Alternative to Practical Skills Worked ExamplesIGCSE Chemistry 0620

(a) A is a conical flask; B is a gas syringe.

(b) The gas syringe collects the carbon dioxide and measures its volume: the plunger is pushed out as gas enters.

(c) Each small division in the enlarged view is $1$ cm³. The front face of the plunger is one division past $35$, so the volume is $36$ cm³.

(d) A stop-watch.

(a) A burette reads downwards. Read the bottom of the meniscus to the nearest $0.1$ cm³ and record every value to one decimal place.

Volume used = final reading minus initial reading, e.g.

(b) Use the concordant results — titrations 1 and 2 agree to within $0.2$ cm³. The rough titration is only a guide and is left out.

(a) Each division is $1$ °C, so read to the nearest half division ($0.5$ °C). The first level is halfway between $19$ and $20$: $19.5$ °C. The second is exactly on $32$: $32.0$ °C (to the same resolution).

(b) Subtract the starting temperature from the highest temperature.

(c) The readings are recorded to different resolutions: $22$ should be written $22.0$ °C, to $0.5$ °C like $34.5$ °C.

(a) Read the small (minutes) dial first, then the large (seconds) dial, and convert to seconds only.

(b) Evaluate the fraction.

(c) The shortest time means the fastest reaction: experiment 1 ($47$ s).

(a) The points lie on a smooth curve that falls steeply at first and then levels off. Draw a single smooth curve through (or close to) all the points — not straight lines joining them, and not a straight line.

(b) Draw a vertical line up from $25$ cm³ to the curve, then a horizontal line across to the time axis. The curve gives about $50$ s.

(c) Water makes up the total to $50$ cm³.

(d) So that the concentration of sodium thiosulfate depends only on the volume of sodium thiosulfate solution used, and the depth of liquid the cross is viewed through is the same each time.

(a) Mean rate = volume of gas ÷ time taken.

(b) Over the first $15$ s:

At the start the zinc and acid are at their highest concentration / amount, so collisions are most frequent. As they are used up, the rate falls, so the average over $45$ s is lower.

(c) Between $60$ s and $75$ s — the volume stays at $48.0$ cm³. The zinc has all reacted (the acid is in excess), so no more gas is made.

(a) Observations are things you see: effervescence (fizzing / bubbles), and the magnesium gets smaller and disappears. (The test-tube also becomes warm.)

(b) Hold a lighted splint at the mouth of the tube: it pops. The gas is hydrogen.

(c) With ethanoic acid there are fewer bubbles / slower fizzing, and the magnesium takes longer to disappear.

Ethanoic acid is a weak acid: it is only partially dissociated, so it has a lower concentration of hydrogen ions than hydrochloric acid of the same concentration.

(a) A burette is more accurate (it reads to $0.1$ cm³) and can deliver different volumes each time.

(b) Polystyrene is a good (thermal) insulator and the lid stops heat escaping from the surface, so less heat is lost to the surroundings and the temperature change measured is closer to the true value.

(c) The temperature change is the thing being measured. A water bath would keep the mixture near $25$ °C, so it would reduce the temperature change and defeat the purpose of the experiment.

(d) A volumetric pipette (with a pipette filler).

(a) The flask and delivery tube were full of air at the start. The first tube collects this air mixed with oxygen, so the gas is not pure oxygen. Later tubes contain almost pure oxygen.

Improvement: discard the first tube of gas before collecting samples.

(b) Keep the reactants apart until the flask is sealed: put the manganese(IV) oxide in a small tube inside the flask (or hang it from the bung), put the bung in, then tip the flask to mix them.

(c) An inverted measuring cylinder filled with water (or a gas syringe connected to the flask).

(a) Methyl orange. It is yellow in the alkaline solution; at the end-point it turns orange (yellow → orange; red if too much acid is added).

(b) Universal indicator changes through many colours gradually, so there is no sharp end-point.

(c) Swirl the flask to mix the acid with the solution. The white tile makes the colour change easier to see.

(d) The volume of acid needed would be larger. Water left in the burette dilutes the acid, so more of it is needed to react with the same amount of sodium carbonate.

(a) Temperature rise:

Substitute into the equation given:

(b) About $16.5$ °C — the same. Doubling both reactants releases twice the energy, but it now has to heat twice the volume of solution, so the temperature rise is unchanged.

(c) Powder has a larger surface area, so the reaction is faster and finishes quickly — less heat is lost to the surroundings before the highest temperature is reached.

Crush one tablet of P and put it in a conical flask with about $25$ cm³ of distilled water; swirl to mix.

Add a few drops of methyl orange indicator.

Fill a burette with the dilute hydrochloric acid and record the initial reading.

Add the acid while swirling, until the indicator turns from yellow to orange (the end-point). Record the final reading and calculate the volume of acid used.

Repeat with one tablet of Q, using the same acid (same concentration) — repeat each to check the results agree.

The tablet that needs the larger volume of acid to reach the end-point neutralises more acid.

(a) Gas A forms at the positive electrode (anode): chlorine. Test: hold damp litmus paper in the gas — it is bleached.

Gas B forms at the negative electrode (cathode): hydrogen. Test: a lighted splint — the gas burns with a pop.

(b) Chlorine is soluble in water, so some of it dissolves in the solution instead of collecting in the tube.

(c) Chlorine is toxic: carry out the experiment in a fume cupboard (or a well-ventilated room).

(d) Sodium hydroxide (aqueous) is left in the solution.

(a) The point at $4.0$ g ($4.2$ °C) is anomalous: the others lie on a straight line through the origin, which predicts $6.0$ °C at $4.0$ g. The decrease is too small — for example, not all the solid had dissolved (not stirred enough) or the lowest temperature was read too early.

(b) Draw a straight line through the origin and the other four points, ignoring the anomaly. The gradient is $1.5$ °C per gram, so extending the line to $6.0$ g:

(c) Endothermic — the temperature decreases, so energy is taken in from the surroundings (the water).

(d) In $50$ cm³, $3.0$ g gives $4.5$ °C. The same energy is now taken from twice the volume of water, so the decrease is halved: