Thursday, 21 November 2013


As requested, here are my top tips on GCSEs.
Some of it is things you will have heard before, but its really just what I found best from first hand experience.

  1. Don't worry about them. The hardest thing to do, but the best: when you come out the other end, you will be thinking 'why did I even feel stressed!?'
  2. Sleep. Early nights, especially before exams.
  3. Treat yourself. Woo! For revising and during the exam period: I got through a lot of flavoured popcorn doing this blog.
  4. Revise.
  5. Practise. Do every past paper you can find.
  6. Teach someone else; the best way to learn is to explain to other people.
  7. Don't be stupid  (go to hospital/do drugs before your exam.)
  8. Bring all the right equipment to the exam- you just feel so prepared and in control.
  9. Start revising really early to lighten the load and get things in you long term memory.
  10. Deep breathing, kept me calm, focused and stress free ;)

Thursday, 16 May 2013


Another one down, every point in the spec should now be covered!

Good luck to all exam takers!

1.30 recall the charges of common ions in this specification

K +
Na +
Li +
H +

Mg 2+
Ca 2+

Al 3+



SO4 2-

CO3 2-

5.31 describe important uses of sodium hydroxide, including the manufacture of bleach, paper and soap; and of chlorine, including sterilising water supplies and in the manufacture of bleach and hydrochloric acid.

Sodium Hydroxide: Bleach; paper; soap
Chlorine: sterilising water; bleach; hydrochloric acid

5.30 write ionic half-equations for the reactions at the electrodes in the diaphragm cell

2 Cl- > Cl2 + 2e-
2 H+ + 2e- > H2

Really useful video:

5.29 describe the manufacture of sodium hydroxide and chlorine by the electrolysis of concentrated sodium chloride solution (brine) in a diaphragm cell

Brine is NaCl solution.
In electrolysis, chlorine is created at the anode, hydrogen is created at the cathode, sodium hydroxide is left in the solution.

2NaCl + 2H2O > 2NaOH + H2 + Cl2

This happens in a diaphragm cell.

5.28 describe the use of sulfuric acid in the manufacture of detergents, fertilisers and paints

Detergents: used to 'sulphonate' products (apperently.)
Fertilisers: reacted to make phosphates soluble to plants: reacted to make ammonia easier to handle.
Paints: reacted with titanium ore to make a main pigment in paint.

5.27 describe the manufacture of sulfuric acid by the contact process, including the essential conditions

(stage 1) S + O2 > SO2 
(stage 2) SO2 + O > SO3
(satge 3) SO3 + H2O > H2SO4


i) a temperature of about 450C

ii) a pressure of about 2 atmospheres

iii) a vanadium(V) oxide catalyst (in stage 2)

Very helpful source:

5.26 recall the raw materials used in the manufacture of sulfuric acid

Sulphur (sulphur is found in rocks and some natural gasses) and oxygen from the air and water.

5.25 describe the use of ammonia in the manufacture of nitric acid and fertilisers

Ammonia is put into fertilisers it contains nitrate ions plants need to make amino acids and so proteins plants need to grow.

Ammonia is also reacted with oxygen to produce nitric acid:
4NH+ 8O2 > 4HNO+ 4H2O

5.24 understand how the cooling of the reaction mixture liquefies the ammonia produced and allows the unused hydrogen and nitrogen to be recirculated

The products from the reactant are sent through a cooling mechanism, this is at a temperature that condenses ammonia, but not hydrogen and nitrogen. Liquid ammonia is then collected but hydrogen and nitrogen float right back into the reactor.


5.23 describe the manufacture of ammonia by the Haber process, including the essential conditions


i) a temperature of about 450°C
ii) a pressure of about 200 atmospheres
iii) an iron catalyst

5.22 understand that nitrogen from air, and hydrogen from natural gas or the cracking of hydrocarbons, are used in the manufacture of ammonia

Ammonia is made by reacting nitrogen from the air and hydrogen (which comes as a natural gas or from cracking hydrocarbons.)

5.21 understand that condensation polymerisation produces a small molecule, such as water, as well as the polymer

Two monomers come together by loosing a molecule. Atoms from each monomer join together to make the molecule: commonly a H atom from one and a OH molecule from another form water. The two monomers then join together, making a polymer.

5.20 understand that some polymers, such as nylon, form by a different process called condensation polymerisation

Two monomers join together when they loose a small molecule made up of atoms from both monomers.
Commonly, an H atom from one and and OH molecule from the other form water (hence condensation reaction) and the two monomers become joined.


This is what happens to make nylon:


5.19 explain that addition polymers are hard to dispose of as their inertness means that they do not easily biodegrade

Polymers are saturated so they don't react. This means they don't decompose easily.

5.18 describe some uses for polymers, including poly(ethene), poly(propene) and poly(chloroethene)

polyethene: plastic carrier bags; plastic bottels
polypropene: crates; ropes
polychlroethene: piping; cable insulation.

5.17 deduce the structure of a monomer from the repeat unit of an addition polymer

A monomer that is repeated in a polymer looks much like the repeat unit; apart from, instead of having an empty bond either end, it has a double bond in the middle.


5.16 draw the repeat unit of addition polymers, including poly(ethene), poly(propene) and poly(chloroethene)

The repeat unit is the structure that is repeated to form a polymer.
(ignore PTFE on this diagram)

5.15 understand that an addition polymer is formed by joining up many small molecules called monomers

monomers are alkenes with a double bond. If this bond is broken there can be other things bonded, if a carbon from another monomer is bonded in then you can create a chain; do this many times can you have a polymer.

5.14 describe how long-chain alkanes are converted to alkenes and shorter-chain alkanes by catalytic cracking, using silica or alumina as the catalyst and a temperature in the range of 600–700C.

Long chain hydrocarbons are passed over a hot catalyst (silica or alumina at 600-700 degrees) this causes them to break down into smaller molecules.

As some atoms are lost from molecules, they become unsaturated and can therefore form a double bond. This is how you get alkenes from the process as well as shorter chain alkanes.

The animation on this page is helpful:

5.13 understand that fractional distillation of crude oil produces more long-chain hydrocarbons than can be used directly and fewer short-chain hydrocarbons than required and explain why this makes cracking necessary

Long chain hydrocarbons are less flammable and more viscous.
Short chain hydrocarbons burn well and flow well.

Long chain hydrocarbons can be cracked which breaks them up into short chain ones.

5.12 understand that nitrogen oxides and sulfur dioxide are pollutant gases which contribute to acid rain, and describe the problems caused by acid rain

NO and SO2 are given off into the atmosphere by some industrial processes.
When they are in the atmosphere they react with rain water to create H+ ions.
When the rain falls the acid can corrode rocks and buildings. Acid can also alter the PH in soil or rivers which can effect an ecosystem.
Also acid rain corrodes limestone, which damages buildings and stuff.

5.11 understand that, in car engines, the temperature reached is high enough to allow nitrogen and oxygen from air to react, forming nitrogen oxides

In car engines there is a high enough temperature to cause a reaction between oxygen and nitrogen in the air.
This makes NO.

5.10 understand that incomplete combustion of fuels may produce carbon monoxide and explain that carbon monoxide is poisonous because it reduces the capacity of the blood to carry oxygen

Hydrocarbons (from crude oil) + oxygen > carbon dioxide + water

Unless there is not enough oxygen around. Then fuels will combust incompletely:

Hydrocarbons + oxygen > carbon monoxide + carbon + water

Carbon monoxide combines with heamaglobin in red blood cells, meaning they can't carry oxygen around the body.

5.9 describe the trend in boiling point and viscosity of the main fractions

Fractions with low boiling points are less viscous.

Fractions with high boiling points are more viscous.

5.8 recall the names and uses of the main fractions obtained from crude oil: refinery gases, gasoline, kerosene, diesel, fuel oil and bitumen


5.7 describe and explain how the industrial process of fractional distillation separates crude oil into fractions

Crude oil is heated until it boils. As a gas it floats upwards.
As the gas goes higher up and further from the heat source the temperature decreases.
When a compound reaches it condensing point it will condense into a liquid and be collected.
This is known as fractional distillation, and groups with similar condensing temperatures are known as fractions; each fraction is a different substance.

5.6 understand that crude oil is a mixture of hydrocarbons

Crude oil contains different molecules made only of hydrogen and carbon.

5.5 explain the uses of aluminium and iron, in terms of their properties

Aluminium is 'low density' and does not corrode so is used for the bodies of planes.

Iron is used in electromagnets as it is a soft magnetic material.

5.4 describe and explain the main reactions involved in the extraction of iron from iron ore (haematite), using coke, limestone and air in a blast furnace

Iron is displaced from its ore (haematite) by carbon (from coke):
2Fe2O3 + 3C > 4Fe + 3CO2
It is also displaced by carbon monoxide
Fe2O3 + 3CO > 2Fe + 3CO2
Limestone reacts with impurities to from 'slag' which is tapped off.
Air allows burning to take place (of coke.)


5.3 write ionic half-equations for the reactions at the electrodes in aluminium extraction

Al3+ + 3e- > Al
  • 2O2- → O2 + 4e-

5.2 describe and explain the extraction of aluminium from purified aluminium oxide by electrolysis

  • Bauxite is purified into aluminium oxide.
  • This is then dissolved in molten cryolite to bring down the boiling point
  • The walls of the tank are the negative electrode; here aluminium is made
  • The aluminium sinks to the bottom and is tapped off
  • Oxygen is formed at the positive electrode
  • The oxygen formed reacts with the graphite anode to from carbon dioxide; so the anode has to be replaced
It is very expensive to supply the electricity needed for this electrolysis.

5.1 explain how the methods of extraction of the metals in this section are related to their positions in the reactivity series

Anything below carbon can be displaced from its ore by carbon.
Anything above carbon can't so is extracted by electrolysis.

4.9 describe experiments to carry out acid-alkali titrations

  • Have a known volume of acid in a beaker with methyl orange, the solution will be red as it is very acidic.
  • Set up a burette with alkali in it. Open the tap very slightly, so that it drips very slowly into the acid.
  • With each drop stir the contents of the beaker.
  • The more alkali that is added the more neutral and closer to orange it gets.
  • When the solution is neutral the it will be completely orange, at this point close the tap on the burette.
  • The level in the burette will have dropped, showing the volume of alkali used.
  • This shows you how much alkali you need to use to neutralise the acid.

4.8 describe experiments to prepare insoluble salts using precipitation reactions

Silver nitrate and sodium chloride are added together, the product, silver chloride is made, this salt is insoluable and so will form a white precipitate in the solution.

AgNO3 + NaCl > AgCl + NaNO3

4.7 describe experiments to prepare soluble salts from acids

Dilute sulphuric acid is added to an excess of magnesium.
Mg + H2SO4 > MgSO4 + H2
The left over magnesium is filtered off and the mixture boiled down slowly to concentrate.
When it is cooled, crystals will form, these can be blotted dry with a piece of paper.

This can be done with other metals and acids:
Nitric, sulphuric and hydrochloric acids make soluble salts with most metals.
ammonium, potassium and sodium make soluble salts with acids

4.25 predict the effects of changing the pressure and temperature on the equilibrium position in reversible reactions.

If you move the equilibrium, you change the rate of reaction.
If the equilibrium moves to the right, you have more products. (the reactants are reacting faster)
If the equilibrium moves to the left, you have more reactants. (the products are reacting faster)

If you increase the pressure: the equilibrium will move to the side with least molecules.
If you decrease the pressure: the equilibrium will move to the side with the most molecules.

If you increase the temperature, there will be more products that are produced by an endothermic reaction.
This is because the reaction is trying to use up the extra heat, and it does so by putting the energy into making bonds.

4.24 understand the concept of dynamic equilibrium

Dynamic equilibrium is when a reversible reaction is happening both ways at the same time, at the same rate.

4.23 describe reversible reactions such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chloride

If you add water to copper sulphate you can make hydrated copper sulphate.
If you remover the water from hydrated copper sulphate you can make copper sulphate.

When heated, ammonium chloride splits into to hydrogen chloride and ammonia.
Hydrogen chloride and ammonia can be reacted to make ammonium chloride.

4.22 understand that some reactions are reversible and are indicated by the symbol ⇌ in equations

Some reactions can happen both ways: the reactants can make the products and the products can make the reactants.

This symbol shows that ⇌.

