Unit 3Chemical Reactions

MenuThe Chemical IndustryHesss LawEquilibriumAcids and BasesRedox ReactionsNuclear Chemistry

The Chemical Industry

The Chemical IndustryMajor contributor to quality of life and economy.

The Chemical IndustryQuality of lifeFuels (eg petrol for cars)Plastics (Polythene etc)Agrochemicals (Fertilisers, pesticides etc)Alloys (Inc. Steel for building)Chemicals (eg Cl2 for water purification)Dyes (for clothing etc)Cosmetics and medicinesSoaps and detergentsEtc!!!!

The Chemical IndustryContributes to National EconomyMajor employer of people at all skill levelsRevenue from taxation on fuels etcRevenue from sales of productRevenue from exports of products

The Chemical IndustryResearch chemists identify a chemical route to make a new product, using available reactants.

The Chemical IndustryFeasibility study produces small amounts of product to see if the process will work

The Chemical IndustryThe process is now scaled up to go into full scale production.Process so far will have taken months.Many problems will have been encountered and will have to be resolved before full scale production commences

The Chemical IndustryChemical plant is built in a suitable siteOperators employedEarly production will allow monitoring of cost, safety, pollution risks, yield and profitability

The Chemical IndustryFeedstockMIXERREACTION VESSELSEPARATORPRODUCTBY-PRODUCTUnreacted feedstocks recycled

The Chemical Industry - Feedstocks

The Chemical IndustryCan be Continuous processOr can be Batch Process

The Chemical IndustryContinuous Process

Used by big industries where large quantities of product are requiredRequires small workforceOften automated / computer controlledQuality of product checked remotelyEnergy efficiency usually goodPlants expensive to buildPlants not flexible

The Chemical IndustryBatch ProcessMake substance which are required in smaller amountsProcess looks more like the initial reactionOverhaul of system needed regularly time and energy lost if plant has to be shut downPlant can be more flexiblePlant is usually less expensive to build initially

The Chemical Industry: The Costs Involved

The Chemical IndustryIndustries can be classed as:Labour intensiveCapital intensive

The Chemical IndustryService industries (Catering, education, healthcare), are labour intensive

The Chemical IndustryChemical industry tends to be more Capital intensive as a large investment is required to buy equipment and build plants

The Chemical IndustryExpectations of work safety and a clean environment increase during the twentieth centuryH & S legislation protects workforce

The Chemical IndustryTradition is important steel making continues in areas where it was set up even if raw materials are no longer available locallyTransport options are important

The Chemical IndustryChoice of a particular chemical route is dependent upon:Cost of raw materialsSuitability of feedstocksYield of productOption to recycle unreacted feedstockMarketability of by productsCosts of getting rid of wastes, and safety considerations for workforce and localsPrevention of pollution

The Chemical IndustryClick here to repeat The Chemical Industry.

Click here to return to the Menu

Click here to End.

Hesss Law

Hesss lawHesss law states that the enthalpy change for a chemical reaction is independent of the route taken.This means that chemical equations can be treated like simultaneous equations.Enthalpy changes can be worked out using Hesss law.

Hesss lawCalculate the enthalpy change for the reaction: C(s) + 2H2(g) CH4(g) using the enthalpies of combustion of carbon, hydrogen and methane.

Hesss lawFirst write the target equation.

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

Then write the given equations.

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJH2 + O2 H2O DH= -286 kJCH4+ 2O2 CO2 + 2H2O DH=-891 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?Build up the target equation from the given equations.If we multiply we must also multiply DH.If we reverse an equation we reverse the sign of DH.

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJH2 + O2 H2O DH= -286 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJCH4 + 2O2 CO2+2H2O DH=-891 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJCO2+2H2O CH4 + 2O2 DH=+891 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJCO2+2H2O CH4 + 2O2 DH=+891 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?We can add all the equations, striking out species that will appear in equal numbers on both sides.

