JAMB UTME - Chemistry - 2006
Tambaya 1 Rahoto
An example of aliphatic unsaturated hydrocarbon is
Bayanin Amsa
An aliphatic unsaturated hydrocarbon is a hydrocarbon that does not contain any aromatic rings and has at least one double or triple bond between its carbon atoms. Pentane is an example of an aliphatic hydrocarbon, but it is not unsaturated, meaning it contains only single bonds between its carbon atoms. Butanol is also aliphatic, but it contains an -OH group and does not have any double or triple bonds. Propene, on the other hand, is an aliphatic unsaturated hydrocarbon because it has one double bond between its two carbon atoms. Finally, benzene is not aliphatic, but rather it is an aromatic hydrocarbon, which contains at least one aromatic ring with alternating double bonds. Therefore, the correct answer is "propene."
Tambaya 2 Rahoto
In the diagram above, mixture Y is
Tambaya 3 Rahoto
The ion that can be used as an oxidizing agent is
Tambaya 4 Rahoto
In the industrial production of H2 from natural gas the CO2 produced along with the H2 is removed by
Bayanin Amsa
In the industrial production of hydrogen gas from natural gas, carbon dioxide is produced as a byproduct. To remove the carbon dioxide from the hydrogen gas, the mixture is washed under pressure. Therefore, the correct option is "washing under pressure."
Tambaya 5 Rahoto
One mole of a hydrocarbon contains 36g of carbon and its vapour density is 20. The structure of the hydrocarbon is
Bayanin Amsa
2VD = Rmm VD = Vapour density; Rmm - Relative
2 * 20 = 40 Molecular mass
Mole of carbon = 36/12 = 3
Mole of H2 = 40 - 36 = 4
The compound contains 3 mole of carbon and 4 hydrogen
Tambaya 6 Rahoto
The products of the combustion of candle wax are
Bayanin Amsa
When a candle is burned, the heat of the flame melts the wax near the wick. This liquid wax then moves up the wick, where the heat of the flame vaporizes the liquid wax molecules. These vaporized molecules react with oxygen from the air to form new compounds, releasing heat and light in the process. The main products of the combustion of candle wax are carbon dioxide (CO2) and water (H2O). Therefore, the correct answer is "carbon (IV) oxide and water."
Tambaya 7 Rahoto
On heating, which of the following compounds will decompose to the free metal, nitrogen (IV) oxide and oxygen?
Tambaya 8 Rahoto
A difference between chemical and physical change is that in a chemical change
Bayanin Amsa
The main difference between chemical and physical change is that in a chemical change a new substance is formed. In other words, the chemical composition of the original substance changes, resulting in the formation of a new substance with different chemical and physical properties. Physical changes, on the other hand, do not involve a change in the chemical composition of the substance. In physical changes, only the physical properties of the substance such as its shape, size, color, or state of matter may change, but the chemical identity of the substance remains the same.
Tambaya 9 Rahoto
5SO2(g) + 2KMnO4(aq) + 2H2O(l) → K2SO4 + 2MnSO4(aq) + 2H2SO4(aq)
In the reaction above, the products give a
Bayanin Amsa
The reaction given is a chemical equation. It represents a chemical change where reactants combine to form products. The equation shows the reactants on the left-hand side and the products on the right-hand side. In this particular equation, the reactants are 5SO2(g), 2KMnO4(aq), and 2H2O(l). The products are K2SO4, 2MnSO4(aq), and 2H2SO4(aq). Now, to answer the question, the equation does not involve any purple solution or purple precipitate, so we can eliminate the first two options. The products of the reaction are K2SO4, 2MnSO4(aq), and 2H2SO4(aq). None of these products is a colourless precipitate, so we can eliminate the third option as well. Therefore, the correct answer is that the products give a colourless solution.
Tambaya 10 Rahoto
If 30cm3 of gas at 50oC is warmed to 80o at a fixed pressure, the fractional increase in volume is
Tambaya 11 Rahoto
Which of the following functional groups will give gas bubbles when treated with a saturated solution of sodium hydrogen trioxocarbonate (IV)?
Bayanin Amsa
Tambaya 12 Rahoto
Which of the following typically represents a photochemical reaction?
