WAEC SSCE - Chemistry - 2007
Ibeere 1 Ìròyìn
from the following ores is iron extracted? I. Haematite ll. Bauxite III. Magnetite IV. Cassiterite
Awọn alaye Idahun
Iron is extracted from haematite (Fe2O3) and magnetite (Fe3O4) ores. Bauxite (Al2O3), on the other hand, is an ore of aluminum, while cassiterite (SnO2) is an ore of tin. Therefore, the correct option is (b) l and lll only.
Ibeere 2 Ìròyìn
What is the oxidation number of chromium in K2CrO4?
Awọn alaye Idahun
The oxidation number of an element is the charge it would have in a compound if electrons were completely transferred. In K2CrO4, the compound is neutral (has no net charge) so we can determine the oxidation number of chromium by adding up the oxidation numbers of all the atoms in the compound and setting the sum equal to zero. The oxidation number of K (potassium) is +1 since it is an alkali metal and forms a +1 ion in compounds. The oxidation number of O (oxygen) is -2 since it usually forms -2 ions in compounds, except in peroxides where it is -1. Let x be the oxidation number of chromium (Cr). So we have: 2(+1) + x + 4(-2) = 0 Simplifying this equation gives: x - 6 = 0 x = +6 Therefore, the oxidation number of chromium in K2CrO4 is +6.
Ibeere 3 Ìròyìn
Alkenes undergo the following reactions except
Awọn alaye Idahun
Alkenes are hydrocarbons that contain at least one carbon-carbon double bond in their structure. The double bond makes alkenes more reactive than alkanes (saturated hydrocarbons) and capable of undergoing several types of reactions, such as addition, hydration, and polymerization. However, alkenes do not undergo substitution reactions, which involve the replacement of one atom or group of atoms in a molecule with another atom or group of atoms. This is because the double bond in alkenes is too strong to break under normal conditions, which makes it difficult to substitute an atom or group of atoms in the molecule. Therefore, the correct answer to the question is substitution. Alkenes do not undergo substitution reactions.
Ibeere 4 Ìròyìn
Which of the following methods can be used to separate a mixture of two miscible liquids with different boiling points?
Awọn alaye Idahun
The method that can be used to separate a mixture of two miscible liquids with different boiling points is distillation. Distillation is a process of separating the components of a mixture based on their different boiling points. In this process, the mixture is heated to its boiling point, and the vapor is condensed and collected. The component with the lower boiling point will vaporize first and will be collected as the distillate, while the component with the higher boiling point will remain in the distillation flask. Therefore, by using distillation, the two liquids can be separated based on their boiling points. The other options, such as decantation, evaporation, and filtration, are not suitable for separating two miscible liquids with different boiling points.
Ibeere 5 Ìròyìn
A solution of a salt X reacts with AgNO3 solution to give white precipitate which dissolves in excess aqueous ammonia. X contains
Awọn alaye Idahun
The white precipitate formed by the reaction of the salt X with AgNO3 indicates the presence of chloride ions, Cl-. The fact that the precipitate dissolves in excess aqueous ammonia confirms the presence of Cl- since silver chloride, AgCl, is soluble in excess aqueous ammonia. Therefore, the salt X contains chloride ions, Cl-.
Ibeere 6 Ìròyìn
Which of the following properties of atoms generally increases down a group in the periodic table?
Awọn alaye Idahun
The ionic radius of an atom generally increases down a group in the periodic table. The reason for this trend is that as you move down a group, the number of electron shells increases, meaning that the atomic size increases. This leads to a decrease in the effective nuclear charge, which reduces the attraction between the nucleus and the outermost electrons, allowing them to spread out and occupy a larger volume. As a result, the ionic radius of the atoms in the group increases, making them larger in size.
Ibeere 7 Ìròyìn
Which of the following compounds reacts readily with sodium to liberate hydrogen?
