Chemistry 4CH1 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
This diagram is part of the ionic arrangement in sodium oxide, a compound used in making some glass and ceramic glazes. A manufacturer compares sodium oxide with magnesium oxide because both compounds contain oxide ions. The diagram is not drawn to scale, but it shows the ratio of the ions in sodium oxide.
(a) State the charge on an oxide ion. [3]
(b) Complete the formula of sodium oxide. [3]
(c) Give the formula of magnesium oxide. [3]
(d) Name the type of bonding in both compounds. [3]
(e) Explain why sodium oxide has a regular arrangement of ions rather than separate molecules. [4]
(a) An oxide ion has charge 2−, written O2-. [3]
(b) Two Na+ ions balance one O2- ion, so sodium oxide is Na2O. [3]
(c) Mg2+ and O2- balance in a 1:1 ratio, so magnesium oxide is MgO. [3]
(d) Both compounds have ionic bonding. [3]
(e) Sodium atoms transfer electrons to oxygen atoms. This forms positive sodium ions and negative oxide ions. Oppositely charged ions attract, and this attraction acts in all directions, producing a giant regular ionic lattice rather than separate molecules. [4]
Answer Details
(a) An oxide ion has charge 2−, written O2-. [3]
(b) Two Na+ ions balance one O2- ion, so sodium oxide is Na2O. [3]
(c) Mg2+ and O2- balance in a 1:1 ratio, so magnesium oxide is MgO. [3]
(d) Both compounds have ionic bonding. [3]
(e) Sodium atoms transfer electrons to oxygen atoms. This forms positive sodium ions and negative oxide ions. Oppositely charged ions attract, and this attraction acts in all directions, producing a giant regular ionic lattice rather than separate molecules. [4]
Question 2 Report
The diagram shows the apparatus used by a technician to remove excess magnesium oxide solid from magnesium sulfate solution. Magnesium oxide was added in small pieces to dilute sulfuric acid until no more of the solid reacted. The mixture was then poured through the apparatus in Fig. 1. The clear liquid collected in the beaker is later heated to obtain magnesium sulfate crystals. This method prevents acid from remaining in the final solution.
(a) Name the parts labelled X, Y and Z. [3]
(b) State the purpose of the filter paper in this process. [1]
(c) Complete the word equation.
magnesium oxide + sulfuric acid → ................................ + ................................ [3]
(d) Give two observations when magnesium oxide is added to sulfuric acid before filtration. [2]
(e) When the filtrate is heated, state two safety precautions. [2]
(f) Which piece of apparatus is used to heat the filtrate gently until crystals begin to form? [1]
(g) Explain why the technician adds excess magnesium oxide before filtering. [3]
(a) X is a filter funnel, Y is filter paper, and Z is the filtrate. [3]
(b) Filter paper separates insoluble magnesium oxide from solution: liquid passes through but solid is retained. [1]
(c) magnesium oxide + sulfuric acid → magnesium sulfate + water. [3]
(d) The white solid dissolves at first. Once excess magnesium oxide has been added, some white solid remains in the colourless solution. [2]
(e) Wear eye protection and heat gently. Using tongs or a heatproof mat, and not heating to dryness, are also suitable precautions. Any two. [2]
(f) Heat the filtrate in an evaporating basin. [1]
(g) Excess magnesium oxide ensures all sulfuric acid reacts. The excess is insoluble, so it remains as a solid and can be removed by filtration. This leaves no acid in the magnesium sulfate solution. [3]
Answer Details
(a) X is a filter funnel, Y is filter paper, and Z is the filtrate. [3]
(b) Filter paper separates insoluble magnesium oxide from solution: liquid passes through but solid is retained. [1]
(c) magnesium oxide + sulfuric acid → magnesium sulfate + water. [3]
(d) The white solid dissolves at first. Once excess magnesium oxide has been added, some white solid remains in the colourless solution. [2]
(e) Wear eye protection and heat gently. Using tongs or a heatproof mat, and not heating to dryness, are also suitable precautions. Any two. [2]
(f) Heat the filtrate in an evaporating basin. [1]
(g) Excess magnesium oxide ensures all sulfuric acid reacts. The excess is insoluble, so it remains as a solid and can be removed by filtration. This leaves no acid in the magnesium sulfate solution. [3]
Question 3 Report
This diagram is from a school investigation of ammonium chloride. A student placed a small mass of solid ammonium chloride at one end of a dry hard-glass tube and heated it gently. After several minutes, a white solid appeared on the cooler part of the tube. The student was told that gaseous ammonia and hydrogen chloride can form near the heated solid and can react again as they move to a cooler region. The tube is not sealed, so the observation is used to discuss a reversible reaction rather than a full equilibrium mixture.
