the correct partial charges and direction of electron-density shift.

(c) Electronegativity trends

Across a period, electronegativity generally increases because nuclear charge increases while shielding changes relatively little. The smaller atomic radius also allows the nucleus to attract bonding electrons more strongly.

Down a group, electronegativity generally decreases because the bonding electrons are farther from the nucleus and experience greater shielding from inner electrons.

Marking guide (2 marks):

• 1 mark for explaining the increase across a period.
• 1 mark for explaining the decrease down a group.

Solution 8

(a) Ionic structures

In solid sodium chloride, the ions occupy fixed positions in a giant ionic lattice. Although the ions carry charge, they cannot move freely through the solid, so the solid does not conduct electricity.

When sodium chloride melts, the lattice is disrupted and the ions become mobile. These moving ions carry electrical charge, allowing the molten substance to conduct.

Marking guide (2 marks):

• 1 mark for explaining why ions cannot move freely in the solid.
• 1 mark for explaining that mobile ions carry charge in the molten substance.

(b) Metallic structures

Copper contains a lattice of positive metal ions surrounded by delocalised electrons.

The delocalised electrons can move through the structure and carry electrical charge, explaining copper’s conductivity. The layers of metal ions can also slide past one another while the attraction between the ions and the delocalised electrons is maintained. This allows copper to be hammered into shape without shattering.

Marking guide (2 marks):

• 1 mark for linking delocalised electrons to electrical conductivity.
• 1 mark for linking the movement of metal-ion layers and maintained metallic attraction to malleability.

(c) Giant covalent and molecular structures

Silicon dioxide forms a giant covalent network. Strong covalent bonds extend throughout the structure, so substantial energy is needed to disrupt the network.

Solid carbon dioxide consists of discrete \(\mathrm{CO_2}\) molecules. The covalent bonds within each molecule are strong, but the attractions between separate molecules are much weaker. During sublimation, the molecules separate without breaking their internal covalent bonds.

Consequently, much less energy is needed to separate the particles in solid carbon dioxide than to disrupt the silicon dioxide network.

Marking guide (2 marks):

• 1 mark for identifying the giant covalent network and its strong extended bonding.
• 1 mark for identifying discrete \(\mathrm{CO_2}\) molecules and the weaker intermolecular attractions.

(d) Allotropy

Allotropy is the existence of two or more different structural forms of the same element in the same physical state.

In diamond, each carbon atom forms four covalent bonds. Its valence electrons are involved in these bonds, so diamond has no freely moving electrons to carry electrical charge.

In graphite, each carbon atom forms three covalent bonds, leaving one electron per atom delocalised within the layered structure. These electrons can move and carry charge, so graphite conducts electricity.

Marking guide (2 marks):

• 1 mark for defining allotropy.
• 1 mark for explaining the conductivity difference using the structures and electron arrangements of diamond and graphite.

Solution 9

(a) Lewis dot diagram

Nitrogen has five valence electrons, and each hydrogen has one. Nitrogen forms three single covalent bonds with three hydrogen atoms, leaving one lone pair on the nitrogen atom.

One correct Lewis diagram is:

..

H – N – H
|
H

The three N-H bonds each represent a shared pair of electrons. The two dots above nitrogen represent its lone pair.

Marking guide (2 marks):

• 1 mark for nitrogen in the centre with three single N-H bonds.
• 1 mark for the lone pair on nitrogen.

(b) Molecular shapes

Ammonia, \(\mathrm{NH_3}\), has three bonding pairs and one lone pair around its central nitrogen atom. The electron pairs repel one another, giving the molecule a trigonal pyramidal shape.

Carbon dioxide, \(\mathrm{CO_2}\), has two regions of electron density around its central carbon atom, one for each double bond. These regions repel one another and arrange themselves on opposite sides of the carbon atom, giving a linear molecule.

Marking guide (2 marks):

• 1 mark for the trigonal pyramidal shape of ammonia, with a suitable electron-pair explanation.
• 1 mark for the linear shape of carbon dioxide, with a suitable electron-pair explanation.

(c) Intermolecular and intramolecular forces

Intramolecular forces are the bonds within a molecule. In water, these are the covalent O-H bonds.

Intermolecular forces act between separate molecules. Water molecules form hydrogen bonds with one another. Hydrogen sulfide does not form comparably strong hydrogen bonding, so its intermolecular attractions are weaker overall.

More energy is needed to overcome the intermolecular attractions between water molecules. Water therefore has a much higher boiling point than hydrogen sulfide. Boiling separates the molecules; it does not normally break the covalent O-H bonds within each water molecule.

Marking guide (3 marks):

• 1 mark for distinguishing intramolecular bonds from intermolecular forces.
• 1 mark for identifying hydrogen bonding in water and comparing it with hydrogen sulfide.
• 1 mark for linking stronger intermolecular attractions to water’s higher boiling point.

Solution 10

(a) Naming compounds

(i) \(\mathrm{FeCl_3}\): iron(III) chloride.