4.21 explain that a catalyst speeds up a reaction by providing an alternative pathway with lower activation energy

A catalyst provides an alternative route for the reaction to start, this route requires less energy to start the reaction.

4.18 describe the effects of changes in surface area of a solid, concentration of solutions, pressure of gases, temperature and the use of a catalyst on the rate of a reaction

Higher temperature, more surface area, higher concentration, higher pressure and use of a catalyst all make a reaction faster.

4.19 understand the term activation energy and represent it on a reaction profile

Activation energy is the amount of energy required for a reaction start happening.


4.20 explain the effects of changes in surface area of a solid, concentration of solutions, pressure of gases and temperature on the rate of a reaction in terms of particle collision theory

Collision theory says that to react particles must:
Collide with enough energy to react
Collide in the right orientation to react (the more frequent the collisions, the more likely this is)

Surface area

  • Particles collide more frequently if there is more surface area, as there is more contact between the reactants. Faster rate of reaction.

Concentration/ pressure

  • There is more chance of particles colliding at a higher concentration/pressure, so they react more often. Faster rate of reaction.


  • Particles move about more and will collide more frequently the higher the temperature; react more often. Increases the rate of reaction.


  • Provides an alternative pathway for the reaction to start which requires a lower activation energy.

4.17 describe experiments to investigate the effects of changes in surface area of a solid, concentration of solutions, temperature and the use of a catalyst on the rate of a reaction

Surface area

  • Put a set mass of magnesium in hydrochloric acid
  • Time the reaction
  • Change the from of magnesium keeping the mass the same (powder, wire, strips)
  • The more surface area (the smaller the pieces of magnesium) the faster the reaction


  • Put a set mass of marble chips into dilute hydrochloric acid
  • Time the reaction
  • Change the ratio of water to hydrochloric acid
  • The more concentrated the hydrochloric acid (the lower the ratio of water) the faster the reaction


  • Put a set mass of magnesium powder into a set mass of hydrochloric acid
  • time the reaction
  • Carry out this reaction at different temperatures
  • The higher the temperature the faster the rate of reaction


  • If you have hydrogen peroxide it will not decompose
  • If you put it with manganese dioxide it will decompose into water and oxygen
  • The manganese dioxide will be unaltered by the reaction
  • The more of the catalyst the faster the reaction

4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction.

To work out the enthalpy change you do:
 the sum of bond energies in the reactants - the sum of bond energies in the products

Given average bond energies, you can easily work out the sum of bond energies.
For example, if we are told that
C-H = 413
O=O= 498
C=O= 746

Work out the enthalpy change for this reaction: CH4 + 2O2 > CO2 + 2H2O

((4 x C-H) + (2 x O=O)) - ((2x C=O) + (4 x H-O))

Then we substitute in:

(4 x 413) + (2 x 498) - (2 x 746) + (4 x 464) = 2648KJ - 3348KJ = 700KJ enthalpy change

Credit to author of this source:,d.d2k

4.15 understand that the breaking of bonds is endothermic and that the making of bonds is exothermic

Breaking bonds requires energy, this takes in heat: endothermic.

Making bonds releases energy, this gives out heat: exothermic.

4.14 represent exothermic and endothermic reactions on a simple energy level diagram

Exothermic: lower energy level at end;
Endothermic: more energy at end;
Activation energy is the energy requires to start the reaction.

4.13 understand the use of ΔH to represent enthalpy change for exothermic and endothermic reactions

ΔH is the symbol that represents the amount of energy lost or gained in a reaction.

+ΔH is endothermic (because it gains heat)
-ΔH is exothermic (because it looses heat)

4.12 calculate molar enthalpy change from heat energy change

ΔH =
T (C) x mass of H2O (g) x 4.2J/g C./
number of moles

ΔH is measured in J/mol or kJ/mol.

4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation in which heat energy changes can be calculated from measured temperature changes

Measure the temperature at the beginning of the experiment, measure the temperature at the end. How ever much heat has gone up or down is the calorimetry of the reaction.

For example if you have a beaker of water and take its temperature, then burn a piece of bread under it, the change in temperature is the calories (energy) of the bread.

4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermic

In an exothermic reaction heat is given out. Because bonds are made which gives out energy.
Think expelled; exit; exo.

In a endothermic reaction heat is taken in. Because bonds are Broken which requires energy.

4.6 understand the general rules for predicting the solubility of salts in water

i) all common sodium, potassium and ammonium salts are soluble

ii) all nitrates are soluble

iii) common chlorides are soluble, except silver chloride

iv) common sulfates are soluble, except those of barium and calcium

v) common carbonates are insoluble, except those of sodium, potassium and ammonium

4.5 predict the products of reactions between dilute hydrochloric, nitric and sulfuric acids; and metals, metal oxides and metal carbonates


  • Hydrochloric acid + metal > metal chloride salt + hydrogen
  • Nitric acid + metal > You don't need to know
  • Sulphuric acid + metal > metal sulphate + hydrogen

Metal oxides

  • Hydrochloric acid + metal oxide > metal chloride salt + water
  • Nitric acid + metal oxide > metal nitrate salt + water
  • Sulphuric acid + metal oxide > metal sulphate + water

Metal carbonates

  • Hydrochloric acid + metal carbonate > metal chloride salt + water + carbon dioxide
  • Nitric acid + metal carbonate > metal nitrate salt + water + carbon dioxide
  • Sulphuric acid + metal carbonate > metal sulphate + water + carbon dioxide

4.4 define acids as sources of hydrogen ions, H+, and alkalis as sources of hydroxide ions, OH¯

Essentially if something is acidic it contains positive hydrogen ions, if something is alkaline it contains negative hydroxide ions.

4.3 describe the use of universal indicator to measure the approximate pH value of a solution

Neutral is green

The more red, the more acidic.

The more purple, the more alkaline.

4.2 understand how the pH scale, from 0–14, can be used to classify solutions as strongly acidic, weakly acidic, neutral, weakly alkaline or strongly alkaline

4.1 describe the use of the indicators litmus, phenolphthalein and methyl orange to distinguish between acidic and alkaline solutions

Red litmus paper turns blue in the presence of alkali.

Blue litmus paper turns red in the presence of acid.