Awọn alaye Idahun
Ibeere 8 Ìròyìn
Glucose reduces Fehling's solution on warming to
Awọn alaye Idahun
Fehling's solution is a chemical test used to detect the presence of reducing sugars such as glucose. It consists of two solutions: Fehling's A (a solution of copper(II) sulfate) and Fehling's B (a solution of sodium hydroxide and potassium sodium tartrate). When glucose is added to Fehling's solution and heated, the blue color of copper(II) ions in Fehling's A turns into reddish-brown precipitate of copper(I) oxide, Cu2O (not copper (ll) oxide). This happens because glucose acts as a reducing agent and donates electrons to the copper(II) ions, which get reduced to copper(I) ions. The copper(I) ions then combine with hydroxide ions from Fehling's B to form the insoluble copper(I) oxide precipitate. Therefore, the correct answer is copper(I) oxide (Cu2O).
Ibeere 9 Ìròyìn
Which of the following gases is light than air?
Awọn alaye Idahun
Ammonia (NH3) is lighter than air. This is because its molar mass is lower than that of air, which is mostly composed of nitrogen and oxygen. When released into the air, ammonia gas rises due to the buoyancy force exerted by the air around it. In contrast, carbon dioxide (CO2), sulfur dioxide (SO2), and hydrogen chloride (HCl) are heavier than air and tend to settle towards the ground when released.
Ibeere 10 Ìròyìn
In which of the following atoms is the ionic radius larger than the atomic radius? [11 Na, 12Mg,13AI,17CI]
Awọn alaye Idahun
An atom becomes an ion by either gaining or losing electrons. When an atom gains an electron, it forms a negative ion (anion) which is larger than its corresponding neutral atom because the added electron increases the electron-electron repulsion and pushes the other electrons further away from the nucleus. Conversely, when an atom loses an electron, it forms a positive ion (cation) which is smaller than its corresponding neutral atom because the lost electron reduces the electron-electron repulsion and the remaining electrons are pulled closer to the nucleus. From the given options, only Chlorine (Cl) gains an electron to form a chloride ion (Cl⁻) which is larger than its atomic radius. The other elements (Na, Mg, and Al) lose electrons to form cations which are smaller than their atomic radii. Therefore, the correct answer is Chlorine (Cl).
Ibeere 11 Ìròyìn
The complete hydrogenation of C 6
H 6 in the presence of nicked catalyst at 200oC gives
Awọn alaye Idahun
The hydrogenation of C6H6 is a reaction in which hydrogen gas is reacted with benzene in the presence of a nickel catalyst. This reaction is an example of an addition reaction. In this reaction, the carbon-carbon double bond present in benzene is broken and replaced by single carbon-carbon bonds, and each carbon atom forms a single bond with a hydrogen atom. The product of the reaction is cyclohexane, which has the molecular formula C6H12. Therefore, the answer is option (C) C6H12.
Ibeere 12 Ìròyìn
What volume will 0.5 g of H2 occupy at s.t.p? [H = 1; 1 mole of a gas occupy 22.4 dm3 at s.t.p.]
Awọn alaye Idahun
To solve this problem, we need to use the formula: number of moles = mass / molar mass From the periodic table, the molar mass of hydrogen (H) is 1 g/mol. Therefore, the number of moles of H2 present in 0.5 g is: number of moles = 0.5 g / 2 g/mol = 0.25 mol At standard temperature and pressure (s.t.p), 1 mole of any gas occupies 22.4 dm^3. Therefore, 0.25 moles of H2 will occupy: volume = number of moles x 22.4 dm^3/mol = 0.25 x 22.4 dm^3 = 5.6 dm^3 Therefore, the correct answer is: 5.6 dm^3.
Ibeere 13 Ìròyìn
Which of these cell notations is the correct representation of a Daniell cell?
Awọn alaye Idahun
Ibeere 14 Ìròyìn
How many atoms are there in one mole of hydrogen gas? [Avogadro's constant = 6.02 x 1023mol-1]
Awọn alaye Idahun
Ibeere 15 Ìròyìn
Which of the following species does not contain a co-ordinate bond?
