JAMB UTME - Chemistry - 2022
Question 1 Report
An organic compound contains 69% carbon, 15.3% hydrogen and 30.7% oxygen. Calculate the the empirical formula [C=12, H = 1, O = 16]
Answer Details
Question 2 Report
The sub-atomic particles located in the nucleus of an atom are?
Answer Details
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.
Question 3 Report
An organic compound decolourized acidified KMnO4 solution but failed to react with ammonical AgNO3 solution. The organic compound is likely?
Answer Details
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.
Question 4 Report
Which of the following conducts electricity
Answer Details
Graphite is the option that conducts electricity.
Question 6 Report
2H2 + O2 → 2H2 O
From the equation above, calculate the volume of unreacted oxygen gas if a mixture of 50cm3 of hydroden and 75cm3 of oxygen are involved
Answer Details
The balanced chemical equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor. Therefore, the stoichiometric ratio of hydrogen to oxygen is 2:1. In this problem, there are 50cm3 of hydrogen gas and 75cm3 of oxygen gas. Since the gases are at the same temperature and pressure, their volumes are directly proportional to the number of moles of gas present. Using the stoichiometric ratio, we can calculate that the amount of oxygen gas required to react completely with 50cm3 of hydrogen gas is (1/2) * 50cm3 = 25cm3. Since there are 75cm3 of oxygen gas present, there must be (75cm3 - 25cm3) = 50cm3 of unreacted oxygen gas remaining. Therefore, the volume of unreacted oxygen gas is 50cm3. Answer: 50cm3
Question 7 Report
The IUPAC nomenclature of the structure is
Answer Details
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".
Question 8 Report
Which of the following substances is not a homogeneous mixture?
Answer Details
The substance that is not a homogeneous mixture is flood water. Flood water is typically a mixture of various substances, such as sediment, dirt, debris, and organic matter, that have been carried along by the water. As such, flood water is usually a heterogeneous mixture, meaning that it does not have a uniform composition throughout. In contrast, filtered sea water, soft drinks, and writing ink are all examples of homogeneous mixtures, where the components are evenly distributed and the mixture has a uniform composition throughout.
Question 9 Report
Which of the following will precipitate in dil. HCl
Answer Details
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.
Question 10 Report
Ethene, when passed into concentrated H2SO4, is rapidly absorbed. The product is diluted with water and then warmed to produce
Answer Details
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.
Question 11 Report
The function of sulphur during the vulcanization of rubber is to
Answer Details
The function of sulphur during the vulcanization of rubber is to form chains which bind rubber molecules together.
Question 12 Report
Which of the following will act as both oxidizing agents and reducing agents?
Answer Details
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.
Question 13 Report
On the basis of the electrochemical series, which of these ions will show the greater tendency to be discharged at the cathode in an electrolytic cell
Answer Details
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.
Question 14 Report
An organic compound which liberate carbon(iv)oxide from trioxocarbonate(iv) solution is likely to be?
Answer Details
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.
Question 15 Report
If the volume of a given mass of a gas at 0ºc is 29.5cm3 . What will be the volume of the gas at 15ºc, given that the pressure remains constant.
Question 16 Report
A piece of radioactive element has initially 8.0×10^22 atoms. The half life of two days after 16 days the number of atom is
Question 17 Report
Answer Details
The addition of charcoal to the filter bed of sand during water treatment for township supply is to remove odors and improve the taste of the water. Charcoal is a porous material that can adsorb impurities and chemicals from the water, such as dissolved organic matter that can contribute to unpleasant tastes and odors. This process helps to produce a better-quality drinking water that is free from unpleasant tastes and odors. It should be noted that while the addition of charcoal can help remove impurities, it does not kill germs or prevent tooth decay or goiter. Other water treatment methods, such as disinfection with chlorine or ultraviolet light, are required to kill harmful microorganisms and ensure the safety of the drinking water.
