JAMB UTME - Chemistry - 2024

Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver.

[Mg = 24.0, Ag = 108 ]

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To solve this problem, we must consider the concept of electrochemistry and Faraday's laws of electrolysis. These laws are crucial for determining the mass of a substance liberated during electrolysis.

Faraday's first law states that the mass of a substance liberated is directly proportional to the quantity of electricity that passes through the electrolyte. The mass can be calculated using the formula:

m = (Q * M) / (n * F)

Where:

• m = mass of substance liberated
• Q = total electric charge (in Coulombs)
• M = molar mass of the isolated element
• n = number of moles of electrons required to reduce 1 mole of ions of the element
• F = Faraday's constant (approximately 96500 C/mol)

For silver (Ag), the chemical reaction at the cathode is:

Ag⁺ + e⁻ → Ag

This shows that **1 mole of electrons** is required to discharge **1 mole** of silver ions.

For magnesium (Mg), the chemical reaction at the cathode is:

Mg²⁺ + 2e⁻ → Mg

This means that **2 moles of electrons** are required to discharge **1 mole** of magnesium ions.

Given:

• Mass of Ag liberated = **16 g**
• Molar mass of Ag (M) = **108 g/mol**
• Molar mass of Mg (M) = **24 g/mol**

First, find the number of moles of Ag liberated:

Number of moles of Ag = 16 g / 108 g/mol = 0.1481 mol

The same quantity of electricity will be used to liberate an equivalent in moles of electrons for Mg.

0.1481 moles of Ag require 0.1481 moles of electrons, equivalent to:

0.1481 moles of electrons for Mg. Since Mg requires 2 moles of electrons for 1 mole of Mg:

Number of moles of Mg = 0.1481 / 2 = 0.07405 mol

Finally, calculate the mass of Mg liberated:

m = 0.07405 mol * 24 g/mol = 1.7772 g

Rounding this to the closest answer provided:

The mass of magnesium that will be liberated is approximately **1.78 g**.

The IUPAC nomenclature of the compound above is

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

The highest isotope of hydrogen is

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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.

From the table above, which of these two compounds can form functional group isomers?

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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

The combustion of candle under limited supply of air forms

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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).

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.

127g of sodium chloride was dissolved in 1.0dm3 ${}^3$ of distilled water at 250 ${}^0$C . Determine the solubility in moldm−3 ${}^{-3}$ of sodium chloride at that temperature. [Na = 23, Cl = 35.5] 

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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.

If 11.0g of a gas occupies 5.6 dm3 ${}^3$ at s.t.p., calculate its vapour density (1 mole of a gas occupies 22.4 dm3 ${}^3$).

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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.

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.

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.

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

The process above produces 

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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 empirical formula of an organic liquid hydrocarbon is XY. If the relative molar masses of X and Y are 72 and 6 respectively, it's vapour density is likely to be

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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**.

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.

Which of the following is an air pollutant?

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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).

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

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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**.

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.

For chemical reaction to be spontaneous, ∆G must be

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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.

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.

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

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

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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 oxide of nitrogen that can rekindle a glowing splint is 

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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.

An example of an amphoteric oxide is

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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.

Heat of solution involves two steps that is accompanied by heat change. The energies involved in this steps are

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

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

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

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

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.

Which of the following is used in forming slag in the blast furnace for the extraction of iron?

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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.

The IUPAC name of the compound above is

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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."

The ions responsible for permanent hardness in water are sulphates of 

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Permanent hardness in water is mainly caused by the presence of certain metal ions, specifically the **sulfates (SO₄²⁻)** and **chlorides (Cl⁻)** of calcium (Ca) and magnesium (Mg). These compounds do not precipitate out when the water is boiled, which means they remain dissolved and continue to contribute to the hardness of the water.

Among the options you provided, the ions responsible for permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. The presence of calcium sulfate (CaSO₄) and magnesium sulfate (MgSO₄) in water keeps it hard.

When compared to temporary hardness, which can be removed by boiling the water to precipitate bicarbonates, **permanent hardness cannot be removed by boiling**. Instead, methods such as ion exchange or the use of water softeners are required to remove these ions from the water.

In summary, the ions causing permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. These ions remain dissolved and continue to make the water hard, despite boiling.

The group VIII elements are the inert gases because they 

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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.

The term that is not associated with petroleum industry is ?

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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.

