chemistry perfect score 2011 module answer

Download Chemistry Perfect Score 2011 module answer

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Chemistry Perfect Score 2011 module answer



2. JAWAPAN SET 1PAPER 2 : STRUCTURED QUESTIONSection ANo.AnswerMark1 (a)The formula that shows the simplest whole number ratio of atoms1 of each element in a compound.(b)H2SO4 + Zn ZnSO4 + H22(c)Heating, cooling and weighing are repeated until a constant mass is1 obtained.(d)Element CopperOxygenMass, g47.70 25.30 53.30 47.70=22.40 =5.60Mole atom22.405.60 6416= 0.35= 0.35Simplest ratio 11 Empirical formula = CuO4(e)H2 + CuO Cu + H2O2(f)To prevent the hot copper from being oxidized again. 1(g) Magnesium ribbon Heat2 TOTAL 13No. AnswerMark2 (a) (i) Al2CO3 1(ii) Al2(CO3)3Al2O3 + 3CO2 2(iii) The number of mole of Al2 (CO3) 3 = 70.2/ 234= 0.3 mol 1 Based on the balanced equation; Al2 (CO3)3 : Al2O3 1: 10.3 : 0.3 1Mass of Ag = 0.4 x 102 = 30.6 g 1(iv)Based on the balanced equationAl2 (CO3)3 : CO21 : 3 0.3 : 0.61Volume of CO2 = 0.9 x 24= 21.6 dm31 = 21600 cm31(b) (i) Zinc carbonate1(ii)Zinc oxide and carbon dioxide 12 3. (iii) ZnCO3 ZnO + CO2 1TOTAL 12No. AnswerMark3 (a) (i)The number of protons found in the nucleus of an atom 1(ii) 7 1(b)133 Q 16(c)P and S // Q and R1(d) (i)Q and R 1(ii) Have same proton number but different nucleon number // 1 Have same number of protons but different number of neutrons(e) (i)Melting point : 63 OC [values & unit must be correct] 1(ii) Section Physical state ABSolid DELiquid and gas 1(iii) the heat energy absorbed by the particles is used1to overcome the forces of attraction between particles 1TOTAL 10No. AnswerMark4 (a)Sodium and magnesium // sodium and aluminium // magnesium and 1 aluminium(b)Halogen 1(c)2.8.3 1(d) (i)Sodium, magnesium, aluminium, chlorine, argon 1 Atomic size decreases(ii) From left to right : The proton number // the positive charge increases from sodium to argon 1 The forces of attraction by the nucleus on the electrons (nuclei attraction)1 in the first three occupied shells become stronger(e) (i)Sodium burnt rapidly and brightly with a yellow flame //1 White fumes liberated // white solid formed(ii)2Na + Cl2 2NaCl[Formula of reactants and product are correct] 1[Balanced equation]1has high melting / boiling point // conduct electricity1(iii) in molten state or aqueous solution // soluble in waterTOTAL103 4. No.AnswerMark5 (a) (i)X1 (ii)8 valence electron // electron arrangement 2.8 // achieve octet electron 1 arrangement(b)Covalent 1(c) (i) VW4 ( (b)1 (a) i i(ii) ) ( 1+1 WWV W W(iii)has low melting / boiling point // cannot conduct electricity1 in molten and solid state . // insoluble in water// soluble in organic solvent.(d) (i)Ionic compound 1 (ii)Atom U donate one electron to form U+ ion1 Atom W accept one electron to form W- ion1 U+ ion and W- ion attracted to each other by strong electrostatic force /1 ionic bond. (iii)11U W [Number of electron each shells are correct] [Number of charge symbol are correct] TOTAL13 4 5. PAPER 2: ESSAY QUESTIONSection BNo. Answer Mark6 (a) Group 171Period 31Has seven valence electrons.1Has three shells occupied with electron 1 (b) (i) Between Y and X 1.Atom Y has 1 valence electron and atom X has 7 valence electron1 2. to achieve octet electron arrangement 1 3. Atom Y loses/donates/transfers 1 electron to form ion Y+1 4. Atom X gains/receives 1 electrons from atom Y to form ion X-1 5 Y+ ion and X- ion are attracted by a strong electrostatic force / ionic bond 1 6. Diagram+Y X1 (ii) Between W and X 1. Atom W has 4 valence electrons and atom X has 7 valence electrons.1 2. Each atom W contributes 4 electrons whereas each atom X contributesone electron for sharing. 