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| Chemistry Past Paper Questions January 2009 | On the front of the card is a chemistry past paper question from January 2009. Try and answer it and then check your answer with the model one on the back of the card. Good luck! |
| (1) Carbon occurs in a wide range of compounds and is essential to living systems. (a) Two isotopes of carbon are 12C and 13C. (i) State what is meant by the term isotopes. 1 mark | Atoms of the same element with different numbers of neutrons. |
| (ii) Isotopes of carbon have the same chemical properties. Explain why. 1 mark | They all have the same number of electrons in the outer shell. |
| (iii) The 12C isotope is used as the standard measurement of relative masses. Define the term relative isotopic mass. 2 marks | Mass of the isotope compared to 1/12th the mass of a carbon-12 atom. |
| (b) One form of naturally occurring carbon is graphite. The table below lists some properties of graphite. Electrical conductivity: good conductor Hardness: soft Melting point: very high • Describe the bonding and structure in graphite. • Explain, in terms of bonding and structure, the properties of graphite shown above. 5 marks | Graphite is a giant covalent lattice and has layers. It is a good conductor because it has delocalised electrons that can move. It has a high melting point because the strong covalent bonds have to be broken and this requires a lot of energy. It is soft because of the weak London forces between layers and so the layers can easily slide over each other. |
| (c) In the sixteenth century, a large deposit of graphite was discovered in the Lake District. People at the time thought that the graphite was a form of lead. Nowadays, graphite is used in pencils but it is still referred to as ‘pencil lead’. A student decided to investigate the number of carbon atoms in a ‘pencil lead’. He found that the mass of the ‘pencil lead’ was 0.321 g. (i) Calculate the amount, in mol, of carbon atoms in the student’s pencil lead. Assume that the ‘pencil lead’ is pure graphite. 1 mark | 0.321/12 = 0.0268 |
| (ii) Using the Avogadro constant, NA, calculate the number of carbon atoms in the student’s ‘pencil lead 1 mark | 0.0268 X (6.022140857747474 x1023) = 1.61X10^22 Total of 11 marks for this section. |
| (2) Chemists have developed models for bonding and structure which are used to explain different properties. (a) Ammonia, NH3, is a covalent compound. (i) Explain what is meant by a covalent bond. 1 mark | The electrostatic attraction between a shared pair of electrons. |
| (ii) Draw a ‘dot-and-cross’ diagram to show the bonding in NH3. Show outer electrons only. 1 mark | |
| (iii) Name the shape of the ammonia molecule. Explain, using your ‘dot-and-cross’ diagram, why ammonia has this shape and has a bond angle of 107°. 3 marks | The shape is trigonal pyramid. There are three bonded pairs and one lone pair of electrons and the lone pair repels more than the bonded pairs. |
| (b) Ammonia reacts with hydrogen chloride, HCl, to form ammonium chloride, NH4Cl. NH4Cl is an ionic compound containing NH4 + and Cl – ions. (i) Complete the electron configuration of the Cl – ion. 1s2 ................................. 1 mark | 1s2 2s2 2p6 3s2 3p6 |
| (ii) Draw a ‘dot-and-cross’ diagram to show the bonding in NH4 +. Show outer electrons only. 1 mark | |
| (iii) State the shape of, and bond angle in, an NH4 + ion. 2 marks | Tetrahedral, 109.5 degrees |
| (iv) A student investigated the conductivity of ammonium chloride. She noticed that when the ammonium chloride was solid it did not conduct electricity. However, when ammonium chloride was dissolved in water, the resulting solution did conduct electricity. Explain these observations. 2 marks | The ions/electrons cannot move in a solid but the ions can move in a solution. |
| (c) Ammonium compounds such as ammonium sulfate, (NH4 )2SO4, can be used as fertilisers. (i) Write a balanced equation to show how ammonium sulfate could be formed by the reaction between aqueous ammonia and sulfuric acid. 1 mark | 2NH3 + H2SO4 -> (NH4)2SO4 |
| (ii) Ammonium sulfate is an example of a salt formed when an acid is neutralised by a base. Explain what is meant by the term salt. 1 mark | When the H+ in an acid is replaced by a metal ion. |
| (iii) Why is ammonia acting as a base in this neutralisation? 1 mark | It accepts a proton (H+). |
| (iv) What is the relative formula mass of (NH4 )2SO4? Give your answer to one decimal place. 1 mark | 132.1 Total of 15 marks for this section. |
| (3) A student used the internet to research chlorine and some of its compounds. (a) He discovered that sea water contains chloride ions. The student added aqueous silver nitrate to a sample of sea water. (i) What would the student see? 1 mark | A white precipitate. |
| (ii) Write an ionic equation, including state symbols, for the reaction that would occur. 2 marks | Ag+(aq) + Cl-(aq) -> AgCl(s) |
