JAMB UTME - Chemistry - 2024

When the **purity of solutes** is of utmost importance, the most preferred separation technique is **recrystallization**. This method is widely used in chemistry for purifying solid compounds.

Here's a simple explanation of **recrystallization**:

1. **Dissolving the Impure Compound**: The impure solid is dissolved in a suitable hot solvent. The choice of solvent is crucial; it should dissolve the compound well at high temperatures but poorly at low temperatures.

2. **Cooling the Solution**: The solution is slowly cooled. As it cools, the solubility of the compound in the solvent decreases, causing the pure compound to form crystals and precipitate out of the solution.

3. **Collection and Drying of Crystals**: The pure crystals are collected through filtration and then allowed to dry, separating them from any remaining impurities that stay dissolved in the solvent.

The **advantage** of recrystallization is that it allows for the **removal of impurities** that are either more soluble than the desired compound at low temperatures or less soluble at high temperatures, resulting in a more purified product. Therefore, when achieving high purity is a priority, **recrystallization** is often the method of choice.

The main constituent of water-glass is sodium trioxosilicate(IV). Water-glass, also known as liquid glass, is common terminology for a mixture of sodium silicate and water. The primary chemical component in water-glass is sodium silicate, which includes sodium ions (Na⁺) bonded with silicate ions (SiO₄^4-).

Essentially, when sodium silicate is dissolved in water, it results in a viscous liquid that can be utilized in various applications such as in cements, passive fire protection, textile and lumber processing, and as a sealant. Sodium trioxosilicate(IV) forms a significant part of this mixture as it reacts with other compounds to create a hardened, glass-like structure when it dries. Therefore, when water-glass is mentioned, it is mostly referring to solutions that have sodium trioxosilicate(IV) as their principal compound.

The principle that states that no two electrons in the same orbitals of an atom can have the same value for all four quantum numbers is the Pauli Exclusion Principle.

To understand this principle, it's important to know a bit about the structure of an atom and what quantum numbers are:

Quantum Numbers:
1. **Principal Quantum Number (n):** This describes the energy level or shell of the electron.
2. **Angular Momentum Quantum Number (l):** This describes the subshell or shape of the orbital (s, p, d, f...).
3. **Magnetic Quantum Number (m_l):** This describes the specific orbital within a subshell where the electron is located.
4. **Spin Quantum Number (m_s):** This describes the spin direction of the electron, which can be either +1/2 or -1/2.

The Pauli Exclusion Principle asserts that each electron in an atom has a unique set of these four quantum numbers. While electrons can share the first three quantum numbers if they are in the same orbital (meaning they share the same energy level, the same subshell, and the same specific orbital within that subshell), they must have different Spin Quantum Numbers. This means that in any given orbital, one electron can have a spin of +1/2 and the other must have a spin of -1/2. This principle is fundamental in explaining the electronic structure of atoms and, consequently, the behavior and properties of elements.

In the process of silver plating a spoon using an electrolytic cell, the correct configuration involves the following:

Cathode: The object to be plated, which in this case is the spoon. In an electrolytic cell, the cathode is where the reduction reaction occurs, and it is the surface on which the metal ions are deposited.

Anode: A rod made of silver. The anode is where oxidation occurs, meaning the silver rod will dissolve into the solution in the form of silver ions. These ions then move towards the cathode to be deposited as a thin layer on the spoon.

Electrolyte: A solution that contains a soluble silver salt (such as silver nitrate, AgNO₃). The silver ions from this salt help in the process of transferring the silver from the anode to the cathode.

Thus, the proper order for silver plating a spoon in an electrolytic cell for industrial production is: "Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt."

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.

The shape of the molecule of Carbon(IV) oxide, also known as carbon dioxide (CO₂), is linear. This is because of the following reasons:

• Carbon dioxide (CO₂) consists of one carbon atom covalently double bonded to two oxygen atoms.
• Carbon, being the central atom, has no lone pairs of electrons. It forms double bonds with each of the two oxygen atoms.
• The double bonds between carbon and the two oxygen atoms are on opposite sides of the carbon atom. This creates a structure where the atoms are aligned in a straight line, producing a linear shape.
• The bond angle between the oxygen-carbon-oxygen atoms is approximately 180 degrees, further confirming its linear nature.

Due to this arrangement, carbon dioxide has a symmetric shape, making it non-polar despite having polar covalent bonds. The pulling forces of the two oxygen atoms on either side of the carbon atom cancel each other out, reinforcing its linear configuration.

