Chemistry

ChemistryEdexcel International GCSE in Chemistry (4CH0)First examination June 201311Principles of chemistryStates of matterAtomsAtomic structureRelative formula masses and molar volumes of gasesChemical formulae and chemical equationsIonic compoundsCovalent substancesMetallic crystalsElectrolysis2States of matter1.1 understand the arrangement, movement and energy of the particles in each of the three states of matter: solid, liquid and gasSolid: Particles in fixed positions, regular repeating pattern, strong inter-particle forces.Liquid: Particles not in fixed positions, can flow, irregular pattern, weaker inter-particle forces.Gas: Particles not in fixed positions, can flow, irregular pattern, non-existent inter-particle forces.

1.2 understand how the interconversions of solids, liquids and gases are achieved and recall the names used for these interconversionsSolid Liquid is MELTING. Heat the solid until it melts.Liquid Sold is FREEZING. Cool the liquid until it freezes.Liquid Gas is EVAPORATION. Heat the liquid until it boils.Liquid Gas is CONDENSATION. Cool the gas until it condenses.Solid Gas is SUBLIMATION. Heat the solid until it sublimes.

3States of matter1.3 explain the changes in arrangement, movement and energy of particles during these interconversions.Melting: The particles gain kinetic energy as they are heated and vibrate faster and faster. This allows the particles to overcome the forces of attraction between them. This allows the particles to break the regular pattern and to flow over one another.Freezing: The particles lose kinetic energy as they are cooled and this allows the forces of attraction to hold them together. The particles arrange themselves into a regular pattern and are no longer able to flow over one another.Evaporation: The particles gain kinetic energy as they are heated and move further apart. Eventually the forces of attraction between the particles are completely broken and they are able to escape.Condensation: The particles lose kinetic energy as they are cooled and this allows the forces of attraction to bring them closer together. The particles gradually clump together to form a liquid.Sublimation: The particles gain kinetic energy as they are heated and vibrate faster and faster. Eventually the forces of attraction between the particles are completely broken and they are able to escape.

4Atoms1.4 describe and explain experiments to investigate the small size of particles and their movement including:i dilution of colored solutionsWhen potassium manganate (VII) crystals are dissolved in water, a purple solution is formed. When this solution is diluted with water several times, the color becomes less intense. However it takes many dilutions for the color to completely fade.This indicates that there is a large number of particles in a small solid, denoting that potassium manganate particles are very small.

ii diffusion experimentsThe bromine gas in the flask on one side will diffuse quickly into the flask on the other side. This is because of large gaps between particles, allowing the air and bromine to easily mix.The ammonium chloride experiment shows that particles in different gases travel at different speeds. The while layer of ammonium chloride is closer to the hydrochloric acid because the ammonia gas travels faster because it is lighter.

5Atoms1.5 understand the terms atom and moleculeAtoms are made up of smaller, sub-atomic particles called protons, neutrons and electrons.Molecules are made up of two ore more atoms covalently bonded together.

1.6 understand the differences between elements, compounds and mixturesElements are substances that cannot be further broken down.Compounds are substances that are created through a chemical reaction from different elements. They have different properties to their elements.Mixtures are substances that are created through a physical action of mixing from different elements. They have properties from each of the elements used.

6Atoms1.7 describe experimental techniques for the separation of mixtures, including simple distillation, fractional distillation, filtration, crystallization and paper chromatographyFiltration is to separate an undissolved solid from a mixture of the solid and a liquid.Pass the mixture through a filter, thus filtering out the solid.Evaporation is to separate a dissolved solid from a solution, when the solid has similar solubilities in both cold and hot solvent.Heat the solution until the liquid evaporates and only the solid remains.Crystallization is to separate a dissolved solid from a solution, when the solid is much more soluble in hot solvent than in cold.Dissolve the solution in a hot solvent, then allow it to rest and crystals will appear in the flask. Using filtration, extract the crystals, then rinse it with cold solution to remove impurities.Simple Distillation is to separate a liquid from a solution.Heat the solution until the liquid evaporates, then pass it through a condenser to condense the gas back into liquid.7AtomsFractional Distillation is to separate two or more liquids that are miscible with one another.Heat the mixture up to a certain boiling point that only one of the liquids evaporates, then pass it through a condenser to condense the gas back into liquid.Paper Chromatography is to separate substances that have different solubilities in a given solvent.The different substances travel up the paper at different speeds due to their different solubilities.

1.8 explain how information from chromatograms can be used to identify the composition of a mixture.The distance they travel on the paper denotes their solubility.

8Atomic Structure1.9 understand that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons, orbiting in shellsNeutronsElectronsProtons

9Atomic Structure1.10 recall the relative mass and relative charge of a proton, neutron and electronProtons: +1 Charge, 1 MassElectrons: -1 Charge, 1/1836 (Basically 0) MassNeutrons: 0 Charge, 1 Mass

1.11 understand the terms atomic number, mass number, isotopes and relative atomic mass (Ar)Atomic Number is the number of protons.Mass Number is the number of protons + the number of neutrons.Isotopes occur when atoms have the same number of protons but different number of neutrons.Relative Atomic Mass is calculated from the masses and relative abundancies of all the isotopes of an element. Given the symbol Ar.10Atomic Structure1.12 calculate the relative atomic mass of an element from the relative abundances of its isotopesChlorine has two isotopes: chlorine-35 and chlorine-37.In a typical sample of chlorine, 75% is chlorine-35 and 25% is chlorine-37.(0.75 X 35) + (0.25 X 37) = 35.5 = Ar of chlorine

1.13 understand that the Periodic Table is an arrangement of elements in order of atomic numberThe periodic table is arranged by the number of protons in the element. The element is defined by its number of protons.

1.14 deduce the electronic configurations of the first 20 elements from their positions in the Periodic TableThe period (row) in the periodic table defines the number of shells it has.

11Atomic Structure1.15 deduce the number of outer electrons in a main group element from its position in the Periodic Table.The group (column) 1-7 in the periodic table defines the number of electrons on the outer shell.

12Relative formula masses and molar volumes of gases1.16 calculate relative formula masses (Mr) from relative atomic masses (Ar)Hydrogen has an Ar of 1.H2 has a Mr of 2 X 1 = 2Oxygen has an Ar of 16.H2O has a Mr of 2 X 1 + 16 = 18Likewise, (NH4)2SO4 has an Mr of 132(2 X 14) + (8 X 1) + 32 + (4 X 16) = 132

1.17 understand the use of the term mole to represent the amount of substanceMole is a measure of the amount of substance.

1.18 understand the term mole as the Avogadro number of particles (atoms, molecules, formulae, ions or electrons) in a substance6.023 X 1023 is known as the Avogrado number.1 mol of an substance contains 6.023 X 1023 of the substance.

13Relative formula masses and molar volumes of gases1.19 carry out mole calculations using relative atomic mass (Ar) and relative formula mass (Mr)Mole calculations can be done with Ar and Mr..e.g H2 has a Mr of 21 mol of H2 is 2g.Likewise, 8 mols of H2 is 16ge.g. Mg has a Ar of 240.5 mols of Mg is 12gMass of Substance (g) = Amount (mols) X Mr

1.20 understand the term molar volume of a gas and use its values (24 dm3 and 24,000 cm3) at room temperature and pressure (rtp) in calculations.One mole of any gas has a volume of 24 dm3.To calculate the volume of gas at rtp: Volume of gas (dm3) = Amount (mols) X 24To calculate the amount of gas at rtp: Amount of gas (mols) = Volume of gas (dm3) / 2414Chemical formulae and chemical equations1.21 write word equations and balanced chemical equations to represent the reactions studied in this specificationUnbalanced Equation: Al + CuO Al2O3 + CuLeft side: 1 Al, 1 Cu, 1 ORight side: 2 Al, 1 Cu, 3 OBalanced Equation: 2Al + 3CuO Al2O3 + 3CuLeft side: 2 Al, 3 Cu, 3 ORight side: 2 Al, 3 Cu, 3 OWord Equation: Aluminum + Copper Oxide Aluminum Oxide + Copper

1.22 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectivelyZn(s) + CuSO4(aq) ZnSO4 (aq) + Cu(s)Tells us that when solid zinc is added to an aqueous Copper Sulfate solution, it forms an aqueous Zinc Sulfate solution and solid Copper.

