JAMB UTME - Chemistry - 2024

Determine the half-life of a first order reaction with constant 4.5 x 10−3 ${}^{-3}$ sec−1 ${}^{-1}$.

Détails de la réponse

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.

An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5?   [ C =12, H = 1, N = 14, O = 16]. 

Détails de la réponse

To find the molecular formula of the compound, follow these steps:

1. Determine the Empirical Formula:

Start by assuming you have 100 grams of the compound. This means you have:

• 53.1 grams of Carbon
• 6.2 grams of Hydrogen
• 12.4 grams of Nitrogen
• 28.3 grams of Oxygen

Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):

• Moles of Carbon = 53.1 g / 12 g/mol = 4.425 moles
• Moles of Hydrogen = 6.2 g / 1 g/mol = 6.2 moles
• Moles of Nitrogen = 12.4 g / 14 g/mol = 0.886 moles
• Moles of Oxygen = 28.3 g / 16 g/mol = 1.769 moles

Next, divide each by the smallest number of moles to get the simplest ratio:

• Carbon: 4.425 / 0.886 ≈ 5
• Hydrogen: 6.2 / 0.886 ≈ 7
• Nitrogen: 0.886 / 0.886 = 1
• Oxygen: 1.769 / 0.886 ≈ 2

This gives us the empirical formula: C₅H₇NO₂.

2. Determine the Molecular Formula:

The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.

Calculate the empirical formula mass:

• 5(C) + 7(H) + 1(N) + 2(O) = 5(12) + 7(1) + 1(14) + 2(16) = 101 g/mol

The molar mass can be calculated from the vapor density:

• Molar Mass = Vapor Density × 2 = 56.5 × 2 = 113 g/mol

Now, find the ratio of the molar mass to the empirical formula mass:

• Ratio = 113 / 101 ≈ 1.12

This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.

Therefore, the molecular formula is C₅H₇NO₂.

Hydrochloric acid is regarded as a strong acid because it

Détails de la réponse

Hydrochloric acid (HCl) is regarded as a strong acid because it ionizes completely in water. This means that when HCl is dissolved in water, it breaks down entirely into hydrogen ions (H⁺) and chloride ions (Cl^-). In a solution, there are no molecules of HCl left; only its ions are present.

This complete ionization results in a high concentration of hydrogen ions, which is a key characteristic of strong acids. Because there are more hydrogen ions available, hydrochloric acid can readily participate in chemical reactions, particularly those involving proton transfers, like neutralization reactions with bases.

In summary, the reason HCl is considered strong is due to its ability to consistently and completely ionize in an aqueous solution, not because of its physical state, source, or reactive nature with bases. Therefore, the property that defines it as a strong acid is that it ionizes completely.

  H2 ${}_2$SO4 ${}_4$

C2 ${}_2$H5 ${}_5$OH      →         C2 ${}_2$H4 ${}_4$

                    1700 ${}^0$C

The reaction above illustrates 

Détails de la réponse

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.

Which of the following is an air pollutant?

Détails de la réponse

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 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

Détails de la réponse

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

Détails de la réponse

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.

Nitrogen obtained from air is not absolutely pure because it contains the following except

Détails de la réponse

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.  

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 ]

Détails de la réponse

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**.

A major effect of oil pollution in coastal water is 

Détails de la réponse

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.

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

Détails de la réponse

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 general molecular formula Cn ${}_n$H2n?2 ${}_{2n?2}$ represents that of an

Détails de la réponse

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.

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

Détails de la réponse

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%.

A typical chemical reaction will be spontaneous if 

Détails de la réponse

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

The criteria for spontaneity is as follows:

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

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

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

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

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

Détails de la réponse

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.

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

Détails de la réponse

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.

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]

Détails de la réponse

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.

If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are

Détails de la réponse

To answer this question, let's break it down step by step:

Mass Number: The mass number is the total number of protons and neutrons in an atom's nucleus. In this case, the mass number is given as 31.

Stable Neutral Atom: A stable neutral atom has no overall electrical charge, meaning the number of protons (positively charged) must equal the number of electrons (negatively charged).

If we symbolize the number of protons by the atomic number (Z), we can say:

1. **Protons = Electrons** in a neutral atom.

2. **Mass Number (A) = Protons + Neutrons**.

Given that the mass number is 31, we have the equation:

A = Protons + Neutrons = 31.

Assuming a commonly known stable element like Phosphorus, which has an atomic number (Z) of 15, it means:

1. **Protons = 15**.

2. **Electrons = 15** (because it's a neutral atom).

