WAEC SSCE - Chemistry - 2006
Ajụjụ 1 Ripọtì
Waste plastics accumulate in the soil and pollute the environment because plastic materials are
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The reason why waste plastics accumulate in the soil and pollute the environment is that plastic materials are non-biodegradable. This means that they cannot be broken down by the natural processes that decompose other organic materials. As a result, plastics persist in the environment for hundreds of years, slowly breaking down into smaller and smaller pieces called microplastics. These microplastics can contaminate soil, water, and air, and harm wildlife and human health. Therefore, it is important to reduce plastic waste by recycling, reusing, and reducing our plastic consumption.
Ajụjụ 2 Ripọtì
An element X has isotopic masses of 6 and 7. if the relative abundance is 1 to 12.5 respectively, what is the relative atomic mass of X?
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Isotopes are atoms of the same element with different numbers of neutrons, and they have the same atomic number but different mass numbers. The relative abundance of an isotope is the proportion of that isotope in a sample of the element. In this question, element X has two isotopes with mass numbers of 6 and 7, and their relative abundances are given as 1 and 12.5, respectively. The relative atomic mass (RAM) of X can be calculated using the following formula: RAM = (mass of isotope 1 x abundance of isotope 1 + mass of isotope 2 x abundance of isotope 2) / total abundance Plugging in the given values, we get: RAM = (6 x 1 + 7 x 12.5) / (1 + 12.5) Simplifying the expression, we get: RAM = (6 + 87.5) / 13.5 RAM = 93.5 / 13.5 RAM ≈ 6.93 Therefore, the relative atomic mass of X is approximately 6.93. Hence, the correct option is "6.9".
Ajụjụ 3 Ripọtì
If 20 cm3 of distilled water is added to 80cm 3 of 0. 50 mol dm-3 hydrochloric acid, the new concentration of the acid will be
Ajụjụ 4 Ripọtì
When aqueous ammonia is added to one of the following solutions,a white precipitate which dissolved in excess ammonia is formed Identify the solution.
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The solution that forms a white precipitate when aqueous ammonia is added, which dissolves in excess ammonia, is the solution of zinc chloride (ZnCl2(aq)). The addition of aqueous ammonia to zinc chloride solution forms a white precipitate of zinc hydroxide (Zn(OH)2(s)), which dissolves in excess ammonia to form a colorless, complex ion called tetraamminezinc(II) (Zn(NH3)42+). This reaction can be represented by the following equation: ZnCl2(aq) + 2NH3(aq) + H2O(l) → Zn(OH)2(s) + 2NH4Cl(aq) The formation of a white precipitate that dissolves in excess ammonia indicates the presence of a metal cation that can form a complex ion with ammonia. Zinc ions (Zn2+) have a small size and high charge density, making them highly polarizing and able to form complex ions with ammonia. In contrast, lead ions (Pb2+), copper ions (Cu2+), and iron ions (Fe2+) can also form white precipitates with aqueous ammonia, but these precipitates do not dissolve in excess ammonia due to the formation of insoluble hydroxides.
Ajụjụ 5 Ripọtì
An atom 23892X decays by alpha particle emission to give an atom Y. The atomic number and mass number of Y are
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An alpha particle consists of two protons and two neutrons, and it is symbolized by the Greek letter alpha (α). When an atom emits an alpha particle, it loses two protons and two neutrons, which means that its atomic number decreases by 2 and its mass number decreases by 4. In the given question, the parent atom is X with a mass number of 238 and atomic number of 92. It decays by alpha particle emission, which means it loses two protons and two neutrons. Therefore, the daughter atom Y will have a mass number of 238-4=234 and an atomic number of 92-2=90. So, the atomic number and mass number of the daughter atom Y are 90 and 234, respectively. Therefore, the correct option is: "90 and 234 respectively".
Ajụjụ 6 Ripọtì
Which of the following compounds absorbs moisture from the atmosphere and dissolves in it?