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJCO2+2H2O CH4 + 2O2 DH=+891 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJCO2+ 2H2O CH4 +2O2 DH=+891 kJ C + 2H2 CH4 DH=(-394 572 + 891)kJ = -75 kJ

Hesss lawC(s) + 2H2(g) CH4(g) DH=?

C + O2 CO2 DH= -394 kJ2H2 + O2 2H2O DH= -572 kJCO2+ 2H2O CH4 +2O2 DH=+891 kJ C + 2H2 CH4 DH=-75 kJ

Hesss LawClick here to repeat Hesss Law.

Click here to return to the Menu

Click here to End.

Equilibrium

Dynamic EquilibriumReversible reactions reach a state of dynamic equilibrium The rates of forward and reverse reactions are equal.At equilibrium, the concentrations of reactants and products remain constant, although not necessarily equal.

Changing the EquilibriumUsing a catalyst does not change the position of the equilibrium.A catalyst speeds up both the forward and back reactions equally and so the equilibrium is reached more quickly.

Changing the EquilibriumChanges in concentration, pressure and temperature can alter the position of equilibrium.Le Chateliers Principle states that when we act on an equilibrium the position of the equilibrium will move to reduce the effect of the change.

ConcentrationConsider the equilibrium:A + B C + DIf we increase the concentration of A, we speed up the forward reaction.This results in more C and D being formed.

ConcentrationConsider the equilibrium:Br2(aq) + H2O(l) 2H+(aq) + Br-(aq) + BrO-(aq)The solution is red-brown, due the Br2 molecules.If we add sodium bromide, increasing the concentration of Br-, we favour the RHS and so the equilibrium moves to the left.The red-brown colour will increase.

PressureRemember: 1 mole of any gas has the same volume (under the same conditions of pressure and temperature).This means that the number of moles of has are the same as the volumes.

PressureIncreasing pressure means putting the same number of moles in a smaller space.This is the same as increasing concentration.To reduce this effect the equilibrium will shift so as to reduce the number of moles of gas.

PressureIncreasing pressure favours the side with the smaller volume of gas. Consider:N2O4(g) 2NO2(g) 1 mole 2 moles 1 volume 2 volumesIf we increase the pressure we favour the forward reaction, so more N2O4 is formed.

TemperatureAn equilibrium involves two opposite reactions. One of these processes must release energy (exothermic).The reverse process must take in energy (endothermic).

TemperatureFirst consider an exothermic reaction.

Exothermic Reaction

Increasing temperature leads to a small increasein the number of molecules with sufficient activation energy

TemperatureNow consider an endothermic reaction.

Endothermic Reaction

Increasing temperature leads to a greater increasein the number of molecules with sufficient activation energy

TemperatureThe percentage increase in the number of molecules with sufficient activation energy is much greater in the endothermic reaction, compared to the exothermic reaction.

Thus both the endothermic and exothermic processes are speeded up by increasing temperature.However an increase in temperature has a greater effect on the endothermic process.

Increasing temperature favours the endothermic side of the equilibrium. Consider:N2O4(g) 2NO2(g) DH= +58 kJIf we increase the temperature we favour the forward reaction so more NO2 is formed.

The Haber ProcessThe Haber process involves the preparation of ammonia from nitrogen and hydrogen.N2 + 3H2 2NH3 DH = -88 kJWe shall look at the factors affecting this equilibrium.

The Haber ProcessN2 + 3H2 2NH3 DH = -88 kJA catalyst of finely divided iron is used to increase the reaction speed and so shorten the time needed to reach the equilibrium.

The Haber ProcessN2 + 3H2 2NH3 DH = -88 kJ1 mole 3 moles 2 moles1 vol3 vols 2 vols4 vols 2 volsSince the RHS has a lower volume of gas than the LHS, higher pressure will favour the production of ammonia.A reaction chamber to withstand the higher pressure will cost much more.

The Haber ProcessN2 + 3H2 2NH3 DH = -88 kJSince the forward reaction is exothermic more ammonia will be produced at low temperatures.At low temperatures the reaction is very slow so the rate of production of ammonia is low.