15Chemical formulae and chemical equations1.23 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallizatione.g. When 20g of hydrated copper sulfate salts are heated, and 7.2g of water evaporates.The mass of the anhydrous salt copper sulfate is 12.8g.The mass of the water is 7.2g.To deduce the mole ratio divide the masses by their respective Mr..Copper Sulfate: 12.8 / 160 = 0.08 Water: 7.2 / 18 = 0.40.08 : 0.4 = 1 : 5The empirical formula of hydrated copper sulfate salts is CuSO4.5H2O16Chemical formulae and chemical equations1.24 calculate empirical and molecular formulae from experimental datae.g. To find the empirical formula of aluminum oxide:From an experiment it is deduced that there is 52.94g of aluminum and 47.06g of oxygen in a sample of aluminum oxide.Aluminum: 52.94 / 27 = 1.96Oxygen: 47.06 / 16 = 2.941.96 : 2.94 = 2 : 3The empirical formula of aluminum oxide is Al2O3e.g. To find the molecular formula of benzene:Empirical formula of benzene is CH.The Mr of Benzene is 78.The empirical formula mass of CH = 1378 / 13 = 6Therefore, the molecular formula of Benzene is C6H6

17Chemical formulae and chemical equations1.25 calculate reacting masses using experimental data and chemical equationse.g. Calculate the mass of magnesium oxide made by completely burning 6g of magnesium in oxygen. (2Mg + O2 2MgO)Ar of magnesium is 24. Therefore, 6 / 24 = 0.25 mols.The equation states that 2 mols of Mg makes 2 mols of MgO, thus 0.25 mols of MgO is created from 0.25 mols of Mg.Mr of MgO is 400.25 X 40 = 10g of MgO formed.

18Chemical formulae and chemical equations1.26 calculate percentage yieldPercentage yield = Yield obtained / theoretical yielde.g. To displace copper from copper sulfate, 6.5g of zinc was added to an excess of copper sulfate solution. The mass of copper obtained was 4.8g.The equation for the reaction is Zn(s) + CuSO4 (aq) ZnSO4(aq) + Cu(s)Ar of zinc is 65. Therefore, 6.5 / 65 = 0.10 mols.Amount of copper = 0.10 mols. Mass of copper = (0.10 X 64) = 6.4gYield obtained = 4.8gTheoretical yield = 6.4g4.8 / 6.4 X 100 = 75%19Chemical formulae and chemical equations1.27 carry out mole calculations using volumes and molar concentrations.e.g. Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm3 that is required to react completely with 2.5g of calcium carbonate. Mr of calcium carbonate is 100. Therefore, 2.5 / 100 = 0.25 mols.The equation states that 1 mol of CaCO3 requires 2 mols of HCl, thus 0.025 mols of CaCo3 requires 0.05 mols of HCl.Volume = (Amount (cm3) X 1000) / concentration (mol / dm3)(0.05 X 1000) / 1.0 = 50 cm3 of hydrochloric acid needed.

20Ionic compounds1.28 describe the formation of ions by the gain or loss of electronsAn ion is an electrically charged atom or group of atoms. Ions are formed by the loss or gain of electrons.

1.29 understand oxidation as the loss of electrons and reduction as the gain of electronsOxidation: Loss of electronsReduction: Gain of electrons

1.30 recall the charges of common ions in this specification

Ag+OH-NH4+NO3-Cr3+SO42-Zn2+CO32-21Ionic compounds1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formedAn atom seeks to achieve a full shell by oxidation or reduction.When oxidation occurs, an atom loses electrons and thus gains a positive charge.When reduction occurs, an atom gains electrons and thus gains a negative charge.e.g. Magnesium is in Group 2, thus has 2 electrons in its outer shell.To achieve a full shell, it has to lose 2 electrons.Magnesium gains a charge of +2, since each electron lost has a -1 charge.

22Ionic compounds1.32 explain, using dot and cross diagrams, the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7e.g. In NaCl, sodium chloride:Na has 1 electron in its outer shell.Cl has 7 electrons in its outer shell. To achieve full shells, Na donates its extra electron to Cl.Now that Na has one less electron, it has a charge of +1.Now that Cl has one more electron, it has a charge of -1.

NaClNaCl[ ]-[ ]+23Ionic compounds1.33 understand ionic bonding as a strong electrostatic attraction between oppositely charged ionsAn ionic compound forms when a oppositely charged (positive / negative) atoms are attracted to each other. It is a bond between a metal and a non-metal.

1.35 understand the relationship between ionic charge and the melting point and boiling point of an ionic compoundStrong ionic charge: Very high melting and boiling point (MgO: 2800 degrees)Weak ionic charge: High melting and boiling point (NaCl: 800 degrees)

1.36 describe an ionic crystal as a giant three-dimensional lattice structure held together by the attraction between oppositely charged ionsWhen many ions arrange themselves in a three dimensional structure called an giant ionic lattice. Ionic compounds often form crystals.

24Ionic compounds1.37 draw a diagram to represent the positions of the ions in a crystal of sodium chloride.25Covalent substances1.38 describe the formation of a covalent bond by the sharing of a pair of electrons between two atomsWhen atoms of non-metallic elements combine together they share electrons between them to make full shells. This electron bonds them together.

1.39 understand covalent bonding as a strong attraction between the bonding pair of electrons and the nuclei of the atoms involved in the bondWhen the electrons are shared, the shared region of the electron shells bond the two nuclei together. This forms a covalent compound.

26Covalent substances1.40 explain, using dot and cross diagrams, the formation of covalent compounds by electron sharing for the following substances:

HHClClHClHCHHHHCHHCHHHCHHCHH27Covalent substances1.41 understand that substances with simple molecular structures are gases or liquids, or solids with low melting pointsSimple molecular structures are formed from covalent bonds because it consists of individual molecules, these structures have low melting points.Often gases (e.g. hydrogen, oxygen, nitrogen, methane).Low boiling point liquids (e.g. water, bromine)Low melting point solids (e.g. iodine)

1.42 explain why substances with simple molecular structures have low melting and boiling points in terms of the relatively weak forces between the moleculesForces of attraction between molecules are relatively weak compared to ionic bonds, therefore, very little energy is required to overcome them.e.g. If it is a diatomic chlorine molecule, it is breaking the intermolecular bond between the chlorine molecules not the covalent bond.

28Covalent structures1.44 draw diagrams representing the positions of the atoms in diamond and graphite

DiamondGraphite29Covalent substances1.45 explain how the uses of diamond and graphite depend on their structures, limited to graphite as a lubricant and diamond in cutting.Diamond can be used in cutting because covalent bonds are strong, and there are many of them in the giant covalent structure. There are no weak forces in diamond.Graphite can be used as a lubricant because the weak intermolecular forces of attraction between the layers are weak so the layers easily slide over one another and can easily be separated. This gives graphite the soft and slippery property.

30Metallic crystals1.46 understand that a metal can be described as a giant structure of positive ions surrounded by a sea of delocalized electronsMetals have a giant, three dimensional lattice structure in which positive ions are arranged in a regular pattern in a sea of electrons. The outer shell electrons are detached from the atoms and are delocalized throughout the entire structure.

