WAEC SSCE - Chemistry - 2021
Tambaya 2 Rahoto
Cathodic protection of metals is based on
Tambaya 3 Rahoto
Which of the following scientists discovered the electrons?
Bayanin Amsa
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.
Tambaya 4 Rahoto
Which of the following half reaction equations represent the reaction at the cathode
Bayanin Amsa
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.
Tambaya 5 Rahoto
Which of the following statements about Group VII elements is correct?
Bayanin Amsa
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.
Tambaya 6 Rahoto
Which of the following gasses is highly soluble in water at room temperature
Bayanin Amsa
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.
Tambaya 7 Rahoto
If humid air is polluted by chlorine discharged, the air can be restored by sprinkling
Bayanin Amsa
Tambaya 8 Rahoto
How many molecules of oxygen would occupy a volume of 2.24 cm\(^{3}\) at s.t.p?
[Molar volume at s.t.p = 22,400 cm\(^{3}\), Avogadro's number = 6.02 x 10\(^{23}\)]
Tambaya 9 Rahoto
Which of the following statements is correct about ionization energy? It
Bayanin Amsa
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.
Tambaya 11 Rahoto
A coordinate covalent bond could be formed between?
Tambaya 13 Rahoto
Potassium trioxonitrate (V) can be obtained from its solution by----------
Tambaya 14 Rahoto
Study the graphs below and use them to answer the question below
Which of the following illustrates the variation of the rate of evolution of gas from a given length of magnesium ribbon (y-axis) with an increase in the concentration of the added (x-axis)?
Tambaya 15 Rahoto
A molecular of phosphorus is
Bayanin Amsa
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.
Tambaya 16 Rahoto
Study the graphs below and use them to answer the question below
Which of the graphs illustrates the variation of the solubility of a salt in water (y-axis), with an increase in temperature (x-axis) if the dissolution process is exothermic?
Tambaya 17 Rahoto
Protons and electrons are called fundamental particles because they----------
Bayanin Amsa
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.
Tambaya 18 Rahoto
The 25°C evaporation of a 100 cm\(^{3}\) solution of K\(_{2}\)CO\(_{3}\) to dryness gave 14g of the salt. What is the solubility of K\(_{2}\)CO\(_{3}\) at 25°25C? [K\(_{2}\)CO\(_{3}\) = 138]
Bayanin Amsa
Tambaya 19 Rahoto
Under which conditions of pressure (p) and temperature (T) would the volume of an inflated ballon increase? When
Bayanin Amsa
Tambaya 20 Rahoto
What would be observed when aqueous ammonia is added in drops and then in excess to a solution of copper(II) ions?
Bayanin Amsa
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.
Tambaya 21 Rahoto
The alkanol represented by the structure below is?
Tambaya 22 Rahoto
Which of the following pairs of compounds would form a precipitate when their aqueous solutions are mixed?
Tambaya 23 Rahoto
The most common method of preparing insoluble salts is by
Bayanin Amsa
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.
Tambaya 24 Rahoto
The basic property of salts used as drying agents is by?
Bayanin Amsa
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.
Tambaya 25 Rahoto
An element, Q, contains 69% of\(^{63}\)Q and 31% of \(^{65}\)Q. What is the relative atomic mass of Q?
Bayanin Amsa
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.
Tambaya 27 Rahoto
The ratio of carbon atoms of hydrogen atoms in a hydrocarbon is 1:2. If its molecular mass is 56, what is its molecular formula?
Bayanin Amsa
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\(_{8}\).
Tambaya 28 Rahoto
The Bohr model of the atom proposed the existence of
Bayanin Amsa
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.
Tambaya 29 Rahoto
Student X titrated 25 cm\(^{3}\) of Na\(_{2}\)CO\(^{3}\) with 0.1 mol dm\(^{-3}\) HCI, using methyl orange as indicator. Student Y carried out the same exercise but used phenolphthalein as an indicator. Which of the following statements about the titration is true?
Tambaya 31 Rahoto
Which of the following oxide causes acid rain?
Bayanin Amsa
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\(_{2}\).
Tambaya 32 Rahoto
The following ions have the same electron configuration except
\(_{8}\)O, \(_{12}\)Mg, \(_{13}\)AI, \(_{17}\)CI
Tambaya 33 Rahoto
What is the empirical formula of a hydrocarbon containing 0.160 moles of carbon and 0.640 moles of hydrogen
Bayanin Amsa
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\(_{4}\).
Tambaya 34 Rahoto
The region around the nucleus where electrons can be located is called?
Bayanin Amsa
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.
Tambaya 35 Rahoto
How many coulombs of electricity would liberate 1.08g of Ag from a solution of silver salt?
[Ag = 108.0; 1F = 96500 C]
Bayanin Amsa
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.
Tambaya 36 Rahoto
The collision between ideal gas molecules are considered to be perfectly elastic because
Bayanin Amsa
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.
Tambaya 37 Rahoto
Consider the structure below
How many carbon atoms does the parent chain contain?
Bayanin Amsa
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.
Tambaya 38 Rahoto
Which of the following species has the largest ionic radius
Bayanin Amsa
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.
Tambaya 39 Rahoto
The reactivity of fluorine is high because of
Bayanin Amsa
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.
Tambaya 40 Rahoto
What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm\(^{2}\) cylinder?
[R = 0.082 atm dm\(^{3}\) mol\(^{-1}\) K\(^{-1}\)]
Tambaya 41 Rahoto
Elements with high ionization energies would-------
Bayanin Amsa
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.
Tambaya 42 Rahoto
Which of the following pairs of molecules form hydrogen bonds?
Bayanin Amsa
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.
Tambaya 43 Rahoto
The formation of ethene from dehydration of ethanol can be described as-------
Bayanin Amsa
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.
Tambaya 44 Rahoto
Which of the following statements about elements in Group VII is correct?
Bayanin Amsa
Tambaya 45 Rahoto
The energy profile diagram illustrates-------------
Bayanin Amsa
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.
Tambaya 46 Rahoto
The hydrolysis of proteins by diluting mineral acids produces
Bayanin Amsa
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.
Tambaya 47 Rahoto
What is the concentration of a solution which contains 0.28g of potassium hydroxide in 100 cm\(^{3}\) of solution [KOH = 56]
Bayanin Amsa
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.
Tambaya 48 Rahoto
Study the graphs below and use them to answer the question below
Which of the graphs illustrates the variation of the pH of a given volume of strong acid solution (y-axis) with the volume of strong base titrated against its (x-axis)?
Bayanin Amsa
Tambaya 49 Rahoto
The isotopes of neon are represented by the symbols \(^{20}\)\(_{x}\) Ne, \(^{21}\) \(_{y}\) Ne, and \(^{22}\) \(_{z}\) Ne. The relationship between x, y, and z is?
Bayanin Amsa
Tambaya 50 Rahoto
The activation energy of the reaction is
Tambaya 51 Rahoto
(a)i. State two characteristics of a homologous series.
(ii) Explain briefly why there are differences in the reaction of ethane and ethene.
(b) When crystals of sodium chloride were warmed with concentrated tetraoxosulphate( VI) acid, a gas was evolved.
(i) Name the gas.
(ii) State two physical properties of the gas.
(iii) Write a balanced chemical equation for the reaction.
(c)i. What are hydrocarbons?
(ii) State two natural sources of hydrocarbons.
(iii) A hydrocarbon contains 83% of carbon by mass. Calculate its empirical formula. [H=1.0, C=12.0]
(d) Draw and label a diagram of a set-up that could be used to electroplate a copper ornament with silver
(a)
(i) Homologous series have members that have similar chemical properties and successive members differ by a -CH2- unit.
(ii) Ethane is a saturated hydrocarbon and cannot undergo addition reactions, whereas ethene is an unsaturated hydrocarbon and can undergo addition reactions due to the presence of a double bond.
(b)
(i) The gas evolved is hydrogen chloride (HCl).
(ii) Hydrogen chloride is a colorless gas with a pungent odor and is highly soluble in water.
(iii) NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g)
(c)
(i) Hydrocarbons are organic compounds made up of hydrogen and carbon atoms.
(ii) Two natural sources of hydrocarbons are crude oil and natural gas.