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Tambaya 13 Rahoto
In the diagram above, gas X is
Tambaya 14 Rahoto
Water has the ability to dissolve almost all substance because its molecules has a structure with
Bayanin Amsa
Water has a unique molecular structure which makes it a universal solvent. A water molecule consists of two hydrogen atoms and one oxygen atom. The oxygen atom has a slightly negative charge, while the hydrogen atoms have a slightly positive charge. This uneven distribution of charge creates a polarity in the molecule, with one end being more positively charged and the other end being more negatively charged. Therefore, the correct answer is: "one positive end and one negative end".
Tambaya 15 Rahoto
Which of the following has an isomer?
Tambaya 16 Rahoto
Which of the following results in the fall of acid rain?
Bayanin Amsa
The presence of certain gases and particles in the air can contribute to the formation of acid rain. Acid rain occurs when rainwater mixes with these gases and particles, forming acidic solutions that fall to the ground. Out of the given options, the oxides of sulfur are known to be the main contributor to acid rain. When sulfur dioxide (SO2) and nitrogen oxides (NOx) are released into the atmosphere, they react with water, oxygen, and other chemicals to form sulfuric acid and nitric acid, respectively. These acids then fall to the ground in the form of acid rain, damaging forests, lakes, and other ecosystems. Therefore, the correct answer is "Oxides of sulphur."
Tambaya 18 Rahoto
In which of the following is the oxidation number of sulphur equal to -2?
Bayanin Amsa
The oxidation number of an element is a number that represents the number of electrons that an atom has gained or lost. In the case of sulfur, it can have different oxidation numbers depending on the compound it is present in. When the oxidation number of sulfur is -2, it means that sulfur has gained two electrons. So, we need to identify the compound in which sulfur has an oxidation number of -2. Out of the given options, only the second option - H2S - contains sulfur with an oxidation number of -2. In H2S, hydrogen has an oxidation number of +1 and sulfur has an oxidation number of -2, which makes the overall charge of the compound neutral. In the other options: - S8 is a form of elemental sulfur and has an oxidation number of 0 since it is uncombined. - SO2 contains sulfur with an oxidation number of +4. - SO32- contains sulfur with an oxidation number of +4. Therefore, the correct answer is the second option - H2S.
Tambaya 19 Rahoto
The idea of planetary electron was that of
Tambaya 20 Rahoto
A decrease in atomic size and an increase in nuclear charge across a period lead to
Tambaya 21 Rahoto
The solution that will conduct the highest amount of electricity is
Tambaya 23 Rahoto
2-methylbut-2-ene has the structure
Bayanin Amsa
The structure of 2-methylbut-2-ene can be determined by identifying the location of the double bond and the position of the methyl group. The name indicates that there are four carbon atoms in the main chain, with a methyl group attached to the second carbon and the double bond between the second and third carbon. Looking at the options, it can be seen that option A and B have the double bond between the first and second carbon, which is incorrect. Option C has the correct position of the double bond, but the methyl group is attached to the first carbon instead of the second carbon. Therefore, the correct structure for 2-methylbut-2-ene is option D, where the double bond is between the second and third carbon and the methyl group is attached to the second carbon.
Tambaya 24 Rahoto
Which of the following pairs of substance will react further with oxygen to form a higher oxide?
Tambaya 25 Rahoto
The oxidation of the compound above produces
Bayanin Amsa
The compound shown in the question is 2-butanone, also known as methyl ethyl ketone. The oxidation of 2-butanone produces 2-butanal, also known as butyraldehyde. The carbonyl group (C=O) in the 2-butanone is converted to an aldehyde group (-CHO) during the oxidation reaction. Therefore, the correct answer is 2-butanal.
Tambaya 26 Rahoto
Dynamic equilibrium describes a situation where
Bayanin Amsa
Dynamic equilibrium describes a situation where both forward and reverse reactions proceed at the same rate, resulting in no net change in the concentration of reactants and products over time. This means that even though the reaction is still ongoing, the concentrations of both reactants and products remain constant. It is important to note that the reactants and products are still being converted into each other, but at the same rate. This state of balance can only be reached under certain conditions, such as constant temperature and pressure.
Tambaya 27 Rahoto
The characteristic colour of the flame shown when a copper sail moistened with concentrated hydrochloric acid on a platinum wire is put on bunsen burner is
Bayanin Amsa
When a copper salt moistened with concentrated hydrochloric acid is placed on a platinum wire and introduced into a flame, it produces a characteristic green color. This is due to the emission of photons by excited copper atoms as they return to their ground state. The photons emitted have a wavelength of about 510 nm, which corresponds to the green part of the visible spectrum. Therefore, the correct answer is green.
