2025 Christ Church Prelim P2 MS

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 1 Marking Scheme1aisulfur dioxide1iicopper (II) chloride1iiizinc sulfate1ivsulfur dioxide1biArrangement: particles closely packed in an unorderly manner Movement: particles able to slide past each other 2iiNa2O + H2O  2NaOH1iiiBlue precipitate formed12aNo. of moles of S2O32 = x 0.1 = 0.005 mol No of moles of I2 = x 0.1 = 0.004 mol [1]Mole ratio of S2O32 : I2  2 : 10.005 mol of S2O32- requires 0.0025 mol of I2 to react [1]Thus, the limiting reactant is S2O32- [1]3bMole ratio of S2O32 : I-  1 : 1No. of moles of I- = 0.005 mol [1]% yield = x 100 % = 26.0% [1][Accept ECF from (a)]2cThe oxidation state of iodine decreases [1] from 0 in I2 to 1 in I- [1] Accept answer where student show thiosulfate being oxidised and link it back to the characteristic of a reducing agent. 2dNo. Different elements are being reduced and oxidised at the same time.13a2Mg(s) + O2(g)  2MgO(s)[Reject: state symbols not written or incorrect]1bNitrogen.1

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 2 cvolume of O2 reacted = 90 – 72= 18 cm3number of moles of O2 reacted= (18/1000) / 24= 0.00075 mol [1]From the balanced chemical equation, 2 mol of Mg reacts with 1 mol of O2.Hence, 0.0015 mol of Mg will react with 0.00075 mol of O2. [1]Mass of Mg reacted= 0.0015 x 24= 0.036 g [1][Accept: ECF from (a)] 3 4aMost LeastR, S, P, Copper, Q1bAnode: R(s)  R+(aq) + eCathode: Cu2+(aq) + 2e  Cu(s)[Reject: state symbols not written or incorrect.]2cAny two:Electrical energy is not supplied but produced through chemical reactions in a simple cell.Electrons move spontaneously from the anode to the cathode.The anode is the negative terminal in a simple cell.25aN2 (g) + 3H2 (g) 2NH3 (g)[1]From the equation, 1 mole / volume of N2 reacts with 3 moles / volumes of H2. [1]2bTemperature: about 450 oCPressure: about 250 atmCatalyst: finely divided iron[1] for all 3 conditions3cMethod: upward delivery [1]Reason: ammonia is less dense than air [1]2diWhen pressure increases, particles come closer together. [1]Frequency of effective collisions increases; rate of reaction increases.[1]2iiSince ammonia and nitrogen monoxide are gases, Volume ratio of NH3 : NO  1 : 1Theoretical volume of NO = 440 dm3 [1]40% = x 100 % x = 176 dm3 [1]2

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 3 6aSilicon(IV) dioxide is a covalent compound and with no mobile charge carriers.1bThey both have a macromolecular structure. [1]They both have atoms in a rigid tetrahedral arrangment. [1]2ciAny twoSilicate contains only covalent bonds while wine glass contains both covalent and ionic bonds.Silicate has a regular arrangement while wine glass has an irregular arrangement.In silicate, each silicon atom is bonded to three oxygen atoms but in wine glass, each silicon atom can be bonded to one or more oxide ions.Silicate contains no ions but wine glass has calcium and sodium ions.2iiThe strong electrostatic forces of attraction between the ions are overcome when the glass is melted. [1] The ions become free to move about / act as mobile charge carriers to conduct electricity [1].27aA strong acid is a substance that ionises/dissociates completely in water to produce high concentration of hydrogen ions/ H+ ions.[1] 1bThe iron stirrer would react with the sulfuric acid immediately.1cBa(OH)2(aq) + H2SO4(aq)  BaSO4(s) + 2H2O(l)1m – correct balanced equation1m – correct state symbols2dThe conductivity was high at the start as there were free moving mobile barium and hydroxide ions [1]. As more sulfuric acid was added, the ions are being removed from the solution when they combine with the hydrogen and sulfate ions. Hence, the number of free moving mobile ions in the solution decreases [1].When a total of 5 cm3 of sulfuric acid was added, only the insoluble barium sulfate salt and water are present. There are no free moving mobile ions in the solution at that time [1], Hence, conductivity of the solution is zero. When excess acid was added, the conductivity of the solution increases as there is an increase in the number of free moving mobile ions from the sulfuric acid [1].4 e = = concentration of barium hydroxide, Mb = =0.100 mol/dm3 [Accept ECF from (c)]1

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 4 8aCH3OCH3 + 3O2  2CO2 + 3H2O 1bMr of diethylether = 74 [1]Enthalpy change of combustion in kJ/g = Enthalpy in kJ/mol / Mr= -2726 / 74 = -36.8 kJ/g [1]2cThe overall enthalpy change is calculated by the difference between the energy absorbed during bond breaking and the energy released during bond forming. [1] In the combustion reaction of these products, there is more energy released from bond forming than energy absorbed for bond breakling. [1]2dIncrease1eThe types of bonds in the molecules that need to be overcome are different so different amount of energy are needed when these bonds are broken.1fButanol Methylproponal [Accept: hydroxyl group on any carbon]1 gThe potassium manganate (VII) solution will be decolourised / turn from purple to colourless in the sample with the alcohol.The potassium manganate (VII) solution will remain purple in the sample with ethoxyethane1h 1m – correct hydroxyl groups1m – correct carbon backbone29a 2

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 5 1m – shared electrons; 1m – unshared electrons[Minus 1m if legend not shown.]bCCl3F is a simple molecule [structure] with weak intermolecular forces of attraction [bonding] [1] that require a low amount of energy to overcome [1]. Hence, it would have a low boiling point. 2diAllotropes1iiThe ClO particle that is produced in step 1 reacts with an ozone molecule in step 2 to produce another Cl atom. [1]This Cl atom is then able to start step 1 again / to be reused in step 1 / used again for step 1 [1] and react with more ozone molecules.2Excessive UV radiation/light that prevents risk of skin cancer will not be absorbed.110aCalcium carbonate reacts with hydrochloric acid to form carbon dioxide gas.The mass loss is due to the carbon dioxide gas being liberated.1bAny 2:Reading on the electronic balance remains unchanged.There would be no more zinc carbonate left in the reaction mixtureNo more effervescence is observed.2ci Experiment 2 1iiIn the same concentration and volume, sulfuric acid has twice the number of hydrogen ions as compared to hydrochloric acid. [1]Concentration of reactant particles increases, increase frequency of effective collision, rate of reaction increases. Since calcium carbonate is the limiting reagent, total mass loss remains at 0.6 g. [1]2

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 6 iii 1diThe lead(II) sulfate produced forms an insoluble layer on the surface of the lead (II) carbonate, preventing further reaction with the acid.1iiPrecipitation method [1]Lead(II) nitrate [1]211aAdd aqueous bromine to the polymers seperately. Aqueous bromine will decolourise in the sample with polymer W. 1b 2cn= 14000 / [(4x12)+(8x1)]= 250 [1]No. of carbon atoms= 250 x 4= 1000 [1]2dNo. Polymer Y is a polyamide which does not require an alcohol / which requires a diacid and diamine. 1ei 1iiMethanol1

Christ Church Secondary SchoolSecondary 4 Express2025 Preliminary ExaminationChemistry Page | 7 fDisadvantage: Polymer can only be disposed off by burying in landfills which willlead to land or air pollution.[1]Advantage: Polymer is durable and resistant to corrosion in daily use. [1]2