(iii) Let's assume we have 100g of the hydrocarbon. Therefore, the mass of carbon present in the hydrocarbon is:
83g = (83/12) mol of carbon
The mass of hydrogen present in the hydrocarbon is:
(100 - 83)g = 17g = (17/1) mol of hydrogen
The empirical formula is the simplest ratio of the atoms in the compound, and it can be calculated by dividing the number of moles of each element by the smallest number of moles. In this case, the smallest number of moles is 1. Therefore, the empirical formula of the hydrocarbon is:
C(83/12)/1 : H(17/1)/1
which simplifies to:
C7H2
(d)
To electroplate a copper ornament with silver, we need a set-up that includes a silver anode, a copper cathode, and a solution of a silver salt such as silver nitrate in water. The ornament is connected to the negative terminal of a power supply, which is also connected to the copper cathode. The silver anode is connected to the positive terminal of the power supply.
When an electric current flows through the set-up, silver ions from the silver anode move towards the copper cathode and gain electrons to form silver atoms, which deposit on the surface of the copper ornament. This process is known as electroplating, and it allows us to coat a less expensive metal with a more expensive metal, such as copper with silver.
Bayanin Amsa
(a)
(i) Homologous series have members that have similar chemical properties and successive members differ by a -CH2- unit.
(ii) Ethane is a saturated hydrocarbon and cannot undergo addition reactions, whereas ethene is an unsaturated hydrocarbon and can undergo addition reactions due to the presence of a double bond.
(b)
(i) The gas evolved is hydrogen chloride (HCl).
(ii) Hydrogen chloride is a colorless gas with a pungent odor and is highly soluble in water.
(iii) NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g)
(c)
(i) Hydrocarbons are organic compounds made up of hydrogen and carbon atoms.
(ii) Two natural sources of hydrocarbons are crude oil and natural gas.
(iii) Let's assume we have 100g of the hydrocarbon. Therefore, the mass of carbon present in the hydrocarbon is:
83g = (83/12) mol of carbon
The mass of hydrogen present in the hydrocarbon is:
(100 - 83)g = 17g = (17/1) mol of hydrogen
The empirical formula is the simplest ratio of the atoms in the compound, and it can be calculated by dividing the number of moles of each element by the smallest number of moles. In this case, the smallest number of moles is 1. Therefore, the empirical formula of the hydrocarbon is:
C(83/12)/1 : H(17/1)/1
which simplifies to:
C7H2
(d)
To electroplate a copper ornament with silver, we need a set-up that includes a silver anode, a copper cathode, and a solution of a silver salt such as silver nitrate in water. The ornament is connected to the negative terminal of a power supply, which is also connected to the copper cathode. The silver anode is connected to the positive terminal of the power supply.
When an electric current flows through the set-up, silver ions from the silver anode move towards the copper cathode and gain electrons to form silver atoms, which deposit on the surface of the copper ornament. This process is known as electroplating, and it allows us to coat a less expensive metal with a more expensive metal, such as copper with silver.
Tambaya 52 Rahoto
(a)i. With the aid of an equation, explain briefly why aluminum metal is not affected by air.
(ii) In the extraction of aluminum from bauxite, state the:
- substance used for purifying the ore;
- composition of the mixture electrolyzed.
(b) ZnO is an amphoteric oxide. Write equations to illustrate this statement.
(c)i) List three uses of sodium trioxocarbonate(IV).
(ii) Explain briefly why a solution of trioxonitrate(V) acid turns yellowish on storage for some time.
(ii) Describe briefly how trioxonitrate(V) ions could be tested for in the laboratory.
(d) Write balanced chemical equations for the preparation of hydrogen chloride.
(i) using concentrated H\(_{2}\)SO\(_{4}\):
(ii) by direct combination of its constituent elements.
(iii) State one use of hydrogen chloride.
(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
Bayanin Amsa
(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
Tambaya 53 Rahoto
(a) Distinguish between molecular formula and structural formula
(b) List three factors that determine the ionization energy of an atom.
(c) State the two conditions necessary for the establishment of a chemical equilibrium
(d) Consider the following table
| Element | A | B | C |
| Ionization energy KJ mol\(^{-1}\) | 619 | 518 | 594 |
(d)(i) State which of the elements is the strongest reducing agent.
(ii) Give a reason for the answer stated in (d)(i)
(e) State Graham's law of diffusion
(f) Consider the following salts: Mg(NO\(_{3}\))\(_{2}\), CaCO\(_{3}\), Na\(_{2}\)SO\(_{4}\). State which of the salts is/are:
(i) readily soluble in water:
(ii) insoluble in water.
