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  Unit F212: Molecules, Biodiversity, Food and Health   Module 1: Biological Molecules (a)    Describe how hydrogen bonding occurs between water molecules, and relate this, and other properties of water, to the roles of water in living organisms (HSW1); Hydrogen bond is a weak interaction that can occur wherever molecules contain a slightly negatively charged atom bonded to a slightly positively charged hydrogen. The bonds between oxygen and hydrogen are covalent, with electrons shared between them. They exist, as oxygen in water is more electronegative than the hydrogen so a permanent dipole forms. Water is polar as oxygen atom pulls the shared electrons towards it.   Property Key Points Role of water Good solvent for charged & uncharged substances Water molecules are attracted to ions and polar molecules e.g. glucose Transport in blood, xylem and phloem High specific heat capacity 4.2kJ are necessary to increase the temperature of 1kg of water by 1oC. The thermal energy absorbed is used to break hydrogen bonds. Helps prevent rapid body temperature changes allowing them to keep their temperature fairly stable High latent heat of vaporization Much thermal energy is used to cause water molecules to change to water vapour – this happens in transpiration in plants, and in sweating and panting in mammals. Coolant – water is used efficiently as a small amount of water absorbs and uses as much thermal energy when water evaporates High cohesion Hydrogen bonds stick water molecules together as they’re polar Helps draw up water in xylem, helps water flow making it great for transporting substances Can be reactive/Metabolic Water reacts with other substances Involved in hydrolysis reactions and in photosynthesis Incompressibility Outside pressure cannot force water into a smaller space Hydrostatic skeleton for some animals e.g. earthworms and provides turgidity in plant cells                (b)    Describe, with the aid of diagrams, the structure of an amino acid; H2N-CHR-COOH. Cells polymerise amino acids into polypeptides by forming peptide bonds. A        peptide bond forms between an amine group and a carboxylic acid group.           (c)    Describe, with the aid of diagrams, the formation and breakage of peptide bonds in the synthesis and hydrolysis of dipeptides and polypeptides; Amino acids are linked together by peptide bonds to form dipeptides (2) and polypeptides (2+). A molecule of water is eliminated in a condensation reaction. The chemical addition of water breaks the bond is hydrolysis. Properties of proteins depend on R group, some are charged (polar) and interact with water hydrophilic and some aren’t – hydrophobic, and can interact with phospholipids within membranes. R groups determine active site of an enzyme.   (d)    Explain, with the aid of diagrams, the term primary structure The sequence of amino acids that make up the protein.       e)    Explain, with the aid of diagrams, the term secondary structure with reference to hydrogen             bonding The coiling or pleating of parts of the polypeptide molecule stabilised by the numerous hydrogen      bonds. Formed when the chain of amino acid coils to form an α-helix or coils to form a β-pleated sheet. Hydrogen bonds between the amino acids hold the coils in place. f) Explain, with the aid of diagrams, the term tertiary structure, with reference to hydrophobic and hydrophilic interactions, disulfide bonds and ionic interactions; Refers to the overall 3D structure of the final polypeptide or protein molecule. It’s the further coiling and folding of a polypeptide to give a more complex 3D shape specific to the function. 3D shape is held in place by a number of different types of bonds and interactions. ·       Hydrogen bonds between polar groups. Wherever slightly positively charged groups are found close to slightly negatively charged groups. ·       Disulfide bonds is when the amino acid cysteine contains sulfur and 2 cysteines are found close to each other a covalent bond can form. ·       Ionic bonds between R groups which ionize to form positively and negatively charged groups that attract each other. ·       Hydrophobic interactions between non-polar R groups. Hydrophobic amino acids will be most stable if they are held together with water excluded. Hydrophilic amino acids tend to be found on the outside in globular proteins, with hydrophobic amino acids in the centre.  (Some bonds break when proteins are heated up or treated with acids and alkalis, the tertiary structure changes and protein does not function. Destruction of shape and function is denaturation)   g) Explain, with the aid of diagrams, the term quaternary structure, with reference to the structure of hemoglobin. Protein structure where a protein consists of more than one polypeptide chain. Hemoglobin and insulin both have a quaternary structure. Hb is made of 4 polypeptide chains bonded together. For proteins made from 1+ polypeptide chain, the quaternary structure is the protein’s final 3D structure.   h) Describe, with the aid of diagrams, the structure of a collagen molecule Made up of three polypeptide chains, each about 1000 amino acids that are tightly coiled into a strong triple helix. The chains are interlinked by strong covalent bonds. Minerals can bind to the triple helix to increase its rigidity.   (i)              Compare the structure and function of haemoglobin (as an example of a globular protein) and collagen (as an example of a fibrous protein); Haemoglobin Collagen Globular transport protein transports oxygen and carbon dioxide  Fibrous structural protein strengthens tissues Soluble in water (good for transport in the blood) Insoluble in water Wide range of amino acid constituents in primary structure Approximately 35% of the molecule’s primary structure is one type of amino acid (glycine) Contains a prosthetic group – haem Doesn’t contain prosthetic group Much of the molecule is wound into alpha helix structures Much of the molecule consists of left-handed helix structures Complex 3D shape Triple helix In red blood cells other globular proteins e.g. enzymes found in all organisms and plasma proteins and antibodies found in the blood of mammals Found in skin, bone, cartilage, teeth, tendons, muscles, ligaments and walls of blood vessels other fibrous protein e.g. keratin which is skin, nails and hair and elastin which is in blood vessel walls and alveoli   (ii)             Describe, with the aid of diagrams, the molecular structure of alpha-glucose as an example of a monosaccharide carbohydrate; Glucose is a source of energy in organisms and is built up into macromolecules. It is a monosaccharide with six carbon atoms in each molecule. It is polar so it interacts with water – soluble and suitable for transport in the blood.             k) State the structural difference between alpha- and beta- glucose; In α-glucose the –OH on carbon 1 is below the plane of the ring. In β-glucose it is above the chain of the ring. Since respiration is powered by enzymes, which need to be complementary to the shape of the substrate- glucose, one enzyme which can bind with α-glucose may not be able to bind with β-glucose meaning that plants and animals are unable to use β-glucose for respiration.   l) Describe, with the aid of diagrams, the formation and breakage of glycosidic bonds in the synthesis and hydrolysis of a disaccharide (maltose) and a polysaccharide (amylose); Two monosaccharide molecules can be joined together in a condensation reaction forming a disaccharide. A new covalent bond called a 1,4-glycosidic bond forms between C1 on one glucose and C4 on another and water is eliminated (released) and the reverse hydrolysis reaction uses a water molecule to break the glycosidic bond. Two α-glucose molecules are joined together by a glycosidic bond to form maltose         Lots of α-glucose molecules are joined together by 1,4-glycosidic bonds to form amylose         m) Compare and contrast the structure and functions of starch (amylose) and cellulose; Amylose Cellulose Monomer α-glucose Monomer β-glucose Unbranched helix so compact Unbranched straight chain Angles of 1,4-glycosidic bonds give it a coiled structure β-glycosidic bond can only be broken down by cellulose enzymes, which humans do not. Because the H and OH groups on the C1 and C4 atoms are on opposite sides of the ring, each alternating molecule is rotated through 180° Energy storage in plants Forms plant cell walls Insoluble in water Many microfibrils are held together by hydrogen bonds to form macrofibrils.   n) Describe, with the aid of diagrams, the structure of glycogen; Made up of α-glucose subunits. Similar to amylopectin but 1-4 linked glucose chains in glycogen is shorter and have more branches extending from chain. Forms glycogen granules in animal cells, in liver and muscle cells, it is the energy store in animals, they have higher energy demand than plants so needs to be able to break down glycogen quickly into many glucose molecules. It is a compact molecule. They can easily be broken off from ends to provide glucose for respiration when required.           o) Explain how the structures of glucose, starch (amylose), glycogen and cellulose molecules relate to their functions in living organisms; Glucose Simplest sugar, small carbohydrates, soluble and crystalline. Used in respiration. Starch (Amylose) Amylose in starch is unbranched and coiled à compact àgood energy storage to break the glycosidic bonds and release glucose quickly. It’s insoluble so It can be stored in cells without causing water to enter by osmosis which would cause them to swell, doesn’t affect water potential of the cell. Glycogen Highly branched so can be broken down into many glucose molecules quickly Cellulose Macrofibrils have great mechanical strength due to large number of hydrogen bonds helps cell walls provide strength to each cell, allows water to easily move through and along macrofibrils and prevents cell from bursting in turgid conditions. p) Compare, with the aid of diagrams, the structure of a triglyceride and a phospholipid Glycerol is a three-carbon compound, each carbon atom ha a hydroxyl group (-OH) which can react with a fatty acid to form an ester bond. It has a glycerol + 3 fatty acids. Phospholipid has a glycerol + 2 fatty acids + a phosphate group.           