The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:

• Van der Waals forces include interactions that occur not from permanent charges (like those in ions) but from temporary or transient dipoles that momentarily exist even in non-polar molecules.
• These forces can be divided into different types, such as London dispersion forces and dipole-dipole interactions:
• London Dispersion Forces: This type of Van der Waals force is found in all molecules, regardless of whether they are polar or non-polar. It arises due to the momentary unequal distribution of electrons in a molecule, causing a temporary dipole which induces a dipole in a neighboring molecule.
• Dipole-Dipole Interactions: These occur between molecules that have permanent dipoles, such as when one end of a molecule is slightly negative and the other end is slightly positive. The positive end of one molecule will be attracted to the negative end of another.

Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.

Fats and oils are types of lipids that belong to the category of esters of fatty acids. These are organic compounds formed when fatty acid molecules react with an alcohol. In the case of fats and oils, the alcohol involved is a trihydric alkanol, meaning it has three hydroxyl (-OH) groups.

The trihydric alkanol commonly found in fats and oils is glycerol. Glycerol, also known as glycerine, has the chemical formula C₃H₈O₃ and has three carbon atoms, each of which is attached to a hydroxyl group, making it a perfect candidate to form esters with three fatty acid molecules.

When these fatty acids react with the hydroxyl groups of glycerol, they form compounds called triglycerides. These triglycerides are the primary constituents of both fats and oils. Therefore, the correct answer is that fats and oils are esters of fatty organic acids combined with glycerol as the trihydric alkanol.

This reaction illustrates dehydration. In chemistry, dehydration refers to the process of removing water (H₂O) from a compound. Let's break down the given reaction to understand this better.

The provided chemical equation is:

C₂H₅OH → C₂H₄ + H₂O

This equation indicates that ethanol (C₂H₅OH) is being transformed into ethylene (C₂H₄) with the production of water (H₂O).

The process involves the breaking of bonds in ethanol and the removal of a water molecule, as follows:

• The ethanol molecule, C₂H₅OH, has the –OH group, which combines with a hydrogen atom from the rest of the molecule to form H₂O (water).
• This results in the formation of C₂H₄ (ethylene), a smaller molecule, and the water molecule is removed.

This reaction is typically carried out under certain conditions, in this case at a high temperature of 1700°C, to facilitate the dehydration process.

Therefore, this is indeed a dehydration reaction as it involves converting ethanol into ethylene by removing water.

To understand the effect of increased pressure on the equilibrium position of the given reaction:

CuO(s) + H₂(g) ⇌ Cu(s) + H₂O(g)

We need to consider Le Chatelier's Principle. According to this principle, if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust itself to counteract the effect of the change and re-establish equilibrium.

For the reaction in question, let's consider the number of gas molecules on each side of the equation:

• Left side (reactants): 1 molecule of H₂(g)
• Right side (products): 1 molecule of H₂O(g)

Since both sides of the equation have the same number of gas molecules, an increase in pressure will not favor a shift to either the left or the right because the number of moles of gas on both sides of the equilibrium is the same.

Therefore, the effect of increased pressure on the equilibrium is that there is no effect. The position of the equilibrium remains unchanged, and pressure changes do not influence the production of more H₂(g) or H₂O(g) in this specific reaction.

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.

The problem requires calculating the **vapor density** of the gas. Vapor density is defined as the mass of a certain volume of a gas compared to the mass of an equal volume of hydrogen, where the hydrogen standard is 2 g/mol (as the molecular weight of hydrogen gas, H₂, is 2).

Here's a step-by-step explanation:

1. Molar Volume: At standard temperature and pressure (s.t.p.), 1 mole of any gas occupies a volume of 22.4 dm³.
2. Volume of Gas Provided: The given gas occupies a volume of 5.6 dm³.
3. Find Moles of Gas: Using the proportion of the molar volume:
   • Moles of Gas = Volume of Gas Provided / Molar Volume
   • Moles of Gas = 5.6 dm³ / 22.4 dm³/mole = 0.25 moles
4. Calculate Molecular Weight: Molecular weight (M) can be calculated using the relation:
   • M = Mass of Gas / Moles of Gas
   • M = 11.0 g / 0.25 moles = 44 g/mol
5. Vapor Density: Vapor density is half of the molecular weight of the gas.
   • Vapor Density = M / 2
   • Vapor Density = 44 g/mol / 2 = **22**

The calculated vapor density of the gas is 22.

To understand the geometrical isomers of butene, we need to explore its structure. Butene has four carbon atoms, and there are various structural forms that butene can take. These structural forms include linear or branched chains, with a double bond present between carbon atoms.

