1. Particles that substances aremade up of

2. Atoms Basic building blocksof matter.Element Mono-atomic elements Metals giant structures Diatomic moleculesDiatomic molecular elementsPolyatomic molecular elements 3. CompoundsTwo or more atoms of different elements: Molecular compound Covalent network structures Ionic compounds Metallic structure Fixed proportionsElectrically neutral 4. Investigating elements andcompounds 5. Hydrogen gas 6. Carbon dioxide gas 7. Dehydration of copper sulfate 8. Dehydration of copper sulfate 9. Electrolysis of water 10. Representation of elements and compounds Molecular formula of a compoundH2O actual number and type of atoms in the molecule. Empirical formula of a compoundNaCl the simplest formula.Strucutural formula of a compoundO = C = O shows arrangement betweenatoms in a compound. 11. Diagrams of elements and compounds 12. Covalent structuresElectrons shared.Non-metals.Form molecules. 13. Covalent molecular solidsSolids = moleculesthat are bonded in amolecular lattice.Low boiling andmelting points.Often sublimates. 14. Covalent network structureWhen bonds extend throughout the structure of the substance to form a giant covalent network.These solids are very stable and have high melting and boiling points.Diamond and graphite are two allotropes of carbon. Allotropes are when the same element occur in different crystal forms. 15. DiamondColourless transparent crystal.Hard strong covalent bonds (4)Very high M.P. and B.P.Cannot conduct electricity no free e- or ions. 16. GraphiteSoft/slippery form layers held together by weak intermolecular forces.Conduct electricity due to one free e-. 17. Silica compoundEvery silicon atom bonds covalently to four oxygen atoms. 18. Boron compoundsBN share properties similar to graphite and diamond, depending on its molecular structure. 19. Uses of diamond, graphite and silicaSUBSTANCE PROPERTIES USESDIAMOND Hardest known substance In tools to cut and drillVery high M.P. and B.P.Precious stonesDoes not conduct JewelleryelectricityGRAPHITESoft and slipperyLubricants for engines and locksHigh M.P. and B.P. Lead of pencilsConducts electricity Electrodes and connectors in generatorsSILICAHard, used to sand On sand paperobjectsHigh M.P. And B.P. Make glass and lensesDoes not conduct In bricks to line furnaceelectricityovens. 20. Properties of covalentcompoundsLow melting and boiling points due to weak intermolecular forces. Covalent network structures have very high M.P.and B.P due to the strong covalent bonds betweenatoms.Covalent structures are poor conductors of electricity and heat they are insulators becausethey have no free electrons. 21. Ionic structuresElectrons transferredMetal and non-metalsForms crystal lattice 22. Complete the ion and metalstuff 23. Chemical bond IonicCovalentStrong electrostatic forces between ionsForces between atoms inside ofare broken. Requires large amounts ofmolecules are brokenenergy to break.Rearrange atoms/molecules. Rearrange ions.Energy is released New bond, newNew bond, new substancesubstanceOnly reversible in certain circumstances. www.docscientia.co.za 24. Physical and chemical change 25. Physical change Composition does not change.Properties appearance, form and state changes.A physical change is a change where no newchemical compounds are formed. The state may change, but the identity does not. 26. PhysicalArrangementReversibilityConservation of mass, atoms and molecules Energy involved 27. Melting EvaporationAbsorbs energy Absorbs energyReleases energy Releases energy Solidification SublimationArrangement of particleswww.docscientia.co.za 28. Physical ReversibilityGenerally easily reversed 29. Physical Conservation of mass, atoms, and moleculesMatter cannot created or destroyed, only changed. Therefore mass stays the same after a physical change, as does the amount of atoms and molecules. 30. Physical changes Energy involvedPhysical changes can be caused by small energy changes.Only intermolecular forces break during physicalchanges. 31. Chemical changes Arrangement of the particlesAtoms, molecules or ions regroup. Decomposition reaction Synthesis reaction 32. Chemical changes ReversibleA few are reversible, but most are difficult toreverse. 33. Chemical changesConservation of mass, atoms and moleculesThe quantity of a specific subsance canchange, but total amount of atoms stay the same. 34. Chemical changes Energy involvedAbsorbed or released during chemical reactions. Much more energy involved. Intramolecular forces have to be broken and re-formed. 