Trial 2010 mara

1 (a) Diagram 1.1 shows the step involved in an industrial process to produce ammonia.

(i) Name the process in the production of ammonia.

Haber

Hydrogen gas

Gas T

Ammonia

Combine to form

Berpadu membentuk

(ii) Name gas T.

Nitrogen

Iii) Write the chemical equation for the reaction between hydrogen and gas T to produce ammonia.

3H2 + N2 2NH3

Types of food additives

Example Uses

USodium benzoate

Food can last longer

Antioxidants VNeutralises free radicalsand reduce risk of cancer

• 1(b)(i) • State the type of food additive U.• preservative• State the example of antioxidant that is

represented by V.• vitamin E/tocopherol

Gliseryl tristearate Glyserol Soap

(i) Name the soap molecules produced.

sodium stearate

• Name the process represented by the equation.

• saponification• Why are the soap molecules not effective in

hard water?• soap ion react with Ca2+ / Mg2+ ion• to produce an insoluble salt / solid / scum

Atom of element

Proton number

Nucleon number

P 8 16

Q 9 19

R 8 17

• Which pair of atoms are isotopes?• P and R• Give the reason for your answer in (a)(i).• Both atoms have same number of protons but

different number of neutrons // Both atoms have same proton number but different nucleon number

Draw the electron arrangement for atom Q.

When metals of Group 1 react with water, bubbles of colourless gas are evolved. Diagram 2 shows the reaction of the first three alkali metals with water in three different beakers labeled as W, X and Y.

X

W X Y

In which beaker shows the reaction of potassium with water?

• Based on Diagram 2, name the gas evolved when metals of

• Group1 react with water.• hydrogen gas• State one observation when a few drops of

phenolphthalein indicator is added into the solution formed in W, X and Y.

• Explain your answer.• Colourless solution turns pink• The reaction produces alkaline solution• a: presence of OH-

• Complete the following chemical equation:• Lengkapkan persamaan kimia berikut:

• …… Na + …… H2O → ………..… + …………..

• Based on Diagram 2, what can you infer about the density of Group 1 metals compared to water?

• density of Group 1 metals are lower than the density of water // less dense than water

Molten lead(II) oxideCarbon F

HeatPanaskanCarbon GCarbon L

Silver nitrate solutionLarutan argentum nitrat

Carbon M

Diagram 3.1Rajah 3.1

Diagram 3.2Rajah 3.2

What is meant by electrolyte?

A compound/substance that can conduct electricity in molten state or aqueous solution and undergoes chemical changes

• Name the product formed at electrode G.• lead• Write the half equation for the reaction that

occurs at electrode F.

• 2 O2- O2 + 4e

• State the ions present in silver nitrate solutio Ag+ , H+ , NO3

- , OH- (a: name of ions)

• n.

• What would you observe at carbon electrode M?

• grey solid/ deposit• Explain your answer in (c)(ii).• Silver is produced// Ag+ is discharged• Because silver ion is lower than hydrogen

ion in the Electrochemical Series

• Describe how you can verify the product formed at carbon electrode L.

• Insert a glowing wooden splinter into the test tube

• Glowing splinter will rekindles / relights / reignites

• State one application of electrolysis in industries. • Extraction of reactive metal / electroplating /

purification of metal

Type of acidJenis asid

ExampleContoh

pH

Strong acidAsid kuat

Hydrochloric acidAsid hidroklorik

1.0

Weak acidAsid lemah

Oxalic acidAsid oksalik

3.0

What is meant by a weak acid?

Acid (substance) which ionises partially// has lower degree of dissociation in water to produce low concentration of H+ ion

• Why is the pH value of hydrochloric acid lower than the pH value of oxalic acid?

• The concentration of hydrogen ions in hydrochloric acid is higher

• The higher the concentration of hydrogen ions, the lower the pH value

Hydrochloric acid

Asid hidroklorik

25 cm3 of 0.1 mol dm-3 sodium hydroxide

25 cm3 natrium hidroksida 0.1 mol dm-3

Write a chemical equation to represent the reaction.

