Model Test 1 Paper 2

PAPER 2

Two hours and thirty minutes

Section A ( 60 marks) Instruction : Answer all questions in this section. The time suggested to answer Section A is 90 minutes 1

Substance Physical condition Melting point/ °C Boiling point/ °C Benzene -55 16 P Iodine 20 20 Q Naphthalene 79 115 Solid Sodium chloride 130 800 Solid Table 1 Table 1 shows four types of substance at room temperature. (a) State the condition of i. P : ii. Q: (2m) (b) Name the particles in i. Benzene : ii. Sodium chloride : (2m) (c) Explain why the boiling point to sodium chloride is high. (2m) (d)i. Write down the chemical formula for iodine at room temperature and pressure. (1m) ii. What is the bond formed between the particles of iodine. (1m) iii. Give one example that can vapourise when heated. (1m) (e) Explain why the temperature becomes even for a short time when the naphthalene melts although the heating process continues. (1m)

Answers: (a)i. gas ii. gas/solid (b)i. molecule ii. ions (c) Strong electrostatic forces of attraction exist between sodium ion and chlorine ion. A lot of energy are needed to break the attraction force. (d)i. I 2 ii. Van der Waals bond iii. dry carbon dioxide (e) The heat energy used to break the bonds between the molecules of naphthalene. 2 Solution pH value

P 1

Q 4

R 10

S 14

Table 2 Four types of solution with the same concentration are poured into four different test tubes. The solutions are tested with universal indicator. The results are recorded in Table 2. (a) Which of the following solutions is i. strong acid? (1m) ii. weak alkali? (1m) (b)Why the pH values of solutions P and Q are different although their concentrations are the same? (2m) (c) If solution R is ammonia, write the equation for the ionization of ammonia in water. (1m) 3 -3 3 (d) 20 cm of 0.5 mol dm reacts with 50 cm of sodium hydroxide completely. i. Name the type of reaction that takes place. (1m) ii. Write down the equation of the reaction. (1m) −3 iii. Calculate the concentration of sodium hydroxide solution in mol dm . (2m)

Answers: (a)i. P ii. R (b) P ionises completely in water and produces a high concentration of H + ion. Q ionises partially in water and produces a low concentration of H + ion. + (c) NH 3 (g) + H 2 O(l) NH 4 (aq) + OH - (aq) (d)i. neutralisation ii. HCl(aq) + NaOH(aq) → NaCl(s) + H 2 O(l) 20 × 0.5 = 0.2 mol dm -3 iii. concentration of NaOH = 50 3.

When 250 cm 3 of 2 mol dm −3 calcium chloride solution is poured into a beaker containing 250 cm 3 of 2 mol dm −3 sodium carbonate solution, a white precipitate formed and the temperature of the solution decreases by 3 °C .

[Heat capacity of solution: 4.2 J g -1 °C -1 ] (a) Name the white precipitate formed. (1m) (b) Write the equation for the reaction that takes place. (1m) (c) Calculate the heat change for this experiment. (1m) (d) Calculate the heat of precipitation. (2m) (e) Draw the energy level diagram for the reaction that takes place. (2m) (f) 25 cm 3 of 2 mol dm −3 calcium chloride solution is mixed with 25 cm 3 of 2 mol dm −3 sodium carbonate solution. Will the change in the volume affects the drop in temperature? Explain your answer. (2m)

Answers: (a) Calcium carbonate (b) CaCl 2 (aq) + Na 2 CO 3 (aq) → CaCO 3 (s) + 2NaCl(aq) (c) ∆H = mcθ = 500 × 4.2 × 3 = 6300 J 2− (d) Number of moles of Ca 2+ ion used = number of moles of CO 3 ion used 250 × 2 = = 0.5 mole 1000

Ca 2+ (aq) + CO 3 (aq) → CaCO 3 (s) 1 mole 1 mole 1 mole 2− 0.5 mole of Ca 2+ ion react with 0.5 mole CO 3 ion to produce 0.5 mole of 2-

CaCO 3 precipitate, where 0.5 mole of CaCO 3 absorbs 6300 J heat energy. ∴ Heat of precipitation of one mole of CaCO 3 precipitate 1 × 6300 = 0.5 × 1000 = 12.5 kJ mol −1 (e) CaCO 3 (s) + 2NaCl(aq) Energy

