WAEC SSCE - Chemistry - 2019
Question 1 Report
consider the following reaction equation: 2NH\(_3 \) .+ 3Cl\(_2\) → N2 + 6HCl.
NH\(_3\) is acting as?
Answer Details
NH\(_3\) is acting as a reducing agent in the given reaction. A reducing agent is a substance that donates electrons, and as a result, it causes the reduction of another substance in a chemical reaction. In the given reaction equation, NH\(_3\) donates electrons to Cl\(_2\), which gets reduced to HCl. At the same time, NH\(_3\) gets oxidized to N\(_2\). Therefore, NH\(_3\) acts as a reducing agent in the given reaction. It is important to note that a reducing agent always gets oxidized in a chemical reaction as it donates electrons to another substance. Similarly, an oxidizing agent always gets reduced as it accepts electrons from another substance.
Question 2 Report
Stoichiometry is based on the law?
Answer Details
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, the total mass of the reactants must equal the total mass of the products. This means that in a chemical reaction, the number of atoms of each element involved must be conserved. This can be used to calculate the amounts of reactants and products needed to produce a given reaction, and to predict the products that will result from a given reaction. The law of conservation of mass is the fundamental principle upon which stoichiometry is based. Therefore, stoichiometry is based on the law of conservation of mass.
Question 3 Report
Which of the following compounds is formed by the oxidation of ethanol?
Answer Details
The compound that is formed by the oxidation of ethanol is acetic acid, which has the chemical formula CH\(_3\)CO\(_2\)H. Ethanol, which has the chemical formula C\(_2\)H\(_5\)OH, can be oxidized by various oxidizing agents, such as potassium dichromate or acidified potassium permanganate, to form acetic acid. During the oxidation process, the two carbon atoms in ethanol are oxidized to form a carbonyl group (C=O) in acetic acid, and two hydrogen atoms are removed to form a carboxyl group (-COOH). The other options listed do not correspond to the oxidation product of ethanol. C\(_2\)H\(_4\)CO\(_2\)H is oxalic acid, which is not derived from ethanol. C\(_2\)H\(_5\)CO\(_2\)H is ethanoic acid, which is another name for acetic acid and is the correct answer. CH\(_3\)OH is methanol, which is not an oxidation product of ethanol. Finally, CH\(_3\)CO\(_2\)H is also acetic acid, but it is not an oxidation product of ethanol since it is already in its fully oxidized form.
Question 4 Report
The process by which alkanoic acid reacts reversibly with alkanols is known as?
Answer Details
The process by which alkanoic acid reacts reversibly with alkanols is known as esterification. During this chemical reaction, a molecule of alkanoic acid and a molecule of an alcohol combine to form an ester and a molecule of water. This reaction is reversible, which means that the products can react with each other to produce the starting materials. Esters are a group of organic compounds that are commonly found in fragrances, flavors, and plastics. They have a characteristic fruity or flowery smell, and they are responsible for the flavor and aroma of many fruits and flowers. Esterification is an important industrial process used to produce a wide variety of products such as perfumes, flavors, and plastics.
Question 5 Report
The method used to collect gas in the laboratory depends on its?
I. boiling point II. density III. smell IV. solubility in water
Answer Details
Question 6 Report
When quicklime dissolves in water?
Answer Details
When quicklime (calcium oxide, CaO) dissolves in water, heat is evolved. This process is known as slaking of lime. When CaO is added to water, it undergoes a vigorous reaction in which it combines with water (H2O) to form calcium hydroxide [Ca(OH)2]. The balanced chemical equation for this reaction is: CaO + H2O -> Ca(OH)2 + heat The heat is released because the reaction is exothermic. This means that the products (Ca(OH)2) have a lower energy than the reactants (CaO and H2O). The excess energy is released in the form of heat. The heat evolved during the slaking of lime makes it useful in a variety of applications, such as in construction and in the treatment of water and wastewater. It can also be used as a source of heat in some industrial processes. Therefore, when quicklime dissolves in water, heat is evolved.
Question 7 Report
If the value of ΔH is positive for a reaction, it means that the reaction is?
Answer Details
If the value of \(\Delta H\) is positive for a reaction, it means that the reaction is endothermic. \(\Delta H\) stands for the change in enthalpy, which is a measure of the amount of heat absorbed or released during a chemical reaction. A positive value of \(\Delta H\) indicates that the reaction absorbs heat from the surroundings. This means that the reaction requires an input of energy in order to proceed. Endothermic reactions are those that absorb heat from the surroundings. In an endothermic reaction, the products have a higher energy than the reactants. The energy is absorbed from the surroundings, so the surroundings become cooler. For example, melting ice is an endothermic process because it absorbs heat from the surroundings. On the other hand, exothermic reactions release heat to the surroundings, and have a negative \(\Delta H\) value. These reactions are often spontaneous and occur without the need for an external energy input. For example, burning wood is an exothermic process because it releases heat to the surroundings. Therefore, if the value of \(\Delta H\) is positive for a reaction, it means that the reaction is endothermic.
Question 8 Report
The group to which elements belong in the periodic table is determined by the number?
Answer Details
The group to which elements belong in the periodic table is determined by the number of valence electrons. Valence electrons are the electrons in the outermost shell (valence shell) of an atom that are involved in chemical bonding. The group number of an element in the periodic table corresponds to the number of valence electrons in its atom. For example, elements in Group 1 (also known as the alkali metals) have one valence electron, elements in Group 2 (also known as the alkaline earth metals) have two valence electrons, and so on. Elements in the same group have similar chemical and physical properties due to their similar valence electron configurations. Therefore, the correct option is "valence electrons".
Question 9 Report
A solid with higher melting and boiling points is likely to be?
Answer Details
A solid with higher melting and boiling points is likely to be an electrovalent compound, also known as an ionic compound. Electrovalent compounds are made up of positively charged ions (cations) and negatively charged ions (anions) held together by electrostatic attractions. These attractions are strong and require a significant amount of energy to break, which is why ionic compounds have higher melting and boiling points compared to covalent compounds, which are held together by weaker intermolecular forces. Covalent compounds consist of molecules with covalent bonds between atoms, where atoms share electrons to form a bond. Covalent compounds typically have low melting and boiling points, as the intermolecular forces are weak. A dative compound, also known as a coordinate compound, contains a coordinate bond where one atom donates both electrons to the bond. These types of compounds can have varying melting and boiling points depending on the strength of the bond. Non-metals can form both covalent and ionic compounds, so their identity alone does not determine the melting and boiling points of a solid. Therefore, the correct option is "An Electrovalent compound".
Question 10 Report
One of the deductions from Rutherford's alpha particle scattering experiment is that?
Answer Details
Rutherford's alpha particle scattering experiment was conducted by bombarding a thin gold foil with alpha particles. The experiment led to the conclusion that most of the alpha particles passed through the gold foil with little or no deflection, but some were scattered at various angles. Rutherford deduced that the positive charge of an atom is concentrated in a tiny nucleus at the center of the atom, which is why most of the alpha particles passed through the gold foil with little or no deflection. Therefore, one of the deductions from Rutherford's alpha particle scattering experiment is that the nuclei of atoms are positively charged.
Question 11 Report
Which of the following methods is NOT used for the separation of mixtures?
Answer Details
Electrolysis is NOT used for the separation of mixtures. Electrolysis is a process in which an electric current is passed through a solution or molten compound to decompose it into its constituent elements or ions. It is used for the extraction of metals from their ores and for the purification of some chemicals. On the other hand, the other three methods mentioned - chromatography, crystallization, and distillation - are commonly used for the separation of mixtures. - Chromatography is a technique used to separate different components of a mixture based on their properties such as size, polarity, and solubility. It is used in fields such as chemistry, biochemistry, and environmental science. - Crystallization is the process of forming solid crystals from a solution, melt, or gas. It is used to purify solids and to separate a mixture of solids into individual components. - Distillation is a process that separates the components of a mixture based on their boiling points. It is used to purify liquids, to separate a mixture of liquids into individual components, and to remove impurities from a liquid. Therefore, electrolysis is NOT used for the separation of mixtures, while the other three methods - chromatography, crystallization, and distillation - are commonly used.
Question 12 Report
Which of the following substances can be classified as heavy chemicals?
Answer Details
Among the options given, H\(_2\)SO\(_4\) (sulfuric acid) can be classified as a heavy chemical. Heavy chemicals refer to chemicals that are produced in large quantities and are used as raw materials in various industries, such as fertilizers, pharmaceuticals, plastics, and dyes. Sulfuric acid is one of the most widely produced chemicals in the world, and it is used in the production of fertilizers, detergents, and other industrial chemicals. Therefore, H\(_2\)SO\(_4\) can be considered a heavy chemical. CaOCl\(_2\) (calcium hypochlorite), AgNO\(_3\) (silver nitrate), and CaO (calcium oxide) are also chemicals but they are not typically classified as heavy chemicals because they are not produced in the same large quantities as sulfuric acid and are not as widely used as raw materials in various industries.
Question 13 Report
The bleaching action of chlorine is through the process?