3. To find Neutrons: Neutrons = Mass Number - Protons = 31 - 15 = 16.

So, in this scenario, the number of electrons is 15 and the number of neutrons is 16. This combination is found in the first option given.

Water gas obtained from the gasification of coke is made up of 

Détails de la réponse

The gasification of coke to produce water gas involves reacting coke, which is primarily composed of carbon, with steam. The main chemical reaction that occurs is:

C (s) + H₂O (g) → CO (g) + H₂ (g)

From this reaction, the main constituents of water gas are hydrogen (H₂) and carbon monoxide (CO), also known as carbon(II) oxide. Therefore, water gas obtained from the gasification of coke is made up of hydrogen and carbon(II) oxide.

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 ${}^3$).

Détails de la réponse

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 percentage of carbon(IV) oxide in air is

Détails de la réponse

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

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

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

Determine the empirical formula of an oxide of sulphur containing 60% of oxygen 

[S = 32, O = 16 ]

Détails de la réponse

To determine the empirical formula of an oxide of sulfur containing 60% of oxygen, we have to understand the concept of empirical formulas, which give the simplest whole-number ratio of atoms of each element in a compound.

Step 1: Assume 100g of the compound. In 100g of the compound:

• Oxygen (O) = 60g
• Sulfur (S) = 100g - 60g = 40g

Step 2: Convert masses to moles. Use the molar mass to find moles.

• Moles of Sulfur (S) = (40g) / (32g/mol) = 1.25 moles
• Moles of Oxygen (O) = (60g) / (16g/mol) = 3.75 moles

Step 3: Determine the simplest whole-number ratio.

To find the ratio, divide each mole value by the smallest number of moles calculated:

• For Sulfur: 1.25 / 1.25 = 1
• For Oxygen: 3.75 / 1.25 = 3

The simplest ratio of S:O is 1:3.

Thus, the empirical formula of the oxide is SO₃.

The IUPAC Nomenclature of CH3 ${}_3$CH2 ${}_2$C(CH3 ${}_3$)=C(CH3 ${}_3$)2 ${}_2$ for the compound is

Détails de la réponse

The compound in question is written as CH₃₃CH₂₂C(CH₃₃)=C(CH₃₃)₂₂, which seems to be intended as (CH₃)₃CH₂CH=C(CH₃)₃. The IUPAC nomenclature of organic compounds follows specific rules to name the compound uniquely such that it is understood universally. Here is a comprehensive breakdown:

1. Select the longest carbon chain that includes the highest-order functional group, which, in this case, is the alkene group (double bond).

2. The longest chain consists of 5 carbons, which gives us the root name "pentene". We choose the carbon chain such that the double bond gets the lowest possible number, starting from the end of the chain closest to the double bond.

3. Number the carbon atoms in the chain from the end closest to the double bond. The numbering direction will determine the position of the double bond and substituents. The double bond starts on carbon 2.

4. Identify and name the substituents attached to the carbon chain. In this case, there are two methyl groups on carbon 3. This means it is dimethyl as there are two of them.

Thus, the complete name of the compound is 2,3-dimethylpent-2-ene. Here, "2,3-dimethyl" indicates the position and quantity of methyl groups, "pent" indicates the longest chain with 5 carbons, and "-2-ene" indicates a double bond starting at the second carbon.

C2 ${}_2$H4(g) ${}_{4(g)}$  +  3O2(g) ${}_{2(g)}$ → 2CO2(g) ${}_{2(g)}$  +  2H2 ${}_2$O(g) ${}_{(g)}$

The above equation represents the combustion of ethene.If 10cm3 ${}^3$ of ethene is burnt in 50cm3 ${}^3$ of oxygen, what would be the volume of oxygen that would remain at the end of the reaction?

Détails de la réponse

Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant. 

                                      C2 ${}_2$H4(g) ${}_{4(g)}$  +  3O2(g) ${}_{2(g)}$ → 2CO2(g) ${}_{2(g)}$  +  2H2 ${}_2$O(g) ${}_{(g)}$

                                         1 mole      :       3 moles

Total volume required:     10 cm3 ${}^3$          50 cm3 ${}^3$

Reacted Volume:              10 cm3 ${}^3$          30 cm3 ${}^3$

Residual volume:               0                    (50 - 30) = 20 cm3

Scandium is not regarded as a transition metal because its ion has 

Détails de la réponse

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.

A gas when mixed with oxygen, it produces a very hot and early controllable flame. What is the name of the flame and where is it used?

