JAMB UTME - Chemistry - 2024

The IUPAC nomenclature of the compound above is

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

How much of 5g of radioactive element whose half life is 50days remains after 200days?

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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.

Fog is a colloid in which

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**Fog** is a type of colloid, which is a mixture where very small particles of one substance are evenly distributed throughout another substance. In the case of fog, it consists of tiny **liquid droplets** that are dispersed in a **gas**. Specifically, these are tiny droplets of water suspended in the air. When you walk through fog, you are essentially walking through air that contains these minute water droplets.

Thus, the correct description of fog as a colloid is that it consists of **liquid particles dispersed in a gas medium**. The liquid here is water, and the gas is air.

Na2 ${}_2$X  ⇌  2Na+ ${}^{+}$  +  X2− ${}^{2-}$

The bond between Na and X is likely to be

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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.

Alkylation of benzene is catalyzed by 

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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.

When the subsidiary quantum numbers (l) equals 1, the shape of the orbital is 

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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.

What accounts for the low melting and boiling points of covalent molecules?

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The low melting and boiling points of covalent molecules are primarily due to the presence of weak intermolecular forces between the molecules. While covalent molecules consist of atoms bonded together by strong covalent bonds, the forces between separate molecules, known as van der Waals forces or London dispersion forces, are much weaker. These weak forces require significantly less energy to overcome, which explains why covalent molecules tend to have lower melting and boiling points.

Although covalent molecules have definite shapes and possess shared electron pairs, these characteristics have little influence on the melting and boiling points. The focus is instead on how much energy is needed to separate the molecules from one another.

Covalent molecules are not typically three-dimensional structures like ionic compounds or metals which form intricate lattices and require more energy to disrupt. Thus, the primary reason for their lower melting and boiling points is the presence of weak intermolecular forces that can be more easily overcome with minimal energy input.

The principle which states that no two electrons in the same orbitals of an atom have same value for all four quantum numbers is the

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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.

The compound of Copper used as a fungicide is

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The compound of copper that is commonly used as a fungicide is **Copper(II) sulfate**, which is represented by the chemical formula **CuSO₄**.

Let's break this down for better understanding:

• Copper(II) sulfate (CuSO₄): This is a blue-colored, crystalline solid that is highly soluble in water, making it easy to apply as a fungicide. It is often used in a mixture called Bordeaux mixture, which is a blend of copper sulfate and slaked lime (calcium hydroxide). This mixture is effective in controlling fungal infections on crops like grapes, melons, and berries.

The other compounds listed do not serve as common fungicides:

• Copper(II) carbonate (CuCO₃): This compound is mainly used as a pigment or for other industrial applications, not as a fungicide.
• Copper(II) nitrate (Cu(NO₃)₂): This compound is primarily used in industrial processes, such as electroplating or as a catalyst, rather than in agriculture.
• Copper(II) chloride (CuCl₂): While it is used in a variety of chemical applications, it is not the standard choice for a fungicide in agricultural practices.

Therefore, the correct and widely used copper compound as a fungicide is Copper(II) sulfate (CuSO₄).

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

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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.

The stability of atomic nucleus is determined by ratio of 

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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.

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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.

A radioactive element of mass 1g has half-life of 2 minutes, what fraction of the substance would have disintegrated after 10 minutes?

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Originalmass2n
  =   Residual mass

Where n = number of activity =  exposuretimehalflife

Given: 

Original mass = 1g, exposure time = 10 minutes , half life = 2 minutes, Residual mass = ?

Substituting all the given parameters appropriately, we have

                     n = 102 $\frac{\mathrm{<br>}}{}$

                     n = 5

Originalmass2n $\frac{\mathrm{<br>}}{}$ =   Residual mass

125
5  =   Residual mass

132 $\frac{\mathrm{<br>}}{}$     =   Residual mass

Residual mass =  132
 or 0.03125g

The chemical formula for potassiumhexacyanoferrate(II) is

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The chemical formula for potassiumhexacyanoferrate(II) is K₄Fe(CN)₆.

Let's break down the name to understand why:

1. Potassium (K): The compound includes potassium ions. In this case, four potassium ions are present, indicated by the subscript 4 in K₄.

2. Hexacyano: The prefix "hexa" means six, which signifies there are six cyanide ions (CN^-) in the complex. This is represented as (CN)₆.

3. Ferrate (II): The word "ferrate" suggests the presence of iron (Fe). The Roman numeral (II) indicates that the iron is in the +2 oxidation state.

Overall, the complex ion is [Fe(CN)₆] with a charge of 4-, so to balance the charge, four potassium ions (each with a charge of +1) are needed, resulting in the formula K₄Fe(CN)₆.

Solubility curve is a plot of solubility against 

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A solubility curve is a plot of solubility against temperature. Let me explain in a simple way:

Solubility refers to the amount of a substance (solute) that can dissolve in a given quantity of solvent to form a homogeneous solution at a specified condition. The most common factor that affects solubility is the temperature.

Here's why a solubility curve typically involves temperature:

• For most solid solutes, as the temperature increases, the solubility also increases. This means that more solute can dissolve in the solvent at higher temperatures.
• This behaviour is because higher temperature generally provides more energy to the solute and solvent molecules, allowing them to interact more effectively and dissolve.
• Temperature can have different effects on the solubility of gases in liquids, where an increase in temperature usually results in decreased solubility.

Therefore, plotting solubility against temperature in a solubility curve allows us to visualize and understand how solubility changes with variations in temperature.

The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in 

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

Here is a simplified explanation:

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

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

Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are

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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.

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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".

CH3 ${}_3$-CH2 ${}_2$-OH and CH3 ${}_3$-O-CH3 ${}_3$

The relationship between the two compounds above, is that they are 

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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.

