JAMB UTME - Chemistry - 2024

When the acidified potassium dichromate (\(K_2Cr_2O_7\)) solution changes from orange to green, it indicates a chemical reaction is occurring where the chromium in the dichromate ion is being reduced. In this context, acidified \(K_2Cr_2O_7\) is commonly used as an oxidizing agent.

The change in color from orange (dichromate ion) to green (chromium ion) suggests that the dichromate ion is being reduced, and something in the person's breath is being oxidized.

The substances that can be oxidized in the breath are organic compounds, typically those containing functional groups with oxidizable hydrogen atoms or structures.

• Alkanol: The presence of alkanols (such as alcohols) can result in this color change because alkanols can be oxidized to aldehydes or carboxylic acids by the dichromate ion. In an alcohol, the oxidation is facilitated by the presence of an -OH group.

Therefore, when the color of acidified potassium dichromate changes from orange to green, it suggests the presence of an alkanol.

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.

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.

Aqueous solution of sodium hydroxide (NaOH) is a versatile reagent in chemistry, often used to test for the presence of metal ions. When sodium hydroxide is added to solutions containing certain metal ions, it forms precipitates that are characteristic of those ions. Here's how it interacts with each of the mentioned ions:

Calcium ions (Ca²⁺): When NaOH is added to a solution containing calcium ions, a white precipitate of calcium hydroxide (Ca(OH)₂) can form. However, the precipitate is only slightly soluble in water, and this reaction is not the most definitive test for calcium ions.

Zinc ions (Zn²⁺): When sodium hydroxide is added to a solution containing zinc ions, a white gelatinous precipitate of zinc hydroxide (Zn(OH)₂) forms. This precipitate is soluble in excess NaOH, leading to a clear, colorless solution. This reaction is used to test for zinc ions.

Copper ions (Cu²⁺): When NaOH is added to a solution containing copper ions, a pale blue precipitate of copper(II) hydroxide (Cu(OH)₂) forms. This precipitate is insoluble even in excess NaOH, and the formation of this blue precipitate is a common test for copper ions.

Therefore, an aqueous solution of sodium hydroxide can be used to test for the presence of all three ions: calcium (Ca²⁺), zinc (Zn²⁺), and copper (Cu²⁺). The reaction and precipitate formation with each ion serve as indicators of their presence. Thus, the correct answer is:

I, II and III.

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).

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.

The electrolysis of copper(II) tetraoxosulphate(VI) involves the deposition of copper at the cathode. To understand how many moles of copper are deposited when 2 Faraday of electricity is passed through, we need to consider Faraday's first law of electrolysis. Faraday's first law states that the mass (or number of moles) of a substance deposited at an electrode is directly proportional to the quantity of electricity that is passed through the electrolyte.

A Faraday (or Faraday constant) is the charge of one mole of electrons, which is approximately **96500 coulombs** (C). During electrolysis, the chemical reaction occurring at the cathode for copper deposition can be represented by the following equation:

Cu²⁺ + 2e^- → Cu

This equation shows that **2 moles of electrons** (represented by 2e^-) are needed to deposit **1 mole of copper (Cu)**.

If we have **2 Faradays** of electricity, it means we have **2 x 96500 C = 193000 C**. Since **1 Faraday (96500 C)** is required to deposit **0.5 mole** of copper, **2 Faradays** will deposit twice that amount:

0.5 mole of copper deposited per Faraday x 2 Faradays = **1.0 mole** of copper

Thus, when **2 Faradays** of electricity are passed through copper(II) tetraoxosulphate(VI) solution, **1.0 mole** of copper will be deposited.

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 terms strong and weak acids are used to indicate the extent of ionization of an acid. This means how completely an acid dissociates into its ions in water.

Strong acids completely dissociate in water. This means that nearly all the acid molecules break down into positive hydrogen ions (H⁺) and their respective anions. Examples include hydrochloric acid (HCl), sulfuric acid (H₂SO₄), and nitric acid (HNO₃).