Tambaya 28 Rahoto
The efficiency of a solid catalyst to increase reaction rates depends on its
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Tambaya 29 Rahoto
The gas obtained as a product of anaerobic action on organic matter buried in the earth is
Bayanin Amsa
The gas obtained as a product of anaerobic action on organic matter buried in the earth is methane. Anaerobic decomposition of organic matter such as dead plants and animals, in the absence of oxygen, leads to the formation of methane gas, which is a major component of natural gas. Methane is also a greenhouse gas that contributes to climate change.
Tambaya 30 Rahoto
Nigerian crude oil is described as light crude because of its
Bayanin Amsa
Nigerian crude oil is described as light crude because of its low sulfur content. "Light crude" refers to the oil's density and refers to crude oil that has low viscosity and low specific gravity. Low sulfur content is also a characteristic of light crude oil, which makes it less polluting when burned.
Tambaya 31 Rahoto
When a crystal was added to the clear solution of its salt, the crystal did not dissolve and the solution remained unchange. This showed that the solution was
Bayanin Amsa
The observation that a crystal added to a clear solution of its salt does not dissolve and the solution remains unchanged indicates that the solution is already holding as much of the salt as it can at that particular temperature. This means that the solution is saturated with the salt. Therefore, the correct option is "saturated".
Tambaya 33 Rahoto
When 50cm2 of a saturated solution of sugar of molar mass 342 g at 40oC was evaporated to dryness, 34.2g of dry solid was obtained. The solubility of sugar at 40oC is
Bayanin Amsa
The question is asking for the solubility of sugar at 40°C given the mass of dry solid obtained after evaporating a known volume of a saturated solution of sugar. To solve this problem, we need to use the formula: Solubility = (mass of solute ÷ molar mass of solute) ÷ (volume of solvent in dm³) Using the given data, we can find the mass of sugar dissolved in 50cm³ of the saturated solution: mass of sugar = 34.2g Next, we need to calculate the number of moles of sugar dissolved: number of moles = mass of sugar ÷ molar mass of sugar = 34.2g ÷ 342g/mol = 0.1mol Finally, we can use the formula to calculate the solubility: solubility = (0.1mol ÷ 0.05dm³) = 2.0 mol dm⁻³ Therefore, the solubility of sugar at 40°C is 2.0 mol dm⁻³. The main labels for this question would be: Chemistry, Solubility, Sugar, Saturated solution, Evaporation, Molar mass, Volume, Mass.
Tambaya 35 Rahoto
The order in which a mixture of sand, ammonium chloride and sodium chloride can be separated is
Bayanin Amsa
To separate a mixture of sand, ammonium chloride, and sodium chloride, we need to use methods that take advantage of their distinct physical properties: ammonium chloride can sublime, sodium chloride is soluble in water, and sand is neither soluble in water nor sublimes.
The correct order of separation steps should be:
- Sublimation: Heat the mixture to sublime the ammonium chloride. Ammonium chloride will transition from solid to gas without passing through the liquid phase, leaving behind sand and sodium chloride.
- Dissolution: Dissolve the remaining mixture of sand and sodium chloride in water. Sodium chloride is soluble in water, while sand is not.
- Filtration: Filter the mixture to separate the insoluble sand from the sodium chloride solution.
- Evaporation: Evaporate the water from the sodium chloride solution to obtain solid sodium chloride.
Given these steps, the correct order is:
C. sublimation → dissolution → filtration → evaporation
Tambaya 36 Rahoto
What is the molar mass of a substance, if 0.4 mole of the substance has a mass of 25.0g?
Bayanin Amsa
The molar mass of a substance is the mass of one mole of the substance. We are given that 0.4 mole of the substance has a mass of 25.0 g. To find the molar mass, we can use the formula: Molar mass = Mass / Number of moles Plugging in the given values, we get: Molar mass = 25.0 g / 0.4 mol Simplifying this expression, we get: Molar mass = 62.5 g/mol Therefore, the molar mass of the substance is 62.5 g/mol. Answer: 62.5 g
Tambaya 37 Rahoto
In the electrolysis of CuSO4(g) using platinum electrodes, the reaction at the anode is
Bayanin Amsa
Tambaya 38 Rahoto
Na2CO3(g) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
How many moles of sodium trioxocarbonate (IV) are there in a 25 cm3 solution which required 10cm3 of 0.05 mol dm-3 hydrochloric acid solution to neutralize it?