3. to achieve octet electron arrangement1 4. Four atoms of X share a pair of electrons with one atom W to form a1WX4 molecule / Diagram1XW X X W W W X WMolecules WX4 5 6. (c) Compound P : ionic bond1Compound Q : Covalent bond 1Melting PointCompound PIons are held by strong electrostatic forces.More energy is needed to overcome these forces.1Compound Q 1Molecules are held by weak intermolecular forces.Only a little energy is required to overcome the forces.Or 1Electrical conductivity 1Compound PIn molten state or aqueous solution , there are free moving ions 1Ions carry charge 1Compound Q 1In molten and solid states , no free moving ions 1exist as moleculeTOTAL20No.Answer Mark7 (a) (i)2.8.7, Chlorine1+1(ii) 2Fe + 3Cl2 2FeCl3 Correct formulae of reactants and product 1 Balanced1(b) (i)Z,Y,X 1 Z more reactive than X1 Atomic size of Z bigger than atomic size X1 Valence electron become further away from nucleus 1 Valence electron to be more weakly pulled by the nucleus1 Valence electron can be released more easily in atom Z1(ii) same/similar1 Same valence electron 1(c)X : 2.4 1 Y : 2.6 1 to achieve octet electron arrangement one X atom contributes four electron and each two Y atoms 1 contributes two electrons for sharing 1 Group 161 Period 21 6 valence electron1 2 shells occupied with electrons1 TOTAL206 7. PAPER 2: ESSAY QUESTIONSection CNo. AnswerMark8 (a) (i)Dilute acid: Hydrochloric acid / Sulphuric acid/ Nitric acid1 Metal N: Magnesium / zinc 1(ii) Anhydrous calcium chloride1 To dry the hydrogen gas 1(iii) Example: Copper(II) oxide1Copper ion is reduced// reduction process1Because oxidation number of copper decrease from +2 to 0Hydrogen is oxidised// oxidation process 1Because oxidation number of hydrogen increase from 0 to +1 1Hydrogen is reducing agent 1Copper(II) ion// Copper(II) oxide is oxidising agent 1(b) (i)Relative Molecular mass of (CH2)n = 561 (12 + 2) n = 56n=4Molecular formula = C4H8 1(ii)Unglasedporcelain chips Glass wool soaked in butanol Heat Water 2 Procedure: 1.A small amount of glass wool soaked in butanol is placed in a boiling tube.1 2.The boiling tube is clamped horizontally1 3.The unglazed porcelain chips are placed in the middle section of the boiling tube.1 4.The boiling tube is closed with a stopper fitted with a delivery tube 1 5.The unglazed porcelain chips are heated strongly. Then, the glass wool is warmed gently to vaporize the propanol. 1 6.The gas released is collected in a test tube. 1TOTAL 20 7 8. No.Answer Mark9 (a) (i)Formula that shows the simplest ratio of the number of atoms for each 1 element in the compound.