WAEC SSCE - Chemistry - 2002
Pergunta 1 Relatório
The gas given off when NH4 Cl is heated with an alkali is
Detalhes da Resposta
When NH4Cl is heated with an alkali such as NaOH, it undergoes a decomposition reaction to form NH3 gas, water, and a salt. The balanced chemical equation for the reaction is: NH4Cl + NaOH → NH3 + H2O + NaCl Therefore, the gas given off is NH3, which is ammonia (option D).
Pergunta 3 Relatório
What will happen if more heat is applied to the following system at equilibrium? X2(g) + 3Y 2(g) ⇌ 2XY3(g); ∆H = xKJmol-1
Detalhes da Resposta
Pergunta 4 Relatório
The atomic number of chlorine is 17. What is the number of electrons in a chloride ion?
Detalhes da Resposta
Chlorine has an atomic number of 17, which means it has 17 protons and 17 electrons in its neutral state. When chlorine gains an electron, it becomes a chloride ion with a charge of -1. The negative charge on the chloride ion indicates that it has one more electron than protons. Therefore, the number of electrons in a chloride ion is 17 + 1 = 18. Hence, the answer is 18.
Pergunta 5 Relatório
If the relative molecular mass of an element is not a whole number, it can be deduced that the element is
Detalhes da Resposta
If the relative molecular mass of an element is not a whole number, it can be deduced that the element is an isotopic mixture. This is because the relative molecular mass of an element is the average of the masses of all the isotopes of that element, taking into account their relative abundances. Isotopic mixture means that the element is composed of different isotopes with different masses and different relative abundances. For example, chlorine has two isotopes, Cl-35 and Cl-37, which have relative abundances of 75.5% and 24.5%, respectively. The average atomic mass of chlorine is therefore not a whole number, but a weighted average of the masses of these isotopes, taking into account their abundances.
Pergunta 6 Relatório
If two metallic ions of the same concentration migrate to a graphite cathode, the one that would be preferentially discharged is the one that has the
Detalhes da Resposta
In an electrolytic cell, when two metallic ions of the same concentration migrate to a graphite cathode, the one that would be preferentially discharged is the one that has the stronger tendency to accept electrons. This is because reduction, which involves the gain of electrons, occurs at the cathode. The ion with a stronger tendency to accept electrons will have a higher reduction potential and will therefore be preferentially reduced at the cathode.
Pergunta 7 Relatório
Which of the following processes takes place during the production of margarine from vegetable oils?
Detalhes da Resposta
The process that takes place during the production of margarine from vegetable oils is hydrogenation. Hydrogenation is a chemical process that involves the addition of hydrogen gas to unsaturated fatty acids in vegetable oils, resulting in the conversion of the unsaturated fatty acids into saturated fatty acids. This process makes the vegetable oils more solid and stable, improving their texture and shelf life. The other options listed are not relevant to the production of margarine from vegetable oils. Esterification involves the formation of esters from carboxylic acids and alcohols, hydrolysis involves the breakdown of molecules into their constituent parts by the addition of water, and saponification involves the reaction of a fat or oil with a strong base to produce soap and glycerol.
Pergunta 8 Relatório
The following compounds are hydrocarbons except
Detalhes da Resposta
The compound that is not a hydrocarbon is methylpropanoate. A hydrocarbon is a compound that contains only carbon and hydrogen atoms. - 2-methylbutane is a hydrocarbon because it only contains carbon and hydrogen atoms. - Benzene is a hydrocarbon with a ring structure made up of carbon atoms and hydrogen atoms. - Cyclohexane is also a hydrocarbon with a ring structure made up of carbon atoms and hydrogen atoms. Methylpropanoate, on the other hand, contains carbon, hydrogen, and oxygen atoms, which means it is not a hydrocarbon.
Pergunta 9 Relatório
The following compounds are hydrocarbons except
Detalhes da Resposta
Among the given options, only "methylpropanoate" is not a hydrocarbon. This is because a hydrocarbon is a compound made up of only hydrogen and carbon atoms. "Methylpropanoate," on the other hand, contains an additional oxygen atom in its structure, making it an ester compound. In contrast, "2-methylbutane," "benzene," and "cyclohexane" are all hydrocarbons as they contain only hydrogen and carbon atoms in their structures.
Pergunta 10 Relatório
Which of the following salt solutions will have a pH greater than 7?
Detalhes da Resposta
Among the given options, Na2CO3(aq) is the only salt solution that can have a pH greater than 7. When Na2CO3 is dissolved in water, it undergoes hydrolysis to produce NaOH and HCO3- ions. The hydroxide ion concentration increases, which makes the solution basic. Hence, the solution will have a pH greater than 7. On the other hand, NaCl, Na2SO4, and NaHSO4 are all salts of strong acids and strong bases, or weak acids and strong bases, and therefore their solutions will be neutral or slightly acidic but never basic.
Pergunta 11 Relatório
One of the methods used for combating air pollution is
Detalhes da Resposta
Legislation against some industrial practices is one of the methods used for combating air pollution. This involves enacting laws and regulations to control or limit the amount of pollutants released into the atmosphere by industries. The laws may require industries to install pollution control devices or use cleaner production methods that generate fewer pollutants. The aim is to reduce the amount of pollutants in the air, thus reducing air pollution and its negative impacts on the environment and human health.
Pergunta 12 Relatório
An element Y has the electronic configuration 1s22s22p6623p4. To what period does it belong in the periodic table?
Detalhes da Resposta
The electronic configuration of an element provides information about its position in the periodic table. The period of an element is determined by the highest energy level occupied by electrons in the ground state. In the given electronic configuration, the highest energy level occupied by electrons is the third energy level. Therefore, element Y belongs to period 3 of the periodic table.
Pergunta 13 Relatório
When alkynes are hydrogenated completely, they produce compounds with the general molecular formula
Detalhes da Resposta
When alkynes are completely hydrogenated, each triple bond (-C≡C-) is converted to a single bond (-C-C-) and adds two hydrogen atoms. Therefore, each alkyne molecule loses two hydrogen atoms and gains two carbon atoms. The general molecular formula of completely hydrogenated alkynes is C nH 2n+2. Thus, is the correct answer.
Pergunta 14 Relatório
Which of the following compounds will leave a metal residue when heated?
Detalhes da Resposta
Pergunta 15 Relatório
Which of the following statements about soapless detergents is correct? They
Detalhes da Resposta
Soapless detergents are also known as synthetic detergents. They are made up of molecules that are not affected by hard water, unlike soap molecules that react with hard water to form scum. Therefore, soapless detergents do not form a curdy precipitate with soft water. So, the correct statement is that they lather readily with hard water.
Pergunta 16 Relatório
Which of the following substances increases in mass when heated in air?
Detalhes da Resposta
Pergunta 17 Relatório
Consider the following equation; Cr2O7(aq)2- + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(I). The oxidation number of chromium changes from
Detalhes da Resposta
In the given equation, the reactant Cr2O72- has chromium in the +6 oxidation state (since each oxygen atom has an oxidation number of -2, and there are seven of them, the total negative charge of the oxygen atoms is -14, and the total negative charge of the ion is -2, so the total positive charge of the chromium atoms must be +12 to balance it out). In the product side, the product Cr3+ has a +3 oxidation state. This means that the oxidation number of chromium has changed from +6 to +3, a decrease of 3. Therefore, the answer is - 2 to + 6, since chromium goes from +6 to +3.
Pergunta 19 Relatório
If two metallic ions of the same concentration migrate to a graphite cathode, the one that would be preferentially discharged is the one that has the
Detalhes da Resposta
When two metallic ions of the same concentration migrate to a graphite cathode, the ion that would be preferentially discharged is the one that has a stronger tendency to accept electrons. This is because during the process of electrolysis, the positively charged ions are reduced (gain electrons) at the cathode. The ion with a stronger tendency to accept electrons is more likely to be reduced at the cathode than the other ion, leading to its preferential discharge. This is known as the selective discharge of ions. The mass, electrical charge, and stability of the ion in solution do not affect its preferential discharge.
Pergunta 20 Relatório
A suitable reagent for distinguishing between ethanoic acid ethanol is
Detalhes da Resposta
Pergunta 21 Relatório
CuSO45H2O can be obtained from an aqueous solution of copper (ll) tetraoxosulphate (VI) by
Detalhes da Resposta
CuSO45H2O can be obtained from an aqueous solution of copper (II) tetraoxosulphate (VI) by **crystallization**. Crystallization is the process of separating a solid substance from a liquid by cooling the liquid or allowing it to evaporate slowly. In this case, by slowly evaporating the water from the solution of copper (II) tetraoxosulphate (VI), CuSO45H2O will crystallize out of the solution as blue crystals. The crystals obtained will have a definite shape and size, and can be filtered and dried to obtain the pure solid CuSO45H2O.
Pergunta 22 Relatório
Which of the following ions require the largest quantity of electricity for discharge at an electrode?
Detalhes da Resposta
Pergunta 23 Relatório
A major factor considered in selecting a suitable method for preparing a simple salt is its
Detalhes da Resposta
One major factor considered in selecting a suitable method for preparing a simple salt is its solubility in water. The method chosen should be able to effectively dissolve the necessary reactants in water to form the salt. If the salt is insoluble in water, a different method such as precipitation or metathesis may be used. Additionally, the solubility of the salt can affect the purification process and the yield of the final product. Therefore, the solubility of the salt in water is an important consideration in selecting a suitable method for preparing it.
Pergunta 24 Relatório
Calculate the mass of chlorine gas which occupies a volume of 1.12 dm\(^3\) a volume of 1.12 dm\(^3\) at s.t.p [Cl = 35.5, 1 mole og gas occupies 22.4 dm\(^3\) at s. t. p ]
Detalhes da Resposta
The given information includes the volume of chlorine gas and the conditions of temperature and pressure (s.t.p). The formula for calculating the mass of a gas at s.t.p is: mass = (volume of gas at s.t.p in dm³) x (molar mass of gas) / (molar volume at s.t.p) The molar mass of chlorine gas (Cl₂) is 2 x atomic mass of Cl = 2 x 35.5 = 71 g/mol. The molar volume of any gas at s.t.p is 22.4 dm³. Plugging in the values, we get: mass = (1.12 dm³) x (71 g/mol) / (22.4 dm³/mol) = 3.55 g Therefore, the mass of chlorine gas which occupies a volume of 1.12 dm³ at s.t.p is 3.55 g. The correct option is (b) 3.55g.
Pergunta 25 Relatório
What is the expression for the equilibrium constant (Kc) for the following reaction? N2(g) + O2(g) ⇌ 2NO(g)
Detalhes da Resposta
Pergunta 26 Relatório
A substance is said to be hygroscopic if it absorbs
Detalhes da Resposta
A substance is said to be hygroscopic if it absorbs moisture from the atmosphere without dissolving. This means that the substance has the ability to attract and hold water molecules from the air around it, but it does not dissolve in the water to form a solution. Hygroscopic substances are commonly used as desiccants to control the moisture content of products such as foods, medicines, and electronic components. Examples of hygroscopic substances include salt, sugar, silica gel, and calcium chloride.
Pergunta 27 Relatório
A hydrogen atom which has lost an electron contains
Detalhes da Resposta
A hydrogen atom which has lost an electron contains one proton only. When an electron is lost from a neutral hydrogen atom, the positively charged proton remains in the nucleus while the electron is lost to the surrounding environment. This results in a positively charged hydrogen ion, also known as a proton. Therefore, the correct answer is "one proton only".
Pergunta 28 Relatório
Which of the following decreases when a given mass of gas is compressed to half its initial volume?
