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Ajụjụ 1 Ripọtì
What is the product Cx Hy in the following equation? C10H12 → C8H18 + CxHy
Ajụjụ 2 Ripọtì
The process of seeding during the extraction of aluminium involves the addition of pure AI(OH)3 crystals in order to
Akọwa Nkọwa
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."
Ajụjụ 3 Ripọtì
Which of the following properties of elements deceases generally across a period in the Periodic Table?
Akọwa Nkọwa
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.
Ajụjụ 4 Ripọtì
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)
Ajụjụ 5 Ripọtì
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
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
WHICH OF THE FOLLOWING SALTS IS READILY SOLUBLE IN WATER?
Akọwa Nkọwa
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.
Ajụjụ 7 Ripọtì
A nuclide emits a beta-particle (e) in the process, that atomic number
Akọwa Nkọwa
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".
Ajụjụ 8 Ripọtì
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)
Akọwa Nkọwa
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.
Ajụjụ 9 Ripọtì
Which of the following subtances will not produce effervescence with dilute hydeochloric acid?
Akọwa Nkọwa
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.
Ajụjụ 10 Ripọtì
What of the following pairs of acid and base will produce a solution with pH less than 7 at equivalence point?
Ajụjụ 11 Ripọtì
Which of the following salts is not prepared by precipitation?
Akọwa Nkọwa
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.
Ajụjụ 12 Ripọtì
The alloy used extensively in the building industry is
Akọwa Nkọwa
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.
Ajụjụ 13 Ripọtì
Which of the following compounds is a soapless detergent?
Akọwa Nkọwa
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.
Ajụjụ 14 Ripọtì
Which of the following reactions will be affected by change in pressure at equilibrium?
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
A metal X forms two chlorides with the formulae XCI2 and XCI3. Where is X in the Periodic Table?
Akọwa Nkọwa
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."
Ajụjụ 16 Ripọtì
The key factor to be considered in siting a chemical industry is
Akọwa Nkọwa
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.
Ajụjụ 17 Ripọtì
The method of collection of gases prepared in the laboratory depend on their
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
Consider the cell represented below: Zn/Zn2+(aq) // Cu2+(aq) / Cu(s). Which of the following statements is correct about the cell
Akọwa Nkọwa
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+".
Ajụjụ 19 Ripọtì
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?
Akọwa Nkọwa
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.
Ajụjụ 20 Ripọtì
The condensation of several amino acid molecules gives
Akọwa Nkọwa
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.
Ajụjụ 21 Ripọtì
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.
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
If 60 cm3 of gas is heated from 27oC TO 50oC, what is the new volume of the gas at constant pressure?
Ajụjụ 23 Ripọtì
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?
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
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]
Akọwa Nkọwa
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.
Ajụjụ 25 Ripọtì
Which of the following measures is not suitable for controlling water pollution?
Akọwa Nkọwa
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.
Ajụjụ 26 Ripọtì
Which of the following electron transition results in the emission of energy?
Akọwa Nkọwa
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.
Ajụjụ 27 Ripọtì
Which of the following pairs of salts undergo hydrolysis?
Ajụjụ 28 Ripọtì
When ammonium trioxocarbonate (IV) solution is added to separate solutions of calcium, zinc and sodium salt precipitate could be formed in the case of
Akọwa Nkọwa
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.
Ajụjụ 29 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 30 Ripọtì
The reaction that takes place at the cathode when an aqueous solution of NaCI is electrolysed using carbon electrodes is
Akọwa Nkọwa
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.
Ajụjụ 31 Ripọtì
A solution of zinc chloride should not be stored in a container made of
Akọwa Nkọwa
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.
Ajụjụ 32 Ripọtì
Which of the following pollutants is biodegradable?
Akọwa Nkọwa
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.
Ajụjụ 33 Ripọtì
What quantity of electrons in moles is needed to discharge two moles of aluminum from aluminum oxide (AI 2O3)
Akọwa Nkọwa
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.
