edisi-jwpn-kys-2015

Chemistry Workshop 2015

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[MRSM14-09] (a) Diagram 9 shows the apparatus set-up of a voltaic cell using solution Tand solution V. The observation is recorded in Table 9.

CopperKuprum

Diagram 9 / Rajah 9

ZincZink

Solution T/VLarutan TN

Solution Deflection of voltmeter's needleT Yesv No

Table 9

The different between structure question and essay question is

NO LINE /NO SPACEto write the answer

(i) Suggest suitable solutions for T and V. Explain the differences in observation for solutions T and V.[4Mj

This is voltaic cell diagram.But the observation is the deflection of voltmeter's needle

The question asking about electrolyte.



Tis electrode, contain free moving ion- show by deflection.

Vis non-electrolyte. No free moving ion - show by no deflection

Answer:

T: any soluble ionic compoundExample : Copper(II) nitrate

V : any covalent compoundExample : Benzene, methanolReject: alcohol - this is homologous series.

Explanation:T solution contains free moving ion that can conduct electricity

V solution exist as molecule, no free moving ion present in the solution.

(ii) Explain the rocess that occurs in the voltaic cell in Diagram 9 when solution Tis ed.

ould include the half equations involved. [6M]

Process that occurs in question is

Refer to redox



At negative terminal

1. Zinc as negative terminal

2. Zinc will be oxidation because zinc lost electronor oxidation number increases from Oto +2

3. Zn "7 zn2+ + 2e

At positive terminal

1. Copper as positive terminal

[T solution is Copper(II)nitrateCopper(II)ion and H+ ion present in the solution. Cu2+ ion was choose to discharge because lower in Electrochemical series/ less electropositive.}

2. Copper(II) ion will be reduction because gain electron to form copper metal or decreases oxidation numberfrom +2 to 0.

3. Cu-" + 2e "7 Cu

(b) A standard solution can be prepared using dilution method.Describe how you would prepare 100 cm-' 0.5 mol dm-3 sodium hydroxide solution from 2.0 mol dm-3sodium hydroxide solution.

Your description should include the followings:• List of materials and apparatus• Calculation involved• Procedure

[lOM]


II

h r

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sodium hydroxide,


--------=- 2moldm-3- - - - - - - - - sodium hydroxide,- - - - - -- - - NaOH solution

Pipette

(a) Calculate the volume ofstock solution required.

2.0 mol dm-3

- - - - - NaOH solution

---Sodium hydroxide, NaOH solution

(b) Use a pipette to draw up the required volume of stock solution.

Distilled waterCalibration mark

Volumetric flask

(c) Transfer the stock solution to a volumetric flask.

(d) Add water to bring the level of the solution to the calibration mark. Shake well to ensure thorough mixing.

List of materials and apparatus

Material2.0 mol dm-3 of sodium hydroxide, distil water

Apparatus100 cm-' volumetric flask, 25 cm-' pipet, 250 cm-' beaker,

Calculation involvedUsed formula

M1V1 = M2V2Before/ original after

Need to calculate the volume, V1V1 = M2V2 = 0.5 X 100 = 25 cm3

M1 2.0



Procedure

Gunakan - Teknik ABC - untuk menulis laporan eksperimenI Used - ABC Technique - to write the laboratory report

Simbol/Symbol

Maksud/Mean

Contoh/Example

A Arahan

Masukkan I add/ pour Bersihkan I clean Timbang /weights Bakar /Burn

B Bahan

Kepingan Magnesium/ magnesium ribbon

Larutan kuprum(II) sulfat I Copper(II)sulphate solution# perlukan MV/needMV

M - Kepekatan I molarityV- Isipadu /volume

Kertas litmus biru I Blue litmus paper

c Container

Bikar [Nyatakan saiz] I beaker [state the size]Tabung uji /Test Tube Kelalang kun I Conical Flask Buret I Burrete

Contoh menggunakan Teknik ABCI Example using ABC Technique

AArahan

B Bah an

cContainer

1. BersihkanClean

pita magnesiummagnesium ribbon

dengan menggunakan kertas pasir.by using sand paper.

2. MasukkanAdd

Kertas litmus biru lembabWet blue litmus paper

ke dalam tabung uji.into the test tube.

3. TimbangWeights

Pita magnesiummagnesium ribbon

dengan menggunakan penimbang elektronikby using weighing balance

4. MasukkanAdd

50 cm3 Larutanku prum(II) sulfat 1. 0moldrrr-'50 cm3 of 1. 0 moldm:" of copper(II)sulphate solution

ke dalam kelalang kuninto the conical flask.



Suggestion answer

1. Calculate the volume of sodium hydroxide need.

2. Prepare 5 cm3 of sodium hydroxide by using pipe.

3. Transfer the sodium hydroxide in the 2i2etl into 100 cm3 volumetric flask.

4. Add the distil wate into the volumetric flask that was filled half full with water.

5. Drop aistil water wiselY.l by using the pipet until reach graduation/ calibration mark

4. Close the volumetric flask and shaking the volumetric flask.

Overall

List of materials and apparatus1. Material 2.0 mol dm-3of sodium hydroxide, distil water2. Apparatus 100 cm-' Volumetric flask, 25 cm-' pipet, 250 cm-' beaker,

Calculation involved

V1 = M2V2 = 0.5 X 100 = 25 cm3

M1 2.0

Procedure1. Calculate the volume of sodium hydroxide need.2. Prepare 25 cm-' of sodium hydroxide by using pipet.3. Transfer the sodium hydroxide in the i etl into 100 cm-' volumetric flask.4. Add the distil wate into the volumetric flask that was filled half full with water.5. Drop aistil water wiselyj by using the pipet until reach graduation/ calibration mark4. Close the volumetric flask and shaking the volumetric flask.



ide(II)

[SPM14-04]

SwitchSuts

Carbon electrodes

.-----1111-----,

Elek trod karbon """"'==~- CrucibleMangkuk pijar

Mo1 ten lead{I ) brom Leburan plumbum bromide

Panaskan

Diagram 4.1

Anode /Anod Cathode/ Katod1. Ion presentIon hadir lead(II), Pb2+and Bromide, Br-

2. Ion attract toIon bergerak ke

{Positive terminal}

Bromide, Br-

{Negative terminal}

leadll), Pb2+3. Ion chooseIon dipilih Bromide, Br- lead(II), Pb2+

4. ReasonAlas an• Position

• Concentration

• electrode

5. ObservationPemerhatian Brown gas released Grey solid deposited

6. Comfirmatory testUjian PengesahanIf gas/ jika gas1. Method I Kaedah2. Result I Keputusan

none none

7. ProductHasil Bromine gas Lead



Ceppertll) sulphate solutionLannan kuprum,(D} suifa:

S'oivitchSuis

~-------•[--~Diagram 4.2

Anode /Anod Cathode/ Katod1. Ion presentIon hadir Copper(II) ion, Cu>, Sulphate, SQ42-

Hydrogen, H+ ion, Hydroxide, OH- ion


{Positive terminal}

Sulphate, S042- Hydroxide, OH- ion

{Negative terminal}

Copper(II) ion, Cu-" Hydrogen, H+ ion

3. Ion chooseIon dipilih Hydroxide, OH- ion Copper(II) ion, Cu-"


• Concentration

• electrode

Less electronegative{Position}

Less electropositive{Position}

5. ObservationPemerhatian

{40H- 7 2H20 + 02 + 4e}

Bubbles gas released

{Cu2+ + 2e "7 Cu}

Brown solid deposited


{test for oxygen}

1. Put the glowing splinter into the test tu be contain the gas.2. Glowing splinter willignite.

None

7. ProductHasil Oxygen gas Copper

es'"""='"-- Crucible

romidem(II)

SwitchPanaskan Suis



[SPM 14-04] Diagram 4.1 and Diagram 4.2 show the apparatus set-up for two electrolytic cells with different electrolytes.

SwitchSuis

Carbon electrod Elektrod karbon

~---1111----,

Copper(ll) sulphate solutionLarutan kuprum(ll) suifat

Moltenlead(TI) b Leburan plumbu bromide

Diagram 4.1

(a) State the meaning of electrolyte.

