SOALAN ULANGKAJI SPM 00 37 Jawapan Chemistry [454/] [454/] [454/3] Chemistry Paper (454/) A A 3 A 4 B 5 D 6 C 7 D 8 B 9 C 0 A B D 3 B 4 A 5 C 6 D 7 C 8 D 9 B 0 A A D 3 D 4 C 5 B 6 C 7 D 8 A 9 A 30 A 3 B 3 C 33 C 34 B 35 C 36 A 37 A 38 C 39 A 40 C 4 B 4 A 43 C 44 B 45 B 46 B 47 B 48 A 49 C 50 C
38 SOALAN ULANGKAJI SPM 00 Chemistry Paper (454/) Q Mark scheme (sample answer) Sub Mark (a) Saponification Total Mark. Detergent. Detergent does not form scum. Detergent more effective in hard water. Detergent more effective in acidic water (d)(i) To prevent the growth of bacteria // to make sure food last longer. To make meat look fresh (e)(i) Salt//sugar//vinegar.(a) (i). absorb water out of the food. bacteria cannot live without water A :,8, //.8. B :,8,6 //.8.6 Ionic bond. Atom A donate electrons to form A+. Atom B receive electron to form B- 0 + (iv) High melting and boiling point // Can conduct electricity in molten and aqueous solution// Soluble in water// Insoluble in organic compound (i) CB 76 0
SOALAN ULANGKAJI SPM 00 39 3. Silver metal // silver Silver nitrate Ag + + e Ag. Intensity increase. Because the concentration of Cu + ions increase (d)(i) Z, Y, X, W. X. Because X less electropositive than Z // vice versa 4. 0.6 v Zinc chloride 0 (i) Lead(II) chloride ZnCO 3 + HCl ZnCl + CO + H O. Mol HCl = (.0)(0) 000 = 0.0. From equation, mol HCl produce mol R So, 0.0 mol HCl produce 0.0 mol R + (i) 5. 3. Volume R = 0.0 x 4 dm 3 = 0.4 dm 3 / 40 cm 3 Precipitate reaction // double decomposition reaction. The mixture is filtered. the residue is rinsed with distilled water Hydrogenation 3 0 C 3 H 6 + H C 3 H 8 +. Put cm 3 of propene and substance X into two difference test tube.. add bromine water / acidified KMnO 4 into both test tube 3. Brownish colour of bromine / purple colour of acidified KMnO 4 is decolourised for propene. No change for X. 3
330 SOALAN ULANGKAJI SPM 00 Bromine water Hydration (d). Propyl methanoate 6. O H H H. H C O C C C H H H H Brown colour. add sodium hydroxide solution. brown precipitate is formed From + to +3 Acidified potassium manganate(vii) solution (d) (e)(i) 7.(a) Acidified potassium dichromate(vi) // Chlorine water // bromine water Reduction X + 4(-) = + x = +7 I - I + e When the number of proton increase, the number of electron also increase. The force attraction between nucleus and valence electron increase The size of atom become smaller Z, Y, X + 0 4 The electron arrangement of X is. and Z is.6 X and Y are hold together by ionic bond Two atom X donate its valence electron to form X+ ions One atom Z receive two electron to form Z - ion X + ions and Z - ion achieve stable duplet and octet electron arrangement 6
SOALAN ULANGKAJI SPM 00 33 (i). right electron arrangement. has nucleus Compound in. high melting point Compound in. low melting point 3. because electrostatic force is a strong force. 5. can conduct electricity in molten state and aqueous solution. 7. because it has freely moving ions. 4. because van der waals force is a weak force 6. cannot conduct electricity in any state. 8. Because it has no ions. 8 0 8.(a) (i). remove the wire gauze.. use a windshield 3. replace beaker with copper can. 4. place the spirit lamp on a wooden block. 4 (iv). some heat loss to the surrounding. incomplete combustion of ethanol any two 3. some of ethanol evaporate. Heat change / release when mol of compound. burnt completely in excess oxygen. C H 5 OH + 3O CO + 3H O. correct reactant and product. balance equation + 30 ++ 3. y-axis with energy label.. correct product, reactant 3. correct value of heat of combustion with correct unit.. Mol ethanol = 0.3 = 0.005 46. mol ethanol release 376 kj heat So, heat release for 0.005 mol ethanol = 0.005 x 376 kj = 6880 J 3. 6880 = 500 x 4. x θ θ = 3.3 O C. 4. final temperature = 8.0 + 3.3 = 3.3 oc (with correct unit) 4
33 SOALAN ULANGKAJI SPM 00 9.. when going down the homologous series the number of carbon atom and hydrogen atom increase. more number of molecules of carbon dioxide and water produced. 3. the heat of combustion also increase. Substance that can change the rate of reaction. 3 0. use in a small amount.. Remain unchanged until the end of reaction. Any two 3. Specific in action. 4. cannot change the product of reaction. +. Harber process // contact process // hydrogenation // Ostwald. Iron // vanadium pentoxide // nickel // platinum. size of reactant (total surface area). smaller size of reactant has the large total surface area. 3. more surface area expose to the collision. 4. affective collision increase, rate of reaction increase 5. the temperature of reactant. 6. the higher the temperature, the higher the rate of reaction 7. kinetic energy of particle increase and particle move faster 8. affective collision increase, rate of reaction increase.. cooking meat in small pieces.. small pieces of meat have large total surface area. 3. more surface area of meat expose to the heat. 4. meat cook faster 5. store food in freezer. 6. in freezer, temperature is low. 7. bacteria become inactive. 8. the decomposition of food by bacteria become slow. 9. food last longer 8 8 0