Detalhes da Resposta
When a given mass of gas is compressed to half its initial volume, the volume of the container holding the gas is reduced, but the number of gas molecules remains constant. This means that the gas molecules have less space to move around, and the average intermolecular distance between them decreases. As the distance between the molecules decreases, the frequency of collisions between them increases. Thus, the correct answer is: - Average intermolecular distance
Pergunta 29 Relatório
Equal amounts of marble chips are reacted separately with 100cm3 of hydrochloric acid of different concentrations. If all the marble chips reacted; which of the following remained the same in each case?
Detalhes da Resposta
The total mass of gas evolved during the reaction remained the same in each case. This is because equal amounts of marble chips were reacted separately with 100cm3 of hydrochloric acid of different concentrations. Since the same amount of marble chips was used in each reaction, the total amount of gas produced by the reaction will be equal, regardless of the concentration of the hydrochloric acid used. The other options, such as the average rate of evolution of gas, time taken for the reaction to reach completion, and initial reaction rates, may differ depending on the concentration of hydrochloric acid used.
Pergunta 30 Relatório
Calculate the mass of sodium hydroxide in 5.00 dm3 of a 0.125 mol dm-3 solution. [NaOH = 40 g mol-1]
Detalhes da Resposta
To calculate the mass of sodium hydroxide in 5.00 dm3 of a 0.125 mol dm-3 solution, we need to use the formula: mass = moles × molar mass First, let's calculate the number of moles of NaOH in 5.00 dm3 of the solution: moles = concentration × volume moles = 0.125 mol dm-3 × 5.00 dm3 moles = 0.625 mol Now, we can use the formula to calculate the mass of NaOH: mass = 0.625 mol × 40 g mol-1 mass = 25.0 g Therefore, the mass of sodium hydroxide in 5.00 dm3 of a 0.125 mol dm-3 solution is 25.0 g. Answer: 25.0 g
Pergunta 31 Relatório
Which of the following processes takes place during the production of margarine from vegetable oils?
Detalhes da Resposta
The correct process that takes place during the production of margarine from vegetable oils is hydrogenation. In this process, vegetable oils, which are unsaturated, are reacted with hydrogen gas in the presence of a catalyst such as nickel. This process converts the unsaturated fatty acids in the vegetable oils into saturated fatty acids, making the oils more solid at room temperature. The hydrogenation process also increases the shelf life of the oils and gives the margarine a desirable texture and consistency. Therefore, is the correct answer.
Pergunta 32 Relatório
If the relative molecular mass of an element is not a whole number, it can be deduced that the element is
Detalhes da Resposta
If the relative molecular mass of an element is not a whole number, it can be deduced that the element is an isotopic mixture. This is because the relative molecular mass is a weighted average of the masses of all the isotopes of the element, taking into account their abundance. If the element has only one isotope, its relative molecular mass will be a whole number. However, if the element has two or more isotopes with different masses and abundances, its relative molecular mass will be a decimal or fraction. Therefore, the non-whole number relative molecular mass indicates that the element is made up of a mixture of isotopes with different masses and abundances.
Pergunta 33 Relatório
A visible change is observed when a strip of iron is placed in an aqueous solution of
Detalhes da Resposta
When a strip of iron is placed in an aqueous solution of copper sulfate (CuSO4), a visible change is observed. The iron strip reacts with the copper ions in the solution and displaces them, forming iron sulfate (FeSO4) and copper metal. This reaction is known as a single displacement or replacement reaction. The copper metal formed will appear as a reddish-brown solid on the iron strip, causing a visible change.
Pergunta 34 Relatório
The reaction between ethene and chlorine to form 1, 2-dichloroethane is
Detalhes da Resposta
The reaction between ethene and chlorine to form 1, 2-dichloroethane is an addition reaction. In this reaction, the double bond between the two carbon atoms in ethene is broken, and each carbon atom bonds to a chlorine atom to form 1,2-dichloroethane. The reaction is called an addition reaction because the two molecules add together to form a single molecule, rather than substituting or removing atoms as in a substitution or oxidation reaction.
Pergunta 35 Relatório
The conclusion from Rutherford's alpha-scattering experiment is that
Detalhes da Resposta
Rutherford's alpha-scattering experiment led to the conclusion that atoms are mostly empty space with a small, dense, positively charged nucleus. This was based on the observation that when alpha particles were fired at a thin gold foil, most of the particles passed straight through the foil, but a few were deflected at large angles. This indicated that the positive charge of the atom was concentrated in a small, dense nucleus, and that most of the atom was empty space.
Pergunta 36 Relatório
Which of the following ions require the largest quantity of electricity for discharge at an electrode?
Detalhes da Resposta
Pergunta 37 Relatório
When concentrated sodium chloride solution is electrolysed using inert electrodes, the products are
Detalhes da Resposta
When a concentrated sodium chloride solution is electrolyzed using inert electrodes, the products are hydrogen gas and chlorine gas. This is because the electrolysis of the sodium chloride solution leads to the formation of chloride ions (Cl-) and hydrogen ions (H+) at the anode and cathode respectively. At the anode, chloride ions (Cl-) are oxidized to form chlorine gas (Cl2) while at the cathode, hydrogen ions (H+) are reduced to form hydrogen gas (H2). The sodium ions (Na+) do not participate in the reaction as the electrodes are inert and do not react with them.
Pergunta 38 Relatório
The reaction of sucrose with dilute HCI produces
Detalhes da Resposta
The reaction of sucrose with dilute HCl produces glucose and fructose. This process is known as hydrolysis, which means the breaking of a bond by the addition of water. The sucrose molecule is a disaccharide made up of one molecule of glucose and one molecule of fructose, held together by a glycosidic bond. When sucrose is hydrolyzed with dilute HCl, the glycosidic bond is broken, and glucose and fructose are formed as the products.
