Chemistry Answers Chapter 12

Answers Chapter 12 Exercises 1 2 3 A Substance χaverage Δχ % ionic character Bonding Cl2O 3.3 0.2 6 (Polar) covalent PbCl2 2.5 1.4 44 Polar covalent Al2O3 2.5 1.8 56 Ionic HBr 2.6 0.8 25 Polar covalent NaBr 1.95 2.1 66 Ionic Substance χaverage Δχ % ionic character Bonding van Arkel-Ketelaar Triangle of Bonding 3.0 electronegativity difference 2.5 ∆ = |a − b| ionic % % covalent ionic 92 8 25 75 50 50 75 25 100 0 Polar covalent CuO 2.65 1.5 47 ∆ eneg 2.0 CuO 1.5 polar covalent 1.0 0.5 metallic 0.0 0.79 1.0 1.5 covalent 2.0 2.5 3.0 3.5 4.0 average electronegativity ∑ = (a + b) 2 4 to aluminium atoms: Al3++ 3e− → Al (d) Aluminium is more reactive than hydrogen. Hydrogen gas would be produced as the hydrogen from the water is reduced in preference to the aluminium. Metal atoms can slide across each other with the metallic bonding not breaking as the delocalized electrons can move to accommodate the changes in the lattice. The ionic and covalent bonds are directional and more rigid in ceramics. They resist changes in the atomic arrangement but will break if the applied forces are too strong. 5 6 (e) Aluminium oxide is only 56% ionic based on electronegativity values. The ions are not completely free in the molten compound. (f) Oxygen is produced at the anode from the oxide ions: Concrete can contain iron or carbon fibres. If these are connected into a network within the concrete the material will conduct electricity along the network. 2O2− → O2 + 4e− The oxygen reacts with the carbon to produce carbon dioxide: (a) Bauxite C + O2 → CO2 (b) Aluminium is more reactive than carbon. (c) Aluminium ions are attracted towards the negative electrode where they are reduced 7 At the cathode Cu2+ + 2e− → Cu 1 mol 2 mol 1 mol 58 10:01 2Fe2O3(s) + 2CH4(g) + O2 → 4Fe(l) + 2CO2(g) + 4H2O(l) Balanced n(Cu)/n(e−) = 1/2 n(Cu) = (1/2)n(e−) 0.100 F = 0.100 mol of e− n(Cu) = 0.0500 mol 10 m(Cu) = 0.0500 mol × 63.55 g mol−1 = 3.18 g 11 (b) TiCl4 + 2Mg → Ti + 2MgCl2 1 mol 2 mol n(Cl2)/n(e−) = 1/2 n(Cl2) = (1/2)n(e−) 0.100 F = 0.100 mol of e− 12 n(Cl2) = 0.0500 mol V(Cl2) = 0.0500 mol × 22.4 dm3 mol−1 = 1.1 dm3 8 The alloy is stronger than the pure metal. Adding atoms of different size disrupts the regular metal lattice so that it is difficult for one layer to slide over another. Alloying can make the metal harder, stronger and more resistant to corrosion. At the anode 2Cl− → Cl2 + 2e− 2 mol (a) TiO2 + 2C + 2Cl2 → TiCl4 + 2CO All of the electron spins are paired in diamagnetic elements. Atoms are paramagnetic if they have unpaired electrons. So, to determine whether the elements are paramagnetic or diamagnetic, we need to consider the electron configuration for each element. n(e−) = 0.0965 A × 1000 s/96500 C mol−1 = 96.5/96500 = 0.00100 mol n(Ti) = 0.011975/47.9 = 0.000250 mol n(e−)/n(Ti) = 4 Electron config. Element Ti is in the +4 state. Formula: TiCl4 9 Fe2O3(s) + CH4(g) + O2 → 2Fe(l) + CO2(g) + H2O(l) Unbalanced Balance the elements only present in a combined state first. Balance the C and H: Fe2O3(s) + CH4(g) + O2 → Fe(l) + CO2(g) + 2H2O(l) Unbalanced Balance the Fe and O: Atom K [Ar]4s1 Electron configuration No. of 1 unpaired electrons Magnetic behaviour Na [Ne]3s1 Mg Al Si P S [Ne]3s2 [Ne]3s23p1 [Ne]3s23p2 [Ne]3s23p3 [Ne]3s23p4 1 0 1 2 3 2 Para Dia Para Para Para Para Cl Ar [Ne]3s23p5 [Ne]3s23p6 1 0 Para Dia Phosphorus has the most unpaired electrons and so is the most paramagnetic. Fe2O3(s) + CH4(g) + 