JAMB UTME - Chemistry - 2024

Akọwa Nkọwa

Sulphur(IV) oxide has many uses including food preservation, refrigeration, laboratory reagent and solvent, sulphuric acid production, fumigant etc.Sulphur(IV) oxide  is a good refrigerant because it has a high heat of evaporation and can be easily condensed.

Akọwa Nkọwa

In the Contact Process, the catalyst used for the conversion of sulphur(IV) oxide (SO₂) to sulphur(VI) oxide (SO₃) is vanadium(V) oxide, also chemically represented as V₂O₅. This catalyst is preferred because it is more cost-effective and significantly more durable under reaction conditions than other catalysts such as platinum. Moreover, while platinum is also an effective catalyst, it is prone to poisoning by impurities that may be present in the reaction mixture. Vanadium(V) oxide, on the other hand, offers a better balance of efficiency, cost, and durability, making it the catalyst of choice in industrial applications of the Contact Process.

Akọwa Nkọwa

The question is asking about the pH of a 0.001 mol dm⁻³ solution of H₂SO₄ (sulfuric acid). To find the pH, we need to understand how sulfuric acid dissociates in water.

Step 1: Dissociation of H₂SO₄
Sulfuric acid, H₂SO₄, is a strong acid and dissociates completely in water in two steps:

1. The first dissociation: H₂SO₄ → H⁺ + HSO₄^-

2. The second dissociation: HSO₄^- → H⁺ + SO₄^2-

For dilute solutions, particularly below 0.1 M, the first dissociation provides the major contribution to the H⁺ concentration. The second dissociation also contributes slightly to the acidity, but for simplicity and due to the dilute nature of this solution, the first step's contribution is primarily considered.

Step 2: Calculate the H⁺ Concentration
Since this is a strong acid and dissociates completely, for every 1 mole of H₂SO₄, we get 2 moles of H⁺. Therefore, for a 0.001 mol dm⁻³ solution of H₂SO₄, the concentration of H⁺ ions will be:

2 x 0.001 = 0.002 mol dm⁻³

Step 3: Calculate the pH
The pH is calculated using the formula: pH = -log[H⁺]

Substitute the H⁺ concentration:

pH = -log(0.002)

We know that log(10^-2) = -2 and log(2) = 0.3 (as provided), so:

pH = -(log(2) + log(10^-3))

pH = -(0.3 - 3)

pH = 3 - 0.3

pH = 2.7

Therefore, the pH of the 0.001 mol dm⁻³ H₂SO₄ solution is 2.7.

Akọwa Nkọwa

In evaluating the factors that affect the rate of a chemical reaction, we can look at each of the possible influences: surface area, temperature, volume, and catalyst.

Surface Area: When you increase the surface area of reactants, it allows more particles to collide with each other per unit of time, which in turn increases the rate of reaction. Imagine smaller particles like powders reacting faster than larger chunks because they have a greater surface exposed to the other reactants.

Temperature: Increasing the temperature usually increases the rate of reaction. Higher temperatures cause particles to move faster, increasing the energy of collisions, and therefore increasing the chance of successful reactions.

Catalyst: A catalyst is a substance that increases the rate of a chemical reaction without being consumed by it. It lowers the activation energy needed for the reaction to occur, thus allowing it to proceed faster.

Volume: The volume of the container or the amount of space in which a reaction occurs generally does not directly affect the rate of the reaction. While changing the volume can alter pressure or concentration in gaseous reactions, which in turn affects the rate, the volume itself is not a direct factor affecting reaction rate.

Therefore, the factor that does not directly affect the rate of a chemical reaction is volume. It indirectly affects reaction rates by altering concentration or pressure in certain reaction conditions, but it is not a direct influencing factor on its own.

Akọwa Nkọwa

The petroleum fractions that distill at 200–250°C are naphtha and kerosene,

• Bitumen: has a boiling point value of 5250 ${}^0$C
• Gasoline: Has a boiling range of 40–200°C
• Diesel: Has a boiling range of 250–350°C
• Fuel oil, paraffin, lubricants: Have a boiling range of 300–370°C
• Residue, crude oil: Has a boiling point of more than 370°C

Akọwa Nkọwa

When a candle burns under a limited supply of air, it doesn't get enough oxygen to completely burn the hydrocarbons in the wax. In complete combustion (with enough air), the candle would ideally produce water (H₂O) and carbon dioxide (CO₂). However, under limited air supply, the process is incomplete and results in the formation of soot and carbon monoxide (CO).