Pergunta 39 Relatório
If a metal R forms a compound with the formula R(NO3)2 , the formula of its tetraoxosulphate (IV) is
Detalhes da Resposta
Pergunta 40 Relatório
When alkynes are hydrogenated completely they produce compounds with the general molecular formula
Detalhes da Resposta
When alkynes are hydrogenated completely, they produce compounds with the general molecular formula CnH2n+2. This is because the double or triple bond between the carbon atoms in the alkyne molecule is converted to a single bond by the addition of hydrogen atoms. As a result, each carbon atom in the original alkyne molecule becomes bonded to two hydrogen atoms, leading to the formula CnH2n+2 for the fully hydrogenated compound.
Pergunta 41 Relatório
Which of the following compounds can be represented by the molecular formula C2H6O?
Detalhes da Resposta
The molecular formula C2H6O indicates that the compound has two carbon atoms, six hydrogen atoms, and one oxygen atom. By checking the options: - Propanal has the molecular formula C3H6O, which means it has three carbon atoms and cannot be represented by the given formula. - Ethanol has the molecular formula C2H6O and can be represented by the given formula. - Methanoic acid has the molecular formula CH2O2, which means it has only one carbon atom and cannot be represented by the given formula. - Glucose has the molecular formula C6H12O6, which means it has six carbon atoms and cannot be represented by the given formula. Therefore, the compound that can be represented by the molecular formula C2H6O is ethanol.
Pergunta 42 Relatório
Iron is often galvanized in order to
Detalhes da Resposta
Iron is often galvanized to protect it against corrosion. Galvanizing is a process of coating iron or steel with a layer of zinc to prevent rusting. The zinc acts as a sacrificial anode, which means that it corrodes before the iron or steel underneath it, providing a protective layer that prevents the underlying metal from coming into contact with moisture and oxygen, which are the main agents of corrosion. This process helps to extend the lifespan of iron or steel objects and structures.
Pergunta 43 Relatório
When concentrated sodium chloride solution is electrolysed using inert electrodes the products are?
Detalhes da Resposta
When concentrated sodium chloride solution is electrolysed using inert electrodes, the products are hydrogen and chlorine gases. This process is known as the electrolysis of brine. At the cathode (negative electrode), hydrogen ions (H+) from water are reduced to hydrogen gas (H2). At the anode (positive electrode), chloride ions (Cl-) from sodium chloride are oxidized to form chlorine gas (Cl2). The sodium ions (Na+) and hydroxide ions (OH-) produced by the electrolysis of water remain in solution and react to form sodium hydroxide (NaOH). Therefore, the correct option is "sodium hydroxide, hydrogen, and chlorine" but it is not listed as an option. The closest option is "hydrogen and chlorine."
Pergunta 44 Relatório
What volume of hydrogen will be produced if 100cm3 of ammonia is completely decomposed at constant temperature and pressure? The equation for reaction is 2NH3(g) → N2(g) + 3H2(g)
Detalhes da Resposta
The balanced chemical equation for the reaction between ammonia and hydrogen is: 2NH3(g) → N2(g) + 3H2(g) From the equation, we can see that 2 moles of ammonia reacts to produce 3 moles of hydrogen. Therefore, the ratio of ammonia to hydrogen is 2:3. This means that for every 2 moles of ammonia, 3 moles of hydrogen will be produced. To determine the volume of hydrogen produced, we need to use the volume of ammonia given in the question. We can assume that the ammonia is measured under the same temperature and pressure as the hydrogen produced. Since we are given the volume of ammonia, we need to convert it to moles using the ideal gas law: PV = nRT Where: P = pressure (constant) V = volume of gas (in liters) n = number of moles R = ideal gas constant T = temperature (constant) Assuming constant temperature and pressure, we can simplify the equation to: V = n Therefore, the volume of ammonia (100 cm3) is equal to its number of moles. Using the ratio from the balanced equation, we can calculate the number of moles of hydrogen produced: 2 moles NH3 : 3 moles H2 100 cm3 NH3 : x cm3 H2 x = (100 cm3 NH3)(3 mol H2/2 mol NH3) = 150 cm3 H2 Therefore, the volume of hydrogen produced is 150 cm3. The correct answer is (B) 150 cm3.
Pergunta 45 Relatório
Which of the following types of alkanols undergo oxidation to produce alkanoic acids? 1. Primary alkanols 11. Secondary alkanols 111. Tertiary alkanols
Detalhes da Resposta
Only primary alkanols undergo oxidation to produce alkanoic acids. This is because primary alkanols have a hydrogen atom attached to the carbon atom with the -OH group, which can be easily oxidized to a carboxylic acid (-COOH) group. Secondary and tertiary alkanols lack this hydrogen atom and cannot be directly oxidized to alkanoic acids.
Pergunta 46 Relatório
Which of the following statements about soapless detergents is correct? They
Detalhes da Resposta
Soapless detergents, also known as synthetic detergents, are different from soaps in that they do not contain fatty acids as their active ingredient. Rather, they contain synthetic surfactants that are more soluble in water and are better at removing dirt and oils. Of the given options, the correct statement is that soapless detergents lather readily with hard water. This is because hard water contains dissolved metal ions, such as calcium and magnesium, which can react with the soap molecules and form an insoluble precipitate, reducing its cleaning ability. However, since synthetic detergents do not contain the same fatty acid molecules as soaps, they do not form this precipitate and can therefore lather well even in hard water.
Pergunta 47 Relatório
Which of the following elements has a ground state electronic configuration with an incomplete penultimate shell?
Detalhes da Resposta
The electronic configuration of an atom refers to the arrangement of electrons in its various shells or energy levels. The penultimate shell is the second-to-last shell of an atom. An incomplete penultimate shell means that the penultimate shell is not completely filled with electrons. Out of the given options, Iron has a ground state electronic configuration with an incomplete penultimate shell. The electronic configuration of Iron (Fe) is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. In this configuration, the 3d sub-shell is incomplete with 6 electrons, and the 4s sub-shell is complete with 2 electrons. Therefore, Iron has an incomplete penultimate shell. Sodium, Aluminium, and Calcium all have complete penultimate shells in their ground state electronic configurations. Sodium (Na) has an electronic configuration of 1s² 2s² 2p⁶ 3s¹, Aluminium (Al) has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p¹, and Calcium (Ca) has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². In all of these configurations, the penultimate shell is complete, and thus none of these elements have an incomplete penultimate shell. Therefore, the correct answer is Iron.