Awọn alaye Idahun
Ibeere 17 Ìròyìn
Consider the reaction represented by the following equation: N2(g) + 3H2(g) ⇌ 2NH3(g); ∆H = - 92kJmol1. Finely divided iron is used as catalyst in the reaction in order to
Awọn alaye Idahun
The finely divided iron catalyst is used to shorten the time for the attainment of equilibrium in the reaction. A catalyst is a substance that increases the rate of a chemical reaction without undergoing any permanent change itself. In the Haber process, which is the industrial process for the production of ammonia, the reaction is slow at room temperature and pressure, even though the overall enthalpy change (∆H) is negative (exothermic). By adding a catalyst such as finely divided iron, the activation energy required for the reaction to proceed is lowered, which speeds up the reaction and allows the system to attain equilibrium faster. This results in a higher yield of ammonia in a shorter amount of time.
Ibeere 18 Ìròyìn
What is the IUPAC name of the compound represented by the molecular formula NaCIO4?
Awọn alaye Idahun
The IUPAC name of the compound represented by the molecular formula NaClO4 is Sodium tetraoxochlorate (VII). This is because the chlorate ion has a charge of -1 and contains four oxygen atoms, which gives it the formula ClO4-. The Roman numeral in the compound's name represents the oxidation state of chlorine in the ion, which is +7 in this case. Therefore, the correct name is Sodium tetraoxochlorate (VII).
Ibeere 19 Ìròyìn
Which of the following substances is suitable solvent for perfumes?
Awọn alaye Idahun
Ethanol is a suitable solvent for perfumes. Ethanol is a type of alcohol that can dissolve various types of compounds, including oils and aromatic compounds, which are the main components of perfumes. This property of ethanol makes it an ideal solvent for perfumes. In contrast, benzene and turpentine are toxic and not safe for use in perfumes. Water, on the other hand, is a polar solvent and is not suitable for dissolving non-polar aromatic compounds present in perfumes.
Ibeere 20 Ìròyìn
14.8 g of a salt (z)dissolved in 250 cm3 of distilled water gives a concentration of 0.80 mol dm-3. Calculate the molar mass of the salt (z).
Awọn alaye Idahun
To find the molar mass of the salt (z), we need to use the formula: concentration (mol dm-3) = number of moles / volume (dm3) We are given the concentration of the solution as 0.80 mol dm-3 and the volume of the solution as 250 cm3 which is equivalent to 0.250 dm3. Therefore, we can rearrange the above formula to find the number of moles of the salt (z): number of moles = concentration x volume number of moles = 0.80 x 0.250 number of moles = 0.200 Now we can use the formula: molar mass (g mol-1) = mass (g) / number of moles We are given the mass of the salt (z) as 14.8 g. Therefore: molar mass = 14.8 / 0.200 molar mass = 74.0 g mol-1 Therefore, the molar mass of the salt (z) is 74.0 g mol-1.
Ibeere 21 Ìròyìn
Consider the reaction represented by the equation: Ca(OH)2(aq) + CO 2(g) → CaCO3(s) + H 2O(l)
in the reaction above, CO 2 acts as
Awọn alaye Idahun
In the given chemical equation, Ca(OH)2 reacts with CO2 to form CaCO3 and H2O. Carbon dioxide (CO2) is a non-metallic oxide and when it reacts with water, it forms carbonic acid (H2CO3). The equation can be written as CO2 + H2O → H2CO3 Carbonic acid is a weak acid that can donate H+ ions to form HCO3- ions. The HCO3- ions formed then react with Ca2+ ions present in Ca(OH)2 to form CaCO3 and H2O. Thus, CO2 acts as an acidic oxide in the given reaction. Therefore, the answer is "an acidic oxide".
Ibeere 22 Ìròyìn
What product is formed at the cathode during the electrolysis of concentrated sodium chloride solution using carbon electrodes?
