JAMB UTME - Chemistry - 2022

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An acidic oxide is an oxide that reacts with water to form an acidic solution. Non-metals have a greater tendency to form acidic oxides than metals. Therefore, among the given options, the non-metal that does not form an acidic oxide with oxygen would be the one that does not react with water to form an acidic solution. Out of the given options, chlorine is the non-metal that does not form acidic oxides with oxygen. Chlorine reacts with oxygen to form a number of oxides such as chlorine monoxide (Cl2O), chlorine dioxide (ClO2), and chlorine trioxide (ClO3), but none of these oxides react with water to form an acidic solution. Instead, they react with water to form oxyacids or oxoacids such as hypochlorous acid (HClO), chlorous acid (HClO2), and chloric acid (HClO3), which are stronger acids than the oxides. Therefore, the correct answer is chlorine.

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The electrochemical series is a list of metals and ions arranged in order of their decreasing tendency to lose or gain electrons, and thus, their ability to act as reducing or oxidizing agents. The higher the position of a metal or ion in the electrochemical series, the greater its tendency to lose electrons and undergo oxidation, while the lower its position, the greater its tendency to gain electrons and undergo reduction. In an electrolytic cell, the cathode is the electrode where reduction occurs, meaning that cations (positively charged ions) are attracted and gain electrons to form neutral atoms or molecules. Based on the electrochemical series, the ion with the higher position in the series will have a greater tendency to gain electrons and be discharged at the cathode, while the ion with the lower position will have a lower tendency and may not be discharged at all. Among the given options, the electrochemical series order is: Cu2+ > Sn2+ > Fe2+ > Zn2+ Therefore, Cu2+ has the highest tendency to be discharged at the cathode and undergo reduction, while Zn2+ has the lowest tendency. So, in an electrolytic cell, Cu2+ will be discharged at the cathode, while Zn2+ may not be discharged at all, depending on the conditions of the cell.

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The organic compound that liberates carbon(iv)oxide from trioxocarbonate(iv) solution is CH3COOH (acetic acid). When acetic acid is added to a solution of trioxocarbonate(iv) (carbonate) it reacts to form carbon(iv)oxide gas, water and a salt. The balanced chemical equation for the reaction is: 2CH3COOH + Na2CO3 → CO2 + 2H2O + 2NaCH3COO The carbon(iv)oxide gas is released as bubbles, causing the solution to fizz. Therefore, CH3COOH is the organic compound that liberates carbon(iv)oxide from trioxocarbonate(iv) solution.

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Among the given options, only CuS will precipitate in dilute HCl. CuS is insoluble in dilute HCl, and hence it will precipitate when added to dilute HCl. However, the other options will dissolve in dilute HCl, and hence they will not precipitate. ZnS will dissolve in dilute HCl to form ZnCl2 and H2S. Na2S will react with dilute HCl to produce H2S and NaCl. FeS will dissolve in dilute HCl to form FeCl2 and H2S. Therefore, the correct answer is (4) CuS.

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The sub-atomic particles located in the nucleus of an atom are neutron and proton. The nucleus is the dense core of an atom that contains most of its mass. Protons are positively charged particles found in the nucleus, and they determine the atomic number of the element. Neutrons are neutral particles found in the nucleus, and they help stabilize the nucleus by balancing the repulsive forces between the positively charged protons. Electrons, on the other hand, are negatively charged particles that are located outside the nucleus in energy levels or shells. They are attracted to the positively charged nucleus by electrostatic forces and are involved in chemical bonding between atoms. The number of protons in the nucleus determines the identity of the element, while the number of neutrons determines its isotopes. Isotopes of an element have the same number of protons but different numbers of neutrons in the nucleus. In summary, the two sub-atomic particles located in the nucleus of an atom are neutron and proton.

