JAMB UTME - Chemistry - 1999
Ajụjụ 1 Ripọtì
Which of the following statement is true of the electrochemical series?
Akọwa Nkọwa
The electrochemical series is a list of metals and non-metals in order of their relative reactivity towards other substances in electrochemical reactions. The statement that is true of the electrochemical series is that electropositivity of metals decreases down the series. This means that the most electropositive metals (such as potassium and sodium) are found at the top of the series, while the least electropositive metals (such as gold and platinum) are found at the bottom. As you go down the series, the metals become less likely to lose electrons and form positive ions, which means that they become less reactive.
Ajụjụ 3 Ripọtì
RS(aq) + HF(aq) → RF(s) + HS(aq)
ΔH = -65.7 kj mol(-1).
From the equation above, it can be deduced that
Akọwa Nkọwa
The given chemical equation represents a reaction between RS and HF to form RF and HS. The value of ΔH (-65.7 kJ/mol) indicates the heat change associated with the reaction. A negative value of ΔH indicates that the reaction is exothermic, meaning that the heat content of the products is lower than the heat content of the reactants. Therefore, the correct answer is "the heat content of the reactants is higher than that of the products." "the heat content of the reactants is lower than that of the products," is incorrect as it contradicts the given value of ΔH. "the reaction is slow," and "a large amount of heat is absorbed," are irrelevant and not supported by the given information.
Ajụjụ 4 Ripọtì
Use the table above to answer this question .The least reactive element is
Akọwa Nkọwa
From the reactivity series in the table, the least reactive element is Z. This is because Z is placed at the bottom of the series and is the only element that does not react with any of the other elements. Elements higher up the series are more reactive because they can displace the ions of elements below them in the series. Therefore, since Z is placed at the bottom of the reactivity series and does not react with any of the other elements, it is the least reactive element in the table.
Ajụjụ 5 Ripọtì
When the two end alkyl groups ethyl ethanoate are interchange, the compound formed is known as
Ajụjụ 6 Ripọtì
The correct order of increasing oxidation number of the transition metal ions for the compounds K2Cr2O7, V2O5 and KMnO4 is
Akọwa Nkọwa
The oxidation state or number is a measure of the degree of oxidation of an atom in a compound, which is determined by counting the number of electrons transferred from or to the element when it forms a compound. In this question, we are given three transition metal compounds, K2Cr2O7, V2O5, and KMnO4, and we are asked to arrange them in order of increasing oxidation number of the metal ion. The oxidation number of the metal ion in each compound can be determined by assigning electrons to the metal and oxygen atoms based on their electronegativity values and the overall charge of the compound. Oxygen is generally assigned an oxidation number of -2, and the sum of the oxidation numbers of all the atoms in the compound should equal the overall charge of the compound. For K2Cr2O7, we can assign an oxidation state of +6 to the chromium ion, since each oxygen atom contributes -2 to the total charge, and the two potassium ions each contribute +1. For V2O5, we can assign an oxidation state of +5 to the vanadium ion, since each oxygen atom contributes -2 to the total charge. For KMnO4, we can assign an oxidation state of +7 to the manganese ion, since each oxygen atom contributes -2 to the total charge, and the potassium ion contributes +1. Therefore, the correct order of increasing oxidation number of the metal ions in these compounds is: V2O5 < K2Cr2O7 < KMnO4.
Ajụjụ 7 Ripọtì
A gaseous metallic chloride MClx consists of 20.22% of M by mass. The formula of the chloride is
[M = 27, Cl = 35.5]
Akọwa Nkọwa
The given percentage of M in the gaseous metallic chloride MCl_x is 20.22%. We need to determine the formula of the chloride. First, we can assume that we have 100g of the compound. This would mean that there are 20.22g of M and 100 - 20.22 = 79.78g of Cl in the compound. Next, we need to determine the ratio of M to Cl in the compound. We can do this by dividing the mass of each element by its atomic weight and then dividing by the smaller result to get the simplest ratio. For M, (20.22 / 27) gives a result of 0.748 and for Cl, (79.78 / 35.5) gives a result of 2.25. Dividing both by 0.748 gives a ratio of approximately 1:3. This means that the formula for the compound is MCl3, which is option (C).