CaCl 2 (aq) + Na 2 CO 3 (aq)

∆H = +12.6 kJ

(f) No, because the volume of solution changes but the concentration does not change. 4

Dropper Bromine water Test tube

Iron(II) chloride solution

Figure 1 Bromine water is added drop by drop into iron(II) chloride solution as shown in Figure 1. The test tube is shaken until there is no more changes. (a) State the chemical change that occurs to i. iron(II) chloride solution (1m) ii. bromine water (1m) (b) Name the chemical process that takes place to iron(II) ion. (1m) (c) What is the change in oxidation number of bromine in this reaction? (1m) (d) In the above reaction, name the substance that acts as i. oxidising agent. (1m) ii. reducing agent. (1m) (e) Write the ionic equation for the overall reaction. (1m) (f)i. Give one substance that can change iron(III) ion to iron(II) ion in the laboratory. (1m) ii. How can you confirm the presence of iron(II) ion? (1m)

Answers: (a)i. The colour change from light green to yellow ii. The colour change from brown to colourless (b) Oxidation (c) 0 to -1 (d)i. Bromine ii. Fe 2+ ion (e) 2Fe 2+ (aq) + Br2 (aq) → 2Fe 3+ (aq) + 2Br - (aq)

(f)i. Zinc powder ii. Add potassium hexacyanoferrate(III) solution and blue precipitate formed. 5

Figure 2 shows the flow chart for the reactions of substances W, X, Y and Z. Liquid W Excess concentrated H 2SO 4 burnt

Gas X

bromine water

sodium blue flame

colour bleached The gas released burns with a ‘pop’ sound

Liquid X pH paper

+W Water

heated

3.5 value Colourless solution

Steam Y with a nice smell

Figure 2 (a) Introduce W, X, Y and Z by writing the names of each of them. W: Y: X : Z: (2m) (b) Write the equation that represents combustion of W. (1m) (c) Write the equation that represents the formation of Y from W and X. (1m) (d) What is type of reaction between Z and bromine? Write a chemical equation to represent this process. (2m) (e) What is the name of the process of the formation of Z from W? Write a chemical equation to represent this process. (2m)

Answers: (a)W : Ethanol X :Ethanoic acid Y :Ethyl ethanoic Z : Ethene (b) C 2 H 5 OH + 3O 2 → 2CO 2 + 3H 2 O CH 3 COOC 2 H 5 + H 2 O (c) CH 3 COOH + C 2 H 5 OH (d) The addition reaction, C 2 H 4 + Br2 → C 2 H 4 Br2 (e) Dehydration C 2 H 5 OH → C 2 H 5 + H 2 O 6 Four homologous series of organic compound can be represented by the following general formulae: C n H 2n C n H 2n +1COOH Series A Series C C n H 2n + 2 C n H 2n +1OH Series B Series D (a) Name an important natural source for the compounds in Series B. (1m) (b) Using chemical equation, show a way where… i. A compound chosen from Series A can be changed into a compound in Series B. (1m) ii. A compound chosen from Series D can be changed into a compound in Series A.(1m) (c) Write the balanced chemical equations to show how a compound chosen from Series C can react with: i. magnesium tape (1m) ii. sodium carbonate (1m) iii. sodium hydroxide (1m) iv. a compound chosen from Series D (1m)

Answers: (a) Petroleum (b)i. C 2 H 4 + H 2  → C 2 H 6 Ni(Catalyst)

2 4 → C 2 H 5 + H 2 O ii. C 2 H 5 OH    

Concentrated H SO Excess

(c)i. 2CH 3 COOH + Mg → (CH 3 COO) 2 Mg + H 2 ii. 2CH 3 COOH + Na 2 CO 3 → 2CH 3 COONa + H 2 O + CO 2 iii. CH 3 COOH + NaOH → CH 3 COONa + H 2 O CH 3 COOC 2 H 5 + H 2 O iv. CH 3 COOH + C 2 H 5 OH 7

(a) Glass and ceramic are two important substances in trading. Name the main raw material to make i. glass. (1m) ii. ceramic. (1m) (b) What extra chemicals are added to glass to obtain i. borosilicate glass? (1m)

ii. lead glass? (c) State two uses of borosilicate glass. (d) Give one main difference between ceramic and glass. (e) State one advantage of glass compared to ceramic.