1---------<III---~

Diagram 4.2

Chemical substance that can conducts electricity in molten or aqueous solution........................................................................................................................ [lM]

(b) Based on Diagram 4.1,(i) Why does the bulb light up when molten lead(II) bromide is used as electrolyte?Contains free moving ion........................................................................................................................ [lM]

(ii) Write all the formulae for the ions present in lead(II) bromide.Pb2+and Br- Sebab molten........................................................................................................................ [lM]

(iii) State the observation at cathode.Grey solid deposited Sebab logam Pb yang terhasil........................................................................................................................ [lM]

(iv) Write the half equation at cathode.Pb2++ 2e 7 Pb........................................................................................................................ [lM]

Based on Diagram 4.2,(i) State which electrode is anode?Carbon electrode X........................................................................................................................ [lM]

(ii) State the observation at the anode.Bubbles gas released Sebab factor kedudukan dan ion OH- yang terpilih........................................................................................................................ [lM]

(iii) Write the half equation at the anode.40H- 7 2H20 + 02 + 4e4 org hutan minum 2 air bersama oren 2 dan 4 epal........................................................................................................................ [lM]

(d) In Diagram 4.2, carbon anode is replaced with copper and electrolysis is carried out for 20 minutes. State the observation at the anode. Give a reason.Copper electrode became thinner.Copper electrode dissolves/ I Copper electrode produce copper(II) ion........................................................................................................................ [2M]



[MRSM 14-03] Diagram 3 shows the apparatus set-up to investigate the electrolysis of1.0 mol dm-3potassium iodide solution using carbon electrodes.Rajah 3 menunjukkan susunan radas untuk mengkaji elektrolisis larutan kalium iodida1. 0 mol dm= menggunakan elektrod karbon.

Carbon TKarban T

Carbon RKarban R

-- - - -1.0 mol dm' potassium iodide solutionLarutan kalium iodida1.0 mol dm'

Anode /Anod Cathode/ Katod1. Ion presentIon hadir Potassium, K+ ion, Iodide, I- ion



{Positive terminal}

Iodide, I- ionHydroxide, OH- ion

{Negative terminal}

Potassium, K+ ionHydrogen, H+ ion

3. Ion chooseIon dipilih Iodide, I- ion Hydrogen, H+ ion


• Concentration

• electrode

Concentration of iodide ion Less electropositive{position}

5. ObservationPemerhatian Brown solution formed Bubbles gas released


1. Put 3 drop of iodine into 2 cm-' of starch in test tube.2. Blue black/dark blue produce

1. Place the burning splinter near the mouth of test tube contain the gas.2. Pop sound heard.

7. ProductHasil Iodine Hydrogen gas



[MRSM14-03] Diagram 3 shows the apparatus set-up to investigate the electrolysis of1.0 mol dm-3potassium iodide solution using carbon electrodes.

~-----<II>------~

0Carbon T Carbon RKarban T Karban R

(a) What is meant by electrolyte?

1.0 mol dm' potassium iodide solutionLarutan kalium iodida1.0 mol dm'

Chemical substance that can conducts electricity in molten or aqueous solution.......................................................................................................................... [lM]

(b) Write the ionic formulae for all ions present in potassium iodide solution.K+,I-, a-, OH-

·························································································································[·lM]

(c) (i) Name the ions attracted to anode.Iodide and hydroxide Reject formula sebab soalan nak nama.......................................................................................................................... [lM]

(ii) Write the half equation for the ion discharged at the anode.21- 7 h + 2e Sebab kepekatan, ion halida terpilih.......................................................................................................................... [lM]

(d) (i) State the observation at electrode R.Bubbles gas released.......................................................................................................................... [lM]

(ii) Explain your answer in (d)(i).Hydrogen ion less electropositive than potassium ionHydrogen ion receive electron and produce hydrogen atom2 hydrogen atom combine then produce molecule of hydrogen.......................................................................................................................... [2M]

(e) In another experiment, 1.0 mol dm-3potassium iodide solution is replaced with 0.001 mol dm-3potassium iodide solution.(i) Name the gas released at the anode.Oxygen gas.......................................................................................................................... [lM]

(ii) The volume of gas collected at anode is 24.0 cm3 at room condition. Calculate the number of molecules of gas collected.[Avogadro's constant: 6.02 X 1023moli: 1 mol of gas occupies 24 dm3 at room condition]

Mol =volume/molar volume= [24/ 1000] /24 = 0.001

Mol = No of Molecules/Na

No of molecules= mol X Na= 0.001 X 6.02 X 102a = 6.02 X 1020Molecules [2M]


[SBPtrialOS-02] Diagram 2 shows the setup of apparatus to investigate the electrolysis ofsilver nitrate solution with carbon electrodes and copper(II) sulphate solution with copper electrodes.

Cell 1 Cell 2Anode /Anod Cathode/ Katod Anode /Anod Cathode/ Katod

1. Ion presentIon hadir Silver, Ag+ ion, nitrate, N03- ion


Copper(II), Cu-" ion, Sulphate, S042- ions



{Positiveterminal}

nitrate, N03- ion,

Hydroxide, OH- ion

{Negativeterminal}

Silver, Ag+ ion, Hydrogen, H+

10n

{Positiveterminal}

Sulphate, S042- ion, Hydroxide,

OH- ion

{Negativeterminal}

Copper(II) , Cu-" ion,

Hydrogen, H+10n

3. Ion chooseIon dipilih Hydroxide,

OH- ionSilver, Ag+ ion, NONE Copper(II) ,

Cu-" ion,4. ReasonAlas an• Position

• Concentration

• electrode

Less electronegative

{Position}

Less electropositive

{Position}

Used active electrode

{Electrode}

Less electropositive

{Position}

5. ObservationPemerhatian Bubbles gas

releasedGrey solid depsited

Copper electrode thinner

Copper electrode thicker

6. Comfirmatory testUjianPengesahanIf gas/ jika gas1. Method/Kaedah2. Result IKeputusan

1. Put the glowing splinter in to the test tu be contain the gas.2. Glowingsplinter will ignite.

7. ProductHasil

Oxygen gas Copper formed Copper(II) ion produce


Carbon Copper

Silver

nitrate

solution

Copper(II) sulphate

solution

Copper formed


[SBPtrialOS-02] Diagram 2 shows the setup of apparatus to investigate the electrolysis ofsilver nitrate solution with carbon electrodes and copper(II) sulphate solution with copper electrodes.


Carbon

electrode

Copper

electrode

Silvernitrate

solutionCopper(II) sulphate

solution

(a) What is the energy change in electrolysis? [lM]Electrical energy to chemical energy

(b) Write the formulae of all the ions present in silver nitrate solution. [lM]Ag+,NOa-,ff+, OH-

(c) In the electrolysis of Cell 1

(i) What is the observation at electrode B? [lM]grey solid deposited

(ii) Write the half equation for the reaction at electrode B. [lM]Ag+ + e 7 Ag

(d) In the electrolysis of Cell 2.

(i) What is the observation at electrode C? [lM]Copper electrode became thinner

(ii) Write the half equation for the reaction at electrode C. [lM]Cu 7 cu2+ + 2e

(e) What are the processes that occur at electrodes A and D? [2M]Electrode A Electrode D

: Oxidation: Reduction

(f) State which cells the concentration of electrolyte remains unchanged. [lM]Cell 2

(g) State one application of electrolysis in industrial. [lM]electroplating of metal/ I purification of metal I I extraction of metal


[MRSM 13-03] Diagram 3.1 shows the apparatus set-up of a voltaic cell.

r::t---

•---


Rajah 3.1 menunjukkan susunan radas bagi suatu sel voltan p.

