Chemistry 0620 | Paper 3 Mock 01 | Theory (Core)
Question 1 Report
Several reactions take place inside the furnace used to extract iron. Fig. 5.1 is a simple outline of the furnace with three regions labelled X, Y and Z. Air enters at point G near the base.
Fig. 5.1
(a) At region Z the coke burns in the hot air entering at G. Write the symbol equation for the burning of carbon in oxygen. [2]
(b) State whether the reaction in (a) is exothermic or endothermic, and give the importance of this reaction to the furnace. [2]
(c) Higher up, at region Y, the gas from part (a) reacts with more hot coke. Write the symbol equation for this reaction. [2]
(d) Name the gas that is the main reducing agent produced at region Y. [1]
(e) State the two products formed when this reducing agent reacts with iron(III) oxide in region X. [2]
This question follows the sequence of reactions down the iron blast furnace, region by region.
(a) At region Z the coke burns in the hot air:
C + O2 → CO2
Award [1] for correct formulae and [1] for balancing (1 C and 2 O on each side).
(b) This reaction is exothermic [1]; it is important because it releases heat that keeps the furnace hot enough for the other reactions to happen [1].
(c) Higher up at region Y, the carbon dioxide from part (a) reacts with more hot coke:
C + CO2 → 2CO
Award [1] for correct formulae and [1] for balancing (2 C and 2 O on each side).
(d) The main reducing agent produced at region Y is carbon monoxide [1].
(e) When carbon monoxide reacts with iron(III) oxide in region X, the two products are iron [1] and carbon dioxide [1] (Fe2O3 + 3CO → 2Fe + 3CO2).
Answer Details
This question follows the sequence of reactions down the iron blast furnace, region by region.
(a) At region Z the coke burns in the hot air:
C + O2 → CO2
Award [1] for correct formulae and [1] for balancing (1 C and 2 O on each side).
(b) This reaction is exothermic [1]; it is important because it releases heat that keeps the furnace hot enough for the other reactions to happen [1].
(c) Higher up at region Y, the carbon dioxide from part (a) reacts with more hot coke:
C + CO2 → 2CO
Award [1] for correct formulae and [1] for balancing (2 C and 2 O on each side).
(d) The main reducing agent produced at region Y is carbon monoxide [1].
(e) When carbon monoxide reacts with iron(III) oxide in region X, the two products are iron [1] and carbon dioxide [1] (Fe2O3 + 3CO → 2Fe + 3CO2).
Question 2 Report
A student adds a few drops of chlorine water to a test-tube of aqueous potassium bromide, as shown in Fig. 2.1.
Fig. 2.1
(a) State the colour of the aqueous potassium bromide before the chlorine water is added. [1]
(b) Describe the colour change seen in the test-tube after the chlorine water is added. [1]
(c) Name the type of reaction taking place. [1]
(d) Name the halogen that is formed in the solution. [1]
(e) Explain why chlorine is able to react in this way with potassium bromide. [1]
(f) Complete the word equation for the reaction:
chlorine + potassium bromide → .............. + .............. [2]
This is a halogen displacement reaction. A more reactive halogen pushes a less reactive halogen out of its salt. Chlorine is above bromine in Group VII, so chlorine is more reactive and displaces bromine from potassium bromide.
(a) Aqueous potassium bromide is colourless before the chlorine water is added [1].
(b) After adding chlorine water the solution turns orange / brown (yellow-brown) [1], the colour of bromine forming in the solution.
(c) The reaction type is displacement [1].
(d) The halogen formed is bromine [1], which gives the orange-brown colour.
(e) Chlorine can react this way because chlorine is more reactive than bromine, so it displaces bromine from the bromide [1]. Reactivity falls down Group VII, and chlorine sits above bromine.
(f) Completed word equation: chlorine + potassium bromide → potassium chloride [1] + bromine [1]. The chlorine takes the place of the bromine, forming potassium chloride, and the freed bromine appears as the coloured product.
Exam tip: to predict a displacement, remember the order of reactivity Cl > Br > I; a halogen can only displace one below it in the group.
Answer Details
This is a halogen displacement reaction. A more reactive halogen pushes a less reactive halogen out of its salt. Chlorine is above bromine in Group VII, so chlorine is more reactive and displaces bromine from potassium bromide.
(a) Aqueous potassium bromide is colourless before the chlorine water is added [1].
