WAEC SSCE - Chemistry - 2021

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To determine the empirical formula of a hydrocarbon, we need to know the number of moles of each element present in the compound. In this case, we are given that the hydrocarbon contains 0.160 moles of carbon and 0.640 moles of hydrogen. To find the empirical formula, we need to determine the simplest whole-number ratio of the atoms in the compound. We can start by dividing the number of moles of each element by the smallest number of moles to obtain a ratio of the atoms in the compound. In this case, the smallest number of moles is 0.160, which corresponds to the number of moles of carbon. Dividing the number of moles of hydrogen by 0.160, we get: 0.640 mol H / 0.160 mol C = 4 mol H/mol C This means that there are four times as many hydrogen atoms as carbon atoms in the compound. Therefore, the empirical formula of the hydrocarbon is CH\(_{4}\), which corresponds to one carbon atom and four hydrogen atoms. So, the answer is option (C) CH\(_{4}\).

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The relative atomic mass (RAM) of an element is the weighted average of the masses of all its isotopes, taking into account their abundance. In this case, we are given the percentage abundance of two isotopes of element Q, and we can use this information to calculate its RAM. To calculate the RAM of Q, we need to multiply the atomic mass of each isotope by its percentage abundance (as a decimal), and then add these values together. The formula for calculating the RAM is: RAM = (mass\(_{63}\)Q x % abundance of\(^{63}\)Q) + (mass\(_{65}\)Q x % abundance of\(^{65}\)Q) The atomic mass of\(^{63}\)Q is 63, and the atomic mass of\(^{65}\)Q is 65. Substituting the given values into the formula, we get: RAM = (63 x 0.69) + (65 x 0.31) RAM = 43.47 + 20.15 RAM = 63.62 Therefore, the relative atomic mass of element Q is 63.62. So, the answer is option (B) 63.6.

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The parent chain in the given structure contains 5 carbon atoms. The parent chain is the longest continuous chain of carbon atoms in a molecule, and it is used as the basis for naming organic compounds. In the given structure, the longest continuous chain of carbon atoms starts from the leftmost carbon atom and extends to the rightmost carbon atom, with a total of 5 carbon atoms. The other carbon atoms in the molecule are attached to the parent chain as substituents. In this case, there are two substituents, each consisting of a single carbon atom with a methyl (-CH3) group attached. Therefore, the parent chain in the given structure contains 5 carbon atoms.

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Joseph J. Thomson discovered electrons. Joseph J. Thomson was a British physicist who conducted a series of experiments in the late 19th century to study the nature of electric discharge in gases. In 1897, he performed a famous experiment involving cathode rays, which are beams of electrons that travel from the negatively charged electrode (cathode) to the positively charged electrode (anode) in a vacuum tube. Thomson observed that cathode rays could be deflected by magnetic and electric fields, indicating that they were negatively charged particles. He proposed that these particles were a fundamental component of all atoms and called them "corpuscles," which later became known as electrons. Thomson's discovery of the electron revolutionized our understanding of atomic structure and led to the development of modern physics. He was awarded the Nobel Prize in Physics in 1906 for his work on the conduction of electricity in gases. Therefore, Joseph J. Thomson discovered electrons.

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Protons and electrons are called fundamental particles because they are considered to be the building blocks of matter. They cannot be divided into smaller particles and still retain their unique properties. Additionally, protons have a positive charge and electrons have a negative charge, which makes them essential for creating stable atoms. Finally, protons and electrons are found in all matter, meaning that they are fundamental to the composition of the universe. Therefore, all of the given options are correct, and each contributes to the overall understanding of why protons and electrons are fundamental particles.