(g) Classily each of the following products as addition polymer or condensation polymer:
(i) protein:
(ii) perspex:
(iii) nylon.
(h) Define atomic radius.
(i) Explain briefly why ethanol has a higher boiling point than propane even though they both have comparable molar masses.
(j) State three significance of the pH value in everyday life.
(a) Molecular formula and structural formula are both used to represent the composition of a molecule, but they differ in their level of detail. A molecular formula simply shows the types and numbers of atoms in a molecule, while a structural formula shows the arrangement of atoms and the bonds between them. (b) The three factors that determine the ionization energy of an atom are: - Nuclear charge: the more protons an atom has, the stronger the pull on the electrons, which makes it harder to remove an electron. - Distance from the nucleus: the further away an electron is from the nucleus, the weaker the attraction to the nucleus, which makes it easier to remove an electron. - Shielding effect: the more inner shells of electrons an atom has, the less the outer electrons feel the pull of the nucleus, which makes it easier to remove an electron. (c) The two conditions necessary for the establishment of a chemical equilibrium are: - A reversible reaction: a reaction that can proceed in both forward and reverse directions. - A closed system: a system in which no reactants or products can escape or be added from the outside. (d)(i) Element B is the strongest reducing agent because it has the lowest ionization energy, which means it is easier for it to lose electrons and be oxidized. (ii) A reducing agent is a substance that donates electrons to another substance and gets oxidized in the process. Element B has the lowest ionization energy, so it is easier for it to lose electrons and become oxidized, making it a stronger reducing agent. (e) 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. (f) (i) Na2SO4 is readily soluble in water. (ii) CaCO3 and Mg(NO3)2 are insoluble in water. (g) (i) Protein is a condensation polymer. (ii) Perspex is an addition polymer. (iii) Nylon is a condensation polymer. (h) Atomic radius is the distance between the nucleus of an atom and its valence electrons. (i) Ethanol has a higher boiling point than propane even though they have comparable molar masses because ethanol has stronger intermolecular forces due to its ability to form hydrogen bonds, while propane can only form weak London dispersion forces. (j) Three significance of the pH value in everyday life are: - Biological systems: the pH of bodily fluids such as blood and stomach acid must be maintained within a narrow range for proper functioning of enzymes and other biological processes. - Household cleaning: certain cleaning products work best at specific pH values, such as acidic cleaners for removing mineral deposits and basic cleaners for removing grease and oils. - Agriculture: the pH of soil can affect the availability of nutrients to plants and the growth of crops.
Bayanin Amsa
(a) Molecular formula and structural formula are both used to represent the composition of a molecule, but they differ in their level of detail. A molecular formula simply shows the types and numbers of atoms in a molecule, while a structural formula shows the arrangement of atoms and the bonds between them. (b) The three factors that determine the ionization energy of an atom are: - Nuclear charge: the more protons an atom has, the stronger the pull on the electrons, which makes it harder to remove an electron. - Distance from the nucleus: the further away an electron is from the nucleus, the weaker the attraction to the nucleus, which makes it easier to remove an electron. - Shielding effect: the more inner shells of electrons an atom has, the less the outer electrons feel the pull of the nucleus, which makes it easier to remove an electron. (c) The two conditions necessary for the establishment of a chemical equilibrium are: - A reversible reaction: a reaction that can proceed in both forward and reverse directions. - A closed system: a system in which no reactants or products can escape or be added from the outside. (d)(i) Element B is the strongest reducing agent because it has the lowest ionization energy, which means it is easier for it to lose electrons and be oxidized. (ii) A reducing agent is a substance that donates electrons to another substance and gets oxidized in the process. Element B has the lowest ionization energy, so it is easier for it to lose electrons and become oxidized, making it a stronger reducing agent. (e) 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. (f) (i) Na2SO4 is readily soluble in water. (ii) CaCO3 and Mg(NO3)2 are insoluble in water. (g) (i) Protein is a condensation polymer. (ii) Perspex is an addition polymer. (iii) Nylon is a condensation polymer. (h) Atomic radius is the distance between the nucleus of an atom and its valence electrons. (i) Ethanol has a higher boiling point than propane even though they have comparable molar masses because ethanol has stronger intermolecular forces due to its ability to form hydrogen bonds, while propane can only form weak London dispersion forces. (j) Three significance of the pH value in everyday life are: - Biological systems: the pH of bodily fluids such as blood and stomach acid must be maintained within a narrow range for proper functioning of enzymes and other biological processes. - Household cleaning: certain cleaning products work best at specific pH values, such as acidic cleaners for removing mineral deposits and basic cleaners for removing grease and oils. - Agriculture: the pH of soil can affect the availability of nutrients to plants and the growth of crops.
