Scheme Mid Year Exam f52012

7/31/2019 Scheme Mid Year Exam f52012

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Question

No

Description Marks

1 ( a ) ( i ) S- 11 ( a ) ( ii ) Atoms of the same element which have same number of

proton but different number of neutron // came proton

number but different nucleon number

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1( a ) ( iii ) 2.8.7 11 ( b ) ( i ) Period 4 11( b ) ( ii ) V atom have four shells filled up with electrons 11 ( c ) ( i ) PS4 11 ( c ) ( ii ) 2

1 ( d ) ( i ) V 11( d ) ( ii ) 2V + 2H2O 2VOH + H2 1

2 ( a ) A formula that shows simplest / smallest / lowest / ratio no

of mole of atoms of each element in the compound

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2 ( b )

( i )

( ii )

[Able to name suitable acid and metal and its equation]

For example:

Hydrochloric acid and zinc

Zn + 2HClZnCl2 + H2

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2 ( c ) To dry hydrogen gas 12 ( d ) Hydrogen gas must be flowed / through / into the

combustion tube for a few minutes before heating // flow of

hydrogen gas must be continuous throughout the

experiment

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2 ( e )Number of mole of copper =

64

62.1= 0.025 mole

Number of mole of oxygen =16

40.0= 0.025 mole

Empirical formula of copper(II) oxide is CuO

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2 ( f ) Iron(II) oxide / Tin(II) oxide / Lead(II) oxide 12 ( g ) Burning of metal in excess oxygen 13 ( a ) Water displacement method 1

3 ( b ) Mg + H2SO4 MgSO4 + H2 13 ( c ) ( i ) Answer must be in range ( 0.317-0.325)cm3s-1 13 ( c ) ( ii )


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3 ( c )

( iii )

Draw a tangent correctly

Answer must be in range ( 0.1375 + 0.001 cm3s-1 )

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13 ( d ) ( i ) The graph shows

- Curve with lower gradient

- Total volume of H2 gas is half of the volume of H2

gas in Experiment I

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3 ( d ) ( ii ) [able to explain in term of collision theory]

The concentration // number of H+ per unit volume in

hydrochloric acid is half

Frequency of collision between magnesium atom and

hydrogen ion decreases

Frequency of effective collision also decreases

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14 ( a ) ( i ) 3H2 + N22NH3 1

4 ( a ) ( ii ) Air 14 ( b ) ( i ) Neutralization 14 ( b ) ( ii ) H2SO4 + 2NH3(NH4)2SO4 //

H2SO4 + 2NH4OH (NH4)2SO4 + H2O

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4 ( b )

( iii )

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4 ( c ) ( i ) Tin 14 ( c ) ( ii )

No overlapping between the atoms

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4 ( c )

( iii )

- In pure copper the layer of atom are easily slide

over each other when external force is applied on

them

- The different size of foreign atoms in alloy are

prevent the layers of atom from slide each other

when external force is applied

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5 ( a ) ( i ) Magnesium oxide / magnesium / magnesium carbonate

Hydrochloric acid

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15 ( a ) ( ii ) 2


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5 ( b ) ( i ) Precipitation reaction // double decomposition reaction 15 ( b ) ( ii ) Ba2+ + SO4

2-BaSO4

[formula of reactants correct]

[formula of product correct]

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5 ( c ) Number of mole of hydrochloric acid = 0.1 mol

Number of mole of zinc chloride = 0.05 mol

Mass of zinc chloride = g8.613605.0 =

1

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6 ( a ) Propanol 16 ( b ) C3H6 +H2OC3H7OH 1

6 ( c ) ( i ) Dehydration 16 ( c ) ( ii ) Functional apparatus

Labelled diagram : glass wool soaked with alcohol P /

propanol , porcelain chips , delivery tube , a test tube

invert in a basin of water to collect propene

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6 ( d ) ( i ) Oxidizing agent 16 ( d ) ( ii ) Oxidation 16 ( d )

