CHEMISTRY WAEC SPECIMEN QUESTIONS AND ANSWERS

1. The collision theory proposes that?

2. The colour of phenolphthalein indicator in dilute HNO3(aq) is?

3. An aqueous solution of CaCl2 is?

4. Which of the following compounds would dissolve in water to give a solution whose pH is less than 7? A. Al(NO3)3 B. KNO3 C. N2O D. NH3

5. Consider the reaction represented by the equation below: KOH(aq) + HCL(aq) KCL(aq)+H2O(l) . What volume of 0.25 mole dm-3 KOH would be required to completely neutralize 40 cm3 of 0.10 mole-3 HCL? 6. When 100 cm3 of saturated solution of KCIO3 at 40oC is evaporated, 14g of the salt is recovered. What is the solubility if KCIO3 at 40oC?

7. The high solubility of ethanol in water is due to its?

8. Consider the equilibrium reaction represented by the equation: 2SO2(g) + O2(g) ↔2SO3(g) ∆H = -197 kjmol-1 At equilibrium, increase in the total pressure of the system brought about?

9. In dry Leclanche cell, manganese (IV) oxide acts as?

10. Oxidation takes place at the anode during electrolysis because the anode?

11. Calculate the quantity of electricity passed when 0.4 A flows for a 1 hour 20 minutes through an electrolytic cell

12. A mixture of kerosene and diesel oil can be separated by?

13. Which of the following compound would not give a precipitate with ammoniacal AgNO3 solution A. CH3C = CCH3 B. HC = CH C. CH3C=CH D. CH3CH2C=CH

14. The hydrolysis of groundnut oil by potassium hydroxide is known as?

15. When ethanol is heated with excess concentrated tetraoxosulphate (VI) acid, the organic product formed is?

16. What is the main type of reaction alkenes undergo?

17. The following compound with the following structure represents CH3 CH3 C C2H5 OH

18. Consider the organic compound X with the following structure. H H H H C C C H H OH H The complete oxidation of X gives?

19. What is the IUPAC name of the compound with the following structure? H H C C = C H H H H C H H 20. A hydrocarbon containing 88.9% carbon has the empirical formula?

21. A solution of salt was acidified with HCL. When a few drop of BaCL2 solution were added, a white precipitate was formed. Which anions is present in the salt?

22. Which substances causes the depletion of the ozone layer in the atmosphere?

23. Chemical that are produced in small quantities and with very high degree of purity are?

24. Aluminum is a good roofing material because it?

25. Which of the following alloys is mainly used in making statues? A. Brass B. Bronze C. Duralumin D. Steel

solution/answers

1. The collision theory proposes that chemical reactions occur when reactant molecules collide with sufficient energy and proper orientation. According to this theory, for a reaction to take place, the reacting species must collide with each other. The collisions should have enough energy to overcome the activation energy barrier, and the molecules should be oriented in a way that allows the necessary bond rearrangements to occur.
2. The color of phenolphthalein indicator in dilute HNO3(aq) is colorless. Phenolphthalein is an organic compound that acts as an acid-base indicator. In an acidic solution, such as dilute HNO3, phenolphthalein remains colorless. However, in a basic solution, it turns pink or magenta.
3. An aqueous solution of CaCl2 is formed when calcium chloride (CaCl2) is dissolved in water. When CaCl2 dissolves in water, it dissociates into calcium ions (Ca2+) and chloride ions (Cl-). Therefore, the resulting solution contains calcium ions and chloride ions dispersed throughout the water.
4. The compound that would dissolve in water to give a solution with a pH less than 7 is D. NH3 (ammonia). NH3 is a weak base that can accept a proton (H+) from water, forming NH4+ (ammonium) and OH- ions. The presence of hydroxide ions increases the concentration of OH- in the solution, making it basic. Therefore, NH3 would give a solution with a pH greater than 7.
5. In the given reaction: KOH(aq) + HCl(aq) → KCl(aq) + H2O(l), the stoichiometry of the reaction tells us that the ratio of moles of KOH to moles of HCl is 1:1. To calculate the volume of 0.25 mol/dm³ KOH required to neutralize 40 cm³ of 0.10 mol/dm³ HCl, we can use the equation:

Volume(KOH) x Concentration(KOH) = Volume(HCl) x Concentration(HCl)

Volume(KOH) = (Volume(HCl) x Concentration(HCl)) / Concentration(KOH) Volume(KOH) = (40 cm³ x 0.10 mol/dm³) / 0.25 mol/dm³ Volume(KOH) = 16 cm³

Therefore, 16 cm³ of 0.25 mol/dm³ KOH would be required to completely neutralize 40 cm³ of 0.10 mol/dm³ HCl.

6. The solubility of KClO3 at 40°C can be determined by dividing the mass of the salt dissolved by the volume of the saturated solution.

Solubility = Mass of KCIO3 / Volume of solution

Given that 14g of KCIO3 is recovered from a 100 cm³ saturated solution, the solubility can be calculated as follows:

Solubility = 14g / 100 cm³ = 0.14 g/cm³

Therefore, the solubility of KCIO3 at 40°C is 0.14 g/cm³.

7. The high solubility of ethanol in water is due to its ability to form hydrogen bonds. Ethanol (C2H5OH) contains a hydroxyl group (-OH) that can form hydrogen bonds with water molecules. These hydrogen bonds allow ethanol to mix homogeneously with water and dissolve easily. The polarity of the hydroxyl group and its ability to form hydrogen bonds make ethanol highly soluble in water.
8. According to Le Chatelier’s principle, an increase in the total pressure of a system at equilibrium will shift the equilibrium in the direction that reduces the pressure. In the given equilibrium reaction: 2SO2(g) + O2(g) ↔ 2SO3(g), an increase in total pressure would cause the equilibrium to shift in the direction that produces fewer moles of gas. In this case, it would shift to the right, favoring the formation of SO3, which results in a decrease in the total number of gas molecules.
9. In a dry Leclanche cell, manganese(IV) oxide (MnO2) acts as the depolarizer. The depolarizer helps to remove hydrogen gas that is produced at the anode during the cell’s operation. It oxidizes the hydrogen gas back to water, preventing the buildup of hydrogen bubbles on the anode and maintaining the cell’s efficiency.
10. Oxidation takes place at the anode during electrolysis because the anode is the positive electrode where electrons are released. The anode attracts negatively charged ions or anions from the electrolyte solution. These anions lose electrons, resulting in oxidation.
11. To calculate the quantity of electricity passed, you can use the equation:

Quantity of electricity (Coulombs) = Current (Amperes) × Time (Seconds)

First, convert 1 hour 20 minutes to seconds: 1 hour = 60 minutes × 60 seconds = 3600 seconds 20 minutes = 20 minutes × 60 seconds = 1200 seconds

Total time = 3600 seconds + 1200 seconds = 4800 seconds

Quantity of electricity = 0.4 A × 4800 s = 1920 Coulombs

Therefore, the quantity of electricity passed is 1920 Coulombs.

12. A mixture of kerosene and diesel oil can be separated by the process of fractional distillation. Fractional distillation takes advantage of the different boiling points of the components in the mixture. Kerosene and diesel oil have different boiling points, so when the mixture is heated, the component with the lower boiling point (kerosene) vaporizes and can be collected separately.
13. The compound that would not give a precipitate with ammoniacal AgNO3 solution is A. CH3C = CCH3 (2-butyne). Ammoniacal silver nitrate (AgNO3) solution is commonly used to test for the presence of alkynes (carbon-carbon triple bonds). Alkynes do not form precipitates with ammoniacal AgNO3 solution, whereas alkenes (C=C double bonds) and aromatic compounds (benzene rings) do form precipitates.
14. The hydrolysis of groundnut oil by potassium hydroxide (KOH) is known as saponification. Saponification is a chemical reaction where a fat or oil reacts with a strong base to form soap and glycerol. In the case of groundnut oil, it contains triglycerides that react with KOH to produce potassium salts of fatty acids (soap) and glycerol.
15. When ethanol is heated with excess concentrated tetraoxosulphate(VI) acid (H2SO4), the organic product formed is ethene. The reaction is a dehydration reaction, where ethanol loses a molecule of water in the presence of an acid catalyst. The resulting product is an alkene, ethene (C2H4).
16. The main type of reaction alkenes undergo is addition reactions. Alkenes contain a carbon-carbon double bond (C=C), which is a reactive site. In an addition reaction, a molecule or an atom adds to the carbon-carbon double bond, breaking the double bond and forming new single bonds. Examples of addition reactions include hydrogenation, halogenation, hydration, and polymerization.
17. The compound represented by the given structure is 2-methylpropan-2-ol (also known as tert-butanol). It has the following structural formula:CH3 CH3 C C2H5 OH
18. The complete oxidation of the organic compound X gives carbon dioxide (CO2) and water (H2O). The structural formula of X is:

H H H | C C C H H OH H

When X undergoes complete oxidation, the alcohol group (-OH) is oxidized to a carbonyl group (C=O), forming a ketone. The carbon chain is fully oxidized to carbon dioxide, and the hydrogen atoms combine with oxygen to form water.