21O2 → 2Fe(l) + CO2(g) + 2H2O(l) Balanced 13 No. of unpaired electrons Sc V Mn Ga As [Ar]3d14s2 [Ar]3d34s2 [Ar]3d54s2 [Ar]3d104s24p1 [Ar]3d104s24p3 1 3 5 1 3 K, Sc, Ga < V, As < V 59 10:01 Answers 14 (a) Positive argon ions and (free) electrons. An activated complex corresponds to a maximum in the energy and a reaction intermediate corresponds to a local minimum in energy. Reaction intermediates can in theory be isolated. (b) Argon, as it is present in the plasma. (c) ICP-MS (d) ICP-AES , ICP-MS is less effective with nonmetals as they have higher ionization energies and so form positive ions less readily. 15 (b) Heterogeneous catalysts are in a different phase from the reactants; they can be easily removed by filtration. (a) Different calibrations are produced for each electron transition so three transitions are analysed. (c) They have a large surface area per unit mass for reactants to be adsorbed and their surface structure can be modified to improve effectiveness. (b) It produced 3.00 × 107 counts in one minute= 0.0500 × 107 c s−1 = 5.00 × 105cs−1 = 500 kc s−1 (d) Toxicity of the nanoparticles is dependent on their size, so need to regulate for type of material and size of particles. [Hg] = 1.95 µg dm−3 (c) 798 kc s−1 (d) Series 2, as it has the steepest gradient; small differences in concentration can be detected with large differences in count rate. 16 0.37% by mass 17 Transition metals are effective heterogeneous catalysts as they form weak bonds to small reactant molecules which allow them to come together with the correct orientation. 20 (a) An activated complex is an unstable combination of reactant molecules that can go on to form products or fall apart to form reactants. A reaction intermediate is a species that is produced and consumed during a reaction but does not occur in the overall equation. disordered disordered Molecular orientation disordered ordered (c) The molecules/ions group together to form a spherical arrangement; the hydrophilic heads are exposed to water, shielding the non-polar tails. 21 (b) Sulfur impurities block the active sites of the catalyst; the impurities are adsorbed on the catalyst surface more strongly than reactant molecules. 19 Molecular arrangement The phase transitions of lyotropic liquid crystals depend on both temperature and concentration. (a) Lower temperatures needed so reduced energy costs. Catalysts act selectively, increasing the yield of the desired product. They are not used up and so can be reused over long periods of time. Liquid crystal (b) The phase transitions of thermotropic liquid crystals depend on temperature. The ability of transition metals to show variable oxidation states allows them to be particularly effective homogeneous catalysts. 18 Liquid (a) Thermotropic liquid crystal materials are pure substances that show liquid crystal behaviour over a temperature range between the solid and liquid states. Example: biphenyl nitriles. Lyotropic liquid crystals are solutions that show the liquid crystal state at certain concentrations. Examples: soap and water, Kevlar® in solution. 22 (a) Low reactivity of C–H bond due to high bond energy and low polarity. (b) Increases polarity. Molecule can change orientation when an electric field is applied. 23 (a) C24H23N 60 10:01 (b) The addition of a benzene ring makes the molecule more rigid and rod-shaped. 24 Atom economy = (b) (a) A CH2 (b) There are strong C–C covalent bonds within the chains and relatively strong intermolecular forces between the large polymer chains. 