| (iii) After carrying out the test in (i), the student added dilute aqueous ammonia to the mixture. What would the student see? 1 mark | The precipitate dissolves and it goes colourless. |
| (b) The student also discovered that chlorine, Cl 2, is used in the large-scale treatment of water. (i) State one benefit of adding chlorine to water. 1 mark | Kill bacteria in the water, making it safe to drink. |
| (ii) Not everyone agrees that chlorine should be added to drinking water. Suggest one possible hazard of adding chlorine to drinking water. 1 mark | It is toxic. |
| (c) The equation for the reaction of chlorine with water is shown below. Cl 2(g) + H2O(l) HCl(aq) + HCl O(aq) (i) State the oxidation number of chlorine in: Cl2 ……… HCl ……… HClO ……… 1 mark | Cl2=0 HCl=-1 HClO=+1 |
| (ii) The reaction of chlorine with water is a disproportionation reaction. Use the oxidation numbers in (i) to explain why. 2 mark | It has been both oxidised and reduced. Oxidised from 0 to +1 and reduced from 0 to -1. |
| (iii) Chlorine reacts with sodium hydroxide to form bleach in another disproportionation reaction. Write an equation for this reaction. 1 mark | Cl2 + 2NaOH -> NaClO + NaCl + H20 |
| (d) Two other chlorine compounds of chlorine are chlorine dioxide and chloric(V) acid. (i) Chlorine dioxide, ClO2, is used as a bleaching agent in both the paper and the flour industry. When dry, ClO2 decomposes explosively to form oxygen and chlorine. Construct an equation for the decomposition of ClO2. 1 mark | 2Cl2 -> Cl2 + 2O2 |
| (ii) Chloric(V) acid has the following percentage composition by mass: H, 1.20%; Cl, 42.0%; O, 56.8%. Using this information, calculate the empirical formula of chloric(V) acid. Show all of your working. 2 marks | H: 1.2/1 = 1.2 Cl: 42/35.5 = 1.183 O: 56.8/16 = 3.55 1.2/1.183 = 1.014 1.183/1.183 = 1 3.55/1.183 = 3 HClO3 |
| (iii) What does (V) represent in chloric(V) acid? 1 mark | The oxidation number of chlorine. Total of 14 marks for this section. |
| (4) The table shows the melting points, atomic radii and other features of the elements in Period 3. 1 pm=1×10–12 m (a) (i) Explain the difference in melting point for the elements Na and Mg. 3 marks | The magnesium ions have a greater charge and magnesium has more delocalised electrons. Therefore, Mg has a greater attractions between ions and electrons and so has stronger metallic bonds. More energy is needed to break these bonds and so Mg has a melting boiling point. |
| (ii) Sulfur exists as S8 molecules and chlorine as Cl 2 molecules. Use this information to explain the difference in their melting points. 2 marks | S8 has more electrons and so it has stronger London forces than Cl2. Therefore, more energy is needed to break the intermolecular forces in S8. |
| (b) Explain the decrease in the atomic radii across the period from Na to Cl. In your answer, you should use appropriate technical terms, spelt correctly. 3 marks | Across the period, the number of protons in each atom increases by one and so the nuclear charge increases. Electrons are added to the same shell, meaning shielding remains the same. Therefore, there is a greater attraction between the nucleus and the outer electrons, so the atom is smaller. Total of 8 marks for this section. |
| (5) The Group 2 element barium, Ba, is silvery white when pure but blackens when exposed to air. The blackening is due to the formation of both barium oxide and barium nitride. The nitride ion is N3–. (a) Predict the formula of: barium oxide …… barium nitride …… | BaO and Ba3N2 |
| (b) A 0.11 g sample of pure barium was added to 100 cm3 of water. Ba(s) + 2H2O(l) Ba(OH)2(aq) + H2(g) (i) Show that 8.0 × 10−4 mol of Ba were added to the water. 1 mark | 0.11/137.3 |
| (ii) Calculate the volume of hydrogen, in cm3, produced at room temperature and pressure 1 mark | 8x(10^-4) X 24000 = 19.2cm3 |
| (iii) Calculate the concentration, in mol dm−3, of the Ba(OH)2(aq) solution formed. 1 mark | 8x10^-3 |
| (iv) State the approximate pH of the Ba(OH)2(aq) solution. 1 mark | >7 but <15 |
| (c) A student repeated the experiment in (b) using a 0.11 g sample of barium that had blackened following exposure to the air. Suggest why the volume of hydrogen produced would be slightly less than the volume collected using pure barium. 1 mark | Less barium to react as sodium barium has already reacted. |
| (d) Describe and explain the trend, down the group, in the reactivity of the Group 2 elements with water. 5 marks | The reactivity increases down the group. The atomic radii increases, as there are more shells. There is also more shielding. Therefore, the nuclear attraction decreases between the outer electron and the nucleus. The increased shielding and distance outweigh the increased nuclear charge. It is easier to remove the outer electrons, as ionisation energy decrease. Total of 12 marks for this section. |
| End of Test. Well done! Now check how you did using the grade boundaries below! A = 46 B = 40 C = 34 D = 28 E = 23 U = 0 Turn over for a quick recap on bonding. |
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