Here's why:

• Soot is composed of tiny particles of carbon that are given off when some of the wax does not fully combust. This happens because there isn't enough oxygen to ensure complete combustion.
• Carbon monoxide (CO) is a colorless, odorless, and toxic gas produced alongside soot when the hydrocarbon in the wax doesn't get enough oxygen to form CO₂. Instead of converting to carbon dioxide, carbon remains only partially oxidized, resulting in CO.

In summary, under limited air conditions, the combustion of a candle primarily forms soot and carbon monoxide (CO).

Akọwa Nkọwa

A titration is a process used to determine the concentration of an unknown solution by adding a solution of known concentration. The indicator used in a titration is a substance that changes color at the specific pH level of the solution, which usually happens at the equivalence point.

For a titration between a strong acid and a weak base, the solution at the equivalence point is slightly acidic. This is because the salt formed as a result of the neutralization reaction can undergo hydrolysis, producing an excess of hydronium ions (H₃O⁺) which makes the solution acidic.

Among the provided indicators, methyl orange is the most suitable for indicating this type of reaction because it changes color in an acidic pH range of about 3.1 to 4.4. It shifts from red at a pH below 3.1 to yellow at a pH above 4.4.

Therefore, for a titration involving a strong acid and a weak base, methyl orange is the appropriate indicator as it can show the end point effectively when the solution is slightly acidic. The pH at the equivalence point falls within the color change range of methyl orange.

Akọwa Nkọwa

The group VIII elements, also known as the noble gases, are called inert gases primarily because they all have completely filled valence shells. In a very simplified explanation:

1. Complete Valence Shells: All the noble gases have their outermost shells completely filled with electrons. This configuration is considered very stable and requires no additional electrons to reach stability, unlike other elements that may gain, lose, or share electrons to achieve a full valence shell.

2. Highly Stable: Due to this completely filled valence shell, the noble gases do not readily react with other elements to form compounds. Their stability comes from the fact that they do not need to bond with other elements to achieve a more stable state.

3. Examples: For instance, Helium (He) has two electrons filling its first shell, Neon (Ne) has eight electrons in its second shell, and similarly, other noble gases also have fully occupied outer shells.

This property is why the noble gases are termed "inert," which means they are largely non-reactive.

Akọwa Nkọwa

One of the major effects of oil pollution in coastal water is the destruction of aquatic life.

When oil spills into a water body, it forms a thin layer called a sheen on the surface of the water. This oil layer blocks sunlight from reaching aquatic plants and phytoplankton, inhibiting their ability to perform photosynthesis. As a result, these plants and microorganisms suffer, impacting the entire food chain.

Moreover, oil can coat the feathers of birds and the fur of marine mammals, which affects their insulation and buoyancy, leading to hypothermia, drowning, or inability to fly. Additionally, the toxic components in oil are harmful if ingested, causing internal damage to fish and other marine organisms. These combined effects can lead to significant mortality in aquatic ecosystems, threatening biodiversity and the natural balance of coastal waters.

Therefore, oil pollution can severely affect the health and survival of aquatic life, creating disruptions that can persist for many years.

Akọwa Nkọwa

The reaction between hydrogen and chlorine to produce hydrogen chloride gas is explosive in sunlight. This is because sunlight contains a broad range of electromagnetic radiation, including ultraviolet (UV) light, which is energetic enough to initiate the reaction.

Here is a simplified explanation:

• When hydrogen and chlorine gases are combined, they can react to form hydrogen chloride gas. The equation for this reaction is: H₂ + Cl₂ → 2HCl.
• This reaction is highly exothermic, meaning it releases a large amount of energy.
• The reaction requires a source of energy to get started. Sunlight provides the necessary energy to break the bonds in the chlorine molecules (Cl₂), forming chlorine radicals. These radicals are highly reactive and can quickly react with hydrogen to form hydrogen chloride.
• Once initiated by sunlight, the reaction can proceed explosively, as the formation of radicals and the subsequent reaction with hydrogen release more energy, which propagates the reaction further.

In contrast, other forms of light like diffused light, infrared light, and Raman light do not provide enough energy to initiate this explosive reaction because they lack the necessary UV component found in sunlight.

