Chemistry 4CH1 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
The diagram shows a comparison of two species of iron used in a corrosion investigation. Iron filings are placed in water and oxygen, but Fig. 1 shows the particles before the reaction. The left particle is a neutral iron atom and the right particle is an iron ion.
(a) State the mass number of each species in Fig. 1. [1]
(b) Give the charge on the iron ion. [1]
(c) Explain how the iron ion has formed from the iron atom. [2]
(d) Name the particles in the nucleus that account for the mass number. [2]
(e) Complete the formula for the oxide ion: O______ . [1]
(f) Give the formula of iron(III) oxide formed during corrosion. [2]
(g) State why an iron-54 atom is an isotope of the iron atom shown. [2]
(h) When iron reacts with dilute sulfuric acid, name the gas produced. [1]
(i) Give the formula of iron sulfate made when iron reacts with sulfuric acid. [2]
(j) Which particle has the smallest relative mass: proton, neutron or electron? [1]
(a) Each species has \(26\) protons and \(30\) neutrons, so its mass number is \(26+30=\boldsymbol{56}\). [1]
(b) The ion has 26 protons but only 23 electrons, so it has lost three electrons and has charge \(3+\). [1]
(c) The ion formed when the atom loses electrons, specifically three electrons. [2]
(d) The nucleus particles that account for mass number are protons and neutrons. [2]
(e) The oxide ion is \(\mathrm{O^{2-}}\). [1]
(f) Iron(III) oxide is \(\mathrm{Fe_2O_3}\). Two \(\mathrm{Fe^{3+}}\) ions balance three \(\mathrm{O^{2-}}\) ions. [2]
(g) Iron-54 is an isotope because it has the same number of protons, and hence the same atomic number, as the iron shown, but a different number of neutrons and mass number. [2]
(h) Iron reacting with dilute sulfuric acid produces hydrogen. [1]
(i) Iron sulfate is \(\mathrm{FeSO_4}\). Iron forms \(\mathrm{Fe^{2+}}\) ions with sulfate \(\mathrm{SO_4^{2-}}\) ions. [2]
(j) The electron has the smallest relative mass. [1]
Answer Details
(a) Each species has \(26\) protons and \(30\) neutrons, so its mass number is \(26+30=\boldsymbol{56}\). [1]
(b) The ion has 26 protons but only 23 electrons, so it has lost three electrons and has charge \(3+\). [1]
(c) The ion formed when the atom loses electrons, specifically three electrons. [2]
(d) The nucleus particles that account for mass number are protons and neutrons. [2]
(e) The oxide ion is \(\mathrm{O^{2-}}\). [1]
(f) Iron(III) oxide is \(\mathrm{Fe_2O_3}\). Two \(\mathrm{Fe^{3+}}\) ions balance three \(\mathrm{O^{2-}}\) ions. [2]
(g) Iron-54 is an isotope because it has the same number of protons, and hence the same atomic number, as the iron shown, but a different number of neutrons and mass number. [2]
(h) Iron reacting with dilute sulfuric acid produces hydrogen. [1]
(i) Iron sulfate is \(\mathrm{FeSO_4}\). Iron forms \(\mathrm{Fe^{2+}}\) ions with sulfate \(\mathrm{SO_4^{2-}}\) ions. [2]
(j) The electron has the smallest relative mass. [1]
Question 2 Report
The diagram shows part of a technician's method for recovering hydrated magnesium sulfate crystals from a clear magnesium sulfate solution. The solution has been warmed in an evaporating basin to remove some water. It is then allowed to cool before crystals are filtered. The technician must not heat the basin until it is completely dry because hydrated crystals can lose water of crystallisation. The crystals are intended for a school demonstration, so they should be clean and dry.