HCl + NaOH NaCl + H2O

• 25 cm3 of 0.1 mol dm-3 sodium hydroxide solution is required to react completely with 12.50 cm3 of dilute hydrochloric acid.

• Calculate the molarity of the dilute hydrochloric acid used.• ) No of mole of NaOH = 0.1 (25)• 1000• = 0.0025 mol• • 1 mol of NaOH react completely with 1 mol of HCl• 0.0025 mol of NaOH react completely with 0.0025 mol HCl• • Molarity of HCl = 0.0025 x 1000• 12.5• = 0.2 mol dm-3

25 cm3 of 0.1 mol dm-3 sodium hydroxide solution is required to react completely with 12.50 cm3 of dilute hydrochloric acid.

Calculate the molarity of the dilute hydrochloric acid used.

Zinc oxide

Zink oksida

+ HCl

Step 1

Langkah 1

Solution N

Larutan N

Zinc carbonate

Zink karbonat

Sodium chloride

Natrium klorida

+

Step 2

Langkah 2

+ S

Name the following solutions: Namakan larutan- larutan berikut:

4(c)(i)

N: ……………………..………

S: ..……………………………

N : Zinc chloride S : Sodium carbonate

r: formula

• State the type of reaction in Step 2.• double decomposition/ precipitation

Burning lithium

Litium yang terbakar

Chlorine gas

Gas klorin

Write a chemical equation for the reaction.

) 2Li + Cl2 2LiCl

State the changes in oxidation number for chlorine

0 to -1

Explain why lithium acts as the reducing agent in terms of electron transfer.

) lithium atom loses // donates electron // is electron donor

Bromine water

Air bromin

Potassium iodide solution

Larutan kalium iodida

A small amount of 1,1,1-trichloroethane liquid is added to the product in the test tube and the mixture is shaken. What is the colour of 1,1,1-trichloroethane layer?

purple

Write the ionic equation for the reaction between bromine and potassium iodide solution. Br2 + 2I- 2Br - + I2

• What is the role of bromine water in this reaction?

• oxidizing agent

Dry hydrogen gas

Gas hidrogen kering

Heat

Panaskan

Metal oxide powder

Serbuk oksida logam

ExperimentEksperimen

ResultKeputusan

Hydrogen + oxide of metal JHidrogen + oksida logam J

Metal oxide powder glows brightly.Black powder turned brown.

Serbuk oksida logam berbara dengan terang.Serbuk hitam menjadi perang.

Hydrogen + oxide of metal THidrogen + oksida logam T

No reaction.Powder turns yellow when hot and white

when cold. Tiada tindak balas.

Serbuk bertukar kuning apabila panas dan putih apabila sejuk.

Hydrogen + magnesium oxideHidrogen + magnesium oksida

No reaction.White powder remained.

Tiada tindak balas.Serbuk putih kekal.

Suggest a name for metal T. ) zinc

• Arrange the reactivity of J, T, Magnesium and Hydrogen in ascending order.

• J, H , T(zinc), magnesium• Based on the observations, explain how you obtain the

arrangement• Reaction occur between H2 and oxide J, so, H is • more reactive than J• 2. No reaction occur between H2 and oxide T and • between H-2 and magnesium oxide• so, H2 is less reactive than Mg and T• 3. Magnesium is more reactive than zinc

Compound K Compound L

Write the general formula for the homologous series of compound K.

CnH2n+1COOH, n= 0,1,2, ….