Akọwa Nkọwa

When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode

At the anode, 2Cl− ${}^{-}$  →  Cl2 ${}_2$ +  2e− ${}^{-}$

At the cathode, 2H+ ${}^{+}$ +  2e− ${}^{-}$ →  H2 ${}_2$

During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.

Akọwa Nkọwa

To determine the condition of the solution containing KNO₃ at 30⁰C, let's start by calculating the molarity of the given solution.

The molecular weight of KNO₃ (Potassium Nitrate) is approximately:

• K = 39 g/mol
• N = 14 g/mol
• O = 16 g/mol

Thus, KNO₃ = 39 + 14 + (16 * 3) = 101 g/mol.

Now, to determine the molarity of the given solution:

• 303 g/dm³ of KNO₃
• Moles of KNO₃ = 303 g / 101 g/mol = 3 mol/dm³

Compare with the solubility at 30⁰C:

• The solubility of KNO₃ at 30⁰C is given as 3.10 mol/dm³
• The calculated molarity of the given solution is 3 mol/dm³.

If we compare the values:

• The solution's molarity (3 mol/dm³) is less than the solubility limit (3.10 mol/dm³)

Hence, the solution is unsaturated because it can still dissolve more KNO₃ until it reaches the solubility limit of 3.10 mol/dm³.

Akọwa Nkọwa

To determine the difference in molecular mass between an alkene and an alkyne, let's first take a look at their general formulas.

Alkene: An alkene is a hydrocarbon with at least one double bond between carbon atoms. For an alkene with six carbon atoms, the general formula is C_nH_2n. Therefore, for 6 carbon atoms, the molecular formula is C₆H₁₂.

Alkyne: An alkyne is a hydrocarbon with at least one triple bond between carbon atoms. For an alkyne with six carbon atoms, the general formula is C_nH_2n-2. Therefore, for 6 carbon atoms, the molecular formula is C₆H₁₀.

Now let's calculate the molecular masses:

Molecular mass of alkene (C₆H₁₂):

1. Carbon (C): 6 atoms x 12 g/mol = 72 g/mol
2. Hydrogen (H): 12 atoms x 1 g/mol = 12 g/mol
3. Total for C₆H₁₂ = 72 g/mol + 12 g/mol = 84 g/mol

Molecular mass of alkyne (C₆H₁₀):

1. Carbon (C): 6 atoms x 12 g/mol = 72 g/mol
2. Hydrogen (H): 10 atoms x 1 g/mol = 10 g/mol
3. Total for C₆H₁₀ = 72 g/mol + 10 g/mol = 82 g/mol

The **difference** in molecular mass between the alkene and alkyne is **84 g/mol - 82 g/mol** = 2 g/mol.

Akọwa Nkọwa

Alkanoates, also known as fatty acid esters, are primarily found in lipids. Lipids are a broad group of naturally occurring molecules that include fats, waxes, sterols, fat-soluble vitamins (such as vitamins A, D, E, and K), and others. One of the main components of lipids is fatty acids and their derivatives, such as alkanoates.

To be more specific, alkanoates can be found in the form of triglycerides, which are the main constituents of body fat in humans and animals, as well as vegetable fat. Triglycerides are composed of glycerol bound to three fatty acids, and these fatty acids are usually present in the form of alkanoates.

Unlike proteins and rubber, which are made up of amino acids and polymers of isoprene respectively, lipids are the primary class of biomolecules where these alkanoate compounds can be found in significant amounts.

Akọwa Nkọwa

The ability to rekindle a glowing splint is an indicator of the presence of an oxidizing agent, typically oxygen or a substance that releases oxygen. Among oxides of nitrogen, only a few are capable of doing this.

Nitrogen(I) oxide, commonly known as nitrous oxide (N₂O), is not a strong enough oxidizer to rekindle a glowing splint.

Nitrogen(II) oxide, known as nitric oxide (NO), is not stable in the presence of oxygen and does not have the ability to rekindle a glowing splint because it does not actively release oxygen.

Nitrogen(IV) oxide or nitrogen dioxide (NO₂), can support combustion by releasing oxygen as it decomposes. It is a brown gas and an effective oxidizer.

Dinitrogen tetraoxide (N₂O₄) is in equilibrium with nitrogen dioxide (NO₂). However, at standard conditions, it is not as effective an oxidizer for rekindling a glowing splint as pure NO₂.

In conclusion, the oxide of nitrogen that can rekindle a glowing splint is nitrogen(IV) oxide or nitrogen dioxide (NO₂) due to its ability to release oxygen and support combustion.

Akọwa Nkọwa

Cx ${}_x$Hy ${}_y$O + 5O2 ${}_2$  →  4CO2 ${}_2$  +  4H2 ${}_2$O

On balancing the equation, we should have

X = 4 , y = 8  and O = 2  ⇒   C4 ${}_4$H8 ${}_8$O2 ${}_2$

Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.

We finally have  C2 ${}_2$H4 ${}_4$O

Akọwa Nkọwa

The amount of water that is chemically combined with a substance is referred to as water of crystallization. This is the water present in the crystalline form of a compound, necessary to maintain the structure of the crystals.

When certain substances crystallize from an aqueous solution, they incorporate a specific amount of water molecules into their crystal lattice structure. These water molecules are an integral part of the crystal and often affect its color, stability, and solubility. The water is combined in stoichiometric amounts, which means it is present in a fixed ratio relative to the rest of the molecule.

An example of this is copper(II) sulfate pentahydrate, which consists of copper(II) sulfate combined with five molecules of water per formula unit, represented as CuSO₄·5H₂O.

Akọwa Nkọwa

An example of an amphoteric oxide is Al₂O₃ (aluminum oxide).

Amphoteric oxides are special because they can act as both acidic and basic oxides. This means they can react with both acids and bases to form salts and water, showcasing their dual behavior.

Here is how it works:

• Reacts with Acids: When aluminum oxide reacts with an acid, such as hydrochloric acid (HCl), it behaves like a base and forms a salt. For example, it will react to form aluminum chloride and water:
  Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O.


• Reacts with Bases: When it reacts with a strong base, like sodium hydroxide (NaOH), it acts like an acid and also forms a salt and water. An example reaction might be:
  Al₂O₃ + 2 NaOH + 3 H₂O → 2 NaAl(OH)₄.

In contrast, oxides like CuO (copper(II) oxide) are basic oxides, and K₂O (potassium oxide) is a basic oxide as well. They don't exhibit both acidic and basic properties.

Therefore, the amphoteric nature of Al₂O₃ is what distinguishes it from common oxides that are strictly acidic or basic. This property is crucial in various chemical processes and applications.

Akọwa Nkọwa

The term basicity of an acid refers to the number of hydrogen ions (H⁺) that an acid can donate when it dissociates in water. In simpler terms, it's the number of replaceable hydrogen ions in one molecule of the acid.

Tetraoxophosphate(V) acid is another name for phosphoric acid, which has the chemical formula H₃PO₄. In this molecule, there are three hydrogen (H) atoms bonded to the phosphate group (PO₄).

When H₃PO₄ dissolves in water, it donates hydrogen ions in three steps:

• In the first step, it donates one H⁺ to form H₂PO₄⁻.
• In the second step, it can donate another H⁺ to form HPO₄²⁻.
• In the third and final step, it can donate the last H⁺ to form PO₄³⁻.

Therefore, phosphoric acid, or tetraoxophosphate(V) acid, can donate a total of three hydrogen ions. Hence, the basicity of tetraoxophosphate(V) acid is 3.

Akọwa Nkọwa

To find the vapour density of a gas, you can use the formula:

Vapour density = (Molar mass of gas) / 2

However, first, we need to determine the molar mass of the gas. One can find the molar mass using the given data:

We know that at standard temperature and pressure (s.t.p.), 1 mole of any gas occupies 22.4 dm³. We need to convert the volume from cm³ to dm³ because the molar volume is given in dm³:

560 cm³ = 0.560 dm³

Now, let's find the number of moles in 0.560 dm³:

The number of moles (n) = Volume of gas (dm³) / Molar volume at s.t.p. (dm³/mol)

n = 0.560 dm³ / 22.4 dm³/mol

n = 0.025 moles

Given that the mass of the gas is 0.4 grams, we can find the molar mass by using the relation:

Molar Mass = Mass / Number of Moles

Molar Mass = 0.4 g / 0.025 moles

Molar Mass = 16 g/mol

Now that we have the molar mass, we can find the vapour density:

Vapour density = Molar mass / 2

Vapour density = 16 g/mol / 2

Vapour density = 8.0

Hence, the vapour density of the gas is 8.0.