The Haber ProcessN2 + 3H2 2NH3 DH = -88 kJTo ensure maximum conversion the unreacted gases are recycled through the reaction chamber after reaction.

The Haber ProcessN2 + 3H2 2NH3 DH = -88 kJTo achieve the most profitable production of ammonia the following conditions are used:iron powder as catalyst250 atmospheres pressuretemperature of 500oC - 600oC

The Haber ProcessN2 + H2MIXER

REACTIONChamber Fe catalyst

SEPARATORNH3Unreacted N2 + H2recycled

EquilibriumClick here to repeat Equilibrium.

Click here to return to the Menu

Click here to End.

Acids and Bases

pHpH is a scale of acidity.It can be measured using: pH paperUniversal Indicator solution A pH meter.

We carry out an experiment where we progressively dilute acid.Tube 110 ml 0.1 mol/l hydrochloric acid

Transfer 1 ml of acid fromTube 1Tube 2

Add 9 mlof waterto Tube 2Tube 20.01 mol/lhydrochloricacid

Repeat this process five more times so you have a series test tubes.

Concentrations are:

Add Universal Indicator:

Look for a relationship between the concentration of acid and the pH.If [H+] = 10-xpH = x

We repeat the experiment but this time we progressively dilute alkali.Tube 110 ml 0.1 mol/l sodium hydroxide

You now have five test tubes, numbered as below.

Concentrations are:

Add Universal Indicator:

Look for a relationship between the concentration of alkali and the pH.If [OH-] = 10-ypH = 14-y

If we look at water, pH=7[H+] = [OH-] = 10-7 mol/l [H+] x [OH-] = 10-14 mol2/l2This is due to the equilibrium in water: H2O(l) H+(aq) + OH-(aq)For any solution [H+] x [OH-] = 10-14 mol2/l2

Thus we can find [H+] for any solution.What is [H+] of a solution with pH 10? [OH-] = 10-4 mol/l [H+] x 10-4 = 10-14 mol2/l2Thus [H+] = 10-10

Strong and Weak AcidsA strong acid is one which completely dissociates in solution:HCl(aq) H+(aq) + Cl-(aq)A weak acid is one which partially dissociates in solution:CH3CO2H(aq)H+(aq) + CH3CO2-(aq)

We can compare eqimolar solutions of strong and weak acids e.g. 0.1 mol/l hydrochloric acid and 0.1 mol/l ethanoic acid.We compare pH, conductivity, reaction rates and stoichiomery.

The differences between the properties of strong and weak acids are caused by the fact that weak acids contain many fewer H+ ions than strong acids.Both acids can produce the same number of H+ ions, its just that weak acids do so more slowly.

Weak AcidsSolutions of ethanoic acid, carbon dioxide and sulphur dioxide are weak acids.CH3CO2H(aq)H+(aq) + CH3CO2-(aq)CO2(g) + H2O(l) H2CO3(aq)H2CO3(aq) 2H+(aq) + CO32-(aq)SO2(g) + H2O(l) H2SO3(aq)H2SO3(aq) 2H+(aq) + SO32-(aq)

Strong and Weak BasesA strong base is one which completely dissociates in solution:NaOH(aq) Na+(aq) + OH-(aq)A weak base is one which partially dissociates in solution:NH4OH(aq)NH4+(aq) + OH-(aq)

We can compare eqimolar solutions of strong and weak bases e.g. 0.1 mol/l sodium hydroxide and 0.1 mol/l ammonium hydroxide.When we compare pH, conductivity, reaction rates and stoichiomery we find similar results to the comparison of weak and strong acids.

Weak BasesA solution of ammonia is a weak base.NH3(g) + H2O(l) NH4OH(aq)NH4OH(aq) NH4+(aq) + OH-(aq)

Acids + BasesA strong acid and a strong base produce a salt which is neutral.A strong acid and a weak base produce a salt which is acidic.A weak acid and a strong base produce a salt which is basic.

