JAMB UTME - Chemistry - 2024

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**

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).

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.

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.

To determine what particle X is, we need to understand the reaction given:

N + X → \^146\\ C + \^11\ \P

The notation in nuclear reactions is important. The numbers on top (superscripts) are the mass numbers, which represent the total number of protons and neutrons. The numbers on the bottom (subscripts) are the atomic numbers, which represent the number of protons.

Here's what we have:

• N: This is likely to represent some isotope of nitrogen. However, we do not have its specific isotope number, so we represent it with N.
• 146/6 C: This represents a carbon isotope with a mass number of 146 and 6 protons.
• 11/1 P: This represents a particle with a mass number of 11 and 1 proton, which is likely a proton since it's positively charged with atomic number 1 and a larger mass compared to typical neutrons.

Let's consider the conservation of mass and charge:

1. **Conservation of Mass Number:** The mass number of the reactants should equal the mass number of the products. If N has a mass number 'a' and X has a mass number 'b', then:

a + b = 146 + 11 = 157

2. **Conservation of Atomic Number:** The total number of protons should also be conserved. If N has an atomic number 'c' and X has an atomic number 'd', then:

c + d = 6 + 1 = 7

To satisfy these rules:

- Option X could be a **neutron**, as neutrons have a mass number of 1 and an atomic number of 0, which means they do not affect the atomic number but contribute to the mass number.

Let's verify:

- Assume X is a neutron with a mass number of 1 and an atomic number of 0, which fits the requirement for conservation of atomic mass:

• a + 1 = 157 (where a is from N mass number)
• c + 0 = 7 (where c is from N atomic number)

Therefore, X is a neutron because it helps conserve both the mass number and the atomic number in the given nuclear reaction.

The compound in question is K₄[Fe(CN)₆]. To name this complex using IUPAC nomenclature, let's break it down into parts:

• Potasium (K₄): This denotes four potassium ions. Potassium is not part of the complex ion; it is a counter ion that balances the charge of the complex ion.
• Fe(CN)₆: This is the complex ion.
• The ligand CN is the cyanide ion, which is named as cyano.
• Hexacyano: The prefix 'hexa-' indicates that there are six cyanide ligands attached to the central metal atom.

Next, consider the oxidation state of Fe:

• The formula of the complex is [Fe(CN)₆]^4-. Because each cyanide ion has a charge of -1, the total charge contributed by the six cyanide ions is -6.
• The overall charge is -4 due to the four potassium ions (K⁺), thus Fe must have a charge of +2 to balance the charge.

Finally, we consider the oxidation state of the iron. Since calculations show that it is +2, the complex ion is named based on its oxidation state.

Hence, the IUPAC name of this compound is potassium hexacyanoferrate(II).

An air pollutant is any substance in the air, introduced by natural or human activity, that causes harm or discomfort to living organisms, or damages the environment. Let's analyze the substances mentioned:

1. O₂ (Oxygen)
Oxygen is the gas we need to breathe. It's not considered an air pollutant because it is essential for human and animal life, as well as many natural processes.

2. CO (Carbon Monoxide)
Carbon Monoxide is a colorless, odorless gas that is produced by burning fuel (like in cars and factories). This gas can be very dangerous if there is a lot of it, as it can prevent oxygen from entering the bloodstream. Because of its harmful effects, it is considered an air pollutant.

3. H₂ (Hydrogen)
Hydrogen, while a flammable gas, is generally not harmful to the air or to organisms when it is released into the environment. Therefore, it is not considered an air pollutant.

4. O₃ (Ozone)
Ozone is a bit tricky because it is both good and bad. Higher up in the atmosphere, it forms a layer that protects us from the sun’s UV radiation. However, at ground level, it is a harmful air pollutant. Ground-level ozone can cause health problems such as respiratory difficulties, so in this context, it is considered an air pollutant.

In conclusion, the substances that are considered air pollutants in this context are Carbon Monoxide (CO) and ground-level Ozone (O₃).

When **ethyne** (also known as acetylene) is passed through a hot tube containing finely divided iron, a process called decomposition occurs. The heat causes the ethyne molecules to break down, and under these conditions, they **re-combine** to form structures that result in more complex molecules.

The key transformation involves the conversion of these ethyne molecules into **aromatic compounds**. Aromatic compounds, such as **benzene**, have a distinct ring structure and are characterized by **stability** due to resonance (a phenomenon where electrons are delocalized over a certain structure, providing extra stability).

Thus, when ethyne is passed through a hot iron tube, it undergoes trimerization to form benzene, an **aromatic** compound. Therefore, the product formed is **aromatic**.

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.

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.

