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SULITNO. KAD PENGENALANANGKA GILIRAN.ry%ffi e@4+'

SIJIL PELAJARAN MALAYSIA 2OO9CHEMISTRYKertas2Nov./Dis.^1l lam

LEMBAGA PEPERIKSAANMALAYSIAKEMENTERIAN PELAJARAN MALAYSIA

4541t2

Dua am tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU1. Tulis nombor kad pengenalan dan angka gitiranandapada ruang y(tng disediakan.2. Kertas soalan ini adalah dalam dwibahasa.3. Soalan dalam bahasa Inggeris mendahului soalanyang sepadandalam bahasaMelayu.4. Calon dibenarkan menjawab keseluruhan atausebahagian soalan sama ada dalam bahasa Inggerisatau bahasaMelayu.

INFORMATION FOR CANDIDATESMAKLUMATUNTAKCALON1. This questionpaperconsists f three sections.Section A, Section B and Section C.Kertassoalan ni mengandungiigabahagian. ahagianA,BahagianB dan BahagianC.2. You are advise o spend90 minutes to answerquestions n Section A, 30 minutes for Section B and30 minutes for Section C.Andadinasihat upayamengambilmasa90minit untukmenjawaboalandalamBahagianA, 30 minituntukBahagianB dan30 minituntukBahagianC.

Untuk KegunaanP emeriksaKod Pemeriksa:Bahagian Soalan MarkahPenuh MarkahDiperoleh

A

I 9) 93 l04 105 116 11

B 208 20C 9 2010 20

Jumlah

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ForExaminer'sUse454112sulrr " section

Bahaginn A160marksl160markahl

Answer all questions n this section'Jowabsemua soalandalam bahagian ni'1 Diagram 1 shows the arrangementof atoms in two types of copper alloy'nawtmenunjukknnSusunanatomdalamduajenisaloibagikuprum.

Zinc atomAtom ink

CopperatomAtomkuprum

Alloy XAloiX

State the meaning of alloY.Nyatakanmaksudaloi.

BronzeGangsa

Diagram 1Rajah I(a)

AtomM

ll marklll markah)(b) State the name of alloY X.Nyatakannamabagi aloi X.

ll mark)ll markahl(c) State the name of atom M.Nyatakannama bagi atom M-

fl marklll markahlSULIT@ 2009 Hak Cipta Kerajaan MalaYsia 184

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SULIT 454U2(d) (i) What is the difference in terms of hardnessbetweenbronze and pure copper?Apakahperbezaandaripada segi kekerasan ntara gangsadengan uprumiilenZ

DifferencePerbezaan BronzeGangsa Pure copperKuprumtulenSize of atomsSaiz atomArrangement of atomsSusunanatom

12marlcsl[2 markah]Describe what happens o the atoms when a force is applied to a bronze and purecopper.Huraikan apayang aknn berlaku kepadaatom-atomapabila satudayadikenakankepadagangsadan kuprumtulen.Bronze/Gangsa:

Pwe copper/Kuprum tulen:

I marklfl markahl(ii) CompleteTbble 1 to show the differences n terms of size and arrangementof atomsin bronze and pure copper.Lengkapkan adual I untuk menunjukkan erbezaandaripada segi saiz dan susunanatom dalam gangsadengankuprumtulen.

Table IJadual 1

(iii)

ForExaminer'sUse1(dxi)

I r---I l l l

r(e)I r---lI l r lTotal Alt-=I le l

[2 marks][2 markah](e) Pewter is also an example of an alloy.Stateone use of pewterPewter uga adalah satucontohbagi aloi.Nyatakansatu kegunaan ewter.

I marnI markahl

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ForExarniner'sUseSULIT 454ll2

2 Diagram 2 sh6ws the standard representatlon for the atoms of two elements' lithium andoxygen. , - . . - ,^. . ." :+Rajah 2 menunjukk,onerwakilanpiawai bagi atom bagi dua unsur, itiurn dan oksig'en.

(a)

Diagratn 2Raiah2What is representedby the number I in7rr;tApakahyang diwakili oleh nombor1 dalam iLi?

ll mark)ll markahl

(b) (D Write the elecffon arrangement for an atom of:Tulis susunanelektronbagi atom:LithiltrnlLitium

OxygenlOksigen

12markslf2 marknhl

(ii) Compare the size of the lithium atom with the oxygen atom'Baniingkan saiz atom litium denganatom oksigen

fl marklll markahl

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SULIT 454U2(c) complete Table 2 to show the position of oxygen in the periodic Table.LengkapkanJadual 2 untuk menunjukkankedudikan oksigendalam Jadual Berkala.

(d)

ElementUnsur PeriodKala GroupKampulanLi 2 Io

Table 2Jadual2U marklfl markahl

Lithium reactswith oxygen to form a compound, lithium,oxide with the formula, Liro.Litium bertindak balas irrgon oksigenuntulimembentuksatu sebatian, litium oksida denganfortnula, Li2O.(i) Write the formulae of all the ions in lithium oxide.Tulis foTmula bagi semua on dalam litium oksida.

