JAMB UTME - Chemistry - 2024

When Sulphur(IV)oxide is passed into solution of acidified tetraoxomanganate(VII), the colour changes from 

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When Sulphur(IV) oxide (SO₂) is passed into a solution of acidified tetraoxomanganate (VII) (KMnO₄), it acts as a reducing agent. This reaction involves the reduction of potassium permanganate (KMnO₄), which is characterized by a distinctive color change.

The tetraoxomanganate (VII) ion (MnO₄^-) is purple in color. During the reaction, SO₂ gets oxidized while the MnO₄^- ion is reduced to Mn²⁺, which is almost colorless or pale pink, depending on the concentration.

Thus, the color of the solution changes from purple to almost colorless as the reaction progresses.

In a chemical reaction, surface area of reactants can affect

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The surface area of reactants affects the rate of a reaction between limestone and hydrochloric acid because it increases the number of collisions between the particles of the reactants. For example, if you have a large marble chip of calcium carbonate and hydrochloric acid, the acid can't reach all the calcium carbonate in the middle of the chip. If you break the marble chip into smaller pieces, you'll have a larger surface area for the acid to react with, and the reaction will happen faster. 

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

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

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

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



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

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



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

Convert ml to liters: 80 ml = 0.080 L



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

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

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

Concentrated sodium chloride solution is electrolyzed using mercury cathode and graphite anode. The products at the anode and the cathode respectively are

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When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode

At the anode, 2Cl− ${}^{-}$  →  Cl2 ${}_2$ +  2e− ${}^{-}$

At the cathode, 2H+ ${}^{+}$ +  2e− ${}^{-}$ →  H2 ${}_2$

During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.

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

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When the acidified potassium dichromate (\(K_2Cr_2O_7\)) solution changes from orange to green, it indicates a chemical reaction is occurring where the chromium in the dichromate ion is being reduced. In this context, acidified \(K_2Cr_2O_7\) is commonly used as an oxidizing agent.

The change in color from orange (dichromate ion) to green (chromium ion) suggests that the dichromate ion is being reduced, and something in the person's breath is being oxidized.

The substances that can be oxidized in the breath are organic compounds, typically those containing functional groups with oxidizable hydrogen atoms or structures.

• Alkanol: The presence of alkanols (such as alcohols) can result in this color change because alkanols can be oxidized to aldehydes or carboxylic acids by the dichromate ion. In an alcohol, the oxidation is facilitated by the presence of an -OH group.

Therefore, when the color of acidified potassium dichromate changes from orange to green, it suggests the presence of an alkanol.

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.

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

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

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

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

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

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

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

This simplifies to:

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

Rearranging the equation, we find:

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

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

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

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.

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

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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.

An example of a physical change is 

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A physical change involves a change in the physical properties of a substance, without a change in its chemical composition. This means that the substance remains the same at the molecular level, despite how it might appear differently.

An example of a physical change from the given options is the liquefaction of liquids. In this process, a substance transitions from a solid or gas to a liquid state. This change is purely physical because the molecular structure of the substance does not change; only its state or form does. Importantly, such a change is usually reversible, meaning the substance can return to its original state. For instance, water can change into ice (frozen) or steam (vapor), and can still revert back to liquid water.

On the other hand, the other options involve chemical changes, where the original substances undergo chemical reactions to form new substances with different properties, thus altering the molecular structure depending on the option.

The highest isotope of hydrogen is

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Hydrogen has three naturally occurring isotopes, and each of them contains the same number of protons but different numbers of neutrons. Let's briefly differentiate them:

• Protium: This is the most common isotope of hydrogen, which has one proton and no neutrons. It's simply referred to as hydrogen.
• Deuterium: This isotope of hydrogen has one proton and one neutron. It is heavier than protium because of the additional neutron.
• Tritium: This is the heaviest naturally occurring isotope of hydrogen, consisting of one proton and two neutrons.

The highest isotope of hydrogen is tritium because it has the most neutrons and, therefore, the greatest atomic mass compared to the other isotopes. It is also noteworthy that tritium is radioactive, while the other hydrogen isotopes are stable.

An example of an amphoteric oxide is

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An example of an amphoteric oxide is Al₂O₃ (aluminum oxide).

Amphoteric oxides are special because they can act as both acidic and basic oxides. This means they can react with both acids and bases to form salts and water, showcasing their dual behavior.

Here is how it works:

• Reacts with Acids: When aluminum oxide reacts with an acid, such as hydrochloric acid (HCl), it behaves like a base and forms a salt. For example, it will react to form aluminum chloride and water:
  Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O.


• Reacts with Bases: When it reacts with a strong base, like sodium hydroxide (NaOH), it acts like an acid and also forms a salt and water. An example reaction might be:
  Al₂O₃ + 2 NaOH + 3 H₂O → 2 NaAl(OH)₄.

In contrast, oxides like CuO (copper(II) oxide) are basic oxides, and K₂O (potassium oxide) is a basic oxide as well. They don't exhibit both acidic and basic properties.

