Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Sample p2 kimia 2013

6,389 views

Published on

Koleksi soalan Trial berjawapan

Published in: Education
  • Be the first to comment

Sample p2 kimia 2013

  1. 1. TARGET SPM P2 2013 1. STRUCTURE OF ATOM/BONDING/BOILING AND MELTING POINT (a) (i) The electron arrangement for argon is 2.8.8. Why is the element very stable and not reactive?[1 mark] Argon has achieved stable octet electron arrangement (Helium only achieved stable duplet electron arrangement) (ii) Name one other element that has the same stability as gas argon. [1 mark] Helium // Neon // other Noble Gases element (b) Atom of both sodium and chlorine are unstable. They react to form an ionic compound which is more stable. Diagram 3.1 shows a sodium chloride compound, NaCl, that is produced by the formation of an ionic bond between a sodium ion, Na+ , and a chloride ion, Cl- Diagram 3.1 (i) How are sodium ion and chloride ion formed from their respective atoms? [2 marks] Sodium ion produced when sodium atom donate/release one electron to chlorine atom. Chloride ion formed when chlorine atom accept/gain/receive one electron from sodium atom. (ii) Name the force that exists between these ions in the compound. Ionic bond // electrovalent bond [1 mark] (iii) The melting point of sodium chloride, NaCl, is 801ºC and its boiling points is 1413ºC. What will happen to the ions in this compound at 900ºC? [1 mark] Ions will moves more faster (iv) Give one reason for your answer in (b)(iii). Ions are absorbed heat energy. (c) [1 mark] Diagram 3.2 shows the proton number and nucleon number for two elements, X and Y. The letters used do not represent the actual symbols of the elements. Diagram 3.2 Draw a diagram to show the bonding formed between elements X and Y. [3 marks] (X : 4 proton number, electron arrangement 2.2; Y : 8 proton number, electron arrangement 2.6; so ions X2+ and Y2- produced XY compound) 2. (a) Table 1 shows four substances and their respective formulae. Substances Chemical formula Iodine I2 Copper Cu Target SPM (P2&P3) 2013 1
  2. 2. Naphthalene Copper(II) sulphate C10H8 CuSO4 Table 1 Use information from Table 1 to answer the following questions. (More than one atom is compound, 2 type of compound are ionic or covalent. Tips : if has metal atom – ion particles; non-metal atom only – molecules particles) (i) State one substance from Table 1 which exists as a molecule. Iodine // Naphthalene [1 mark] (ii) Which substance has the highest melting point, iodine, copper or naphthalene? Copper (because it is metal element) [1 mark] (iii) What is the state of matter of copper(II) sulphate at room temperature? [1 mark] Solid (because properties of ionic compound exist as solid state at room condition) (iv) State the substance in Table 1 which can conduct electricity in the solid state. Copper (conductor because metal) [1 mark] (v) Draw the arrangement of particles in the substance in (a)(iv). (solid particles – more than 3 layers of atom and not overlap) [1 mark] (vi) Write the ionic formula for the substance in (a)(iv). Cu+ // Cu2+ (b) [1 mark] Graph below shows the temperature against time when solid naphthalene is heated. (i) State the melting point of naphthalene. T1 º C [1 mark] (ii) Explain why there is no change in temperature from Q to R. [2 mark] Heat energy absorbed by naphthalene particles are used to overcome the forces of attraction between particles so that the solid can turn to liquid. (iii) State how the movement of naphthalene particles changes between R and S during heating. [1 mark] Particles will moves more faster Target SPM (P2&P3) 2013 2
  3. 3. 3. Graph 8 shows the heating curve of element X. Describe Graph 8 in terms of states of matter, particle arrangements and changes in energy.[10 marks] Stage to – t1 State of matter Solid t1 - t 2 Solid and Liquid t2 – t3 Liquid Particles arrangement The particles are close to each other. The particles arrangement is orderly. Some of particles are close to each other in orderly arrangement and some particles are close to each other but arrangement is not orderly. The particles are close to each other. The particles arrangement is not orderly. Changes in energy The kinetic energy increases The kinetic energy is constant The kinetic energy increases 4. Formation of ionic compound (metal [Group 1, 2 & 13] and non metal [ Group 14, 15, 16& 17]) Sample answer: 1. Electron arrangement of atoms ( Na , 28.1 ; Cl 2.8.7 ) // valence electrons 2. To achieve stable / octet electron arrangement 3. Atom ( Na) releases one / valence electron to form sodium ion, Na+ 4. Half equation ( Na  Na+ + e) 5. Atom (Cl) gain / accept electron to form chloride ion, Cl6. Half equation ( Cl + e  Cl- ) 7. Oppositely charged ion, Na+ & Cl- are attracted to one another by strong electrostatic force of attraction to form ionic compound, NaCl 8. Diagram Formation of covalent compound (nonmetal) 1. electron arrangement of the atom /valence electrons 2. to achieve duplet /octet electron arrangement 3. Atom (Carbon) contributes 4 electrons while (H) atom contributes 1 electron (for sharing). 4. one ( Carbon ) atom share 4 pairs of electrons with 4 (H) atoms to form covalent compound , CH4 / ratio 5. diagram Compare the physical properties of covalent and ionic compound Properties Covalent compound ( naphthalene) Ionic compound ( sodium chloride) Melting and - low - high boiling - consist of molecules - consist of oppositely charged ions - weak inter molecular forces - ions are held together by strong electrostatic between molecules forces . Target SPM (P2&P3) 2013 3
  4. 4. - less energy needed to overcome the weak forces Electrical - consist of molecules conductivity - does not conduct electricity in any state (molten or aqueous). - more heat energy needed to overcome the strong forces - consist of oppositely charged ions - conduct electricity in molten or aqueous solution. - in molten or aqueous solution, ions can move freely. 2. EMPIRICAL FORMULA a) Unreactive metal – reaction oxide metal with hydrogen gas, (CuO, PbO ,SnO ) b) Diagram Hydrogen c) Procedure - Weigh and record the mass of combustion tube with porcelain dish - Add a spatula of copper (II) oxide on the porcelain dish. Weigh the tube again. - Allow hydrogen gas flow into the tube for 5 – 10 minutes. - Burn the excess hydrogen. - Heat copper (II) oxide strongly. - Turn off the flame when black solid turns brown completely. - Continue the flow of hydrogen until the set of apparatus cool down to room temperature. - Weigh the combustion tube with its content. - -Repeat the process heating, cooling and weighing until a constant mass is obtained and record. Result : - combustion tube with porcelain dish = a g - combustion tube with porcelain dish + copper (II) oxide = b g -combustion tube with porcelain dish + copper = c g - mass of copper = ( c- a) g , Mass of oxygen = ( b- c ) g Target SPM (P2&P3) 2013 4 Reactive metal ( Mg, Zn – burn in excess oxygen / air ) – more reactive than H2 - Weigh and record a crucible with its lid - Clean Mg ribbon with sand paper then coil the Mg ribbon and place into the crucible. Weigh and record. - Heat strongly - When Mg ribbon start to burn, cover the crucible with lid. - Lift / raise the lid at intervals. - When the burning is complete, remove the lid and heat strongly. -Allow the crucible to cool down. -Weigh and record the crucible with content and lid. -Repeat the process heating, cooling and weighing until a constant mass is obtained and record. - Observation : White fume / solid formed - mass of crucible + lid = a g - mass of crucible + lid + Mg = b g - mass of crucible + lid + magnesium oxide = c g - mass of Mg = ( b – a ) g - mass of oxygen = ( c – b) g