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An energy profile diagram is a graphical representation of the energy changes that occur during a chemical reaction. It shows the energy of the reactants, the energy of the products, and the energy changes that occur during the course of the reaction. The x-axis of the energy profile diagram represents the reaction coordinate, which is a measure of the progress of the reaction from the reactants to the products. The y-axis represents the energy of the system, usually in units of joules or kilojoules per mole. An endothermic reaction is a reaction that absorbs heat from the surroundings, resulting in a positive change in enthalpy (?H>0). In an energy profile diagram for an endothermic reaction, the energy of the products is higher than the energy of the reactants, and the reaction requires an input of energy to proceed. The activation energy, which is the energy required to initiate the reaction, is represented by a peak in the diagram. In contrast, an exothermic reaction is a reaction that releases heat to the surroundings, resulting in a negative change in enthalpy (?H<0). In an energy profile diagram for an exothermic reaction, the energy of the products is lower than the energy of the reactants, and the reaction releases energy as it proceeds. The activation energy is also represented by a peak in the diagram. A spontaneous reaction is a reaction that occurs naturally without the need for an input of energy. In an energy profile diagram for a spontaneous reaction, the energy of the products is lower than the energy of the reactants, and the reaction releases energy as it proceeds. However, the activation energy may still be required to initiate the reaction. A redox reaction is a reaction that involves a transfer of electrons from one reactant to another. An energy profile diagram for a redox reaction may show the energy changes associated with the transfer of electrons, but it is not specific to redox reactions. In summary, an energy profile diagram illustrates the energy changes that occur during a chemical reaction, and can be used to identify whether the reaction is endothermic or exothermic, spontaneous or non-spontaneous. Therefore, the answer to this question is either (a) an endothermic reaction or (b) an exothermic reaction, depending on the specific diagram being referred to.

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Elements with high ionization energies would have high effective nuclear charges. This means that the positively charged nucleus of the atom has a strong hold on its electrons, making it difficult to remove them. The ionization energy is the energy required to remove an electron from an atom. Elements with high ionization energies are therefore less likely to lose electrons easily and become positively charged ions. This can also result in elements having higher electronegativities, which is a measure of an atom's tendency to attract electrons towards it.

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The pair of molecules that form hydrogen bonds are C\(_{2}\)H\(_{5}\)OH and CH\(_{3}\)OH. A hydrogen bond is a type of chemical bond that forms between a partially positively charged hydrogen atom and a partially negatively charged atom, such as oxygen or nitrogen. In the case of C\(_{2}\)H\(_{5}\)OH and CH\(_{3}\)OH, both molecules contain oxygen atoms, which have a partial negative charge. Additionally, the hydrogen atoms in these molecules are partially positively charged. This creates an opportunity for hydrogen bonding to occur between the oxygen atom in one molecule and the hydrogen atom in another molecule. In contrast, the other pairs of molecules listed do not contain the necessary atoms with partial charges for hydrogen bonding to occur. CH\(_{3}\)OH and H\(_{2}\) do not have any partially charged atoms, while H\(_{2}\)S and CH\(_{4}\) do not contain atoms with the necessary partial charges to form hydrogen bonds. NH\(_{3}\) and SO\(_{2}\) have partially charged atoms, but these atoms do not interact with each other in a way that allows for hydrogen bonding to occur. Therefore, only C\(_{2}\)H\(_{5}\)OH and CH\(_{3}\)OH can form hydrogen bonds.

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The statement that is correct about Group VII elements (also known as the halogens) is that their reactivity decreases down the group. Group VII elements are located in the periodic table in the 17th column and include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). These elements have similar chemical and physical properties, including high reactivity. However, as you move down the group, the reactivity of the halogens decreases. This is due to the increase in atomic size and the decrease in effective nuclear charge. The larger size of the atoms makes it harder for the halogens to gain an electron, reducing their reactivity. This trend can be seen in the reactivity of the halogens in different chemical reactions.

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Among the given species, Ca\(^{2+}\) has the largest ionic radius. Ionic radius is defined as half the distance between the nuclei of two adjacent ions of the same element in a crystal lattice. The ionic radius increases as we move down a group in the periodic table and decreases as we move across a period. Ca\(^{2+}\) has the largest ionic radius among the given species because it has the least nuclear charge compared to the other ions. As we move from left to right in a period, the number of protons in the nucleus increases, resulting in a greater nuclear charge. The greater nuclear charge attracts the electrons more strongly, causing the ionic radius to decrease. In contrast, as we move down a group, the number of electron shells increases, leading to an increase in the size of the ion. Therefore, among the given species, Ca\(^{2+}\) has the largest ionic radius because it has more electron shells and the least nuclear charge, while S\(^{2-}\) has the smallest ionic radius because it has the greatest nuclear charge and the least electron shells.