Détails de la réponse

The Oxy-ethylene flame is a type of flame produced when oxygen is mixed with a gas called ethylene. This mixture results in a flame that is extremely hot and can be easily controlled. Such a flame is often used in industrial applications related to cutting and welding metals. The heat generated by an oxy-ethylene flame is sufficient to melt metals, allowing them to be welded together or cut apart efficiently.

Alkylation of benzene is catalyzed by 

Détails de la réponse

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.

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

Détails de la réponse

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**.

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

Détails de la réponse

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

The element which can combine with oxygen to form an acid anhydride of the form XO2 ${}_2$ is

Détails de la réponse

An Acid anhydride can be defined as a non-metal oxide which forms an acidic solution when reacted with water. 

Sulphur is the element that can combine with oxygen to form an acid anhydride of the form XO2 ${}_2$.

An acid oxide is a compound that forms an acid when it reacts with water. Non-metals in groups 4–7 form acidic oxides.

Détails de la réponse

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.

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 ${}^3$ ]

Détails de la réponse

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.

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

The bond between Na and X is likely to be

Détails de la réponse

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.

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

Détails de la réponse

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 difference in molecular mass between an alkene and alkyne with six carbon per mole is

Détails de la réponse

To determine the difference in molecular mass between an alkene and an alkyne, let's first take a look at their general formulas.

Alkene: An alkene is a hydrocarbon with at least one double bond between carbon atoms. For an alkene with six carbon atoms, the general formula is C_nH_2n. Therefore, for 6 carbon atoms, the molecular formula is C₆H₁₂.

Alkyne: An alkyne is a hydrocarbon with at least one triple bond between carbon atoms. For an alkyne with six carbon atoms, the general formula is C_nH_2n-2. Therefore, for 6 carbon atoms, the molecular formula is C₆H₁₀.

Now let's calculate the molecular masses:

Molecular mass of alkene (C₆H₁₂):

1. Carbon (C): 6 atoms x 12 g/mol = 72 g/mol
2. Hydrogen (H): 12 atoms x 1 g/mol = 12 g/mol
3. Total for C₆H₁₂ = 72 g/mol + 12 g/mol = 84 g/mol

Molecular mass of alkyne (C₆H₁₀):

1. Carbon (C): 6 atoms x 12 g/mol = 72 g/mol
2. Hydrogen (H): 10 atoms x 1 g/mol = 10 g/mol
3. Total for C₆H₁₀ = 72 g/mol + 10 g/mol = 82 g/mol

The **difference** in molecular mass between the alkene and alkyne is **84 g/mol - 82 g/mol** = 2 g/mol.

A liquid hydrocarbon obtained from fractional distillation of coal tar that is used in the pharmaceutical industry is

Détails de la réponse

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

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

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

If the solubility of KNO3 ${}_3$ at 300 ${}^0$C is 3.10 mol/dm3 ${}^3$ a solution containing 303g/dm3 ${}^3$ KNO3 ${}_3$ is likely to be

Détails de la réponse

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.

The amount of Faraday required to discharge 4.5 moles of Al3+ ${}^{3+}$ is

Détails de la réponse

To determine the amount of Faraday required to discharge 4.5 moles of Al³⁺ ions, it is essential to understand Faraday's laws of electrolysis and the concept of moles in chemistry.

When discharging Al³⁺ ions to form aluminum metal (Al), the reduction half-reaction involved is:

Al³⁺ + 3e^- → Al

From this equation, it can be seen that 3 moles of electrons (e^-) are required to discharge 1 mole of Al³⁺ ions to form 1 mole of aluminum metal.

A Faraday is the amount of electric charge carried by one mole of electrons. Therefore, 1 Faraday corresponds to the charge needed to discharge 1 mole of electrons.

Now, to discharge 4.5 moles of Al³⁺, we need:

4.5 moles of Al³⁺ × 3 moles of electrons (e^-)/mole of Al³⁺ = 13.5 moles of electrons

Since each Faraday discharges 1 mole of electrons, 13.5 moles of electrons correspond to 13.5 Faradays of charge.

Hence, the amount of Faraday required to discharge 4.5 moles of Al³⁺ ions is 13.5 Faradays.

CuOs ${}_s$   +  H2 ${}_2$(g ${}_g$)  ⇌ Cus ${}_s$   +  H2 ${}_2$O(g ${}_g$)

In the equation above, the effect of increased pressure on the equilibrium position is that

Détails de la réponse

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

CuO(s) + H₂(g) ⇌ Cu(s) + H₂O(g)

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

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

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

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

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

The combustion of candle under limited supply of air forms

Détails de la réponse

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).