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.

The molecular formula C_nH_2n-2 represents an alkyne.

To understand this, let's take a look at the characteristics of hydrocarbons, which are compounds made up of hydrogen and carbon:

• Alkanes are saturated hydrocarbons (only single bonds) and have the general formula C_nH_2n+2.
• Alkenes are unsaturated hydrocarbons with at least one double bond and have the general formula C_nH_2n.
• Alkynes are also unsaturated hydrocarbons, but they contain at least one triple bond and follow the general formula C_nH_2n-2.
• Alkanals are not hydrocarbons; they are aldehydes with at least one -CHO group.

The formula C_nH_2n-2 indicates the presence of two fewer hydrogen atoms than in an alkene. This deficiency of hydrogen atoms is characteristic of a triple bond, which is a key feature of alkynes. Therefore, hydrocarbons with this formula must contain at least one triple carbon-carbon bond.

Biodegradable pollutants are substances that can be broken down by natural processes and microorganisms. However, when they are present in water systems, they can lead to several environmental and health issues. One of the main concerns is their potential to cause ill health. Here's why:

When biodegradable pollutants such as organic waste are introduced into water bodies, they are decomposed by bacteria and other microorganisms. This process consumes dissolved oxygen in the water. As the oxygen levels decrease, aquatic life such as fish and plants may suffer or die due to a lack of oxygen, disrupting the entire aquatic ecosystem.

This situation is known as eutrophication, which can lead to the excessive growth of algae, commonly referred to as algal blooms. These blooms often produce toxins that are harmful to both aquatic life and humans. Furthermore, when this polluted water is used for drinking, agriculture, or recreational purposes, it poses serious health risks to humans. These risks may include gastrointestinal infections, neurological disorders, and skin problems.

In addition, as the pollutants decompose, foul smells may be released, which can affect air quality in the vicinity, although the primary concern with biodegradable pollutants in water is related to how they affect water quality and health.

Therefore, it is crucial to properly manage and treat biodegradable pollutants before they enter water systems to prevent these health hazards. Failure to do so can result in significant environmental and health issues.

To determine the IUPAC name of a compound, follow these steps:

1. Identify the longest carbon chain: The compound should be named based on the longest continuous chain of carbon atoms. In this scenario, consider a butane chain (which has 4 carbon atoms) as it appears to feature prominently.
2. Number the carbon atoms: Number the chain in such a way to give the lowest possible numbers to the substituents. The substituents here are a bromo group (Br) and a methyl group (CH₃).
3. Assign locants and organize substituents alphabetically: There seem to be two groups attached to the main chain: a bromo group and a methyl group. When writing the full name, organize them alphabetically (i.e., bromo comes before methyl).
4. Write the complete IUPAC name: After numbering, identify the position of the bromo and methyl groups. If they are both on the second carbon, the name is "2-bromo, 2-methylbutane." But if they are on the third carbon, the name would be "3-bromo, 3-methylbutane."

Hence, by following these steps, if the bromo and methyl groups are both attached to the second carbon (lowest numbering possible), the IUPAC name of the compound is "2-bromo, 2-methyl butane."

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."

Among the options listed, the constituent of petroleum used in surfacing roads is bitumen. Bitumen, also known as asphalt, is a sticky, black, and highly viscous liquid or semi-solid form of petroleum. It is the last fraction obtained when crude oil is distilled and is often left over after the lighter components are extracted.

Reasons why bitumen is used for road surfacing:

• Binding property: Bitumen acts as a strong binding agent that holds together the crushed stone aggregates, forming a solid road surface.
• Water resistance: It is water-resistant, which helps protect the road from rain and moisture, preventing damage and erosion.
• Durability: Roads made with bitumen can withstand heavy traffic and adverse weather conditions, making them long-lasting.
• Flexibility: Bitumen has the ability to flex and adapt to the movements of the road, reducing the occurrence of cracks.

Due to these properties, bitumen is extensively used in road construction and surfacing, ensuring roads are durable, smooth, and safe for travel.

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.

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

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.

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.

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.

Cracking, saponification and polymerization are all terminologies associated with the petroleum industry but fermentation is associated with the brewery industry.

Cracking is a chemical process that breaks down heavy hydrocarbon molecules into lighter, more useful ones.

Saponification is a chemical reaction that converts fats and oils into soap. It's not directly involved in petroleum, but it can be used to analyze petroleum products.

Polymerization is a process in the petroleum industry that converts light olefin gases into higher molecular weight hydrocarbons.

Fermentation is the process in which a substance breaks down into a simpler substance. Microorganisms like yeast and bacteria usually play a role in the fermentation process, creating beer, wine, bread,yogurt and other foods.