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Ajụjụ 7 Ripọtì
Which of the following compounds is an alkanoate?
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The compound that is an alkanoate is CH3COOCH3. An alkanoate is a salt or an ester of an organic acid that contains a carbonyl group (-CO-) bonded to an oxygen atom (-O-) that is in turn bonded to an alkyl group. CH3COOCH3 is an ester of acetic acid (CH3COOH) and methanol (CH3OH), which is formed by the condensation reaction between the carboxylic acid and alcohol. This ester has the characteristic -COO- functional group and an alkyl group, which makes it an alkanoate. CH3COOH is acetic acid, a carboxylic acid. CH3CH2OH is ethanol, a primary alcohol. CH3CH2COOH is propanoic acid, a carboxylic acid.
Ajụjụ 8 Ripọtì
Considered the reaction: H+(aq) + OH - (aq) → H2O
(l). The energy change taking place in the reaction above is enthalpy of
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The given reaction involves the combination of hydrogen ions (H+) and hydroxide ions (OH-) to form water (H2O). The reaction releases energy in the form of heat, which is an indication that the reaction is exothermic. When a substance is formed from its constituent elements, the energy change taking place is called the enthalpy of formation. However, in this case, no new substance is being formed from its constituent elements. When an ionic compound dissolves in water, the energy change taking place is called the enthalpy of solution. However, in this case, no ionic compound is dissolving in water. When an acid and a base react to form a salt and water, the energy change taking place is called the enthalpy of neutralization. Since the given reaction is an acid-base reaction, and water is formed as one of the products, the energy change taking place is the enthalpy of neutralization. Thus, the answer is neutralization.
Ajụjụ 9 Ripọtì
Consider the reaction: 2AI(s) + 6H+ (aq) → 2AI3+ (aq) + 3H2(g). What is the total number of moles of electrons transferred from the aluminium atoms to the hydrogen ions?
Ajụjụ 10 Ripọtì
The indicator used in neutralizing CH3COOH and NaOH solution has pH range of
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The indicator used to neutralize CH3COOH and NaOH solution is called phenolphthalein, and its pH range is 8-10. When an acid, like CH3COOH, is mixed with a base, like NaOH, a neutralization reaction occurs, and the resulting solution is neither acidic nor basic but neutral. Phenolphthalein is a chemical compound that changes color depending on the pH of the solution it is in. It is colorless in acidic solutions with a pH below 8 and pink in basic solutions with a pH above 10. Therefore, when phenolphthalein is added to the CH3COOH and NaOH solution, it will remain colorless as long as the solution is acidic but will turn pink once the solution becomes basic. Since the pH range of phenolphthalein is 8-10, it is suitable for use in neutralizing the CH3COOH and NaOH solution, as the resulting solution will have a pH within that range, indicating that the acid and base have reacted completely to form a neutral solution.
Ajụjụ 12 Ripọtì
Which of the following aqueous methods cannot be used to distinguish between a strong acid and a a weak acid?
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The aqueous method that cannot be used to distinguish between a strong acid and a weak acid is the action on starch iodide paper. The reason for this is that both strong acids and weak acids are capable of reacting with the starch-iodide paper to produce a blue-black color. This reaction occurs because the iodide ions in the paper react with the hydrogen ions released by the acids to form hydrogen iodide gas. The hydrogen iodide gas then reacts with the starch in the paper to produce the blue-black color. Conductivity measurement and measurement of pH are useful methods to distinguish between strong acids and weak acids. Strong acids completely dissociate in water, producing a high concentration of ions that conduct electricity well and have a very low pH. Weak acids only partially dissociate in water, producing a lower concentration of ions that conduct electricity less efficiently and have a higher pH. Measurement of heat of neutralization is also a useful method to distinguish between strong acids and weak acids. Strong acids release more heat during neutralization than weak acids because they produce more hydrogen ions when they react with a base.
Ajụjụ 13 Ripọtì
What type of reaction occurs between vegetable oil and plant ash extract?
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The reaction between vegetable oil and plant ash extract is a saponification reaction. Saponification is a process that involves the reaction of an ester, such as vegetable oil, with an alkali, such as plant ash extract, to form soap and glycerol. In the saponification reaction, the ester bond in the vegetable oil is broken, and the alkali reacts with the fatty acids in the oil to form soap molecules and glycerol. The plant ash extract provides the necessary alkali, usually in the form of potassium or sodium hydroxide. The reaction is exothermic, meaning it releases heat, and it is typically carried out in a heated vessel. The resulting mixture is a crude soap that contains impurities such as excess alkali, glycerol, and unsaponified fats. To produce a purer soap, the crude soap is dissolved in water and the impurities are removed through a process called salting out. The purified soap can then be used for cleaning or other purposes. In conclusion, the reaction between vegetable oil and plant ash extract is a saponification reaction, which produces soap and glycerol. This reaction has been used for centuries to produce soap and is still used today in traditional soap-making processes.
Ajụjụ 14 Ripọtì
Which of the following industrial processes is chlorines not used?
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Chlorine is not used in the manufacturing of common salt. Common salt, also known as table salt or sodium chloride, is typically obtained through the evaporation of saltwater or brine. Chlorine is not needed in this process because saltwater and brine naturally contain high concentrations of sodium chloride. Chlorine is used in the other three processes listed: - Production of polyvinylchloride (PVC): Chlorine is used to create vinyl chloride monomer, which is then polymerized to form PVC. - Manufacturing of hydrochloric acid: Chlorine is used to react with hydrogen gas to produce hydrochloric acid. - Manufacturing of domestic bleach: Chlorine is used as the active ingredient in many types of bleach, including household bleach.
Ajụjụ 16 Ripọtì
The ammonium compound used in the manufacture of dry cells is
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The ammonium compound used in the manufacture of dry cells is (NH4)Cl, which is also known as Ammonium Chloride. Dry cells are a type of electrochemical cell that converts chemical energy into electrical energy. They are commonly used in household appliances such as flashlights and remote controls. Ammonium chloride is used in the manufacture of dry cells as an electrolyte. An electrolyte is a substance that conducts electricity when dissolved in water. In dry cells, the ammonium chloride electrolyte helps to facilitate the flow of electrons between the electrodes, allowing the cell to generate an electric current. Out of the options provided, (NH4)Cl is the only ammonium compound. NH4NO3 is ammonium nitrate, which is commonly used as a fertilizer and explosive. (NH4)SO4 is ammonium sulfate, which is also used as a fertilizer. (NH4)2CO3 is ammonium carbonate, which is used in the manufacturing of ceramics and as a cleaning agent. NH4Cl is the only option that is commonly used in the manufacture of dry cells as an electrolyte.
Ajụjụ 17 Ripọtì
The oxidation number of iodine in the iodate ion (IO -3) is
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In the iodate ion (IO3-), the overall charge of the ion is -1. Since oxygen atoms typically have an oxidation state of -2 in ionic compounds, and there are three oxygen atoms in the ion, their total charge contribution is -6. Therefore, the oxidation state of iodine can be calculated by equating the total charge of the ion (-1) with the sum of the oxidation states of iodine and oxygen (-6 + x, where x is the oxidation state of iodine). Solving for x, we get: -6 + x = -1 x = +5 Therefore, the oxidation state of iodine in the iodate ion (IO3-) is +5.
Ajụjụ 18 Ripọtì
Consider the reaction represented by the equation: 2SO\(_{2(g)}\) + O\(_{2(g)}\) ⇌ 2SO\(_{3(g)}\); ∆H = - 197 kJmol\(^{-1}\). Which of the following conditions will not increase the yield of sulphur (VI) oxide?
Ajụjụ 19 Ripọtì
What is CaHb in the following equation? C a
H b + 5O2 → 3CO 2 + 4H2O
Ajụjụ 20 Ripọtì
Metals can be extracted from their ores by a process involving
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Metals can be extracted from their ores by a process called reduction. In this process, a chemical reaction is used to remove the metal from its ore. Reduction involves adding a reducing agent to the ore, which reacts with the metal to form a pure, metallic substance. The reducing agent takes away the oxygen or other elements from the ore, leaving behind the pure metal. This process is commonly used for metals like iron, copper, and aluminum, which are all extracted from their ores through reduction. Oxidation, hydrolysis, and decomposition are not commonly used to extract metals from their ores.
Ajụjụ 21 Ripọtì
Greenhouse effect can be reduced by controlling
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The greenhouse effect is a natural process that occurs when certain gases in the Earth's atmosphere, such as carbon dioxide, methane, and water vapor, trap heat from the sun and keep the planet warm. However, human activities, such as burning fossil fuels, deforestation, and industrial processes, have increased the concentration of these greenhouse gases in the atmosphere, leading to an enhanced greenhouse effect and causing global warming and climate change. To reduce the greenhouse effect, it is necessary to reduce the concentration of these greenhouse gases in the atmosphere. This can be achieved by controlling the human activities that contribute to their emissions. Some effective ways to control the greenhouse effect are: - Reducing the use of fossil fuels: Fossil fuels are the main source of carbon dioxide emissions. Therefore, reducing the use of fossil fuels, such as oil, coal, and gas, can help reduce the concentration of carbon dioxide in the atmosphere. - Planting more trees: Trees absorb carbon dioxide from the atmosphere through photosynthesis, so planting more trees can help reduce the concentration of carbon dioxide in the atmosphere. - Improving energy efficiency: Using energy-efficient appliances, insulating homes and buildings, and reducing energy waste can help reduce the use of fossil fuels and, therefore, reduce greenhouse gas emissions. - Promoting the use of renewable energy: Renewable energy sources, such as solar, wind, and hydropower, do not emit greenhouse gases and can help reduce greenhouse gas emissions. - Reducing waste and promoting recycling: Landfills and incineration of waste produce methane and carbon dioxide, which are greenhouse gases. Therefore, reducing waste and promoting recycling can help reduce greenhouse gas emissions. In conclusion, the greenhouse effect can be reduced by controlling human activities that contribute to the emissions of greenhouse gases. By implementing measures such as reducing the use of fossil fuels, planting more trees, improving energy efficiency, promoting renewable energy, and reducing waste, we can reduce the concentration of greenhouse gases in the atmosphere and mitigate the effects of global warming and climate change.
Ajụjụ 22 Ripọtì
Which of the following equations represents a substitution reaction?
Ajụjụ 23 Ripọtì
If 1 mole of sodium contains 6 x 1023 atoms, how many atoms are contained in 0.6 g of sodium? [Na = 23]
Ajụjụ 24 Ripọtì
Which of the following substance is an ore of iron?
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The substance that is an ore of iron is Haematite. Haematite (Fe2O3) is a mineral form of iron oxide and is a primary source of iron ore. It is found in abundance throughout the world and is mined to extract iron metal. When iron is needed for manufacturing purposes, haematite is processed into pig iron, which is the basic building block of steel. Bauxite is an ore of aluminium, Cassiterite is an ore of tin, and steel is not an ore but an alloy composed mainly of iron with carbon and other elements.
Ajụjụ 25 Ripọtì
In the reaction represented by the equation. 5Fe 2+ (aq) + MnO- 4(aq) + 8H+ (aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) which species is reduced?
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In the given chemical reaction, the oxidation state of Fe changes from +2 to +3, which means that Fe has lost an electron. Similarly, the oxidation state of Mn changes from +7 in MnO-4 to +2 in Mn2+, which means that Mn has gained five electrons. Therefore, Fe is undergoing oxidation and MnO-4 is undergoing reduction. Reduction is a process in which a species gains electrons, leading to a decrease in its oxidation state. In this case, MnO-4 is gaining electrons, so it is being reduced. Thus, the answer is MnO-4.