1.47 explain the electrical conductivity and malleability of a metal in terms of its structure and bonding.Metallic bonds are strong and there are many of them in a giant structure, hence a lot of heat energy is required, so melting points are high.Metals can conduct electricity because the delocalized electrons are free to move when a potential difference is applied across the metal.Metals are malleable and ductile because the layers of positive ions can easily slide over one another and take up different positions. The electrons move with them so no bonds are broken.31Electrolysis1.48 understand that an electric current is a flow of electrons or ionsAn electric current is a flow of electrons or ions within a substance, this is why metals are such good conductors.

1.49 understand why covalent compounds do not conduct electricityCovalent compounds do not conduct electricity because they do not have free delocalized electrons that enables this property.

1.50 understand why ionic compounds conduct electricity only when molten or in solutionIonic compounds conduct electricity when molten or in aqueous form because the ions are free to move in these forms and thus allow it to conduct a charge.

32Electrolysis1.51 describe experiments to distinguish between electrolytes and nonelectrolytesElectrolytes are substances that conduct electricity in the molten state or when dissolved in water.Non-electrolytes are substances that dont conduct electricity in these statesTo distinguish between these two, simply place two electrodes in the water connected to a power source and a light bulb, and if the bulb lights up, electrolytes are present.

1.52 understand that electrolysis involves the formation of new substances when ionic compounds conduct electricityPassing a charge through the electrolytes cause chemical changes, thus the chemical reactions produce new products.

33Electrolysis1.53 describe experiments to investigate electrolysis, using inert electrodes, of molten salts such as lead(II) bromide and predict the productsThe electrodes are inert which means they do not react.Cathode: The negative electrode, when ions get to the electrode, they gain electrons.Anode: The positive electrode, when ions get to the electrode, the lose electrons.e.g. In the electrolysis of lead(II) bromide:Pb2+ is attracted towards the cathode, where they gain electrons to become Pb.Pb2+ + 2e- PbBr- is attracted towards the anode, where they lose electrons to become Br.2Br- Br2 + 2e-34Electrolysis1.54 describe experiments to investigate electrolysis, using inert electrodes, of aqueous solutions such as sodium chloride, copper(II) sulfate and dilute sulfuric acid and predict the productsSelective discharge occurs:At the cathode, hydrogen is preferably discharged over all cations unless the cation is less reactive than it.At the anode, oxygen is preferably discharged over all anions unless there is a high concentration of anions. e.g. in the electrolysis of sodium chloride:There are three particles present in the solution, Na+, Cl-, H2O.At the cathode, the water gains electrons to produce hydrogen gas and hydroxide ions.2H2O + 2e- H2 + 2OH-At the anode, the chloride loses electrons to produce chlorine gas.2Cl- Cl2 + 2e-2NaCl(aq) + H2O(l) Cl2(g) + H2(g) + 2NaOH(aq)

35Electrolysise.g. in the electrolysis of copper(II) sulfate:There are three particles present in the solution, Cu2+, SO42-, H2O.At the cathode, copper ions gain electrons to produce copper deposits. (Copper is lower than hydrogen on the reactivity series).Cu2+ + 2e- CuAt the anode, water loses electrons to produce oxygen gas and hydrogen ions.2H2O O2 + 4H+ +4e-2CuSO4(aq) + 2H2O(l) 2Cu(s) + O2(g) + 2H2SO4(aq) e.g. in the electrolysis of dilute sulfuric acid:There are three particles present in the solution, H+, SO42-, H2O.At the cathode, hydrogen ions gain electrons to produce hydrogen gas. 2H+ + 2e- H2At the anode, water loses electrons to produce oxygen gas and hydrogen ions.2H2O O2 + 4H- + 4e- and then 2H2O O2 + 2H2OThe water is decomposed leaving a more concentrated solution of sulfuric acid.

36Electrolysis1.55 write ionic half-equations representing the reactions at the electrodes during electrolysisTypically ionic half equations show the gain and loss of electrons at the electrodes:At the cathode: cation + e- [hydrogen or metal]At the anion: anion [non-metal] + e-

1.56 recall that one faraday represents one mole of electronsOne faraday is one mole of electrons, which equals 96500 coulombs of charge.

37Electrolysis1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions.e.g. When copper is deposited, the cathode reaction is Cu2+ + 2e- CuThus, for every 1 mol of copper deposited, 2 mols of electrons are needed.e.g. Calculate the masses of metals deposited when one faraday of electricity flows through molten lead bromide.Pb2+ + 2e- Pb1 mol of electrons deposit 0.5 mols of lead = 103.5ge.g. A current of 0.010 A passes for 4 hours through a solution of gold (III) nitrate. What mass of metal is deposited.Charge = Current X Time0.010 X (4 X 60 X 60) = 144 CAu3+ + 3e- Au3 x 96500C deposit 1 mol (197g) of gold.144C deposits (197 / (3 X 96500)) X 144 = 0.098g of gold.

38Electrolysise.g. A metal of Ar 48.0 is deposited by electrolysis. If 0.239g of the metal is deposited when 0.100 A flow for 4 hours, what is the charge on the ion of this element.0.100 X (4 X 60 X 60) = 1440 C1440 C deposits 0.239g of metal.48g of metal is deposited by 1440/0.239 X 48 = 289205 C289205 / 96500 = 3 mols of electrons.3 mols of electrons produces 1 mol of metal.The metal has the charge of 3+.

39Chemistry of the elementsThe Periodic TableGroup 1 elements lithium, sodium and potassiumGroup 7 elements chlorine, bromine and iodineOxygen and oxidesHydrogen and waterReactivity seriesTests for ions and gases40The Periodic Table2.1 understand the terms group and periodGroups are the vertical columns of the periodic table. The elements have the same number of electrons in their outermost shell.Periods are the horizontal columns of the periodic table. The elements have the same number of electron shells.

2.2 recall the positions of metals and non-metals in the Periodic TableThe metals are on the left and the righter-most metals are Beryllium, Aluminum, Gallium, Tin, Bismuth.

2.3 explain the classification of elements as metals or non-metals on the basis of their electrical conductivity and the acid-base character of their oxidesMetals conduct electricity, are malleable and ductile, often shiny, solid at room temperature and form oxides that are basic.Non metals do not conduct electricity, are brittle and form oxides that are acidic.

41The Periodic Table2.4 understand why elements in the same group of the Periodic Table have similar chemical propertiesThey have similar chemical properties because they have the same number of electrons in their outer shell.

2.5 understand that the noble gases (Group 0) are a family of inert gases and explain their lack of reactivity in terms of their electronic configurations.Noble gases have full outer shells, thus they do not lose or gain electrons easily, because doing so would either require breaking a full shell, or creating a new one. Therefore, they are inert.

42Group 1 elements lithium, sodium and potassium2.6 describe the reactions of these elements with water and understand that the reactions provide a basis for their recognition as a family of elementsThe alkali metals are recognized as a family due to their reactions with water:Lithium: Moves around the surface of the water, hissing sound, effervescence, deteriorates over time.Sodium: All of the above, melts into a shiny ball.Potassium: All of the above, burns with a lilac-colored flame.

2.7 describe the relative reactivities of the elements in Group 1Increases down the group: Potassium Sodium Lithium

2.8 explain the relative reactivities of the elements in Group 1 in terms of distance between the outer electrons and the nucleus.The reactivity goes up going down the group because less energy is required to remove the one electron on the outer shell. This is because the electron is further away from the nucleus, thus it is less strongly attached.

43Group 7 elements chlorine, bromine and iodine2.9 recall the colors and physical states of the elements at room temperature

2.10 make predictions about the properties of other halogens in this groupFluorine is a pale-yellow gas.Astatine is a black solid.