  5. 5. Calculation: Element / atom Cu O Mass (g) x y Number of mole x / 64 y / 16 Simplest ratio of mole Precaution : 1. The flow of H2 must be continuous during cooling – to prevent hot copper metal from oxidized. 2. Allow hydrogen gas flow into the tube for 5 – 10 minutes to unsure air totally removed. The mixture H2 and air may cause an explosion. 3. To determine all air totally removed, collect the air and place lighted splinter, the gas burn quietly. [To prepare H2] 4. Zn + 2HCl  ZnCl2 + H2 5. Anhydrous calcium chloride – to dry the H2 gas. 6. Chemical equation : CuO + H2  Cu + H2O Target SPM (P2&P3) 2013 5 Element / atom Mg O Mass (g) x y Number of mole x / 24 y / 16 Simplest ratio of mole Precaution : 1. Clean Mg ribbon with sand paper to remove the layer of oxide on its surface. 2. Lift / raise the lid at intervals to allow air in 3. When Mg ribbon start to burn, cover the crucible with lid to avoid the white fume produced from being escape to the air. 4. Repeat the process heating, cooling and weighing to make sure all magnesium is completely reacted with oxygen. 5. Chemical equation : 2Mg + O2  2MgO
  6. 6. Target SPM (P2&P3) 2013 6
  7. 7. 8. Diagram 3 shows the apparatus set-up to determine the empirical formula of oxide metal M. Rajah 3 menunjukkan susunan radas untuk menentukan formula empirik bagi oksida logam M. Oxide metal M Dry hydrogen gas Oksida logam M Gas hidrogen kering Heat (a) (i) State the name of two reactants to prepare hydrogen gas in the laboratory. Panaskan Nyatakan nama dua bahan tindak balas untuk menyediakan gas hidrogen dalam makmal. Zinc , hydrochloric acid / suphuric acid (ii) Write the chemical equation for the reaction in (a)(i). Tuliskan persamaan kimia bagi tindak balas di (a)(i). Zn + 2HCl  ZnCl2 + H2 (b) State one precaution that must be taken when carrying out the experiment. Nyatakan satu langkah berjaga-jaga yang mesti diambil semasa menjalankan eksperimen itu. The air in the combustion tube must be displaced before lighting the hydrogen gas// The heating, cooling and weighing is repeated until a constant mass is obtained (c) Table 3 shows the results of the experiment: Jadual 3 menunjukkan keputusan eksperimen itu: Mass of combustion tube + asbestos paper Jisim tiub pembakaran + kertas asbestos Mass of combustion tube + asbestos paper + M oxide Jisim tiub pembakaran + kertas asbestos + oksida M Mass of combustion tube + asbestos paper + M Jisim tiub pembakaran + kertas asbestos + M 36.50 g 37.30 g 37.14 g (i) Based on the results in Table 3, determine the empirical formula of M oxide. Berdasarkan keputusan dalam Jadual 3, tentukan formula empirik bagi oksida M. [Relative atomic mass ; O=16, M=64] Element Mass Number of mole Simplest ratio M 0.64 0.64  0.01 64 1 O 0.16 0.16  0.01 16 1 Empirical formula is MO (ii) Write the chemical equation for the reaction between M oxide and hydrogen gas Tulis persamaan kimia bagi tindak balas antara oksida M dengan gas hidrogen. MO + H2  M + H2O (d) (i) The empirical formula of magnesium oxide cannot be determined by the above method. Explain why. Target SPM (P2&P3) 2013 7
  8. 8. Formula empirik bagi magnesium oksida tidak boleh ditentukan melalui kaedah di atas. Terangkan mengapa. Magnesium is more reactive than hydrogen. (ii) Draw a suitable set up of apparatus for the experiment to determine the empirical formula of magnesium oxide. Lukiskan susunan radas yang sesuai untuk eksperimen bagi menentukan formula empirik magnesium oksida. 3. PERIODIC TABLE OF ELEMENTS 1. Diagram 2 shows part of the Periodic Table of Elements. Diagram 2 (a) Based on Diagram 2, answer the following questions. What is the element represented by the symbol Fe? [1 mark] Ferum // Iron // Besi (b)In Diagram 2, mark ‘X’ in the boxes for all the transition elements. Cross all boxes between group 2 and 13 include Fe [1 mark] (c) State one specific characteristic of transition elements. [1 mark] Form coloured compounds //Have more than one oxidation number//Act as catalyst //Form complex ions / compounds. (d)Write the electron configuration for the Mg atom. [1 mark] 2.8.2 (period 3 and group 2, has 3 shell filled with electrons and 2 valence electron) (e) Write the chemical equation for the reaction between aluminium and oxygen gas. [1 mark] 4Al + 3O2 → 2Al2O3 (f) Briefly state the electron transfer in the bond formation between aluminium and oxygen. [ 2 mark] Target SPM (P2&P3) 2013 8
  9. 9. Aluminium atom donate 3 electron to oxygen atom to achieved stable octet electron arrangement. Oxygen atom will accepted 2 electron to achieved stable octet electron arrangement also. (g)State one gas is more suitable to be used in meteorological balloons? Give a reason. [1 mark] Helium because it is not flammable gas. 2. Diagram below shows part of the Periodic Table of the Elements. Q,R,T,X and Y do not represent the actual symbol of the elements. (a) Using the letters in the Periodic Table of the Elements in Diagram above, answer the following questions. You may use the Periodic Table of the Elements at the back. (i) Choose an element that is a halogen. Y [1 mark] (ii) Which element is monoatomic? R [1 mark] (iii) Which element forms an amphoteric oxide? X [1 mark] (b) Arrange Q,R,T,X and Y according to the increase in size of the atoms. R, Q, Y, X, T [1 mark] (c ) Write the electron arrangement for an atom of element Q. 2.4 (Group 14 Period 2) [1 mark] (d) Write the formula for the ion formed from an atom of element Y. Y- [1 mark] (e) Why are elements Q and R placed in the same period? Have same number of shells occupied/filled with electrons [1 mark] (f) When a small piece of element T is put into water, TOH solution is formed and hydrogen gas is released. State one observation when red litmus paper is put into the solution. [1 mark] Red litmus paper turns blue (g) State the common name of the elements between group 2 and group 13. Transition elements Target SPM (P2&P3) 2013 9 [1 mark]
  10. 10. 3. Diagram below shows part of the periodic table of elements. Based on diagram above: (a) (i) name one element which is a metal. Sodium, Na // Magnesium, Mg // Aluminium, Al [1 mark] (ii) which group and period is the metal in (a)(i) found in? [1 mark] Group 1 Period 3 // Group 2 Period 3 // Group 13 Period 3 (b)(i) Name the element that exists a monoatomic gas. Argon, Ar [1 mark] (ii) Explain why this gas is monoatomic. [2 marks] Has achieved stable octet electron arrangement // Valence electron is full with electrons. Do not need to donate or lose or share their electron with other elements. (c ) Sodium reacts with oxygen gas to form sodium oxide, Na2O. (i) write a balanced chemical equation for this reaction [2 marks] 4Na + O2 → 2Na2O (ii) sodium oxide reacts with water to produced a solution. In table below, mark (√) in the box which shows the value of pH of the solution. [1 mark] pH value 4 7 11 √ (iii) State which is more electronegative, sodium or chlorine. Explain your answer. Chlorine because has stronger nuclei attraction to attract electrons towards its nucleus. 4. ELECTROCHEMISTRY 1 Diagram 6 shows two types of cell. Diagram 6 (a) Compare and contrast cell P and cell Q. Include in your answer the observation and half equations for the reactions of the electrodes in both cells. [8 marks] Target SPM (P2&P3) 2013 10 [2 marks]
  11. 11. Cell P Electrical chemical (ec) +ve / anode: copper (OXIDATION) -ve / cathode: copper Cu2+ , H+ OH- , SO42Anode :Cu  Cu2+ + 2e (type of electrode) Cathode: Cu2+ + 2e  Cu ( ECS) Anode: copper electrode become thinner Cathode: brown solid formed/ becomes thicker. Electrolyte: intensity blue solution / concentration of Cu2+ solution remain. Rate of ionized of copper atom to form copper (II) ion at the anode same as rate of discharged copper (II) ion at the cathode. Characteristics Cell Q Energy change Chemical  electrical (ce) Electrode +ve/cathode: copper -ve/ anode: zinc (OXIDATION) Ions present in Cu2+ , H+ the electrolyte OH- , SO42Anode: Zn  Zn2+ + 2e Half equation Cathode: Cu2+ + 2e  Cu (ECS) Anode: becomes thinner Cathode: becomes thicker / brown solid Observation formed Electrolyte: intensity blue solution decrease / blue becomes paler 2. Figure 1 shows an experiment to construct an electrochemical series by measuring the voltage of a pair of metals in a simple voltaic cell. Rajah 1 menunjukkan satu eksperimen untuk membentuk satu siri e elektrokimia dengan cara mengukur nilai voltan pasangan logam dalam satu sel kimia. e Logam lain Kuprum Larutan natrium nitrat Figure 2/ Rajah 2 Target SPM (P2&P3) 2013 11
  12. 12. Figure 2 shows the reading on the voltmeter in different sets of experiment. Rajah 2 menunjukkan bacaan voltmeter dalam set eksperimen yang berbeza. (a) Write the voltage for each experiment in the spaces provided in the table below Tuliskan nilai voltan untuk setiap eksperimen di dalam jadual yang diberikan Experiment Eksperimen 1 2 3 4 Positive terminal Terminal positif Copper/Kuprum Copper/Kuprum Copper/Kuprum Copper/Kuprum Negative terminal Terminal negatif Magnesium/Magnesium Iron/Ferum Lead/Plumbum Zinc/Zinc Voltage/V Voltan/V 2.5 0.8 0.4 1.1 (b) Show the direction of electron flow on figure 1. Tunjukkan arah pergerakan elektron pada gambarajah 1. (c) What is the function of voltmeter in voltaic cell. Apakah fungsi voltmeter di dalam sel kimia. To measure the reading of voltage (d) Arrange the metals in (a) in the electrochemical series in ascending order. Susun logam dalam soalan (a) dalam siri elektrokimia mengikut tertib menaik. Mg, Zn, Fe, Pb, Cu (d) Explain why copper be a positive terminal compare to it’s metal pair base on the arrangement of metal in electrochemical series. Jelaskan mengapa kuprum bertindak sebagai terminal positif berbanding pasangan logamnya berpandukan kepada aspek susunan logam di dalam siri elektrokimia. Because Copper is place lower than other metals in electrochemical series/Copper is less electropositive metal than its pair of metal (e) Another voltaic cell is formed by using magnesium and iron as electrodes. Which electrode will be the positive terminal? Satu sel kimia yang lain dibina dengan menggunakan magnesium dan ferum sebagai elektrod.Elektrod yang manakah akan bertindak sebagai terminal positif? Iron 3. Electroplating Target SPM (P2&P3) 2013 12
  13. 13. 5. ACIDS AND BASES 4 (a) Table 4.1 shows the degree of ionisation and the colour of phenolphthalein in the solution P, Q and R. Jadual 4.1 menunjukkan darjah penceraian dan warna larutan fenolftalein dalam larutan P, Q dan R. Solution Larutan Degree of ionisation Darjah penceraian Colour of phenolphthalein in the solution Warna fenolftalein dalam larutan itu P Ionises completely Mengion lengkap Colourless Tanpa warna Q Ionises partially Mengion separa Colourless Tanpa warna R Ionises completely Mengion lengkap Pink Merah jambu Table / Jadual 4.1 (i) Which solution has the lowest pH value? Larutan manakah yang mempunyai nilai pH paling rendah? [1 mark] (ii) Give a reason for your answer in (a) (i). Beri satu sebab bagi jawapan anda di (a) (i).[1 mark] (iii) Solution P, Q and R might be acid or alkali. Classify the solutions into acid or alkali. Larutan P, Q dan R mungkin asid atau alkali. Kelaskan larutan itu kepada asid atau alkali. [2 marks] (b) Diagram 4.2 shows the observations in test tube I and test tube II when hydrogen chloride in tetrachloromethane and hydrogen chloride in solvent X are reacted with zinc. Rajah 4.2 menunjukkan pemerhatian dalam tabung uji I dan tabung uji II apabila hidrogen klorida dalam tetraklorometana dan hidrogen klorida dalam pelarut X bertindak balas dengan zink. Test tube I II Zinc Zinc Apparatus set-up Susunan radas Hydrogen chloride in tetrachloromethane Observation Target SPM (P2&P3) 2013 Zink Zink Hidrogen klorida dalam tetraklorometaa No change 13 Hydrogen chloride in solvent X Hidrogen klorida dalam pelarut of Bubbles X gas are produced
  14. 14. (i) State the name of solvent X. Nyatakan nama pelarut X. [1 mark] (ii) Write the formula of ion that causes an acid shows its acidic properties. Tuliskan formula ion yang menyebabkan asid menunjukkan sifat asid. [1 mark] (iii) Explain the differences in observation in test tube I and II. Terangkan perbezaan pemerhatian dalam tabung uji I dan II. (c) [2 marks] Vinegar consists of an ethanoic acid. Describe briefly a chemical test to verify the acid without using an indicator. Cuka mengandungi asid etanoik. Huraikan secara ringkas satu ujian kimia untuk mengenal pasti asid tanpa menggunakan penunjuk. [2 marks] (a) (b) (i) (ii) P Concentration of H+ ion in P highest 1 1 (iii) 4 Acid : P & Q Alkali : R 1 1 (i) (ii) (iii) Water H+ Test tube I : HCl exist as molecule / No H+ ion Test tube II : HCl ionise to produces H+ ion 1 1 1 1 Add magnesium // calcium carbonate //[suitable metal//metal carbonate] Bubble gas release 1 1 (c) TOTAL 10 6. SALTS 1. Diagram 7.1 shows a series of reaction for salt S which is a green colour compound. Salt S is heated strongly to produce black residue of compound T and gas U. Compound T reacts with sulphuric acid to form blue solution of compound W. Rajah 7.1 menunjukkan satu siri tindak balas bagi garam S yang merupakan sebatian berwarna hijau. Garam S dipanaskan dengan kuat untuk menghasilkan baki hitam sebatian T dan gas U. Sebatian T bertindak balas dengan asid sulfurik membentuk larutan biru sebatian W. Salt S Garam S Heat Panaskan Compound T Sebatian T + H2SO4 Compound W Sebatian W + Gas U Gas U Pass through lime water Alir ke dalam air kapur Lime water turns cloudy Air kapur keruh Based on Diargram 7.1: Berdasarkan Rajah 7.1: (a) (i) Suggest one formula of the anion in salt S. Cadangkan satu formula bagi anion dalam garam S.[1 mark] Target SPM (P2&P3) 2013 14