(b) After adding chlorine water the solution turns orange / brown (yellow-brown) [1], the colour of bromine forming in the solution.
(c) The reaction type is displacement [1].
(d) The halogen formed is bromine [1], which gives the orange-brown colour.
(e) Chlorine can react this way because chlorine is more reactive than bromine, so it displaces bromine from the bromide [1]. Reactivity falls down Group VII, and chlorine sits above bromine.
(f) Completed word equation: chlorine + potassium bromide → potassium chloride [1] + bromine [1]. The chlorine takes the place of the bromine, forming potassium chloride, and the freed bromine appears as the coloured product.
Exam tip: to predict a displacement, remember the order of reactivity Cl > Br > I; a halogen can only displace one below it in the group.
Question 3 Report
Fig. 2.1 shows the apparatus a student set up to make ethanol from a warm sugar solution. The flask was left in a warm place for four days.
Fig. 2.1
(a) Name the substance that must be added to solution G so that the reaction takes place. [1]
(b) The reaction works best at about 30 °C. Explain why the reaction is very slow at 5 °C and why it stops completely at 80 °C. [2]
(c) A gas bubbles through liquid L. Name this gas and state the change you would see in L. [2]
(d) Explain why air must be kept out of the flask during the reaction. [1]
(e) Write a word equation for the reaction taking place inside the flask. [1]
This question is about making ethanol by fermentation, the effect of temperature on the enzymes in yeast, and testing the gas given off.
(a) The substance that must be added to sugar solution G is yeast [1]. Yeast contains the enzymes that ferment the sugar.
(b) Fermentation is catalysed by enzymes, which are temperature sensitive. At 5°C the particles/enzymes have too little energy, so the reaction is very slow [1]; at 80°C the enzymes in the yeast are denatured (their shape is destroyed), so the reaction stops completely [1]. This is why around 30°C is ideal.
(c) The gas that bubbles through liquid L is carbon dioxide [1]; the limewater L turns milky / cloudy / white [1], which is the confirmatory test for carbon dioxide.
(d) Air must be kept out to stop the ethanol being oxidised (to ethanoic acid / vinegar) by oxygen in the air [1].
(e) The word equation for fermentation is:
glucose → ethanol + carbon dioxide [1].
Exam tip: "denatured" is the correct word at 80°C, not "killed" or "melted"; heat changes the enzyme's shape so it can no longer work.
Answer Details
This question is about making ethanol by fermentation, the effect of temperature on the enzymes in yeast, and testing the gas given off.
(a) The substance that must be added to sugar solution G is yeast [1]. Yeast contains the enzymes that ferment the sugar.
(b) Fermentation is catalysed by enzymes, which are temperature sensitive. At 5°C the particles/enzymes have too little energy, so the reaction is very slow [1]; at 80°C the enzymes in the yeast are denatured (their shape is destroyed), so the reaction stops completely [1]. This is why around 30°C is ideal.
(c) The gas that bubbles through liquid L is carbon dioxide [1]; the limewater L turns milky / cloudy / white [1], which is the confirmatory test for carbon dioxide.
(d) Air must be kept out to stop the ethanol being oxidised (to ethanoic acid / vinegar) by oxygen in the air [1].
(e) The word equation for fermentation is:
glucose → ethanol + carbon dioxide [1].
Exam tip: "denatured" is the correct word at 80°C, not "killed" or "melted"; heat changes the enzyme's shape so it can no longer work.
Question 4 Report
Table 8.1 shows some aqueous solutions electrolysed using inert electrodes. Some products are missing.
| Aqueous electrolyte | Product at cathode | Product at anode |
|---|---|---|
| dilute sulfuric acid | ||
| concentrated hydrochloric acid | hydrogen | |
| copper(II) sulfate | oxygen |
(a) Complete the four empty boxes in Table 8.1. [4]
(b) State the rule that decides whether a metal or hydrogen is formed at the cathode. [2]
(c) Explain why chlorine is formed at the anode with concentrated hydrochloric acid, but oxygen is formed with dilute sulfuric acid. [2]
(d) State one observation that shows chlorine is being formed at the anode. [1]
(e) Name one material used to make an inert electrode. [1]
This question tests the electrolysis of aqueous solutions with inert electrodes, where water competes with the dissolved ions. At the cathode either the metal or hydrogen is discharged; at the anode either a halogen (if concentrated halide) or oxygen is discharged.
