# Triple

Triple-Science-only spec point

## 1.25 know and use the relationship between momentum, mass and velocity: P=m x v

momentum (kgm/s)= mass (kg) x velocity (m/s)

## 1.26 use the idea of momentum to explain safety features

To reduce the force experienced by the passenger you need to extend the time for a passenger to stop in a collision. As force is the change in momentum divided by time.

## 1.28 use the relationship between force, change in momentum and time take

Force is the rate of change of momentum. So Force (N) = change in momentum (kgm/s) / time (s)

## 1.29 demonstrate an understanding of Newton’s third law

Every action has an equal and opposite reaction.

Book pushes down on table, table pushes up on book. So book doesn’t accelerate.

Table pushes down on floor, floor pushes up on table. So table doesn’t accelerate.

## 1.30 know and use the relationship between the moment of a force and its perpendicular distance from the pivot

moment = force x perpendicular distance from the pivot

## 1.32 use the principle of moments for a simple system of parallel forces acting in one plane

The principle of moments states that when the clockwise moments are equal to the anticlockwise moments a body will be in equilibrium.

## 1.33 understand how the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam

when moments are taken from the right hand side as the block is a greater distance the force from the left hand pivot must be bigger to counteract it. The opposite is true for the left hand side.

## 5.10 describe the arrangement and motion of particles in solids, liquids and gases

solids:

• Tightly packed
• Held in fixed pattern

liquids:

• Tightly packed
• Can slide over each other

gasses:

• Move with rapid, random motion

## 5.11 practical: obtain a temperature–time graph to show the constant temperature during a change of state

1. Remove the boiling tube of stearic acid from
the water bath
2. Place the tube into a beaker of room
temperature water
3. Add a separate thermometer to the water
4. Take readings from the thermometer in the
stearic acid and the water every minute
[Make sure to avoid parallax error while doing so]
5. Note readings in the table below
6. Note on the table when you observe the stearic
acid change from a liquid to a solid.
7. Plot your results in a graph

## 5.12 know that specific heat capacity is the energy required to change the temperature of an object by one degree Celsius per kilogram of mass (J/kg °C)

Specific heat capacity:

• Amount of heat energy required to increase the temperature of 1kg of a substance by 10
• Unit J/kg 0C

## 5.13 use the equation: change in thermal energy: ΔQ = m × c × ΔT

Change in thermal energy [J] = Mass [kg] x Specific heat capacity [J/kg 0C] x Change in temperature [0C]

## 5.14 practical: investigate the specific heat capacity of materials including water and some solids

1. Set up the apparatus as shown the diagram.
2. Make note of all measurements: current (A), potential difference (V), mass (kg).
3. Use the electronic balance to measure the mass of your
4. Record the initial temperature of you block.
5. Switch on the heater and start your stopwatch.
[You will now leave the heater on for 10 minutes]
6. While the heater is switched on take readings from the
Ammeter and the Voltmeter.
7. Use these to calculate the Thermal Energy that will be
supplied to the block in 10 minutes
8. Record the temperature of your block after 10 minutes.
9. Calculate the Change in Temperature

## 6.09 describe the construction of electromagnets

A soft iron core wrapped in wire. When current flows through the coil of wire it becomes magnetic.

## 6.11 know that there is a force on a charged particle when it moves in a magnetic field as long as its motion is not parallel to the field

 The movement of the charged particle is a current so it produces a magnetic field. This magnetic field interacts with the permanent magnetic field to create a force. The force is perpendicular to the direction of motion and the permanent magnetic field.

## 6.17 describe the structure of a transformer, and understand that a transformer changes the size of an alternating voltage by having different numbers of turns on the input and output sides

 AC current in the primary coil produces a changing magnetic field around the primary coil. The iron core channels the changing field through the secondary coil. The changing magnetic field induces a voltage in the secondary coil.

