CHEMISTRY ANSWERING TECHNIQUES

GUIDELINE FOR ANSWERING ESSAY PAPER 2 ( SECTION B AND C)

SECTION B

Consists of RESTRICTED RESPONSE (close end) Normally FACTS / CONCEPT Require explanation on concepts/facts

SECTION C

Consists of OPEN RESPONSE (open end) Normally FACTS / CONCEPT / EXPERIMENTS / PROCESS Multiple answers for FACTS / CONCEPTS / EXPERIMENTS Require explanation

SECTION B [CLOSE END]

QUESTION 1

Diagram shows the chemical symbols which represent three element X, Y and Z. These letters are not the actual symbols of the elements.

(a) Explain the formation of the bond that is formed between(i) element X and element Y(ii) element Y and element Z [12 marks]

(b) Compare the melting points of compound formed in (a). Explain your answer.

[6 marks]ANSWER

(a) (i) X and Y The electron arrangement of atom X is 2.8.2 and atom Y is 2.8.7. Atom X release 2 electron to achieve a stable octet electron

arrangement X2+ ion is formed/ Atom Y accept 1 electron to achieve a stable octet electron

arrangement Y¯ ion is formed. Two electrons are transferred from one atom X to two atom Y. X2+ ion and Y¯ ion are attracted to each other by strong electrostatic

force. An ionic compound with the formula of XY2 is formed.

(ii) Y and Z The electron arrangement of atom Z is 2.4 and atom Y is 2.8.7. Atom Z contributes 4 electrons for sharing Atom Y contributes 1 electron for sharing One atom Z share 4 pairs of electrons with four Y atoms to achieve a

stable octet electron arrangement. A covalent compound with the formula of ZY4 is formed.

(b) Compound XY2 (ionic) Compound ZY4 (covalent)

1. High melting point

2. Ions (X 2+ ion and Y ¯ ion) are attracted to each other by strong electrostatic force

3. More heat energy is needed to overcome the forces of attraction

4. Low melting point

5. Molecules (ZY4 molecules) are attracted to each other by weak intermolecular forces

6. Less heat energy is needed to overcome the forces of attraction

QUESTION 2

The following information is about hydrochloric acid solution and ethanoic acid solution.

Acid Concentration (mol dm-3)

pH

Ethanoic acid solution 1.0 4Hydrochloric acid solution

1.0 1

Explain why these two solutions have different pH values. [4 marks]

ANSWER

Hydrochloric acid solution is a strong acid and ethanoic acid is a weak acid.

Hydrochloric acid ionize completely in water to form high concentration of hydrogen ion// ethanoic acid ionize partially in water to form low concentration of hydrogen ion.

The concentration of hydrogen ion in hydrochloric acid is higher than ethanoic acid.

Therefore, the higher the concentration of hydrogen ion, the lower the pH value.

QUESTION 3

The following information is about sodium hydroxide solution and ammonia solution.

Alkali Concentration (mol dm-3)

pH

Sodium hydroxide solution

1.0 14

Ammonia solution 1.0 11

Explain why these two solutions have different pH values. [4 marks]

ANSWER

Potassium hydroxide solution is a strong alkali and aqueous ammonia is a weak alkali.

Potassium hydroxide ionize completely in water to form high concentration of hydroxide ion// Ammonia ionize partially in water to form low concentration of hydroxide ion.

The concentration of hydroxide ion in potassium hydroxide is higher than ammonia.

Therefore, the higher the concentration of hydroxide ion, the higher the pH value.

QUESTION 4

A group of students carried out three experiments to investigate the factors affecting the rate of reaction. The table shows information about the reactants and the temperature used in each experiment.

Experiment Reactants Temperature / °CI Large marble chips and 25 cm3 of 0.5

mol dm-3 hydrochloric acid30

II Large marble chips and 25 cm3 of 0.5 mol dm-3 hydrochloric acid

40

III Small marble chips and 25 cm3 of 0.5 mol dm-3 hydrochloric acid

40

Based on the table, compare the rate of reaction between

Experiment I and II Experiment II and III

In each case, explain the difference in the rate of reaction with reference to the collision theory.