Phenolphthalein goes pink in alkalis.

Methyl orange is yellow for alkali, red for acid.


3.12 describe the dehydration of ethanol to ethene, using aluminium oxide.

C2H5OH > C2H4 + H2O

ethanol > ethene + water

aluminium oxide is the catalyst for this reaction.

3.11 evaluate the factors relevant to the choice of method used in the manufacture of ethanol, for example the relative availability of sugar cane and crude oil


  • Cane sugar widely avalible/ cheap/ renewable
  • Slow process
  • Impurities in the product
  • Done in batches

Hydrating (ethene and steam)

  • Crude oil (cracked to make ethene) expensive/ non-renewable
  • Fast process
  • Pure product
  • Continuous reaction

3.10 describe the manufacture of ethanol by the fermentation of sugars, for example glucose, at a temperature of about 30°C

Ethanol can be made by the anaerobic respiration of microorganisms.

glucose > ethanol + carbon dioxide

This happens at 30 degrees.

3.9 describe the manufacture of ethanol by passing ethene and steam over a phosphoric acid catalyst at a temperature of about 300°C and a pressure of about 60–70 atm

C2H4 (ethene) + H2O (steam) > C2H5OH (ethanol)

This reaction takes place at a high pressure (60-70 atm) and a high temperature (300) to make the reaction happen very quickly. Phosphoric acid also speeds up the reaction as it is a catalyst.

2.39 describe tests for the gases

i hydrogen
  • burns with a 'squeaky pop' sound

ii oxygen
  • will relight a glowing splint

iii carbon dioxide
  • Turns lime water cloudy

iv ammonia
  • Damp red litmus paper blue
  • Damp universal indicator purple

v chlorine
  • bleaches damp litmus paper white

2.38 describe tests for the anions

i) Cl-, Br- and I-, using dilute nitric acid and silver nitrate solution
  • Chloride ions + nitric acid + silver nitrate > white precipitate (silver chloride)
  • Bromide ions + nitric acid + silver nitrate > cream precipitate (silver bromide)
  • Iodide ions + nitric acid + silver nitrate > yellow precipitate (silver iodide)
ii) SO2- (sulphate ions) using dilute hydrochloric acid and barium chloride solution
  • SO4(2-) + HCl + Ba(2+) > white precipitate (barium sulphate)
iii) CO2-, using dilute hydrochloric acid and identifying the carbon dioxide evolved
  • Carbonate + acid > salt + water + carbon dioxide
  • Carbon dioxide produced will turn lime water cloudy

2.37 describe tests for the cations

i) Li+, Na+, K+, Ca2+ using flame tests
  • Lithum: red
  • Sodium: orange (so strong can mask other colours)
  • Potassium: lilac
  • Calcium: brick red

ii) NH4+, using sodium hydroxide solution and identifying the ammonia evolved
  • NH4 + OH > NH3 + H2O
  • ammonium ions + hydroxide ions > ammonia + water
  • ammonia (pungent smelling gas) turns red litmus paper blue

iii) Cu2+, Fe2+ and Fe3+, using sodium hydroxide solution
  • Copper(ii) sulphate + sodium hydroxide > blue precipitate
  • Iron(ii) sulphate + sodium hydroxide > green precipitate
  • Iron(iii) sulphate + sodium hydroxide > brown precipitate

Wednesday, 15 May 2013

3.8 describe the addition reaction of alkenes with bromine, including the decolourising of bromine water as a test for alkenes

An alkene will make its double bond into a single bond, to bond to two bromines. Bromine is added to the molecule. The product made is colourless. When alkenes are put in bromine water it turns from brown to colourless (a good way of testing for alkenes.)
For example:

C2H4(g) + Br2 (aq) → C2H4Br2 (aq)

3.7 draw displayed formulae for alkenes with up to four carbon atoms in a molecule, and name the straight-chain isomers

In every alkene there is one double bond between two carbons. Bearing in mind that carbons can only make four bonds, then the double bonded carbons will either be joined to: another carbon and a hydrogen; or two hydrogens.

3.6 recall that alkenes have the general formula CnH2n

All compounds in the homologous group alkenes have the general formula CnH2n.

3.5 describe the substitution reaction of methane with bromine to form bromomethane in the presence of UV light.

In UV light bromine and methane will form bromomethane:

CH4  +  Br2CH3Br  +  HBr

What has happened in this reaction is a bromine has taken the place of a hydrogen (substitution.)

3.4 recall the products of the complete and incomplete combustion of alkanes

Complete combustion gives carbon dioxide and water.

Incomplete combustion gives carbon monoxide and water.

3.3 draw displayed formulae for alkanes with up to five carbon atoms in a molecule, and name the straight-chain isomers

The formula for alkanes is CnH2n+2. This means that every carbon is bonded to two hydrogens, and there is also one hydrogen on each end.
The isomers you are likely to encounter will just be this formula in a straight line (you only need up to pentane):


2.36 understand the sacrificial protection of iron in terms of the reactivity series.

Sacrificial is covering a metal with a more reactive metal. What this means is water and/or air will react with the more reactive metal instead of the one underneath.

2.35 describe how the rusting of iron may be prevented by grease, oil, paint, plastic and galvanising

Grease, oil, paint and plastic prevent air and/or water from coming into contact with iron. This means the reaction that rusts iron can't occur.

Galvanising is coating in zinc. This Zinc react in the air to form ZnCO3 which prevents air and/or water from coming into contact with the iron.

3.2 recall that alkanes have the general formula CnH2n +2

Alkanes is a homologous series with the formula CnH2n +2.

What this means is that for every one carbon there are two times the amount of hydrogens plus two more hydrogens.

3.1 explain the terms homologous series, hydrocarbon, saturated, unsaturated, general formula and isomerism.

Compounds in the same homologous series have the same general formula and similar chemical properties.

A hydrocarbon is a compound made up only of hydrogen and carbon.

Saturated means something has bonded as many time as possible. Unsaturated means that more bonds can be made.

General formula is the most simplified the ratio of molecules can be.

Isomers have the same general formula but different structures.

2.34 describe the conditions under which iron rusts

Water and oxygen are needed to rust iron: iron that reacts with these becomes hydrated iron(iii) oxide.

