JAMB UTME - Chemistry - 2024

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

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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.

Hydrochloric acid is regarded as a strong acid because it

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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.

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

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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.

What would be the order of the electrolytic cell in an industry intending the production of silver plated spoons?

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In the process of silver plating a spoon using an electrolytic cell, the correct configuration involves the following:

Cathode: The object to be plated, which in this case is the spoon. In an electrolytic cell, the cathode is where the reduction reaction occurs, and it is the surface on which the metal ions are deposited.

Anode: A rod made of silver. The anode is where oxidation occurs, meaning the silver rod will dissolve into the solution in the form of silver ions. These ions then move towards the cathode to be deposited as a thin layer on the spoon.

Electrolyte: A solution that contains a soluble silver salt (such as silver nitrate, AgNO₃). The silver ions from this salt help in the process of transferring the silver from the anode to the cathode.

Thus, the proper order for silver plating a spoon in an electrolytic cell for industrial production is: "Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt."

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

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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 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**.

The basicity of tetraoxophosphate(V) acid is

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The term basicity of an acid refers to the number of hydrogen ions (H⁺) that an acid can donate when it dissociates in water. In simpler terms, it's the number of replaceable hydrogen ions in one molecule of the acid.

Tetraoxophosphate(V) acid is another name for phosphoric acid, which has the chemical formula H₃PO₄. In this molecule, there are three hydrogen (H) atoms bonded to the phosphate group (PO₄).

When H₃PO₄ dissolves in water, it donates hydrogen ions in three steps:

• In the first step, it donates one H⁺ to form H₂PO₄⁻.
• In the second step, it can donate another H⁺ to form HPO₄²⁻.
• In the third and final step, it can donate the last H⁺ to form PO₄³⁻.

Therefore, phosphoric acid, or tetraoxophosphate(V) acid, can donate a total of three hydrogen ions. Hence, the basicity of tetraoxophosphate(V) acid is 3.

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.

Strong acids can be distinguished from weak acids by any of the following methods, EXCEPT

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To distinguish between strong acids and weak acids, we can employ several methods based on their chemical properties:

Conductivity Measurement: Strong acids dissociate completely in water, releasing more ions. Because ion concentration is directly related to electrical conductivity, strong acids exhibit higher conductivity than weak acids, which only partially dissociate.

Litmus Paper: This method helps determine if a solution is acidic or basic but does not provide detailed information about the strength (strong or weak) of an acid. Both strong and weak acids turn blue litmus red. Therefore, **litmus paper cannot effectively distinguish between a strong and a weak acid.**

Measurement of pH: Strong acids have a lower pH because they fully dissociate to release more hydrogen ions (H+), whereas weak acids have a relatively higher pH as they do not dissociate completely. Thus, pH measurement can distinguish the extent of acidity.

Measurement of Heat of Reaction: The heat of reaction can give insights into the strength of an acid because it involves the degree of ionization and the energetics associated with it. A strong acid will exhibit a different calorimetric response compared to a weak acid.

In summary, **litmus paper is not suitable for distinguishing between a strong and a weak acid**, as it only indicates acidity but does not reveal the strength of the acid.

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.

The volume in cm3 ${}^3$ of a 0.12 moldm−3 ${}^{-3}$ HCl required to completely neutralize a 20cm3 ${}^3$ of 0.20 moldm−3 ${}^{-3}$ of NaOH is 

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

In this particular reaction, the balanced chemical equation is:

HCl + NaOH → NaCl + H₂O

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

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

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

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

= 0.20 × 0.020

= 0.004 moles

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

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

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

= 0.004 moles / 0.12 mol/dm³

= 0.03333 dm³

Convert this volume from dm³ to cm³:

0.03333 dm³ × 1000 cm³ / dm³ = 33.33 cm³

Therefore, the volume of HCl required is 33.33 cm³.

The 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.

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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".

The percentage of carbon(IV) oxide in air is

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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%.

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**.

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 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

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 electronic configuration of an atom of Nitrogen is 1s2 ${}^2$ 2s2 ${}^2$ 2p1x ${}_x^1$ 2p1y ${}_y^1$ 2p1z ${}_z^1$ because the atom is

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

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

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

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

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

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%.

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.

Biuret test is a chemical test used for detecting the presence of 

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The Biuret test is a chemical test used for detecting the presence of proteins. When you perform a Biuret test, you are looking for peptide bonds, which are the connections between the amino acids in a protein. This is how it works:

• When a solution containing proteins is mixed with Biuret reagent, which contains copper sulfate, a chemical reaction occurs.
• In this reaction, the copper ions form a complex with the peptide bonds present in the protein.
• This complex gives the solution a characteristic purple color if proteins are present.
• The intensity of the purple color can often indicate the amount of protein present in the sample.

The test is specifically tailored to proteins because carbohydrates, amines, and alkanoates do not exhibit the required peptide bonds necessary for this color change. Therefore, the Biuret test is not suitable for detecting these compounds.

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

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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.

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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.

The hybridization scheme in ethyne is

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

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

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

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

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

In conclusion, the hybridization scheme in ethyne is sp.

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 amount of water a substance chemically combined with is called water of

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The amount of water that is chemically combined with a substance is referred to as water of crystallization. This is the water present in the crystalline form of a compound, necessary to maintain the structure of the crystals.

When certain substances crystallize from an aqueous solution, they incorporate a specific amount of water molecules into their crystal lattice structure. These water molecules are an integral part of the crystal and often affect its color, stability, and solubility. The water is combined in stoichiometric amounts, which means it is present in a fixed ratio relative to the rest of the molecule.

An example of this is copper(II) sulfate pentahydrate, which consists of copper(II) sulfate combined with five molecules of water per formula unit, represented as CuSO₄·5H₂O.