Pergunta 48 Relatório
Which of the following compounds can be represented by the molecular formula C2H6O?
Detalhes da Resposta
The molecular formula C2H6O indicates that the compound contains 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. Among the options given, the only compound that fits this molecular formula is Ethanol (C2H6O), which has the chemical formula CH3CH2OH. Propanol (Propanol), has the molecular formula C3H8O, Methanoic acid (Formic acid) has the molecular formula HCOOH and Glucose has the molecular formula C6H12O6.
Pergunta 49 Relatório
Determine the mass of sulphur (IV) oxide obtained when 91.0g of oxygen reacts completely with sulphur according to the following equation: S(s) + O2(g) → SO2(g) [S = 32; O = 16]
Detalhes da Resposta
The balanced chemical equation for the reaction between sulfur and oxygen to form sulfur dioxide is: S + O2 → SO2 This equation tells us that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide. We are given the mass of oxygen that reacts completely with sulfur, which is 91.0 g. We can use this information to calculate the number of moles of oxygen: Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen Number of moles of oxygen = 91.0 g / 32 g/mol Number of moles of oxygen = 2.84 mol From the balanced equation, we can see that the number of moles of sulfur dioxide produced is equal to the number of moles of oxygen that reacts. Therefore, we have: Number of moles of sulfur dioxide = 2.84 mol Now we can use the molar mass of sulfur dioxide to calculate its mass: Mass of sulfur dioxide = Number of moles of sulfur dioxide × Molar mass of sulfur dioxide Mass of sulfur dioxide = 2.84 mol × (32 g/mol + 2 × 16 g/mol) Mass of sulfur dioxide = 2.84 mol × 64 g/mol Mass of sulfur dioxide = 182 g Therefore, the mass of sulfur dioxide produced when 91.0 g of oxygen reacts completely with sulfur is 182 g. So, the correct option is (d) 182g.
Pergunta 50 Relatório
Consider the following equation: Cr2O2-7(aq) + 14H+(aq) + 6e- → 2Cr3+(ag) + 7H2O(1). The oxidation number of chromium changes from
Detalhes da Resposta
The oxidation number is a number that represents the charge of an atom in a compound. In the given equation, the chromium atom changes from a +6 oxidation state on the reactant side (Cr2O2-(aq)) to a +3 oxidation state on the product side (2Cr3+(aq)). This means that the oxidation number of chromium has decreased by 3, indicating that it has undergone reduction. Therefore, the correct answer is +6 to +3.
Pergunta 51 Relatório
The atom of an element X is represented as \(^y_z X\). The basic chemical properties of X depend on the value of
Detalhes da Resposta
In the symbol of an element, the number of protons (Z) in its nucleus is called its atomic number. The atomic number determines the element's identity and its basic chemical properties. Elements with the same atomic number have the same number of protons, so they have the same electronic configuration, meaning they have similar chemical behavior. Therefore, the value of Z in the symbol \(^y_z X\) determines the element's basic chemical properties.
Pergunta 52 Relatório
The atomic number of chlorine is 17. What is the number of electrons in a chloride ion?
Detalhes da Resposta
Chlorine has 17 electrons because it has 17 protons, which is its atomic number. A chloride ion (Cl-) has gained one electron to form a negative ion. Therefore, it now has one more electron than the number of protons (which is still 17). This means that a chloride ion has 18 electrons. So, the answer is (c) 18.
Pergunta 53 Relatório
(a) Giving different examples, mention one metal in each case which produces hydrogen on reacting with
(i) dilute mineral acid
(ii) cold water;
(iii) steam;
(iv) hot, concentrated alkali.
(b) In an experiment, excess 0.50 mol dm\(^{-3}\) HCI was added to 1Og of granulated zinc in a beaker. Other conditions remaining constant, state how the reaction rate would be affected in each case, if the experiment was repeated using:
(i) 1.0 mol dm\(^{-3}\) HCI;
(ii) 8.0g of granulated zinc;
(iii) 10g of zinc dust;
(iv) a higher volume of 0.50 mol dm HCI;
(v) a reaction vessel dipped in crushed ice;
(vi) equal volumes of water and 0.50 mol dm\(^3\) HCI.
(c) Aluminium is extracted from its ore by electrolysis.
(i) Name the ore from which the metal is extracted.
(ii) State the role of molten cryolite in the extraction.
(iii) Describe in outline how the ore is purified before electrolysis
(iv) Calculate the current in amperes required to produce 18.0g of aluminium in 1.50 hours. [Al = 27.0; F = 96500C]
(d) Give the reason why
(i) aluminium, which is a reactive metal, is resistant to corrosion.
(ii) metals are generally good reducing agents.
Pergunta 54 Relatório
(a) Explain in terms of the kinetic theory why a tyre should not be overinflated.
(b)The following results were obtained at room temperature in an experiment to verify one of the gas laws using a glass syringe:
| Pressure (P) of air in syringe (atm) | Volume (V) of air in syringe (\(cm^3\)) | \(\frac{I}{V}\) |
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(i) Plot a graph of P against \(\frac{1}{v}\), using 1 cm to represent 0.01 atm on the vertical axis and 1cm to represent 0.02 unit on the horizontal axis.
(ii) Which of the gas laws is in agreement with the results?
(c) The flow chart below represents the stages involved in the manufacture of H\(_2\)SO\(_4\).
| S + O\(_2\) \(\to\) SO\(_2\) | SO\(_2\) +x \(\to\) SO\(_3\) | SO\(_3\) +Conc. H\(_2\)SO\(_4\) \(\to\) Y | Y +H\(_2\)O \(\to\) Conc H\(_2\)SO\(_4\) |
| stage I | stage II | stage III | stage IV |
(i) Name the process represented by the chart.
(ii) Identify reactant X and product Y.
