JAMB UTME - Chemistry - 2024

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

Awọn alaye Idahun

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.

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?

Awọn alaye Idahun

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 scientist that performed the experiment on discharged tubes that led to the discovery of the cathode rays as a sub-atomic particle is 

Awọn alaye Idahun

The **scientist who performed the experiment on discharge tubes that led to the discovery of cathode rays as a sub-atomic particle** is J.J. Thomson.

In the late 19th century, J.J. Thomson conducted experiments using a cathode ray tube. This device involved an evacuated glass tube with electrodes at each end, through which an electric current was passed. **When a high voltage was applied, Thomson observed a stream of particles traveling from the negative electrode (cathode) to the positive electrode (anode).** These streams of particles were what he called "cathode rays."

Through his experiments, J.J. Thomson discovered that these cathode rays were composed of negatively charged particles. **He concluded that these particles were much smaller than atoms, and named them "electrons," which are now known to be sub-atomic particles.** His work was fundamental in advancing the atomic model and in understanding the structure of the atom.

Thomson's discovery was pivotal because it provided the first evidence that atoms are not indivisible, but rather consist of smaller subatomic particles. This **challenged the then-prevailing notion of atoms as indivisible units**, thus marking the birth of modern particle physics.

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

Awọn alaye Idahun

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

Esterification reaction is analogous to

Awọn alaye Idahun

The **esterification reaction** is analogous to a **condensation reaction**. In chemistry, a **condensation reaction** is a type of chemical reaction where two molecules or functional groups combine to form a larger molecule, with the simultaneous loss of a small molecule, usually water. **Esterification** specifically involves the reaction between an acid (often a carboxylic acid) and an alcohol, resulting in the formation of an **ester** and the release of a molecule of water.

To explain this further, in an esterification reaction:

• An **alcohol** provides one part of the ester, containing the oxygen and part of the carbon chain.
• A **carboxylic acid** provides the other part, which includes the carbon double-bonded to an oxygen atom (the carbonyl group).
• The reaction produces the **ester** and water (**H₂O**) as byproducts.

Conversely, the other types of reactions you've mentioned have different mechanisms:

• An **oxidation reaction** involves the losing of electrons by a molecule, atom, or ion.
• A **neutralization reaction** involves the reaction of an acid and a base to form salt and water.
• A **hydrolysis reaction** essentially does the reverse of esterification, breaking an ester down into its constituent acid and alcohol with the addition of water.

Therefore, given the nature of how molecules join and release water, it's clear that the **esterification reaction** is analogous to a **condensation reaction**.

The number of molecules of helium gas contained in 11.5g of the gas is

Awọn alaye Idahun

To find the number of molecules of helium gas in a given mass, we can use Avogadro's number and the molar mass of helium.

Step 1: Determine the molar mass of helium.

Helium is a noble gas with an atomic mass of approximately 4 grams per mole (g/mol).

Step 2: Calculate the number of moles in 11.5 grams of helium.

The formula to find the number of moles is:

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

So for helium:

Number of moles = 11.5 g / 4 g/mol = 2.875 moles

Step 3: Use Avogadro's number to find the number of molecules.

Avogadro's number is 6.022 x 10²³ molecules per mole.

The formula to find the number of molecules is:

Number of molecules = Number of moles x Avogadro's Number

Number of molecules = 2.875 moles x 6.022 x 10²³ molecules/mole

Number of molecules ≈ 1.73 x 10²⁴ molecules

Therefore, the number of molecules of helium gas in 11.5g of helium is approximately 1.73 x 10²⁴.

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

Awọn alaye Idahun

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.

Sulphur(IV)oxide can be used as a

Awọn alaye Idahun

Sulphur(IV) oxide has many uses including food preservation, refrigeration, laboratory reagent and solvent, sulphuric acid production, fumigant etc.Sulphur(IV) oxide  is a good refrigerant because it has a high heat of evaporation and can be easily condensed.

The ions responsible for permanent hardness in water are sulphates of 

Awọn alaye Idahun

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 IUPAC name of the compound above is

Awọn alaye Idahun

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."

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

Awọn alaye Idahun

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.

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]

Awọn alaye Idahun

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.

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?

Awọn alaye Idahun

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**.

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

Awọn alaye Idahun

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.

An organic compound with general formula RCOR' is an

Awọn alaye Idahun

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.

An example of highly unsaturated hydrocarbon is

Awọn alaye Idahun

To determine a highly unsaturated hydrocarbon, we must first understand the concept of saturation in hydrocarbons. **Saturated hydrocarbons** are compounds that contain the maximum possible number of hydrogen atoms, single-bonded to carbon atoms, and they are alkanes. **Unsaturated hydrocarbons** have one or more double or triple bonds between carbon atoms, which reduces the number of hydrogen atoms that can be bonded.