25 (a) ( ) H NH N CH3 C C H H HN CH2 H CH3 H CH3 H CH3 H CH3 isotactic polypropene H CH3 (c) Isotactic polypropene has a regular structure with the methyl groups pointing in the same direction and so is crystalline and toug h. n = 2.1 × 10 /42.09 = 50000 6 (e) The chains in a polymer are not all the same length. 27 (a) Relative molar mass of reactant = (6 × 12.01) + (24 × 1.01) + (12 × 14.01) + ( 6 × 16.00) = 360.42 Relative molar mass of desired product = (3 × 12.01) + (6 × 1.01) + (6 × 14.01) = 126.15 NH CH2 Elastomers have very limited cross-links which knot some chains together and prevent molecules slipping across each other without restricting the freedom of the molecules to coil and uncoil. 28 (b) The non-expanded form of polystyrene is a colourless, transparent, brittle plastic. The expanded form is opaque with a lower density. It is a better insulator and shock absorber. The expanded form is produced by heating polystyrene beads with a volatile hydrocarbon such as pentane. The pentane evaporates and causes bubbles to form in the plastic. N (d) There is extensive cross-linking in thermosetting plastics which means they cannot be reshaped and do not melt when heated. (d) Mr = (3 × 12.01) + (6 × 1.01) = 42.09 (a) The pure form, which has strong diplole– dipole interactions between the chains, is hard and brittle. The addition of plasticizers allows the chains to slip across each other and makes the plastic more flexible. N (c) There are relatively weak intermolecular forces between the chains in polyethene. These forces are broken when the solid melts but are reformed when the liquid is cooled. The crosslinks between the chains in the thermosetting resin are made from strong covalent bonds. When heated to high temperature the resin does not melt but burns. n (b) 26 126.15 × 100% = 35.0% 360.42 (a) Polystyrene can act as a good shock absorber. Its low density will reduce transport costs and make it easier to handle. (b) A volatile hydrocarbon is added during the polymerization process. The volatile hydrocarbon turns into a gas, forming bubbles that force the surrounding polymer to expand and take the shape of the mould. 29 (a) No. of diameters = 10 × 10−6 m/1 × 10−9 m = 10−5/10−9 = 104 = 10 000 (b) Strong covalent C–C bonds must be broken. (c) Range of tube lengths with different structures lead to a less regular structure in the solid, which reduces strength. As properties are sensitive to tube length, it is difficult to produce tubes with required properties. 61 10:02 Answers (d) Large surface area for reactants to be adsorbed; the shape and size of the tubes make them shape-selective catalysts, only reactants of the appropriate geometry are able to interact effectively with the active sites. (e) Quantum effects predominate and the electrons behave like waves; the length of the tube affects the behaviour of electrons; the tubes are conductors or semiconductors depending on the length. 30 33 (a) The size of the nanoparticles is similar to the wavelength of harmful UV radiation. UV is scattered and not absorbed. (b) Toxicity of the nanoparticles is dependent on their size, so need to regulate for type of material and size of particles. 31 (d) Natural polymers (e.g. starch, cellulose or protein) can be added. Bacteria can break down the natural polymers and so the bag is broken down into smaller pieces. 34 35 36 (b) Scanning tunnelling microscope (STM) or atomic force microscope (AFM). (a) Plastics are easily moulded; they are nonbiodegradable; they have low density. Method Advantages Disadvantages landfill simple method to deal with large volumes plastics are not biodegradable; limited sites incineration reduces volume; plastics are concentrated energy source CO2 is a greenhouse gas; CO is poisonous; HCl produced from combustion of PVC causes acid rain conserves natural resources plastics need to be sorted recycling Disadvantages efficient method to deal with large volumes not popular with locals; needs to be maintained and monitored after use incineration reduces volume; energy source can cause pollutants such as greenhouse gases and dioxins Advantages: saves natural resources; saves energy; reduces pollution (c) Bacteria do not have the enzymes needed to break the C–C bonds present. (a) 1–5 (b) 1–4 Approximate length = 25 × 2 × 70 × 10−12 m = 3.5 × 10−9 m (b) Method Disadvantages: materials need to be sorted (a) Approx. 25 atoms high Each C atom has a diameter of 2 × 75 × 10−12 m and each O atom has a diameter of 2 × 64 × 10−12 m 32 Advantages landfill Both molecules have C–H bonds so they have strong absorptions between 2850 and 3090 cm−1. The monomer has a C=C bond not found in the polymer so it will have a weak absorption at 1620–1680 cm−1. 37 (a) A (b) A: The resistance is zero at very low temperature due to the formation of Cooper electron pairs. The crystal structure is distorted at low vibrational energies by the presence of electrons, and pairs are more difficult to impede than single electrons. B: The resistance decreases as the reduced vibrational energies of the ions in the lattice offer reduced resistance to the passage of single electrons. 38 The electrons are given energy as they are accelerated by the power source. They collide with the ions in the lattice and pass on some of their kinetic energy. The average vibrational energy and thus temperature of the ions increases. 62 10:02 39 Type 1 superconductors are metals or alloys. Density = (4 × 196.97/6.02 × 1023)/(407 × 10−12)3 g m−3 They only exhibit superconductivity at very low temperatures (<20 K). They lose their superconductivity suddenly as the temperature or magnetic field strength is increased. = 19.4 × 103 kg m−3 43 d = λ/2 sin θ Type 2 superconductors are ceramic metal compounds. d = 150 × 10−12 m/(2 × sin(17.9°)) = 244 × 10−12m Some can exhibit superconductivity at higher temperatures (<100 K). They lose their superconductivity gradually as the temperature or magnetic field strength is increased. nλ = 2d sin θ With n = 1 (θ = 17.9°) Assuming the atoms are touching R = d/2 = 122 × 10−12 m 44 –(–NH–(CH2)6–NH–CO–(CH2)8–CO–)n– 45 40 Location of atoms Corner Number of atoms Contribution 8 8 × (1/8) Total atoms HO H 46 (a) The diagonal of the cube = 4rM The length of a unit cell = 4rM / • √3 = (4 × 220/ • √3) × 10−12 m = 508 × 10−12 m (a) [O (b) [ HN (c) 47 [O (b) H2O Mass of individual atom = 39.10/6.02 × 1023g (c) CO Mass of unit cell = 2 × 39.10/6.02 × 1023 g = 9.91 × 105 g m−3 CH2 CH2 (CH2)4 O CO CH2 CH2 CO CH2 NH CO CH2 CH2 CH2 CH2 CH2 O C ] n + H 2O CO ] n + HCl O NH (CH2)3 NH C O ] n [ NH CONH CONH CO ] n NH2 (e) The C, H, N and O atoms have a lower relative atomic mass than Fe. The atoms in Kevlar® are not close packed, unlike in a metal. = 991 kg m−3 (a) The diagonal a face = 4rM 48 (a) H = 407 × 10−12 m (b) The volume of a cube = (407 × 10−12)3 m3 CH3 C C H H (b) Mass of