Model Test 1 Paper 2

8/14/2019 Model Test 1 Paper 2

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PAPER 2 Two hours and thirty minutes

Section A ( 60 marks)

Instruction :Answerallquestions in this section. The time suggested to answerSection A is 90

minutes

1

Substance Melting point/ C Boiling point/ C Physical condition

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)


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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 2ii. Van der Waals bond

iii. dry carbon dioxide

(e) The heat energy used to break the bonds between the molecules of naphthalene.

2

Solution P Q R SpH value 1 4 10 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)(d) 20 cm -33 dmmol0.5of reacts with 50 cm 3 of sodium hydroxide completely.

i. Name the type of reaction that takes place. (1m)ii. Write down the equation of the reaction. (1m)

iii. Calculate the concentration of sodium hydroxide solution in mol dm 3 . (2m)


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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) O(l)H(g)NH 23 + (aq)OH(aq)NH-

4 ++

(d)i. neutralisation

ii. O(l)HNaCl(s)NaOH(aq)HCl(aq) 2++

iii. concentration of NaOH =3-dmmol2.0

50

5.020=

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 .


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[Heat capacity of solution: 4.2 J-1-1 Cg ]

(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 intemperature? Explain your answer. (2m)

Answers:

(a) Calcium carbonate

(b) 2NaCl(aq)(s)CaCO(aq)CONa(aq)CaCl 3322 ++

(c) 32.4500 == mcH

= 6300 J

(d) Number of moles of Ca +2 ion used = number of moles of CO2

3 ion used

=1000

2250= 0.5 mole


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(s)CaCO(aq)CO(aq)Ca 3-2

3

2+

+

1 mole 1 mole 1 mole

0.5 mole of Ca +2 ion react with 0.5 mole CO2

3 ion to produce 0.5 mole of

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

=10005.0

63001

= 12.5 kJ mol 1

(e)

kJ6.12+=H

(aq)CONa(aq)CaCl 322 +

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

4

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 asi. 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)

Dropper

Iron(II) chloride solution

Bromine water

Test tube

Energy

2NaCl(aq)(s)CaCO3 +


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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) (aq)2Br(aq)2Fe(aq)Br(aq)2Fe-3

2

2++

++


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(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.

Excess concentrated42

SOH

burnt

bromine watersodium

blue flame colo

pH paper +W Water heated 3.5 value

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)

Liquid W

The gas released burns with a

pop sound

Gas X

Colourless solution

Liquid X

Steam Y with a nice smell


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Answers:

(a)W : Ethanol

X :Ethanoic acidY :Ethyl ethanoic

Z : Ethene

(b) O3H2CO3OOHHC 22252 ++

(c) OHHCCOOHCH 523 + OHHCOOCCH 2523 +

(d) The addition reaction, 242242 BrHCBrHC +

(e) Dehydration OHHCOHHC 25252 +

6 Four homologous series of organic compound can be represented by the following general

formulae:

Series A 2nn HC Series C COOHHC 12nn +

Series B 22nn HC + Series D OHHC 12nn +(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 SeriesC can react with:

i. magnesium tape (1m)ii. sodium carbonate (1m)

iii. sodium hydroxide (1m)

iv. a compound chosen from Series D (1m)


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Answers:

(a) Petroleum

(b)i. 62t)Ni(Catalys

242 HCHHC +

ii. OHHCOHHC 252SOHedConcentrat

5242

+

Excess

(c)i. 2233 HMgCOO)(CHMgCOOH2CH ++

ii. 223323 COOHCOONa2CHCONaCOOH2CH +++

iii. OHCOONaCHNaOHCOOHCH 233 ++

iv. OHHCCOOHCH 523 + OHHCOOCCH 2523 +

7 (a) Glass and ceramic are two important substances in trading. Name the main raw

material to makei. glass. (1m)

ii. ceramic. (1m)

(b) What extra chemicals are added to glass to obtain

i. borosilicate glass? (1m)


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ii. lead glass? (1m)

(c) State two uses of borosilicate glass. (2m)(d) Give one main difference between ceramic and glass. (1m)

(e) State one advantage of glass compared to ceramic. (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 30minutes.

1The rusting of steel is an example of chemical corrosion that creates a lot of problem

in the metal industries today.


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(a) Explain the meaning of metal corrosion. (2m)

(b) Explain in detail how the corrosion of steel occurs. Your explanation must includei. a suitable diagram (2m)

ii. the condition for the formation of rust (2m)

iii. the rusting process (8m)

(c)Experiment I II

Condition

Observation after

two days

No dark blue colour.Steel nail does not rust.

Dark blue colour formed.Steel nail rusts.

Table 1

A student carried out two experiments to investigate the effect of metals R and Ttowards rusting. The observations obtained are shown in Table 1.