[10 marks]

ANSWER

Experiment I and II

The rate of reaction in Experiment II is higher.

The temperature of solution in Experiment II is higher.

The kinetic energy of hydrogen ions in Experiment II is higher.

The frequency of collision between carbonate ions and hydrogen ions

in Experiment II is higher.

The frequency of effective collision between carbonate ions and

hydrogen ions in Experiment II is also higher.

Experiment II and III

The rate of reaction in Experiment III is higher.

The particle size of marble chips in Experiment III is smaller.

The total surface area of marble chips exposed to collision in

Experiment III is higher.

The frequency of collision between carbonate ions and hydrogen ions

in Experiment III is higher.

The frequency of effective collision between carbonate ions and

hydrogen ions in Experiment III is also higher.

QUESTION 5

A group of students carried out three experiments to investigate the factors affecting the rate of reaction. The table shows information about the reactants used in each experiment.

Experiment

Reactants Time taken to collect 25 cm3 hydrogen gas (s)

I Excess zinc granules and 25 cm3

of 0.5 mol dm-3 hydrochloric acid50

II Excess zinc granules and 25 cm3

of 1.0 mol dm-3 hydrochloric acid20

III Excess zinc granules and 25 cm3

of 0.5 mol dm-3 hydrochloric acid with 2 cm3 of copper(II) sulphatesolution

15

Based on the table, compare the rate of reaction between Experiment I and II Experiment I and III

In each case, explain the difference in the rate of reaction with reference to the collision theory.

[10 marks]

ANSWER

Experiment I and II

The rate of reaction in Experiment II is higher. The concentration of hydrochloric acid in Experiment II is higher. The number of hydrogen ions per unit volume hydrochloric acid in

Experiment II is higher. The frequency of collision between zinc (atom) and hydrogen ions in

Experiment II is higher. The frequency of effective collision between zinc (atom) and hydrogen

ions in Experiment II is also higher.

Experiment I and III

The rate of reaction in Experiment III is higher. Copper(II) sulphate catalyst present in Experiment III provide an

alternative path with lower activation energy. More particles that collide be able to achieve this lower activation

energy. The frequency of collision between zinc (atom) and hydrogen ions in

Experiment III is higher. The frequency of effective collision between zinc (atom) and hydrogen

ions in Experiment III is also higher.

ANSWER

Experiment I Experiment IIObservation at anode

Gas bubbles are released

Copper anode becomes thinner

Observation at cathode

Brown solid deposited Copper cathode becomes thicker // Brown solid deposited

Product at anode Oxygen gas Copper (II) ionsHalf equation at anode

4OHˉ → O2 +2H2O + 4e Cu → Cu2+ + 2e

QUESTION 6

Diagram shows the apparatus set- up of two experiments to investigate the electrolysis of copper(II) sulphate solution using different electrodes.

Compare Experiment I and Experiment II in terms of the observations at the anode and cathode, the name of the products formed at the anode half equations for the reactions at the anode.

[10 marks]

QUESTION 7

(a) Describe how the process of iron rusting occurs naturally in daily life.

(b) Why metal easily corrodes in the area closer to the sea? [9marks]

ANSWER(a)

Rusting of iron occurs in the presence of water and oxygen.

Iron becomes anode which is in contact with water and oxygen.

Iron atoms release electrons to form iron(II) ions// Fe → Fe2+ + 2e

The electron flows to the edge of the water circle at cathode where a lot of dissolve oxygen can be found.

Oxygen and water accept electron to form OH- ion //O2 + 2H2O +4e →

4OH ̄ Hydroxide ions, OH- ions combined with iron(II) ions to form iron(II)

hydroxide// Fe2+ + 2OH ̄ → Fe(OH)2

Iron(II) hydroxide is then further oxidized to form hydrated iron(III) oxide known as rust.

