9701/11Paper 1 Multiple Choice May/June 2013Cl-1Cl-

Group 70Cl2

+1ClO-

+3ClO2-

+4ClO2

+5ClO3-

+7ClO4-

1. for ions likeClO4-thecharge on the ion is equal to the sum of the oxidation numbers so Cl is +7, O is -2, so +7 + (4 x -2) = -1 =- for overall charge on ion Note the convention: for oxidation state, the sign comes before the number for charge on an ion, the sign comes after the number======================================================ExtractionofAluminium(continued) -Electrolysis Cell.

Thesteelcontaineriscoatedwithcarbon(graphite) and this is used as thenegative electrode(cathode).Aluminiumoxide(Al2O3) is anioniccompound.When it ismeltedtheAl3+andO2-ionsarefreetomove andconductelectricity.Electrolysisof thealumina/cryolitesolution givesaluminiumat thecathodeandoxygenat theanode.4Al3++ 12e- 4Al(aluminium metalat the (-)cathode)reduction.6O2- - 12e- 3O2 (oxygen gasat the (+)anode)oxidation.Aluminiumismoredensethan thealumina/cryolitesolution and so itfallsto thebottomof thecellwhere it can betapped offaspureliquid metal.Theoverall reactionisaluminiumoxidealuminium+oxygen.2Al2O3(l) 4Al(l) + 3O2(g)Oxygenisgivenoffat thepositivecarbonanode.Carbon dioxideis alsogivenoffat thecarbonanodebecause hotoxygenreactswith hecarbonanodeto formcarbon dioxidegas.carbon+oxygencarbon dioxide.C(s) +O2(g)CO2(g)Thecarbonanodesslowlydisappear becauseeach moleculeofcarbon dioxidewhich isgivenofftakes alittle pieceofcarbonaway with it.Thecarbonanodesneed to bereplacedwhen they becometoo small.A. -Aluminium ions are reduced in this processB. -Aluminium is liberated at the cathode by the reaction Al3+ + 3e Al.C. Cryolite is not a purified aluminium oxide instead a white or colourless mineral consisting of a fluoride of sodium and aluminium. Cryolite(Na3AlF6, sodium hexafluoroaluminate).

-What is cryolite's purpose in the extraction of aluminum?

The alluminium oxide has to be molten for its electrolytic extraction. To melt the electrolyte, heat has to be provided upto temperatures of 2200 decree celsius. The provision of this energy can be very expensive and cost-ineffective.

Cryolite is added as an impurity to the electrolyte. The cryolite, lowers the melting point of alluminium oxide to 980. The rule is as it is for all impurities they lower the melting point and raise the boiling point, so the reduction of the melting point allows the alluminium to melt at lower temperatures, making electrolysis work at lower temperatures, hence requires lesser energy, making the process cheaper, and more cost effective.================================================================== there are more protons in each nucleus so thenuclear chargein each element increases ... therefore theforce of attractionbetween the nucleus and outer electron is increased, and ...

there is a negligible increase inshieldingbecause each successive electron enters the same energy level ... so more energy is needed to remove the outer electron. the first ionization of an atom is always an endothermic processThe law of conservation of energy dictates that the difference

between the energy of the ionizing radiation and the energy of the ejected electron equals the energy required for ionization. Ionization energies are typically reported either in kJ/mol or in eV (1eV= 96.485 kJ/mol). ionization energy generally increases to the right across a period and decreases down a given group.=================================6. hydrolysis reactions of an organic compound like eter require a catalyst - an acid (H+ions) or alkali (OH-ions) as reaction with water is very slow. The dilute acid provides both the acid catalyst and the water.============================================================7. If the pressure is higher, the chances of collision are greater.

==============================================9.As molecules get larger, then dispersion forces will increase, and you may get other intermolecular forces such as dipole-dipole attractions as well. Gases made of molecules such as these will be much less ideal. Like helium, a hydrogen molecule also has two electrons, and so the intermolecular forces are going to be small - but not as small as helium. In the hydrogen molecule, you have two atoms that you can distribute the charges over. A helium molecule consists of a single small atom, and the van der Waals dispersion forces are as low as it is possible for them to be.====================================================== The ideal gas equation is: pV = nRTPressure, p: Pressure is measured in pascals, Pa - sometimes expressed as newtons per square metre, N m-2. These mean exactly the same thing.Be careful if you are given pressures in kPa (kilopascals). For example, 150 kPa is 150,000 Pa. You must make that conversion before you use the ideal gas equation.Should you want to convert from other pressure measurements: 1 atmosphere = 101,325 Pa 1 bar = 100 kPa = 100,000 Pa

Volume, VThis is the most likely place for you to go wrong when you use this equation. That's because the SI unit of volume is the cubic metre, m3-notcm3or dm3.1 m3= 1000 dm3= 1,000,000 cm3

Number of moles, nThis is easy, of course - it is just a number. You already know that you work it out by dividing the mass in grams by the mass of one mole in grams.You will most often use the ideal gas equation by first making the substitution to give:

Therefore Mr = mRT ---------------- pV

=================================================================these quations are given in data booklet so no need to memorize itSn2+ Sn4++ 2e- equation 1(MnO4)+ (8H+)+ 5e (Mn2+)+ 4H2O) equation 2balance the number of electrons so that each equation have same number of electrons multiply equation 1 by 5 and equation 2 by 5 and then add these equations we get(16H+)+(5Sn2+)+2MnO4- yields (5sn4+)+2Mn2+ignore the liquid we are concerned with ions and ions only2moles of mno4- reacts with 5moles of sn2+2/5moles of mn 1 moles sn(9.5/190)*2/5 9.5/190The amount of SnCl2 in the question is 0.05 moles. Because of thestoichiometry of the reaction (worked out using oxidation states) this will produce 0.02 moles of Mn2+.============================================ The hydroxides becomemore solubleas you go down the Group. The sulphates becomeless solubleas you go down the Group The carbonates tend to becomeless solubleas you go down the Group.====================================because iodine is less reactive than bromine so it's 1- ions do not displace the bromine ions in sodium bromide. (It is lower down in the group).================================================================Astatide ion, At-, is a reducing agent. So when it's oxidised, the product is astatine:

2At- ------> At2+ 2e-

HAt is not a redox product, and the others are reduction or neutralisation products of the H2SO4.=================================================17.It can be either Nacl or NaBr uptill the colorless solution but what is happeneing on the addition of excess HNO3.??

HNO3 simply removes the NH3 ligands by protonating them, regenerating the silver halide ppt.The key is the "excess NH3(aq)" used, instead of "excess concentrated NH3(aq)" that isrequired to dissolve AgBr(s). Hence it can be deduced that the ppt is AgCl(s), rather than AgBr(s).======================================================18.Equilibrium constantsaren't changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature.==========================================19.Ammonium compunds ALWAYS react with alkalis to form salt,water and AMMONIA gas.So the answer would be C since limewater i.e. calcium hydroxide would react with ammonium sulphate to liberate ammonia.===========================================================21.B is correct because the secondary alcohol group in lactic acid will form anester with methanoic acid. C is incorrect because the secondary alcohol group in lactic acid is not acidic enough to liberate CO2 from NaHCO3. It is possible that some candidates did not select B because the given formula of the organic product, CH3CH(O2CH)CO2H, proved difficult to comprehend. In this situationcandidates may make the question easier for themselves by drawing out fully displayed formulae on their question paper.

=======================

22.

Electrophile================================