pecutan akhir kimia spm 2015

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Rate of ReactionA group of students carried out three experiments to investigate the factors affecting the rate of reaction between hydrochloric acid and zinc. Table shows the results of the experiments.

i) Calculate the average rate of reaction for Experiment I and Experiment II.[2 marks]ii) On the same axis , sketch the graph for the three sets of experiments for the liberation of 40 cm of hydrogen gas [3 marks]iii) Write the ionic equation for the reaction between zinc and hydrochloric acid. [2 marks]iv) Based on Table above, compare the rate of reaction between Experiment I and Experiment II Experiment II and Experiment III Explain the difference in the rate of reaction based on the Collision Theory. [10 marks]

Answeri) Average rate of reaction experiment I = 40/90 = 0.44 cm s Average rate of reaction experiment II = 40/150 = 0.27 cm s


iii) Zn + 2H Zn + H

iv) Experiment I and II Rate of reaction of Experiment I is higher than Experiment IICopper(II) sulphate solution acts as catalyst Which provide an alternative path with lower activation energy More colliding particles able to achieve the activation energy The frequency of effective collision between zinc atom and hydrogen ions is higher in Experiment I than in Experiment II Experiment II and III Rate of reaction in experiment II is higher than Experiment IIIThe concentration of hydrochloric acid in experiment II is higher than in Experiment IIIThe number of hydrogen ions per unit volume is higher in Experiment IIThe frequency of collision between zinc atom and hydrogen ions is higher in Experiment II than Experiment IIIThe frequency of effective collision between zinc atom and hydrogen ions is higher in Experiment II than in Experiment III

Manufactured of Substances in IndustryTable shows three substance, example and their components respectively

Name substance P and suggest how the strength of concrete can be increased to be used as pillars of building. [2 marks] ii) Polymer R is formed through polymerization process. Write the chemical equation to produce the polymer R and give a name for the polymer. [3 marks]iii) Q is one type of alloy. Compare and explain the hardness between alloy Q and its pure metal. [5 marks]

SubstanceExampleComponentPReinforced concretecement, sand, small pebbles and steelAlloyQcopper and zincPolymerRethene

AnswerComposite material - Adding steel rod into the concrete

nC2H4 [C2H4 ] Polyethene

iii) - Alloy Q/brass is harder than its pure metal/copper - the presence of zinc atom in alloy Q disrupts the orderly arrangement of copper atom - These make the atomic layers of atoms harder to slide over on another - in pure metal/copper the atoms are arranged packed closely and in orderly manner. - this allow the layers of atoms are easily to slide one another


Chemical For ConsumersDiagram shows the apparatus used to investigate the cleaning action of cleaning agent X and Y to remove oily stain from the cloth.

Based on Diagram compare and explain the effectiveness of cleaning action between experiment I and II. Identify the cleaning agent X and Y.

Answer- Cleaning agent Y in experiment II is more effective than cleaning agent XCleaning agent Y do not form scum in hard water therefore it can remove oily stain from the clothCleaning agent X in experiment I is not effective in hard water because hard water contain high calcium ion and magnesium ionThese ions will react with cleaning agent X to formed an insoluble precipitate/scum- The formation of scum will reduces the number of cleaning agent A Cleaning agent X is soapCleaning agent Y is detergent

Chemical Bond

By using the element in Diagram above, explain the formation of compound between element R and Q , also formation of compound between element L and T. The two compounds should have different bond types. [10 marks]

AnswerR and Q Electron arrangement of atom R is 2.8.2 Electron arrangement of atom Q is 2.6.Atom R donates 2 electrons to achieve stable octet electron arrangement to form R ionAtom Q receive 2 electrons to achieve stable octet electron arrangement to form Q ionR ion and Q ion are attracted by a strong electrostatic force to form ionic bond

L and T Electron arrangement of atom L is 2.4 Electron arrangement of atom T is 2.8.7 Atom L contribute 4 electron and atom T contribute 1 electron for sharing To achieve stable octet electron arrangement1 atom L share 4 pairs of electron with 4 atom T to form covalent bond

ElectrochemistryDiagram shows two type of cells.

Compare and contrast cell X and cell Y in terms of:Type of cellThe energy changeThe terminals of the cellsIons presence in the electrolyteObservationHalf equation for both electrodesName of the processes occurred at the positive terminal of each cell [ 10 marks]

Cell X Cell Y

AnswerCell XCell YType of cellElectrolytic cellVoltaic cellThe energy changeElectrical energy to chemical energyChemical energy to electrical energyThe terminal of the cellPositive terminal / anode: CopperNegative terminal / cathode: copper Positive terminal / cathode: copperNegative terminal / anode: aluminiumIons present in the electrolyteCu 2+, H+SO4 2- , OH-ObservationAnode: Electrode become thinnerCathode: Brown solid is deposited//thickerNegative terminal/Aluminium plate: Electrode become thinnerPositive terminal/Copper plate:Brown solid is deposited//copper plate become thickerHalf equation for both electrodesAnode:Cu Cu 2+ + 2eCathode:Cu 2+ +2e CuAl plate/- terminal:Al Al 3+ + 3eCu plate//+ terminal:Cu 2+ +2e CuName of the process occurred at both electrodes/terminalAnode/Al plate: OxidationCathode/Copper plate//negative terminal: Reduction

Empirical FormulaMetal X is more reactive than hydrogen.[Relative atomic mass: O = 16 ; X = 24 , ionic formula : X ]

Describe a laboratory experiment to determine the empirical formula of oxide X. Your answer should consist of the following: Procedure of the experiment Calculation involved [ 10 marks ]

AnswerProcedure:A crucible and its lid are weighed 10 cm of cleaned X ribbon is coiled loosely and placed in the crucible. The crucible with its lid and content are weighed again. The crucible is heated strongly without its lid. Using a pair of tongs, the lid is lifted at intervals. When the burning is completed, the lid is removed and the crucible is heated strongly for 2 minutes. The crucible is allowed to cool to room temperature. The crucible and its lid and content are weighed again

AnswerCalculation:Mass of crucible + lid = a g Mass of crucible + lid + magnesium = b g Mass of crucible + lid + X oxide = c g

Empirical formula is XpOq


Diagram 10.2 shows the reactions involving Fe2+ ion and Fe3+ ion

i) State the suitable examples for reagent A and reagent B. ii) Explain the oxidation and reduction in terms of electron transfer in Reaction I.iii) Write the half equation involved in Reaction II and state the observation occurred. [7 marks]

Answeri) A : Bromine/chlorine water // acidified KMnO4 solution B : Zincii) Fe2+ undergoes oxidation reaction because Fe2+ releases electron. Bromine water undergoes reduction because Bromine water gains electroniii) Fe3+ + e Fe2+ Zn Zn2+ + 2e

Observation : Brown solution turns to green