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The chromatogram of pure dyes A B C and a dye mixture D is shown below Determine the pure dyes present in D On the diagram show
ithe solvent front
ii Baseline
Iii the most soluble pure dye
i Solvent extraction
Solvent extraction is a method of separating oil from nutsseeds Most nuts contain oil First the nuts are crushed to reduce their size and increase the surface area A suitable volatile solvent is added The mixture is filtered The filtrate solvent is then allowed to crystallize leaving the oilfat If a filter paper is rubbedsmeared with the oilfat it becomes translucent This is the test for the presence of oilfat
Experiment To extract oil from Macadamia nut seeds
Procedure
Crush Macadamia nut seeds form the hard outer cover Place the inner soft seed into a mortar Crush add a little sand to assist in crushing
Add a little propanone and continue crushing Continue crushing and adding a little propanone until there is more liquid mixture than the solid Decantfilter Put the filtrate into an evaporating dish Vapourize the solvent using solar energy sunlight Smearrub a portion of the residue left after evaporation on a clean dry filter paper
Observation Explanation
Propanone dissolve fatoil in the macadamia nuts Propanone is more volatile lower boiling point than oilfat In sunlightsolar energy propanone evaporatevaporize leaving oilfathas a higher boiling pointAny seed like corn wheat rice soya bean may be used instead of macadamia seed When oilfat is rubbed smeared on an opaque paper it becomes translucent
j Crystallization
Crystallization is the process of using solubility of a solutesolid to obtain the solutesolid crystals from a saturated solution by cooling or heating the solution
A crystal is the smallest regular shaped particle of a solute Every solute has unique shape of its crystals
Some solutions form crystals when heated This is because less solute dissolves at higher temperature Some other solutions form crystals when cooled This is because less solute dissolves at lower temperature
Experiment To crystallize copper II sulphate VI solution
Procedure
Place about one spatula full of hydrated copper sulphate VI crystals into 200cm3 of distilled water in a beaker Stir Continue adding a little more of the hydrated copper sulphate VI crystals and stirring until no more dissolve Decantfilter Cover the filtrate with a filter paper Pierce and make small holes on the filter paper cover Preserve the experiment for about seven days
ObservationExplanation
Large blue crystals formed
When hydrated copper II sulphate crystals are placed in water they dissolve to form copper II sulphate solution After some days water slowly evaporate leaving large crystals of copper II sulphate If the mixture is heated to dryness small crystals are formed
PhysicalTemporary and Chemical changes
A physicaltemporary change is one which no new substance is formed and is reversible back to original
A chemicalpermanent change is one which a new substance is formed and is irreversible back to original
The following experiments illustrates physical and chemical changes
aHeating ice
Place about 10g of pure ice in a beaker Determine its temperature Record it at time 00 in the table below Heat the ice on a strong Bunsen flame and determine its temperature after every 60seconds1minute to complete the table below
Plot a graph of time against Temperature yaxes
Explain the shape of your graph
Meltingfreezingfusionsolidification and boiling vaporization evaporation are the two physical processes
Melting freezing point of pure substances is fixed constant
The boiling point of pure substance depends on external atmospheric pressure
Meltingfusion is the physical change of a solid to liquid
Freezing is the physical change of a liquid to solid
Meltingfreezingfusionsolidification is therefore two opposite but same reversible physical processes ie
A s A l
Boilingvaporizationevaporation is the physical change of a liquid to gas
Condensation liquidification is the physical change of gas to liquid
Boilingvaporizationevaporation and condensation liquidification are therefore two opposite but same reversible physical processes ie
B l Bg
Practically
i Meltingliquidificationfusion involves heating a solid to weaken the strong bonds holding the solid particles together
Solids are made up of very strong bonds holding the particles very close to each other Kinetic Theory of matter
On heating these particles gain energyheat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom
iiFreezingfusionsolidification involves cooling a liquid to reform rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom Kinetic Theory of matter
Freezing fusion solidification is an exothermic H process that require particles holding the liquid together to lose energy to the surrounding
iiiBoilingvaporizationevaporation involves heating a liquid to completely breakfree the bonds holding the liquid particles together
Gaseous particles have high degree of freedom Kinetic Theory of matter
Boiling vaporization evaporation is an endothermic H process that requireabsorb energy from the surrounding
ivCondensationliquidification is reverse process of boiling vaporization evaporation
It involves gaseous particles losing energy to the surrounding to form a liquid
AIR OXYGEN AND COMBUSTION
ATHE ATMOSPHERE
1 The atmosphere is made up of air Air is a mixture of colourless odorless gases which is felt as wind air in motionAll living things breath in air for respiration Plants use air for respiration and photosynthesis
2 The main gases present in the atmosphereair
3 The following experiments below shows the presence and composition of the gases in airatmosphere
aTo find the composition of air supporting combustion using a candle stick
Procedure
Measure the length of an empty gas jar M1 Place a candle stick on a Petri dish Float it on water in basintrough Cover it with the gas jar Mark the level of the water in the gas jar M2 Remove the gas jar Light the candle sick Carefully cover it with the gas jar Observe for two minutes Mark the new level of the water M3
Set up of apparatus
Sample observations
Candle continues to burn then extinguishedgoes off
Level of water in the gas jar rises after igniting the candle
Length of empty gas jar M1 14cm
Length of gas jar without water before igniting candle M2 10 cm
Length of gas jar with water before igniting candle M1 M2 14 10 4 cm
Length of gas jar with water after igniting candle M3 8 cm
Length of gas jar without water after igniting candle M1 M3 10 8 2 cm
Explanation
Candle burns in air In a closed system vessel the candle continues to burn using the part of air that support burningcombustion This is called the active part of air The candle goes offextinguished when all the active part of air is used up The level of the water rises to occupy the space volume occupied by the used active part of air
The experiment is better when very dilute sodiumpotassium hydroxide is used instead of water Dilute Potassium sodium hydroxide absorb Carbon IV oxide gas that comes out from burningcombustion of candle stick
From the experiment above the composition of the
i Active part of air can be calculated
M2 M3 x 100 10 8 x 100 20
M2 10cm
ii Inactive part of air can be calculated
100 20 80 M3 8 x 100 80
M2 10cm
bTo find the composition of active part of air using heated copper turnings
Procedure
Clamp a completely packedfilled open ended glass tube with copper turnings Seal the ends with glasscotton wool
Label two graduated syringes as A and B Push out air from syringe A Pull in air into syringe B
Attach both syringe A and B on opposite ends of the glass tube
Determine and record the volume of air in syringe B V1
Heat the glass tube strongly for about three minutes
Push all the air slowly from syringe B to syringe A as heating continues Push all the air slowly from syringe A back to syringe B and repeatedly back and forth
After about ten minutes determine the new volume of air in syringe B V2
Set up of apparatus
Sample observations
Colour change from brown to black
Volume of air in syringe B before heating V1 1580cm3
Volume of air in syringe B after heating V2 1272cm3
Volume of air in syringe B used by copper V1 V2 308cm3
Sample questions
1 What is the purpose of
i glasscotton wool
To preventstop copper turnings from being blown into the syringeout of the glass tube
ii Passing air through the glass tube repeatedly
To ensure all the active part of air is used up