2.33 understand the terms redox, oxidising agent, reducing agent

In a redox reaction, a more reactive metal gains an oxygen from a less reactive metal which looses it.
i.e. a more reactive metal is oxidised and a less reactive metal is reduced.

The reducing agent is the more reactive metal which reduces the other metal.
The oxidising agent is the less reactive metal which allows the other metal to be oxidised.

2.32 understand oxidation and reduction as the addition and removal of oxygen respectively

oxidation is the gain of oxygen,
reduction is the loss of oxygen.

2.31 deduce the position of a metal within the reactivity series using displacement reactions between metals and their oxides, and between metals and their salts in aqueous solutions

A metal oxide or a metal salt dissolved in water:

  • introduce a more reactive metal and it will displace the current one
  • introduce a less reactive metal and no displacement will take place
From this you can deduce which metals are more and less reactive.

2.30 describe how reactions with water and dilute acids can be used to deduce the following order of reactivity: potassium, sodium, lithium, calcium, magnesium, zinc, iron and copper

potassium, sodium, lithium and calcium all react with water and acids
magnesium, zinc and iron all react with acids (and very slowly with water.)
copper doesn't react with either.

The more vigorous the reaction the more reactive the metal. The more things a metal will react with, the more reactive the metal.

2.29 understand that metals can be arranged in a reactivity series based on the reactions of the metals and their compounds

Inline images 1

Tuesday, 7 May 2013

2.28 describe a physical test to show whether water is pure.

If water is pure it will boil at exactly 100° and freeze at exactly 0°

2.27 describe the use of anhydrous copper(II) sulfate in the chemical test for water

anhydrous copper sulphate will become hydrous copper sulphate when it is reacted with water.
So if anhydrous copper sulphate goes from white to blue in the presence of a liquid it will be water.

2.25 describe the reactions of dilute hydrochloric and dilute sulfuric acids with magnesium, aluminium, zinc and iron

acid + metal > salt + hydrogen

For example:
magnesium + hydrochloric acid > magnesium chloride + Hydrogen
Mg + 2HCl > MgCl2 + H2

2.26 describe the combustion of hydrogen

The combustion of hydrogen is its reaction with oxygen.
Water is created. and a lot of energy.
2H2 + O2 > 2H2O

2.24 understand that carbon dioxide is a greenhouse gas and may contribute to climate change

Carbon dioxide prevents heat leaving the earth's atmosphere in rays that the earth emits.
Significant amounts of green house gasses will warm up the earth, changing the climate.

2.23 explain the use of carbon dioxide in carbonating drinks and in fire extinguishers, in terms of its solubility and density

Carbon dioxide is dissolved into drinks at a high pressure, this makes CO2 bubbles in fizzy drinks.

Some fire extinguishers have CO2 in, because it is denser than air it will fall over the fire creating a barrier between the air and fire: the fire can't burn with out the oxygen in the air.

2.22 describe the properties of carbon dioxide, limited to its solubility and density

It is denser than air.
It is soluble in water at a high pressure.

2.21 describe the formation of carbon dioxide from the thermal decomposition of metal carbonates such as copper(II) carbonate

When metal carbonates are heated they become carbon dioxide and a metal.
For example:
copper carbonate > copper oxide + carbon dioxide
CuCO3 > CuO + CO2

2.20 describe the laboratory preparation of carbon dioxide from calcium carbonate and dilute hydrochloric acid

calcium carbonate + hydrochloric acid → calcium chloride + water + carbon dioxide

CaCO3 + 2HCl → CaCl2 + H2O + CO2

2.19 describe the reactions of magnesium, carbon and sulfur with oxygen in air, and the acid-base character of the oxides produced

The two non-metals burn in air- giving out heat and light- to bond with oxygen.
They become non-metal oxides, which are, by nature, acids.

Magnesium will burn in air to from a metal-oxide, these are always basic.

2.18 describe the laboratory preparation of oxygen from hydrogen peroxide, using manganese(IV) oxide as a catalyst

hydrogen peroxide is put in a conical flask with manganese(IV) oxide as a catalyst. Plus there could be some water to diloute the hydrogen peroxide.
Hydrogen peroxide > water + oxygen
2H2O2 > 2H2O + O2
You can then collect oxygen by downwards displacement method.

This video is helpful:

2.17 explain how experiments involving the reactions of elements such as copper, iron and phosphorus with air can be used to investigate the percentage by volume of oxygen in air

Copper, iron and phosphorus all react with air.
If you know the volume of air that you have, then react it with on of these, then re measure the volume of air; what has been lost is all oxygen that reacted.
This page describes it well:

2.16 recall the gases present in air and their approximate percentage by volume


2.15 understand these displacement reactions as redox reactions.

When a more reactive halogen displaces a less reactive one this is a redox reaction.
This means that one element has been reduced (gained electrons) and one has been oxidised (lost electrons.)
Helpful hint!
Oilrig helps to remind you how redox reactions go:


2.14 describe experiments to demonstrate that a more reactive halogen will displace a less reactive halogen from a solution of one of its salts

A more reactive halogen will displace a less reactive one that is bonded as a salt. This will only happen if the salt is dissolved in water or a gas.

This page has a really good example:

2.12 explain, in terms of dissociation, why hydrogen chloride is acidic in water but not in methylbenzene

Water (H2O) is a polar molecule.
Methylbenzene is a non-polar molecule.
Hydrogen chloride is a polar molecule.

Polars only dissolve in polars.
When hydrogen chloride is dissolved you get +H ions.
These are acidic.

Wednesday, 1 May 2013

2.13 describe the relative reactivities of the elements in Group 7

Group 7 elements become less reactive as you go down the group.

At the top, the positive charge of the proton in the nucleus is close to the surface (as there are few shells) this makes it easy for them to pull in the one electron they need to become stable, meaning they are very reactive.

Lower down where there are more shells the pull of the proton is further from the surface making it less easy to pull in another electron.

2.11 understand the difference between hydrogen chloride gas and hydrochloric acid

Hydrogen chloride gas is HCl

Hydrochloric acid is hydrogen chloride dissolved in water. The two ions become detached leaving Cl- and H+ ions. H+ is acidic, hence the term acid.

2.10 make predictions about the properties of other halogens in this group

We would expect the colour to keep getting darker and the melting and boiling points to keep getting higher further down the group.