(a) The completed table is:
| Aqueous electrolyte | Product at cathode | Product at anode |
|---|---|---|
| dilute sulfuric acid | hydrogen [1] | oxygen [1] |
| concentrated hydrochloric acid | hydrogen | chlorine [1] |
| copper(II) sulfate | copper [1] | oxygen |
(b) The rule at the cathode: if the metal is more reactive than hydrogen, hydrogen is formed at the cathode [1]; if the metal is less reactive than hydrogen (for example copper), the metal is formed [1].
(c) At the anode: a concentrated halide solution releases the halogen (chlorine) [1], whereas a dilute solution or a non-halide (like sulfate) releases oxygen from the water/hydroxide ions [1].
(d) An observation for chlorine: a pale green/yellow gas is seen that bleaches damp litmus paper [1].
(e) A material for an inert electrode: carbon (graphite) or platinum [1].
Exam tip: sulfuric acid electrolysis is really the electrolysis of water (hydrogen and oxygen in a 2:1 volume ratio); the sulfate ion is too stable to be discharged, so oxygen comes off instead.
Answer Details
This question tests the electrolysis of aqueous solutions with inert electrodes, where water competes with the dissolved ions. At the cathode either the metal or hydrogen is discharged; at the anode either a halogen (if concentrated halide) or oxygen is discharged.
(a) The completed table is:
| Aqueous electrolyte | Product at cathode | Product at anode |
|---|---|---|
| dilute sulfuric acid | hydrogen [1] | oxygen [1] |
| concentrated hydrochloric acid | hydrogen | chlorine [1] |
| copper(II) sulfate | copper [1] | oxygen |
(b) The rule at the cathode: if the metal is more reactive than hydrogen, hydrogen is formed at the cathode [1]; if the metal is less reactive than hydrogen (for example copper), the metal is formed [1].
(c) At the anode: a concentrated halide solution releases the halogen (chlorine) [1], whereas a dilute solution or a non-halide (like sulfate) releases oxygen from the water/hydroxide ions [1].
(d) An observation for chlorine: a pale green/yellow gas is seen that bleaches damp litmus paper [1].
(e) A material for an inert electrode: carbon (graphite) or platinum [1].
Exam tip: sulfuric acid electrolysis is really the electrolysis of water (hydrogen and oxygen in a 2:1 volume ratio); the sulfate ion is too stable to be discharged, so oxygen comes off instead.
Question 5 Report
Fig. 8.1 shows apparatus used to identify the products formed when an alkane fuel is burned in a good supply of air.
Fig. 8.1
Tube X contains anhydrous copper(II) sulfate. Tube Y contains limewater.
(a) When the alkane burns completely, two products pass through the apparatus. Name both products. [2]
(b) State the colour change seen in tube X and name the substance that causes this change. [2]
(c) State what would be observed in tube Y and name the gas responsible. [2]
(d) The alkane used is hexane, C6H14. Write the balanced symbol equation for its complete combustion. [2]
(e) The pump draws the gases through the apparatus. Suggest why the apparatus is arranged so that tube X comes before tube Y. [1]
(f) When hexane burns in a limited supply of air, incomplete combustion occurs. Name the two carbon-containing products that may form. [2]
(g) Explain, using ideas about oxygen supply, why incomplete combustion produces these products instead of carbon dioxide. [2]
This question is about identifying the products of burning a hydrocarbon and testing for them. The gases from the flame are drawn first through anhydrous copper(II) sulfate (tube X, tests for water) and then through limewater (tube Y, tests for carbon dioxide).
(a) The two products of complete combustion [2]. A hydrocarbon burned in plenty of air gives carbon dioxide [1] and water (vapour) [1]. The carbon in the fuel ends up as CO2 and the hydrogen ends up as H2O.
(b) Colour change in tube X [2]. Anhydrous copper(II) sulfate turns from white to blue [1]. This change is caused by water [1], which is being condensed and absorbed from the gas stream. This is the standard chemical test for the presence of water.
(c) Observation in tube Y [2]. The limewater turns milky (cloudy) [1]. The gas responsible is carbon dioxide [1], which reacts with the calcium hydroxide in limewater to form insoluble calcium carbonate.
(d) Balanced equation for complete combustion of hexane [2].
2C6H14 + 19O2 → 12CO2 + 14H2O
Award [1] for the correct products (CO2 and H2O) and [1] for the whole equation being balanced. Check the balance: 12 C on each side, 28 H on each side, and \(12\times2 + 14 = 38\) O atoms on the right, matching \(19\times2 = 38\) on the left. Using 2 molecules of hexane avoids a fractional coefficient in front of O2.
(e) Why tube X comes before tube Y [1]. So that the water vapour is removed and tested before the gas reaches the limewater [1]. If limewater came first, water from the flame would mix with it and you could not test the two products separately.
(f) Carbon-containing products of incomplete combustion [2]. In a limited air supply the fuel forms carbon monoxide [1] and carbon (soot) [1].
(g) Why these form instead of carbon dioxide [2]. There is not enough oxygen for the carbon to be fully oxidised [1], so the carbon is only partly oxidised to carbon monoxide, or not oxidised at all and left as solid carbon [1]. Full oxidation to CO2 needs the greatest amount of oxygen, so it is the first thing to be lost when air is restricted.
Exam tip: remember the pairing of tests, white to blue anhydrous copper(II) sulfate for water and milky limewater for carbon dioxide, and always balance combustion equations by doubling the hydrocarbon if an odd number of oxygen atoms appears.
Answer Details
This question is about identifying the products of burning a hydrocarbon and testing for them. The gases from the flame are drawn first through anhydrous copper(II) sulfate (tube X, tests for water) and then through limewater (tube Y, tests for carbon dioxide).
(a) The two products of complete combustion [2]. A hydrocarbon burned in plenty of air gives carbon dioxide [1] and water (vapour) [1]. The carbon in the fuel ends up as CO2 and the hydrogen ends up as H2O.
(b) Colour change in tube X [2]. Anhydrous copper(II) sulfate turns from white to blue [1]. This change is caused by water [1], which is being condensed and absorbed from the gas stream. This is the standard chemical test for the presence of water.
(c) Observation in tube Y [2]. The limewater turns milky (cloudy) [1]. The gas responsible is carbon dioxide [1], which reacts with the calcium hydroxide in limewater to form insoluble calcium carbonate.
(d) Balanced equation for complete combustion of hexane [2].
2C6H14 + 19O2 → 12CO2 + 14H2O
Award [1] for the correct products (CO2 and H2O) and [1] for the whole equation being balanced. Check the balance: 12 C on each side, 28 H on each side, and \(12\times2 + 14 = 38\) O atoms on the right, matching \(19\times2 = 38\) on the left. Using 2 molecules of hexane avoids a fractional coefficient in front of O2.
(e) Why tube X comes before tube Y [1]. So that the water vapour is removed and tested before the gas reaches the limewater [1]. If limewater came first, water from the flame would mix with it and you could not test the two products separately.
(f) Carbon-containing products of incomplete combustion [2]. In a limited air supply the fuel forms carbon monoxide [1] and carbon (soot) [1].
(g) Why these form instead of carbon dioxide [2]. There is not enough oxygen for the carbon to be fully oxidised [1], so the carbon is only partly oxidised to carbon monoxide, or not oxidised at all and left as solid carbon [1]. Full oxidation to CO2 needs the greatest amount of oxygen, so it is the first thing to be lost when air is restricted.
Exam tip: remember the pairing of tests, white to blue anhydrous copper(II) sulfate for water and milky limewater for carbon dioxide, and always balance combustion equations by doubling the hydrocarbon if an odd number of oxygen atoms appears.
Question 6 Report
The method chosen to extract a metal depends on the position of the metal in the reactivity series. Fig. 2.1 shows part of the reactivity series; carbon is included in brackets. Bracket A and bracket B mark two ranges of metals.
Fig. 2.1
Table 2.1 lists four metals and how each is obtained.
| Metal | Main extraction method |
|---|---|
| potassium | |
| zinc | |
| iron | reduction in a furnace |
| gold |
(a) State the general method used to extract the metals in bracket A. [1]
(b) Explain, in terms of reactivity, why the metals in bracket A cannot be extracted by heating their oxides with carbon. [2]
(c) Complete Table 2.1 by giving the main extraction method for potassium, zinc and gold. [3]
(d) Name the substance normally used to reduce the metals in bracket B from their oxides. [1]
(e) Zinc oxide is reduced by this substance. Write the symbol equation for the reaction of zinc oxide with carbon to form zinc and carbon monoxide. [2]
(f) State the meaning of the term ore. [1]
(g) Suggest why gold is found in the ground as the uncombined metal. [2]
This question links the reactivity series to the method used to extract each metal.
