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  1. Chemistry Edexcel International GCSE in Chemistry (4CH0) First examination June 2013

  2. Principles of chemistry • States of matter • Atoms • Atomic structure • Relative formula masses and molar volumes of gases • Chemical formulae and chemical equations • Ionic compounds • Covalent substances • Metallic crystals • Electrolysis

  3. States of matter 1.1 understand the arrangement, movement and energy of the particles in each of the three states of matter: solid, liquid and gas • Solid: 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 interconversions • Solid  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.

  4. States of matter 1.3 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.

  5. Atoms 1.4 describe and explain experiments to investigate the small size of particles and their movement including: idilution of colored solutions • When 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 experiments • The 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.

  6. Atoms 1.5 understand the terms atom and molecule • Atoms 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 mixtures • Elements 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.

  7. Atoms 1.7 describe experimental techniques for the separation of mixtures, including simple distillation, fractional distillation, filtration, crystallization and paper chromatography • Filtration 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. • Evaporationis 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.

  8. Atoms • Fractional 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.

  9. Atomic Structure 1.9 understand that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons, orbiting in shells • Neutrons • Electrons • Protons

  10. Atomic Structure 1.10 recall the relative mass and relative charge of a proton, neutron and electron • Protons: +1 Charge, 1 Mass • Electrons: -1 Charge, 1/1836 (Basically 0) Mass • Neutrons: 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.

  11. Atomic Structure 1.12 calculate the relative atomic mass of an element from the relative abundances of its isotopes • Chlorine 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 number • The 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 Table • The period (row) in the periodic table defines the number of shells it has.

  12. Atomic Structure 1.15 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.

  13. Relative formula masses and molar volumes of gases 1.16 calculate relative formula masses (Mr) from relative atomic masses (Ar) • Hydrogen has an Ar of 1. • H2has a Mr of 2 X 1 = 2 • Oxygen has an Ar of 16. • H2O has a Mr of 2 X 1 + 16 = 18 • Likewise, (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 substance • Mole 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 substance • 6.023 X 1023 is known as the Avogrado number. • 1 mol of an substance contains 6.023 X 1023 of the substance.

  14. Relative formula masses and molar volumes of gases 1.19 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 H2has a Mr of 2 • 1 mol of H2 is 2g. • Likewise, 8 mols of H2 is 16g • e.g. Mg has a Ar of 24 • 0.5 mols of Mg is 12g • Mass 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 24 • To calculate the amount of gas at rtp: Amount of gas (mols) = Volume of gas (dm3) / 24

  15. Chemical formulae and chemical equations 1.21 write word equations and balanced chemical equations to represent the reactions studied in this specification • Unbalanced Equation: Al + CuO Al2O3 + Cu • Left side: 1 Al, 1 Cu, 1 O • Right side: 2 Al, 1 Cu, 3 O • Balanced Equation: 2Al + 3CuO  Al2O3 + 3Cu • Left side: 2 Al, 3 Cu, 3 O • Right side: 2 Al, 3 Cu, 3 O • Word 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 respectively • Zn(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.

  16. Chemical formulae and chemical equations 1.23 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallization • e.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 = 0.4 • 0.08 : 0.4 = 1 : 5 • The empirical formula of hydrated copper sulfate salts is CuSO4.5H2O

  17. Chemical formulae and chemical equations 1.24 calculate empirical and molecular formulae from experimental data • e.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 = 1.96 • Oxygen: 47.06 / 16 = 2.94 • 1.96 : 2.94 = 2 : 3 • The empirical formula of aluminum oxide is Al2O3 • e.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 = 13 • 78 / 13 = 6 • Therefore, the molecular formula of Benzene is C6H6

  18. Chemical formulae and chemical equations 1.25 calculate reacting masses using experimental data and chemical equations • e.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 40 • 0.25 X 40 = 10g of MgO formed.

  19. Chemical formulae and chemical equations 1.26 calculate percentage yield • Percentage yield = Yield obtained / theoretical yield • e.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) = 6.4g • Yield obtained = 4.8g • Theoretical yield = 6.4g • 4.8 / 6.4 X 100 = 75%

  20. Chemical formulae and chemical equations 1.27 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.

  21. Ionic compounds 1.28 describe the formation of ions by the gain or loss of electrons • An 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 electrons • Oxidation: Loss of electrons • Reduction: Gain of electrons 1.30 recall the charges of common ions in this specification

  22. Ionic compounds 1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formed • An 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.

  23. Ionic compounds 1.32 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, 7 • e.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. [ ]+ Na Na [ ]- Cl Cl

  24. Ionic compounds 1.33 understand ionic bonding as a strong electrostatic attraction between oppositely charged ions • An 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 compound • Strong 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 ions • When many ions arrange themselves in a three dimensional structure called an giant ionic lattice. Ionic compounds often form crystals.

  25. Ionic compounds 1.37 draw a diagram to represent the positions of the ions in a crystal of sodium chloride.

  26. Covalent substances 1.38 describe the formation of a covalent bond by the sharing of a pair of electrons between two atoms • When 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 bond • When the electrons are shared, the shared region of the electron shells bond the two nuclei together. This forms a covalent compound.

