2022 CCEA AS-Level Chemistry 1110 模擬試題連答案詳解

A clean nichrome wire is dipped into concentrated hydrochloric acid, then into a sample of solid Y, and held in the edge of a roaring blue Bunsen flame. The flame colour observed identifies the metal cation present. Solid Y produces a brick-red (orange-red) flame colouration, which is the characteristic flame test result for calcium ions, Ca2+. Answer: brick-red flame.

1 mark: correct observation stated as brick-red or orange-red flame colouration. No mark for stating the ion (Ca2+) without the correct colour, as only the observation is credited here.

Solid Y is calcium bromide, CaBr2. Concentrated sulfuric acid first protonates the bromide ion to give steamy white (misty) fumes of hydrogen bromide gas: \( \text{Br}^- + \text{H}_2\text{SO}_4 \rightarrow \text{HBr} + \text{HSO}_4^- \). Because Br– is a moderately strong reducing agent, concentrated sulfuric acid also partially oxidises some of the HBr produced to orange-brown bromine vapour, being itself reduced to sulfur dioxide: \( 2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \). Answer: steamy white fumes of HBr, together with orange-brown fumes of Br2 (redox by-product).

1 mark: steamy/misty white fumes observed (HBr gas). 1 mark: orange or orange-brown fumes also observed (Br2, formed by partial oxidation of bromide by concentrated H2SO4). Max 2 marks.

Silver ions react with the halide ion present in solid Y to form an insoluble silver halide precipitate: \( \text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr} \). Silver bromide is a cream-coloured precipitate, distinguishable from the white precipitate given by chloride (AgCl) and the yellow precipitate given by iodide (AgI). Answer: a cream precipitate (AgBr) forms.

1 mark: correct observation — a cream precipitate forms. (Do not accept 'white' or 'yellow'.)

Silver bromide is only sparingly soluble in dilute ammonia (unlike silver chloride, which readily dissolves in dilute ammonia to form the soluble complex ion [Ag(NH3)2]+). The cream precipitate therefore remains essentially undissolved when dilute ammonia is added. Answer: the cream precipitate does not dissolve (remains largely insoluble) in dilute ammonia.

1 mark: correct observation — the precipitate does not dissolve (remains insoluble/only very slightly soluble) in dilute ammonia.

Unlike silver iodide (which remains insoluble even in concentrated ammonia), silver bromide dissolves in excess concentrated ammonia, forming the soluble complex ion \( [\text{Ag}(\text{NH}_3)_2]^+ \): \( \text{AgBr} + 2\text{NH}_3 \rightarrow [\text{Ag}(\text{NH}_3)_2]^+ + \text{Br}^- \). The precipitate dissolves to give a colourless solution. Because the precipitate was cream (ruling out chloride, which is white and dissolves even in dilute ammonia) and dissolved in concentrated (but not dilute) ammonia (ruling out iodide, which never dissolves), this confirms the halide ion present is bromide. Answer: precipitate dissolves in concentrated ammonia to give a colourless solution, confirming Br– is present.

1 mark: correct observation — the precipitate dissolves in concentrated ammonia to give a colourless solution. 1 mark: correct conclusion that this confirms the halide is bromide (with reference to the cream precipitate that was insoluble in dilute but soluble in concentrated ammonia). Max 2 marks.

Chlorine is a more powerful oxidising agent than bromine, because oxidising power decreases down Group 7. Chlorine therefore oxidises bromide ions to bromine, itself being reduced to chloride ions: \( \text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2 \). This is observed as the colourless solution turning pale yellow/orange, the colour of aqueous bromine. Answer: solution turns pale yellow/orange (Br2 formed), because Cl2 is a stronger oxidising agent than Br2 and displaces bromide from solution.

1 mark: correct observation — solution turns (pale) yellow/orange. 1 mark: correct explanation that chlorine, being a stronger oxidising agent than bromine (oxidising power decreases down the group), displaces/oxidises bromide to bromine. Max 2 marks.

Hexane is a non-aqueous, non-polar solvent that does not mix with water, so two distinct layers form, with the less dense hexane layer on top. Bromine is more soluble in hexane than in water, so it is extracted preferentially into the upper hexane layer, which is observed to turn orange (a more intense orange than the pale yellow/orange of the aqueous bromine solution), while the lower aqueous layer becomes noticeably paler as bromine leaves it. Answer: two layers form; the upper hexane layer is orange, the lower aqueous layer is paler/almost colourless.

1 mark: two distinct layers form, with the hexane layer on top. 1 mark: the upper (hexane) layer turns orange (bromine extracted into the non-aqueous layer) while the aqueous layer becomes paler. Max 2 marks.

Chlorine also oxidises iodide ions to iodine: \( \text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2 \). Iodine dissolves in hexane to give a distinctive purple/violet colour, quite different from the orange colour seen with bromine. Because chlorine is able to displace both bromide and iodide, this confirms that chlorine is a stronger oxidising agent than both bromine and iodine, consistent with the trend that oxidising power decreases down Group 7. Answer: the hexane layer turns purple/violet (I2 extracted), confirming chlorine is a stronger oxidising agent than iodine, in line with the trend of decreasing oxidising power down the group.

