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

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

(a) Molecular formula gives the actual number of each kind of atom in one molecule while structural formula is a graphical representation of a molecule of a compound which shows how the atoms are linked/bonded together OR Molecular formula gives the total number of atoms of each element in one molecule of a compound. while structural formula shows li arrangements/relative position of atoms/atoms and bonds in one molecule of a compound. (b) Factors that Determine the ionization Energy of an Atom: - Atomic radius/size of atom Nuclear charge. - Shielding effect/screening effect. - Electron configuration,-Penetration effect. (c) Conditions Necessary for the Establishment of a Chemical Equilibrium - The rates of the forward and backward reactions must be equal. - \(\bigtriangleup\)G must be zero. - The temperature must be constant - System must be dynamic/ all reactant and product must be present. - System must be closed. (d) Consider the following table: Element A B C Ionization energy KJ mol\(^{-1}\) 619 518 594 (i) Element B Reason: Element B has the lowest ionization energy/ it loses electrons most easily. (e) Graham's Law of Diffusion: Graham's law of diffusion states that the rate of diffusion of gas at the same temperature and pressure is inversely proportional to the square root of its vapor density/molar mass (f)i. Salts that is/are readily soluble in water: - Mg(N\(O_3\))\(_{2}\), N\(a_2\)S\(O_4\) (ii) Insoluble in Water - CaC\(O_3\) (g) Product classification as addition polymer or condensation polymer: (i) Protein - Condensation. (ii) Perspex - Addition. (iii) Nylon - Condensation. (h) Define Atomic Radius: - lt is defined as half the distance between the nuclei of any two close atoms in a molecule/or is a measure of the size of its atoms, the mean distance from the center of the nucleus to the boundary of the surrounding shells of electrons. (i) Why ethanol has a higher boiling point than propane even though they both have comparable molar mass: - The intermolecular forces in ethanol are hydrogen bonds which are stronger than the van der Waal's forces in propane. (j) Three Significance of pH Value in Everyday Life: - Digestion of food. - Determination of pH of soil/used in agriculture. - Important role in the survival of animals (self-defense) existence of living beings. - pH change as the cause of tooth decay. - Treatment of water for public supply. - reatment of industrial efluent/ liquid waste. - ln pharmaceuticals. - In medicine (to determine pH of body fluid/urine /blood).

(a)i. Characteristics of homologouS series - All members can be represented by a general formula. - Each member of the series differs from the next by a constant/fixed number of atoms - C\(H_2\)/ 14 amu - All members of the series show a gradual change in physical properties such as density. - AII members have similar chemical properties/same functional groups. - All members of the group are prepared using a similar chemical method (ii) Differences in the Reaction of Ethane and Ethene - Ethane is a saturated hydrocarbon containing a single bond and therefore undergoes mainly substitution reaction while ethene is an unsaturated hydrocarbon containing a double bond and hence undergoes an addition reaction (b)i. The gas is hydrogen chloride. (ii) Physical properties of the gas: - Colourless - Has choking smell - Denser than air - Gas is soluble in water. - Gas fumes in moist air. (iii) Write a balanced equation reaction: 2HaCl\(_{(s)}\) + H,SO\(_{4(aq)}\), \(\implies\) Na2SO\(_{4(aq)}\) + 2HC\(_{(g)}\) OR HaCI\(_{(s)}\) + \(H_2\)SO\(_{4(aq)}\) \(\implies\) NaHSO\(_{4(aq)}\) + HCl\(_{(g)}\) (c)i. Hydrocarbons are compounds containing only hydrogen and carbon (ii) Two Natural Sources of Hydrocarbons: - Coal - Petroleum/crude oil Natural gas. (iii) A hydrocarbon contains 83.0 % of carbon by mass. calculate its emprical formular (H =1.0; C = 12.0). Solution: C = 83.0 %. \(\therefore\) % \(H_2\) = 100 - 83 17% (d) Draw and label a diagram of a set-up that could be used to electroplate a copper ornament with silver.