Awọn alaye Idahun
During the electrolysis of concentrated sodium chloride solution using carbon electrodes, the product formed at the cathode is hydrogen. In this process, the chloride ions (Cl⁻) in the solution are attracted to the anode (positive electrode) and undergo oxidation to form chlorine gas (Cl₂) while water (H₂O) molecules are reduced at the cathode (negative electrode) to form hydrogen gas (H₂) and hydroxide ions (OH⁻). The overall reaction can be represented as: 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂ Since the electrolysis is carried out using carbon electrodes, neither sodium nor oxygen is produced at the electrodes as these reactions would require a more reactive electrode material than carbon. Thus, the product formed at the cathode is hydrogen gas.
Ibeere 23 Ìròyìn
Which of the following compounds is a basic salt?
Awọn alaye Idahun
A basic salt is a salt that contains a metal cation and the conjugate base of a weak acid. The conjugate base of the weak acid behaves as a base and hence the salt shows basic properties. Among the given options, Zn(OH)CI is the only compound that can be a basic salt. It contains the metal ion zinc (Zn) and the anion of a weak acid, hydroxide (OH-). The hydroxide ion can act as a base and accept a proton from water molecules, increasing the concentration of hydroxide ions in solution and making the solution basic. Therefore, the correct answer is Zn(OH)CI.
Ibeere 24 Ìròyìn
Which of the following pairs of substances can be distinguished by use of NaHCO3 solution?
Awọn alaye Idahun
NaHCO3 solution is commonly used as a test for carboxylic acids. When a carboxylic acid is added to NaHCO3 solution, it reacts to produce carbon dioxide gas, water, and a sodium carboxylate salt. Therefore, the pair of substances that can be distinguished by use of NaHCO3 solution is: - CH3CH2OH and CH3COOH CH3CH2OH is ethanol, which is not a carboxylic acid, and therefore does not react with NaHCO3 solution. CH3COOH is ethanoic acid, which is a carboxylic acid, and reacts with NaHCO3 solution to produce carbon dioxide gas. The other options cannot be distinguished using NaHCO3 solution because they do not involve carboxylic acids.
Ibeere 26 Ìròyìn
Which of the following statements about sodium is not true?
Awọn alaye Idahun
The statement that is not true is "Na is smaller in size than Na+". In reality, the ionic radius of Na+ is smaller than the atomic radius of Na. When sodium loses one electron to form Na+, the number of electrons in the atom decreases from 11 to 10 while the number of protons in the nucleus remains the same at 11. This leads to an increase in the effective nuclear charge and a decrease in the size of the ion. Thus, Na+ is smaller than Na. The other statements are true. Na+ is smaller in size than Na, has fewer electrons than Na, and has a greater effective nuclear charge than Na.
Ibeere 27 Ìròyìn
Pig-iron is brittle because it contains
Awọn alaye Idahun
Pig-iron is brittle because it contains a high percentage of carbon as impurity. Pig-iron is the product obtained from the smelting of iron ores in a blast furnace. It is a crude form of iron and contains about 4% carbon, along with other impurities such as silicon, sulfur, and phosphorus. The high percentage of carbon in pig-iron makes it brittle and weak, as it tends to form carbon inclusions in the iron matrix, leading to cracks and fractures under stress. Therefore, pig-iron needs to be refined further to remove impurities, including carbon, to make it stronger and more ductile.
Ibeere 28 Ìròyìn
Which of the these metals will not liberate hydrogen from dilute HCI?
Awọn alaye Idahun
Copper will not liberate hydrogen from dilute HCl. This is because copper is below hydrogen in the reactivity series of metals. Metals below hydrogen in the reactivity series are unable to displace hydrogen from dilute acids. Copper, therefore, cannot react with dilute HCl to liberate hydrogen gas.