Ajụjụ 8 Ripọtì
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
From the equation above, the mass of magnesium required to react with 250cm3 of 0.5M HCl is
[Mg = 24]
Akọwa Nkọwa
The given equation shows a reaction between magnesium and hydrochloric acid which produces magnesium chloride and hydrogen gas. The question asks for the mass of magnesium required to react with a given amount of hydrochloric acid. To solve this problem, we need to use the following formula: moles = concentration x volume / 1000 where moles is the number of moles of hydrochloric acid, concentration is the molarity of hydrochloric acid, and volume is the volume of hydrochloric acid in liters. First, we need to calculate the number of moles of hydrochloric acid in 250 cm³ of 0.5 M HCl: moles HCl = 0.5 x 250 / 1000 = 0.125 moles From the balanced chemical equation, we know that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. Therefore, the number of moles of magnesium required is half the number of moles of hydrochloric acid: moles Mg = 0.125 / 2 = 0.0625 moles Finally, we can calculate the mass of magnesium required using its molar mass: mass = moles x molar mass = 0.0625 x 24 = 1.5 g Therefore, the answer is option B: 1.5g.
Ajụjụ 10 Ripọtì
The entropy and enthalpy of a system are a measure of
Akọwa Nkọwa
The entropy and enthalpy of a system are both thermodynamic properties used to describe the state of a system. Entropy is a measure of the degree of disorderliness or randomness of a system, while enthalpy is a measure of the heat content of a system at constant pressure. In other words, entropy tells us how much energy in a system is unavailable to do work, while enthalpy tells us how much energy is stored in the chemical bonds of the system. Therefore, the correct option is: - degree of disorderlinesss and heat content respectively
Ajụjụ 11 Ripọtì
Given that electronegativity increases across a period and decreases down a group in the periodic table, in which of the following compounds will the molecules be held together by the strongest hydrogen bond?
Akọwa Nkọwa
Hydrogen bonding is a type of intermolecular force that occurs between a hydrogen atom bonded to a highly electronegative atom (such as nitrogen, oxygen or fluorine) and another highly electronegative atom in a separate molecule. The strength of a hydrogen bond depends on the electronegativity of the atoms involved and the distance between them. In the given compounds, HF has the strongest hydrogen bond because it has the highest electronegativity difference between the hydrogen and fluorine atoms. Fluorine is the most electronegative element in the periodic table, so the bond between hydrogen and fluorine is highly polarized. This results in a strong partial positive charge on the hydrogen atom, which attracts the highly electronegative fluorine atom of another molecule, forming a strong hydrogen bond. In contrast, NH3 and CH4 have weaker hydrogen bonds because nitrogen and carbon are less electronegative than fluorine. HCl has the weakest hydrogen bond because hydrogen and chlorine have a smaller electronegativity difference than hydrogen and fluorine.
Ajụjụ 12 Ripọtì
The solution with the lowest pH value is
Ajụjụ 13 Ripọtì
In order to remove one electron from a 3S orbital of gaseous sodium atom, about 496KJMol-1 of energy is required. This energy is referred to as
Akọwa Nkọwa
The energy required to remove an electron from an atom is called ionization energy. In this case, the question specifically states that the energy required to remove one electron from a 3S orbital of gaseous sodium atom is 496 kJ/mol. Therefore, the correct answer is ionization energy. Ionization energy is a measure of how strongly an atom holds onto its electrons. The higher the ionization energy, the more difficult it is to remove an electron from the atom. The ionization energy is often used to describe the reactivity of an element and its ability to form chemical bonds.
Ajụjụ 14 Ripọtì
Chlorine, bromine and iodine resemble one another in that they
Akọwa Nkọwa
Chlorine, bromine, and iodine resemble one another in that they displace one another from solutions of their salts. This is due to the fact that they are all halogens and have similar chemical properties. Halogens are highly reactive non-metals that have seven electrons in their outermost energy level. Because they all have a similar electronic configuration, they react similarly with other elements and compounds. In the case of displacing one another from solutions of their salts, this is because they are all capable of oxidizing and reducing one another. As a result, if a more reactive halogen is added to a solution containing a less reactive halogen, the more reactive halogen will displace the less reactive halogen from its salt. This displacement reaction is a common property of halogens and is used in various chemical applications.
