Ppt kimia ting 5 paper 2 (2018) answer scheme

ANSWER SCHEME

1

PPT KIMIA P2 TING. 5 (2018)

a: accept ; r: reject ; adp: avoid double penalty ; wcr: wrong cancels right ; ecf: error carried forward ; bod: benefit of doubt Question 1

Answer

Marks

(a)

A mixture of two or more elements with a certain fixed composition in which the major component is a metal.

Suatu campuran dua unsur atau lebih mengikut peratusan tertentu dengan unsur utamanya ialah logam.

1

(b)

Brass / Loyang

1

(c)

Tin / stanum / timah r : Sn

1

(d)

Any three points 1. Bronze is harder than pure copper Gangsa lebih keras daripada kuprum tulen 2. Size of atoms in bronze is different whereas size of atoms in pure copper is the same Saiz atom dalam gangsa adalah berbeza manakala saiz atom dalam kuprum tulen adalah sama 3. In bronze, atoms are not arranged in orderly manner whereas in pure copper, atoms are arranged in orderly manner Dalam gangsa, atom-atom tidak tersusun dengan teratur manakala dalam kuprum tulen, atom-atom tersusun dengan teratur 4. Atoms in bronze do not slide easily when external force is applied whereas in pure copper, atoms easily slide over one another Atom-atom dalam gangsa tidak mudah menggelongsor apabila daya dikenakan manakala dalam kuprum tulen, atom-atom mudah menggelongsor antara satu sama lain.

1 1

1

1

Max. = 3 (e)

Making souvenirs / membuat cenderamata

1

(f)

1. Fused glass / Kaca silika terlakur 2. Borosilicate glass / Kaca borosilikat

1 1 TOTAL

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

9

SULIT

ANSWER SCHEME

2

Question 2

(a)

(b)

PPT KIMIA P2 TING. 5 (2018) Answer

Marks

(i)

Diffusion / resapan r : wrong spelling

1

(ii)

1. Particles of fried chicken smell are tiny Zarah-zarah bau ayam goreng adalah halus 2. Moves randomly in between air particles Bergerak secara rawak di antara zara-zarah udara 3. From higher concentration to lower concentration region Dari kawasan berkepekatan tinggi ke kawasan berkepekatan rendah

1

(i)

1 1

1 1

(ii) 1+1

AB

CD

(iii) 1. Heat absorbed (by the particles) is used Haba yang diserap (oleh zarah-zarah) digunakan untuk 2. To overcome the forces (of attraction) between particles Mengatasi daya antara (tarikan) zarah-zarah

1 1 TOTAL

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

9

SULIT

ANSWER SCHEME

3

Question 3

PPT KIMIA P2 TING. 5 (2018) Answer

Marks

(a)

17

1

(b)

Atom P : 2.7 Atom Q : 2.8.7

1 1

(c)

Element P : (Period / Kala) 2 Element Q : (Period / Kala) 3

1 1

(d)

(e)

1

(i)

Increases / meningkat // becomes bigger / semakin besar

1

(ii)

Decreases / berkurang

1

(iii) 1. Correct formulae of reactants and product 2. Balanced equation Answer: 2Fe + 3Q2 → 2FeQ3

1 1

TOTAL

4541/2

© 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

10

SULIT

ANSWER SCHEME

4

Question 4

(a)

(b)

(c)

PPT KIMIA P2 TING. 5 (2018) Answer

Marks

(i)

2+

Lead(II) ion / ion plumbum(II) / Pb , Bromide ion / ion bromida / Br-

1

(ii)

Sodium ion / ion natrium / Na+ , Sulphate ion / ion sulfat / SO42- , Hydrogen ion / ion hidrogen / H+ , Hydroxide ion / ion hidroksida / OH-

1

(i)

Plumbum

1

(ii)

Pb2+ + 2e → Pb

1

(iii) Brown gas bubbles released // gelembung gas perang terbebas

1

(i)

Hydroxide ion / ion hidroksida r : OH-

1

(ii)

Anode : oxygen / oksigen / O2 Cathode : hydrogen / hidrogen / H2

1 1

(iii) 1. Insert / put a glowing wooden splinter into a test tube containing the gas Masukkan / letakkan kayu uji berbara ke dalam kayu uji yang mengandungi gas itu 2. Wooden splinter lights up, (confirming the presence of oxygen gas) Kayu uji menyala, (mengesahkan kehadiran gas oksigen) TOTAL