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In an electrochemical cell, the cathode is the electrode where reduction occurs. Reduction is the gain of electrons by a species. So, the half-reaction equation that involves the gain of electrons is the reaction at the cathode. Option B represents the reaction at the cathode because it involves the gain of two electrons, which is a reduction half-reaction. The species B\(^{2+}\) is being reduced to B\(_{(s)}\), which means that B\(^{2+}\) is gaining two electrons to become a neutral atom of B. Option A is not the reaction at the cathode because it involves the oxidation of A\(_{(s)}\) to A\(^{3+}\). This is an oxidation half-reaction, and it occurs at the anode. Option C is not the reaction at the cathode because it involves the reduction of A\(^{3+}\) to A\(_{(s)}\). This is a reduction half-reaction, but it occurs at the anode in an electrolytic cell. Option D is not the reaction at the cathode because it involves the oxidation of B\(_{(s)}\) to B\(^{2+}\). This is an oxidation half-reaction, and it occurs at the anode. In summary, option B represents the reaction at the cathode because it involves the gain of two electrons, which is a reduction half-reaction.

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To calculate the number of coulombs of electricity required to liberate 1.08g of Ag from a silver salt solution, we need to use Faraday's laws of electrolysis, which state that the amount of a substance produced by electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The formula for calculating the amount of electricity required to liberate a certain amount of a substance is: Quantity of electricity = (Mass of substance / Equivalent weight) × 96500 The equivalent weight of Ag is equal to its atomic weight divided by its valency. Since the valency of Ag is 1, the equivalent weight is equal to its atomic weight, which is 108.0 g/mol. Using the above formula, we get: Quantity of electricity = (1.08 g / 108.0 g/mol) × 96500 C/mol Quantity of electricity = 965 C Therefore, the number of coulombs of electricity required to liberate 1.08g of Ag from a silver salt solution is 965 C. In summary, option C is the correct answer.

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A molecule is a group of two or more atoms held together by chemical bonds. A monoatomic molecule is a molecule made up of a single atom. Phosphorus, which has the chemical symbol P and atomic number 15, can exist in several allotropes, including white, red, and black phosphorus. At standard temperature and pressure, white phosphorus is the most stable allotrope and exists as P\(_{4}\) molecules, which are tetraatomic. Therefore, the correct answer is tetraatomic.

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The collision between ideal gas molecules are considered to be perfectly elastic because they collide without losing energy. In an ideal gas, the molecules are considered to be point masses that move randomly in a straight line and collide with each other and the walls of the container. The collisions are perfectly elastic, meaning that no energy is lost during the collision. During a collision between ideal gas molecules, energy can be transferred from one molecule to another, but the total energy of the system remains constant. This is because the molecules are not considered to have any internal energy, such as rotational or vibrational energy, that could be lost during a collision. The lack of energy loss during collisions is a fundamental assumption of the kinetic theory of gases, which describes the behavior of gases at the molecular level. This assumption allows us to calculate the properties of gases, such as pressure, volume, and temperature, using simple equations based on the average kinetic energy of the gas molecules. Therefore, the collision between ideal gas molecules are considered to be perfectly elastic because they collide without losing energy.

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When aqueous ammonia is added in drops to a solution of copper(II) ions, a pale blue precipitate is formed. This is due to the formation of a complex ion known as tetraamminecopper(II) [Cu(NH3)4]2+. This complex ion has a blue color and is responsible for the color change observed in the solution. If more aqueous ammonia is added to the solution, the blue precipitate dissolves due to the formation of a soluble complex ion known as pentamminecopper(II) [Cu(NH3)5]2+. This solution will appear deep blue in color. Therefore, when excess aqueous ammonia is added to a solution of copper(II) ions, a deep blue solution of pentamminecopper(II) complex ion is formed, and there will be no precipitate. It is important to note that the color changes observed are due to the formation of complex ions and not due to the formation of a new substance.

