Skema Pecutan Akhir Chemistry p2 p3 Spm 2013

18
8/10/2019 Skema Pecutan Akhir Chemistry p2 p3 Spm 2013 http://slidepdf.com/reader/full/skema-pecutan-akhir-chemistry-p2-p3-spm-2013 1/18 SECTIO 1. (a) an (b) Zn (c) copp (d) To e (e) Repe (f) Mass No of m Ratio Empirical  2. (a)(i) (ii) U (iii) W (b) (i) 2. (ii) X +  (c)(i) Th (ii) X, Y, (d) (i) 4 (ii) 2X +  3. (a)(i) (ii) Grou (iii) Ato (b)(i) 2. (ii) (iii) Q  (c) P an (d)(i) ion (ii) (e)(i) Co N A hydrous cal  2HCl  r(II) oxide sure that t at the heati X 5.32 l 0.76 1 formula = .1 y have the Z + O 2   2H 2 O  .4 14 and pe  P has 4 val  Q 2+  + 2e R. They ha ic bond alent bond ium chlorid ZnCl 2  + H 2 e combusti g, cooling  – 4.560 =  / 64 = 0.0 XO same 3 she 2X 2 O 2XOH + iod 2 lence electr ve the sam  n tube is t nd weighin 0.768 g 12 lls occupied 2  ns and has proton nu tally filled process u O 5.520 – 5 0.192 / 1 1 with electro 2 shells occ ber but dif  ith hydroge til a consta .328 = 0.19  = 0.012 ns upied with ferent nucle n gas nt mass is p 2 g lectrons on number. roduced

Transcript of Skema Pecutan Akhir Chemistry p2 p3 Spm 2013

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SECTIO

1. (a) an

(b) Zn

(c) copp

(d) To e(e) Repe

(f)

MassNo of mRatioEmpirical

 

2. (a)(i)

(ii) U

(iii) W

(b) (i) 2.

(ii) X+ 

(c)(i) Th

(ii) X, Y,

(d) (i) 4

(ii) 2X +

 

3. (a)(i)

(ii) Grou

(iii) Ato

(b)(i) 2.

(ii)

(iii) Q→ 

(c) P an

(d)(i) ion(ii)

(e)(i) Co

N A

hydrous cal

  2HCl → 

r(II) oxide

sure that tat the heati

X5.32

l 0.761

formula =

.1

y have the

Z

+ O2  → 

2H2O → 

.4

14 and pe

 P has 4 val

 

Q2+  + 2eR. They ha

ic bond

alent bond

ium chlorid

ZnCl2  + H2

e combustig, cooling

 – 4.560 = / 64 = 0.0

XO

same 3 she

2X2O

2XOH +

iod 2

lence electr

ve the sam

 

n tube is tnd weighin

0.768 g12

lls occupied

ns and has

proton nu

tally filledprocess u

O5.520 – 50.192 / 11

with electro

2 shells occ

ber but dif 

 

ith hydrogetil a consta

.328 = 0.19 = 0.012

ns

upied with

ferent nucle

n gasnt mass is p

2 g

lectrons

on number.

roduced

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(ii)

4. (a) Al

(b)

  p

  s

c

  T

h

(c) Maki

(d) M1 V1

(2.0)(25

  M2 

(e) 2Na

(f) (i)

(ii) HClNo of m

From the

 

No of m

  0.0

  M

ali which io

H value for

dium hydr

ncentratio

he concentr

igher the co

g soap

= M2 V2 

= M2(75)

= 0.67 m

  2H2O

NaOHl NaOH =

equation,

0.

l HCl = MV

= M(2

  = 2.5

nizes compl

odium hyd

xide is stro

 of hydroxi

ation of hyd

ncentration

l dm-3 

2NaOH

NaCl + H V / 1000 =

mol NaOH

5 mol NaO

 /1000

0) / 1000

mol dm-3 

tely in wat

oxide is 14

g alkali whi

e ions

roxide ion i

of hydroxid

+ H2 

O2.0(25) / 1

: 1 mol HCl

 : 0.05 mol

r to form hi

and pH valu

ch ionizes c

 sodium hy

 ion, the hi

00 = 0.05

HCl

igh concent

e for ammo

ompletely i

droxide is hi

gher the pH

 

ol

ation of hy

nia is 11

water to f 

igher than a

 value.

roxide ions

rm higher

mmonia. T

.