[1 mark]ft markahl(ii) Explain how each of these ons is formed,Ielaskan bagaimanasetiap ion ini terbentuk.

2k\[=I l r l

ForExaminer'sUse

Total .{2

f2 marksl[2 markah](iii) State one physical property of lithium oxide.Nyatakan satu sifat fizik litium oksida.

Hak Cipta Kerajaan Mataysia

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ForExaminer'sUse4s4ll2SULIT

3 Diagram 3 ,t o*, the apparatus set-uplto determine the empirical formula of copper oxide'Rajah 3 menuniukkan iinon radas uituk rnenentukanormula empirik bagi kaprum olesida'Burningof excess Ydrogen asPembalcaran lebihan gas hidrogen

HydrogengasGas hidrogen

AnhYdrous alciumchlorideKalsiumklorida erhidratDiagram3Rajah3

Table 3 shows the result of this experiment'Jadual 3 menunjukkan eputusaneksperimen

ni'

t \HeatPanasCopperoxideKuprumoksida

PorcelaindishPiringporselin

(a) (i) "l!)k"hat is the meaning of empirical formula?Apakahmaksudormula etnPirik?

DescriptionPeneranganMass (g)Jisim (si)

Combustion tube + Porcelain dishTiubpembakaran+ Piring Porselin 32.25Combustion tube + Porcelain dish + Copper oxideTiubpembakaran+ Pirtng porselin + Kuprum oksida 42.25Combustion tube + Porcelain dish + CopperTiubpembakaran+ Piring porselin + Kuprum 40.25

ll marklU markahl(iD State the function of the anhydrous calcium chloride'Nyatakanungsi knlsiumklorida terhidrat'

ll marlcllI markahl(b) (i) Basedon Table 3, calculate the massof:Berdasarkan adual 3, hitung isim bagi:

CoPPerlKuPrum:

OxYgenlOksigen:12marks\12markahlSULIT@ 2009 Hak CiPta Keraiaan MalaYsia r88

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SULIT 4 4U2

[7 mark][l markah]

(ii) Calculate the ratio of moles of copper atoms to oxygen atoms.[Relative atomic mass: Cu = 64, O = 16]Hitung nisbah mol bagi atom kuprum kepadaatom oksigen.fJisim atom relatif: Cu = 64,O = 16]

(iii) Determine the empirical formula of copper oxide.Tentukanformula empirik bagi kuprum olcsida.

ForExaminer'sUse

Total A3

(c) (i) Why is hydrogengas passed hrough thestopped?Mengapakah gas hidrogen dialirkan melaluitamat?

combustion tubetiub pembakaran

after

selepas

[1 mark]ft markah)heating has

pemanasan

(ii)ft marklfl markahl

State how to determine that the reaction betweencopper oxide with hydrogen hascompleted.Nyatakan bagaimana untuk menentukanbahawa tindak balas yang berlaku antarakuprum oksida denganhidrogen telah lengkap.

fl mark]I markahl(d) (i) Statewhy the empirical formula of magnesiumoxide cannotbe determinedby usingthe sametechnique.Nyatakan nxengapaformula empirik bagi magnesiumoksida tidak dapat ditentukandenganmenggunakan eknik yang sarna.

[1 mark]ll markah]State the name of another metal oxide whose empirical formula can be determinedusing the same echnique.Nyatakan nama suatu logam oksida rain yang formula empiriknya boleh ditentukanmenggunakan eloik yang sama.

I mark)ll markahl

[Lihat halaman sebelah]i} SULIT

(ii)

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SULIT(d) Compound P undergoespolymerisation.The structural formula for compound P is given below.SebatianP menjalaniprosespempolimeran.Formula struktur basi sebatianP diberikandi bawah.

HHWrite the equation for the polymerisation of compound P.Tulispersamaanbagi prosespempolimeransebatianP.

4541t2

HHt t

ForExaminer'sUse

4(d)t-=I l2l

4(e)[=I | r lTotal A4t-:I l10l

[2 marksl12markahl(e) Draw the structural formula for Q.Lukis ormula struktur bagi Q.

ll marklfl markahl

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ForExarniner'sUse

s(a)[=I l2 l

s(b)[=I l1 l

SULIT 454u25 Diagram 5 shows two sets of experiments to study the factor affecting the rate of reactionbetween hydrochloric acid, HCl and calcium carbonate,CaCO3.Rajah 5 menunjukkandua set eksperimenuntuk mengkaji aktor yang rnempengaruhikadar tindikbalas antara asid hidroklorik, HCI dengankalsiurnkarbonat, CaCO3.