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

1 10

SULIT

ANSWER SCHEME

5

5

PPT KIMIA P2 TING. 5 (2018)

Question

Answer

Marks

(a)

(i)

Blue coloured solution forms Larutan berwarna biru terbentuk // Colourless gas bubbles which turns lime water chalky / milky / cloudy forms Gelembung gas tidak berwarna yang mengeruhkan air kapur terbentuk a : colourless gas bubbles released gelembung gas tidak berwarna terbebas

1

(ii)

1. Copper(II) sulphate / Kuprum(II) sulfat 2. Carbon dioxide / Karbon dioksida

1 1

(i)

Double decomposition reaction / Penguraian ganda dua

1

(ii)

Ba2+ + SO42- → BaSO4

1

(b)

(iii) 1. Ratio of moles From the chemical equation, 1 mol of BaSO4 : 1 mol of Ba(NO3)2 0.02 mol of BaSO4 : 0.02 mol of Ba(NO3)2 2. Volume of Ba(NO3)2 = 0.02 mol // 0.04 dm3 // 40 cm3 0.5 mol dm-3

1 1

(c)

1. Correct formulae of reactants and product 2. Balanced equation Answer: Zn + CuSO4 → ZnSO4 + Cu

1 1

(d)

1. Put 2 cm3 of dilute hydrochloric acid and 2 cm3 of barium chloride solution into the copper(II) salt solution / anion Masukkan 2 cm3 asid hidroklorik cair dan larutan barium nitrat ke dalam larutan garam kuprum(II) / anion 2. White precipitate forms, (confirming the presence of sulphate ion) Mendakan putih terbentuk (mengesahkan kehadiran ion sulfat)

1

TOTAL

11

4541/2

© 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

SULIT

ANSWER SCHEME Question 6

(a)

(b)

6

PPT KIMIA P2 TING. 5 (2018) Answer

1. Functional diagram – rubber stopper above the mouth of conical flask, burette clamped, rubber hose inserted into burette, dashed line for acid and water 2. Label – zinc / zink, sulphuric acid / asid sulfurik, water / air

Marks 1

1

1. Average rate of reaction for Experiment I = 20 cm3 // 10 cm3 min-1 // 20 cm3 // 0.167 cm3 s-1 2 min 120 s

1

2. Average rate of reaction for Experiment II = 32 cm3 // 16 cm3 min-1 // 32 cm3 // 0.267 cm3 s-1 2 min 120 s

1

*adp P2 for unit (c)

(d)

4541/2

1. Number of moles of H2SO4 = 0.1 mol dm-3 × 0.02 dm3 // 0.002 mol 2. Ratio of moles 1 mol of H2SO4 : 1 mol of H2 0.002 mol of H2SO4 : 0.002 mol of H2 3. Volume of H2 gas = 0.002 mol × 24 dm3 mol-1 // 0.048 dm3 // 48 cm3 1. Rate of reaction in Experiment II is higher than Experiment I Kadar tindak balas Eksperimen II lebih tinggi berbanding Eksperimen I 2. Catalyst / copper(II) sulphate lowers activation energy Mangkin / kuprum(II) sulfat merendahkan tenaga pengaktifan © 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

1 1 1 1 1

SULIT

ANSWER SCHEME

7

PPT KIMIA P2 TING. 5 (2018)

3. More colliding particles achieve the lower activation energy Lebih banyak zarah yang berlanggar dapat mencapai tenaga pengaktifan yang lebih rendah 4. Frequency of effective collision between hydrogen ion and zinc atom is higher in Experiment II Frekuensi perlanggaran berkesan antara ion hidrogen dan atom zink lebih tinggi dalam Eksperimen II TOTAL

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

11

SULIT

ANSWER SCHEME

8

Question 7

(a)

PPT KIMIA P2 TING. 5 (2018) Answer

Marks

(i)

As catalyst / sebagai mangkin

1

(ii)

Esterification / Pengesteran r : esterifikasi

1

(iii) 1. Correct formulae of reactants and product 2. Balanced equation Answer: CH3COOH + C3H7OH → CH3COOC3H7 + H2O

1 1

(iv) 1. Correct drawing of structural formula Answer:

1

2. Name : propyl ethanoate / propil etanoat (v)

Any two answers 1. Colourless liquid Cecair tidak berwarna 2. Fruity / sweet / fragrant smell Berbau buah / manis / harum 3. Low boiling point Takat didih rendah 4. Volatile Mudah meruap 5. Does not dissolve in water Tidak larut dalam air 6. Dissolves in organic solvent Larut dalam pelarut organic 7. Less dense than water Kurang tumpat berbanding air

1 Total = 2 1 1 1 1 1 1 1 Max. = 2

(vi)

1+1 Total = 2

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

SULIT

ANSWER SCHEME (b)

(i)

(ii)

9

PPT KIMIA P2 TING. 5 (2018)

1. Bacteria produces (lactic) acid Bakteria menghasilkan asid (laktik) 2. Acid / H+ (ions) neutralize negative charges of the (protein) membrane Asid / ion H+ meneutralkan cas negatif membran (protein) 3. Rubber particles collide with one another causing the breakage of (protein) membranes Zarah getah berlanggar antara satu sama lain menyebabkan (protein) membran pecah 4. Rubber molecules / polymer entangle / combine together Molekul / polimer getah bergabung antara satu sama lain r : menggumpal 5. Hydroxide ion / OH- from ammonia will neutralize H+ from the acid Ion hidroksida / OH- daripada ammonia akan meneutralkan H+ daripada asid 6. (Protein) membrane remain negatively charged Membran (protein) kekal bercas negatif 7. Rubber particles repel each other Zarah getah menolak antara satu sama lain // Rubber particles cannot combine with each other Zarah getah tidak boleh bergabung antara satu sama lain 1. The presence of cross-linkage of sulphur atoms between rubber molecules improves the weakness of natural rubber Kehadiran rangkai silang atom sulfur antara molekul getah memperbaiki kelemahan getah asli 2. When vulcanized rubber is stretched and released, cross-linkage will pull the chain back to its original arrangement 3.

4. Elasticity and strength of rubber is improved Kekenyalan dan kekuatan getah dapat diperbaiki // Vulcanised rubber is more resistant to heat / organic solvent Getah tervulkan lebih tahan haba / pelarut organik OR

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1 1

1

1

1

1 1

Max. = 6 1

1 1

1

Total = 4 OR SULIT

ANSWER SCHEME

10

PPT KIMIA P2 TING. 5 (2018)

1. The presence of cross-linkage of sulphur atoms between rubber molecules improves the weakness of natural rubber Kehadiran rangkai silang atom sulfur antara molekul getah memperbaiki kelemahan getah asli 2.

1

1

3. C = C in vulcanized rubber is less than C = C in natural rubber C = C dalam getah tervulkan kurang berbanding C = C dalam getah asli 4. So, vulcanized rubber is more resistant to oxidation Maka, getah tervulkan lebih tahan terhadap pengoksidaan TOTAL

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

1 Total = 4 20

SULIT

ANSWER SCHEME

11

Question 8

PPT KIMIA P2 TING. 5 (2018)

Answer

Marks

(a) Compound Sebatian

Structural formula Formula struktur

Name Nama

1.

2. Propene Propena

X

1+1

3.

4. Propanoic acid Asid propanoik

Y

1+1

5.

Z

6. Propyl ethanoate Propil etanoat 1+1 Total = 6

(b)

1. Reaction I : Dehydration / Pendehidratan 2. Put 2 cm3 of bromine water / acidified potassium manganate(VII) solution into a test tube containing compound X Masukkan 2 cm3 air bromin / larutan kalium manganat(VII) berasid ke dalam tabung uji yang mengandungi sebatian X // Flow gas / compound X into a test tube containing 2 cm3 of bromine water / acidified potassium manganate(VII) solution Alirkan gas / sebatian X ke dalam tabung uji yang mengandungi air bromin / larutan kalium manganat(VII) berasid 3. Brown bromine water becomes colourless Warna perang air bromin menjadi tidak berwarna // Purple acidified potassium manganate(VII) solution becomes colourless Warna ungu larutan kalium manganat(VII) berasid menjadi tidak berwarna

1 1

1

Total = 3

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

SULIT

ANSWER SCHEME (c)

12

PPT KIMIA P2 TING. 5 (2018)

