NYGH 2014-S3EOY-Chem P1 & P2 Ans

1 2014 Secondary Three End-of-Year Examination (Chemistry) Paper 1 1 D 6 B 11 C 16 A 21 A 26 D 2 B 7 A 12 A 17 A 22 B 27 B 3 B 8 C 13 C 18 C 23 A 28 B 4 C 9 B 14 C 19 C 24 B 29 D 5 B 10 D 15 B 20 D 25 A 30 C Paper 2 Section A No Answers Marks A1(a) Hexane and Heptane are miscible [1]. They have different boiling points that are close to each other/ less than 25oC [1]. [2] (b) temperature/oC 100 90 80 70 total volume of distillate axes – 1m graph – 1 m [2] A2 (a) XO2 [1] (b) 32 32 162 162XO in oxygen of ncompositio Percentage 2 +=×+ ×= xx ( ) [1] 0.28 944.14533.0 32056.17533.0 3232533.0 [1] 533.032 32 = = =+ =+ =+ x x x x x Practise Error Carry Forw ard (E.C.F.) if the formula of the oxide from (a) is incorrect. Example: X4O : 56.1, X 2O : 7.00, X 3O2: 9.35, X 8O9: 53.3, X 2O4: 28.0, X 4O2: 7.00, X9O8: 12.5, X2O3: 21 [1] A3(a)(i) Reaction 1: CaCO3 → CaO + CO2 [1] Reaction 2: CaO + H2O → Ca(OH)2 [1] [2] (a)(ii) Lime is used in agriculture to reduce the acidity of the soil so as to obtain appropriate pH for growth of plants./neutralise/ increase the pH of the soil. [1] (a)(iii) Sodium hydroxide is a stronger alkali which ionises completely to give a higher concentration of OH-. [1]

2 No Answers Marks (b)(i) CaMg(CO3)2 + 4HCl → CaMgCl4 + 2CO2 + 2H2O Accept this answer : CaMg(CO3)2 + 4HCl → MgCl2 + CaCl2 + H2O + 2CO2 [1] (b)(ii) mol 01023.0 21612 450.0CO of moles of No. 2 = ×+= ( ) .CO of mole 2 forms COCaMg of mole 1 equation, the From 223 ( ) [1] mol 0.005115 2 01023.0COCaMg of moles of No. 23 = = ( ) ( )[ ] g 94116.0 1631222440005115.0COCaMg of Mass 23 = ×+×++×= [1] %1.94 %100000.1 94116.0Purity Percentage = ×= Practise Error Carry Forward (E.C.F.) if the equation from (b)(i) is not balanced. [2] A4(a) J: van der waals forces / intermolecular forces of attraction. (intermolecular forces and hydrogen bond is not accepted. ) [1] (b) Greater energy is required to overcome the strong ionic bonds/electrostatic forces of attraction between oppositely charged ions [1] in K than the weak van der waals forces between molecules of J [1]. [2] (c) A large amount of energy [1] is required to overcome the strong electrostatic forces of attraction between the positive ions and the delocalised electrons/ overcome the strong metallic bond. [1] [2] A5(a)(i) sulfuric acid chemical formula is not accepted. [1] (a)(ii) Reaction of aqueous sodium hydroxide and sulfuric acid will yield sodium sulfate and water according to the equation 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) [1] With the use of 0.100 mol/dm3 of NaOH, half the volume H2SO4 with the same concentration must be reacted with it to form Na 2SO4. /Titrate fixed volume of NaOH with H2SO4. [1] The salt solution is then heated until a saturated solution is formed and crystallised [1]. The crystals are dried by pressing between pieces of filter paper [1]. [4] (b) White precipitates are formed. [1] A6(a) Element X: nitrogen (or any Group V element) [1] Element Z: hydrogen/ any Group VII element [1] [2] (b)(i) H3AsO3 / Na2HAsO3 / NaH2AsO3 [1]