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The correct statement about ionization energy is that it decreases down the group. Ionization energy is defined as the energy required to remove an electron from an atom or a positive ion in the gaseous state. The ionization energy of an atom depends on the nuclear charge, the distance between the nucleus and the outermost electron, and the shielding effect of inner electrons. As we move down a group in the periodic table, the number of electron shells increases, and the shielding effect of inner electrons also increases. This means that the outermost electrons are held less tightly by the nucleus and are easier to remove. Therefore, the ionization energy decreases down the group. Conversely, as we move across a period, the number of electron shells remains the same, but the nuclear charge increases. This means that the outermost electrons are held more tightly by the nucleus and require more energy to remove. Therefore, the ionization energy increases across the period. The other options in the question are incorrect. Ionization energy does not result in the formation of an anion, nor does it cause metallic nuclei to disintegrate.

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The solubility of a gas in water is determined by several factors, including temperature, pressure, and the nature of the gas and the solvent. At room temperature, the gas that is highly soluble in water is ammonia (NH\(_{3}\)). Ammonia is a polar molecule, meaning it has regions of positive and negative charge, and it readily dissolves in water due to the strong hydrogen bonding interactions between the ammonia molecule and the water molecules. In contrast, carbon (IV) oxide (CO\(_{2}\)) is only slightly soluble in water, and it is often used as a reference for measuring the solubility of other gases. Chlorine (Cl\(_{2}\)) is highly reactive and toxic, and it is not very soluble in water at room temperature. Nitrogen (N\(_{2}\)) is an unreactive gas and is only slightly soluble in water. So, of the gases listed, ammonia is the most soluble in water at room temperature.

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The property of salts that makes them effective drying agents is their ability to absorb moisture from their surroundings, a characteristic called hygroscopy. When exposed to an environment with high humidity, the salt molecules attract and hold onto water molecules, effectively removing moisture from the surrounding area. This makes them useful for drying out materials such as solvents or wet surfaces. Salts with high solubility, low melting point, or efflorescence are not typically effective as drying agents because they tend to absorb water and dissolve rather than removing it from the environment. Salts with low solubility, on the other hand, tend to be more effective drying agents as they can absorb more moisture without dissolving.

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The hydrolysis of proteins by diluting mineral acids breaks down proteins into their constituent building blocks, which are called amino acids. Proteins are large, complex molecules made up of long chains of amino acids that are linked together by peptide bonds. Hydrolysis is a process in which a molecule is broken down by the addition of a water molecule. In the case of proteins, hydrolysis occurs when the peptide bonds between the amino acids are broken by the addition of water. This process is catalyzed by dilute mineral acids, such as hydrochloric acid (HCl). The result of this reaction is the breakdown of proteins into individual amino acids, which are then available for use by the body to synthesize new proteins or to be used as a source of energy. Therefore, the correct answer is (C) amino acids. Sucrose, glucose, and fatty acids are not the products of protein hydrolysis. Sucrose is a disaccharide made up of glucose and fructose, while glucose is a simple sugar that is a component of many carbohydrates. Fatty acids are the building blocks of lipids, which are a different class of biomolecules.

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The dehydration of ethanol to form ethene is a chemical reaction in which water is removed from ethanol, resulting in the formation of ethene. The reaction can be represented as: CH\(_{3}\)CH\(_{2}\)OH → CH\(_{2}\)=CH\(_{2}\) + H\(_{2}\)O This reaction involves the elimination of a molecule of water (H\(_{2}\)O) from ethanol (CH\(_{3}\)CH\(_{2}\)OH) to form ethene (CH\(_{2}\)=CH\(_{2}\)). Such reactions are known as elimination reactions. In elimination reactions, a molecule is removed from a reactant, resulting in the formation of a new product. This is different from addition reactions, in which two or more reactants combine to form a single product, and substitution reactions, in which one atom or group is replaced by another. Therefore, the correct answer is (B) an elimination reaction.