(a) Name the white solid initially heated in the tube. [1]
(b) Complete the equation for the change near the heated end.
NH4Cl(s) ⇆ ................. + ................. [2]
(c) State the two gases that diffuse towards the cooler part of the tube. [2]
(d) Explain why a white solid forms in the cool region. [3]
(e) Give one change needed to allow a true equilibrium to become established in this experiment. [2]
(a) The solid initially heated is ammonium chloride. [1]
(b) Near the heated end:
\[\mathrm{NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)}\]
Ammonia, \(\mathrm{NH_3(g)}\), gains one mark and hydrogen chloride, \(\mathrm{HCl(g)}\), gains one mark. [2]
(c) The gases diffusing to the cooler region are ammonia [1] and hydrogen chloride. [1] [2]
(d) The ammonia and hydrogen chloride gases meet in the cooler region. [1] They react together [1] to form solid ammonium chloride, which deposits as a white solid. [1] [3]
(e) The tube must be sealed, or a closed container used. [1] Then no gas can escape and both the forward and reverse reactions can continue, allowing a true equilibrium to form. [1] [2]
Answer Details
(a) The solid initially heated is ammonium chloride. [1]
(b) Near the heated end:
\[\mathrm{NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)}\]
Ammonia, \(\mathrm{NH_3(g)}\), gains one mark and hydrogen chloride, \(\mathrm{HCl(g)}\), gains one mark. [2]
(c) The gases diffusing to the cooler region are ammonia [1] and hydrogen chloride. [1] [2]
(d) The ammonia and hydrogen chloride gases meet in the cooler region. [1] They react together [1] to form solid ammonium chloride, which deposits as a white solid. [1] [3]
(e) The tube must be sealed, or a closed container used. [1] Then no gas can escape and both the forward and reverse reactions can continue, allowing a true equilibrium to form. [1] [2]
Question 4 Report
Table 1 shows the results of a school investigation into two packaging films. Equal masses of film were placed in separate compost containers containing damp soil. The students kept the temperature, amount of water and mass of soil the same in each container. Film A was poly ethene. Film B was a polymer designed to break down more readily after use. Both films were initially solids with a mass of 12.0 g.