Basic Salts

Basic SaltsSodium carbonate is completely ionised.

Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)

Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present.

Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present. H2O(l) H+(aq) + OH-(aq)

Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present. H2O(l) H+(aq) + OH-(aq)The ions set up an equilibrium.

Basic SaltsSodium carbonate is completely ionised.Na2CO3(aq) 2Na+(aq) + CO32-(aq)Water is also present. H2O(l) H+(aq) + OH-(aq)The ions set up an equilibrium.2H+(aq) + CO32-(aq) H2CO3(aq)

Basic SaltsThis removes of H+(aq) from water. H2O(l) H+(aq) + OH-(aq)

Basic SaltsThis removes of H+(aq) from water. H2O(l) H+(aq) + OH-(aq)The OH-(aq) left behind make the resulting solution basic.

Acid Salts

Acid SaltsAmmonium chloride is completely ionised.

Acid SaltsAmmonium chloride is completely ionised.NH4Cl(aq) NH4+(aq) + Cl-(aq)

Acid SaltsAmmonium chloride is completely ionised.NH4Cl(aq) NH4+(aq) + Cl-(aq)Water is also present.

Acid SaltsAmmonium chloride is completely ionised.NH4Cl(aq) NH4+(aq) + Cl-(aq)Water is also present. H2O(l) H+(aq) + OH-(aq)

Acid SaltsAmmonium chloride is completely ionised.NH4Cl(aq) NH4+(aq) + Cl-(aq)Water is also present. H2O(l) H+(aq) + OH-(aq)The ions set up an equilibrium.

Acid SaltsAmmonium chloride is completely ionised.NH4Cl(aq) NH4+(aq) + Cl-(aq)Water is also present. H2O(l) H+(aq) + OH-(aq)The ions set up an equilibrium. NH4+(aq) + OH-(aq) NH4 OH(aq)

Acid SaltsThis removes of OH-(aq) from water. H2O(l) H+(aq) + OH-(aq)

Acid SaltsThis removes of OH-(aq) from water. H2O(l) H+(aq) + OH-(aq)The H+(aq) left behind make the resulting solution acidic.

Acids and BasesClick here to repeat Acids and Bases.

Click here to return to the Menu

Click here to End.

Redox Reactions

RedoxAn oxidation reaction is one where electrons are lost.Zn(s) Zn2+(aq) + 2eA reduction reaction is one where electrons are gained.Cu2+(aq) + 2e Cu(s)A redox reaction is one in which both oxidation and reduction are occurring.Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

RedoxAn oxidising agent is a substance which accepts electrons.This means that an oxidising agent must itself be reduced.

RedoxA reducing agent is a substance which donates electrons. This means that a reducing agent must itself be oxidised.

RedoxWe should be able to recognise oxidising and reducing agents from the reaction equation.5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn 2+ + 4H2OFe2+ is oxidised to Fe3+ so MnO4 acts as an oxidising agent.

Writing Ion-Electron Equations.Simple equations can be obtained from the data booklet.More complex equations are written using the following routine.

Writing Ion-Electron Equations.Write the reactants and products.2IO3- I2 Add H2O to the side with less oxygen.2IO3- I2 + 6H2OAdd H+ to the other side.2IO3- + 12H+ I2 + 6H2OBalance charge by adding electrons.2IO3- + 12H+ + 10e- I2 + 6H2O

Combining Oxidation and Reduction Equations.Combining the ion-electron half equations produces the overall reaction equation.This must be done so that the number of electrons on opposie sides are equal, and so cancel each other out.

