CHEMISTRY PAPER 3 4541/3 - asiskl.orgasiskl.org/v4/wp-content/uploads/2014/07/ACE-KIMIA-B.pdf · CHEMISTRY PAPER 3 4541/3 Q1 An experiment is carried out to study the relationship

CHEMISTRY PAPER 3 4541/3 Q1

An experiment is carried out to study the relationship between the

concentration of H+ ions and pH value of hydrochloric acid. The

following diagram shows the pH value of 5 solutions of hydrochloric

acid, a strong acid with concentrations of 1.0, 0.1, 0.01, 0.001 and

0.0001 moldm-3.i


(a) Classify the ions in hydrochloric acid solution into anion and cation.

Anions Cations

Cl- H+

OH- - Q1

(b) Complete the following table based on the experiment.

Manipulated variable Method to manipulate the variable

Concentration of hydrochloric

acid

Use different concentration of HCl in each set of

experiment

Responding variable What to measure

Acidity of solution/pH value

pH values

How the variable is responding

pH values increases as the concentration of hydrochloric

acid increases

Controlled variable Method to maintain the controlled variable

50 cm3 of HCl Use 50 cm3 of HCl in each set of experiment


(c) State the hypothesis for the experiment.

The higher the concentration of H+ ions the lower the pH value

As the concentration of H+ ions increases, the pH value

decreases as well

Q1

(c) (i) Construct a table and record the concentration and the pH

value for this experiment.

Concentration of

HCl/moldm-3

pH value

1.0 0.0

0.1 1.0

0.01 2.0

0.001 3.0

0.0001 4.0


(d) (ii) Calculate the number of moles of hydrogen ions in 50 cm3 of

0.01 moldm-3 hydrochloric acid

(0.01X50)/(1000)

=0.0005 mol

Q1

(e) State the operational definition of a strong acid

Is HCl with concentration of 0.001 to 1.0 mol/dm3

That has pH value of 4 to 0

Why not…….

Is an acid

That dissolves and dissociates in water completely to produce high

concentration of hydrogen, H+ ions.

CHEMISTRY PAPER 3 4541/3

Q1

(f) If the experiment is repeated by using 0.01 moldm-3 of ethanoic

acid, predict the reading of the pH meter.

4 to 5//equals or more than 4 and equals or less than 5

Operational definition:

Follow template – classification is definite and

specific – usually the variable used and being

manipulated (HCl with various concentrations) –

characteristic obtain from What to measure for RV

(pH values) as well as How RV changes (pH

decreases as the concentration of HCl increases)

Define:

a. Melting point

b. Neutralization

c. Electrolysis

d. Acid

e. Alkali

f. Proton number

g. Isotopes

Definition


Q4

The following table shows the proton number of two elements X and

Y.

Element Proton number

X 6

Y 11

(a) Draw the atomic structure of atom X.

Write Draw


1st 2nd

Number of Proton

Electron

Neutron

1st BUT 2nd

Proton Number

Nucleon Number

Q4

(b) (i) Atom of element X has isotopes. What is meant by isotopes?

a. Isotopes are atoms (of the same element)

b. That has the same number of protons/proton number

but different number of neutrons/neutron number.

Spot the mistake?...............


Definition

i. Meaning of words

ii. Operational definition

•Classification

•Characteristic


1. What do you mean by empirical

formula?

2. What is the meaning of molecular

formula?

3. What is an acid?

4. What is strong alkali?


What is meant by acid?

A substanceclass that yields

hydrogen ions when dissolved in

watercharacteristic


What is meant by ‘melting point’?

The temperatureclass at which a solid

becomes a liquid at standard atmospheric

pressurecharacteristic.

What is meant by proton number?

The numberclass of protons in an atomic

nucleuscharacteristic


What is empirical formula?

A chemical formulaclass that indicates the relative

proportions of the elements in a molecule rather than the

actual number of atoms of the elementscharacteristic.

What is molecular formula?