  15. 15. (ii) Identify salt S, compound T, gas U and compound W. Kenal pasti garam S, sebatian T, gas U dan sebatian W. [4 marks] (b) Write a chemical equation for the reaction between compound T and sulphuric acid. Tuliskan persamaan kimia bagi tindak balas antara sebatian T dan asid sulfurik. [2 marks] (c) Compound W is a soluble salt. Describe chemical test to verify the cation and anion in compound W. Sebatian W adalan garam terlarutkan. Huraikan ujian kimia untuk menentu sahkan kation dan anion dalam sebatian W. [5 marks] (d) Diagram 7.2 shows the chemical equation for the reaction between blue solution of compound W and barium nitrate solution. Rajah 7.2 menunjukkan persamaan kimia bagi tindak balas antara larutan biru sebatian W dan larutan barium nitrat. Compound W (aq) + Ba(NO3)2(aq) Sebatian W (ak) + Ba(NO3)2(ak) → → Salt X(s) + Salt Y(aq) Garam X(p) + Garam Y(ak) Diagram 7.2 Based on Diagram 7.2: Berdasarkan Rajah 7.2: (i) State the name of salt X and salt Y. Nyatakan nama bagi garam X dan garam Y. [2 marks] (ii) State one observation and name the type of reaction occurred. Nyatakan satu pemerhatian dan namakan jenis tindak balas yang berlaku. [2 marks] (iii) Compound W reacts with 50 cm3 of 0.1 mol dm-3 barium nitrate solution. Write the ionic equation for the reaction and calculate the mass of salt X produced. [Relative molecular mass of salt X: 233] Sebatian W bertindak balas dengan 50 cm3 larutan barium nitrat 0.1 mol dm-3. Tuliskan persamaan ion bagi tindak balas itu dan hitung jisim garam X yang dihasilkan. [Jisim molekul relatif garam X: 233] [4 marks] No. 1(a)(i) (ii) (b) Carbonate ion // CO32Salt S : Copper(II) carbonate // CuCO3 Compound T: Copper(II) oxide // CuO Gas U: Carbon dioxide // CO2 Compound W: Copper(II) sulphate // CuSO4 1 1 1 1 1. Correct formulae of reactants and products 2. Balanced equation 1 1 CuO + H2SO4 → CuSO4 (c) Target SPM (P2&P3) 2013 Sub Mark 1 Answer 1. 2. 3. 4. 5. 1 4 2 + H2O Add sodium hydroxide solution Blue precipitate formed indicate the presence of Cu+ ion Add hydrochloric acid Add barium chloride solution White precipitate formed indicate the presence of SO42- ion 15 Mark 1 1 1 1 1 5
  16. 16. Salt X : Barium sulphate Salt Y: Copper(II) nitrate 1 1 2 (ii) White precipitate Double decomposition reaction 1 1 2 1 1 (iii) Ba2+ + SO42- → BaSO4 Number of mol Ba2+ = // 0.005 2+ 1 mol Ba produce 1 mol BaSO4 // 0.005 mol Ba2+ produce 0.005 mol BaSO4 Mass BaSO4 = 0.005 x 233 g // 1.165 g (d)(i) 1 1 TOTAL 2. Preparation of lead(II)sulphate. Procedure: 1. pour ( 25 – 50cm3) of soluble salt Pb(NO3)2 into a beaker 2. add ( 25 – 50cm3) of soluble salt (Na2SO4) 3. stir 4. filter the mixture 5. rinse residue / solid / precipitate 6. dry between sheets of filter paper Observation Chemical equation Ionic equation (a) Test for anion (Cl-) 1. pour 2 cm3 the solution into a test tube 2. add 1 cm3 of dilute nitric acid and silver nitrate solution. 3. white precipitate formed 4. confirm the presence of chloride ions Target SPM (P2&P3) 2013 16 4 20
  17. 17. 7. RATE OF REACTIONS 10. Experiments I and II are carried out to investigate the effect of different sizes of solid X on the rate of reaction. Table 10 shows the reactants and time taken to collect 30 cm3 of colourless gas. Eksperimen I dan II telah dijalankan untuk mengkaji kesan perbezaan saiz pepejal X ke atas kadar tindak balas. Jadual 10 menunjukkan bahan tindak balas dan masa yang diambil untuk mengumpul 30cm3 gas yang tidak berwarna. Target SPM (P2&P3) 2013 17
  18. 18. Experiment Eksperimen Reactants Bahan Tindak balas Time taken to collect 30 cm3 of gas /s Masa yang diambil untuk mengumpul 30 cm3 gas /s 2 g of solid X + 40 cm3 0.1 mol dm-3 of hydrochloric acid 2 g pepejal X + 40 cm3 asid hidroklorik 0.1 mol dm-3 2 g of solid X + 40 cm3 0.1 mol dm-3 of hydrochloric acid 2 g pepejal X + 40 cm3 asid hidroklorik 0.1 mol dm-3 Table 10 I II 120 90 Based on the information given in table 10, Berdasarkan maklumat yang diberikan di dalam Jadual 10, (a) (i) Suggest the size of solid X in experiment I and experiment II. Cadangkan saiz bagi solid X dalam eksperimen I dan eksperimen II. [2 marks] (ii) Suggest the name of solid X. By using the suggested chemical substance of solid X, write the chemical equation for the reaction of solid X and hydrochloric acid. Cadangkan nama pepejal. Dengan menggunakan bahan kimia yang dicadangkan bagi pepejal X, tuliskan persamaan kimia bagi tindak balas pepejal X dan asid hidroklorik. [3 marks] (iii) Calculate the maximum volume of gas released in the experiment. [Molar volume of gas at room conditions =24 dm3 mol-1]. Based on the volume of gas obtained, sketch the graph volume of gas liberated againts time for both experiment on the same axes. Hitungkan isipadu maksimum bagi gas yang telah dibebaskan dalam eksperimen ini.[Isipadu molar gas pada keadaan bilik = 24 dm3 mol-1] [5 marks] (iv) Draw the apparatus set-up for the experiment. Lukiskan gambarajah susunan radas bagi eksperimen ini.[3 marks] (b) Calculate the average rate of reaction for experiment I and experiment II. Hitungkan kadar tindak balas purata bagi kedua-dua eksperimen 1 dan eksperimen II. [2 marks] (c) By using the name of solid X in (a)(ii), compare the rate of reaction between experiment I and experiment II. Explain the different in rate of reaction with reference to the collision theory. Dengan menggunakan nama bagi pepejal X di dalam (a)(ii), bandingkan kadar tindak balas di antara eksperimen I dan eksperimen II. Terangkan perbezaan dalam kadar tindak balas dengan merujuk kepada teori perlanggaran. [5 marks] No. 10 (a) Answer Experiment I : larger size // granulated// large crushed Experiment II : smaller size // powder// small creushed Target SPM (P2&P3) 2013 18 Sub Mark 1 1 Mark 2
  19. 19. No. 10 (a) (ii) 10 (a) (iii) Answer Sub Mark -Name of solid X : any suitable metal that higher than H in ECS// any suitable carbonate metal. Sample answer : Zinc // calcium carbonate -correct formula for reactans and products -correct balanced Sample answer : Zn + 2HCl  ZnCl2 + H2 // CaCO3 + 2HCl  CaCl2 + H2O + CO2 1. mol of HCl 40 x 0.1 // 0.004 1000 2. ratio 2 mol HCl : 1 mol H2 // 0.004 mol HCl : 0.002 mol H2 3. Volume of gas with unit 0.002 x 24 dm3 // 0.048 dm3 // 48 cm3 Sketch graph -both axes label & unit -correct curve & label for experiment I and II Mark 1 1 1 3 1 1 1 1 1 Volume of gas / cm3 II I 5 Time /s 10 (a)(iv) -functional diagram for the reaction between solid X and HCl -label (solid X/substance given eg. Zn/Mg/CaCO3, hydrochloric acid, water) -functional diagram to collect the gas 1 1 1 3 HCl water Zn 10 (b) -average rate of reactions for experiment I with unit -average rate of reactions for experiment II with unit Exp I : 30/120 cm3s-1 // 0.25 cm3s-1 Exp II : 30/90 cm3s-1 // 0.333 cm3s-1 Target SPM (P2&P3) 2013 19 1 1 2
  20. 20. No. 10 (c) Answer Sub Mark -Experiment II use smaller size of named solid X compare to experiment I -Named Solid X in experiment II has larger total surface area -frequency of collision higher -frequency of effective collision between named X particles (eg :Zn atom) and hydrogen ions higher -rate of reaction in experiment I is higher than experiment II. Mark 1 1 1 1 1 TOTAL 5 20 10. Diagram 10.1 shows the time taken for meat to cook using different size of beef. Rajah 10.1 menunjukkan masa yang diambil untuk memasak daging menggunakan saiz daging lembu yang berbeza. 120 minutes Diagram / Rajah 10.1 20 minutes (a) Explain why different size of meat takes different times to cook? [2 marks] Terangkan mengapa saiz daging yang berbeza mengambil masa yang berbeza untuk masak? (b) Two experiments were carried out to study the effect size of magnesium on the rate of reaction between magnesium and an acid Graph in Diagram 10.2 shows the results of Experiment I and Experiment II. Dua eksperimen dijalankan untuk mengkaji kesan saiz magnesium ke atas kadar tindak balas antara magnesium dengan suatu asid. Graf dalam Rajah 10.2 menunjukkan keputusan Eksperimen I dan Eksperimen II. Volume of gas (cm3 ) Experiment Isipadu gas (cm3 ) II : Acid and magnesium powder 20 15 Eksperimen II : Asid dan serbuk magnesium 10 Experiment I : Acid and magnesium granules 5 10 Target SPM (P2&P3) 2013 20 30 Eksperimen I : Asid dan ketulan magnesium 40 60 70 80 50 Diagram / Rajah 10.2 20 Time (s) Masa (s)