2.11 understand the difference between hydrogen chloride gas and hydrochloric acidHydrochloric acid is hydrogen chloride dissolved in water, as a solution. It forms H+ ions which makes it acidic.HalogenStateColorChlorineGasPale greenBromineLiquidRed-brownIodineSolidBlack44Group 7 elements chlorine, bromine and iodine2.12 explain, in terms of dissociation, why hydrogen chloride is acidic in water but not in methylbenzeneWhen hydrogen chloride is dissolved in water, it dissociates (splits up) into H+ ions and Cl- ions. The H+ ions are the ones responsible for its acidic properties.When hydrogen chloride is dissolved in methylbenzene, it only dissolves and does not dissociate, thus not forming ions. This means it only exists as HCl molecules, not H+ and Cl- ions.

2.13 describe the relative reactivities of the elements in Group 7Decreases down the group: Fluorine Chlorine Bromine Iodine Astatine

45Group 7 elements chlorine, bromine and iodine2.14 describe experiments to demonstrate that a more reactive halogen will displace a less reactive halogen from a solution of one of its saltse.g. Adding chlorine to a potassium bromide solution will displace the bromide from the potassium.Cl2 + 2KBr 2KCl + Br2The color of aqueous chlorine is very pale green.The color of aqueous bromine is orange.The color of aqueous iodine is brown.

2.15 understand these displacement reactions as redox reactions.In the above reaction, chlorine oxidizes bromine and gains an electron each, this makes Cl- which is attracted to the K+ ion. There is now two bromine atoms, and thus forms a covalent bond to make the diatomic molecule Br2.Cl2 + 2e- 2Cl-2Br- Br2 + 2e-46Oxygen and oxides2.16 recall the gases present in air and their approximate percentage by volumeComposition of air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% carbon dioxide.

2.17 explain how experiments involving the reactions of elements such as copper, iron and phosphorus with air can be used to investigate the percentage by volume of oxygen in airIn the reaction of copper and oxygen: 2Cu + O2 2CuOUsing syringes to pass air over copper over a Bunsen burner. Measure how much air there is left after it stops decreasing. The difference is the volume of oxygen.Placing wet iron filings in the end of the burette, and allowing the water to rise up the burette to show the volume of oxygen that reacted with the iron. The volume of oxygen used divided by the initial volume of gas is the percentage of oxygen in air.47Oxygen and oxides2.18 describe the laboratory preparation of oxygen from hydrogen peroxide, using manganese(IV) oxide as a catalyst2H2O2(aq) 2H2O(l) + O2(g)Oxygen is not very soluble in water, so very little is lost.The speed of hydrogen peroxide decomposition can be catalyzed by MnO2.

2.19 describe the reactions of magnesium, carbon and sulfur with oxygen in air, and the acid-base character of the oxides producedMagnesium burns with a bright, white flame to form a white powder.In solution it forms Mg(OH)2, a basic solution with a pH of 10.Carbon burns with a yellow-orange flame to form a colorless gas.In solution it forms H2CO3 a acidic solution with a pH of 5-6.Sulfur burns with a blue flame to form a colorless gas.In solution it forms H2SO3 (sulfurous acid), a acidic solution with a pH of 3-4.

48Oxygen and oxides2.20 describe the laboratory preparation of carbon dioxide from calcium carbonate and dilute hydrochloric acidCaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)Calcium carbonate is used in the form of marble chips. They are easy to handle and the rate of reaction is not too fast.

2.21 describe the formation of carbon dioxide from the thermal decomposition of metal carbonates such as copper(II) carbonateCuCO3(s) CuO(s) + CO2(s)Green copper carbonate Black copper oxideMgCO3(s) MgO(s) + CO2(g)CaCO3(s) CaO(s) + CO2(g)ZnCO3(s) ZnO(s) + CO2(g)White to yellow when hot, then yellow again when cold.

49Oxygen and oxides2.22 describe the properties of carbon dioxide, limited to its solubility and densityCarbon dioxide is slightly soluble in water.It is more dense than air.

2.23 explain the use of carbon dioxide in carbonating drinks and in fire extinguishers, in terms of its solubility and densityIt is used as a fire extinguisher as it sinks below other gases onto the fire and prevents oxygen from getting to it.Under high pressure it becomes much more soluble, hence its use to make carbonated drinks.

2.24 understand that carbon dioxide is a greenhouse gas and may contribute to climate change.As a greenhouse gas, it traps heat inside the atmosphere, thus heating up the earth.

50Hydrogen and water2.25 describe the reactions of dilute hydrochloric and dilute sulfuric acids with magnesium, aluminum, zinc and ironMetals above hydrogen in the reactivity series will react with acids to form a salt.Metal + hydrochloric acid metal chloride + hydrogenMetal + sulfuric acid metal sulfate + hydrogenCommon observations:Metal disappearsEffervescenceColorless solution formed (except green iron)

2.26 describe the combustion of hydrogenHydrogen combusts violently to give water.To test for hydrogen, place the flame of a lighted splint at the mouth of the test tube, a squeaky pop will be heard.

51Hydrogen and water2.27 describe the use of anhydrous copper(II) sulfate in the chemical test for waterAdd the liquid to anhydrous copper(II) sulfate. The white power will turn blue if water is present.CuSO4(s) + 5H2O(l) CuSO4.5H2O(l)Anhydrous copper(II) sulfate is white.Hydrated copper (II) sulfate is blue.

2.28 describe a physical test to show whether water is pure.Boil the water and measure its boiling point.If it is 100 degrees, then the water is pure.

52Reactivity series2.29 understand that metals can be arranged in a reactivity series based on the reactions of the metals and their compounds: potassium, sodium, lithium, calcium, magnesium, aluminum, zinc, iron, copper, silver and goldPotassium Sodium Lithium Calcium Magnesium Aluminum Carbon Zinc Iron Tin Lead Copper Silver Gold PlatinumHydrogen is between lead and copper.

2.30 describe how reactions with water and dilute acids can be used to deduce the following order of reactivity: potassium, sodium, lithium, calcium, magnesium, zinc, iron and copperFirst four react vigorously with cold water to form a solution of metal hydroxide and hydrogen. Magnesium reacts very slowly with cold water but burns in steam. Zinc, iron do not react with cold water but react with steam, without burning. They react with steam to form metal oxide and hydrogen, and react with decreasing vigor with dilute acids to form the metal salt and hydrogen.Copper does not react with anything53Reactivity series2.31 deduce the position of a metal within the reactivity series using displacement reactions between metals and their oxides, and between metals and their salts in aqueous solutionsA metal oxide is a compound of metal and oxygen. When the oxygen is removed from, the oxide is reduced. The substance that carries out the reduction is the reducing agent.A metal that is higher in the reactivity series will displace a metal from its oxide.e.g. CuO(s) + Mg(s) Cu(s) + MgO(s)In this case, magnesium is the reducing agent.A metal that is higher in the reactivity series will displace a metal from a solution of its salts.e.g. Mg(s) + FeSO4(aq) MgSO4(aq) + Fe(s)In this case, magnesium is the reducing agent.Pale green of FeSO4 replaced by colorless of MgSO4.

54Reactivity series2.32 understand oxidation and reduction as the addition and removal of oxygen respectivelyOxidation: addition of oxygen.Reduction: removal of oxygen.

2.33 understand the terms redox, oxidizing agent, reducing agentRedox reaction: A reaction in which both reduction and oxidation are taking place.Oxidizing agent: A substance that is capable of oxidizing another substance.Reducing agent: A substance that is capable of reducing another substance.

2.34 describe the conditions under which iron rustsIron rusts in the presence of oxygen and water.It is catalyzed by the presence of electrolytes such as salt.