19. The IUPAC name of the compound with the given structure is 2,2-dimethylpropane. The structural formula is:

H H | C C = C H H H H H

The longest continuous carbon chain in the molecule is three carbons, and the carbon-carbon double bond is located between the second and third carbons. Therefore, the IUPAC name is based on the parent chain propane, with the double bond indicated by the suffix “-ene.” The two methyl groups attached to the second carbon are designated by the prefix “2,2-dimethyl-.”

20. To determine the empirical formula of a hydrocarbon containing 88.9% carbon, we assume a 100g sample of the hydrocarbon. This means that the sample contains 88.9g of carbon.

The molar mass of carbon is 12 g/mol, so we can calculate the number of moles of carbon: Number of moles of carbon = mass of carbon / molar mass of carbon Number of moles of carbon = 88.9g / 12 g/mol ≈ 7.41 mol

Since the empirical formula represents the simplest ratio of atoms in a compound, we divide the number of moles by the smallest value to find the simplest ratio. In this case, the smallest value is approximately 7.41 mol.

Empirical formula ratio = Number of moles of carbon / 7.41 ≈ 1

Therefore, the empirical formula of the hydrocarbon is CH.

21. When a white precipitate is formed upon adding BaCl2 solution to an acidified salt solution, it indicates the presence of sulfate ions (SO4^2-) in the salt. Barium chloride (BaCl2) reacts with sulfate ions to form insoluble barium sulfate (BaSO4), which appears as a white precipitate.
22. Substances that cause depletion of the ozone layer in the atmosphere are known as ozone-depleting substances (ODS). The primary ODS are chlorofluorocarbons (CFCs), halons, carbon tetrachloride, and methyl chloroform. These substances contain chlorine and bromine atoms that can catalytically destroy ozone molecules in the stratosphere, leading to the thinning of the ozone layer.
23. Chemicals that are produced in small quantities and with very high purity are known as specialty chemicals. Specialty chemicals are typically manufactured for specific applications and industries. They undergo extensive purification and quality control processes to ensure high purity levels and meet the strict requirements of their intended uses.
24. Aluminum is a good roofing material because it has several desirable properties. Firstly, aluminum is lightweight, which makes it easier to install and reduces the overall load on the building structure. Secondly, aluminum is highly corrosion-resistant, forming a protective oxide layer that prevents rusting and degradation. Additionally, aluminum is durable, long-lasting, and can withstand various weather conditions without significant deterioration. These properties make aluminum a preferred choice for roofing applications.
25. The alloy mainly used in making statues is B. Bronze. Bronze is an alloy primarily composed of copper and tin. It is widely used in sculpture and statue making due to its desirable properties, including excellent casting ability, durability, and resistance to corrosion. Bronze statues often have a distinctive golden-brown color and can retain their appearance and structural integrity for long periods, making them suitable for outdoor and artistic applications.

part B

1. The structure of benzene consists of a ring of six carbon atoms, with each carbon atom bonded to a hydrogen atom. The structure is often represented as a hexagon, with a circle inside to indicate the delocalized electrons. The type of bonding in benzene is known as aromatic bonding or resonance bonding. It involves the overlapping of p orbitals of the carbon atoms, resulting in a continuous delocalization of electrons throughout the ring.
2. Benzene and alkenes undergo several types of reactions under ordinary conditions, including:

• Substitution reactions: In a substitution reaction, one or more atoms or groups are replaced by other atoms or groups. An example of a substitution reaction in benzene is the nitration reaction, where a nitro group (-NO2) is substituted for a hydrogen atom. The reaction is typically carried out in the presence of a nitric acid and sulfuric acid catalyst:C6H6 + HNO3 → C6H5NO2 + H2O
• Addition reactions: In an addition reaction, new atoms or groups are added to the carbon-carbon double bond of an alkene or the benzene ring. An example is the hydrogenation of benzene, where hydrogen gas is added in the presence of a catalyst (e.g., platinum) to form cyclohexane:C6H6 + 3H2 → C6H12
• Electrophilic aromatic substitution: This type of reaction involves the substitution of a hydrogen atom in the benzene ring with an electrophilic species. An example is the Friedel-Crafts acylation reaction, where an acyl group (-COR) is added to the benzene ring in the presence of a Lewis acid catalyst (e.g., aluminum chloride):C6H6 + RCOCl → C6H5COR + HCl