• Name compound L.• Methyl propanoate• Compare two physical properties of

compounds K and L by completing the following table:

Physical propertySifat fizik

Compound KSebatian K

Compound LSebatian L

OdourBau

SolubilityKeterlarutan

Properties Compound K Compound L

Odour pungent smell sweet/fruity smell

Solubility soluble in water //soluble in organic solvent

not soluble in water //soluble in organic solvent

• Compound L can be produced from K in the laboratory.• Sebatian L boleh dihasilkan dari K di dalam

makmal.• • (i) Describe briefly how this process can be carried

out.• Add methanol to compound K (propanoic acid) in a

boiling tube • Add a few drops of concentrated sulphuric acid • Heat gently / warm the mixture

• Write the equation for the reaction

• C2H5COOH + CH3OH C2H5COOCH3 + H2O

• Compound K is produced from the oxidation of propanol.

• Draw all possible isomers for propanol.

t1 t2 Time,s

Masa

Volume of gas, cm3

Isipadu gas

Compare the rate of reaction at t1 and t2.Based on the graph, explain your answer.

Rate of reaction at t1 is higher than t2

2. because gradient at t1 is greater than t2

3. The concentration of hydrochloric acid decreases with time

2HCl + CaCO3 CaCl2 + CO2 + H2O

The following chemical equation shows the reaction between calcium carbonate and hydrochloric acid. Persamaan kimia berikut menunjukkan tindak balas antara kalsium karbonat dengan asid hidroklorik.

Determine the mass of calcium carbonate needed in the reaction if 480 cm3 of gas is released at room condition. (The molar volume of gas at room condition: 24 dm3 mol-1, Relative atomic mass: C = 12 g, O= 16 g, Ca = 40 g)

No of mole of gas = 480 / 24000 = 0.02 molFrom equation,1 mol of CO2 is produced from 1 mol of CaCO3

0.02 mol of CO2 is produced from 0.02 mol of CaCO3

Mass of CaCO3 = 0.02 [40 + 12 + (16)(3)] g mol-1

= 2.0 g

Two experiments are carried out to study the effect of the size of calcium carbonate on the rate of reaction.Dua eksperimen dijalankan untuk mengkaji kesan saiz kalsium karbonat ke atas kadar tindak balas .

Experiment I : 1 g of calcium carbonate chips react with 20.0 cm3 of 0.2 mol dm-3 hydrochloric acid.Eksperimen I : 1 g ketulan kalsium karbonat bertindak balas dengan 20.0 cm3 asid hidroklorik 0.2 mol dm-3

.

Experiment II : 1 g of calcium carbonate powder react with 20.0 cm3 of 0.2 mol dm-3 hydrochloric acid.

Time / sMasa/ s 0 60 120 180 240 300 360

Volumeof gas / cm3

Isipadugas / cm3

ExperimentI 0.00 25.9

033.0

037.0

040.5

042.0

042.0

0Experiment

II 0.00 28.00

36.50

41.00

42.00

42.00

42.00

The volume of gas released is recorded in Table 7.1

Isipadu gas yang terhasil direkodkan dalam Jadual 7.1

•Plot a graph of volume of gas against time for both experiments in the graph paper provided on page 28.

Axis label & unitAll points transferred correctlyCurve – smooth curve, correct shapeConsistent scaleLabel curve for both experiment

• Based on the graph, determine the rate of reaction at 90 sec for Experiment I.

• Correct gradient drawn at t=90s• Calculation, answer and unit (0.11 ± 0.03)

cm3s-1

• Another set of experiment was carried out to study the effect of temperature on the rate of reaction between sodium thiosulphate solution and sulphuric acid as shown in Table 7.2

ExperimentEksperimen

I II

Temperature of sodium thiosulphate / oC

Suhu natrium tiosulfat 40 50

Write the ionic equation for the reaction.

S2O32- + 2H+ S + SO2 + H2O

Correct formulae Balance

• Compare the rate of reaction between Experiment I and II by using Collision Theory.

• The rate of reaction in Experiment II is higher than Experiment I

• 2. Temperature of sodium thiosulphate solution in Experiment II is higher

• 3. Kinetic energy of particles in Experiment II is higher,

• 4. frequency of collision between thiosulphate ions and hydrogen ions higher in Experiment II,

• [a: ions if the name of the ions is given in point 3]

• 5. frequency of effective collision higher.