| Time in compost / weeks | Mass of film A / g | Mass of film B / g |
|---|---|---|
| 0 | 12.0 | 12.0 |
| 4 | 11.9 | 9.8 |
| 8 | 11.8 | 7.2 |
(a) State one variable that the students controlled in this investigation. [1]
(b) Calculate the percentage decrease in mass of film B after 8 weeks. [2]
(c) Which film has changed least in mass after 8 weeks? [1]
(d) Explain why the mass of poly ethene changes very little in the compost. [2]
(e) Give two environmental problems caused when used poly ethene bags are not collected or recycled. [2]
(a) One controlled variable was the temperature. The amount of water, mass or type of soil, initial film mass, or measurement time were also controlled. [1]
(b)
\[\text{mass decrease}=12.0-7.2=4.8\text{ g}\]
\[\text{percentage decrease}=\frac{4.8}{12.0}\times100=40\%\]
The percentage decrease is 40%. [2]
(c) Film A, poly ethene, changed least in mass: it decreased by only 0.2 g. [1]
(d) Poly ethene is chemically unreactive because of its strong covalent bonds. Microorganisms do not readily break its polymer chains, so it is not biodegradable and its mass changes very little. [2]
(e) Two valid environmental problems are that poly ethene can persist in landfill and that litter can harm animals. It can also block drains or waterways, form microplastics, or require finite crude-oil resources to replace. [2]
Answer Details
(a) One controlled variable was the temperature. The amount of water, mass or type of soil, initial film mass, or measurement time were also controlled. [1]
(b)
\[\text{mass decrease}=12.0-7.2=4.8\text{ g}\]
\[\text{percentage decrease}=\frac{4.8}{12.0}\times100=40\%\]
The percentage decrease is 40%. [2]
(c) Film A, poly ethene, changed least in mass: it decreased by only 0.2 g. [1]
(d) Poly ethene is chemically unreactive because of its strong covalent bonds. Microorganisms do not readily break its polymer chains, so it is not biodegradable and its mass changes very little. [2]
(e) Two valid environmental problems are that poly ethene can persist in landfill and that litter can harm animals. It can also block drains or waterways, form microplastics, or require finite crude-oil resources to replace. [2]
Question 5 Report
A technician checks the purity of magnesium metal used to make lightweight bicycle components. Table 1 gives information for the three naturally occurring magnesium isotopes. Each species is a neutral atom.
| Isotope | Number of protons | Number of neutrons | Relative abundance / % |
|---|---|---|---|
| magnesium-24 | 12 | 12 | 78.99 |
| magnesium-25 | 12 | 13 | 10.00 |
| magnesium-26 | 12 | 14 | 11.01 |
The metal is later reacted with dilute acid to produce hydrogen gas. The isotope does not change which element the atom is.
(a) Name the particles in an atom that have a positive charge. [1]
(b) Give the mass number of a magnesium-25 atom. [1]
(c) Complete the electronic configuration of a magnesium atom: 2, 8, ________. [1]
(d) Calculate the relative atomic mass of magnesium using Table 1. Give your answer to two decimal places. [3]
(e) Explain why magnesium-24 and magnesium-26 have similar chemical properties. [2]
(f) State the charge on a magnesium ion formed when a magnesium atom loses two electrons. [1]
(g) Give the formula of magnesium chloride. [1]
(a) Protons have a positive charge. [1]
(b) Magnesium-25 has mass number 25. [1]
(c) Magnesium has 12 electrons arranged \(2,8,\boldsymbol{2}\). [1]
(d) \[\frac{(24\times78.99)+(25\times10.00)+(26\times11.01)}{100}=24.3202\] To two decimal places, the relative atomic mass is 24.32. [3]
(e) Magnesium-24 and magnesium-26 have the same number of electrons and the same outer-shell electron arrangement. Chemical properties depend mainly on outer-shell electrons, not neutron number. [2]
(f) Losing two electrons forms an ion with charge \(2+\). [1]
(g) Magnesium chloride is \(\mathrm{MgCl_2}\). One \(\mathrm{Mg^{2+}}\) ion needs two \(\mathrm{Cl^-}\) ions for an overall neutral compound. [1]
Answer Details
(a) Protons have a positive charge. [1]
(b) Magnesium-25 has mass number 25. [1]
(c) Magnesium has 12 electrons arranged \(2,8,\boldsymbol{2}\). [1]
(d) \[\frac{(24\times78.99)+(25\times10.00)+(26\times11.01)}{100}=24.3202\] To two decimal places, the relative atomic mass is 24.32. [3]
(e) Magnesium-24 and magnesium-26 have the same number of electrons and the same outer-shell electron arrangement. Chemical properties depend mainly on outer-shell electrons, not neutron number. [2]
(f) Losing two electrons forms an ion with charge \(2+\). [1]
(g) Magnesium chloride is \(\mathrm{MgCl_2}\). One \(\mathrm{Mg^{2+}}\) ion needs two \(\mathrm{Cl^-}\) ions for an overall neutral compound. [1]
Question 6 Report
A technician at a metal-finishing factory receives an unlabelled pale-green solution. The factory uses iron compounds and sodium compounds in different treatment tanks, so the technician carries out cation tests on separate 2 cm3 portions. With sodium hydroxide solution, a green precipitate forms. With aqueous ammonia, a green precipitate also forms and does not dissolve when excess ammonia is added. A fresh portion gives no flame colour apart from the usual blue Bunsen flame.