  27. Covalent substances 1.40 explain, using dot and cross diagrams, the formation of covalent compounds by electron sharing for the following substances: H H H H H H H C C H C H C H H C Cl Cl H H H H H H Cl

  28. Covalent substances 1.41 understand that substances with simple molecular structures are gases or liquids, or solids with low melting points • Simple 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 molecules • Forces 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.

  29. Covalent structures 1.44 draw diagrams representing the positions of the atoms in diamond and graphite Graphite Diamond

  30. Covalent substances 1.45 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.

  31. Metallic crystals 1.46 understand that a metal can be described as a giant structure of positive ions surrounded by a sea of delocalized electrons • Metals 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.

  32. Electrolysis 1.48 understand that an electric current is a flow of electrons or ions • An 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 electricity • Covalent 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 solution • Ionic 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.

  33. Electrolysis 1.51 describe experiments to distinguish between electrolytes and nonelectrolytes • Electrolytes are substances that conduct electricity in the molten state or when dissolved in water. • Non-electrolytes are substances that don’t conduct electricity in these states • To 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 electricity • Passing a charge through the electrolytes cause chemical changes, thus the chemical reactions produce new products.

  34. Electrolysis 1.53 describe experiments to investigate electrolysis, using inert electrodes, of molten salts such as lead(II) bromide and predict the products • The 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- Pb • Br- is attracted towards the anode, where they lose electrons to become Br. • 2Br-  Br2 + 2e-

  35. Electrolysis 1.54 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 products • Selective discharge occurs: • At the cathode, hydrogen is preferably discharged over all cations unless the cationis 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)

  36. Electrolysis • e.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- Cu • At 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-  H2 • At the anode, water loses electrons to produce oxygen gas and hydrogen ions. • 2H2O  O2 + 4H- + 4e-and then 2H2O  O2 + 2H2O • The water is decomposed leaving a more concentrated solution of sulfuric acid.

  37. Electrolysis 1.55 write ionic half-equations representing the reactions at the electrodes during electrolysis • Typically 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 electrons • One faraday is one mole of electrons, which equals 96500 coulombs of charge.

  38. Electrolysis 1.57 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- Cu • Thus, 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-  Pb • 1 mol of electrons deposit 0.5 mols of lead = 103.5g • e.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 Time • 0.010 X (4 X 60 X 60) = 144 C • Au3+ + 3e-  Au • 3 x 96500C deposit 1 mol (197g) of gold. • 144C deposits (197 / (3 X 96500)) X 144 = 0.098g of gold.

  39. Electrolysis • e.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 C • 1440 C deposits 0.239g of metal. • 48g of metal is deposited by 1440/0.239 X 48 = 289205 C • 289205 / 96500 = 3 mols of electrons. • 3 mols of electrons produces 1 mol of metal. • The metal has the charge of 3+.

  40. Chemistry of the elements • The Periodic Table • Group 1 elements – lithium, sodium and potassium • Group 7 elements – chlorine, bromine and iodine • Oxygen and oxides • Hydrogen and water • Reactivity series • Tests for ions and gases

  41. The Periodic Table 2.1 understand the terms group and period • Groups 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 Table • The 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 oxides • Metals 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.

  42. The Periodic Table 2.4 understand why elements in the same group of the Periodic Table have similar chemical properties • They 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.

  43. Group 1 elements – lithium, sodium and potassium 2.6 describe the reactions of these elements with water and understand that the reactions provide a basis for their recognition as a family of elements • The 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 1 • Increases 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.

  44. Group 7 elements – chlorine, bromine and iodine 2.9 recall the colors and physical states of the elements at room temperature 2.10 make predictions about the properties of other halogens in this group • Fluorine is a pale-yellow gas. • Astatine is a black solid. 2.11 understand the difference between hydrogen chloride gas and hydrochloric acid • Hydrochloric acid is hydrogen chloride dissolved in water, as a solution. It forms H+ ions which makes it acidic.

  45. Group 7 elements – chlorine, bromine and iodine 2.12 explain, in terms of dissociation, why hydrogen chloride is acidic in water but not in methylbenzene • When 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 7 • Decreases down the group: Fluorine  Chlorine  Bromine  Iodine  Astatine

  46. Group 7 elements – chlorine, bromine and iodine 2.14 describe experiments to demonstrate that a more reactive halogen will displace a less reactive halogen from a solution of one of its salts • e.g. Adding chlorine to a potassium bromide solution will displace the bromide from the potassium. • Cl2 + 2KBr  2KCl + Br2 • The 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-

  47. Oxygen and oxides 2.16 recall the gases present in air and their approximate percentage by volume • Composition 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 air • In the reaction of copper and oxygen: 2Cu + O2 2CuO • Using 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.

  48. Oxygen and oxides 2.18 describe the laboratory preparation of oxygen from hydrogen peroxide, using manganese(IV) oxide as a catalyst • 2H2O2(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 produced • Magnesium 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.

  49. Oxygen and oxides 2.20 describe the laboratory preparation of carbon dioxide from calcium carbonate and dilute hydrochloric acid • CaCO3(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) carbonate • CuCO3(s)  CuO(s) + CO2(s) • Green copper carbonate  Black copper oxide • MgCO3(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.

  50. Oxygen and oxides 2.22 describe the properties of carbon dioxide, limited to its solubility and density • Carbon 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 density • It 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.