Magnesi nn---+•

Magnesium

Magnesium nitrate----+- solutionLarutan magnesium nitrat

m---4---CopperKuprum

Porous potPasu berliang

Copper(II) sulphate solutionLarutan kuprum(II) su/fat

Anode /AnodNegative Terminal

Cathode/ KatodPositive Terminal

1. Determine TerminalKenalpasti terminal Magnesium Copper

2. ReasonAlas an• more electropositive

• less electropositive

More electropositive Less electropositive

3. The direction of movement of electron Arah pergerakan elektron

Magnesium to copper

4. Ion presentIon hadir

Mg2+, N03- H+, OH-

Cu2+,S042- H+, OH-

5. Ion chooseIon dipilih Cu-"

6. Half EquationPersamaan setengah Mg 7 Mg2+ + 2e Cu-" + 2e "'7 Cu

7. ObservationPemerhatian Magnesium electrode thinner Copper electrode thicker

...,,..

---


[MRSM 13-03] Diagram 3.1 shows the apparatus set-up of a voltaic cell.


-----! v , _

Magnesiun---H•Magnesium

Magnesiumnitrate--+'" solutionLarutan magnesium nitrat

Based on Diagram 3 .1,

(a) What is the function of porous pot?

----1--CopperKuprum

PorouspotPasu ber/iang

Copper(II)sulphate solutionLarutan kuprum(II) sulfat

Allow ion through it to complete the circuit........................................................................................................................ [lM]

(ii) State the negative terminal of the voltaic cell.Magnesium I /Mg

State - boleh NAMA/ / FORMULA........................................................................................................................ [lM]

(iii) Explain your answer in (a)(ii).Magnesium more electropositive/ IMagnesium higher than Copper in Electrochemical Series

Jawapan yg kedua - mesti sebut ECS........................................................................................................................ [lM]

(iv) Name the type of reaction that occurred at copper electrode.Reduction

Sebab -Terima elektron / /berlaku penurunan no pengoksidaan dari +2 ke O........................................................................................................................ [lM]

(b) Diagram 3.2 shows an incomplete cell Q to electroplate an iron ring with silver.


Fm1--11--=--c1 Iron


Cell Qsa»

Mag:111:t">imrr-· 1Afag1n1:1irm1

~---- ..-. -- ~

solution f.'<troJ,an wgcnlum nilNU

ring

Magne.siwn mitrate Copper.oolution Kfipn,m C0pper(fil) sulphateLtmit(II! nlfl'SJleSilll'l:I K~m(II)lulfatmfnlt-~

(i) Complete the diagram of Cell Qin the space provided.

(ii) State one observation in Cell Q after 30 minutes.Silver electrode became thinner I IIron ring coated with grey metal/ silvery

Iron ring became thicker - kurang sesuai........................................................................................................................ [lM]

(iii) Write the half equation for the reaction occurred at the iron ringAg++ e 7 Ag

Sebab ion Ag+yang terendah dan dipilih untuk dinyahcas........................................................................................................................ [lM]

(iv) What happens to the concentration of silver ions in Cell Q. Explain your answer.

Concentration of silver ion is equal/ I unchangedRate to produce silver ion at anode equal to rate discharge silver ion at cathode........................................................................................................................ [lM]

....,,..... CK

-·=+------ PoPas

rate

otng



[SPM12-06] Diagram 6 shows the apparatus set-up for the combination of cell I and cellII. Cell I supplies electrical energy for cell II.

iagne mmMagne tum opper

uprum

m1.1 s pu

berlia

Magnesium niti solution

Larutan magne firmnit at

CelJ I oppe ( 11 ulphat ohnionell Larutan hprn.m(il si1lfa1

C HHSe/II

(a) State all the ions present in the copper(II) sulphate solution.Cu2+ , SQ42-, ff+ , OH-

State - Boleh bagi nama atau formula......................................................................................................................... [lM]

(b) State the negative terminal in cell I.Magnesium

Sebab - Mg higher than Cu in ECS......................................................................................................................... [lM]

(c) After twenty minutes,

(i) State the observation at magnesium electrode in cell I.Magnesium became thinner

......................................................................................................................... [lM]

(ii) Write half equations for the reaction occurred at the magnesium and copper electrodes in cell I.

Magnesium electrode : Mg 7 Mg2+ + 2e

Copper electrode : cu2+ + 2e 7 Cu [2M]

(d) State the change in colour of copper(II) sulphate solution in cell I and cell II.

Cell I : Blue CuS04 turn to colourless



Cell II : Blue CuS04 unchanged I I remain blue [2M]

(e) A metal Z is found containing some impurities. Z is located below copper in the electrochemical series.

(i) State the method used to purify the metal Z.Purification of metal

......................................................................................................................... [lM]

(ii) Draw a labelled diagram for the apparatus set-up for 6(e)(i).

------11+-

Impure Z i,----- Pure Z

----Z .soJutio.D

Lukis dan berfungsi - 1 MarkahLabel - 1 Markah

[3M]

Chapter 12 - RedOX - Reduction and Oxidation ChemQuest 2015

Adura Azlin Ishak I Update O 1 April 2015 I Since 2007www.cikguadura.wordpress.com 7

__-.-,,,•

-i-">lr

--=t

[SPM03-05] Diagram 5 shows the setup of apparatus to investigate the reactions that take place in test tu bes P and Q.

Displacement Reaction - of metal

Magnesium Dissolve/ thinner Mg7 Mg2+ + 2e Oxidation Reducing agent Oto+ 2

Copperf Il) sulphate -- solution

Magnesiumribbon -- .

~ Brominewater

0

Ferumtl l)--- sulphate

solution

Conversion of iron(II) to iron(III)

Bromine water Br2 +2e7 2Br• O to -1ReductionOxidising agent

Ferum(II) sulphateFe2+ "'7 Fe3+ + e

Copper(II)sulphate

Test tube P Test tube Q

Diagram 5

Oxidation Redcuting agent Green "'7 brown

( a) State the observation for the reaction

(i) is test tube P. [ lM]Brown solid is formed.//Blue solution turns colourless.

(ii) is test tube Q. [ lM]Green solution turns brown.//Brown colour of bromine decolorises.

(b) Write the ionic equation for the reaction in (a)(i).[lM]Mg + Cu 2+ --+ Mg2+ + Cu

(c) State what is meant by oxidizing agent in terms of electron transfer. [lM]A substance that receives electron.

(d) referring to the reaction that takes place in test tube p.

(i) What is the change in the oxidation number of magnesium? [lM]Oto +2

(ii) State the oxidation number of bromine in bromine water. [lM]0

(iii) What is the function of bromine water? [lM]oxidising agent

(iv) Name another reagent that can replace bromine water. [lM]chlorine water

--~

___.,.._



[MRSM14-06] Diagram 6 shows the apparatus set- up to investigate the reaction between iron(II) sulphate solution and bromine water through the transfer of electrons at a distance.

Iron(II) sulphate Fe 2+ "'7 F3+ + e Green "'7 brown Oxidation Reducing agent

rbon electrode Lktrod karbon L

Iron(II) sulphate solutionLannan ferum(II) - sulfat

Dilute sulphuric acid _Asid sulfurik cair

Bromine water Br2 +2e "'7 2Br• O to -1Reduction

Carbon electrode Oxidising agentElektrod karbon M

Bromine water Airbromin

(a) What is the function of dilute sulphuric acid? To allow the movement of ions complete the circuit.......................................................................................................................... [lM]

(b) Show the direction of electron flow in Diagram 6. [ 1 M] Shown in the diagram (from electrode L to electrode M through external circuit).[r: answers in words]

(c) Referring to the reaction that take place at electrode L:(i) Name the product formed.Iron(III)ion / Ferum(III) ion / Iron(III)sulphate[r: formula].......................................................................................................................... [lM]

(ii) Describe a chemical test to determine the product formed in (c)(i). Alternative 11. Add sodium hydroxide/ammonia solution.2. Brown precipitate is formed.

Alternative 21. Add potassium hexacyanoferrate(II) solution.2. Dark blue precipitate is formed.

Alternative 31. Add potassium thiocyanate solution.2. Blood red colouration is formed.