Ibeere 29 Ìròyìn
Which of the following statements is true of the molecules of a gas under ideal conditions? The molecules
Awọn alaye Idahun
The statement that is true of the molecules of a gas under ideal conditions is that they move at random. In the kinetic theory of gases, it is assumed that gas molecules are in constant random motion, colliding with each other and the walls of their container. This movement is what gives gases their pressure, as the molecules collide with the walls of the container and exert a force. The other options are not true of ideal gases. Gas molecules under ideal conditions undergo perfectly elastic collisions, meaning they do not lose any kinetic energy during their collisions. They do not attract each other, as they are assumed to be point masses with no intermolecular forces, and they occupy a small volume, as they are assumed to have no volume themselves, but only occupy the space in the container they are in.
Ibeere 30 Ìròyìn
Which of the following statements about rare gases are correct?
Awọn alaye Idahun
The correct statement about rare gases is: - Their outermost shells are fully filled Explanation: Rare gases, also known as noble gases, are elements in Group 18 of the periodic table. They have a complete outermost electron shell, which makes them stable and unreactive. The filled outermost shell is the reason behind their low reactivity with other elements. Therefore, the statement that their outermost shells are fully filled is correct. The other options are incorrect as they do not apply to the properties of rare gases.
Ibeere 31 Ìròyìn
Consider the reaction below: Vegetable oil \(\frac{H2;NI catalyst}{High temperature}\) product. The reaction is applied in the manufacture of?
Awọn alaye Idahun
The given reaction describes the conversion of vegetable oil into some other product in the presence of a catalyst and high temperature. Among the options, the only product that can be obtained from vegetable oil using such a process is margarine. Margarine is made from vegetable oil by a process called hydrogenation, which involves the addition of hydrogen gas in the presence of a catalyst (such as nickel) and high temperature. This process converts the unsaturated fatty acids in vegetable oil to saturated ones, making the oil more solid and suitable for use as a spread. Therefore, the correct answer is margarine.
Ibeere 32 Ìròyìn
In which of the following compound (s) will heat be absorbed on adding water?
l. KNO\(_3\) II. Concentrated H\(_2\)SO\(_4\) III. NH\(_4\)CI
Awọn alaye Idahun
Heat is absorbed when a substance dissolves in water and the process is called endothermic. In this case, KNO3 and NH4CI are ionic compounds that dissolve in water to form ions, and this process is endothermic. On the other hand, concentrated H2SO4 is a molecular compound and does not dissociate into ions when mixed with water, so no heat is absorbed. Therefore, the compounds that will absorb heat when mixed with water are option I and III only, which is answer choice C.
Ibeere 34 Ìròyìn
Hydrocarbons which react with ammoniacal copper (l) chloride solution conform to the general molecular formula
Awọn alaye Idahun
Ibeere 35 Ìròyìn
Which of the following properties is characteristic of the halogens?
Awọn alaye Idahun
The property that is characteristic of the halogens is their ability to accept electrons readily. The halogens are a group of nonmetal elements found in Group 17 (VIIA) of the periodic table. They have seven valence electrons, which means they require only one more electron to achieve a noble gas configuration. Therefore, they have a high tendency to gain an electron and form negatively charged ions, such as Cl-, Br-, and I-. This ability to accept electrons readily is what makes halogens highly reactive and forms the basis for many of their chemical properties. Therefore, the correct answer is "Ability to accept electrons readily."
Ibeere 36 Ìròyìn
If the difference between the electronegativities of two elements is large, the type of bond that can be formed between them is
Awọn alaye Idahun
The electronegativity difference between two elements determines the type of bond that can be formed between them. When the difference in electronegativity is large, one element attracts the electrons in the bond more strongly than the other, resulting in the formation of an ionic bond. Therefore, if the difference between the electronegativities of two elements is large, the type of bond that can be formed between them is ionic.
Ibeere 37 Ìròyìn
What process is illustrated below? S(s) ⇌H2O⇌ S (aq)
Awọn alaye Idahun
The process illustrated below is the formation of a saturated solution of a solid. The solid S is in equilibrium with its dissolved form in water, represented by the double-headed arrow. The symbol (s) indicates that S is a solid, while (aq) represents the aqueous or dissolved form of S. When the maximum amount of S has been dissolved in the water, the solution is saturated. This process is often used in chemistry to determine the solubility of a solid in water or other solvents.