Geometric isomerism is a type of stereoisomerism. It occurs due to restricted rotation around the double bond, leading to different spatial arrangements of groups attached to the carbons forming the double bond. The geometric isomerism primarily occurs in alkenes like butene where the positions of substituents can vary.

Let's consider the different types of butene, focusing on the possibility of geometrical isomerism:

• 1-butene: This is a straight-chain isomer with the double bond starting at the first carbon. It doesn't show geometrical isomerism because the double bond is at the end of the chain, so there are no different spatial arrangements of groups possible.
• 2-butene: Here the double bond is between the second and third carbon. In this case, geometrical isomerism is possible because different groups (or atoms) attached to the carbon atoms can have different spatial arrangements. Specifically:
  • cis-2-butene: Both methyl groups (CH₃) are on the same side of the double bond.
  • trans-2-butene: The methyl groups are on opposite sides of the double bond.
• Isobutene (or 2-methylpropene): This is a branched isomer, and it does not have geometric isomers because the substituents around the double bond do not allow for different spatial configurations on either side of the double bond.

In conclusion, for butene, only 2-butene has geometrical isomers (cis and trans). Therefore, the number of geometric isomers is 2.

Alnico is a type of alloy that is known for its strong magnetic properties. The name "Alnico" comes from the elements it is primarily composed of: Aluminum (Al), Nickel (Ni), and Cobalt (Co). These elements are combined to form an alloy that retains its magnetism well and can operate at high temperatures, making it ideal for applications like electric motors, sensors, and various electronic devices.

While there are different variations of Alnico, the presence of Cobalt (Co) is essential for enhancing the magnetic properties of the alloy. The other elements listed, such as Magnesium (Mg), Manganese (Mn), and Copper (Cu), are not typical core constituents of Alnico. Although trace amounts of other elements like copper may sometimes be included in specific formulations, the primary and most significant component responsible for Alnico's powerful magnetic characteristics is Cobalt (Co).

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.

Quantum numbers are a set of numbers that describe the position and energy of an electron in an atom. 

When the quantum number is equal to 3, the possible values for the azimuthal quantum number  are 0, 1, and 2:

• l=0, The resulting subshell is an "s" subshell.
• l =1, The resulting subshell is a "p" subshell.
• l =2: The resulting subshell is a "d" subshell. 

The three possible sub-shells when n=3 are 3s, 3p, and 3d.

In the fountain experiment, hydrogen chloride gas (HCl) and ammonia (NH₃) are used to demonstrate the creation of a visible 'fountain' due to their high solubility in water. Here's a simple explanation:

When hydrogen chloride gas and ammonia gas come into contact with water, they dissolve very quickly and react vigorously. This is because both gases are very soluble in water. As they dissolve, a vacuum-like pressure is created inside the container where the gases are held, pulling water up into it, creating the 'fountain' effect.

Moreover, when HCl and NH₃ gases react with each other, they form a white, solid product known as ammonium chloride (NH₄Cl), which is a demonstration of how both gases can effectively dissolve and react with not just water, but also with each other.

Thus, the ability of these gases to create a fountain effect is primarily because they are very soluble in water, which allows them to dissolve rapidly and create the pressure differential necessary for the water to be pulled into the container dynamically.

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.

Benzene is a liquid hydrocarbon that is obtained from the fractional distillation of coal tar, and it is extensively used in the pharmaceutical industry. Let me break this down for you:

• **Fractional Distillation:** This is a process used to separate a mixture into its component parts, or fractions, based on differences in boiling points. Coal tar, a byproduct of coal processing, contains a myriad of compounds, and fractional distillation helps isolate specific hydrocarbons like benzene.
• **Benzene:** It is a simple aromatic hydrocarbon with the chemical formula C₆H₆. Benzene has a unique ring structure that makes it a crucial starting material or intermediate in **pharmaceutical synthesis.**
• **Uses in Pharmaceuticals:** Benzene is used to synthesize various medicinal compounds. It acts as a building block to create more complex molecules necessary for **drug manufacturing.**

That's why benzene plays an important role in the pharmaceutical industry, making it a highly valued product obtained through the distillation of coal tar.

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.

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.

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.

Hydrogen has three naturally occurring isotopes, and each of them contains the same number of protons but different numbers of neutrons. Let's briefly differentiate them:

• Protium: This is the most common isotope of hydrogen, which has one proton and no neutrons. It's simply referred to as hydrogen.
• Deuterium: This isotope of hydrogen has one proton and one neutron. It is heavier than protium because of the additional neutron.
• Tritium: This is the heaviest naturally occurring isotope of hydrogen, consisting of one proton and two neutrons.