3ttcIAfter 3 minutesSelepas minit-------)ExcessCaCO, chipsKe ulan CzCO,berlebihan

100cm30.50 mol dm-3 HCIAfter 3 minutesSelepas 3 minit..------>

ExcessCaCO, chipsKetulanCaCO,be leb han

60 cm3of gas60 cm3 as

,il;o,ru,120 cm3gas

Diagram5Rajah5(a) Write a balanced chemical equation for the reaction in these experiments.Tulis persam.aankimia seimbang bagi tindak balas dalam eksperimen ni.

[2 marksl[2 markah]What is the reading needed o be recorded in both experiments to determinethe rate ofreaction in 3 minutes?Apakah bacaan yang perlu dicatat dalam kedua-dua elcsperimenuntuk menentukaru adartindak balas dalam masa3 minit?

(b)

Calculate the average rate of reaction in set 1.Hitung kadar tindak balaspurata bagi set L(c)

II mark]ll marknhl

II mark]ll markahlSULIT

5(c)[=I l r l

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454u2Compare the rate of reaction in set I and set 2.Explain your answerbasedon the factor affecting the rate of reaction.Bandingkan ka"dar indak balas bagi set I denganset Z.Jelaskan awapan anda berdasarkanfaktor yang mempengaruhikadar tindak balas.

[2 marksf[2 markah](ii) Explain the answer n 5(d)(i) wirh reference o the collision theory.Jelaskan awapan di 5(d)(i) denganmerujuk kepadateori perlanggaran.

[3 marksf[3 markah]Sketch the graph of volume of carbon dioxide gas producedagainst ime for both setsofexperiment in the first 3 minutes.I^akar graf isi padu gas karbon dioksida yang dihasilkan melawanmasabagi kedua-duasetelcsperimenalam rnasa3 minit yangpertanw.

Volume of carbondioxide (cm3)Isipaduknrbondioksida cm3)

Time (min)Masa(min)

[2 marksf[2 markah]

ll.ihat halamansebetahl', SULIT

SULIT(d) (i)

(e)

Hak Cipta Kerajaan Malaysia r93

ForExaminer'sUse

5(e)t-=I l2lTotal A5i-=I l l r I

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ForEmminer'sUse454U2ULIT ". :,

6 Diagram 6 shows the apparatus set-up to determine the heat of neutralisation between nitricacid and sodium hydroxide solution.Raiah 6 menunjukkansusunan adasuntuk menentukanhabapeneutralan antara asid nitrik deiganlarutan natrium hidroksida.Polystyrene upCawan polistirena

Mixing andreactingBercampuranbertindak balns__>

25 cm3 f 1.0 mol dm-3 25 cm3of 1.0mol dm-3nitric acid of sodiumhydroxide25cm31.0moldm-3 25 cm31.0mol dm-3asidnitrik natriumhidroksi"dnDiagram 6

Rajah 6Table 6 shows the result of this experiment.Jadual 6 menunjukkankeputusaneksperimen ni.

Mixture of nitric acidand sodiumhydroxideCampuransidnitrikdannatriumhidroksi&t

Table 6Jadual 6

(a) What is the meaning of heat of neutralisation?Apakahmaksudhaba peneutralan?

DescriptionPenerangan

Temperature("C)Suha 'C\

Initial temperatureof nitric acidSuhuawal asid nitrik 30.0Initial temperatureof sodium hydroxideSuhu awal natrium hidroksida 30.0Highest temperature of the mixtureSuhu maksimum ampuran 36.8

I marklfl markahl

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SULIT(b) Calculate:Hitung:

(i) The heat releasedduring the reaction,[Specific heat capacity of solution, c = 4.2 J g-toc-t;Densi tyof solut ion= I gcm-31Haba yang dibebaskansemasa indak balas,lMuatan haba tentu bagi larutan, c = 4.2 J g-to6-t'Keturnpatanarutan = I g cm-'l

4 4u2

[1 mark]lI markahl(ii) The number of moles of nitric acid reacting,Bilangan mol asid nitrik yang bertindak,

ll marklI markahl(iii) The heat of neutralisation.Haba peneutralan.

ll marklll markahl@2009Hak cipta KerajaanMalaysia 195 [Lihat halaman sebelah]SULIT

ForExaminer'sUse

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ForExaminer'sUje

6(e)r-:I l1 lTotat A6r-;I l11l

454112SULIT(c) Draw an energy level diagram for ttiis reaction'Lukis rajah aras tenaga bagi tindak balas ini'

13marksl13markshl(d) The experiment is repeatedusing 25 cm3 of 1.0 mol dm-3 of ethanoic acid to replace thenitric acid.The heat of neutralisation using ethanoic acid is 55'0 kJ mol IExplain the difference of the heat of neufralisation'Eksperimen iulangdenganmenggunakansid. tanoik25cm31.0mol dmabagimenggantil

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SULIT 414ll2SectionBBahaginn Bl2Omarksl[20markah]

Answerany onequestionrom this section.Jawab mana-manasatu soalan daripa.dabahagian ini.7 (a) Table 7 shows some nformation about three membersof a homologous series.Jadual 7 menunjukkanbeberapamaklumat mengenaitiga ahli satu siri homolog.