(ii) Copper(II)oxide // lead(II)oxide 1 CuO + H2 Cu + H2O // PbO + H2 Pb + H2O1+1(b) (i)Magnesium oxide / zinc oxide 1(ii)Procedure: 1. Clean magnesium / zinc ribbon with sand paper 1 2. Weigh crucible and its lid1 3. Put magnesium ribbon into the crucible and weigh the crucible withits lid 1 4. Heat strongly the crucible without its lid1 5. Cover the crucible when the magnesium starts to burn and lift/raise 1 the lid a little at intervals 6.Remove the lid when the magnesium burnt completely 1 7.Heat strongly the crucible for a few minutes 1 8.Cool and weigh the crucible with its lid and the content 1 9.Repeat the processes of heating, cooling and weighing until a constant 1 mass is obtained 10. Record all the mass1 Result: DescriptionMass/g1Crucible + lidxCrucible + lid + magnesiumyCrucible + lid + magnesium oxidez Calculation: Element Mg O Mass, g y-x z-y Moley-x z-y24161 =0.1=0.1 Simplest ratio1 11 Empirical formula: MgO Max 10(c)Element CHMass (%) 84.6 15.4Number of moles84.6/12 15.4/11=7.05=15.41Mole ratio12 Empirical formula : CH2 1 RMM of (CH2)n = 70 1 [ 12 + 2]n = 7014 n = 70n = 5 1 Molecular formula : C5H10120 8 9. JAWAPAN SET 2PAPER 2 : STRUCTURED QUESTIONSection ANo.AnswerMark1 (a)Cell II1(b) (i)Magnesium electrode1(ii) eV 1 MagnesiumCopper electrode electrode (iii) Copper electrode thicker // Brown solid deposited1 (c) 1. Correct formulae of reactant and product1 2. Balanced equation 1Cu2+ + 2e Cu (d) (i) Electrical energy to chemical energy 1 (ii)Blue colour remain unchange1 (iii) 1. Concentration / Number of mole of Cu2+ ion remain unchanged 1 2. Rate of Cu2+ ion discharge at cathode is the same as rate of Cu atom ionize at anode 1 TOTAL10No. Answer Mark2 (a) (i)Iodine 1 r: formula/iodide/iodine gas (ii)MnO4 - + 8 H+ + 5 e Mn2+ + 4 H2O 1 (iii) +7 +21 reduction1 (iv)Potassium chloride // iron(II) sulphate // [any reducing agent]1(b)(i) Zinc 1 (ii)1. Correct formulae of reactant and product1 2. Balanced equation 12 Zn + O2 2 ZnOa: 2 J + O2 2 JO (iii) K,J, L 1 (iv)Predict : no changes 1 r: no reaction Reason : L is more reactive than J/zinc1 r: more electropositive TOTAL11 9 10. PAPER 2 : ESSAY QUESTIONSection BNo. AnswerMark3 (a)1.Propanone is a covalent compound1 2.Propanone exist as molecule // No freely moving ion in propanone1 3.Sodium chloride is an ionic compound 4.Sodium chloride solution has freely moving ion1 1(b) (i)Properties Cell XCell Y1. Type of cellVoltaic cellElectrolytic cell2. EnergyChemical electrical Electrical chemical change 13. ElectrodesAnode: APositive terminal: C // Copper 1 Cathode: BNegative terminal: D // Zinc4. Ions in Cu2+, SO42-, H+ and OH- Cu2+, SO42-, H+ and OH- ions 1 electrolyte ions5. HalfAnode:Positive terminal: 1 equationCu Cu2+ + 2eCu2+ + 2e Cu Cathode:Negative terminal1 Cu2+ + 2e CuZn Zn2+ + 2e6. Observation Anode:Positive terminal: 1 Copper ecomes thinner Copper plate becomes thicker1 Cathode:Negative terminal: 1.6 Copper becomesZinc becomes thinner thicker(c)1. Ag, M, L1 2. L is more electropositive than silver 1 3. L displace silver from silver nitrate solution1 4. M is more electropositive than silver 1 5. M displace silver from silver nitrate solution1 6. M is less electropositive than L1 7. M cannot displace L from L nitrate solution 1(i Copper // Cu 1i)TOTAL20No. AnswerMark4 (a) (i) 1.Correct formulae of reactant and product 12.Balanced equation1Zn+ 2e Zn2+3.Correct formulae of reactant and product 14.Balanced equation1Pb2+ + 2e Pb(ii)1.Zinc is oxidized 12.Zinc atom donates / losses electrons 13.Lead(II) nitrate / Pb2+ is reduced 14.Lead(II) nitrate / Pb2+ receives electrons 1(b) (i) 1. Green colour of iron(II) sulphate change to brown 12. Correct formulae of reactant and