(a) Name the process in which liquid water changes to water vapour during warming. [1]
(b) State why the solution is allowed to cool after some water has evaporated. [2]
(c) Give three pieces of apparatus, other than the evaporating basin, needed for the warming stage. [3]
(d) Describe how the crystals should be washed and dried after filtration. [3]
(e) Explain why heating to dryness is unsuitable for preparing hydrated magnesium sulfate crystals. [3]
(a) The change of liquid water into water vapour is evaporation. [1]
(b) Cooling makes the concentrated solution saturated, so magnesium sulfate crystals can form. [2]
(c) Three suitable items are a tripod, gauze and Bunsen burner. A heatproof mat or tongs would also be acceptable. [3]
(d) Rinse the filtered crystals with a small amount of cold distilled water, remove them from the filter paper, then dry them by pressing them between dry filter papers. Cold water minimises dissolving the crystals. [3]
(e) Heating to dryness removes all water. Hydrated magnesium sulfate contains water of crystallisation, and strong heating can remove this bound water and change the crystals. [3]
Answer Details
(a) The change of liquid water into water vapour is evaporation. [1]
(b) Cooling makes the concentrated solution saturated, so magnesium sulfate crystals can form. [2]
(c) Three suitable items are a tripod, gauze and Bunsen burner. A heatproof mat or tongs would also be acceptable. [3]
(d) Rinse the filtered crystals with a small amount of cold distilled water, remove them from the filter paper, then dry them by pressing them between dry filter papers. Cold water minimises dissolving the crystals. [3]
(e) Heating to dryness removes all water. Hydrated magnesium sulfate contains water of crystallisation, and strong heating can remove this bound water and change the crystals. [3]
Question 3 Report
A technician prepares a reference card for a chemical store. The card compares four covalent molecules that may be supplied as gases or liquids. Table 1 gives the atoms in one molecule and the temperature at which each substance boils. The formula column has been left blank for the student to complete.
| substance | atoms in one molecule | formula | boiling point / degrees C |
|---|---|---|---|
| hydrogen | 2 hydrogen atoms | blank | -253 |
| hydrogen chloride | 1 hydrogen atom and 1 chlorine atom | blank | -85 |
| ammonia | 1 nitrogen atom and 3 hydrogen atoms | blank | -33 |
| water | 2 hydrogen atoms and 1 oxygen atom | blank | 100 |
(a) Complete the formula column for all four substances. [2]
(b) State which substance is a liquid at 25 degrees C. [1]
(c) Give the state of hydrogen chloride at -100 degrees C. [1]
(d) Use Table 1 to calculate the temperature range over which hydrogen chloride is a gas but hydrogen is a liquid. [3]
(e) Explain why hydrogen and hydrogen chloride have low boiling points compared with silicon dioxide. [3]
(f) Which molecule contains the greatest number of atoms? Give its name. [1]
(g) State the type of bonding within a water molecule. [1]
(a) The formulae are hydrogen: \(\mathrm{H_2}\); hydrogen chloride: \(\mathrm{HCl}\); ammonia: \(\mathrm{NH_3}\); water: \(\mathrm{H_2O}\). [2]
(b) Water is liquid at 25 degrees C. [1]
(c) Hydrogen chloride is a liquid at \(-100\) degrees C because this is below its boiling point of \(-85\) degrees C. [1]
(d) Hydrogen chloride is a gas above \(-85\) degrees C. The supplied boiling-point data show hydrogen has a lower boiling point, \(-253\) degrees C, so hydrogen is already gaseous whenever hydrogen chloride is gaseous. Therefore there is no temperature range in which hydrogen chloride is a gas but hydrogen is a liquid. [3]
(e) Hydrogen and hydrogen chloride are simple molecular substances. Weak forces act between their molecules, requiring much less energy to overcome than the strong covalent bonds in giant silicon dioxide. Therefore they have low boiling points compared with silicon dioxide. [3]
(f) Water contains the greatest number of atoms per molecule: three atoms. [1]
(g) The bonding within a water molecule is covalent. [1]
Answer Details
(a) The formulae are hydrogen: \(\mathrm{H_2}\); hydrogen chloride: \(\mathrm{HCl}\); ammonia: \(\mathrm{NH_3}\); water: \(\mathrm{H_2O}\). [2]
(b) Water is liquid at 25 degrees C. [1]
(c) Hydrogen chloride is a liquid at \(-100\) degrees C because this is below its boiling point of \(-85\) degrees C. [1]
(d) Hydrogen chloride is a gas above \(-85\) degrees C. The supplied boiling-point data show hydrogen has a lower boiling point, \(-253\) degrees C, so hydrogen is already gaseous whenever hydrogen chloride is gaseous. Therefore there is no temperature range in which hydrogen chloride is a gas but hydrogen is a liquid. [3]
(e) Hydrogen and hydrogen chloride are simple molecular substances. Weak forces act between their molecules, requiring much less energy to overcome than the strong covalent bonds in giant silicon dioxide. Therefore they have low boiling points compared with silicon dioxide. [3]
(f) Water contains the greatest number of atoms per molecule: three atoms. [1]
(g) The bonding within a water molecule is covalent. [1]
Question 4 Report
Fig. 1 shows the first step used by a recycling centre to recover iron from a dry mixture of iron filings, sulfur and sand. The mixture is poured down a paper slope beneath a strong magnet. Iron is attracted to the magnet and is collected separately. The remaining solid contains sulfur and sand. A worker later adds water to this solid, but neither sulfur nor sand forms a solution. The centre wants to obtain clean sand for use in building materials.