Each chloride ion has a charge of \(-1\). Three chloride ions give a total charge of \(-3\), so iron must have oxidation state \(+3\).

(ii) \(\mathrm{N_2O_5}\): dinitrogen pentoxide.

This is a molecular compound, so prefixes indicate the numbers of nitrogen and oxygen atoms.

Marking guide (2 marks):

• 1 mark for iron(III) chloride.
• 1 mark for dinitrogen pentoxide.

(b) Writing formulae

(i) Calcium nitrate

Calcium forms \(\mathrm{Ca^{2+}}\), and nitrate is \(\mathrm{NO_3^-}\). Two nitrate ions are needed to balance one calcium ion.

\[ \boxed{\mathrm{Ca(NO_3)_2}} \]

(ii) Sulfur dioxide

The prefix di- indicates two oxygen atoms.

\[ \boxed{\mathrm{SO_2}} \]

Marking guide (2 marks):

• 1 mark for \(\mathrm{Ca(NO_3)_2}\).
• 1 mark for \(\mathrm{SO_2}\).

(c) Oxidation states and nomenclature

The compound \(\mathrm{CuO}\) is copper(II) oxide.

Oxygen has oxidation state \(-2\) in this compound. Since the compound is neutral, the oxidation states must sum to zero:

\[ x+(-2)=0 \]

\[ x=+2 \]

Copper therefore has oxidation state \(+2\), which is indicated by the Roman numeral II.

Marking guide (2 marks):

• 1 mark for the name copper(II) oxide.
• 1 mark for calculating the \(+2\) oxidation state using charge balance.

Solution 11

(a) Classifying substances and mixtures

(i) Distilled water is a pure substance, assuming it contains only \(\mathrm{H_2O}\).

(ii) Sodium chloride solution is a homogeneous mixture. Its composition is uniform throughout the solution, but it contains more than one substance.

(iii) Settled oil and water form a heterogeneous mixture. The composition is not uniform throughout, and two distinct liquid phases are present.

Marking guide (3 marks):

• 1 mark for classifying distilled water as a pure substance.
• 1 mark for classifying salt solution as a homogeneous mixture.
• 1 mark for classifying settled oil and water as a heterogeneous mixture.

(b) Separation of a mixture

A suitable procedure is:

1. Filter the mixture. The insoluble sand remains as the residue, while the salt solution passes through as the filtrate.
2. Rinse the sand residue with a small amount of distilled water to remove adhering salt solution, then dry it to obtain dry sand.
3. Distil the filtrate, including the rinse water. Collect and condense the water vapour to obtain pure water, while the dissolved sodium chloride remains in the distillation flask.

Filtration works because the insoluble sand particles are retained by the filter, while dissolved ions pass through. Distillation separates water from dissolved salt by using their different volatility.

Marking guide (3 marks):

• 1 mark for filtration to separate sand from the solution.
• 1 mark for rinsing and drying the sand to obtain a clean, dry sample.
• 1 mark for distillation and collection of the condensed water.

(c) Crystallisation

Crystallisation is a suitable technique for recovering solid sodium chloride from its solution. Some water is evaporated to concentrate the solution, allowing salt crystals to form as the solution cools or loses more water.

Marking guide (1 mark):

• 1 mark for identifying crystallisation as a suitable technique.

(d) Chromatography

Paper chromatography can separate the coloured dyes in ink. The dyes travel different distances because they differ in their attraction to the stationary phase and their solubility in the moving solvent.

Marking guide (1 mark):

• 0.5 marks for naming paper chromatography.
• 0.5 marks for explaining that differences in solubility or attraction to the two phases cause separation.

Solution 12

(a) Percentage composition of sodium sulfate

The formula is \(\mathrm{Na_2SO_4}\).

First calculate its relative formula mass:

\[ \begin{aligned} M_r &= 2(23.0)+32.1+4(16.0)\\ &=46.0+32.1+64.0\\ &=142.1 \end{aligned} \]

The total mass contribution from oxygen is:

\[ 4(16.0)=64.0 \]

The percentage by mass of oxygen is therefore:

\[ \begin{aligned} \%\text{ O} &=\frac{64.0}{142.1}\times100\\ &=45.038\ldots\%\\ &=\boxed{45.0\%} \end{aligned} \]

Marking guide (2 marks):

• 1 mark for the correct relative formula mass and oxygen mass contribution.
• 1 mark for the correct percentage calculation and final answer.

(b) Percentage composition of a mixture

The total mass of the solution is:

\[ m_{\text{solution}}=18.0+42.0=60.0\text{ g} \]

The percentage by mass of sodium chloride is:

\[ \begin{aligned} \%\text{ NaCl} &=\frac{18.0}{60.0}\times100\\ &=\boxed{30.0\%} \end{aligned} \]

Marking guide (1 mark):

• 1 mark for the correct mass-percentage calculation and final answer.

End of practice paper – 80 marks

When checking your work, focus on the reasoning behind each answer. In this unit, the most useful habit is to connect what you observe or calculate to the underlying particles, electron arrangements, structures, and forces.