Revision questions

Water, H₂O

• Symbols use an uppercase first letter and, where present, a lowercase second letter: Co is cobalt, whereas CO is carbon monoxide. For the first twenty elements learn H hydrogen, He helium, Li lithium, Be beryllium, B boron, C carbon, N nitrogen, O oxygen, F fluorine, Ne neon, Na sodium, Mg magnesium, Al aluminium, Si silicon, P phosphorus, S sulfur, Cl chlorine, Ar argon, K potassium and Ca calcium. Group 1 also includes Rb rubidium, Cs caesium and Fr francium; Group 7 includes F fluorine, Cl chlorine, Br bromine, I iodine and At astatine. Use the supplied periodic table for other specified elements, including Fe iron, Cu copper and Zn zinc. Very radioactive francium and astatine are names to interpret from the table, not school reaction specimens.

8 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

2Mg + O₂ → 2MgO

• Begin with the actual substance formulae and make an element-count table. Change coefficients to equalize the count for each element. Never change MgO to MgO₂ just to balance oxygen: that would identify a different substance. For hydrogen burning, 2H₂ + O₂ → 2H₂O conserves four hydrogen atoms and two oxygen atoms. State symbols, when required, distinguish solid (s), liquid (l), gas (g) and aqueous solution (aq).

6 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Mg → Mg²⁺ + 2e⁻

• In the magnesium half equation, the right-hand total charge is +2 − 2 = 0, matching neutral Mg. In copper reduction, +2 − 2 = 0 on the left, matching neutral copper. Neutralization can be represented as H⁺(aq) + OH⁻(aq) → H₂O(l); the net charge is zero on each side. Electron numbers must match when combining half equations so no free electrons remain in the overall reaction.

9 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

In the filtrate

• For sand mixed with salt, add water and stir to dissolve the salt, filter the insoluble sand, then gently evaporate some water from the filtrate. Stop concentration before boiling dry and allow the solution to cool; filter and dry the crystals. Washing the sand removes adhering solution, and washing crystals with a little cold suitable solvent removes some surface impurity while limiting dissolution.

75 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Simple distillation

• Paper chromatography separates soluble components according to their different distributions between the moving solvent and the paper. Draw the baseline in pencil above the solvent level, place a small sample spot and let solvent rise. Components that spend more time in the mobile phase generally travel farther under these conditions. Compare spots with known standards using the same solvent and paper; a single spot alone does not prove purity under every possible method.

0.6

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A tiny dense positively charged nucleus

• The nuclear model replaced diffuse positive charge with a dense nucleus and electrons outside it. Bohr added electrons at specific distances or energy levels; his theoretical predictions agreed with observations. Later work identified positively charged protons. Chadwick provided evidence for uncharged neutrons in the nucleus about twenty years after the nuclear idea became accepted. A scientific model can develop further without every earlier observation becoming wrong.

0.05 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Protons

• Net relative charge = proton number − electron number. Losing electrons produces a positive ion; gaining electrons produces a negative ion. Ordinary chemical ion formation changes electrons, not the nucleus. An atom with eleven protons is sodium even when it has ten electrons; its net charge is +1. A model with ten protons is neon, regardless of a similar electron arrangement.

2

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Neutron number

• In nuclide notation the upper-left number is A and the lower-left is Z. Protons=Z; neutrons=A−Z. A neutral atom has Z electrons; a positive ion has lost electrons and a negative ion has gained them. Chlorine-35 and chlorine-37 both have Z=17, with 18 and 20 neutrons respectively. Their mass numbers differ without a change in proton number.

13 neutrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

That isotope contributes more atoms to the sample

• For a model chlorine sample containing 75% chlorine-35 and 25% chlorine-37, Ar=(35×75+37×25)/100=35.5. The mean is closer to 35 because that isotope is more abundant. An ordinary unweighted mean would give 36 and incorrectly treat both abundances as equal. Ar is a relative quantity without a gram unit; it is not the mass number of an individual atom.

10.4

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

2,8,8,2

• The first twenty arrangements are H 1; He 2; Li 2,1; Be 2,2; B 2,3; C 2,4; N 2,5; O 2,6; F 2,7; Ne 2,8; Na 2,8,1; Mg 2,8,2; Al 2,8,3; Si 2,8,4; P 2,8,5; S 2,8,6; Cl 2,8,7; Ar 2,8,8; K 2,8,8,1; Ca 2,8,8,2. Verify every sum against the neutral atom’s proton number. This simplified sequence should not be extrapolated to every later element or used to claim the third shell can never contain more than eight.

17 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Period 3, Group 7

• Group 1 atoms tend to lose one outer electron and form +1 ions; Group 7 atoms can gain one to form −1 ions. Predict similar reaction types within a group, then use the group-specific trend for relative reactivity. Helium has two outer electrons in a full first shell and belongs with Group 0, not Group 2. Hydrogen is an unusual non-metal and should not be described as an alkali metal simply from its printed position.

4 shells

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Later elements matched predicted properties

• Later discoveries filled gaps with elements whose properties agreed with the predictions, supporting the classification. Modern order follows atomic number rather than relative atomic mass. Isotopes explain why average mass need not rise strictly with proton number: different isotope masses and abundances influence the mean. The historical improvement was evidence-based prediction, not knowledge of electron shells that had yet to be discovered.

5 g/cm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It loses one electron

• Sodium, 2,8,1, loses its outer electron to form Na⁺. Chlorine, 2,8,7, gains one to form Cl⁻. Oppositely charged ions can form a compound. Non-metal atoms also share electrons in covalent molecules, rather than always becoming isolated negative ions. Characteristic property patterns have exceptions: graphite conducts electricity and mercury is a liquid metal. Use more than one line of evidence.

6 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Its first shell is full with two electrons

• Neon is 2,8 and argon 2,8,8. Compare these stable structures with an alkali metal’s one outer electron or a halogen’s seven. Unreactivity and boiling point are different properties: a rising boiling point does not imply rising chemical reactivity. Given trend data, predict an ordering; a trend alone rarely determines an exact numerical value for an unmeasured element.

60 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The outer electron is farther away and more shielded, so is lost more easily

• With oxygen, these metals tarnish and burn to form oxygen-containing solids; lithium burns crimson, sodium yellow and potassium lilac. The simple oxide model uses 4Li + O₂ → 2Li₂O; sodium and potassium burning products can include peroxide or superoxide, so do not invent a single oxide formula for every demonstration. Reactivity increases Li→Na→K because the outer electron is farther from the nucleus and more shielded by inner shells, so it is lost more easily despite increased nuclear charge.

5 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Chlorine is more reactive than iodine, while iodine has the higher boiling point

• A halogen gains an electron to form a −1 halide ion when reacting with a metal, giving an ionic salt such as NaCl. With non-metals it generally forms covalent compounds through shared electron pairs, such as HCl. Down the group the outer shell is farther from the nucleus and more shielded, reducing attraction for an incoming electron despite greater nuclear charge. This is the opposite reactivity trend to Group 1, which loses an electron.

254

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Chlorine with potassium iodide solution

• A displacement table should identify the added halogen, starting halide and expected product before recording colour. Bromine in water is orange/brown and iodine solutions can appear brown; observation must be compared with suitable blanks, because the added halogen also has colour. Same-halogen/halide combinations have no net displacement. A missing colour change alone may also reflect low concentration or observation limits.