Question 18 Report
Hard water is water with high concentrations of dissolved ions, in particular calcium and
Answer Details
Hard water is water that contains high amounts of dissolved minerals, specifically calcium and magnesium ions. These minerals come from the rocks and soil that the water flows through and can accumulate in the water as it travels to your home. When you use hard water, it can leave mineral deposits on your pipes, fixtures, and appliances, which can reduce their efficiency and lifespan. It can also make soap less effective and leave your skin feeling dry and itchy. Therefore, it is important to treat hard water if it is a problem in your area.
Question 19 Report
2-methylprop-1-ene is a structural isomer of?
Question 20 Report
Mixing aqueos solution of barium hydroxide and sodium tetraoxocarbonate (iv) yields a white precipitate of
Answer Details
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.
Question 21 Report
An organic functional group which can likely decolorize ammoniacal silver nitrate is?
Answer Details
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.
Question 22 Report
A sample of gas exerts a pressure of 8.2 atm when confined in a 2.93 dm3 container at 20c. The number of moles of gas in the sample is
Question 23 Report
An organic compound with fishy smell is likely to have a general formula?
Answer Details
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.
Question 24 Report
Complete dehydrogenation of ethyne yields
Question 25 Report
Which of the following is stable to heat
Answer Details
Out of the given options, K2CO3 is stable to heat.
Question 26 Report
Chlorine is a common bleaching agent. This is not true with
Answer Details
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.
Question 27 Report
In the preparation of salts, the method employed will depend on the?
Answer Details
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.
Question 28 Report
Copper (II) tetraoxosulphate (IV) is widely used as
Answer Details
Copper (II) tetraoxosulphate (IV), also known as copper sulfate or CuSO4, is widely used as a fungicide and a disinfectant. As a fungicide, copper sulfate is effective in controlling fungal diseases in plants, including mildew, leaf spots, and blights. It is also used as a fungicide in swimming pools to prevent the growth of algae. As a disinfectant, copper sulfate is effective in killing bacteria and viruses. It is used in a variety of applications, including in the production of animal feed, as a preservative for wood, and in water treatment to kill bacteria and algae. While copper sulfate has been used as a fertilizer in the past, its use in this capacity has largely been replaced by other compounds. It is not commonly used as a purifier.
Question 29 Report
When marble is heated to 1473K, another whiter solid is obtained which reacts vigoriously with water to give an alkaline solution. The solution contains
Answer Details
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).
Question 30 Report
Zn + 2HCL → ZnCl2 + H2
What happens to zinc in the above reaction?
Answer Details
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."
Question 31 Report
N2 O4 ? 2NO2 (? = -ve)
From the reaction above, which of these conditions would produce the highest equilibrium yield for N2 O4 ?
Answer Details
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.
Question 32 Report
H+ + OH− → H2 O
The equation above illustrates
Question 33 Report
Wrought iron is obtained by heating cast iron in a furnace with?
Answer Details
Wrought iron is a type of iron that is very malleable and ductile, meaning it can be easily shaped and formed into various objects. It is obtained by heating cast iron in a furnace with haematite, also known as iron(III) oxide. When cast iron is heated with haematite in a furnace, a chemical reaction takes place where the haematite reacts with the carbon in the cast iron to produce carbon dioxide gas. This reaction also produces molten iron, which is then further heated to remove any impurities like sulfur and phosphorus. This molten iron is then poured into molds to form ingots of wrought iron. Therefore, haematite is essential in the process of obtaining wrought iron from cast iron.
Question 34 Report
H2SO4 is used to remove rust on the surface of iron (picking) before electroplating. The type of reaction involved is
Answer Details
The type of reaction involved when using H2SO4 to remove rust on the surface of iron is a redox reaction. This is because the sulfuric acid oxidizes the iron in the rust, converting it into iron(II) sulfate, while the acid itself is reduced to sulfur dioxide. The overall reaction can be written as follows: Fe2O3(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3H2O(l) In this reaction, the iron in Fe2O3 is oxidized from a +3 to a +2 oxidation state, while the sulfur in H2SO4 is reduced from a +6 to a +4 oxidation state. This transfer of electrons between the reactants is what defines a redox reaction.
Question 35 Report
Which of the following roles does sodium chloride play in preparation? It
Answer Details
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.