Akọwa Nkọwa

To determine the percentage of hydrogen in the sixth member of aliphatic alkanes, we first need to understand the general formula for alkanes. Aliphatic alkanes are a class of hydrocarbons with the general formula C_nH_2n+2, where 'n' is the number of carbon atoms.

The sixth member of this series will have n = 6. Therefore, the molecular formula for the sixth member is C₆H₁₄.

To find the percentage of hydrogen, we first calculate the molar mass of C₆H₁₄:

• Molar mass of 1 carbon atom = 12
• Molar mass of 6 carbon atoms = 6 × 12 = 72
• Molar mass of 1 hydrogen atom = 1
• Molar mass of 14 hydrogen atoms = 14 × 1 = 14

Total molar mass of C₆H₁₄ = 72 + 14 = 86

Next, we calculate the percentage of hydrogen:

Percentage of hydrogen = (Molar mass of hydrogen atoms / Total molar mass) × 100

Percentage of hydrogen = (14 / 86) × 100 = 16.28%

Therefore, the percentage of hydrogen in the sixth member of the aliphatic alkanes is 16.28%.

Akọwa Nkọwa

The observation that a salt initially weighs 2g, but reduces to 1.5g after exposure to the atmosphere is primarily due to the process called efflorescence.

Efflorescence occurs when a salt loses water molecules from its crystal structure when exposed to air, which is why the weight of the salt decreases over time. This loss of water is because some salts contain water of crystallization, and when such salts are exposed to the atmosphere, they can release this water, leading to a reduction in weight.

In this specific case, the salt has lost 0.5g of water, leading to the weight change from 2g to 1.5g. This process is different from hygroscopy, which involves absorbing moisture from the atmosphere, or deliquescence, where a substance absorbs moisture and eventually dissolves in it. It's also not related to effervescence, which is the escape of gas from an aqueous solution.

Akọwa Nkọwa

When a strong acid reacts with a strong base, the result is the formation of a neutral salt. This reaction is a part of a chemical process known as neutralization.

Let's break it down further:

• A strong acid is an acid that completely dissociates in solution to release hydrogen ions (H⁺). An example of a strong acid is hydrochloric acid (HCl).
• A strong base is a base that completely dissociates in solution to release hydroxide ions (OH⁻). An example of a strong base is sodium hydroxide (NaOH).

During a neutralization reaction, the hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O). Meanwhile, the remaining ions (for example, Na⁺ from NaOH and Cl⁻ from HCl) come together to form a compound known as a salt. This salt does not affect the acidity or basicity of the solution, hence it is considered neutral.

Therefore, the salt formed in such a reaction is a neutral salt, which is what is referred to as a normal salt in the options provided.

Akọwa Nkọwa

To distinguish between strong acids and weak acids, we can employ several methods based on their chemical properties:

Conductivity Measurement: Strong acids dissociate completely in water, releasing more ions. Because ion concentration is directly related to electrical conductivity, strong acids exhibit higher conductivity than weak acids, which only partially dissociate.

Litmus Paper: This method helps determine if a solution is acidic or basic but does not provide detailed information about the strength (strong or weak) of an acid. Both strong and weak acids turn blue litmus red. Therefore, **litmus paper cannot effectively distinguish between a strong and a weak acid.**

Measurement of pH: Strong acids have a lower pH because they fully dissociate to release more hydrogen ions (H+), whereas weak acids have a relatively higher pH as they do not dissociate completely. Thus, pH measurement can distinguish the extent of acidity.

Measurement of Heat of Reaction: The heat of reaction can give insights into the strength of an acid because it involves the degree of ionization and the energetics associated with it. A strong acid will exhibit a different calorimetric response compared to a weak acid.

In summary, **litmus paper is not suitable for distinguishing between a strong and a weak acid**, as it only indicates acidity but does not reveal the strength of the acid.

Akọwa Nkọwa

The stability of an atomic nucleus is primarily determined by the neutron/proton ratio. This refers to the number of neutrons in relation to the number of protons within the nucleus. Let's break down why this ratio is crucial for nuclear stability:

• Protons: These are positively charged particles found in the nucleus. Since like charges repel each other, having too many protons can cause the nucleus to become unstable due to the electrostatic repulsion between them.
• Neutrons: These are neutral particles that help to provide a binding force within the nucleus. Neutrons play a vital role in offsetting the repulsive forces between protons by providing additional nuclear forces, which are attractive in nature.

The right balance between the number of neutrons and protons helps in achieving nuclear stability.

• For light elements (like helium or carbon), a neutron/proton ratio near 1:1 is generally stable.
• For heavier elements, more neutrons are needed to stabilize the nucleus, leading to a higher neutron/proton ratio.

An imbalance in this ratio often results in an unstable nucleus, leading to radioactive decay as the nucleus attempts to reach a more stable form. This is why the neutron/proton ratio is a fundamental factor in the stability of the atomic nucleus.

Akọwa Nkọwa

To determine the change in oxidation state of sulfur, follow these steps:

In the given reaction:

H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)

We observe:

• **Hydrogen sulfide (H₂S)** is a molecule where hydrogen (H) with an oxidation state of **+1** bonds with sulfur (S). Since the molecule is neutral, the oxidation state of sulfur in H₂S must balance out the positive oxidation states of the two hydrogen atoms, which is **+1** each. Therefore, the oxidation state of sulfur in H₂S is **-2**.
• **Sulfur (S) in elemental form** at the products is just sulfur solid (S_(s)), and for elements in their natural state, the oxidation state is **0**.

Thus, the change in oxidation state of sulfur when moving from the reactants to the products is from **-2** to **0**. This indicates that sulfur is being oxidized.

The correct answer is that the oxidation state of sulfur changes from **-2 to 0**.

Akọwa Nkọwa

When Sulphur(IV) oxide (SO₂) is passed into a solution of acidified tetraoxomanganate (VII) (KMnO₄), it acts as a reducing agent. This reaction involves the reduction of potassium permanganate (KMnO₄), which is characterized by a distinctive color change.

The tetraoxomanganate (VII) ion (MnO₄^-) is purple in color. During the reaction, SO₂ gets oxidized while the MnO₄^- ion is reduced to Mn²⁺, which is almost colorless or pale pink, depending on the concentration.

Thus, the color of the solution changes from purple to almost colorless as the reaction progresses.

Akọwa Nkọwa

To solve the problem, we can use **Graham's law of effusion**. This law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is represented as:

Rate of diffusion of Gas X / Rate of diffusion of Gas Y = sqrt(Molar mass of Gas Y / Molar mass of Gas X)

According to the given information, gas X diffuses **twice as fast** as gas Y. This implies:

2 = sqrt(Molar mass of Gas Y / Molar mass of Gas X)

To eliminate the square root, square both sides of the equation:

(2)^2 = Molar mass of Gas Y / Molar mass of Gas X

This simplifies to:

4 = Molar mass of Gas Y / Molar mass of Gas X

Rearranging the equation, we find:

Molar mass of Gas Y = 4 * Molar mass of Gas X

This means that **Gas Y is four times as heavy as Gas X**. Therefore, the correct statement is:

• **Gas Y is four times as heavy as Gas X**

Akọwa Nkọwa

The gasification of coke to produce water gas involves reacting coke, which is primarily composed of carbon, with steam. The main chemical reaction that occurs is:

C (s) + H₂O (g) → CO (g) + H₂ (g)

From this reaction, the main constituents of water gas are hydrogen (H₂) and carbon monoxide (CO), also known as carbon(II) oxide. Therefore, water gas obtained from the gasification of coke is made up of hydrogen and carbon(II) oxide.

Akọwa Nkọwa

When discussing the process of substances changing states, some substances can transition directly from a solid to a gas without passing through a liquid state. This process is called sublimation. However, not all substances exhibit this behavior. Let's examine the substances provided:

• Ammonium chloride (NH₄Cl): This substance can sublimate, meaning it can go directly from a solid state to a gaseous state when heated.
• Iodine (I₂): Iodine is known for sublimating as well. When iodine crystals are heated, they turn directly into a violet gas without becoming a liquid.
• Calcium carbonate (CaCO₃): This substance does not sublimate. Instead, when heated, it decomposes, releasing carbon dioxide gas and leaving calcium oxide as a residue. Therefore, it does not directly transition from a solid to a gas.
• Sulfur (S₈): This substance melts into a liquid before it evaporates, so it doesn't sublimate.

In conclusion, calcium carbonate (CaCO₃) is the substance that does not change directly from a solid to a gas when heated, as it undergoes a decomposition process instead.

Akọwa Nkọwa

Kerosene is commonly used as a solvent for paints. Let me explain why in a simple way:

Kerosene is a type of fuel that is composed of hydrocarbons, which are molecules made up of hydrogen and carbon atoms. These hydrocarbons give kerosene the ability to dissolve other similar substances.

Paints often contain oils and other hydrocarbon-based compounds. Since kerosene is also hydrocarbon-based, it can effectively dissolve and thin these compounds. This makes it suitable for use as a solvent in paints, allowing the paint to be thinned or cleaned up after use. This property makes kerosene a good choice for cleaning brushes and other painting tools or for dissolving dried paint.

On the other hand, sulphur, gums, and fats are typically not dissolved effectively by kerosene because of their different chemical properties. Therefore, kerosene as a solvent is primarily useful in the context of working with paints and similar hydrocarbon-based materials.

Akọwa Nkọwa

The heat of solution refers to the overall energy change that occurs when a solute dissolves in a solvent. This process involves breaking and making of intermolecular forces, and it can be broken down into two main steps that are each accompanied by heat change. The energies involved in these steps are:

Lattice energy: This is the energy required to break the bonds between the ions in the solid crystal lattice of the solute. Breaking these bonds requires energy, and this step is usually endothermic, meaning it absorbs heat from the surroundings. The more energy needed to break the lattice, the higher the lattice energy.

Hydration energy: Once the lattice is broken, the ions are surrounded by solvent molecules, typically water, in a process known as hydration. The energy released when the solvent molecules interact with and stabilize the ions is called the hydration energy. This step is usually exothermic, meaning it releases heat into the surroundings.

In conclusion, the two energies involved in the heat of solution are lattice energy and hydration energy. The balance between these two energies determines whether the overall process of dissolving a solute in a solvent is endothermic or exothermic.

Akọwa Nkọwa

The gas that turns lime water milky is **Carbon Dioxide**. This is because carbon dioxide reacts with calcium hydroxide, which is the main component of lime water, to form calcium carbonate. This chemical reaction can be represented by the equation:

Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) + H₂O (l)

In this equation, calcium hydroxide ({Ca(OH)₂}) in the lime water reacts with carbon dioxide ({CO₂}) to produce calcium carbonate ({CaCO₃}) and water ({H₂O}).

The result is a milky or cloudy appearance due to the formation of insoluble calcium carbonate precipitate in the lime water. This reaction is a common test for the presence of carbon dioxide gas.

Among the options given, **Trioxocarbonate(IV)** is another name for the Carbonate group involving the gas carbon dioxide ({CO₂}). Hence, the gas related to Trioxocarbonate(IV) is the one that turns lime water milky.

Akọwa Nkọwa

To determine the half-life of a first-order reaction, you can use the formula:

Half-life (\(t_{1/2}\)) = \(\frac{0.693}{k}\)

where \(k\) is the rate constant of the reaction. For the given problem, the rate constant (\(k\)) is 4.5 x 10^-3 s^-1.

Substituting the value of \(k\) into the formula, we have:

\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}}\)

Perform the division:

\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}} \approx 154\) s

Therefore, the half-life of the reaction is 154 seconds.

Akọwa Nkọwa

The IUPAC nomenclature of the compound above is 2-methylpropan-2-ol.

Akọwa Nkọwa

To understand what y represents in the graph, we need to think about what graphs in chemistry, specifically regarding energy changes in reactions, generally show.

Chemical reaction energy diagrams often depict a reaction's energy change as a curve from the reactants to the products, showing different energy levels throughout the process. The energy required to start a reaction or to transform the reactants into an activated complex (also known as the transition state) is crucial.

The height of this energy barrier is called the activation energy. This is the minimum amount of energy required to start a chemical reaction. The activation energy is represented by the peak in the energy graph between the reactant energy level and the top of the curve.

Therefore, in this context, y represents the activation energy needed for the reaction to proceed. Understanding activation energy is vital as it determines how quickly a reaction will occur. Reactions with a high activation energy tend to happen more slowly because it is less probable that the necessary energy for the reaction to occur spontaneously will be present.