(a) Name the metal ions indicated by the green precipitate tests. [2]
(b) State the formula of the green precipitate. [2]
(c) Give the name of the reagent used in the first test. [2]
(d) Which observation shows that sodium ions are unlikely to be present? [2]
(e) Explain why separate portions of the original solution are used for the two tests. [4]
(a) The green precipitates indicate iron(II) ions, \(\mathrm{Fe^{2+}}\). [2]
(b) The precipitate is iron(II) hydroxide, \(\mathrm{Fe(OH)_2}\). [2]
(c) The reagent is sodium hydroxide solution, \(\mathrm{NaOH(aq)}\). [2]
(d) No yellow flame colour is seen. Sodium ions produce an intense yellow flame, so sodium ions are unlikely to be present. [2]
(e) If one sample were used, the first reagent would remain in it and could react with the next reagent or change the ions present. This could alter the observation. Separate portions give independent, reliable tests. [4]
Answer Details
(a) The green precipitates indicate iron(II) ions, \(\mathrm{Fe^{2+}}\). [2]
(b) The precipitate is iron(II) hydroxide, \(\mathrm{Fe(OH)_2}\). [2]
(c) The reagent is sodium hydroxide solution, \(\mathrm{NaOH(aq)}\). [2]
(d) No yellow flame colour is seen. Sodium ions produce an intense yellow flame, so sodium ions are unlikely to be present. [2]
(e) If one sample were used, the first reagent would remain in it and could react with the next reagent or change the ions present. This could alter the observation. Separate portions give independent, reliable tests. [4]
Question 7 Report
Fig. 1 shows the output from a mass spectrometer used to check a cylinder of neon gas before it is used in a laboratory reaction. The sample was vaporised, ionised and accelerated. Each atom formed an ion with a charge of +1. The peaks are caused by atoms with different numbers of neutrons, not by different elements. The horizontal scale gives mass-to-charge ratio, m/z. The diagram is not to scale. A student also has a separate sodium chloride solution for another investigation, but it must not be mixed with the gas sample.
(a) State the number of neutrons in ion A. [2]
(b) Give the meaning of the term isotope. [2]
(c) Use Fig. 1 to state which ion, A, B or C, is least abundant and explain your choice. [3]
(d) Calculate the relative atomic mass of the neon sample, using relative abundances of 30 for A, 50 for B and 5 for C. Show your working. [4]
(a) Neon has atomic number 10, so every neon atom has 10 protons. For ion A, the mass number is 20. \(20-10=10\), so ion A has 10 neutrons. [2]
(b) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. The differing neutrons give different mass numbers. [2]
(c) Ion C is least abundant. Its peak is the shortest, and peak height represents relative abundance: fewer atoms of that isotope produce a lower peak. [3]
(d) Use a weighted mean, because each isotope contributes according to its abundance.
\[\frac{(20\times30)+(21\times50)+(22\times5)}{30+50+5}=\frac{1760}{85}=20.7\]
The relative atomic mass of this neon sample is 20.7 (or 20.71). [4]
Exam reminder: For a \(+1\) ion, \(m/z\) is equal to its mass number.