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.

To determine the vapor density of the organic liquid hydrocarbon with the empirical formula XY, we first need to determine the **molecular formula** of the compound, which represents the actual number of atoms of each element in a molecule.

The **relative molar masses** of X and Y are given as 72 and 6, respectively. To find the molar mass of XY, we can add these values together:

Molar mass of XY = Molar mass of X + Molar mass of Y = 72 + 6 = 78 g/mol

Vapor density is defined as half of the molar mass of the compound, since vapor density is often compared to hydrogen, where hydrogen is taken as the standard with a molar mass of 2 g/mol. Therefore, vapor density can be calculated using the formula:

Vapor Density = (Molar Mass of the Compound) / 2

Substituting the molar mass of XY:

Vapor Density of XY = 78 / 2 = 39

Therefore, the vapor density of the hydrocarbon with the empirical formula XY is **39**.

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.

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.

Nitrogen obtained from air is not absolutely pure because it contains other gases, including:

• Carbon dioxide: Makes up less than 0.04% of the atmosphere, but is important environmentally
• Argon: A component of air - one of the inert gases (0.93%)
• Water vapour.  

ROH' and ROR' can form functional group isomers because they are the functional groups of alcohols and ethers, respectively.

Ethers have a pair of alkyl or aromatic groups attached to a linking oxygen atom. ROH is the functional group of alcohols, which are derivatives of water with one hydrogen atom replaced by an alkyl group.

Alcohols (ROH) and ethers (ROR') can form functional group isomers because they have the same chemical formula but different functional groups. E.g CH3 ${}_3$CH2 ${}_2$OH and CH3 ${}_3$OCH3

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.

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.

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.

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.

The bond between Na and X is most likely to be ionic. Let's break this down simply:

In the equation provided:

Na₂X ⇌ 2Na⁺ + X²⁻

The sodium (Na) atoms become positively charged ions (Na⁺), while X becomes a negatively charged ion (X²⁻). This change in charge occurs because sodium atoms donate electrons to the X atom. The donation of electrons by sodium to X indicates a transfer of electrons, which is a hallmark of an ionic bond.

In an ionic bond, electrons are transferred from one atom to another, resulting in a positively charged ion and a negatively charged ion. These oppositely charged ions attract each other, forming a strong ionic bond.

In summary, since sodium (Na) donates electrons to X forming ions, the bond between Na and X is most likely to be ionic.

The subsidiary quantum number, often referred to as the azimuthal quantum number or angular momentum quantum number, is denoted by l. This quantum number defines the shape of the atomic orbital. The value of l determines the type of orbital as follows:

• l = 0: This corresponds to an s orbital, which has a spherical shape.
• l = 1: This describes a p orbital, which has a dumb-bell shape.
• l = 2: This corresponds to a d orbital, which has a more complex shape, often described as a cloverleaf.
• l = 3: This describes an f orbital, which has an even more complex structure.

For l = 1, the atomic orbital is a p orbital, which is characterized by its dumb-bell shape. This means that the electron density is concentrated in two lobes on opposite sides of the nucleus, resembling a dumb-bell.

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.

To determine the solubility of sodium chloride (NaCl) in mol/dm³ at the given temperature, you need to first calculate the number of moles of NaCl dissolved.

Step 1: Calculate the molar mass of NaCl.

The molar mass of a compound is found by adding the atomic masses of its constituent elements:

- Sodium (Na) has an atomic mass of 23.

- Chlorine (Cl) has an atomic mass of 35.5.

Thus, the molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

Step 2: Calculate the number of moles of NaCl.

The formula to calculate moles is:

Number of moles = Mass (g) / Molar mass (g/mol)

Given mass of NaCl = 127 g,

Number of moles = 127 g / 58.5 g/mol ≈ 2.17 mol

Step 3: Calculate the solubility in mol/dm³.

Since the sodium chloride is dissolved in 1.0 dm³ of water, the solubility is the same as the number of moles, since the volume is already 1.0 dm³.

Therefore, the solubility of sodium chloride at that temperature is 2.17 mol/dm³.

Rounded to the options given, 2.17 mol/dm³ is approximately equal to 2.2 mol/dm³.

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.

Rust on the surface of a metal, specifically on **iron**, is primarily composed of **hydrated iron(III) oxide**. The rusting process occurs when **iron** reacts with **oxygen** and **water** from the environment. This chemical reaction typically produces a compound called **iron(III) oxide**, which is then combined with water molecules, resulting in **hydrated iron(III) oxide**. This hydrated state gives rust its characteristic flaky and reddish-brown appearance.

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.