GCSE_Chemistry_Single Archives - TutorMyself Chemistry

1:01 understand the three states of matter in terms of the arrangement, movement and energy of the particles

Solid

Arrangement: Particles are close together and regularly packed.

Movement: Particles vibrate around a fixed point.

Energy: Particles have less kinetic energy than both liquids and gasses.

Liquid

Arrangement: Particles are close together but irregular.

Movement: Particles are free to move.

Energy: Particles have less kinetic energy than gasses but more than solids.

Gas

Arrangement: Particles are far apart and there are no forces between them.

Movement: Particles are free to move.

Energy: Particles have more kinetic energy than liquids and solids.

Hydr0Gen2022-11-29T13:49:09+00:00Categories: (a) States of matter, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:01, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Fundamentals|

1:02 understand the interconversions between the three states of matter in terms of: the names of the interconversions, how they are achieved and the changes in arrangement, movement and energy of the particles

Melting: When a solid is heated, the energy makes the particles vibrate fast enough so that the forces of attraction between the particles break. For example H₂O(s) –> H₂O(l)

Freezing: When a liquid is cooled, the particles move slow enough so that the forces of attraction between them will hold them into a solid. For example H₂O(l) –> H₂O(s)

Boiling: When a liquid is heated strongly, the energy makes the particles move fast enough so that all forces of attraction are broken. For example H₂O(l) –> H₂O(g)

Condensing: When a gas is cooled, the particles move slow enough so that the forces of attraction between them will hold them as a liquid. For example H₂O(g) –> H₂O(l)

Sublimation: A small number of substances have the ability to change directly from a solid to a gas when heated. For example CO₂(s) –> CO₂(g)

Hydr0Gen2022-11-30T13:57:13+00:00Categories: (a) States of matter, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:02, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Fundamentals|

1:03 understand how the results of experiments involving the dilution of coloured solutions and diffusion of gases can be explained

Diffusion is the spreading out of particles in a gas or liquid. There is a net movement of particles from areas of high concentration to areas of low concentration until a uniform concentration is achieved.

i) dilution of coloured solutions

Dissolving potassium manganate(VII) in water demonstrates that the diffusion in liquids is very slow because there are only small gaps between the liquid particles into which other particles diffuse.

The random motion of particles cause the purple colour to eventually be evenly spread out throughout the water.

Adding more water to the solution causes the potassium manganate(VII) particles to spread out further apart therefore the solutions becomes less purple. This is called dilution.

ii) diffusion experiments

When ammonia gas and hydrogen chloride gas mix, they react together to form a white solid called ammonium chloride.

ammonia + hydrogen chloride –> ammonium chloride

NH₃(g) + HCl(g) –> NH₄Cl(s)

A cotton wool pad was soaked in ammonia solution and another was soaked in hydrogen chloride solution. The two pads were then put into opposite ends of a dry glass tube at the same time.

The white ring of ammonium chloride forms closer to the hydrochloric acid end because ammonia particles are lighter than hydrogen chloride particles and therefore travel faster.

Even though these particles travel at several hundred metres per second, it takes about 5 min for the ring to form. This is because the particles move in random directions and will collide with air particles in the tube.

Hydr0Gen2023-09-18T10:54:46+00:00Categories: (a) States of matter, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:03, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Equilibria, Topic: Reversible Reactions (D), Topic: Simple Molecules & Covalent Bonding, Topic: Specification experiments|

1:08 understand how to classify a substance as an element, a compound or a mixture

Element: The simplest type of substances made up of only one type of atom.

Compound: A substance that contains two or more elements chemically joined together in fixed proportions.

Mixture: Different substances in the same space, but not chemically combined.

Note: elements such as oxygen (O₂) are described as diatomic because they contain two atoms.

The full list of elements that are diatomic is:

Hydrogen (H₂)
Nitrogen (N₂)
Fluorine (F₂)
Oxygen (O₂)
Iodine (I₂)
Chlorine (Cl₂)
Bromine (Br₂)

Hydr0Gen2022-11-30T13:57:56+00:00Categories: (b) Elements, compounds and mixtures, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:08, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Fundamentals|

1:09 understand that a pure substance has a fixed melting and boiling point, but that a mixture may melt or boil over a range of temperatures

Pure substances, such as an element or a compound, melt and boil at fixed temperatures.