  21. 21. (i) State a suitable example of the acid used and write the chemical equation for the reaction between this acid and magnesium. Nyatakan satu contoh asid yang sesuai digunakan dan tuliskan persamaan kimia bagi tindak balas antara asid ini dengan magnesium. [3 marks] (ii) Calculate the average rate of reaction for Experiment I and Experiment II and compare the rate of reaction for both experiments. Hitung kadar tindak balas purata untuk Eksperimen I dan Eksperimen II dan bandingkan kadar tindak balas bagi kedua-dua eksperimen.[3 marks] (iii) Describe how to carry out the experiment in the laboratory. Huraikan bagaimana untuk menjalankan eksperimen ini di dalam makmal.[12 marks] 10 (a) (b) 1. The smaller the size of beef the larger the total surface area of beef exposed to heat 2. More heat absorbed (i) (ii) (c) Example of acid Sample answer : Hydrochloric acid / HCl Correct formula of reactant and product Balance Sample answer 2HCl + Mg → MgCl2 + H2 1. Experiment I : 20 cm3 / 60 s // 0.33 cm3s-1 2. Experiment II : 20 cm3 / 50 s // 0.4 cm3s-1 3.Rate of reaction Experiment II higher than Experiment I Apparatus Conical flask , burette , basin , measuring cylinder , stopwatch retort stand, stopper with delivery tube Procedure 1. A burette is filled with water and inverted over a basin of water and burette is clamped vertically using retort stand. 2. Initial burette reading is recorded 3. [ 1.0 – 2.0 ] g large pieces of magnesium is weigh and put into conical flask 4. [ 20 – 50 ] cm3 of [ 0.5 – 2.0 ] moldm-3 an acid is pour into conical flask 5. The conical flask is closed immediately with stopper and delivery tube 6. Quickly the stopwatch is started 7. Conical flask is swirled throughout the experiment 8. The burette reading is recorded at 30 s intervals 9. Step 1 to 9 are repeated using powder magnesium Results 1. Table contain time , burette reading and volume of gas 2. Correct unit for time , burette reading and volume of gas Sample answer Time (s) Burette reading (cm3) Volume of gas (cm3) Target SPM (P2&P3) 2013 21 1 1..…..2 1 1 1 . …3 1 1 1 ……3 1 1 1 1 1 1 1 1 1 1 1 1…...12
  22. 22. 8. CARBON COMPOUNDS Homologous series Alkane General formula Functional group Member , example CnH2n + 2 , n = 1,2.. Single covalent bond between carbon atoms. C- C Ethane Alkene CnH2n , n = 2.. Double covalent bond between carbon atoms. C=C Ethene Alcohols CnH2n + 1 OH, n = 1,2.. Hydroxyl group / - OH Ethanol Carboxylic acid CnH2n + 1 COOH, n = 0,1,2.. Carboxyl group , -COOH Ethanoic acid CH3COOH 1. Your are required to prepare one namely ester by using ethanoic acid is one of the reactants. By using a namely alcohol, describe one experiment to prepare the ester. In your description include the chemical equation and observation involved. Ester: ethylethanoate Material: ethanol, etahanoic acid, water, concentrated sulphuric acid Apparatus: Boiling tube / test tube, Bunsen burner, test tube holder, beaker Procedure: 1. Pour 2 cm3 of ethanol into a boiling tube / test tube 2. Add 1 cm3 of ethanoic acid 3. Add 2 to 4 drops of concentrated sulphuric acid 4. Heat the mixture gently for about two minutes 5. Pour the mixture into a beaker containing water. Observation: Sweet/ pleasant / fruity smell // insoluble in water Chemical equation: CH3COOH + C2H5OH  CH3COO C2H5 + H2O 2. Dehydration of alcohol Diagram of set up of apparatus 1. Complete and functional 2. Labels of set up of apparatus correct Target SPM (P2&P3) 2013 22
  23. 23. Procedure: a) Place some glass wool in a boiling tube b) Use a dropper to add propan-1-ol to wet the glass wool. c) Clamp the boiling tube horizontally and placed unglazed porcelain chips in the mid section of the boiling tube. d) Heat the unglazed porcelain chips strongly. e) Then heat the glass wool gently to vaporize the propanol. f) [Description of the chemical test to the gas collected in the test tube.] Add 1 cm3 of bromine water and shake well. Observation: Reddish brown colour of bromine decolourised Or, add 1 cm3 of acidified potassium manganate (VII) solution and shake well. Observation: Purple colour of potassium manganate(VII) solution decolourised Chemical equation: C3H7OH  C3H6 + H2O 3. Diagram 9 shows how compound Y is formed from an alkene W. Then compound Y react with alcohol X to produce ester Z. Rajah 9 menunjukkan bagaimana sebatian Y terbentuk daripada alkena W.Kemudian sebatian Y bertindak balas dengan alkohol X menghasilkan ester Z. Alkene W Alcohol X + H2O K2Cr2O7 Sebatian Y Alkohol X Alkena W Compound Y H3PO4 , 3000C,60 atm Ester Z Ester Z Diagram / Rajah 9 (a)(i) Name one alkene that has less than five carbon atoms.Write its molecular formula Namakan satu alkena yang mempunyai kurang daripada lima atom karbon.Tulis formula molekulnya. [2 marks] (ii) Based on the answer in 9 (a) (i), state the name of alcohol X and compound Y. Berdasarkan jawapan di 9 (a) (i), nyatakan nama bagi alkohol X dan sebatian Y. [2 marks] (iii) Write the chemical equation and state the observation for the reaction between alcohol X and potassium dichromate, K2Cr2O7 to produce compound Y that you named in (a) (ii). Tuliskan persamaan kimia dan nyatakan pemerhatian bagi tindak balas antara alkohol X dan kalium dikromat, K2Cr2O7 untuk menghasilkan sebatian Y yang anda namakan di (a) (ii). [3 marks] (b) (i) By using alcohol X and compound Y that you named in (a) (ii), describe the preparation of ester Z in the laboratory. In your description, include the chemical equation for the reaction. Dengan menggunakan alkohol X dan sebatian Y yang anda namakan di (a) (ii), huraikan penyediaan ester Z di dalam makmal. Dalam huraian anda, sertakan persamaan kimia bagi tindak balas itu. [5 marks] (ii) Alkene W can be prepared from alcohol X. Draw the set –up of apparatus for the preparation of the alkene W. Alkena W boleh disediakan daripada alkohol X. Lukiskan susunan radas bagi penyediaan alkena W itu. [2 marks] (c) Table 9 shows the results of latex coagulation. Jadual 9 menunjukkan keputusan pengumpalan getah. Target SPM (P2&P3) 2013 23
  24. 24. Procedure Prosedur Compound Y is added to latex Sebatian Y ditambah kepada susu getah Observations Pemerhatian The latex coagulates immediately Susu getah menggumpal dengan serta merta. Compound T is added to latex Sebatian T ditambah kepada suhu getah The latex does not coagulate within a longer period. Susu getah tidak menggumpal dalam suatu tempoh yang lebih lama.. The latex coagulates slowly. Susu getah menggumpal dengan perlahan. Latex is left under natural conditions Susu getah dibiarkan pada keadaan semulajadi Table / Jadual 9 Explain why there is a difference in these observations. Suggest the compound T. Terangkan mengapa terdapat perbezaan pemerhatian itu. Cadangkan sebatian T itu. [6 marks] (a) ethene / propene / butene C2H4 /C3H6 / C4H8 1 1..….2 ethanol and ethanoic acid // propanol and propanoic acid // butanol and butanoic acid 1+1 …2 (iii) (b) (i) (ii) 3 Correct formula of reactants and products Balanced Sample answer C2H5OH + 2[O] → CH3COOH + H2O Orange to green Sample answer Pour [ 2-5 cm3] ethanoic acid into a boiling tube Add [ 2-5 cm3] ethanol into the acid Add a few drops of concentrated sulphuric acid Heat the mixture CH3COOH + C2H5OH→ CH3COOC2H5 + H2O 1 1 Functional diagram Label Compound Y contains a lot of hydrogen ions H+ ions neutralize the negative charge on the protein membranes The particles collide and the protein membranes break 1 1……2 1 1 1 (i) 1 ……3 1 1 1 1 1 ……5 (ii) (c) Target SPM (P2&P3) 2013 24