3 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Higher density and melting point, lower reactivity

• Iron can react with steam under suitable conditions and can rust slowly in oxygen and water; copper does not react with cold water in the Group 1 manner. Many transition metals react with oxygen or halogens when heated, so less reactive does not mean incapable of reaction. Compare the same conditions and specify whether an observation is rapid burning, slow corrosion or no visible change.

24 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Fe²⁺ and Fe³⁺ require different chloride ratios

• Catalysts increase reaction rate without being used up overall. Iron catalyses the Haber process; nickel is used in hydrogenation; manganese dioxide catalyses hydrogen peroxide decomposition. A catalyst changes the route and rate, not the balanced reaction’s overall atom count. Variable charges require matching negative-ion charge: FeCl₂ contains Fe²⁺ and FeCl₃ contains Fe³⁺ with Cl⁻ in each case.

6 ions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Its ions can move

• To explain a bulk property, name the structure, particles, forces and mobile charge carriers. Simple molecular substances can have strong covalent bonds inside molecules but weak attractions between molecules.

18

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Attraction between positive metal ions and delocalised electrons

• Most non-metal elements and compounds of non-metals have covalent bonding, but their structures can be small molecules, very large molecules or giant networks. An ionic bond is the attraction after ion formation, not the electron transfer itself. A covalent pair is shared, not transferred completely to one nucleus. Metal electrons are free to move through the structure rather than assigned to one pair of atoms.

3 ions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

One Mg²⁺ and two Cl⁻ ions with completed outer shells

• Draw final ions in separate square brackets, with charges outside. A dot-and-cross diagram uses different marks for electrons originating from different atoms; transferred electrons keep their original marks. Mg²⁺ has no electrons left in its original third shell, while O²⁻ has six original outer electrons plus two transferred ones. If only outer electrons are shown, state whether the remaining filled shell of the cation is omitted by that convention.

10 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A simplest 1:1 ratio of sodium and chloride ions

• To infer an empirical formula, count ions in a stated representative region and reduce the ratio to the smallest whole numbers. A model with six Mg²⁺ and twelve Cl⁻ gives MgCl₂ after reducing 6:12 to 1:2. Counts in an arbitrary boundary fragment need not themselves give the bulk ratio, so use the representative information or account for shared positions as specified. Knowledge of other named ionic crystal structures is not required.

2

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Three

• Oxygen has six outer electrons, so O₂ has two shared pairs, a double bond, and two lone pairs on each atom. Nitrogen has five outer electrons, so N₂ has three shared pairs, a triple bond, and one lone pair on each atom. A line diagram writes H–H, Cl–Cl, H–Cl, O=O or N≡N. One line represents one shared pair, not one electron.

4 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Three shared pairs and one lone pair on nitrogen

• Covalent substances can consist of small molecules, very large molecules such as polymers, or giant networks such as diamond and silicon dioxide. A polymer has many repeated units linked along each molecule: [–CH₂–CH₂–]ₙ represents poly(ethene), with bonds passing through brackets and n large. A giant network diagram shows only a small part of continuing covalent connections, not a separate molecule with that exact pictured atom count.

60 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Delocalised electrons

• The shared electrons extend throughout the structure, so shifting layers can preserve attraction and allow bending. When a potential difference is applied, mobile electrons carry charge; the positive ions remain in their positions in a solid rather than drifting through the wire. The drawing’s ions and electron symbols are a model of the bulk bonding, not isolated metal ions mixed with a separate substance.

20 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Liquid

• Stronger relevant forces require more energy to overcome and commonly give higher melting and boiling points. Identify the particles and forces before explaining: boiling a molecular liquid overcomes intermolecular attractions, while melting an ionic lattice overcomes strong ion attractions. Bulk properties belong to large collections, not an individual atom that is itself a tiny piece of solid or liquid. At a transition temperature more than one state can coexist.

80 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Forces between particles

• Without attractions, the drawing cannot explain energy needed to separate particles or the different melting and boiling temperatures of substances. Treat the balls as symbols for particles, not literal hard miniature pieces of bulk material. A model may succeed at showing close liquid particles while needing an extended force account to explain boiling. Individual particles do not acquire the substance’s bulk hardness or melting point.

6 particles

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

NaCl(aq)

• For an aqueous precipitation reaction, AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) distinguishes the new solid from dissolved substances. Magnesium reacting with acid can be written Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Correct state labels add evidence about the reaction while the coefficients still conserve atoms.

2 species

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Its ions can move and carry charge

• The carrier is an ion, not a free electron travelling through the salt. Dissolution separates ions into the solution; it does not simply melt the salt. Not every ionic compound is readily soluble, so explain aqueous conductivity only when a dissolved sample is specified. High melting point supports a strong giant structure, while state-dependent conductivity strengthens the ionic interpretation.

3

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Attractions between molecules

• For comparable molecules, intermolecular forces generally increase with molecular size, so larger molecules tend to have higher melting and boiling points. Molecular shape and other interactions also matter; a size trend is not an exact universal temperature rule. A compound such as hydrogen chloride can form ions on dissolving, so the pure molecular substance and its reacting aqueous solution must be distinguished.

35 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Many repeated units linked along a very large molecule

• Distinguish bonds along a chain from attractions between separate chains. The size of polymer molecules gives many opportunities for intermolecular attraction. A simplified chain picture helps explain the solid material, but real polymer properties also depend on chain arrangement and other structural features. Do not replace intermolecular attractions with a claim that every neighbouring chain is necessarily joined by covalent bonds.

100 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Many strong covalent connections in its giant structure must be overcome

• In silicon dioxide, each silicon is linked to four oxygens and each oxygen bridges two silicons, producing the overall Si:O ratio 1:2. A local drawing may show four surrounding oxygens because each is shared with another silicon elsewhere in the network. Graphite has strongly bonded layers but weaker attractions between layers; its high-temperature behaviour and layer sliding concern different connections.

36 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Different-sized atoms disrupt layers and hinder sliding

• Different-sized atoms distort the regular layers and make sliding more difficult, so many alloys are harder than the corresponding pure metal. Hardness is resistance to indentation or scratching, while strength concerns resisting deformation or failure under a load; do not treat the words as identical measurements. An alloy still has metallic bonding and need not have one fixed compound formula.

60 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Mobile delocalised electrons carry charge

• Electrical conduction is transfer of charge; thermal conduction is transfer of energy from a hotter region toward a cooler region. Do not confuse electron drift with the whole wire moving or with electrons being permanently consumed. A metal can conduct thermal energy without being part of an electric circuit. The two explanations share mobile electrons but refer to different observations.

0.3 C/s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It has no delocalised electrons available to carry charge

• Explain one property with the appropriate connection: resistance to deformation follows the rigid many-direction network; high melting point follows energy needed to overcome many strong covalent bonds; non-conductivity follows the absence of mobile charged carriers. A drawing with a carbon surrounded by four neighbours shows local coordination, not a complete diamond molecule.

80 bonds

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Weak attractions between layers without covalent bonds joining them

• Do not use weak interlayer forces to explain the high melting point: that property involves strong covalent bonding. Conversely, naming strong covalent bonds alone does not explain easy layer sliding. Graphite resembles metals in having delocalised electrons, but its bonding network is layered covalent rather than metallic. Conductivity directions in real graphite are not represented fully by one simple flat drawing.