Answer Details
The bleaching action of chlorine is through the process of oxidation. Chlorine is a strong oxidizing agent, which means that it is able to remove electrons from other molecules. In the presence of water, chlorine reacts with water molecules to form hypochlorous acid, which then breaks down into hypochlorite ions and hydrogen ions. The hypochlorite ions are responsible for the bleaching action. When hypochlorite ions come into contact with a colored molecule, such as a dye or a pigment, they oxidize the molecule by removing electrons. This changes the chemical structure of the molecule, making it colorless. This is why chlorine is such an effective bleaching agent. Oxidation is a chemical process in which a molecule loses electrons. This process is opposite to reduction, in which a molecule gains electrons. Chlorine is a strong oxidizing agent because it has a high tendency to attract electrons from other molecules. This makes it useful in a variety of applications, including water treatment and bleaching.
Question 14 Report
The atoms of \(^{64}_{29}\)Cu and \(^{65}_{30}\)Zn have the same number of?
Answer Details
Question 15 Report
If 6.5g of magnesium ribbon reacts completely with Hydrochloric acid, how many grams of hydrogen gas would be liberated?
Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)
[ H = 1.0, Mg = 24.0 ]
Question 16 Report
An organic compound contains 72% carbon, 12% hydrogen and 16% oxygen by mass. What is the empirical formula of the compound?
[ H = 1.0. C = 12.0, O = 16.0 ]
Answer Details
The empirical formula of a compound is the simplest whole-number ratio of atoms in the compound. To find the empirical formula, we need to convert the percent composition by mass of each element to its corresponding number of moles, and then divide each number by the smallest number of moles to find the smallest whole-number ratio. Given the following percent composition by mass: - Carbon: 72% = 72 g / (12.0 g/mol) = 6.0 mol - Hydrogen: 12% = 12 g / (1.0 g/mol) = 12.0 mol - Oxygen: 16% = 16 g / (16.0 g/mol) = 1.0 mol Dividing each number of moles by the smallest number of moles (1.0 mol for oxygen), we get: - Carbon: 6.0 mol / 1.0 mol = 6.0 - Hydrogen: 12.0 mol / 1.0 mol = 12.0 - Oxygen: 1.0 mol / 1.0 mol = 1.0 Rounding each number to the nearest whole number, we get: - Carbon: 6.0 - Hydrogen: 12.0 - Oxygen: 1.0 The smallest whole-number ratio of atoms in the compound is C\(_6\)H\(_12\)O, which is the empirical formula of the compound.
Question 18 Report
If the oxidation potential of Mg and Al are +2.37 volts and + 1.66 volts respectively, then the Mg would?
Answer Details
Oxidation potential is a measure of the tendency of a substance to lose electrons and get oxidized (or become positively charged). A higher oxidation potential means that the substance is more likely to lose electrons, while a lower oxidation potential means that it is less likely to lose electrons. In the case of Mg and Al, Mg has a higher oxidation potential of +2.37 volts compared to Al's +1.66 volts. This means that Mg is more likely to lose electrons and get oxidized, while Al is less likely to do so. Therefore, Mg will replace Al in a reaction. In other words, Mg will take the place of Al, meaning that Mg will be oxidized while Al will remain in the solution.
Question 19 Report
Which of the substances is non-polar?
Answer Details
Non-polar substances are those that have an equal distribution of electrons and therefore have no positive or negative poles. An example of a non-polar substance is Hydrogen Bromide. On the other hand, polar substances have a positive and negative pole, meaning that the electrons are not evenly distributed. Examples of polar substances are Water and Ammonia. Bromine, like Hydrogen Bromide, is a non-polar substance.
Question 20 Report
Calcium and magnesium belong to the group of the periodic table because both?
Answer Details
Calcium and magnesium belong to the same group in the periodic table because they have the same number of valence electrons in their outermost shell. They both have two valence electrons, which makes them very similar in terms of their chemical behavior. This similarity is also reflected in their properties, such as their ability to form cations and their tendency to form colorless salts. Therefore, the correct option is "have the same number of valence electrons".
Question 21 Report
Which of the following compounds has coordinate bonds in its structure?
Answer Details
The compound that has coordinate bonds in its structure is Al\(_2\)Cl\(_6\). A coordinate bond (also known as a dative bond) is a type of covalent bond in which both electrons in the bond are donated by one atom. In Al\(_2\)Cl\(_6\), each aluminum atom shares its three valence electrons with three chlorine atoms, and each chlorine atom shares one of its electrons with the aluminum atom. This results in a compound with two aluminum atoms that are both surrounded by six chlorine atoms, and each aluminum atom has six shared electrons, three from its own valence electrons and three from the chlorine atoms' lone pairs of electrons. MgCl\(_2\), AgCl, and NaCl do not have coordinate bonds in their structures. MgCl\(_2\) and NaCl have ionic bonds, where the electrons are transferred from one atom to another. AgCl has a polar covalent bond, where the electrons are shared unequally between the two atoms.
Question 22 Report
which of the following elements has the highest ionization energy?
Answer Details
Question 23 Report
How many moles of carbon(iv)oxide contains 16.0g of oxygen?
[ C = 12.0, O = 16.0 ]
Question 24 Report
which of the following solutions containing one mole per dm\(^3\) of the compound would have the lowest PH?
Answer Details
The pH of a solution is a measure of its acidity, with a lower pH indicating a more acidic solution and a higher pH indicating a more basic solution. The pH scale ranges from 0 to 14, with 7 being neutral. Of the options given, hydrochloric acid has the lowest pH because it is a strong acid. Ethanoic acid (also known as acetic acid) is a weak acid, so its pH is higher than that of hydrochloric acid. Sodium chloride is a salt and does not contribute to the acidity of a solution, so its pH would be neutral (7). Sodium hydrogen trioxocarbonate (IV), also known as sodium carbonate, is a basic compound, so its pH would be higher than 7, meaning it is not as acidic as the other options. Therefore, hydrochloric acid would have the lowest pH of the solutions listed.
Question 25 Report
Vinegar is an aqueous solution of?
Answer Details
Vinegar is an aqueous solution of ethanoic acid.
Question 26 Report
Kipps apparatus can be used to prepare?
Answer Details
Kipps apparatus can be used to prepare H2S. Kipps apparatus is a type of apparatus used in chemistry laboratories to prepare and store gases. It consists of a glass bulb that is connected to a tap and a gas generator. The bulb can be filled with a solution of a substance that will react with a base to produce a gas, such as H2S. To prepare H2S, the bulb is filled with a solution of sulfur and an acid, which will react to produce H2S gas. The gas can then be collected and stored in the bulb, and the tap can be used to release the gas as needed. Overall, Kipps apparatus is a useful tool for preparing and storing gases, and is commonly used in chemistry and other scientific fields to produce and store gases like H2S for research and experimentation.
Question 27 Report
A sample of air was bubbled into water. The PH of the water slowly changed from 7 to 6. Which of the following gases in the samples caused the change?
Answer Details
The pH of water is a measure of its acidity, with a lower pH indicating a more acidic solution and a higher pH indicating a more basic solution. When the pH of water changes from 7 to 6, it means that the water has become more acidic. Carbon dioxide (CO\(_2\)) is known to dissolve in water and form carbonic acid, which is a weak acid. This weak acid will increase the acidity of the water, causing its pH to decrease. Oxygen and nitrogen are not acidic gases and would not cause the pH of the water to change. Carbon(II) oxide, also known as carbon monoxide, is a toxic gas that can interfere with the ability of the blood to transport oxygen, but it is not an acidic gas and would not cause the pH of the water to change. Therefore, the gas that caused the change in the pH of the water from 7 to 6 is carbon dioxide (CO\(_2\)).
Question 28 Report
The number of electrons in the 3d orbital of \(_24\)Cr is
Answer Details
The number of electrons in the 3d orbital of \(_{24}\)Cr is 5. In the electronic configuration of chromium (Cr), the atomic number 24 indicates that there are 24 electrons present in the atom. The electronic configuration of Cr is: 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^1\) 3d\(^5\). The electronic configuration tells us that there are five electrons present in the 3d orbital of Cr. This is because the 3d orbital can hold up to 10 electrons, and in the case of Cr, only five of those slots are occupied by electrons. Therefore, the number of electrons in the 3d orbital of \(_{24}\)Cr is 5.
Question 29 Report
Which of the following pairs of components have the same molecular mass?
[ H = 1.0, C = 12.0, O = 16.0 ]
Answer Details
Molecular mass is the sum of the atomic masses of all the atoms in a molecule. To determine which pair of components have the same molecular mass, we need to calculate the molecular mass of each component and compare them. The molecular mass of each component is calculated by adding up the atomic masses of all the atoms in the molecule. - Ethene: C\(_2\)H\(_4\) = (2 x 12.0) + (4 x 1.0) = 28.0 - Ethyne: C\(_2\)H\(_2\) = (2 x 12.0) + (2 x 1.0) = 26.0 - Ethane: C\(_2\)H\(_6\) = (2 x 12.0) + (6 x 1.0) = 30.0 - Carbon (ii) Oxide: CO = (1 x 12.0) + (1 x 16.0) = 28.0 - Carbon (iv) Oxide: CO\(_2\) = (1 x 12.0) + (2 x 16.0) = 44.0 From the calculations above, we can see that ethene and carbon (ii) oxide have the same molecular mass of 28.0.
