Question Library

Caustic soda, \(NaOH\) , serves to remove gaseous impurities from the gas evolved during the laboratory preparation of ethene. Basically, the major impurities are Sulphur (IV) oxide and Carbon (IV) oxide.

(a) Two differences in the chemical properties of metals and non-metals Metals Non-Metals Metals Non-Metals 1. it can form basic oxides 2. It can form electrovalent compounds 1. It can form acidic oxides 2. It can form covalent compounds (b) Two general methods of extracting metals from their ores are:- (i) Reduction by electrolysis and (ii) By reduction with carbon II oxide.

(a) Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. (b)(i) Gaseous state (ii) Solid state (iii) Gaseous state.

(a) (i) E (ii) D (iii) A (iv) B (b)(i) Electrovalent bond (ii) DO.

(a) (i) Glucose (ii) Citric acid (b)(i) The process of obtaining paraffin from crude oil is called fractional distillation (ii) The process of obtaining benzene from ethyne is called polymerization.

(a) Oxidation is defined as the loss or removal of electron. (b)(i) The species that is reduced is Ag\(^+\) (ii) Cu \(\rightleftharpoons\) Cu\(^{2+}\) + 2e\(^-\) 2Ag\(^+\) + 2e\(^-\) \(\rightleftharpoons\) 2Ag

(a)(i) The name of the phenomenon is Efflorescence (ii) Another example is Na\(_2\)CO\(_3\).10H\(_2\)O. (b) Molar mass of MgSO\(_4\) x H\(_2\)O = 2x 123 = 246 i.e. 24 + 32 + (4 x 16) + x (2X 1 + 16) = 246 120 + 18x = 246; 18x = 246 - 120; 18x = 126 x = \(\frac{126}{18}\) x = 7

(a)(i) 2KNO\(_3\) \(\to\) 2KNO\(_2\) + O\(_2\) (ii) C\(_{12}\)H\(_{22}\)O\(_{11}\) + 11H\(_2\)O (b) Two observations include: (i) gas is given off (ii) The zinc disappears to form a colourless solution.

(a) The type of process involved in the reaction is called dehydration (b) The basicity of the acid is 2 (c) Carbon (ii) oxide (CO) Carbon (iv) oxide (CO\(_2\)) 1. It burns in air 2. It is a reducing agent 3. No action on lime water 1. Does not burn in air 2. Has no reducing properties 3. It turns lime water milky.

(a) (i) The entropy decreases (ii) The entropy increases (iii) The entropy increases (iv) The entropy decreases. (b)(i) The reaction is at equilibrium. (iii) The reaction is feasible.

(a) (i) The substance that can be used is manganese (iv) oxide (ii) The substance used is alkaline pyrogallol. (b) The precipitate is silver chloride. In KCI, there is free chloride ion to form the precipithte but in KCIO\(_3\), the chlorine atom is a complex. Therefore, it is not available to react with silver ion to form the precipitate.

(a)(i) Diagram for the laboratory preparation of a dry sample of chlorine. Reactants are conc. HCI and MnO\(_2\), drying agent is conc. H\(_2\)SO\(_4\) removal of impurities is brine or KHCO\(_3\) (i) When moist starch iodide paper is applied, the colour will change from white to blue black indicating the presence of chlorine (b)(i) CI\(_2\) + 2FeCl\(_2\) \(\to\) 2FeCl\(_3\) (ii) Cl\(_2\) + 2Kl \(\to\) 2Kcl + l\(_2\) (iii) 3Cl\(_2\) + 6NaOH -> NaCIO\(_3\) + 5NaCI + 3H\(_2\)O (c)(i) White precipitate is formed. (ii) The precipitate disappears on heating and re-appears on cooling. (d)(i) A white precipitate or suspension would form. The solution turn out to be milky. The reason is that the addition of water would drive the equilibrium to the left with the result that BiCI\(_3\) and H\(_2\)O would react to produce more BiOCI which is insoluble.

i) Hydrocarbons are compounds consisting of carbon and hydrogen only. (ii) H = (100 - 92.3)% = 7.7% C : H \(\frac{92.3}{12}\) : \(\frac{7.7}{1}\) \(\frac{7.7}{1}\) : \(\frac{7.7}{1}\) Therefore, the empirical formula is CH, hence the molecular formula is (CH)n. Molar mass = 2 x vapour Density = 2 x 39 = 78; (CH)n = 13n = 78, :. n = \(\frac{78}{13}\) = 6 13 Therefore, the molecular formula is C\(_6\)H\(_6\). (b)(i) The cassava is crushed and pressure cook to extract starch granules. Treat with malt which contains diastase to convert the starch to maltose. Add yeast and leave to ferment for about two days. The maltose in yeast coriverts the maltose into glucose. The zymase also in yeast converts the glucose into ethanol and carbon (iv) oxide. Distil to obtain ethanol from the "wash" (ii) Acidified potassium tetraoxomanganate (VII) and acidified potassium heptaoxodichromate (VI) (c)(i) Esterification (ii) Hydrolysis (iii) Hydrogenation (iv) Saponification

(a)(i) For M, n = 3. Substituting into 2n\(^2\), 2 x 3 x 3 = 18 is obtained. (ii) Befor Combination after Combination Sodium 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^1\) (Na) Fluorine 1s\(^2\)2s\(^{2}\)2p\(^5\) (F) s\(^2\)2s\(^2\)2p\(^6\) (Na\(^+\)) 1s\(^2\)2s\(^2\)2p\(^6\) (F\(^-\)) (iii) (1) High melting and high boiling points. (2) Conduction of electricity in solution (3) Solubility in water (b) 4Na + O\(_2\) -> 2Na\(_2\)O Na\(_2\)O + H\(_2\)O -> 2NaOH (c) Place 100cm\(^3\) of each (or equal volumes) in separate identical beakers and connect as shown in the diagram below. Record the ammeter reading using one solution as the test solution. Using one solution as the test solution, connect again as shown in the diagram and take the reading. The higher reading indicates the electrolyte which has the higher conductance.

(a)(i) Heat of combustion is the heat evolved when one mole of a substance is completely burnt in oxygen (ii) The name given is calorimeter. (b)(i) (ii) C + O\(_2\) -> CO\(_2\); H = —3935KJ\(^{-1}\). Mole of carbon =12g of carbon will evolve — 3935 KJ :. 60g of C will evolve — \(\frac{3935 \times 60}{12}\) Heat change = -19675 KJ (iii) Neutralization reaction can be represented generally as H\(_2\)O\(^+\) + OH\(^-\) —> 2H\(_2\)O. Strong acids and strong bases lame completely in water to furnish H\(_3\)O\(^{+}\) and OH\(^{-}\) respective Thus in all cases, one mole of H\(_3\)O\(^{+}\) of a strong acid completely es one mole of OH\(^-\) of a strong base. Therefore the enthalpy of neutralization is constant. (c)(i) Both rusting and combustion involve combination with oxygen to produce the oxide. Both reactants are exothermic. Rusting takes a longer period of time than combustion. (ii) iron filings possess greater net surface area and reaction rate increases with increased surface area. (iii) Zinc is more reactive which is more reactive than tin. Consequently, when tin - plated iron is exposed, the iron goes into solution and rusts but in the case of galvanized iron, it is the zinc that dissolves thus protecting the iron better.