55 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A single hexagonal carbon layer with delocalised electrons

• A composite combines materials so their useful properties can contribute to a designed product. Graphene can reinforce a material, but strength claims need actual data for the particular composite, loading and manufacturing method. Conductivity supports electronic applications without proving every graphene-containing product is automatically a good conductor. A drawn lattice edge is a clipped boundary, not the full bulk bonding environment.

25 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A long hollow cylindrical fullerene

• A hollow molecular cage can be considered for carrying other substances in proposed delivery systems; nanotubes can be used as reinforcing fibres and in electronic materials. A use must be linked to supplied properties and tested for suitability. Do not claim every nanotube has identical conductivity or that a proposed medical use is automatically safe or effective. Graphene is a sheet, graphite stacked layers and a fullerene a hollow shape.

500

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It becomes ten times larger

• For a cube side L, area=6L², volume=L³ and area/volume=6/L. Reducing side length tenfold raises area/volume tenfold. Nanoparticles can have different properties from the same bulk material because a much larger fraction interacts at the surface. More exposed area can improve catalytic effectiveness per mass, but particle shape, aggregation and surface chemistry also affect performance.

0.3 per nm

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Lower quantity was needed in this stated trial; safety still needs evidence

• Consider useful performance, required quantity, cost and durability alongside possible human exposure, environmental release and uncertain long-term effects. Risk depends on the material, particle form, exposure route and amount. A contained particle in a product and an inhalable loose powder need separate exposure evidence. A proposed benefit is not proof of safety, while a possible risk is not proof that harm occurs at every dose.

75 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Use a balanced equation and convert mass to amount

• Calculate the amount available for each reactant and divide by its coefficient. The smaller ratio limits the reaction. Use that reactant to calculate the maximum product before comparing actual yield.

0.05 mol

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Two whole water molecules or formula amounts

• For 2H₂ + O₂ → 2H₂O, the coefficient 2 multiplies an entire formula. The subscript 2 belongs to the preceding element, so one water molecule contains two H atoms and one O atom. Both sides contain four H atoms and two O atoms. Using supplied relative atomic masses H=1 and O=16 gives reactant totals 2×2 + 32 =36 and product total 2×18=36. The coefficients 2:1:2 are not gram ratios.

4 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

40+2×(16+1)

• An element’s percentage by mass is its total relative mass contribution divided by Mr, multiplied by 100. Calcium contributes 40 of the 74 units in Ca(OH)₂, giving 40/74×100=54.1% to three significant figures. Oxygen contributes 32 and hydrogen 2. These percentages describe the compound’s fixed mass proportions, not the proportion of different atoms counted equally.

72.7 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Oxygen from the surroundings becomes part of the oxide

• Identify which material is on the balance at each reading. In the metal example, oxygen joins the weighed solid; in the carbonate example, gas leaves the weighed apparatus. In particle terms the same atoms are redistributed among substances, with some crossing the selected system boundary. A sealed-system total includes all products; an open-residue measurement does not.

11 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Highest reading minus lowest reading

• For 2.40, 2.42 and 2.44 g, the mean is 2.42 g and the full range is 0.04 g. Half-range is 0.02 g, so the stated estimate is 2.42±0.02 g. Here the mean lies midway between the extremes. For an asymmetric distribution the extreme deviations from the mean differ; display the actual results as well as the chosen uncertainty estimate.

0.2 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The number of stated particles

• State what is being counted. A mole of CO₂ molecules contains one mole of carbon atoms and two moles of oxygen atoms, three moles of atoms altogether. A mole of NaCl formula units corresponds to one mole of Na⁺ ions and one mole of Cl⁻ ions in the ionic lattice. Calling either sample simply a mole of particles without specifying which particles can make an answer ambiguous.

0.5 mol

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Convert Mg mass into moles using its molar mass

• For 2Mg + O₂ → 2MgO, supplied molar masses Mg=24, O₂=32 and MgO=40 g per mol give mass proportions 48:32:80. These differ from coefficients 2:1:2. A calculation from one reactant assumes sufficient other reactant and complete reaction unless the question supplies a limiting amount. If the known quantity is product mass, work backwards through the same ratio.

20 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Multiply every term by two to get 4:3:2

• For Mg, O₂ and MgO with molar masses 24, 32 and 40 g per mol, supplied masses 2.4, 1.6 and 4.0 g give 0.10, 0.050 and 0.10 mol. Dividing by 0.050 gives 2:1:2, hence 2Mg + O₂ → 2MgO. The oxygen formula is O₂, so its molar mass is 32, not the atomic value 16.

3

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

HCl

• For Mg + 2HCl → MgCl₂ + H₂, 0.20 mol Mg would require 0.40 mol HCl. If only 0.30 mol HCl is supplied, acid limits and only 0.15 mol Mg can react. Maximum H₂ amount is 0.15 mol, and 0.05 mol Mg remains. This assumes reaction proceeds as written and does not describe incomplete conversion caused by slow rate or equilibrium.

0.25 mol

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

250 cm³ converted to 0.250 dm³

• For a uniformly mixed solution, the amount of solute taken is proportional to the sample volume. A 0.250 dm³ sample of an 8.0 grams per dm³ solution contains 2.0 g solute. The unit calculation cancels dm³ and leaves grams. This shared-tier numerical task is distinct from the embedded Higher-only explanation of how changing mass or solution volume changes concentration.

3 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Concentration halves

• During simple dilution with pure solvent and no losses or reaction, the solute mass stays constant. Therefore c₁V₁=c₂V₂ when both concentrations use the same mass units and both volumes the same volume units. The final volume includes the original sample and added solvent; it is not the volume of water added alone. Dissolution and dilution should not be confused with a chemical transformation of solute.

8 g per dm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Product remains dissolved during separation

• Recovery may be below the maximum because a reversible reaction does not go to completion, product is lost during separation, or reactants undergo side reactions. Some atoms can remain in reactants or enter unwanted products; some desired product can remain dissolved or on apparatus. Conservation concerns all atoms and substances, while yield concerns the recovered desired product.

80 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Multiply by the product molar mass

• For CaCO₃ → CaO + CO₂, the amount ratio CaCO₃:CaO is 1:1. Supplied molar masses are 100 and 56 g per mol. A 10.0 g CaCO₃ sample contains 0.100 mol and can give 0.100 mol CaO, mass 5.60 g. The other 4.40 g is theoretical carbon dioxide, so the product-residue mass is lower than the original carbonate mass even at complete conversion.

14 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Coefficient-weighted relative masses from the balanced equation

• For CaCO₃ → CaO + CO₂ with relative masses 100, 56 and 44, choosing CaO as desired product gives 56/100×100=56%. Choosing CO₂ as the desired product for a different purpose gives 44%. Conservation still accounts for all 100 mass units. A high atom economy can reduce waste and raw-material cost, but it does not alone establish safety, energy demand or commercial suitability.

70 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Comparing supplied criteria against the stated production priority

• Imagine route A with atom economy 90%, yield 50% and a slow rate, and route B with atom economy 70%, yield 90% and a faster rate under supplied comparable conditions. Neither percentage alone decides every use. If equal starting mass could theoretically enter product according to these fractions, recovered desired mass per 100 g starting material is 45 g for A and 63 g for B under this simplified comparison.

120 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

n=cV, then m=nM

• Convert cm³ to dm³ by dividing by 1,000. A 250 cm³ portion of 0.20 moles per dm³ solution contains 0.20×0.250=0.050 mol solute. If solute molar mass is 40 g per mol, that amount weighs 2.0 g. At fixed amount, greater final solution volume means lower molar concentration. At fixed final volume, more solute moles means higher concentration.