Question 30 Report
What is the name of the gas liberated when diluted hydrochloric acid is added to iron(ii)sulphide?
Answer Details
Question 31 Report
in Boyle's law?
Question 32 Report
which of the following reagents could be used to distinguish between propan-1-ol and propan-2-ol
Answer Details
Question 33 Report
which of the following pairs of substances has the highest heat of neutralization?
Answer Details
The pair of substances that has the highest heat of neutralization is the one composed of a strong acid and a strong base. When an acid reacts with a base, they undergo a neutralization reaction, in which an acid donates a proton (H+) to a base to form water and a salt. The heat released during this reaction is called the heat of neutralization. Strong acids and bases are highly reactive, meaning that they will release a large amount of heat when they neutralize each other. This is because they completely dissociate in water to form ions, making their reaction highly exothermic. On the other hand, weak acids and bases have a lower degree of dissociation and are less reactive, meaning that they will release less heat when they neutralize each other. This is because only a small fraction of the weak acid or base molecules will react, and the reaction will not be as exothermic as that of strong acids and bases. Therefore, the highest heat of neutralization will be observed in the reaction between a strong acid and a strong base.
Question 34 Report
If a catalyst is added to a system, the temperature and the pressure remain constant, there would be no effect on the?
Answer Details
A catalyst is a substance that speeds up the rate of a chemical reaction without being consumed in the process. When a catalyst is added to a system, the activation energy of the reaction remains unchanged, but the rate of the reaction increases. The activation energy of a reaction is the minimum amount of energy that must be supplied to the reactants for the reaction to occur. The addition of a catalyst does not change this energy requirement, but it can provide an alternative reaction pathway with a lower activation energy, allowing the reaction to proceed more easily and quickly. The rate of the reverse reaction, which is the rate at which the products of the reaction revert back to the reactants, may also be affected by the addition of a catalyst, but this depends on the specific reaction and the type of catalyst being used. The heat of reaction, which is the change in enthalpy (heat content) that occurs during a chemical reaction, is also not affected by the addition of a catalyst. This is because a catalyst only speeds up the reaction rate, but does not change the energy required for the reaction to occur. Therefore, the addition of a catalyst to a system with constant temperature and pressure would affect the rate of the reaction, but would not affect the activation energy, rate of the reverse reaction, or heat of reaction.
Question 35 Report
Consider the following reaction represented by the following equation: C\(_2\)H\(_2\) \(^{H\(_2\)}_{→}\) X \(^{H\(_2\)}_{→}\) Y. X and Y respectively are?
Answer Details
The reactant in the equation is C\(_2\)H\(_2\), which is the chemical formula for ethyne (also known as acetylene). The products are formed by adding hydrogen gas (H\(_2\)) to the ethyne molecule. When ethyne reacts with hydrogen, it undergoes a reduction reaction and forms ethene (also known as ethylene) as the first product, which still has a carbon-carbon double bond, and then ethane, which is a saturated hydrocarbon and has only single bonds between carbon atoms, as the second product. Therefore, the products X and Y, respectively, are ethene and ethane. So, the correct option is "ethene and ethane".
Question 36 Report
The most suitable indicator for the reaction between NaOH and HCOOH is?
Answer Details
The best indicator for the reaction between NaOH (sodium hydroxide) and HCOOH (formic acid) is phenolphthalein. Phenolphthalein is a pH indicator that changes color based on the acidity of the solution. In an acidic solution, it is colorless, while in a basic (alkaline) solution, it turns pink or red. This makes it ideal for monitoring the progress of a reaction between an acid and a base. As the reaction between NaOH and HCOOH proceeds and the concentration of hydroxide ions increases, the solution will become more basic, causing the phenolphthalein to change color, indicating that the reaction is complete.
Question 37 Report
When excess chlorine is mixed with ethane at room temperature, the product formed is?
Answer Details
When excess chlorine is mixed with ethane at room temperature, the product formed is 1,2-dichloroethane. This is because ethane reacts with chlorine in the presence of ultraviolet light or heat to form a mixture of 1,2-dichloroethane and 1,1,2,2-tetrachloroethane. However, in the absence of light or heat, only 1,2-dichloroethane is produced as the major product.
Question 38 Report
Which of the following metals reacts slowly with cold water?
Answer Details
Of the given options, calcium is the metal that reacts slowly with cold water. Calcium reacts with water, producing hydrogen gas and calcium hydroxide. The reaction is exothermic but the heat produced is not sufficient to ignite the hydrogen gas. The other metals listed in the options - iron, silver, and potassium - are more reactive with water than calcium. Iron reacts with steam, while silver and potassium react vigorously with cold water. Therefore, the correct option is "Calcium".
Question 39 Report
The properties of a good primary standard solution include the following except?
Answer Details
Question 40 Report
When an equilibrium is established between dissolved and undissolved solutes, the solution is said to be?
Answer Details
Question 41 Report
Which of the following apparatus is NOT used in volumetric analysis?
Answer Details
A desiccator is not used in volumetric analysis. A desiccator is a container used for preserving moisture-sensitive items by creating a dry and airtight environment. It is commonly used in laboratory settings to store chemicals and other materials that are sensitive to moisture. On the other hand, a pipette and a burette are essential apparatus in volumetric analysis. A pipette is a laboratory tool used to measure and dispense precise volumes of liquid, while a burette is a long, graduated glass tube with a precisely adjustable valve used for adding a solution of known volume to a reaction mixture. A conical flask is also commonly used in volumetric analysis. It is a type of laboratory flask with a cone-shaped body and a narrow neck. It is used to mix and store solutions, as well as to hold reactions while they are being observed.
Question 42 Report
which of the following organic compounds has the lowest boiling point?
Answer Details
The organic compound with the lowest boiling point is (CH\(_3\))\(_3\). This is because it is a gas at room temperature and has a very low molecular weight. As the molecular weight of the organic compound increases, the boiling point also increases. Therefore, C\(_4\)H\(_10\) has a higher boiling point than (CH\(_3\))\(_3\), C\(_5\)H\(_12\) has a higher boiling point than C\(_4\)H\(_10\), and C\(_6\)H\(_14\) has the highest boiling point of all.
Question 43 Report
The periodic table is an arrangement of elements according to their?
Answer Details
The periodic table is an arrangement of elements according to their atomic number, which is the number of protons in the nucleus of an atom. The elements are arranged in order of increasing atomic number in rows and columns, with elements in the same row (period) having the same number of electron shells and elements in the same column (group/family) having similar chemical and physical properties due to their similar electronic configurations. The periodic table provides a systematic way to organize and study the properties of elements, allowing chemists and scientists to predict the behavior of elements and their compounds. The atomic number is a fundamental property of an element that determines its identity and properties, so it is the most important factor in arranging the elements in the periodic table. Therefore, the correct option is "Atomic number".
Question 44 Report
consider the following reaction equation: 2Al(OH)\(_3\) + 3H\(_2\)SO\(_4\) →Al\(_2\)(SO4)\(_3\)(aq)+6H\(_2\)O. How many moles of tetraoxosulphate(vi)acid is needed to produce 0.5mol of aluminium tetraoxosulphate(vi)?
Question 45 Report
consider the reaction represented by the following equation: 2SO\(_2\) + O\(_2\) ⇌ 2SO\(_3\)
A decrease in the pressure of the reaction would
Answer Details
Question 46 Report
What is the solubility of a salt X at 25ºC if the saturated solution of the salt contains 0.28g in 100cm\(^3\) of the solution in mol\dm\(^3\)?
[ X = 56.0g ]
Answer Details
The solubility of a salt in a solution is the maximum amount of the salt that can dissolve in the solution at a given temperature to form a homogeneous mixture. It is typically expressed as the concentration of the salt in the solution, in units of moles per liter (mol/L) or grams per liter (g/L). To determine the solubility of a salt, you need to know the mass of the salt that dissolves in a given volume of solution at a given temperature. In this case, the saturated solution of the salt contains 0.28 g of the salt in 100 cm\(^3\) of the solution, which means that 0.28 g of the salt dissolves in 100 cm\(^3\) of the solution. To convert the mass of the salt to moles, you need to know the molecular weight of the salt. Given that the molecular weight of the salt is 56.0 g, 0.28 g of the salt is equal to 0.28/56.0 = 0.005 moles of the salt. The concentration of the salt in the solution is then given by the number of moles of the salt divided by the volume of the solution, which is 0.005 moles/100 cm\(^3\) = 0.05 mol/L. Therefore, the solubility of the salt X at 25ºC is 0.05 mol/L or 0.05 M.
Question 47 Report
Biotechnology is applied in the?
Answer Details
Biotechnology is applied in the treatment of sewage, treatment of water, and purification of chemicals. Biotechnology is the application of living organisms or their products to improve human health and the environment. In the case of sewage treatment, biotechnology is used to degrade organic matter and remove pathogens from wastewater, making it safe for discharge into the environment. This can be done using microbes such as bacteria and fungi that break down organic matter through processes such as aerobic and anaerobic digestion. Similarly, in water treatment, biotechnology can be used to remove pollutants and pathogens from water sources to make them safe for drinking or other uses. This can involve the use of living organisms such as bacteria or algae to remove contaminants or the use of biological processes such as bioremediation to degrade pollutants. In addition, biotechnology can be applied in the purification of chemicals. This can involve the use of enzymes or other biological molecules to catalyze chemical reactions or the use of microbes to produce chemicals such as antibiotics or enzymes. Purification of metals, on the other hand, typically involves physical and chemical methods such as smelting, leaching, and electrolysis, and does not typically involve biotechnology.