A chemical formulaclass that shows the number and kinds

of atoms in a moleculecharacteristic.

What is a structural formula?

A chemical formulaclass that shows how the atoms and

bonds in a molecule are arrangedcharacteristic


Q3

Polymers are long chained molecules made by joining together

thousands of smaller molecules called monomers.

(a) Polypropene and polyvinyl chloride are examples of polymers.

State the name of their monomers.

Polypropene : Propene

Polyvinyl chloride : Chloroethane//Vinyl chloride


Structural formula

of monomer

Structural formula

of polymer

Polyethene

Polyvinyl Chloride

Polypropene


Task:

a. Name the monomer for each of the polimer

b. Draw the structural formula for each of the monomer


C C -

H

-

H H - - -

H -

C C - -

H

- H

H H - - -


C C -

H

-

H H - -

H -

n


C C - -

Cl

- H

H H - - -


C C -

Cl

-

H H - -

H -

n


C C - -

CH3

- H

H H - - -


C C - -

H H - - -

H -

n CH3


POLYETHENE


H

C C -

H

-

H

H H - -

- C C -

H -

H

H H - -

-

C C - -

H

H H - -

-

- - - -


POLYVINYL CHLORIDE

Cl

C C -

Cl

-

H

H H - -

- C C -

Cl -

H

H H - -

-

C C - -

H

H H - -

-

- - - -


POLYPROPENE

CH3

C C -

CH3

-

H

H H - -

- C C -

CH3

- H

H H - -

-

C C - -

H

H H - -

-

- - - -

Drawing the structural formula:

-C-O-H

C C C - -

O H -

H

H H

H

H H H - -

- - -

- - -

C C C - -

O H

- H

H H

H

H H H - -

- - -

- -

-

Careful when you connect H and hydroxyl group to

any of the atom from the top or bottom

i. Calculate the mass of magnesium oxide, MgO, which is formed when 1.2

g of magnesium, Mg, is burnt in oxygen, O2. (RAM; O,16;Mg,24). The

chemical equation of the reaction is as follows: 2Mg + O2 2MgO. ii. What is the volume of carbon dioxide, CO2, released when 10 g of

calcium carbonate, CaCO3, reacts with excess dilute hydrochloric acid,

HCl, at room conditions? (RAM; C,12;O,16;Ca,40; Molar volume: 24

dm3 at room temperature)

iii. Calculate the mass of zinc, Zn, reacting with dilute nitric acid, HNO3, if

360 cm3 of hydrogen, H2, gas is released in this reaction. (RAM: Zn,65;

Molar volume: 24 dm3 at room temperature)

iv. The following reaction can be used to prepare copper(II)chloride salt.

CuCO3 + 2HCl CuCl2 + H2O + CO2. Excess copper(II)carbonate is

added to react with 50 cm3 of 2.0 moldm-3 hydrochloric acid to form the

salt. Calculate the mass of the salt formed. ( RMM CuCl2 ,135 )


4 STEPS PROBLEM SOLVING:

Step 1.

Write balanced chemical equation

2Mg + O2 2MgO

Step 2 and 3.

Write exact number of mol of each reactants Step 2 and products Step 3

[mol=mass/ram or mass/rmm or MV M[V/1000] or volume ofgas/molar

volume of gas]

Mol=1.2/24=0.05

Step 4.

Answer the question

[ mass-volume-molarity or volume of solution ]

Mass of MgO=0.05X40=2 g

2Mg + O2 2MgO

0.05 005/2

=0.025 0.05


Calculate the mass of saltProduct formed when 2.3 g sodium metalReactant reacts completely with excess chlorine gas. The chemical equation of the reaction is as follows: 2Na + Cl2 NaCl. ( Relative Atomic Mass: Na,23;Cl,35.5 ) Reactant