(1m) (2m) (1m) (1m)

Answers: (a)i. sand ii. clay (b)i. Boron oxide ii. Lead(II) oxide (c)1. Make plates and bowls 2. Make flask and test tube (d) The glass is transparent, whereas ceramic is opaque (e) Glass can be melted to other shapes but ceramic cannot. Section B ( 20 marks) Instruction : Answer any one question in this section. The time suggested for Section B is 30 minutes. 1

The rusting of steel is an example of chemical corrosion that creates a lot of problem in the metal industries today.

(a) Explain the meaning of metal corrosion. (2m) (b) Explain in detail how the corrosion of steel occurs. Your explanation must include i. a suitable diagram (2m) ii. the condition for the formation of rust (2m) iii. the rusting process (8m) (c) Experiment I II Condition

Steel nail Metal R wire Sodium nitrate solution+ potassium hexacyanoferrate solution

Steel nail Metal T wire Sodium nitrate solution+ a little potassium hexacyanoferrate solution

Observation after No dark blue colour. Dark blue colour formed. two days Steel nail does not rust. Steel nail rusts. Table 1 A student carried out two experiments to investigate the effect of metals R and T towards rusting. The observations obtained are shown in Table 1. Compare both experiments and explain the observations. (6m)

Answers: (a) A process of changing metal atoms into their positive ions by losing of electrons in the presence of air(oxygen), electrolyte and water, and metals that are less electropositive than the metal that corrode. (2m) (b)i. Water droplet Cathode

iron

O 2 (g) + 2H 2 O(l) + 4e - → 4OH Anode

Fe(s) → Fe 2 + (aq) + 2e

(2m) ii. Steel rusts in the presence of water and air. (2m) iii.- The air layer at the edge of the water droplet is thinner than the center causing the concentration of dissolved oxygen higher at the edge. (1m) - The edge of the water droplet acts as cathode and the center of the water droplet acts as anode. - At anode : Fe(s) → Fe 2 + (aq) + 2e (1m) 2+ Iron metal corrodes to form Fe ion through oxidation by losing electron. (1m) - The electron released at anode flows to cathode. (1m) - The water and oxygen molecule receive electron and produce OH − ion by reduction. (1m) - At cathode: O 2 (g) + 2H 2 O(l) + 4e → 4OH (aq) (1m) − 2+ - OH ion combines with Fe ion to form iron(II) hydroxide which is a green solid. (1m) - Then iron(II) hydroxide is oxidised by oxygen in the air to hydrated iron(III) oxide ( Fe 2 O 3 .2H 2 O ) or rust. Fe(OH) 2 (s) Oxidised  → Fe 2 O 3 .2H 2 O(s) (1m) rust

(c) In experiment I, metal R is more electropositive than steel. - This causes metal R corrode through oxidation by losing electron to become metal R ion. (1m) - Therefore, the steel nail will be protected from corroding. (1m) - Hence iron(II) ion will not be produced and steel will not rust. (1m) - In experiment II, steel is more electropositive than metal T. (1m) - This causes steel nail corrode through oxidation by releasing electron to become iron(II) ion. (1m) - Iron(II) ion react with potassium hexacyanoferrate(III) to produce dark blue precipitate and changed to rust by the air. (1m) - Metal T is protected from corrosion. (1m) Maximum: 6 marks

2 Salt is an ionic compound prepared from acid. (a) Based on the statement above, state four methods of preparing soluble salt from acid. Each method must be accompanied with suitable examples. (8m) (b) A student wants to determine the pH value of two types of acidic solution with the same concentration. Table 2 shows two results obtained. Type of acid

Hydrochloric acid 0.1 mol dm −3 2

Ethanoic acid 0.1 mol dm −3 6

Concentration pH Table 2 i. What is represented by the pH value? (2m) ii. Compare the type of acid, explain why there is a difference in pH value for every solution. (6m) 3 −3 (c) 5.0 g of zinc powder is left to react with 50 cm of 0.2 mol dm sulphuric acid. Calculate the maximum volume of hydrogen gas produced? [Relative atomic mass: Zn,65; Molar volume of gas: 24 dm 3 mol -1 at room condition] (4m)

