Question Library

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option A.

Orange juice is an acidic beverage with a typical pH value ranging between 3.3 and 4.2. However, in the context of the specific multiple-choice question provided by the examiner, the intended answer or best option to be considered is D. between 5.5 and 6.5.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option C.

no of mole in 250\(cm^3\) of its solution is about molar concentration. using n = cv c = \(\frac{n}{v}\) where c = molar concentration, n = number of moles and v = volume in \(dm^3\) number of mole of Na\(_H\)CO\(_3\) = mass of Na\(_H\)CO\(_3\) / molar mass of Na\(_H\)CO\(_3\) = \(\frac{8.4}{84}\) = 0.1moles but v = 250\(cm^3\) = 250/1000 = 0.25\(dm^3\) therefore c = 0.25 x 0.1 = 0.025mol\(dm^3\)

(i) On bubbling SO\(_2\) into acidified KMnO\(_4\) solution the solution becomes decolorized/ the solution disappears/discharged. (ii) On mixing zinc dust with CuSO\(_{4(aq)}\) solution, the blue colour of the CuSO\(_{(aq)}\) gradually disappears/the solution becomes colourless. A reddish-brown or pink deposit is obtained. (iii) On adding concentrated HNO\(_3\) to freshly prepared FeSO\(_4\) the solution, the solution changes colour from green to brown. (b) Substances which, if added to dilute H\(_2\)SO\(_4\) would give; (i) Hydrogen-magnesium/iron/zinc (ii) ZnSO\(_{4(aq)}\) - ZnO/ZnCO\(_3\)/Zn (c) (i) I - Filtration II - Chromatography III - Use of separating funnel/solvent extraction IV - Evaporation (ii) The set-up labelled IV is used for concentrating salt solutions. (iii) The set-up labelled I is used in obtaining clear water from muddy water. (iv) The set up labelled lIl is used for the separation of a polar solvent from a non-polar solvent. (v) In order to use the set-up labelled (IV) for separating a mixture of NaCl and NH\(_4\)CI, cover the evaporating dish with an inverted funnel.

(a)(i) Q belongs to group VI in the periodic table. (ii) S is strongly metallic in character (iii) T readily ionizes by gaining one electrons (iv) P and Q contain two unpaired electrons in the ground state atom; (v) S readily loses two electrons during chemical bonding (vi) R does not participate readily in chemical reactions (vii) S is an s-block element. (b) Particle Number of Protons Number of Electrons Number of Neutrons \(^1_1H\) 1 1 0 \(^{27}_{13}\)Al\(^{3+}\) 13 10 14 \(^{16}_{8}O^{2+}\) 8 10 8 (ii) Atomic radius-increases down a group in the periodic table but decreases from left to right across a period, because as atomic number/ number of electrons increases down a group, number of shells increases, hence atomic size increases. Across a period the number of electrons and protons also increases but there is no increase in the number of shells/ the electrons are added to the same shell. The resultant effect is that the electrons in the orbit experience a greater pull towards the nucleus/ effective nuclear charge increases resulting in decrease in size. (c)(i) The half-life of a radioactive element is the time taken for half of the total number of atoms in a given sample of the element to decay. (ii) \(^{210}_{84}PO\) \(\to\) \(^{206}_{82}PO\) + \(^{4}_{2}He\) (d) The type of chemical bonding that accounts for the following observations: (i) Chlorine exists as discrete molecules-covalent (ii) Sodium chlorine dissolves readily in water-ionic/electrovalent (iii) CuSO\(_{4(aq)}\) forms a deep blue complexion with excess NH\(_3\) –coordinate covalent/dative.

(a)(i) Characteristic properties of transition metals. – Variable oxidation states – Complex ion. formation – Formation of coloured ions/compounds – Paramagnetism – Catalytic ability. (ii) The oxidation state of manganese in I. MnCl\(_2\) is +2 II. MnO\(_2\) is +4 III. MnO\(^-_4\) is +7 (iii) Manganese conducts electricity in the solid state because it consists of metal nuclei and a sea of mobile (valence) electrons which conduct the current. Manganese chloride conducts only when molten or solution because conduction is by mobile ions. (b)(i) Factors that determine whether or not the collision of reactant particles would result in the formation of products –Energy of collision/colliding particles (temperature or activation energy) –Alignment of the particles. (ii) (c) Fe\(^{3+}_{(aq)}\) + 3SCN\(^-_{(aq)}\) \(\rightleftharpoons\) (SCN)\(_{(aq)}\) yellow colourless deep red Equilibrium mixture has a pale red colour (i) On adding more of KSCN\(_{(aq)}\) to the equilibrium mixture, the red colour will deepen. Addition of KSCN would increase the concentration of SCN\(^-\) causing equilibrium to shift to Products the right. (ii) Fe\(^{3+}\) reacts with NaOH ; Fe\(^{3+}_{(aq)}\) + 3OH\(^-\) --> Fe(OH)\(_3\). (iii) On adding NaOH\(_{(aq)}\) to the equilibrium mixture, a brown precipitate is formed, a colourless solution is obtained/the red colourpale disappears.