(a)i. 2KOH\(_{aq}\) + \(H_2\)SO\(_{4(aq)}\) \(\implies\) \(K_2\)SO\(_{4(aq)}\) + 2\(H_2\)O\(_{(l)}\) ii. Amount of KOH used = \(\frac{24}{1000} \times \frac{0.150}{1}\) = 0.0036mol Mole ratio of acid to base = 1:2 Amount of \(H_2\)S\(O_4\) used \(\frac{0.0036}{2}\) = 0.0018mol \(\therefore\) Concentration of \(H_2\)S\(O_4\) = \(\frac{0.0018}{0.025}\) = 0.072mol dm\(^{-3}\) (b)(i) 2Mg(s) + CO2(g) >>>>>> 2MgO(s) + C(s) (ii) Magnesium is above carbon in the reactivity series and so can remove oxygen from carbon dioxide and continue to burn. (iii) Molar mass Mg(N\(O_3\))\(_{2}\) = 24 x 1 + 2 x [14 + (16 x 3)] 24 x 1 + 2 x [14 + 48] = 24 + 28 + 96 = 148 gmol\(^{-1}\) Mass of \(N_2\) = 14 x 2- 28 \(\therefore\) %Nitrogen = \(\frac{28}{149} \times \frac{100}{1}\) = 18.9% (c)i. Reaction with brominei n CC\(l_4\)/bromine water/H\(^{+}\) / K Mn\(O_4\)/ H\(^{+}\) / \(K_2\)C\(r_2\)\(O_7\) - Reaction with aqueous NaHC\(O_3\)/ N\(a_2\)C\(O_3\). OR - Reaction with alkanol + Conc. H\(+{2}\)S\(O_4\) (ii) - Compound will decolourize bromine in CC\(l_4\)/ bromine water - Change H\(^{+}\)/K\\(_{2}\)C\(r_2\)\(O_7\) from orange to green. - Change H\(^{+}\) /KMn\(O_4\) from purple to colourless. - There will be effervescence/evolution of gas in reaction with NaCO\(_{3(aq)}\) / Production of sweet-smelling/fruity odour (d) Add sufficient water to dissolve the pellets of NaOH. The solution is then added to the vegetable oil, stirred, and boiled. It is cooled to precipitate the soap/salt out the soap/salt. Filter off the soap takes off the soap crust. (e) Define the Term Electronegativity: The tendency of an atom to pull to itself the shared pair of electrons in a covalent bond. The ability of an atom to attract shared electrons between It and another atom in a chemical bond

(a) Graphite:-hexagonal: Diamond:-letrahedral/octahedral (ii) Explain briefly why diamond is hard and a non-conductor of electricity while graphite is soft and an electrical conductor: n diamond each carbon atom is covalently bonded to four other carbon atoms forming an interlocking three dimens1onal network hence the hardness. Each of the four valence electrons is localized in a covalent bond hence there are no free mobile electrons, while in graphite each carbon atom is bonded to three other carbon atoms forming large sheets with van der Waal's force between the sheets which can easily slide over one another. One of the valence electrons of each atom is delocalized in a covalent bond and can act as a conductor of electricity OR Diamond is hard because it has a rigid/three-dimensional structure. It has no free/ mobile electrons hence it does not conduct electricity. Graphite has a layered structure held together by weak van der Waal's forces hence it is soft. It conducts electricity because it has free/ mobile electrons. (bi) State what is Achieved at each of the Following Stages in the Purification of Water for Town Supply: - Aeration- removes odor/kills bacteria. - Screening- removal of suspended dirt - Sedimentary- removal of suspended dirt particles after coagulation. (ii) Name two Substances Responsible for Hardness in Water: - Calcium hydrogen trioxocarbonate (IV). - MagnesIum hydrogen triOxocarbonate(IV). - Magnesium chloride. - Magnesium tetraOxOSulphate( Vl). - Calcium tetraoxosulphate (VI). - Iron (lI) hydrogen trioxocarbonate(IV). - Calcium chloride. - Iron(Il) tetraoxosulphate(VI) (iii) State two Methods for the Removal of Hardness in Water: - By adding washing soda (N\(a_2\)C\(O_3\)). Adding (CaCOH)\(_{2}\) /lime water. - Using ion-exchange resin / permutit/ zeolite. - Boiling/ distillation (iv) Give one Disadvantage of Hard Water: - Furring of kettle/pipes causes blockage leading to power wastage/ scales in the boiler. - Soap is wasted/ forms scums - Cannot be used in dyeing/ tanning - Not suitable for laundry. - Not suitable for textile industry (c) Describe Briefly the Extraction of Tin from its Ores: (i) The ore is concentrated/crushed and washed with water, it is then roasted in air to remove impurities. The product is mixed with charcoal/carbon /coke and heated to reduce the oxide. it is then refined to obtain pure tin. (ii) SnO\(_{2(s)}\) + 2C\(_{(s)\) \(\implies\) Sn\(_{(l)}\) + 2CO\(_{(g)}\) (iii) (I) Sn\(_{(s)}\) + O2\(_{(g)}\) \(\implies\) Sn\(_{2(s)}\) (II) Sn\(_{(s)\) + 2Cl\(_{2(g)}\) \(\implies\) SnC\(l_4\)