Question 36 Report
6g of Mg was to 100cm3 of 1 moldm3 H2 SO4 . What mass of Mg remained undissolved? (Mg = 24)
Answer Details
The balanced chemical equation for the reaction between magnesium (Mg) and sulfuric acid (H2SO4) is: Mg + H2SO4 -> MgSO4 + H2 According to the equation, one mole of Mg reacts with one mole of H2SO4 to produce one mole of magnesium sulfate (MgSO4) and one mole of hydrogen gas (H2). Since the concentration of the sulfuric acid is 1 moldm3, this means that there is one mole of H2SO4 in every 1 liter (1000 cm3) of solution. To determine the amount of Mg that reacts with the H2SO4, we need to use stoichiometry. One mole of Mg reacts with one mole of H2SO4, so the amount of Mg that reacts with 1 moldm3 of H2SO4 is given by: 6g / 24g/mol = 0.25 mol Since the reaction is 1:1, this means that 0.25 mol of H2SO4 is consumed in the reaction. The volume of the solution is 100cm3 (0.1 dm3), so the amount of H2SO4 in the solution is: 1 mol/dm3 x 0.1 dm3 = 0.1 mol The amount of H2SO4 that remains after the reaction is: 0.1 mol - 0.25 mol = -0.15 mol This negative value means that all of the H2SO4 was consumed in the reaction, and there is excess Mg left over. The mass of Mg that remains undissolved is given by: 0.15 mol x 24g/mol = 3.6g Therefore, the correct answer is 3.6g.
Question 37 Report
A certain volume of gas at 298k is heated such that its volume and pressure are now four times the original values. What is the new temperature?
Answer Details
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.
Question 38 Report
30 cm3 of oxygen at 10 atmosphere pressure is placed in a 20 dm3 container. Calculate the new pressure if the temperature is kept constant.
Answer Details
Given:
- Initial pressure, 𝑃1=10P1=10 atmospheres
- Initial volume, 𝑉1=30 cm3V1=30cm3
- Final volume, 𝑉2=20 dm3V2=20dm3
First, convert all volumes to the same units. Since 1 dm3dm3 is 1000 cm3cm3:
𝑉2=20 dm3=20×1000 cm3=20000 cm3V2=20dm3=20×1000cm3=20000cm3
Now, using Boyle's Law:
𝑃1𝑉1=𝑃2𝑉2P1V1=P2V2
Substitute the known values into the equation:
10×30=𝑃2×2000010×30=P2×20000
300=𝑃2×20000300=P2×20000
Solve for 𝑃2P2:
𝑃2=30020000P2=20000300
𝑃2=0.015 atmospheresP2=0.015atmospheres
Therefore, the new pressure if the temperature is kept constant is:
0.015 atmospheres0.015atmospheres
Question 39 Report
The sulphide that is commonly used in coating electric fluorescent tubes is?
Answer Details
The sulphide commonly used in coating electric fluorescent tubes is Zinc Sulphide. Zinc Sulphide is a type of material that glows when it is exposed to ultraviolet light. When ultraviolet light is generated inside a fluorescent tube, it excites the Zinc Sulphide particles, causing them to emit visible light. This visible light is what we see as the bright light coming from the tube. So, Zinc Sulphide acts as a phosphor and helps in producing the bright light in fluorescent tubes.
Question 40 Report
The IUPAC name for CICH2-CH2-CH2-OH is
Answer Details
The IUPAC name for CICH2-CH2-CH2-OH is 3-chloropropan-1-ol. To name the compound using the IUPAC nomenclature system, we start by identifying the longest continuous chain of carbon atoms that contains the functional group (-OH). In this case, the longest chain contains three carbon atoms, so the root name is propane. Next, we identify the position of the substituent (-Cl) on the chain. The substituent is attached to the third carbon atom in the chain, so the name of the compound becomes 3-chloropropane. Finally, we add the suffix -ol to indicate that the compound contains an alcohol functional group (-OH), so the complete name of the compound is 3-chloropropan-1-ol. Therefore, the correct answer is 3-chloropropan-1-ol.