The chromatogram of pure dyes A B C and a dye mixture D is shown below Determine the pure dyes present in D On the diagram show

ithe solvent front

ii Baseline

Iii the most soluble pure dye

i Solvent extraction

Solvent extraction is a method of separating oil from nutsseeds Most nuts contain oil First the nuts are crushed to reduce their size and increase the surface area A suitable volatile solvent is added The mixture is filtered The filtrate solvent is then allowed to crystallize leaving the oilfat If a filter paper is rubbedsmeared with the oilfat it becomes translucent This is the test for the presence of oilfat

Experiment To extract oil from Macadamia nut seeds

Procedure

Crush Macadamia nut seeds form the hard outer cover Place the inner soft seed into a mortar Crush add a little sand to assist in crushing

Add a little propanone and continue crushing Continue crushing and adding a little propanone until there is more liquid mixture than the solid Decantfilter Put the filtrate into an evaporating dish Vapourize the solvent using solar energy sunlight Smearrub a portion of the residue left after evaporation on a clean dry filter paper

Observation Explanation

Propanone dissolve fatoil in the macadamia nuts Propanone is more volatile lower boiling point than oilfat In sunlightsolar energy propanone evaporatevaporize leaving oilfathas a higher boiling pointAny seed like corn wheat rice soya bean may be used instead of macadamia seed When oilfat is rubbed smeared on an opaque paper it becomes translucent

j Crystallization

Crystallization is the process of using solubility of a solutesolid to obtain the solutesolid crystals from a saturated solution by cooling or heating the solution

A crystal is the smallest regular shaped particle of a solute Every solute has unique shape of its crystals

Some solutions form crystals when heated This is because less solute dissolves at higher temperature Some other solutions form crystals when cooled This is because less solute dissolves at lower temperature

Experiment To crystallize copper II sulphate VI solution

Procedure

Place about one spatula full of hydrated copper sulphate VI crystals into 200cm3 of distilled water in a beaker Stir Continue adding a little more of the hydrated copper sulphate VI crystals and stirring until no more dissolve Decantfilter Cover the filtrate with a filter paper Pierce and make small holes on the filter paper cover Preserve the experiment for about seven days

ObservationExplanation

Large blue crystals formed

When hydrated copper II sulphate crystals are placed in water they dissolve to form copper II sulphate solution After some days water slowly evaporate leaving large crystals of copper II sulphate If the mixture is heated to dryness small crystals are formed