  25. 25. Rubber molecules / polymers are released and combined Compound T contains OH- ion The existent of bacteria in natural conditions Bacteria produce weak acid /little H+ ions 1 1 1 1 Max 5 1 …...6 20 Compound T : Example: Ammonia TOTAL 4. Alkane and alkene are two homologous series for hydrocarbon compounds. (a) Explain the meaning of hydrocarbon compounds. Hydrocarbon compounds are made up of carbon and hydrogen elements only. (b) Butane and butene are member of alkane and alkene series respectively. Write down the equation which represents complete combustion in air of i) butane : 2C4H10 + 13O2 → 8CO2 + 10H2O ii)butene : C4H8 + 6O2 → 4CO2 + 4H2O (c) Draw the structural formula for all butene isomers and name the isomers you have drawn. H H H H H H H H │ │ │ │ │ │ │ │ H─ C ─ C ─ C ═ C ─ H H─C─ C ═ C─C─H │ │ │ │ H H H H But-1-ene But-2-ene (d) Methane is the simplest member in the alkane series but ethene is the simplest member in the alkene series. Explain why. General formula of alkane is CnH2n+2 . In methane, carbon atom forms four single covalent bond with hydrogen atoms which is stable. General formula of alkene is CnH2n. Ethene is the first member in alkene because they must have at least one double covalent bond between carbon atoms in the molecules. (e) Ethane and ethene can be differentiated by using aqueous bromine. State how this can take place. Ethene reacts with aqueous bromine through addition reaction which decolourises aqueous bromine. There is no reaction between ethane and aqueous bromine. 5. Figure 1 show changes of a carbon compound involving a series of reactions. Process I Propane Process II Alkene Y a. Write the general formula for homologous series of alkene Y. CnH2n b. i) Based on the flow chart name process II Hydration (Addition of water) ii) Write the chemical equation for process II C3H6 + H2O  C3H7OH iii) State the condition needed for process I and process II Target SPM (P2&P3) 2013 25 Propanol
  26. 26. Process I Process II c. - Hidrogen, temperature 1800C, catalyst Nickel/Platinum (Addition of hydrogen/ hydrogenation) - water steam, catalyst concentrated phosphoric acid, H3PO4, temperature 3000C and 60 atm. Draw the structural formulae of two propanol isomers. Name both isomers H OH  H H H    H  C  C C  OH    H H H Propan-1-ol  H  C  C  CH3   H H Propan-2-ol d. Table 1 shows the results of a test to differentiate between alkene Y and propane. Procedure Observations Bromine water is added to alkene Y Brown solution is decolourised Bromine water is added to propane Brown colour remains Explain why there is difference in these observations. Because alkene Y is unsaturated hydrocarbon which undergo addition reaction with bromine. Propane is saturated hydrocarbon which does not undergo addition reaction but substitution reaction. 6. Figure below shows the set-up of apparatus for the preparation of ethyl ethanoate from the reaction of ethanol with ethanoic acid. Y Liebig condenser X Mixture of ethanol, ethanoic acid, and concentrated sulphuric acid Water bath a b On the Liebig condenser in figure above, mark ‘X’ to indicate the place where water flows in and Y’ where water flows out. Why is the mixture heated using a water bath ? To give a uniform supply of heat to prevent the mixture from burning. c (i) Name the reaction for the preparation of ethyl ethanoate. Esterification Target SPM (P2&P3) 2013 26
  27. 27. (ii) Name the catalyst used for the preparation of ethyl ethanoate. Concentrated sulphuric acid (iii) Write the chemical equation for the reaction in c(i). CH 3COOH  C 2 H 5 OH  CH 3COOC 2 H 5  H 2 O d. The experiment is repeated by replacing ethanol with propanol. (i) Name the ester formed. Prophyl ethanoate (ii) State one physical property of the ester. Sweet smell (iii) State one usage of ester in everyday life. To produce perfume// flavour agent. 7 Figure above shows structural formula of monomer of natural rubber. a Give the IUPAC name of the monomer. 2-methylbuta-1,3-diene//isoprene b What type of polymerization takes place between this monomer to form natural rubber? Additional polymerisation c Give two properties of natural rubber. Soft, sticky d The properties of natural rubber can be improved by treating rubber with sulphur. (i) Name the process for this reaction. Vulcanisation reaction (ii) Draw the structure formula for the rubber produced. S S S S S S (iii ) Give two properties of the rubber produced. Elastic, strong (iv) What happens to the rubber when excess of sulphur is used. The rubber becomes brittle 9. REDOXS Target SPM (P2&P3) 2013 27
  28. 28. A. Rusting of iron 1. When iron exposed to water and oxygen 2. Iron atom releases 2 electrons to form iron (II) ion, Fe2+ / is oxidized to form iron (II) ion, Fe2+ 3. Fe  Fe2+ + 2e // (anode) [oxidation] 4. Iron acts as reducing agent 5. Oxygen and water receives /gain electrons to form hydroxide ions. 6. O2 + 2H2O + 4e  4OH- (cathode) [reduction] 7. Oxygen acts as oxidizing agent. 8. Iron (II) ion, Fe2+ combine with hydroxide ion, OH- to form iron (II) hydroxide, Fe(OH)2. 9. Iron (II) hydroxide, Fe(OH)2 oxidized by oxygen to form iron (III) oxide, brown solid/precipitate, Fe2O3.x H2O. // Fe2+  Fe3+ + e B. Describe an experiment to investigate oxidation and reduction in the change of iron(II) ions to iron(III) ions and vice versa. (i) Changing of Fe2+ ions to Fe3+ ions Procedure: 1. Pour 2 cm3 of freshly prepared iron(II)sulphate solution into a test tube. 2. Using dropper, add bromine water drop by drop until no further changes are observed. 3. Heat slowly / gently 4. Add 3 drops of potassium hexacyanoferrate (II) solution / sodium hydroxide solution. 5. Dark blue precipitate // brown precipitate formed. (ii) Changing of Fe3+ ions to Fe2+ ions Procedure: 1. Pour 2 cm3 of iron(III)sulphate solution into a test tube. 2. Add half spatula of zinc / Mg powder to the solution. 3. Shake the mixture until no further changes are observed. 4. Filter the mixture. 5. Add 3 drops of potassium hexacyanoferrate (III) solution / sodium hydroxide solution into the filtrate. 6. Dark blue precipitate // green precipitate formed. C Diagram 6 shows an apparatus set up to investigate the effect of three different metals A,B and C on the rusting of iron. Rajah 6 menunjukkan susunan radas untuk mengkaji kesan tiga logam berbeza A,B, C yang berlainan ke atas pengaratan besi. 1 2 Iron nail Paku besi 3 Iron nail Paku besi Metal A Logam A Metal C Logam C Iron nail Paku besi Metal B Logam B Jelly solution + potassium hexacyanoferrate(III) solution + phenolphthalein Larutan agar + larutan potassium heksasianoferrat(III) + fenolftalein Diagram 6 Target SPM (P2&P3) 2013 28