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To calculate the concentration of a solution, we need to know the amount of solute and the volume of the solution. The given amount of potassium hydroxide is 0.28 g and the volume of the solution is 100 cm\(^{3}\) or 0.1 dm\(^{3}\). First, we need to calculate the number of moles of potassium hydroxide in the solution by dividing its mass by its molar mass: Number of moles of KOH = Mass of KOH / Molar mass of KOH = 0.28 g / 56 g/mol = 0.005 mol Then, we can calculate the concentration of the solution in mol dm\(^{-3}\) by dividing the number of moles by the volume of the solution in dm\(^{3}\): Concentration of KOH solution = Number of moles / Volume of solution = 0.005 mol / 0.1 dm\(^{3}\) = 0.05 mol dm\(^{-3}\) Therefore, the concentration of the solution is 0.05 mol dm\(^{-3}\). is the correct answer.

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The reactivity of fluorine is high due to its high electronegativity. Electronegativity is the ability of an atom to attract electrons towards itself. Fluorine has the highest electronegativity of all elements, which means that it attracts electrons towards itself more strongly than any other element. This makes fluorine highly reactive, as it tends to form strong bonds with other atoms by either gaining or sharing electrons. In addition to its high electronegativity, the small size of the fluorine atom also contributes to its high reactivity. The small size allows fluorine atoms to approach other atoms very closely, which enhances the strength of the interactions between them. The availability of d-orbitals and the strong F-F bond are not responsible for the high reactivity of fluorine. Therefore, the correct answer is option A: its high electronegativity.

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The most common method of preparing insoluble salts is by double decomposition. Double decomposition is a chemical reaction that involves the exchange of ions between two ionic compounds. In this reaction, the cations and anions of two different compounds switch places, resulting in the formation of two new compounds. When one of the products of this reaction is an insoluble salt, it can be separated from the solution by filtration. For example, suppose we want to prepare the insoluble salt lead(II) chloride (PbCl\(_2\)). We can do this by reacting lead(II) nitrate (Pb(NO\(_3\))\(_2\)) with sodium chloride (NaCl): Pb(NO\(_3\))\(_2\) + 2 NaCl → PbCl\(_2\) + 2 NaNO\(_3\) In this reaction, the lead(II) cation (Pb\(^{2+}\)) and the chloride anion (Cl\(^-\)) switch places, resulting in the formation of lead(II) chloride and sodium nitrate (NaNO\(_3\)), which remains soluble in water. The lead(II) chloride is insoluble in water and will precipitate out of the solution. We can then isolate the solid PbCl\(_2\) by filtration. Therefore, double decomposition is the most common method of preparing insoluble salts, as it allows us to precisely control the reaction conditions and obtain the desired product with high purity.

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The region around the nucleus where electrons can be located is called an orbital. An orbital is a three-dimensional region in space where an electron is most likely to be found. The electrons in an atom occupy specific orbitals, which are determined by the energy of the electron and its distance from the nucleus. The electrons in an orbital are said to be in a specific energy state and occupy a particular energy level. The shape and size of an orbital depend on the energy of the electron and the number of electrons in the orbital. In simple terms, an orbital is like a cloud around the nucleus that defines the region where an electron can be found.

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In order to determine the molecular formula of the hydrocarbon, we need to use the information given in the question to calculate the number of carbon and hydrogen atoms in the molecule. The ratio of carbon atoms to hydrogen atoms in the hydrocarbon is 1:2, which means that for every one carbon atom, there are two hydrogen atoms. We can represent this ratio using the general formula C\(_{n}\)H\(_{2n}\), where n is the number of carbon atoms in the molecule. Next, we are given the molecular mass of the hydrocarbon, which is 56. To find the value of n, we need to use the molecular formula to calculate the molecular mass. For the general formula C\(_{n}\)H\(_{2n}\), the molecular mass can be calculated using the formula: Molecular mass = n x (atomic mass of carbon) + 2n x (atomic mass of hydrogen) Substituting the values of atomic mass for carbon and hydrogen (12 and 1, respectively), and the given molecular mass of 56, we get: 56 = n x 12 + 2n x 1 Simplifying this equation, we get: 56 = 14n n = 4 Therefore, the molecular formula of the hydrocarbon is C\(_{4}\)H\(_{8}\), which is option B in the question. In summary, the hydrocarbon in the question has a ratio of carbon atoms to hydrogen atoms of 1:2, and a molecular mass of 56. Using this information, we can calculate that its molecular formula is C\(_{4}\)H\(_{8}\).