Weak acids, on the other hand, only partially dissociate in water. This means that only a small fraction of the acid molecules break down into ions. Most of the acid remains in its molecular form. An example of a weak acid is acetic acid (CH₃COOH), which is found in vinegar.

Therefore, the strength of an acid in terms of its classification as strong or weak is about how fully it dissociates into ions in an aqueous solution, not about the number of H⁺ ions or the strength of its action on substances.

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.

Ethyne, also known as acetylene, is a simple alkyne with the chemical formula C₂H₂. In ethyne, each carbon atom is bonded to two other atoms: one hydrogen atom and the other carbon atom. The molecular structure of ethyne is linear, with a triple bond between the two carbon atoms.

To determine the hybridization scheme in ethyne, we need to examine the arrangement of the electron pairs around each carbon atom. In ethyne, each carbon atom is forming two sigma (σ) bonds and two pi (π) bonds. Let's explain:

• The two carbon atoms are linked by a triple bond, consisting of one sigma bond and two pi bonds. The sigma bond is formed by the overlap of orbitals directly between the two carbon nuclei. The pi bonds are formed by the overlap of p orbitals above and below the plane of the molecule.
• The carbon atoms in ethyne are bonded to only two atoms, which is indicative of a linear geometry.

When we consider the hybridization of the carbon atoms, we focus on the formation of sigma bonds and lone pairs. In ethyne, each carbon atom utilizes two orbitals to form sigma bonds: one with the hydrogen atom and one with the other carbon atom. This implies that each carbon atom in ethyne must use two hybrid orbitals.

The two hybrid orbitals formed by each carbon atom in ethyne are a result of mixing one s orbital with one p orbital. This hybridization is referred to as sp hybridization, characterized by a linear electron geometry. The remaining two unhybridized p orbitals on each carbon atom are responsible for forming the two pi bonds in the triple bond.

In conclusion, the hybridization scheme in ethyne is sp.

Silver and Gold are classified as noble metals. These metals are known for their resistance to corrosion and oxidation in moist air, unlike most other base metals. They can be found in the earth's crust as free, uncombined elements because they do not easily react with oxygen and other elements to form compounds. This property is what distinguishes noble metals from more reactive or corrosive ones. While the term "natural metals" seems applicable in that they occur naturally, the more precise and widely accepted term for metals like Silver and Gold is "noble metals".

The **scientist who performed the experiment on discharge tubes that led to the discovery of cathode rays as a sub-atomic particle** is J.J. Thomson.

In the late 19th century, J.J. Thomson conducted experiments using a cathode ray tube. This device involved an evacuated glass tube with electrodes at each end, through which an electric current was passed. **When a high voltage was applied, Thomson observed a stream of particles traveling from the negative electrode (cathode) to the positive electrode (anode).** These streams of particles were what he called "cathode rays."

Through his experiments, J.J. Thomson discovered that these cathode rays were composed of negatively charged particles. **He concluded that these particles were much smaller than atoms, and named them "electrons," which are now known to be sub-atomic particles.** His work was fundamental in advancing the atomic model and in understanding the structure of the atom.

Thomson's discovery was pivotal because it provided the first evidence that atoms are not indivisible, but rather consist of smaller subatomic particles. This **challenged the then-prevailing notion of atoms as indivisible units**, thus marking the birth of modern particle physics.

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.

The law that states a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is the law of definite proportion.

To explain this simply, let's consider water as an example. Water is made up of hydrogen and oxygen. According to the law of definite proportion, a sample of pure water taken from anywhere in the world will always contain the same ratio of hydrogen to oxygen by mass. Specifically, water will always have approximately 88.8% oxygen and 11.2% hydrogen by mass.

This is because a chemical compound has a fixed composition, regardless of the process used to create it or the source from which it is derived. The law of definite proportion, also known as the law of constant composition, is fundamental in chemistry because it supports the idea that chemical compounds are composed of elements in specific and fixed ratios. This does not change regardless of how the compound is prepared or where it is found.