Combining Oxidation and Reduction Equations.Oxidation

Combining Oxidation and Reduction Equations.Oxidation2IO3- + 12H+ + 10e- I2 + 6H2O

Combining Oxidation and Reduction Equations.Oxidation2IO3- + 12H+ + 10e- I2 + 6H2OReduction

Combining Oxidation and Reduction Equations.Oxidation2IO3- + 12H+ + 10e- I2 + 6H2OReductionSO32- + H2O SO42- +2H+ + 2e-

Combining Oxidation and Reduction Equations.Oxidation2IO3- + 12H+ + 10e I2 + 6H2OReduction multiplied by 55SO32- + 5H2O 5SO42- +10H++ 10e

Combining Oxidation and Reduction Equations.Add the equations2IO3- + 12H+ + 10e I2 + 6H2O5SO32- + 5H2O 5SO42- +10H++ 10e

2IO3- + 2H+ + 5SO32- I2 + H2O + 5SO42- We now can extract the mole relationship 2 moles iodate react with 5 moles of sulphite

Redox Titrations.These can be carried out to calculate concentration.Many use permanganate or starch/iodine reactions which are self-indicating the colour change of the reaction tells you when the end point is reached.

Redox Titrations.It was found that 12.5 ml of of 0.1 mol/l acidified potassium dichromate was required to oxidise the alcohol in a sample of 1 ml of wine.Calculate the mass of alcohol in 1 ml of wine.

Redox Titrations.Equations:Cr2O72-+14H+ + 6e 2Cr 3+ + 7H2OC2H5OH + H2O CH3COOH +4H++4eMole Relationship2 moles dichromate react with 3 moles ethanol1 mole dichromate react with 1.5 moles ethanol

Redox Titrations.12.5 ml of 0.1 mol/l dichromate contain 0.0125x0.1 moles dichromate.1.25x10-3 molesMoles of alcohol = 1.25x10-3 x1.5 = 1.875x10-3Mass of alcohol = 46 x 1.875x10-3 g = 0.08625 g

ElectrolysisElectrolysis takes place when electricity is passed through an ionic liquid.Chemical reaction take place at the electrodes reduction at the negative electrode and oxidation at the positive electrode.

ElectrolysisThe electrode reactions can be represented by ion electron equations.In the electrolysis of nickel(II) chloride the reactions are:+ electrode 2Cl- Cl2 + 2e- electrode Ni2+ + 2e Ni

Electrolysis+ electrode 2Cl- Cl2 + 2e- electrode Ni2+ + 2e NiIn both of these ion electron equations one mole of product is produced by two moles of electrons.

The FaradayTo find the value for one mole of electrons multiply Avogadros number by the charge on the electron (1.6x10-19 coulombs)One mole of electrons is called a Faraday and is 96,500 coulombs.

The FaradayUsing the value for the Faraday and the equation:Charge = Current x Time (Coulombs) (Amps) (Seconds) we can carry out many calculations.

Redox ReactionsClick here to repeat Redox Reactions.

Click here to return to the Menu

Click here to End.

Nuclear Chemistry

Stable nucleiNuclei contain protons and neutrons.Energy is needed to hold these particles together.We can plot the number of protons against the number of neutrons.