However, mixtures melt and boil over a range of temperatures.

Example: although pure water boils at 100⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will raise the boiling point to 100.2⁰C.

Example: although pure water melts at 0⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will lower the melting point to -0.6⁰C.

Hydr0Gen2023-05-11T16:37:43+00:00Categories: (b) Elements, compounds and mixtures, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:09, Chemistry, ChemistrySpecPointsNew2019, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Fundamentals|

1:10 describe these experimental techniques for the separation of mixtures: simple distillation, fractional distillation, filtration, crystallisation, paper chromatography

Simple distillation

This method is used to separate a liquid from a solution. For example: separating water from salt water.

The salt water is boiled. The water vapour condenses back into a liquid when passed through the condenser. The salt is left behind in the flask.

Note: cold water is passed into the bottom of the condenser and out through the top so that the condenser completely fills up with water.

Fractional distillation

This method is used to separate a mixture of different liquids that have different boiling points. For example, separating alcohol from a mixture of alcohol and water.

Water boils at 100^oC and alcohol boils at 78^oC. By using the thermometer to carefully control of temperature of the column, keeping it at 78^oC, only the alcohol remains as vapour all the way up to the top of the column and passes into the condenser.

The alcohol vapours then condense back into a liquid.

Filtration

This method is used to separate an insoluble solid from a liquid. For example: separating sand from a mixture of sand and water.

The mixture is poured into the filter paper. The sand does not pass through and is left behind (residue) but the water passes through the filter paper and is collected in the conical flask (filtrate).

Crystallisation

This method is used to obtain a salt which contains water of crystallisation from a salt solution. For example: hydrated copper sulfate crystals (CuSO₄.5H₂O(s)) from copper sulfate solution (CuSO₄(aq)).

Gently heat the solution in an evaporating basin to evaporate some of the water
until crystals form on a glass rod (which shows that a hot saturated solution has formed).
Leave to cool and crystallise.
Filter to remove the crystals.
Dry by leaving in a warm place.

If instead the solution is heated until all the water evaporates, you would produce a powder of anhydrous copper sulfate (CuSO₄(s)).

Paper chromatography

This method can be used to separate the parts of a mixture into their components. For example, the different dyes in ink can all be separated and identified.

The coloured mixture to be separated (e.g. a food dye) is dissolved in a solvent like water or ethanol and carefully spotted onto the chromatography paper on the baseline, which is drawn in pencil so it doesn’t ‘run or smudge’.

The paper is carefully dipped into the solvent and suspended so the baseline is above the liquid solvent, otherwise all the spots would dissolve in the solvent. The solvent is absorbed into the paper and rises up it as it soaks into the paper. The choice of solvent depends on the solubility of the dye. If the dye does not dissolve in water then normally an organic solvent (e.g. ethanol) is used.

As the solvent rises up the paper it will carry the dyes with it. Each different dye will move up the paper at different rates depending on how strongly they stick to the paper and how soluble they are in the solvent.

Hydr0Gen2023-09-18T11:04:24+00:00Categories: (b) Elements, compounds and mixtures, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:10, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Reactions, Topic: Specification experiments|

1:11 understand how a chromatogram provides information about the composition of a mixture

Paper chromatography can be used to investigate the composition of a mixture.

A baseline is drawn on the paper. The mixture is spotted onto the baseline alongside known or standard reference materials. The end of the paper is then put into a solvent which runs up the paper and through the spots, taking some or all of the dyes with it.

Different dyes will travel different heights up the paper.

The resulting pattern of dyes is called a chromatogram.

In the example shown, the mixture is shown to contain the red, blue and yellow dyes. This can be seen because these dots which resulted from the mixture have travelled the same distance up the paper as have the red, blue and yellow standard reference materials.