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 table above shows the formulae of some ions. In which of these compounds is the formula not correct?

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To assess the correctness of the chemical formulae for the given compounds, let's break down each compound:

1. Aluminium Tetraoxosulphate(VI), Al₂(SO₄)₃:

   Aluminium ion is denoted as Al³⁺, and the sulphate ion is SO₄^2-. To balance the charges between the positive and negative ions:

   2 x (+3) from aluminium ions = +6

   3 x (-2) from sulphate ions = -6

   Thus, the charges balance out, making the formula correct.

2. Calcium Trioxonitrate(V), Ca(NO₃)₂:

   Calcium ion is Ca²⁺, and the nitrate ion is NO₃^-. To balance the charges:

   1 x (+2) from calcium ion = +2

   2 x (-1) from nitrate ions = -2

   The charges balance out, therefore, this formula is also correct.

3. Iron(III) Bromide, Fe₃Br:

   Iron(III) ion is Fe³⁺, and bromide ion is Br^-. Each iron ion would pair with three bromide ions to balance the charges:

   FeBr₃, where:

   1 x (+3) from iron = +3

   3 x (-1) from bromide = -3

   The charges balance out in the correct formula which should be FeBr₃, making the given formula Fe₃Br incorrect.

4. Potassium Sulphide, K₂S:

   Potassium ion is K⁺, and sulphide ion is S^2-. To balance the charges:

   2 x (+1) from potassium ions = +2

   1 x (-2) from sulphide ion = -2

   The charges balance out, making this formula correct.

Therefore, the compound with the incorrect formula is Iron(III) Bromide where the proper chemical formula should be FeBr₃, not Fe₃Br.

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 ${}^3$ ]

Maelezo ya Majibu

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.

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

Maelezo ya Majibu

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.

Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca2+ ${}^{2+}$ ,   II. Zn2+ ${}^{2+}$ ,  III. Cu2+

Maelezo ya Majibu

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

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

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

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

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

I, II and III.

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?

Maelezo ya Majibu

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

The pH of a 0.001 mol dm−3 ${}^{-3}$ of  H2 ${}_2$SO4 ${}_4$ is 

[Log10 ${}_{10}$2 = 0.3]

Maelezo ya Majibu

The question is asking about the pH of a 0.001 mol dm⁻³ solution of H₂SO₄ (sulfuric acid). To find the pH, we need to understand how sulfuric acid dissociates in water.

Step 1: Dissociation of H₂SO₄
Sulfuric acid, H₂SO₄, is a strong acid and dissociates completely in water in two steps:

1. The first dissociation: H₂SO₄ → H⁺ + HSO₄^-

2. The second dissociation: HSO₄^- → H⁺ + SO₄^2-

For dilute solutions, particularly below 0.1 M, the first dissociation provides the major contribution to the H⁺ concentration. The second dissociation also contributes slightly to the acidity, but for simplicity and due to the dilute nature of this solution, the first step's contribution is primarily considered.

Step 2: Calculate the H⁺ Concentration
Since this is a strong acid and dissociates completely, for every 1 mole of H₂SO₄, we get 2 moles of H⁺. Therefore, for a 0.001 mol dm⁻³ solution of H₂SO₄, the concentration of H⁺ ions will be:

2 x 0.001 = 0.002 mol dm⁻³

Step 3: Calculate the pH
The pH is calculated using the formula: pH = -log[H⁺]

Substitute the H⁺ concentration:

pH = -log(0.002)

We know that log(10^-2) = -2 and log(2) = 0.3 (as provided), so:

pH = -(log(2) + log(10^-3))

pH = -(0.3 - 3)

pH = 3 - 0.3

pH = 2.7

Therefore, the pH of the 0.001 mol dm⁻³ H₂SO₄ solution is 2.7.

The Van der waals forces of attraction operates between

Maelezo ya Majibu

The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:

• Van der Waals forces include interactions that occur not from permanent charges (like those in ions) but from temporary or transient dipoles that momentarily exist even in non-polar molecules.
• These forces can be divided into different types, such as London dispersion forces and dipole-dipole interactions:
• London Dispersion Forces: This type of Van der Waals force is found in all molecules, regardless of whether they are polar or non-polar. It arises due to the momentary unequal distribution of electrons in a molecule, causing a temporary dipole which induces a dipole in a neighboring molecule.
• Dipole-Dipole Interactions: These occur between molecules that have permanent dipoles, such as when one end of a molecule is slightly negative and the other end is slightly positive. The positive end of one molecule will be attracted to the negative end of another.

Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.

A major effect of oil pollution in coastal water is 

Maelezo ya Majibu

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.

What method is suitable for the separation of gases present in air?

Maelezo ya Majibu

The suitable method for the separation of gases present in air is the fractional distillation of liquid air. This method is used due to the differing boiling points of the gases present in the air. Let me explain this in simple terms:

Air is a mixture of different gases, primarily nitrogen, oxygen, and argon, along with small amounts of other gases like carbon dioxide, neon, and krypton. Each of these gases turns into a liquid at different temperatures.

The process begins by cooling the air until it becomes a liquid. This is done at very low temperatures (around -200 degrees Celsius). Once the air is in liquid form, it is slowly warmed up in a distillation column. As it heats up, each gas boils off or evaporates at its respective boiling point and can be collected separately.

For example, nitrogen, which has a boiling point of about -196 degrees Celsius, will evaporate first and can be collected at the top of the distillation column. Following nitrogen, oxygen will evaporate at its boiling point of around -183 degrees Celsius. Finally, argon and other gases will do so at their respective temperatures.

In summary, fractional distillation of liquid air is effective because it takes advantage of the different boiling points to separate each gas from the air mixture.

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

Maelezo ya Majibu

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.

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 ]

Maelezo ya Majibu

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**.