JAMB UTME - Chemistry - 2013
Pergunta 2 Relatório
The constituent of air that acts as a diluent is
Detalhes da Resposta
The constituent of air that acts as a diluent is nitrogen. Diluent refers to a substance that is added to a gas or liquid to reduce its concentration. In the case of air, nitrogen is the most abundant gas and makes up about 78% of the air we breathe. The other gases present in air such as oxygen, carbon dioxide, and noble gases make up the remaining 22%. Nitrogen, being an inert gas, does not react with other elements, which makes it an ideal diluent in air. It helps to lower the concentration of reactive gases, making it easier for us to breathe and for combustion reactions to occur safely.
Pergunta 3 Relatório
the minimum amount of energy required for a reaction to take place is
Detalhes da Resposta
The minimum amount of energy required for a reaction to take place is called the activation energy. Activation energy is the energy required to initiate a chemical reaction by breaking the bonds in the reactant molecules. It is the energy barrier that must be overcome for the reaction to occur. Once the activation energy is surpassed, the reaction will proceed on its own without the need for additional energy input. The activation energy is often provided in the form of heat or light. Lattice energy refers to the energy released when ions combine to form a crystal lattice structure. Ionization energy is the energy required to remove an electron from an atom or ion. Kinetic energy is the energy of motion. None of these terms are directly related to the minimum amount of energy required for a reaction to take place.
Pergunta 4 Relatório
The shape of the carbon (lV) oxide molecule is
Detalhes da Resposta
The shape of the carbon dioxide (CO2) molecule is linear. This is because the molecule consists of two oxygen atoms bonded to a central carbon atom in a straight line. The bond angles in CO2 are all 180 degrees, resulting in a linear shape. This shape is determined by the electron geometry and molecular geometry of the molecule, which are both linear.
Pergunta 5 Relatório
In a water body, too much sewage leads to
Detalhes da Resposta
When there is too much sewage in a water body, it can lead to an increase in the bacterial population. This is because sewage contains organic matter and nutrients that can provide food for bacteria to grow. However, as the bacteria grow, they use up oxygen from the water. If there is too much bacteria, it can lead to a decrease in the level of oxygen in the water. This is called "eutrophication." A decrease in oxygen levels in water can be harmful for aquatic animals. Many fish and other aquatic creatures need a certain level of oxygen to survive, and if there's not enough, they can become sick or die. This can also have a ripple effect throughout the ecosystem, as the loss of certain species can affect the food chain and balance of the environment.
Pergunta 6 Relatório
Detalhes da Resposta
Copper is displaced from the solution of its salts by most metals because it is at the bottom of the activity series. The activity series is a list of metals arranged in order of their decreasing chemical reactivity. Metals at the top of the activity series are highly reactive and can displace metals below them from their solutions. Copper, on the other hand, is at the bottom of the activity series and is less reactive than most other metals, so it can be displaced by them from its salts. Therefore, when a more reactive metal is added to a solution containing copper ions, the more reactive metal will displace copper ions from the solution, forming its own ions and causing copper to precipitate out of the solution as a solid.
Pergunta 7 Relatório
Detalhes da Resposta
The presence of an impurity in a substance will reduce its melting point. The melting point of a pure substance is a fixed temperature at which the solid and liquid phases coexist in equilibrium. However, when an impurity is added to the substance, the impurity molecules become incorporated into the crystal lattice of the substance. This incorporation of impurity molecules disrupts the regular arrangement of the substance's molecules, making it more difficult for the substance to maintain its crystal structure as the temperature increases. As a result, the melting point of the substance is lowered, and it melts at a lower temperature than the pure substance. This is why impurities are often added to substances like ice on roads to reduce their melting point and make them melt more quickly.
Pergunta 8 Relatório
Which of the following fraction is used as raw material for the cracking process?
Detalhes da Resposta
The fraction used as raw material for the cracking process is typically a heavier fraction of crude oil, such as lubricating oil or diesel oils. In the cracking process, these heavier fractions are broken down into lighter fractions, such as gasoline and kerosene. The goal of cracking is to produce more valuable, in-demand products from the heavier, less valuable fractions of crude oil.