3 No Answers Marks Chemical formula of Arsenic is As and NOT Ar or AS No award of mark if Ar or AS was given as the answer. (b)(ii) Na3AsO3 / Na2HAsO3 / NaH2AsO3 [1] A7(a) Oxidising agent: Ag2O or H2O2 [1] Reducing agent: H2O2 [1] Need to have both answers correct to score 1 mark. [1] (b)(i) mol 1.0 24 10002400 oxidenitrogen of moles of No. = = [1] (b)(ii) [1] mol 400.0 1612 2.7OH of moles of No. 2 = +×= [1] mol 200.0 1000 2001NH of moles of No. 3 = ×= [2] (b)(iii) [1] 41.0 4.0y 21.0 2.0 == ==x OR No of mole of NxOy : No of mole of H2O : No of mole of NH3 0.1 : 0.4 : 0.2 1 : 4 : 2 [1] Formula of nitrogen oxide = N2O4 [1] [2] (b)(iv) ( ) [1] 4 [1] 2 420N of number Oxidation += ×−−= Error Carry Forward (E.C.F.) if the formula of the oxide from (b)(iii) is incorrect. [2] (c) N2O5 [1] When nitric acid reacts with a reducing agent (zinc), it is reduced and the oxidation number of nitrogen is decreased[1] . The oxidation number of nitrogen remains unchanged at +5 [1] if N2O5 is produced and hence, N2O5 is least likely to be produced. [3] Paper 2 Section B B8(a) D: zinc hydroxide/ Zn(OH)2 E: barium sulfate/ BaSO4 F: carbon dioxide/ CO2 [4]

4 G: calcium chloride/ CaCl2 H: calcium carbonate/ CaCO3 I: calcium hydroxide/ Ca(OH)2 J: silver chloride/ AgCl 1M for each correct answer, maximum 4m (b) Observation: The white precipitate dissolves to give a colourless solution/ solution turns clear [1]. Not accepted: 1. Effervescence (this will occur during bubbling of F , cannot tell that it is due to reaction of H and F.) 2. Same products F and G will be formed. (Is this still called a reaction if nothing changed?) Explanation: Carbon dioxide is acidic/ H2CO3, carbonic acid is formed when F dissolves in water / H is basic/ Reaction between F and H forms a soluble salt, calcium hydrogen carbonate. [1] Not accepted: F (CO2) reacts with limewater (This is not H.). F and H will react. (Repeating the question) No e.c.f for this question. [2] (c) ZnSO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + Zn(NO3)2(aq) Only e.c.f. from identitiy of D. Equation must still be chemically correct and sound to obtain any e.c.f. [1] (d) CaCO3 (s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) [1] mol 00250.0 1000 0.25100.0HC of moles of No. = ×=l [1] mol 0500.0 1631240 00.5CaCO of moles of No. 3 = ×++= reactant. limiting the is HCl [1] mol 00125.0 2 00250.0CaC of moles of No. 2 = =l Those who did not check for limiting reactant, maximum 2m. Error Carry Forward (E.C.F.) if the G is wrongly identified and only if the reaction given is a carbonate acid reaction. For e.c.f, will still deduct marks if working is wrong. [3]

5 B9(a)(i) [1] dm 000500.0 0.1100 05.0octane unburnt of Volume 3= ×= [1] (a)(ii) [1] mol 102.08mol/ 0000208.0 24 000500.0octane unburnt of moles of No. 5-×= = E.c.f from (a)(i) [1] (a)(iii) [1] dm 133.0 0.1100 3.13CO and CO of Volume 3 2 = ×= [1] (a)(iv) [1] mol 00554.0 24 133.0CO and CO of moles of No. 2 = = E.c.f from (a)(iii) [1] (a)(v) [1] mol 000693.0 00554.08 1octane of moles of No. = ×= E.c.f from (a)(iv) [1] (b)(i) Observations: A brown solution / black solid of iodine is formed./ Purple KMnO4 decolourised. [1] Not accepted: 1. Brown solid. 2. Solution/ Mixture turned colourless.(Solution/ Mixture appears as brown at the end due to presence of iodine. 3. KMnO4 turned colourless. (without mentioning that it was originally purple) Explanation: The oxidation state of manganese decreases from +7 to +2 while the oxidation state of iodine increases from -1 to 0 [1]. (If student wrote explanat ion for change in oxidation state of either iodine/ manganese only, explanation must match observation.) Not accepted: 1. Iodine increased/ oxidized from -1 to 0. 2. Manganese decreased/ reduced from +7 to +2. 3. Oxidation state of manganate (MnO4 -) decreased from +7 to +2. 4. Oxidation state of iodide (I-) increased from -1 to 0. 5. Writing oxidation states as 7+, 2+, 1- 6. Described changes in charge instead of oxidation states. [2]

6 (b)(ii) Concentration of K2Cr2O7 Alternatively, method No. of moles of KMnO 4 : No. of moles of K2Cr2O7 is 6:5 [1] No. of moles of K 2Cr2O7 = (0.0300 x 0.025) ÷ 6 x 5 = 0.000625 mol [1] [K2Cr2O7] = 0.001875 ÷0.025 = 0.0250mol/dm3 [1] [3] 10 Either (a) i. Li2CO3 + 2HCl → 2LiCl + H2O + CO2 [1] ii Moles of pure lithium carbonate = (0.635 x 5.00) ÷ Mr of Li2CO3 = 0.04290 mol Li2CO3