Hydr0Gen2023-09-18T11:02:13+00:00Categories: (b) Elements, compounds and mixtures, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:11, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding|

1:12 understand how to use the calculation of Rf values to identify the components of a mixture

When analysing a chromatogram, the mixture being analysed is compared to standard reference materials by measuring how far the various dyes have travelled up the paper from the baseline where they started.

For each dye, the R_f value is calculated. To do this, 2 distances are measured:

The distance between the baseline and the dye
The distance between the baseline and the solvent front, which is how far the solvent has travelled from the baseline

The R_f value is calculated as follows:

$R_f=\frac{distance\:of\:dye\:from\:baseline }{distance\:of\:solvent\:front\:from\:baseline}$

If the R_f value of one of the components of the mixture equals the R_f value of one of the standard reference materials then that component is know to be that reference material.

Note that because the solvent always travels at least as far as the highest dye, the R_f value is always between 0 and 1.

Dyes which are more soluble will have higher R_f values than less soluble dyes. In other words, more soluble dyes move further up the paper. The extreme case of this is for insoluble dyes which don’t move at all (R_f value = 0). The other aspect affecting how far a dye travels is the affinity that dye has for the paper (how well it ‘sticks’ to the paper).

Hydr0Gen2023-09-18T11:02:34+00:00Categories: (b) Elements, compounds and mixtures, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:12, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding|

1:13 practical: investigate paper chromatography using inks/food colourings

A pencil line (baseline) is drawn 1cm from the bottom of the paper. Pencil will not dissolve in the solvent, but if ink were used instead it might dissolve and interfere with the results of the chromatography.
A spot of each sample of dye is dropped at different points along the baseline.
The paper is suspended in a beaker which contains a small amount of solvent. The bottom of the paper should be touching the solvent, but the baseline with the dyes should be above the level of the solvent. This is important so the dyes don’t simply dissolve into the solvent in the beaker.
A lid should cover the beaker so the atomosphere becomes saturated with the solvent. This is so the solvent does not evaporate from the surface of the paper.
When the solvent has travelled to near the top of the paper, the paper is removed from the solvent and a pencil line drawn (and labelled) to show the level the solvent reached up the paper. This is called the solvent front.
The chromatogram is then left to dry so that all the solvent evaporates.

Common solvents are water or ethanol. The choice of solvent depends on whether most of the dyes are soluble in that solvent.

Hydr0Gen2023-09-18T11:03:01+00:00Categories: (b) Elements, compounds and mixtures, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:13, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding, Topic: Specification experiments|

1:14 know what is meant by the terms atom and molecule

Atom: An atom is the smallest part of an element.

Molecule: A molecule is made of a fixed number of atoms covalently bonded together.

Hydr0Gen2022-11-30T14:10:40+00:00Categories: (c) Atomic structure, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:14, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding|

1:15 know the structure of an atom in terms of the positions, relative masses and relative charges of sub-atomic particles

An atom consists of a central nucleus, composed of protons and neutrons.

This is surrounded by electrons, orbiting in shells (energy levels).

Atoms are neutral because the numbers of electrons and protons are equal.

	Mass	Charge
Proton	1	+1
Neutron	1	0
Electron	negligible (1/1836)	-1

Hydr0Gen2022-11-30T14:11:04+00:00Categories: (c) Atomic structure, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:15, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Fundamentals, Topic: Simple Molecules & Covalent Bonding|

1:16 know what is meant by the terms atomic number, mass number, isotopes and relative atomic mass (Aᵣ)

Atomic number: The number of protons in an atom.

Mass number: The number of protons and neutrons in an atom.

Isotopes: Atoms of the same element (same number of protons) but with a different number of neutrons.

Relative atomic mass (A_r): The average mass of an atom compared to 1/12th the mass of carbon-12.

