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Showing posts from April, 2022

Full CBT Rollout for SSCE Postponed to 2027 by WAEC

 Abuja — The West African Examinations Council (WAEC) has postponed the full implementation of the computer-based test (CBT) format for the 2026 Senior School Certificate Examination (SSCE) until 2027, according to a statement released at the council’s 63rd annual meeting in Umuahia, Abia State.  Hajia Binta Abdulkadir, Chairman of WAEC’s Nigeria National Committee, explained that the 2026 May/June edition will now serve as a pilot phase — allowing candidates to sit for either the traditional paper-and-pen test or the digital format. She emphasized that no student will be disadvantaged by the transition.  The delay follows widespread concern over the readiness of many schools, particularly in rural areas, for full digital exams. Many institutions lack adequate computers, reliable electricity, internet connectivity, and trained personnel—factors WAEC and the Federal Ministry of Education say are essential for a successful rollout.  To facilitate the switch, WAEC co...
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Stoichiometry of Chemical Reaction involving Volume of Gases part two

    In this second part, we are going to be looking at the stoichiometry of reaction involving the volume of gases. Just like what we saw in mass and mole, there is a relationship which also exists between mass and volume or mole and volume that is what we are going to look at.    A close look at a balanced chemical equation shows the ratio in which the reactants react and the ratio in which the products are produced. However, an experiment has shown that at standard temperature and pressure, gases have an equal volume which is numerically equal to 22.4 dm 3 or 22400cm 3 . This volume is known as the molar volume of gases. 1mole of any gas at STP = 22.4dm 3 32g of oxygen at stp =    22.4dm 3 Let us consider the decomposition of potassium trioxochlorate (V) compound.   2KClO 3(s) ---------- 2KCl (s) + 3O 2(g)    2 moles of KClO 3 will decompose to give 2 moles of KCl and 3 moles of oxygen gas. Since the molar volume gas is 22.4dm...
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Stoichiometry of Chemical Reaction involving Mass and Mole Part one

       Stoichiometry is the study of the relationships which exist between reactants and products in a chemical reaction. This relationship can easily be seen in a balanced chemical equation. For example,   K 2 SO 4 ------ 2K + + SO 4 2-       1              2            1 In the ionization reaction above, notice that one-mole potassium tetraoxosulphate (VI) gives two moles of potassium ions and one mole of tetraoxosulphate (VI) ion in solution. This ionization equation also shows the mole ratio of the reactants and the products, which is known as the stoichiometry of the chemical reaction. It is necessary to note that the mole of a substance is not affected if the reactants or products are ions, atoms or molecules etc. in a chemical reaction. CaCO 3(s) + 2HCl (aq)   ------ CaCl 2(s)   + H 2 O (l) + CO (g) ...
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How to apply Gay-Lussac’s Law in chemistry calculations

    Gay-Lussac’s Law explains the conditions that govern the combination of gases in a chemical reaction. This law states that gases react in volumes which are in simple ratio to each other and to the volumes of the products, provided that the temperature and pressure are constant. For instance, 4NH 3(g) + 3O 2(g) =  6H 2 O (g) + 2N 2(g) In the reaction above, according to Gay-Lussac’s Law, it can be said that 4cm 3 of ammonia reacts with 3cm 3 of oxygen to give 6cm 3 of water vapour and 2cm 3 of nitrogen. This is in line with his observation in his experiment with oxygen and hydrogen and also with hydrogen and chlorine. 2H 2(g) + O 2(g) = 2H 2 O (g) 2cm 3     1cm 3    2cm 3 The ratios of the reactants and product are 2:1:2 and the moles are 2 moles, 1 mole, and 2 moles respectively. He also noticed the same pattern in the case of the reaction with hydrogen and chlorine. H 2(g) + Cl 2(g) = 2HCl (g) 1:    1...
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How to prepare a standard solution of an acid

   A standard solution is a solution which has a known concentration, which means the amount in mole or mass concentration of the substance which is dissolved in a known volume of a solvent is known.    If you want to prepare a standard solution of an acid, the knowledge of the relationship between amount, molar concentration and volume is very necessary                                      C = n/V Where: n = amount (mole)               V = volume (dm 3 )              C = molar concentration (mole per dm 3 ) Experiment has shown that        C 1 V 1 = C 2 V 2 Since the amount of solute in a solution is not affected by incr...
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