55Reactivity series2.35 describe how the rusting of iron may be prevented by grease, oil, paint, plastic and galvanizingOne can stop rusting by preventing contact with oxygen and water, this is done by:Covering it with grease / oil.Painting the iron.Covering it with plastic.However, one it is scratched, the iron will rust very quickly.Galvanizing.Coating it with a more reactive metal.

2.36 understand the sacrificial protection of iron in terms of the reactivity series.Zinc is more reactive than iron, thus, the zinc will react with the air before it can corrode the iron. Therefore, the zinc sacrifices itself to stop iron from rusting. Galvanization is only used to describe being coated by zinc, for other metals it is sacrificial protection.

56Tests for ions and gases2.37 describe tests for the cations:i Li+, Na+, K+, Ca2+ using flame testsClean the end of a piece of platinum wire by dipping it into clean HCl and then placing it into a roaring flame until no color is produced.The end of the wire should be dipped into hydrochloric acid and then into the sample.Lithium (Li+) burns red.Sodium (Na+) burns yellow / orange.Potassium (K+) burns lilac.Calcium (Ca2+) burns brick red.

ii NH4+, using sodium hydroxide solution and identifying the ammonia evolvedAdd aqueous sodium hydroxide to the solid, or solution, under test and warm it.If ammonium is present, pungent-smelling gas is produced. The gas produced turns damp red litmus paper blue. It is ammonia (NH3).NH4+(aq) + OH- NH3 + H2O

57Tests for ions and gasesiii Cu2+, Fe2+ and Fe3+, using sodium hydroxide solutionMost metal hydroxides are insoluble and hence can be precipitated by adding sodium hydroxide solution drop by drop.Copper(II) (Cu2+): Blue precipitateCu2+(aq) + 2OH-(aq) Cu(OH)2(s)Iron(II) (Fe2+): Green precipitateFe2+(aq) + 2OH-(aq) Fe(OH)2(s)Iron(III) (Fe3+): Brown precipitateFe3+(aq) + 3OH-(aq) Fe(OH)3(s)

58Tests for ions and gases2.38 describe tests for the anions:i Cl, Br and I, using dilute nitric acid and silver nitrate solutionTo an aqueous solution of the solid under test, add some dilute nitric acid followed by a few drops of silver nitrate solution.Chloride (Cl-): White precipitateAg+(aq) + Cl-(aq) AgCl(s)Bromide (Br-): Cream precipitateAg+(aq) + Br-(aq) AgBr(s)Iodide (I-): Yellow precipitateAg+(aq) + I-(aq) AgI(s)The precipitate goes from white to yellow down the halogen group.

59Tests for ions and gasesii SO42-, using dilute hydrochloric acid and barium chloride solutionTo an aqueous solution of the solid under test, add dilute hydrochloric acid followed by a few drops of barium chloride solution.Sulfate (SO42-): White precipitateBa2+(aq) + SO42-(aq) BaSO4(s)

iii CO32-, using dilute hydrochloric acid and identifying the carbon dioxide evolvedTo either the solid, or an aqueous solution of it, under test add dilute hydrochloric acid.Carbonate (CO3-): Bubbles of gas. The gas produced turns limewater milky.CO32-(aq) + 2H+(aq) CO2(g) + H2O(l)

2.39 describe tests for the gases:i hydrogenMix with air and ignite. If positive, burns with a squeaky pop.2H2(g) + O2(g) 2H2O(l)60Tests for ions and gasesii oxygenInsert glowing splint into test tube of oxygen. If positive, split relights.

iii carbon dioxideBubble through limewater. If positive, limewater turns milky colored.Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)The CaCO3 solid turns limewater milky.

iv ammoniaDamp red litmus paper. If positive, red litmus paper turns blue.NH3(g) + H2O(l) NH4+(aq) + OH-(aq)OH- responsible for basic properties. OH- present.H+ responsible for acidic properties. H+ not present.

61Tests for ions and gasesv chlorineDamp litmus paper. If positive, bleaches litmus paper white.

62Organic chemistryIntroductionAlkanesAlkenesEthanol

63Introduction3.1 explain the terms homologous series, hydrocarbon, saturated, unsaturated, general formula and isomerism.Homologous series: a series of organic compounds that have the same general formula, similar chemical reactions, and where each member differs from the next by a -CH2- group.Hydrocarbon: a compound containing only the elements hydrogen and carbon.Saturated: an organic compound in which all the bonds are single bonds.Unsaturated: an organic compound that contains a carbon-carbon double bond.General formula: a formula that states the ratio of atoms of element in the formula of every compound in a particular homologous series.Isomerism: compounds that have the same molecular formula but different displayed formulae. The different compounds are called isomers.

64Alkanes3.2 recall that alkanes have the general formula CnH2n+2The general formula CnH2n+2 produces CH4, C2H6, etc

3.3 draw displayed formulae for alkanes with up to five carbon atoms in a molecule, and name the straight-chain isomersMethaneEthanePropaneButanePentane65Alkanes3.4 recall the products of the complete and incomplete combustion of alkanesIn complete combustion, the products are CO2 and water:CH4(g) + 2O2(g) CO2(g) + 2H2O(l)C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)In incomplete combustion, the products are CO and water:CH4(g) + 1.5O2(g) CO(g) + 2H2O(l)

3.5 describe the substitution reaction of methane with bromine to form bromomethane in the presence of UV light.Methane and bromine react together in the presence of ultra-violet radiation to form bromomethane.CH4(g) + Br2(g) CH3Br(g) + HBr(g)One hydrogen atom in methane has been replaced by a bromine atom, this reaction is called substitution reaction.

66Alkenes3.6 recall that alkenes have the general formula CnH2nThe general formula CnH2n produces C2H4, C3H6, etc

3.7 draw displayed formulae for alkenes with up to four carbon atoms in a molecule, and name the straight-chain isomers (knowledge of cis- and trans- isomers is not required)EthenePropeneButeneBut-1-eneBut-2-eneMethylpropene67Alkenes3.8 describe the addition reaction of alkenes with bromine, including the decolorizing of bromine water as a test for alkenes.A bromine molecule will perform an addition reaction across the double bond of ethene to form 1,2-dibromoethane:1.2-dibromoethane is colorless, so when bromine or bromine is shaken with any alkenes, it will decolorize bromine because of the carbon-carbon double bond.This reaction can be used as a test for unsaturated compounds (hydrocarbons).68Ethanol3.9 describe the manufacture of ethanol by passing ethene and steam over a phosphoric acid catalyst at a temperature of about 300 degrees and a pressure of about 6070 (atm)Direct hydration of ethene:C2H4(g) + H2O(g) C2H5OH(g)The ethanol is condensed as a liquid.Process described in statement.

3.10 describe the manufacture of ethanol by the fermentation of sugars, for example glucose, at a temperature of about 30 degreesFermentation:C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)Dissolve sugar in water and add yeast.Leave mixture to ferment at 25-40 degrees for several days in the absence of air.Filter off the excess yeast to obtain a dilute solution of ethanol.69Ethanol3.11 evaluate the factors relevant to the choice of method used in the manufacture of ethanol, for example the relative availability of sugar cane and crude oil

3.12 describe the dehydration of ethanol to ethene, using aluminum oxide.C2H5OH(g) C2H4(g) + H2O(g)Aluminum oxide is used as a catalyst to the reaction.MethodFermentationHydration of etheneRaw MaterialsUses renewable resourcesUses non-renewable resourcesTypeBatch processContinuous processRateSlow (several days)FastQualityDilute ethanol solutionPure ethanolConditionsLow temp.High temp. and pressure70Physical chemistryAcids, alkalis and saltsEnergeticsRates of reactionEquilibria71Acids, alkalis and salts4.1 describe the use of the indicators litmus, phenolphthalein and methyl orange to distinguish between acidic and alkaline solutionsLitmus: 1-5 red, 5-8 purple, 8-14 blueMethyl orange: 0-3 red, 3-5 orange, 5-14 yellowPhenolphthalein: 0-10 colorless, 10-14 red

4.2 understand how the pH scale, from 014, can be used to classify solutions as strongly acidic, weakly acidic, neutral, weakly alkaline or strongly alkaline0-3 pH is strongly acidic.4-6 pH is weakly acidic.7 is neutral.8-10 is weakly basic.11-14 is strongly basic.