Pergunta 9 Relatório
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Pergunta 10 Relatório
Choose the correct option from the graph above.
3Fe(S) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g). The equilibrium constant, K, of the reaction above is represented as
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Pergunta 11 Relatório
The bond formed between two elements with electron configurations 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4 is
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Pergunta 12 Relatório
The rate of diffusion of a gas Y is twice that of Z If the relative molecular mass of Y is 64 and the two gases diffuse under the same conditions, find the relative molecular mass of Z
Detalhes da Resposta
√y
= 2x
√Z
= x
rzry
= √mymz
2xx
= √64mz
41
= 64mz
mz = 644
mz = 16
Pergunta 13 Relatório
A primary amide is generally represented by the formula
Detalhes da Resposta
A primary amide is generally represented by the formula RCONH2. This means that it is a compound made up of an R group (which can be any alkyl or aryl group) attached to a carbon atom, which is then connected to a nitrogen atom by a double bond, and that nitrogen atom is also bonded to two hydrogen atoms. This arrangement of atoms is characteristic of a primary amide, which is an organic compound that contains the functional group -CONH2. The other options listed are not primary amides. RCOOR is an ester, RCONHR is a secondary amide (with one hydrogen atom replaced by an R group), and RCONR2 is a tertiary amide (with both hydrogen atoms replaced by R groups).
Pergunta 15 Relatório
In an endothermic reaction, if there is a loss in entropy the reaction will
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Pergunta 16 Relatório
Which of the following substances is not a salt?
Detalhes da Resposta
The substance that is not a salt among the options given is Aluminium oxide. A salt is a type of chemical compound that is formed when an acid reacts with a base. It is typically made up of a metal cation and a non-metal anion. Sodium hydrogentrioxosulphate (V), sodium trioxocarbonate (V), and zinc chloride are all examples of salts. Aluminium oxide, on the other hand, is not formed from the reaction of an acid and a base. It is an oxide of the metal aluminium and is therefore not a salt. In summary, of the substances listed, Aluminium oxide is not a salt, while sodium hydrogentrioxosulphate (V), sodium trioxocarbonate (V), and zinc chloride are all examples of salts.
Pergunta 17 Relatório
If 24.4 g of lead (ll) trioxonitrate (V) were dissolved in 42 g of distilled water at 20oC; calculate the solubility of the solute in gdm-3.
Detalhes da Resposta
To calculate the solubility of lead (II) trioxonitrate (V) in water, we need to use the formula: Solubility (g/dm³) = (mass of solute ÷ volume of solvent) First, let's convert the temperature to Kelvin (K) using the formula K = °C + 273.15: 20°C + 273.15 = 293.15 K Next, we need to calculate the volume of water that the lead (II) trioxonitrate (V) was dissolved in. We can assume that the density of water is 1 g/cm³, so: Volume of water = mass of water ÷ density of water Volume of water = 42 g ÷ 1 g/cm³ Volume of water = 42 cm³ Now, we can calculate the solubility of lead (II) trioxonitrate (V) in water: Solubility (g/dm³) = (mass of solute ÷ volume of solvent) Solubility (g/dm³) = 24.4 g ÷ (42 cm³ ÷ 1000) Solubility (g/dm³) = 581 g/dm³ Therefore, the solubility of lead (II) trioxonitrate (V) in water is 581 g/dm³. The correct option is the one that reads "581.000".
Pergunta 18 Relatório
2- methylbutan- 2ol is an example of a
Detalhes da Resposta
2-methylbutan-2-ol is an example of a tertiary alkanol. An alkanol is a type of organic compound that contains a hydroxyl (-OH) group attached to a carbon atom in an alkane chain. In this case, the -OH group is attached to a tertiary carbon atom, which means that the carbon is bonded to three other carbon atoms. Tertiary alkanols have the hydroxyl group attached to a tertiary carbon atom, while primary and secondary alkanols have the hydroxyl group attached to a primary or secondary carbon atom, respectively.
Pergunta 19 Relatório
Three drops of a 1.0 mol dm-3 solution of HCl was added to 20 cm3 of a solution of pH6.4. The pH of the resulting solution will be
Detalhes da Resposta
The addition of HCl (hydrochloric acid) to a solution of pH 6.4 will result in a decrease in pH. Hydrochloric acid is a strong acid and will release hydrogen ions (H+) into the solution, thereby lowering the pH. The exact pH will depend on the concentration of HCl added, but it will be less than 6.4.
Pergunta 20 Relatório
In the laboratory preparation of ammonia, the flask is placed in a slanting position so as to
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Pergunta 21 Relatório
2SO2(g) + O2(g) ? 2SO3(g)
?H=?395.7kJmol?1
In the reaction above, the concentration of 2SO3(g) can be in creased by
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Pergunta 22 Relatório
Aluminium containers are frequently used to transport trioxonitrate (V) acid because aluminium
Detalhes da Resposta
Aluminium containers are used to transport trioxonitrate (V) acid because aluminium does not react with the acid and does not corrode, which makes it a safe material for containing the acid. This is important because if the container were to react with the acid or corrode, it could cause leaks or spills, which would be dangerous. Additionally, aluminium has a low density and is lightweight, which makes it easy to transport and handle.
Pergunta 23 Relatório
The radioisotope used in industrial radiography for the rapid checking of faults in welds and casting is
Detalhes da Resposta
The radioisotope used in industrial radiography for the rapid checking of faults in welds and casting is cobalt-60. Cobalt-60 is a radioactive isotope that emits gamma rays, which are high-energy electromagnetic radiation. These gamma rays can penetrate through solid materials such as metal and concrete, allowing technicians to see the internal structure of the material without having to physically cut it open. This makes it ideal for checking faults in welds and castings, as it can quickly reveal any cracks or voids that might otherwise be difficult to detect. Once the radiography is done, the cobalt-60 source is removed and the material is safe to handle. The use of cobalt-60 in industrial radiography is a well-established and widely used technique that has been safely and effectively used for many decades.
Pergunta 25 Relatório
Sulphur (IV)oxide is a strong reducing agent in the presence of water due to the formation of
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Pergunta 27 Relatório
How many unpaired electron are in the p-orbitals of a fluorine atom?
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Pergunta 28 Relatório
10.0 dm3 of water was added to 2.0 mol dm-3 of 2.5dm3 solution of HCl. What is the concentration of the final solution in mol dm-3?
Detalhes da Resposta
C1V1 = C2V2
C2 = C1V1V2
= 2.0×2.512.5
= 0.4mol/dm-3
Pergunta 29 Relatório
Use the above option to answer this question. The dehydration of ammonium salt of alkanoic acids produces a compound with the general formula
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Pergunta 30 Relatório
An organic compound contains 60% carbon, 13.3% hydrogen and 26.7% oxygen. Calculate the empirical formula
rn(C=12,H =1, O=16)
Detalhes da Resposta
To calculate the empirical formula, we need to determine the simplest whole number ratio of atoms in the compound. Given that the compound contains 60% carbon, we can assume that the compound contains 60 grams of carbon in a 100-gram sample of the compound. Similarly, we can assume that the compound contains 13.3 grams of hydrogen and 26.7 grams of oxygen in a 100-gram sample. Next, we need to convert the masses of each element to the number of moles of each element using their respective atomic masses. The atomic masses are: - Carbon (C) = 12 - Hydrogen (H) = 1 - Oxygen (O) = 16 So, the number of moles of each element is: - Carbon (C) = 60g / 12 g/mol = 5 moles - Hydrogen (H) = 13.3g / 1 g/mol = 13.3 moles - Oxygen (O) = 26.7g / 16 g/mol = 1.67 moles The next step is to divide each of the mole values by the smallest mole value to obtain the simplest whole number ratio of atoms. In this case, the smallest mole value is 1.67 moles of oxygen. Dividing by 1.67 gives: - Carbon (C) = 5 / 1.67 = 2.99 ≈ 3 - Hydrogen (H) = 13.3 / 1.67 = 7.95 ≈ 8 - Oxygen (O) = 1.67 / 1.67 = 1 Therefore, the empirical formula of the compound is C3H8O. is the correct formula as it matches the empirical formula determined above.