Which of the following represents an order of increasing reactivity?

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To determine the order of increasing reactivity of the elements listed, it's important to understand the general trends in metal reactivity. Metals react by losing electrons, and their reactivity is often influenced by their ability to lose these electrons easily. In many cases, generally, alkali metals are the most reactive, and noble metals are the least reactive. Here's a basic description of the reactivity of the given metals:

• Calcium (Ca): An alkaline earth metal, it is highly reactive and will readily react with water and acids.
• Iron (Fe): A transition metal, less reactive than calcium, but it will rust (react with oxygen) and react with acids.
• Tin (Sn): Also a metal, less reactive than iron, but it can still react with acids.
• Copper (Cu): Another transition metal, copper is less reactive than tin and does not react with water easily, though it will slowly react with oxygen over time.
• Gold (Au): A noble metal, gold is very unreactive, which is why it's used for jewelry and electronics.

With these considerations in mind, the order of increasing reactivity from the given options would be:

Gold (Au) < Copper (Cu) < Tin (Sn) < Iron (Fe) < Calcium (Ca)

This is the order where the least reactive element is first (gold), and the most reactive element is last (calcium). Hence, the correct option represents the order: Au < Cu < Sn < Fe < Ca.

The main constituent of water-glass is

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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.

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

Cx ${}_x$Hy ${}_y$O in the equation is

Answer Details

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

On balancing the equation, we should have

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

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

We finally have  C2 ${}_2$H4 ${}_4$O

An example of a substance that does not change directly from solid to gas when heated is

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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.

When Calcium ethynide is decomposed by water, the gas produced is 

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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.

What is the vapour density of 560cm3 ${}^3$ of a gas that weighs 0.4g at s.t.p?

[Molar Volume of gas at s.t.p = 22.4 dm3 ${}^3$ ]

Answer Details

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

Vapour density = (Molar mass of gas) / 2

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

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

560 cm³ = 0.560 dm³

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

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

n = 0.560 dm³ / 22.4 dm³/mol

n = 0.025 moles

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

Molar Mass = Mass / Number of Moles

Molar Mass = 0.4 g / 0.025 moles

Molar Mass = 16 g/mol

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

Vapour density = Molar mass / 2

Vapour density = 16 g/mol / 2

Vapour density = 8.0

Hence, the vapour density of the gas is 8.0.

After breathing in a test tube that contains acidified K2 ${}_2$Cr2 ${}_2$O7 ${}_7$, a man noticed the change in the colour of K2 ${}_2$Cr2 ${}_2$O7 ${}_7$ from orange to green. This suggests the presence of

Answer Details

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.

How many moles of CO2 ${}_2$ are produced when ethanol is burnt with 6g of oxygen

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To determine how many moles of carbon dioxide (CO₂) are produced when ethanol is burnt with 6g of oxygen, we need to understand the balanced chemical equation for the combustion of ethanol. The reaction is as follows:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

This equation tells us that 1 mole of ethanol (C₂H₅OH) reacts with 3 moles of oxygen (O₂) to produce 2 moles of carbon dioxide (CO₂).

First, let's calculate how many moles of oxygen 6 g represents. The molecular weight of oxygen (O₂) is approximately 32 g/mol. Therefore, the number of moles of oxygen is:

Number of moles of O₂ = 6 g / 32 g/mol = 0.1875 moles

According to the balanced equation, 3 moles of O₂ produce 2 moles of CO₂. Hence, the relationship between moles of O₂ and moles of CO₂ is:

2 moles of CO₂ / 3 moles of O₂ = x moles of CO₂ / 0.1875 moles of O₂

Solving for x, we have:

x = (2/3) * 0.1875 = 0.125

Therefore, 0.125 moles of CO₂ are produced when 6g of oxygen is used to burn ethanol.

Answer Details

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.

One of the following is not a water pollutant?

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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.

In the graph above, y represents 

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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.

At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that

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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**

An example of highly unsaturated hydrocarbon is

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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).

An example of a physical change is 

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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.

A major effect of oil pollution in coastal water is 

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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.

The law which states that 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

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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.

23892 ${}_{92}^{238}$U  + 10 ${}_0^1$n → 23992 ${}_{92}^{239}$U 

The process above produces 

Answer Details

The process described appears to depict a nuclear reaction involving a nuclear transmutation. Let's break down the process:

1. The starting element is initially denoted as "23892", which represents Uranium-238. In nuclear notation, "23892" indicates an atomic mass number of 238 and an atomic number of 92.

2. The next step so happens with the element "238"; however, the numbers remain: "92" indicates that the atomic number is unchanged, suggesting no change in the element. This often means a step in between of hypothetical notation.

3. Then there's the occurrence of adding a "U + 10", which again leaves the original atomic number "92".

4. In subsequent steps, it seems that the number "n" transitions to become "23992". The mass number has increased by one unit, turning the initial isotope into "23992", which represents Uranium-239.

The key point here is the transition from Uranium-238 to Uranium-239, which typically happens through the process of a neutron absorption in which a neutron is added, resulting in a change of the mass number. Such a process often leads to the creation of a radioactive isotope.

Therefore, the process described is indicative of producing a radioactive isotope, specifically Uranium-239.

The product formed when ethyne is passed through a hot tube containing finely divided iron is

Answer Details

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 IUPAC nomenclature of the complex K4 ${}_4$Fe(CN)6 ${}_6$ is

Answer Details

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).

The term that is not associated with petroleum industry is ?

Answer Details

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.

Fats and oils are esters of fatty organic acids combined with a trihydric alkanol commonly referred to as 

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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.