72Acids, alkalis and salts4.3 describe the use of universal indicator to measure the approximate pH value of a solutionFrom strongly acidic to neutral to strongly alkali: Red Yellow Green Blue Purple.

4.4 define acids as sources of hydrogen ions, H+, and alkalis as sources of hydroxide ions, OHHydrogen (H+) ions give an acidic nature.Hydroxide (OH-) ions give an alkaline nature.

73Acids, alkalis and salts4.5 predict the products of reactions between dilute hydrochloric, nitric and sulfuric acids; and metals, metal oxides and metal carbonates (excluding the reactions between nitric acid and metals)Metal + acid salt + hydrogenMg(s) + 2HCl(aq) MgCl2(aq) + H2(g)Acids + base salt + waterNaOH(aq) + HCl(aq) NaCl (aq) + H2OAcids + metal carbonates salt + carbon dioxide + waterNa2CO3(s) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l)74Acids, alkalis and salts4.6 understand the general rules for predicting the solubility of salts in water:SolubleAll common sodium, potassium and ammonium salts (includes carbonates)All nitratesAll common chlorides (except silver and lead(II) chloride)All common sulfates (except barium, calcium and lead(II) sulfate)InsolubleAll carbonates (except sodium, potassium and ammonium carbonate)

4.7 describe experiments to prepare soluble salts from acidsAcid + insoluble basePut some dilute acids into a beaker and heat it using a Bunsen burner flame.Add the insoluble base, a little at the time and stir until there is excess.Filter the mixture into an evaporation basin to remove the excess base.Leave the filtrate for the water to evaporate and to form the crystals.

75Acids, alkalis and saltsAcid + soluble basePut an aqueous solution of the alkali into a conical flask and add a suitable indicator. Add dilute acid from a burette until the indicator just changes color.Add powdered charcoal and shake the mixture to remove the color of the indicator.Filter to remove the charcoal and then obtain crystals from the filtrate in through crystallization. Dilute acid + metal carbonateFor most carbonates, follow acid + insoluble base, except for sodium and potassium carbonates where following the acid + soluble technique is suitable.Dilute acid + metalFollow acid + insoluble base procedure.

76Acids, alkalis and salts4.8 describe experiments to prepare insoluble salts using precipitation reactionsAgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)Silver nitrate provides Ag+Sodium chloride provides Cl-Ag+ + Cl- AgClThe precipitate of silver chloride is removed by filtration, washed with a little distilled water, and left to dry in a warm place.

77Acids, alkalis and salts4.9 describe experiments to carry out acid-alkali titrations.Titrations are used to find the volume of acids required to react exactly with a given volume of alkali.Using a pipette, put 25.0 cm3 of the alkali solution into a conical flask.Add a few drops of an indicator, such as methyl orange.Put the acid into a burette and note the initial reading.Add the acid into the alkali until the indicator just changes color. Methyl orange turns from yellow to orangeNote the final reading of acid in the burette.Subtract the initial reading from the final reading to obtain the volume of acid added.

78Energetics4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermicExothermic: Energy is evolved. (Rise in temperature of surroundings)Reactants have more energy than products.Water IceCombustion of fuelsNegative heat energy changeEndothermic: Energy is absorbed. (Fall in temperature of surroundings)Reactants have less energy than products.Ice WaterPhotosynthesisPositive heat energy change

79Energetics4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralization in which heat energy changes can be calculated from measured temperature changese.g. for reactions in solutionUsing a measuring cylinder, place 50 cm3 of 0.1 mol / dm3 HCl into a polystyrene cup supported in a beaker. Measure and record the temperature of the acid.Tip approximately 0.15g of magnesium power into the acid and stir the mixture.Measure and record the highest temperature reached by the mixture.Heat given out = mass of solution x heat capacity x temperature riseTemp. rise = 10 degreesMoles of acid = 50 / 1000 X 0.1 = 0.005 molsHeat given out = 50 X 4.2 X 10J = 2.1 kJ (Density = 1g / cm3, capacity = 4.2)0.0005 mol of acid produces 2.1 kJ1 mol of acid produces = 420 kJMolar enthalpy change = H = -420 kJ / mol80Energeticse.g. combustion reactionsUsing a measuring cylinder, put 100 cm3 of water into a copper can.Measure and record the initial temperature of the water.Fill the spirit burner with the substance that is being measured and record its mass.Place the burner under the copper can and light the wick.Stir the water constantly with the thermometer and measure the highest temperature of the water.Measure and record the mass of the empty spirit burner.Temp. rise = 24.5 degreesMass of ethanol burnt = 0.46gHeat given out = 100 X 4.2 X 24.5J = 10.29 kJ0.46g of ethanol produces 10.29 kJ1 mol of ethanol = 46g10.29 X 46 / 0.46 = 1029 kJMolar enthalpy change = H = -1029 kJ / mol81Energetics4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction.Breaking bonds takes in energy. (endothermic / positive)Making bonds gives out energy. (exothermic / negative)e.g. H-H = 436, Cl-Cl = 242, H-Cl = 432In the reaction H2 + Cl2 = 2HCl:Breaking H-H = +436Breaking Cl-Cl = +242Breaking bonds total = +678Forming 2 X H-Cl = 2 X -432 = -864H = +678 864 = -186kJ / mol

82Rates of reaction4.17 describe experiments to investigate the effects of changes in surface area of a solid, concentration of solutions, temperature and the use of a catalyst on the rate of a reactione.g. In the laboratory preparation of CO2:CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)The size of the marble chips (surface area).Concentration of HCl.Temp.e.g. In the laboratory preparation of O2:2H2O2(aq) 2H2O(l) + O2(g)Use of manganese(IV) oxide as a catalyst.

83Rates of reaction4.18 describe the effects of changes in surface area of a solid, concentration of solutions, pressure of gases, temperature and the use of a catalyst on the rate of a reactionIncrease concentration: Increase rate.Increase temperature: Increase rate.Increase surface area: Increase rate.Catalysts: Increase rate.84Rate of reaction4.19 understand the term activation energy and represent it on a reaction profileWith catalystWithout catalystReactantsProductsActivation Energy85Rate of reaction4.20 explain the effects of changes in surface area of a solid, concentration of solutions, pressure of gases and temperature on the rate of a reaction in terms of particle collision theoryConcentration: More particles in a given volume. Particles collide more often, more successful collisions per second.Temperature: Particles have more energy. More collisions have required activation energy, more successful collisions per second. Particles move faster, particles collide more often.Surface area: More particles in contact with other reactant. Particles collide more often, more successful collisions per second.

4.21 explain that a catalyst speeds up a reaction by providing an alternative pathway with lower activation energy.Catalysts provide an alternate pathway for the reaction. This pathway has a lower activation energy. This means that although there are the same number of collisions, there are more collisions that have enough activation energy.

86Equilibria4.22 understand that some reactions are reversible and are indicated by the symbol in equationsThe symbol indicates that the reaction occurs in both directions. Such reactions are said to be reversible.