(a) The metals in bracket A (the most reactive, above carbon) are extracted by electrolysis of the molten compound / oxide [1].
(b) Carbon cannot be used for bracket A because [2]: these metals are more reactive than carbon [1], so carbon cannot remove the oxygen from (cannot reduce) their oxides [1]. A more reactive metal holds onto its oxygen too strongly for carbon to take it.
(c) Completing Table 2.1, [1] each: potassium is above carbon so electrolysis; zinc is below carbon so reduction (heating) with carbon; gold is so unreactive it is found native and simply dug up (no chemical extraction).
| Metal | Main extraction method |
|---|---|
| potassium | electrolysis |
| zinc | reduction (heating) with carbon |
| iron | reduction in a furnace |
| gold | found native / dug up as the metal |
(d) The substance normally used to reduce the bracket B metals from their oxides is carbon (coke) [1].
(e) Zinc oxide reacts with carbon to give zinc and carbon monoxide:
ZnO + C → Zn + CO
Award [1] for correct formulae and [1] for balancing (1 Zn, 1 C and 1 O on each side).
(f) An ore is a naturally occurring rock or mineral from which a metal can be extracted [1].
(g) Gold is found as the uncombined (native) metal because [2]: it is very unreactive and low in the reactivity series [1], so it does not react with other substances and stays as the free metal rather than forming a compound [1].
Answer Details
This question links the reactivity series to the method used to extract each metal.
(a) The metals in bracket A (the most reactive, above carbon) are extracted by electrolysis of the molten compound / oxide [1].
(b) Carbon cannot be used for bracket A because [2]: these metals are more reactive than carbon [1], so carbon cannot remove the oxygen from (cannot reduce) their oxides [1]. A more reactive metal holds onto its oxygen too strongly for carbon to take it.
(c) Completing Table 2.1, [1] each: potassium is above carbon so electrolysis; zinc is below carbon so reduction (heating) with carbon; gold is so unreactive it is found native and simply dug up (no chemical extraction).
| Metal | Main extraction method |
|---|---|
| potassium | electrolysis |
| zinc | reduction (heating) with carbon |
| iron | reduction in a furnace |
| gold | found native / dug up as the metal |
(d) The substance normally used to reduce the bracket B metals from their oxides is carbon (coke) [1].
(e) Zinc oxide reacts with carbon to give zinc and carbon monoxide:
ZnO + C → Zn + CO
Award [1] for correct formulae and [1] for balancing (1 Zn, 1 C and 1 O on each side).
(f) An ore is a naturally occurring rock or mineral from which a metal can be extracted [1].
(g) Gold is found as the uncombined (native) metal because [2]: it is very unreactive and low in the reactivity series [1], so it does not react with other substances and stays as the free metal rather than forming a compound [1].
Question 7 Report
Bronze was one of the first alloys ever made and is still used today for bells, statues and medals.
(a) Bronze is an alloy of copper. Name the two metals present in bronze. [2]
(b) State one use of bronze. [1]
(c) Bronze is harder than pure copper. Explain why, in terms of the arrangement of the atoms. [2]
(d) Brass also contains copper. Name the second metal present in brass. [1]
(e) Give one property, other than hardness, that copper alloys such as brass and bronze have that makes them useful. [1]
(f) State whether an alloy is an element, a compound or a mixture. [1]
This question is about bronze, one of the oldest copper alloys, and tests the same layer-sliding model of hardness plus the basic classification of an alloy.
(a) Bronze is an alloy of copper [1] and tin [1].
(b) Any one correct use of bronze, for example bells, statues, medals or springs. [1]
(c) Bronze is harder than pure copper because the tin atoms are a different size from the copper atoms and distort the regular layers [1]; the distorted layers cannot slide over each other easily, so the alloy is harder [1].
(d) Brass is copper mixed with zinc. [1]
(e) Any one useful property of copper alloys other than hardness, for example resistance to corrosion, good appearance, or good electrical conductivity. [1]
(f) An alloy is a mixture. [1] The atoms are only mixed together, not chemically combined in a fixed ratio, so it is not a compound; and because it contains more than one type of atom it is not an element.