(a) Aluminum first reacts With oxygen in the air to form a film of oxide/oxide layer on the Surface. Al\(_{(s)}\) + 3O2\(_{(g)}\) \(\implies\) 2AlO3\({(s)}\). The film of oxide formed is not porous and insoluble in water (in the air) and hence protects the metal underneath from further attack. OR The film of oxide coats the metal and prevents it from further attack. (ii) The composition of the ore:- Cryolite / N\(a_3\)Al\(F_6\) -Aluminà / A\(l_2\)\(O_3\) (b) ZnO is an amphoteric oxide. Write an equation to illustrate this statement. ZnO\(_{(s)}\) + 2HCl\(_{(aq)}\) \(\implies\) ZnCl\(_{2(aq)}\) + \(H_2\)O\(_{(l)}\); ZnO\(_{(s)}\) + 2NaOH\(_{(aq)}\) + N\(a_2\)ZnO\(_{2(aq)}\) + \(H_2\)\(O_l\) (c)i. List three uses of Sodium Trioxocarbonate(IV): - Glass production. - Water treatment/ softening hard water - Preparation of baking powder. - Removal of sulfur (IV) Oxide in power stations. - As a standard alkali in acid-base titration. - As an analytical reagent. - Production of rayon polymer. - As a household cleaning agent used in laundry. - As food additive. - Used in textile industry. - In petroleum refining. - Manufacturing of other sodium compounds. - production of soaps and detergents - paper production - production/ preparation of NaHC\(O_3\) (ii) Explain briefly why a solution of trioxonitrate( V) acid turns yellowish on storage for sometime:- The yellowish color is caused by dissolved nitrogen(IV) oxide which results from the slow decomposition of the acid. (iii) Describe briefly how trioxonitrate (V) ions could be tested for in the laboratory:- To the solution, add freshly prepared iron (ll) tetraoXoSulphate (Vl) solution and followed by Conc \(H_2\)S\(O_4\), a brown ring is observed or warm the solid with ( copper turnings and ) Conc. \(H_2\)S\(O_4\) in a test tube. Brown fumes are observed. (d)i. \(H_2\)SO\(_{4(aq)}\) + 2NaCl\({s}\) \(\implies\) N\(a_2\)SO\(_{4(aq)}\) + 2HC\(l_g\). OR \(H_2\)S\(O_4\) + 2KCl \(\implies\) \(K_2\)SO\(_{4(aq)}\) + 2HCI (ll) H\(_{2(g)}\) + Cl\(_{2(g)}\) \(\implies\) 2HCl\(_{(g)}\) (ii) State one use of hydrogen chloride: - Manufacturing of drug. - Cleaning - Pickling - Electroplating metals - Tanning leather - Refining -Production of C\(H_3\)Cl/PVC/plastics.

- Hydrolysis is any chemical reaction in which a molecule of water breaks one or more chemical bonds. -Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution.