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Heavy water (D2O) is used as a moderator to control nuclear fission. A moderator is a substance that is used to slow down the neutrons produced in a nuclear reaction, making them more likely to be captured by the fuel nuclei and causing further fission. Heavy water is a type of water that contains a larger amount of the isotope deuterium (D) than regular water. Deuterium has an extra neutron compared to the more common hydrogen isotope, and this makes heavy water more effective at slowing down neutrons than regular water. Lead, iron, and chromium are not typically used as moderators in nuclear reactors. Lead can be used as a shield to absorb radiation, while iron and chromium are used in the construction of the reactor vessel and other components.

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The method employed in the preparation of salts will depend on the composition of the salt. Different salts have different chemical properties, and the method used to prepare them will depend on these properties. For example, some salts can be easily dissolved in water, while others are not very soluble and may require the use of a different solvent or special conditions to dissolve. The dissociating ability, stability to heat, and precipitating ability of the salt may also play a role in determining the preparation method, but the most important factor is the composition of the salt.

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The highest equilibrium yield of N2O4 would be produced at low temperature and low pressure. In a chemical reaction, the position of the equilibrium can be influenced by changing the temperature or pressure. A decrease in temperature or an increase in pressure favors the side of the reaction with the fewer moles of gas (in this case, N2O4). This means that, if the temperature is low and the pressure is low, there will be more N2O4 at equilibrium, as the reaction will shift to the right to counteract the reduction in the concentration of N2O4. So, low temperature and low pressure would produce the highest equilibrium yield of N2O4.

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When ethene is passed into concentrated H2SO4, it undergoes electrophilic addition reaction to form ethyl hydrogen sulfate as the product. The reaction mixture is then diluted with water and warmed to produce ethanol as the main product. Therefore, the answer is ethanol.

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The process of converting crude petroleum into useful products is known as fractional distillation. Crude petroleum is a mixture of different hydrocarbons, and fractional distillation separates these hydrocarbons based on their boiling points. During the process of fractional distillation, crude petroleum is heated to a high temperature, and the resulting vapors are passed through a tower called a fractionating column. This column contains a series of trays, and each tray contains a specific temperature range. As the vapors rise up the column, they cool and condense into liquids on the tray with a temperature that matches their boiling point. The liquids are then collected and further refined into useful products like gasoline, diesel, jet fuel, and heating oil. Fractional distillation is an important process because it allows us to separate and purify the different components of crude petroleum, which have different properties and uses. For example, gasoline has a lower boiling point and is more volatile than diesel fuel, which makes it ideal for use in cars. By separating these components, we can create products that meet specific needs and requirements.

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The given information suggests that the organic compound is an unsaturated compound (because it decolorized the acidified KMnO4 solution), but it does not contain a functional group that reacts with ammonical AgNO3 solution. Therefore, the likely organic compound is an alkene or an alkyne. Carboxylic acids can also react with acidified KMnO4 solution, but they would also react with ammonical AgNO3 solution to form a silver carboxylate salt. Alkanes are saturated compounds and do not react with either reagent, so they would not decolorize the acidified KMnO4 solution. Therefore, based on the given information, the most likely option is either an alkene or an alkyne.

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The compound that is NOT correctly formed when the parent metal is heated in air is: tri-iron tetraoxide (Fe2O). This is because the correct compound formed from the heating of iron in air is iron (III) oxide or Fe2O3. The formula for tri-iron tetraoxide is incorrect, as it implies that there are only three iron atoms in the compound when there should be four.

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We can use the ideal gas law to solve this problem, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in kelvin. If the volume and pressure are both increased by a factor of 4, then the new volume V' and new pressure P' are given by: V' = 4V P' = 4P Substituting these values into the ideal gas law, we get: (4P)(4V) = nR(T') Simplifying this equation, we get: 16PV = nRT' Dividing both sides by PV, we get: 16 = nRT' / PV Since n, R, and P are constant, we can simplify this to: 16 = T' / T Solving for T', we get: T' = 16T Therefore, the new temperature is 16 times the original temperature. Substituting T = 298 K, we get: T' = 16 x 298 K = 4768 K So the correct answer is 4768.0K.