• Based on the statement, explain the meaning of ecorrect elements [ie: carbon and hydrogen]

• 2. The simplest ratio of mole/atom of carbon to hydrogen is 2:5

• 3. The molecular formula shows the actual number of carbon and hydrogen atoms in a molecule

• 4. 1 molecule of butane contains 4 carbon atoms and 10 hydrogen atoms.

• mpirical formula and molecular formula

The molecular formula of butane is C4H10

and its empirical formula is C2H5.

Formula molekul bagi butana ialah C4H10

dan formula empiriknya ialah C2H5

The decomposition of copper(II) nitrate is shown in the following equation:

2 Cu(NO3)2 2 CuO + 4 NO2 + O2

(Relative atomic mass: N=14, O=16, Cu=64, molar volume of gas at room condition; 24 dm3 mol-1)

Determine the percentage composition by mass of oxygen in copper(II) nitrate.

% of oxygen = 6(16) X 100 = 51.06 %

64 + 14(2) + 16(6)Correct RMM shownCalculation and answer

• If 3.2 g of copper(II) oxide is produced during the heating process, calculate the volume of oxygen gas evolved at room condition.

• Number of mole of CuO = 3.2 = 0.04 mol

• 64 + 16

• 2 mol of CuO is released with 1 mol of O2

• 0.04 mol of CuO is released with 0.02 mol of O2

• Volume of O2 evolved = 0.02 (24) dm3 mol-1

• = 0.48 dm3

• A student carried out two experiments to determine the empirical formulae for magnesium oxide and copper(II) oxide.

• Diagram 8 shows the apparatus set-up for both experiments.

Experiment IEksperimen I

HeatPanaskan

Magnesium coilPita magnesium

Experiment IIEksperimen II

Copper(II) oxideKuprum(II) oksida

HeatPanaskan

Dry hydrogen

Hidrogen kering

Explain the differences in the method used for the determination of the empirical formulae for both oxides.

• Magnesium is a reactive metal• 2. Magnesium can combine / react [readily]

with oxygen to form magnesium oxide• 3. Copper is less reactive than hydrogen //

copper situated lower than hydrogen in Reactivity Series

• 4. Hydrogen can reduce copper(II) oxide to copper.

• The following equations show two redox reactions involving iron(II) ion, Fe2+.

• I 2 Fe2+ + Br2 → 2 Fe3+ + 2 Br-

• II Fe2+ + Zn → Fe + Zn2+

• Compare the role of Fe2+ ion in both reactions. • Explain your answer.

Reaction I Reaction IIrole Reducing agent1 Oxidizing agent1

The oxidation number of iron increases from

+2 to +3// Fe2+ loss electron to form Fe3+1

oxidation number of iron decreases from +2

to 0 // Fe2+ gain electron to form Fe

atom1

Fe2+ undergoes oxidation1

Fe2+ undergoes reduction1

• The reaction between sodium and chlorine forms a compound with a high melting point.

• Determine the mass of the compound formed when 2.3 g sodium reacts with excess chlorine.

• [Relative atomic mass Na= 23, Cl= 35.5]

• 2Na + Cl2 2NaCl

• Number of mole of Na = 2.3 = 0.1 mol 23

• • 2 mol of Na produce 2 mol of NaCl

– mol of Na produce 0.1 mol of NaCl

• • Mass of NaCl = 0.1[23 + 35.5] = 5.85 g•

ReactionTindak balas

ReactantsBahan Tindak balas

ObservationPemerhatian

I Iron + Chlorine gasFerum + Gas klorin

The hot iron wool ignites rapidly with a bright flame. A brown solid is formed.Wul besi panas menyala dengan cepat dan terang.Pepejal perang terhasil.

II Iron + Bromine gasFerum + Gas bromin

The hot iron wool glows moderately bright and moderately fast. A brown solid is formed.Wul besi panas berbara sederhana terang dan sederhana cepat. Pepejal perang terhasil.

