WAEC SSCE - Chemistry - 2004
Question 1 Report
The condensation of several amino acid molecules gives
Answer Details
The condensation of several amino acid molecules gives proteins. A protein is a macromolecule composed of one or more chains of amino acid residues. When two amino acids combine, a peptide bond is formed, and the resulting molecule is called a dipeptide. By continued condensation reactions involving many amino acids, a long chain of amino acids is formed, which is a protein. Therefore, option D, "proteins," is the correct answer.
Question 2 Report
A metal X forms two chlorides with the formulae XCI2 and XCI3. Where is X in the Periodic Table?
What type of bonding exists between X and chlorine in each of the chlorides?
Answer Details
Question 3 Report
Which of the following is required in an experiment to determine the enthalpy of solution of anhydrous CuSO? l. Mass of the anhydrous CuSO 4 II. Volume of water in which the CuSO4 is dissolved lll. lnitial temperature of water lV. Final temperature of the solution
Answer Details
To determine the enthalpy of solution of anhydrous CuSO4, we need to dissolve a known mass of CuSO4 in a known volume of water, measure the initial temperature of the water and the final temperature of the solution. Therefore, we need all the four options: i.e., l, ll, lll, and IV. We need to know the mass of CuSO4 used to determine the amount of solute in the solution. The volume of water is required to calculate the concentration of the solution. The initial temperature of the water is required to determine the amount of heat absorbed by the solution, and the final temperature of the solution is required to determine the amount of heat released by the solution.
Question 4 Report
The reaction that takes place at the cathode when an aqueous solution of NaCI is electrolysed using carbon electrodes is
Answer Details
The reaction that takes place at the cathode when an aqueous solution of NaCl is electrolyzed using carbon electrodes is: - 2H+2(g) + 2e- → H2(g) During the electrolysis of an aqueous solution of NaCl using carbon electrodes, the NaCl molecules dissociate into Na+ and Cl- ions. The Na+ ions are attracted towards the cathode where they gain electrons and become sodium atoms which are deposited on the cathode. However, water molecules are also present in the solution and can be reduced instead of Na+ ions. The reduction of water molecules results in the production of hydrogen gas. The hydrogen gas bubbles out of the solution, and the overall reaction at the cathode can be represented as: 2H+(aq) + 2e- → H2(g) Therefore, option A is the correct answer.
Question 5 Report
If 60 cm3 of gas is heated from 27oC TO 50oC, what is the new volume of the gas at constant pressure?
Question 6 Report
Which of the following statements is true of an endothermic reaction? I. Heat is absorbed from the surrounding II. The heat content of the products is more than that of the reactants III. The enthalpy change is positive IV. The surrounding is at a lower temperature than the system.
Answer Details
Question 7 Report
Consider the cell represented below: Zn/Zn2+(aq) // Cu2+(aq) / Cu(s). Which of the following statements is correct about the cell
Answer Details
The correct statement about the cell represented as Zn/Zn2+(aq) // Cu2+(aq) / Cu(s) is that Zn is oxidized to Zn2+. In this cell, the zinc metal (Zn) is the anode, where oxidation occurs. Zn atoms lose electrons to form Zn2+ ions, which go into solution as Zn2+(aq). The electrons released in the oxidation of Zn are transferred through the external circuit to the cathode, which is the copper metal (Cu) electrode. The Cu2+ ions in solution gain electrons and are reduced to form copper metal, which deposits on the electrode as Cu(s). The double line in the cell notation represents the salt bridge, which completes the circuit by allowing the flow of ions between the two half-cells. It prevents the mixing of the solutions in the half-cells, which could disrupt the electrochemical reaction. Therefore, the correct option is "Zn is oxidized to Zn2+".
Question 8 Report
Which of the following pollutants is biodegradable?
Answer Details
Domestic sewage is biodegradable. Biodegradable means that a substance can be broken down by living organisms such as bacteria, fungi, and other microorganisms. Domestic sewage consists of organic matter such as food waste and human waste, which can be decomposed by microorganisms. The process of decomposition breaks down the organic matter into simpler compounds that can be absorbed by the soil, making domestic sewage a natural fertilizer. Therefore, domestic sewage can be biodegraded by microorganisms in the soil or in wastewater treatment plants.
Question 9 Report
Which of the following compounds is a soapless detergent?
Answer Details
The compound that is a soapless detergent is C 12H 25OSO3Na. Soapless detergents are also known as synthetic detergents, and they are produced from organic compounds. They do not contain any soap, hence the name soapless. They are usually made by combining an alkyl benzene or a long-chain alcohol with a sulfonic acid or a sulfuric acid, which creates a compound that is highly effective at removing dirt and grease from surfaces. This compound is a sodium alkyl benzene sulfonate, which is a type of synthetic detergent commonly used in household cleaning products.
Question 10 Report
WHICH OF THE FOLLOWING PETROLEUM FRACTIONS YIELDS VASELINE AND PARAFFIN WAX ON REDISTILLATION?
Answer Details
The petroleum fractions that yield vaseline and paraffin wax on redistillation is kerosene. When crude oil is distilled, it is separated into different fractions based on their boiling points. Kerosene is one of the fractions that is obtained from the distillation of crude oil. Upon redistillation, the kerosene fraction further separates into different components, including vaseline and paraffin wax. Therefore, the correct option is kerosene.
Question 11 Report
What of the following pairs of acid and base will produce a solution with pH less than 7 at equivalence point?
Answer Details
Question 12 Report
Which of the following measures is not suitable for controlling water pollution?
Answer Details
Burning industrial and agricultural wastes before discharge into waterways is not a suitable measure for controlling water pollution. Burning of waste can release harmful chemicals and gases into the air, which can contribute to air pollution. Moreover, burning the waste before discharge into water can increase the temperature of the water, which can have negative effects on aquatic life. Therefore, this measure can actually worsen water pollution and is not a recommended solution. The other options such as treating industrial effluent and domestic sewage before discharge into water, controlling the use of agrochemicals, and recycling industrial and agricultural wastes are effective measures for controlling water pollution.