Member ofhomologous eriesAhli bagi siri homologBoiling point ('C)Takatdidih ("C) PreparationPenyediaan Oxidation productHasilpengoksidaan

EthanolEtanol 78 CrHo + HrO -+ C2H5OH Ethanoic acidAsid etanoikPropanolPropanol 97 CrHu + HrO -+ CaHTOH Propanoic acidAsid propanoikButanolButanol 118 CoH, + HrO -+ C4HeOH Butanoic acidAsid butanoik

Table 7Jadual7Based on Table 7, state and explain five characteristicsof a homologous series.BerdasarkanJadual 7, nyataknn dan terangkan lima ciri siri homolog.

[I0 markslll0 markahj(b) The following information is about an organic compound X.Maklumatberikut adalah mengenaisuatusebatianorganik X.. Empirical formula is CH2OFormula empirik ialah CHrO

. Relative molecular mass s 60Jisim molekul relatif ialah 60' Reacts with calcium carbonate to produce a type of gas that turns lime water chalkyBertindakbalas dengankalsium karbonatuntuk menghasilkan ejenisgas yang mengeruhkan ir kapurBased on the information given:Berdasarkanmaklumatyang diberi:(i) Determine the molecular formula of X.(Relativeatomic mass:C = l2,H= 1, O = 16)Tentukanormula molekul X.(Jisim alom relatif: C = lZ,H = l, O = 16)

[2 marlcs][2 markah](ii) State the name of the homologous series or X and explain your answer.Nyatakan nama siri homolog bagi X dan terangkan awapan anda.

12marksl[2 markah]

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ST]LIT

CompoundQSebatian

454u2

f4 marksff4 markahl

(iii) Write a balanced chemical equation for the reaction of compound X with calcium carbonate.Tulispersamaankimia seimbanguntuk tindak balas sebatianX dengankalsiumkarbonat.12mqrksl12markahl(c) Diagram 7 shows the structural formulae of hydrocarbon of compoundsP and Q.Rajah 7 menuniakkanormula struktur bagi sebatianhidrokarbon P dan Q.

HHl lH-C:C-HCompoundPSebatianP

Diagram7Raiah 7Compare and contrast thesetwo hydrocarbonsbasedon their structures'Banding dan bezakankedua-duahidrokarbon ini berdasarkanstrukturnya.

HHl lH-C-C-Ht lHH

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SULIT 4541t2

Pair of metalsPasangan lagam Positive terminalTerminal positif Potential diffference (V)Bezakeupayaan (Y)W, CU Cu 3.1X,Y Y 0.3w,x x 1.8X, CU

Tiilif,:i(i) Based on the values of the potential differences, ilrange the metals in descendingorder in the. electrochemical series.Berdasarkannilai beza keupayaan,susun ogam-logam ersebutdalam tertib menurundalam sirielektrokimia.ll marlclI markahl(ii) Predict the value of the positive terminal and the potential difference for the pair of metals X andCu.Explain your answer.Ramal nilai terminarpositif dan bezakeupayaan agi pasangan ogamx dan cu.Ielaskan jawapan anda.

[3 marks]13markahl(b) Diagram 8 shows a voltaic cell. Metal Q is situatedbelow copper in the electrochemicalseries.Raiah 8 menuniukkan uatusel voltan.Lr)gamQ terletakdi bawahkuprumdalam sirt elektrokimia.

Meral QLogamQ

SolutionRLarutanR Copper(Il)nitratKuprum(Il)nitrat

"';ffffrtstate the positive terminal and the negative terminal of this cen.Suggesta metal that is suitable as melal Q and a solution that is suitable as solution R.p"!"tu! terminalpositif dan terminal negiif bagi sel ini.cadangkan ogamyang sesuaisebagai ogam Q dan larutan yang sesuaisebagai arutanR.

[4 marks][4 markah]

ll,ihat halamansebelahlSULIT

8 (a) Table 8'1 shows the results of a seriesof experimentscarried out to construct the electrochemical series.The positive terminal and value for the potential difference for the pair of metals X and copper, cuare not given. W, X and y are not the actual symbols of the metals.Jadual 8'l menuniukkankeputusan satu siri iksperimen yang telah dijalankan untuk membina sirilektrokimia.Terminalpositif dan nitai bezakeupayaan agi pasangan ogamx dan kuprum,Cu tidak diberi. .vt, x danbukansimbol sebenar ogam_logamtu.

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4s41"1ulrr 1{ ,l(c) Experiment I and experiment II are carried out to investigate the factors affecting the discharge of ionsat the electrodes.Table 8.2 shows the apparatusset-up and the observations or experiment I and experiment II.Eksperimen dan el

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SULIT 4s4lt2SectionCBahagianC120marksl[20 markah]

Answer any one question from this section.Jawab mana-mana satu soalan daripada bahagian ini.9 (a) Diagram 9 shows the apparatus and observations for a redox reaction between iron(Itr) chloridesolution and a.metal.Rajah 9 menunjuklan ra.dasdanpemerhatian bagi satu tindak balasredoks antara larutan ferum(Ill) kloridadan sekeping ogam.