0.06 mol

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Multiply acid moles by two

• A pipette supplies the measured sample volume and a burette delivers the titre. The titre is final minus initial burette reading, not the final reading alone. Use consistent close results under the school’s stated concordance rule and exclude a rough trial from the accurate mean. A suitable indicator end-point approximates the required neutralisation condition; it is not evidence that every solution must be pH 7.

0.24 mol per dm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

For gases compared at the same temperature and pressure

• Equal gas amounts occupy equal volumes at identical conditions, so balanced coefficients give gaseous volume ratios. N₂(g) + 3H₂(g) → 2NH₃(g) gives ratio 1:3:2 for gas volumes measured at the same temperature and pressure. This ratio does not apply to a liquid or solid volume. A predicted product volume assumes sufficient reactants and the stated complete-conversion model; real equilibrium can reduce conversion.

4.8 dm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It gains electrons

• Predict products using the specified electrolyte and electrode material. In an aqueous solution, hydrogen or oxygen may form because water-related species compete. Molten salts contain only the ions of the salt.

0.4 mol

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

To avoid dilution by residual water

• Calculate the known amount first, apply the stoichiometric ratio, then divide by the unknown solution volume in cubic decimetres. Use concordant titres as required by the school method and report the accepted values.

22.25 cm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Copper oxide, because it loses oxygen

• For 2CuO + C → 2Cu + CO₂, copper oxide loses oxygen and is reduced, while carbon gains oxygen and is oxidised. The same oxygen atoms leave the oxide and appear in carbon dioxide. The two changes occur together in this reaction. Oxygen is an element; oxide is a compound containing oxygen with another element, so the names are not interchangeable.

3 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Copper

• Potassium, sodium and lithium react vigorously with cold water to produce hydroxides and hydrogen, with potassium generally most vigorous. Calcium also reacts, producing hydrogen and calcium hydroxide. Magnesium reacts very slowly with cold water, while zinc, iron and copper show no appreciable reaction under this school comparison. With dilute hydrochloric or sulfuric acid, Mg, Zn and Fe produce hydrogen, with Mg more vigorous and Fe slower; Cu does not. Very reactive alkali metals are not student acid-test samples.

0.6 cm³ per s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Zn > Fe > Cu

• For supplied results, translate each successful displacement into a comparison arrow: A displaces B implies A>B. Combine comparisons into an order and check for contradictions. If A displaces B and B displaces C, A>B>C is consistent. Lack of visible reaction alone needs care because an unsuitable method, protective layer or short observation time can hide a reaction.

80 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

An oxide of a metal less reactive than carbon

• For 2CuO + C → 2Cu + CO₂, copper oxide is reduced and carbon oxidised in oxygen terms. Carbon cannot be assumed to reduce every metal oxide under useful extraction conditions. Aluminium is more reactive than carbon and its extraction uses electrolysis, not simply this copper-oxide route. Detailed blast-furnace process recall is not required by this section’s stated limits.

18 kg

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Cu²⁺

• Check both atom counts and total charge. The combined left charge is +2 and the right charge is +2. Zinc atoms become positive ions by losing electrons; copper ions become neutral metal atoms by accepting them. Identify the actual species reduced, Cu²⁺, rather than saying the already-neutral copper product gains further electrons.

12 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Zinc sulfate and hydrogen

• The metal must be suitable relative to hydrogen. Copper does not normally release hydrogen from these dilute acids. Do not extend this acid–metal pattern to nitric acid or concentrated acid reactions, which are outside the stated scope and can produce different outcomes. With a metal oxide, the usual products are salt and water instead of hydrogen.

12 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

H⁺

• Charge is +2 on both sides of the net equation. Two H⁺ ions are needed to form one H₂ molecule; gaining one electron each makes their total gain two electrons. The metal product is an ion, while the hydrogen product is a neutral diatomic molecule. Electron gain must be assigned to hydrogen ions, not to chloride ions just because hydrochloric acid was used.

12 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Calcium chloride, water and carbon dioxide

• Use charges to make a neutral formula: Cu²⁺ and SO₄²⁻ give CuSO₄; Mg²⁺ and NO₃⁻ give Mg(NO₃)₂. CuO + H₂SO₄ → CuSO₄ + H₂O is balanced. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ has two chloride ions per calcium ion and one carbon dioxide molecule per carbonate formula unit.

8 ions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

To remove unused insoluble oxide while salt solution passes through

• Allow cooling and filter to remove excess insoluble oxide. The filtrate contains dissolved copper sulfate, so it passes through filter paper. Gently concentrate it in an evaporating basin using a water bath or electric heater, then allow cooling to form crystals. Separate the crystals and dry gently as directed. Evaporating completely to dryness is not the crystal-preparation endpoint.

4.55 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

11

• Neutralisation between acid and alkali forms water: H⁺(aq) + OH⁻(aq) → H₂O(l). The positive and negative charges cancel, and atoms are conserved. Excess acid after mixing leaves the solution acidic; excess alkali leaves it alkaline. Neutralisation does not automatically mean that equal volumes of every pair of solutions will produce pH 7.

6 ions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Final burette reading minus initial reading

• Titre equals final burette reading minus initial reading. Read the appropriate meniscus at eye level, with the burette vertical and the filling funnel removed. A rough trial locates the endpoint; subsequent careful trials supply an accurate mean under the school’s stated concordance rule. Phenolphthalein is pink in alkali and becomes permanently colourless at the handbook’s acid-into-alkali endpoint. Universal indicator is unsuitable for a sharply defined accurate endpoint.

20.35 cm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Acid amount is half alkali amount

• The acquired handbook PDF page 35 correctly states the equation and the 2:1 ratio, but prints a reversed half-mole statement afterwards. The correct inference is acid moles = half alkali moles. The balanced equation gives the correct amount relationship. The common-tier RP2 task still concerns accurate reacting volumes, while this numerical determination is Higher-only.

14.7 g per dm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Strength concerns ionisation; concentration concerns amount per volume

• At a given concentration under comparable conditions, the stronger acid has lower pH because it produces greater hydrogen-ion concentration. Each decrease of one pH unit corresponds to a tenfold increase in H⁺ concentration. A decrease of two units means a hundredfold increase. Use whole-number pH comparisons in this source scope; do not treat pH as a linear concentration scale.

1000 times

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Towards the negative cathode

• Charge travels through the electrolyte by moving ions and through the external circuit by electrons. Positive ions are not positive electrons, and solid salt ions do not become mobile simply because a wire touches the solid. Electrolysis uses electrical energy to drive chemical change. The polarity names here apply to an externally powered electrolytic cell, not a blanket rule for every electrochemical cell.

14 ions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Lead metal

• Identify the positive metal ion and negative non-metal ion from the formula. The cathode product is the metal element, while a halogen anode product is diatomic, such as Br₂ or Cl₂. Do not write bromide as the elemental product. The formula’s ratio is determined by charge, and the product equation must preserve every atom.

16 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

They react with oxygen products and are consumed

• Aluminium forms at the negative cathode. Oxygen associated with oxide discharge at the positive carbon anode reacts with carbon, forming carbon dioxide in the GCSE account. The carbon anode is consumed and needs continual replacement. It is therefore not an inert electrode like the one assumed in a simple product-prediction exercise.

25 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Copper at cathode, oxygen at anode

• Aqueous copper(II) chloride therefore gives copper at the cathode and chlorine at the anode. Sodium chloride solution gives hydrogen and chlorine under the approved handbook conditions. Copper(II) sulfate gives copper and oxygen, while sodium sulfate gives hydrogen and oxygen. State that these are the specified single-solute, inert-electrode predictions; concentration and electrode material can affect actual competing reactions outside this simple rule.

2 solutions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A brown/red deposit on the negative electrode with appropriate supporting evidence

• The acquired handbook uses carbon rods, low-voltage direct current and approved CuCl₂/NaCl solutions. A brown/red deposit supports copper at the negative electrode; gas at the positive electrode bleaches damp blue litmus under the approved test. A bubble observation alone does not identify a gas. Label each observation with electrode sign and keep the products separate.

3 cm³ per min

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

2Cl⁻ → Cl₂ + 2e⁻

• For 4OH⁻ → O₂ + 2H₂O + 4e⁻, oxygen count is four on each side and hydrogen count four on each side. Charge is −4 on both sides. For 2Cl⁻ → Cl₂ + 2e⁻, chlorine is diatomic and the electron total balances the two negative charges. A correct formula is needed before balancing; changing Cl₂ to Cl would misidentify the product.

15 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Endothermic: energy is taken from the surroundings

• Distinguish the reacting chemicals from their surroundings. A thermometer in a reacting solution measures the temperature of that solution, which receives or supplies energy during the chemical change. A temperature rise supports an exothermic interpretation under the stated conditions; a fall supports endothermic behaviour. Heating a vessel externally can obscure this evidence. Not every process involving cooling is a chemical reaction: new substances must also be formed.

7.5 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Endothermic: energy is taken from the surroundings

• Distinguish the reacting chemicals from their surroundings. A thermometer in a reacting solution measures the temperature of that solution, which receives or supplies energy during the chemical change. A temperature rise supports an exothermic interpretation under the stated conditions; a fall supports endothermic behaviour. Heating a vessel externally can obscure this evidence. Not every process involving cooling is a chemical reaction: new substances must also be formed.

7.5 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Extra cooler liquid and heat loss can lower it without further neutralisation warming

• After each addition, replace the lid, stir gently and record the highest temperature. Repeat the whole investigation with fresh starting solutions and calculate means at each added volume. Plot mean temperature against total alkali volume added. Draw appropriate best-fit lines through the rising and falling regions and estimate their intersection. It estimates the peak between discrete readings; do not join every noisy point and call the highest measured point exact.

29.5 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

An endothermic overall change

• In an exothermic profile products lie below reactants, showing an overall energy decrease of the reacting chemicals and transfer to surroundings. In an endothermic profile products lie above reactants, showing energy taken from surroundings. Both profiles can have an activation barrier. The vertical difference between reactants and products is the overall change, distinct from the larger climb to the peak. Horizontal distance is not elapsed time or a measured reaction rate.

65 relative units

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

−250 kJ, exothermic

• Use the balanced equation and actual bond types. A coefficient multiplies all bonds in that molecule. For H₂ + Cl₂ → 2HCl, break one H–H and one Cl–Cl and form two H–Cl bonds. For 2H₂ + O₂ → 2H₂O, break two H–H bonds and one O=O bond, and form four O–H bonds. An O=O double bond uses its supplied double-bond value once; do not double a single-bond value.

-170 kJ

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It produces a potential difference from chemical reactions

• Use supplied readings or reactivity information to compare cells under specified conditions. Do not promise a universal voltage from a metal pair alone: electrolyte choice and conditions also matter. Connecting identical cells in series adds their potential differences in the ideal stated model; reversed orientation subtracts one cell’s contribution. Voltage measures potential difference, not the amount of chemical reactant remaining or the rate of all reactions.

3 V

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

An external current reverses its chemical reaction

• Choose a cell for its intended use. Compare required voltage, available operating time, mass, initial cost, replacement frequency, charging access and disposal information supplied. Repeated use can make a rechargeable option cheaper over time despite greater initial cost. A rarely used emergency device might place more weight on storage performance and immediate availability. These conclusions depend on stated data, not a universal rule that one cell type is always best.

19 £

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Water is the product of the local hydrogen–oxygen reaction

• Compare both devices against a purpose. Supplied information may include mass, operating time, refuelling or charging time, infrastructure, fuel storage and cost. Water is the direct product of the hydrogen–oxygen reaction; hydrogen production and transport may have other environmental effects. A conclusion about total emissions needs data for those stages. Do not infer zero lifecycle carbon emissions merely from the absence of carbon in the local equation.

5 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

H₂ → 2H⁺ + 2e⁻

• Check atoms and charges separately. The hydrogen half equation has four hydrogen atoms on each side and zero net charge because +4 from H⁺ balances −4 from electrons. The oxygen half equation has two oxygen atoms and four hydrogen atoms on each side, again zero net charge. In this generating cell electrons leave the hydrogen anode through the external circuit and reach the oxygen cathode. The anode is negative and the cathode positive while the cell supplies electricity.

16 electrons

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Their rates

• A product-time graph has a steeper gradient where rate is larger. A tangent estimates instantaneous rate; a secant gives average rate over an interval. The final plateau reflects the total collected product under the stated conditions.

0.6 cm³/s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Forward and reverse reaction rates

• Beginning with mostly reactants, the forward rate may be high while the reverse rate is initially low. As products accumulate, reverse reaction becomes possible more often until the rates match under the stated model. A reaction-rate graph should show convergence to the same non-zero rate. A composition graph can level off at different reactant and product amounts. Identify which quantity is on the axis before claiming two curves must meet.

0 particles

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

(30−10)/(60−20)

• Plot time on the horizontal axis and measured quantity on the vertical axis with labelled units and sensible scales. A steeper product-time curve means faster formation; a horizontal plateau means no further measured product forms. A tangent touches the local curve direction at the chosen point and its steepness indicates rate there. Drawing and interpreting tangents is both-tier; calculating their numerical gradients is in the Higher lesson. A straight chord between distant points gives an interval mean, not the local rate.

0.6 cm³ per s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

On the tangent line itself

• First mark the requested time, draw a tangent following the smooth curve’s local direction, then select two readable points on that line. They need not be measured data points or lie on the curve elsewhere. A larger triangle reduces the relative effect of reading uncertainty. Tangent estimates can differ slightly between reasonable drawings. A chord linking two curve points measures an interval mean and must not be presented as the tangent gradient.

0.7 cm³ per s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Rate, without necessarily changing total product

• State exactly what changes and what is controlled. For equal masses of the same carbonate with acid in excess, smaller pieces can reach the same final carbon dioxide amount sooner. If increasing acid concentration also changes the limiting reactant amount, both speed and final amount can change; this is a different comparison. Higher gas pressure is relevant to reacting gases, not automatically a general explanation for every liquid reaction.

1.5 cm³ per s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Using different magnesium masses in the two trials

• Fit the collector and check for unobstructed movement and an open gas path into the collector. Add magnesium, replace the bung promptly and start timing consistently. Record gas volume at suitable intervals, for example every 10 s, until it becomes constant. Plot both volume–time curves on the same axes and compare steepness at comparable times. Identical final volumes require equal limiting magnesium and enough acid; real collection losses or sample differences can change recorded plateaux.

0.9 cm³ per s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

To avoid also changing liquid depth and total mixture volume

• The concentration labels refer to the 50 cm³ pre-acid dilution; adding acid further dilutes the mixture consistently. Maintain total volumes, flask, viewing depth, lighting, temperature and endpoint judgement. Repeat and calculate mean endpoint time. For a comparable fixed turbidity threshold, 1/mean time is a relative rate proxy: a shorter time means a larger proxy. It does not measure a known mass of sulfur per second or the exact completion time of the entire reaction.

0.025 per s

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Frequency of reactant collisions

• Keep frequency distinct from collision energy. At unchanged temperature, raising concentration alone does not mean particles move faster or each has more energy. In a simple stated model with one collision partner held fixed, doubling the other partner’s particle concentration can double encounter opportunities. It is not a universal measured rate law for every mechanism or for simultaneous changes to both reactants. Different reactions require suitable evidence for exact proportional relationships.

3

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Collisions become more frequent and a greater fraction are energetic enough

• A cube model makes the size comparison explicit. A cube side L has area 6L² and volume L³. Cutting it into eight half-side cubes preserves total volume but doubles total area. This model assumes newly exposed faces are accessible and not clumped or coated. Real powders need appropriate containment; aggregate formation can reduce accessible area. Do not claim every ten-degree temperature rise always doubles rate: exact effects depend on reaction and conditions.

48 cm²

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The activation barrier becomes lower

• At the same temperature more colliding particles can meet the smaller activation requirement. This does not mean the catalyst adds energy to every particle or increases temperature by definition. Evidence for catalytic behaviour combines rate increase and regeneration/no net consumption. An unchanged small sample alone does not prove catalysis unless the reaction-rate effect is shown. The overall equation omits a regenerated catalyst from its stoichiometric reactant and product lists.

3

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The products can react to reform ammonium chloride

• The symbol does not mean that all substances must be present in equal quantities or that the reaction continually switches completely from one side to the other. A reversible reaction may reach equilibrium when substances are contained under suitable fixed conditions, but reversibility alone is not proof that an open apparatus has reached equilibrium. Melting/freezing is a reversible physical change; it does not by itself demonstrate this chemical product-to-reactant relationship.

20 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Exothermic with the same transferred amount

• Products with greater energy in the forward endothermic reaction become the higher-energy starting materials for the reverse exothermic reaction. Reversing the equation swaps the levels. The equal transferred amounts do not mean equal activation energies for the two directions: the climb to the same peak starts from different levels. That distinction can be read qualitatively without introducing an advanced thermodynamic calculation.

6 packets

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Forward and reverse reaction rates

• Beginning with mostly reactants, the forward rate may be high while the reverse rate is initially low. As products accumulate, reverse reaction becomes possible more often until the rates match under the stated model. A reaction-rate graph should show convergence to the same non-zero rate. A composition graph can level off at different reactant and product amounts. Identify which quantity is on the axis before claiming two curves must meet.

0 particles

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The forward and reverse rates become equal again

• Begin with the system already at equilibrium, identify one imposed change and select the appropriate response rule. Increased reactant concentration favours its consumption; removing product favours forming more. Warming favours the endothermic direction. Compression of a gaseous equilibrium favours the side with fewer gas molecules. These are composition responses, not the same as saying every faster reaction makes a larger final yield.

13 particles

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A reacts to form more B

• Separate the immediate imposed change from the response. If product is removed, its amount first falls; a subsequent net forward change can raise it from that lowered value without necessarily restoring the original level. At the new equilibrium both directions again occur at equal rates. A plot may show a sudden jump in the changed species followed by slower adjustment; the other species need not have the same immediate jump.

62 particles

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A shift toward reactants

• Both forward and reverse reactions can become faster on warming, yet their relative rate balance changes, causing a net shift before a new equilibrium is reached. For an exothermic forward industrial reaction, a low temperature may favour yield while slowing production; an appropriate temperature choice can be a compromise. A catalyst can help rate without turning that exothermic forward direction into an endothermic one or increasing its equilibrium yield at fixed conditions.

15 percentage points

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Ammonia: two gas molecules rather than four

• If the gas-molecule counts are equal on both sides, changing pressure by compression gives no pressure-driven composition shift in this model. H₂(g) + I₂(g) ⇌ 2HI(g) has two on each side. If solids or liquids occur in a supplied equation, do not count their coefficients as gas molecules. Compression can also affect rates by increasing gas particle concentration; a rate effect alone is not proof of an equilibrium shift.

4

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It remains two carbons long

• Distinguish addition, substitution, oxidation and polymerization by tracing bonds before and after reaction. Conditions and reagents belong to the reaction arrow; they are not interchangeable labels.

0.15 mol

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

C₄H₁₀

• A mixture contains different substances together without a fixed bonding ratio between those substances. Each individual compound still has its own definite formula. C₃H₈ is a hydrocarbon; C₂H₅OH contains oxygen and therefore is not a hydrocarbon, even though it contains carbon and hydrogen. Carbon dioxide is not a hydrocarbon either. Finite describes a limited resource, not a prediction that every deposit will disappear on one exact date.

60 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

C₃H₈

• For a straight chain, each end carbon has three hydrogen atoms, while an internal carbon has two. Methane has no carbon neighbour and so has four hydrogens. Count the bond lines at each carbon to check a displayed structure rather than adding hydrogen randomly. The general formula identifies the expected molecular composition of this series; molecular composition alone need not establish a unique arrangement of atoms.

16 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The lower region is hotter and it condenses before reaching cooler regions

• Evaporation and condensation are physical changes: molecules are separated without breaking their carbon chains. A sufficiently volatile fraction can remain gaseous at the top; heavy material can remain near the base. Fuels include petrol, diesel oil, kerosene, heavy fuel oil and liquefied petroleum gases. Fractions also provide feedstock for the petrochemical industry, producing materials such as solvents, lubricants, polymers and detergents. Carbon atoms can bond to each other in families of related structures; this helps explain the wide variety of natural and synthetic carbon compounds.

140 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Boiling point and viscosity increase; flammability decreases

• Keep the three comparisons separate. Boiling point concerns liquid becoming gas, viscosity concerns flow, and flammability concerns ignition and burning. Stronger intermolecular attractions in larger comparable hydrocarbon molecules help explain higher boiling points. Boiling does not normally break the covalent bonds within the molecules. The acquired section limits recalled property trends to these three; density or a precise energy-per-molecule trend is not required here.

3.5

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Carbon dioxide and water

• One carbon atom requires one carbon dioxide molecule, while each pair of hydrogen atoms forms one water molecule. Oxygen atoms on the product side come from both products, so count both before finding O₂. If an intermediate oxygen coefficient is a half number, multiply every coefficient by two for smallest whole-number coefficients. Never alter the given hydrocarbon’s subscripts to force balance.

15 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Cracking makes new smaller molecules; distillation separates existing compounds

• High demand for small-molecule fuels makes some cracking products useful as fuels. Alkenes can be used to make polymers and other chemicals. A cracking equation is an atom-conservation example; different products can form depending on conditions. A formula supplied in the equation must be preserved. The full named alkene structures and addition reactions are separate Chemistry-only 4.7.2 requirements, not evidence that this introductory lesson completes them.