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The role that sodium chloride (NaCl) plays in soap preparation is to separate soap from glycerol. When fats or oils are hydrolyzed with an alkali, such as sodium hydroxide (NaOH), the result is a mixture of soap and glycerol. Adding NaCl to this mixture helps to induce the precipitation of the soap, allowing it to be separated from the glycerol. This process is known as "salting out" and is used to purify the soap and remove impurities. Sodium chloride does not react with glycerol or accelerate the decomposition of fat and oil. Also, it does not convert the fatty acid to its sodium salt as this conversion is done by the alkali (such as NaOH) during the saponification process.

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The oxidizing and reducing properties of a substance depend on its ability to gain or lose electrons. A substance that can gain electrons acts as an oxidizing agent, while a substance that can lose electrons acts as a reducing agent. Among the given options, both Cl2 (chlorine gas) and SO2 (sulfur dioxide) can act as both oxidizing and reducing agents depending on the reaction conditions. - Cl2 can act as an oxidizing agent when it gains electrons to form Cl- ions, and it can act as a reducing agent when it loses electrons to form Cl+ ions. For example, in the reaction Cl2 + 2KBr → 2KCl + Br2, chlorine gas is acting as an oxidizing agent since it is gaining electrons from bromide ions to form bromine gas. However, in the reaction 2Cl- + Cl2 → 2Cl2-, chlorine gas is acting as a reducing agent since it is losing electrons to form chloride ions. - SO2 can act as an oxidizing agent when it gains electrons to form sulfite ions (SO32-), and it can act as a reducing agent when it loses electrons to form sulfur trioxide (SO3). For example, in the reaction SO2 + 2H2S → 3S + 2H2O, sulfur dioxide is acting as a reducing agent since it is losing electrons to form elemental sulfur. However, in the reaction 2SO32- + O2 → 2SO42-, sulfur dioxide is acting as an oxidizing agent since it is gaining electrons to form sulfate ions. H2S (hydrogen sulfide) and NH3 (ammonia) are not likely to act as both oxidizing and reducing agents under normal conditions. H2S tends to act as a reducing agent by donating electrons to oxidizing agents, while NH3 tends to act as a reducing agent by donating electrons to oxidizing agents or as a base by accepting protons.

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Chlorine is not a common bleaching agent for wet litmus paper, wet pawpaw leaf, and most wet fabric dyes. It is commonly used as a bleaching agent for printer's ink.

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The organic compound with a fishy smell is most likely to have the general formula RNH2, which represents a primary amine. Amines are organic compounds that contain a nitrogen atom bonded to one or more carbon atoms. Primary amines have one alkyl or aryl group and two hydrogen atoms bonded to the nitrogen atom. Some primary amines have a fishy smell, which is caused by the presence of volatile amines. These amines are small molecules that can easily evaporate and have a strong odor, similar to that of fish. Examples of compounds that have a fishy smell include trimethylamine, which is found in fish, and butylamine, which is used in the production of rubber and pharmaceuticals. In summary, the organic compound with a fishy smell is likely to have the general formula RNH2, which represents a primary amine.

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The reason why calcium forms complexes with ammonia is that it has empty d-orbitals.

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The organic functional group that can likely decolorize ammoniacal silver nitrate is an alkyne. When ammoniacal silver nitrate is added to a solution containing an alkyne functional group, a white or yellowish precipitate of silver acetylide is formed. Silver acetylide is a highly explosive compound and is sparingly soluble in water, causing it to appear as a white or yellowish solid precipitate. This reaction is used as a test to detect the presence of an alkyne functional group in an organic compound. In contrast, alkanes, alkenes, and alkanols do not react with ammoniacal silver nitrate, so they cannot decolorize it. Therefore, an organic functional group that can likely decolorize ammoniacal silver nitrate is an alkyne.

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The function of sulphur during the vulcanization of rubber is to form chains which bind rubber molecules together.

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The compound that exhibits resonance is benzene.

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When air is passed through alkaline pyrogallol, the oxygen in the air is absorbed by the pyrogallol, resulting in an increase in the weight of the pyrogallol. The other gases in air, namely nitrogen, neon, and argon, do not react with pyrogallol under these conditions. Therefore, the answer is oxygen.