e

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5. (a) Th

between

(b) (i) H

(ii) no of

No of m

(iii) 1 m

0.025 m

(iv) 0.02

1 mol Ag

Heat of

(c)

(d) Ag+ 

6. (a) To

(b) (i) H

(ii) no ofno of mo

0.2 mol

0.2 mol

1 mol H2

Heat of

e heat chan

sodium chl

= mcӨ = (

mol Ag+ =

l Cl¯ = MV

l Ag+ + 1 m

l Ag+ + 0.0

 mol AgCl

Cl release 5

recipitation

  Cl¯ →  A

 reduce the

at change,

mol NaOHl HCl = MV

Cl + 0.2 m

2O release

→ 5460

eutralizatio

ge(release)

ride solutio

0 + 50) (4.

 V / 1000

 / 1000 = (

ol Cl¯ → 1

25 mol Cl¯

elease 147

8800 J mol-

,  ∆ H = - 58.

gCl

heat loss to

H = mcӨ 

= MV / 100 / 1000 = (2

ol NaOH→ 

10920 J of

J mol-1 

,  ∆ H = - 5

when 1 mol

n and silver

) ( 30.5 –

(0.5)(50) /

.5)(50) / 1

mol AgCl

→ 0.025 m

 J of heat

of heat

8 kJ mol-1 

the surrou

 (100 + 10

 = (2.0)(10.0)(100) / 1

 0.2 mol H2

eat

.6 kJ mol-1

 of silver ch

nitrate solu

7) = 1470

1000 = 0.0

00 = 0.025

l AgCl

ding

) (4.2) (41

0) / 1000 =00 = 0.2

loride is for

ion.

25 mol

mol

 – 28) = 10

 0.2 molol

ed from th

20 J

e reaction

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(c)

(d) The

hydrochl

(e)(i) les

(ii) Etha

complete

 

7. (a)

(b) Zn

(c) Exp I

Exp II:

eat change

ric acid an

 than - 54.

oic acid is a

ly.

H2SO4 → 

: Average r

verage rate

 / released

sodium hy

 kJ mol-1 //

weak acid.

ZnSO4  +

te of reacti

of reaction

hen 1 mol

roxide solu

any value l

Some of he

H2 

n = 32 cm3

= 20 cm3 /

of water is

tion.

wer than -

at released

 / 120 s =

120 s = 0.1

ormed fro

54.6 kJ mol

is used to i

.27 cm3s-1 

7 cm3s-1 

the reactio

-1 

nize the et

 

n between

anoic acid

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(c)

9. (a) (i)

(ii) body

(iii) PVC

(b) (i) Sa

(ii) To pr

(iii) Diss

Dissolve

(c)(i) as

(ii) To tr

(iii) To c

 

zinc

of an aerop

does not ru

ponification

oduce soap

lves in wat

 in oils and

irin // para

at diabetes

lm the pati

lane

t

precipitate

r : Part B

grease: Hy

etamol

nt.

 / To reduc

rophobic

solubility of soap

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SECTIO

10. (a) C

Tempera

Pressure

Catalyst

(b)

  P

 

 

  B

 

 

(c)

  S

S

  S

2

C

T

P

  S

S

 

 

H

N B

onditions fo

ture : 450°

: 200 atm

: Iron

ure copper

hich are ar

hen force i

ronze consi

hich disrup

hen force i

ulphur is bu

 + O2  → 

ulphur dioxi

SO2  + O2 

onditions re

emperature

ressure: 1

atalyst: Van

ulphur trioxi

O3  + H2S

leum is dilu

2S2O7  + H

r Haber Pro

 

onsist of at

anged in or

 applied, th

ts of tin ato

 the orderly

 applied, th

rned in oxy

SO2 

e is burne

 →  2SO3 

quired:

: 450°C

tm

adium(V) o

de is dissol

4 →  H2S2O

ed in water

O →  2H2

ess:

oms of the

derly mann

e layer of a

ms of differ

 arrangeme

e layers of

en to form

in oxygen

ide

ed in conce

7

to form sul

O4 

ame size

r

oms easily

ent size

nt of pure c

ure copper

sulphur dio

o form sulp

ntrated sulp

phuric acid

lide over o

opper atom

not easily s

ide

hur trioxide

huric acid t

e another

s

lide over on

 form oleu

e another

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(d)

  Sulphur dioxide gas is released to the atmosphere and cause air pollution.

  The air pollution can cause respiratory problems and asthma.

  Sulphur dioxide released also react with rain water to form acid rain.

   Acid rain can corrode buildings, cars, increase acidity of rivers and lakes and increase

acidity of soil.

(e)(i) Procedure

1. Fill a burette with 1 mol dm-3 sulphuric acid. Record the initial burette reading, V1 .

2. Pour 25 cm3 of 1 mol dm-3 ammonia / ammonium hydroxide solution into a conical flask. Add

three drops of phenolphthalein into the conical flask.

3. Add sulphuric acid solution from the burette into the conical flask slowly. Swirl the conical

flask.

4. Stop adding sulphuric acid when the solution in conical flask turn from pink to colourless.

Record the final burette reading, V2.

5. Calculate the volume of sulphuric acid used, V = V2 – V1 6. Pour 25 cm3 of ammonia / ammonium hydroxide solution into a beaker.

7. Add V cm3 of sulphuric acid into the beaker and stir the mixture.

8. Heat and evaporate the salt solution until 1/3 its original volume.

9. Cool the saturated salt solution at room temperature.

10. The ammonium sulphate crystals are filtered.

11. The ammonium sulphate crystals are pressed between two filter papers.

(ii)

 

Pour 2 cm3

 of ammonium sulphate solution into a test tube   Add 2 cm3 of dilute hydrochloric acid followed by 2 cm3 of barium chloride solution into

the test tube

  Shake the mixture

  White precipitate is formed

  Sulphate ion is confirmed present.

11. (a) (i) Haber process

(ii) N2  + 3H2  →  2NH3 

(iii) Catalyst : IronPressure: 200 atm

Temperature : 450°C

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(b) (i) ethanol and propanoic acid

(ii) Ethanol:

  Ethanol undergo dehydration process when with heated porcelain chips to produce

ethene and water

  Ethanol undergo oxidation process when heated with acidified potassium

manganate(VII) to produce ethanoic acid and waterPropanoic acid:

  Propanoic acid react with calcium carbonate to produce calcium ethanoate, carbon

dioxide and water

  Propanoic acid react with zinc to produce zinc ethanoate and hydrogen gas

(c) Reaction with bromine water:

1. Pour 2 cm3 of hexane into a test tube.

2. Add 3 drops of bromine water into the test tube

3. Shake the mixture

4. Repeat steps 1 to 3 using hexene.5. If bromine water change from brown to colourless, the solution is hexene.

6. If no change, the solution is hexane.

Reaction with acidified potassium manganate(VII) :

1. Pour 2 cm3 of hexane into a test tube.

2. Add 3 drops of acidified potassium manganate(VII) into the test tube

3. Shake the mixture

4. Repeat steps 1 to 3 using hexene.

5. If acidified potassium manganate(VII) change from purple to colourless, the solution is

hexene.6. If no change, the solution is hexane.

13. (a)

  Chlorine is more reactive than bromine

  Chlorine can displace bromide ion from potassium bromide solution

  Bromide ion release electron to form bromine molecule. Oxidation occur.

2Br¯ → Br2 + 2e

  Chlorine molecule accept electron to form chloride ion. Reduction occur.

Cl2  + 2e → 2Cl¯

 

Oxidising agent: Chlorine  The product is bromine molecule and potassium chloride

(i)   Add 2 cm3 of 1,1,1-trichloroethane into the test tube

(ii)  Shake the test tube

(iii)  Record the observation on top and bottom layer of the solution

(iv)  The bottom layer becomes orange. Therefore, bromine is displaced from

potassium bromide solution by chlorine.

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(b)

Procedur

1. Clamp2. PourU-tube.3. Add iruntil the4. Add clayer of5. Imme

6. Conne7. Deterpointer.8. Left t 

Transfer

Negative

  I

  I

 

  C

Positive  C

C

  R 

  C

 

  T

e:

a U-tube tilute sulphu

n(II) sulphlayer of thelorine watehe solutionse carbon

ct the electine the ne

e apparatu

of electrons

 terminal:

on(II) sulp

on(II) ion r

xidation oc

arbon electr

erminal:hlorine wat

hlorine acce

eduction oc

arbon electr

he electron

a retort staric acid into

ate solutionsolution rear carefully ireaches thelectrodes in

odes to a gative and p

 aside for 3

:

ate is reduc

lease elect

ur

ode immers

r is oxidizin

pt electron

ur

ode immers

flow from ir

nd.the U-tube

carefully inches the heto the rightheight of 3to each of t

lvanometeositive term

0 minutes a

ing agent.