Ajụjụ 26 Ripọtì
How many orbitals are in the d-sub shell?
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The d-sub shell is one of the electron shells in an atom. It can hold a maximum of 10 electrons, which are distributed among its orbitals. Each orbital in the d-sub shell can hold 2 electrons. There are 5 different orbitals in the d-sub shell, labeled as dxy, dxz, dyz, dz2, and dx2-y2. These orbitals have different shapes and orientations in space, and they can accommodate electrons with different energies. Therefore, the answer is 5.
Ajụjụ 27 Ripọtì
Consider the reaction represented by the equation. 2NaHCO3(s) \(\rightarrow\) 2Na2CO 3(S) + CO 2(g) + H2O(g) what volume of carbon (IV)oxide at s.t.p is evolved when 0. 5 moles of NaHCO3 is heated? [Molar volume = 22.4 dm3 at s.t. p]
Ajụjụ 28 Ripọtì
a) Define each of the following terms and indicate one use of each:
(i) Nuclear fission; (ii) Nuclear fusion.
(b) Alpha particle emission by \(^{293}_{25}U\) proceduces an element A. Beta particle emission by the particle A produces another element B. Element B also undergoes alpha particle emission to produce \(^{227}_{89}AC\). Write balanced equations to represent the above statement.
(c) The models below represent the filling of orbitals in an atom.
State which rule(s) is/are violated or obeyed by each model.
(d) Explain why the boiling point of H\(_2\)S with relative molecular mass of 34 is lower than that of H\(_2\)O with relative molecular mass of 18.
(e) HCI is passed into each of the following solvents:
(i) water;
(ii) methylbenzene. I. State the effect of each solution on blue litmus paper II. Compare the electrical conductivities of the two solutions.
(f) Zinc dust is added to copper (II) tetraoxosulphate (VI) solution. State;
(i) what is observed; (ii) the type of reaction that occurs.
Ajụjụ 29 Ripọtì
TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette reading (initial and final) must be given to two decimal places. Volume of pipette used must be recorded but no account of experimental procedure is required. All calculations must be done in you answer book.
A is a solution containing 6.3 g dm\(^{-3}\) of HNO\(_3\), B is a solution Na\(_2\)CO\(_3\)
(a) Put A Into the burette and titrate it against 20.0 cm\(^3\) portions of B using menyl Indicator. Record the volume of your pipette. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is:
2HNO\(_{3(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) 2NaNO\(_{3(aq)}\) + CO\(_{2(g)}\)
(b) From your results and information provided above, calculate the;
(i) concentration of B in mol dm\(^{-3}\)
(ii) concentration of B in g dm\(^{-3}\):
(iii) mass of sodium ions in 1:0 dm\(^{-3}\) of B
[H = 1; C = 1; O = 16; N = 14; Na = 23]
Ajụjụ 30 Ripọtì
(a) State the following laws of chemical combination: (i) Law of constant composition (ii) Law of multiple' proportion.
(b) Copper reacts with oxygen to form two oxides X and Y. On analysis, 1.535 g of X yielded 1.365 g copper and 1.450 g of Y yielded 1.160 g of cooper.
i) Determine the chemical formula of X and Y.
(ii) Calculate the mass of copper which can react with 0.500 g of oxygen to yield I. X II. Y.
(iii) Which of the laws of chemical combination is illustrated by the result in (b)(i) above. [ = 16, Cu = 63.51]
(c) Write the structure of the product responsible for the observation in each of the following reactions:
(i) A mixture of butanoic acid and ethanol warmed in the presence of concentrated H\(_2\)SO\(_4\) gives off a fragrant odour.
(ii) Sodium dissolves in propan-2-ol with effervescence to give a solution which on evaporation to dryness leaves a white precipitate.
(d) Consider the compound CH\(_3\)CH\(_2\)COOCH\(_2\)CH\(_3\).