Chemical equation

Product

2Na + Cl2 NaCl (2.3/23)=0.1 (2.3/23)=0.1/2=0.05 (2.3/23)=0.1/2=0.05

Mol=(2.3/23)=0.1

Mass=0.05X58.5=2.925 g

Mass

Mole Relative Atomic Mass/ Relative Molecular Mass


Calculate the volume of alkaliReactant used to react completely with Hydrochloric acid to produce 1.4625 g saltProduct is formed . ( Relative Atomic Mass: H,1;O,16;Na,23;Cl,35.5 )

Hydrochloric acid

Sodium hydroxide solution, 1.0 mol/dm3 Reactant

Chemical equation

Product

Mol=1.4625/58.5=0.025

HCl + NaOH NaCl + H2O

0.025 0.025 0.025 0.025

Volume=Mol/Molarity=(0.025/1)X1000=25 cm3

Mole

Volume Molarity


x H Cl


Covalent

molecular compound

Explain the formation of the above mentioned compound. ( 8 points )

Na

+

2.8

O

2.8

2- Na

+

2.8

Na2O


Ionic compound

Explain the formation of the above compound.

( 5 + 5 + 3 points )


1. Electron arrangement of atom of sodium, Na is

2.8.1

2. Achieve stable electron arrangement

3. Atom of sodium, Na releases one valence electron

to atom of chlorine to form ion of sodium, Na+

4. Electorn arrangement of ion of sodium, Na+ is 2.8

5. Ionic equation for the formation of ion of sodium,

Na+ is Na e + Na+


1. Electron arrangement of atom of chlorine, Cl is

2.8.7

2. Achieve stable electron arrangement

3. Atom of chlorine, Cl gain one electron from atom of

sodium, Na to form ion of chlorine, Cl-

4. Electorn arrangement of ion of chlorine, Cl- is 2.8

5. Ionic equation for the formation of ion of chlorine,

Cl- is Cl + e Cl-

A

B


1 2

4 3

6 7

5

Arrangement of

atoms of a pure

metal Arrangement of

atoms of an alloy

The following data shows the differences in pH value

for samples of acids provided. Explain.


All acids have the same concentration

Semua asid mempunyai kepekatan sama 0.5 mol/dm3

Ethanoic Acid

Asid etanoik

Hydrochloric Acid

Asid hidroklorik

Hydrochloric Acid

Asid hidroklorik

Sulphuric Acid

Asid Sulfurik

pH 6 pH 4 pH 4 pH 2

Aspects

Aspek

Ethanoic Acid

Asid etanoik

Hydrochloric Acid

Asid hidroklorik

pH 6 pH 4

Acidity

Keasidan

Concentration of hydrogen ions

Kepekatan ion-ion hidrogen

Asid molecule that dissociates

Molekul asid yang bercerai

Degree of dissociation

Darjah penceraian

Name of solvent

Nama pelarut

Strength of acid

Kekuatan asid


Aspects

Aspek

Hydrochloric Acid

Asid hidroklorik

Sulphuric Acid

Asid Sulfurik

pH 4 pH 2

Acidity

Keasidan

Concentration of hydrogen ions

Kepekatan ion-ion hidrogen

Asid molecule that dissociates

Molekul asid yang bercerai

Degree of dissociation

Darjah penceraian

Name of solvent

Nama pelarut

Basicity of acid

Kebesan asid


CHEMISTRY PAPER 2 4541/2 3 sets of medium are prepared. Set 1: hydrogen chloride gas is

dissolved in water. Set 2: hydrogen chloride gas mixed with

methylbenzene. Set 3: hydrogen chloride gas is left dry without any

solvent. Zinc powder is added into each of the medium. Predict your

observations and explain why for each of the sets of experiment.

Set 1: hydrogen chloride gas is dissolved in water.