Yellow solution ofiron(IID chlorideLarutan aningferum(fr) klori.da

GreensolutionLarutanhijauMetalIngamAt the beginning of the experiment' Di awal el

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SULIT 4 41t210 (a) In an experiment, 50 cm3 of 0.1 mol dm-3of ethanoic acid, CH3COOH reactscompletely with x g ofsodium hydroxide, NaOH, and is dissolved in 100 cm3 of solution.Calculate the value of x.Dalam satu eksperimen,50 mJ 0.1 mol dmj asid etanoik,CH3COOHbenindak balas engkapdenganx gnatrium hidrolcsida. NaOH. dan dilarutkan dalam 100 cm' larutan.Hitung nilai x.

14marksl14markahl

(b) Table 10 shows the results when zinc reacts with hydrogen chloride in solvent L and solvent M.Jadual l0 menunjukkan eputusan pabila zink bertindakbalas denganhidrogen klorida di dalampelarutL danpelarut M.SubstanceBahan SolventPelarut ObservationPetnerhatian

Zinc + Hydrogen chlorideZink + Hidrogen klorida L Bubbles of gasGelembung asZinc + Hydrogen chlorideZink + Hidrogen klorida M No bubble of gasTiadagelembunggas

Table 10Jadual lO

Based on Table 10, suggest he name of solvent L and solvent M.Explain the observations.Write the equation for the reaction that occurs in solvent L.Berdasarkan Jadual 10, cadangkan namapelarut L dan pelarut M.Jelaskanpemerhatian.Tulispersamaanbagi tinlak balasyang berlaku dalampelarut L. f6 marksl[6 markah]

(c) You are given a solution that contains a mixture of iron(trI) nitrate and iron(III) chlorideDescribe the confirmatory tests to determine the presence of cation and anion in the solution. Yourdescriptionmust include all the materials used, observations and conclusion.Anda diberikan larutan yang mengandungi campuran erum(lll) nitrat danferum(IlI) klorida.Huraikan ujianpengesahan ntukmenentukan ehadirankation dan anion dalam arutan tersebut.Huraiananda mesti mengandungi emuabahanyang digunakan,pemerhatiandan kesimpulan.ll0 mar*sl

[I0 marknh]END OF QUESTION PAPERKBRTASSOALANTAMAT

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KERTAS 2Section NBahagian AI (a) An alloy is a mixture of two or more elementswith a certain composition in which the majorcomponent is a metal.Suatualoi ialah campurandua atau lebih unsurdengan komposisi tertentu dengan komponenutamanyaalah logam.(b) Brass/Loyang

(c) Tin/Timah(d) (D Bronze is harder than pure copper.Gangsaadalah lebih keras daripadakuprum.(ii)

DifferencePerbezaan BronzeGangsa Fure copperKuprum ulenSize ofatomsSaiz atom

Some arebiggerSesetengahnyalebihbesar

All of thesame sizeSemuanyasamasaiz

Arrangementof atomsSusunan tom

Randomarrangementof atom oftwo typesSusunanawakdua enis atom

Regulararrangementof atom ofone typeSusunantersusun atujenis atomsahaja

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Atom oksigen adalah lebih kecil dari atomlitium.

(c) Oxygen: Period 2 group 16Oksigen: Kala 2, kumpulan 16

(d) (i) Li+ and 02-(ii) Li, electron configuration: 2.1

Li, susunan elektron: 2.1It can donates 1 electron to form octet.This forms a more stable particle Li*Ia menderma I elektron untuk membentukoktet. Ini membentuk zarah Li* yang lebihstabil.O, electron configuration: 2.6.O, susunan elektron: 2.6It can take in 2 electrons to form octet.This forms a more stable particle 02-Ia mengambil 2 elektron untuk membentukol

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The higher the concentration, the higheris the reaction will be.Kadnr tindak balas di dalam set 2 adalnhlebih cepat dari kadar tindak balas di dalamset l. Ini disebabkan kepekatan HCI dalamtindak balas dalam set 2 adalah lebih tinggidari kepekatanHCl dalam tindak balas dalamset l. Kadar tindak balas akan bertambah ikakepekatanbertambah.