Examining the given options:

• C₂H₂ - also known as ethyne (or acetylene), has one carbon-carbon triple bond. A triple bond means it is highly unsaturated because it reduces the number of hydrogen atoms per carbon.
• CH₄ - known as methane, is a saturated hydrocarbon with single bonds only.
• C₂H₄ - known as ethene (or ethylene), has one carbon-carbon double bond, making it unsaturated.
• C₄H₁₀ - known as butane or isobutane, is another saturated hydrocarbon with single bonds only.

Based on this analysis, **C₂H₂** (ethyne) is a highly unsaturated hydrocarbon due to the presence of a **triple bond**. The triple bond signifies a greater level of unsaturation compared to double bonds in hydrocarbons like ethene (C₂H₄).

The indicator used in a titration between strong acid and weak base is

Awọn alaye Idahun

A titration is a process used to determine the concentration of an unknown solution by adding a solution of known concentration. The indicator used in a titration is a substance that changes color at the specific pH level of the solution, which usually happens at the equivalence point.

For a titration between a strong acid and a weak base, the solution at the equivalence point is slightly acidic. This is because the salt formed as a result of the neutralization reaction can undergo hydrolysis, producing an excess of hydronium ions (H₃O⁺) which makes the solution acidic.

Among the provided indicators, methyl orange is the most suitable for indicating this type of reaction because it changes color in an acidic pH range of about 3.1 to 4.4. It shifts from red at a pH below 3.1 to yellow at a pH above 4.4.

Therefore, for a titration involving a strong acid and a weak base, methyl orange is the appropriate indicator as it can show the end point effectively when the solution is slightly acidic. The pH at the equivalence point falls within the color change range of methyl orange.

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]. 

Awọn alaye Idahun

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.

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

Awọn alaye Idahun

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.

In the extraction of Aluminium, the silica impurity is removed by 

Awọn alaye Idahun

Aluminum is extracted from bauxite by electrolysis. The extraction proceeds in two stages;

1. Purification of the Bauxite: The impure bauxite is heated with sodium hydroxide solution to form soluble sodium tetrahydroxy aluminate (iii). The impurities in the ore which are iron (iii) oxide and trioxosilicate (iv) compounds are not soluble in the alkali. They are therefore filtered off as a sludge. 

Aluminum hydroxide crystals is then added to filtrate, NaAl(OH)4 ${}_4$ solution to induce the precipitation of Aluminum hydroxide.

2. The electrolysis of the pure alumina

Which of the following represents an order of increasing reactivity?

Awọn alaye Idahun

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

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

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

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

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

One of the following is not a water pollutant?

Awọn alaye Idahun

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.

A type of isomerism that ClCH=CHCl can exhibit is 

Awọn alaye Idahun

ClCH=CHCl can exhibit geometrical isomerism and positional isomerism. ClCH=CHCl can exhibit positional isomerism because the positions of the functional groups or substituent atoms are different. Positional isomerism occurs when compounds with the same molecular formula have different properties due to the difference in the position of a functional group, multiple bond, or branched chain. 

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 ${}^{-1}$ ]

Awọn alaye Idahun

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

In the graph above, y represents 

Awọn alaye Idahun

To understand what y represents in the graph, we need to think about what graphs in chemistry, specifically regarding energy changes in reactions, generally show.

Chemical reaction energy diagrams often depict a reaction's energy change as a curve from the reactants to the products, showing different energy levels throughout the process. The energy required to start a reaction or to transform the reactants into an activated complex (also known as the transition state) is crucial.

The height of this energy barrier is called the activation energy. This is the minimum amount of energy required to start a chemical reaction. The activation energy is represented by the peak in the energy graph between the reactant energy level and the top of the curve.

Therefore, in this context, y represents the activation energy needed for the reaction to proceed. Understanding activation energy is vital as it determines how quickly a reaction will occur. Reactions with a high activation energy tend to happen more slowly because it is less probable that the necessary energy for the reaction to occur spontaneously will be present.

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

Awọn alaye Idahun

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.

The group VIII elements are the inert gases because they 

Awọn alaye Idahun

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.

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

Awọn alaye Idahun

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 constituents of Alnico are Aluminium, Nickel and 

Awọn alaye Idahun

Alnico is a type of alloy that is known for its strong magnetic properties. The name "Alnico" comes from the elements it is primarily composed of: Aluminum (Al), Nickel (Ni), and Cobalt (Co). These elements are combined to form an alloy that retains its magnetism well and can operate at high temperatures, making it ideal for applications like electric motors, sensors, and various electronic devices.