SOALAN ULANGKAJI SPM 00 333 0.(a). hydrogen gas. H + + e H Cell P. electrolytic cell // electric energy change to chemical energy 3. use the same type of electrode // both using copper electrode. 5. Cathode become thicker, anode become thinner 7. cathode : Cu + + e Cu Anode : Cu Cu + + e Cell Q. voltaic cell // chemical energy change to electric energy 4. use two difference electrode // difference metals. 6. cathode become thicker, anode become thinner and 8. cathode : Cu + + e Cu Anode : Zn Zn + + e 8 Example : electroplate iron key with silver.. Chemical : silver metal, 0.5 moldm -3 silver nitrate solution. Procedure:. iron key is placed at the anode and silver metal at cathode of cell. 3. iron key and silver metal are dipped in the silver nitrate solution. 4. the switch is on. 5. diagram. 6. cathode : Ag + + e Ag 7. anode : Ag Ag + + e 8. to get best result : use a small current (0.5A) Rotate the iron key continuously Clean iron key using sand paper any one total 0 0
334 SOALAN ULANGKAJI SPM 00 Chemistry Paper (453/). (a) [able to record three burette reading correctly with two decimal places and correct unit] Exp. :Initial burette reading = 5.60 cm3 Final burette reading = 30.60 cm3 Exp. : Initial burette reading =.60 cm3 Final burette reading = 5.0 cm3 [able to construct table with three columns and three rows with correct unit] Exp. Exp. Initial burette reading / cm3 5.60.60 Final burette reading / cm3 30.60 5.0 Total volume of acid used / cm3 5.00.50 - sulphuric acid is diprotic acid, hydrochloric acid is mono protic acid. - mole of sulphuric acid produce mol of H+ ions, - the volume of sulphuric acid used half than volume of hydrochloric acid (d) [able to state all three variable correctly] (i) Manipulated Responding Constants : type of acids used : Volume of acid used ; concentration of sodium hydroxide used, size of conical flask (e) [able to state hypothesis correctly] The higher the concentration of H+ ions in acid, the lower the volume of acid used for the neutralisation process. (f) [able to state the observation correctly] The pink colour of phenolphthalein turn to colourless. (g) [able to state operation definition correctly] The volume of acid needed to neutralise sodium hydroxide accurately from the titration of acid-base. (h) (i) [able to write the reactants and products correctly also able to balance correctly] H SO 4 + NaOH Na SO 4 + H O [able to calculate concentration of acid with correct unit for the answer] Mol of NaOH = 5.0 x.0 000 = 0.05
SOALAN ULANGKAJI SPM 00 335 From equation, mol of H SO 4 need mol of NaOH, So, mol of HSO4 needed = 0.05 = 0.05 Concentration of hydrochloric acid = 0.05 x 000 5 =.0 moldm -3 (i) (i) [able to predict the volume correctly with correct unit. ] 5.00 cm 3 - because nitric acid also monoprotic acid same as hydrochloric acid - mol of nitric acid produce mol of H + ions. (j) [able to classify the acid given correctly] Strong acid Hydrochloric acid Sulphuric acid Nitric acid Weak acid Ethanoic acid Methanoic acid. (a) Problem statement. Does the brass harder than pure copper? All variables Manipulated : Brass and copper block Responding : Diameter of dent produced Constant : steel ball, mass of weighed, distance of weighed from the block. Hypothesis. The harder the substance, the smaller the diameter of dent produced. (d) List of materials and apparatus. Brass block, copper block, meter ruler, kg of weighed, retort stand, steel ball, thread and cellophane tape. (e) Procedure of the experiment. A steel ball is taped on the copper block using a cellophane tape.. kg of weighed is suspended about meter from the copper block. 3. The weighed is released to the steel ball on the copper block 4. The dent produced is measured using the ruler. 5. Step until 4 are repeated twice to get the average of dent produced. 6. Experiment was repeated by replace the copper block with brass block. (f) Tabulation of data Copper block Brass block Diameter of dent 3 Average