(ii) A higher concentration of HCl containsmore particles of H+ and Cl- per unitvolume. The more particles there are, thehigher the effective rate of collision of H+on CO32 will be. This increases the rate offormation of CO2.Kepekatan HCl yang lebih tinggi akanmempunyai ebih banyakzarahH* dan Cl- perunit isi padu. Zarah-zarah yang lebih banyakakan menyebabkan perlanggaran berkesanyang lebih banyak antara H+ dan COr2-. Iniakan meningkatkan kadar pembentukan CO2.(e)

I' Volume of CO2(cm3)Isi padu COr(cm3)

Time/minMasalmin(a) The heat of neutralisation is the heat energyproduced when I mol of OH- ions combinewith one mole of H+ ions to form one mole ofHzo.Haba peneutralan alah haba yang dihasilkanapabilaI mol ion OH- berpadudengansatu molion H+ untukmembentukatumol of H2O.(b) (D Total massof the liquid (25 + 25) gJumlahisim cecatr=50g

Total rise in temperature(36.8 - 30)'CJumlahkenaikan uhu= 6.8oCTotal amount of heat releasedJumlahhabayangdibebaskan= mcT = 50 x 4.2 x 6.8 = 1428J(ii) 1000cm3-ofnitric acid contain 1.0 mol ofnitric acidTherefore 25 cm3 nitric acid contain25 + 1000mol = 0.025 mol1000cm3 asid nitrik mengandungi .0 molasidnitrik.Jadi25 cm3 sidnitrikmengandungi

12060

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25 - 1000mol = 0.025mol(iii) I mol of HNO3 will produce I mol ofHzo.I zol HNO3akan menghasilkan mol H2O.Total amountof water formed = 0.025 molJumlahair yang dibentuk0.025 mol of water causes he releaseoft428 J0.025mol air akanmenyebabknnembebasan1428JTherefore I mol of water will requireJadi I mol air akan memerlukan1428 = _57120I.0.025Heat of neutralisation s -5ll2O J mol I or57 12 kI mol1Habapeneutralan alah -57120J mol-r atau-57.12 J mol '

EnergyTenagaHNO3(aq)+ NaOH(aq)HNOr(ck)+ NaOH(aft)

= -57.12 kJ mol NaNO3(aq) HrO(aq)NaNOr(aft) H2O(ak)(d) Ethanoic acid is a weak acid. Weak acids haveweak ionisation. Before neutralisation cantakeplace,energy s needed o breakthe bondsof the unionised weak acid molecules. Thisenergy s takenfrom the heatof neutralisation.Therefore the final amount of heat released isless.Asid etanoik ialah asid lemnh. Asid lemahmempunyaiengionanemah.Sebelumeneutralanboleh berlaku, tenaga diperluknn untuk memecahikatan ikatan molekul asid lemah yang tidakmengion. enagani diambildarihaba eneutralan.Jadi nilai akhir habayang erbebas kanmenjadilebihrendah.(e) A copper container is a good conductor ofheat. It absorbs a lot more heat. Thus, thetemperaturerise will be smaller and the heatof neutralisationwill be lower.Bekaskuprum alah konduktorhabayang baik. Iamenyerapebih banyak haba.Jadi kenaikansuhu

akan menjadi lebih rendahdan habapeneutralanaknn menjadi ebih rendah.Section BlBahagian B7 (a) The five characteristicsare as follows:Lima ciri-ciri adalah:

(c)

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. All themembers ontain he sameunctionalgroup,Semuaahli mempunyai kumpul.anberfungsiyangsama. Can be represented by a general formulaBoleh diwakili dengan satu formuln am. The difference between one member and thenext member is -CH;Perbezaanantara satu ahli dengan ahli berikutialah -CH;. All members have similar chemicalpropertiesSemw ahli mempunyai sifat-sifa kiminyang sarna. All can be prepared by using a similarmethod of preparationSemua ahli boleh disediakan denganmenggurakan knednhpeniryediaanang sama(b) ( i) (12 +2 + 16)x= 60.Thereforex=2(12+2 + l6)x =60.Jadix=2

Formula (CH2O)2 or C2H4O2 orcH3cooHFormula (CH2O)2 atau C2H4O2 ataucH3cooH(ii) The name of the homologous series iscarboxylic acids.This is because all the members containthe functionalgroup -COOH.Namasiri homolog alah asidknrbosilik.Ini disebabkansemua ahli mengandungikumpulan e fungs -COOH(iii) 2CH3COOH + CaCO3 -r(CH3COO)2Ca+HzO COz

(c) Both have two carbon atoms per molecule(ethyl group). However, P is an unsaturatedcompound (alkene) with one double bond andQ is a saturated compound (alkane) with nodouble bond.Kedua-duanyamempunyai ua atomkarbon setiapmolekul (kumpulanetil). Akan tetapi P ialahsebatian tak tepu (alkena) dengan satu ikatanganda d.uadan Q ialah sebatian tepu (alkana)dengan iad.a katangandadua.8(a)( i ) w x Y

Tendency of cations o accptelecfronsncreasesKecenderunganation meneima elelaronbertambah

(ii) Positive terminal is Cu.. Terminal positif ialah Cu.The potential difference is 1.3 V

Perbezaan keupayaan alah 1.3 Y(b) The positive terminal is Q and the negativeterminal is Cu.