3. A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. It consists of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. The weak acid/base component of the buffer system can neutralize the added acid or base, preventing a significant change in the solution’s pH.
4. One example of a buffer solution is a mixture of acetic acid (CH3COOH) and sodium acetate (CH3COONa). Acetic acid is a weak acid that partially dissociates in water to release hydrogen ions (H+). Sodium acetate is the conjugate base of acetic acid. When a small amount of acid (HCl) is added to the buffer solution, the acetate ions (CH3COO-) react with the added H+ ions to form acetic acid, thereby preventing a large increase in the solution’s acidity:CH3COO- + H+ → CH3COOHAs a result, the buffer solution maintains its pH relatively constant even with the addition of an acid.
5. (i) In a 0.025 dm-3 aqueous solution of HCl, the concentration of H3O+ ions is equal to the concentration of HCl, which is 0.025 mol/dm³. Therefore, the concentration of H3O+ is 0.025 mol/dm³.(ii) The pH of a solution can be calculated using the equation: pH = -log[H3O+]. In this case, since the concentration of H3O+ is 0.025 mol/dm³, the pH is calculated as:pH = -log(0.025) ≈ 1.60The pOH of a solution can be calculated using the equation: pOH = -log[OH-]. Since HCl is a strong acid and fully dissociates in water, the concentration of OH- ions is negligible. Therefore, the pOH is essentially 0.
6. A covalent bond is a type of chemical bond that forms between atoms by the sharing of electrons. It occurs between non-metal atoms and is characterized by the overlap of atomic orbitals to create a shared electron pair. Covalent bonds are typically strong and result in the formation of stable molecules.
7. Two properties of covalent bonds are:

• Strong bond strength: Covalent bonds are generally strong, holding atoms together in a molecule. The strength of the bond arises from the sharing of electrons between atoms, resulting in the formation of a stable electron configuration.
• Directional bonding: Covalent bonds have a directional nature, meaning that the atoms involved in the bond have specific spatial arrangements. This directional bonding allows molecules to have unique shapes and determines the arrangement of atoms in the molecule.

8. The formation of an ammonia (NH3) molecule involves the sharing of electrons between nitrogen (N) and three hydrogen (H) atoms. Here is a diagram showing the bonding in an ammonia molecule:H | H – N – H | HThe nitrogen atom has five valence electrons, while each hydrogen atom has one valence electron. The nitrogen atom shares three of its electrons with the three hydrogen atoms, resulting in the formation of three covalent bonds.
9. The formation of the ammonium ion (NH4+) occurs when ammonia (NH3) accepts a proton (H+). Here is a diagram illustrating the formation of the ammonium ion:H | H – N – H | H diff + H
10. In the presence of an acidic solution or a proton donor, one of the lone pairs of electrons on the nitrogen atom in ammonia accepts a proton, forming a coordinate covalent bond. The resulting species is the ammonium ion (NH4+), which has a positive charge due to the addition of the extra proton.
11. The three subatomic particles are:

• Proton: It has a positive charge (+1) and a relative mass of approximately 1 atomic mass unit (1 amu). Protons are found in the nucleus of an atom.
• Neutron: It has no charge (neutral) and a relative mass of approximately 1 amu. Neutrons are also found in the nucleus of an atom.
• Electron: It has a negative charge (-1) and a negligible mass compared to protons and neutrons. Electrons orbit around the nucleus in specific energy levels.

11. Water can exist in three states: solid (ice), liquid (water), and gas (water vapor). These different states are determined by the temperature and pressure conditions. At low temperatures (below 0°C), water freezes and becomes solid ice. At temperatures between 0°C and 100°C, water is in its liquid state. At temperatures above 100°C, water boils and changes into water vapor, which is its gaseous state.
12. Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. In other words, lighter gases diffuse faster than heavier gases. The law can be expressed using the following equation:

Rate1/Rate2 = √(M2/M1)

Where Rate1 and Rate2 are the rates of diffusion of two gases, and M1 and M2 are their molar masses, respectively.