2.9 recall the colours and physical states of the elements at room temperature

Fluorine.............. Gas........... yellow
Chlorine............. Gas........... yellow-green
Bromine............. Liquid........ red-brown
Iodine................ Solid.......... purple
Astatine............. Solid.......... black

2.7 describe the relative reactivities of the elements in Group 1

Group one elements get more reactive the further down the group.

2.8 explain the relative reactivities of the elements in Group 1 in terms of distance between the outer electrons and the nucleus.

Group one elements are more reactive further down the group.

Group one elements need to loose an electron- the one on the outer shell- to react. Electrons are held to an atom by the protons in the nucleus. If an electron is close to the nucleus the force holding it in will be very strong, if it is further away it will be weaker.

So bigger atoms (towards the bottom of the group) with the outer orbital far from the nucleus will loose their electron more easily: this means they react more easily/quickly/more/vigorously.

Smaller atoms with the electron closer to the pull of the nucleus (at the top of group one) will be less reactive as it takes more to lose the electron.

2.6 describe the reactions of these elements with water and understand that the reactions provide a basis for their recognition as a family of elements

The group one elements- lithium, sodium, potassium- are easily identifiable as the same group due to the fact that they all react vigorously with water (clearly due to the fact they have similar electronic configurations.)

The reactions that occur are huge- they get bigger further down the group. Hydrogen gas is produced as well as metal hydroxide.

Search on youtube; there are plenty of videos of there reactions.

2.5 understand that the noble gases (Group 0) are a family of inert gases and explain their lack of reactivity in terms of their electronic configurations.

Nobel gasses are inert, this means they do not react. The reason for this is because they are stable: meaning they have a full outer shell, so they do not need to loose or gain electrons.

2.4 understand why elements in the same group of the Periodic Table have similar chemical properties

Elements in the same group have the same number of electrons on their outer shell.
This means they will behave in a similar way; they will react and bond similarly.

For example in group one, the elements all form +1 ions as they each loose one electron to become stable (to have a full outer shell).

2.3 explain the classification of elements as metals or non-metals on the basis of their electrical conductivity and the acid-base character of their oxides

Metals are all conductors. Metals form metal-oxides which are alkaline.

Nonmetals don't conduct. They form nonmetal-oxides which are acidic.

2.2 recall the positions of metals and non-metals in the Periodic Table


To the left are the metals, to the right are the nonmetals

2.1 understand the terms group and period

Groups are columns in the periodic table, the number of a group represents the number of electrons on an atom's outer shell.
Periods are rows on the periodic table, the row represents the number of orbitals that a atom has.

1.56 recall that one faraday represents one mole of electrons

One Faraday is 96500 coulombs. That is the amount of coulombs in one mole of electrons.

Tuesday, 30 April 2013

1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions

One faraday is 96500 coulombs. It is also one mole of electrons.

If current of 0.2 Apms is passed through copper(ll) sulphate for tow hours, how much copper do you get?

  • Write out the half equation
Cu2+ + 2e > Cu
  • Work out coulombs of electrons flowing
Coulombs= current x time
time is 2x60x60 (times 60 makes minutes, times 60 again makes it seconds)
Q= 0.2 x 7200= 1440 coulombs
  • Convert C into moles of electrons
Moles= C/Faraday
Mol= 1440/96500
Mol= 0.015
  • Work out scale factor
Cu2+ + 2e > Cu
For every 2 moles of electrons, there will be one Cu
Sf= Moles of product/ moles of electrons
Sf= 1/2
Sf= a half
  • Work out moles of product using Scale factor
so we do the moles of electrons times the scale factor
0.015x1/2= 0.0075 Moles of Cu
  • Convert moles into mass
Moles x Mr
0.0075 x 63.5= 0.48g of copper

Sunday, 28 April 2013

1.55 write ionic half-equations representing the reactions at the electrodes during electrolysis

At the positive electrode, electrons will be lost: to show this we write the lost electrons as products:
2Br- > Br2 + 2e-
Make sure the charges are equal on both sides: 1- > 1-.

At the negative electrode, electrons will be gained so we write them as reactants:
2H+ + 2e- > H2
And to make sure the charges are the same on both sides: 0 > 0.

1.54 describe experiments to investigate electrolysis, using inert electrodes, of aqueous solutions such as sodium chloride, copper(II) sulfate and dilute sulfuric acid and predict the products

Place inert electrodes (ones that wont react) into an aqueous solution.

At the positive electrode the negatively charged ion from the compound will form an atom. At the negative electrode the atom of the positive ion will form.

sodium chloride: Hydrogen at the negative; chlorine at the positive
copper(II) sulfate: copper at the negative; oxygen at the positive
dilute sulfuric acid: Hydrogen at the negative; oxygen at the positive

If the metal in the solution is more reactive then hydrogen, the hydrogen from the water will be a product, as the metal will bond with the oxygen.

Test the products using known methods: eg damp blue litmus paper turned red by chlorine.

1.53 describe experiments to investigate electrolysis, using inert electrodes, of molten salts such as lead(II) bromide and predict the products

Inert electrodes are ones that don't react with any other substances, but only play a role in the transfer of electrons.

To describe an experiment you need to be able to draw and label an electrolysis experiment. As an example:

Lead bromide will make lead and bromine, you can use chemical tests to see is you got these products from electrolysis.

1.52 understand that electrolysis involves the formation of new substances when ionic compounds conduct electricity

In electrolysis ionic compounds conduct electricity. Positively charged ions move to one end, negatively to the other, these are then turned into atoms (by losing their charge) and so new substances are formed.

1.51 describe experiments to distinguish between electrolytes and nonelectrolytes

Set up an electric circuit with an LED and a break in the wire, put both ends of wire into a solution/molten substance. If the LED lights up then there is a current flowing, this will only be able to happen if the solution is conducting: so it must be an electrolyte. Conversely if the LED does not light up then there is no current flowing, and so the solution has not conducted electricity meaning it must be a nonelectrolyte.

1.50 understand why ionic compounds conduct electricity only when molten or in solution

When ionic compounds are molten or in solution, the positive and negative ions separate  this means that there are ions free to flow, and so they can conduct electricity.