Terminal positif ialah Q dan terminal negatif ialahCu.

Cu

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Suitable metal for Q = silverLogamsesuaiuntukQ = argentumSuitable Solution for R = silver nitrate solution.LarutansesuaiuntukR = larutanLrRentum irat(c) (i) Experiment I:Eksperimen:

Product formed at the anode s oxygen.Hasil yang dibentuk pada anod ialahoksigen.The anions presenceare OH- and trAnion-anion angada ialah OH- dnnl-OH- ions are preferentially to bedischarged as shown in the equationbelow.Ion OH diutamalmnuntuk dinyahcassepertiyangditunjukkan alampersamaan erikut.4OH -+ zHzO + 02 + 4e'This happens because the concentrationof I- is too low.Ini berlaku kerana kepekatanrendah.The product formed at the

I terlalucathode is

hydrogen.The cationspresentare H+ andK+.Hasil yang dibentuk pada katod ialahhidrogen.Kation-kation anghadir ialah H+dan Kt.Potassiumoccupies oo high a position inthe electrochemical eries.Thus, H+ ionsare preferentially discharged as shown inthe equation below.2H* + 2e- -s H2Kalium mengambilkedudukan ang terlalutinggi dalam siri elektrokimia. adi ion H+diutamakan untuk dinyahcas seperti yangditunjukkan alampersamaan i bawah.2H+ + 2e- -->H2Experiment II:Eksperimen I:The product formed at the anode is iodin.Hasil yangdibentakpada anod alah iodin.The electrolyte used is a concentratedsolution of potassium iodide. Thus, theconcentrationof I- is very high, Due to itshigh concentration it is, preferentially tobe dischargedcompared to OH-.2I- 112+2eElektrolit yang digunakan mempunyailarutanpekat kalium iodida. Jadi larutanpekat l- turut menjadi inggi. Disebabkankepekatan angtinggi maka a diutamakanuntukmenyahcas erbanding enganOH-.2I - -+12+2e

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The product formed at the cathode ishydrogen.Hasil yangdibentuk adakatod nl"ah idrogen.Both H+ ions and K+ ions are attracted othe cathodebut H+ ions are preferentiallydischargedbecauseof the lower positionin the electrochemicalseries.2H++ 2e -+H2Kedua-duaon H+ dan ion K+ adalahditarikkekatod.Akan etapi utnaonH+dinyahc skankerana kedudukanyang lebih rendnhdalamsiri elektrokimia.2H++ 2e-+H,(ii) At the cathode: 2H+ + 2e -+ H2Pada kntod : 2I{+ + 2e -+ Hc

Section ClBahagian C9 (a) Suggestedmetal is magnesium.The ions present n the greensolution are Fe2+- and (H+, OH- from water)Logamyangdicadangkanalah magnesium.Ion-ionyangberada alam nrutanhijau alahFez+dan (H+, OH- daripadaair)Change in the oxidation number for ...Perubahan umborpengoksidaanntuk:. Fe3* to Fe2+ s from +3 to +2 (reductionreaction occur)Fe3*ke Fe2* alah dari +3 to +2 (tindakbalaspenurunan erlaku). Mg to Mgz*: is from 0 to +2 (oxidationreaction occur)Mg ke Mgz+: alah dari 0 to +2 (tindak balaspengolcsidaane laku)Fe3* s an oxidising agent. Mg is a reducingagent.Fe3+alah agenpengoksidaan. g ialah agenpenurunan.Fe3*+ e -->Fe2* reduction)/(penurunan)Mg -+ Mg2++ 2e (oxidation)l(pengoksidaan)(b) Galvanometer

Potassiumiodidasolution

CarbonelectrodeElektrodkarbonAcidified potassiummanganate(VII)solutionI-arutankaliummanganatYII) berasid

LarutankaliumiodidaU-tubeTiub-U SulphuricacidAsidsulfurik

24s

Procedute/prosedur:1. Dilute sulphuric acid is poured into a U-tube and clamped to a retort stand.Asid sulfurikcair dituangkan e daktmsebuahtiub-U dan diapitkan kepada sebuah kakiretort.2. A dropper is used to fill one arm of theU-tube with potassium odide solution.

Satupenitisdigunakan ntukmengisi alahsatulengan iub-Udenganarutankalium odida.3. Acidified potassium manganate(Vll) isadded carefully to the other arm of the U-tube by using a different dropper.Larutan kalium manganat(YIl) dimasukkandengan eliti ke dalam engan ain pada tiub-lJdenganmeng unakan enitisberlainan.4. Both arm of the U-tube is fitted with acarbon electrode each as shown in thediagram.Kedua-duaengan iub-U dipasangkan enganelektrod karbonsepertiyang ditunjukkm padarajah.