1. Correct formulae of reactants and product 2. Balanced equation C3H6 + H2O → C3H7OH 3. Phosphoric acid / Asid fosforik / H3PO4 300°C 60 atm 1. Pour [2 – 5] cm ethanoic acid into a boiling tube Tuang [2 – 5] cm3 asid etanoik ke dalam tabung didih 2. Add [2 – 5] cm3 propan-1-ol into the acid Tambah [2 – 5] cm3 propan-1-ol ke dalam asid itu 3. Add [2 – 5] drops of concentrated sulphuric acid Tambah [2 – 5] titis asid sulfurik pekat 4. Heat the mixture Panaskan campuran 3

(d)

1 1 1 Total = 3 1 1 1 1 Total = 4

(e)

(i)

Carboxyl / karboksil

1

(ii)

CnH2n+1COOH n = 0, 1, 2…

1

(iii) 1. Correct formulae of reactants and product 2. Balanced equation Answer: 2C2H5COOH + CaCO3 → Ca(C2H5COO)2 + CO2 + H2O TOTAL

4541/2

© 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1 1

20

SULIT

ANSWER SCHEME Question 9

Any two equations involving : Acid + metal → salt + hydrogen gas Acid + base / alkali → salt + water Acid + carbonate metal → salt + water + carbon dioxide gas Sample answer: Zn + 2HNO3 → Zn(NO3)2 + H2 ZnO + 2HCl → ZnCl2 + H2O ZnCO3 + H2SO4 → ZnSO4 + CO2 + H2O (i)

(ii)

(c)

4541/2

PPT KIMIA P2 TING. 5 (2018)

Answer

(a)

(b)

13

Marks

1 1

1. Methanoic acid ionizes partially in water Asid metanoik mengion separa dalam air 2. Producing low concentration of hydrogen ion / H+ Menghasilkan kepekatan ion hidrogen / H+ yang rendah

1

1. Glacial methanoic acid consists of neutral molecule Asid metanoik glasial terdiri daripada molekul neutral // Glacial methanoic acid does not contain free moving ions Asid etanoik glasial tidak mengandungi ion yang bergerak bebas 2. Aqueous methanoic acid contains free moving ions Asid metanoik akueus mengandungi ion-ion yang bergerak bebas 3. Which carry electrical charges to conduct electricity Yang membawa cas elektrik untuk mengkonduksikan elektrik

1

1. pH value of ethanoic acid is higher than nitric acid Nilai pH asid etanoik lebih tinggi berbanding asid nitrik 2. Ethanoic acid is a weak acid while nitric acid is a strong acid Asid etanoik ialah asid lemah manakala asid nitrik ialah asid kuat 3. Ethanoic acid ionizes partially in water to produce low concentration of hydrogen ion / H+ Asid etanoik mengion separa dalam air menghasilkan kepekatan ion hidrogen / H+ yang rendah 4. Nitric acid ionizes completely in water to produce high concentration of hydrogen ion / H+ Asid nitrik mengion sepenuhnya dalam air menghasilkan kepekatan ion hidrogen / H+ yang tinggi

1

© 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

1 1

1 1

1

Total = 4

SULIT

ANSWER SCHEME (d)

14

PPT KIMIA P2 TING. 5 (2018)

1. Pour [50 – 250 cm3] of [0.1 – 2.0 mol dm-3] sulphuric acid into a beaker Tuang [50 – 250 cm3] asid sulfurik [0.1 – 2.0 mol dm-3] ke dalam sebuah bikar 2. Warm the acid Hangatkan asid itu 3. Add magnesium / Mg / magnesium oxide / MgO / magnesium carbonate / MgCO3 powder little by little into the acid Tambah serbuk magnesium / Mg / magnesium oksida / MgO / magnesium karbonat / MgCO3 sedikit demi sedikit ke dalam asid itu 4. Stir the mixture Kacau campuran itu 5. Add magnesium / Mg / magnesium oxide / MgO / magnesium carbonate / MgCO3 powder continuously until in excess / some of it no longer dissolve Tambah serbuk magnesium / Mg / magnesium oksida / MgO / magnesium karbonat / MgCO3 berterusan sehingga berlebihan / tidak larut lagi 6. Filter the mixture to remove excess / unreacted magnesium / Mg / magnesium oxide / MgO / magnesium carbonate / MgCO3 Turas campuran itu untuk mengasingkan magnesium / Mg / magnesium oksida / MgO / magnesium karbonat / MgCO3 berlebihan / yang tidak bertindak balas 7. Transfer / pour the filtrate into an evaporating dish Pindahkan / tuang hasil turasan ke dalam mangkuk penyejat 8. Heat gently to produce a saturated solution // one-third / onequarter of the original volume Panaskan perlahan-lahan untuk menghasilkan larutan tepu // satu pertiga / satu perempat daripada isi padu asal 9. Cool (to room temperature) Sejukkan (pada suhu bilik) 10. Filter the crystals Turaskan hablur 11. Dry the crystals between filter papers Keringkan hablur antara kertas turas