Question 14 Report
The alloy used extensively in the building industry is
Answer Details
The alloy used extensively in the building industry is steel. Steel is an alloy of iron and carbon, with small amounts of other elements such as manganese, silicon, and phosphorus. Steel is known for its strength and durability, making it an ideal material for use in the construction of buildings and other structures. It is also a versatile material that can be easily fabricated and formed into various shapes and sizes, making it an ideal choice for a wide range of construction applications.
Question 15 Report
Glucose gives a brick-red precipitate with Fehling's solution because it is
Answer Details
Glucose gives a brick-red precipitate with Fehling's solution because it is a reducing sugar. Fehling's solution contains copper ions (Cu2+) which are reduced to copper (I) oxide (Cu2O) by reducing sugars. The brick-red precipitate is the copper (I) oxide formed in the reaction. Glucose is a reducing sugar because it has a free aldehyde or ketone functional group that can donate electrons and reduce other substances like the copper ions in Fehling's solution.
Question 16 Report
What volume of hydrogen would be left over when 300cm3 of oxygen and 1000cm of hydrogen are exploded in a sealed tube? [2H2(g) + O2(g) → 2H2 Og]
Question 17 Report
What is the IUPAC name of CH3CH2COOCH 3?
Answer Details
The IUPAC name of CH3CH2COOCH 3 is methyl propanoate. This is because the compound has a methyl group (-CH3) attached to the third carbon of a propanoic acid (-COOH) molecule, which results in the name "methyl propanoate". The prefix "methyl" indicates the presence of a -CH3 group, and the suffix "-oate" indicates the presence of an ester functional group (-COO-).
Question 18 Report
Which of the following equipment is not used to detect radioactivity?
Answer Details
The equipment that is not used to detect radioactivity is the mass spectrometer. A Wilson cloud chamber is a device used to visualize the path of ionizing radiation by producing condensation trails. A Geiger-Muller counter is an instrument used to detect and measure ionizing radiation. A photographic plate can also be used to detect ionizing radiation. However, a mass spectrometer is an analytical instrument used to determine the composition of a sample by separating its components based on their mass-to-charge ratio. It is not specifically designed to detect radioactivity.
Question 19 Report
When ammonium trioxocarbonate (IV) solution is added to separate solutions of calcium, zinc and sodium salt precipitate could be formed in the case of
Answer Details
When ammonium trioxocarbonate (IV) solution is added to separate solutions of calcium, zinc and sodium salt, precipitates could be formed in the case of calcium and zinc only. The ammonium trioxocarbonate (IV) solution reacts with the calcium and zinc ions in the solutions to form insoluble carbonates as follows: Ca2+ + CO32- → CaCO3(s) Zn2+ + CO32- → ZnCO3(s) These carbonates are insoluble in water and will form precipitates in the solutions. The sodium ion does not react with the ammonium trioxocarbonate (IV) solution and therefore, no precipitate will be formed in its solution.
Question 20 Report
Which of the following factors that affect the rate of chemical reactions can be explained in terms of the collision theory? l. concentration ll. pressure lll. temperature lV catalyst
Answer Details
Question 21 Report
A nuclide emits a beta-particle (e) in the process, that atomic number
Answer Details
When a nuclide emits a beta-particle (e), a neutron inside the nucleus is converted into a proton and an electron is emitted. The mass number of the nuclide remains constant because the total number of protons and neutrons in the nucleus remains the same. However, since a neutron is converted into a proton, the atomic number of the nuclide increases by 1, because the atomic number of an element is equal to the number of protons in its nucleus. Therefore, the correct option is: "the atomic number increases by 1 while the mass number remains constant".
Question 22 Report
The key factor to be considered in siting a chemical industry is
Answer Details
The key factor to be considered in siting a chemical industry is its nearness to the source of raw materials. This is because chemical industries rely heavily on the availability and accessibility of raw materials for production. Therefore, the cost of transportation and logistics of raw materials to the industry should be minimized to increase efficiency and profitability. Additionally, the proximity to raw materials reduces the risks of shortages, delays and damages in transit, and fluctuations in prices. Other factors such as favourable climatic conditions, the availability of space to store raw materials, and nearness to other industrial establishments may be important but are usually secondary.
Question 23 Report
Which of the following electron transition results in the emission of energy?
Answer Details
When an electron transitions from a higher energy level to a lower energy level, it releases energy in the form of electromagnetic radiation. The energy of the emitted radiation corresponds to the difference in energy between the two levels involved in the transition. Based on this information, the correct answer to the question is the electron transition from the 3p orbital to the 3s orbital, as this involves a higher energy level (3p) transitioning to a lower energy level (3s), resulting in the emission of energy. The other options involve transitions from lower to higher energy levels, and therefore do not result in the emission of energy.
Question 24 Report
The process of seeding during the extraction of aluminium involves the addition of pure AI(OH)3 crystals in order to
Answer Details
The addition of pure AI(OH)3 crystals during the extraction of aluminium is called seeding. The purpose of seeding is to facilitate the precipitation of solid Al(OH)3. When the pure AI(OH)3 crystals are added, they act as nucleation sites where additional Al(OH)3 molecules can deposit and grow. This results in the formation of larger and more uniform particles of Al(OH)3, which can be easily separated from the solution. Therefore, the correct option is "facilitate the precipitation of solid."
Question 25 Report
The method of collection of gases prepared in the laboratory depend on their
Answer Details
Question 26 Report
What is the change in the oxidation number of 1- in the reaction represented by the following equation? 51 -1(aq) + 6H+(aq) + 1O-3(aq) → 312(g) + 3H2O(l)
Answer Details
The oxidation number of an ion is the charge of the ion. The oxidation number of 1- is -1. In the equation, the reactant 51 -1(aq) is being oxidized because its oxidation number is changing from -1 to 0. Thus, the change in the oxidation number of 1- is from -1 to 0, which means an increase in oxidation number or a loss of electrons. Therefore, the answer is -1 to 0.
Question 27 Report
The mass of an element is 27 and its atomic number is 13. What is the composition of the nucleus of its atom?
Answer Details
The atomic number of an element is equal to the number of protons in its nucleus. In this case, the atomic number is 13. This means the nucleus of the atom contains 13 protons. The mass number of an element is equal to the number of protons plus the number of neutrons in its nucleus. The mass of the element is given as 27. Therefore, the number of neutrons in the nucleus can be calculated as follows: Number of neutrons = Mass number - Atomic number Number of neutrons = 27 - 13 Number of neutrons = 14 So, the composition of the nucleus of the atom is 13 protons and 14 neutrons. Therefore, the correct option is "13 protons and 14 neutrons".