(a) Aluminum first reacts With oxygen in the air to form a film of oxide/oxide layer on the Surface. Al\(_{(s)}\) + 3O2\(_{(g)}\) \(\implies\) 2AlO3\({(s)}\). The film of oxide formed is not porous and insoluble in water (in the air) and hence protects the metal underneath from further attack. OR The film of oxide coats the metal and prevents it from further attack. (ii) The composition of the ore:- Cryolite / N\(a_3\)Al\(F_6\) -Aluminà / A\(l_2\)\(O_3\) (b) ZnO is an amphoteric oxide. Write an equation to illustrate this statement. ZnO\(_{(s)}\) + 2HCl\(_{(aq)}\) \(\implies\) ZnCl\(_{2(aq)}\) + \(H_2\)O\(_{(l)}\); ZnO\(_{(s)}\) + 2NaOH\(_{(aq)}\) + N\(a_2\)ZnO\(_{2(aq)}\) + \(H_2\)\(O_l\) (c)i. List three uses of Sodium Trioxocarbonate(IV): - Glass production. - Water treatment/ softening hard water - Preparation of baking powder. - Removal of sulfur (IV) Oxide in power stations. - As a standard alkali in acid-base titration. - As an analytical reagent. - Production of rayon polymer. - As a household cleaning agent used in laundry. - As food additive. - Used in textile industry. - In petroleum refining. - Manufacturing of other sodium compounds. - production of soaps and detergents - paper production - production/ preparation of NaHC\(O_3\) (ii) Explain briefly why a solution of trioxonitrate( V) acid turns yellowish on storage for sometime:- The yellowish color is caused by dissolved nitrogen(IV) oxide which results from the slow decomposition of the acid. (iii) Describe briefly how trioxonitrate (V) ions could be tested for in the laboratory:- To the solution, add freshly prepared iron (ll) tetraoXoSulphate (Vl) solution and followed by Conc \(H_2\)S\(O_4\), a brown ring is observed or warm the solid with ( copper turnings and ) Conc. \(H_2\)S\(O_4\) in a test tube. Brown fumes are observed. (d)i. \(H_2\)SO\(_{4(aq)}\) + 2NaCl\({s}\) \(\implies\) N\(a_2\)SO\(_{4(aq)}\) + 2HC\(l_g\). OR \(H_2\)S\(O_4\) + 2KCl \(\implies\) \(K_2\)SO\(_{4(aq)}\) + 2HCI (ll) H\(_{2(g)}\) + Cl\(_{2(g)}\) \(\implies\) 2HCl\(_{(g)}\) (ii) State one use of hydrogen chloride: - Manufacturing of drug. - Cleaning - Pickling - Electroplating metals - Tanning leather - Refining -Production of C\(H_3\)Cl/PVC/plastics.

1 mole of CaCO\(_3\) = 100g or 6.02 X 10\(^{23}\) liberated 1 mole of CO\(_2\) or 44g or 6.02 X 10\(^{23}\) :100g = 6.02 X 10\(^{23}\) 10g of CaCO\(_3\) = x cross multiply [10g X 6.02 X 10\(^{23}\)] / 100g = x ⇒ x = 6.02 X 10\(^{22}\) Since 6.02 X 10\(^{23}\) of CaCO\(_3\) liberated 6.02 X 10\(^{23}\) of CO\(_2\) ⇒ 6.02 X 10\(^{22}\) of CaCO\(_3\) will liberate 6.02 X 10\(^{22}\) of CO\(_2\)

COMPOUNDS BOILING POINTS Butane -1 ^\circC Ethanoic acid 118 ^\circC Water 100 ^\circC Ethanol 78.37 ^\circC

Halide: A binary compound of a halogen with another element or group.

No. of moles = \(\frac{molar conc \times volume}{1000}\) No. of moles = \(\frac{0.108 \times 50}{1000}\) : No. of moles = 0.0054 moles = \( 5.4 \times 10^{-3}\)

Once you figure out the electron configuration, you fill up the corresponding orbitals with electrons, any left with one is considered unpaired. Since 1s can only hold 2 electrons, and P has 15, that's obviously filled and has no unpaired electrons. The same is for 2s which holds 2, 2p which holds 6, 3s which holds 2. However 3p can hold 6 electrons and in order for that to be filled up you would need to have an element of 18 electrons. So you fill up as much as you can in 3p by first adding 1 electron to each energy level. 3p has 3 energy levels and there are only 3 electrons left to distribute, so each of those energy levels only gets 1, because you have to fill them all with one before you can start adding a second. So since you are only able to fill one electron in each of the three energy levels of the 3p orbital, that leaves the orbital open for 1 more electron in each of its energy levels. So there are 3 unpaired electrons in P.