( iii )

Orange to green 1

6 ( e )( i ) Esterification 16 ( e ) ( ii ) CH3COOC3H7OH 1

Section B

7 ( a ) 1. Hydrogen gas

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2. 2H+ + 2e H2

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( b )

Properties Cell A Cell B1. Type of cell Voltaic cell Electrolytic cell

12. Energy change Chemical electrical Electrical chemical

13. Electrodes Positive terminal : Copper

Negative terminal :

Magnesium

Anode : Copper

Cathode : Copper

14 Ions in

electrolytes

Cu2+ , SO42- , H+ , OH- Cu2+ , SO4

2- , H+ , OH- 1

5. Half equation Positive terminal :

Cu2+ + 2e Cu

Negative terminal :

Mg Mg2+ + 2e

Anode :

CuCu2+ + 2e

1

Cathode :

Cu2+ + 2e Cu

16. Observation Positive terminal :

Copper plate becomes

thicker

Negative terminal :

Magnesium becomes thinner

/ dissolve

Anode :

Copper dissolves // becomes

thinner

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Cathode :

Copper becomes thicker

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( c ) ( i ) improve the appearance // to make it more attractive

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To prevent / reduce corrosion / rusting

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( ii ) Procedure :

1. iron ring is then connected to the negative plate on the battery while the

silver plate is connected to the positive terminal of the battery // iron ring is

made as cathode while silver plate is made as anode

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2. both plates are immersed into the silver nitrate solution

13. the circuit is completed

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Functional apparatus set-up

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Label correctly : silver plate . silver nitrate solution , iron ring

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Cathode : Ag+ +e Ag

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Observation : Grey / silvery solid is deposited

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Anode : Ag Ag+ + e

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Observation : Anode / silver become thinner // dissolve

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Question Description Marks8 ( a ) ( i ) Acid A : Hydrochloric acid // nitric acid // sulphuric acid

[accept any strong acid]

Acid B : Ethanoic acid

[accept any weak acid]

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8 ( a ) ( ii ) 1. pH value of acid A is lower than pH value of acid B

2. Acid A ionises completely in water to produce a higher

concentration of hydrogen ions.

3. Acid B ionises partially in water to produce a lower

concentration of hydrogen ion

4. the concentration of hydrogen ions in acid A is higher than

in acid B

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Max : 38 ( b ) ( i ) Solvent X : Water

Solvent Y : Benzene [accept any other suitable organic solvent]

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18 ( b ) ( ii ) In Beaker I:

1. Ethanoic acid ionises in water to form hydrogen ion

2. hydrogen ion in water react with calcium carbonate toproduce carbon dioxide gas // CO32- + 2H+ H2O + CO2

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3. Ethanoic acid does not ionises in Beaker II //

Etahnoic acid exists as molecules in Beaker II //

Hydrogen ions are not present in Beaker II to react with

calcium carbonate in Beaker II8 ( c ) ( i ) Standard solution is the solution with known concentration 1

8 ( c ) ( ii ) Preparation of 250 cm3

of 1 mol dm-3

of sodium hydroxide

solution

Material : solid sodium hydroxide , distilled water //

Apparatus : Electronic balance , beaker , 250cm3 volumetric

flask , filter funnel , glass rod

[accept from labelled diagram / description]

Calculation :

No of moles of NaOH =1000

2501

= 0.25 mol

RFM of NaOH = 23 + 16 + 1 = 40

Mass of NaOH = 4025.0

= 10 g

Procedure :

1. Weigh 10g of solid sodium hydroxide and dissolve in 100

cm3 of distilled water in a beaker.

2. Stir the solution using a glass rod

3. Pour the solution into 250cm3 volumetric flask using a filter

funnel

4. Rinse the beaker , filter funnel with distilled water and

transfer the solution into volumetric flask.