(iii) What are the operating temperature and pressure at stage II?
(iv) Mention the stage which requires a catalyst and state the catalyst used.
(v) Give the reason why the SO\(_3\) produced in stage II is not dissolved directly in water to form the acid
(d) When K\(_4\)Cr\(_2\)C\(_7\) dissolves in water, the following equilibrium is established:
\(\mathrm{Cr_2O_{7(aq)}^{2-} + H_2O_{(l)} \to 2CrO_{4(aq)}^{2-} + 2H_{aq}}\)
(i) State the colour observed on adding a few drops of dilute H\(_2\)SO\(_4\) to the system.
(ii) Explain your answer in (d)(1).
(iii) What principle is applicable to this explanation?
(a) Why a tyre should not be overinflated (kinetic theory)
According to the kinetic theory, the air inside a tyre consists of tiny particles in continuous rapid random motion. These particles constantly collide with the inner walls of the tyre, and the total force of these collisions per unit area is the pressure the gas exerts. When the tyre is overinflated, more particles are forced into the same fixed volume, so the frequency of collisions with the walls increases and the pressure rises sharply. In addition, as the vehicle moves, friction warms the trapped air; the particles gain kinetic energy and move faster, striking the walls harder and more often, which raises the pressure still further. If the internal pressure exceeds the strength of the tyre material, the tyre bursts. The tyre should therefore not be overinflated.
(b) Verifying a gas law with a glass syringe
The recorded readings, with \(\tfrac{1}{V}\) computed to three significant figures, are:
| Pressure, \(P\) (atm) | Volume, \(V\) (cm\(^3\)) | \(\dfrac{1}{V}\) (cm\(^{-3}\)) | \(P\times V\) (atm cm\(^3\)) |
|---|---|---|---|
| 0.100 | 10.00 | 0.100 | 1.00 |
| 0.125 | 8.00 | 0.125 | 1.00 |
| 0.150 | 6.60 | 0.152 | 0.99 |
| 0.175 | 5.60 | 0.179 | 0.98 |
| 0.200 | 4.80 | 0.208 | 0.96 |
| 0.225 | 4.40 | 0.227 | 0.99 |
(i) Plotting \(P\) (vertical axis) against \(\tfrac{1}{V}\) (horizontal axis) gives a straight line passing through the origin:
The graph is a straight line through the origin, which shows that \(P\) is directly proportional to \(\tfrac{1}{V}\). The gradient of the line,
\[\text{slope}=\frac{\Delta P}{\Delta(1/V)}=\frac{0.225-0.100}{0.227-0.100}=\frac{0.125}{0.127}\approx 1.0\ \text{atm cm}^3,\]which equals the constant product \(P\times V\) in the last column of the table. Hence \(PV=\text{constant}\).
(ii) The result that agrees with the data is Boyle's Law, which states that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. \(P\propto\tfrac{1}{V}\), so \(PV=\text{constant}\).
(c) Manufacture of H\(_2\)SO\(_4\)
(i) The process represented by the chart is the Contact Process.
(ii) Reactant X is oxygen, O\(_2\) (air); product Y is oleum (fuming sulphuric acid), H\(_2\)S\(_2\)O\(_7\).
(iii) At Stage II the operating temperature is about 450 °C (in the range 400-500 °C) and the pressure is about 1-2 atm (approximately atmospheric).
(iv) The stage that requires a catalyst is Stage II (the conversion of SO\(_2\) to SO\(_3\)), and the catalyst used is vanadium(V) oxide, V\(_2\)O\(_5\).
(v) The SO\(_3\) is not dissolved directly in water because the reaction is strongly exothermic; the large amount of heat released would vaporise the water and create a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense and would escape the plant. Instead SO\(_3\) is absorbed in concentrated H\(_2\)SO\(_4\) to form oleum, which is then safely diluted with water.
(d) The chromate-dichromate equilibrium
The equilibrium established is
\[\text{Cr}_2\text{O}_7^{2-}{}_{(aq)}+\text{H}_2\text{O}_{(l)}\rightleftharpoons 2\text{CrO}_4^{2-}{}_{(aq)}+2\text{H}^+{}_{(aq)}\]where \(\text{Cr}_2\text{O}_7^{2-}\) (dichromate) is orange and \(\text{CrO}_4^{2-}\) (chromate) is yellow.
(i) On adding a few drops of dilute H\(_2\)SO\(_4\), the colour changes from yellow to orange (the solution becomes orange).
(ii) Dilute H\(_2\)SO\(_4\) supplies H\(^+\) ions, increasing their concentration on the right-hand side of the equilibrium. The system responds by shifting the position of equilibrium to the left so as to remove some of the added H\(^+\). This produces more of the orange dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\), and less of the yellow chromate ion, so the solution turns orange.
(iii) The principle applicable to this explanation is Le Chatelier's Principle.
Detalhes da Resposta
(a) Why a tyre should not be overinflated (kinetic theory)
According to the kinetic theory, the air inside a tyre consists of tiny particles in continuous rapid random motion. These particles constantly collide with the inner walls of the tyre, and the total force of these collisions per unit area is the pressure the gas exerts. When the tyre is overinflated, more particles are forced into the same fixed volume, so the frequency of collisions with the walls increases and the pressure rises sharply. In addition, as the vehicle moves, friction warms the trapped air; the particles gain kinetic energy and move faster, striking the walls harder and more often, which raises the pressure still further. If the internal pressure exceeds the strength of the tyre material, the tyre bursts. The tyre should therefore not be overinflated.