individual atom = 196.97/6.02 × 1023 g CH2 (d) The polymer formed has straight chains. Hydrogen bonds can form between the closely packed chains. Density = (2 × 39.10/6.02 × 1023)/ (508 × 10−12)3 g m−3 No. of atoms = 4 CH2 (a) The primary amine and carboxylic acid groups. No. of atoms = 2 The length of a unit cell = 4rM / • √2 = (4 × 144/ • √2) × 10−12 m OH CH3 O (b) The volume of a cube = (508 × 10−12)3 m3 42 C C 1 There is one atom. 41 O HO CH3 O CH C OH 23 Mass of unit cell = 4 × 196.97/6.02 ×10 g 63 10:02 Answers (c) 49 50 H CH3 C C H CO2CH3 2Fe2+ + H2O2 → 2Fe3+ + 2OH− Heavy metal ions interfere with the function of key enzymes. They bind to the proteins instead of essential ions such as Mg2+ or Ca2+ and so interfere with key biochemical processes. (c) Oxidation number = −12 One electron is transferred to Fe3+, which is reduced to Fe2+. (a) The ligand has two O atoms which can form dative coordinate bonds. It is a bidentate ligand. (b) Each ox forms two bonds. Coordination number = 6. Oxidation state = +3 51 NiS(s) Ni2+(aq) + S2−(aq) Challenge yourself 1 Ksp = [Ni (aq)] [S (aq)] = s = 2.0 × 10 2+ s = √2.0 × 10 2− 2 −26 2 −26 = 1.4 × 10−13 mol dm−3 52 (c) In regions where there is no applied voltage the liquid crystal rotates the plane of polarization so light is now transmitted by the analyser. In regions where there is voltage applied the liquid crystal molecules align with the electric field and no longer rotate the plane of polarization. The observer would see a dark circle surrounded by light. (c) Ksp = [Al3+(aq)] [PO43−(aq)] (d) Ksp = [Ni2+(aq)] [OH−(aq)]2 (a) 1.3 × 10−5 mol dm−3 (b) [Ag+(aq)] [Cl−(aq)] = 1.6 × 10−10 1.6 × 10−10 [Ag+(aq)] = 0.100 = 1.6 × 10−9 mol dm−3 54 (a) s2 (b) 4s3 (c) 4s3 (d) 108s5 (e) 27s4 55 3 The halogen atoms have a larger mass and so C–X bonds vibrate at a lower frequency. 4 ON(Ca) = +2 ON(O) = −2 (a) Ksp = [Pb2+] [S2−] = 1.30 × 10−28 (b) [Pb2+] = [S2−] [Pb2+] = √1.30 × 10−28 = 1.14 × 10−14 mol dm−3 (c) As the product of the concentrations is constant, an increase in [S2−] will lead to a decrease in [Pb2+] and Pb2+ will be precipitated out of solution. (a) The observer can see nothing. The whole area would appear dark as the polarizer and analyser are crossed. (b) The whole area would appear light as the liquid crystal rotates the plane of polarization, so light is now transmitted by the analyser. (a) Ksp = [Pb2+(aq)] [S2−(aq)] (b) Ksp = [Cu+(aq)]2 [S2−(aq)] 53 (a) The molecule is rigid and rod-shaped. (b) Thermotropic. If the solubility is s: [Ni2+(aq)] = s; [S2−(aq)] = s ON(Ti) = 6 − 2 = +4 5 ON(Ba) = +2 ON(O) = −2 ON(Y) = +3 3 × ON(Cu) = 14 – 4 – 3 = +7 ON = +7/3 = +213 Two of the Cu ions are Cu2+ and one is Cu3+. 64 10:02 6 There are four atoms in the unit cell so the volume occupied = 4 × volume of one atom 12 Assuming the atoms are spherical: Consider a 1 dm3 sample of H2O with a mass of 1.000 × 103 g Volume = 4 × (4/3)πr3 n(H2O) = m(H2O)/M(H2O) Edge of a unit cell = (4 / √‾2)r = 1.000 × 103/(16.00 + 2(1.01)) Volume of unit cell = ((4 / √‾2)r) 3 = 55.49 mol dm−3 Volume occupied = [(16 / 3)πr / ((4 / √‾2)r) ] × 100% 3 3 13 If the solubility is s: [Mm+(aq)] = ps; [Xn−(aq)] = qs Ksp = [Mm+(aq)]p [Xn−(aq)]q = [(√‾2)3 ×π / 12] × 100% = (ps)p(qs)q = 74% 7 [H2O] = n(H2O)/V(H2O) = ppqqsp+q There are two atoms in the unit cell so the volume occupied = 2 × volume of one atom Assuming the atoms are spherical: Practice questions Volume occupied = 2 × (4/3)πr3 1 Edge of a unit cell = (4 / √‾3)r Volume of unit cell = ((4 / √‾3)r) 3 Volume occupied = [(8 / 3)πr3 / ((4 / √‾3) r)3] × 100% = [(√‾3)3 ×π / 24] × 100% = 68% 8 9 10 Element No. of atoms Allow lowers melting point or lowers melting point of aluminium oxide. Position Total number of atoms Ca 1 Centre 1×1=1 O 12 Edges 12/4 = 3 Ti 8 Corners 8 × 1/8 = 1 Formula: CaTiO3 Do not allow lowers melting point of aluminium. (ii) Positive electrode: 2O2− → O2 + 4e− / O2− → 12O2 + 2e− Negative electrode: Al3+ + 3e− → Al Polyesters have similar linkages to biopolymers and so can be broken down by bacteria. They are broken down by hydrolysis. Allow e instead of e−. (b) use of fossil fuels (to provide energy) Hg binds to sulfur 2+ There are two particles on the left and seven particles on the right. There will be an increase in the number of ways the energy can be distributed and an increase in entropy. This means ∆G (= ∆H − T∆S) is more likely to be negative and the reaction more likely to be spontaneous. [2] Award [1] for correct equations but wrong electrodes. oxidation of the (graphite) positive electrode / anode [2] Cysteine 11 (a) (i) melting point of the cryolite solution is much lower than the melting point of alumina / Al2O3 / it lowers the melting point of the mixture / cell operates at lower temperature [1] 2 (a) Al is more reactive than Fe / Al is higher than Fe in the reactivity series / it is harder to reduce aluminium ores compared to iron ores / Fe3+ is a better oxidizing agent than [2] Al3+ / OWTTE (b) (i) Fe3O4 + 4CO → 3Fe + 4CO2 (ii) Fe3O4 + 4H2 → 3Fe + 4H2O [1] [1] 65 10:02 Answers 3 (a) homogeneous mixture of metals / a metal and non-metal [1] 6 (a) C–Cl bond / molecule is polar stronger intermolecular / van der Waals / London / dispersion forces / dipole–dipole attraction [2] (b) alloying element(s) disrupts regular / repeating (metal) lattice difficult for one layer to slide over another / atoms smaller than the metal cations can fit into the (holes of) metal lattice disrupting bonding (b) addition of plasticizers Allow misspelling within reason. gets between polymer chains / keeps chains further apart and reduces attraction (between the chains) [2] can make metal harder / stronger / more corrosion resistant / brittle [2 max] 4 (c) Mode of action of homogeneous catalysis: catalyst reacts in one step (of the mechanism) and is regenerated at a later step /ability to form a range of oxidation states (for transition metals) / reaction steps with catalyst have lower activation energies than for reaction without catalyst / OWTTE 8 Example using chemical equation: 2SO3(g) / N2(g) + 3H2(g) 2NH3(g) / C2H4(g) + H2(g) C2H6(g) / other suitable example Reversible sign not required for mark. Catalyst and states must be specified to score mark. [4] 5 H H C C C C C C H Cl H Cl H Cl (a) biphenyl nitriles / cyanobiphenyls [1] (b) nitrile groups make molecule polar intermolecular forces are strong enough to align in a common direction biphenyl groups make molecules more rigid/ rod-shaped (long) alkane chain ensures that molecules cannot pack together closely (to maintain liquid-crystal state) [4] [1] makes the polymer low density / good thermal insulator / expanded / softer / better shock absorber packaging/insulation catalyst provides the reactive surface / presence of active sites / adsorb reactant molecule(s) (on surface) Ni(s) H Hydrogen atoms must be included. 