5. Add distilled water drop by drop into the volumetric flask

until reaches the graduation mark

6. stopper the volumetric flask and shake the volumetric flask

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Max :10

9 ( a ) ( i ) Copper chloride 19 ( a ) ( ii ) Silver chloride 19 ( a ) ( iii ) Anion : chloride ion // Cl-

Cation : Copper (II) ion , Cu2+

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19 ( b ) Confirmatory test for zinc ion

1. zinc nitrate crystals is dissolved in distilled water

2. the solution obtained is poured into two different test tubes

3. for the first test tube, ammonia solution is added drop by

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drop until excess

4. a white precipitate soluble in excess ammonia solution is

formed. This confirms the presence of zinc ions, Zn2+

Confirmatory test for NO3- ions

1. Add dilute sulphuric acid followed by iron(II) sulphatesolution

2. shake the test tube to mix well

3. slowly add concentrated sulphuric acid

4. brown ring is formed

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Max: 6

9 ( c ) ( i ) All soluble salt carbonate. eg : sodium carbonate 19 ( c ) ( ii ) 1. ( 25-50 ) cm3 of ( 0.5-2.0 ) moldm-3 magnesium sulphate

solution is measured and is put into a beaker.

2. ( 25-50 ) cm3 of solution X is added into the beaker

3. the mixture is stirred by using a glass rod

4. the mixture is filtered to get the precipitate

5. the precipitate is rinsed with distilled water to remove

impurities

6. the precipitate is then dried between 2 filter papers

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Max : 49 ( c ) ( iii ) 1. ( 25-50 ) cm3 of dilute / ( 0.5-2.0 ) moldm-3 nitric acid is

measured and poured into a beaker and is heated gently

2. the precipitate obtained from ( c ) ( ii ) is added into nitric

acid a little at time until excess

3. the mixture is stirred by using a glass rod and filtered

4. the filtrate is poured into an evaporating dish and then is

heated until saturated

5. the hot saturated solution is cooled

6. the crystal form is filtered

7. the crystal is pressed between filter papers to be dried

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1Max : 5

10 ( a ) ( i ) P : [Any metal situated above Cu in the ECS]

Reject : potassium / sodium

Q : Any acid either strong or weak acid

[ chemical equation ]

1. correct formula of reactant and product

2. balance chemical equations

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110 ( a ) ( ii ) Experiment I : 30/10 // 3 cm3s-1

Experiment II : 30/20 // 1.5 cm3s-1

[ unit must be correct ]

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10 ( a ) ( iii ) 1. rate of reaction in experiment I is higher than Experiment II

2. the concentration of acid in Experiment I more than in

Experiment II // no of hydrogen ions per unit volume in

Experiment I more than in Experiment II

3. Frequency of collision between hydrogen ion and metal P in

Experiment I is higher than in Experiment II

4. Frequency of effective collision between particles in

Experiment I is higher than in Experiment II

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10 ( b ) Factor : Size of reactant

1. [ name the reactants used ]

Example :

Zinc / magnesium / calcium carbonate and hydrochloric acid

2. pour ( 20-50 ) cm3 and acid* 1.0 mol dm-3 into a conical

flask.

3. filled a burette with a water and inverted it over a basin of

water and clamp a burette vertically using retort stand

4. initial burette reading is recorded

5. granulated / pieces of metal / metal carbonate is added into

a conical flask. The conical flask is closed immediately with

stopper and delivery tube

6. start the stopwatch7. the volume of gas collected is recorded at 30 seconds

intervals

8. step 1 to 8 is repeated by using a powder of metal / metal

carbonate

9. results :

Exp 1 : using a large piece of metal / metal carbonate

Time (s) 0 30 60 90

Volume of gas (cm3

)

Exp II : using a powder of metal / metal carbonate

Time (s) 0 30 60 90Volume of gas (cm3)

10. sketch the graph of volume of gas against time for both

experiment at same axes.

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11. [gradient graph using powder is higher than large pieces]

12. rate of reaction using powder is higher than large pieces

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Scheme Mid Year Exam f52012