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The question is asking about the Bohr model of the atom and what it proposed. The Bohr model of the atom was proposed by Niels Bohr in 1913. It is a model that describes the structure of the atom, with the nucleus at the center and electrons orbiting the nucleus in specific energy levels or shells. Therefore, the answer to the question is "electron shells". The Bohr model proposed the existence of electron shells, which are specific energy levels that electrons occupy around the nucleus of an atom. These shells are quantized, meaning that they can only have certain energy values, and electrons can move between them by absorbing or releasing energy. Option A, "the nucleus", is not specific to the Bohr model and was already known to exist before the development of the Bohr model. Option C, "nucleons", refers to the collective term for the particles in the nucleus (protons and neutrons), which are not part of the Bohr model's description of the electron shells. Option D, "neutrons", is a type of nucleon, but again, it is not part of the Bohr model's description of the electron shells.

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The oxide that causes acid rain is NO\(_{2}\). Acid rain is a type of rain that has a low pH value (below 5.6) due to the presence of certain pollutants in the air. These pollutants, such as sulfur dioxide and nitrogen oxides, can react with water in the atmosphere to form acids that fall to the ground as acid rain. Among the options given, NO\(_{2}\) is the only oxide that contains nitrogen, which is a major contributor to acid rain. When NO\(_{2}\) is released into the atmosphere, it reacts with water to form nitric acid, which is a strong acid that can lower the pH of rainwater. Therefore, the correct answer is option D: NO\(_{2}\).

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(a)(i) Characteristics of a homologous series

• All members have the same functional group and can be represented by the same general formula.
• Successive members differ by a \(-CH_2-\) unit, that is, by 14 in relative molecular mass.
• They have similar chemical properties and show a gradual change in physical properties such as boiling point.

(a)(ii) Ethane is a saturated hydrocarbon containing only carbon-carbon single bonds. It undergoes mainly substitution reactions. Ethene is unsaturated and contains a carbon-carbon double bond, \(C=C\), which is reactive; it therefore readily undergoes addition reactions.

(b)(i) The gas evolved is hydrogen chloride, \(HCl\).

(b)(ii) Hydrogen chloride is a colourless gas with a pungent, choking smell. It is denser than air and is very soluble in water.

(b)(iii)

\[\mathrm{NaCl_{(s)} + H_2SO_{4(l)} \rightarrow NaHSO_{4(s)} + HCl_{(g)}}\]

(c)(i) Hydrocarbons are organic compounds that contain carbon and hydrogen only.

(c)(ii) Natural sources of hydrocarbons include crude oil (petroleum) and natural gas.

(c)(iii) Empirical formula

Assume 100 g of the hydrocarbon.

Element	Mass (g)	Moles	Mole ratio
C	83	\(\frac{83}{12}=6.92\)	\(\frac{6.92}{6.92}=1\)
H	\(100-83=17\)	\(\frac{17}{1}=17\)	\(\frac{17}{6.92}=2.46\approx2.5\)

The simplest ratio is \(\mathrm{C:H}=1:2.5\). Multiplying throughout by 2 gives \(2:5\).

\[\boxed{\text{Empirical formula }=\mathrm{C_2H_5}}\]

(d) Electroplating a copper ornament with silver

The silver electrode is connected to the positive terminal and is the anode. The copper ornament is connected to the negative terminal and is the cathode. Aqueous silver trioxonitrate(V), \(\mathrm{AgNO_3(aq)}\), is used as the electrolyte. Silver dissolves from the anode and is deposited on the copper ornament.