Stable nuclei

Chart2

0

1

1

2

3

4

5

5

6

6

7

7

8

8

9

10

10

10

11

12

12

12

13

14

14

14

15

16

16

16

17

18

20

18

20

18

20

22

20

22

20

22

23

24

26

24

24

25

26

27

28

28

26

28

29

30

30

28

30

31

32

32

30

32

33

34

36

34

36

34

36

37

38

40

38

40

38

40

41

42

44

42

40

42

43

44

46

48

44

46

42

44

46

47

48

50

48

50

46

48

49

50

50

50

51

52

54

56

52

50

52

53

54

55

56

58

52

54

55

56

57

58

60

58

56

58

59

60

62

64

60

62

58

60

62

63

64

65

66

68

64

66

62

64

65

66

67

68

69

70

72

74

70

72

68

70

71

72

73

74

76

78

74

70

72

74

75

76

77

78

80

82

78

74

76

78

79

80

81

82

82

78

80

82

84

82

82

83

85

86

88

90

82

86

87

88

90

92

88

90

88

90

91

92

93

94

96

94

90

92

94

95

96

97

98

98

94

96

98

99

100

102

100

98

100

101

102

103

104

106

104

105

102

104

105

106

107

108

108

106

108

109

110

112

110

108

110

111

112

113

114

116

114

116

114

116

118

119

120

118

116

118

119

120

121

122

124

122

124

122

124

125

126

126

protons

neutrons

Number of protons v number of neutrons

Sheet1

10

11

21

22

33

34

45

55

56

66

67

77

78

88

89

810

910

1010

1011

1012

1112

1212

1213

1214

1314

1414

1415

1416

1516

1616

1617

1618

1620

1718

1720

1818

1820

1822

1920

1922

2020

2022

2023

2024

2026

2124

2224

2225

2226

2227

2228

2328

2426

2428

2429

2430

2530

2628

2630

2631

2632

2732

2830

2832

2833

2834

2836

2934

2936

3034

3036

3037

3038

3040

3138

3140

3238

3240

3241

3242

3244

3342

3440

3442

3443

3444

3446

3448

3544

3546

3642

3644

3646

3647

3648

3650

3748

3750

3846

3848

3849

3850

3950

4050

4051

4052

4054

4056

4152

4250

4252

4253

4254

4255

4256

4258

4452

4454

4455

4456

4457

4458

4460

4558

4656

4658

4659

4660

4662

4664

4760

4762

4858

4860

4862

4863

4864

4865

4866

4868

4964

4966

5062

5064

5065

5066

5067

5068

5069

5070

5072

5074

5170

5172

5268

5270

5271

5272

5273

5274

5276

5278

5374

5470

5472

5474

5475

5476

5477

5478

5480

5482

5578

5674

5676

5678

5679

5680

5681

5682

5782

5878

5880

5882

5884

5982

6082

6083

6085

6086

6088

6090

6282

6286

6287

6288

6290

6292

6388

6390

6488

6490

6491

6492

6493

6494

6496

6594

6690

6692

6694

6695

6696

6697

6698

6798

6894

6896

6898

6899

68100

68102

69100

7098

70100

70101

70102

70103

70104

70106

71104

71105

72102

72104

72105

72106

72107

72108

73108

74106

74108

74109

74110

74112

75110

76108

76110

76111

76112

76113

76114

76116

77114

77116

78114

116

118

119

120

79118

80116

80118

80119

80120

80121

80122

80124

81122

81124

82122

82124

82125

82126

83126

Sheet1

protons

neutrons

Number of protons v number of neutrons

Sheet2

Sheet3

Stable nucleiAll stable nuclei fit in a narrow bandSome nuclei are unstable because they need too much energy to hold them together.Thus they split apart, sending out some small particles.

Radioactive decay

a decaya decay takes place when the nucleus ejects a helium nucleus.This causes a change in the nucleus.

b decayb decay takes place when the nucleus ejects an electron.This causes a change in the nucleus.

g decayg decay takes place when the nucleus loses energy.This is the extra energy which is no longer needed to hold the nucleus together.

Nuclear EquationsWhen we write a nuclear equation the sum of the mass numbers and atomic numbers on each side must be equal.

Half lifeHalf life is the time which it takes for the radioactivity to half.For any radioactive substance this time is constant.

Half lifeThe decay of individual nuclei within a sample is random and is does not depend of chemical or physical state of the element.Half lives of individual elements may vary from seconds to thousands of years.

Half lifeCalculations involving half life usually involve precise fractions e.g.3H is a b-emitting isotope with a half life of 12.3 years. How long will it take for the radioactivity of a sample to drop to 1/8 of its original value?

Half life3H is a b-emitting isotope with a half life of 12.3 years. How long will it take for the radioactivity of a sample to drop to 1/8 of its original value?Time12.3y24.6y36.9yFraction1/8

Half lifeFor examples where the numbers are more complex the quantity of radioactive material against time is best estimated from a graph.

Half lifeTimeActivity

The Nuclear ChemistryClick here to repeat Nuclear Chemistry.

Click here to return to the Menu

Click here to End.

The EndHope you found the revision useful.

Come back soon!!

Fossil fuelsMetallic ores