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

When condensation polymerisation occurs, a small molecule (such as water) is produces, as well as the monomer.

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

Most often, condensation polymerisation involves two different types of monomer. These monomers react together forming bonds between them, making polymer chains. However, for each new bond that forms, a molecule (e.g. water) is lost.

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

Polyethene is light and stretchy making it ideal for packaging such as plastic bags, water bottles/food containers and carpets.

Polypropene is a tough polymer but is quite flexible and heat resistant too, this makes it ideal for making things like crates

Polychloroethene is used to make clothes, drainpipes and insulating cables

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

To find the repeating unit, just find the section of the polymer that is repeated and draw it out, writing the number of monomers there are next to it (if you don't know how many monomers there are, you must put an 'n')...

Polyethene


Polypropene

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Polychloroethene

image credit: wiki // quizlet

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

Each type of bond (for example O-O or C-H) has a particular bond energy... we will be given these in the exam so don't worry about learning them. These energies can be used to calculate enthalpy change, for example...

Using bond energies, calculate the enthalpy change for the following reaction
H2 + Cl2 ---> 2HCl

Bond energies...
H-H: +436 kJ/mol
Cl-Cl: +242 kJ/mol
H-Cl: +431kJ/mol

1. Work out what bonds are broken & the energy made... 

1 mole of H-H is broken and 1 mole of Cl-Cl is broken. Therefore, +436 + +242 = +678 kJ/mol is required to break the bonds in this reaction

2. Work out what bonds are being made & the energy released...

Forming 2 moles of H-Cl bonds. This released 2 x +431 = 862kJ/mol


3. Use the formula 'ΔH = total energy absorbed to break bonds - total energy released in making bonds' to find out the enthalpy change

ΔH  = 678 - 862 = -184 kJ/mol


4. ΔH  is negative which means the reaction must be exothermic

4.12 Calculate molar enthalpy change from heat energy change

Okay so you have calculated the amount of energy produced , this can be used to work out the molar enthalpy change (this is basically the enthalpy change given out by one mole of the reactant).

NOTE:

To calculate the molar enthalpy change, you need to know the equations moles = mass / Mr (if unsure of this equation, click here) and molar enthalpy change = energy produced / moles. For example...


0.9g of methylated spirit produces 6510J of heat energy, work out the heat produced per mole. (The Mr of methylated spirit is 44.6)

- First, work out the amount of energy transferred...

We know that 6510J of heat energy was produced, this means 6510J of energy was transferred... this needs to be converted into kJ (as the unit or enthalpy change is kJ/mol)... 6.510kJ

- Next, find out how many moles of fuel produced this heat...

moles = mass / Mr

= 0.9 / 44.6

= 0.02 moles

- Now, divide the amount of heat energy produced by the amount of moles...


6.510 / 0.02 = 325.5 kJ/mol

The end (:

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

If you have ethanol and you want ethene, you can dehydrate it. This is done by removing water from the ethanol and is known as a dehydration reaction. 

Method
Ethanol vapour is passed over a hot catalyst (aluminium oxide, Al2O3)


NOTE: The catalyst is used as it provides a large surface area for the reaction to take place, meaning the reaction will be quick(ish)

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

Producing ethanol by reacting ethene and steam is relatively cheap (as, right now, ethane is quite cheap and not much is wasted). However, ethene is produced from crude oil which is a finite source, soon it will become fairly expensive as it gets rarer and in the end it will run out, meaning ethanol will no longer be able to be made using ethene and steam.

An advantage of making ethanol by fermenting sugars is that all 'reactants' are renewable (sugar and yeast). It also can be produced at a much lower temperature. However, the ethanol you get from fermenting sugars is not as concentrated as when you react steam with ethene (basically, its super weak). This needs to be distilled to increase its strength and it also needs to be purified. So, although its simpler, its also a lot more hassle.

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

Another method of producing ethanol is by fermentation. The raw material for fermentation is sugar (e.g. glucose), which is converted into ethanol using yeast. This process is done at 30ºC.

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

Ethene will react with steam to produce ethanol. This reaction will take place at 300ºC with a pressure of 60-70 atm. This process is very slow, so a catalyst of phosphoric acid is used.

NOTE: atm stands for atmospheres

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

All elements in Group 1 have 1 electron in their outer shell. 