  29. 29. Table 6 shows the results of the experiment after three days Jadual 6 menunjukkan keputusan eksperimen itu 6 selepas tiga hari Test tube Tabung uji 1 2 Observation Pemerhatian High intensity of blue colour Keamatan warna biru yang tinggi Low intensity of pink colour. Keamatan warna merah jambu yang rendah Table 6 3 Low intensity of blue colour Keamatan warna biru yang rendah (a) (i) Rusting is a redoxs reaction. What is meant by redoxs reaction? [1 mark] Pengaratan adalah satu tindak balas redoks. Apakah yang dimaksudkan dengan tindak balas redoks? (ii) Write the half equation for the oxidation reaction in Test tube 1. [2 marks] Tuliskan setengah persamaan bagi tindak balas pengoksidaan dalam Tabung uji 1. (b) Based on the observation in Table 6: Berdasarkan pemerhatian dalam Jadual 6: (i) Identify metals A, B and C. Kenal pasti logam A, B dan C. [3 marks] Metal A Logam A Metal B Logam B Metal C Logam C (ii) Compare why there is different in the observation in Test tubes 2 and 3. [1 mark] Bandingkan mengapa terdapat perbezaan pemerhatian dalam Tabung uji 2 dan 3. (c) (i) Describe your answer in (b)(ii). Huraikan jawapan anda di (b)(ii). [3 marks] (c) (ii)What is the change of oxidation number of iron in Test tube 3. [1 mark] Apakah perubahan nombor pengoksidaan besi dalam Tabung uji 3. C No (a)(i) (a)(ii) (b) (i) (b) ii) (c) (i) (c) (ii) Target SPM (P2&P3) 2013 Explanation Oxidation reaction and reduction reaction that occur simultaneously // at same time 1. Correct formulae of reactant and products 2. Balanced equation Fe → Fe2+ + 2e metal A= Siver / Ag // Copper / Cu metal B = Zinc / Zn metal C= Lead / Pb // Tin / Sn Iron in test tube does 2 not rust but iron in test tube 3 rust 1. Metal B / Zn is more electropositive than iron 2. Metal C / Pb / Sn is less electropositive than iron 3. Iron in test tube 3 is oxidized // Fe2+ formed 0 → +2 29 Mark  mark 1 1 1 1 1 1 1 1 1 1 1 1 Total 2 3 1 3 1 11
  30. 30. D. (a) Diagram shows the apparatus set-up of an experiment to investigate the transfer of electrons at a distance. Rajah menunjukkan susunan radas eksperimen untuk mengkaji pemindahan elektron pada suatu jarak. Carbon X Carbon Y Karbon X Karbon Y Acidified potassium manganate(VII) solution Iron (II) sulphate solution Larutan ferum(II) sulfat Larutan kalium manganat (VII) berasid Dilute sulphuric acid (i) State the name of the oxidizing agent in this Asid sulfurikNyatakan nama agen pengoksidaan dalam reaction. cair tindak balas ini. Acidified potassium manganate(VII) solution (ii) Referring to the reaction that takes place at carbon X : Merujuk pada tindak balas yang berlaku di karbon X : Write the half equation for the reaction. Tuliskan persamaan setengah untuk tindak balas itu. Fe2+ ------------- Fe3+ + e(iii) State one observation that occurred. Nyatakan satu pemerhatian yang berlaku. Green solution turns to brown solution (purple to colourless) (iv) Show the direction of the electron flow in Diagram. Rajah From electrod carbon X to Y Tunjukkan arah pengaliran elektron dalam (v) Referring to the reaction that takes place at carbon Y, calculate the oxidation number of manganese in MnO4-. Merujuk pada tindak balas yang berlaku pada karbon Y, hitungkan nombor pengoksidaan bagi mangan dalam MnO4X + 4 (-2) = -1 X + (-8) = -1 X = +7 (b) Diagram 6.2 shows the apparatus set-up to investigate the displacement of halogen from its halide solution. Chlorine water was added to a test tube containing a potassium iodide solution and organic solvent, 1,1,1-trichloroethane. Rajah 6.2 menunjukkan susunan radas untuk mengkaji penyesaran halogen daripada larutan halidanya. Air klorin ditambah ke dalam tabung uji yang mengandungi larutan kalium iodida dan pelarut organik 1,1,1-trikloroetana. Chlorine water Air klorin Potassium iodide solution Larutan kalium iodida 1,1,1-trichloroethane 1,1,1-trikloroetana Target SPM (P2&P3) 2013 Shake Goncang Brown solution Larutan perang 30 Purple organic layer Lapisan organik berwarna ungu
  31. 31. (i) Write the ionic equation for the reaction. Tuliskan persamaan ion bagi tindak balas itu. Cl2 + 2I- ----------- I2 + 2Cl(ii) What is the function of chlorine water? Apakah fungsi air klorin? An oxidizing agent (iii) State the change of oxidation number for iodine. Nyatakan perubahan nombor pengoksidaan bagi iodin. -1 to 0 (iv) State the name of another reagent that can replace chlorine water. Nyatakan nama satu bahan uji lain yang boleh menggantikan air klorin. Bromine water 10. THERMOCHEMISTRY 10. A student carried out an experiment to determine the heat of displacement of copper for the reaction between zinc and copper(ll) sulphate solution. In this experiment, excess zinc powder is added to 25cm 3 of 0.2mol/dm3 copper(ll) sulphate solution in a plastic cup. The temperatures of copper(ll) sulphate solution before and after the addition of zinc powder are recorded Initial temperature of copper sulphate solution = 28oC The highest temperature of the solution after the addition of zinc = 37.5oC (specific heat capacity of the solution, 4.2 Jg-1oC; the density of the solution, 1 gcm-3) Seorang pelajar menjalankan satu eksperimen untuk menentukan haba penyesaran kuprum bagi tindakbalas di antara zinc dan larutan kuprum sulfat. Dalam eksperimen ini serbuk zink berlebihan ditambahkan kepada 25cm3 of 0.2mol/dm3 larutan kuprum sulfat dalam cawan plastik. Suhu larutan kuprum sulfat sebelum dan selepas penambahan serbuk zink dicatatkan; Suhu awal larutan kuprum sulfat = 28oC Suhu tertinggi larutan selepas penambahan zinc = 37.5oC ( Muatan haba tentu bagi larutan, 4.2 Jg-1oC; ketumpatan larutan, 1 gcm-3 ) (a) What is meant by heat of displacement ? [1M] Apakah yang dimaksudkan dengan haba penyesaran? Heat of displacement is the heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal. (b) What is the colour change of the copper(ll) sulphate solution in this reaction? Apakah perubahan warna larutan kuprum sulfat dalam tindakbalas ini? [1M] The intensity of blue solution ( copper(ll) sulphate solution ) in this reaction decreases until they become colourless. (c) Write an ionic equation for the reaction. Tuliskan persamaan ion untuk tindakbalas ini [1M] Zn + Cu 2+  Zn 2+ + Cu (d) Based on the results of this experiment, calculate Berdasarkan keputusan eksperimen ini, kirakan (i) The number of mole of copper(ll) ions displaced. Bilangan ion kuprum yang dinyahcaskan. Number of mole of copper(ll) ions = MV 1000 = 2(25) 1000 = 0.005 mole. [2 marks] Target SPM (P2&P3) 2013 31
  32. 32. (ii) The heat given out in this reaction. Haba yang dibebaskan dalam eksperimen ini. mcθ = 25 x 4.2 x 9.5 = 997.5 joule [2 marks] (iii) The heat of displacement of copper in this reaction. Haba penyesaran kuprum dalam tindakbalas ini. [2 marks] (iii) The heat of displacement of copper by zinc 997.5 = ---------0.005 = 199500 J = 199.5 kJ/mol (e) Draw an energy level diagram for the reaction. Lukis gambarajah aras tenaga untuk tindakbalas ini. [2 marks] Energy Zn + Cu 2+ ∆H = -199. 5 kJ/mol Zn 2+ + Cu Target SPM (P2&P3) 2013 32
  33. 33. Target SPM (P2&P3) 2013 33
  34. 34. 11. Chemicals for consumer 1. You are given liquid soap, sample of hard water, sample of soft water and other materials. Describe an experiment to investigate the effect of cleaning action of the soap in different types of water. You description must include example of hard water and soft water, observation and conclusion. [10 marks] Sample answer: 1. hard water : sea water 2. soft water : distilled water Materials: liquid soap, sea water, distilled water, pieces of cloth with oil stain. Apparatus: beaker (suitable container), glass rod, measuring cylinder Procedure: 1. pour (100 – 200) cm3 sea water into a beaker/ suitable container 2. Add (10 – 20 ) cm3 liquid soap into the beaker. 3. stir the mixture 4. Place a piece of cloth with oil stain into the beaker. 5. Record the observation. 6. Repeat step 1 – 4 using distilled water. Observation: 1. The oil stain in hard water remained but removed in soft water. Conclusion: 1. Hard water contains Mg2+ or Ca2+. Soap anion formed scum (insoluble salt) when react with Mg2+ or Ca2+. 2. Soap is not an effective cleansing agent in hard water but only effective in soft water. 2. (a) Table 5.1 shows the results obtained when cleaning agent L and cleaning agent M are used to wash cloth in hard water and soft water. Jadual 5.1 menunjukkan keputusan yang diperolehi apabila agen pencuci L dan agen pencuci M digunakan untuk mencuci kain dalam air liat dan air lembut. Target SPM (P2&P3) 2013 34