Question 48 Report
The oxidation of chromium in Cr\(_2\)O\(_7)\(-2\) ?
Answer Details
The oxidation state of chromium in the Cr\(_2\)O\(_7)\(-2\) ion is +6. The oxidation state of an element refers to the number of electrons it has lost or gained in a chemical reaction. In the Cr\(_2\)O\(_7)\(-2\) ion, there are two chromium atoms, each with a different oxidation state. To determine the oxidation state of chromium in the Cr\(_2\)O\(_7)\(-2\) ion, we need to consider the overall charge of the ion, which is -2. Oxygen has an oxidation state of -2, so the seven oxygen atoms in the ion contribute a total oxidation state of -14. We also know that the oxidation state of the entire Cr\(_2\)O\(_7)\(-2\) ion is -2, so the two chromium atoms must contribute a total oxidation state of +12 to balance out the negative charge. Since there are two chromium atoms, each chromium atom must have an oxidation state of +6. This is because +6 + +6 = +12, which balances out the -2 charge on the ion. Therefore, the oxidation state of chromium in the Cr\(_2\)O\(_7)\(-2\) ion is +6.
Question 49 Report
In a mixture of gases which do not react chemically together, the pressure of the individual gas is?
Answer Details
In a mixture of gases that do not react chemically together, the pressure of each individual gas is called its partial pressure. When different gases are mixed together in a container, each gas behaves independently of the others. The total pressure exerted by the mixture of gases is equal to the sum of the partial pressures of each gas present in the mixture. This is known as Dalton's law of partial pressures. Mathematically, this can be represented as: Total pressure = Partial pressure of gas A + Partial pressure of gas B + ... + Partial pressure of gas N The partial pressure of each gas in the mixture depends on its mole fraction, which is the ratio of the number of moles of that gas to the total number of moles of all gases present in the mixture. The partial pressure of a gas can be calculated using the following formula: Partial pressure of gas A = Mole fraction of gas A x Total pressure Therefore, in a mixture of gases that do not react chemically together, the pressure of the individual gas is known as its partial pressure, and it can be calculated using the above formula.
Question 50 Report
The name of the compound CH\(_3\)C(CH\(_3\))CH\(_2\)CH\(_2\)CH\(_3\)
Answer Details
The name of a organic compound is based on its molecular structure, which includes the arrangement of its carbon and hydrogen atoms. The compound CH\(_3\)C(CH\(_3\))CH\(_2\)CH\(_2\)CH\(_3\) has a total of seven carbon atoms, which form a straight chain. The first carbon atom has three methyl groups (CH\(_3\)) attached to it, while the second carbon atom has only one methyl group attached to it. Based on these characteristics, the name of the compound is 2,2-dimethylpentane. The prefix "2,2-dimethyl" indicates that there are two methyl groups attached to the second carbon atom, and the suffix "pentane" indicates that there are five carbon atoms in the chain.
Question 51 Report
(a) (i) What is meant by hardness of water?
(ii) Describe briefly how you would determine what proportion of hardness in a given sample of water is due to permanent hardness.
(iii) Give two reasons why hardness of water is an undesirable property.
(b) State the:
(i) reagents;
(ii) conditions for the laboratory preparation of trioxonitrate (V) acid.
(iii) How does concentrated trioxonitrate (V) acid reacts with:
(I) sulphur;
(II) aluminium.
(c) Name one amphoteric oxide.
(a)
(i) Hardness of water refers to the presence of dissolved metal ions, mainly calcium and magnesium, in water. These ions can form insoluble salts with soap, reducing its effectiveness, and can also cause scale buildup in pipes and appliances.
(ii) To determine the proportion of hardness due to permanent hardness in a sample of water, we can first determine the total hardness by titrating the water sample with a solution of ethylenediaminetetraacetic acid (EDTA) using an indicator such as Eriochrome Black T. The EDTA will react with all the metal ions present in the water. Then, we can remove the temporary hardness by boiling the water sample to precipitate out the calcium and magnesium bicarbonates. The difference between the total hardness and the hardness after boiling will give us the proportion of permanent hardness.
(iii) Hardness of water is undesirable because it can cause scale buildup in pipes and appliances, leading to decreased efficiency and increased maintenance costs. It can also reduce the effectiveness of soaps and detergents, requiring more of these products to achieve the desired cleaning results.
(b)
(i) Reagents for the laboratory preparation of trioxonitrate (V) acid include concentrated sulfuric acid and a nitrate salt, such as potassium nitrate or sodium nitrate.
(ii) Trioxonitrate (V) acid can be prepared in the laboratory by adding a nitrate salt to concentrated sulfuric acid while cooling the mixture in an ice bath. The mixture is then gently warmed until trioxonitrate (V) acid fumes appear.
(iii)
(I) When concentrated trioxonitrate (V) acid reacts with sulfur, it produces nitrogen dioxide gas, sulfur dioxide gas, and oxygen gas.
(II) When concentrated trioxonitrate (V) acid reacts with aluminum, it produces aluminum nitrate, water, and nitrogen dioxide gas.
(c)
One example of an amphoteric oxide is aluminum oxide (Al2O3). An amphoteric oxide is a type of oxide that can react with both acids and bases to form salts and water. Aluminum oxide can react with acids to form salts such as aluminum nitrate, and can also react with bases to form salts such as sodium aluminate.
Answer Details
(a)
(i) Hardness of water refers to the presence of dissolved metal ions, mainly calcium and magnesium, in water. These ions can form insoluble salts with soap, reducing its effectiveness, and can also cause scale buildup in pipes and appliances.
(ii) To determine the proportion of hardness due to permanent hardness in a sample of water, we can first determine the total hardness by titrating the water sample with a solution of ethylenediaminetetraacetic acid (EDTA) using an indicator such as Eriochrome Black T. The EDTA will react with all the metal ions present in the water. Then, we can remove the temporary hardness by boiling the water sample to precipitate out the calcium and magnesium bicarbonates. The difference between the total hardness and the hardness after boiling will give us the proportion of permanent hardness.
(iii) Hardness of water is undesirable because it can cause scale buildup in pipes and appliances, leading to decreased efficiency and increased maintenance costs. It can also reduce the effectiveness of soaps and detergents, requiring more of these products to achieve the desired cleaning results.
(b)
(i) Reagents for the laboratory preparation of trioxonitrate (V) acid include concentrated sulfuric acid and a nitrate salt, such as potassium nitrate or sodium nitrate.
(ii) Trioxonitrate (V) acid can be prepared in the laboratory by adding a nitrate salt to concentrated sulfuric acid while cooling the mixture in an ice bath. The mixture is then gently warmed until trioxonitrate (V) acid fumes appear.
(iii)
(I) When concentrated trioxonitrate (V) acid reacts with sulfur, it produces nitrogen dioxide gas, sulfur dioxide gas, and oxygen gas.
(II) When concentrated trioxonitrate (V) acid reacts with aluminum, it produces aluminum nitrate, water, and nitrogen dioxide gas.
(c)
One example of an amphoteric oxide is aluminum oxide (Al2O3). An amphoteric oxide is a type of oxide that can react with both acids and bases to form salts and water. Aluminum oxide can react with acids to form salts such as aluminum nitrate, and can also react with bases to form salts such as sodium aluminate.
Question 52 Report
(a) State the conditions necessary for the cracking of long-chain hydrocarbons to produce more gasoline.
(b) State two reasons why metallic objects are electroplated
(c) (i) Explain briefly why calcium oxide cannot be used to dry hydrogen chloride gas.
(ii) State one drying agent for hydrogen chloride gas.
(d) Concentrated trioxonitrate (V) acid was added to a solution of iron (II) tetraoxosulphate (VI) and the mixture heated. The mixture turned from pale green to yellow with the evolution of a brown gas. Explain briefly these observations.
(e) (i) Write the equation for the reaction between zinc oxide and
- dilute tetraoxosulphate (VI) acid,
- sodium hydroxide solution.
(ii) State which property of zinc oxide is shown by the reaction in (e)(i).
(f) Two isotopes of chlorine are \(^{35}_{17}\)Cl and \(^{37}_{17}\)Cl State one:
- similarity;
- difference between the isotopes.
(g) State the two products formed when chlorine water is exposed to sunlight.
(h) Consider the reaction represented by the following equation:
State the:
- species that is undergoing oxidation;
- oxidizing agent.
(i) What is meant by carbon-12 scale?
(j) State two properties of a chemical system in equilibrium.
(a) The cracking of long-chain hydrocarbons to produce more gasoline requires the following conditions:
- High temperature, typically between 500 and 900 °C.
- Presence of a catalyst, such as silica or alumina.
- High pressure, usually around 1 atmosphere or higher.
(b) Metallic objects are electroplated for two main reasons:
- To improve their appearance or to provide a decorative finish.
- To protect them from corrosion or wear and tear.