Ibeere 38 Ìròyìn
Which of the following compounds has hydrogen bonds between its molecules?
Awọn alaye Idahun
Among the given compounds, only HF has hydrogen bonds between its molecules. Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative element like fluorine, oxygen, or nitrogen, and this hydrogen atom interacts with another electronegative atom from another molecule. In the case of HF, the hydrogen atom is bonded to the highly electronegative fluorine atom, and this hydrogen atom forms a hydrogen bond with the fluorine atom of another HF molecule. HBr, HCl, and HI have covalent bonds between their molecules, but they don't have hydrogen bonds because the hydrogen atoms are not bonded to highly electronegative atoms like fluorine, oxygen, or nitrogen.
Ibeere 39 Ìròyìn
When air is successively passed through sodium hydroxide solution, alkaline pyrogallol and then concentrated tetraoxosulphate (VI) acid, its remaining components are
Awọn alaye Idahun
When air is passed through sodium hydroxide solution, carbon (IV) oxide and water vapour are removed as they react with the alkali to form sodium carbonate and sodium hydrogencarbonate respectively. The remaining gases, which are mainly nitrogen, are passed through alkaline pyrogallol to remove any remaining oxygen, as it reacts with the pyrogallol to form brown precipitate. The nitrogen gas is then passed through concentrated tetraoxosulphate (VI) acid to remove any remaining traces of water vapour. Therefore, the remaining components of air after the above process are nitrogen gas and noble gases such as argon, helium, neon and krypton, which are present in very small amounts. Hence, the correct option is nitrogen and noble gases.
Ibeere 40 Ìròyìn
(a) Define the following terms:(i) Saturated solution; (ii) Solubility.
(b) In an experiment to determine the solubility of a given salt Y, the following data were provided:
Mass of dry empty dish = 7.16 g
Mass of dish + saturated solution of salt Y = 17.85 g
Mass of dish + salt Y = 9.30 g Temperature of solution = °C
Molar mass of salt Y = 100
Density of solution Y = 1.00 g cm\(^{-3}\) Calculate the solubility of salt Y in
(i) g dm\(^{-3}\) of solution, (ii) mol dm\(^{-3}\) of solution.
(c) State the type of bond broken on melting each of the following substances: (i) NaCI\(_{(s)}\) (ii) CO\(_{2(s)}\) (iii) SiO\(_{2(s)}\) (iv) Al\(_{(s)}\)
(d) Explain the following observations: (i) the chemical reactivity of alkali metals increases down the group; (ii) Mg has higher melting point than Na; (iii) K is a better reducing agent than Na.
(e)(i) What are isotopes? (ii) Lithium exists as \(^6_3\)Li and \(^7_3\)Li in the ratio 2:25. Calculate the relative atomic mass of the lithium.
None
Awọn alaye Idahun
None
Ibeere 41 Ìròyìn
(a)(i) What is 2 homologous series?
(ii) Give two homologous series present in petroleum
(iii) Give one example of a compound belonging to each of the homologous series in (a)(ii).
(iv) Name two fractions obtained from the fractional distillation of petroleum
(v) Why is there a gradual change in the physical properties of petroleum fractions?
b) Write a two-step balanced chemical equation for the reaction of (i) ethanol with excess concentration traoxosulphate (VI) acid at high temperature
(ii) excess ethanol with concentrated tetraoxosulphate (VI) acid at lower temperature.
(c) An organic compound of relative molecular mass 46, on analysis vas ound to contain 52.0% carbon, 13.3% hydrogen and 34.7% oxygen.
(i) Determine its I. empirical formula, II. molecular formula.
(ii) Draw two possible structures of the compound and name one of them [O = 16; C = 12; H = 1 ]
Ibeere 42 Ìròyìn
a) Define the following in term:. of electron transfer: (i) oxidation; (ii) reduction.
(b)(i) Determina the oxidation stale of phosphorus in each of the following structures: I. POCI\(_3\) II. PH\(_3\).