5. The appa.ratuss left for about 20 minutesand the changesale recorded.Susunan adasdibiarkan untuk 20 minit danpe ubahan- e ubahandireko kan.Verification of statement:Pengesahanemyataan:Oxidising agent: acidified KMnOaAgenpengoksidaan: MnOaberasidReducing agent: KIAgenpenurunan:KlAt the negative electrode, iodide ions areoxidised to iodine as shown in the equation.Pada elektrod negatif, ion iodida dioksidakankepada iodin seperti yang ditunjukkan dalampersanqaani bawah.2T(aq) + I2(aq) + 2e (oxidation)2l(ak) -+ I2@k)+ 2e (pengoksidaan)The colourlessKI solution becomesyellow 12.InrutanKlyang tidakberwamamenjadi ekuningl2.At the positive terminal, the permanganateions, MnO; are reduced o Mn2+by acceptingelectronsas shown in the equation below.Pada erminal ositif, on-ionpermanganat nOa-diturunkan epadaMn2*secaramenerima lektronsepeni yang ditunjukknndalam persamaandibawah.MnO;(aq) + 8H+(aq) + 5e --)4H2O0)MnO;(ck) + 8H+(aft) + 5e -+4H2O(ce)

Mn2*(aq) +Mn2*1ak1+

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10

Here MnO;(aq) is reduced by I- to formMn2*. The purple colour layer of potassiummanganate(Vll) slowly chariges o a colourlesssolution.Di sini MnO4 ak) diturunkankepada - untukmembentukMnz*. Warna ungu.larutan kaliummanganat(Vllperlahan-lahan ertukarke arutantak berwarna.Thus, the iodide ions from KI are a goodreducing agent.Jadi ion-ion odida dari Kl ialah agenpenurunanyangbaik.(a) CH:COOH (aq) + NaOH(aq) -+CH3COONa(aq) + H2O(l)CH3COOHaft)+ NaOH(ak) >CH3COONa(ak) H2O(ce)1000 crn3 of CH3COOH contain 0.1 mol1000cm3CH3COOHmengandungi .1 molThus, 50 cm3 of CH3COOH contain0'1 * 50 = 0.005mol1000

Jadi , 5Ocm3CHTCOOHmengandungiolff i x 50=0.005 olI mol of ethanoic acid reacts with I mol ofNaOHI mol asid etanoikbertindakbalasdengan molNaOHThus, 0.005 mol of ethanoic acid reacts with0.005 mol of NaOHJadi,0.005mol asidetanoik ertindak alasdengan0.005molo/NaOAMolar massof NaOH = 23 + i6 + 1 - 40 gJisim molar NaOHThus, 0.005 mol NaOH = 40 x 0.005= 0.2 sJadi, 0.005mol NaOH(b) Solvent L is water. M is Methylbenzene(propanonecan also be used)PelarutL alahair. M ialahmetilbenzenapropanonjuga bolehdigunaknn)Hydrogen is evolved when zinc reacts withacid in aqueoussolution.Without water i.e. inmethylbenzene, no hydrogen is evolvedbecause he acid will not behave ike an acidwithout the presenceof water.Zn(s) + HCI(aq) -->ZnCl2@q) + HzG)Hidrogendibebaskan pabilazinkbertindakbalasdengan asid dalam larutan akueus. Tanpa airseperti dalam metilbenzena, iada hidrogendibebaskan erana asid tidak akan bersifat asid

246

tanpa kehadiran air.Zn(p) + HCI(ak) -+ ZnCl2@k) + Hz(s)(c) Materialsused:Test tubes, Iron(II) sulphate solution, dilutesulphuric acid, silver nitrate solution, hexa-cyanoferate(tr) !Fe(CN)6 solution,concentratedsulphuric acidBahanyangdigunakan:Tabunguji, larutan besi(Il) sulfut, asid sulfurikcair, larutan argentum nitrat, larutanheksasianofera(Il) aFe(CN)6,sid sulfurikpekatConfirmatory test for nitrate: Brown ring testUjianpengesahanitrat: Ujian cincinperang

ConcentratedsulphuricacidAsid sulfurikpekat

Nitrate ion + dilute sulphuric acid+ iron(Il) sulphate olutionIonnitrat + asidsulfurikcair+ larutanbesi(lI)sulfat

Procedure:Prosedur:Pour about 2 cm3 of the mixture into a testtube. Some dilute sulphuric acid and iron(Il)sulphate solution are added. Concentratedsulphuric acid is added slowly down the sideof the test fube by using dropper.Tuangkanebihkurang2 cm3 ampuran e dalamsebuah abunguji. Sedikitasid sulfurik cair danlarutan besi(Il) sulfat dicampurkan. sid sulfurikpekat ditambahkan erlahan-lahanmelalui tepitabunguji menggunakanenitik.Observation:Pemerhatian;A brown ring will be formed.Satucincinperangaknn erbentuk.Conclusion:Kesimpulan;The presence of a brown ring confirms thepresenceof nitrate ions.Kehadirancincinperangmengesahkanehadiranion-ionnitrat.

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