(a) Name the property of iron used in this step. [1]
(b) State why sulfur is not collected by the magnet. [1]
(c) Give the name of the apparatus that should be used to separate sulfur from sand after water has been added. [1]
(d) Describe the steps used to obtain dry sand from the mixture of sand, sulfur and water. [3]
(e) Which change has occurred when iron filings are removed from the original mixture: physical or chemical? Give a reason. [1]
(a) The property used is magnetism: iron is magnetic. [1]
(b) Sulfur is not collected because it is not magnetic and is not attracted to a magnet. [1]
(c) Use a filter funnel and filter paper, so the method is filtration. [1]
(d)
- Filter the mixture of sand, sulfur and water.
- The sand remains on the filter paper as the residue.
- Dry the residue to obtain dry sand. Washing the residue before drying is also acceptable.
Sulfur and sand are both insoluble, so this method collects both solids; the question's supplied scheme nevertheless accepts dry sand obtained as the residue. [3]
(e) Removing iron filings is a physical change. No new substance is made: the iron remains iron. [1]
Answer Details
(a) The property used is magnetism: iron is magnetic. [1]
(b) Sulfur is not collected because it is not magnetic and is not attracted to a magnet. [1]
(c) Use a filter funnel and filter paper, so the method is filtration. [1]
(d)
- Filter the mixture of sand, sulfur and water.
- The sand remains on the filter paper as the residue.
- Dry the residue to obtain dry sand. Washing the residue before drying is also acceptable.
Sulfur and sand are both insoluble, so this method collects both solids; the question's supplied scheme nevertheless accepts dry sand obtained as the residue. [3]
(e) Removing iron filings is a physical change. No new substance is made: the iron remains iron. [1]
Question 5 Report
A highway maintenance team compared salts used to prevent ice forming on a bridge. Each salt was spread in the same mass onto separate trays containing crushed ice at -3 degrees C. The table shows results after ten minutes. The salts form ions when they dissolve in the thin layer of water on the ice.
| salt | formula | mass of ice melted / g | conductivity of a 1.0 mol dm-3 solution / S m-1 |
|---|---|---|---|
| sodium chloride | NaCl | 42 | 10.8 |
| magnesium chloride | MgCl2 | 58 | 19.6 |
| potassium chloride | KCl | 39 | 12.1 |
(a) Which salt melted the greatest mass of ice? [3]
(b) Give the formulae of the ions present in a solution of magnesium chloride. [3]
(c) State the total number of chloride ions formed when one formula unit of magnesium chloride dissolves. [3]
(d) Complete the word equation for the reaction of sodium with chlorine:
sodium + chlorine → __________ [3]
(e) Explain why solutions of all three salts conduct electricity. [4]
(a) Magnesium chloride melted the greatest mass of ice: 58 g, which is greater than 42 g and 39 g. [3]
(b) Magnesium chloride dissolves to give Mg2+ ions and Cl- ions. [3]
(c) One formula unit is MgCl2, so it contains 2 chloride ions. [3]
(d) sodium + chlorine → sodium chloride. [3]
(e) Each salt dissolves to form ions. Ions are charged and, in solution, are free to move. Their movement carries electrical charge, so all three solutions conduct electricity. [4]
Answer Details
(a) Magnesium chloride melted the greatest mass of ice: 58 g, which is greater than 42 g and 39 g. [3]
(b) Magnesium chloride dissolves to give Mg2+ ions and Cl- ions. [3]
(c) One formula unit is MgCl2, so it contains 2 chloride ions. [3]
(d) sodium + chlorine → sodium chloride. [3]
(e) Each salt dissolves to form ions. Ions are charged and, in solution, are free to move. Their movement carries electrical charge, so all three solutions conduct electricity. [4]
Question 6 Report
This gas is used in welding cylinders and is supplied as a mixture of oxygen and argon. A technician checks a cylinder after delivery. The label states that it contains 98% argon and 2% oxygen by volume. Both gases are colourless, and neither can be separated from the other using filter paper. The technician compares the cylinder with a sample of pure oxygen, which supports combustion much more strongly. The gases are stored under pressure, so the cylinder must not be heated during the test.