Hydr0Gen2022-11-30T14:11:18+00:00Categories: (c) Atomic structure, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:16, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations|

1:17 be able to calculate the relative atomic mass of an element (Aᵣ) from isotopic abundances

75% of chlorine atoms are the type ³⁵Cl (have a mass number of 35)

25% of chlorine atoms are of the type ³⁷Cl (have a mass number of 37)

In order to calculate the relative atomic mass (A_r) of chlorine, the following steps are used:

Multiply the mass of each isotope by its relative abundance
Add those together
Divide by the sum of the relative abundances (normally 100)

$A_r = \frac{( (35 \times 75) + (37 \times 25) )}{100}$

$A_r = 35.5$

Example question:

A sample of bromine contained the two isotopes in the following proportions: bromine-79 = 50.7% and bromine-81 = 49.3%.

Calculate the relative atomic mass (A_r) of bromine.

$A_r = \frac{( (79 \times 50.7) + (81 \times 49.3) )}{100}$

$A_r = 79.99$

Hydr0Gen2022-11-30T14:11:27+00:00Categories: (c) Atomic structure, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:17, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Key Calculations|

1:18 understand how elements are arranged in the Periodic Table: in order of atomic number, in groups and periods

The elements in the Periodic Table are arranged in order of increasing atomic number.

Columns are called Groups. They indicate the number of electrons in the outer shell of an atom.

Rows are called Periods. They indicate the number of shells (energy levels) in an atom.

Hydr0Gen2023-01-10T12:19:15+00:00Categories: (d) The Periodic Table, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:18, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Fundamentals, Topic: Halogens (D), Topic: Metal Reactivity & Halogens, Topic: Metals (D), Topic: Simple Molecules & Covalent Bonding|

1:21 identify an element as a metal or a non-metal according to its position in the Periodic Table

Metals on the left of the Periodic Table.

Non-Metals on the top-right, plus Hydrogen.

Hydr0Gen2023-01-10T12:26:21+00:00Categories: (d) The Periodic Table, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:21, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Fundamentals, Topic: Halogens (D), Topic: Metal Reactivity & Halogens, Topic: Metals (D), Topic: Simple Molecules & Covalent Bonding|

1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided

Example:

Sodium (Na) reacts with water (H₂O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).

Word equation:

sodium + water –> sodium hydroxide + hydrogen

Writing the chemical equation

A chemical equation represents what happens in terms of atoms in a chemical reaction.

Step 1: To write a chemical equation we need to know the chemical formulae of the substances.

Na + H₂O –> NaOH + H₂

Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.

During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.

2Na + 2H₂O –> 2NaOH + H₂

If asked for an equation, the chemical equation must be given.

State symbols are used to show what physical state the reactants and products are in.

State symbols	Physical state
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous solution (dissolved in water)

Example:

A solid piece of sodium (Na) reacts with water (H₂O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).

2Na(s) + 2H₂O(l) –> 2NaOH(aq) + H₂(g)

Hydr0Gen2023-09-18T11:05:10+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:25, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil, Topic: Fundamentals, Topic: Ions, Topic: Key Reactions, Topic: Moles|

1:26 calculate relative formula masses (including relative molecular masses) (Mᵣ) from relative atomic masses (Aᵣ)

Relative formula mass (M_r) is mass of a molecule or compound (on a scale compared to carbon-12).

It is calculated by adding up the relative atomic masses (A_r) of all the atoms present in the formula.

Example:

The relative formula mass (M_r) for water (H₂O) is 18.

Water = H₂O

Atoms present = (2 x H) + (1 x O)

M_r = (2 x 1) + (1 x 16) = 18

Hydr0Gen2022-11-30T14:12:46+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:26, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Fundamentals, Topic: Key Calculations, Topic: Moles|

1:37 understand how ions are formed by electron loss or gain

Ions are electrically charged particles formed when atoms lose or gain electrons.

They have the same electronic structures as noble gases.

Metal atoms form positive ions (cations).

Non-metal atoms form negative ions (anions).

