Topic 5: Separate chemistry 1

Question 1

What are most metals?

Answer 1

Most metals are transition metals

Question 2

properties of transition metals:

Answer 2

● High melting point (due to electrostatic forces of attraction between positively charged metal ions and ‘sea’ of electrons)
● High density
● They have ions with many different charges
● Form coloured compounds
● Are useful as catalysts.
o Shown by iron and its use in the Haber process
as a catalyst

Question 3

How does the oxidation of metals result in corrosion?

Answer 3

● Corrosion = destruction of materials by chemical reactions with substances in the
environment
o E.g. rusting
▪ Both air and water are necessary for iron to rust – i.e. oxidation – gain of oxygen results in corrosion

Question 4

How can rusting of iron can be prevented by:

exclusion of oxygen, exclusion of water,

Answer 4

● rusting can be prevented by excluding oxygen and water e.g. by:
o painting
o coating with plastic
o using oil or grease
● Aluminium has an oxide coating that protects the metal from further corrosion –
exclusion of oxygen and water
● water can be kept away using a desiccant in the container (absorbs water
vapour)
● oxygen can be kept away by storing the metal in a vacuum container

Question 5

How can rusting of iron can be prevented by:

sacrificial protection

Answer 5

● Sacrificial protection: where the metal you want to be protected from rusting is galvanised with a more reactive metal, which will rust first and prevent water
and oxygen reaching the layer underneath
o E.g. zinc is used to galvanise iron

Question 6

How can electroplating be used to improve the appearance and/or the resistance to corrosion of metal objects?

Answer 6

● Electroplating acts as a barrier in order to exclude oxygen and water
● Also improves appearance as you can electroplate a metal with an unreactive
metal such as gold that is more attractive and will not corrode
● it is done using the metal to be plated as the cathode and the metal you’re
plating it with as the anode, then have a solution containing ions of the metal being used to do the plating

Question 7

What are most metals in every day uses?

Answer 7

● Most metals in everyday uses are alloys.

Question 8

why converting pure metals into alloys often

increases the strength of the product

Answer 8

● In a pure metal, all the + metal ions are the same size and in a regular arrangement, allowing the layers to slide over each other relatively
easily, making the metal soft and malleable.
●In an alloy, you have + ions of different metals, which have different sized ions. This disrupts the regular
structure and prevents the ions being able to slide as easily, leaving a much harder, stronger metal.

Question 9

Why is pure copper,gold, iron and aluminium mixed with small amounts of similar metals?

Answer 9

● Pure copper, gold, iron and aluminium are all too soft for everyday uses and so are mixed with small amounts of similar metals to make them harder for everyday use.

Question 10

Explain why iron is alloyed with other metals to produce alloy steels?

Answer 10

Steels are alloys since they used mixtures of carbon and iron
o Some steels contain other metals. Alloys can be designed to specific uses.
o Low-carbon steels are easily shaped - used for sheeting (malleable)
o High carbon steels are hard - used for cutting tools
o Stainless steels (containing chromium and nickel) are resistant to corrosion - used for cutlery

Question 11

Uses of metals related to their properties (and vice

versa), including aluminium, copper and gold and their alloys includingmagnalium and brass

Answer 11

● aluminium: low density, used for aircraft
● copper: good conductor, used in electrical cables
● gold: good resistance to corrosion, used in jewelry
● magnalium (aluminum + magnesium): low density, used in cars and planes
● brass (copper + zinc): hard, resistant to corrosion, used in coins

Question 12

What is the concentration of a solution measured in?

Answer 12

● Concentration of a solution can be measured in moles per given volume of solution e.g. moles per dm3 (mol/dm3)

Question 13

How can you calculate moles of a solute in a given volume of a known concentration?

Answer 13

● To calculate moles of solute in a given volume of a known concentration use:
moles = conc x vol i.e. mol = mol/dm3 x dm3

Question 14

What does a smaller volume or larger number of moles of solute give?

Answer 14

● a smaller volume or larger number of moles of solute gives a higher concentration

Question 15

What does a larger volume or smaller number of moles of solute give?

Answer 15

● a larger volume or smaller number of moles of solute gives a lower concentration

Question 16

Core Practical: Carry out an accurate acid-alkali titration, using burette, pipette and a suitable indicator
● method:

Answer 16

○ add acid to burette using a funnel, record the volume in the burette to start
○ add known volume of alkali to a conical flask and add some indicator
○ place conical flask on white tile (so you can see colour change clearly)
○ add acid to alkali until you reach the end point
○ calculate how much acid has been added (titre)
○ repeat until you get concordant titres

Question 17

How can you use the results of titrations

to calculate an unknown concentration or an unknown volume of a solution

Answer 17

● if you only knew the concentration of the acid and wanted to calculate the concentration of the alkali:
○ calculate moles of acid using moles = concentration x volume
○ calculate the mole ratio of acid to alkali using the equation for the reaction
○ work out how many moles of alkali you have using the mole ratio and moles of acid (e.g. if you have 5 moles of acid and the ratio of acid to alkali is 1:2, you will have 10 moles of alkali)
○ calculate the concentration of the alkali using concentration = mol volume
● follow the same method for if you have both concentrations but only one volume

Question 18

Yield

Answer 18

● Amount of product obtained is known as yield

Question 19

Percentage yield

Answer 19

Percentage​ ​yield​ ​=​ ​Amount​ ​of​ ​product​ ​produced/ Maximum​ ​amount​ ​of​ ​product​ ​possible X 100

Question 20

causes of actual yield being less than theoretical yield:

Answer 20

○ incomplete reactions (not all of the reactants have reacted)
○ practical losses during the experiment (some product has been left in the weighing boat etc)
○ side reactions (some of the products react to form other products than those you wanted)

Question 21

Atom economy

Answer 21

● atom economy- a measure of the amount of starting materials that end up as useful products
● Important for sustainable development and for economic reasons to use reactions with high atom economy

Question 22

How to calculate atom economy

Answer 22

● atom economy = (Mr of desired product from reaction / sum of Mr of all reactants) x 100

Question 23

How can you explain why a particular reaction pathway is chosen to produce a specified product?

Answer 23

● look for a high atom economy, high yield, fast rate, equilibrium position to the right (towards products) and useful by-products – be prepared to look for these
within given information for the question and present them as an answer

Question 24

Describe the molar volume, of any gas at room temperature & pressure, as the volume occupied by one mole of molecules of any gas

Answer 24

● Equal amounts in mol. of gases occupy the same volume under the same conditions of temperature and pressure (e.g. RTP)
● Volume of 1 mol. of any gas at RTP (room temperature and pressure: 20 degreesC and 1 atmosphere pressure) is 24 dm

Question 25

How to calculate the volumes of gaseous reactants and products at RTP

Answer 25

Volume (dm3) of gas at RTP = Mol. x 24

Volume (cm3) of gas at RTP = mol x 24,000

Question 26

How to calculate moles of reactant with volume or mass

Answer 26

if given a mass: moles=mass ÷ molar mass

if given a volume: moles = volume ÷ 24

Question 27

How to calculate mass/volume using moles

Answer 27

for calculating mass, mass=moles x molar mass

for calculating volume, volume=moles x 24

Question 28

avogadro’s law

Answer 28

• one mole of a substance contains 6.02 x 10^23 particles

- this means if you had 10 moles of a substance, you would have 10 x 6.02 x 10^23 particles = 6.02 x 10^24

Question 29

Haber process

Answer 29

Haber process is a reversible reaction between nitrogen and hydrogen to form ammonia
●The reaction is reversible so ammonia breaks down again into nitrogen and hydrogen.
nitrogen​ ​+​ ​hydrogen​ ​⇌​ ​ammonia

Question 30

equilibrium is reached at a faster rate when:

Answer 30

● a higher temperature is used (particles have more kinetic energy so collide more frequently so have more successful collisions)
● a higher pressure/concentration is used (more particles in a given volume, so more frequent successful collisions)
● a catalyst is used

Question 31

What conditions are needed for the haber process?

Answer 31

● For the Haber process - the purified gases are passed over a catalyst of iron at a high temperature (about 450 °C) and a high pressure (about 200 atmospheres).

Question 32

In industrial reactions how are conditions related to:
The availability & cost of raw materials and energy supplies, control of temperature, pressure and
catalyst used

Answer 32

o High temperatures and pressures are desired for industrial reactions toincrease the rate of reaction
o But, a higher temperature shifts equilibrium towards the reactants (as the forwards reaction is exothermic), therefore a compromise is required to ensure a fast rate of reaction and a high yield of products
o Catalyst is used because of the effect of having an increased rate of reaction, however they are also expensive
o High temperatures and pressures can be expensive and dangerous as well (particularly pressures) and the equipment required for them can be very expensive

Question 33

What may fertilizers contain?

Answer 33

-fertilisers may contain nitrogen, phosphorus and

potassium compounds to promote plant growth

Question 34

How does ammonia reacts with nitric acid to produce a salt that is used as a fertiliser?

Answer 34

● Ammonia can be used to manufacture ammonium salts with nitric acid
o Ammonia acts as a base
o ammonia + nitric acid → ammonium nitrate
o NH3 + HNO3 → NH4NO3

Question 35

Laboratory preparation of ammonium sulfate from ammonia solution and dilute sulfuric acid on a small scale

Answer 35

● In the lab:
○ reactants: ammonia solution and dilute sulfuric acid (bought from
chemical manufacturers)
○ SMALL scale (very little is produced)
○ only involves a few stages (titration then crystallisation)

Question 36

Industrial production of ammonium sulfate, used as a fertiliser

Answer 36

● In industry:
○ reactants: natural gas, air, water (to make ammonia) and sulfur, air, water (to make sulfuric acid)
○ LARGE scale (produces a lot)
○ Many stages required (need to make ammonia and sulfuric acid, react accurate volumes then evaporate)

Question 37

What does a chemical cell produce?

Answer 37

● a chemical cell produces a voltage until one of the
reactants is used up
● Chemical reactions stop when one of the reactants has been used up

Question 38

In a hydrogen-oxygen fuel cell where hydrogen and oxygen are used to produce a voltage, what is the product?

Answer 38

● Supplied by an external source of fuel (eg hydrogen) and oxygen or air
o The reaction takes place within the fuel cell to produce a potential difference
o Overall reaction in a hydrogen fuel cell involves the oxidation of hydrogen to produce water
o 2H2 + O2 → 2H2O

Question 39

Strengths of fuel cells

Answer 39

• produce only water as waste

- keep producing fuel if fuel keeps being supplied

Question 40

Weaknesses of fuel cells

Answer 40

• difficult to transport/store hydrogen so aren’t suitable for portable devices
• expensive to make