C5

Question 1

Recall the formula of the following substances: • hydrochloric acid and ethanoic acid • carbon dioxide, hydrogen and water • sodium chloride and potassium chloride • ammonia and calcium carbonate.

Answer 1

Hydrochloric Acid : HCL 
Ethanoic Acid: CH3COOH
Carbon Dioxide: CO2 
Hydrogen : H2
Water: H20
Sodium Chloride: NaCl
Potassium Chloride: KCl
Ammonia: NH3 
Calcium Carbonate: CaCO3

Question 2

Recall the formula of the following substances: • sulfuric acid and nitric acid • sodium hydroxide, potassium hydroxide and magnesium carbonate • sodium sulfate, potassium sulfate, magnesium sulfate and barium sulfate • lead(II) nitrate and lead iodide • potassium iodide and potassium nitrate

Answer 2

Sulfuric Acid: H2SO4 
Nitric Acid: HNO3
Sodium Hydroxide: NaOH
Potassium Hydroxide: KOH
Magnesium Carbonate: MgNO3
Sodium Sulfate: Na2SO4
Potassium Sulfate: K2SO4
Magnesium Sulfate: MgSO4
Barium Sulfate: BaSO4
lead(II)nitrate: Pb(NO3)2
Lead iodide: PbI2
Potassium Iodide: KI
Potassium nitrate: KNO3

Question 3

Recall and use the relationship between molar mass, number of moles and mass: • number of moles = mass ÷ molar mass • determine the number of moles of an element from the mass of that element • determine the number of moles of a compound from the mass of that compound • determine the masses of the different elements present in a given number of moles of a compound.

Answer 3

Number of Mass= Mass/Molar Mass

Question 4

Recall that the relative atomic mass of an element is the average mass of an atom of the element compared to the mass of 1/12th of an atom of carbon-12.

Answer 4

The relative atomic mass of an element is the average mass of an atom of that element , compared with the mass of 1/12th of an atom of carbon-12

Question 5

Calculate mass of products and/or reactants using the mole concept from a given balanced equation and the appropriate relative atomic masses.

Answer 5

2Mg + O2 -> 2MgO
224 + 216 -> 2(16+24)
48 + 32 -> 80

Question 6

Recall and use the relationship between molar mass, number of moles and mass: • number of moles = mass ÷ molar mass. Determine the number of moles of an element from the mass of that element.

Answer 6

What mass of water can be made from 6g of Hydrogen Gas.

54g

Question 7

Calculate the percentage by mass of an element in a compound given its formula and the appropriate atomic masses.

Answer 7

Percentage of element= (mass of element/mass of compound)*100

Question 8

Calculate empirical formula of a compound from the: • percentage composition by mass • mass of each element in a sample of the compound.

Answer 8

MgO=>. Mg. O
Each mass. 0.72. 0.48
Each Ar. 24. 16
No. of moles. 0.72/24=0.03. 0.48/16=0.03
Divide by 0.03/0.03= 1. 0.03/0.03=1
Smallest

                                    1:1
                                   MgO

Question 9

Convert volume in cm3 into dm3 or vice versa.

Answer 9

To go from cm^3 to dm^3 , divide by 1000

To go from dm^3 to cm^3, multiply by 1000

Question 10

Recall and use the relationship between the amount in moles, concentration in mol/dm3 and volume in dm3: • amount in moles = concentration × volume • concentration = amount in moles ÷ volume • volume = amount in moles ÷ concentration.

Answer 10

Moles = concentration * volume

Question 11

Interpret information on food packaging about guideline daily amounts (GDA) for example: • percentage of GDA in a portion.

Answer 11

A portion of food containing 5g of saturated fat will provide 25% of an adults GDA for saturated fat.

Question 12

Interpret more complex food packaging information and its limitations for example: • convert amounts of sodium to amounts of salt. Explain why the above conversion may be inaccurate, to include sodium ions coming from other sources.

Answer 12

Percentage mass of sodium in NaCl is 39.3%. So 1g of salt is the same as 1*39.3/100=0.393g of sodium. However sodium ions can come from other sources in food, so this conversion may be inaccurate. For eg. Monosodiumglutamate, commonly used to enhance the flavours of food, contains sodium ions.

Question 13

Explain the need for several consistent titre readings in titrations.

Answer 13

Several consistent titre readings are needed to ensure that the titre is accurate.

Question 14

Describe the difference in colour change during a titration using a single indicator, such as litmus or phenolphthalein, compared to a mixed indicator, such as universal.

Answer 14

Phenolphthalein is pink in alkaline solutions and colourless in acidic solutions. Litmus is blue in alkaline solutions and red in acidic solutions. A mixed indicator changes colour over a different range of pH values, it gives a continuous change and gives an approximation.

Question 15

Explain why an acid-base titration should use a single indicator rather than a mixed indicator.

Answer 15

A mixed indicator gives an approximate pH value, whereas a single indicator gives a sudden colour change between pink and colourless at the end point.

Question 16

Describe an experimental method to measure the volume of gas produced in a reaction given appropriate details about the reaction.

Answer 16

A gas syringe is a gas tight glass syringe used for measuring the volume of gas. Gas syringes can be used to measure any volume of gas. The syringe is connected to the reaction container by tubing. As the reaction occurs , gas fills the gas syringe and pushes the plunger out.

Question 17

Describe an experimental method to measure the mass of gas produced in a reaction given appropriate details about the reaction.

Answer 17

A weighing balance can be used to measure the gas (loss of mass due to gas escaping) a reaction container is placed on the balance, the reaction takes place and the mass lost is the mass of the gas.

Question 18

Recall that in a reversible reaction at equilibrium: • the rate of the forward reaction equals the rate of the backward reaction • the concentrations of the reactants and the products do not change. Understand how the position of equilibrium is related to the ratio of the concentration of the products to the concentration of the reactants.

Answer 18

When a reversible reaction reaches equilibrium, the rate of the forward reaction will be equal to the rate of the backward reaction. The concentration of the products do not change. The position of the equilibrium is in the left if the concentration of the reactants is greater than the concentration of the products. It is on the right if the concentration of products is greater than the concentration of reactants.

Question 19

Explain why a reversible reaction may reach an equilibrium: • importance of a closed system • initially rate of forward reaction decreases • initially rate of backward reaction increases • eventually rate of forward equals rate of backward reaction.

Answer 19

Reversible reaction starting off with just the substances on the left of the equilibrium, these will react quickly because they are at a high concentration. The rate of the forward reaction will then decrease as these substances are used up. However, as they are used up, the concentration on the right will increase. They will react increasingly quick as their concentration increases. Eventually the rate of forward reaction will equal the rate of the backward reaction. Equilibrium has been reached. This works in a closed system, such as stoppered flask or a beaker of liquid in which all of these reactants stay in the solution.

Question 20

Understand in simple qualitative terms factors that affect the position of equilibrium: • removing a product moves the position of equilibrium to the right or vice versa • adding extra reactant moves the position of equilibrium to the right or vice versa • increasing the temperature moves the position of equilibrium in the direction of the endothermic reaction or vice versa • increasing the pressure moves the position of equilibrium to the side with the least number of moles of gas molecules or vice versa. Explain the effect of changing product concentration, reactant concentration, temperature or pressure on the position of equilibrium given appropriate details about a reaction.

Answer 20

If the pressure is increased: the position of the equilibrium moves to the side with the lower number of moles of gas molecules.

If the concentration of the substance is increased: the position of equilibrium moves to the opposite side.

If the concentration of the substances decreases: the position of the equilibrium moves to other side.

If the temperature is increase: the position of the equilibrium moves in the direction of the endothermic reaction.

Removing a product moves the position of the equilibrium to the right or vice versa. Adding on extra reactants moves the position of equilibrium it the right or vice versa, the change in position of the equilibrium will affect the composition of the minute of substance

Question 21

Understand that the reaction between sulfur dioxide and oxygen is reversible: • sulfur dioxide + oxygen sulfur trioxide • 2SO2(g) + O2(g) 2SO3(g) Describe the conditions used in the Contact Process: • V2O5 catalyst • around 450°C • atmospheric pressure.

Answer 21

2SO2 + O2 ->. 2SO3

Question 22

Explain the conditions used in the Contact Process: • increasing the temperature moves the position of equilibrium to the left and increases rate of reaction so a compromise temperature is used • addition of catalyst increases rate but does not change position of equilibrium • even at low pressure, the position of equilibrium is already on right so expensive high pressure is not needed.

Answer 22

An optimum temperature of around 450 degrees is used. It is high enough to give a reasonable rate of reaction without the yield decreasing too much. At atmospheric pressure, the position of the equilibrium moves to the right. A high pressure would increase the yield however the extra cost involved would not be worth it as it will be a small increase to the yield. The vandium Oxide catalyst does not change the position of the equilibrium, but it increase the rate of reaction.

Question 23

Understand that an acid ionises in water to produce H+ ions. Understand that a strong acid completely ionises in water and a weak acid does not fully ionise and forms an equilibrium mixture.

Answer 23

Acid ionise in water to produce hydrogen ions, H+. Strong Acid such as HCl completely ionises in solution. Weak acids such as ethanoic Acid only partially ionises in water. Many of their molecule do not release hydrogen ions. The rest ion is reversible and an equilibrium mixture is produced.

Question 24

Explain why the pH of a weak acid is much higher than the pH of a strong acid of the same concentration. Explain the difference between acid strength and acid concentration: • acid strength (strong or weak) is a measure of the degree of ionisation of the acid • acid concentration is a measure of the number of moles of acid in one dm3

Answer 24

The concentration or an Acid is a measure of the number of moles of Acid in 1dm^3. The more moles of Acid dissolved in the same volume of Acid, the higher the concentration. The stronger the Acid, the more ionise (lower pH)

Question 25

Explain why ethanoic acid reacts slower than hydrochloric acid of the same concentration: • ethanoic acid has a lower concentration of hydrogen ions • in ethanoic acid the hydrogen ions have a lower collision frequency with reactant particles.

Answer 25

The concentration of hydrogen ions is greater in HCl that it is in ethanoic Acid. Ethanoic Acid has a lower concentration of hydrogen ions(higher pH), meaning that the hydrogen ions will have a lower collision frequency rate with react particles.

Question 26

Explain why ethanoic acid is less conductive than hydrochloric acid of the same concentration: • lower concentration of hydrogen ions to carry the charge in ethanoic acid.

Answer 26

Ethanoic is less conductive because it is a weak Acid, there is a lower concentration of H+ ions to carry the charge through ethanoic acid.

Question 27

Explain in terms of collisions between ions, why most precipitation reactions are extremely fast.

Answer 27

The collision frequency between ions in solution is very large, there is a high chance that different ions will collide with each other and cause a reaction, so precipitation reaction are extremely fast.

Question 28

Construct word equations for simple precipitation reactions e.g. for the reaction between solutions of barium chloride and sodium sulfate (products not given)

Answer 28

Lead nitrate + potassium iodide -> potassium nitrate + lead iodide

Barium Chloride + sodium Sulfate -> sodium chloride + barium Chloride.

Question 29

Construct ionic equations, with state symbols, for simple precipitation reactions, given the formulae of the ions that react. Explain the concept of ‘spectator ions’

Answer 29

Ba2+(aq) + SO4 2-(aq) -> BaSO4(s)
Pb2+(aq) + 2Cl-(aq) -> PbCl2(s)

The other ions, Na+ and NO3- are called spectator ions. They do not take part in the reaction, but instead form the other product, aqueous sodium nitrate.

Question 30

Describe the stages involved in the preparation of a dry sample of an insoluble compound by precipitation given the names of the reactants: • mix solutions of reactants • filtration • wash and dry residue

Answer 30

1. A suitable combination of solutions are mixed together to form a precipitate of the required insoluble compound.
2. The mixture is filtered to separate the precipitate from the other reactants and products
3. The precipitate is washed with water while it is on the filter paper. It is then dried (in a warm oven)