Write the chemical equation for either of the reactions.

) 2 Fe + 3Cl2 2FeCl3 or 2 Fe + 3Br2 2FeBr3

• Compare the reactivity of both reactions.• Explain your answer.• The reactivity of reaction I is higher than

reaction II.• 2. The atomic size of chlorine is smaller

than bromine• 3. The forces of attraction of the nucleus

toward the • electrons is stronger in chlorine atom

than in bromine atom• 4. It is easier for chlorine atom to attract

electron

• Diagram 9 shows the standard representation for the atoms of three elements; Li, C and Cl.

Using the given information, describe the formation of two compounds with different types of bonding.

• . Electron arrangement of atom Li: 2.1, Cl: 2.8.7, C: 2.4

• Formation of compound between Li and Cl• 2. To achieve stable electron arrangement• 3. Lithium atom need to lose / donate 1

valence electron to form lithium ion // Li Li+ + e while

• 4. Chlorine atom need to gain 1 electron to form chloride ion // Cl2 + 2e 2Cl-

• 5. Li+ ion attracted to Cl- ion by electrostatic force to form ionic bond

• 6. Diagram - correct number of electrons and shells - shows Li+ ion and Cl- ion

• Formation of compound between C and Cl• 7. Carbon atom share valence electron with

chlorine atom• 8. Each C atom contributes 4 electrons and

each Cl atom contributes 1 electron // each C atom share electron with 4 Cl atom

• 9. To form #4 single# covalent bond• 10. diagram - correct number of

electrons and shells• - shows sharing electron

Mg + 2 HCl

ΔH = -50.4 kJ mol -1

Mg Cl2 + H2

Determine the temperature change when 50 cm3 of 1.0 mol dm-3 of hydrochloric acid reacts with excess magnesium.

[Specific heat capacity of solution: 4.2 J g-1 o C-1, density of solution: 1 g cm-3 ]

• Mol of hydrochloric acid = 1 x 50• 1000• = 0.05 mol• 2 mol HCl react to produce 50.4 kJ• 0. 05 mol of HCl react to produce = 0. 05 x

50.4• 2 = 1.26 kJ• θ = H• mc• = 1.26 x 1000• (50)(4.2)• = 6 0C

AlcoholAlkohol

Molecular FormulaFormula molekul

Heat of combustion/ kJ mol-1

Haba pembakaran

Propanol C3H7OH -2100Butanol C4H9OH -2877

•Write the equation for the complete combustion of either one of the alcohol.

C3H7OH + 9/2O2 3CO2 + 4H2O

or C4H9OH + 6 O2 4CO2 + 5H2O

• Compare the heat of combustion between propanol and butanol.

• Explain your answer.• Heat of combustion of butanol is higher than

propanol.• 2. The number of carbon atom per molecule

butanol is bigger than propanol• 3. Butanol produce more carbon dioxide and

water molecules than propanol // more bonds are formed // energy content is higher in butanol.

• 4. Released more heat energy

• Describe a laboratory experiment to determine the heat of combustion of a named alcohol.

• Your answer should include: • a labeled diagram • procedure

• Diagram:• Arrangement of apparatus is functional• can with water, thermometer, spirit lamp with

alcohol• no wire gauze, thermometer not touching

surface, flame touching can• Labeled water, alcohol, [metal] can•

• Procedure:• 1. (100-250 cm3) of water is measured and

poured into a copper can and the copper can is placed on a tripod stand.

• 2. The initial temperature of the water is measured and recorded.

• 3. A spirit lamp with ethanol is weighed and its mass is recorded.

• 4. The lamp is then placed under the copper can and the wick of the lamp is lighted up immediately.

• 5. The water in the can is stirred continuously until the temperature of the water increases by about 20- 50 oC.

• 6. The flame is put off and the highest temperature reached by the water is recorded.

• 7. The lamp and its content are weighed and the mass is recorded.