7 Mode of action of heterogeneous catalysis: V2O3(g) H Continuation bonds at end of structure not needed. H+ (aq) CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O(l) / other suitable example Fe(s) H Accept any structure with all the Cl atoms shown on the same side. Example using chemical equation: 2SO2(g) + O2(g) H Award [1 max] if thermal insulation given for both answers. [2] (a) 1 nm to 100 nm [1] (b) physical techniques move atoms to a specific position chemical techniques involve chemical reactions to position atoms (in molecules) Accept suitable examples for chemical techniques. [2] (c) reference to effect on human health (e.g. unknown, immune system may not cope, unsatisfactory toxicity regulations) reference to effect on employment (e.g. increased/decreased job opportunities, adverse effect on traditional industries) reference to effect on quality of life (e.g. medical advances, faster computers, improved performance of electronic equipment) 66 10:02 reference to public opinion (e.g. need to improve information, encourage discussion, seek approval) ends have half a buckyball (fullerene) / carbons in pentagons (and hexagons) reference to nanotechnology being developed in wealthier nations hence increasing the divide between different nations [2 max] 9 Do not accept ‘reactive surface’. high selectivity related to dimensions of tube [2] Advantage: (ii) unknown health effects Accept potentially harmful as easily ingested/inhaled. Disadvantage: expensive to build and operate / can form dioxins/toxic gases / requires energy / adds to greenhouse effect [2] Accept difficulty of preparing nanotubes in required amounts. [1] 13 (a) (i) CN makes molecule polar, ensures common orientation which can be changed by electric field (a) Landfill: can be used to deal with large volumes/ amounts / filled ground can be re-used / low cost (ii) C5H11 prevents close packing of molecules Do not accept ‘no air pollution’. (iii) Incineration: reduces volume / requires minimal space / source of energy molecules rigid and rod-shaped Accept chemical stability for second or third mark not both. [3] Do not accept ‘no land pollution’ (b) liquid crystal between two glass plates which have scratches at 90° to each other Apply list principle, i.e. award [0] when onecorrect and one incorrect advantage given. [2] molecules form a twisted arrangement between plates due to intermolecular bonds (b) limited supply of oxygen (prevents the bacteria from acting) Do not accept air. 11 [1] when polarizers are aligned with scratches, light will pass through film and pixel will appear bright (a) Positive electrode: graphite / carbon applied voltage aligns polar molecules and pixel appears dark [4] Negative electrode: graphite / carbon (on a steel liner) [2] (b) much less energy required to recycle than to produce Al from ore / OWTTE 14 (a) walls have rolled/single sheets of graphite / carbons bonded in hexagons (rod-shaped) molecules aligned in the same direction increasing temperature causes arrangement to lose its directional order / molecules to become more randomly arranged until normal liquid state occurs [3] less production of CO2/greenhouse gases (graphite used in the electrolysis is converted into CO2) / the more that is recycled the less there will be in landfill sites / OWTTE [2] 12 [1] (c) (i) large surface area reduce volume / stable odour-free residue / source of energy 10 [2] (b) covalent bonds are very strong 15 (a) (i) CH2CHCH3 [1] 67 10:02 Answers (ii) H CH3 H CH3 H CH3 Do not accept boiling point. C C C C