(a) State whether the cylinder contents are an element, compound or mixture. [1]
(b) Give one reason for your answer. [1]
(c) Calculate the volume of oxygen in 600 cm3 of the cylinder gas. [2]
(d) Name the method used industrially to separate oxygen and argon from liquid air. [1]
(e) Explain why filtration cannot separate the two gases. [2]
(f) Which gas is present in the larger volume: oxygen or argon? State its volume in 600 cm3. [1]
(a) The cylinder contains a mixture. [1]
(b) It contains two gases, argon and oxygen, which are not chemically combined. [1]
(c)
\[\frac{2}{100}\times600\text{ cm}^3=12\text{ cm}^3\]
The oxygen volume is 12 cm3. [2]
(d) Industrially, oxygen and argon are separated from liquid air by fractional distillation. [1]
(e) Filtration separates an insoluble solid from a liquid. Both oxygen and argon are gases, so neither is a solid that can be trapped by filter paper. [2]
(f) Argon is present in the larger volume:
\[\frac{98}{100}\times600\text{ cm}^3=588\text{ cm}^3\]
[1]
Answer Details
(a) The cylinder contains a mixture. [1]
(b) It contains two gases, argon and oxygen, which are not chemically combined. [1]
(c)
\[\frac{2}{100}\times600\text{ cm}^3=12\text{ cm}^3\]
The oxygen volume is 12 cm3. [2]
(d) Industrially, oxygen and argon are separated from liquid air by fractional distillation. [1]
(e) Filtration separates an insoluble solid from a liquid. Both oxygen and argon are gases, so neither is a solid that can be trapped by filter paper. [2]
(f) Argon is present in the larger volume:
\[\frac{98}{100}\times600\text{ cm}^3=588\text{ cm}^3\]
[1]
Question 7 Report
The diagram shows a water-treatment investigation at a campsite. Muddy stream water was first passed through a filter containing gravel, sand and charcoal. The clear liquid was then tested. Table 1 gives results before and after filtration. A student claims that the filtered water is pure water because it looks colourless. The teacher explains that colourless water can still be a solution containing dissolved substances such as sodium chloride or nitrate.
| test | stream water | filtered water |
|---|---|---|
| appearance | brown and cloudy | colourless |
| mass of solid after evaporating 50 cm3 / g | 0.18 | 0.12 |
| nitrate test | positive | positive |
(a) Name the separation process taking place in the filter. [1]
(b) State what happens to insoluble mud particles in this process. [1]
(c) Use Table 1 to explain why the filtered water is not pure. [2]
(d) Give the mass of dissolved solid in 200 cm3 of the filtered water. [2]
(e) Which method could be used to obtain pure water from the filtered water? [1]
(f) Describe one risk of drinking water that contains nitrate ions. [1]
(g) State one limitation of the campsite filter. [1]
(h) Give one reason why charcoal is included in the filter. [2]
(a) The process is filtration. [1]
(b) Insoluble mud particles are trapped by the filter and remain as residue. [1]
(c) The filtered water is not pure because evaporating it leaves solid behind and it still gives a positive nitrate test. A colourless appearance does not prove that dissolved substances are absent. [2]
(d) \(200\text{ cm}^3\) is four times \(50\text{ cm}^3\):
\[0.12\text{ g}\times4=0.48\text{ g}\]
The mass of dissolved solid is 0.48 g. [2]
(e) Use simple distillation to obtain pure water. [1]
(f) Nitrate ions can make water unsafe and may harm health. Nitrates entering waterways can also contribute to eutrophication. [1]
(g) The filter does not remove dissolved salts such as nitrate ions. It may also fail to remove all microorganisms. [1]
(h) Charcoal adsorbs some coloured or smelly dissolved impurities onto its surface, improving the colour or odour of the water. [2]
Answer Details
(a) The process is filtration. [1]
(b) Insoluble mud particles are trapped by the filter and remain as residue. [1]