·······································································································[·2·M··]·······

(d) Write the half equation for the reaction that takes place at electrode M. Br2 + 2e ~ 2Br-· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · [lM]

(e) The experiment is repeated by replacing bromine water with acidified potassium dichromate(VI) solution. Predict the observation at electrode Mand explain your answer.1. Orange solution turns green2. Cr2012- ion undergoes reduction to form Cr3+ ion

l-i-oiiiiiiiiiiiapiiiiiiiiiiiiiiiiiiii c

swlplhate solution



(f) Without using U Tube, draw another apparatus set up to investigate the transfer of electron at a distance, using the same materials as in Diagram 6. Mark in the diagram the positive and negative terminal of the cell.

arbon electrode

lmn(II) ·· orol!l.s po

Bromine wafer

suis Carbon

Iron(II) sulphate solution

Functional apparatusLabel apparatus and materialLabel positive and negative terminal

[3M]

e

SA

----r



[MRSM09-06] ( a) Diagram 6.1 shows the apparatus set up to investigate the transfer of electrons at a distance between potassium iodide solution and acidified potassium manganate(VII) solution. After a few minutes, colourless solution turns brown at electrode R.

Or---"'"·Potassium iodide

KMn04

Potassium icdidKali1.1m iodida

Acidified potassium man.ganate (YKalium manganal{Vli) berasid

ulphuric addstd suifank

(i) Name the product formed at electrode R. [lM]Iodine (solution)

(ii) Complete the half equation for the reaction at electrode S. [lM]...... H+ +

.. 8... H+ +..... e --+ Mn2+ + H20

.. 5.. e ----+ Mn2+ + .4 .. H20

(iii) State the change in oxidation number of manganese and name the process that occurs atS. [2M]

Change in oxidation number : +7 to +2

Name of process : reduction

(iv) Suggest a substance that can replace potassium iodide solution in order to obtain the same reaction. [lM]sodium iodide

(b) Diagram 6.2 shows the setup of the apparatus to investigate the reactivity of metals J, Kand L. The different metals are heated consecutively.

Metal powder Glass woolSerbuk logam Kapas kaca

Solid potassium manganat (VII)Pepejal kalium manganat (VII)

He at stronglyPanaskan kuat

HeatPanaskan

Table 6.1 shows the observation of the experiment.

MetalNameMetal

ObservationsColour of residue

Hot ColdJ Zinc Burns brightly - 2 Yellow WhiteK Copper Glows dimly -3 Black BlackL Mg Burns with a very bright flame -1 White White

Table 6.1



(i) Name metal J. [lM]Zinc

(ii) Write a chemical equation for the reaction between metal J and oxygen. [lM]2 Zn + 02 ~ 2 ZnO

(iii) Based on the observation in Table 6.1, arrange metals J, Kand Lin ascending order of reactivity towards oxygen. [lM]K,J,L

iv) A mixture of metal J and oxide of metal Lis heated strongly. Predict an observation and explain your answer. [2M]No changes.Because Metal Lis more reactive with oxygen than metal J



[SPM 14-05] Table 3 shows the information for four sets of experiment to construct the reactivity series of metals.

Set Reactants Observation Reaction Happen positionI Carbon + Iron(III) oxide Grey solid is formed yes C> feII Carbon + Oxide of X Brown solid is formed yes C>X

III Carbon + Magnesium oxide

No change No C< Mg

IV X + Iron(III) oxide No change X< Fe

(a) Set I is a redox reaction.Table 3

(i) What is the meaning of redox reaction?reduction and oxidation happen same time........................................................................................................................ [lM]

(ii) Write the chemical equation for the reaction.3C + 2Fe203 7 3C02 + 4Fe........................................................................................................................ [2M]

(iii) State the change in the oxidation number of iron.+3 to O........................................................................................................................ [lM]

(iv) Which substance undergoes reduction? Iron(III) oxide........................................................................................................................ [lM]

(b) Based on set II and set III, explain the difference in the observations. In set II, carbon is more reactive than metal XIn set III, carbon is less reactive than Magnesium........................................................................................................................ [2M]

(c) (i) Arrange X, carbon, magnesium and iron in descending order of reactivity. Mg, C, Fe, X........................................................................................................................ [lM]

(ii) Suggest X. Copper........................................................................................................................ [lM]

(d) Draw a labelled diagram for the apparatus set-up used in set II.Carbon powder+oxide of motal x

serbuk karbon +logamoksida X

farapor,uinigdi I,

mangkuk pijar

[2M]



[MRSMlO-OSc] (c) Diagram 5.3 shows the set-up of apparatus for an experiment to compare the reactivity of reactions between metal oxides and hydrogen gas.

Metal oxide powderSerbuk oksida logam

Dry hydrogen gaGas hidrogen kering

Table 5 shows the result of the experiment:

Experiment Result Metal

Hydrogen + oxide of metal J Metal oxide powder glows brightly.Black powder turned brown.

Copper

Hydrogen + oxide of metal T No reaction. Powder turns yellow when hot and white when cold.

Zinc

Hydrogen+ magnesium oxide No reaction.White powder remained. Magnesium

(i) Suggest a name for metal T. [lM]Zinc

(ii) Arrange the reactivity of J, T, Magnesium and Hydrogen in ascending order. [lM]J, H, T, Mg

(iii) Based on the observations, explain how you obtain the arrangement in (c)(ii).[3M]1. Metal J lowest in arrangement because metal J can displace from it Metal Oxide by hydrogen2. Metal T and Metal Mg is higher than Hydrogen, because Hydrogen cannot react with their metal oxide.3. Mg is higher because, metal Tis Zn, it less reactive than Mg.

-



[SPM07-06] Iron is metal that rust easily.

(a)(i)State the condition for the rusting of iron. [lM]water and oxygen

(ii)Draw a labelled diagram to show the conditions for the rusting of iron involve the ionization of iron and the flow o! el~ctron. [3 M]

Water

Cathode:Anode:

(b)(i)Describe the reactions that take place at the edge of water droplet (positive terminal)during the rusting of iron after the Fe2+ and OH- ions are formed. [3M]Fe2+ and OH- ions combine to form iron(II) hydroxide. Iron(II) hydroxide is oxidised to iron(III)hydroxide.Iron(III)hydroxide form hydrated iron(III) oxide/ rust.

(ii)State the change in the oxidation number of iron in 6(b)(i). [1 M]+2 -----+ +3

(c) Diagram 6 shows the use of zinc plates on an iron ship to prevent rusting.

--------------------------

Diagram 6

Iron shipK.apal' besi

Zinc pleteK.epmgcm zi1;1t

(i) Explain how the zinc plates protect the iron ship from rusting. [2 M]Zinc is more electropositive than iron.Zinc atoms lose electrons more easily than iron. Zinc corrodes but iron does not.

(ii) Write the half equation for the reaction in 6(c)(i). [lM]Zn -----+ Zn2+ + 2e

Seminar SKORA+ SPM 2015 - Kimia Bersama Cikgu Adura


[SBPTrial14-06] Experiments I, II and III are carried out to investigate the factors affecting the rate of reaction. Table 6.1 shows the reactants and temperature used in each experiment.Eksperimeri I, II dan III dijalankan untuk mengkajifaktor-faktor yang mempengaruhi kadar tindak balas. Jadual 6.1 menunjukkan bahan tindak balas dan suhu yangdigunakan dalam setiap eksperimen. Experiment

ExperimentEksperimeri

ReactantsBahan tindak balas

Temperature (0C) Suhu (0C)

IExcess zinc powder + 25 cm-' of 0.1 mol dm-3hydrochloric acidSerbuk zink berlebihan + 25 ems asid hidroklorik: 0.1 mol dm=

30

II

Excess zinc powder + 25 cm., of 0.1 mol dm-3hydrochloric acidSerbuk zink berlebihan + 25 ems asid hidroklorik: 0.1mol dm=

40

IIIExcess zinc powder + 25 cm-' of 0.1 mol dm-3 sulphuric acidSerbuk zink berlebihan + 25 cm3 asid sulfurik: 0.1 mol dm=

30

Table 6.1 / Jadual 6.1

a) Write the ionic equation for the reaction in Experiment I. [2M]Tuliskan persamaan ion untuk tindak balas dalam Eksperimeri I.