Hydr0Gen2022-11-30T14:13:00+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:37, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Ions|

1:38b know the charges of these ions: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)

Name of Ion	Formula	Charge
Sulfate	SO₄^2-	-2
Carbonate	CO₃^2-	-2
Nitrate	NO₃^-	-1
Hydroxide	OH^-	-1
Ammonium	NH₄⁺	+1
Hydrogen ion	H⁺	+1

Ion charges on the periodic table

Sam Mahal2022-12-08T10:46:44+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:38b, GCSE, GCSE_Chemistry_Single, SpecPoint, Topic: Acids & Salts|

1:39 write formulae for compounds formed between the ions listed in 1:38

Writing the electron configuration of an atom allows you to work out the electron configuration of the ion and therefore the charge on the ion.

Examples:

Atom = Mg

Electron configuration = 2,8,2

remove the two electrons from the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)

Ion = Mg²⁺

[2,8]²⁺

Atom = O

Electron configuration = 2,6

add two electrons to the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)

Ion = O^2- [2,8]^2-

Hydr0Gen2022-11-30T14:13:29+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:39, Chemistry, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Ions, Topic: Key Reactions|

1:41 understand ionic bonding in terms of electrostatic attractions

Ionic bonding: a strong electrostatic attraction between oppositely charged ions.

Hydr0Gen2023-01-10T11:24:42+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:41, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Ions, Topic: Structure & Bonding|

1:42 understand why compounds with giant ionic lattices have high melting and boiling points

Ionic compounds have high melting and boiling points because they have a giant structure with strong electrostatic forces between oppositely charged ions that require a lot of energy to break.

Giant 3D lattice of sodium chloride (NaCl)

Hydr0Gen2023-01-10T11:24:55+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:42, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Ions, Topic: Structure & Bonding|

1:44 know that a covalent bond is formed between atoms by the sharing of a pair of electrons

A covalent bond is formed between two non-metal atoms by sharing a pair of electrons in order to fill the outer shell.

Hydr0Gen2023-01-10T11:26:44+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:44, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:47 explain why substances with a simple molecular structures are gases or liquids, or solids with low melting and boiling points. The term intermolecular forces of attraction can be used to represent all forces between molecules

Carbon dioxide (CO₂) has a simple molecular structure. This just means that it is made up of molecules.

Within each molecule are atoms bonded to each other covalently. These covalent bonds inside the molecules are strong.

However, between the molecules are weak forces of attraction that require little energy to break. These forces are not covalent bonds. This is why simple molecular substances such as carbon dioxide have a low boiling point.

So when carbon dioxide changes from a solid to a gas, for example, that process can be represented as:

CO₂ (s) → CO₂ (g)

Notice that even though there has been a dramatic change of state from solid to gas, the substance before and after the change is always made up of carbon dioxide molecules. During the change of the state the covalent bonds within each molecule remain unbroken. Instead it is the weak forces of attraction between the molecules which have been overcome.

Hydr0Gen2023-01-10T11:29:01+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:47, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil, Topic: Bonding & Structure (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:49 explain why substances with giant covalent structures are solids with high melting and boiling points

Diamond has a high melting point because it is a giant covalent structure with many strong covalent bonds that require a lot of energy to break.

Hydr0Gen2023-01-10T11:30:25+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:49, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Structure & Bonding|

2:01 understand how the similarities in the reactions of lithium, sodium and potassium with water provide evidence for their recognition as a family of elements

Group 1 metals such as potassium, sodium and lithium, react with water to produce a metal hydroxide and hydrogen. For example:

lithium + water → lithium hydroxide + hydrogen

2Li (s) + 2H₂O (l) → 2LiOH (aq) + H₂ (g)

The observations for the reaction of water with either potassium or sodium or lithium have the following similarities:

fizzing (hydrogen is produced)
metal floats and moves around on the water
metal disappears

In each case a metal hydroxide solution is produced.

These similarities in the reactions provide evidence that the 3 metals are in the same group of the Periodic Table (i.e. have the same number of electrons in their outer shell).

Hydr0Gen2023-01-10T12:29:10+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:01, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:02 understand how the differences between the reactions of lithium, sodium and potassium with air and water provide evidence for the trend in reactivity in Group 1

Lithium is the first element in group 1 of the Periodic Table. The observations for the reaction of lithium and water are:

fizzing (hydrogen gas is released)
lithium floats and moves around on the water
lithium disappears

Sodium is the second alkali metal in the group. The reaction of sodium and water is more vigorous than lithium’s:

fizzing (hydrogen gas is released)
sodium floats and moves around on the water
sodium melts into a silver-coloured ball
sodium disappears

Potassium is the third alkali metal in the group. The reaction of potassium and water is more vigorous than sodium’s:

fizzing (hydrogen gas is released)
potassium floats and moves around on the water
catches fire with a LILAC flame
potassium disappears

When the group 1 metals react with air they oxidise, showing a similar trend in reactivity as we go down the group of the Periodic Table.

Therefore, as we go down group 1 (increasing atomic number), the elements become more reactive: Li<Na<K<Rb<Cs<Fr

Hydr0Gen2023-01-10T12:29:20+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:02, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:03 use knowledge of trends in Group 1 to predict the properties of other alkali metals

From the data in the table, it is possible to deduce the properties of francium from the trends in the other group 1 metals.

For example, we can predict that francium will have a melting point around 20⁰C and a density of just over 2g/cm³.

We can also predict that francium will react violently with water, producing francium hydroxide and hydrogen.

Alkali metal	Melting point (⁰C)	Density (g/cm³)	Reaction with water	Products
lithium (Li)	181	0.53	fizzing	lithium hydroxide + hydrogen
sodium (Na)	98	0.97	rapid fizzing	sodium hydroxide + hydrogen
potassium (K)	63	0.86	vigorous fizzing and lilac flame	potassium hydroxide + hydrogen
rubidium (Rb)	39	1.53	?	rubidium hydroxide + hydrogen
caesium(Cs)	29	1.88	?	caesium hydroxide + hydrogen
francium (Fr)	?	?	?	?

Hydr0Gen2023-01-10T12:29:32+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:03, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:05 know the colours, physical states (at room temperature) and trends in physical properties of chlorine, bromine and iodine

Element	Colour	State at room temp
Chlorine (Cl₂)	Green	Gas
Bromine (Br₂)	Red-brown	Liquid
Iodine (l₂)	Grey	Solid

Chlorine is a toxic gas, so should be handled in a fume cupboard.

Hydr0Gen2023-01-10T11:50:40+00:00Categories: (b) Group 7 (halogens) – chlorine, bromine and iodine, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:05, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Halogens (D), Topic: Metal Reactivity & Halogens|

2:06 use knowledge of trends in Group 7 to predict the properties of other halogens

If you look at the trends in the physical properties of the halogens, Cl₂, Br₂, I₂ you can make predictions about the properties of the other halogens.

Element	Colour	State at room temp
Fluorine (F₂)	Yellow	Gas
Astatine (At₂)	Black	Solid

Hydr0Gen2023-01-10T11:51:01+00:00Categories: (b) Group 7 (halogens) – chlorine, bromine and iodine, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:06, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Halogens (D), Topic: Metal Reactivity & Halogens|

2:09 know the approximate percentages by volume of the four most abundant gases in dry air

Air is a mixture of different gases.

The abundance of gases in the air is as follows:

Gas	% by volume
Nitrogen, N₂	78.1
Oxygen, O₂	21.0
Argon, Ar	0.9
Carbon dioxide, CO₂	0.04

Hydr0Gen2023-09-18T11:05:32+00:00Categories: (c) Gases in the atmosphere, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:09, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding|

2:10 understand how to determine the percentage by volume of oxygen in air using experiments involving the reactions of metals (e.g. iron) and non-metals (e.g. phosphorus) with air

The following 3 experiments can be used to determine that oxygen (O₂) makes up approximately 20% by volume of the composition of air.

Copper

The copper is in excess and uses up the oxygen to form copper oxide (CuO).

All the oxygen in the air is therefore used up, and so the volume of the air decreases by about 20% (the percentage of oxygen in air).

Iron

The iron reacts with the oxygen in the air (rusting).

As long as the iron and water are in excess, the total volume of air enclosed by the apparatus decreases by about a fifth (20%) over several days.

Phosphorus

The phosphorus is lit with a hot wire.