Ajụjụ 34 Ripọtì
What is the IUPAC name of CH3CH2COOCH 3?
Akọwa Nkọwa
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-).
Ajụjụ 35 Ripọtì
WHICH OF THE FOLLOWING PETROLEUM FRACTIONS YIELDS VASELINE AND PARAFFIN WAX ON REDISTILLATION?
Akọwa Nkọwa
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.
Ajụjụ 36 Ripọtì
Which of the following equipment is not used to detect radioactivity?
Akọwa Nkọwa
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.
Ajụjụ 37 Ripọtì
One of the characteristics of transition metals is
Akọwa Nkọwa
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.
Ajụjụ 38 Ripọtì
Two solutions containing Pb2+ and AI3+ respectively can be distingusihed using
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
Glucose gives a brick-red precipitate with Fehling's solution because it is
Akọwa Nkọwa
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.
Ajụjụ 40 Ripọtì
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]
Ajụjụ 41 Ripọtì
The mass of an element is 27 and its atomic number is 13. What is the composition of the nucleus of its atom?
Akọwa Nkọwa
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".
Ajụjụ 42 Ripọtì
Which of the electrons in the following orbitals will experience the greatest nuclear attraction?
Akọwa Nkọwa
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.
Ajụjụ 43 Ripọtì
(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.
Ajụjụ 44 Ripọtì
TEST OF PRACTICAL KNOWLEDGE QUESTION
Credit will be given for strict adherere 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 sample of iron (ii) tetraoxosulphate (VI). D is a sample of zinc trioxocarbonate (IV). Carry out the following exercises on C and D. Record your observations and identify any gases evolved. State the conclusion you draw from the result of each test.
(a)(i) Put all of C in a test tube and add about 5 cm\(^3\) of distilled water. Stir and test with litmus paper. Divide the solution into two portions
(ii) To the first portion, add sodium hydroxide solution in drops and then in excess.
(iii) To the second portion, add few drops of barium chloride solution, followed by dilute hydrochloric acid in excess.
(b)(i) Put half of D in a dry test tube and heat strongly. Allow to cool.
(ii) Add about 5 cm\(^3\) of dilute hydrochloric acid to the residue and divide the solution into two portions.
(iii) To the first portion, add sodium hydroxide solution in drops and then in excess.
(iv) To the second portion, add aqueous ammonia in drops and then in excess.
None
Akọwa Nkọwa
None
Ajụjụ 45 Ripọtì
(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.
None
Akọwa Nkọwa
None
Ajụjụ 46 Ripọtì
(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
Akọwa Nkọwa
None
Ajụjụ 47 Ripọtì
TEST OF PRACTICAL KNOWLEDGE QUESTION
(a) Give the reason in each case why it is necessary to:
(i) moisten a piece of litmus paper used in testing for the acidity or alkalinity of a gas.
(i) acidify the test solution with dilute hydrochloric acid in the confirmatory test for SO\(_4^2\)
(b)(i) List two gases that must not be prepared in the open laboratory
(ii) Mention one precaution that should be taken in the laboratory to prevent excessive inhalation of these gases during their preparation.
(iii) State one use of each of the following pieces of apparatus in the laboratory. I. Wash bottle II. Tripod stand
(c) State what would be observed when:
(i) Pb(NO\(_3\))\(_2\) is needed
(ii) concentrated HCI is added to MnO\(_2\)
Akọwa Nkọwa
None
Ajụjụ 48 Ripọtì
TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution of HCI containing 5.0g dm\(^{-3}\). B is a solution of impure KOH containing 6.50g dm\(^{-3}\).
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 Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is: HCI\(_{(aq)}\) KOH\(_{(aq)}\) \(\to\) Cl\(_{(aq)}\) H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided above, calculate the:
(i) concentration of A in mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) percentage purity of KOH in B [H= 1; CI = 35.5; KOH = 56.0g mol\(^{-1}\)]
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