Question 28 Report
One of the characteristics of transition metals is
Answer Details
One of the characteristics of transition metals is the formation of coloured ions. Transition metals have partially filled d-orbitals, which allow for the absorption and reflection of certain wavelengths of light. When transition metals form ions, the arrangement of electrons in the d-orbitals changes, causing the ion to absorb and reflect different wavelengths of light. This results in the ion having a different colour than the original metal. Therefore, the ability of transition metals to form coloured ions is one of their defining characteristics.
Question 29 Report
WHICH OF THE FOLLOWING SALTS IS READILY SOLUBLE IN WATER?
Answer Details
Pb(NO3)2 is readily soluble in water because it is a nitrate salt. Nitrate salts are generally soluble in water. On the other hand, PbCl2, PbSO4 and PbCO3 are insoluble or only slightly soluble in water.
Question 30 Report
Which of the following subtances will not produce effervescence with dilute hydeochloric acid?
Answer Details
Sodium chloride will not produce effervescence with dilute hydrochloric acid. When a substance reacts with dilute hydrochloric acid, it produces bubbles of carbon dioxide gas, which causes effervescence. Copper (II) trioxocarbonate (IV), potassium hydrogen trioxocarbonate (IV), and zinc granules are all carbonates that react with hydrochloric acid to produce carbon dioxide gas and effervescence. Sodium chloride, on the other hand, is a halide that does not react with hydrochloric acid, and therefore does not produce effervescence.
Question 31 Report
Which of the following properties of elements deceases generally across a period in the Periodic Table?
Answer Details
The correct answer is "Atomic radius". Across a period in the Periodic Table, the atomic radius generally decreases. This is because, as you move from left to right across a period, the number of protons in the nucleus increases, and the number of energy levels or shells remains the same. This results in a greater attractive force between the positively charged nucleus and the negatively charged electrons in the outermost shell, pulling them closer to the nucleus and hence decreasing the atomic radius.
Question 32 Report
What is the product Cx Hy in the following equation? C10H12 → C8H18 + CxHy
Question 33 Report
What are the values of x, y and z respectively in the following in the following equation? MnO 2(s) + xHCI(aq) → MnCI2(g) + yHO(l) + zCI2(g)
Question 34 Report
A metal X forms two chlorides with the formulae XCI2 and XCI3. Where is X in the Periodic Table?
Answer Details
Metal X forms two chlorides with formulae XCI2 and XCI3. The Roman numeral in the name of a metal chloride indicates the charge of the metal ion. The chlorine atom has a fixed charge of -1. Since the formula of XCI2 has two chlorine atoms, the total charge contributed by the chlorine atoms is -2. Therefore, the charge of the metal ion in XCI2 must be +2 in order for the compound to be electrically neutral. Likewise, the charge of the metal ion in XCI3 must be +3 in order for the compound to be electrically neutral. From the above information, we can conclude that the metal X has variable oxidation states of +2 and +3. This means that the metal X must be a transition metal and located in the d-block of the periodic table. Therefore, the correct answer is "d-block."
Question 35 Report
Two solutions containing Pb2+ and AI3+ respectively can be distingusihed using
Answer Details
Question 36 Report
Which of the electrons in the following orbitals will experience the greatest nuclear attraction?
Answer Details
The nuclear attraction experienced by an electron depends on the distance between the electron and the nucleus and the number of protons in the nucleus. The greater the number of protons in the nucleus, the greater the nuclear attraction experienced by an electron. Therefore, out of the given options, the 1s electron of helium will experience the greatest nuclear attraction since it has the smallest atomic number and the least distance from the nucleus. The 1s electron of potassium will have the least nuclear attraction since it has the largest atomic number and the greatest distance from the nucleus. The 2s electron of carbon and the 2p electron of boron will have intermediate nuclear attraction.
Question 37 Report
A solution of zinc chloride should not be stored in a container made of
Answer Details
A solution of zinc chloride should not be stored in a container made of aluminium. This is because zinc chloride is a corrosive substance and reacts with aluminium to form aluminium chloride and zinc metal. This reaction can lead to the weakening and eventual failure of the container, potentially causing a hazardous spill. Therefore, it is important to store zinc chloride in containers made of materials that are resistant to its corrosive properties, such as glass, plastic, or certain types of metals.
Question 38 Report
Which of the following reactions will be affected by change in pressure at equilibrium?
Answer Details
Question 39 Report
What quantity of electrons in moles is needed to discharge two moles of aluminum from aluminum oxide (AI 2O3)
Answer Details
In order to discharge two moles of aluminum from aluminum oxide (Al2O3), each Al3+ ion in the Al2O3 must be reduced to Al metal, which requires the gain of three electrons per Al3+ ion. Therefore, for two moles of aluminum, there are 2 x Avogadro's number (6.022 x 1023) of Al3+ ions that need to be reduced, and each of these requires 3 electrons. Thus, the total number of electrons needed is 2 x Avogadro's number x 3 = 3.612 x 1024 electrons, which can be expressed as 6 moles of electrons. Therefore, the correct answer is 6.
Question 40 Report
Which of the following salts is not prepared by precipitation?
Answer Details
Sodium trioxocarbonate (IV) is not prepared by precipitation. Precipitation is a process that involves the formation of an insoluble solid from a solution. Lead (II) trioxocarbonate (IV), barium trioxocarbonate (IV), and calcium trioxocarbonate (IV) are all prepared by precipitation reactions. In the case of sodium trioxocarbonate (IV), it is a soluble salt, and it can be obtained by the reaction between sodium hydroxide and carbon dioxide. Thus, it is not prepared by precipitation, and this is the correct answer.