The highest isotope of hydrogen is tritium because it has the most neutrons and, therefore, the greatest atomic mass compared to the other isotopes. It is also noteworthy that tritium is radioactive, while the other hydrogen isotopes are stable.

Water pollutants are substances that, when introduced into the water, cause harm to ecosystems, human health, and the overall quality of the water. Each of the options provided has the potential to be considered a water pollutant, except for one. Let's explain them:

1. Inorganic fertilizers: These are substances mainly composed of synthetic chemicals, including nitrates and phosphates. When these fertilizers enter water bodies, they can lead to nutrient pollution, which causes excessive growth of algae (eutrophication), leading to a decrease in oxygen levels in the water, harming aquatic life.

2. Warm water affluent: This refers to the discharge of heated water into natural water bodies. This heat contamination can change the temperature of the water, affecting the metabolism of aquatic life and leading to thermal pollution.

3. Oxygen gas: Oxygen gas is a fundamental component of the Earth's atmosphere and is not considered a water pollutant. In fact, dissolved oxygen is crucial for the survival of aquatic organisms. Rather than causing any harm, adequate levels of dissolved oxygen in water bodies are essential for maintaining healthy aquatic ecosystems.

4. Biodegradable waste: These are organic materials that decompose in the environment. When introduced in large quantities into water bodies, they can consume a significant amount of dissolved oxygen as they decompose, which can lead to depletion of oxygen levels and cause harm to aquatic life, making them pollutants in aquatic ecosystems.

Given the explanations above, oxygen gas is the option that is not a water pollutant. It is vital for the health of aquatic ecosystems, unlike the other options, which can all lead to some form of pollution in water bodies.

Aluminum is extracted from bauxite by electrolysis. The extraction proceeds in two stages;

1. Purification of the Bauxite: The impure bauxite is heated with sodium hydroxide solution to form soluble sodium tetrahydroxy aluminate (iii). The impurities in the ore which are iron (iii) oxide and trioxosilicate (iv) compounds are not soluble in the alkali. They are therefore filtered off as a sludge. 

Aluminum hydroxide crystals is then added to filtrate, NaAl(OH)4 ${}_4$ solution to induce the precipitation of Aluminum hydroxide.

2. The electrolysis of the pure alumina

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.

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.

To determine a highly unsaturated hydrocarbon, we must first understand the concept of saturation in hydrocarbons. **Saturated hydrocarbons** are compounds that contain the maximum possible number of hydrogen atoms, single-bonded to carbon atoms, and they are alkanes. **Unsaturated hydrocarbons** have one or more double or triple bonds between carbon atoms, which reduces the number of hydrogen atoms that can be bonded.

Examining the given options:

• C₂H₂ - also known as ethyne (or acetylene), has one carbon-carbon triple bond. A triple bond means it is highly unsaturated because it reduces the number of hydrogen atoms per carbon.
• CH₄ - known as methane, is a saturated hydrocarbon with single bonds only.
• C₂H₄ - known as ethene (or ethylene), has one carbon-carbon double bond, making it unsaturated.
• C₄H₁₀ - known as butane or isobutane, is another saturated hydrocarbon with single bonds only.

Based on this analysis, **C₂H₂** (ethyne) is a highly unsaturated hydrocarbon due to the presence of a **triple bond**. The triple bond signifies a greater level of unsaturation compared to double bonds in hydrocarbons like ethene (C₂H₄).

In thermodynamics, a chemical reaction is considered spontaneous when it occurs naturally under a given set of conditions without needing to be driven by an external force. The spontaneity of a reaction is best determined by the Gibbs Free Energy change, denoted as ΔG.

The criteria for spontaneity is as follows:

• A reaction is spontaneous if the Gibbs Free Energy change (ΔG) is negative. This indicates that the process can perform work on its surroundings, which means it is energetically favorable.

Now, let's relate this to the given options:

• An enthalpy change is negative (exothermic reaction) can contribute to the decrease in Free Energy. However, this alone does not determine spontaneity as it also depends on the entropy change and temperature.
• A free energy change is greater than one is not relevant in terms of spontaneity. The key is whether ΔG is less than zero (negative), not its magnitude.
• An entropy change is zero implies no change in disorder, but again, spontaneity would depend on other factors, such as enthalpy changes and the temperature.
• If enthalpy and free energy change are equal, it doesn't imply anything directly regarding spontaneity. Spontaneity is determined by ΔG, not the comparison between enthalpy and free energy.

Thus, a chemical reaction is spontaneous when the Gibbs Free Energy change (ΔG) is negative.

Scandium is not regarded as a transition metal because its ion has no electron in the d-orbital.