(b) Verifying a gas law with a glass syringe
The recorded readings, with \(\tfrac{1}{V}\) computed to three significant figures, are:
| Pressure, \(P\) (atm) | Volume, \(V\) (cm\(^3\)) | \(\dfrac{1}{V}\) (cm\(^{-3}\)) | \(P\times V\) (atm cm\(^3\)) |
|---|---|---|---|
| 0.100 | 10.00 | 0.100 | 1.00 |
| 0.125 | 8.00 | 0.125 | 1.00 |
| 0.150 | 6.60 | 0.152 | 0.99 |
| 0.175 | 5.60 | 0.179 | 0.98 |
| 0.200 | 4.80 | 0.208 | 0.96 |
| 0.225 | 4.40 | 0.227 | 0.99 |
(i) Plotting \(P\) (vertical axis) against \(\tfrac{1}{V}\) (horizontal axis) gives a straight line passing through the origin:
The graph is a straight line through the origin, which shows that \(P\) is directly proportional to \(\tfrac{1}{V}\). The gradient of the line,
\[\text{slope}=\frac{\Delta P}{\Delta(1/V)}=\frac{0.225-0.100}{0.227-0.100}=\frac{0.125}{0.127}\approx 1.0\ \text{atm cm}^3,\]which equals the constant product \(P\times V\) in the last column of the table. Hence \(PV=\text{constant}\).
(ii) The result that agrees with the data is Boyle's Law, which states that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. \(P\propto\tfrac{1}{V}\), so \(PV=\text{constant}\).
(c) Manufacture of H\(_2\)SO\(_4\)
(i) The process represented by the chart is the Contact Process.
(ii) Reactant X is oxygen, O\(_2\) (air); product Y is oleum (fuming sulphuric acid), H\(_2\)S\(_2\)O\(_7\).
(iii) At Stage II the operating temperature is about 450 °C (in the range 400-500 °C) and the pressure is about 1-2 atm (approximately atmospheric).
(iv) The stage that requires a catalyst is Stage II (the conversion of SO\(_2\) to SO\(_3\)), and the catalyst used is vanadium(V) oxide, V\(_2\)O\(_5\).
(v) The SO\(_3\) is not dissolved directly in water because the reaction is strongly exothermic; the large amount of heat released would vaporise the water and create a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense and would escape the plant. Instead SO\(_3\) is absorbed in concentrated H\(_2\)SO\(_4\) to form oleum, which is then safely diluted with water.
(d) The chromate-dichromate equilibrium
The equilibrium established is
\[\text{Cr}_2\text{O}_7^{2-}{}_{(aq)}+\text{H}_2\text{O}_{(l)}\rightleftharpoons 2\text{CrO}_4^{2-}{}_{(aq)}+2\text{H}^+{}_{(aq)}\]where \(\text{Cr}_2\text{O}_7^{2-}\) (dichromate) is orange and \(\text{CrO}_4^{2-}\) (chromate) is yellow.
(i) On adding a few drops of dilute H\(_2\)SO\(_4\), the colour changes from yellow to orange (the solution becomes orange).
(ii) Dilute H\(_2\)SO\(_4\) supplies H\(^+\) ions, increasing their concentration on the right-hand side of the equilibrium. The system responds by shifting the position of equilibrium to the left so as to remove some of the added H\(^+\). This produces more of the orange dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\), and less of the yellow chromate ion, so the solution turns orange.
(iii) The principle applicable to this explanation is Le Chatelier's Principle.
Pergunta 55 Relatório
(a) What is the shape of (i) p - orbital; (ii) a molecule of methane; (iii) a molecule of carbon (IV) oxide?
(b) Consider the following elements: Ne, S, CI, 0, Fe, Mg. State which of them
(i) exhibit(s) allotropy;
(ii) form(s) coloured ions;
(iii) is/are malleable;
(iv) consist(s) of molecules that are far apart at room temperature;
(v) form(s) hydrides by sharing electrons with hydrogen;
(vi) has/have complete outermost shell.
(c)(i) List three applications of radioactivity in different fields.
(ii) Explain clearly the difference between the following reactions involving electron loss from lead.
\(^{211} pb\) \(\to\) \(^{ 211}Bi\) + \(^0_{-1}\); Pb \(\to\) pb\(^{3+}\) _ 2e\(^-\)
(iii) Give one advantage and one disadvantage of nuclear power generation over the use of fossil fuels.
Pergunta 56 Relatório
a)(i) Give two uses of ammonia.
(ii) Name the process by which ammoniacal liquor can be obtained from coal and list two other products of the reaction
(iii) What type of reaction is involved in the conversion of ammoniacal liquor to (NH\(_4\))\(_2\)SO\(_4\) by dilute H\(_2\)SO\(_4\)?
(iv) Sketch and label an energy profile diagram to show the effect of presence of Pt/Rh on the reaction represented by the following equation: 4NH\(_3\) + 5O\(_2\) \(\to\) 6H\(_2\)O + 4NO; \(\Delta\)H = —907 kJmol\(^1\)
(b) Rock salt is an impure form of sodium chloride.
(i) Outline a suitable procedure for preparing a pure sample of sodium chloride from rock salt.
(ii) State two methods that can be used to prepare chlorine from rock salt. Write an appropriate equation in each case.
(c) Lead pigments were used in a water colour painting which turned black after prolonged exposure to an air pollutant. The original colour was restored by using H\(_2\)O\(_2\) which converted the black substance to a simple, white lead (II) salt.
(i) Which pollutant turned the painting black?
(ii) Write the formula of the black substance
(iii) What is the white salt?
(iv) State the role of H\(_2\)O in the restoration process.
(a)(i) Two uses of ammonia are:
- Manufacture of fertilizers such as ammonium sulphate and ammonium nitrate.
- Manufacture of trioxonitrate(V) acid, i.e. nitric acid.
(a)(ii) Ammoniacal liquor is obtained by the destructive distillation of coal.
Two other products are:
- Coke
- Coal tar
Coal gas is also produced.
(a)(iii) The reaction is a neutralisation (acid-base) reaction.
\(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(a)(iv) Energy profile diagram for the reaction:
The reaction is exothermic because the products are at a lower energy level than the reactants. Pt/Rh lowers the activation energy but does not change the value of \(\Delta H\).
(b)(i) Procedure for preparing pure sodium chloride from rock salt:
- Crush the rock salt and dissolve it in water.
- Filter the mixture to remove insoluble impurities such as sand.