Ajụjụ 15 Ripọtì
Nitrogen obtained from the liquefaction of air has a higher density than that obtained from nitrogen containing compounds because the former contains
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Ajụjụ 16 Ripọtì
In the diagram above, the mixture of the solids P and Q can be separated by
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The diagram shows two solids, P and Q, mixed together. The question is asking how these two solids can be separated from each other. Crystallization involves dissolving a solid in a solvent to create a solution, then allowing the solvent to evaporate slowly to produce crystals of the dissolved solid. This method can be used when one of the solids is soluble in a particular solvent and the other is not. However, from the diagram, it is not clear whether either of the solids is soluble in a particular solvent, so we can eliminate options C and D. Distillation is a process used to separate liquids based on differences in their boiling points. This method would not be effective for separating solids, so we can eliminate option A. Fractional distillation is similar to distillation, but is used to separate a mixture of liquids with similar boiling points. This method also would not be effective for separating solids, so we can eliminate option B. Therefore, the correct answer is option C, crystallization.
Ajụjụ 17 Ripọtì
2-methylprop-1-ene is an isomer of
Ajụjụ 18 Ripọtì
Which of the following metals is passive to concentrated trioxonitrate (V) acid?
Ajụjụ 19 Ripọtì
The pair of metals in the reactivity series that are usually extracted by the electrolysis of their ores is
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Ajụjụ 20 Ripọtì
An organic compound that does not undergo a reaction with both hydrogen cyanide and hydroxylamine can be
Akọwa Nkọwa
An organic compound that does not undergo a reaction with both hydrogen cyanide and hydroxylamine can be an alkane. Hydrogen cyanide (HCN) and hydroxylamine (NH2OH) are chemical reagents that are commonly used to test for the presence of a carbonyl group (-C=O) in organic compounds. When these reagents are added to a carbonyl-containing compound, a chemical reaction occurs, and a new product is formed. However, alkanes do not contain a carbonyl group and are therefore unreactive towards both HCN and NH2OH. Hence, an organic compound that does not undergo a reaction with both HCN and NH2OH can be an alkane. Alkenes contain a carbon-carbon double bond (-C=C-) and are generally more reactive than alkanes. Alkanals, alkanones, and alkanoc acids all contain a carbonyl group and are therefore reactive towards HCN and NH2OH.
Ajụjụ 21 Ripọtì
X + Y → Z. The rate equation for the chemical reaction above is ((-Δ[x])/Δt) = K[X]2[Y].
The overall order of the reaction is
Ajụjụ 22 Ripọtì
2SO2(g) + O2(g) ↔ 2SO3(g). In the chemical reaction above, the substance that will increase the rate of production of sulphur (VI) oxide is
Ajụjụ 23 Ripọtì
200cm3 each of 0.1M solutions of lead (II) trioxonitrate (V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated.
[Pb = 207, Cl= 35.5, N = 14, O = 16]
Akọwa Nkọwa
The balanced chemical equation for the reaction between lead (II) trioxonitrate (V) and hydrochloric acid is: Pb(NO3)2(aq) + 2HCl(aq) → PbCl2(s) + 2HNO3(aq) From the equation, we can see that 1 mole of lead (II) trioxonitrate (V) reacts with 2 moles of hydrochloric acid to produce 1 mole of lead (II) chloride. Therefore, we need to first calculate the number of moles of lead (II) trioxonitrate (V) and hydrochloric acid present in the solution: Number of moles of Pb(NO3)2 = concentration × volume = 0.1 × (200/1000) (convert cm³ to dm³) = 0.02 moles Number of moles of HCl = concentration × volume = 0.1 × (200/1000) = 0.02 moles From the equation, we can see that 1 mole of Pb(NO3)2 produces 1 mole of PbCl2. Therefore, the number of moles of PbCl2 that will be formed is also 0.02 moles. Now, we can calculate the mass of PbCl2 formed: Mass of PbCl2 = number of moles × molar mass = 0.02 × (207 + 2(35.5)) = 0.02 × 278 = 5.56 g Therefore, the mass of lead (II) chloride that will be precipitated is 5.56g. The correct option is: 5.56g.
Ajụjụ 24 Ripọtì
POlyvinyl chloride is used to produce
Akọwa Nkọwa
Polyvinyl chloride (PVC) is a synthetic plastic material that is widely used in various applications due to its versatility, durability, and affordability. One of the most common uses of PVC is in the production of pipes. PVC pipes are used in various industries, including construction, agriculture, and plumbing, because they are strong, lightweight, and easy to install. PVC is not used in the production of bread, pencils or ink. Bread is made from flour, water, yeast, and other ingredients, while pencils are made from graphite and wood or plastic. Ink is typically made from a mixture of pigments or dyes, a solvent, and a binder. PVC is not a common component in any of these products.
Ajụjụ 25 Ripọtì
The method that can be used to convert hard water to soft water is
Akọwa Nkọwa
Hard water contains high levels of dissolved minerals like calcium and magnesium, which can cause problems like soap scum, scale buildup, and decreased efficiency of water heaters and pipes. Soft water, on the other hand, contains very low levels of these minerals and is more desirable for household use. One effective method of converting hard water to soft water is the use of an ion-exchange resin. This resin contains sodium ions that can exchange with the calcium and magnesium ions in the water, effectively removing them and replacing them with sodium ions. This process is called ion exchange, and it can be done using a device called a water softener. Chlorination, passage over activated charcoal, and aeration are not effective methods of converting hard water to soft water.
Ajụjụ 26 Ripọtì
0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm3. If 15.0 cm3 of the solution requires 12.50 cm3 of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution
Akọwa Nkọwa
M1 V1 = M2 V2
∴ Mn Vn represent volume and molar concentration of HCL also MyVy represent volume and molar concentration of Na2CO3
MnVn = MyVy
0.25 * 15 = 12.50 * My
My = (0.25 * 15)/12.50
My = 0.3m/dm3
Ajụjụ 27 Ripọtì
What quantity of electricity will liberate 0.125 mole of oxygen molecules during the electrolysis of dilute sodium chloride solution?
[ F = 96 5000 C mol-1 ]
Akọwa Nkọwa
4 Faradays of electricity is required to liberate one mole of oxygen gas
∴ quantity of electricity required
= 0.125 * 4 * 96500 = 48250 coulombs
Ajụjụ 28 Ripọtì
A pair of compounds that can be used to generate a gas which has physiological effect on human beings is
Ajụjụ 29 Ripọtì
Which of the compounds above would react to take up two molecules of bromine during bromination?
Akọwa Nkọwa
The compound that would react to take up two molecules of bromine during bromination is l only. This is because l has a double bond between carbon and carbon, which allows it to undergo addition reactions, such as bromination. The double bond is broken, and a bromine atom is added to each carbon atom, resulting in the product with two bromine atoms attached to it. On the other hand, lll has no double bonds and cannot undergo addition reactions like bromination. Therefore, l and ll only have double bonds and can react to take up two molecules of bromine during bromination.
Ajụjụ 30 Ripọtì
The set of pollutants that is most likely to be produced when petrol is accidentally spilled on plastic materials and ignited is
Ajụjụ 31 Ripọtì
1s2, 2s2, 2p6, 3s2, 3p6, 3d7, 4s2
An element with the electronic configuration above is a
Akọwa Nkọwa
The electronic configuration given in the question represents an atom with 26 electrons. To determine the identity of the element, we can count the number of electrons in each energy level and sub-level. The highest energy level, 4, has 2 electrons in the s sub-level and 0 electrons in the d sub-level, which means that it is a Group 2 element. The 3d sub-level has 7 electrons, which confirms that the element is a transition element. Therefore, the correct answer is "transition element."
Ajụjụ 32 Ripọtì
When a current I was passed through an electrolyte solution for 40 minutes, a mass X g of a univalent metal was deposited at the cathode. What mass of the metal will be deposited when a current 2I is passed through the solution for 10 minutes?