Type of activity Zinc/Magnesium Copper(II)

carbonate

Electrolysis of

solution

Blue litmus

paper

Observations

Acidity of solution

Presence or absence

hydrogen ions or freely

moving ions

Name of solution

What dissociates/not

dissociates

Name of solvent

CHEMISTRY PAPER 2 4541/2 Name of reactants

Nama bahan bertindak balas Magnesium and dan Hydrochloric acid Asid hidroklorik

Balanced chemical equation

Persamaan kimia tindak balas seimbang

……………………………………………………………

Ionic equation

Persamaan ion ……………………………………………………………

Name of reactants

Nama bahan bertindak balas

Copper(II) carbonate Kuprum(II) karbonat and dan Nitric acid

Asid nitrik

Balanced chemical equation

Persamaan kimia tindak balas seimbang

……………………………………………………………

Ionic equation

Persamaan ion

……………………………………………………………

a Mg + HCl

b CuCO3 + HNO3

a Mg + H+

b CO32- + H+

a HCl

b H2SO4

c HNO3

d CH3COOH


Electrolyte

– NaCl (aq) 0.001 moldm-3

Electrodes – Carbon

Electrolyte

– CuSO4 (aq)

Electrodes – Carbon

Electrolyte

– CuSO4 (aq)

Electrodes – Copper

Draw a labled diagram for the following set up of apparatus of

electrolysis Lukiskan rajah berlabel bagi susunan radas untuk menjalankan elektrolisis berikut:

CHEMISTRY PAPER 2 4541/2 Electrolyte – CuSO4 (aq) - Electrodes – Carbon

Name/Write the formula of all ions present in the electrolyte

Name/Write the formula of ions that move to the :

Anode: Cathode:

Direction of flow of electrons

Lable the anode and cathode for the cell

At which electrode does the OiL/RiG occur

At which electrode does lost/gain of electrons occur

Write the half ionic equation of oxidation and reduction occurring at the anode and cathode

respectively


Electrolyte – CuSO4 (aq) - Electrodes – Carbon

State the observations at the anode and cathode

Name the products at the anode and cathode

Anode: Cathode:

Give reasons why to the answers above

Anode:

Cathode:

CHEMISTRY PAPER 2 4541/2 The 3 factors influencing the selectively discharged ions at cathode and anode

1. Position of ions in the electrochemical series (always at cathode and sometimes at anode )

2. Concentration of ions in the electrolyte ( always at anode )

3. Types of electrodes used ( always at anode )

Factors influencing Explanation

Cell Electrolyte Name of electrode Anode Cathode Anode Cathode

Electrolytic cell

Sodium chloride solution Carbon Position Position

OH- is lower to Cl- in the electrochemical series

H+ is lower to Na+ in the electrochemical series

Electrolytic cell

Copper(II)sulphate solution Carbon Position Position

OH- is lower to SO4

2- in the electrochemical series

Cu2+ is lower to H + in the electrochemical series

Electrolytic cell

Copper(II)sulphate solution Copper

Types of electrode Position

Copper is reactive and it reacts

Cu2+ is lower to H+ in the electrochemical series

Electrolytic cell

Hydrochloric acid /Sodium chloride 0.001 moldm-3

Carbon Concentration Position Conc of Cl- is less than OH-


Electrolytic cell

Hydrochloric acid/Sodium chloride 1.0 moldm-3

Carbon Concentration Position Conc of OH- is less than Cl-



The following diagram shows the set up of the apparatus with the

combination of an electrolytic cell and a chemical cell.



1.

Describe an experiment to show that chemical energy can be

transformed into electrical energy. Huraikan satu eksperimen untuk

menunjukkan bahawa tenaga kimia boleh ditukarkan menjadi

tenaga elektrik.

( 12 marks markah )

Electrolytic cell Chemical cell

Anode Cathode Negative terminal Positive terminal

Name of electrodes Copper Copper Zinc Copper

Half ionic equation

Observations

Change of energy

Blue copper(II)sulphate solution

intensity

Explain the change in colour

intensity of the electrolyte

Summary of the differences between an electrolytic cell and a chemical cell


Group 17

a. Non-metal elements – halogens

b. Relationship between electrons arrangement of an atom and its

grouping

i. Chlorine atom – 2.8.7

ii. 7 valence electron

iii. group 17

c. Physical properties changes progressively down the group

i. Diatomic molecules

ii. Very low melting/boiling points


a. Explain in terms of electron arrangement why…

(i) Elements of group 18 are chemically inert/unreactive

compared to elements of group 17

Atom has achieved the stable octet/duplet electron

arrangement – Outermost shell fully filled – Need not share,

gain or lose electrons – Thus unreactive


a. Explain in terms of electron arrangement why

i. Going down group 1 the elements become more

reactive.