Akọwa Nkọwa

Alkylation of benzene is a part of a reaction class called **Friedel-Crafts alkylation**. In this reaction, an alkyl group is transferred to the aromatic benzene ring, making it a more complex molecule. The catalyst used in this process is **aluminium chloride (AlCl₃)**.

Here's how the reaction typically works:

• Role of Aluminium Chloride: Aluminium chloride acts as a Lewis acid catalyst in this process. It helps in generating a carbocation from the alkyl halide. This carbocation is highly electrophilic, which means it is positively charged and seeks electrons.
• Interaction with Benzene: Benzene has a high electron density due to its conjugated π-electron system, making it nucleophilic, or electron-rich. The carbocation readily attacks the benzene ring to form a new carbon-carbon bond, attaching the alkyl group to the ring.
• Completion of the Reaction: Finally, the aluminium chloride can be regenerated, returning to its initial form, and can be reused in the catalytic cycle. This aspect is crucial as it underscores the functioning of aluminium chloride as a catalyst.

In contrast, the other options wouldn't effectively catalyze alkylation of benzene for the following reasons:

• Palladium: This metal is often used in hydrogenation and coupling reactions but is not suitable for Friedel-Crafts alkylation.
• Nickel: Similar to palladium, nickel is used in hydrogenation but not in this type of reaction.
• Sunlight: It can provide energy for photochemical reactions but does not play the role of a catalyst in the alkylation of benzene.

Therefore, **aluminium chloride** is the catalyst used for the alkylation of benzene in Friedel-Crafts reactions.

Akọwa Nkọwa

To determine how much of a radioactive element remains after a certain period, we use the concept of half-life. The half-life of a substance is the time it takes for half of the initial amount of a radioactive element to decay. In this example, the half-life is given as 50 days.

We want to know how much of a 5g sample remains after 200 days. First, calculate how many half-lives occur in 200 days:

Number of half-lives = Total time elapsed / Half-life
                      = 200 days / 50 days
                      = 4 half-lives

Next, we calculate the remaining amount after each half-life period:

• Initial amount: 5g
• After first half-life (50 days): 5g / 2 = 2.5g
• After second half-life (100 days): 2.5g / 2 = 1.25g
• After third half-life (150 days): 1.25g / 2 = 0.625g
• After fourth half-life (200 days): 0.625g / 2 = 0.3125g

After 200 days, 0.31g of the radioactive element remains.

Akọwa Nkọwa

In the process of extracting iron in a blast furnace, CaCO3, or calcium carbonate, plays a crucial role in forming slag. Here is a simple and comprehensive explanation of how it works:

1. Role of Calcium Carbonate (CaCO3):

Calcium carbonate is commonly used as a flux in the blast furnace. When it is introduced into the furnace, it undergoes a decomposition reaction due to the high temperatures, breaking down into calcium oxide (CaO) and carbon dioxide (CO2).

2. Formation of Slag:

The calcium oxide (CaO) produced then reacts with silicon dioxide (SiO2) present in the iron ore. This reaction forms a liquid slag of calcium silicate. The slag serves two main functions:

• Removal of Impurities: It captures impurities such as silica from the ore and other materials, thereby purifying the iron.
• Protection and Insulation: The slag forms a layer over the molten iron, protecting it from oxidation and acting as an insulating layer to retain heat.

Thus, calcium carbonate (CaCO3) is crucial for forming slag by providing the necessary calcium oxide (CaO) that reacts with impurities to form slag during the extraction of iron in a blast furnace.

Akọwa Nkọwa

To determine the empirical formula of an oxide of sulfur containing 60% of oxygen, we have to understand the concept of empirical formulas, which give the simplest whole-number ratio of atoms of each element in a compound.

Step 1: Assume 100g of the compound. In 100g of the compound:

• Oxygen (O) = 60g
• Sulfur (S) = 100g - 60g = 40g

Step 2: Convert masses to moles. Use the molar mass to find moles.

• Moles of Sulfur (S) = (40g) / (32g/mol) = 1.25 moles
• Moles of Oxygen (O) = (60g) / (16g/mol) = 3.75 moles

Step 3: Determine the simplest whole-number ratio.

To find the ratio, divide each mole value by the smallest number of moles calculated:

• For Sulfur: 1.25 / 1.25 = 1
• For Oxygen: 3.75 / 1.25 = 3

The simplest ratio of S:O is 1:3.

Thus, the empirical formula of the oxide is SO₃.