d strucutred question answers

STRUCTURED QUESTION ANSWERS

(2003-2008)

Answer To Score Chemistry Structured Question Answers

FORM 4 CHAPTER 2 THE STRUCTURE OF THE ATOM & CHAPTER 3 CHEMICAL FORMULAE AND EQUATION

ANSWER

1 SPM 2003/P2/Q1(a) It is a formula which shows the simplest ratio of atoms of the elements present

in a compound.b(i) Mass of magnesium = (26.4 – 24.0)g = 2.4g

Mass of oxygen = (28.0 – 26.4)g = 1.6g

(ii) Number of moles of magnesium atom = 2.4/24 = 0.1

Number of moles of oxygen atom = 1.6 / 16 = 0.1

0.1 mole of magnesium atom combine with 0.1 mole of oxygen atom. Therefore 1 mole of magnesium atom combines with 1 mole of oxygen atom.

(iii) Empirical formula of magnesium oxide is MgO(iv) 2Mg + O2 2MgO

(c) To allow oxygen to enter into the crucible. This is to make sure that all magnesium reacts completely with oxygen.

d(i)

(ii) 1. collect some gas into the test tube

2. place a burning splint near the mouth of the test tube3. no ‘pop sound

2 SPM 2004/ P2/Q1a(i) Iodine / naphthalene(ii) Copper(iii) Solid(iv) Copper (v)

(vi) Cu2+ / Cu+

b(i) T1 °C

(ii) Heat is absorbed to overcome the intermolecular forces of attraction between naphthalene particles. Therefore, the temperature remains constant even though heating continues.

(iii) Naphthalene particles will move faster.

2

dry hydrogen gas

metal X oxide

combustion excess hydrogen gas

Answer To Score Chemistry Structured Question Answers3 SPM 2006/ P2/Q2

ai) H2O

ii) Isotope Carbon-12

bi) Number of mole = 6.0 dm3 / 24 dm3 = 0.25 molMass = 11g

(ii) 0.25 x 6.02 x 1023 = 1.505 x 1023

(iii) 6.0 dm3 of carbon dioxide with the mass of 11g contains of 1.505 x 1023 molecules.

4 SPM 2007/ P2/ Q3(a) It is a formula which shows the simplest ratio of atoms of the elements present

in a compound.

(b) = number of moles of atom

c(i) Method I

(ii) Magnesium is an active element which burns completely in air.

(iii) To allow oxygen to flow into the crucible. This is to make sure that all magnesium combines completely with oxygen.

d(i) Mass of lead = (113.68 – 64.00 )g = 49.68g

(ii) Number of moles of lead = 49.68 / 207 = 0.24 mole

(iii) Mass of oxygen = (117. 52 – 113.68)g = 3.84g

(iv) Number of moles of oxygen = 3.84 / 16 = 0.24 mole

(v) PbO

5 SPM 2007/ P2/ Q5(a) 11(bi) In Group 1 and Period 3b(ii) Contains 1 valence electron and 3 shells filled with electron

c(i) Nucleus attractions towards valence electron in atom X is stronger ,so valence electron in X is difficult to be released orNucleus attraction towards valence electron in atom Y is weaker so valence electron in Y is easier to be released.

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c(ii)

Oxygen gas

Substance X / Y

d(i) 2.3/23 = 0.1

d(ii) 4 mol of X 2 mol of X2O0.1 mol of X 0.05 mol of X2OMass of X2O = 0.05 x 62g = 3.1g

6 SPM 2008/ P2/ Q3a(i) Diffusion (ii) Molecule (iii) The particles are made up of tiny or discrete particles. They are randomly

moving. The particles fill the space between the air particles

(iv) Less than 10 minutesb(i) 0.5 x 32g = 16.0g

(ii) 0.5 x 24 dm3 = 12 dm3

(iii) The number of gas molecules in balloon A and in balloon B is the same because the number of moles is the same

7 SPM 2004/P3/Q1

(a) Observation Inference (i) white fumes released(ii) the mass of crucible and its contents increases