## 6.18 explain the use of step-up and step-down transformers in the large-scale generation and transmission of electrical energy

 Step Up transformers increase the voltage – more secondary turns than primary Step Down transformers decrease the voltage – more primary turns than secondary

## 6.19 know and use the relationship between input (primary) and output (secondary) voltages and the turns ratio for a transformer:

 Worked example:

## 1:05 (Triple only) know what is meant by the term solubility in the units g per 100g of solvent

Solubility is defined in terms of the maximum mass of a solute that dissolves in 100g of solvent. The mass depends on the temperature.

For example, the solubility of sodium chloride (NaCl) in water at 25⁰C is about 36g per 100g of water.

## 1:06 (Triple only) understand how to plot and interpret solubility curves

The solubility of solids changes as temperature changes. This can be plotted on a solubility curve.

The salts shown on this graph are typical: the solubility increases as temperature increases.

For example, the graph above shows that in 100g of water at 50⁰C the maximum mass of potassium nitrate (KNO₃) which will dissolve is 80g.

However, if the temperature were 80⁰C a mass of 160g of potassium nitrate (KNO₃) would dissolve in 100g of water.

## 1:07 (Triple only) practical: investigate the solubility of a solid in water at a specific temperature

At a chosen temperature (e.g. 40⁰C) a saturated solution is created of potassium nitrate (KNO₃) for example.

Some of this solution (not any residual solid) is poured off and weighed. The water is then evaporated from this solution to leave a residue of potassium nitrate which is then weighed.

The difference between the two measured masses is the mass of evaporated water.

The solubility, in grams per 100g of water, is equal to 100 times the mass of potassium nitrate residue divided by the mass of evaporated water.

## 1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

Concentration is a measurement of the amount of substance per unit volume.

In Chemistry, concentration is measured in mol/dm3 (read as moles per cubic decimetre).

The following formula allows for the interconversion between a concentration (in mol/dm3), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm3).

or

Note: 1 dm3 = 1000 cm3

For example, to convert 250 cm3 into dm3:

Example 1:     0.03 mol of sodium carbonate (Na2CO3) is dissolved in 300 cm3 of water. Calculate the concentration of the solution.

Step 1:  Convert the volume of water from cm3 into dm3.

Step 2: Use the molar concentration formula to calculate of the concentration of the solution.

Example 2:     Calculate the amount, in moles, of 25cm3 of hydrochloric acid (HCl) with a concentration of 2 mol/dm3.

Step 1: Convert the volume of HCl from cm3 into dm3.

Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.

Titration calculations, example 1

The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.

For example, a titration problem will look like this:

A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution (NaOH) required 23.50 cm3 of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm3 of the acid.

The chemical equation for this reaction is:

NaOH(aq)               +              HCl(aq)      –>    NaCl(aq)     +     H2O(l)

volume                    25.0cm3                                  23.50cm3

concentration         0.100mol/dm3

It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.

Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).

First convert the volume of NaOH from cm3 into dm3.

Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.

Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).

From the chemical equation the ratio of NaOH to HCl is 1:1

Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.

Step 3: Calculate the concentration of the hydrochloric acid (HCl).

Convert the volume of HCl from cm3 into dm3.

Use the molar concentration formula to calculate of the concentration of the HCl.

Titration calculations, example 2

25.0 cm3 of sodium carbonate solution (Na2CO3) of unknown concentration was neutralised by 30.0 cm3 of 0.100 mol/dm3 nitric acid (HNO3).

Na2CO3(aq)      +      2HNO3(aq)      –>      2NaNO3(aq)      +      CO2(g)      +      H2O(l)

Calculate the concentration, in mol/dm3 of the sodium carbonate solution.

Step 1: Calculate the amount, in moles, of nitric acid (HNO3).

Step 2: Deduce the amount, in moles of sodium carbonate (Na2CO3).

Using the equation:

Na2CO3(aq)      +      2HNO3(aq)      –>      2NaNO3(aq)      +      CO2(g)      +      H2O(l)

Ratio Na2CO3:HNO3 = 1:2

Step 3: Calculate the concentration of sodium carbonate (Na2CO3).

