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(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\) Burette Readings (cm\(^3\)) Rough 1st titre 2nd titre Final burette readings 18.20 18.50 38.40 Initial burette readings 0.00 0.40 20.20 Volume of A used (KMnO\(_4\)) 18.20 18.10 18.20 Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\) = \(\frac{18.20 + 18.10 + 18.20}{3}\) = 18.166 V\(_A\) = 18.17 cm\(^3\) (i) Concentration of B in moldm\(^{-3}\) From the balanced equation : MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\) C\(_A\) = 0.020 moldm\(^{-3}\) N\(_A\) = 1 V\(_A\) = 18.17cm\(^3\) N\(_B\) = 5 C\(_B\) = ? V\(_B\) = 25.0cm\(^3\) Using \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\) C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\) C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\) C\(_B\) = 0.07268 C\(_B\) = 0.070 moldm\(^{-3}\) (ii) Concentration of B in gdm\(^{-3}\) If 3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B X = \(\frac{ 3.8 X 1000}{250}\) = 15.2 gdm\(^{-3}\) Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\) (iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B. To get the mass, we'll use the formula n = \(\frac{mass}{Molar mass}\) However, since do not know the number of moles, n , hence we'll need to obtain n, using n = \(\frac{CV}{1000}\) since we have C and V So, n = \(\frac{{0.070}\times{250}}{1000}\) = 0.025 mole Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56 n = \(\frac{mass}{Molar mass}\) m = n x M = 0.018 x 56 = 1.02 g Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B. (iv) Percentage of Fe\(^{2+}\) in the granules = \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\) = \(\frac{1.02}{3.8}\times 100\) = 26.89% = 26.9%

Test Observation Inference (a) C + 10cm\(^3\) of H\(_2\)O C partially dissolves C is a mixture of a soluble and insoluble salt. b(i) 2cm\(^3\) of filtrate + heat Colourless and odourless gas in form of bubbles CO\(_3\)\(^{2-}\) suspected (ii) residue + t.t + dil. HCl No visible reaction NO\(_3\)\(^-\), SO\(_4\)\(^{2-}\) suspected (iii) Solution from b(ii) + NaOH in drops and then in excess White ppt which is dissolves in excess NaOH Zn\(^{2+}\) , Pb\(^{2+}\) , Al\(^{3+}\) suspected (iv) Solution from b(ii) + NH\(_3\) in drops and then in excess White ppt insoluble in excess NH\(_3\) Pb\(^{2+}\) confirmed.

(a) Gases that are soluble in water are: - Ammonia - Hydrogen chloride - oxygen, - carbon dioxide, - Sulphur dioxide, - chlorine, and - hydrogen sulfide. (b) Apparatus that could be used to measure volume of solutions accurately: - volumetric flask, - graduated cylinder, - burette, or - volumetric pipette. (c) In a laboratory, fume cupboards are essential safety equipment used to capture and remove airborne hazardous or poisonous substances like gases, vapours, aerosols, and dust, protecting users from exposure to harmful chemicals. (d) Personal protective equipment (PPE) used in the laboratory. - Lab coats - safety glasses - respiratory protector - Hand gloves - Face shield