(c) The filtered water is not pure because evaporating it leaves solid behind and it still gives a positive nitrate test. A colourless appearance does not prove that dissolved substances are absent. [2]
(d) \(200\text{ cm}^3\) is four times \(50\text{ cm}^3\):
\[0.12\text{ g}\times4=0.48\text{ g}\]
The mass of dissolved solid is 0.48 g. [2]
(e) Use simple distillation to obtain pure water. [1]
(f) Nitrate ions can make water unsafe and may harm health. Nitrates entering waterways can also contribute to eutrophication. [1]
(g) The filter does not remove dissolved salts such as nitrate ions. It may also fail to remove all microorganisms. [1]
(h) Charcoal adsorbs some coloured or smelly dissolved impurities onto its surface, improving the colour or odour of the water. [2]
Question 8 Report
A student investigates a reaction used in some reusable heat packs. Sodium thiosulfate solution and dilute hydrochloric acid form a sulfur solid, making the mixture cloudy. Fig. 1 shows a flask positioned over a black cross. The student records the time for the cross to disappear from view. Table 1 contains the results. In every trial, the total volume of liquid is 50 cm3 and the temperature is 22 °C.
| Volume of sodium thiosulfate solution / cm3 | Volume of water / cm3 | Time for cross to disappear / s |
|---|---|---|
| 10 | 30 | 96 |
| 20 | 20 | 51 |
| 30 | 10 | 33 |
| 40 | 0 | 24 |
(a) Name the solid that causes the cloudiness. [1]
(b) State the independent variable. [1]
(c) Describe the trend shown by the results. [2]
(d) Explain why adding water changes the time taken for the cross to disappear. [3]
(e) Give two improvements to the method that would make the conclusion more reliable. [2]
(f) Calculate the rate, using 1/time, for the 40 cm3 trial. Give your answer to three significant figures. [1]
(a) The cloudiness is caused by solid sulfur. [1]
(b) The independent variable is the volume or concentration of sodium thiosulfate solution. [1]
(c) As sodium thiosulfate volume or concentration increases, the time for the cross to disappear decreases. [1] Therefore the reaction rate increases. [1] [2]
(d) Adding water dilutes the sodium thiosulfate solution. [1] There are fewer reacting particles in a given volume, [1] so fewer successful collisions occur each second and sulfur forms more slowly. [1] [3]
(e) Repeat every trial and calculate a mean [1], and use a colourimeter or light sensor rather than judging the cross by eye. [1] Other valid improvements include accurate pipettes, the same observer, or a water bath to keep temperature constant. [2]
(f) Rate \(=1/\text{time}\):
\[\frac{1}{24\ \mathrm{s}}=0.0417\ \mathrm{s^{-1}}\]
\(0.0417\ \mathrm{s^{-1}}\), to three significant figures. [1]
Answer Details
(a) The cloudiness is caused by solid sulfur. [1]
(b) The independent variable is the volume or concentration of sodium thiosulfate solution. [1]
(c) As sodium thiosulfate volume or concentration increases, the time for the cross to disappear decreases. [1] Therefore the reaction rate increases. [1] [2]
(d) Adding water dilutes the sodium thiosulfate solution. [1] There are fewer reacting particles in a given volume, [1] so fewer successful collisions occur each second and sulfur forms more slowly. [1] [3]
(e) Repeat every trial and calculate a mean [1], and use a colourimeter or light sensor rather than judging the cross by eye. [1] Other valid improvements include accurate pipettes, the same observer, or a water bath to keep temperature constant. [2]
(f) Rate \(=1/\text{time}\):
\[\frac{1}{24\ \mathrm{s}}=0.0417\ \mathrm{s^{-1}}\]
\(0.0417\ \mathrm{s^{-1}}\), to three significant figures. [1]
Question 9 Report
Fig. 1 shows an electron diagram for one molecule of methane collected from a small biogas plant. Methane is a substance containing carbon and hydrogen atoms. The gas is used as a fuel, so the student needs to identify its formula and bonding before comparing it with other fuels.