Converting mol/dm3 into g/dm3

Concentration can also be expressed in g/dm3 (grams per cubic decimetre).

Therefore mol/dm3 can be converted into g/dm3.

Example:     Convert 0.06 mol/dm3 of sodium carbonate (Na2CO3) into g/dm3

Step 1: calculate the relative formula mass (Mr) of sodium carbonate (Na2CO3).

Step 2: Recall the formula giving the relationship between mass, amount and formula mass.

therefore

## 1:35 (Triple only) understand how to carry out calculations involving gas volumes and the molar volume of a gas (24dm³ and 24,000cm³ at room temperature and pressure (rtp))

The molar volume of a gas is the volume that one mole of any gas will occupy.

1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm3 or 24,000 cm3.

Note: 1 dm3 = 1000 cm3

The following formulae allows for the interconversion between a volume in dm3 or cm3 and a number of moles for a given gas:

or

Example 1:

Calculate the amount, in moles, of 12 dm3 of carbon dioxide (CO2).

Example 2:

Calculate the volume at rtp in cubic centimetres (cm3), of 3 mol of oxygen, (O2).

## 1:52 (Triple only) know how to represent a metallic lattice by a 2-D diagram

When metal atoms join together the outer electrons become ‘delocalised’ which means they are free to move throughout the whole structure.

Metals have a giant regular arrangement of layers of positive ions surrounded by a sea of delocalised electrons.

## 1:53 (Triple only) understand metallic bonding in terms of electrostatic attractions

Metallic bonding is the strong electrostatic attraction between positive metal ions and a sea of delocalised electrons.

## 1:54 (Triple only) explain typical physical properties of metals, including electrical conductivity and malleability

Metals are good conductors because they have delocalised electrons which are free to move.

Metals are malleable (can be hammered into shape) because they have layers of ions that can slide over each other.

## 1:55 (Triple only) understand why covalent compounds do not conduct electricity

Electrical conductivity is the movement of charged particles.

In this case, charged particles means either delocalised electrons or ions.

These particles need to be free to move in a substance for that substance to be conductive.

Covalent compounds do not conduct electricity because there are no charged particles that are free to move.

## 1:56 (Triple only) understand why ionic compounds conduct electricity only when molten or in aqueous solution

Ionic compounds only conduct electricity only when molten or in solution.

When solid the ions are not free to move.

When molten or in solution the ions are free to move.

## 1:57 (Triple only) know that anion and cation are terms used to refer to negative and positive ions respectively

A negative ion is called an anion. Examples are the bromide ion (Br⁻) and the oxide ion (O²⁻).

A positive ion is called a cation. Examples are the sodium ion (Na⁺) and the aluminium ion (Al³⁺).

A trick to remember this is to write the ‘t’ as a ‘+’ in the word cation: ca+ion

## 1:58 (Triple only) describe experiments to investigate electrolysis, using inert electrodes, of molten compounds (including lead(II) bromide) and aqueous solutions (including sodium chloride, dilute sulfuric acid and copper(II) sulfate) and to predict the products

Electrolysis: The breaking down of a substance caused by passing an electric current through an ionic compound which is molten or in solution. New substances are formed.

ELECTROLYSIS OF MOLTEN IONIC COMPOUNDS

Example: The electrolysis of molten lead bromide (PbBr2)

• Solid lead bromide is heated and becomes molten. Explanation: ions become free to move.

• Electrodes attached to a power source are placed in the molten lead bromide. Explanation: these electrodes are made of either graphite or platinum because both conduct electricity and are fairly unreactive.
• From the diagram, the left-hand electrode becomes positively charged, this is called the anode. The right-hand becomes negatively charged, this is called the cathode. Explanation: delocalised electrons flow from the anode to the cathode.
• At the anode a brown gas is given off. This is bromine gas (Br2(g)). Explanation: Negatively charged bromide ions are attracted to the anode (positive electrode). At the anode, bromide ions lose electrons (oxidation) and become bromine molecules.
• At the cathode a shiny substance is formed. This is molten lead (Pb(l) ). Explanation: Positively charged lead ions are attracted to the cathode (negative electrode). At the cathode, lead ions gain electrons (reduction) and become lead atoms.