Pergunta 31 Relatório
Which of the gases is employed as an anaesthesia?
Detalhes da Resposta
The gas that is commonly used as an anaesthetic is Nitrous Oxide (N2O). Nitrous oxide is a colorless, non-flammable gas with a slightly sweet taste and odor. It is also known as laughing gas due to its euphoric and pain-relieving effects. Nitrous oxide works by decreasing the sensation of pain and inducing a state of relaxation and sedation. It is commonly used in dental procedures, minor surgeries, and obstetrics. Nitrous oxide is preferred over other anaesthetic gases because it is fast-acting, has a rapid onset and offset of action, and is easily administered. It is also a relatively safe gas to use as an anaesthetic, with a low risk of side effects or complications. In summary, Nitrous Oxide is the gas that is commonly used as an anaesthetic due to its fast-acting, safe, and easily administered properties.
Pergunta 32 Relatório
Calculate the volume in cm3 of oxygen evolved as s.t.p. when a current of 5 A is passed through acidified water for 193s
{F = 96500 Cmol-1, Molar volume of a gas at s.t.p. = 22.4 dm3}
Detalhes da Resposta
The volume of oxygen evolved can be calculated using the formula: Volume of gas = (Current × Time × Molar volume of gas at s.t.p.) / (2 × Faraday constant) where current is in amperes, time is in seconds, the molar volume of gas at s.t.p. is 22.4 dm³/mol, and the Faraday constant is 96500 C/mol. Plugging in the given values, we get: Volume of gas = (5 A × 193 s × 22.4 dm³/mol) / (2 × 96500 C/mol) = 0.056 dm³ Therefore, the volume of oxygen evolved is 0.056 dm³ or 56 cm³. So, the correct answer is option (B) 0.056 dm³.
Pergunta 33 Relatório
Detalhes da Resposta
The soluble trioxosulphate (IV) ion is also known as the sulphate (IV) ion or the sulphite ion, SO32-. This ion is formed when sulphur dioxide gas (SO2) dissolves in water, and reacts with oxygen in the air. Out of the given metals, only one metal can form a soluble trioxosulphate (IV) ion, and that is potassium. Potassium sulphite (K2SO3) is a soluble salt that can dissociate into potassium ions (K+) and sulphite ions (SO32-) in water. Barium can form a sulphite ion, but it is not soluble in water. Manganese can form a sulphate ion, but not a sulphite ion. Aluminium can also form a sulphate ion, but it cannot form a sulphite ion. Therefore, the metal that forms a soluble trioxosulphate (IV) ion is potassium.
Pergunta 34 Relatório
2H2O (l) + 2F2(g) → 4HF(aq) + O2(g).
In the reaction above, the substance that is being reduced is
Detalhes da Resposta
The substance that is being reduced in the reaction is F2(g). Reduction is a chemical reaction in which electrons are gained by a species, leading to an increase in its oxidation state. In the reaction above, F2(g) is gaining electrons and its oxidation state is decreasing, which indicates that it is being reduced. On the other hand, oxidation is a chemical reaction in which electrons are lost by a species, leading to a decrease in its oxidation state. In the reaction, O2(g) is losing electrons and its oxidation state is increasing, which indicates that it is being oxidized.
Pergunta 38 Relatório
The coloured nature of transition metal ions are associated with their partially filled
Detalhes da Resposta
The colored nature of transition metal ions is associated with their partially filled d-orbital. When an electron in a transition metal ion absorbs a certain amount of energy, it gets excited and jumps to a higher energy level. When this electron falls back down to its original energy level, it emits a certain amount of energy in the form of light. This emitted light has a characteristic color that is determined by the energy difference between the excited and ground states of the electron. The d-orbitals in transition metal ions are partially filled with electrons, and these electrons can easily move between different energy levels. When light is absorbed by a transition metal ion, the electrons in the partially filled d-orbitals can get excited to higher energy levels. When these electrons fall back down to lower energy levels, they emit light with a specific wavelength and color that corresponds to the energy difference between the excited and ground states of the electron. Therefore, the color of transition metal ions is directly related to the presence of partially filled d-orbitals, which allow electrons to absorb and emit light in a way that produces characteristic colors.
Pergunta 39 Relatório
Steam changes the colour of anhydrous cobalt (ll) chloride from
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Pergunta 40 Relatório
Detalhes da Resposta
The reaction between ammonia and ethyl ethanoate produces ethanol and ethanamide. When ammonia reacts with ethyl ethanoate, it undergoes a reaction called a nucleophilic substitution reaction. In this reaction, the nitrogen atom of the ammonia molecule acts as a nucleophile, or a molecule with a strong tendency to donate electrons, and attacks the carbon atom of the ethyl ethanoate molecule, which is bonded to the carbonyl group. This results in the formation of ethanol and ethanamide.
Pergunta 42 Relatório
What volume of carbon (ll) oxide is produced by reacting excess carbon with 10 dm3 of oxygen?
Detalhes da Resposta
The volume of carbon dioxide produced by reacting excess carbon with 10 dm^3 of oxygen can be calculated using the balanced chemical equation for the reaction. The balanced chemical equation for the reaction between carbon and oxygen to form carbon dioxide is: C + O2 → CO2 From this equation, it can be seen that one molecule of carbon reacts with one molecule of oxygen to form one molecule of carbon dioxide. So, if we have 10 dm^3 of oxygen, we would need 10 dm^3 of carbon to react with it. The volume of carbon dioxide produced would be equal to the volume of the reactants, which in this case is 10 dm^3 + 10 dm^3 = 20 dm^3. Therefore, the answer is 20 dm^3 of carbon dioxide would be produced.
Pergunta 43 Relatório
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
In the reaction above, the oxidizing agent is
Detalhes da Resposta
In the given reaction, Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s), zinc (Zn) is oxidized to form Zn2+ ions and copper ions (Cu2+) are reduced to form solid copper (Cu). Therefore, the oxidizing agent in this reaction is CuSO4 (aq), which causes the oxidation of Zn by accepting electrons from it. The reducing agent, on the other hand, is Zn (s), which causes the reduction of Cu2+ ions to form solid copper (Cu).
Pergunta 44 Relatório
Which of the following molecules is held together by hydrogen bond?
Detalhes da Resposta
Hydrogen bonds are a type of chemical bond that forms between a hydrogen atom and an electronegative atom, such as nitrogen, oxygen, or fluorine. Out of the given options, only "HF" molecule is held together by hydrogen bonds. In this molecule, the hydrogen atom is bonded to the electronegative fluorine atom through a hydrogen bond. This bond results from the attraction between the positively charged hydrogen atom and the negatively charged fluorine atom. The other molecules listed, "CH4," "HBr," and "H2SO4," do not contain hydrogen bonds as they do not involve the interaction between a hydrogen atom and an electronegative atom.
Pergunta 45 Relatório
In the graph above, the activation energy of the catalyzed reaction is
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Pergunta 46 Relatório
The radioactive emission with the least ionization power is
Detalhes da Resposta
The radioactive emission with the least ionization power is gamma (γ) radiation. Gamma radiation is a type of electromagnetic radiation that has the highest frequency and shortest wavelength among all types of radiation. While it is highly penetrating and can cause damage to biological tissues, it has the least ionization power because it carries no electric charge and does not interact strongly with matter. This means that gamma radiation can easily pass through materials without being deflected or slowed down, and it takes a lot of energy to ionize atoms or molecules through gamma radiation. In comparison, alpha and beta radiation are particles that carry an electric charge and interact more strongly with matter, making them more ionizing than gamma radiation.
Pergunta 47 Relatório
Use the following option above to answer this question. Which of the following compounds in solution will turn red litmus paper blue?
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