Question 41 Report
The mass of 800cm3 of a gas X at s.t.p. is 1.0g. What is the molar mass of X? [1 mole of a gas at s.t.p occupies 22.4dm 3]
Answer Details
The molar mass of gas X can be determined by using the formula: molar mass = mass / (volume of gas at stp / molar volume of gas at stp) The mass of the gas X is given as 1.0g and the volume of 800cm3 can be converted to dm3 by dividing by 1000: volume of gas at stp = 800cm3 / 1000 = 0.8dm3 The molar volume of a gas at stp is given as 22.4dm3/mol. Substituting these values into the formula, we get: molar mass = 1.0g / (0.8dm3 / 22.4dm3/mol) molar mass = 1.0g / (0.0357mol) molar mass = 28.0g/mol Therefore, the molar mass of gas X is 28.0g/mol. The correct option is: 28.0gmol-1.
Question 42 Report
An orange drink concentrate is suspected to contain traces of poisonous green dye and a harmless dye having the same boiling point. Which of the following techniques is most suitable for isolating the dyes?
Answer Details
Paper chromatography is the most suitable technique for isolating the dyes in the orange drink concentrate. This is because paper chromatography can separate the components of a mixture based on their solubility and polarity, which is exactly what is needed to separate the two dyes. The technique involves placing a small amount of the mixture on a piece of paper and allowing the components to move up the paper as a solvent travels up the paper. The dyes will move at different rates depending on their solubility and polarity, allowing them to be separated and identified. Fractional distillation separates components based on their boiling points, which is not useful in this case since the dyes have the same boiling point. Coagulation involves the clumping of particles in a liquid, which is not applicable in this scenario. Recrystallization is used to purify solids and is not applicable for isolating dyes from a liquid mixture.
Question 43 Report
(i) State Two assumptions of the kinetic theory of gases
(ii) When some solids are heated, they change directly into the gaseous state. What narne is given to this phenomenon?
(iii) List two substances which exhibit the phenomenon referred to in (a)(ii) above
(iv) Write an expression to show the mathematical relationship between the rate of diffusion of a gas and its vapour density.
(b) Consider the following equilibrium reaction:
3Fe\(_{(s)}\) + 4H\(_2\)O\(_{(g)}\) \(\rightleftharpoons\) FeO\(_3\)O\(_{4(s)}\) + 4H\(_{2(g)}\), \(\Delta\)H = - 150KJ mol\(^{-1}\)
Explain the effect of the following factors on the position of equilibrium: (i) tecrease in temperature; (ii) Increase in pressure; (iii) Removal of hydrogen.
(c) . Three beakers labelled P, Q and S each contained zinc metal of the same mass but in different forms. P contained a length of zinc rod, Q contained zinc dust while S contained zinc foil. 100cm\(^3\) of 5.0 mol dm\(^{-3}\) hydrochloric acid was added to each beaker to react with all the zinc.
(i) State the order in which the reaction came to completion in beakers P,Q and S starting with the fastest.
(ii) Give reason for your answer in (c)(i) above
(iii) Write an equation to represent the reaction between zinc rid the hydrochloric acid.
(d) (i) What is meant by pH of a solution?
(ii)(I) State with reason in each case whether the pH would increase, decrease or remain constant if the following experiments were carried out Neutralizing bench HNO\(_3\);
II. Diluting 25.0 cm\(^3\) of a given NaOH solution to 100.0cm\(^3\) Concentrating a solution of NaCI.
(a)(i) Two assumptions of the kinetic theory of gases: gas molecules are in constant, rapid, random motion; the actual volume of the molecules is negligible compared with the volume of the container (and there are negligible forces between molecules).
(ii) The phenomenon is sublimation.
(iii) Substances that sublime: iodine and ammonium chloride (also solid CO2, naphthalene, camphor).
(iv) Rate of diffusion is inversely proportional to the square root of the vapour density:
\[R \propto \frac{1}{\sqrt{d}}\]
(b) For \(3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_4_{(s)} + 4H_{2(g)}\), \(\Delta H = -150\ \text{kJ mol}^{-1}\) (exothermic):
- (i) Decrease in temperature: favours the exothermic forward reaction, so the equilibrium shifts to the right (more hydrogen).
- (ii) Increase in pressure: there are 4 moles of gas on each side, so pressure has no effect on the position of equilibrium.
- (iii) Removal of hydrogen: the equilibrium shifts to the right to replace the hydrogen removed.
(c)(i) Order (fastest first): Q (dust) > S (foil) > P (rod).
(ii) The rate depends on surface area. Zinc dust has the largest surface area exposed to the acid (fastest), the foil an intermediate area, and the rod the smallest surface area (slowest).
(iii) \[Zn + 2HCl \rightarrow ZnCl_2 + H_2\]
(d)(i) The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii)
- I. Neutralizing bench HNO3 with a base: the pH increases as H+ ions are removed.
- II. Diluting 25.0 cm3 of NaOH to 100.0 cm3: the pH decreases as the alkali becomes more dilute (lower [OH-]).
- III. Concentrating a solution of NaCl: the pH remains constant, because NaCl is a neutral salt and does not change [H+].
Answer Details
(a)(i) Two assumptions of the kinetic theory of gases: gas molecules are in constant, rapid, random motion; the actual volume of the molecules is negligible compared with the volume of the container (and there are negligible forces between molecules).
(ii) The phenomenon is sublimation.
(iii) Substances that sublime: iodine and ammonium chloride (also solid CO2, naphthalene, camphor).
(iv) Rate of diffusion is inversely proportional to the square root of the vapour density:
\[R \propto \frac{1}{\sqrt{d}}\]
(b) For \(3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_4_{(s)} + 4H_{2(g)}\), \(\Delta H = -150\ \text{kJ mol}^{-1}\) (exothermic):
- (i) Decrease in temperature: favours the exothermic forward reaction, so the equilibrium shifts to the right (more hydrogen).
- (ii) Increase in pressure: there are 4 moles of gas on each side, so pressure has no effect on the position of equilibrium.
- (iii) Removal of hydrogen: the equilibrium shifts to the right to replace the hydrogen removed.
(c)(i) Order (fastest first): Q (dust) > S (foil) > P (rod).
(ii) The rate depends on surface area. Zinc dust has the largest surface area exposed to the acid (fastest), the foil an intermediate area, and the rod the smallest surface area (slowest).
(iii) \[Zn + 2HCl \rightarrow ZnCl_2 + H_2\]
(d)(i) The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii)
- I. Neutralizing bench HNO3 with a base: the pH increases as H+ ions are removed.