(i) magnesium oxide is formed(ii) magnesium reacts with oxygen

(b) Mass of crucible and lid = 25.35gMass of crucible, lid and magnesium ribbon = 27.75gMass of crucible, lid and magnesium oxide when cooled = 29.35g

c(i) Mass of Mg = (27.75 – 25.35)g = 2.4g

(ii) Mass of O = ( 29.35 – 27.75)g = 1.6g

(iii) Number of moles Mg = 0.1moleNumber of moles O = 0.1moleRatio of Mg : O = 1 : 1Empirical formula is MgO

(d) 0.1 mole of Mg reacts with 0.1 mole of oxygen

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8 SPM 2005/P3/Q1(a) Time/ s 0 30 60 90 120 150 180 210

Temperature/°C 95.0 85.0 82.0 80.0 80.0 80.0 78.0 70.0(b)

c(i) 80°C(ii) There is no temperature change during the cooling process(d) The heat released during the formation of bonds balances the heat loss to the

surroundings(e)

(f) Covalent compound Glucose, ethanol, tetrachloromethane

Ionic compound Potassium iodide, copper (II) sulphate, aluminium oxide

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6

Answer To Score Chemistry Structured Question AnswersCHAPTER 4 : PERIODIC TABLE OF THE ELEMENTS

1 SPM 2003/P2/Q2

a) Iron/Ferumb)

c) 1. Formed compound with different oxidation numbers2. Formed colored ions or compound3. Formed complex ions4. As a catalyst

d) 2.8.2e) i) 4Al + 3O2 2 Al2O3

ii) Aluminium atom releases 3 electrons to form aluminium ion, (Al3+). Oxygen atom receive 2 electrons to form oxide ion, (O2-). Two aluminium atoms release a total of 6 electrons while 3 oxygen atoms receive 2 electrons each. The strong electrostatic force between Al3+ and O2- form aluminium oxide compound (Al2O3)

f) Helium gas because helium is inert gas

2 SPM 2005/P2/Q1

a) i) Yii) Riii) X

b) R, Q, Y, X and Tc) 2.4

d) Y-

e) Both Q and R have the same number of electron shells filled with electronsf) Red litmus paper turns blueg) Transition elements

3 SPM 2007/P2/Q5

a) 11b) i) Period 3, Group 1

ii) The electron arrangement is 2.8.1. X has three electron shells and one valence electron.

c) i) The size of atom Y is larger than the size of atom X. The distance between the valence electron and nucleus increases. The force of attraction is smaller. It is easier for Y to release its valence electron.ii)

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+ -

Answer To Score Chemistry Structured Question Answersd) i) Number of moles of element X

= 2.3/23 = 0.1

ii) Relative molecular mass of X2O = 46 + 16 = 62 4 moles of X 2 moles of X2O 0.1 mole of X 0.05 mole of X2O

Maximum mass of X2O = 0.05 x 62 = 3.1 g

4 SPM 2008/P2/Q2

a) Group 17b) 2.7c) The outermost occupied shell of a fluorine atom is nearer to the nucleus. The

strength of the fluorine nucleus to attract electrons is higher. Fluorine atom can accept electron easily to form negative ions.

d) Covalent bonde) i) Ionic bond

ii)

f)

Chemicals

KI (aq) + Cl2 (aq) /

KCl (aq) + Br2 (aq)

KBr (aq) + KCl (aq)

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Na+ Cl-


CHAPTER 5 : CHEMICAL BOND

1 SPM 2006/P2/Q3

a) i) Because argon has 8 valence electron.ii) Neon (Group 18 elements)

b) i) Sodium ion : Sodium atom donates one electron (Na Na+ + e)

Chloride ion : Chlorine atom receives one electron from the sodium atom. (Cl + e Cl-) ii) Electrostatic forceiii) The ions can move freely nowiv) Energy or heat is used to break the ionic bond in sodium chloride

c) Q1

SPM 2007/P2/Q4a) i) Structure : Molecule

Bonding : Covalentii) The molecules in P are held together by weak intermolecular forces. A small amount of heat energy is needed to overcome the weak intermolecular forces.

b) i) Substance Q : By sharing of electronsii) Substance R : By transfer of electrons

c) Solid state : The negative and positive ions are in fixed positions and they cannot move freely.

Molten and aqueous state :The negative and positive ions can move freely. This enables them to conduct electricity.

d) Substance R is soluble in waterSubstance P and Q are insoluble in water

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F4 Chapter 6 ElectrochemistryF5 Chapter 3 Oxidation & Reduction

1 SPM 2003/P2/Q5

(a) (i) Brown solid is formed.Blue solution turns colourless.

(ii) Green solution turns brown.Brown colour of bromine decolorizes.

(b) Mg + Cu2+ → Mg2+ + Cu(c) A substance that receives electron.

(d) (i) 0 → +2(ii) copper(II) ion // copper(II) sulphate

(e) (i) redox(ii) 0(iii) oxidising agent(iv) chlorine water

2 SPM 2004/P2/Q3

(a) (i) A positively charged ion.(ii) Electrical → Chemical

(b) (i) Cu2+, SO42-, H+ , OH-

(ii) In the table below, write the ions in (b)(i) which moved to electrodes X and Y.

Electrode X Electrode Y SO4

2-

OH- Cu2+

H+

(iii)Electrode X : Oxidation

Electrode Y : Reduction

(iv) Brown solid(v) Blue to colorless.

(c) (i) Oxygen(ii)

20___ = 0.0008 mol 24000

(iii) 0.0008 x 6.02 x 1023

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3 SPM 2005/P2/Q6

(a) (i)

(ii) A burning splinter gives a ‘pop’ sound.(iii) Na+ and H+ ions are attracted towards the cathode. H+ ions are selected to be

discharged as hydrogen gas.

(b) (i) MgO(ii) 2Mg + O2 → 2MgO(iii) Oxidation number for:

Magnesium = +2Oxygen = -2

4 SPM 2007/P2/Q6

(a) (i) water and oxygen

(ii)

(b) (i) 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) +2 → +3

(c) (i) Zinc is more electropositive than iron.Zinc atoms lose electrons more easily than iron.Zinc corrodes but iron does not.

(ii) Zn → Zn2+ + 2e

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5 SPM 2004/P3/Q2

(a)Metal ObservationsZincMagnesiumLead

Moderately bright flameVery bright flameBright flame

(b)Name of variables Action to be taken

(i) Type of metal (i) Replace the metal with different metals

(ii) Intensity of the flame (ii) Observe the intensity of the flame

(iii) Quantity of metal (iii) Use a constant mass of metal

(c) The higher the metal in the reactivity series, the brighter the intensity of the flame.

(d) (i)

Descending order of reactivity of metal towards oxygen.

(ii) Mg, ↓ , Zn, Pb, Cu

(e) Metals more reactivethan carbon

Metals less reactive than carbon

MagnesiumSodium

LeadCopper

6 SPM 2005/P3/Q2

(a) The blue colour of copper (II) sulphate solution fades.

(b) Manipulated variable:The metal as negative electrode

Method to manipulate the variable:Replace the negative electrode with different metals.

Responding variable:Voltmeter reading

How the variable is responding:The voltmeter reading changes.

Controlled variable:

Concentration of copper(II) sulphate solution, copper metal

Method to maintain the controlled variable:Use copper(II) sulphate solution of the same concentration.Use copper as the positive electrode.

(c) The greater the distance between two metals in the electrochemical series, the higher the voltage reading.

Zn Pb Cu

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Mg

Answer To Score Chemistry Structured Question AnswersFORM 4 CHAPTER ACIDS AND BASES

FORM 4 CHAPTER 8 SALTS2004/P2/Q5

1 (a) From pink to colourless or intensity of the pink colour decreases or the container of the mixture becomes warm or hot

(b) (i) To ensure all the the acid react completely with copper(II) oxide(ii) By filtering (iii) CuO + H2SO4 CuSO4 + H2O or

CuO + 2H+ Cu2+ + H2O(iv) n(H2SO4) = 0.1 x 50/1000 Mass of Cu SO4 = 0.005 x 160

= 0.005 = 8.0 g(c) 20.0 cm3 or twice the volume of sulphuric acid(d)