Tulis <lulu persamaan penuh

Zn + 2HC1 Zn Cb +

Kemudian, potong unsur yang tidak berubah,

Zn + 2H+ zn2+ +

b) Based on the experiments, state two factors that affect the rate of reaction. [2M] Merujuk kepada eksperimen, nyatakan dua faktor yang mempengaruhi kadar tindak balas.

Exp 1 dan 3 : Kepekatan/ I concentrationExp 1 dan 2: suhu // temperature



(c) Compare the rate of reaction between Experiment I and II. Explain the difference by using collision theory. [4M]Bandingkan kadar tindak: balas antara Eksperimen I dan II. Terangkan perbezaan itu dengan menggunakan teori perlanggaran.

Mesti bagi seperti ini

FaktorApa factor tu buatFOC di antara apa dengan apa bertambah/meningkat FOEC di antara apa dengan apa bertambah/meningkat Kesimpulan - ROR bertambah

1. Experiment II has higher temperature than experiment 12. The particles in experiment II has high kinetic energy3. Frequency of collision between Zn and H+ ion increases4. Frequency of effective collision between Zn and H+ ion increases5. Rate of reaction experiment II is higher than experiment I

1. Eksperimen II mempunyai suhu yang tinggi daripada eksperimen 12. zarah-zarah dalam eksperimen II mempunyai tenaga kinetic yang tinggi3. Bilangan perlanggaran di antara Zn dan ion H+ bertambah4. Bilangan perlanggaran berkesan di antara Zn dan ion H+ bertambah5. Kadar tindak balas eksperimen II lebih tinggi daripada eksperimen I

(d) Diagram 6.2 shows the curve of the graph of total volume against time for ExperimentI. Sketch the curve obtained for Experiment III on the same axes.Rajah 6.2 menunjukkan garis lengkung bagi graf jumlah isi padu gas melawan masa bagi Eksperimen I. Lakarkan garis lengkung yang diperolehi bagi Eksperimen IIIpada paksi yang sama.



Volume of gas ( crn3)

III

I

Time (s)

• Correct curve which shows the volume is double[lM]

e) During a master chef competition, an apprentice found that a piece of meat is still not tender after cooking for one hour.Semasa satu pertandingan 'master chef', seorang pelatih mendapati ketulan daging yangdimasak masih tidak lembut selepas satu jam.



State one method that should be taken to make the meat become tender in a shorter time. Explain you answer. [2M]Nyatakan satu kaedah yang boleh diambil supaya daging itu menjadi lembut dalam masa lebih singkat. Terangkan jawapan anda.

1. Cut the meat into smaller size I I potong daging kepada ketulan kecil2. Larger total surface area of meat will absorb more heat

OR

1. Cook in pressure cooker2. High pressure in pressure cooker increase the temperature

,._



[SPM 14-06] Diagram 5.1 shows the graph of the mass of zinc against time for the reaction between zinc and hydrochloric acid. In this experiment, 5.00 g of zinc is added to 100 cm-' of 1.0 mol dm-3hydrochloric acid to study the rate of reaction at the temperature of 30 °C.Rajah 5.1 menunjukkan graf jisim zink melawan masa bagi tindak: balas antara zink danasidhidroklorik: Dalam eksperimeri ini, 5. 00 g zink ditambahkan kepada 1 00 ems asidhidroklorik: 1. 0 mol dm= untuk mengkaji kadar tindak: balas pada suhu 30 °C.

Mass of Zinc (g) / Jisim. zink (g)5·0

The L~l"\'C at the temperature of30 "'CLengk1111g pada suh,~ 30 "C

t,

Dia.g1·M.115. IRtyah -.1

'firne(s)M"asa(s)

(a) Write the chemical equation for the reaction. [2M]Tulis persamaan kimia bagi tindak: balas ini.

Zn + HCl 7 ZnCh +

(b) Based on Diagram 5.1, / Berdasarkan Rajah 5.1,

(i) Why is the curve in the graph remains constant after t, second? [lM]mengapakah lengkung bagi graf kekal mendatar selepas t, saat?

All hydrochloric acid was reacted completely with ZnSemua asid hidroklorik telah bertindak balas lengkap dengan Zn

(ii) determine the mass of unreacted zinc in the experiment. [lM]tentukan jisim zink yang tidak: bertindak: balas dalam eksperimeri itu.

---The c urvLe11gkmig



Kira semua mol: Zn dan HCl

Mol Zn= 5/65 = 0.077

Mol HCl = MV/1000 = 1.0 X 100/1000 = 0.1

Used Zn: Used HCl

1 mol Zn bertindak balas dengan 2 mol HCl0.077 0.077 X 2/1 = 0.1540.1 x 1/2 = 0.05 0.1

Mass zinc reacted = mol X mass = 0.05 X 65 = 3.25 gMass zinc not reacted= 5 - 3.25 = 1.75 g

Atau

Daripada graf,

at the tern p,;;:ml\Ln: of3Q ~p.ad1 ulru 30 ac

Diagram:5.1Ro.juh 5.1

imc( Masa(s)

1.75 g

(c) In this experiment, the rate of reaction can also be determined by measuring the volume of hydrogen gas produced at regular intervals of time. Draw the apparatus set-up for the experiment.Dalam eksperimeri ini, kadar tindak: balas boleh ditentukan dengan mengukur isi padn gas hidrogen yang dihasilkan pada sela masa yang tetap. Lukis rajah susunan radas untuk eksperimeri ini.

id

----The cLeng

\\\\

'-.. "-



h drochlori -

Zinc powder

(d) The experiment is repeated at the temperature of 40 °C with other factors remain unchanged.Eksperimeri diulangi pada suhn 40 °C dengan semua faktor lain kekal tidak berubah.

(i) Sketch the curve obtained for this experiment on the same axis in Diagram 5.2. Lakarkan lengkungyang diperoleh dalam eksperimen pada paksiyang sama dalam

Mass of Zinc (g)Jisim. zink (g)

5·00

[2M]

ur e at the temperature of 30 °I kung' pada uhu 30 °C

l ·7

Diagram 5.2Rajah 5.2

ime()Ma a()

[lM](ii) Based on your answer in 6(d)(i),explain how temperature affects the rate of reaction by using collision theory. [3M]Berdasarkan jawapan anda di 6(d)(i),terangkan bagaimana suhu mempengaruhi kadar tindak balas dengan menggunakan teori perlanggaran.



1. The higher temperature, the higher kinetic energy of particles2. the frequency of collision between Zn and Hydrogen ion increases3. The frequency of effective collision between Zn and Hydrogen ion increases4. rate of reaction is higher

1. Suhu tinggi, menyebabkan kandungan tenaga kinetic zarah bertambah2. Bilangan perlanggaran di antara Zn dan ion H+ bertambah3. Bilangan perlanggaran berkesan di antara Zn dan ion H+ bertambah4. Kadar tindak balas lebih tinggi

(e) Apart from temperature, state one other factor that will also affect the rate of reaction in this experiment. [lM]Selain daripada suhu, nyatakan satu faktor lain yang juga boleh mempengaruhi kadar tindak balas dalam eksperimeri ini.