- II. Diluting 25.0 cm3 of NaOH to 100.0 cm3: the pH decreases as the alkali becomes more dilute (lower [OH-]).
- III. Concentrating a solution of NaCl: the pH remains constant, because NaCl is a neutral salt and does not change [H+].
Question 44 Report
(a) Methane reacts with chlorine under certain condition to produce tetrachloromethane.
(i) State the condition for the reaction
(ii) Name the type of reaction
(iii) Give two uses of methane
(iv) Name one major natural source of methane
(b)(i) Mention one similarity between the reaction of ethanol with sodium and that of sodium with water
(ii) Write the structure of two isomers of C\(_3\)H\(_8\)O
(iii) Differentiate between a fine chemical and a heavy chemical
(iv) Give one example of each in (b)(iii) above
(c) Two compounds X and Y have the same percentage composition by mass of 92.3% carbon and 7.7% hydrogen. Calculate the:
(i) empirical formula of X and Y;
(ii) molecular formula of each compound if molar mass of X is 26 g and Y is 78g.
(d) A protein is boiled for a long time with dilute HCI and a reaction occurred.
(i) State the type of reaction that occurred
(ii) Name the major product formed
(iii) Give the functional groups present in (d)(ii) above.
(a)(i) Condition: presence of (ultraviolet) sunlight / diffused light.
(ii) Type of reaction: substitution (photochemical, free-radical substitution).
(iii) Uses of methane: as a domestic and industrial fuel (natural gas); manufacture of hydrogen; manufacture of carbon black / methanol.
(iv) Major natural source: natural gas (also biogas/marsh gas from decaying organic matter).
(b)(i) Similarity: both reactions liberate hydrogen gas. (\(2Na + 2C_2H_5OH \rightarrow 2C_2H_5ONa + H_2\); \(2Na + 2H_2O \rightarrow 2NaOH + H_2\)).
(ii) Two isomers of C3H8O
Propan-1-ol: CH3-CH2-CH2-OH
Propan-2-ol: CH3-CH(OH)-CH3(iii) A fine chemical is made in small quantities to a high degree of purity and is costly; a heavy chemical is made in large (bulk) quantities and is of lower purity.
(iv) Fine chemical: a drug (for example aspirin) or a dye. Heavy chemical: tetraoxosulphate(VI) acid (or sodium hydroxide, ammonia).
(c) C : H by mass = 92.3% : 7.7%.
C: \(\dfrac{92.3}{12} = 7.69\); H: \(\dfrac{7.7}{1} = 7.7\). Ratio \(\approx 1:1\).
(i) Empirical formula of X and Y = CH (empirical mass = 13).
(ii) X: \(\dfrac{26}{13} = 2\), so molecular formula = C2H2 (ethyne). Y: \(\dfrac{78}{13} = 6\), so molecular formula = C6H6 (benzene).
(d)(i) Type of reaction: hydrolysis.
(ii) Major product: amino acids.
(iii) Functional groups present: the amino group (-NH2) and the carboxyl group (-COOH).
Answer Details
(a)(i) Condition: presence of (ultraviolet) sunlight / diffused light.
(ii) Type of reaction: substitution (photochemical, free-radical substitution).
(iii) Uses of methane: as a domestic and industrial fuel (natural gas); manufacture of hydrogen; manufacture of carbon black / methanol.
(iv) Major natural source: natural gas (also biogas/marsh gas from decaying organic matter).
(b)(i) Similarity: both reactions liberate hydrogen gas. (\(2Na + 2C_2H_5OH \rightarrow 2C_2H_5ONa + H_2\); \(2Na + 2H_2O \rightarrow 2NaOH + H_2\)).
(ii) Two isomers of C3H8O
Propan-1-ol: CH3-CH2-CH2-OH
Propan-2-ol: CH3-CH(OH)-CH3(iii) A fine chemical is made in small quantities to a high degree of purity and is costly; a heavy chemical is made in large (bulk) quantities and is of lower purity.
(iv) Fine chemical: a drug (for example aspirin) or a dye. Heavy chemical: tetraoxosulphate(VI) acid (or sodium hydroxide, ammonia).
(c) C : H by mass = 92.3% : 7.7%.
C: \(\dfrac{92.3}{12} = 7.69\); H: \(\dfrac{7.7}{1} = 7.7\). Ratio \(\approx 1:1\).
(i) Empirical formula of X and Y = CH (empirical mass = 13).
(ii) X: \(\dfrac{26}{13} = 2\), so molecular formula = C2H2 (ethyne). Y: \(\dfrac{78}{13} = 6\), so molecular formula = C6H6 (benzene).
(d)(i) Type of reaction: hydrolysis.
(ii) Major product: amino acids.
(iii) Functional groups present: the amino group (-NH2) and the carboxyl group (-COOH).
Question 45 Report
(a)(i) State one physical method and one chemical method in each case by which the following can be removed:
I. Permanent hardness in water
II. A suspension of CaCO\(_3\) in water.
(ii) Give one disadvantage of hard water.
(b) Explain the following observations: (i) Crystals of washing soda become powdery on exposure to air for a long time
(ii) The concentration of chloride ions in 0.02 mol dm\(^{-3}\) calcium chloride solution is not the same s in 0.02 mol dm\(^{-3}\) sodium chloride solution.
(iii) Iron filings corrode faster than iron nails of the same mass.
(c)(i) Classify each of the following oxides as acidic, basic, neutral or amphoteric. I. ZnO II. CO III. NO\(_2\)
(ii) Give the formula of the acid anhydride of each of the following: I. H\(_2\)CO\(_3\) II. H\(_2\)SO\(_4\)
(iii) Give the IUPAC name of the following salts: I. COCl\(_2\) II. Mg(NO\(_3\))\(_2\).
(d)(i) Mention one pollutant associated with depletion of ozone layer in the atmosphere
(ii) Calculate the volume occupied by 0.125 mole of oxygen at 27°C and a pressure of 2.02 x 105 Nm\(^{-2}\) [I mole of gas occupies 22.4 dm\(^3\) at s.t.p; standard pressure = 1.01 x 105 Nm\(^{-2}\)]
(iii) State one process used for the industrial preparation of oxygen.