Overall Reaction

chemical equation:              PbBr2(l)              –>   Pb(l) +   Br2(g)

Remember

OILRIG : Oxidation Is the Loss of electrons and Reduction Is the Gain of electrons

PANCAKE : Positive Anode, Negative Cathode

ELECTROLYSIS OF IONIC SOLUTIONS

Rules for working out elements formed from electrolysis of solutions

Follow these rules to decide which ions in solution will react at the electrodes:

At the cathode

Metal ions and hydrogen ions are positively charged. Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series:

• The metal will be produced if it is less reactive than hydrogen
• Hydrogen gas (H2) will be produced if the metal is more reactive than hydrogen

At the anode

At the anode, the product of electrolysis is always oxygen gas (O2) unless the solution contains a high concentration of Cl, Br­- or I ions, in which case a halogen is produced, e.g. chlorine gas (Cl2), bromine gas (Br2), and iodine gas (I2).

The electrolysis of sodium chloride solution (NaCl(aq))

• Solid sodium chloride is dissolved in water. Explanation: The sodium ions and chloride ions become free to move.

• The solution also contains hydrogen ions (H+) and hydroxide ions (OH). Explanation: Water is a very weak electrolyte. It ionises very slightly to give hydrogen ions and hydroxide ions:

H2O(l) ⇋ H+(aq) + OH(aq)

• Chloride ions (Cl) and hydroxide ions (OH) are attracted to the anode.
• Sodium ions (Na+) and hydrogen ions (H+) are attracted to the cathode.
• At the anode a green gas is given off. This is chlorine gas (Cl2(g)). Explanation: chloride ions lose electrons (oxidation) and form molecules of chlorine. The chloride ions react at the anode instead of the hydroxide ions because the chloride ions are in higher concentration. The amount of chlorine gas produced might be lower than expected because chlorine is slightly soluble in water.

Electron half equation:   2Cl(aq)    –>  Cl2 (g) + 2e

• At the cathode a colourless gas is given off. This is hydrogen gas (H2(g)). Explanation: hydrogen ions gain electrons (reduction) and form molecules of hydrogen. The hydrogen ions react at the cathode because hydrogen is below sodium in the reactivity series.

Electron half equation:   2H+(aq) + 2e  –>  H2 (g)

• The solution at the end is sodium hydroxide (NaOH(aq)).

The electrolysis of copper sulfate solution (CuSO4(aq))

• Copper sulfate solution is composed of copper ions (Cu2+), sulfate ions (SO42-), hydrogen ions (H+) and hydroxide ions (OH).
• At the cathode a brown layer is formed. This is copper. Explanation: copper ions gain electrons (reduction) and form atoms of copper. The copper ions react at the cathode instead of hydrogen ions because copper is below hydrogen in the reactivity series.

Electron half-equation:   Cu2+(aq) + 2e  –>  Cu (s)

• At the anode, bubbles of gas are given off. This is oxygen gas (O2(g)). Explanation: hydroxide ions lose electrons (oxidation) and form molecules of oxygen and water. The hydroxide ions react at the anode instead of the sulfate ions because the hydroxide ions are less stable.

Electron half-equation:   4OH(aq)     –>  O2 (g) + 2H2O(l) + 4e

The electrolysis of sulfuric acid (H2SO4(aq))

• Sulfuric acid is composed of sulfate ions (SO42-), hydrogen ions (H+) and hydroxide ions (OH).
• At the cathode bubbles of gas are formed. This is hydrogen gas (H2(g)). Explanation: hydrogen ions gain electrons (reduction) and form molecules of hydrogen.

Electron half-equation:   2H+(aq) + 2e  –>  H2(g)

• At the anode, bubbles of gas are given off. This is oxygen gas (O2(g)). Explanation: hydroxide ions lose electrons (oxidation) and form molecules of oxygen and water. The hydroxide ions react at the anode instead of the sulfate ions because the hydroxide ions are less stable.

Electron half-equation:   4OH(aq)     –>  O2 (g) + 2H2O(l) + 4e

• Twice the volume of hydrogen gas is produce compared to oxygen gas. Explanation: from the two half equations, O2 needs 4ebut H2 only needs 2e– as can be seen from the equation

2H2O(l)  –>  2H2(g)  +  O2(g)

There are twice the amount (in moles) of H2 compared to O2

## 1:59 (Triple only) write ionic half-equations representing the reactions at the electrodes during electrolysis and understand why these reactions are classified as oxidation or reduction

Oxidation: the loss of electrons or the gain of oxygen

Reduction: the gain of electrons or the loss of oxygen

Example: The electrolysis of lead (II) bromide, PbBr2

At the cathode (negative electrode):   Pb2+ (l) + 2e  →  Pb (l)         (reduction)

At the anode (positive electrode):       2Br(l)      →  Br2 (g) + 2e       (oxidation)

Example: The electrolysis of aluminium oxide, Al2O3

At the cathode:   Al3+ + 3e    →    Al         (reduction)

At the anode:      2O2-    →    O2 + 4e       (oxidation)

Example: The electrolysis of sodium chloride solution (NaCl (aq))

At the cathode:   2H(aq) + 2e  →  H2 (g)       (reduction)

At the anode:      2Cl– (aq)    →  Cl2 (g) + 2e       (oxidation)

Example: The electrolysis of copper sulfate solution (CuSO(aq))

At the cathode:   Cu2+ (aq) + 2e  →  Cu (s)      (reduction)

At the anode:      4OH– (aq)     →  O2 (g) + 2H2O (l) + 4e       (oxidation)

## 1:60 (Triple only) practical: investigate the electrolysis of aqueous solutions

The diagram shows an electrolytic cell.

The electrolyte is an aqueous solution. For example it might be concentrated sodium chloride, NaCl (aq).

The test tubes over the electrodes must not completely cover them to make sure the ions are free to move throughout the solution.

In the case of NaCl (aq) bubbles of gas will be seen forming at the electrodes. These float up and collect in the test tubes when each gas can be tested to assess its identity.

## 2:04 (Triple only) explain the trend in reactivity in Group 1 in terms of electronic configurations

As you go down the group the outer electron lost from the group 1 metal is further from the nucleus therefore the electron is less attracted by the nucleus and therefore more easily lost.

## 2:08 (Triple only) explain the trend in reactivity in Group 7 in terms of electronic configurations

The higher up we go in group 7 (halogens) of the periodic table, the more reactive the element. The explanation concerns how readily these elements form ions, by attracting a passing electron to fill the outer shell.

In fluorine the outer electron shell is very close to the positively charged nucleus, so the attraction between this nucleus and the negatively charged electrons is very strong. This means fluorine is very reactive indeed.

However, for iodine the outer electron shell is much further from the nucleus so the attraction is weaker. This means iodine is less reactive.

## 2:22 (Triple only) know that most metals are extracted from ores found in the Earth’s crust and that unreactive metals are often found as the uncombined element

Most metals are found in the Earth’s crust combined with other elements. Such compounds are found in rocks called ore, rocks from which it is worthwhile to extract a metal.

A few very unreactive metals, such as gold, are found native which means they are found in the Earth’s crust as the uncombined element.

## 2:23 (Triple only) explain how the method of extraction of a metal is related to its position in the reactivity series, illustrated by carbon extraction for iron and electrolysis for aluminium

Extraction of a metal from its ore typically involves removing oxygen from metal oxides.

If the ore contains a metal which is below carbon in the reactivity series then the metal is extracted by reaction with carbon in a displacement reaction.

If the ore contains a metal which is above carbon in the reactivity series then electrolysis (or reaction with a more reactive metal) is used to extract the metal.

## 2:24 (Triple only) be able to comment on a metal extraction process, given appropriate information

Extraction of a metal from its ore typically involves removing oxygen from metal oxides.

If the ore contains a metal which is below carbon in the reactivity series then the metal is extracted by reaction with carbon in a displacement reaction.

If the ore contains a metal which is above carbon in the reactivity series then electrolysis (or reaction with a more reactive metal) is used to extract the metal.

## 2:25 (Triple only) explain the uses of aluminium, copper, iron and steel in terms of their properties the types of steel will be limited to low-carbon (mild), high-carbon and stainless

Aluminium
UseProperty
Aircrafts and cansLow density / resists corrosion
Power cablesConducts electricity / ductile
Pots and pansLow density / strong (when alloyed) / good conductor of electricity and heat

Aluminium resists corrosion because it has a very thin, but very strong, layer of aluminium oxide on the surface.

Copper
UseProperty
Electrical wiresvery good conductor of electricity and ductile
Pots and pansvery good conductor of heat / very unreactive / malleable
Water pipesunreactive / malleable
Surfaces in hospitalsantimicrobial properties / malleable
Iron
UseProperty
BuildingsStrong
SaucepansConducts heat / high melting point / malleable
Steel
Type of steelIron mixed withSome uses
Mild steelup to 0.25% carbonnails, car bodies, ship building, girders
High-carbon steel0.6%-1.2% carboncutting tools, masonry nails
Stainless steelChromium (and nickel)cutlery, cooking utensils, kitchen sinks

Mild steel is a strong material that can easily be hammered into various shapes (malleable). It rusts easily.

High-carbon steel is harder than mild steel but more brittle (not as malleable).

Stainless steel forms a strong, protective oxide layer so is very resistant to corrosion.

## 2:26 (Triple only) know that an alloy is a mixture of a metal and one or more elements, usually other metals or carbon

An alloy is a mixture  of a metal with, usually, other metals or carbon.

For example, brass is a alloy of copper and zinc, and steel is an alloy of iron and carbon.

## 2:27 (Triple only) explain why alloys are harder than pure metals

Alloys are harder than the individual pure metals from which they are made.

In an alloy, the different elements have slightly different sized atoms. This breaks up the regular lattice arrangement and makes it more difficult for layers of ions to slide over each other.

## 2:33 (Triple only) describe how to carry out an acid-alkali titration

Titration is used to find out precisely how much acid neutralises a certain volume of alkali (or vice versa).

The diagram shows the titration method for a neutralisation reaction between hydrochloric acid and sodium hydroxide, using phenolphthalein as an indicator. The indicator changes colour when neutralisation occurs.

The conical flask is swirled to mix the solutions each time alkali is added. When reading the burette it is important to be aware that the numbers on the scale increase from top to bottom. Readings are usually recorded to the nearest 0.05cm³ so all readings should be written down with 2 decimal places. The second decimal place is given as a ‘0’ if the level of the solution is on a line, or ‘5’ if it is between the lines. The volume of alkali added is calculated by subtracting the final reading from the initial reading. Various indicators can be used such as phenolphthalein or methyl orange. However universal indicator should not be used since it has a wide range of colours rather than one specific colour change so it would be unclear when the precise endpoint of titration was achieved.

This process is repeated a number of times. The first time it is done roughly to get a good approximation of how much alkali needs to be added. On subsequent attempts, the alkali is added very slowly when approaching the correct volume.

## 2:40 (Triple only) describe an experiment to prepare a pure, dry sample of a soluble salt, starting from an acid and alkali

Titration Method:

Preparing pure dry crystals of sodium chloride (NaCl) from hydrochloric acid (HCl) and sodium hydroxide (NaOH)

Before the salt preparation is carried out using the below method, the volume of acid that exactly reacts with 25cm3 of the alkali is found by titration using methyl orange indicator.

StepExplanation
Pipette 25cm3 of alkali (NaOH) into a conical flaskAccurately measures the alkali (NaOH)
Do not add indicatorPrevents contamination of the pure crystals with indicator
Using the titration values, titrate the known volume acid (HCl) into conical flask containing alkaliExactly neutralises all of the alkali (NaOH)
Transfer to an evaporating basin & heat the solutionForms a hot saturated solution (NaCl(aq))
Allow the solution to cool so that hydrated crystals formSodium chloride is less soluble in cold water
Remove the crystals by filtration and wash with distilled waterRemoves any impurities
Dry by leaving in a warm placeEvaporates the water

(Note – This process could be reversed with the acid in the pipette and the alkali in the burette)

How to select the right method for preparing a salt:

## 2:41 (Triple only) describe an experiment to prepare a pure, dry sample of an insoluble salt, starting from two soluble reactants

Precipitation Method:

Preparing pure dry crystals of silver chloride (AgCl) from silver nitrate solution (AgNO3) and potassium chloride solution (KCl)

StepExplanation
Mix the two salt solutions together in a beakerForms a precipitate of an insoluble salt (AgCl)
Stir with glass rodMake sure all reactants have reacted
Filter using filter paper and funnelCollect the precipitate (AgCl)
Wash with distilled waterRemoves any the other soluble salts (KNO3)
Dry by leaving in a warm placeEvaporates the water

## 2:43 (Triple only) practical: prepare a sample of pure, dry lead(II) sulfate

Objective: prepare a pure, dry sample of lead (II) sulfate (PbSO₄).

Preparing a pure, dry sample of lead (II) sulfate (PbSO₄) from lead (II) nitrate solution (Pb(NO₃)₂) and sodium sulfate solution (Na₂SO₄).

Pb(NO₃)₂ (aq)      +      Na₂SO₄ (aq)      →        PbSO₄ (s)      +      2NaNO₃ (aq)

1. Mix similar volumes lead nitrate solution and sodium sulfate solution in a beaker. The precise volumes do not matter since any excess will be removed later.
2. A white precipitate of lead (II) sulfate will form.
3. The reaction mixture is filtered.
4. The residue left on the filter paper is washed with distilled water several times to remove impurities.
5. The residue is then moved to a warm oven to dry.

## 3:05 (Triple only) draw and explain energy level diagrams to represent exothermic and endothermic reactions

The symbol ΔH is used to represent the change in heat (or enthalpy change) of a reaction.

ΔH is measured in kJ/mol (kilojoules per mole).

The change in heat (enthalpy change) can be represented on an energy level diagram. ΔH must also labelled.

In an exothermic reaction, the reactants have more energy than the products.

Energy is given out in the form of heat which warms the surroundings.

ΔH is given a negative sign, because the reactants are losing energy as heat, e.g  ΔH = -211 kJ/mol.

In an endothermic reaction, the reactants have less energy than the products.

Energy is taken in which cools the surroundings.

ΔH is given a positive sign, because the reactants are gaining energy, e.g  ΔH = +211 kJ/mol.

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Section 1: Principles of chemistry

a) States of matter

b) Atoms

c) Atomic structure

d) Relative formula masses and molar volumes of gases

e) Chemical formulae and chemical equations

f) Ionic compounds

g) Covalent substances

h) Metallic crystals

i) Electrolysis

Section 2: Chemistry of the elements

a) The Periodic Table

b) Group 1 elements: lithium, sodium and potassium

c) Group 7 elements: chlorine, bromine and iodine

d) Oxygen and oxides

e) Hydrogen and water

f) Reactivity series

g) Tests for ions and gases

Section 3: Organic chemistry

a) Introduction

b) Alkanes

c) Alkenes

d) Ethanol

Section 4: Physical chemistry

a) Acids, alkalis and salts

b) Energetics

c) Rates of reaction

d) Equilibria

Section 5: Chemistry in industry

a) Extraction and uses of metals

b) Crude oil

c) Synthetic polymers

d) The industrial manufacture of chemicals

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