Experiment I Experiment II needs filtering a mixture between two solutions

need to add an indicator experiment is repeated without

using an indicator volume of sulphuric acid is added

accurately

no need filtering a mixture between a solid and

solution no need to add an indicator no need to repeat the experiment

solute or CuO is added in excess

2005/P2/Q42 (a) cation

(b) Cu2+ ions & SO42- ions , H+ ions & OH- ions

(c) Na2SO4

(d) (i) Na2SO4 + Pb (NO3)2 2NaNO3 + PbSO4

(ii) I mol of lead(II) nitrate reacts with 1 mol of sodium nitrate to produce 1 mol of lead(II) sulphate and 2 moles of sodium nitrate

(iii) Lead(II) sulphate(iv) Number of mole = 10/1000 x 0.5 = 0.005(v) Mass = 0.005 x (207 + 32 + 16 x 4) or 0.005 x 303 = 1.515 g

2006/P2/Q43 (a) (i) Concentration – the quantity or amount of solute (grams) dissolves in a

given volume(1 dm3) of solution

(Ii) Molarity – the number of moles of solutes that are present in 1 dm3 of solution.(iii) n = MV(cm3) /1000 or n = MV(dm3)(iv) n = 8/40 = 0.2 mole , M = 0.2 x 1000/1000, M = 0.2 mol dm-3

(b) (i) Parameter I : mass / moles of NaOH Parameter II : volume of solution or distilled water

(ii) No traces of sodium hydroxide is left on the filter funnel or beaker for accurate concentration or amount of solute used is accurate and not less

(iii) Add distilled water drop by drop until the meniscus is at the calibration mark(iv) Measures the volume accurately(v) To prevent evaporation or evaporation of water can cause the changes in

concentration or easy to swirl the solution

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Answer To Score Chemistry Structured Question Answers2003/P3/Q2

4 (a) Test Tube P Q R S T U VVolume of lead(II) nitrate solution 1.0 mol dm-3/cm3

0.5 1.0 1.5 2.0 2.5 3.0 3.5

Height of lead(II) iodide precipitate /cm

1.1 2.2 3.4 4.4 5.5 5.5 5.5

Table 1(The diagram given is not to scale - please refer to the original SPM question paper for accuracy of readings)

(b) 1. label of the x-axis – volume(cm3)2. label of the y-axis – height of precipitate(cm)3. uniform scale4. size of graph more than 50%5. all points are transferred correctly6. smooth graph

(c) (i) Suggested answer:

Height of precipitate(cm)

Volume of lead(II) nitrate(cm3 )

(ii) No. of mole (Pb2+) = 2.5 X 1.0 = 0.0025 1000

No. of mole (I-) = 5 X 1.0 = 0.005 1000

No. of mole of I- reacted with 1 mol of Pb2+ = 2 mole(iii) Pb2+ + 2I- → PbI2

(d) The height of precipitate increases gradually from test tubes P to SThe height of precipitate in test tubes T,U,V are the same

(e) In test tubes P, Q, R, and S, more and more yellow precipitate of lead(II) iodide is formed due to the increasing amount of lead(II) nitrate added to the test tubes or potassium iodide has not completely reacted with lead(II) nitrate solution

In test tubes T,U,and V, potassium iodide/iodide ions have reacted completely or a complete reaction has taken place

(f)Solution Positive ions Negative ions

Lead(II) nitrate Pb2+ , H+ NO3- ,OH-

Potassium iodide K+ , H+ I-, OH-

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Minimum volume- must be written on the graph (2.5cm3)

Answer To Score Chemistry Structured Question AnswersFORM 4 CHAPTER 9 MANUFACTURED SUBSTANCES IN INDUSTRY

Question 1 2003/P2/Q3(a) Contact Process(b) Sulphur trioxide(c) The reaction gives off a lot of heat. (or Large cloud of sulphuric acid formed)(d) H2S2O7 + H2O 2H2SO4

(e) Substance Y: ammoniaFertilizer Z : ammonium sulphate

(f) 2SO2 + O2 + 2H2O 2H2SO4 or SO2 + H2O H2SO3

Question 2 2006/P2/Q5(a) N2 + 3H2 2NH3

(b) (i) Percentage of ammonia produced from factory Q is higher than factory P.

