(a)(i) What is an acid-base indicator?
(ii) Give one example of an acid-base indicator.
(b) State the property exhibited by nitrogen(IV) oxide in each of the following equations:,
(i) 4Cu + 2NO\(_2\) -> 4CuO + N\(_2\) (ii) H\(_2\)O + 2NO\(_2\) --> HNO\(_3\) + HNO\(_2\)
(c)(i) Define enthalpy of combustion..
(ii) State why the enthalpy of combustion is always negative.
(d)(i) Distinguish between a primary cell and a secondary cell.
(ii) Give an example of each of the cells stated in I (d)(i).
(e) Define the term mole.
(f) Calculate the amount of hydrochloric acid in 40.0 cm\(^3\) of 0.40 moldm\(^{-3}\) dilute HCl.
(g) Name two substances which can be used as electrodes during the electroylsis of acidified water.
(h) List two forces of attraction that can exist between covalent molecules.
(i) Name the products formed when butane undergoes incomplete combustion.
(j) Write the electron configuration of \(_{26}\)Fe\(^{3+}\)
(a)(i) An acid-base indicator is a substance (usually a weak acid or weak base) that shows different colours in acidic and alkaline media and is used to detect the end point of a neutralization.
(ii) Example: litmus (or methyl orange, or phenolphthalein).
(b)(i) In \(4Cu + 2NO_2 \rightarrow 4CuO + N_2\), nitrogen(IV) oxide acts as an oxidizing agent (it is reduced and oxidizes the copper).
(ii) In \(H_2O + 2NO_2 \rightarrow HNO_3 + HNO_2\), nitrogen(IV) oxide undergoes disproportionation, acting as both an oxidizing and a reducing agent (N goes from +4 to +5 and to +3).
(c)(i) Enthalpy of combustion is the heat change when one mole of a substance is completely burnt in excess oxygen under standard conditions.
(ii) It is always negative because combustion is exothermic - heat is released to the surroundings.
(d)(i) A primary cell cannot be recharged; its chemical reaction is irreversible. A secondary cell can be recharged because its reaction is reversible.
(ii) Primary cell: dry (Leclanche) cell. Secondary cell: lead-acid accumulator.
(e) A mole is the amount of a substance that contains as many elementary entities as there are atoms in 12 g of carbon-12 (that is, \(6.02 \times 10^{23}\) entities).
(f) Amount of HCl \(= C \times V = 0.40 \times \dfrac{40.0}{1000} = 0.016\ \text{mol}\).
(g) Platinum and carbon (graphite) - inert electrodes.
(h) Van der Waals (dispersion) forces and hydrogen bonds.
(i) Carbon(II) oxide (carbon monoxide) and water (with some carbon/soot).
(j) \(_{26}\)Fe is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\); Fe3+ loses the two 4s and one 3d electron:
\[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\]
(a)(i) An acid-base indicator is a substance (usually a weak acid or weak base) that shows different colours in acidic and alkaline media and is used to detect the end point of a neutralization.
(ii) Example: litmus (or methyl orange, or phenolphthalein).
(b)(i) In \(4Cu + 2NO_2 \rightarrow 4CuO + N_2\), nitrogen(IV) oxide acts as an oxidizing agent (it is reduced and oxidizes the copper).
(ii) In \(H_2O + 2NO_2 \rightarrow HNO_3 + HNO_2\), nitrogen(IV) oxide undergoes disproportionation, acting as both an oxidizing and a reducing agent (N goes from +4 to +5 and to +3).
(c)(i) Enthalpy of combustion is the heat change when one mole of a substance is completely burnt in excess oxygen under standard conditions.
(ii) It is always negative because combustion is exothermic - heat is released to the surroundings.
(d)(i) A primary cell cannot be recharged; its chemical reaction is irreversible. A secondary cell can be recharged because its reaction is reversible.
(ii) Primary cell: dry (Leclanche) cell. Secondary cell: lead-acid accumulator.
(e) A mole is the amount of a substance that contains as many elementary entities as there are atoms in 12 g of carbon-12 (that is, \(6.02 \times 10^{23}\) entities).
(f) Amount of HCl \(= C \times V = 0.40 \times \dfrac{40.0}{1000} = 0.016\ \text{mol}\).
(g) Platinum and carbon (graphite) - inert electrodes.
(h) Van der Waals (dispersion) forces and hydrogen bonds.
(i) Carbon(II) oxide (carbon monoxide) and water (with some carbon/soot).
(j) \(_{26}\)Fe is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\); Fe3+ loses the two 4s and one 3d electron:
\[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\]