(c)
- (i) Calcium oxide cannot be used to dry hydrogen chloride gas because it reacts with it to form calcium chloride, which is a hygroscopic compound and will not remove water vapor from the gas.
- (ii) Phosphorus pentoxide (P2O5) can be used as a drying agent for hydrogen chloride gas.
(d) The pale green color of the iron (II) tetraoxosulphate (VI) solution indicates the presence of Fe2+ ions. Upon adding concentrated trioxonitrate (V) acid and heating, the Fe2+ ions are oxidized to Fe3+ ions, which give the yellow color to the solution. The brown gas evolved is likely to be nitrogen dioxide (NO2), which is a product of the reaction between the trioxonitrate (V) acid and the Fe2+ ions.
(e)
- (i) The equation for the reaction between zinc oxide and dilute tetraoxosulphate (VI) acid is:
ZnO + H2SO4 → ZnSO4 + H2O - (ii) The property of zinc oxide shown by the reaction is its basic nature, as it reacts with an acid to form a salt and water.
(f) Similarity: Both isotopes of chlorine have the same number of electrons and protons, and hence have the same chemical properties.
Difference: The isotopes have different atomic masses due to the presence of different numbers of neutrons in their nuclei.
(g) When chlorine water is exposed to sunlight, the two products formed are:
- Chlorine free radicals (Cl*) which are highly reactive and can participate in further reactions.
- Oxygen gas (O2) which is a byproduct of the reaction between the Cl* radicals and water molecules.
(h) The reaction can be represented as follows:
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Species undergoing oxidation: Fe2+ ions
Oxidizing agent: Cr2O72- ions
(i) The carbon-12 scale is a standard used to define the relative atomic masses of elements. It is based on the isotope carbon-12
Answer Details
(a) The cracking of long-chain hydrocarbons to produce more gasoline requires the following conditions:
- High temperature, typically between 500 and 900 °C.
- Presence of a catalyst, such as silica or alumina.
- High pressure, usually around 1 atmosphere or higher.
(b) Metallic objects are electroplated for two main reasons:
- To improve their appearance or to provide a decorative finish.
- To protect them from corrosion or wear and tear.
(c)
- (i) Calcium oxide cannot be used to dry hydrogen chloride gas because it reacts with it to form calcium chloride, which is a hygroscopic compound and will not remove water vapor from the gas.
- (ii) Phosphorus pentoxide (P2O5) can be used as a drying agent for hydrogen chloride gas.
(d) The pale green color of the iron (II) tetraoxosulphate (VI) solution indicates the presence of Fe2+ ions. Upon adding concentrated trioxonitrate (V) acid and heating, the Fe2+ ions are oxidized to Fe3+ ions, which give the yellow color to the solution. The brown gas evolved is likely to be nitrogen dioxide (NO2), which is a product of the reaction between the trioxonitrate (V) acid and the Fe2+ ions.
(e)
- (i) The equation for the reaction between zinc oxide and dilute tetraoxosulphate (VI) acid is:
ZnO + H2SO4 → ZnSO4 + H2O - (ii) The property of zinc oxide shown by the reaction is its basic nature, as it reacts with an acid to form a salt and water.
(f) Similarity: Both isotopes of chlorine have the same number of electrons and protons, and hence have the same chemical properties.
Difference: The isotopes have different atomic masses due to the presence of different numbers of neutrons in their nuclei.
(g) When chlorine water is exposed to sunlight, the two products formed are:
- Chlorine free radicals (Cl*) which are highly reactive and can participate in further reactions.
- Oxygen gas (O2) which is a byproduct of the reaction between the Cl* radicals and water molecules.
(h) The reaction can be represented as follows:
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Species undergoing oxidation: Fe2+ ions
Oxidizing agent: Cr2O72- ions
(i) The carbon-12 scale is a standard used to define the relative atomic masses of elements. It is based on the isotope carbon-12
Question 53 Report
(a) In the Solvay process, explain briefly with equations the functions of the following substances:
- limestone;
- ammonia.
(b) (i) Write a chemical equation for the fermentation of glucose.
(ii) Explain briefly why a tightly-corked glass bottle filled to the brim with fresh palm-wine shatters on standing for some time.
(c) Consider the following metals: Na, Fe, K and Cu.
(i) Arrange the metals in order of increasing reactivity.
(ii) Which of the metals will react with cold water?
(ii) Which of the metals could form coloured salts?
(d)(i) What is a redox reaction?
(ii) Identify which of the following reaction equations are redox.
(i) 2Na + Cl\(_2\) ? 2NaCl
(ii) AgCl + 2NH\(_3\) ? [Ag(NH\(_3\))\(_2\)]Cl
(iii) C\(_2\)H\(_2\) + H\(_2\) ? C\(_2\)H\(_4\)
(IV) HCl + KOH ? KCl + H\(_2\)O
(V) 2FeCl\(_3\) + 2KI ? 2FeCl\(_2\) + 2KCl + I\(_2\)
(iii) Give a reason for each of the answers in (d)(ii).
(iv) Write balanced equations of the half reactions for any two of the redox reactions in (d)(ii).
Solvay Process and Chemical Reactions
(a) Functions in the Solvay Process:
- Limestone (CaCO3):
- Function: Source of calcium oxide (CaO) for the regeneration of ammonia.
- Equation: CaCO3 (g) -> CaO (s) + CO2 (g) (Limestone decomposes to lime and carbon dioxide)
- Ammonia (NH3):
- Function:
- Reacts with calcium and chloride ions in brine to form ammonium chloride (NH4Cl).
- Makes the solution alkaline, favoring the precipitation of calcium carbonate.
- Equation: Not directly involved in a reaction, but participates in the overall process.
- Function:
(b) Fermentation and Shattered Bottle:
(i) Glucose Fermentation:
C6H12O6 (aq) -> 2C2H5OH (aq) + 2CO2 (g) (Glucose ferments to ethanol and carbon dioxide)
(ii) Shattered Bottle:
Fresh palm wine undergoes fermentation by yeast. During fermentation, yeast consumes sugar and produces carbon dioxide gas. In a tightly sealed bottle, the CO2 gas gets trapped and builds up pressure. As pressure increases, the glass bottle can no longer withstand the force and shatters.
(c) Metal Reactivity:
(i) Increasing Reactivity: K > Na > Fe > Cu
(ii) Reacting with Cold Water: K, Na (These alkali metals react vigorously with water)
(iii) Colored Salts: Cu (Copper forms characteristic colored salts, like blue copper sulfate)
(d) Redox Reactions and Half Reactions:
(i) Redox Reaction:
A redox reaction is a type of chemical reaction where there is a transfer of electrons between the participating atoms or ions. Oxidation is the loss of electrons, and reduction is the gain of electrons.
(ii) Identifying Redox Reactions:
- Redox: (II), (V)
- Not Redox: (I), (III), (IV)
(iii) Explanation for Redox Reactions:
- (II) AgCl + 2NH3 -> [Ag(NH3)2]Cl:
- Silver (Ag) loses an electron to become Ag+ (oxidation).
- Nitrogen (N) in ammonia gains an electron to form the complex ion [Ag(NH3)2]+ (reduction).
- (V) 2FeCl3 + 2KI -> 2FeCl2 + 2KCl + I2:
- Iron (Fe) in FeCl3 loses an electron to become Fe2+ in FeCl2 (oxidation).
- Iodine (I) gains an electron to form I2 (reduction).
(iv) Balanced Half Reactions (Examples):
- Oxidation Half Reaction (from reaction II):
Ag -> Ag+ + e-
- Reduction Half Reaction (from reaction V):
I- -> I2 + e- (Note: I- needs to be balanced with Cl- from the reactant side)
Answer Details
Solvay Process and Chemical Reactions
(a) Functions in the Solvay Process:
- Limestone (CaCO3):
- Function: Source of calcium oxide (CaO) for the regeneration of ammonia.
- Equation: CaCO3 (g) -> CaO (s) + CO2 (g) (Limestone decomposes to lime and carbon dioxide)
- Ammonia (NH3):
- Function:
- Reacts with calcium and chloride ions in brine to form ammonium chloride (NH4Cl).
- Makes the solution alkaline, favoring the precipitation of calcium carbonate.
- Equation: Not directly involved in a reaction, but participates in the overall process.
- Function:
(b) Fermentation and Shattered Bottle:
(i) Glucose Fermentation:
C6H12O6 (aq) -> 2C2H5OH (aq) + 2CO2 (g) (Glucose ferments to ethanol and carbon dioxide)
(ii) Shattered Bottle:
Fresh palm wine undergoes fermentation by yeast. During fermentation, yeast consumes sugar and produces carbon dioxide gas. In a tightly sealed bottle, the CO2 gas gets trapped and builds up pressure. As pressure increases, the glass bottle can no longer withstand the force and shatters.
(c) Metal Reactivity:
(i) Increasing Reactivity: K > Na > Fe > Cu
(ii) Reacting with Cold Water: K, Na (These alkali metals react vigorously with water)
(iii) Colored Salts: Cu (Copper forms characteristic colored salts, like blue copper sulfate)
(d) Redox Reactions and Half Reactions:
(i) Redox Reaction:
A redox reaction is a type of chemical reaction where there is a transfer of electrons between the participating atoms or ions. Oxidation is the loss of electrons, and reduction is the gain of electrons.