More proton number – More number of electrons –

More occupied shells – Bigger distance1 between

valence electrons and nucleus – Weaker nuclear

charge2 – Greater ease of electron lost3 – Easier to

form positive ion4 – Thus more reactive


a. Explain in terms of electron arrangement why

i. Going down group 17 the elements become less

reactive.

More proton number – More number of electrons –

More occupied shells – Bigger distance between

valence electrons and nucleus – Weaker nuclear

charge – Less ease of electron gain – More difficult to

form negative ion – Thus less reactive


In a reaction, 2.7 g of element R combines with 2.4 g of

element S. What is the formula of the compound

produced? ( RAM of S,16; R,27 )

Elements R S

Relative atomic mass 27 16

Mass 2.7 2.4

Mole (2.7/27)=0.1 (2.4/16)=0.15√1

Mole ratio 0.1/0.1=1 0.15/0.1=1.5

Simplest mole ratio 1X2=2 1.5X2=3√2

Empirical formula of compound is R2S3 √3


0.36 g of carbon combines with 0.06 g of hydrogen to

form a hydrocarbon. The relative molecular mass of the

hydrocarbon is 42. Calculate:

a. The empirical formula of the hydrocarbon

b. The molecular formula of the hydrocarbon

( Relative atomic mass: C,12; O,16 )


Elements Carbon Hydrogen

Relative atomic mass 12 1

Mass 0.36 0.06

Mole (0.36/12)=0.03 (0.06/1)=0.06√1

Mole ratio 0.03/0.03=1 0.06/0.03=2√2

Empirical formula of compound is CH2 √3


(CH2)n = 42

(12 + 2(1))n = 42

(14)n = 42

n = 42/14

n = 3 √4

Molecular formula of the compound is

(CH2)3 = C3H6√5


Describe a laboratory activity to prepare a dry and pure crystals of

copper(II) sulphate using copper(II) carbonate solid and sulphuric acid.

You are also provided with appropriate apparatus to carry out the

activity. [ 6 + 3 + 1 points ]

Describe a laboratory activity to prepare a dry and pure magnesium

carbonate using magnesium nitrate solution and potasium carbonate

solution. You are also provided with appropriate apparatus to carry out

the activity. [ 5 + 3 + 1 + 1 points ]


Describe a laboratory activity to prepare a dry and pure crystals of zinc

carbonate using zinc oxide, nitric acid and sodium carbonate solution.

You are also provided with appropriate apparatus to carry out the

activity. [ 9 marks ]

Remember:

a. Zinc carbonate is an insoluble salt which can be prepared using the precipitation

method [ double decomposition reaction ]

b. There are 2 common reactions used: (i) Metal oxide (ZnO) and Acid (Nitric acid)

or (ii) Metal carbonate and Acid.

c. Sodium carbonate and potassium carbonate are two common soluble salts used to

prepare a typical insoluble salt when it is reacted to another soluble salt that

provides the intended metal ion

d. The later should be prepared from (b)

e. Describe the Soluble salt method of preparation followed by Insoluble salt method

of preparation

f. THIS STEP APPLIES VISE VERSA IF YOU ARE ASKED TO PREPARE A

TYPICAL SOLUBLE SALT

Documents

CHEMISTRY PAPER 3 4541/3 - asiskl.orgasiskl.org/v4/wp-content/uploads/2014/07/ACE-KIMIA-B.pdf · CHEMISTRY PAPER 3 4541/3 Q1 An experiment is carried out to study the relationship