1.49 understand why covalent compounds do not conduct electricity

In covalent compounds there are no electrons free to move, this means there can be no transfer of electricity through a covalent compound

1.48 understand that an electric current is a flow of electrons or ions

An electric current is a flow of electrons, although it can also be a flow of ions (as they have a charge.)

1.47 explain the electrical conductivity and malleability of a metal in terms of its structure and bonding.

Metals have delocalised electrons, electrons carry electricity; so because there are free electrons charge can pass easily through a metal.

The structure of a metal is with rows of atoms on top of one another, in pure metals as all the atoms will be the same size, the layers can slide easily over one another making them easy to bend.

1.46 understand that a metal can be described as a giant structure of positive ions surrounded by a sea of delocalised electrons

In a metal atoms come together into a lattice, the electrons become detached from their atoms- delocalised- making the atoms positive ions.

1.36 describe an ionic crystal as a giant three-dimensional lattice structure held together by the attraction between oppositely charged ions

An ionic crystal is a lattice of electrons in a 3D structure, the ions are alternate positive and negative and their opposing charges hold the structure together.

1.37 draw a diagram to represent the positions of the ions in a crystal of sodium chloride.

In a lattice structure with negative touching positive and visa versa.

Wednesday, 24 April 2013

1.45 explain how the uses of diamond and graphite depend on their structures, limited to graphite as a lubricant and diamond in cutting.

In Graphite the atoms from layers, these layers can slide over each other, this makes it very slippery and so can be used as a lubricant.

Diamond is extremely hard because it has a many bonds in it, this means it is great for cutting as it can cut anything.

1.44 draw diagrams representing the positions of the atoms in diamond and graphite
Carbon atoms in Graphite (left), are each bonded to three other atoms.

Diamond (right) is formed of carbon atoms each joined to four others.

1.43 explain the high melting and boiling points of substances with giant covalent structures in terms of the breaking of many strong covalent bonds

A giant covalent structure is one with many atoms bonded together. To melt or boil them you are not separating intermolecular bonds (between molecules), you are separating intramolecular bonds that keep the molecule together. These bonds are strong covalent bonds which take a lot of energy to break, so a lot of heat energy is required before the bonds will break to boil or melt; meaning they have high melting and boiling points.

1.42 explain why substances with simple molecular structures have low melting and boiling points in terms of the relatively weak forces between the molecules

A substance with a simple molecular structure is one that contains only a few atoms in a molecule.
The intermolecular forces (between the molecules) are weak, so it doesn't take much energy- or heat- to break them- this means they will melt and boil under low heats, as even small amounts of heat energy are enough to break the bonds.

1.41 understand that substances with simple molecular structures are gases or liquids, or solids with low melting points

A simple molecule (one with only a few atoms) will have a low melting point.

1.40 explain, using dot and cross diagrams, the formation of covalent compounds by electron sharing

To draw a dot and cross diagram for a covalent bond, you need to draw the outer shells of the two atoms involved with an overlap, in this overlap should be the electrons they share. Half the electrons in the overlap should be dots and half crosses, because one electron in every pair comes from each atom. The rest of the electrons for the outer shell should be drawn on, one atom with dots, the other with crosses.

For example, H + Cl = HCl:
Inline images 1

You need to be able to do dot and cross for the following substances (comment if you need help with any):
i hydrogen
ii chlorine
iii hydrogen chloride
iv water
v methane
vi ammonia
vii oxygen
viii nitrogen
ix carbon dioxide
x ethane
xi ethene

1.39 understand covalent bonding as a strong attraction between the bonding pair of electrons and the nuclei of the atoms involved in the bond

Electrons, being shared by atoms in a covalent bond, are attracted to the nucleus of each atom in the bond. Remember that electrons are negative and protons- in the nucleus- are positive.

1.38 describe the formation of a covalent bond by the sharing of a pair of electrons between two atoms

A covalent bond is a bond formed between atoms by sharing a pair of electrons (one from each atom.)

1.20 understand the term molar volume of a gas and use its values (24 dm3 and 24,000 cm3) at room temperature and pressure (rtp) in calculations.

At standard temperature and pressure, one mole of any gas will occupy 24000 cm3; also known as 24dm3.

A really good way to do calculations with this information is by using this triangle:

In a recent past paper I did this was one of the questions: calculate the amount, in moles, of carbon dioxide gas collected if you collect 144cm3.

The correct way to find this out is by doing 144 / 24000 which gives you 0.006 Mol

Sunday, 21 April 2013

1.35 understand the relationship between ionic charge and the melting point and boiling point of an ionic compound

The bigger the difference in charge, the stronger the attraction: if you have + and - ions they will have a weaker bond than +3 and -3 ions. The stronger the attraction, the harder it is to break the bonds, this means that the melting and boiling points will be higher. So the bigger the charge of an ion, the higher the melting and boiling point.

1.34 understand that ionic compounds have high melting and boiling points because of strong electrostatic forces between oppositely charged ions

To melt or boil anything, heat is used to break bonds. The stronger the bonds, the more heat needed. Ionic compounds have strong bonds, so they don't melt or boil unless there is a considerable amount of heat, this means the have high melting and boiling points.

1.33 understand ionic bonding as a strong electrostatic attraction between oppositely charged ions

Ionic bonding happens between two ions: they are attracted to each other due to their opposite charges, so we say the ions have electrostatic attraction. This attraction bonds them together into an ionic compound.

1.32 explain, using dot and cross diagrams, the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7

Dot and cross diagrams represent electron transfer. One atom will have dots as electrons, the other crosses.

Here we begin with sodium and chlorine. Sodium looses one electron, so is drawn on the right with one less. Chlorine gains the electron that sodium looses so is drawn with one extra electron: because the electron came from sodium it is a cross instead of a dot.

The ions are drawn in brackets with their charge written outside.

Sorry the picture is hard to see.