(ii) Identifying Redox Reactions:
- Redox: (II), (V)
- Not Redox: (I), (III), (IV)
(iii) Explanation for Redox Reactions:
- (II) AgCl + 2NH3 -> [Ag(NH3)2]Cl:
- Silver (Ag) loses an electron to become Ag+ (oxidation).
- Nitrogen (N) in ammonia gains an electron to form the complex ion [Ag(NH3)2]+ (reduction).
- (V) 2FeCl3 + 2KI -> 2FeCl2 + 2KCl + I2:
- Iron (Fe) in FeCl3 loses an electron to become Fe2+ in FeCl2 (oxidation).
- Iodine (I) gains an electron to form I2 (reduction).
(iv) Balanced Half Reactions (Examples):
- Oxidation Half Reaction (from reaction II):
Ag -> Ag+ + e-
- Reduction Half Reaction (from reaction V):
I- -> I2 + e- (Note: I- needs to be balanced with Cl- from the reactant side)
Question 54 Report
(a) The following reaction scheme is an illustration of the contact process. Study the scheme and answer the questions that follow.
(i) Name X and Y
(ii) Write a balanced chemical equation for each of the processes I, II, III and IV
(iii) Name the catalyst used in process II
(iv) Using Le Chatelier's principle, explain briefly why increasing the temperature would not favour the reaction in II
(v) State two uses of SO\(_2\)
(b) Consider the following equation: 2H\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2H\(_2\)O\(_{(g)}\)
Calculate the volume of unused oxygen gas when 40 cm\(^3\) of hydrogen gas is sparked with 30cm\(^3\) of oxygen gas
(c) Calcium carbonate of mass 1.0 g was heated until there was no further change.
- Write an equation for the reaction which took place.
- Calculate the mass of the residue.
- Calculate the volume of the gas evolved at s.t.p.
- What would be the volume of the gas measured at 15 and 760 mm Hg? [C= 12.0, O = 16.0, Ca = 40.0, molar volume of a gas at s.t.p. = 22.4 dm\(^3\) ]
a)
(i) X is sulfur dioxide (SO2) and Y is sulfur trioxide (SO3).
(ii)
Process I: Sulfur (S) is burned to form sulfur dioxide (SO2).
S (s) + O2 (g) → SO2 (g)
Process II: Sulfur dioxide (SO2) is oxidized to form sulfur trioxide (SO3) using oxygen (O2) in the presence of a catalyst, vanadium(V) oxide (V2O5).
2SO2 (g) + O2 (g) → 2SO3 (g)
Process III: Sulfur trioxide (SO3) is dissolved in concentrated sulfuric acid (H2SO4) to form oleum (H2S2O7).
SO3 (g) + H2SO4 (l) → H2S2O7 (l)
Process IV: Oleum (H2S2O7) is diluted with water (H2O) to form concentrated sulfuric acid (H2SO4).
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
(iii) The catalyst used in process II is vanadium(V) oxide (V2O5).
(iv) In process II, the reaction is exothermic (it releases heat). According to Le Chatelier's principle, increasing the temperature would shift the equilibrium in the opposite direction, favoring the reactants (SO2 and O2). Therefore, increasing the temperature would not favor the forward reaction in process II.
(v) Two uses of SO2 are:
- - It is used in the production of sulfuric acid (H2SO4), which is an important industrial chemical.
- - It is used as a preservative for dried fruits, such as apricots, to prevent them from turning brown.
b)
The equation given is:
2H2 (g) + O2 (g) → 2H2O (g)
To solve the problem, we can use the mole ratio
Answer Details
a)
(i) X is sulfur dioxide (SO2) and Y is sulfur trioxide (SO3).
(ii)
Process I: Sulfur (S) is burned to form sulfur dioxide (SO2).
S (s) + O2 (g) → SO2 (g)
Process II: Sulfur dioxide (SO2) is oxidized to form sulfur trioxide (SO3) using oxygen (O2) in the presence of a catalyst, vanadium(V) oxide (V2O5).
2SO2 (g) + O2 (g) → 2SO3 (g)
Process III: Sulfur trioxide (SO3) is dissolved in concentrated sulfuric acid (H2SO4) to form oleum (H2S2O7).
SO3 (g) + H2SO4 (l) → H2S2O7 (l)
Process IV: Oleum (H2S2O7) is diluted with water (H2O) to form concentrated sulfuric acid (H2SO4).
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
(iii) The catalyst used in process II is vanadium(V) oxide (V2O5).
(iv) In process II, the reaction is exothermic (it releases heat). According to Le Chatelier's principle, increasing the temperature would shift the equilibrium in the opposite direction, favoring the reactants (SO2 and O2). Therefore, increasing the temperature would not favor the forward reaction in process II.
(v) Two uses of SO2 are:
- - It is used in the production of sulfuric acid (H2SO4), which is an important industrial chemical.
- - It is used as a preservative for dried fruits, such as apricots, to prevent them from turning brown.
b)
The equation given is:
2H2 (g) + O2 (g) → 2H2O (g)
To solve the problem, we can use the mole ratio
Question 55 Report
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
(a) State what would be observed when BaCl, the solution is a portion of a saturated Na\(_2\)CO\(_2\) followed by dilute HCI in excess.
(ii) A gas Q decolorized acidified kM\(_n\)O\(_4\) Solution. Suggest what Q could be
(b) Name one substance used in the laboratory for drying each of the following substances:
(i) ammonia gas:
(ii) carbon (IV) oxide.
(c) Give a reason why a given mass of sodium hydroxide pellets cannot be used to prepare a standard solution.
(a)
BaCl2(aq) + Na2 CO3(aq) followed saturated by dilute HCl in excess
- A white precipitate is formed.
- Effervescence occurs and a gas evolved and the white precipitate dissolves in excess dil HCl
(ii)
Gas Q that decolourized acidified KMnO2 solution could be any of these.
SO2(g) , H2 S(g) , unsaturated hydrocarbon (C2 H2 , C2 H4 ....)
(b)
One substance used in the laboratory for drying each of the following:
(i) ammonia gas - Calcium oxide
(ii) carbon (IV) Oxide - Calcium chloride
- Concentrated tetraoxosulphate( VI) acid.
(c)
Sodium hydroxide pellets cannot be used to prepare a standard solution because it is deliquescent, it absorbs water vapour (moisture) and carbon (IV) oxide from the air, and this would add to its mass.
Answer Details
(a)
BaCl2(aq) + Na2 CO3(aq) followed saturated by dilute HCl in excess
- A white precipitate is formed.
- Effervescence occurs and a gas evolved and the white precipitate dissolves in excess dil HCl
(ii)
Gas Q that decolourized acidified KMnO2 solution could be any of these.
SO2(g) , H2 S(g) , unsaturated hydrocarbon (C2 H2 , C2 H4 ....)
(b)
One substance used in the laboratory for drying each of the following:
(i) ammonia gas - Calcium oxide
(ii) carbon (IV) Oxide - Calcium chloride
- Concentrated tetraoxosulphate( VI) acid.
(c)
Sodium hydroxide pellets cannot be used to prepare a standard solution because it is deliquescent, it absorbs water vapour (moisture) and carbon (IV) oxide from the air, and this would add to its mass.
Question 56 Report
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
C is a mixture of two inorganic compounds. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusions you draw from the result of each test.
(a) Put all of C in a boiling tube and add about 10 cm\(^3\) of distilled water. Shake thoroughly and filter. Keep both the residue and the filtrate.
(b)(i) To about 2 cm\(^3\) of the filtrate add a few drops of Silver trioxoitrate (V) followed by dilute HNO\(_{3}\)
(ii) Add excess NH\(_3\) solution to the resulting mixture in (b)(i).
(C)(i) Put the residue in a test tube, add about 2 cm\(^3\) of dilute HCl and shake.
(ii) Add NH\(_3\) Solution in drops to the mixture from (c)(i) and then in excess.
Experiment:
This experiment involves analyzing a mixture (C) containing two inorganic compounds. You will perform tests on the filtrate (soluble part) and residue (insoluble part) after filtration with water.
Observations and Conclusions:
(a) Filtration:
- Observation: The mixture might separate into a clear liquid (filtrate) and a solid residue.
- Conclusion: The mixture contains at least one soluble and one insoluble compound in water.
(b) Tests on Filtrate (Solution):
(i) Silver Nitrate (AgNO3) and Nitric Acid (HNO3):
- Observation: A white precipitate might form.
- Conclusion (Possible): Presence of chloride (Cl-) ions. Silver nitrate in acidic solution reacts with chloride ions to form a white silver chloride (AgCl) precipitate.
(ii) Excess Ammonia (NH3) Solution:
- Observation: The white precipitate (if formed in step (i)) might dissolve.
- Conclusion (Possible): The white precipitate from step (i) could be AgCl. Ammonia forms a complex ion with silver ions, dissolving the AgCl precipitate.
(c) Tests on Residue:
(i) Dilute Hydrochloric Acid (HCl):
- Observation: There might be fizzing or gas evolution.
- Conclusion (Possible): Presence of a carbonate (CO3²⁻) or hydrogencarbonate (HCO3⁻) compound. These react with acid to release carbon dioxide (CO2) gas.