C C softness / hardness / flexibility / strength / rigidity / density [2] H H H H H H (c) atactic harder / more rigid / higher melting point / stronger / denser crystalline / chains closer together [3] methyl groups arranged randomly / OWTTE 19 (b) polystyrene beads contain pentane / volatile hydrocarbon to produce heat C + O2 → CO2 [4] OR Any two properties, [1] each. to act as a reducing agent / to produce carbon monoxide (c) carbon dioxide is a greenhouse gas / CO2 causes global warming, climate change etc. produces toxic chlorine compounds / causes acid rain due to HCl 16 Fe2O3 + 3C → 2Fe + 3CO / 2Fe2O3 + 3C → 4Fe + 3CO2 / C + CO2 → 2CO [2] (limestone) HDPE: to remove impurities / silica no / very few branches CaCO3 → CaO + CO2 and CaO + SiO2 → CaSiO3 / CaCO3 + SiO2 → CaSiO3 + CO2 chains pack closer together stronger intermolecular forces 20 Allow converse argument, e.g. LDPE has more branches, so its chains are further apart and the intermolecular forces are weaker. [4] 17 coke / carbon / C and limestone / calcium carbonate / CaCO3 (coke) heating causes pentane to evaporate white / opaque / lower density / better insulator / (better) shock absorber [5] (metal) does not rust / corrode low density thermal insulator / poor conductor of heat (a) addition of plasticizers more flexible / flexibility [2] electrical insulator / poor conductor of electricity [2] Accept any answer above for [1]. (b) polymer disadvantages (wood) difficult to dispose of polymer properly easily moulded fills up landfill sites non-biodegradable / does not rot litter low density lack of biodegradability Accept any answer above for [1]. use of natural resources Do not accept reference to cost. [2] Award [1] each for any two. 21 PVC disadvantages burning produces toxic gases / HCl [3 max] 18 (a) CH3 C [1] H C / H H (b) melting point CH3 − CH = CH2 (a) lyotropic liquid crystal rigid rod-shaped molecules alignment of molecules depends on concentration of solution [3] (b) (Kevlar has) strong hydrogen bonds between chains creating a very ordered / strong structure [2] 68 10:02 (c) acid donates a proton to the O and N atoms, breaking the hydrogen bonds 22 = 3.82 × 10−2 g [2] more Ni(OH)2 will precipitate due to common ion effect / OWTTE [2] [Cl−] = 2.3 × 10−8 mol dm−3 Ksp = [Pb2+][Cl−]2 / 1.7 × 10−5 = 1.9 × 10−2 × [Cl−]2 [Cl−] = 3.0 × 10−2 mol dm−3 AgCl will precipitate first (because it is less soluble) 23 24 [5] (a) The diagonal of a face = 4rM and the length of a unit cell = 4rM/√‾2 = 4 × 174/√‾2 × 10−12 m = 492 × 10−12 m (b) The volume of a cube = (492 × 10−12)3 m3 (a) Ksp = [Ni2+] × [OH−]2 [OH−] = 2[Ni2+], hence Ksp = 4[Ni2+]3 6.50 × 10−18 [Ni2+] = = 1.18 × 10−6 mol dm−3 4 Mass of Ni2+ in 1 dm3 = 58.71 × 1.18 × 10−6 = 6.90 × 10−5 g ( ) No. of unit cells = 1 × 10−6/(492 × 10−12)3 = 8.4 × 1021 25 = 4 × 130/√‾2 × 10−12 m = 368 × 10−12 m Accept Ksp = [Ni2+] × [OH−]2 The volume of a cube = (368 × 10−12)3 m3 No. of atoms = 4 pH = pOH = 7 → [OH−] = 10−7 mol dm−3 Mass of individual atom = 195.08/6.02 × 1023 g [Ni2+] 6.50 × 10−18 = 6.50 × 10−4 mol dm−3 (10−7)2 2+ Mass of Ni in 1 dm3 = 6.50 × 10−4 × 58.71 2 [4] The length of a unit cell = 4rM/√‾2 Award [4] for correct final answer. = [4] Award [4] for correct final answer. (b) add excess hydroxide ions / increase the pH Ksp = [Ag+][Cl−]/1.8 × 10−10 = 8.0 × 10−3 × [Cl−] Mass of unit cell = 4 × 195.08/6.02 × 1023 g Density = (4 × 195.08/6.02 × 1023)/ (368 × 10−12)3 g m−3 = 26.00 × 103 kg m−3 [5]  10:02