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.

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.

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.

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.

In the treatment of water for municipal supply, chlorine is used to kill germs. This process is known as chlorination. Chlorine is a very effective disinfectant and is used to eliminate harmful microorganisms such as bacteria, viruses, and protozoans that may be present in the water. By doing so, chlorine helps to ensure that the water is safe for human consumption and protects public health by preventing waterborne diseases. It is important to note that **chlorine is not used to prevent tooth decay, prevent goitre, or to remove colour or odour** in water treatment for municipal supply.

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.

The electronic configuration of nitrogen is given as: 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹.

This configuration suggests that nitrogen has 7 electrons, as follows:

• 2 electrons in the 1s orbital (1s²).
• 2 electrons in the 2s orbital (2s²).
• 1 electron in the 2p_x orbital (2p_x¹).
• 1 electron in the 2p_y orbital (2p_y¹).
• 1 electron in the 2p_z orbital (2p_z¹).

This is the **ground state** electron configuration of nitrogen, meaning that the atoms have electrons in the **lowest possible energy levels**. It demonstrates nitrogen's **stable configuration**, where it has half-filled p orbitals, each with a single electron. This configuration obeys Hund's Rule, which states that every orbital in a subshell gets one electron before any one orbital gets two (due to electron repulsion). It also obeys the Aufbau principle which suggests electrons fill orbitals starting from the lowest energy level.

Therefore, this configuration indicates that the atom is simply obeying rules governing electron configuration. The electrons are in their lowest energy orbitals, consistent with the principles that direct electron arrangement in an atom, ensuring stability without being excited or unstable. There are no **energy changes** being depicted nor is the atom in an **excited state**—it is showing the normal ground state.

When a species undergoes oxidation, it experiences an increase in its oxidation number. Oxidation is a chemical process where a species loses electrons. In terms of oxidation number, electrons have a negative charge, so losing them results in an increase in charge. Thus, the oxidation number of the species becomes more positive or less negative.

To help understand, consider sodium (Na) reacting with chlorine (Cl₂) to form sodium chloride (NaCl):

• Initially, the oxidation number of Na is 0 (it's in a pure element form).
• In NaCl, sodium now has an oxidation number of +1 (because it loses an electron).

This change clearly shows that when sodium is oxidized, its oxidation number increases.

Therefore, the correct explanation is: a species undergoing oxidation will have its oxidation number increase.

The air we breathe is made up of a mixture of gases. The most abundant gases in the atmosphere are nitrogen and oxygen, but there are other gases present in smaller amounts, one of which is carbon dioxide, chemically known as carbon(IV) oxide.

Carbon dioxide makes up approximately 0.03% of the Earth's atmosphere by volume. This value can also be expressed in different terms, such as 300 parts per million (ppm). Even though it is a small percentage, carbon dioxide plays a significant role in maintaining the Earth's temperature through the greenhouse effect.

In summary, the percentage of carbon(IV) oxide in air is 0.03%.

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).

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.

The relationship between the two compounds is that they are isomers.

To understand why these compounds are isomers, let's break down their structures and definitions:

1. Structures of the Compounds:

• The first compound, CH₃-CH₂-OH, is known as ethanol. This compound is composed of two carbon atoms (C), six hydrogen atoms (H), and one oxygen atom (O).
• The second compound, CH₃-O-CH₃, is known as dimethyl ether. This compound also consists of two carbon atoms (C), six hydrogen atoms (H), and one oxygen atom (O).

2. Definitions:

• Isomers: Compounds that have the same molecular formula (same types and numbers of atoms) but different structural arrangements and properties.
• Allotropes: Different physical forms of the same element (e.g., carbon can exist as diamond, graphite, etc.).
• Isotopes: Atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei.
• Isobars: Atoms of different elements that have the same atomic weight (mass number) but different atomic numbers.

Both compounds have the same molecular formula: C₂H₆O. However, they have different arrangements of their atoms. Ethanol has a hydroxyl group (-OH) attached to an ethyl group (CH₃-CH₂-), while dimethyl ether involves two methyl groups (CH₃-) bonded to an oxygen atom (O). This difference in structure leads to different chemical and physical properties, despite having the same molecular formula. Hence, these two compounds are classified as isomers.

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.

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.

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.

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.

When water reacts with calcium ethynide, the gas produced is ethyne (also known as acetylene), which is represented by the chemical formula C₂H₂.

The chemical reaction involved is as follows:

CaC₂ + 2 H₂O → C₂H₂ + Ca(OH)₂

Let's break down this process to make it understandable:

• Calcium ethynide (CaC₂), often called calcium carbide, is a compound composed of calcium and the acetylide ion (C₂^2-).
• When water (H₂O) is added to calcium carbide, a chemical reaction occurs.
• The reaction produces **ethyne (C₂H₂)**, a colorless gas with a distinctive odor.
• Another product of this reaction is calcium hydroxide (Ca(OH)₂), which is a solid.

The key point to remember here is that the gas produced is **ethyne (C₂H₂)**, which is useful in various industrial applications, such as welding and as a precursor for other chemicals.

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.

To find the volume of HCl that is required to completely neutralize the NaOH solution, we need to use the concept of a neutralization reaction. A neutralization reaction occurs when an acid and a base react to form water and a salt, thus neutralizing each other.

In this particular reaction, the balanced chemical equation is:

HCl + NaOH → NaCl + H₂O

Here, the equation tells us that one mole of HCl reacts with one mole of NaOH. Therefore, the molar ratio of HCl to NaOH is 1:1.

First, let's determine the number of moles of NaOH present in 20 cm³ solution:

Number of moles of NaOH = Concentration (mol/dm³) × Volume (dm³)

= 0.20 mol/dm³ × 20 cm³ × (1 dm³ / 1000 cm³)

= 0.20 × 0.020

= 0.004 moles

Since the reaction is in a 1:1 ratio, the number of moles of HCl required is also 0.004 moles.

Now, let's determine the volume of HCl solution required:

Volume of HCl (dm³) = Number of moles / Concentration

= 0.004 moles / 0.12 mol/dm³

= 0.03333 dm³

Convert this volume from dm³ to cm³:

0.03333 dm³ × 1000 cm³ / dm³ = 33.33 cm³

Therefore, the volume of HCl required is 33.33 cm³.

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

In the context of chemical reactions, the spontaneity of a reaction is determined by the Gibbs Free Energy change, represented by the symbol ΔG. A chemical reaction is considered to be spontaneous if it proceeds on its own without needing continuous external input of energy.

For a reaction to be spontaneous, the value of ∆G must be negative. This is based on the Gibbs Free Energy equation:

ΔG = ΔH - TΔS

Where:

• ΔG is the change in Gibbs Free Energy.
• ΔH is the change in enthalpy (heat content).
• T is the temperature in Kelvin.
• ΔS is the change in entropy (degree of disorder).

A negative value for ΔG indicates that the process releases energy and will proceed spontaneously. This means the system is moving towards a lower energy and more stable state, naturally favoring the products over the reactants.

In contrast, a positive ΔG indicates that the reaction is non-spontaneous and requires energy input. If ΔG is zero, the system is at equilibrium, meaning there is no net change taking place, but this doesn't indicate spontaneity.

Therefore, in summary, for a reaction to be spontaneous, ∆G must be 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.

According to the Ideal gas equation, PV = nRT

Given: P = 1.58 atm, V = ?,  n = 1 mole, R = 0.082,  T= 13 + 273K = 286K

Substituting all the given parameters, 

                   V  = nRTP $\frac{nRT}{P}$

                   V =  1×0.082×2861.58 $\frac{\mathrm{<br>}}{}$

                   V = 14.84 dm3