(ii) Ammonia Solution (NH3):
- Observation: No significant change expected.
- Conclusion: Ammonia is usually not reactive with the residue components in this type of experiment.
Important Note:
These are general observations and conclusions based on common inorganic compounds. The actual behavior depends on the specific components present in mixture C.
Answer Details
Experiment:
This experiment involves analyzing a mixture (C) containing two inorganic compounds. You will perform tests on the filtrate (soluble part) and residue (insoluble part) after filtration with water.
Observations and Conclusions:
(a) Filtration:
- Observation: The mixture might separate into a clear liquid (filtrate) and a solid residue.
- Conclusion: The mixture contains at least one soluble and one insoluble compound in water.
(b) Tests on Filtrate (Solution):
(i) Silver Nitrate (AgNO3) and Nitric Acid (HNO3):
- Observation: A white precipitate might form.
- Conclusion (Possible): Presence of chloride (Cl-) ions. Silver nitrate in acidic solution reacts with chloride ions to form a white silver chloride (AgCl) precipitate.
(ii) Excess Ammonia (NH3) Solution:
- Observation: The white precipitate (if formed in step (i)) might dissolve.
- Conclusion (Possible): The white precipitate from step (i) could be AgCl. Ammonia forms a complex ion with silver ions, dissolving the AgCl precipitate.
(c) Tests on Residue:
(i) Dilute Hydrochloric Acid (HCl):
- Observation: There might be fizzing or gas evolution.
- Conclusion (Possible): Presence of a carbonate (CO3²⁻) or hydrogencarbonate (HCO3⁻) compound. These react with acid to release carbon dioxide (CO2) gas.
(ii) Ammonia Solution (NH3):
- Observation: No significant change expected.
- Conclusion: Ammonia is usually not reactive with the residue components in this type of experiment.
Important Note:
These are general observations and conclusions based on common inorganic compounds. The actual behavior depends on the specific components present in mixture C.
Question 57 Report
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
A is a solution containing 15.8 g dm\(^3\) of Na\(_2\)S\(_2\)O\(_3\). B was obtained by dissolving 9.0 g of an impure sample of I\(_2\) in aqueous Kl and the solution made up to 1 dm\(^3\).
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B. Use starch solution as indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of A used. The equation for the reaction involved in the titration is I\(_2\) + 2S\(_2\)O\(_3\) \(\to\) 2I\(^-\) + S\(_4\)O\(_6^{2-}\).
(b) From your results and the information provided, calculate the:
(i) concentration of A in mol dm\(^{-3}\)
(ii) concentration of I\(_2\) in B in mol dm\(^{-3}\);
(iii) percentage by mass of I\(_2\) in the sample
(c) Give reasons why the starch indicator was not added to the titration mixture at the beginning of the titration. [O = 16.0, Na = 23.0, S = 32.0, 1 = 127.0] Credit will be given for strict adherence to the instructions for observations precisely recorded and for accurate inferences. AIl tests, observations and inferences must be clearly entered in the booklet in ink at the time they are made.
Titration Experiment with Sodium Thiosulfate and Iodine
Experiment:
You are titrating a solution of sodium thiosulfate (Na2S2O3, solution A) against a solution containing iodine (I2, solution B) obtained from an impure sample.
Requirements:
- Record initial and final burette readings for titrating solution A against B (20.0 cm3 or 25.0 cm3 portions).
- Use starch solution as an indicator (not added at the beginning).
- Repeat the titration for concordant titers (consistent volumes used).
(a) Average Volume of Solution A:
- From your recorded burette readings, calculate the volume of solution A used in each titration.
- Perform multiple titrations and discard any outliers.
- Calculate the average volume of solution A used for all concordant titrations.
(b) Concentration Calculations:
i) Concentration of Solution A (Na2S2O3):
- Use the provided concentration of Na2S2O3 in solution A (15.8 g dm-3).
- Convert the concentration from g dm-3 to mol dm-3 using the molar mass of Na2S2O3 (248.2 g mol-1).
ii) Concentration of Iodine (I2) in Solution B:
- Use the balanced equation for the reaction: I2 + 2S2O3- → 2I- + S4O6-2
- The moles of Na2S2O3 used are equal to the moles of I2 reacted (from stoichiometry).
- Calculate the moles of Na2S2O3 used from the concentration of solution A and the average volume used.
- Divide the moles of Na2S2O3 by the volume of solution B used (0.0200 dm3 or 0.0250 dm3 depending on your chosen volume) to get the concentration of I2 in mol dm-3.
iii) Percentage by Mass of I2 in the Sample:
- Calculate the total moles of I2 present in 1 dm3 of solution B using the concentration obtained in (b)(ii).
- Multiply the moles of I2 by the molar mass of I2 (254.0 g mol-1) to get the total mass of I2 in 1 dm3 of solution B.
- Divide the mass of I2 by the mass of the impure sample used (9.0 g) and multiply by 100% to get the percentage by mass of I2 in the sample.
(c) Why Not Add Starch Indicator at the Beginning?
Starch forms a blue-black complex with free iodine molecules. If added at the beginning, the solution would appear blue even before the endpoint. By adding starch near the endpoint (when free I2 concentration is low), a faint blue color appears, indicating the complete reaction of I2 with Na2S2O3. This allows for a more accurate determination of the titration endpoint.
Answer Details
Titration Experiment with Sodium Thiosulfate and Iodine
Experiment:
You are titrating a solution of sodium thiosulfate (Na2S2O3, solution A) against a solution containing iodine (I2, solution B) obtained from an impure sample.
Requirements:
- Record initial and final burette readings for titrating solution A against B (20.0 cm3 or 25.0 cm3 portions).
- Use starch solution as an indicator (not added at the beginning).
- Repeat the titration for concordant titers (consistent volumes used).
(a) Average Volume of Solution A:
- From your recorded burette readings, calculate the volume of solution A used in each titration.
- Perform multiple titrations and discard any outliers.
- Calculate the average volume of solution A used for all concordant titrations.
(b) Concentration Calculations:
i) Concentration of Solution A (Na2S2O3):
- Use the provided concentration of Na2S2O3 in solution A (15.8 g dm-3).
- Convert the concentration from g dm-3 to mol dm-3 using the molar mass of Na2S2O3 (248.2 g mol-1).
ii) Concentration of Iodine (I2) in Solution B:
- Use the balanced equation for the reaction: I2 + 2S2O3- → 2I- + S4O6-2
- The moles of Na2S2O3 used are equal to the moles of I2 reacted (from stoichiometry).
- Calculate the moles of Na2S2O3 used from the concentration of solution A and the average volume used.
- Divide the moles of Na2S2O3 by the volume of solution B used (0.0200 dm3 or 0.0250 dm3 depending on your chosen volume) to get the concentration of I2 in mol dm-3.
iii) Percentage by Mass of I2 in the Sample:
- Calculate the total moles of I2 present in 1 dm3 of solution B using the concentration obtained in (b)(ii).
- Multiply the moles of I2 by the molar mass of I2 (254.0 g mol-1) to get the total mass of I2 in 1 dm3 of solution B.
- Divide the mass of I2 by the mass of the impure sample used (9.0 g) and multiply by 100% to get the percentage by mass of I2 in the sample.
(c) Why Not Add Starch Indicator at the Beginning?
Starch forms a blue-black complex with free iodine molecules. If added at the beginning, the solution would appear blue even before the endpoint. By adding starch near the endpoint (when free I2 concentration is low), a faint blue color appears, indicating the complete reaction of I2 with Na2S2O3. This allows for a more accurate determination of the titration endpoint.
Question 58 Report
(a) (i) Name three different methods for preparing salts.
(ii) Give one example of a balanced equation for each of the methods named in (a)(i).
(iii) State two uses of sodium trioxocarbonate (IV).
(b) If you were given some impure copper, describe how you would
obtain a specimen of the pure metal by electrolysis.
(c) Given that sodium chloride has a solubility of 36.3 at 30 and 39.0 at
100 and that of silver nitrate is 297.0 at 30 and 952.0 at 100.
(i) Calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30
(ii) Deduce which of the two salts can be purified more efficiently by crystallization.
a)
- (i) Three different methods for preparing salts are precipitation, neutralization, and evaporation.
- (ii) Examples of balanced equations for each method:
- Precipitation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
- Neutralization: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
- Evaporation: NaCl(aq) → NaCl(s)
- (iii) Two uses of sodium trioxocarbonate (IV) are:
- As a cleaning agent: Sodium trioxocarbonate (IV) is used as a cleaning agent in many household and industrial cleaning products.
- As a water softener: Sodium trioxocarbonate (IV) can be used to soften hard water by removing excess calcium and magnesium ions.
b)
- To obtain a specimen of pure copper by electrolysis, you would need to follow these steps:
- Obtain a supply of direct current (DC) and two electrodes (anode and cathode).
- Place the impure copper in a solution that conducts electricity (such as copper sulfate solution).
- Attach the impure copper to the anode and a pure copper plate to the cathode.
- Turn on the DC and allow the copper ions to be attracted to the cathode, where they will deposit and form pure copper.
c)
To calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30:
- For sodium chloride: (36.3/39.0) x 100 = 92.56%
- For silver nitrate: (297.0/952.0) x 100 = 31.22%
Based on the percentage of each substance that is deposited, sodium chloride can be purified more efficiently by crystallization.