Size of Zinc/ I concentration of HCl I I Add catalyst (CuS04)Saiz zink/ / kepekatan HCl / / penambahan mangkin

Seminar SKORA+ SPM 2015 @Jr[MUCy'berjaya - Kimia Bersama Cikgu Adura


1

[SBPtrial04-05] One experiment to determine heat of precipitate of lead(II) sulphate was done by mix 25 cm-' of 0.5 mol dm-3 lead(II) nitrate solution and 25 cm-' of 0.5 mol dm-1

V M V Mpotassium sulphate solution. The result of the experiment in the table below:

Initial of temperature of lead(II) nitrate = 28.0 °C Initial of temperature of potassium sulphate = 29.0 °C The higher temperature of mixture = 33.0 °C

(a) What mean by the heat of precipitate of that reaction. [lM]Heat change when 1 mol of precipitate /lead(II) sulphate formed from its ion

(b) Write the ionic equation for the reaction. [lM] Tunjuk <luluPb(N03)2+ K2S04 7 PbS04 + 2KN03

Solid

(c) Calculate the number of mol of lead(II) ions and sulphate ions that exist in every solution. [2M]

(i) Lead(II) ions (II) Sulphate ionsMol sulphate ion = mol K2S04

Mol lead(II) ion= mol Pb(N03)2= MV/1000= 0.5 X 25/ 1000= 0.0125 mol

= MV/1000= 0.5 X 25/ 1000= 0.0125 mol

(d) Calculate the change of heat in the experiment. [2M] [specific heat capacity of the solution=4.2 Jg-1°c-1J

Q=mcO= [25 + 25] x 4.2 X [33.0 - (28.0 + 29.0)/2]= 50 X 4.2 X 4.5= 945J= 0.945 kJ

(e) Calculate the heat of precipitate of this reaction. [2M]

i::1H = Q/ mole, n= 0.945/ 0.0125= - 75.6 kJ mol-1

(f) Draw the energy level diagram for the reaction. [2M]Energy

L'.1H = - 75.6 kJ m olPbSQ4



[SBPTrial13-04] Diagram 4 shows the apparatus set-up for an experiment to determine the heat of displacement.Rajah 4 menunjukkan susunan radas bagi satu eksperimeri untuk menentukan haba penyesaran.

Polystyrene cupCmvan polistirena

Excess zinc powderSerbuk zink berlebihan

Based on the experiment, Berdasarkan eksperimen di atas,

O cm3 of 0.2 mol chu-3copper (II) sulphate50 cm'kuprunull) sulfat 0.2 mo/ dm"

Diagram 4 Rajah 4

(a) State the meaning of heat of displacement.Nyatakan maksud haba penyesaran.Heat change when 1 mol of copper was displace from opper(II) sulphate by zinc that more electropositive metal

(b) Give one reason why polystyrene cup is used in the experiment.Berikari satu sebab mengapa cawan polistirena digunakan dalam eksperimeri ini.Polystyrene is a heat insulator I I to reduce heat loss to surroundings

[lM]

(c) The thermochemical equation below represents the displacement reaction.Persamaan termokimia di bawah mewakili tindak balas penyesaran itu.

Zn + CuS04 7 ZnS04 + Cu ; 11H = - 210 kJ mol-!

Calculate / Hitung:

(i) the number of moles of copper(II) ion.bilangan mol ion kuprum.

No of moles of Copper(II) ion= 0.2 x 50 I I 0.01 mol1000

[lM]

(ii) the heat released during the reaction.haba yang dibebaskan semasa tindak balas.

Q = 210 000 x 0.01= 2100 J

1. 1 mol of Cu is displaced produce 210 kJ heat

2. 0.01 mol of Cu = 0.01 x 210 kJI I 2.1 kJ I 2100 J

[lM]

[2M]



(iii) the change of temperature in the experiment.[Specific heat capacity of solution= 4.2 J g-1"C:"; Density of solution= 1 gcm-"]Perubahan suhu dalam eksperimeri ini.[Muatan haba tentu larutan = 4.2 J g-1°c-1; Ketumpatan larutan =1 qcm:"]

Q =mcO

0 = Q/mc = 2100/[50 x 4.2] = 10 C

2100 J = so x 4.2 x e 11 e = 10 °c

[lM]

(d) The experiment is repeated using magnesium powder to replace zinc powder. The volume and concentration of copper (II) sulphate used is remained the same.Eksperimeri diulang dengan menggunakan serbuk magnesium menggantikan serbuk zink. Isi padu dan larutan kuprum(II) sulfat yang digunakan adalah sama.

(i) Predict the heat of displacement for the reaction. Ramalkan haba penyesaran bagi tindak balas itu. More than -210 kJ mol-1 I Higher I Increases

[lM]

(ii) Give a reason for your answer in 4(d)(i).Ben sebab bagi jawapan di 4(d(i).Magnesium is more electropositive than zinc I I magnesium is higher than zinc in electrochemical series I I distance between Mg - Cu is further than Zn-Cu inelectrochemical series

(e) Draw the energy level diagram for the reaction.Lukis gambar rajah aras tenaga bagi tindak balas ini.

Energy

[lM]

Zn + CuS04 IZn-Cu2+

Aff = - 10 kJ mor1

ZnSO .. + Cu/Zn2+ + Cu

1. Label energy and diagram has 2 different energy levels for exothermic reaction2. Balanced chemical I ionic equation, LiH is written



[SBPTriallS-06] Table 6 shows the heats of combustion of some common fuels.Jadual 6 menunjukkan haba pembakaran beberapa bahan api yang biasa digunakan.

Fuel / Bahan api Heat of combustion (kJ moli] Haba pembakaran (kJ mol-1)

Methane / Metana -890Propane / Propana -2 230Ethanol / Etanol -1 376Propanol / Propanol -2 016

Table 6 / Jadual 6

(a) The combustion of the fuels is an exothermic reaction. What is meant by exothermic reaction?Pembakaran bahan api adalah tindak balas eksotermik. Apakah yang dimaksudkan dengantindak balas eksotermik? [ 1 M]Reaction that gives out I released heat to the surroundings.

(b) Diagram 6 shows the energy profile for the combustion of ethanol. Rajah 6 menunjukkan profil tenaga bagi pembakaran etanol.

Mark i::1H for the reaction in Diagram 6.Diagram 6 / Rajah 6

Tandakan LJH bagi tindak balas itu dalam Rajah 6. [IM}

(c) (i)Compare the heat of combustion of methane and propane. [lM] Bandingkan haba pembakaran metana dan propana.Heat of combustion of propane is higher than methane

(ii) Explain your answer in (c)(i)./ Terangkanjawapan anda di (c)(i) [3M]1. The number of carbon atom per molecule propane is higher2. The More carbon dioxide I water produced when propane is burnt3. More heat energy released

(d) Calculate the fuel value of propanol. / Hitungkan nilai bahan api bagi propanol. [2M] [Molar mass of propanol, C3H70H = 60 g mol-1J[Jisim Molarpropanol, C3H70H = 60 g mol-1]Molar mass of propanol, C3H70H = 60 g mol-160 g of C3H70H burnt released 2016 kJ //1 g C3H70H burnt released 1 x 2016 kJ60= 33.6 kJ g-1(correct answer with correct unit)



(e) During a football game, a player found that his knee was swollen after being hit by the opponent player.Semasa perlawanan bola sepak, seorang pemain mendapati lututnya bengkak selepas berlanggardengan pemain lawan.

A physiotherapy put ice cubes on his knee to relieve the pain. As a chemistry student, suggest another method to help the player. Explain how the method you choose will help the player. [3M]Seorang ahli fisioterapi meletakkan ketulan ais pada lutut pemain itu untukmengurangkan kesakitan. Sebagai seorang pelajar kimia, cadangkan kaedah lain untuk membantu pemain itu. Terangkan bagaimana kaedah yang dipilih dapat membantu pemain itu.

1. place the cold packs on his swollen knee2. to absorbs heat from his swollen knee3. constrict blood vessels and slows down blood flow I reduce the formation of fluid in the affected area.



SPMl0-01 Rajah 1 menunjukkanperubahan keadaan tiqa jirim, X, Y dan Z bagi air. Diagram 1 shows the inter-conversion of the three states of matter, X, Y and Z of water.

0 000

0x y z

SolidPepejal

Freezing condesationliquid gascecair gas

Melting boiling

(a) Apakahjenis zarah yang terdapat dalam air? What type of particle is found is water?

Ada 3 jenis zarah

Iaitu

Atom, ion, molekul

Molekul/ I molecule........................................................................................................................ [lM]

(b) Di bawah suhu bilik; pada suhu berapakah ais berubah kepada air? Under the room temperature, at what temperature does ice change to water?

Jawapan MESTI bersama dengan unit yang BETUL

0 °C........................................................................................................................ [lM]



(c) Apakah keadaanfizikal yang berlabel ZWhat is the physical state labelled Z?