(b) Sea water contains salt . Salt solution acts as an electrolyte.

SECTION C [OPEN END]

QUESTION 1

You are required to prepare dry copper (II) sulphate salt.

By using suitable chemicals, describe a laboratory experiment to prepare the salt. In your description, include the chemical equations involved.

[10 marks]

ANSWER

Chemicals/Materials: Copper(II) oxide powder, 0.5 mol dm-3 sulphuric acid, distilled water.

Procedure:1. Measure 100 cm3 of 0.5 mol dm-3 sulphuric acid solution by using measuring cylinder and then pour into a beaker. Heat the solution gently.2. Add copper(II) oxide powder little by little until in excess.3. Stir the mixture with glass rod.4. The heating is stopped when copper(II) oxide is no longer dissolve in acid.5. Filter the mixture. 6. Heat the filtrate of copper(II) sulphate solution until one-third its original volume.7. Cool the saturated copper(II) sulphate solution to room temperature.8. Filter the copper(II) sulphate crystals.9. Rinse the copper(II) sulphate crystals.10. Press the copper(II) sulphate crystals between two filter papers.

Equation:CuO + H2SO4 → CuSO4 + H2O

QUESTION 2

Silver chloride is insoluble salt. You are required to prepare dry silver chloride salt.

Describe a laboratory experiment to prepare the salt. In your description, include the procedure, observation and ionic equation.

[10 marks]

ANSWER

Materials: 0.5 mol dm-3 silver nitrate, 0.5 mol dm-3 sodium chloride, distilled water.

Procedure:1. Measure 20 cm3 of 0.5 mol dm-3 silver nitrate solution using measuring cylinder and pour into a beaker.2. Measure 20 cm3 of 0.5 mol dm-3 sodium chloride solution using measuring cylinder and pour into the same beaker.3. Stir the mixture with glass rod.4. Filter the mixture.5. Rinse the residue of silver chloride with distilled water.6. Press the silver chloride salt between two filter papers.

Observation:White precipitate is formed. Silver chloride is white solid.

Ionic equation: Ag+ + Cl¯ → AgCl

QUESTION 3

You are given a solution that contains a mixture of iron(III) nitrate and iron(III) chloride.

Describe the confirmatory tests to determine the presence of cation and anion in the solution. Your description must include all reagents used, observation and conclusion.

[10 marks]

ANSWERTest for iron (III) ion:1. Pour 2 cm3 of solution into a test tube.2. Add 3 drops of potassium hexacyanoferrate(II) solution into the test tube.3. Shake the test tube.4. Observation: Dark blue colouration is formed.5. Iron(III) ion is confirmed present.

ORTest for iron (III) ion:1. Pour 2 cm3 of solution into a test tube.2. Add 3 drops of potassium thiocyanate solution into the test tube.3. Shake the test tube.4. Observation: Blood red colouration is formed.5. Iron(III) ion is confirmed present.

Test for nitrate ion:1. Pour 2 cm3 of solution into a test tube.2. Add 2 cm3 of dilute sulphuric acid followed by 2 cm3 of iron(II) sulphate solution into the test tube.3. Add concentrated sulphuric acid drop by drop slowly into a slanted test tube using a dropper.4. Observation: Brown ring is formed.5. Nitrate ion is confirmed present.

Test for chloride ion:1. Pour 2 cm3 of solution into a test tube.2. Add 2 cm3 of dilute hydrochloric acid followed by 2 cm3 of silver nitrate solution into the test tube.3. Shake the test tube.4. Observation: White precipitate is formed.5. Chloride ion is confirmed present.

QUESTION 4

Describe two chemical properties to verify the solution is an acid.In your answer, include example of an acid, procedure, observatiom and chemical equation for the reaction involved.