Question 46 Report
(a) Draw an energy profile diagram to illustrate a catalysed exothermic reaction and label parts of the curves representing the following:
(i) activated complex (without catalyst);
(ii) activated energy (with catalyst)
(iii) enthalpy change
(b) Give the reasons for the following observations:
(i) A balloon filled with liyilrogen becomes deflated faster than a balloon filled with air under the same conditions.
(ii) Hydrogen peroxide decomposes slowly at room temperature but when a pinch of MnO, is added, bubbles form rapidly.
(iii) A solution of hydrogen chloride as in methylbenzene has no effect on `litmus but a solution of the gas in water turns blue litmus paper red.
(c) Consider the reaction represented by the following equation: 2MnO\(^-_{4(aq)}\) + 5C\(_2\)O\(^{2-}_4\) + 16H\(^+\) \(\to\) 2Mn\(^{2+}_{(aq)}\) + 8H\(_2\)O\(_{(l)}\) + 10C\(_{2(g)}\) .
Write down: (i) the species undergoing reduction giving reasons;
(ii) the reducing agent giving reasons;
(iii) the reduction half equation;
(iv) one observation made during the reaction.
(d)(i) What is an electrochemical cell?
(ii) State three differences between an electrochemical cell and an electrolytic cell.
(a) Energy profile for a catalysed exothermic reaction
In an exothermic reaction the products lie at a lower energy than the reactants, so the enthalpy change \(\Delta H\) is negative. A catalyst provides an alternative reaction pathway of lower activation energy, so its energy hump (and the activated complex on it) is lower than that of the uncatalysed path. The catalyst changes neither the energy of the reactants nor that of the products, so \(\Delta H\) is exactly the same for both paths.
Reading of the labelled parts:
- (i) Activated complex (without catalyst) - the highest point (top of the upper, uncatalysed curve). It is the unstable transition state through which the reactants pass on the uncatalysed path.
- (ii) Activation energy (with catalyst), \(E_{a(cat)}\) - the smaller energy gap from the reactant level up to the top of the lower (catalysed) curve.
- (iii) Enthalpy change, \(\Delta H\) - the vertical drop from the reactant level down to the product level; it is negative because the reaction is exothermic.
(b) Reasons for the observations
(i) Hydrogen gas is much lighter (less dense) than air. By Graham's law the rate of diffusion of a gas is inversely proportional to the square root of its density, so the light hydrogen molecules diffuse (escape) through the tiny pores of the balloon wall much faster than the heavier oxygen and nitrogen molecules of air. The hydrogen balloon therefore loses its gas and deflates faster.
(ii) Manganese(IV) oxide, \(MnO_2\), acts as a catalyst. It provides an alternative pathway of lower activation energy for the decomposition of hydrogen peroxide, so the reaction that is very slow on its own now proceeds rapidly, giving off oxygen gas quickly as bubbles:
\[2H_2O_2\;\xrightarrow{\;MnO_2\;}\;2H_2O\;+\;O_2\]
The \(MnO_2\) is recovered chemically unchanged at the end.
(iii) Hydrogen chloride is a covalent molecule. In a non-polar solvent such as methylbenzene it stays as neutral \(HCl\) molecules and produces no ions, so there are no free \(H^{+}\) ions and litmus is unaffected. In water it ionises completely, giving hydrogen ions (hydroxonium ions):
\[HCl(g)\;+\;H_2O(l)\;\to\;H_3O^{+}(aq)\;+\;Cl^{-}(aq)\]
These \(H^{+}/H_3O^{+}\) ions make the solution acidic, so it turns blue litmus paper red.
(c) The reaction \(2MnO_4^{-}(aq) + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+}(aq) + 8H_2O(l) + 10CO_2(g)\)
(i) Species undergoing reduction: the permanganate ion, \(MnO_4^{-}\). Reason: the oxidation number of manganese decreases from \(+7\) in \(MnO_4^{-}\) to \(+2\) in \(Mn^{2+}\); a decrease in oxidation number is reduction (gain of electrons).
(ii) Reducing agent: the oxalate ion, \(C_2O_4^{2-}\). Reason: it is itself oxidised (the oxidation number of carbon increases from \(+3\) in \(C_2O_4^{2-}\) to \(+4\) in \(CO_2\)), and in being oxidised it supplies the electrons that reduce the \(MnO_4^{-}\).
(iii) Reduction half equation:
\[MnO_4^{-}\;+\;8H^{+}\;+\;5e^{-}\;\to\;Mn^{2+}\;+\;4H_2O\]
(iv) One observation: the purple (pink) colour of the permanganate is discharged, leaving an almost colourless (very pale pink) solution, and effervescence of a colourless gas (\(CO_2\)) is seen.
(d) Electrochemical cell
(i) An electrochemical cell is a device in which a spontaneous chemical (redox) reaction is used to produce electrical energy; that is, it converts chemical energy into electrical energy.
(ii) Three differences between an electrochemical cell and an electrolytic cell:
| Electrochemical (galvanic) cell | Electrolytic cell |
|---|---|
| A spontaneous chemical reaction produces electricity. | An external electric current is supplied to drive a non-spontaneous reaction. |
| Converts chemical energy into electrical energy. | Converts electrical energy into chemical energy. |
| The anode is the negative electrode and the cathode is the positive electrode. | The anode is the positive electrode and the cathode is the negative electrode. |
Answer Details
(a) Energy profile for a catalysed exothermic reaction
In an exothermic reaction the products lie at a lower energy than the reactants, so the enthalpy change \(\Delta H\) is negative. A catalyst provides an alternative reaction pathway of lower activation energy, so its energy hump (and the activated complex on it) is lower than that of the uncatalysed path. The catalyst changes neither the energy of the reactants nor that of the products, so \(\Delta H\) is exactly the same for both paths.
Reading of the labelled parts:
- (i) Activated complex (without catalyst) - the highest point (top of the upper, uncatalysed curve). It is the unstable transition state through which the reactants pass on the uncatalysed path.
- (ii) Activation energy (with catalyst), \(E_{a(cat)}\) - the smaller energy gap from the reactant level up to the top of the lower (catalysed) curve.
- (iii) Enthalpy change, \(\Delta H\) - the vertical drop from the reactant level down to the product level; it is negative because the reaction is exothermic.