1

1 1

1

1

1

1 1

1 1 1 Max. = 9

TOTAL

4541/2

© 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

20

SULIT

ANSWER SCHEME Question 10 (a)

15

PPT KIMIA P2 TING. 5 (2018)

Answer Formation of ionic compound 1. P atom donates / releases 1 electron Atom P menderma / melepaskan 1 elektron 2. to achieve stable octet electron arrangement untuk mencapai susunan elektron oktet yang stabil 3. P+ / P ion is formed P+ / ion P terbentuk 4. Q atom receives / accepts 1 electron Atom Q menerima 1 elektron 5. Q- / Q ion is formed Q- / ion Q terbentuk 6. P+ and Q- are attracted together by electrostatic force, to form an ionic compound PQ P+ dan Q- ditarik bersama oleh daya elektrostatik, membentuk sebatian ion PQ Formation of covalent compound 7. R atom needs 4 electrons (to achieve stable octet electron arrangement) Atom R memerlukan 4 elektron (untuk mencapai susunan elektron oktet yang stabil) 8. Q atom needs 1 electron (to achieve stable octet electron arrangement) Atom Q memerlukan 1 elektron (untuk mencapai susunan elektron oktet yang stabil) 9. One R atom shares 4 pairs of electrons with 4 Q atoms Satu atom R berkongsi 4 pasang elektron dengan 4 atom Q 10. Forming a covalent compound, RQ4 Membentuk sebatian kovalen, RQ4 11. One R atom contributes 4 electrons and each of the 4 Q atoms contribute one electron each for sharing Satu atom R menyumbang 4 elektron dan setiap satu atom Q menyumbang satu elektron untuk dikongsi 12. One R atom forms 4 single covalent bonds with 4 Q atoms Satu atom R membentuk 4 ikatan kovalen tunggal dengan 4 atom Q

Marks 1 1 1 1 1 1

1

1

1 1 1

1 Max. = 10

(b)

4541/2

1. Named ionic compound : sodium chloride / natrium klorida // copper(II) sulphate / kuprum(II) sulfat // [any suitable ionic compound] 2. Named covalent compound : Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

1

1 SULIT

ANSWER SCHEME

16

PPT KIMIA P2 TING. 5 (2018)

naphthalene / naftalena // sugar / gula / glukosa // [any suitable covalent compound] 3. Apparatus : crucible / mangkuk pijar, spatula / spatula, pipe clay triangle / segi tiga tanah liat, tripod stand / tungku kaki tiga, Bunsen burner / penunu Bunsen 4. Put one spatula of [sodium chloride] into a crucible Letakkan satu spatula [natrium klorida] ke dalam mangkuk pijar 5. Heat the [sodium chloride] Panaskan [natrium klorida] 6. Observe whether [sodium chloride] melts Perhatikan sama ada [natrium klorida] melebur 7. Repeat the experiment using [naphthalene] to replace [sodium chloride] Ulang eksperimen menggunakan [naftalena] untuk menggantikan [natrium klorida] 8. Sodium chloride does not melt (easily) [Natrium klorida] sukar untuk / tidak melebur // No change for sodium chloride Tiada perubahan bagi natrium klorida 9. Sodium chloride has high melting point [Natrium klorida] mempunyai takat lebur yang tinggi 10. Naphthalene melts easily [Naftalena] melebur dengan mudah 11. Naphthalene has low melting point [Naftalena] mempunyai takat lebur yang rendah TOTAL

1

1 1 1 1

1

1 1 1 Max. = 10 20

END OF MARKING SCHEME SKEMA JAWAPAN TAMAT

4541/2

Š 2018 Hak Cipta Panitia Kimia SMK Sultan Yussuf

SULIT