[8 marks]

ANSWER

Example of acid: Hydrochloric acid

Reaction of hydrochloric acid with Magnesium:1. Pour 3 cm3 of 0.5 mol dm-3 hydrochloric acid into a test tube.2. Add 1 cm of magnesium ribbon into the test tube.3. Place a lighted wooden splinter near the mouth of the test tube.4. Observation: Gas bubbles are released. A “pop” sound is produced.5. Equation: 2HCl + Mg → MgCl2 + H2

Reaction of hydrochloric acid with calcium carbonate:1. Pour 3 cm3 of 0.5 mol dm-3 hydrochloric acid into a test tube.2. Add 1 g of calcium carbonate powder into the test tube.3. Close the test tube with stopper fitted with delivery tube. The delivery tube is immersed into lime water.4. Observation: Gas bubbles are released. Lime water turns chalky.5. Equation: 2HCl + CaCO3 → CaCl2 + CO2 + H2O

QUESTION 5Substance Molecular

formulaV C6H14T C6H12

Table above shows two hydrocarbon compounds V and T which are colourless liquids at room temperature. Describe one chemical test used in the laboratory to differentiate liquid V and T. In your explanation, include observation, conclusion and chemical equation.

[8 marks]

ANSWER

Procedure:1. Pour 3 cm3 of liquid V into a test tube.2. Add 3 drops of bromine water into the test tube.3. Shake the test tube.4. Steps 1 to 3 are repeated using liquid T.

Observation:Compound

Observation

Liquid V No change Liquid T Brown solution turns

colourlessEquation:C6H12 + Br2 → C6H12Br2

Conclusion: Liquid V is alkane and liquid T is alkene.OR

Procedure:1. Pour 3 cm3 of liquid V into a test tube.2. Add 3 drops of acidified potassium manganate(VII) into the test tube.3. Shake the test tube.4. Steps 1 to 3 are repeated using liquid T.

Observation:Compound

Observation

Liquid V No change Liquid T Purple solution turns

colourlessEquation:C6H12 + [O] + H2O → C6H12(OH)2

Conclusion: Liquid V is alkane and liquid T is alkene.QUESTION 6

A student intends to prepare an ester Q from the reaction between named alcohol and named carboxylic acid. Describe a laboratory experiment to prepare the ester. Your answer should include the following :

A list of material Procedure of the experiment Observation and chemical equation Name of the ester produced [10 marks]

ANSWER

Materials: Ethanol, ethanoic acid, concentrated sulphuric acid, distilled water.

Procedure:1. Pour 3 cm3 of ethanol into a boiling tube.2. Add 3 cm3 of ethanoic acid into the boiling tube and shake the boiling tube.3. Add 2 cm3 of concentrated sulphuric acid slowly into the boiling tube.4. Heat the mixture until it boils. 5. Boils the mixture for 5 minutes.6. Pour the mixture into a beaker containing distilled water. Record the observation.

Observation:Colourless solution is formed.The sweet smell product is produced.

Equation: CH3COOH + C2H5OH → CH3COOC2H5 + H2O

Name of ester: ethyl ethanoate

QUESTION 7

Ionic compounds can conduct electricity in the molten state but cannot conduct electricity in the solid state.

Name one example of the ionic compound. Describe an experiment to prove the above statement. Your description should include the following:

List of apparatus Procedure Observation [8

marks]

ANSWER

Ionic compound: lead(II) bromide

Apparatus: Crucible, pipe clay triangle, tripod stand, Bunsen burner, glass rod, two carbon electrodes, two dry cells, ammeter, switch, connecting wire.

Procedure:1. Fill the crucible with solid lead(II) bromide solid until it is half full.2. Dip two carbon electrodes into the crucible.3. Connect the electrodes to dry cells, ammeter and switch by using connecting wire.4. Turn on the switch and the reading of ammeter is recorded.5. Heat the crucible strongly until it melts.6. Turn on the switch and the reading of ammeter is recorded.

Observation:Lead(II) bromide solid: No deflection of the ammeter pointerLead(II) bromide molten: Ammeter pointer shows deflection