Keadaan fizikal

Adalah

Pepejal, cecair dan gas

gas........................................................................................................................ [lM]

(d) Namakan proses apabila air berubah daripada keadaan X kepada keadaan Y.Name the process when water changes from state X to state Y.

X (Pepejal)kepada Y (cecair)

Takat lebur/melting point

SAL AH

Sebab soalan tanya proses

Peleburan/ /Melting........................................................................................................................ [lM]

(e) Apabila air berubah daripada keadaan Y kepada keadaan Z, nyatakan perubahan bagi:When water changes from state Y to state Z, state the changes in:

Y (cecair) kepada Z (Gas)



(i) Tenaga zarahthe energy of the particles.

Jawapan MESTI menunjukkan Perubahan

High

Reject: Strong [Salah term]High [tidak menunjukkan perubahan

Bertambah/ I Increases/ I Higher........................................................................................................................ [lM]

(ii) Daya tarikari antara zarah-zarah ituThe forces of attraction between the particles.

Soalan menyatakan air,

Sebatian kovalen

Daya tarikan Van der Waals - antara molekul dengan molekul

Strong

Reject : tidak menunjukkan perubahan

Berkurang I weaker........................................................................................................................ [lM]



(f) Bau masakan kari di dapur merebak hingga ke ruang tamu.Berdasarkan teori kinetik jirim, nyatakan proses yang terlibat. Terangkan jawapan anda. The smell of curry cooking in the kitchen spreads to the living room. Based on the kinetic theory of matter, state the process involved. Explain your answer.

Soalan nak Proses yang terlibat

Kemudian terangkan.

Proses : Resapan/ I diffusion

Penerangan / / explanation1. The curry particles is tiny and discreteZarah kari adalah kecil dan diskrit.

2. move in between air particle from high concentration to lower concentrationbergerak di antara zarah-zarah udara dari kepekatan tinggi ke kepekatan rendah........................................................................................................................ [3M]

Sm

~--



SBP13-01 (a) Diagram 1 shows the set-up of the apparatus an experiment to studyProcess I.Rajah 1 menunjukkan susunan radas bagi satu eksperimeri bagi mengkaji Proses I.

Solid agarAgar-agar peja 1

olid potassium anganate(VII)

Pepejal kalium mangana t(VII)

(i) Name the process involved in this experiment?Namakan proses yang terlibat?

Resapan/ I diffusion

-Whole agar turns purple. Seluruh agar menjadi ungu

[lM]

(ii) State the type of particles present in potassium manganate(VII).Nyatakan jenis zarah dalam kalium manganat(VII).

Potassium manganate(VII) adalah sebatian

Ionik , maka zarah yang hadir ialah

ion[lM]

(iii) Explain the observation in this experiment based on the kinetic theory of matter.Terangkan pemerhatian dalam eksperimeri ini berdasarkan teori kinetik jirim.

Berdasarkan teori kinetik jirim

1. The solid potassium manganate particles is tiny and discreteZarah pepejal kalium manganate adalah kecil dan diskrit.

2. The particles are constantly moving/vibrate and rotateZarah sentiasa bergerak/bergetar dan berputar

3. move in between agar particle from high concentration to lower concentrationbergerak di antara zarah-zarah agar-agar dari kepekatan tinggi ke kepekatan rendah

proton+ n

c c c



SPM14-03 (a) Diagram 3 shows standard representation for three isotopes of carbon which are carbon-12, carbon-13 and carbon-14.Rajah 3 menunjukkan peruiakilari piawai bagi tiqa isotop karbon iaitu karbon-12, karbon-13 dan karbon-14

12 13 14

6 6 6

(i) What is the meaning of isotope?Apakah yang dimaksudkan dengan isotop?

Atom of same element,Atom unsur yang sama

Has same number of proton/ I proton numberMempunyai bilangan proton yang sama/ / nombor proton

Different number of neutron or nucleon numberBerbeza bilangan neutron// nombor nucleon

Reject: number of nucleon// neutron number[lM]

(ii) Determine the number of neutrons in carbon-13.Tentukan bilangan neutron dalam karbon- . 3

Nucleon number=

Neutron = n eon number - proton= 13 - 6

7[lM]

(iii) State one use of carbon-14 in daily life.Nyatakan satu kegunaan karbon-14 dalam kehidupan harian

To estimate age of fossilMenentukan tarikh umur fosil

[lM]



(iv) Draw the atomic structure of carbon-12 and label the subatomic particles.Lukis struktur atom karbon-12 dan label zarah subatomnya.

Biasa kita melukis

Susunan electron

Soalan nak sekarang adalah

Atomic structure

Mesti melukis adanya nucleus yang dikelilingi oleh elektron

[2M]

(b) Table 2 shows the physical properties of substance X and substance Y.Jadual 2 menunjukkan sifat fizik bagi bahan X dan bahan Y.

SubstanceBahan

Melting point (0C)Takat lebur ( 0C)

Boiling point (0C)Takat didih ( 0C)

Electrical conductivityKekonduksian elektrik

Solid /Pepejal Molten / Leburan

x -23 77 CannotTidak boleh

Cannot/Tidak boleh

y 801 1413Cannot

Tidak bolehCan

Boleh

0C)



Based on Table 2,Berdasarkan Jadual 2,

solid Solid + liquid liquid Liquid+ gas gas

Melting point Boiling point

Substance (OC)Bahan

(0C) Takat Takat didih (lebur ( oC}

x -23y 801

(i) what are the physical states of substance X and substance Y at room temperature?apakah keadaan fizik bagi bahan X dan bahan Y pada suhu bilik?

X: liquid/ I cecair

Y: pepejal/ /solid [2M]

(ii) Explain the difference in melting point of substance X and subtance Y.Terangkan perbezaan takat lebur bagi bahan X dan bahan Y

SubstanceBahan

Electrical conductivityKekonduksian elektrik

CompoundSebatian

Solid /Pepejal Molten / Leburan

x CannotTidak boleh

Cannot/Tidak boleh

kovalen

y CannotTidak boleh

CanBoleh

ionik

1. Melting point Y higher than XTakat lebur Y lebih tinggi daripada X

2. Compound Y has strong electrostatic force compare to X that has weak Van derWaals forcesSebatian Y mempunyai daya tarikan elektrostatik yang kuat berbanding dengan X yang mempunyai daya tarikan Van der Waals yang lemah.

3. more heat energy needed by Y to overcome the attractive force between its particlesLebih banyak tenaga haba yang diperlukan oleh Y untuk memutuskan ikatan di zarah•zarahnya........................................................................................................................ [2M]



(iii) state the type of particles in substance X.Nyatakan jenis zarah dalam bahan X.

X ialah sebatian kovalen,

Maka,

molecule[lM]



SPM2013-01 (a) Table 1 shows four substances and their chemical formulae. Jadual 1 menunjukkan empat bahan dan formula kimianya.

Substance /Bahan Chemical formula /Formula KimiaArgon / Argon ArBromine / Bromin Br2Napthalene /Naftalena C10HsSodium chloride /Natrium klorida NaCl

Based on table 1: Berdasarkanjadual 1:

Substance /Bahan Chemical formula /Formula Kimia ZarahnyaArgon / Argon Ar atomBromine / Bromin Br2 MolekulNapthalene /Naftalena C10Hs MolekulSodium chloride /Natrium klorida

NaCl 10n

(i) State one substance which exists as atom. Nyatakan satu bahan yang wujud sebagai atom.

Argon[lM]

(ii) which substances has the highest melting point?Bahan manakah mempunyai takat lebur yang paling tinggi?

Sebatian ion

Sodium chloride /Natrium klorida[lM]

(iii) What is the physical state of bromine at room conditions? Apakah keadaan fizik bromin pada keadaan bilik?

Sebatian kovalen

Cecair [Ikut buku teks F4 - F & Cl - gas, Br - cecair, I- solid][lM]



(iv) Sodium chloride cannot conduct electricity in solid state but can conduct electricity in aqueous solution. Explain why.Natrium klorida tidak boleh mengkonduksi elektrik dalam keadaan pepejal tetapi boleh mengkonduksi elektrik dalam larutan akues. Terangkan mengapa.