10 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A carbon–carbon double bond

• A displayed formula shows all atoms and all bond lines. Count a double bond as two of carbon’s four bonds. In the terminal-double-bond structures shown, the first carbon has two hydrogens and the next has one, except ethene where each has two. A longer chain finishes with CH₃. Butene and pentene can have different atom arrangements; the diagrams show one stated straight-chain arrangement, rather than claiming their molecular formula uniquely fixes a structure.

18 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

One H attaches to each affected carbon and C=C becomes C–C

• An addition reaction joins atoms to the molecule across the multiple bond without removing the existing carbon skeleton. Check the new bond counts: each carbon still makes four bonds. Separately, alkenes burn in oxygen like other hydrocarbons. With sufficient oxygen, complete combustion gives carbon dioxide and water. In air they tend to burn with smoky flames because incomplete combustion produces carbon particles; smoke does not establish complete combustion.

8 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Its carbon skeleton and original hydrogen atoms

• For ethene plus bromine, CH₂=CH₂ becomes BrCH₂–CH₂Br. Each carbon retains its two hydrogen atoms and gains one bromine, giving four bond contributions. Bromine water changes orange to colourless in the specified reaction; an alkane control does not give the same rapid addition under these conditions. A carbon skeleton and hydrogen count are retained while two halogen atoms are added. The displayed panels distinguish chlorine, bromine and iodine products for all four named starting alkenes.

10 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Steam, high temperature/pressure and phosphoric-acid catalyst

• The product is an alcohol, not a hydrocarbon, because it now contains oxygen. Count the H inside OH as well as those bonded directly to carbon. Ethene’s two carbons are equivalent in this simple addition. Unsymmetrical starting alkenes can give different OH positions: the longer terminal-chain panels show the usual secondary-alcohol arrangement with OH on the second carbon. These drawings identify stated products without adding a required advanced mechanism or demanding positional product names beyond the course scope.

15 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

CH₃CH₂OH

• The first-four alcohols are used as fuels, solvents and starting materials for making other chemicals. Methanol is an industrial fuel/feedstock and solvent; ethanol is used as a fuel and solvent in suitable formulations; propanol and butanol serve as solvents and chemical feedstock. The selected use must be tied to actual product information, concentration and purpose. None of these industrial uses implies a sample is suitable to drink or apply to skin.

27 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Sodium reaction releases hydrogen; complete combustion makes carbon dioxide and water

• For the displayed primary examples, methanol can give methanoic acid, ethanol ethanoic acid, propanol propanoic acid and butanol butanoic acid. The OH carbon’s position matters; a different alcohol arrangement cannot automatically be assigned the same oxidation product. In an approved test, acidified potassium dichromate(VI) can change orange to green as it oxidises the primary alcohol. The acquired GCSE requirement asks for observations/descriptions, not aldehyde mechanisms or balanced equations for these non-combustion reactions.

12 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Sugar solution, yeast, warm conditions and oxygen excluded

• Maintain the school-selected temperature and sugar concentration, use the prepared yeast and record actual gas or mass observations over time. An airlock can let gas leave without freely admitting outside air; it is not a licence to tightly seal a gas-producing bottle. The rate can change as substrate runs low or ethanol inhibits yeast. Gas bubbling is qualitative unless quantity is measured, and a silent vessel is not by itself proof that no ethanol is present.

26 cm³

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The carbon in COOH as well as the two in CH₃CH₂

• These first-four acids dissolve in water to make acidic solutions. A suitable indicator or measured pH demonstrates acidity under the stated conditions. They react with carbonates to form a salt, carbon dioxide and water. For example, ethanoic acid with sodium carbonate forms sodium ethanoate plus the gas and water. Carbon dioxide can turn limewater milky in an approved product test. Reaction with alcohols forms an ester and water, developed in the separate ester lesson.

4 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Ethyl ethanoate and water

• The illustrated product contains a C(=O)–O–C connection. The acid-derived carbon framework and alcohol-derived carbon framework both remain. It is different from simply mixing an acid with water, from sodium carbonate reaction and from alkene addition. Formation need not be complete in one trial. A characteristic smell may be supplied as reference evidence but is not a unique identification test or a reason to inhale concentrated vapour.

42 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Only some acid particles ionise, producing H⁺

• Concentration describes the amount of acid in a volume; strength describes its extent of ionisation. These are independent labels. Comparing a dilute strong acid with a concentrated weak acid does not establish the simple same-concentration pH ordering. For whole-number pH differences, a decrease of one pH unit means ten times the hydrogen-ion concentration. No equilibrium constant, pKa calculation or advanced weak-acid approximation is required here.

100 times

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A two-carbon single-bond backbone with continuation bonds

• Draw the two-carbon backbone horizontally, keep both hydrogens on each carbon and show a single bond extending out of each side of the brackets. Put n outside the brackets to represent many repeats. A continuation bond joins the neighbouring unit; it is not an extra hydrogen or a separate monomer still containing C=C. Carbon has four bonds when the continuation bonds are counted. The bracketed unit represents the repeating interior of the chain, not its complete end-group structure.

60 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

It remains a side group on the corresponding backbone carbon

• Trace the two C=C carbons first and label their attached atoms or groups before changing any bond. In the repeat, the CH₂ carbon bonds to two hydrogens and two backbone neighbours. The CH carbon bonds to one hydrogen, the CH₃ group and two backbone neighbours. Both have four bonds. A supplied CH₂=CHCl example similarly gives –CH₂–CHCl–: chlorine stays attached. This supplied structure is for transferring the model, not an extra list of commercial names to memorise.

48 atoms

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Each has two reactive functional groups

• Read the ideal repeating interior as –O–CH₂–CH₂–O–C(=O)–(CH₂)₄–C(=O)–. The CH₂–CH₂ segment came from the diol, and the four-CH₂ segment and carbonyl carbons came from the diacid. At each joining event, H from an alcohol OH and OH from an acid COOH make water. The remaining oxygen links the carbonyl carbon to the diol segment. Unlike addition polymerisation, the repeat therefore does not retain every atom of the original monomer pair.

7 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

C(=O)–NH

• Each joining event loses OH from one carboxylic-acid group and H from one amino group as H₂O. A chain can extend because functional groups remain at its ends. Different amino acids can combine in the same chain to make proteins; their order and folding affect the resulting protein. Glycine-only repetition is a simple model, not a claim that every natural protein contains one repeating amino-acid type. Detailed zwitterion, stereochemistry and protein-structure classifications are not required here.

303

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Nucleotides

• The chains provide a repeated structural framework while the sequence of nucleotide types varies. A ladder model can distinguish the two continuous chains from paired positions across them. In the familiar double-stranded DNA model, A pairs with T and C with G. Twisting that ladder models the double helix. Colour-coded blocks are a representation: they are not actual nucleotide shapes, accurate atom arrangements or a diagram of separate bases joined without a backbone.

48 nucleotides

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Starch and cellulose both use glucose

• Proteins can use different amino acids in the same chain, with sequence and folding contributing to their functions. Starch and cellulose use glucose but differ in how their building blocks are joined and arranged, producing different properties. Starch stores carbohydrate in plants; cellulose gives strength to plant cell walls. The required Chemistry comparison is monomer identity and polymer recognition, not memorising advanced alpha/beta linkage names or complete structural formulae of carbohydrates.

32 units

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Graphite is less likely to dissolve and move with the solvent

• Rf is distance travelled by a component divided by distance travelled by the solvent front, both measured from the baseline. Compare under the same conditions; an Rf value alone does not establish identity across different solvents.

0.6

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Water containing only H₂O

• Compare the measured transition with a reference at the same pressure and with calibrated apparatus. A sharp matching melting point supports purity and identity, while a broadened or shifted interval can indicate an impurity. Boiling temperature depends on pressure and the mixture; do not claim every impurity must shift every boiling point in the same direction. A single matching temperature is supporting evidence rather than unique proof of identity, because different substances can have similar values.

6 °C

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Its measured components are selected for useful properties

• Use the supplied information to connect component, quantity and intended property. Too much solvent may make a paint spread easily but reduce coating thickness, while the right binder/pigment balance affects coverage and adhesion. These are product-specific trade-offs rather than a universal recipe. A formulation’s measured composition is different from the fixed chemical proportions inside a compound. Proprietary component names are not required knowledge.

30 g

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Spot centre and solvent front, both from the origin

• Rf is distance moved by the substance divided by distance moved by the solvent. Measure both from the origin line, using the centre of the spot for the substance. The ratio has no unit and ordinarily lies between zero and one in this model. Compare unknown and reference spots using the same paper, solvent and conditions. A different solvent can change the separation and Rf; it is not a universal identifying number for one substance.

0.7

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

To avoid washing the samples directly into the reservoir

• Draw a horizontal pencil origin 2 cm above the lower paper edge. Mark positions away from the edges, place small concentrated spots using separate clean tubes and label in pencil. The handbook aims for 2–3 mm spots and no more than 1 cm water depth, keeping the origin above the water while the paper bottom dips into it. Support the paper without touching the beaker walls and avoid moving the vessel during the run.

0.55

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Burning splint at the open end; pop

• A reaction prediction and a gas test answer different questions. A known reactive metal with dilute acid can produce hydrogen, but bubbles alone could be another gas in a different system. A positive pop supports hydrogen under the approved test conditions. A weak or absent observation can reflect a small or diluted sample as well as absence of hydrogen, so an inconclusive trial should not be rewritten as a definite negative.

80 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A glowing splint relights

• Oxygen supports combustion but is not itself described as a flammable fuel in this test. Increased burning is evidence about combustion conditions; the school identification relies on the characteristic relighting observation. Colourless gas or a known source reaction alone is weaker evidence than a correctly performed test. A sample mixed with air or contaminated during collection may give less clear observations, so retain uncertain results rather than inventing an identification.

75 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Limewater turns milky

• State the reagent and the positive result together. A flame going out is not a unique carbon-dioxide test because several gases fail to support combustion. Under continued excess carbon dioxide the cloudiness can eventually disappear by further reaction; that optional observation does not invalidate the initial specified milky result. Do not substitute that extension for the basic test or make students memorise an unrequired second equation.

80 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Damp litmus is bleached white

• Chlorine is Cl₂, a molecular element, while chloride is Cl⁻, a negative ion in compounds and solutions. Their tests are different: gas bleaching identifies chlorine under suitable conditions, whereas acidified silver nitrate precipitates chloride ions. Pale green gas appearance can support the description but does not justify smelling a sample. Not every substance containing chlorine atoms releases chlorine gas or bleaches paper in the same way.

75 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Lilac

• A sample is introduced into a suitable flame using the school-approved wire or soaked-splint method. Compare its emission with known references under the same conditions. A mixture may have masked colours; intense sodium emission can obscure another response. Yellow therefore supports sodium in the tested conditions but does not prove no other ions are present. Contamination of the wire, sample or apparatus can produce a misleading result.

40 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

To avoid reagents from one test contaminating the next

• Establish reference results with labelled known cation compounds and carbonate/sulfate/chloride/bromide/iodide salts. Use fresh unknown portions for different tests. Keep flame-wire cleaning, dilute-acid identities and precipitating reagents exact. Record carbonate gas and limewater, sulfate with dilute HCl/barium chloride, and halides with dilute nitric acid/silver nitrate. Compare subtle silver-halide colours side by side. The handbook suggests potassium sulfate as a convenient unknown; the school can choose another suitable single compound.

14 ions

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

White precipitate dissolving in excess sodium hydroxide

• Record initial precipitation separately from what happens on adding excess reagent. A white precipitate that dissolves in excess supports aluminium; one remaining supports calcium or magnesium within the supplied set. Use additional suitable evidence, such as calcium’s orange-red flame, when a single observation leaves alternatives. The Roman numerals in iron(II) and iron(III) distinguish ion charges and their different hydroxides. Colour should be compared with known results in similar conditions.

75 %

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Fe(OH)₃

• The divalent examples have form MCl₂(aq) + 2NaOH(aq) → M(OH)₂(s) + 2NaCl(aq), with M the stated Mg, Ca, Cu or Fe(II). For Al and Fe(III), MCl₃(aq) + 3NaOH(aq) → M(OH)₃(s) + 3NaCl(aq). These are full symbol equations, suitable for the both-tier requirement. Separate HT ionic-equation work may simplify them, but it is not needed to recognise the common-tier formation equations here.

12 units

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Dilute acid gives gas that clouds limewater

• For sodium carbonate and hydrochloric acid, Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O illustrates the source of the gas. The CO₂ test uses aqueous calcium hydroxide and forms insoluble calcium carbonate. Effervescence means gas bubbles, not a unique gas identity. A flame being extinguished is also non-specific: several gases do not support combustion, so it is weaker than the specified limewater result.

6 molecules

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Dilute nitric acid followed by silver nitrate

• Nitric acid does not introduce a halide ion that would contaminate this test. Using hydrochloric acid instead introduces chloride and can produce a misleading white precipitate with silver nitrate. Acidification also removes carbonate interference in the standard comparison. The school method governs quantities and handling. Silver-halide colours are subtle, so compare known results side by side under the same light rather than relying on a tiny colour difference from memory.

5 units

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

White precipitate after dilute HCl and barium chloride

• Dilute hydrochloric acid helps remove carbonate interference before barium chloride is added. Barium salts can otherwise precipitate with carbonate as well. Do not use sulfuric acid to acidify the sample: it introduces sulfate and could produce a false positive. This requirement differs from the halide test, where nitric acid avoids introducing chloride. The acid choice is therefore part of the chemical evidence, not a interchangeable label.

8 units

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

A lower detectable concentration

• Compare an instrument and a chemical test against the same task. Detecting a small metal-ion concentration is different from identifying a high-concentration salt by flame colour. Instrumental outputs can be numerical and easier to compare objectively than a subtle cream/white judgement. Instruments still need appropriate references, calibration, clean samples and correct operation. A fast result can be wrong if contamination or an unsuitable reference invalidates the method.

0.3 units

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

Several matching line positions in the reference

• Separate line position from signal strength. Position supports identity; intensity at a suitable line, compared with known concentrations, supports quantity. Use the supplied wavelength scale or reference table, not a memorised list of invented atomic wavelengths. The original figure’s reference X/Y positions are explicitly fictional teaching data. Check several lines and overlapping signals; one shared line alone can leave alternatives. No electron-transition or advanced instrument mechanism is required here.

3.5 mg/L

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.

The same volume delivered the same number of times

• Define the functional unit before comparing products. The same delivered service, such as carrying one litre of water a hundred times, is fairer than comparing one object with another regardless of lifetime.

20 uses

• Appropriate relation; consistent substitution; correct result; meaningful unit and interpretation.