(ii) At a lower temperature, percentage of ammonia produced is higher.(iii) The pipes might explode or break.(iv) By using tanks or pipes that can withstand high pressure

(c) (i) Use as explosives or cold pack(ii) NH 3 + HNO3 NH4NO3(s)(iii)

Question 3 2007/P2/Q2(a) X : Contact process

Y : Haber process(b) 1. Sulphur

2. Air (or oxygen)3. Water

(c) (i) H2SO4 + 2NH3 (NH4)2SO4

(ii) Sulphuric acid: 1 moleAmmonia : 2 moles

(d) Ammonium sulphate is use as fertilizer

Question 4 2008/P2/Q1(a) Contact process(b) Vanadium(V) oxide (or vanadium pentoxide)(c) (i) Oleum

(ii) SO3 + H2SO4 H2S2O7

(d) Waste gas dissolve in rain water and produces acid rain (or waste gas react with water vapour in the air and produces acid rain.)

(e) (i) Ammonium sulphate (or potassium sulphate)[For other accepted answers, please refer to the text book page 152]

(ii) Use as electrolyte in car batteries (lead-acid accumulators) or making of detergent..[For other accepted answers, please to the text book page 152]

CHAPTER 1 : RATE OF REACTION

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1. SPM 2006/P2/Q6

(a) Total surface area of reactant

(b) (i) Carbon dioxide, CO2.(ii) Because the changes in the volume of CO2 can be measured easily.

(c) 1. Concentration of hydrochloric acid.2. Temperature of the reactants.3. Mass of calcium carbonate.* any two

(d) (i) Gradient of the curve of Experiment II is steeper than the gradient of the curve of experiment I.

(ii) All the reactants have reacted completely within time x.

(iii) Both experiments produced the same amount of products.

(e) The greater the total surface area the higher is the rate of reaction.

(f)

- descending pattern of curve- best fit curve

2. SPM 2008/P2/Q5(a) Change of quantity of reactant/product

Time taken

(b) (i) Draw the tangent at 120 sRate of reaction = (0.12- 0.18) cm3s-1

(ii) 0.267 cm3s-1

(c) Because the concentration of hydrochloric acid decreases(d) (i) Curve II : catalyst/ temperature

Curve III : concentration

(ii) Repeat the experiment as in set IReduce the concentration of hydrochloric acidThe volume of hydrochloric acid remains unchanged

(iii) Because the number of mole of hydrochloric acid used is half of set I

3. SPM 2003/P3/Q1

16

Concentration of dilute hydrochloric acid

Time taken to collect a fixed quantity of product


(a) t1 = 55.0 st2 = 48.0 st3 = 42.0 st4 = 37.0 st5 = 33.0 s

(b)Temperature/ oC Time/ s 1/time/ s-1

3035404550

55.048.042.037.033.0

0.0180.0210.0240.0270.030

(c) (i) Graph :- X and Y axes labeled and unit- Correct scale, size more than 50%- All points transferred correctly- Smooth graph

(ii) Rate of reaction is directly proportional to the temperature

(d) Time = 30.3 s

(e) (i) Manipulated variable : temperature of sodium thiosulphateResponding variable : time for the ‘X’ mark disappear from sight or rate of reactionControlled variable : volume and concentration of acid

(ii) Heat the sodium thiosulphate solution with different temperature while the volume and concentration of sodium thiosulphate and acid remains constant.

(f) The higher the temperature the higher the rate of reaction

(g) The lower the temperature the lower the rate of food turns bad

3. SPM 2003/P3/Q1

(a) t1 = 55.0 st2 = 48.0 st3 = 42.0 st4 = 37.0 st5 = 33.0 s

(b)Temperature/ oC Time/ s 1/time/ s-1

3035404550

55.048.042.037.033.0

0.0180.0210.0240.0270.030

(c) (i) Graph :- X and Y axes labeled and unit

17

Answer To Score Chemistry Structured Question Answers- Correct scale, size more than 50%- All points transferred correctly- Smooth graph

(ii) Rate of reaction is directly proportional to the temperature

(d) Time = 30.3 s

(e) (i) Manipulated variable : temperature of sodium thiosulphateResponding variable : time for the ‘X’ mark disappear from sight or rate of reactionControlled variable : volume and concentration of acid

(ii) Heat the sodium thiosulphate solution with different temperature while the volume and concentration of sodium thiosulphate and acid remains constant.