It reacts with the oxygen in the air and causes the water level in the bell jar to rise by about 20%.

Hydr0Gen2023-09-18T11:06:04+00:00Categories: (c) Gases in the atmosphere, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:10, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding, Topic: Specification experiments|

2:11 describe the combustion of elements in oxygen, including magnesium, hydrogen and sulfur

Magnesium reacts with oxygen producing a bright white flame leaving behind a white ash of magnesium oxide.

magnesium + oxygen → magnesium oxide

2Mg (s) + O₂ (g) → 2MgO

MgO is a base, which can react with an acid to give a salt and water.

Hydrogen reacts with oxygen in an explosive reaction. This is the basis of the ‘squeak pop’ test for hydrogen in test tube. With larger quantities of hydrogen this explosion can be dangerous.

hydrogen + oxygen → water

2H₂ (g) + O₂ (g) → 2H₂O (l)

Sulfur reacts with oxygen producing a blue flame.

sulfur + oxygen → sulfur dioxide

S (s) + O₂ (g) → SO₂ (g)

When sulfur dioxide (SO₂) dissolves in water it forms an acidic solution of sulfurous acid:

SO₂ (g) + H₂O (l) → H₂SO₃ (aq)

Hydr0Gen2022-11-30T14:20:27+00:00Categories: (c) Gases in the atmosphere, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:11, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Reactions|

2:13 know that carbon dioxide is a greenhouse gas and that increasing amounts in the atmosphere may contribute to climate change

Carbon dioxide (CO₂) is a greenhouse gas.

It absorbs infra-red radiation and therefore warms the atmosphere. This leads to global warming.

This may cause climate change.

Hydr0Gen2022-11-30T14:20:45+00:00Categories: (c) Gases in the atmosphere, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:13, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil|

2:14 Practical: determine the approximate percentage by volume of oxygen in air using a metal or a non-metal

The following 3 experiments can be used to determine that oxygen (O₂) makes up approximately 20% by volume of air.

Copper

The copper is in excess and uses up the oxygen to form copper oxide (CuO).

All the oxygen in the air is therefore used up, and so the volume of the air decreases by about 20% (the percentage of oxygen in air).

Iron

The iron reacts with the oxygen in the air (rusting).

As long as the iron, oxygen and water are all in excess, the total volume of air enclosed by the apparatus decreases by about a fifth (20%) over several days.

Phosphorus

The phosphorus is lit with a hot wire.

It reacts with the oxygen in the air and causes the water level in the bell jar to rise by about 20%.

Hydr0Gen2023-09-18T11:06:27+00:00Categories: (c) Gases in the atmosphere, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:14, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Simple Molecules & Covalent Bonding, Topic: Specification experiments|

2:15 understand how metals can be arranged in a reactivity series based on their reactions with: water and dilute hydrochloric or sulfuric acid

Some metals are more reactive than others.

The order of reactivity can be determined by adding acid to different metals and observing the rate of reaction.

For example, when hydrochloric acid is added to iron (Fe) then bubbles of hydrogen are produced slowly. However, if the same acid is added to zinc (Zn) then bubbles will be produced more quickly. This tells us that zinc is more reactive than iron.

Instead of using acid, water can be used to test the relative reactivity of metals. However, many metals are too low in the reactivity series to react with water

Hydr0Gen2023-01-10T12:30:26+00:00Categories: (d) Reactivity series, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:15, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:17 know the order of reactivity of these metals: potassium, sodium, lithium, calcium, magnesium, aluminium, zinc, iron, copper, silver, gold

A more reactive metal will displace a less reactive metal.

In addition a more reactive metal will react more vigorously than a less reactive metal.

For example, potassium takes a shorter time to react than sodium:

Hydr0Gen2023-01-10T12:31:04+00:00Categories: (d) Reactivity series, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:17, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:18 know the conditions under which iron rusts

Iron rusts when oxygen and water are present.

Reaction:

Hydr0Gen2023-01-10T12:31:16+00:00Categories: (d) Reactivity series, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:18, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:19a understand how the rusting of iron may be prevented by: barrier methods, galvanising

Barrier Methods: Rusting may be prevented by stopping the water and oxygen getting to the iron with a barrier of grease, oil, paint or plastic.