(i) Name the compound (ii) Write the structural formula of the compound (iii) State the reagents and conditions for the formation of the compound.
Akọwa Nkọwa
None
Ajụjụ 31 Ripọtì
(a)(i) Draw the energy profile diagram for the reaction
H\(_{2(g)}\) + I\(_{2(g)}\) ---> 2HI\(_{(g)}\) \(\Delta\) = —13 kJmol\(^3\)
(ii) If the concentration of HI increases from 0 to 0.001 mol dm\(^3}\) in 50 seconds, what is the rate of the reaction?
(b) State the type of salt represented by each of the following compounds:
(i) K\(_4\)Fe(CN)\(_6\) (ii) (NH\(_4\))\(_2\)Fe(SO\(_4\))\(_2\)6H\(_2\)O (iii) Mg(OH)NO\(_3\) (iv) NaH\(_2\)PO\(_4\).
(c) Explain, giving equations, the following observation: When carbon (IV) oxide is passed into lime water, it turns milky initially but turns clear with excess carbon (IV) oxide.
(d)(i) Give one use for each of the following compounds: CaCO\(_3\), CaSO\(_4\), NaHCO\(_3\).
(ii) State a drying agent for each of the following gases: i. NH\(_3\), II. HCI Ill. SO\(_4\).
(iii) Write an equation to illustrate the reaction of ammonia as a reducing agent.
(e) An industrial raw material has the following composition by mass:
Iron = 28.1%
Chlorine = 35.7%
Water of crystallization = 36.2%
Calculate the formula for the material. [ H = 1, 0 = 16, Cl = 35.5, Fe = 56 ].
Ajụjụ 32 Ripọtì
(a) A solution of CuSO\(_4\) was electrolyzed between pure copper electrodes and the following results were obtained:
Mass of copper anode before experiment = 7.20 g
Mass of copper anode after experiment = 4.00 g
Mass of copper cathode before experiment = 5.75 g
From the information provided,
(i) calculate the mass of the cathode, after the experiment.
(ii) write an equation for the reaction at the I. anode, II. cathode.
(iii) state whether the colour of the solution would change during the electrolysis. Give a reason for your answer.
(iv) if the electrolysis was carried out for 1 hour 20 minutes with a current of 2.0 amperes, determine the value of the Faraday.
(b) Consider the reaction represented by the following equation:
MnO\(^-_4\) + I\(^-\) + H\(^+\) \(\to\) I\(_2\) + H\(_2\)O + Mn\(^{2+}\)
Write balanced half equation for the (i) oxidation reaction, (ii) reduction reaction.
(c)(i) Describe briefly how tin can be extracted from its ore.
(ii) State one use of tin.
(iii) Mention one property that makes tin suitable for the use stated in (c)(ii)
(d)(i) What is meant by the term pollution?
(ii) Explain why it is dangerous to run a generator in a closed room.
Akọwa Nkọwa
None
Ajụjụ 33 Ripọtì
(a)(i) State two differences betwecii the properties of solids and gases
(ii) What process does each of X, Y and Z represent in the changes shown below?
(b)(i) State Charles' Law (ii) Draw a sketch to graphically illustrate Charles' Law.
(c) 60 cm of hydrogen diffused through a porous membrane in 10 minutes. The same volume of a gas G diffused through the same membrane in 37.4 minutes. Determine the relative molecular mass of G. [ H = I ]
(d)(i) State two assumptions X of the kinetic theory.
(ii) Consider the reaction represented by the Solid of Liquid following equation:
H\(_{2(g)}\) + Cl\(_{2(g)}\) \(\to\) 2HCI\(_{(g)}\)
Use the kinetic theory to explain how the rate of formation of HCI\(_{(g)}\) would be affected by I. increase in temperature; II. decrease in pressure.
(e) Given different examples, mention one metal in each case vihich produces hydrogen on reacting with (i) dilute mineral acid; (ii) cold water; (iii) steam; (iv) hot, concentrated alkali.
Akọwa Nkọwa
None