Bayanin Amsa
CAVACBVB=ab
0.05×10CB×25=21
50CB = 0.5
CB = 0.550=5100
= 0.01
Tambaya 39 Rahoto
The ion that will give a white precipitate with aqueous NaOH and soluble in excess of the base is
Tambaya 40 Rahoto
Carbon exist naturally in the combined state as
Bayanin Amsa
Carbon exists naturally in the combined state as all of the options provided: graphite, coal, diamond, and wood. Graphite is a form of carbon that is used in pencils, while coal is a fossil fuel that is formed from the remains of ancient plants and animals. Diamond is a precious gemstone made entirely of carbon atoms arranged in a crystal lattice structure, while wood is a natural composite material made up of cellulose, hemicellulose, and lignin, all of which contain carbon. Therefore, the correct answer is that carbon exists naturally in the combined state as graphite, coal, diamond, and wood.
Tambaya 41 Rahoto
If the heat of combustion of hydrogen is -285.8kJ,
What is the heat of formation of water?
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Tambaya 42 Rahoto
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
In the reaction above, the mass of carbon(IV)oxide produced on burning 78 g of ethyne is
[C =12, O = 16, H = 1]
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Tambaya 43 Rahoto
The element with the electron configuration 1s22s22p3 is
Bayanin Amsa
The electron configuration provided in the question indicates that the element has two electrons in its 1s orbital, two electrons in its 2s orbital, two electrons in its 2p orbital, and three electrons in its 3p orbital. Based on this electron configuration, the element is "nitrogen."
Tambaya 44 Rahoto
The volume of 2.5 moles of oxygen at s.t.p is
[Molar volume of a gas at s.t.p = 22.4 dm3]
Bayanin Amsa
The question is asking for the volume of 2.5 moles of oxygen at standard temperature and pressure (s.t.p), given that the molar volume of a gas at s.t.p is 22.4 dm^3. To solve this problem, we can use the following formula: Volume = (number of moles) x (molar volume at s.t.p) Substituting the given values, we get: Volume = 2.5 mol x 22.4 dm^3/mol Volume = 56.0 dm^3 Therefore, the volume of 2.5 moles of oxygen at s.t.p is 56.0 dm^3. So, the answer is the last option: 56.0 dm^3.
Tambaya 45 Rahoto
H2(g) + Br2(g) → 2HBr(g)
The above reaction is carried out at 25oC. ΔH is -72 kJ mol-1 and ΔS is - 106 J mol-1K-1, the reaction will
Tambaya 46 Rahoto
A particle that contains 11 protons, 12 neutrons and 10 electrons is probably a
Bayanin Amsa
The particle described in the question has 11 protons, which makes it an atom of the element sodium (Na) since the atomic number of sodium is 11. The particle also has 12 neutrons, which means it has a mass number of 23 (11 protons + 12 neutrons = 23). However, the particle has 10 electrons, which is not the same as the number of protons. Therefore, the particle has a charge of +1 since it has one less electron than protons. This means that the particle is a metallic ion. So, the correct option is "metallic ion".
Tambaya 47 Rahoto
The raw materials used for the smelting of iron ore in a furnace is
Tambaya 48 Rahoto
Nuclear fission involves the splitting of a heavy element into two nuclei of
Bayanin Amsa
Nuclear fission is a process in which the nucleus of a heavy atom is split into two nuclei of smaller atoms. In this process, a large amount of energy is released, which can be harnessed for various applications. When the nucleus is split, the two resulting nuclei are not of nearly equal masses, but rather of unequal masses. Additionally, the process results in the release of energy and radiation. Therefore, the correct option is "unequal masses with a release of energy and radiation."
Tambaya 49 Rahoto
The group IA metals are not found free in nature because they
Bayanin Amsa
The question is asking why the group IA metals are not found free in nature. The correct answer is "are very reactive and unstable." This means that the group IA metals, such as sodium and potassium, react very easily with other elements in the environment and are not stable enough to exist on their own. They are typically found in combination with other elements, such as sodium chloride (table salt), rather than in their pure, free form. The other options are not correct because they do not explain why the group IA metals are not found free in nature.
