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To investigate the effect of concentration on the rate of reaction:

1. Apparatus: Conical flask, measuring cylinders (for measuring volumes of liquid), white paper marked with a black cross, stopwatch (for measuring time taken for reaction).

2. Method:
- Place the conical flask on the paper marked with a cross.
- Add a measured volume (e.g., 50 cm³) of sodium thiosulfate solution to the flask.
- Add a measured volume (e.g., 10 cm³) of dilute hydrochloric acid and start the stopwatch immediately.
- Swirl the flask once and look vertically down through the neck of the flask. Stop the stopwatch when the cross is completely obscured by the yellow precipitate of sulfur.
- Record the time taken (\(t\)).
- Repeat the experiment using different concentrations of sodium thiosulfate. This is achieved by mixing different ratios of sodium thiosulfate and distilled water (e.g., 40 cm³ thiosulfate + 10 cm³ water, 30 cm³ thiosulfate + 20 cm³ water, etc.) ensuring the total volume of the mixture remains constant.

3. Variables to control: Temperature of the reactants, total volume of the reaction mixture, and the concentration/volume of hydrochloric acid used.

4. Analyzing results: The rate is inversely proportional to the time taken (\(\text{Rate} \propto 1/t\)). Compare the times or calculate \(1/t\) for each concentration. A shorter time (higher \(1/t\)) indicates a faster rate of reaction.

(a) Pencil lead (graphite) is insoluble in chromatography solvents (such as water or ethanol), so it will not dissolve, run, or interfere with the chromatogram, whereas ink is soluble and would separate along with the samples.

(b) If the solvent level is above the start line, the spots of dyes will dissolve directly into the solvent in the bottom of the beaker instead of traveling up the paper.

(c) Calculation of \(R_f\) value:
\(R_f = \frac{\text{Distance traveled by substance}}{\text{Distance traveled by solvent front}}\)
\(R_f = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\)

(d) Placing a lid on the beaker prevents the solvent from evaporating and ensures that the atmosphere inside the beaker is saturated with solvent vapor, which helps the solvent travel up the paper evenly.

(e) It is a pure substance / single dye because it did not separate into multiple spots.

(f) (i) A locating agent (or ninhydrin) is sprayed on the paper.
(ii) Spraying should be done in a fume cupboard because the locating agent may be toxic/harmful/flammable, or safety goggles/gloves should be worn.
(iii) Calculate the \(R_f\) values of the spots and compare them to standard/known reference \(R_f\) values of amino acids (or run known samples of amino acids alongside on the same chromatogram and match their heights).

(a) (i) Volume of acid added = Final reading - Initial reading = \(27.0\text{ cm}^3 - 1.2\text{ cm}^3 = 25.8\text{ cm}^3\).
(ii) Swirling ensures thorough mixing of the acid and alkali so that the reaction is complete and the end-point is detected accurately.
(iii) The white tile provides a neutral background, making the color change of the indicator much easier to see clearly.

(b) In alkaline solution (sodium hydroxide), methyl orange is yellow. At the neutralisation end-point, it changes to orange (or pink/red if excess acid is added). Thus, the color change is from yellow to orange (or pink/red).

(c) (i) The first titration volume (\(25.8\text{ cm}^3\)) is anomalous because it is not within \(\pm 0.2\text{ cm}^3\) of the other results (concordant results are \(24.1\), \(24.5\), and \(24.3\text{ cm}^3\)).
(ii) Average of concordant results:
\(\text{Average} = \frac{24.1 + 24.5 + 24.3}{3} = \frac{72.9}{3} = 24.3\text{ cm}^3\).

(d) To obtain pure sodium chloride crystals:
- Repeat the titration using the same volumes (\(25.0\text{ cm}^3\) of \(\text{NaOH}\) and \(24.3\text{ cm}^3\) of \(\text{HCl}\)) but without adding any indicator.
- Heat the resulting sodium chloride solution to evaporate most of the water until the crystallization point is reached.
- Allow the hot, saturated solution to cool slowly so crystals form, then filter and dry the crystals with filter paper.

(a) (i) The gas that turns limewater cloudy is carbon dioxide (\(\text{CO}_2\)).
(ii) The presence of carbon dioxide indicates that the anion in Compound E is carbonate (\(\text{CO}_3^{2-}\)).

(b) The absence of a precipitate when silver nitrate is added means that halide ions (chloride, bromide, and iodide) are absent.

(c) (i) Iron(II) ions (\(\text{Fe}^{2+}\)) form a green precipitate with sodium hydroxide which is insoluble in excess. (Note: Chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to give a green solution, so E must contain \(\text{Fe}^{2+}\)).
(ii) The green precipitate is iron(II) hydroxide, \(\text{Fe(OH)}_2\). It is a basic oxide/hydroxide and is not amphoteric, so it does not react with or dissolve in excess sodium hydroxide (unlike amphoteric hydroxides like zinc or chromium hydroxide).

(d) (i) With aqueous ammonia, iron(II) ions form a green precipitate that is insoluble in excess ammonia.
(ii) Compound E contains \(\text{Fe}^{2+}\) and \(\text{CO}_3^{2-}\), so its formula is \(\text{FeCO}_3\).

(e) To test for sulfate ions, add dilute hydrochloric acid (or nitric acid) followed by aqueous barium chloride (or barium nitrate). If sulfate ions are absent, no white precipitate is formed (the solution remains clear).