To understand this, let's first define a transition metal. A transition metal is defined as an element that has an incomplete d-subshell in either its elemental form or in any of its common oxidation states.

When Scandium (Sc) loses electrons to form its most common ion (Sc³⁺), it loses three electrons. These electrons are removed from the 4s and 3d orbitals. The electron configuration for Scandium is [Ar] 3d¹ 4s². Upon losing three electrons to form Sc³⁺, the resulting electron configuration is [Ar], which means there are:

• No electrons in the 3d sublevel
• No electrons in the 4s sublevel

As a result, there are no electrons in the d-orbital of the Scandium ion, which does not meet the criteria for a transition metal.

Boyle's Law describes the relationship between the volume and pressure of a given amount of gas held at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume. In simpler terms, if you decrease the volume of a gas, its pressure increases, provided the temperature remains constant, and vice versa.

The mathematical expression of Boyle's Law is PV = K, where:

• P represents the pressure of the gas.
• V represents the volume of the gas.
• K is a constant value when the temperature and the amount of gas are kept constant.

This relationship implies that if you multiply the pressure by the volume, the result will always be the same constant as long as no other variables are changed. This is the classic formulation of Boyle's Law, illustrating the inverse relationship between pressure and volume for a gas at constant temperature.

The Biuret test is a chemical test used for detecting the presence of proteins. When you perform a Biuret test, you are looking for peptide bonds, which are the connections between the amino acids in a protein. This is how it works:

• When a solution containing proteins is mixed with Biuret reagent, which contains copper sulfate, a chemical reaction occurs.
• In this reaction, the copper ions form a complex with the peptide bonds present in the protein.
• This complex gives the solution a characteristic purple color if proteins are present.
• The intensity of the purple color can often indicate the amount of protein present in the sample.

The test is specifically tailored to proteins because carbohydrates, amines, and alkanoates do not exhibit the required peptide bonds necessary for this color change. Therefore, the Biuret test is not suitable for detecting these compounds.

The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as esterification.

An alkanoic acid, also known as a carboxylic acid, is a type of organic acid that contains a carboxyl group (-COOH). An alkanol, commonly referred to as an alcohol, contains a hydroxyl group (-OH).

When an alkanoic acid reacts with an alkanol in the presence of an acid catalyst (commonly sulfuric acid), they combine to form an ester and water. This particular reaction is termed esterification. The acid catalyst speeds up the reaction by donating protons, which helps in breaking and forming new bonds.

Here's a simplified view of the reaction:

1. Alkanoic Acid (R-COOH) + Alkanol (R'-OH) -> Ester (R-COOR') + Water (H2O)

The key characteristics of esterification are:

• Combination: It involves combining two different types of organic molecules, an acid, and an alcohol.
• Production of Water: Water is produced as a by-product of the reaction.
• Aromaticity: Esters often have a pleasant aroma, which is why many artificial flavors and fragrances are esters.

Therefore, in summary, the process described is esterification.

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.

When a metal reacts with an acid, a chemical reaction takes place in which the metal displaces the hydrogen in the acid. This reaction produces a salt and hydrogen gas is liberated in the process.

Let's break it down further:

• Metal: Metals are elements that can react with acids to produce salts.
• Acid: Acids typically contain hydrogen ions (H⁺) which are replaced by metal ions during the reaction.
• Salt: A salt is an ionic compound that forms when the hydrogen ions of the acid are replaced by metal ions.
• Hydrogen Gas: This is the gas that is liberated during the reaction. You may observe bubbles forming as the gas escapes.

The general equation for the reaction is:

Metal + Acid → Salt + Hydrogen Gas

For example, when zinc (a metal) reacts with hydrochloric acid (an acid), the reaction is as follows:

Zn + 2HCl → ZnCl₂ + H₂

Here, zinc chloride (a salt) and hydrogen gas are produced. This illustrates that salt and hydrogen gas are formed when a metal reacts with an acid.

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.

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**

A physical change involves a change in the physical properties of a substance, without a change in its chemical composition. This means that the substance remains the same at the molecular level, despite how it might appear differently.

An example of a physical change from the given options is the liquefaction of liquids. In this process, a substance transitions from a solid or gas to a liquid state. This change is purely physical because the molecular structure of the substance does not change; only its state or form does. Importantly, such a change is usually reversible, meaning the substance can return to its original state. For instance, water can change into ice (frozen) or steam (vapor), and can still revert back to liquid water.

On the other hand, the other options involve chemical changes, where the original substances undergo chemical reactions to form new substances with different properties, thus altering the molecular structure depending on the option.

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.