(b) Reasons for the observations
(i) Hydrogen gas is much lighter (less dense) than air. By Graham's law the rate of diffusion of a gas is inversely proportional to the square root of its density, so the light hydrogen molecules diffuse (escape) through the tiny pores of the balloon wall much faster than the heavier oxygen and nitrogen molecules of air. The hydrogen balloon therefore loses its gas and deflates faster.
(ii) Manganese(IV) oxide, \(MnO_2\), acts as a catalyst. It provides an alternative pathway of lower activation energy for the decomposition of hydrogen peroxide, so the reaction that is very slow on its own now proceeds rapidly, giving off oxygen gas quickly as bubbles:
\[2H_2O_2\;\xrightarrow{\;MnO_2\;}\;2H_2O\;+\;O_2\]
The \(MnO_2\) is recovered chemically unchanged at the end.
(iii) Hydrogen chloride is a covalent molecule. In a non-polar solvent such as methylbenzene it stays as neutral \(HCl\) molecules and produces no ions, so there are no free \(H^{+}\) ions and litmus is unaffected. In water it ionises completely, giving hydrogen ions (hydroxonium ions):
\[HCl(g)\;+\;H_2O(l)\;\to\;H_3O^{+}(aq)\;+\;Cl^{-}(aq)\]
These \(H^{+}/H_3O^{+}\) ions make the solution acidic, so it turns blue litmus paper red.
(c) The reaction \(2MnO_4^{-}(aq) + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+}(aq) + 8H_2O(l) + 10CO_2(g)\)
(i) Species undergoing reduction: the permanganate ion, \(MnO_4^{-}\). Reason: the oxidation number of manganese decreases from \(+7\) in \(MnO_4^{-}\) to \(+2\) in \(Mn^{2+}\); a decrease in oxidation number is reduction (gain of electrons).
(ii) Reducing agent: the oxalate ion, \(C_2O_4^{2-}\). Reason: it is itself oxidised (the oxidation number of carbon increases from \(+3\) in \(C_2O_4^{2-}\) to \(+4\) in \(CO_2\)), and in being oxidised it supplies the electrons that reduce the \(MnO_4^{-}\).
(iii) Reduction half equation:
\[MnO_4^{-}\;+\;8H^{+}\;+\;5e^{-}\;\to\;Mn^{2+}\;+\;4H_2O\]
(iv) One observation: the purple (pink) colour of the permanganate is discharged, leaving an almost colourless (very pale pink) solution, and effervescence of a colourless gas (\(CO_2\)) is seen.
(d) Electrochemical cell
(i) An electrochemical cell is a device in which a spontaneous chemical (redox) reaction is used to produce electrical energy; that is, it converts chemical energy into electrical energy.
(ii) Three differences between an electrochemical cell and an electrolytic cell:
| Electrochemical (galvanic) cell | Electrolytic cell |
|---|---|
| A spontaneous chemical reaction produces electricity. | An external electric current is supplied to drive a non-spontaneous reaction. |
| Converts chemical energy into electrical energy. | Converts electrical energy into chemical energy. |
| The anode is the negative electrode and the cathode is the positive electrode. | The anode is the positive electrode and the cathode is the negative electrode. |
Question 47 Report
(a)(i) Mention two types of bond present in the ammonium ion
(ii) Give three characteristic properties of electrovalent compounds
(iii) State. two differences between chemical reactions and nuclear reactions
(b) Two elements represented by the letters and Y have atomic numbers 9 and 12 respectively.
(i) Write the electronic configuration of X using the s,p,d, notation
(ii) To what group does Y belong in the periodic table?
(iii) Write the formula of the compound formed when X copibines with Y
(iv) Explain wily X is a good oxidizing agent
(v) State with reason, whether Y would be expected to form acidic or basic oxide
(c) Balance the following nuclear equations and identify the particles represented by X and Y.
(i) \(^{14}_6C\) \(\to\) X + \(^{14}_7N\)
(ii) \(^{14}_7C\)
Y \(\to\) \(^1_1H\) + \(^{17}_8O\)
(d) Consider the following list of substances: Carbon (IV) oxide, hydrogen, zinc, sulphier, methane, potassium and mercury. From the list above, state the:
(i) elements that are metals
(ii) compounds that are gases at room temperature
(iii) non-metals that are solids at room temperature
(a)(i) The ammonium ion contains a covalent bond and a coordinate (dative) bond.
(ii) Three properties of electrovalent compounds: high melting and boiling points; conduct electricity when molten or in aqueous solution; soluble in water (polar solvents); exist as hard crystalline solids.
(iii) Differences between chemical and nuclear reactions:
- A chemical reaction involves only the outer electrons (rearrangement of atoms); a nuclear reaction involves changes in the nucleus (protons and neutrons).
- A chemical reaction forms no new element; a nuclear reaction forms new elements or isotopes and releases far more energy.
(b) X (Z = 9) is fluorine; Y (Z = 12) is magnesium.
(i) X: \(1s^2 2s^2 2p^5\).
(ii) Y belongs to Group II.
(iii) Compound of X with Y: MgF2.
(iv) X (fluorine) is a good oxidizing agent because it readily gains one electron to complete its octet, owing to its small size and high effective nuclear charge (high electronegativity); in doing so it oxidizes other species.
(v) Y is a metal (electropositive), so it forms a basic oxide (MgO), as metallic oxides are basic.
(c)(i) \(^{14}_{6}C \rightarrow X + ^{14}_{7}N\): mass \(14 = 14 + A\) so A = 0; charge \(6 = 7 + Z\) so Z = -1. X = \(^{0}_{-1}e\) (a beta particle).
(ii) \(^{14}_{7}N + Y \rightarrow ^{1}_{1}H + ^{17}_{8}O\): mass \(14 + A = 18\) so A = 4; charge \(7 + Z = 9\) so Z = 2. Y = \(^{4}_{2}He\) (an alpha particle).
(d) From the list:
- (i) Metals: zinc, potassium, mercury.
- (ii) Compounds that are gases at room temperature: carbon(IV) oxide and methane.
- (iii) Non-metals that are solids at room temperature: sulphur.
Answer Details
(a)(i) The ammonium ion contains a covalent bond and a coordinate (dative) bond.