Answers: (a) The method of preparation of soluble salt by acid : - Reaction between acid and alkali. (1m) Example: The reaction between hydrochloric acid and sodium hydroxide produces sodium chloride solution. - HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) (1m) - Reaction between acid and metal oxide. (1m) Example: The reaction between copper(II) oxide and sulphuric acid to produce copper sulphate solution. CuO(s) + H 2 SO 4 (aq) → CuSO 4 (aq) + H 2 O(l) (1m) - Reaction between acid and reactive metal. (1m) Example: The reaction between zinc and sulphuric acid produces zinc sulphate solution. - Zn(s) + H 2 SO 4 (aq) → ZnSO 4 (aq) + H 2 (g) (1m) - Reaction between acid and metal carbonate. (1m) Example: The reaction between nitric acid and copper carbonate produces copper(II) nitrate salt. - 2HNO 3 (aq) + CuCO 3 (aq) → Cu(NO 3 ) 2 (aq) + CO 2 (g) + H 2 O(l) (1m) (b)i.- The pH value represents the acidity strength for any acidic solution that is closely + related to the concentration of H + ion or hydroxonium ion, H 3 O which exist in solution. (1m) + - The concentration of H ion depends on the degree of ionisation of acid when dissolves in water. (1m) ii.- Dilute hydrochloric acid is a strong acid which has a complete ionisation in water when dissolves in water. (1m) + - This is because the HCl molecule will dissociate completely to form H and Cl − ions. HCl(aq) → H + (aq) + Cl - (aq) (1m) - Therefore, the concentration of H + ion in HCl solution is higher and the pH value is low. (1m) + - Ethanoic acid is a weak acid that ionises partially to produce H ions and its anions. (1m) - This causes the pH value for weak acid is high. (1m) - Strong acid has a lower pH value than weak acid because it has a higher concentration of H + ion than the concentration of H + ion in weak acid of the same concentration. (1m) (c)- Zn(s) + H 2 SO 4 (aq) → ZnSO 4 (aq) + H 2 (g) (1m) - Number of moles of H 2 SO 4 reacted MV 50 × 0.2 = = = 0.01 mole (1m) 1000 1000 - From the equation, 1 mole of sulphuric acid reacted will release hydrogen gas of 0.01 x 24 dm 3 =0.24 dm 3 (1m)

Section C ( 20 marks) Instruction : Answer any one question in this section. The time suggested for Section C is 30 minutes 3

(a) Salt plays an important role in our daily lives. Give two uses of salt in our daily lives. (2m) (b) Figure 1 shows an incomplete flow chart for the anion and cation tests of salt X. Salt X

Cation test Cu 2+

Anion test NO 3

−

Figure 1 Use suitable reagents to confirm salt X in copper(II) nitrate solution. State your observations. (8m) (c) Your are required to prepare dry copper(II) sulphate salt. The substance provided are • Copper(II) sulphate • Suitable acid solution Explain one laboratory experiment on how you prepare the salt. In your explanation, include the equations involved. (10m)

Answers: (a)1. Fertiliser 2. Flavouring (b) Experiment to confirm the presence of cation and anion in salt solution M:

(1m) (1m)

1 g of solid copper(II) nitrate is dissolved with 5 cm 3 of water to produce aqueous solution. (1m) 2+ i. Confirmation test of cation, Cu - 2 cm 3 of aqueous copper(II) sulphate solution is poured into a test tube. (1m) - Sodium hydroxide solution is added drop by drop and shaken until precipitate formed. (1m) - Observation: Aqueous copper(II) ion solution produces blue copper(II) hydroxide precipitate which is insoluble in excess sodium hydroxide. (1m) 2+ + Cu (aq) + 2NaOH(aq) → Cu(OH) 2 (s) + 2Na (aq) (1m) −

ii. Confirmation test of anion, NO 3 - 2 cm 3 of aqueous copper(II) sulphate solution is poured into a test tube. (1m) - 2 cm 3 of dilute sulphuric acid is added, followed by 2 cm 3 of iron(II) sulphate solution and shaken. (1m) - Concentrated sulphuric acid is dropped slowly on the wall of the tilted test tube without shaking the test tube. (1m) - Observation: A brown ring formed between concentrated sulphuric acid and solution. (1m) - The reaction between nitrate ion and concentrated sulphuric acid is reduced by iron(II) sulphate to nitrogen monoxide. Then, nitrogen monoxide will combine with iron(II) sulphate to produce brown FeSO 4 .NO compound (brown ring). (1m) Maximum: 8 marks (c) The experiment to prepare copper(II) sulphate crystals through the following equation: CuO(s) + H 2 SO 4 (aq) → CuSO 4 (aq) + H 2 O(l) (1m) Glass rod

Copper(II) oxide powder

Beaker Wire gauze Tripod stand

Retort stand

Dilute sulphuric acid solution Heat

Glass rod

Evaporating dish mixture Filter paper Filter funnel Residue, excess copper(II) oxide powder Filtrate, copper(II) sulphate

(1m) Copper(II) sulphate solution

Heat

(1m)

(1m)

Method: - 50 cm 3 of dilute hydrochloric acid solution is measured and poured into a beaker. The solution is heated. Copper(II) oxide powder is added until saturated. (1m) - The mixture is stirred a glass rod so that the reaction takes place perfectly. Heating is done to fasten the reaction. (1m) - The mixture is filtered to eliminate the excess copper(II) oxide powder. (1m) - The filtrate is copper(II) sulphate solution which is soluble salt. (1m) - The filtrate is transferred to an evaporating dish. Heating is continued until the solution left is 1/3 of the initial volume of the solution. (1m) - The concentrated solution is left to cool at room temperature. (1m) - The copper(II) sulphate crystals is filtered, rinsed with distilled water, and dried between two filter papers. (1m) Maximum: 10 marks 4

(a) Define heat of neutralisation. (b) The following figure shows two energy profile levels for endothermic and exothermic reactions. Energy

(1m)

Energy

Figure 2(a) Figure 2(b) Compare both energy levels in Figure 2, then explain how both of them is related to the breaking and the formation of bonds. (9m) (c) Explain one experiment to determine the heat of neutralisation between nitric acid and potassium hydroxide. In your answer, include • Labelled diagram • Precaution steps while carrying out the experiment • An energy level diagram for the neutralisation reaction • Explain why the heat of neutralisation obtained from this experiment is less than the real heat of neutralisation. (10m)

Answers: (a) Heat of neutralisation is the heat released when one mole of H + ion from the acid neutralises one mole of OH − ion from the alkali to form one mole of water. (1m) (b)- Figure 2(a) is an exothermic reaction that releases heat. (1m) - Heat is released during the formation of bond. (1m) - The total energy released depends on the strength of bond (1m) - In an exothermic reaction, the total energy released for the formation of bond is higher than the total energy absorbed for the breaking of bond. (1m) - Hence heat is released to the surrounding. (1m) - Figure 2(b) is an endothermic reaction that absorbs heat. (1m) - Heat is absorbed during the breaking of bond (1m) -In an endothermic reaction, the total energy absorbed for the breaking of bond is higher than total energy released for the formation of bond. (1m) - Hence heat is absorbed from the surrounding. (1m) (c) 1 mol dm −3 nitric acid

Thermometer

1 mol dm −3 potassium hydroxide solution Plastic cup

(2m) - 10 cm of 1 mol dm KOH is poured into a plastic cup and the initial temperature of KOH and HNO 3 are recorded. (1m) 3

−3

- 10 cm 3 of 1 mol dm −3 HNO 3 is measured and poured into KOH and the mixture is stirred with thermometer. (1m) - Record the highest temperature reached by the solution. (1m) - Calculation step: If T1 = initial temperature of KOH, T2 = initial temperature of HNO 3 Average initial temperature = T1 + T2 = T3 (1m) T4 = highest temperature reached by the mixture Temperature change, θ = T4 – T3 =T Mass of solution = 10 + 10 = 20 g (1m) - Assume the density of solution = 1 g cm −3 Heat capacity of solution = 4.2 J g -1 °C -1 Using the heat change formula, ∆H = mcθ where m = mass of solution c = heat capacity of solution θ = change in temperature (1m) - Precaution steps during the experiment: • Use plastic cup so that less heat is absorbed (1m) • Mixture of solutions is stirred frequently and left for a while until the highest temperature recorded. (1m) • When taking the initial reading of KOH and HNO 3 , stir the solution with thermometer so that an even temperature is obtained. (1m) - Energy level diagram for the reaction (1m) HNO 3 (aq) + KOH(aq) Energy

∆H = negative

H 2 O(l) + KNO3 (aq) - The value obtained from the experiment is less than the standard value because heat is absorbed by the apparatus used and also released into the air. (1m) Maximum:10 marks