Galvanising: (coating in zinc) also prevents water and oxygen getting to the iron, but with galvanising even if the barrier is broken the more reactive zinc corrodes before the less reactive iron. During the process, the zinc loses electrons to form zinc ions.

Sam Mahal2022-12-12T11:13:19+00:00Categories: (d) Reactivity series, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:19a, Chemistry, GCSE, GCSE_Chemistry_Single, SpecPoint, Topic: Metal Reactivity & Halogens|

2:28a describe the use of litmus to distinguish between acidic and alkaline solutions

Indicator	Colour in acidic solution	Colour in alkaline solution
Litmus	Red	Blue

Sam Mahal2022-12-12T11:16:42+00:00Categories: (f) Acids, alkalis and titrations, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:28a, Chemistry, GCSE, GCSE_Chemistry_Single, SpecPoint, Topic: Key Reactions|

2:29 understand how to use the pH scale, from 0–14, can be used to classify solutions as strongly acidic (0–3), weakly acidic (4–6), neutral (7), weakly alkaline (8–10) and strongly alkaline (11–14)

The pH scale ranges from 0 to 14, and tells you how acidic or how alkaline a solution is.

	strongly acidic	weakly acidic	neutral	weakly alkaline	strongly alkaline
pH	0-3	4-6	7	8-10	11-14

Hydr0Gen2023-01-10T11:38:25+00:00Categories: (f) Acids, alkalis and titrations, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:29, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Key Reactions|

2:30 describe the use of Universal Indicator to measure the approximate pH value of an aqueous solution

An indicator is a substance that has more than one colour form depending on the pH.

Universal indicator is a mixture of different dyes which change colour in a gradual way over a range of pH.

Hydr0Gen2023-01-10T11:39:19+00:00Categories: (f) Acids, alkalis and titrations, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:30, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Key Reactions|

2:31 know that acids in aqueous solution are a source of hydrogen ions and alkalis in a aqueous solution are a source of hydroxide ions

An acid is source of hydrogen ions (H⁺).

An alkali is source of hydroxide ions (OH^–).

Hydr0Gen2023-01-10T11:39:28+00:00Categories: (f) Acids, alkalis and titrations, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:31, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Key Reactions|

2:32 know that bases can neutralise acids

Metal oxides, metal hydroxides and ammonia (NH₃) are called bases.

Bases neutralise acids by combining with the hydrogen ions in them.

The key reaction is:

acid + base → salt + water

An example of this is:

sulfuric acid + copper oxide → copper sulfate + water

H₂SO₄ + CuO → CuSO₄ + H₂O

Hydr0Gen2023-09-18T11:07:11+00:00Categories: (f) Acids, alkalis and titrations, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:32, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Key Reactions|

2:44 describe tests for these gases: hydrogen, oxygen, carbon dioxide, ammonia, chlorine

Tests for gases

Gas	Test	Result if gas present
hydrogen (H₂)	Use a lit splint	Gas pops
oxygen (O₂)	Use a glowing splint	Glowing splint relights
carbon dioxide (CO₂)	Bubble the gas through limewater	Limewater turns cloudy
ammonia (NH₃)	Use red litmus paper	Turns damp red litmus paper blue
chlorine (Cl₂)	Use damp litmus paper	Turns damp litmus paper white (bleaches)

Hydr0Gen2023-01-10T11:43:33+00:00Categories: (h) Chemical tests, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:44, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Metal Reactivity & Halogens, Topic: Simple Molecules & Covalent Bonding, Topic: Specification experiments|

2:45 describe how to carry out a flame test

A flame test is used to show the presence of certain metal ions (cations) in a compound.

A platinum or nichrome wire is dipped into concentrated hydrochloric acid to remove any impurities.
The wire is dipped into the salt being tested so some salt sticks to the end.
The wire and salt are held in a non-luminous (roaring) bunsen burner flame.
The colour is observed.

Properties of the platinum or nichrome wire is: