Chem Past Paper

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Question 19 H6, H7

Identifies neutron bombardment, bombardment with nuclei, the apparatus used and the reasons for this, and gives an example of each.

3

Outlines neutron bombardment, and gives an example

OR

Outlines bombardment with other nuclei, and gives an example

OR

Outlines neutron bombardment and bombardment with nuclei, without correct examples.

2

Identifies neutron bombardment, or bombardment with nuclei.

1

There are two general methods for producing transuranic and commercial radioisotopes: bombardment of target nuclei with neutrons, and with other nuclei. Because neutrons are uncharged, there is no repulsive force to be overcome between them and the target nucleus, hence this process can be done in a nuclear reactor. An example is the production of Fe-59 by neutron bombardment ofFe-58. Bombarding target nuclei with other nuclei, for example alpha particles, requires a cyclotron. This is because alpha particles are positively charged, and in order to combine them with the target nuclei, they must be accelerated to high velocity, in order to overcome the repulsive force between them and the target nuclei, both being positive charged. This is done in a particle accelerator (cyclotron). An example of this is the production of P-30 from Al-27 by bombardment with alpha particles.

Question 20 H10, H11, H14

Identifies that using a pH probe is more effective and justifies their choice using characteristics of both acids and both tests.

3

Identifies that using a pH probe is more effective and provides some justification.

2

Identifies that using a pH probe is more effective.

1

Using a pH probe is more effective. This is because both acids are monoprotic and the same concentration. Thus they will both require the same volume to reach an endpoint in a titration. As a result the titration will not be able to distinguish between the two acids. Using a pH probe will determine that although both acids are the same concentration, one will have a higher pH than the other. This is because ethanoic acid is weak whereas HCl is strong. The pH of 0.1M HCl is 1, whereas that of ethanoic acid is greater than 1. The pH probe will be able to distinguish them whereas the titration will not.

Question 21 (a) H12[pic 5]

Identifies the volumetric (or bulb) pipette.

1

volumetric (or bulb) pipette.

Question 21 (b) H9, H10, H14

Correctly calculates the mass (in mg) of ascorbic acid in the 25 mL of orange juice.

2

Answer contains one calculation error.

1

n (iodine) = CxV = 5.00x10-3 x 0.00915 = 4.575x10-5 mol.

n (ascorbic acid) = 4.575x10-5.

m (ascorbic acid) = 4.575x10-5 x MM = 0.008046 g = 8.05 mg.

[pic 6]

Question 21 (c) H13

Identifies that the claim is valid and shows working to support this.

1

m(ascorbic acid) in 100 mL = 100/25 x 8.05 = 32.2 mg.

This is greater than half the recommended daily intake.

Therefore the claim is valid.

Question 22 H7, H8, H13

Identifies THREE conditions which affect the solubility of CO2 gas in water. AND

Explains how a change in both conditions causes the effect in solubility. AND

Supports their answer with a balanced equation.

4-5

Identifies at least TWO conditions which affect the solubility of CO2 gas in water.

Supports their answer with a balanced equation.

OR

Explains how a change in ONE identified condition affects the solubility of CO2

Supports their answer with a balanced equation

2-3

Identifies a condition which affects the solubility of CO2 gas OR

Writes a balanced equation representing the solubility of CO2

1

Carbon dioxide is partially soluble in water. When it is bubbled through water, the following reversible reaction occurs:

CO2 (g) + H2O (l) ↔ H2CO3 (aq) ↔ H+ (aq) + HCO3- (aq) ΔH = -ve

Once sealed, a carbonated beverage represents an equilibrium system as long as the conditions it is placed under are held constant. If conditions such as temperature and pressure are altered, they disturb the equilibrium and this affects the solubility of the CO2 (g).

- Increasing the temperature of the beverage will disturb the equilibrium and favour the reverse, endothermic process. This releases some heat and thereby minimises the disturbance. As a result of this shift the solubility of the gas decreases. Decreasing the temperature on the system does the reverse and increases the solubility of the gas.

- Keeping the system at a higher pressure favours the reaction to the right, because this produces less moles of gas. Thus during production of the beverage, CO2 is pumped into the drink at high pressure and the drink is then sealed. A decrease in pressure (eg by opening the bottle)