(a) Volume of pipette used = 25.00 cm\(^3\) Burette readings (cm\(^3\)) Rough 1st titre 2nd titre Final burette reading 20.90 40.90 21.00 Initial burette reading 0.00 20.10 0.00 Volume of acid, A used 20.90 20.80 21.00 Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\) = \(\frac{20.90 + 20.80 + 21.00}{3}\) = 20.90 cm\(^3\) (b) 2HNO\(_3\)\(_{aq}\) + Na\(_2\)CO\(_3\).xH\(_2\)O\(_{aq}\) → 2NaNO\(_3\) + CO\(_2\) + (X+1)H\(_2\)0(l) c(i) concentration of E in moldm\(^{-3}\) Since we do not know the concentration of D in moldm\(^{-3}\), but in g per cm\(^3\), it will be difficult using the fomula \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\). Thus, we must first convert the Conc. of D to moldm\(^{-3}\). We were told from the question that D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\), So, if D 6.30 g of HNO\(_3\) was dissolved in 500 cm\(_3\) then, X g of HNO\(_3\) will be dissolved in 1000 cm\(_3\) X = \(\frac{{6.30}\times{1000}}{500}\) = 12.6 gdm\(^{-3}\) Recall that Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) Molar mass od HNO\(_3\) = 1 + 14 + (16 X 3) = 63 g/mol concentration of D(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) C\(_D\) = \(\frac{12.6}{63}\) = 0.20 moldm\(^{-3}\) Now we can use the formula \(\frac{C_D V_D}{C_E V_E}\) = \(\frac{N_D}{N_E}\). Given that C\(_D\) = 0.20 moldm\(^{-3}\) N\(_D\) = 2 V\(_D\) = 20.90cm\(^3\) N\(_E\) = 1 C\(_E\) = ? V\(_E\) = 25.0cm\(^3\) C\(_B\) = \(\frac{C_D V_D N_E}{N_D V_E}\) C\(_B\) = \(\frac{ 0.2 X 20.90 X 1}{25 X 2}\) = 0.0836 moldm\(^{-3}\) = 0.084 moldm\(^{-3}\) (ii) To get the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O First, from the question, if 8.20 g of Na\(_2\)CO\(_3\).xH\(_2\)O was dissolved in 250 cm\(^3\) of distilled water. then, Xg of Na\(_2\)CO\(_3\).xH\(_2\)O will be dissolved in 1000 cm\(^3\) of distilled water. So, X = \(\frac{{8.20}\times{1000}}{250}\) = 32.8 gdm\(^{-3}\) Mass concentration of E = 32.8 gdm\(^{-3}\) Recall that concentration of E(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) Then, Molar mass of E ( Na\(_2\)CO\(_3\).xH\(_2\)O ) = \(\frac{Mass concentration}{Molar concentration}\) = \(\frac{32.8}{0.0836}\) = 392.34 g/mol But Molar mass of Na\(_2\)CO\(_3\).xH\(_2\)O = Na\(_2\)CO\(_3\) + x(H\(_2\)O) 392.34 = (23 x 2) + 12 + (16 x 3) + 18x 392.34 = 46 + 12 + 48 + 18x 392.34 = 106 + 18x 18x = 392.34 - 106 18x = 286.34 x = \(\frac{286.34}{18}\) x = 15.9 x ≈ 16

F = (NH\(_4\))\(_2\)CO\(_3\) TEST OBSERVATION INFERENCE (a) F + t.t + dil. HCl Effervescence of colourless, odourless gas. Evolved gas turned lime water milky CO\(_2\) from CO\(_3\)\(^{2-}\) or HCO\(_3\) confirmed. b(i) 1st portion + t.t + NaOH in drops and then in excess + heat to warm No visible reaction, No ppt was also formed Effervescence of a colourless gas with an irritating smell. Gas turned red litmus to blue,also produced white fumes with conc. HCl NH\(_3\) from NH\(_4\)\(^+\) confirmed. (ii) 2nd portion + t.t + few drops of potassium hexacyano ferrate(II) solution No ppt formed and no visible reaction was observed Fe\(^{2+}\)/ Fe\(^{3+}\) absent