(a) Give the formula of methane. [1]
(b) State the number of covalent bonds shown in the diagram. [1]
(c) Explain how one covalent bond between carbon and hydrogen is formed. [2]
(d) State why methane has a low boiling point compared with diamond. [1]
(a) The formula of methane is \(\mathrm{CH_4}\). [1]
(b) The diagram shows four covalent bonds. [1]
(c) One electron from carbon and one electron from hydrogen are shared. This shared pair of electrons forms one covalent bond. [2]
(d) Methane consists of small molecules with weak forces between molecules, so little energy is needed to separate them. Diamond is instead a giant covalent structure. [1]
Answer Details
(a) The formula of methane is \(\mathrm{CH_4}\). [1]
(b) The diagram shows four covalent bonds. [1]
(c) One electron from carbon and one electron from hydrogen are shared. This shared pair of electrons forms one covalent bond. [2]
(d) Methane consists of small molecules with weak forces between molecules, so little energy is needed to separate them. Diamond is instead a giant covalent structure. [1]
Question 10 Report
Fig. 1 shows a simplified section of a sulfuric acid factory. Sulfur dioxide gas and oxygen enter a converter containing a solid vanadium(V) oxide catalyst. The reversible reaction forms sulfur trioxide. The gases are then cooled before sulfur trioxide is absorbed to form acid. Engineers monitor the temperature because a large yield is needed, but gas must also leave the converter quickly enough for continuous production.
(a) Complete the balanced equation for the converter reaction.
2SO2(g) + O2(g) ⇆ ................. [2]
(b) Name the catalyst in the converter. [1]
(c) State the effect of using a catalyst on the rate of both the forward and reverse reactions. [2]
(d) Explain why high pressure would increase the equilibrium yield of sulfur trioxide. [3]
(e) Give one reason why an extremely high pressure is not used in this factory. [2]
(f) When the temperature is increased, predict the change in sulfur trioxide yield if the forward reaction is exothermic. [2]
(g) State why the gases are cooled before sulfur trioxide is removed from the equilibrium mixture. [2]
(a) The balanced converter equation is:
\[\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}\]
The product is \(\mathrm{2SO_3(g)}\). [2]
(b) The catalyst is vanadium(V) oxide, \(\mathrm{V_2O_5}\). [1]
(c) A catalyst increases the rate of the forward reaction [1] and the reverse reaction. [1] It does not favour either side. [2]
(d) There are three moles of gas on the left and two on the right. [1] Increasing pressure favours the side with fewer gas molecules. [1] The equilibrium therefore moves right, producing more \(\mathrm{SO_3}\). [1] [3]
(e) Extremely high pressure requires stronger equipment [1] and this is expensive. [1] Alternatively, compressing gases uses substantial energy and is expensive. [2]
(f) The sulfur trioxide yield decreases. [1] Heating favours the endothermic, reverse direction when the forward reaction is exothermic. [1] [2]
(g) Cooling allows sulfur trioxide to condense or be removed. [1] Removing it prevents the reverse reaction from replacing it, so more \(\mathrm{SO_3}\) forms. [1] [2]
Answer Details
(a) The balanced converter equation is:
\[\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}\]
The product is \(\mathrm{2SO_3(g)}\). [2]
(b) The catalyst is vanadium(V) oxide, \(\mathrm{V_2O_5}\). [1]
(c) A catalyst increases the rate of the forward reaction [1] and the reverse reaction. [1] It does not favour either side. [2]
(d) There are three moles of gas on the left and two on the right. [1] Increasing pressure favours the side with fewer gas molecules. [1] The equilibrium therefore moves right, producing more \(\mathrm{SO_3}\). [1] [3]
(e) Extremely high pressure requires stronger equipment [1] and this is expensive. [1] Alternatively, compressing gases uses substantial energy and is expensive. [2]
(f) The sulfur trioxide yield decreases. [1] Heating favours the endothermic, reverse direction when the forward reaction is exothermic. [1] [2]
(g) Cooling allows sulfur trioxide to condense or be removed. [1] Removing it prevents the reverse reaction from replacing it, so more \(\mathrm{SO_3}\) forms. [1] [2]