1 mark: correct observation (hexane layer turns purple/violet) linked to the correct conclusion (chlorine displaces iodide too, confirming it is a stronger oxidising agent, consistent with the trend down the group). Accept the observation alone or the conclusion alone if only one is given, but the mark point requires both to be broadly linked for full credit.

\( \Delta T = T_{max} - 18 \). For each volume: 5 cm3: \(21-18=3\); 10 cm3: \(24-18=6\); 15 cm3: \(27-18=9\); 20 cm3: \(30-18=12\); 30 cm3: \(31-18=13\); 35 cm3: \(29-18=11\); 40 cm3: \(27-18=9\). Answer: ΔT / °C = 3, 6, 9, 12, 13, 11, 9 (in that order).

Completeness (up to 3 marks): 1 mark for all four ΔT values in the rising region (3, 6, 9, 12) correct; 1 mark for all three ΔT values in the falling region (13, 11, 9) correct; 1 mark for a fully and clearly completed table with a correctly labelled ΔT / °C column. Precision (1 mark): all ΔT values given to the nearest whole number, consistent with the raw data. Decimal/unit consistency (1 mark): correct unit (°C) used consistently for all entries. Max 5 marks; OFR applies if the initial temperature is misread.

The x-axis should be labelled 'Volume of HCl added / cm3' and the y-axis 'Maximum temperature / °C', each axis fully labelled with quantity and unit and a sensible, evenly spaced scale that uses most of the grid. The four readings taken before the peak, (5, 21), (10, 24), (15, 27) and (20, 30), rise in an approximately straight line and should be used to draw the first best-fit line. The three readings taken after the peak, (30, 31), (35, 29) and (40, 27), fall in an approximately straight line and should be used to draw the second best-fit line. Because no reliable reading could be taken in the region where the temperature was changing fastest (between 20 and 30 cm3), both straight lines are extended (extrapolated) with a ruler beyond the plotted points until they cross. The volume and temperature at this point of intersection give the best estimate of the true (heat-loss-corrected) maximum temperature that would have been reached at the exact point of neutralisation, correcting for the cooling that occurred while readings were being taken. Answer: axes as above; rising-line points (5,21)–(20,30), falling-line points (30,31)–(40,27); the two lines are extrapolated to their intersection to obtain the corrected maximum temperature (and the corresponding neutralisation volume).

1 mark: both axes fully and correctly labelled with quantity and unit (Volume of HCl added / cm3; Maximum temperature / °C). 1 mark: correct identification of the four points used for the rising best-fit line. 1 mark: correct identification of the three points used for the falling best-fit line. 1 mark: correct statement that both lines are extrapolated until they intersect, and that this intersection gives the corrected maximum temperature (accounting for heat loss / the gap in readings). Max 4 marks.

(i) Rising line gradient: \( \dfrac{30 - 21}{20 - 5} = \dfrac{9}{15} = 0.6 \ °C\,cm^{-3} \). The rising line passes through (5, 21), so its equation is \( T = 0.6X + 18 \) (since \(21 = 0.6(5) + c \Rightarrow c = 18\), consistent with the initial temperature). (ii) Falling line gradient: \( \dfrac{27 - 31}{40 - 30} = \dfrac{-4}{10} = -0.4 \ °C\,cm^{-3} \). The falling line passes through (30, 31), so its equation is \( T = -0.4(X-30) + 31 = -0.4X + 43 \). (iii) At the intersection, \( 0.6X + 18 = -0.4X + 43 \Rightarrow 1.0X = 25 \Rightarrow X = 25 \ cm^3 \); \( T = 0.6(25) + 18 = 15 + 18 = 33\ °C \). Answer: rising gradient = +0.6 °C cm-3, falling gradient = −0.4 °C cm-3, intersection at 25 cm3 of HCl added and a corrected maximum temperature of 33 °C.

1 mark: correct rising-line gradient (+0.6 °C cm-3) with points at least 10 cm3 apart and working shown. 1 mark: correct falling-line gradient (−0.4 °C cm-3) with working shown. 1 mark: correct intersection coordinates (25 cm3, 33 °C) found either graphically or algebraically from the two line equations. OFR/ECF applies throughout from the table in the previous part. Max 3 marks.

Total mass of solution, \( m = 25.0 + 25.0 = 50.0\ g \) (density 1.00 g cm-3). Corrected temperature rise from the previous part: \( \Delta T = 33 - 18 = 15\ °C \). \( q = mc\Delta T = 50.0 \times 4.18 \times 15 = 3135\ J = 3.135\ kJ \). \( \Delta H = -\dfrac{q}{n} = -\dfrac{3.135}{0.0500} = -62.7\ kJ\,mol^{-1} \). This is exothermic (negative sign), of the same order of magnitude as the accepted enthalpy of neutralisation for a strong acid and strong base (data-book value approximately −57 kJ mol-1); the somewhat larger magnitude obtained here is a normal feature of a simple polystyrene-cup experiment. Answer: ΔH = −62.7 kJ mol-1.

1 mark: correct final ΔH value with correct negative sign and unit (−62.7 kJ mol-1, ECF from the temperature rise found in the previous part).