4.23 describe reversible reactions such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chlorideWhen hydrated copper(II) sulfate crystals are heated, it turns from blue to white because the water of crystallization is evaporated.CuSO4.5H2O(s) CuSO4(s) + 5H2OWhen water is added to anhydrous copper(II) sulfate, it turns from white to blue as it turns back into hydrated copper(II) sulfate.CuSO4(s) + 5H2O CuSO4.5H2O(s) These two reactions can be represented by a single equation:CuSO4.5H2O(s) CuSO4(s) + 5H2O

87Equilibria4.24 understand the concept of dynamic equilibriumIf a reversible reaction is carried out in a closed reaction container, then it is possible for the reaction to reach a position of dynamic equilibrium.e.g. 3H2(g) + N2(g) 2NH3At the beginning, when hydrogen and nitrogen are in high concentrations, the rate of the forward reaction is fast.However, as N2 and H2 concentration decreases, the rate of reaction decreases.Meanwhile, as the concentration of NH3 increases, the rate increases.This occurs until the rate of forward reaction and rate of backward reaction becomes equal. Thus the reactions reach a dynamic equilibrium.The reaction looks to have stopped, but it is just that the rates are the same.Amount of reactants remain the same.Any property, such as color or pressure, which depends on the amounts of reactions and products that are present remains constant.

88Equilibria4.25 predict the effects of changing the pressure and temperature on the equilibrium position in reversible reactions.e.g. for 3H2(g) + N2(g) 2NH3If the equilibrium is altered so that more ammonia is produced, and less hydrogen and nitrogen, the equilibrium shifts to the right.If the equilibrium is altered so that less ammonia is produced, and more hydrogen and nitrogen, the equilibrium shifts to the left.Increase in pressure: Equilibrium to the right, more ammonia is produced.Smaller number of molecules of gas.Decrease in pressure: Equilibrium to the left, less ammonia is produced.Greater number of molecules of gas.Increase in temperature: Equilibrium to the left, less ammonia is produced.Shifts towards the endothermic (backward reaction) direction.Decrease in temperature: Equilibrium to the right, more ammonia producedShifts towards the exothermic (forward reaction) direction.

89Chemistry in industryExtraction and uses of metalsCrude oilSynthetic polymersThe industrial manufacture of chemicals90Extraction and uses of metals5.1 explain how the methods of extraction of the metals in this section are related to their positions in the reactivity seriesThe more reactive the element, the more stable their compounds.All metals above carbon have to be extracted through electrolysis of the molten chloride or oxide. This is because it cannot use carbon as a reducing agent.All metals below carbon can use heat with a reducing agent such as carbon or carbon monoxide.Silver and gold occur naturally.

5.2 describe and explain the extraction of aluminum from purified aluminum oxide by electrolysis, including:i the use of molten cryolite as a solvent and to decrease the required operating temperaturePurified aluminum oxide (after purification from bauxite) has a very high melting point and hence it is dissolved in molten cryolite to make the electrolyte. This mixture has a much lower melting point and is also a better conduct of electricity than molten aluminum oxide.91Extraction and uses of metalsii the need to replace the positive electrodesPositive electrodes need to be replaced because the oxygen produced at the anode reacts with the carbon used as the material. The anode gradually burns away and needs to be replaced.

iii the cost of the electricity as a major factorElectricity is required in high amounts in the extraction of aluminum oxide, thus molten cryolite is needed to lower the melting temperature to reduce operating costs.

5.3 write ionic half-equations for the reactions at the electrodes in aluminum extractionAt the cathode: Al3+ + 3e- AlAt the anode: 2O2- O2 + 4e-The O2 reacts with the carbon in anode to form CO2:C(s) + O2(g) CO2(g)

92Extraction and uses of metals5.4 describe and explain the main reactions involved in the extraction of iron from iron ore (hematite), using coke, limestone and air in a blast furnaceThe raw materials are iron ore (hematite), coke (carbon), limestone (calcium carbonate) and air.The materials are mixed together and fed into the top of the blast furnace.Hot air is blasted into the bottom of the furnace.Oxygen in the air reacts with the coke to form carbon dioxide:C(s) + O2(g) CO2(g)Carbon dioxide reacts with coke to form carbon monoxide:CO2(g) + C(s) 2CO(g)Carbon monoxide reduces the iron(III)oxide in the iron ore, which melts:Fe2O3(s) + 3CO(g) 2Fe(l) + 3CO2(g)The molten iron collects at the bottom and is tapped off.Calcium carbonate in the limestone decomposes to form calcium oxide:CaCO3(s) CaO(s) + CO2(g)

93Extraction and uses of metalsThe calcium oxide reacts with silicon dioxide, an impurity in the ore to form calcium silicate, which melts:CaO(s) + SiO2(s) CaSiO3(l)The molten slag collects on top of the molten iron, and is tapped off.

5.5 explain the uses of aluminum and iron, in terms of their properties.Aluminum can be used for:Aeroplane bodies: high strength to weight ratioOverhead power cables: good conductor of electricitySaucepans: Good conductor of heatFood cans: Non-toxicWindows frames: Corrosion-resistantIron is used in car bodies, iron nails, ships and bridges because of its strength.94Crude oil5.7 describe and explain how the industrial process of fractional distillation separates crude oil into fractionsCrude oil is refined through fractional distillation, this is done in a fractionating column.Crude oil is split into various fractions, each of these fractions have different boiling points.Crude oil is heated to vapor and fed into the bottom of a fractionating column.The hydrocarbons with very high boiling points (bitumen, fuel oil) immediately turn back into liquids and are tapped off at the bottom of the column.The hydrocarbons that have higher boiling points lower than 400 degrees remains as gases and rise up the column, as they rise they cool down.The hydrocarbons condense at different heights and are tapped off individually.The fraction with the lowest boiling point remains as a gas.

95Crude oil400 degrees40 degreesFuel oilDiesel oilKeroseneGasolineRefinery GasesBitumen96Crude oil5.8 recall the names and uses of the main fractions obtained from crude oil: refinery gases, gasoline, kerosene, diesel, fuel oil and bitumenRefinery gases: bottled gas for campingGasoline: petrol for carsKerosene: fuel for airplanes, paraffin for small heaters and lampsDiesel oil: diesel fuel for buses, lorries, trains and cars.Fuel oil: fuel for ships and industrial heatingBitumen: road surfaces

5.6 understand that crude oil is a mixture of hydrocarbonsCrude oil is a thick, sticky black liquid that is found under the ground and sea, it is a mixture of hydrocarbons.

97Crude oil5.9 describe the trend in boiling point and viscosity of the main fractionsAs boiling point increases, viscosity increases as well.Refinery gases have the lowest boiling point, and the lowest viscosity, which means it flows well.Bitumen gases have the highest boiling point, and the highest viscosity, which means it is sticky and cannot flow well.

5.10 understand that incomplete combustion of fuels may produce carbon monoxide and explain that carbon monoxide is poisonous because it reduces the capacity of the blood to carry oxygenIncomplete combustion of hydrocarbons produces carbon monoxide, water and soot. Carbon monoxide combines with the hemoglobin in the red blood cells, hindering its capability to carry oxygen.

98Crude oil5.11 understand that, in car engines, the temperature reached is high enough toallow nitrogen and oxygen from air to react, forming nitrogen oxidesIn high temperatures such as car engines, the nitrogen and oxygen from the air react, forming nitrogen oxides.

5.12 understand that nitrogen oxides and sulfur dioxide are pollutant gases which contribute to acid rain, and describe the problems caused by acid rainNitrogen dioxide dissolves in the water to become nitric acidSulfur dioxide dissolves in the water to become sulfuric acidThe problems caused by acid rain are:Stunted and sickly tree growthRelease of heavy metals into the food chainLakes are acidic and they can no longer support aquatic life

99Crude oil5.13 understand that fractional distillation of crude oil produces more long-chain hydrocarbons than can be used directly and fewer short-chain hydrocarbons than required and explain why this makes cracking necessaryThe amount of long-chain hydrocarbons (high boiling point) is more than necessary, and a lot of shorter-chain hydrocarbons such as gasoline is needed. Therefore cracking is necessary to convert long-chain into short chain.

5.14 describe how long-chain alkanes are converted to alkenes and shorter-chain alkanes by catalytic cracking, using silica or alumina as the catalyst and a temperature in the range of 600700 degrees.The long-chain hydrocarbon molecules are passed over a catalyst (silica or aluminum oxide) and heated to about 600-700 degrees. The long-chain hydrocarbons break down into short-chain alkane molecules and at least one alkene molecule.C10H22(g) C8H18(g) + C2H4C8H18 (octane) is used to make petrol.C2H4 is the alkene that can be used to make poly(ethene).

100Synthetic polymers5.15 understand that an addition polymer is formed by joining up many small molecules called monomersAn addition polymer is a long-chain molecule that has been formed when many small molecules (monomers) have been joined together.The process of making this is called addition polymerization.The monomer molecule contains a carbon-carbon double bond.One of the bonds breaks during addition polymerization and this allows the monomer molecules to join together to form a long chain of carbon atoms.For example, when ethene undergoes addition polymerization it forms poly(ethene):CH2 = CH2 -(-CH2-CH2-)n-The letter n represents how many times this structure repeats itself in the molecule. 101Synthetic polymers5.16 draw the repeat unit of addition polymers, including poly(ethene), poly(propene) and poly(chloroethene)

CCH3HHCHHpoly(ethene)CHCHHpoly(propene)HCHCHClpoly(chloroethene)102Synthetic polymers5.17 deduce the structure of a monomer from the repeat unit of an addition polymer

PropeneEtheneEthene103Synthetic polymers5.18 describe some uses for polymers, including poly(ethene), poly(propene) and poly(chloroethene)Low density poly(ethene) is used for plastic bags. High density poly(ethene) is used for plastic bottles. Poly(propene) is used to make ropes and crates.Poly(chloroethene) or PVC is strong and rigid, used in pipes and electric insulation.

5.19 explain that addition polymers are hard to dispose of as their inertness means that they do not easily biodegradeTheir inertness means that they do not easily react with other substances. This means that they are hard to corrode.

104Synthetic polymers5.20 understand that some polymers, such as nylon, form by a different process called condensation polymerizationNot all polymers are made with addition polymerization, some are formed by a different process called condensation polymerization. This is because they are made from monomers which are saturated as they do not have double bonds between the carbon atoms. This means that an addition polymerization is not possible.

5.21 understand that condensation polymerization produces a small molecule, such as water, as well as the polymer.Groups of atoms need to be removed from the monomer structures to free up electrons to allow it to connect to other monomers.The atoms which have been removed make a new small molecule such as water or hydrogen chloride.This is called condensation polymerization.

105Synthetic polymersNHCHHdiamine monomerNHHHCOCHHdicarboxylic acid monomerCOOOHHNHCHHNHHHCOCHHCOOOHHwater106Synthetic polymersIn the end, a section of the nylon polymer would have the following structure:NHCHHNHCOCHHCOCOCHHCONHCHHNH107The industrial manufacture of chemicals5.22 understand that nitrogen from air, and hydrogen from natural gas or the cracking of hydrocarbons, are used in the manufacture of ammonia3H2(g) + N2(g) 2NH3(g) H = -98kJ / mol

5.23 describe the manufacture of ammonia by the Haber process, including the essential conditions:The Haber process reaction is reversible and can reach a position of equilibrium. The equilibrium position can be shifted to the right to achieve a high yield of ammonia by using high pressures (less ammonia molecules) and low temperatures (exothermic).i a temperature of about 450 degreesThe rate of reaction is very slow at low temperatures, so a compromise of 450 degrees is used. This is a compromise between rate and yield.

ii a pressure of about 200 atmospheresCarrying out a reaction at very high pressures is expensive since more energy is used to compress the gases. Therefore a pressure of 200 atmospheres is used.

108The industrial manufacture of chemicalsiii an iron catalystThe reaction is still fairly slow at 450 degrees unless a catalyst is present. There are a number of metals that will catalyze this reaction, but the most economical one is iron.

5.24 understand how the cooling of the reaction mixture liquefies the ammonia produced and allows the unused hydrogen and nitrogen to be recirculatedUnder these conditions about 15% of the hydrogen and nitrogen is converted into ammonia. The mixture leaves the reaction vessel and is then cooled. The ammonia liquefies and is tapped off. The hydrogen and nitrogen remain as gases and are mixed with more hydrogen and nitrogen and passed again into the reaction chamber. It is recycled.

5.25 describe the use of ammonia in the manufacture of nitric acid and fertilizersManufacture of fertilizersManufacture of nitric acidManufacture of nylon109The industrial manufacture of chemicals5.26 recall the raw materials used in the manufacture of sulfuric acidSulfurAir Water

5.27 describe the manufacture of sulfuric acid by the contact process, including the essential conditions:Stage 1: Producing sulfur dioxideS + O2 SO2Stage 2: Producing sulfur trioxide2SO2 + O2 2SO3 H = -196 kJ / molStage 3: Making sulfuric acidH2SO4(l) + SO3(g) H2S2O7(l)H2S2O7(l) + H2O(l) 2H2SO4(aq)Final product is a concentrated solution of sulfuric acid.110The industrial manufacture of chemicalsThe reaction is reversible and can reach a position of equilibrium. The equilibrium position can be shifted to the right to achieve a high yield of ammonia by using high pressure (less sulfur trioxide molecules) and low temperatures (exothermic).i a temperature of about 450 degreesThe rate of reaction is very slow at low temperatures, so a compromise of 450 degrees is used. This is a compromise between rate and yield.

ii a pressure of about 2 atmospheresThe yield would be increase with high pressure but the yield is already very high (98%) at a pressure of 2 atmospheres, so using higher pressures would be uneconomical.

iii a vanadium(V) oxide catalystVanadium(V) oxide is used to speed up the reaction by giving a alternate pathway with lower activation energy. This is because at 450 degrees, the reaction is still not quite slow.

111The industrial manufacture of chemicals5.28 describe the use of sulfuric acid in the manufacture of detergents, fertilizers and paintsManufacture of detergentsManufacture of fertilizersManufacture of paints

5.29 describe the manufacture of sodium hydroxide and chlorine by the electrolysis of concentrated sodium chloride solution (brine) in a diaphragm cellSodium hydroxide and chlorine are manufactured by the electrolysis of a concentrated solution of sodium chloride.At the anode the chloride ions lose electrons to form chlorine molecules.This is because the chloride ions are in high concentration, thus preferably electrolyzed over oxygen.At the cathode, water molecules gain electrons to form hydroxide ions and hydrogen molecules.The diaphragm is important to keep apart the sodium hydroxide and chlorine.112The industrial manufacture of chemicals5.30 write ionic half-equations for the reactions at the electrodes in the diaphragm cellAt the cathode: 2H2O(l) + 2e- 2OH-(aq) + H2(g)At the anode: 2Cl-(aq) Cl2(g) + 2e-Overall equation: 2NaCl(aq) + 2H2O(l) NaOH(aq) + H2(g) + Cl2(g)

5.31 describe important uses of sodium hydroxide, including the manufacture of bleach, paper and soap; and of chlorine, including sterilizing water supplies and in the manufacture of bleach and hydrochloric acid.Sodium hydroxideManufacture of bleach (If sodium hydroxide and chlorine mix)Manufacture of paperManufacture of soapManufacture of detergents

113The industrial manufacture of chemicalsChlorineSterilizing water suppliesManufacture of bleachManufacture of hydrochloric acidManufacture of PVC114