Therefore, the amphoteric nature of Al₂O₃ is what distinguishes it from common oxides that are strictly acidic or basic. This property is crucial in various chemical processes and applications.

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

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To determine how much of a radioactive element remains after a certain period, we use the concept of half-life. The half-life of a substance is the time it takes for half of the initial amount of a radioactive element to decay. In this example, the half-life is given as 50 days.

We want to know how much of a 5g sample remains after 200 days. First, calculate how many half-lives occur in 200 days:

Number of half-lives = Total time elapsed / Half-life
                      = 200 days / 50 days
                      = 4 half-lives

Next, we calculate the remaining amount after each half-life period:

• Initial amount: 5g
• After first half-life (50 days): 5g / 2 = 2.5g
• After second half-life (100 days): 2.5g / 2 = 1.25g
• After third half-life (150 days): 1.25g / 2 = 0.625g
• After fourth half-life (200 days): 0.625g / 2 = 0.3125g

After 200 days, 0.31g of the radioactive element remains.

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

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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.

The group VIII elements are the inert gases because they 

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The group VIII elements, also known as the noble gases, are called inert gases primarily because they all have completely filled valence shells. In a very simplified explanation:

1. Complete Valence Shells: All the noble gases have their outermost shells completely filled with electrons. This configuration is considered very stable and requires no additional electrons to reach stability, unlike other elements that may gain, lose, or share electrons to achieve a full valence shell.

2. Highly Stable: Due to this completely filled valence shell, the noble gases do not readily react with other elements to form compounds. Their stability comes from the fact that they do not need to bond with other elements to achieve a more stable state.

3. Examples: For instance, Helium (He) has two electrons filling its first shell, Neon (Ne) has eight electrons in its second shell, and similarly, other noble gases also have fully occupied outer shells.

This property is why the noble gases are termed "inert," which means they are largely non-reactive.

The 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 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.

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

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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.

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**.

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

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

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

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

An example of a physical change is 

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An example of a physical change is the boiling of water. Let me explain why this is considered a physical change:

A physical change is a change where the substances involved do not change their chemical composition, meaning they remain the same substance, just in a different form or appearance. In the case of boiling water, when water is heated to its boiling point, it changes from a liquid to a gas (steam), but it is still comprised of water molecules (H₂O). The change is reversible, so the gas can condense back into liquid water without any new substance being formed.

On the other hand:

• Exposing sodium metal to air: This is a chemical change as sodium reacts with components in the air, such as oxygen, forming new substances like sodium oxide.
• Dissolving calcium metal in water: This is a chemical change because calcium reacts with water to produce calcium hydroxide and hydrogen gas, forming new substances.
• Burning of kerosene: This is a chemical change as kerosene reacts with oxygen in the air to form new substances like carbon dioxide and water, and this transformation is irreversible.

Thus, boiling water is an excellent example of a physical change as it involves only the change in the state of matter without altering the substance's identity.

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 ${}^3$ ]

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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 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.

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?

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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

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.

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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.

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When a metal reacts with an acid, a chemical reaction takes place in which the metal displaces the hydrogen in the acid. This reaction produces a salt and hydrogen gas is liberated in the process.

Let's break it down further:

• Metal: Metals are elements that can react with acids to produce salts.
• Acid: Acids typically contain hydrogen ions (H⁺) which are replaced by metal ions during the reaction.
• Salt: A salt is an ionic compound that forms when the hydrogen ions of the acid are replaced by metal ions.
• Hydrogen Gas: This is the gas that is liberated during the reaction. You may observe bubbles forming as the gas escapes.

The general equation for the reaction is:

Metal + Acid → Salt + Hydrogen Gas

For example, when zinc (a metal) reacts with hydrochloric acid (an acid), the reaction is as follows:

Zn + 2HCl → ZnCl₂ + H₂

Here, zinc chloride (a salt) and hydrogen gas are produced. This illustrates that salt and hydrogen gas are formed when a metal reacts with an acid.

Kerosene is used as solvent for 

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Kerosene is commonly used as a solvent for paints. Let me explain why in a simple way:

Kerosene is a type of fuel that is composed of hydrocarbons, which are molecules made up of hydrogen and carbon atoms. These hydrocarbons give kerosene the ability to dissolve other similar substances.

Paints often contain oils and other hydrocarbon-based compounds. Since kerosene is also hydrocarbon-based, it can effectively dissolve and thin these compounds. This makes it suitable for use as a solvent in paints, allowing the paint to be thinned or cleaned up after use. This property makes kerosene a good choice for cleaning brushes and other painting tools or for dissolving dried paint.

On the other hand, sulphur, gums, and fats are typically not dissolved effectively by kerosene because of their different chemical properties. Therefore, kerosene as a solvent is primarily useful in the context of working with paints and similar hydrocarbon-based materials.

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

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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.

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

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When **ethyne** (also known as acetylene) is passed through a hot tube containing finely divided iron, a process called decomposition occurs. The heat causes the ethyne molecules to break down, and under these conditions, they **re-combine** to form structures that result in more complex molecules.