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To determine the oxidation number of oxygen in BaO2, we can use the fact that the overall charge of a compound must be zero. Barium (Ba) is a Group 2 element and has an oxidation state of +2. The compound BaO2 has no net charge, so the sum of the oxidation states of all the atoms must be zero. Let x be the oxidation state of oxygen in BaO2. Therefore, we have: (+2) + 2(x) = 0 Solving for x, we get: x = -1 Therefore, the oxidation number of oxygen in BaO2 is -1.

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In the above reaction, zinc (Zn) reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl2) and hydrogen gas (H2). The chemical equation for the reaction is: Zn + 2HCl → ZnCl2 + H2 During the reaction, zinc atoms lose two electrons each and get oxidized to form positively charged zinc ions (Zn2+), as they react with the hydrogen ions (H+) from the hydrochloric acid to form zinc chloride. The hydrogen ions, on the other hand, gain an electron each and get reduced to form hydrogen gas molecules (H2). Therefore, in the given reaction, zinc is getting oxidized, as it loses electrons and forms a positively charged ion. Hence, the correct option is "oxidized."

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When heat is absorbed during a chemical reaction, the reaction is said to be endothermic. Endothermic reactions are characterized by the absorption of heat energy from the surroundings. In other words, the reactants absorb energy from the environment, usually in the form of heat, to form the products. As a result, the temperature of the surroundings decreases, and the reaction feels cold to the touch. Endothermic reactions can be found in many natural processes, such as photosynthesis, melting of ice, and the evaporation of liquids. These processes require energy to occur, and they absorb heat from the surroundings to power the reaction.

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The white solid obtained when marble (calcium carbonate, CaCO3) is heated to 1473K is calcium oxide (CaO), also known as quicklime. When quicklime reacts vigorously with water, it forms calcium hydroxide (Ca(OH)2), which is an alkaline solution. Therefore, the solution obtained from the reaction of quicklime with water contains calcium hydroxide (Ca(OH)2).

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Mixing aqueous solutions of barium hydroxide and sodium tetraoxocarbonate (IV) would result in a chemical reaction that produces a white precipitate of barium tetraoxocarbonate (IV). The balanced chemical equation for this reaction is: Ba(OH)2(aq) + Na2CO3(aq) → BaCO3(s) + 2NaOH(aq) In the above equation, the barium hydroxide (Ba(OH)2) reacts with sodium tetraoxocarbonate (IV) (Na2CO3) to form barium tetraoxocarbonate (IV) (BaCO3), which is a white precipitate, and sodium hydroxide (NaOH). Therefore, the correct option is 4) Barium tetraoxocarbonate.

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The correct answer is: "melting does not break the metallic bond but boiling does." The metallic bond is the force of attraction between metal atoms, which holds them together to form a solid. When a metal is heated, its temperature increases, and at a certain point, the energy provided by the heat is enough to overcome the metallic bond and cause the metal to melt. However, even in the liquid state, the metallic bond remains intact, which is why metals have a very high melting point. On the other hand, when the temperature is further increased, the energy provided by the heat becomes enough to break the metallic bond, and the metal atoms become completely detached from one another. This results in the metal boiling and turning into a gas. Because the metallic bond is much stronger than other types of intermolecular forces, such as van der Waals forces, it requires a lot of energy to break, resulting in a large temperature interval between the melting point and boiling point of metal.

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The IUPAC nomenclature of the structure is "2-chloro-2-methylbutane". The name is derived by first identifying the longest carbon chain, which in this case contains four carbon atoms (butane). The carbon chain is numbered from one end to the other, giving the substituents the lowest possible numbers. Starting from either end, we can see that the first carbon atom has a chlorine atom attached to it, which is represented by the prefix "chloro-". Moving along the chain, the second carbon atom has a methyl group attached to it, which is represented by the prefix "methyl-". Since the substituents are in the second position from each other, we use the prefix "di-" to indicate two substituents in this position. Finally, we use the suffix "-ane" to indicate that the molecule is an alkane. Therefore, the correct name for this molecule is "2-chloro-2-methylbutane".