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(a)(i) Structures

• Graphite: a giant covalent structure made of flat layers of carbon atoms. Within each layer every carbon is covalently bonded to three others in a hexagonal (honeycomb) arrangement, leaving one delocalized electron per atom. The layers are held together only by weak van der Waals forces.
• Diamond: a giant covalent structure in which every carbon atom is covalently bonded tetrahedrally to four other carbon atoms, forming a rigid three-dimensional network.

(a)(iii) Hardness and conductivity

Diamond is hard because its carbon atoms are held in a rigid three-dimensional network of strong covalent bonds, and it is a non-conductor because all four valence electrons of each carbon are used in bonding, leaving no free electrons to carry charge. Graphite is soft because its layers are held by weak forces and can slide over one another, and it conducts electricity because each carbon uses only three electrons in bonding, leaving one delocalized electron per atom free to move and carry charge.

(b)(i) Water purification stages

• Aeration: water is sprayed into air to add dissolved oxygen and drive off dissolved gases and unpleasant odours; it also oxidizes dissolved iron and manganese.
• Screening: large floating and suspended solids (leaves, sticks, debris) are removed by passing water through screens/gratings.
• Sedimentation: the water is left to stand (often with a coagulant such as alum) so that fine suspended particles settle out to the bottom.

(b)(ii) Substances causing hardness

Dissolved calcium and magnesium salts, that is, the hydrogencarbonates and sulphates of calcium and magnesium (\(Ca^{2+}\) and \(Mg^{2+}\) ions).

(b)(iii) Methods of removing hardness

Boiling (removes temporary hardness), adding washing soda (\(Na_2CO_3\)), ion exchange, or distillation.

(b)(iv) One disadvantage of hard water

It wastes soap by forming an insoluble scum, and it deposits scale (fur) in kettles, boilers and pipes.

(c)(i) Extraction of tin

Tin is obtained from its ore cassiterite, \(SnO_2\). The ore is concentrated by washing and froth flotation to remove earthy impurities, then roasted in air to remove sulphur and arsenic. The concentrated oxide is then reduced with carbon (coke) in a reverberatory or blast furnace, and the molten tin is run off and refined.

(c)(ii) Equation for the reduction

(c)(iii) Reactions of tin

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(a) (i) Aluminum metal is not affected by air because it is covered with a thin layer of aluminum oxide (Al2O3) that protects the metal from further oxidation. The equation for the formation of aluminum oxide is: 2Al + 3O2 → 2Al2O3 The aluminum oxide layer is very stable and prevents oxygen in the air from reaching the aluminum metal, thus protecting it from further reaction. (ii) In the extraction of aluminum from bauxite: - The substance used for purifying the ore is concentrated sodium hydroxide (NaOH) solution, which is used to dissolve the aluminum oxide (Al2O3) present in bauxite, leaving behind impurities. - The composition of the mixture electrolyzed is a molten mixture of aluminum oxide (Al2O3) and cryolite (Na3AlF6), which is used as a solvent to lower the melting point of Al2O3 and make it easier to extract aluminum by electrolysis. (b) ZnO is an amphoteric oxide, which means it can react with both acids and bases. The equations to illustrate this statement are: ZnO + 2HCl → ZnCl2 + H2O ZnO + 2NaOH + H2O → Na2Zn(OH)4 In the first equation, ZnO reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl2) and water (H2O). In the second equation, ZnO reacts with sodium hydroxide (NaOH) and water (H2O) to form sodium zincate (Na2Zn(OH)4). (c) (i) Three uses of sodium trioxocarbonate(IV) (sodium carbonate, Na2CO3) are: - It is used in the production of glass, where it acts as a flux to lower the melting point of silica (SiO2) and help it fuse into glass. - It is used as a water softener in laundry detergents, where it reacts with hard water ions (such as calcium and magnesium) to form insoluble precipitates that can be removed from clothing. - It is used as a food additive, where it acts as a leavening agent in baked goods to produce carbon dioxide (CO2) gas and make the dough rise. (ii) A solution of trioxonitrate(V) acid (HNO3) turns yellowish on storage for some time because it slowly decomposes to form nitrogen dioxide (NO2) gas, which is a brownish color. The equation for this decomposition is: 4HNO3 → 2H2O + 4NO2 + O2 (iii) Trioxonitrate(V) ions (NO3-) can be tested for in the laboratory using the Brown Ring Test. In this test, a solution of the sample containing NO3- ions is mixed with freshly prepared iron(II) sulfate (FeSO4) solution and concentrated sulfuric acid (H2SO4). A brown ring will form at the junction of the two layers if NO3- ions are present, due to the formation of a complex between Fe2+ ions, NO3- ions, and H2O molecules. (d) (i) The balanced chemical equation for the preparation of hydrogen chloride (HCl) using concentrated sulfuric acid (H2SO4) is: NaCl + H2SO4 → NaHSO4 + HCl (ii) The balanced chemical equation for the preparation of hydrogen chloride (HCl) by direct combination of its constituent elements is: H2 + Cl2 → 2HCl (iii) One use