The key transformation involves the conversion of these ethyne molecules into **aromatic compounds**. Aromatic compounds, such as **benzene**, have a distinct ring structure and are characterized by **stability** due to resonance (a phenomenon where electrons are delocalized over a certain structure, providing extra stability).

Thus, when ethyne is passed through a hot iron tube, it undergoes trimerization to form benzene, an **aromatic** compound. Therefore, the product formed is **aromatic**.

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.

Benzene formed nitrobenzene at temperature of 600 ${}^0$C when it reacts with mixture of concentrated trioxonitrate(V) acid and concentrated 

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The reaction described is the nitration of benzene to form nitrobenzene. This is an example of an electrophilic aromatic substitution reaction. **Nitration** involves replacing a hydrogen atom on a benzene ring with a nitro group (NO2). This reaction requires a nitrating mixture composed of concentrated nitric acid (trioxonitrate(V) acid) and concentrated sulfuric acid (tetraoxosulphate(VI) acid). Let me explain why:

Nitration is typically carried out using a mixture of **concentrated nitric acid and concentrated sulfuric acid** at a temperature of around **60°C**. The role of sulfuric acid in this mixture is to act as a catalyst and a dehydrating agent. It helps generate the nitronium ion (NO2⁺), which is the active electrophile that attacks the benzene ring.

Here's a simplified mechanism for this reaction:

• The concentrated sulfuric acid protonates the nitric acid, forming nitronium ion (NO2⁺).
• This powerful electrophile (NO2⁺) then attacks the electron-rich benzene ring, leading to the formation of an arenium ion.
• Lastly, the arenium ion loses a proton to regenerate the aromatic character of the benzene ring, resulting in the formation of nitrobenzene.

None of the other options listed (hydrochloric acid, phosphoric acid, and hydrogen iodide) contain the necessary combination of properties to generate the nitronium ion and facilitate the nitration of benzene.

Therefore, the correct mixture to carry out the nitration of benzene, forming nitrobenzene at a temperature of 60°C, is a combination of **concentrated nitric acid and concentrated sulfuric acid (tetraoxosulphate(VI) acid)**.

A type of isomerism that ClCH=CHCl can exhibit is 

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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. 

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

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To determine the solubility of sodium chloride (NaCl) in mol/dm³ at the given temperature, you need to first calculate the number of moles of NaCl dissolved.

Step 1: Calculate the molar mass of NaCl.

The molar mass of a compound is found by adding the atomic masses of its constituent elements:

- Sodium (Na) has an atomic mass of 23.

- Chlorine (Cl) has an atomic mass of 35.5.

Thus, the molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

Step 2: Calculate the number of moles of NaCl.

The formula to calculate moles is:

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

Given mass of NaCl = 127 g,

Number of moles = 127 g / 58.5 g/mol ≈ 2.17 mol

Step 3: Calculate the solubility in mol/dm³.

Since the sodium chloride is dissolved in 1.0 dm³ of water, the solubility is the same as the number of moles, since the volume is already 1.0 dm³.

Therefore, the solubility of sodium chloride at that temperature is 2.17 mol/dm³.

Rounded to the options given, 2.17 mol/dm³ is approximately equal to 2.2 mol/dm³.

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.

Which of these is the most preferred separation technique for the isolation of solutes where the purity of the constituent is of utmost importance?

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When the **purity of solutes** is of utmost importance, the most preferred separation technique is **recrystallization**. This method is widely used in chemistry for purifying solid compounds.

Here's a simple explanation of **recrystallization**:

1. **Dissolving the Impure Compound**: The impure solid is dissolved in a suitable hot solvent. The choice of solvent is crucial; it should dissolve the compound well at high temperatures but poorly at low temperatures.

2. **Cooling the Solution**: The solution is slowly cooled. As it cools, the solubility of the compound in the solvent decreases, causing the pure compound to form crystals and precipitate out of the solution.

3. **Collection and Drying of Crystals**: The pure crystals are collected through filtration and then allowed to dry, separating them from any remaining impurities that stay dissolved in the solvent.

The **advantage** of recrystallization is that it allows for the **removal of impurities** that are either more soluble than the desired compound at low temperatures or less soluble at high temperatures, resulting in a more purified product. Therefore, when achieving high purity is a priority, **recrystallization** is often the method of choice.

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

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

Here is a simplified explanation:

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

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

In the treatment of water for municipal supply, chlorine is used to 

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In the treatment of water for municipal supply, chlorine is used to kill germs. This process is known as chlorination. Chlorine is a very effective disinfectant and is used to eliminate harmful microorganisms such as bacteria, viruses, and protozoans that may be present in the water. By doing so, chlorine helps to ensure that the water is safe for human consumption and protects public health by preventing waterborne diseases. It is important to note that **chlorine is not used to prevent tooth decay, prevent goitre, or to remove colour or odour** in water treatment for municipal supply.