Compare both experiments and explain the observations. (6m)

Steel nailMetal R wire

Sodium nitrate

solution+ potassium

hexacyanoferrate

solution

Steel nail

Metal T wire

Sodium nitrate

solution+ a little

potassiumhexacyanoferrate

solution


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Answers:

(a) A process of changing metal atoms into their positive ions by losing of electrons inthe presence of air(oxygen), electrolyte and water, and metals that are less

electropositive than the metal that corrode. (2m)

(b)i. Water droplet

(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 dropletacts as anode.

- At anode :-2 2e(aq)FeFe(s) +

+(1m)

Iron metal corrodes to form Fe +2 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: (aq)4OH4eO(l)2H(g)O--

22 ++ (1m)

- OH ion combines with Fe +2 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

( O.2HOFe 232 ) or rust.

O(s).2HOFe(s)Fe(OH) 232Oxidised

2 (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 metalR 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

iron

Anode

2e(aq)FeFe(s) 2 + +

Cathode--

22 4OH4eO(l)2H(g)O ++


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2

Salt is an ionic compound prepared from acid. (a) Based on the

statement above, state four methods of preparing soluble salt fromacid. 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 thesame concentration. Table 2 shows two results obtained.

Type of acid

Hydrochloric acid Ethanoic acid

Concentration 0.1 mol dm 3 0.1 mol dm 3

pH 2 6 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)

(c) 5.0 g of zinc powder is left to react with 50 cm 3 of 0.2 mol dm 3 sulphuric acid.

Calculate the maximum volume of hydrogen gas produced?

[Relative atomic mass: Zn,65; Molar volume of gas: 24 dm -13 mol at room

condition] (4m)


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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.

- O(l)HNaCl(aq)NaOH(aq)HCl(aq) 2++ (1m)- Reaction between acid and metal oxide. (1m)

Example: The reaction between copper(II) oxide and sulphuric acid to produce

copper sulphate solution.

- O(l)H(aq)CuSO(aq)SOHCuO(s) 2442 ++ (1m)

- Reaction between acid and reactive metal. (1m)

Example: The reaction between zinc and sulphuric acid produces zinc sulphatesolution.

- (g)H(aq)ZnSO(aq)SOHZn(s) 2442 ++ (1m)

- Reaction between acid and metal carbonate. (1m)

Example: The reaction between nitric acid and copper carbonate produces

copper(II) nitrate salt.

- O(l)H(g)CO(aq))Cu(NO(aq)CuCO(aq)2HNO 222333 +++ (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,+OH3 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 waterwhen dissolves in water. (1m)

- This is because the HCl molecule will dissociate completely to form H + and Cl

ions.(aq)Cl(aq)HHCl(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)- (g)H(aq)ZnSO(aq)SOHZn(s) 2442++

(1m)- Number of moles of 42SOH reacted

=1000

2.050

1000

MV = = 0.01 mole (1m)

- 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)


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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.

Figure 1Use suitable reagents to confirm salt X in copper(II) nitrate solution. State yourobservations. (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 (1m)

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

Salt X

Cation test Anion test

Cu +2 3NO


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1 g of solid copper(II) nitrate is dissolved with 5 cm 3 of water to produce aqueous

solution. (1m)

i. Confirmation test of cation, Cu +2

- 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)

(aq)2Na(s)Cu(OH)2NaOH(aq)(aq)Cu 22 ++

++ (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 cm3 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 andsolution. (1m)

- The reaction between nitrate ion and concentrated sulphuric acid is reduced byiron(II) sulphate to nitrogen monoxide. Then, nitrogen monoxide will combine with

iron(II) sulphate to produce brown .NOFeSO4 compound (brown ring). (1m)

Maximum: 8 marks

(c) The experiment to prepare copper(II) sulphate crystals through the following

equation:

O(l)H(aq)CuSO(aq)SOHCuO(s) 2442 ++ (1m)

(1m)

(1m) (1m)

Glass rod

Beaker

Wire gauze

Tripod stand Dilute sulphuric acid

solution

Copper(II) oxide powder

HeatRetort

stand

mixture

Filter paperFilter funnel

Residue, excess

copper(II) oxide

powder

Filtrate, copper(II)

sulphate

Heat

Evaporating dishCopper(II)

sulphate

solution

Glass rod


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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. (1m)

(b) The following figure shows two energy profile levels for endothermic andexothermic reactions.

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)

Energy Energy


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- 10 3cm 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 1T = initial temperature of KOH, 2T = initial temperature of HNO 3

Average initial temperature = 1T + 2T

= 3T (1m)

4T = highest temperature reached by the mixture

Temperature change, = 4T 3T

= 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-1-1 Cg

Using the heat change formula, mcH=

where m = mass of solutionc = heat capacity of solution = 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)

KOH(aq)(aq)HNO3 +

Energy H = negative

(aq)KNOO(l)H 32 +

- The value obtained from the experiment is less than the standard value because heatis absorbed by the apparatus used and also released into the air. (1m)

Maximum:10 marks


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Model Test 1 Paper 2