on to form

ed in iron(II

g agent.

o form chlo

ed in chlori

n(II) sulph

until its lev

o the left aight of 3 c arm of thecm.e arms.

by using cinals based

nd record t

iron(III) ion

) sulphate s

ride ion. Cl2

e water be

ate solution

l are 6 cm

m of the U-.U-tube by

nnecting won the defl

e observati

. Fe2+ →  F

olution bec

  + 2e → 2

omes positi

to chlorine

way from t

tube by usi

sing a drop

ires.ction of gal

n at both e

3+ + e

mes negati

Cl¯

ive terminal

water throu

he mouths

g a droppe

er until the

anometer

lectrodes.

ve terminal

gh external

f the

wire.

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14. (a)

   A

  C

  E

Half equ

 

 A  C

Redox re

   A

  C

 

(b) Mate

 Apparat

Procedur1. Clean

2. Pour 1

3. Coppe

4. Imme

5. Conne

6. Turn

7. Recor

 

Observat

 Anode:Cathode:

 

Half equ

 Anode:

Cathode:

 

bservation:

node P: Ga

athode Q: b

lectrolyte: B

tions:

node P : 4Oathode Q:

action:

node P: hy

athode Q:

ials: iron s

s: Dry cells,

e:the iron sp

50 cm3 of 0

r plate is m

se the iron

ct both elec

n the switc

 the observ

ion:

opper beco Brown soli

tions:

u →  Cu2+

 Cu2+  + 2

 

bubbles ar

rown solid

lue solution

H¯ → O2  +u2+ + 2e →

roxide ion r

opper(II) io

oon, coppe

 switch, am

on and cop

.5 mol dm-3 

de anode a

spoon and

trode to dry

 and allow

ations at el

mes thinnerdeposited

+ 2e

→  Cu

released

eposited

turns light

2H2O + 4 Cu

lease elect

n accept ele

 plate, 0.5

meter, conn

er plate wi

copper(II)

nd iron spo

opper plate

 cells, amm

the current

ctrodes

on iron spo

lue // inten

 

on to form

ctron to for

ol dm-3 co

ecting wire,

h sand pap

ulphate sol

n is made

into copper

ter and swi

to flow for

n

sity of blue

oxygen and

m copper a

per(II) sul

 beaker

r

tion into a

athode

(II) sulphat

itch using c

0 minutes

solution de

 water. Oxi

om. Reduct

hate soluti

beaker

e solution.

nnecting wi

reases

ation occur

ion occur.

n, sand pa

re

.

er

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(c)

  Zinc is more electropositive than copper in electrochemical series

  Zinc becomes negative terminal and copper becomes positive terminal

Negative terminal:

 

Zinc release electron to form zinc ion. Oxidation occur.  Half equation: Zn →  Zn2+ + 2e

  Observation: Zinc becomes thinner

  The electron move from zinc electrode to copper electrode through external circuit /

external wire

Positive terminal:

  Copper(II) ion accept electron to form copper atom. Reduction occur.

  Half equation: Cu2+ + 2e →  Cu

  Observation: Copper becomes thicker

15. (a)(i)  Lead(II) nitrate: neutralization reaction through the reaction between lead(II) oxide and

nitric acid

  Lead(II) sulphate: Double decomposition through the reaction between lead(II) nitrate

solution and sodium sulphate solution

(ii) lead(II) nitrate solution and sodium sulphate solution

(b) Preparation of lead(II) nitrate crystal:

1. Pour 100 cm3 of 1 mol dm-3 nitric acid into a beaker and the solution is heated gently.

2. Add solid lead(II) oxide little by little until in excess.

3. Stir the mixture using glass rod.4. The heating is stopped when lead(II) oxide is no longer dissolve in nitric acid.

5. Filter the mixture

6. The filtrate is heated/evaporated until 1/3 its original volume

7. The saturated lead(II) nitrate solution is cooled to room temperature.

8. The lead(II) nitrate crystals are filtered

9. The lead(II) nitrate crystals are pressed between two filter paper

10. Chemical equation: PbO + 2HNO3 →  Pb(NO3)2  + H2O

(c) Test for chloride ion:

1. Pour 2 cm3 sodium chloride solution into a test tube2. Add 2 cm3 dilute hydrochloric acid followed by 2 cm3 silver nitrate solution into the test tube

3. Shake the mixture

4. White precipitate is formed

5. Chloride ion is confirmed present

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Test for iron(II) ion:

1. Pour 2 cm3 iron(II) nitrate solution into a test tube

2. Add 2 cm3 of potassium hexacyanoferrate (III) solution into the test tube

3. Shake the mixture

4. Dark blue colouration is formed

5. Iron(II) ion is confirmed present

PAPER 3

16. (a)

Pink colouration Iron does not rust Very high intensity of dark blue colouration Iron rust very quicklyLow intensity of dark blue colouration Iron rust quickly

(b) (i) Type of metal in contact with iron

(ii) Rusting of iron / Intensity of dark blue colouration

(iii) Iron nails

(c) Fe → Fe2+ + 2e

(d) The further the distance of metal from iron in electrochemical series, the iron rust very

quickly

(e)

Metal more electropositive than iron Metal less electropositive than ironZinc Copper

tin

(f) When the iron nail coiled with less electropositive metal is immersed in jelly solutioncontaining phenolphthalein and potassium hexacyanoferrate(III) solutions, dark bluecolouration is formed.

17. (a) Reading in 2 d.p with correct unit e.g 0.80 cm

(b)

Type of metal Diameter of dent (cm) Average diameter of dent(cm)1 2 3

Pure copper Alloy X(c) Alloy X is harder than pure copper // The smaller the diameter of dent, the metal is harder

(d) Bronze // Brass

18. (a)

2.8 V 0.8 V1.4 V 0.4 V

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(b)

Pairs of metals Potential difference (V)M and Cu 2.8N and Cu 0.8P and Cu 1.4

Q and Cu 0.4

(c) The further the distance between two metals in electrochemical series, the higher the

potential difference

(d) (i) electrode M becomes thinner

(ii) Copper becomes thicker

(iii) Blue solution turns pale blue / Intensity of blue solution decreases

(e) At copper electrode, copper(II) ion accept electron to form copper atom.

The concentration of copper(II) ion in the solution decreases.

(f)(i) Pairs of metals(ii) Potential difference

(iii) copper electrode

(g) Cu, Q, N, P, M

(h)

Pairs of metals Positive terminal Voltage/ VM and N N 2.8-0.8 = 2.0N and P N 1.4-0.8 = 0.6M and P P 2.8-1.4 = 1.4(i)

Electrolyte Non-electrolyteSodium chlorideZinc sulphate

Silver chlorideLead(II) sulphate

19. (a) How to identify the solubility of salt in water? // How to identify the soluble salt and

insoluble salt when dissolve in water?

(b) Manipulated : Type of salts // Salt A and salt BResponding : Solubility of salt in waterFixed : Type of solvent // distilled water // water(c) When the salt dissolves in water, it is soluble salt whereas when the salt not dissolves inwater, it is insoluble salt(d) Apparatus : Beaker, glass rod, spatula, measuring cylinder.Materials : Salt A, salt B, distilled water

(e) 1. Measure 50 cm3 of distilled water (using measuring cylinder) and pour into a beaker.2. Add one spatula of salt A into the beaker.3. Stir the mixture (using glass rod)4. Record the solubility of salt in water.5. Repeat steps 1 to 4 using salt B.

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(f)

Type of salts ObservationSalt ASalt B

20.

 Aim of experiment To investigate the effect of rusting of iron when in contact with metal

X[Magnesium] and metal Y[copper]

 All variablesinvolved

Manipulated variable: Types of metal in contact with iron//Magnesium and

copper

Responding variable: Rusting or iron

Controlled variable: Iron nails// jelly solution

Statement ofhypothesis

When iron in contact with magnesium, iron does not rust /

When iron in contact with copper, the iron rust

List of materialsand apparatus:

 Apparatus : Test tube, test tube rack, dropper, glass rodMaterials : Iron nails, sandpaper, magnesium ribbon, copper strip, hot jelly,

potassium hexacyanoferrate (III) and phenolphthalein

Experimentalprocedure:

1. Clean three iron nails, Magnesium ribbon and Copper strip with

sandpaper.