Answer Details
(a) Neon has atomic number 10, so every neon atom has 10 protons. For ion A, the mass number is 20. \(20-10=10\), so ion A has 10 neutrons. [2]
(b) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. The differing neutrons give different mass numbers. [2]
(c) Ion C is least abundant. Its peak is the shortest, and peak height represents relative abundance: fewer atoms of that isotope produce a lower peak. [3]
(d) Use a weighted mean, because each isotope contributes according to its abundance.
\[\frac{(20\times30)+(21\times50)+(22\times5)}{30+50+5}=\frac{1760}{85}=20.7\]
The relative atomic mass of this neon sample is 20.7 (or 20.71). [4]
Exam reminder: For a \(+1\) ion, \(m/z\) is equal to its mass number.
Question 8 Report
Fig. 1 shows part of a teaching display showing the first four members of a homologous series. The cards have been placed in order of increasing number of carbon atoms. The formulae are molecular formulae, not displayed formulae. A student notices that each successive alkane has one extra CH2 group. The display is used to predict the formula of a later compound before a sample is made in the laboratory. Alkanes are hydrocarbons and are often obtained from crude oil, then separated for fuels and other reactions.
(a) State the general formula of an alkane. [1]
(b) Complete the formula of pentane: C5H__. [1]
(c) Give the difference in molecular formula between ethane and propane. [1]
(d) Name the alkane with molecular formula C6H14. [1]
(e) Explain why all compounds in Fig. 1 have similar chemical reactions. [2]
(f) State two ways in which the physical properties change as the carbon chain becomes longer. [3]
(a) The alkane general formula is \(\mathrm{C_nH_{2n+2}}\). [1]
(b) For pentane, \(H=2(5)+2=12\). The completed formula is \(\mathrm{C_5H_{12}}\). [1]
(c) Propane has one extra \(\mathrm{CH_2}\) unit compared with ethane. [1]
(d) \(\mathrm{C_6H_{14}}\) is hexane. [1]
(e) The compounds belong to the same homologous series, so they have the same functional-group characteristics. They are all saturated hydrocarbons with the same type of bonding, and therefore have similar chemical reactions. [2]
(f) As the carbon chain becomes longer, any three valid changes are: boiling temperature increases, melting temperature generally increases, viscosity increases, and volatility decreases. [3]
Answer Details
(a) The alkane general formula is \(\mathrm{C_nH_{2n+2}}\). [1]
(b) For pentane, \(H=2(5)+2=12\). The completed formula is \(\mathrm{C_5H_{12}}\). [1]
(c) Propane has one extra \(\mathrm{CH_2}\) unit compared with ethane. [1]
(d) \(\mathrm{C_6H_{14}}\) is hexane. [1]
(e) The compounds belong to the same homologous series, so they have the same functional-group characteristics. They are all saturated hydrocarbons with the same type of bonding, and therefore have similar chemical reactions. [2]
(f) As the carbon chain becomes longer, any three valid changes are: boiling temperature increases, melting temperature generally increases, viscosity increases, and volatility decreases. [3]
Question 9 Report
Fig. 1 shows an investigation into rusting of iron used for outdoor railings. Four identical iron nails were left for seven days in sealed tubes. Tube A contained dry air with calcium chloride. Tube B contained boiled water covered by a layer of oil. Tube C contained tap water and air. Tube D contained salt water and air. Each nail began with the same mass and had no protective oxide coating. The student recorded the amount of rust visible.
(a) State the two substances needed for iron to rust. [2]
(b) Which tube should show the greatest amount of rust? [1]
(c) Explain why the nail in tube B has little or no rust. [2]
(d) Give one method used to prevent rusting of an iron gate. [1]
(e) Explain why coating iron with zinc can still protect it when the coating is scratched. [2]
(f) Name the orange-brown substance formed when iron rusts. [1]
(a) Iron needs water and oxygen to rust. [2]
(b) Tube D should have the most rust because it contains both air and salt water; salt water speeds rusting. [1]
(c) Boiling removes most of the dissolved oxygen from the water. The oil layer then prevents oxygen from the air entering the water, so both conditions for rusting are not available. [2]
(d) One method is to paint the gate. Oil or grease, galvanising, and electroplating with tin are also acceptable. [1]
(e) Zinc is more reactive than iron. At a scratch, zinc oxidises in preference to iron, providing sacrificial protection even where the barrier is broken. [2]
(f) The orange-brown substance is rust, hydrated iron(III) oxide. [1]
Answer Details
(a) Iron needs water and oxygen to rust. [2]
(b) Tube D should have the most rust because it contains both air and salt water; salt water speeds rusting. [1]
(c) Boiling removes most of the dissolved oxygen from the water. The oil layer then prevents oxygen from the air entering the water, so both conditions for rusting are not available. [2]
(d) One method is to paint the gate. Oil or grease, galvanising, and electroplating with tin are also acceptable. [1]
(e) Zinc is more reactive than iron. At a scratch, zinc oxidises in preference to iron, providing sacrificial protection even where the barrier is broken. [2]
(f) The orange-brown substance is rust, hydrated iron(III) oxide. [1]
Question 10 Report
Table 1 records results from a corrosion study on magnesium. Each test tube contained 25.0 cm3 of dilute sulfuric acid at 22 degrees C. A student placed a 0.60 g piece of magnesium ribbon into each tube and measured the mass of the tube and its contents at intervals. Hydrogen gas escaped through a small plug of cotton wool, so the measured mass decreased as the reaction occurred. The reaction produces magnesium sulfate solution and hydrogen gas.
| Time / s | Mass of tube and contents / g |
|---|---|
| 0 | 48.60 |
| 20 | 48.48 |
| 40 | 48.40 |
| 60 | 48.35 |
| 80 | 48.33 |
| 100 | 48.33 |
Fig. 1 shows the type of apparatus used. The student repeats the investigation using powdered magnesium, while keeping the mass of magnesium, acid volume and acid concentration the same.
(a) State the gas that causes the mass to decrease. [1]
(b) Calculate the mean rate of mass loss during the first 40 s, in g/s. [2]
(c) Explain why the mass stays constant from 80 s to 100 s. [2]
(d) Give the formula of the salt in the final solution. [1]
(e) Explain why powdered magnesium is likely to give a faster reaction than magnesium ribbon. [3]
(f) State one safety precaution, other than wearing eye protection, when carrying out this reaction with acid. [1]
(a) The escaping gas is hydrogen. [1]
(b) Mass loss in the first 40 s:
\[48.60\text{ g}-48.40\text{ g}=0.20\text{ g}\]
\[\text{mean rate}=\frac{0.20\text{ g}}{40\text{ s}}=0.0050\text{ g/s}\]
[2]
(c) The reaction is complete, so no more hydrogen is produced and escapes. All the magnesium or all the acid has been used up. [2]
(d) Magnesium sulfate is \(\mathrm{MgSO_4}\). [1]
(e) Powdered magnesium has a larger surface area than ribbon. More acid particles can collide with magnesium each second, giving more successful collisions per second and therefore a faster reaction. [3]
(f) Use a small volume of dilute acid, keep the tube pointed away from people, clean up spills immediately, or tie back long hair. Any one. [1]
Answer Details
(a) The escaping gas is hydrogen. [1]
(b) Mass loss in the first 40 s:
\[48.60\text{ g}-48.40\text{ g}=0.20\text{ g}\]
\[\text{mean rate}=\frac{0.20\text{ g}}{40\text{ s}}=0.0050\text{ g/s}\]
[2]
(c) The reaction is complete, so no more hydrogen is produced and escapes. All the magnesium or all the acid has been used up. [2]
(d) Magnesium sulfate is \(\mathrm{MgSO_4}\). [1]
(e) Powdered magnesium has a larger surface area than ribbon. More acid particles can collide with magnesium each second, giving more successful collisions per second and therefore a faster reaction. [3]
(f) Use a small volume of dilute acid, keep the tube pointed away from people, clean up spills immediately, or tie back long hair. Any one. [1]