(ii) Three properties of electrovalent compounds: high melting and boiling points; conduct electricity when molten or in aqueous solution; soluble in water (polar solvents); exist as hard crystalline solids.
(iii) Differences between chemical and nuclear reactions:
- A chemical reaction involves only the outer electrons (rearrangement of atoms); a nuclear reaction involves changes in the nucleus (protons and neutrons).
- A chemical reaction forms no new element; a nuclear reaction forms new elements or isotopes and releases far more energy.
(b) X (Z = 9) is fluorine; Y (Z = 12) is magnesium.
(i) X: \(1s^2 2s^2 2p^5\).
(ii) Y belongs to Group II.
(iii) Compound of X with Y: MgF2.
(iv) X (fluorine) is a good oxidizing agent because it readily gains one electron to complete its octet, owing to its small size and high effective nuclear charge (high electronegativity); in doing so it oxidizes other species.
(v) Y is a metal (electropositive), so it forms a basic oxide (MgO), as metallic oxides are basic.
(c)(i) \(^{14}_{6}C \rightarrow X + ^{14}_{7}N\): mass \(14 = 14 + A\) so A = 0; charge \(6 = 7 + Z\) so Z = -1. X = \(^{0}_{-1}e\) (a beta particle).
(ii) \(^{14}_{7}N + Y \rightarrow ^{1}_{1}H + ^{17}_{8}O\): mass \(14 + A = 18\) so A = 4; charge \(7 + Z = 9\) so Z = 2. Y = \(^{4}_{2}He\) (an alpha particle).
(d) From the list:
- (i) Metals: zinc, potassium, mercury.
- (ii) Compounds that are gases at room temperature: carbon(IV) oxide and methane.
- (iii) Non-metals that are solids at room temperature: sulphur.
Question 48 Report
(a)(i) Give the reason why copper turnings dissolve in AgNO\(_3\) solution but remain insoluble in Pb(NO\(_3\) )\(_2\) solution.
(ii) Copper turnings of mass 1.06g were placed in 250 cm\(^3\) of 0.20 mol dm\(^{-3}\) AgNO\(_3\). Calculate the amourt of silver ions present. [Cu = 63.5]
(iii) Determine whether all the copper in (a)(ii) above will discolvo in the solution. The equation for the reaction is CU\(_{(S)}\) + 2Ag\(^+_{(aq)}\) --> Ag\(_{(aq)}\) + Cu\(^{2+}_{(aq)}\)
(b)(i) List cheicil properties of acids
(ii) Give two large scale uses of HNO\(_3\)
(iii) Write an equation for the action of heat on each of the following compounds: I. Pb(NO\(_3\))\(_2\)
Il. AgNO\(_3\).
(c)(i) State what would be observed if a piece of damp blue litmus paper is dropped into a glass jar of chlorine.
(ii) Name the type of reaction which occurs in (c)(i) above.
(iii) Give the property of chlorine which is exhibited in the reaction in (c)(i) above.
(i) Name two products obtained directly from the destructive distillation of coal.
(ii) Give one use of each product named in (d)(i) above.
(a)(i) Copper is more reactive than (above) silver in the activity series, so it displaces silver from AgNO3 solution (\(Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag\)). Copper is less reactive than (below) lead, so it cannot displace lead from Pb(NO3)2 solution and stays insoluble.
(ii) Amount of Ag+ \(= 0.20 \times \dfrac{250}{1000} = 0.050\ \text{mol}\).
(iii) Amount of Cu \(= \dfrac{1.06}{63.5} = 0.0167\ \text{mol}\).
From \(Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag\), 0.0167 mol Cu needs \(2 \times 0.0167 = 0.0334\ \text{mol}\) Ag+.
Only 0.050 mol Ag+ is available, which is more than 0.0334 mol, so Ag+ is in excess and all the copper dissolves.
(b)(i) Chemical properties of acids: turn blue litmus red; react with metals above hydrogen to liberate hydrogen; react with bases (neutralization) to form salt and water; react with trioxocarbonates(IV) to give carbon(IV) oxide.
(ii) Large-scale uses of HNO3: manufacture of fertilizers (ammonium trioxonitrate(V)); manufacture of explosives (for example TNT).
(iii) Action of heat:
I. \[2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) The damp blue litmus paper first turns red and is then bleached white.
(ii) The reaction is oxidation (bleaching).
(iii) The property shown is that chlorine is a bleaching (oxidizing) agent.
(d)(i) Products from destructive distillation of coal: coke and coal tar (also coal gas, ammoniacal liquor).
(ii) Coke is used as a fuel and as a reducing agent in the blast furnace; coal tar is used as a source of drugs and dyes (and for surfacing roads).
Answer Details
(a)(i) Copper is more reactive than (above) silver in the activity series, so it displaces silver from AgNO3 solution (\(Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag\)). Copper is less reactive than (below) lead, so it cannot displace lead from Pb(NO3)2 solution and stays insoluble.
(ii) Amount of Ag+ \(= 0.20 \times \dfrac{250}{1000} = 0.050\ \text{mol}\).
(iii) Amount of Cu \(= \dfrac{1.06}{63.5} = 0.0167\ \text{mol}\).
From \(Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag\), 0.0167 mol Cu needs \(2 \times 0.0167 = 0.0334\ \text{mol}\) Ag+.
Only 0.050 mol Ag+ is available, which is more than 0.0334 mol, so Ag+ is in excess and all the copper dissolves.
(b)(i) Chemical properties of acids: turn blue litmus red; react with metals above hydrogen to liberate hydrogen; react with bases (neutralization) to form salt and water; react with trioxocarbonates(IV) to give carbon(IV) oxide.
(ii) Large-scale uses of HNO3: manufacture of fertilizers (ammonium trioxonitrate(V)); manufacture of explosives (for example TNT).
(iii) Action of heat:
I. \[2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) The damp blue litmus paper first turns red and is then bleached white.
(ii) The reaction is oxidation (bleaching).
(iii) The property shown is that chlorine is a bleaching (oxidizing) agent.
(d)(i) Products from destructive distillation of coal: coke and coal tar (also coal gas, ammoniacal liquor).
(ii) Coke is used as a fuel and as a reducing agent in the blast furnace; coal tar is used as a source of drugs and dyes (and for surfacing roads).