While there are different variations of Alnico, the presence of Cobalt (Co) is essential for enhancing the magnetic properties of the alloy. The other elements listed, such as Magnesium (Mg), Manganese (Mn), and Copper (Cu), are not typical core constituents of Alnico. Although trace amounts of other elements like copper may sometimes be included in specific formulations, the primary and most significant component responsible for Alnico's powerful magnetic characteristics is Cobalt (Co).

The IUPAC nomenclature of the compound above is

Awọn alaye Idahun

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

Boyle's law can be expressed mathematically as

Awọn alaye Idahun

Boyle's Law describes the relationship between the volume and pressure of a given amount of gas held at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume. In simpler terms, if you decrease the volume of a gas, its pressure increases, provided the temperature remains constant, and vice versa.

The mathematical expression of Boyle's Law is PV = K, where:

• P represents the pressure of the gas.
• V represents the volume of the gas.
• K is a constant value when the temperature and the amount of gas are kept constant.

This relationship implies that if you multiply the pressure by the volume, the result will always be the same constant as long as no other variables are changed. This is the classic formulation of Boyle's Law, illustrating the inverse relationship between pressure and volume for a gas at constant temperature.

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

Awọn alaye Idahun

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%.

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

Awọn alaye Idahun

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 constituent of petroleum fraction used in surfacing road is

Awọn alaye Idahun

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

Reasons why bitumen is used for road surfacing:

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

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

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

Awọn alaye Idahun

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.

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

Awọn alaye Idahun

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.

The shape of the molecule of Carbon(IV) oxide is

Awọn alaye Idahun

The shape of the molecule of Carbon(IV) oxide, also known as carbon dioxide (CO₂), is linear. This is because of the following reasons:

• Carbon dioxide (CO₂) consists of one carbon atom covalently double bonded to two oxygen atoms.
• Carbon, being the central atom, has no lone pairs of electrons. It forms double bonds with each of the two oxygen atoms.
• The double bonds between carbon and the two oxygen atoms are on opposite sides of the carbon atom. This creates a structure where the atoms are aligned in a straight line, producing a linear shape.
• The bond angle between the oxygen-carbon-oxygen atoms is approximately 180 degrees, further confirming its linear nature.

Due to this arrangement, carbon dioxide has a symmetric shape, making it non-polar despite having polar covalent bonds. The pulling forces of the two oxygen atoms on either side of the carbon atom cancel each other out, reinforcing its linear configuration.

If a salt weighs 2g and upon exposure to the atmosphere weighs 1.5g, this is as a result of 

Awọn alaye Idahun

The observation that a salt initially weighs 2g, but reduces to 1.5g after exposure to the atmosphere is primarily due to the process called efflorescence.

Efflorescence occurs when a salt loses water molecules from its crystal structure when exposed to air, which is why the weight of the salt decreases over time. This loss of water is because some salts contain water of crystallization, and when such salts are exposed to the atmosphere, they can release this water, leading to a reduction in weight.

In this specific case, the salt has lost 0.5g of water, leading to the weight change from 2g to 1.5g. This process is different from hygroscopy, which involves absorbing moisture from the atmosphere, or deliquescence, where a substance absorbs moisture and eventually dissolves in it. It's also not related to effervescence, which is the escape of gas from an aqueous solution.

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

Awọn alaye Idahun

To solve the problem, we can use **Graham's law of effusion**. This law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is represented as:

Rate of diffusion of Gas X / Rate of diffusion of Gas Y = sqrt(Molar mass of Gas Y / Molar mass of Gas X)

According to the given information, gas X diffuses **twice as fast** as gas Y. This implies:

2 = sqrt(Molar mass of Gas Y / Molar mass of Gas X)

To eliminate the square root, square both sides of the equation:

(2)^2 = Molar mass of Gas Y / Molar mass of Gas X

This simplifies to:

4 = Molar mass of Gas Y / Molar mass of Gas X

Rearranging the equation, we find:

Molar mass of Gas Y = 4 * Molar mass of Gas X

This means that **Gas Y is four times as heavy as Gas X**. Therefore, the correct statement is:

• **Gas Y is four times as heavy as Gas X**

The law which states that a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is

Awọn alaye Idahun

The law that states a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is the law of definite proportion.

To explain this simply, let's consider water as an example. Water is made up of hydrogen and oxygen. According to the law of definite proportion, a sample of pure water taken from anywhere in the world will always contain the same ratio of hydrogen to oxygen by mass. Specifically, water will always have approximately 88.8% oxygen and 11.2% hydrogen by mass.

This is because a chemical compound has a fixed composition, regardless of the process used to create it or the source from which it is derived. The law of definite proportion, also known as the law of constant composition, is fundamental in chemistry because it supports the idea that chemical compounds are composed of elements in specific and fixed ratios. This does not change regardless of how the compound is prepared or where it is found.