1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formed

This is a way of working it out:
  • How many electrons are on the outer shell?
  • How many shells does it have?
  • To fill up its outer shell how many electrons will it take?
  • Now see weather it will take more transferring to loose electrons (to go down an shell,) or gain electrons (to fill the shell)
  • Which ever one takes the least transferring will be the route that was taken
  • If it lost, it will have a positive charge of the number of electrons lost to empty the shell
  • If it gained, it will have a negative charge of the number of electrons it gained to fill the shell

Most of the time the atoms will be on the second orbital in which case its simpler to think of it like this:

An atom with less then four electrons on its outer shell will want to loose electrons because that is the quickest way for it to have a full outer shell: if all the current electrons on the outer shell go, then the next shell- which will be full- will become the outer one. So we know they will loose electrons to make positive ions. If you have a group 2 atom (2 electrons on outer shell) it will loose two electrons when it becomes an ion, so you know it will have a +2 charge.

Similarly atoms with more than four electrons will gain electrons to fill their outer shell. This means they will make negative ions: as an example a group 7 electron has to gain one electron to fill its outer shell and so will become -1.

1.29 understand oxidation as the loss of electrons and reduction as the gain of electrons

The name given to loosing electrons is oxidation; the name given to gaining electrons is reduction.

One way to remember this is oil rig:

1.28 describe the formation of ions by the gain or loss of electrons

Electrons are transferred from one atom to another (this is in an effort to either fill or empty the outer shell to become stable.) An atom has no charge because the electrons and protons have equal and opposite charges. But an ion will have a charge: an electron has a charge of -1, so loosing an electron looses one negative charge, making the ion +1. So gaining one electron will make an atom a -1, gaining two will make it -2.

The atoms the gain or loose electrons to each other will have opposite charges: for example if a gives away 1 atom and b gains it, a is +1 and b is -1. These charges mean that the ions are attracted to each other (ionic bond), so they form an ionic compound.

1.27 carry out mole calculations using volumes and molar concentrations.

Moles / volume = concentration

1.26 calculate percentage yield

Percentage yield is: (actual yield / theoretical yield) x 100

Theoretical yield is what you expect to get, if a reaction doesn't finish you may end up with a lower yield than the expected theoretical yeild.

To work out theoretical yield it is important to understand that an equation gives you a ratio of moles, e.g.
Fe2O3 > 2Fe
tells us that every one mole of iron oxide makes two moles of iron. In this equation the weight of Fe will be your yield.

If we are told the weight of the Fe2O3 is 100g we can easily work out the theoretical yield:
Work out the moles of Fe2O3 by doing the weight (g) divided by the atomic mass: 100/160= 0.625Mol
We know that for every one mole of Fe2O3 there are two of Fe so we do: 0.625 x 2= 1.25 Mol
Now we have moles of Fe we can work out weight by Mol x Ar: 1.25 x 56= 70g

If you in fact got 62g of Fe you'd do (62/80)x100= 77.5%

Inline images 2

Saturday, 30 March 2013

1.25 calculate reacting masses using experimental data and chemical equations

This means knowing that the mass of A and the mass of B will equal the mass of AB.
Bearing in mind that the mass of one mole of an element is its atomic mass.
If we have 2 moles of A(24) and react it with 5 moles of B(25)
then we have (2x24)+(5x25) and can calculate that A2B5 will weigh 173g

1.24 calculate empirical and molecular formulae from experimental data

Find the masses of the elements in the compound (by weight the compound then weighing it with elements taken out). Make these masses into percentages, divide each percentage by the ar of that element and you will have the number of atoms of it in a molecule.

Carbon= % mass 80, ar 12
C= 80/12= 6.7
Hydrogen= % mass 20, ar 1
H= 20/1= 20

To find the empirical formula divide the molecular formula by the smallest number.

C6.7H20 /6.7

1.23 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation

Weigh you compound, remove one element of the compound through a reaction (break up a metal oxide or salts), then weigh again.
The first weight is AB the second weight is A so doing AB-A= B.
Now that you have the weight of A and B you can work out the formulae by doing the weight divided by the Ar.
A= 22g with an ar of 11
A= 22/11= 2
B=72g with an ar of 18
B=72/18= 4

A2B4 and to simplify to empirical formula AB2

1.22 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively

(S) solid
(L) liquid
(G) gas
(Aq) aqueous/ solid dissolved in liquid

In balanced equations these symbols go after a element or compound to show what state it is in.

1.21 write word equations and balanced chemical equations to represent the reactions studied in this specification

Word equations have just the names of the reactants and products involved:
Hydrogen + Oxygen > Water
Balanced equations are the symbols of the products and reactants including the numbers of each, there must be an equal number of each element on both sides of the equation, if there are not you can alter this by putting the right number infront of a symbol:
2H + O > H2O

1.19 carry out mole calculations using relative atomic mass (Ar) and relative formula mass (Mr)

The way I like to do this is to have a triangle with Mass (g) on top, with Moles (mol) bottom left and Mr/Ar bottom right.

1.18 understand the term mole as the Avogadro number of particles (atoms, molecules, formulae, ions or electrons) in a substance

A mole is Avogadro's number simpy because if you have 1 mole of an element its weight in grams will be its atomic mass.

1.17 understand the use of the term mole to represent the amount of substance

A mole is an amount, in the same way that you can have a dozen buns, you can have a mole of carbon.
Having a mole of something is having 6.022x1023 of it.

n.b the 1023 thing is meant to be 10 to the power of 23.

1.16 calculate relative formula masses (Mr) from relative atomic masses (Ar)

Mr is relative formula mass, it is the mass of a molecule. Its calculation is (number of that element in the molecule x relative atomic mass) + (number of that element in the molecule x relative atomic mass)

Ar is relative atomic mass, it is the average mass of an atom of a specific element. It is calculated by doing (% of isotope x mass of isotope) + (% of isotope x mass of isotope) over 100.

Wednesday, 13 February 2013

1.15 deduce the number of outer electrons in a main group element from its position in the Periodic Table.

The groups (that's going down the periodic table) tell you how many electrons are on the outer shell!

Also note that the periods (going across) tell you the number of orbitals the atom has.

1.14 deduce the electronic configurations of the first 20 elements from their positions in the Periodic Table

Groups are colmns in the periodic table; the group number represents the number of electrons on the outer shell.

Periods are rows on the periodic table; the row corresponds with number of shell.

So if you have an element on the third row down in group seven: it will have three shell and 7 electrons on its outer shell.

1.13 understand that the Periodic Table is an arrangement of elements in order of atomic number

As you can see above the periodic table is in order of atomic number.
Every atom in an element has the same atomic number.