Answer Details
This question is about bronze, one of the oldest copper alloys, and tests the same layer-sliding model of hardness plus the basic classification of an alloy.
(a) Bronze is an alloy of copper [1] and tin [1].
(b) Any one correct use of bronze, for example bells, statues, medals or springs. [1]
(c) Bronze is harder than pure copper because the tin atoms are a different size from the copper atoms and distort the regular layers [1]; the distorted layers cannot slide over each other easily, so the alloy is harder [1].
(d) Brass is copper mixed with zinc. [1]
(e) Any one useful property of copper alloys other than hardness, for example resistance to corrosion, good appearance, or good electrical conductivity. [1]
(f) An alloy is a mixture. [1] The atoms are only mixed together, not chemically combined in a fixed ratio, so it is not a compound; and because it contains more than one type of atom it is not an element.
Question 8 Report
Water for a town is treated in four stages, A, B, C and D, before it is supplied to homes. At stage A large floating objects such as leaves and twigs are trapped by a metal grid.
(a) Name stages B, C and D. [3]
(b) Suggest a name for stage A. [1]
(c) Complete the table to describe what happens to the water at stages B and C. [2]
| Stage | What happens to the water |
|---|---|
| B | |
| C |
(e) Describe a chemical test, with its result, to show that the treated liquid contains water. [2]
(f) The purity of the water is checked by measuring its boiling point. State the boiling point of pure water and explain how a dissolved impurity would change it. [2]
(g) Give one use of large amounts of water in industry. [1]
This question extends water treatment to four stages: an initial screening to catch large debris, then the familiar sedimentation, filtration and chlorination. Each stage targets a progressively finer or different type of impurity.
(a) Naming the stages: B is sedimentation (settling) [1], C is filtration [1], and D is chlorination (sterilisation). [1]
(b) Stage A, where large floating objects are trapped by a metal grid, is screening. [1]
(c) Completing the table:
| Stage | What happens to the water |
|---|---|
| B | heavy insoluble particles settle to the bottom and are removed [1] |
| C | the water passes through sand and smaller insoluble particles are trapped [1] |
(d)(i) The substance added at stage D is chlorine. [1] (ii) Stage D is necessary because chlorine kills bacteria / microorganisms [1], making the water safe to drink and preventing waterborne disease. [1]
(e) To show the treated liquid contains water, add anhydrous copper(II) sulfate; [1] it changes from white to blue. [1]
(f) Pure water boils at 100°C [1] (at normal atmospheric pressure); a dissolved impurity raises the boiling point above 100°C. [1]
(g) One industrial use of large amounts of water: cooling (as a coolant), or as a solvent, for cleaning, or for raising steam. [1]
Exam tip: match each stage to its purpose: screening removes big debris, sedimentation removes heavy solids, filtration removes fine solids, chlorination kills microbes. Only chlorination affects living organisms.
Answer Details
This question extends water treatment to four stages: an initial screening to catch large debris, then the familiar sedimentation, filtration and chlorination. Each stage targets a progressively finer or different type of impurity.
(a) Naming the stages: B is sedimentation (settling) [1], C is filtration [1], and D is chlorination (sterilisation). [1]
(b) Stage A, where large floating objects are trapped by a metal grid, is screening. [1]
(c) Completing the table:
| Stage | What happens to the water |
|---|---|
| B | heavy insoluble particles settle to the bottom and are removed [1] |
| C | the water passes through sand and smaller insoluble particles are trapped [1] |
(d)(i) The substance added at stage D is chlorine. [1] (ii) Stage D is necessary because chlorine kills bacteria / microorganisms [1], making the water safe to drink and preventing waterborne disease. [1]
(e) To show the treated liquid contains water, add anhydrous copper(II) sulfate; [1] it changes from white to blue. [1]
(f) Pure water boils at 100°C [1] (at normal atmospheric pressure); a dissolved impurity raises the boiling point above 100°C. [1]
(g) One industrial use of large amounts of water: cooling (as a coolant), or as a solvent, for cleaning, or for raising steam. [1]
Exam tip: match each stage to its purpose: screening removes big debris, sedimentation removes heavy solids, filtration removes fine solids, chlorination kills microbes. Only chlorination affects living organisms.