The elements further down the group are more reactive. This is because as you go down the group the outer shell (with the single electron) gets further away from the nucleus (as there are complete shells inbetween them). This means that, because the electron is further away, it is more easily lost (given away).

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

In aqueous solutions, as well as ions from the ionic compound, there will be hydrogen ions (H+) and hydroxide ions (OH-) from the water.

Products

At the cathode, if H+ ions and metal ions are present, hydrogen gas will be produced if the metal ions are more reactive than H+ ions (for example, sodium ions). If the metal ions are less reactive than the H+ ions (for example, copper ions), a solid layer of the pure metal will be produced.

At the anode, if OH- and halide ions (Cl-, Br-, I-) are present, then molecules of chlorine, bromine or iodine will be formed. If no halide ions are present, then oxygen gas and water will be formed.

Electrolysis of sulphuric acid
A solution of sulphuric acid (H2SO4) contains three different ions: SO₄²⁻ , H+ and OH-.

At the cathode: as sulphur (SO₄²⁻) is more reactive than hydrogen, hydrogen gas is produced...

2H+ + 2e- ---> H2


At the anode: as there no halide ions present, oxygen and water is produced.

4OH- ---> O2 + 2H2O + 4e-


Electrolysis of sodium chloride
A solution of sodium chloride (NaCl) contains four different ions: Na+, Cl-, OH- and H+

At the cathode: as sodium is more reactive than hydrogen, hydrogen gas is produced...

2H+ + 2e- ---> H2

At the anode: as chlorine ions are present (halide), then chlorine atoms will be produced (as chlorine gas)...

2Cl- ---> Cl2 + 2e-


Electrolysis of copper(II) sulfate
A solution of copper(II) sulphate (CuSO4) contains four different ions: Cu2+, SO₄²⁻, H+ and OH-.

At the cathode: as copper is less reactive than hydrogen, copper metal is produced...

Cu2+ + 2e- ---> Cu

At the anode: as there are no halide ions present, oxygen and water are produced...

4OH- ---> O2 + 2H2O + 4e-

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

Diamond
In diamond, each carbon atom is joined to four other carbon atoms, forming a giant covalent structure. As a result, diamond is very hard and has a high melting point. This explains why it is used in cutting tools

Graphite

In graphite, each carbon atom is joined to only three other carbon atoms, this results in carbon sheets that are 'stacked' on top of each other. These layers can slide over each other, this means that graphite is much softer than diamond. It is used in pencils, and as a lubricant

NOTE: Diamond does not conduct electricity but graphite does

1.44 draw diagrams representing the positions of the atoms in diamond and graphite

Diamond

Graphite

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

Sodium chloride has a typical ionic structure, it has alternating positive  (Na+) and negative (Cl-) ions, this is how it should be drawn...

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

Compounds with ionic bonding always have giant ionic structures, the ions are held closely together by the attraction between oppositely charged ions in a 3-D lattice arrangement.

1.26 calculate percentage yield

Percentage yield is calculated with the equation...

Percentage yield = (actual yield (grams) / theoretical yield (grams)) x 100

NOTE: a 100% yield means you got all the product you expected to get; a 0% yield means that no reactants were converted into product

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 in calculations.

The space that one mole of gas takes up is called its molar volume.

At room temperature and pressure, one mole of any gas always occupies 24 dm3 (equivalent to 24,000 cm3) - room temperature is 25 ºc and room pressure is 1 atmosphere (atm).

In calculations, use the equations below to convert the number of moles or mass of any gas to a volume…

Volume (dm3) = moles of gas x 24

Volume (dm3) = (mass of gas / Mr of gas) x 24

For example: Whats the volume of 4.5 moles of chlorine at RTP?

Volume (dm3) = 4.5 x 24 = 108 dm3

Example two: How many moles are there in 8280cm3 of hydrogen gas at RTP?

Volume of gas = moles of gas x 24 therefore, moles of gas = volume of gas / 24

One more thing, we need to convert 8280cm3 into dm3. To do this, just divide by 1000.

8280 / 1000 = 8.28

Now substitute into the rearranged equation…

moles of gas = volume of gas / 24 = 8.28 / 24 = 0.345 moles

NOTE: RTP just means room temperature and pressure

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

Much like 1 million is 1,000,000; 1 mole is 6.023x10²³

6.023x10²³ is called Avogadro's number (or the Avogadro constant). One mole of a substance is equal to 6.023x10²³ particles, whether the particles are atoms, molecules, ions or electrons.