35. Decompostion and synthesis reactions 36. Decompostion reactions AB A + B 37. Decompostion reactions 38. Synthesis reactions A + B AB 39. Synthesis reactions 40. Energy changes in chemical reactionsTotal energy absorbed to break bonds are less than the energy required to form new bonds. Heat is given off to the environment. EXOTHERMIC REACTION 41. Energy changes in chemicalreactions 42. Energy changes in chemical reactionsTotal energy absorbed to break bonds are more than the energy released to form new bonds.Heat is absorbed from the environment. ENDOTHERMIC REACTION 43. Total energy Total energy released Energy transferredabsorbed to breakwhen new bonds formbondsProduct:warmer =exothermic reaction -colder endothermic reaction Chemical reaction Exothermic EndothermicreactionreactionTo calculate the energy required, you need the following: Amount of Ep that is necessary to break bonds, as well as Amount of Ep that is released after bonds form.Energy values shown as kJmol-1.www.docscientia.co.za 44. Conservation of atoms andmass 45. Law of conservation of massLaviosier: Matter cannot be created or destroyed. Reactant mass = Product mass 46. During a chemical reaction or a physical change the sum of the reactants areequal to the sum of the mass of the products.Atoms in reactants= Atoms in productsAmount of atoms (reactants) = Amounts of atoms(products)Mass before= Mass afterConsider the following balanced equation:2SO22(32 + 16 + 16)+ O2+ (16 + 16)2SO3 2(32 + 16 + 16 + 16)Ar128+ 32 160Voor: 160 = Na: 160www.docscientia.co.za 47. Law of conservation of atoms but non-conservation of molecule in a chemical reactionDuring a chemical reaction the number of atoms ofeach element stays intact. 48. Law of constant compositionBalanced equation:Everyting in front of the arrow, is still to be found at the back of the arrow.Left side has the same amount of matter as the right side. A specific chemical reaction always has the same elements in the same ratios.Any amount:Carbon dioxide: carbon and oxygen contains 1 C for every 2 Os.Ammonia: nitrogen and hydrogen contains 1 N for every 3 Hs.Dinitrogen tetraoxide: nitrogen and oxygen contains 2 Ns for every 4 Os.Multiple compounds are sometimes possible:For example: : H2O en H2O2CO en CO2www.docscientia.co.za 49. A specific amount of particles of any gas occupies thesame volume at a fixed temperature and pressure.2SO2(g) + O2(g) 2SO3(g) 2SO2+O2 2SO32 particles SO2+ 1 particle O2 2 particle SO32 volume-units SO2 + 1 volume-units 2 volume-units O2 SO32 dm3 SO2+ 1 dm3 O2 2 dm3 SO32 cm3 SO2+ 1 cm3 O2 2 cm3 SO36 cm3 SO2+ 3 cm3 O2 6 cm3 SO3www.docscientia.co.za 50. Balancing of equations 51. Balancing of equations 52. Calculating massSo, we have learned that:Mass stays the same before and after achemical reaction.The amount of atoms on the left hand side isthe same as the amount of atoms on theright hand side they are simply arrangeddifferently.All of this led us to balance our chemicalequations. 53. Calculating massThe balanced equations show us that:Mass stays the same on the left ans right.The volumes of gas reactants are the same asthe balanced equations.All of these facts can help us calculate theactual mass of the reactants. 54. Calculating massSo, we can say that:Mass of reactants and products can becalculated.Volume of gas reactions can be calculated.Actual mass of reactants or products can becalculated. 55. Calculating mass proving thelaw of conservation of massStep 1: Balanced equation Pb(NO3)2(aq) + 2NaI(ag) PbI2 + 2NaNO3(aq) Step 2: Relative mass REACTANTS Mr[Pb(NO3)2] = 207 + 2(14 + (163)) = 331Mr(NaI) = 23 + 127 = 150PRODUCTS Mr(PbI2) = 207 + (2127) = 461 Mr(NaNO3) = 23 + 14 + (316) = 85 56. Calculating mass proving the law of conservation of massStep 3: Determine total mass before and after Pb(NO3)2(aq) + 2NaI(ag) PbI2 + 2NaNO3(aq) 331+ 150 461 + 85 631631 57. Calculating mass Using the law of conservation of mass to determine the actual mass of substances.QUESTION:Determine the mass of oxygen that is releasedWhen 29,4 g potassium chlorate is heated and Completely decomposed into potassium and Oxygen.Step 1: What is asked, and what is given? 29,4 g KClO3 ? g O2 58. Calculating mass Using the law ofconservation of mass to determine the actual mass of substances.Step 2: Balanced equation2KClO3 2KCl + 3O2Step 3: Relative mass of substances in step 12[39 + 35,5 + (316) 3(216) 24596 59. Calculating mass Using the law ofconservation of mass to determine the actual mass of substances.Step 4: Use ratios to caculate the mass of O2245 g KClO3 : 96 g O2 245 g/245 KClO3 : 96 g/245 O2Which is the same as...1 g KClO3 : 96/245 g O2So, we can say that 29,4 g KClO3 : (29,496)/245 g O2= 11,52 g O2A mass of 11,52 g O2 is formed. 60. Calculating mass Using the law of conservation of mass to determine the actual mass of subst ances.QUESTION:Determine what mass of sodium will fully react withchlorine to form 25 g of table salt. 61. Calculating mass Using the law ofconservation of mass to determine the actual mass of subst ances.BALANCED EQUATION:2Na + Cl2 2NaCl WHAT DO WE HAVE? Na - ?25 g NaClRELATIVE MOLAR MASS Na 46: NaCl 117 62. Calculating mass Using the law ofconservation of mass to determine the actual mass of subst ances. Na 46 : NaCl 11746 g Na : 117 g NaCl 46/117 g Na : 1 g NaCl2546/117 g Na : 25 g NaCl9,83 g Na will yield 25 g NaCl. 63. Calculating volumeCalculate the volume of nitrogen oxide that forms,should 3 dm3 nitrogen react with oxygen. Step 1: Balanced equationN2 + O2 2NO Step 2: substitute particles for volume units 1 dm3 N2 + 1 dm3 O2 2 dm3 NO 64. Calculating volumeCalculate the volume of nitrogen oxide that forms,should 3 dm3 nitrogen react with oxygen. Step 1: Balanced equationN2 + O2 2NO Step 2: substitute particles for volume units 1 dm3 N2 + 1 dm3 O2 2 dm3 NO