- Evaporate the filtrate until it is nearly saturated.
- Allow the solution to cool so that sodium chloride crystals form.
- Filter, wash the crystals with a little cold distilled water and dry them.
(b)(ii) Two methods of preparing chlorine from rock salt are:
-
Electrolysis of concentrated sodium chloride solution (brine):
\(2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{electrolysis}} 2NaOH(aq) + H_2(g) + Cl_2(g)\) -
Oxidation of sodium chloride with manganese(IV) oxide and concentrated tetraoxosulphate(VI) acid:
\(2NaCl + MnO_2 + 2H_2SO_4 \rightarrow Cl_2 + MnSO_4 + Na_2SO_4 + 2H_2O\)
(c)(i) The air pollutant is hydrogen sulphide, \(H_2S\).
(c)(ii) The black substance is lead(II) sulphide, \(PbS\).
(c)(iii) The white salt is lead(II) tetraoxosulphate(VI), \(PbSO_4\).
(c)(iv) Hydrogen peroxide, \(H_2O_2\), acts as an oxidising agent. It oxidises black lead(II) sulphide to white lead(II) sulphate:
\(PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O\)
Detalhes da Resposta
(a)(i) Two uses of ammonia are:
- Manufacture of fertilizers such as ammonium sulphate and ammonium nitrate.
- Manufacture of trioxonitrate(V) acid, i.e. nitric acid.
(a)(ii) Ammoniacal liquor is obtained by the destructive distillation of coal.
Two other products are:
- Coke
- Coal tar
Coal gas is also produced.
(a)(iii) The reaction is a neutralisation (acid-base) reaction.
\(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(a)(iv) Energy profile diagram for the reaction:
The reaction is exothermic because the products are at a lower energy level than the reactants. Pt/Rh lowers the activation energy but does not change the value of \(\Delta H\).
(b)(i) Procedure for preparing pure sodium chloride from rock salt:
- Crush the rock salt and dissolve it in water.
- Filter the mixture to remove insoluble impurities such as sand.
- Evaporate the filtrate until it is nearly saturated.
- Allow the solution to cool so that sodium chloride crystals form.
- Filter, wash the crystals with a little cold distilled water and dry them.
(b)(ii) Two methods of preparing chlorine from rock salt are:
-
Electrolysis of concentrated sodium chloride solution (brine):
\(2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{electrolysis}} 2NaOH(aq) + H_2(g) + Cl_2(g)\) -
Oxidation of sodium chloride with manganese(IV) oxide and concentrated tetraoxosulphate(VI) acid:
\(2NaCl + MnO_2 + 2H_2SO_4 \rightarrow Cl_2 + MnSO_4 + Na_2SO_4 + 2H_2O\)
(c)(i) The air pollutant is hydrogen sulphide, \(H_2S\).
(c)(ii) The black substance is lead(II) sulphide, \(PbS\).
(c)(iii) The white salt is lead(II) tetraoxosulphate(VI), \(PbSO_4\).
(c)(iv) Hydrogen peroxide, \(H_2O_2\), acts as an oxidising agent. It oxidises black lead(II) sulphide to white lead(II) sulphate:
\(PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O\)
Pergunta 57 Relatório
(a)(i) State three methods of preparing salts, giving one example in each case of a salt so prepared.
(ii) What type of salt is each of the following? NaH\(_2\)PO\(_4\); (CH\(_3\)COO)\(_2\)Pb; KAI(SO\(_4\))\(_2\). 12H\(_2\)O.
(b)(i) Write an equation for the reaction between dilute HCI and a solution of AgNO\(_3\).
(ii) Explain why NaNO\(_3\) is preferred to AgNO\(_3\) in the preparation of oxygen by thermal decomposition of trioxonitrate (V) salts.
(iii) When silver wire was dipped into an aqueous solution of CuSO\(_4\), the wire remained intact but when the wire was replaced with zinc rod, the rod decreased in size. Give an explanation for this observation.
(c) When a sample of a crystalline salt X was exposed to air, there was a loss in mass.
(i) What phenomenon was exhibited by X?
(ii) Suggest two substances which X could be.
(iii) On heating 5.00 g of a fresh sample of X to constant mass, 1.80g was lost in the form of water vapour. Calculate the number of molecules of water of crystallization in one molecule of X. [H = 1.00; O = 16.00; Anhydrous form of X = 160 g mol\(^{-1}\)
Detalhes da Resposta
None
Pergunta 58 Relatório
(a) What term is used to describe each of the following processes?
(i) Alkaline hydrolysis of fats and oils;
(ii) The conversion of glucose into ethanol by enzymatic action;
(iii) Thermal decomposition of higher petroleum fractions into lower molecular mass hydrocarbons in the presence of catalyst.
(b)(i) Write the structure and IUPAC name for one alkanoic acid with the molecular formula C\(_4\)H\(_8\)0\(_2\).
(ii) Arrange the following compounds in order of increasing boiling point: Butane; Butanoic acid; Methylpropane.
(iii) Give an explanation for your answer in (b)(ii).
(c)(i) Ethanol was used for preparing a gas X which decolorized bromine water. Identify X and describe briefly its laboratory preparation.
(ii) Write an equation to show how ethanol reacts with sodium
(iii) Give the reagent and reaction conditions for the conversion of ethanol into C\(_2\)H\(_5\)COOC\(_2\)H\(_5\).
(d) State the type of recction involved in each of the conversions indicated below:
(i)C\(_6\)H\(_6\)C\(_6\)H\(_5\)CH\(_3\)
(ii) nC\(_2\)H\(_4\) \(\to\) (CH\(_2\) - CH\(_2\)),
(iii) CH\(_3\)CH\(_2\)CH(OH)CH\(_3\) -> CH\(_3\)CH\(_2\)CCH\(_3\)
(iv) (C\(_6\)H\(_{10}\)O\(_5\)) -> C\(_6\)H\(_{12}\)O\(_6\).
(iv) 
Detalhes da Resposta
None