q) Explain how the structures of triglycerides, phospholipid and cholesterol molecules relate to their functions in living organisms;   Triglycerides Phospholipid Cholesterol Long hydrocarbon fatty acid tails contain lots of chemical energy so releases lots of energy when respired. Insoluble so doesn’t affect water potential of cell Hydrophilic phosphate head and hydrophobic tail repelled forms stable bilayers when in water – an ideal structure for cell membranes as it creates barrier, as water-soluble substances can’t easily pass through. Small size and flattened shape allows cholesterol to fit in between phospholipid molecules. Bind to hydrophobic tails causing them to pack more closely. Thermal insulators, buoyancy, protecting vital organs, energy store Maintain membrane fluidity even in low temperatures, no. of unsaturated fatty acid tails is increased Controls membrane fluidity. Gives membrane strength and stability.   r) Describe how to carry out chemical tests to identify the presence of the following molecules: ·       Protein (biuret test) – if present turns from pale blue à lilac/violet/purple ·       Reducing sugars (Benedict’s test) – Add Benedict’s solution (Cu2+), heat to 80oC from blue à orange-red with a precipitate (of red copper (l) oxide). If not the blue colour will have no change. ·       Non-reducing sugars (Benedict’s test) – If reducing sugars test is negative boil with hydrochloric acid, cool and neutralise with sodium hydrogencarbonate/alkali before doing Benedict’s test. Repeat Benedict’s test. ·       Starch (Iodine solution) – turns from yellow à blue black ·       Lipids (emulsion test) – Add ethanol to dissolve lipid and pour alcohol into water in another test tube. White emulsion forms near top of water if lipid is present.   s) Describe how the concentration of glucose in a solution may be determined using colorimetry Colorimeter measures the degree of ‘blueness’ of the solution remaining after the reducing sugar has reacted with Benedict’s solution. After boiling a small sample of test solution with a larger volume of Benedict’s solution, filter the test solution to remove the precipitate. Solution that remains blue has no reducing sugar in it and gives a high reading for absorbance in colorimeter. A solution that has no blue colour has a high concentration of reducing sugar and gives a low reading for absorbance.   Nucleic Acids a) State that deoxyribonucleic acid (DNA) is a polynucleotide, usually double stranded, made up of nucleotides containing the bases adenine (A), thymine (T), cytosine (C) and guanine (G);   b) State that ribonucleic acid (RNA) is a polynucleotide, usually single stranded, made up of nucleotides containing the bases adenine (A), uracil (U), cytosine (C) and guanine (G); Each nucleotide consists of pentose sugar (with 5 carbon atoms), phosphate and nitrogen-containing (nitrogenous) base.           c) Describe, with the aid of diagrams, How hydrogen bonding between complementary base pairs (A to T, G to C) on two antiparallel DNA polynucleotides leads to the formation of a DNA molecule,  There are two types of nucleotide bases – pyrimidines and purines. The bases always arranged so that a purine is opposite pyrimidine: A-T, G-C. There are 2 hydrogen bonds between A and T and three between G and C. The bases are complementary in size and shape so that only the pairings A-T and G-C fit into the space between the sugar-phosphate backbone of DNA.   And how the twisting of DNA produces its ‘double-helix’ shape; The antiparallel chains twist like a rope ladder to form the final structure   d) Outline, with the aid of diagrams, how DNA replicates semi-conservatively, with reference to the role of DNA polymerase; ·       Double helix unwinds and DNA ‘unzips’ as hydrogen bonds between the polynucleotides are broken. ·       Existing polynucleotides act as templates for assembly of nucleotides ·       Free nucleotides, which have been made in the cytoplasm, move towards the exposed bases of DNA ·       Base pairing occurs between free nucleotides and exposed bases. A matches T and C matches G. Hydrogen bonds form between complementary bases. ·       Enzyme DNA polymerase forms covalent bonds between free nucleotides attached to each template ·       Two daughter DNA molecules form separate double helices. Proofreading enzyme DNA polymerase prevents mistakes, the ‘wrong’ nucleotide is cut out and the correct one is inserted.   e) State that a gene is a sequence of DNA nucleotides that codes of a polypeptide;   f) Outline the roles of DNA and RNA in living organisms (the concept of protein synthesis must be considered in outline only). Protein synthesis Vital for living organisms to produce proteins in order to grow and develop. DNA carries instructions to make proteins found in the nucleus. Organelles that make proteins are found in cytoplasm but DNA molecules are too large to move out of nucleus so sections of DNA are copied into RNA. RNA leaves nucleus and joins with a ribosome used to synthesise a protein. ·       (Transcription) Sequence coding for a particular protein (a gene) can be exposed by splitting the hydrogen bonds that hold the double helix together in that region ·       RNA nucleotides form a complementary strand (mRNA). This mRNA molecule is a copy of the DNA coding strand or gene ·       mRNA peels away from DNA and leaves the nucleus through a nuclear pore ·       mRNA attaches to a ribosome ·       Then tRNA molecules bring amino acids to the ribosome in the correct order, according to the base sequence on the mRNA ·       (Translation) Amino acids are joined together by peptide bonds to give a protein with a specific primary structure (which then gives rise to the secondary and tertiary structures)   Types of RNA Structure of polynucleotide Function Messenger RNA (mRNA) Variable length, no base pairing Transfers genetic information from DNA to ribosomes, after which it is broken down Transfer RNA (tRNA) Clover-leaf structure with some base pairing Carries amino acids to ribosomes Ribosomal RNA (rRNA) Folded and attached to proteins to make ribosomes Provides site for assembly of amino acids to make proteins       Enzymes a) State that enzymes are globular proteins, with a specific tertiary structure, which catalyse metabolic reactions in living organisms;   b) State that enzyme action may be intracellular or extracellular;   c) Describe, with the aid of diagrams, the mechanism of action of enzyme molecules, with reference to specificity: The active site of an enzyme is a specific shape, depending on the reaction that it catalyses, meaning that other molecules won’t fit into the active site.   Active site: The area on an enzyme to which the substrate binds   Lock and key hypothesis: The active site has a complementary shape to the substrate molecule so that binding is much easier as substrate fits into enzyme, this is described as being like a lock and key. The substrate is then held in place so that the reaction can occur.   Induced-fit hypothesis: 

AS Unit F322: Chains, Energy and Resources   Basic Concepts a)     Interpret and use the terms: (i)              Empirical formula is the simplest whole number ratio of atoms of each element present in a compound. (ii)             Molecular formula is the actual number of atoms of each element in a molecule Relative molecular mass/Relative empirical mass=n then empirical formula xn (iii)            General formula is the simplest algebraic formula of a member of a homologous series, i.e. for alkanes: CnH2n + 2 ,       Alkenes: CnH2n ,      Alcohols: CnH2n+1OH (iv)            Structural formula is the minimal detail that shows the arrangement of atoms in a molecule. E.g. for butane: CH3CH2CH2CH3 or CH3(CH2)2CH3 (v)             Displayed formula is the relative positioning of atoms and the bonds between them. (vi)            Skeletal formula is the simplified organic formula, shown by removing hydrogen atoms from alkyl chains, leaving just a carbon skeleton and associated functional groups.   b)     Interpret, and use, the terms: (i)              Homologous series is a series of organic compounds having the same functional group but with each successive member differing by CH2. (ii)             Functional group as a group of atoms responsible for the characteristic reactions of a compound. c)      Use the general formula of a homologous series to predict the formula of any member of the series. They contain same functional group, have similar properties but each successive member of a homologous series differs by one carbon and two hydrogen atoms: a –CH2- group.   d)     State the names of the first ten members of the alkanes homologous series.   No. Of carbons Name Formula 1 Methane CH4 2 Ethane C2H6 3 Propane C3H8 4 Butane C4H10 5 Pentane C5H12 6 Hexane C6H14 7 Heptane C7H16 8 Octane C8H18 9 Nonane C9H20 10 Decane C10H22   e)     Use IUPAC rules of nomenclature for systematically naming organic compounds: Nomenclature is a system of naming compounds.   -Identifying parent chain (longest) and assign appropriate stem -Add a suffix or prefix to stem to indicate the functional group -Add a prefix to show the identity of any alkyl groups and side chains and use multipliers such as di, tri, tetra when more than one of the same type of side chain is used -Number the position of all side chains and functional group, if required If the suffix starts with a vowel, remove the final e from the alkane suffix e.g. methaneal à methanal   A stem tells us the longest no. of carbon atoms in a chain. A prefix or suffix is used for different functional groups, alkyl groups are also given stems. An Alkyl group is an alkane with a hydrogen atom removed.         No. Of carbons Stem Alkyl group 1 meth- methyl 2 eth- ethyl 3 prop- propyl 4 but- butyl 5 pent- pentyl 6 hex- hexyl 7 hept- heptyl 8 oct- octyl 9 non- nonyl 10 dec- decyl   Type of compound Functional group Prefix Suffix Comments Alkane C-C   -ane   Alkene C=C   -ene   Halogenoalkane -F -CL -Br -I Fluoro- Chloro- Bromo- Iodo-     Alcohol  -OH Hydroxyl- -ol Hydroxyl used if there are other functional groups also present Aldehyde                                     -CHO   -al No no. ever required as the functional groups is always on carbon 1 Ketone                       C-CO-C     -one   Carboxylic acid                       -COOH     -oic acid     f)      Describe and explain the terms: (i)              Structural isomers as compounds with the same molecular formula but different structural formulae (ii)             Stereoisomers as compounds with the same structural formula but with a different arrangement in space (iii)            E/Z isomerism as an example of stereoisomerism, in terms of restricted rotation about a double bond and the requirement for two different groups to be attached to each carbon atom of the C=C group (iv)            Cis-trans isomerism as a special case of E/Z isomerism in which two of the substituent groups are the same;   g)     Determine the possible structural formulae and/or stereoisomers of an organic molecule, given its molecular formula; Example: Molecular formula Structural isomers   C4H10 CH3CH2CH2CH3 CH3CH(CH3)CH3 butane 2-methylpropane C3H7Cl CH3CH2CH2Cl CH3CHClCH3 1-Chloropropane 2-chloropropane C4H9OH CH3CH2CH2CH2OH CH3CH2CH(OH)CH3 (CH3) 3COH (CH3) 2CHCH2OH butan-1-ol butan-2-ol 2-methylpropan-2-ol 2-methylpropan-1-ol   h)     Describe the different types of covalent bond fission: (i)              Homolytic fission forming two radicals: Each bonded atom takes one of the shared pair of electrons. Each atom has a radical. Two species of the same type are produced.    