a(i) The chemical formula for alumina,(aluminium oxide) is Al\(_2\)O\(_3\) (ii) In the extraction of aluminium,the compound that is used; I. to obtain alumina from Bauxite - Sodium hydroxide, NaOH II. as solvent for alumina - Cryolite, Na\(_3\)AlF\(_6\) (iii) The electrodes used in the extraction of aluminium are - cathode (negative electrode) graphite, - anode (positive electrode) carbon rods or slabs. (iv)The extraction of aluminium is expensive due to the high energy requirements of the electrolysis process, which involves purifying bauxite, melting aluminium oxide, and using electricity to break down electrolytes. (v) In the extraction of aluminium: I. the substance liberated at the anode is oxygen gas (O₂) II. the equation for the reaction that leads to decrease in the size of the anode is: C + O\(_2\) → CO\(_2\) b(i) Ions that cause hardness of water are Ca\(^{2+}\) and Mg\(^{2+}\) (ii) Advantages of hard water - strengthening bones and teeth, - potentially aiding digestion, and - reducing the risk of cardiovascular disease. - improves the taste of water due to the presence of the ions. Disadvantages of hard water - skin and hair dryness, - reduced soap effectiveness, - potential appliance damage, and plumbing issues. c(i) Destructive distillation of coal is a process of heating coal in the absence of air (or with limited oxygen) to break down its complex molecules into simpler, useful products like coke, coal gas, coal tar, and ammoniacal liquor. (ii) products obtained from the destructive distillation of coal are: - coke - coal gas - coal tar and - ammoniacal liquor d(i) Gases which can be used to demonstrate the fountain experiment are; - Ammonia, NH\(_3\) - Hydrogen chloride gas, HCl (ii) Ammonia (NH\(_3\)) and hydrogen chloride (HCl) are used in the fountain experiment to demonstrate the high solubility of gases in water. When water is introduced into a flask containing either gas, the gas dissolves rapidly, creating a pressure drop that forces water upwards, forming a "fountain".

a(i) Salt solution is acidic - Reaction is exothermic. - White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn. - Prepared solution contained soluble salt of Al, Ca and Zn. b(i)PbSO\(_4\) (ii) NaHCO\(_3\) (iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\). (iv) NaHCO\(_3\) (v) NH\(_4\)Cl (c) Precautions to be taken during titration include: - Wear a lab coat to protect your clothing from spills. - Ensure to rinse the burette with an acid solution - Also ensure the funnel is removed immediately after filling the acid into the burette. - Avoid error due to parallax while taking the reading to determine the end point. - Ensure all glassware is clean and free of bubbles.

The dehydration of ethanol is a chemical reaction that produces ethene and water from ethanol: The chemical equation for the dehydration of ethanol is: CH\(_3\)CH\(_2\)OH + H\(_2\)SO\(_4\) → CH\(_2\)=CH\(_2\) + H\(_2\)O The reaction is performed by heating ethanol with concentrated sulfuric acid at around 170^\circC. Sulfuric acid is a dehydrating agent, meaning it absorbs water.

a(i) Salt solution is acidic - Reaction is exothermic. - White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn. - Prepared solution contained soluble salt of Al, Ca and Zn. b(i)PbSO\(_4\) (ii) NaHCO\(_3\) (iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\). (iv) NaHCO\(_3\) (v) NH\(_4\)Cl (c) Precautions to be taken during titration include: - Wear a lab coat to protect your clothing from spills. - Ensure to rinse the burette with an acid solution - Also ensure the funnel is removed immediately after filling the acid into the burette. - Avoid error due to parallax while taking the reading to determine the end point. - Ensure all glassware is clean and free of bubbles.

In the novel, 'The Life Changer', the 'Notes of Salvation' is a euphemism for the cheat notes that Kolawole Abdul passed to Salma. This can be found on page 52 of the novel where Salma convinced her seat partner, Kola, during her final examination to scribble and slip her these "notes of salvation" as she had nothing to write.

archbishop - /ˌɑːtʃˈbɪʃ.əp/ Christian - /ˈkrɪs.tʃən/ chemical - /ˈkem.ɪ.kəl/ archangel - /ˈɑː.keɪn.dʒəl/ sachet - /ˈsæʃ.eɪ/

permit - /pəˈmɪt/ (verb) or /ˈpɜː.mɪt/ (noun) enlist - /ɪnˈlɪst/ commit - /kəˈmɪt/ meat - /mi:t/ profit - /ˈprɒf.ɪt/

boys - /bɔɪz/ moist - /mɔɪst/ elbows - /ˈel.bəʊz/ noise - /nɔɪz/ stays - /steɪz/

For polysyllabic words that end in "ity", the stress is usually placed on the ante-penultimate syllable i.e. the third syllable from the last. Therefore, the most appropriate stress placement is "producTIvity".

busy - /ˈbɪz.i/ vineyard - /ˈvɪn.jɑːd/ automobile - /ˈɔː.tə.mə.biːl/ twilight - /ˈtwaɪ.laɪt/ island - /ˈaɪ.lənd/