  35. 35. Cleaning Agent Agen Pencuci L     M     Hard water Air Liat Scum is formed Cloth is not clean Kekat dihasilkan kain tidak bersih     Scum is not formed  Cloth is clean  Kekat tidak dihasilkan  Kain bersih  Table 5.1 Soft water Air lembut Scum is not formed Cloth is clean Kekat tidak dihasilkan Kain bersih Scum is not formed Cloth is clean Kekat tidak dihasilkan Kain bersih (i) State the type of cleaning agent L and cleaning agent M Nyatakan jenis agen pencuci L dan agen pencuci M (ii) State two ions in hard water that cause the formation of scum. Nyatakan dua ion dalam air liat yang menyebabkan pembentukan kekat (iii)Compare the effectiveness of cleansing action between cleaning agent L and cleaning agent M. Explain your answer. Bandingkan keberkesanan tindakan pencucian antara agen pencuci L dan agen pencuci M. Terangkan jawapan anda. (b) Diagram 5.2 shows label of ice cream wrapper. Rajah 5.2 menunjukkan label pembungkus ais krim ICE CREAM Ingredients: Non-dairy milk, sugar, ethyl butanoate, ascorbic acid, cellulose gum and tartrazine. Product by: Shukrov Company, Bangkong City. KM 23, Rompin Muadzam Road Exp. Date : 02/09/13 Net weight : 450g Diagram 5.2 (i) State one food additive used as thickening agent in the ice cream. Nyatakan satu bahan tambah makanan yang digunakan sebagai agen pemekat dalam ais krim itu. (ii) Determine the food aditive that cause hyper activities among children. Tentukan bahan tambah makanan yang menyebabkan hiper aktif di kalangan kanak-kanak (iii)What is the function of ethyl butanoate in the ice cream? Apakah fungsi etil butanoat dalam ais krim itu? (iv)Draw the structural formula of ethyl butanoate. Lukiskan formula struktur bagi etil butanoat. Target SPM (P2&P3) 2013 35
  36. 36. 2 No (a) (i) Explanation L: Soap M : Detergent (ii) magnesium ion, calcium ion // Mg2+, Ca2+ (iii) Cleaning agent M is more effective then cleaning agent L Cleaning agent L form insoluble salt does not clean the cloth Cleaning agent M does not form insoluble salt and clean the cloth b(i) cellulose gum (ii) tartrazine (iii) Flavouring agent (iv) Mark 1 1 1 +1 1 1  mark 1 7 1 1 1 1 Total 3. 4 11 (a) A patients suffered from tuberculosis that caused by a certain bacteria. What type of medicine that can be used to treat the patient ? Seorang pesakit menghidap batuk kering yang disebabkan oleh bakteria tertentu. Apakah jenis ubat yang boleh digunakan untuk merawat pesakit itu? Explain why the patient must complete the whole course of the medicine that his doctor prescribes to him even if he feels better. Terangkan mengapa pesakit tersebut mesti menghabiskan semua ubat yang dipreskripsikan oleh doktor walaupun jika dia berasa sihat. [5 marks] 1. antibiotic 2. This is to make sure that all the bacteria are killed. 3. Otherwise, patient B may become ill again 4. bacteria may become more resistant to the antibiotic. 5. As a result, the antibiotic is no longer effective. (b) Diagram 2 shows the structure of anion parts of a soap and a detergent. These anions consist of part A and part B as shown in the diagram 2. Diagram 2 menunjukan struktur bagi bahagian anion bagi satu sabun dan satu detergen. Anion-anion ini terdiri daripada bahagian A dan bahagian B seperti ditunjukan dalam Rajah 2 Anion of a soap Anion of a detergent Part A Target SPM (P2&P3) 2013 Part B Diagram 2 36
  37. 37. Name part A and part B of the anions. State the solubility of part A and part B in the cleansing action. Namakan bahagian A dan bahagian B dalam anion itu Nyatakan keterlarutan bahagian A dan bahagian B dalam tindakan pembersihan. [4 marks] A: Hydrophobic part B: Hydrophilic part Part A is dissolved in oil / grease Part B is dissolved in water (c) The statement below is about soap and detergent. detergen. Pernyataan di bawah adalah tentang sabun dan The cleaning action of detergent is more effective than soap in hard water. Tindakan pencucian detergen lebih berkesan daripada pencucian sabun dalam air liat. You are given a piece of cloth stained with grease. Anda diberi sehelai kain yang terkena gris. (i) Describe briefly the experimental procedure, observation and conclusion to prove the above statement. Huraikan dengan ringkas prosedur experiment, pemerhatian dan kesimpulan untuk membuktikan pernyataan di atas. [5 marks] Procedure: 1. Two basins/beakers are filled with hard water. 2. Soap is added to one beaker and detergent is added to another beaker. 3. A piece of dirty cloth is placed into each beaker and agitated. 4. Observations: Soap in hard water Detergent in hard water 1. cloth not clean 1. cloth clean 2. scum forms 2. no scum 3 water less dirty 3. water turns dirty (any one pair) 5. Conclusion: detergent cleans stains more effectively than soap. (ii) Explain how soap works in removing grease from a piece of cloth. Terangkan bagaimana sabun bertindak dalam menanggalkan gris daripada sehelai kain. [6 marks] 1. soap particles dissolve in water and (reduce the surface tension of water) /(water wet the cloth thoroughly) 2. the hydrophilic part dissolve in water while the hydrophobic part dissolves in grease//diagram 3. During cleaning/rubbing/heating/scrubbing, grease is lifted off the surface 4. Hydrophobic part / soap anion surrounded the grease//diagram 5. Grease is broken up into small droplets/forming an emulsion//diagram 6. When rinsed, the grease droplet will be removed 4. (a) Ammonia which is used to produce ammonium fertilizer can be obtained through the Haber process in industry. Ammonia yang digunakan dalam penghasilan baja ammonium boleh diperolehi melalui proses Haber dalam industri. Target SPM (P2&P3) 2013 37
  38. 38. Balance the chemical equation to produce ammonia Imbangkan persamaan kimia untuk penghasilan ammonia. N2 (g) + H2(g)  NH3(g) 2NH3 (g) + 3H2 (g)  2NH3 (g) [1 mark] (b) Ammonium fertilizer can be prepared in the laboratory by adding ammonia solution and certain acids as shown in the Table 1. Baja ammonium boleh disediakan dalam makmal dengan menambahkan larutan ammonia dan asid seperti yang ditunjukkan dalam Jadual 1. Neutralisation reactions Table 1 Tindakbalas peneutralan Name of ammonium salts (fertilizer) Jadual Nama bagi garam ammonium (baja) Alkali + Acid 1 Alkali + Asid Ammonium aqueous + nitric acid Ammonium nitrate Ammonium aqueous (i) + sulphuric acid Ammonium sulphate Complete the Table 1 by writing the name of ammonium salts. Lengkapkan Jadual 1 dengan menuliskan nama bagi garam ammonium. [2 marks] (ii) Write the chemical formula of the product in the chemical equation below. Tuliskan formula kimia bagi hasil yang terbentuk dalam persamaan kimia di bawah. [2 marks] NH3 (aq) + HNO3 (aq)  NH4NO3 NH3 (aq) + H2SO4 (aq)  NH4SO4 Target SPM (P2&P3) 2013 38

×