(f) The higher the temperature the higher the rate of reaction

(g) The lower the temperature the lower the rate of food turns bad

FORM 5 CHAPTERR 2 CARBON COMPOUNDS

Question 1 2003/P2/Q4(a)

(b) Heat supplied is uniform (or prevents the solution from evaporating too fast) (c) (i) Esterification

(ii) C2H5OH + CH3COOH CH3COOC2H5 + H2O(d) (i) Propyl ethanoate

(ii) [Note: Choose any one of the following answers]1. Fruity smell 2. Colourless liquid 3. Low boiling point (volatile) 4. Insoluble in water

(e) (i) C2H5OH C2H4 + H2O(ii) C2H5OH + 2[O] CH3COOH + H2O

(f) Butane

Question 2 2004/P2/Q6(a) (i) Hydrogenation

(ii) Vegetable oil changes from liquid to solid (or changes vegetable oil from unsaturated fats to saturated fats or changes double bond in vegetable oil molecule to single bond)

(b) (i) Catalyst X : Nickel (or platinum)Temperature : 100oC (or 200oC)

(ii) 1. Presence of catalyst reduces the activation energy 2. At a high temperature, particles possess high kinetic energy and3. hence effective collision increases and rate of reaction increases

(iii)

or

(c) Palm oil (or coconut oil or corn oil or other named oil example)

Question 3 2008/P2/Q4

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Answer To Score Chemistry Structured Question Answers(a) Ethanol

(b) CnH2n+1OH(c) (i) Carbon dioxide

(ii) m = 3 , n = 2(d) (i) Ethyl ethanoate

(ii)

(e) (i) [Choose any one of the following answers]Porcelain chips or phosphoric acid or aluminium oxide

(ii)

Question 4 2008/P3/Q1(a) Acid coagulates latex while alkali does not coagulate latex.

(b) Set I: 5 minutes Set II: 6 hours or 300 minutes

(c)

(d) Set I: A white solid lump is formed. (or latex coagulated)Set II: There is no visible change observed. (or latex does not coagulate)Set III: A white solid lump is formed. (or latex coagulated)

(e) Operational definition for the coagulation of latex is latex becomes solid when acid is added or when exposed to air.

(f) (i) The manipulated variable : Ethanoic acid and ammonia solution(ii) The responding variable : Latex coagulates or does not coagulate. (or time taken

for coagulation of latex)(iii) The constant variable : Volume of latex, volume and concentration of acid and

ammonia solution. Temperature(g) (i) Latex coagulates after excess hydrochloric acid is added. (or latex becomes solid

after excess hydrochloric acid is added.)(ii) [For more detailed pictorial explanation, please refer to text book page 93]

When acid is added, the H+ ions from the acid neutralized the ammonia in the latex. The excess H+ ions then neutralized the negative charge on the protein membrane. The protein membrane then breaks during collision and rubber molecules combine and become entangled casing latex to coagulate.

(h) [For more explanation, please refer to text book page 93]This is because the bacteria from the air enter the latex. The growth and spread of bacteria produce lactic acid that causes the coagulation of latex.

(i) (i) Latex in Set I coagulates faster than the latex in Set III because the concentration of H+ ions is higher in Set I.

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Set Time taken / minute

I 5

III 300

Answer To Score Chemistry Structured Question Answers(ii)

CHAPTER 4 : THERMOCHEMISTRY

1. SPM 2003/P2/Q6

(a) Heat change when 1 mol of metal is displaced from its salt solution by a more electropositive metal.

(b) Initial temperature and highest temperature.

(c) 1. Stir the mixture.2. Add the two solutions as quickly as possible.3. Use polystyrene or plastic cup

(any one)

(d) (i) 1. Grey solid is deposited2. Colourless solution turns blue3. The thermometer reading rises or the container becomes hot or warm.