The percentage of hydrogen in the sixth member of the class of the aliphatic alkanes is  [H =1, C =12 ]

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

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

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

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

Total molar mass of C₆H₁₄ = 72 + 14 = 86

Next, we calculate the percentage of hydrogen:

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

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

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

An organic compound with general formula RCOR' is an

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The general formula RCOR' represents a class of organic compounds known as ketones. In this formula, R and R' are alkyl groups, which are chains of carbon and hydrogen atoms. The CO in the middle is a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom. Therefore, with the presence of two alkyl groups on either side of the carbonyl group, the compound is categorized as a ketone, scientifically referred to as an alkanone.

Here is a simple breakdown of the terms:

• Alkanone: A term used to describe ketones, which have the general formula RCOR'.
• Alkanal: Represents aldehydes, which contain the functional group RCHO with at least one hydrogen attached to the carbon atom of the carbonyl group.
• Alkanoate: Typically refers to esters, which have the general form RCOOR'.
• Alkanoic acid: Refers to carboxylic acids, which are denoted by the functional group RCOOH.

Hence, by looking at the general formula RCOR', the organic compound in question is undoubtedly an alkanone.

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

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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.

25.0g of potassium chloride were dissolved in 80g of distilled water at 300 ${}^0$C. Calculate the solubility of the solute in mol dm3 ${}^3$. [K =39, Cl = 35.5]

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To calculate the solubility of potassium chloride (KCl) in mol dm³, we need to follow these steps:

1. First, determine the molar mass of potassium chloride (KCl).
• Molar mass of K (Potassium) = 39 g/mol
• Molar mass of Cl (Chlorine) = 35.5 g/mol
2. Add these to get the molar mass of KCl:

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol



3. Calculate moles of KCl:
• We have 25.0 g of KCl dissolved.

Moles of KCl = Mass of KCl / Molar mass of KCl = 25.0 g / 74.5 g/mol = 0.3356 mol



4. Next, convert the mass of water to liters.
• 80 g of distilled water corresponds approximately to 80 ml, because the density of water is roughly 1 g/ml.

Convert ml to liters: 80 ml = 0.080 L



5. Calculate the concentration (solubility) in mol dm³ (which is equivalent to mol/L):

Concentration = Moles of solute / Volume of solvent in liters = 0.3356 mol / 0.080 L = 4.195 mol/dm³

The solubility of potassium chloride at 30°C in mol/dm³ is therefore approximately 4.2 mol/dm³.

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.

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.

The shape of ammonia molecule is

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The shape of the ammonia molecule (NH₃) is trigonal pyramidal. To understand why, let's explore the electron and molecular geometry using a simple explanation:

Ammonia consists of one nitrogen (N) atom bonded to three hydrogen (H) atoms. The nitrogen atom has five valence electrons requiring three more electrons to complete its octet. These are acquired by forming covalent bonds with three hydrogen atoms. In addition to the three bonding pairs, there is one lone pair of electrons on the nitrogen atom.

According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, electron pairs, including bonding pairs and lone pairs, repel each other and arrange themselves as far apart as possible to minimize repulsion. In ammonia:

• There are four electron pairs around the central nitrogen atom (three bonding pairs and one lone pair).
• These electron pairs arrange themselves in a tetrahedral electron pair geometry.

The presence of the lone pair on nitrogen creates a slight distortion, causing the molecule's shape to be trigonal pyramidal rather than perfectly tetrahedral. The lone pair occupies more space and pushes the hydrogen atoms slightly closer together. This results in a pyramidal shape, with nitrogen at the apex, and the three hydrogen atoms forming the base of the pyramid.

The trigonal pyramidal shape of ammonia is a result of this molecular geometry, not to be confused with any of the other options like V-shaped, tetrahedral, or co-planar.

The volume occupied by 1 mole of an ideal gas at a temperature of 130 ${}^0$C and a pressure of 1.58 atm is 

[ R = 0.082 atm dm3 ${}^3$ K−1 ${}^{-1}$mol−1 ${}^{-1}$ ]

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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

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

During the fractional distillation of crude oil, the fraction that distills at 200 - 2500 ${}^0$ C is

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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

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.

H2 ${}_2$S(g) ${}_{(g)}$  +  Cl2 ${}_2$(g) ${}_{(g)}$ → 2HCl(g) ${}_{(g)}$ + S(s) ${}_{(s)}$

What is the change in oxidation state of sulphur from reactant to product?

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

In the given reaction:

H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)

We observe:

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

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

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