Akọwa Nkọwa
From M = zit
Where M = to the mass of the metal liberated
z = Is the electrochemical equivalent
∴ X g = 1 * 1 * 40 * 60
X g = 2400 I
Let Y represent the new mass
Y = 1*2I * 10 * 60
Y = 1200 I
Y/(X g) = 1200 I/2400 I
Y = X/2 g
Ajụjụ 33 Ripọtì
The gas that will form a white precipitate with acidified silver trioxonitrate (v) is
Akọwa Nkọwa
The gas that will form a white precipitate with acidified silver trioxonitrate (v) is hydrogen chloride (HCl). When HCl is reacted with silver trioxonitrate (v) in the presence of water, it produces a white precipitate of silver chloride (AgCl), which is insoluble in water and forms a white solid. The other options, NH3, SO2, and CO2, do not react with silver trioxonitrate (v) to form a white precipitate.
Ajụjụ 34 Ripọtì
Which of the following will have its mass increased when heated in air?
Akọwa Nkọwa
When materials are heated in air, they can undergo a variety of chemical reactions such as combustion, oxidation, or decomposition. These reactions can result in a change in the mass of the material. Out of the options provided, the material that is most likely to have its mass increased when heated in air is "Magnesium". This is because magnesium is a reactive metal that undergoes combustion in air, which means it reacts with oxygen in the air to form magnesium oxide. The magnesium oxide has a greater mass than the original magnesium metal, so the overall mass of the system increases. Helium is a noble gas and is chemically inert, which means it does not react with other substances under normal conditions, including when heated in air. Therefore, its mass remains the same. Copper pyrites is a mineral composed of various compounds, including copper and sulfur. When heated in air, it can undergo various chemical reactions, but none of these reactions result in an overall increase in mass. Glass is a non-crystalline solid that is formed by heating a mixture of silica, soda, and lime. When heated further in air, it may undergo some minor chemical reactions, but these do not result in a significant increase in mass. Therefore, out of the options provided, magnesium is the material that is most likely to have its mass increased when heated in air.
Ajụjụ 35 Ripọtì
The hydrocarbon that burns in air with a sooty flames is
Ajụjụ 36 Ripọtì
From the diagram above the mass of crystals deposited when 1 dm 3 of a saturated solution of NaCl is cooled from 80oC to 60oC is
[Na = 23, Cl = 35.5]
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
What properties of duralumin make it more useful than its constituent metals?
Akọwa Nkọwa
Duralumin is an alloy made from aluminum, copper, and magnesium. The properties of duralumin that make it more useful than its constituent metals are its strength and lightness. Duralumin is much stronger than pure aluminum, which makes it suitable for use in applications where high strength-to-weight ratios are important, such as in aircraft construction. Additionally, duralumin is resistant to corrosion and has a high level of ductility, which means it can be easily shaped without breaking. Therefore, duralumin has a combination of properties that make it an excellent material for many industrial applications, including aircraft construction, automotive parts, and structural components.
Ajụjụ 38 Ripọtì
The salt that reacts with dilute hydrochloric acid to produce a pungent smelling gas which decolourizes acidified purple potassium tetraoxomanganate (VII) solution is
Ajụjụ 39 Ripọtì
What is observed when aqueous solutions of each of tetraoxosulphate (VI) acid, potassium trioxoiodate (V) and potassium iodide are mixed together?
Ajụjụ 41 Ripọtì
Hydrogen is used in oxy-hydrogen flames for melting metals because it
Akọwa Nkọwa
Hydrogen is used in oxy-hydrogen flames for melting metals because it evolves a lot of heat when burnt. When hydrogen gas is burned in the presence of oxygen, it undergoes a highly exothermic reaction to form water vapor. The heat released during this reaction is intense and can be used to melt metals, such as gold, silver, and copper. This process is known as oxy-hydrogen welding, and it is commonly used in jewelry making, metal sculpture, and other metalworking applications. While hydrogen can be explosive when mixed with air or oxygen in certain conditions, oxy-hydrogen welding relies on the controlled burning of hydrogen in the presence of oxygen to generate heat for melting metals. Hydrogen is also a very light gas, but this property is not relevant to its use in oxy-hydrogen flames for melting metals. Additionally, while hydrogen can be used as a rocket fuel, this is not related to its use in oxy-hydrogen welding.