Tambaya 54 Rahoto
(a) In an experiment, 25.0 cm\(^{3}\) of H\(_{2}\)SO\(_{4}\) completely neutralized 24.0 cm\(^{3}\) of a 0.1 50 mol dm\(^{-3}\) aqueous KOH using a suitable indicator.
(i) Write a balanced chemical equation for the reaction.
(ii) Calculate the concentration of the acid solution.
(b)i. A burning magnesium ribbon was placed in a gas jar containing Carbon (IV) oxide.
- Write an equation for the reaction.
- Explain briefly why the magnesium ribbon burns in carbon(lV) oxide although the gas does not support combustion.
- Calculate the percentage mass of nitrogen in magnesium trioxonitrate (V). |N = 140. O = 16.0. Mg = 24.01]
(c) Consider the following organic compound: CH\(_{3}\)CH\(_{2}\)CH = CHCOOH.
(i) State two chemical reactions which could he sed te identify the compound.
(ii) What would be observed in each of the reactions stated in c(i)
(d) Describe briefly how soap is manufactured using pellets of sodium hydroxide and vegetable oil
(e) Define the term electronegativity
(a)
i. 2KOHaq + H2 SO4(aq) ⟹ K2 SO4(aq) + 2H2 O(l)
ii. Amount of KOH used = 241000×0.1501 = 0.0036mol
Mole ratio of acid to base = 1:2
Amount of H2 SO4 used 0.00362 = 0.0018mol
Concentration of H2 SO4 = 0.00180.025 = 0.072mol dm−3
(b)
(i)
Molar mass Mg(NO
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 N2 = 14 x 2- 28
%Nitrogen = 28149×1001 = 18.9%
(c)
i) Reaction with brominei n CCl4 /bromine water/H+ / K MnO4 / H+ / K2 Cr2 O7
- Reaction with aqueous NaHCO3
/ Na2
CO3
. OR
- Reaction with alkanol + Conc. H+2
SO4
(ii)
- Compound will decolourize bromine in CCl4
/ bromine water
- Change H+
/K\2
Cr2
O7
from orange to green.
- Change H+
/KMnO4
from purple to colourless.
- There will be effervescence/evolution of gas in reaction with NaCO3(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
Bayanin Amsa
(a)
i. 2KOHaq + H2 SO4(aq) ⟹ K2 SO4(aq) + 2H2 O(l)
ii. Amount of KOH used = 241000×0.1501 = 0.0036mol
Mole ratio of acid to base = 1:2
Amount of H2 SO4 used 0.00362 = 0.0018mol
Concentration of H2 SO4 = 0.00180.025 = 0.072mol dm−3
(b)
(i)
Molar mass Mg(NO
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 N2 = 14 x 2- 28
%Nitrogen = 28149×1001 = 18.9%
(c)
i) Reaction with brominei n CCl4 /bromine water/H+ / K MnO4 / H+ / K2 Cr2 O7
- Reaction with aqueous NaHCO3
/ Na2
CO3
. OR
- Reaction with alkanol + Conc. H+2
SO4
(ii)
- Compound will decolourize bromine in CCl4
/ bromine water
- Change H+
/K\2
Cr2
O7
from orange to green.
- Change H+
/KMnO4
from purple to colourless.
- There will be effervescence/evolution of gas in reaction with NaCO3(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
Tambaya 55 Rahoto
(a) What is the structure of:
(i) graphite:
(ii) diamond
(ii) Explain briefly why diamond is hard and a non-conductor of electricity while graphite is soft and an electrical conductor,
(b)i. State what is achieved at each of the following stages in the purification of town water supply:
- aeration;
- screening
- sedimentation.
(ii) Name two substances responsible for hardness in water.
(iii) State two methods for the removal of hardness in water.
(iv) Give one disadvantage of hard water
(c)i). Describe briefly the extraction of tin from its ore.