X-Y à X + Y (2 radicals)   Homolytic fission is the breaking of a covalent bond, with one of the bonded electrons going to each atom, forming 2 radicals.   (ii)             Heterolytic fission forming a cation and an anion; Heterolytic fission is the breaking of a covalent bond with both of the bonded electrons going to one of the atoms, forming a cation (+ion) and an anion (-ion). One of the bonded atoms takes both of the shared paid of electrons. Two ions are produced: Atom that takes both shared electrons becomes negatively charged ion (anion), atom that doesn’t take the shared electrons becomes a positively chared ion (cation).      X-Y à X+ + Y- (2 ions)   (j)            Describe a ‘curly arrow’ as the movement of an electron pair, showing either breaking or           formation of a covalent bond A curly arrow will be used in both addition and substitution mechanisms, arrow goes to the electrophile in addition reactions and arrow goes away from nucleophile in substitution reactions.   An electrophile is an atom (or group of atoms) that is attracted to an electron-rich centre or atom, where it accepts a pair of electrons to form a new covalent bond. An atom with a δ+ partial charge. E.g. Br2, HBr, NO2   An addition reaction is where a reactant is added to an unsaturated molecule to make a saturated molecule. 2 reactants à 1 product E.g. Ethene + Bromine (electrophile) à 1,2-dibromoethane   A nucleophile is an atom (or group of atoms) that is attracted to an electron-deficient centre of atom, where it donates a pair of electrons to form a new covalent bond. δ- partial charge. E.g. :Br-, :OH-, H2O:, :NH3   A substitution reaction is a reaction in which an atom or group of atoms is replaced with a different atom or group of atoms. 2 reactants à 2 products.   Elimination reactions is the removal of a molecule from a saturated molecule to make an unsaturated molecule. 1 reactant à 2 products.   Oxidation is the gain of oxygen or loss of hydrogen. Also the loss of electrons from the organic compound. [O] is used to show an oxidizing agent. E.g. C2H5OH + [O] à CH3CHO + H2O   Reduction is the loss of oxygen or gain of hydrogen. Also gain of electrons to the organic compound.   Neutralisation: Acid + base à salt + water         (k)           Carry out calculations to determine the percentage yield of a reaction.  %Yield = (Actual amount, in mol, of product/theoretical amount in mol, of product) x 100 A 100% percentage yield is when all of the reactants turn into products (rare). Lower atom economies result because of: ·       Equilibrium reactions preventing the reaction reaching completion ·       Side reactions can occur, forming by-products ·       Reactants are not pure ·       Reactants and products left behind on the apparatus ·       Separation and purification results in loss of product Other e.g.: When distilling, not all the distillate was evaporated from the mixture, when recrystallising, not all the crystals crystallised from solution, the reaction did not go to completion, the receiver was not cooled so not all the distillate condensed.   (l)     Explain the atom economy of a reaction as: (Molecular mass of the desired products/sum of molecular masses of all products) x 100% A reaction may produce by-products, as well as the product we want to make. The by-products may be useless waste and disposed of – costly and wasting resources posing environmental problems. By-products may also be sold on or used elsewhere in the chemical plant. (minimising waste) Atom economy describes the efficiency of a reaction in terms of all the atoms involved   (m)  Explain that addition reactions have an atom economy of 100% whereas substitution reactions are less efficient. ·       Addition reactions will theoretically have a 100% atom economy. ·       Substitution and elimination reactions have lower atom economies.   (n)    Carry out calculations to determine the atom economy of a reaction;                       (o)    Describe the benefits of developing chemical processes with a high atom economy in terms of fewer waste materials; A higher atom economy reduces the waste of the Earth’s resources.   (p)    Explain that a reaction may have a high percentage yield but a low atom economy. A reaction can have a high percentage yield but a low atom economy.     Hydrocarbons from crude oil (a)    Explain that a hydrocarbon is a compound of hydrogen and carbon only; They are extracted from crude oil   (b)    Explain the use of crude oil as a source of hydrocarbons, separated as fractions with different boiling points by fractional distillation, which can be used as fuels or for processing into petrochemicals; It is a fossil fuel made naturally from decayed plant & animal material. Takes millions of years to form. Mixture of about 150 hydrocarbons, most are straight chained and unbranched.   ·       Takes place in a fractionating column. Crude oil is heated and passed into the column ·        The column is hottest at the bottom and coolest at the top ·       The crude oil gasses pass through bubble caps into a series of compartments ·       Eventually, each hydrocarbon reaches a compartment with a temperature below its boiling point ·       The particular hydrocarbon then condenses in that compartment ·       The liquid fractions are tapped off into storage compartments ·       Short-chained hydrocarbons with lower boiling points condense near the top of the column ·       Long-chained hydrocarbons with higher boiling points condense near to the bottom ·       Gasses do not condense and pass through an outlet at the top of the column as ‘petroleum gas’ ·       The residue from the process is bitumen, which is removed from the bottom of the column. It can be used in road surfacing ·       Fractions obtained can be used as fuels or may be further separated in a dedicated fractionating column, to obtain pure hydrocarbons. Processed further to make petrochemicals – chemicals made from natural gas and oil, distilled further to form pure liquids in other fractionating columns, operating over a narrower range of temperatures.   (c)    State that alkanes and cycloalkanes are saturated hydrocarbons Alkanes are a homologous series of hydrocarbons. Alkanes and cycloalkanes have no multiple bonds in their molecules and are therefore saturated hydrocarbons.   (d)    State and explain the tetrahedral shape around each carbon atom in alkanes All bon angles are 109.5o  - tetrahedral shape gives the maximum separation of bond-pair electrons. Carbon atom in centre with the other atoms at each of the four corners.   The only bonds present in the molecules are either C-C or C-H bonds that are non-polar making alkanes and cycloalkanes unreactive – they have few reactions. Since they are non-polar molecules, the forces between molecules are weak van der Waals’ forces. Uses: fuels and feedstock for production of other chemicals. Longer-chain alkanes are cracked to give alkenes and shorter alkanes that are used as fuels.   (e)    Explain, in terms of van der Waals’ forces, the variations in the boiling points of alkanes with different carbon-chain length and branching; ·       The longer the carbon chain length, the greater the surface area of the molecule that can take part in intermolecular bonding, and the greater the van der Waals’ forces. ·       When alkanes are branched, the surface area is less and consequently the van der Waals’ forces are weaker.   Hydrocarbons as fuels (f)     Describe the combustion of alkanes, leading to their use as fuels in industry, in the home and in transport;   (g)    Explain, using equations, the incomplete combustion of alkanes in a limited supply of oxygen and outline the potential dangers arising from production of CO in the home and from car use;                

AS Unit G482: Electrons, Waves and Photons   Electric current Candidates should be able to:   (a) Explain that electric current is a net flow of charged particles;  An electric current is a flow of charge. Current is measured in amps. One amp is one coulomb per second. Charge cannot get used up therefore current is conserved.   (b) Explain that electric current in a metal is due to the movement of electrons, whereas in an electrolyte the current is due to the movement of ions; In metals charge is carried by electrons, which are negatively charged. In an electrolyte, the mobile charges are the positive and negative ions.   (c) Explain what is meant by conventional current and electron flow; Conventional current is from positive to negative, but in fact electrons flow around the circuit from negative to the positive terminal. Electron flow is in the opposite direction to conventional current.   (d) Select and use the equation △Q = I△t; Charge = Current x time Q = I x t   (e) Define the coulomb; One coulomb is the total charge supplied by a current of one ampere in a time of one second.   (f) Describe how an ammeter may be used to measure the current in a circuit; To measure current, an ammeter is placed in series in a circuit. Connected in series as the current it is measuring must flow through it)   (g) Recall and use the elementary charge e=1.6 x10-19 C;   (h) Describe Kirchhoff’s first law and appreciate that this is a consequence of conservation of charge; The sum of the currents entering any junction is always equal to the sum of the currents leaving the junction.   (i) State what is meant by the term mean drift velocity of charge carriers; Drift velocity (v) is the average velocity of an electron as it travels through a wire due to a p.d across it. The conduction of electrons in a metal move around rapidly in random directions. When a voltage is applied, electrons gain an additional velocity along the wire. The greater the current (I), the greater the value of v. Current also depends on cross-sectional area A of the conductor and on the number density n of electron sin the metal.   (j) Select and use the equation I = Anev; Current = Cross-sectional area x number density of electrons in the metal per m3 x electron charge x mean drift velocity   (k) Describe the difference between conductors, semiconductors and insulators in terms of the number density n. Metals have high number densities of electrons so drift velocities are low – of the order of mm s-1v. Semiconductors have far fewer free electrons (n is much smaller) so drift velocities are much higher. For insulators, n is close to zero.     Resistance (a) Recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995);   (b) Interpret and draw circuit diagrams using these symbols.   Candidates should be able to:   a) Define potential difference (p.d.);  The p.d. (V) across a component is the energy transferred per unit charge when electrical energy is converted into some other form of energy.   b) Select and use the equation W = VQ; Energy transferred (J) = Voltage x Charge   c) define the volt; Volt is the unit of potential difference and e.m.f (V) One volt is one joule per coulomb. 1 V = 1 J C-1   d) Describe how a voltmeter may be used to determine the p.d. across a component; A voltmeter must be connected across the terminals of a component to measure the p.d. across it/it must be connected in parallel with it. Recall: Voltmeter is measuring the p.d. across the component.   e) Define electromotive force (e.m.f.) of a source such as a cell or a power supply; E.m.f is the energy transferred per unit charge when one other type of energy is converted into electrical energy. Chemical energy to electrical energy in a cell. E.m.f=electrical energy transferred/charge W=VQ (Think as Work, J)   Combing e.m.f.s – when two or more cells or other sources of e.m.f. are connected in series, their voltages add up. They must be connected +-to- or the e.m.f.s subtract.   f) Describe the difference between e.m.f. and p.d. in terms of energy transfer. Potential difference is the electrical energy transferred per unit charge when electrical energy is converted into another form e.g. electrical energy to light energy whereas e.m.f is the energy transferred when a type of energy is being converted into electrical energy.   (a) Define resistance; A property of a component that regulates the electric current through it.   (b) Select and use the equation for resistance Resistance = Voltage / Current                R(Ω) V / I     (c) Define the ohm; Ohm is the unit of resistance 1 Ω = 1VA-1 One volt per amp. Kilo-ohm is 103, mega-ohm is x106   (d) State and use Ohm’s law; Ohm’s law states that the current through a conductor is proportional to the potential difference across it, provided physical conditions, such as temperature, remain constant. Greater the p.d., the greater the current. Ohmic conductor is one in which the current that flows is proportional to the p.d. that pushes it, it obeys ohm’s law. V=IR so resistance of conductor is constant, doesn’t depend on the p.d. across it.   (e) Describe the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); For a metallic conductor at constant temperature, an ohmic conductor gives a straight line through the origin. Metals obey Ohm’s law – provided the temperature remains constant. Non-ohmic conductors has an I-V characteristic graph with no straight line. Filament lamp (non-ohmic conductor): As metal filament gets hotter with increasing current and voltage, its resistance increases. The current is less than it would be if it remained proportional to the voltage. Semiconductor diode: In the forward direction allow current to flow when the p.d. is above 0.7V. In the reverse direction, only a very small current flows. A light-emitting diode (LED) shows similar pattern, it emits light when a sufficiently large current flows through it.   (f) Describe an experiment to obtain the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); ·   An ammeter connected in series measures the current through the resistor ·   A voltmeter connected in parallel measures the p.d. across the resistor ·   Reversing the connection to the resistor makes the current ·   Use long thing wire to keep current low so that heating effect is negligible ·   Take readings from voltmeter and ammeter and increase the no. of cells and take readings for each added cell ·   Plot a graph of current against potential difference/I-V characteristics   (g) Describe the uses and benefits of using light- emitting diodes (LEDs); Advantages of LEDs as light sources: ·   Requires only low p.d. ·   Have a long working life ·   Are very robust ·   Can be used singly as indicators or in large arrays to illuminate ·   Switch on instantly   Candidates should be able to: (a) Define resistivity of a material; ρ of a wire of length, l, resistance R and area of cross-section A is given by ρ=RA/l Resistivity is the ratio of the product of resistance and cross-sectional area of a component and its length. Unit of resistivity is Ωm, ohm metres. Resistivity is constant for all materials at a set temperature. A typical value for a good conductor: resistivity of copper = 1.6 x 10-8 Ωm   (b) Select and use the equation Rpl/A R= resistance p= resistivity l=length A= cross-sectional area. R ∝ l   : The longer the wire, the greater its resistance R ∝1/A : The fatter the wire, the less its resistance So doubling the length of a wire doubles its resistance; doubling its cross-sectional area halves its resistance.     (c) Describe how the resistivities of metals and semiconductors are affected by temperature; A semiconductor has many fewer conduction electrons than a metal (number density is less), when a semiconductor is heated, more electrons break free from their atoms increasing the no. of free electrons that can carry the current (the number density increases rapidly). So resistivity of the semiconductor decreases as it is heated – the opposite of a metal.   (d) Describe how the resistance of a pure metal wire and of a negative temperature coefficient (NTC) thermistor is affected by temperature. When a metal is heated, its resistivity increases as temperature rises atoms of the metal vibrate with increasing amplitude making it more likely that conduction electrons will collide with the atoms as they try to travel through the metal. Electrons lose energy when they collide and so we observe an increase in the metal’s resistivity. Metal’s resistance increases gradually as temperature is increased. Negative Temperature Coefficient (NTC) thermistors - Resistance decreases rapidly over a narrow range of temperature. Resistance goes down as temperature goes up.   Candidates should be able to: (a) Describe power as the rate of energy transfer; The rate of doing work, measured in Watts, W.   Power (W) = Energy transferred (J)/time taken (s)      P = W/t   W = Pt  Energy transferred = power x time 1 watt = 1Js-1  1 kilowatt = 103W = 1000W 1 megawatt = 1MW = 106W = 1000000W 1 gigawatt = 1GW = 109W = 1000000000W   (b) Select and use power equations P = VI, P=I2R and P=V2/R The greater the current and p.d V that is flowing through, the greater the power P. Power = current x p.d. P=IV   combing this with V = IR gives P=I2R and P=V2/R the equation for energy transferred W becomes W=current x p.d. x time. W=IVt   (c) Explain how a fuse works as a safety device (HSW 6a); Protects electrical appliances from dangerous overloading by fuse in plug. Fuse is made of thin copper wire in a ceramic casing. Above a certain value of current, wire becomes so hot that it melts breaking circuit. Fuse should be chosen so that its rating is a little higher than the maximum current drawn by the device when it is operating correctly.   (d) Determine the correct fuse for an electrical device; Most household appliances uses fuses of 3A, 5A or 13A and household mains uses 230V. To work out correct fuse needed use P=IV so I=P/V e.g. if current worked out is 4.3A, so a 5A fuse is needed.   (e) Select and use the equation W = IVt; (f)              define the kilowatt-hour (kW h) as a unit of energy; Unit of energy (kWh) used by electricity companies when charging for electricity. 1kWh=1000W for 3600s=3.6MJ.  1kWh= 3.6x106J  (as 1hour = 3600s and 1kW=1000Js-1)   (g) Calculate energy in kW h and the cost of this energy when solving problems (HSW 6a). Energy transferred (kWh) = power (kW) x time (h)       DC Circuits (a) State Kirchhoff’s second law and appreciate that this is a consequence of conservation of energy; The sum of the e.m.f.s around any circuit loop is equal to the sum of the p.d.s: ΣE=ΣIR (E.g. by the time the charge has completed the circuit, it has lost as much energy as it gained – example of conservation of energy as energy cannot be created or destroyed in a closed loop)   (b) Apply Kirchhoff’s first and second laws to circuits; By applying Kirchhoff’s first law, we know that each resistor has the same current flowing through it, current is always conserved.   (c) Select and use the equation for the total resistance of two or more resistors in series; Rtotal=R1 + R2 +R3 + …. Resistors in series must have the same current through them and the p.d. of the supply is shared between them. So Rtotal = R + r and hence E=I(R+r) where I is current supplied. This value is given by V=IR. The voltmeter reading is less than the e.m.f. of the battery so E=I(R + r) = IR + Ir E=V + Ir   (d) Select and use the equation for the total resistance of two or more resistors in parallel; I=I1 + I2 + I3 = E/R1 + E/R1 + E/R3 since I=E/R where R is the total resistance, we get: E/R=E/R1 + E/R1 +E/3 and cancelling throughout by E gives 1/R=1/R1 + 1/R1 + 1/R3 The battery has negligible internal resistance using Kirchhoff’s second law for the complete loop from the battery to R1 and back again, we find that the p.d. across the resistor equals the e.m.f. of the battery. This will be true for each resistor.   Resistors in parallel have the same p.d. across them but the current flowing from supply is shared between them.   (e) Solve circuit problems involving series and parallel circuits with one or more sources of e.m.f.;       (f) Explain that all sources of e.m.f. have an internal resistance; Same current flows through the interior of the supply (from – to +). Interior of a supply is made up of chemicals or metal wire and must have resistance – this is the internal resistance r of the supple.   (g) Explain the meaning of the term terminal p.d.; Terminal p.d. is the p.d, across the load (external) resistance. Terminal p.d. is the energy transferred when 1 coulomb of charge flows through the load resistance. If there was no internal resistance the terminal p.d. would be the same as the e.m.f.   (h) Select and use the equations e.m.f. = I (R + r), and             e.m.f. = V + Ir .   (a) draw a simple potential divider circuit; (b) explain how a potential divider circuit can be used to produce a variable p.d.; (c) selectandusethepotentialdividerequation Vout          R2         Vin ;