1. Sodium chloride in solid, its ion not free to move. Hold by strong electrostatic force.Natrium klorida di dalam pepejal, tiada ion-ion yang bebas bergerak. Di tarik oleh daya elektrostatik yang kuat2. in liquid, its ions is free to moveDi dalam cecair, ion-ionnya bebas bergerak.

[2M]

(v) Why argon is an unreactive element?Mengapakah argon adalah unsur yang tidak reaktif? [lM]

Group 18

Already achieve octet electron arrangementTelah mencapai susunan electron oktet

[lM]

(b) Diagram 1 shows the graph of temperature againts time when liquid naphthalene is cooled.Rajah 1 menunjukkan graf suhu melawan masa apabila cecair naftalene disejukkan.

Temperature (°C)

Suhu (0C)

R Q

p

Time(s)Masa (s)

Berdasarkan graf diatas, 3 soalan lazim yang akan ditanya

1. Takat lebur/ takat beku2. keadaan fizikal pada graf yang malar3. penerangan pada graf yang malar



Based on diagram 1Berdasarkan Rajah 1:

(i) State the freezing point of napthalene. Nyatakan takat beku bagi naftalena. [lM]

[lM]

(ii) Why there is no change in temperature from R to Q? Mengapakah tidak terdapat perubahan suhu dari R ke Q? [lM]

Heat releasedHaba di bebaskan

Was balance by heat released during the formation of bond between the particles Diseimbangkan oleh haba yang dibebaskan semasa pembentukkan ikatan di antara zarah-zarah........................................................................................................................ [lM]

(iii) What are the states of matter from R to Q? Apakah keadaanjirim dari R ke Q? [lM]

Liquid and solid/ I liquid + solid I I cecair dan pepejal

Reject : liquid, solid[lM]



[SBPTrial14-02] (a) (i) One mole of a substance is defined as the quantity of a substance that contains the same number of particles as in n g of element A.What are n and A?Satu mol bahan ditakrifkari sebagai kuantiti bahan yang mengandungi bilangan zarah yang sama seperti yang terdapat dalam n g unsur A.Apakah n dan A?

n = 12 A = carbon-12/ I karbon-12 2M]

(ii) What is the number of atoms in 0.5 mole of methane gas, CH4? [2M]Berapakah bilangan atom dalam 0.5 mol gas methane, CH4?[Avogadro constant= 6.02 x 1023moli] [Pemalar Avogadro= 6.02 x 102s mol:i]

Telah diberikan: 0.5 mol

Soalan nak atom,

Diberikan molekul CH4

Maka

Ada 1 atom C dan 4 atom H

Maka ada 5~ membentuk CH,

= 0.5 x 6.02 x 1023x 5= 1.505 x 1024atom

(b) Diagram 2 shows an experiment to determine the empirical formula of magnesium oxide.Rajah 2 menunjukkan satu eksperimeri untuk menentukan formula empirik: magnesium oksida.



Iagne ium ribbonPita 111ng11 i11111

Diagram 2 / Rajah 2

(i) When carrying out the experiment, why does the crucible lid need to be opened once a while? [lM]Semasa menjalankan eksperimeri itu, mengapakah penutup mangkuk pijar perlu dibukasekali sekala?

To let oxygen enter to continue the combnustionMembenarkan oksigen masuk dan meneruskan pembakaran

(ii) Why this method not suitable to determine the empirical formula of lead(II) oxide?[lM] Mengapakah kaedah ini tidak: sesuai untuk menentukan formula empirik:plumbum(II) oksida?

Bila melibatkan pertukaran oksigen atau pembakaran dengan oksigen

Gunalah Siri Kereaktifan

Lead less reactive with oxygen than MagnesiumPlum bum kurang reaktif dengan oksigen berbanding dengan magnesium

Reject : plumbum(II) oksida

(c) Copper(II)carbonate, CuC03 is heated strongly to produce copper(II) oxide and carbon dioxide gas. The reaction is given by chemical equation below;Kuprum(II) karbonat, CuC03 dipanaskan dengan kuat menghasilkan kuprum(II) oksida dan gas karbon dioksida. Tindakbalas ditunjukkan oleh persamaan kimia di bawah;

6.2 g copper(II) carbonate, CuC03 is heated during an experiment. Calculate the volume of gas released. [3M]6.2 g kuprum(II) karbonat, CuCOs di panaskan dalam suatu eksperimen. Hitungkan isipadu gas yang dibebaskan.

CuCOa = 6.2 = 6.2 =64 + 12 + 16(3) 124



[Relative atomic mass: Cu=64; C=12, 0=16; 1 mole of gas occupies 24 dm3 at room conditions][Jisim. atom relatif: Cu=64; C=12, 0=16; 1 mol gas menempati 24 dm3 pada keadaan bilikj

0.051 mol CuCOa : 1 mol C020.05 mol CuCOa : 0.05 mol C02

Volume of C02 gas = 0.05 x 24 I 1.2 dm3

HeatPana ka«

-«~



SPM14-02 Diagram 2 shows the apparatus set-up to determine the empirical formula for an oxide of copper.Rajah 2 menunjukkan susunan radas untuk menentukan formula empirik: bagi suatu oksida kuprum.

Oxide of copperOb·lda htprum

Hidrogen gasGas hidrogen

.A. nhydrou · ealei am chloride.Kai.· i'urn klorida ko'i"jumg --

Diagram 2 / Rajah 2

(a) State the name of two reactants for the preparation of hydrogen gas. [2M]Nyatakan nama dua bahan tindak balas bagi penyediaan gas hidrogen.

Tindak balas di antara logam dengan asid kuat

Zinc/magnesium/aluminium I I hydrochloric acid/nitric acid/ sulphuric acidZink/magnesium/aluminium / / asid hidroklorik/asid nitrik/ asid sulfurik

(b) What is the function of anhydrous calcium chloride? [lM]Apakahfungsi kalsium klorida kontang?

Absorb water vapour I I to dry hydrogen gasMenyerap wap air// mengeringkan gas hidrogen

(c) Table 1 shows the data obtained from the experiment.Jadual 1 menunjukkan data diperoleh daripada eksperimeri itu.

Description / Penerangan Mass (g) / Jisim (g)Combustion tube+ porcelain dishTiub pembakaran + pirinq porselin

24.60

Combustion tube+ porcelain dish+ oxide of copperTiub pembakaran + pirinq porselin + oksida kuprum

27.00

Combustion tube+ porcelain dish+ copperTiub pembakaran + piring porselin + kuprum

26.52

Table 1 / Jadual 1



Based on Table 1, calculate the empirical formula for the oxide of copper.Berdasarkan Jadual 1, hitung formula empirik bagi oksida kuprum.

[Relative atomic mass: Cu= 64; 0 = 16][Jisim.atom relatif: Cu = 64; 0 = 16]

Buatjadual

Element/unsur Cu 0Mass/ jisim (g) 26.52 - 24.60 = 1.92 27.00 - 26.52 = 0.48Mol 1.92 = 0.03

640.48 = 0.0316

Ratio 0.03= 10.03

0.03= 10.03

Simplest ratio 1 1Empirical formula/ formula empiric= CuO

[3M]

(d) How do you ensure all oxide of copper is reduced to copper?Bagaimanakah anda memastikan semua oksida kuprum diturunkan kepada kuprum?

Repeat heating, cooling and weighing until constant reading achievedUlang panas, sejuk dan timbang sehingga bacaan malar di capai

[lM]

(e)(i)Can the empirical formula for magnesium oxide be determined by using this method?Bolehkah formula empirik bagi magnesium oksida ditentukan dengan menggunakan kaedah ini?

NO[lM]

(ii) Give one reason for your answer in 2(e)(i).Berikan satu alasan bagi jawapan anda di 2(e)(i).

Magnesium more reactive with oxygen than hydrogenMagnesium lebih reaktif dengan oksigen berbanding dengan hidrogen

[lM]

edisi-jwpn-kys-2015

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