(any one)

(ii) 1. Silver metal is produced 2. copper(II) ion is produced3. exothermic reaction/ heat is released to the surroundings

(e) (i) = 0.5 x 100 = 0.05 mol 1000

(ii) = 0.05 x 105 kJ = 5250 J

(iii) Ө = 5.25 x 1000 = 12.5 oC 100 x 4.2

(f)

(g) 1. Mol of Ag+ = 1 x 100 = 0.1 mol

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Can coagulate latex Cannot coagulate latex

Nitric acidMethanoic acid

Potassium hydroxideSodium hydroxide

Energy

Cu + Ag+

Cu2+ + Ag

∆H = -105 kJ mol-1

Answer To Score Chemistry Structured Question Answers 1000

2. Heat change, Ө = 0.1 x 10500 = 25 oC 100 x 4.2

3. Number of mol of Ag+ is double or concentration of silver nitrate is double.

2. SPM 2004/P2/Q4(a) To reduce the heat loss to the surroundings.

(b) (i) Exothermic reaction

(ii) Total energy of products is less than total energy of reactants

(c) Mix the solutions quickly and stir the reaction mixture.

(d) (i) Number of moles Ag+ = 25 x 0.5 100 = 0.0125 mol

(ii) The heat change = mcө = 50 x 4.2 x (31.5-29.0) = 525 J

(iii) 0.0125 mol of Ag+ ions that reacted with Cl- ions released 525 J

1 mol of Ag+ ions that reacted with Cl- ions released = J

= 42000 J

Heat of precipitation = -42 kJmol-1

(e) Heat is released to surroundings.

3. SPM 2005/P2/Q3

(a) Zn2+ + Cu

(b) (i) ∆H = 100 x 4.2 x 20 = 8400 J

(ii) Number of moles CuSO4 reacted = 0.5 x 100 = 0.05 mol 1000

Heat of displacement = ___ m cө_______ Number of moles = 8400 0.05 = -168 000 J mol-1

= -168 kJ mol-1

(c)

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Energy

Zn2+ + Cu

Zn2+ + Cu

∆H = -168 kJ mol-1


(d) 1. Use a plastic / polystyrene cup 2. add the zinc powder quickly.3. stir the solution

(any one)

(e) The heat released when 1 mole of copper is displaced from its solution.(f) Tin (Sn)

4. SPM 2005/P2/Q5

(a) The heat released when 1 mole of alcohol is completely burnt in excess oxygen.

(b) (i) 1. all points are transferred correctly2. draw a straight line

(ii) The greater the number of carbon dioxide molecules, more products are formed which causes more heat to be released during the formation of bonds.

(iii) Relative molecular mass of ethanol= (12 x 2) + (1 x 6) + 16 = 46

Number of moles ethanol = 2.3 = 0.05 mol 46Heat released = 0.05 x 1376 = 68.8 kJ = 68 800 J

(c)

(d) - Ethanol- The freezing point of ethanol is -117 oC, which is lower than -100 oC.

5. SPM 2006/P3/Q1(a) Initial temperature of mixture : 28.0 oC

Highest temperature of mixture : 40.0 oCChange in temperature : 12.0 oC

(b)

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Answer To Score Chemistry Structured Question AnswersExperiment Experiment I Experiment II

Initial temperature of mixture/ oC

28.0 T1

Highest l temperature of mixture/ oC

40.0 T2

Change in temperature/ oC 12.0 T3

(c) Strong acid produces higher heat of neutralization than weak acid.

(d) 12.5 oC - 15.0 oC(e) To enable us to obtain the change in temperature for both experiments.

(f) Change in temperature = Highest temperature of mixture - Initial temperature of mixture

(g) 1. A colourless mixture of solution is obtained.2. The vinegar smell of ethanoic acid disappears.3. The polysterene cup becomes hot.4. Thermometer reading is rises

(h) 1. The volumes of the acid and the alkali.2. The concentrations of the acid and the alkali.3. The type of cup used in the experiment.

(i) (i) The heat of neutralization is defined as the amount of heat released when 1 mole of water is produced.

(ii) Experiment II uses a strong acid whereas Experiment I uses a weak acid.(j)

Name of acid Type of acidEthanoic acid Weak acidHydrochloric acid Strong acidMethanoic acid Weak acid

6. SPM 2007/P3/Q1(a) (i)

Initial temperature (oC)

Highest temperature (oC)

Experiment I 28.0 36.0Experiment II 29.0 25.0Experiment III 27.0 32.0Experiment IV 30.0 27.0

(ii)Experiment Initial temperature

(oC)Highest

temperature (oC)I 28.0 36.0II 29.0 25.0III 27.0 32.0IV 30.0 27.0

(iii) Exothermic reaction Endothermic reaction

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Answer To Score Chemistry Structured Question AnswersExperiment I Experiment II

Experiment III Experiment IV

(b) (i) 1. The mass of sodium hydroxide.2. the volume of water in the cup.3. The size of the polystyrene cup.

