Matter and materials (II) Paticles that substances are made up of

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Information about Matter and materials (II) Paticles that substances are made up of
Education

Published on March 30, 2013

Author: LKOTZE

Source: slideshare.net

Description

Gr. 10 Physical Science based upon Andries Oliviers' Textbook with input from DocScientia

Particles that substances are made up of

Atoms Basic building blocks of matter. Element Mono-atomic elements Metals – giant structures Diatomic molecules Diatomic molecular elementsPolyatomic molecular elements

CompoundsTwo or more atoms of different elements: Molecular compound Covalent network structures Ionic compounds Metallic structure Fixed proportions Electrically neutral

Investigating elements and compounds

Hydrogen gas

Carbon dioxide gas

Dehydration of copper sulfate

Dehydration of copper sulfate

Electrolysis of water

Representation of elements and compounds Molecular formula of a compound H2O – actual number and type of atoms in the molecule. Empirical formula of a compound NaCl – the simplest formula. Strucutural formula of a compoundO = C = O – shows arrangement between atoms in a compound.

Diagrams of elements and compounds

Covalent structuresElectrons shared.Non-metals.Form molecules.

Covalent molecular solidsSolids = molecules that are bonded in a molecular lattice.Low boiling and melting points. Often sublimates.

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.

DiamondColourless transparent crystal.Hard – strong covalent bonds (4)Very high M.P. and B.P.Cannot conduct electricity – no free e- or ions.

GraphiteSoft/slippery – form layers held together by weak intermolecular forces.Conduct electricity due to one free e-.

Silica compoundEvery silicon atom bonds covalently to four oxygen atoms.

Boron compoundsBN share properties similar to graphite and diamond, depending on its molecular structure.

Uses of diamond, graphite and silicaSUBSTANCE PROPERTIES USESDIAMOND Hardest known substance In tools to cut and drill Very high M.P. and B.P. Precious stones Does not conduct Jewellery electricityGRAPHITE Soft and slippery Lubricants for engines and locks High M.P. and B.P. Lead of pencils Conducts electricity Electrodes and connectors in generatorsSILICA Hard, used to sand On sand paper objects High M.P. And B.P. Make glass and lenses Does not conduct In bricks to line furnace electricity ovens.

Properties of covalent compounds Low 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 between atoms. Covalent structures are poor conductors of electricity and heat – they are insulators because they have no free electrons.

Ionic structuresElectrons transferredMetal and non-metalsForms crystal lattice

Complete the ion and metal stuff

Chemical bond Ionic Covalent Strong electrostatic forces between ionsForces between atoms inside of are broken. Requires large amounts ofmolecules are broken energy to break. Rearrange atoms/molecules. Rearrange ions. Energy is released New bond, new New bond, new substance substanceOnly reversible in certain circumstances. www.docscientia.co.za

Physical and chemical change

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.

Physical Arrangement ReversibilityConservation of mass, atoms and molecules Energy involved

Melting EvaporationAbsorbs energy Absorbs energyReleases energy Releases energy Solidification SublimationArrangement of particles www.docscientia.co.za

Physical ReversibilityGenerally easily reversed

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.

Physical changes Energy involved Physical changes can be caused by small energy changes.Only intermolecular forces break during physical changes.

Chemical changes Arrangement of the particlesAtoms, molecules or ions regroup. Decomposition reaction Synthesis reaction

Chemical changes ReversibleA few are reversible, but most are difficult to reverse.

Chemical changesConservation of mass, atoms and molecules The quantity of a specific subsance canchange, but total amount of atoms stay the same.

Chemical changes Energy involvedAbsorbed or released during chemical reactions. Much more energy involved. Intramolecular forces have to be broken and re-formed.

Decompostion and synthesis reactions

Decompostion reactions AB → A + B

Decompostion reactions

Synthesis reactions A + B → AB

Synthesis reactions

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

Energy changes in chemical reactions

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

Total energy Total energy released Energy transferred absorbed to break when new bonds form bondsProduct: warmer –=exothermic reaction - colder – endothermic reaction Chemical reaction Exothermic Endothermic reaction reactionTo 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 kJ⋅mol-1. www.docscientia.co.za

Conservation of atoms and mass

Law of conservation of massLaviosier: Matter cannot be created or destroyed. Reactant mass = Product mass

 During a chemical reaction or a physical change the sum of the reactants are equal 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 after Consider the following balanced equation: →  2SO2 2(32 + 16 + 16) + O2 + (16 + 16) 2SO3 → 2(32 + 16 + 16 + 16)Ar 128 + 32 → 160 Voor:   160 = Na: 160 www.docscientia.co.za

Law of conservation of atoms but non-conservation of molecule in a chemical reactionDuring a chemical reaction the number of atoms of each element stays intact.

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 H’s.• Dinitrogen tetraoxide: nitrogen and oxygen contains 2 Ns for every 4 Os.Multiple compounds are sometimes possible:For example: : H2O en H2O2 CO en CO2 www.docscientia.co.za

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 SO3 www.docscientia.co.za

Balancing of equations

Balancing of equations

Calculating mass So, 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.

Calculating mass The 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.

Calculating mass So, 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.

Calculating mass – proving the law of conservation of mass Step 1: Balanced equation Pb(NO3)2(aq) + 2NaI(ag) → PbI2 + 2NaNO3(aq) Step 2: Relative mass REACTANTS Mr[Pb(NO3)2] = 207 + 2(14 + (16·3)) = 331 Mr(NaI) = 23 + 127 = 150 PRODUCTS Mr(PbI2) = 207 + (2·127) = 461 Mr(NaNO3) = 23 + 14 + (3·16) = 85

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 + 85 631 → 631

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

Calculating mass – Using the law ofconservation of mass to determine the actual mass of substances. Step 2: Balanced equation 2KClO3 → 2KCl + 3O2Step 3: Relative mass of substances in step 12[39 + 35,5 + (3·16) → 3(2·16) 245 → 96

Calculating mass – Using the law ofconservation of mass to determine the actual mass of substances.Step 4: Use ratios to caculate the mass of O2 245 g KClO3 : 96 g O2 245 g/245 KClO3 : 96 g/245 O2 Which is the same as... 1 g KClO3 : 96/245 g O2 So, we can say that 29,4 g KClO3 : (29,4·96)/245 g O2 = 11,52 g O2 A mass of 11,52 g O2 is formed.

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 with chlorine to form 25 g of table salt.

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 NaCl RELATIVE MOLAR MASS Na – 46 : NaCl – 117

Calculating mass – Using the law ofconservation of mass to determine the actual mass of subst ances. Na – 46 : NaCl – 117 46 g Na : 117 g NaCl 46/117 g Na : 1 g NaCl 25·46/117 g Na : 25 g NaCl 9,83 g Na will yield 25 g NaCl.

Calculating volumeCalculate the volume of nitrogen oxide that forms, should 3 dm3 nitrogen react with oxygen. Step 1: Balanced equation N2 + O2 → 2NO Step 2: substitute particles for volume units 1 dm3 N2 + 1 dm3 O2 → 2 dm3 NO

Calculating volumeCalculate the volume of nitrogen oxide that forms, should 3 dm3 nitrogen react with oxygen. Step 1: Balanced equation N2 + O2 → 2NO Step 2: substitute particles for volume units 1 dm3 N2 + 1 dm3 O2 → 2 dm3 NO

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