PhysicalTemporary and Chemical changes

A physicaltemporary change is one which no new substance is formed and is reversible back to original

A chemicalpermanent change is one which a new substance is formed and is irreversible back to original

The following experiments illustrates physical and chemical changes

aHeating ice

Place about 10g of pure ice in a beaker Determine its temperature Record it at time 00 in the table below Heat the ice on a strong Bunsen flame and determine its temperature after every 60seconds1minute to complete the table below

Plot a graph of time against Temperature yaxes

Explain the shape of your graph

Meltingfreezingfusionsolidification and boiling vaporization evaporation are the two physical processes

Melting freezing point of pure substances is fixed constant

The boiling point of pure substance depends on external atmospheric pressure

Meltingfusion is the physical change of a solid to liquid

Freezing is the physical change of a liquid to solid

Meltingfreezingfusionsolidification is therefore two opposite but same reversible physical processes ie

A s A l

Boilingvaporizationevaporation is the physical change of a liquid to gas

Condensation liquidification is the physical change of gas to liquid

Boilingvaporizationevaporation and condensation liquidification are therefore two opposite but same reversible physical processes ie

B l Bg

Practically

i Meltingliquidificationfusion involves heating a solid to weaken the strong bonds holding the solid particles together

Solids are made up of very strong bonds holding the particles very close to each other Kinetic Theory of matter

On heating these particles gain energyheat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom

iiFreezingfusionsolidification involves cooling a liquid to reform rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom Kinetic Theory of matter

Freezing fusion solidification is an exothermic H process that require particles holding the liquid together to lose energy to the surrounding

iiiBoilingvaporizationevaporation involves heating a liquid to completely breakfree the bonds holding the liquid particles together

Gaseous particles have high degree of freedom Kinetic Theory of matter

Boiling vaporization evaporation is an endothermic H process that requireabsorb energy from the surrounding

ivCondensationliquidification is reverse process of boiling vaporization evaporation

It involves gaseous particles losing energy to the surrounding to form a liquid

AIR OXYGEN AND COMBUSTION

ATHE ATMOSPHERE

1 The atmosphere is made up of air Air is a mixture of colourless odorless gases which is felt as wind air in motionAll living things breath in air for respiration Plants use air for respiration and photosynthesis

2 The main gases present in the atmosphereair

3 The following experiments below shows the presence and composition of the gases in airatmosphere

aTo find the composition of air supporting combustion using a candle stick

Procedure

Measure the length of an empty gas jar M1 Place a candle stick on a Petri dish Float it on water in basintrough Cover it with the gas jar Mark the level of the water in the gas jar M2 Remove the gas jar Light the candle sick Carefully cover it with the gas jar Observe for two minutes Mark the new level of the water M3

Set up of apparatus

Sample observations

Candle continues to burn then extinguishedgoes off

Level of water in the gas jar rises after igniting the candle

Length of empty gas jar M1 14cm

Length of gas jar without water before igniting candle M2 10 cm

Length of gas jar with water before igniting candle M1 M2 14 10 4 cm

Length of gas jar with water after igniting candle M3 8 cm

Length of gas jar without water after igniting candle M1 M3 10 8 2 cm

Explanation

Candle burns in air In a closed system vessel the candle continues to burn using the part of air that support burningcombustion This is called the active part of air The candle goes offextinguished when all the active part of air is used up The level of the water rises to occupy the space volume occupied by the used active part of air

The experiment is better when very dilute sodiumpotassium hydroxide is used instead of water Dilute Potassium sodium hydroxide absorb Carbon IV oxide gas that comes out from burningcombustion of candle stick

From the experiment above the composition of the

i Active part of air can be calculated

M2 M3 x 100 10 8 x 100 20

M2 10cm

ii Inactive part of air can be calculated

100 20 80 M3 8 x 100 80

M2 10cm

bTo find the composition of active part of air using heated copper turnings

Procedure

Clamp a completely packedfilled open ended glass tube with copper turnings Seal the ends with glasscotton wool

Label two graduated syringes as A and B Push out air from syringe A Pull in air into syringe B

Attach both syringe A and B on opposite ends of the glass tube

Determine and record the volume of air in syringe B V1

Heat the glass tube strongly for about three minutes

Push all the air slowly from syringe B to syringe A as heating continues Push all the air slowly from syringe A back to syringe B and repeatedly back and forth

After about ten minutes determine the new volume of air in syringe B V2

Set up of apparatus

Sample observations

Colour change from brown to black

Volume of air in syringe B before heating V1 1580cm3

Volume of air in syringe B after heating V2 1272cm3

Volume of air in syringe B used by copper V1 V2 308cm3

Sample questions

1 What is the purpose of

i glasscotton wool

To preventstop copper turnings from being blown into the syringeout of the glass tube

ii Passing air through the glass tube repeatedly

To ensure all the active part of air is used up