Ajụjụ 42 Ripọtì
When excess ethanol is heated to 145oC in the presence of concentrated H2SO4, the product is
Akọwa Nkọwa
When excess ethanol is heated to 145°C in the presence of concentrated H2SO4, the product formed is diethyl ether. The reaction is known as the Williamson ether synthesis. In this reaction, ethanol acts as a nucleophile and attacks the protonated form of ethanol to form an intermediate compound. The intermediate compound loses a molecule of water to form diethyl ether. The other products mentioned in the options are not formed during this reaction. Ethyne is formed by the dehydration of ethanol in the presence of sulfuric acid, but the reaction conditions mentioned in the question are not suitable for the formation of ethyne. Diethyl sulphate is formed when ethanol is heated with excess concentrated H2SO4 at higher temperatures, while acetone is not formed by this reaction.
Ajụjụ 43 Ripọtì
Which of the following gases will diffuse fastest when passed through a porous plug?
[H = 1, C = 12, N = 14, O = 16]
Akọwa Nkọwa
The gas that will diffuse fastest is the gas with the lowest molecular mass which in this case is methane, CH4.
Ajụjụ 44 Ripọtì
A metal that can be extracted from cassitertite is
Akọwa Nkọwa
The metal that can be extracted from cassiterite is tin. Cassiterite is a mineral containing tin oxide (SnO2) which is the main source of tin. The process of extracting tin from cassiterite involves several stages including concentration, smelting, and refining. Tin is used in many applications including the manufacturing of tinplate, which is used for packaging food and drinks, and in the production of various alloys such as bronze and pewter.
Ajụjụ 45 Ripọtì
In which of the following are water molecules in the most disorderly arrangement?
Akọwa Nkọwa
Water molecules are the most disorderly arranged in the gaseous state, i.e., as steam at 100°C. In the gaseous state, water molecules have the highest amount of kinetic energy, and they move randomly in all directions at high speeds, colliding with each other and other molecules in the surroundings. This random motion results in a highly disordered arrangement of water molecules. In contrast, at lower temperatures and higher pressures, water molecules tend to have less kinetic energy and move less randomly, resulting in more orderly arrangements, such as in ice at 0°C or even -10°C.
Ajụjụ 46 Ripọtì
The solubility product of Cu(IO3)2 is 1.08 * 10-7. Assuming that neither ions react appreciable with water to form H+ and OH-, what is the solubility of this salt?
Ajụjụ 47 Ripọtì
Which is the temperature of a given mass of a gas initially at 0°C and 9 atm, if the pressure is reduced to 3 atm at constant volume?
Akọwa Nkọwa
We can solve this problem using the combined gas law, which relates the pressure, volume, and temperature of a gas: (P1 V1) / T1 = (P2 V2) / T2 where P1 and T1 are the initial pressure and temperature, P2 and T2 are the final pressure and temperature, and V1 and V2 are the initial and final volumes (which are held constant in this problem). We can plug in the values we know and solve for the unknown temperature: (9 atm)(V1) / 273 K = (3 atm)(V1) / T2 Cross-multiplying and simplifying gives: T2 = (3 atm / 9 atm) * 273 K = 91 K Therefore, the temperature of the gas is 91K. Answer is correct.
Ajụjụ 48 Ripọtì
N2O4(g) ↔ 2NO2(g). Increase in total pressure of the equilibrium reaction above will
Ajụjụ 49 Ripọtì
56.00cm3 of a gas at S.T.P. weighed 0.11g. What is the vapour density of the gas?
[Molar volume of a gas at S.T.P = 22.4dm3]
Akọwa Nkọwa
56.00cm3 of a gas at S.T.P weighed 0.11g
Molar volume of the gas = 22.4dm3 = 22400cm3
molar mass = (22400 x 0.11)/56
molar mass = 2464/56 = 44
Vapour density = Molar mass/2 = 44/2 = 22