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(a)(i) Equation for the neutralization

(a)(ii) Concentration of the acid

Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).

From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).

(b) Magnesium burning in carbon(IV) oxide

Equation:

Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.

Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):

(c) The compound \(CH_3CH_2CH=CHCOOH\)

This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.

• (i) Reaction 1: add bromine water. Reaction 2: add sodium trioxocarbonate(IV) (\(Na_2CO_3\)) solution.
• (ii) With bromine water: the red-brown bromine water is decolourized (turns colourless), confirming the \(C=C\) double bond. With \(Na_2CO_3\): brisk effervescence occurs, giving off a colourless gas (\(CO_2\)) that turns limewater milky, confirming the acidic \(-COOH\) group.

(d) Manufacture of soap

Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.

(e) Electronegativity

Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.

Akọwa Nkọwa

(a) Molecular formula and structural formula

A molecular formula shows the actual number of atoms of each element present in one molecule of a compound (for example ethanol \( C_2H_6O \)). A structural formula shows how those atoms are arranged and joined together by bonds (for example \( CH_3CH_2OH \)).

(b) Three factors that determine ionization energy

• The atomic radius (distance of the outermost electron from the nucleus).
• The nuclear charge (number of protons).
• The shielding (screening) effect of the inner electrons.

(c) Two conditions for chemical equilibrium

• The reaction must be reversible.
• The system must be closed (no reactant or product enters or leaves) and kept at constant temperature.

(d) Ionization energies: A = 619, B = 518, C = 594 \( \text{kJ mol}^{-1} \).

(i) The strongest reducing agent is B.

(ii) B has the lowest ionization energy, so it loses (donates) its outermost electron most easily; a good reducing agent is one that readily gives up electrons.

(e) Graham's law of diffusion

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass): \( \text{rate} \propto \dfrac{1}{\sqrt{M}} \).

(f) (i) Readily soluble in water: \( Mg(NO_3)_2 \) and \( Na_2SO_4 \) (all nitrates and all sodium salts are soluble). (ii) Insoluble in water: \( CaCO_3 \).

(g) (i) Protein: condensation polymer. (ii) Perspex: addition polymer. (iii) Nylon: condensation polymer.

(h) Atomic radius

Atomic radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond (half the internuclear distance).

(i) Ethanol versus propane

Ethanol molecules contain an \( -OH \) group and form hydrogen bonds between neighbouring molecules. These hydrogen bonds are much stronger than the weak van der Waals (dispersion) forces between propane molecules, so more energy is needed to separate ethanol molecules, giving it the higher boiling point even though the molar masses are comparable.

(j) Three significances of pH in everyday life

• Control of soil pH so that crops grow well (agriculture).
• Maintaining the correct pH of blood and other body fluids for good health (medicine).
• Quality control in the manufacture of foods, drinks, drugs and cosmetics, and in the treatment of drinking water.