Answer Details
a)
- (i) Three different methods for preparing salts are precipitation, neutralization, and evaporation.
- (ii) Examples of balanced equations for each method:
- Precipitation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
- Neutralization: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
- Evaporation: NaCl(aq) → NaCl(s)
- (iii) Two uses of sodium trioxocarbonate (IV) are:
- As a cleaning agent: Sodium trioxocarbonate (IV) is used as a cleaning agent in many household and industrial cleaning products.
- As a water softener: Sodium trioxocarbonate (IV) can be used to soften hard water by removing excess calcium and magnesium ions.
b)
- To obtain a specimen of pure copper by electrolysis, you would need to follow these steps:
- Obtain a supply of direct current (DC) and two electrodes (anode and cathode).
- Place the impure copper in a solution that conducts electricity (such as copper sulfate solution).
- Attach the impure copper to the anode and a pure copper plate to the cathode.
- Turn on the DC and allow the copper ions to be attracted to the cathode, where they will deposit and form pure copper.
c)
To calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30:
- For sodium chloride: (36.3/39.0) x 100 = 92.56%
- For silver nitrate: (297.0/952.0) x 100 = 31.22%
Based on the percentage of each substance that is deposited, sodium chloride can be purified more efficiently by crystallization.
Question 59 Report
(a) (i) State the important points put forward in Dalton’s atomic theory.
(ii) How does the theory explain the law of multiple proportion?
(b) (i) Name three commercially useful products that may be obtained by chemical transformation of vegetable oils.
(ii) With the aid of equations, describe the chemical reaction involved in the transformations stated in (b)(i).
(c) Certain properties of Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resembles those of Aluminium and its compounds. Explain briefly why this is so.
(d) Write balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI).
a)
(i) The important points put forward in Dalton's atomic theory are:
- All matter is made up of tiny, indivisible particles called atoms.
- Atoms of the same element are identical in size, mass, and other properties.
- Atoms of different elements have different properties.
- Atoms can neither be created nor destroyed in a chemical reaction, only rearranged.
- Compounds are formed when atoms of different elements combine in fixed ratios.
(ii) The theory explains the law of multiple proportion by stating that when two elements combine to form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other element are ratios of small whole numbers.
b)
(i) Three commercially useful products that may be obtained by chemical transformation of vegetable oils are soap, biodiesel, and margarine.
(ii) The chemical reactions involved in the transformations stated in (b)(i) are:
- Soap: Vegetable oil + Sodium hydroxide → Soap + Glycerol
- Biodiesel: Vegetable oil + Methanol → Biodiesel + Glycerol
- Margarine: Vegetable oil + Hydrogen gas → Margarine + Water
c) Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resemble those of Aluminium and its compounds because:
- Beryllium has a smaller atomic size and higher ionization energy than magnesium, which makes it more similar to aluminium in terms of electronic structure and chemical behavior.
- Beryllium forms covalent bonds more readily than magnesium, while magnesium forms primarily ionic bonds. This makes beryllium more similar to aluminium, which also tends to form covalent bonds.
d) The balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI) is:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
This reaction is a redox reaction, in which iodine is reduced to iodide ions and trioxothiosulphate(VI) is oxidized to tetrathionate ions. Sodium ions are spectator ions and do not participate in the reaction. The reaction is commonly used as a titration to determine the concentration of iodine in a sample.
Answer Details
a)
(i) The important points put forward in Dalton's atomic theory are:
- All matter is made up of tiny, indivisible particles called atoms.
- Atoms of the same element are identical in size, mass, and other properties.
- Atoms of different elements have different properties.
- Atoms can neither be created nor destroyed in a chemical reaction, only rearranged.
- Compounds are formed when atoms of different elements combine in fixed ratios.
(ii) The theory explains the law of multiple proportion by stating that when two elements combine to form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other element are ratios of small whole numbers.
b)
(i) Three commercially useful products that may be obtained by chemical transformation of vegetable oils are soap, biodiesel, and margarine.
(ii) The chemical reactions involved in the transformations stated in (b)(i) are:
- Soap: Vegetable oil + Sodium hydroxide → Soap + Glycerol
- Biodiesel: Vegetable oil + Methanol → Biodiesel + Glycerol
- Margarine: Vegetable oil + Hydrogen gas → Margarine + Water
c) Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resemble those of Aluminium and its compounds because:
- Beryllium has a smaller atomic size and higher ionization energy than magnesium, which makes it more similar to aluminium in terms of electronic structure and chemical behavior.
- Beryllium forms covalent bonds more readily than magnesium, while magnesium forms primarily ionic bonds. This makes beryllium more similar to aluminium, which also tends to form covalent bonds.
d) The balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI) is:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
This reaction is a redox reaction, in which iodine is reduced to iodide ions and trioxothiosulphate(VI) is oxidized to tetrathionate ions. Sodium ions are spectator ions and do not participate in the reaction. The reaction is commonly used as a titration to determine the concentration of iodine in a sample.
Question 60 Report
(a) A hydrocarbon having the formula C\(_{10}\)H\(_{22}\) was cracked to produce C\(_6\)H\(_{14}\) and another hydrocarbon P.
(i) Give the molecular formula of P.
(ii) Draw the structures of two isomers of P.
(ii) Give a reason why P could be polymerized.
(b) State the guiding principles which are used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals.
(c)Consider each of the following substances: NaH, H\(_2\), H\(_2\)S, NH\(_4\)Cl.
(i) Describe the nature of the intermolecular forces holding the units or molecules together in the condensed (liquid or solid) state.
(ii) Explain briefly what happens when a sample of each of the substances is added to water.
(iii) Write the chemical equations of any reactions occurring or of any equilibria established.
(d) Element J has the following electron configuration: Is\(^2\)2s\(^2\)2p\(^6\)3s\(^2\).
(i) How many unpaired electrons can be found in J?
(ii) State whether J would be a good oxidizing or reducing agent.
(iii) Give a reason for the answer in (d)(ii).
(a)
(i) The molecular formula of hydrocarbon P can be determined by balancing the carbon and hydrogen atoms in the reactant and products of the cracking reaction. The balanced equation is:
C10H22 → C6H14 + C4H8
Therefore, the molecular formula of hydrocarbon P is C4H8.
(ii) Two isomers of hydrocarbon P are:
1. But-1-ene:
H H | | H-C=C-C-C-H | | H H
2. But-2-ene:
H H | | H-C-C=C-C-H | | H H
(iii) Hydrocarbon P could be polymerized because it contains a carbon-carbon double bond, which can undergo addition polymerization to form a long chain polymer.
(b)
The guiding principles used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals are the Aufbau principle, the Pauli exclusion principle, and Hund's rule.
The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy level, starting with the lowest energy level.
The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.
Hund's rule states that for a set of orbitals of equal energy (degenerate), the lowest energy is attained when the number of electrons with the same spin is maximized.
(c)
(i) NaH has ionic bonds holding the units together, while H2, H2S, and NH4Cl have covalent bonds holding the molecules together.
(ii) When NaH is added to water, it reacts to form hydrogen gas and sodium hydroxide:
NaH + H2O → NaOH + H2
When H2 is added to water, it dissolves and forms a homogeneous mixture.
When H2S is added to water, it reacts to form hydrosulfuric acid:
H2S + H2O ⇌ H3O+ + HS-
When NH4Cl is added to water, it dissolves and dissociates into ammonium cations and chloride anions:
NH4Cl → NH4+ + Cl-
(iii) The chemical equations for the reactions are:
NaH + H2O → NaOH + H2
H2S +
Answer Details
(a)
(i) The molecular formula of hydrocarbon P can be determined by balancing the carbon and hydrogen atoms in the reactant and products of the cracking reaction. The balanced equation is:
C10H22 → C6H14 + C4H8
Therefore, the molecular formula of hydrocarbon P is C4H8.
(ii) Two isomers of hydrocarbon P are:
1. But-1-ene:
H H | | H-C=C-C-C-H | | H H
2. But-2-ene:
H H | | H-C-C=C-C-H | | H H
(iii) Hydrocarbon P could be polymerized because it contains a carbon-carbon double bond, which can undergo addition polymerization to form a long chain polymer.
(b)
The guiding principles used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals are the Aufbau principle, the Pauli exclusion principle, and Hund's rule.
The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy level, starting with the lowest energy level.
The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers.
Hund's rule states that for a set of orbitals of equal energy (degenerate), the lowest energy is attained when the number of electrons with the same spin is maximized.
(c)
(i) NaH has ionic bonds holding the units together, while H2, H2S, and NH4Cl have covalent bonds holding the molecules together.
(ii) When NaH is added to water, it reacts to form hydrogen gas and sodium hydroxide:
NaH + H2O → NaOH + H2
When H2 is added to water, it dissolves and forms a homogeneous mixture.
When H2S is added to water, it reacts to form hydrosulfuric acid:
H2S + H2O ⇌ H3O+ + HS-
When NH4Cl is added to water, it dissolves and dissociates into ammonium cations and chloride anions:
NH4Cl → NH4+ + Cl-
(iii) The chemical equations for the reactions are:
NaH + H2O → NaOH + H2
H2S +