2. Two iron nails are coiled with magnesium and copper each

3. Place the three nails into three different test tubes

4. Add 4 drops of potassium hexacyanoferrate (III) solution followed by 4drops of phenolphthalein into hot jelly solution. Stir the mixture.5. Pour hot jelly solutions into all the test tubes until it covers the entire nail

6. Left the test tube aside for one day7. Any observation are recordedTabulation of data Pairs of metal Intensity of blue

colour

Intensity of pink

colour

Iron Nail

Iron Nail +

magnesium

Iron Nail +copper

21.

Problem statement How does the concentration of ions affect the selective discharge of ions at theanode?

 All variables involved Manipulated: concentration of chloride ionsResponding: Ions discharged at anode/product at anodeConstant: carbon electrodes

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Statement ofhypothesis

When the concentration of chloride ions is higher, chloride ions will be selectivelydischarged at the anode.When the concentration of chloride ions is lower, hydroxide ions will be selectivelydischarged at the anode. //When the concentration of chloride ions is higher, the product at anode is chlorinegas.

When the concentration of chloride ions is lower, the product at anode is oxygengas

List of materials andapparatus:

Materials: 0.001 mol dm-3  hydrochloric acid, 1 mol dm-3  hydrochloric acid, bluelitmus, wooden splinter Apparatus: electrolytic cell, test tubes, measuring cylinder carbon electrodes, twodry cells, connecting wire, switch

Experimentalprocedure:

1. Measure 100 cm3 of 0.001 mol dm-3 hydrochloric acid solution using measuringcylinder and pour into an electrolytic cell.2. Dip two carbon electrodes into the solution. Invert the two test tubes containinghydrochloric acid solution onto both electrodes.3. Connect the electrodes to a switch, ammeter and two dry cells by usingconnecting wires.4. Close the switch and the observation at anode is recorded.

5. Collect the gas in a test tube at anode. Place a glowing wooden splinter into themouth of the test tube. Record the observation.6. Repeat steps 1 to 4 using 1 mol dm-3 hydrochloric acid solution to replace 0.001mol dm-3 hydrochloric acid solution.7. Collect the gas in a test tube at anode. Place a moist blue litmus paper into themouth of the test tube. Record the observation.

Tabulation of data Concentration of hydrochloric acid(mol dm-3)

Observation at Anode

0.0011.0

22.

 Aim of experiment To investigate the cleansing action of soap and detergent in hard water

 All variables involved Manipulated variable: Soap and detergentResponding variable: The presence of greasy stain on clothFixed variable: Volume of hard water// volume of cleaning agent

Statement ofhypothesis

The cleansing action of a detergent is more effective than soap in hard water.

List of materials andapparatus:

Materials: 5% soap solution, 5% detergent solution, hard water, two small pieces

of cloth with greasy stains.

 Apparatus: 100 cm3 beaker, 50 cm3 measuring cylinder, glass rod

Experimentalprocedure:

1. Measure 50 cm3 of hard water using measuring cylinder and pour into a beaker.2. Add 50 cm3 of soap solution into the beaker. Stir the mixture.

3. Immerse a small piece of cloth with greasy stains into the mixture.4. Rub the cloth gently.5. Record the presence of greasy stain on the cloth.6. Repeat steps 1 to 5 using detergent solution.

Tabulation of data Type of cleaning agent ObservationSoap + hard waterDetergent + hard water

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23.

Problem statement How to compare the electrical conductivity between molten naphthalene andmolten lead(II) bromide?

 All variables involved Manipulated: naphthalene and lead(II) bromide // Types of compoundResponding: Reading of ammeter/electrical conductivityConstant: mass of compound// carbon electrodes

Statement ofhypothesis

Molten lead(II) bromide can conduct electricity whereas molten naphthalenecannot conduct electricity.

List of materials andapparatus:

Materials: naphthalene and lead(II) bromide Apparatus: crucible, carbon electrodes, two dry cells, connecting wire, ammeter,switch, Bunsen burner, electronic balance

Experimentalprocedure:

1. Measure 50 g of naphthalene and pour into a crucible.2. Dip two carbon electrodes into the naphthalene.3. Connect the electrodes to a switch, two dry cells and ammeter usingconnecting wires.4. Heat the naphthalene until its melt.5. Close the switch and the observation is recorded.6. Repeat steps 1 to 5 using lead(II) bromide to replace naphthalene.

Tabulation of data Type of compound ObservationNaphthalene moltenLead(II) bromide molten