(ii) The reaction between sodium hydroxide and water is an exothermic reaction.

(c) (i) Temperature change = 4 oC

Reason 1 :Heat energy is absorbed by the reactants from the surroundings.

Reason 2 :The energy of the products is more than the energy of the reactants.

(ii) The decrease in temperature shows that endothermic reaction happens where heat energy is absorbed from the surroundings.

(d) 1. 37 oC2. 32 oC3. 30 oC

(e) (i) 1. Final temperature is lower than the initial temperature.2. The temperature reading decreases.3. Bubbles of gas are released.

(ii) Heat energy is absorbed when hydrochloric acid reacts with sodium hydrogen carbonate to produce sodium chloride, carbon dioxide and water.

(iii)

7. SPM 2008/P2/Q6

(a) Heat change when 1 mole of hydrogen ions reacts with 1 mole of hydroxide ions to form 1 mole of water.

(b) Observation : the mixture becomes hot or temperature increase

Explanation : the reaction is exothermic

(c) (i) No. of moles of NaOH = 100 x 2 = 0.2 mol 1000

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Volume of carbon dioxide gas,/cm3

Time /minute


Energy released = 0.2 x 57.3 = 11.46 kJ

(ii) Temperature change = 11.46 x 1000 200 x 4.2 = 13.6 oC

(d)

(e) 1. Ethanoic acid is a weak acid which partially ionize in water, nitric acid is strong acid that ionize completely in water.

2. energy is used to ionize/dissociate weak acid.

CHAPTER 5 : CHEMICAL FOR CONSUMERS

1 SPM 2004/P2/Q2

a) i) Sodium hydroxide/potassium hydroxideii) To reduce the solubility of soap in water or to precipitate the soap formed.

b) Procedure of the experiment :1. Two beakers are filled with hard water2. Soap is added to one beaker and detergent is added to another beaker3. The socks are dipped into each of the beakers and wash by scrubbing or

agitated

Observation :

Detergent in hard water Soap in hard water1. Socks is clean easily 1. Socks still dirty2. No formation of scum 2. Scum forms3. The water turns dirty 3. Water is less dirty

Conclusion :Detergent cleans stains more effectively compared to soap in hard water. Detergent can still performed its cleansing action and more effective than soap in hard water.

c) i) To calm down patients so that they can sleep easily. As a sedative or to calm or relax.ii) Paracetamoliii) C9H8O4

iv) Molecular mass = (12 x 9) + (8 x 1)(4 x 16) = 108 + 8 + 64 = 180

2 SPM 2005/P2/Q2a) Saponification

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Energy

NaOH + HNO3

Na NO3 + H2O

∆H = -57.3 kJ mol-1

Answer To Score Chemistry Structured Question Answersb) i) Ester

ii) COO-

c) Concentrated potassium hydroxide solutiond) i) Hydrophobic part or hydrocarbon part

ii) - Detergent ions reduce the surface tension of water - Hydrophilic dissolves in water

- Hydrophobic dissolves in grease- Mechanical agitation or scrubbing helps pull the grease free

(Refer to Chemistry Text Book Pg 185)

iii)

3 SPM 2006/P2/Q1a) i)

ii) Stomach pain due to wind in stomachiii) Extract the juice from the rhizome and drink

b) i) X : Analgesics Y : Antibiotics Z : Psychotherapeutic medicineii) Can cause bleeding in the stomachiii) The bacteria becomes immune to medicineiv) Get rid of anxiety

4 SPM 2007/P2/Q1a) i) Saponification

ii) Glyceroliii) To reduce the solubility of soap in water or to precipitate the soap

b) i) J : Soap K : Detergentii) The insoluble precipitate formed when soap react with magnesium and calcium ions (hard water)iii) 1. Magnesium ion 2. Calcium ioniv) J is biodegradable

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d strucutred question answers

Documents

g mass of oxygen

atom x

mole of oxygen atom

g number of moles

mole mass of oxygen

mole of magnesium atom

bi mass of magnesium

mol mass