(ii) State with reasons whether the following compounds will form acidic, neutral or basic aqueous solutions: I. NaNO\(_3\) II. Na\(_2\)H\(_4\)CI; Ill. Na\(_2\)CO\(_3\).
(c) Consider the set-up
(i) What is the gas produced in the experiment illustrated by the set-up above?
(ii) Name the method of collection of gas
(iii) Give a reason for your answer in (c)(ii) above
(iv) State the function of the concentrated H\(_2\)SO\(_4\) in the conical flask
(v) Give of collection of the gas one I. physical property; II. chemical property of the gas
(vi) State one chemical test to identify the gas.
(d) A 4.3 g hydrated sodium tetraoxosulphate (VI) (Na\(_2\)SO\(_4\).xH\(_2\)O) was heated to remove the water of crystallization. The remaining anhydrous salt had a mass of 2.12 g. Calculate the value of x in the t I hydrated salt. [H = 1; O = 16; Na = 23; S = 32 ]
Ibeere 43 Ìròyìn
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initial and final) as well as the size size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution containing 1.04 g HCl per 500 cm\(^3\) of solution. B was prepared by diluting 50.0 cm\(^3\) of a saturated solution of Na\(2\)CO\(_3\) at room temperature to 1000 cm\(^3\)
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^{3}\) portions of B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your results and calculate the average volume of acid used.
9b) From your results and information provided above, calculate the;
(i) concentration of A in moldm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) solubility of Na\(_2\)CO\(_3\) in mol dm\(^{-3}\)
(iv) volume of CO\(_2\) that would be liberated from 1 dm\(^3\) of B if the titration were carried out at s.t.p.
The equation for the reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
[H = 1; C = 12; O = 16; Na = 23; Cl = 35.5; Molar volume of gas at s.t.p = 22.4 dm\(^3\)]
Ibeere 44 Ìròyìn
i) What is the name of the process used for the industrial preparation of tetraoxosulphate (VI) acid?
(ii) State the catalyst used in (a)(i)
(iii) Show by means of bdanced chemical equations only, the industrial preparation of tetraoxosulphate (VI) acid from sulphur (IV) oxide.
(b)(i) Distinguish between dehydration and drying
(ii) Explain why concentrated tetraoxosulphate (VI) acid cannot be used to dry ammonia
(iii) What is the drying agent for ammonia?
(c)(i) Give one example of I. a chloride which is soluble in hot water, II. a trioxocarbonate (IV) which does not decompose on heating, Ill. an amphoteric oxide
(ii) List three methods for the preparation of salts
(iii) State one method for the recovery of salt from its solution.
(d)(i) State Gay Lussac's law of combining volumes
(ii) Consider the reaction represented by the following equation: C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) \(\to\) 2H\(_2\)O\(_{(g)}\) + 2CO\(_{2(g)}\). What is the volume of oxygen required for the complete combustion of 12.5 cm\(^{3}\) of ethene?
Ibeere 45 Ìròyìn
TEST OF PRACTICAL KNOWLEDGE QUESTION
Credit will be given for strict adherence to the instructions, for observations precisely recorded, and for accurate inferences. All tests, Observations, and inferences must be clearly entered in your answer book, in ink, at the time they are made.
C is a mixture of two inorganic salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion you draw from the result of each test.
(a) Put all of C in a test tube and add about 10 cm\(^3\) of distilled water. Stir, filter, and keep the filtrate and the residue.
(b) Put the residue into a test tube and add about 5 cm\(^3\) of dilute HCl. Shake to dissolve
(i) To about 2 cm\(^3\) of the solution, add NaOH\(_{(aq)}\) in drops and then in excess
(ii) To another 2 cm\(^3\) portion of the solution, add NH\(_{3(aq)}\) in drops and then in excess.
(c) To about 2 cm\(^3\) portion of the filtrate, add few drops of dilute HNO\(_3\), and then AgNO\(_{3(aq)}\) a followed by aqueous NH\(_3\) in excess.