(ii) Write a balanced chemical equation for the reaction.
(iii) Write an equation for the reaction of tin with:
- oxygen;
- chlorine.
(a)
(i) Graphite is made up of layers of carbon atoms arranged in a hexagonal pattern. Each carbon atom is bonded covalently to three neighboring carbon atoms, forming a flat layer. These layers are stacked on top of each other, held together by weak van der Waals forces.
(ii) Diamond, on the other hand, is made up of carbon atoms arranged in a three-dimensional crystal lattice structure, with each carbon atom bonded covalently to four other carbon atoms. The strong covalent bonds between the carbon atoms give diamond its hardness and high melting point.
Diamond is hard because of its tightly packed, three-dimensional crystal structure, which makes it difficult for other materials to scratch or break the bonds between the carbon atoms. Graphite, in comparison, is soft because its layers are held together by weak van der Waals forces, which allows the layers to slide easily over one another.
Diamond is a non-conductor of electricity because it is an insulator, meaning that electrons cannot flow freely through its crystal structure. Graphite, on the other hand, is a conductor of electricity because its layers can act as a pathway for electrons to flow through.
(b) (i) In the purification of town water supply:
- Aeration involves adding air to the water, which helps to remove dissolved gases and improve the taste and odor of the water.
- Screening involves removing large particles, such as debris or leaves, from the water.
- Sedimentation involves allowing the water to stand in a large tank, allowing suspended particles to settle to the bottom.
(ii) Calcium and magnesium are two substances responsible for hardness in water.
(iii) Two methods for the removal of hardness in water are ion exchange and reverse osmosis.
(iv) One disadvantage of hard water is that it can cause scale buildup in pipes and appliances, reducing their efficiency and lifespan.
(c) (i) Tin is extracted from its ore, typically cassiterite (SnO2), by heating the ore in a furnace with carbon to form tin metal and carbon dioxide gas.
(ii) The balanced chemical equation for the reaction is:
2SnO2 + 5C ? 2Sn + 4CO2
(iii) The reaction of tin with:
- Oxygen: Tin reacts with oxygen to form tin dioxide (SnO2).
- Chlorine: Tin reacts with chlorine to form tin(IV) chloride (SnCl4).
Bayanin Amsa
(a)
(i) Graphite is made up of layers of carbon atoms arranged in a hexagonal pattern. Each carbon atom is bonded covalently to three neighboring carbon atoms, forming a flat layer. These layers are stacked on top of each other, held together by weak van der Waals forces.
(ii) Diamond, on the other hand, is made up of carbon atoms arranged in a three-dimensional crystal lattice structure, with each carbon atom bonded covalently to four other carbon atoms. The strong covalent bonds between the carbon atoms give diamond its hardness and high melting point.
Diamond is hard because of its tightly packed, three-dimensional crystal structure, which makes it difficult for other materials to scratch or break the bonds between the carbon atoms. Graphite, in comparison, is soft because its layers are held together by weak van der Waals forces, which allows the layers to slide easily over one another.
Diamond is a non-conductor of electricity because it is an insulator, meaning that electrons cannot flow freely through its crystal structure. Graphite, on the other hand, is a conductor of electricity because its layers can act as a pathway for electrons to flow through.
(b) (i) In the purification of town water supply:
- Aeration involves adding air to the water, which helps to remove dissolved gases and improve the taste and odor of the water.
- Screening involves removing large particles, such as debris or leaves, from the water.
- Sedimentation involves allowing the water to stand in a large tank, allowing suspended particles to settle to the bottom.
(ii) Calcium and magnesium are two substances responsible for hardness in water.
(iii) Two methods for the removal of hardness in water are ion exchange and reverse osmosis.
(iv) One disadvantage of hard water is that it can cause scale buildup in pipes and appliances, reducing their efficiency and lifespan.
(c) (i) Tin is extracted from its ore, typically cassiterite (SnO2), by heating the ore in a furnace with carbon to form tin metal and carbon dioxide gas.
(ii) The balanced chemical equation for the reaction is:
2SnO2 + 5C ? 2Sn + 4CO2
(iii) The reaction of tin with:
- Oxygen: Tin reacts with oxygen to form tin dioxide (SnO2).
- Chlorine: Tin reacts with chlorine to form tin(IV) chloride (SnCl4).