(a)(i) X contains 40% carbon, 6.67% hydrogen, the rest being oxygen. To determine the empirical formula of X (i) Percentage oxygen = 100 – (40 + 6.67) = 53.33% C : H O 40 6.67 53.33 12 1 16 3.33 6.67 3.33 1 2 1 Ratio Empirical formula is CH\(_2\)O (ii) Molecular formula of X if its relative molecular mass is 60. (CH\(_2\)O)\(_n\) = 60; (12 + 2 + 16)\(_n\) = 60 ; 30n = 60 ; n = 2 Molecular formula is C\(_2\)H\(_4\)O\(_2\) (b) Y is an alkanoic acid with relative molecular mass is 74. (i) The functional group in Y is –COOH/carboxyl group. (ii) Esterification reaction (iii) If Y is represented by C\(_n\)H\(_{2n+1}\)COOH; then 14n + 46 = 74, giving n = 2 (iv) 2C\(_2\)H\(_5\)COOH + 2Na –> 2C\(_2\)H\(_5\)COONa + H\(_2\). (v) Compound Y in the same homologous series as X has a CH\(_3\)CH\(_2\)COOH relative molecular mass of 74. The boiling point of Y will be N higher. than 118^\circ because in a homologous series boiling OH point increases with increase molar mass. (c)(i) Fermentation is the breakdown of carbohydrates (sugars) to form ethanol (alcohol) and CO\(^2\) by the action of enzymes. (iii) Yeast is added to glucose solution to make glucose ferment. (iv) A tightly caked glass bottle filled to the brim with fresh palm wine shatters.on standing because CO\(^2\) is produced as the palm wine ferments. The gas causes pressure build-up which causes the bottle to shatter.

(a)(i) Aluminium is preferred to copper for making overhead electric cables because aluminium: is lighter/has a lower relative density -is more resistant to corrosion. (ii) Extraction of aluminium from purified bauxite. A solution of the pure alumina in molten cryolite (Na\(_3\)AIF\(_6\)) is placed in the electrolytic cell made of iron lined with graphite which serves as the cathode. The anode consists of graphite rods dipping into the electrolyte (solution of alumina in molten cryolite). The cell is maintained at high temperature 900 \pm 100^\circC. At the cathode: Al\(^{3+}\) + 3e\(^-\)m ---> Al ; At the anode: 2O\(^{2-}\) ---> O\(_2\) + 4e\(^-\). The molten aluminium collects at the bottom of the cell and is tapped off at intervals. (b) Illustration of the cell used for the purification of copper. (i) Electrode I increases in mass during the electrolysis, because Cu\(^{2+}\) migrate to it. The Cu\(^{2+}\) is reduced to copper/Cu\(^{2+}_{(aq)}\)+ 2e --> Cu\(_{(s)}\) and the copper is deposited, leading to an increase in its mass. (ii) Electrode II is the site of oxidation. Oxidation is electron loss and the reaction at this electrode is Cu -> Cu\(^{2+}\) + 2e\(^{2+}\) (iii) Ill is CuSO\(_{4(aq)}\)/any soluble copper (II) salt. Its colour does not change in intensity because Cu\(^{2+}\) which is responsible for the colour is removed by the cathode (i) from the solution and is replaced by the anode. (c) To calculate the current that will deposit 8.00g of calcium from fused CaCl\(_2\) in 1 hr. 15mins. Ca\(^{2+}\) + 2e\(^{-}\) -> Ca. 40g of calcium will be deposited by (2 x 96500)C 8.00g of calcium will be deposited by \(\frac{8 \times 38600}{40}\) = 38600 C ; Let x represent 386000C Q = It xc = 1 x 75 x 60 ; Current (I) = \(\frac{38600}{75 \times 60}\) ; Current = 8.58A.

(a) An oxide which fits the following descriptions: (i) Used in bleaching - SO\(_2\) (ii) Is a reddish-brown gas NO\(_2\) (iii) Reacts with NaOH and also with HCI - ZnO/Al\(_2\)O\(_3\)/SiO\(_2\)/BeO/PbO/SnO\(_2\) (iv) Dissolves readily in water to give a solution with pH greater than 7 - Na\(_2\)O/K\(_2\)O or CaO and MgO. (v) Oxidizes hot concentrated HCI to chlorine -MnO\(_2\) or PbO\(_4\) or Pb\(_3\)O\(_2\) (b)(i) Removal of hardness in a sample of water that contains Ca(CO\(_3\))\(_2\). - By boiling/distillation - Adding washing soda (N\(_2\)CO\(_3\).10H\(_2\)O) -Passing through ion-exchange resin/permutit process -Adding a calculated amount of slacked lime -Adding sodium trioxophosphate (V) NaPO\(_3\). (ii) It is not advisable to build a house with limestone in an environment polluted by SO\(_2\) for the following reason. SO\(_2\) forms acid rain which reacts.with the limestone and wears away the building. CaCO\(_3\) + SO\(_2\) + H\(_2\)O -> CaSO\(_3\) + H\(_2\)O + CO\(_2\) (c)(i) KNO\(_3\)/KCIO\(_3\)/K\(_2\)O\(_2\) yield oxygen when heated strongly. (ii) To calculate the cmount oT gas which occupies 250cm\(^3\) at s.t.p. 1 mole occupies 22400cm\(^3\) at s.t.p.Hence \(\frac{250}{22400}\) mole occupies 0.00112 moles (iii) \(\frac{V_1}{T_1} = \frac{V_2}{T_2}; \frac{250}{273} = \frac{V_2}{300}; v_2 = \frac{250 \times 300}{273} = 275cm^3\) (iv) When a gas is heated at constant pressure, the molecules acquire more kinetic energy and move faster. They collide with one another and with the walls of the container more frequently. To maintain the Same number of collisions on the walls of the container (i.e keep pressure constant) the volume of the gas increases.