JAMB UTME - Chemistry - 2024

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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.

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The stability of an atomic nucleus is primarily determined by the neutron/proton ratio. This refers to the number of neutrons in relation to the number of protons within the nucleus. Let's break down why this ratio is crucial for nuclear stability:

• Protons: These are positively charged particles found in the nucleus. Since like charges repel each other, having too many protons can cause the nucleus to become unstable due to the electrostatic repulsion between them.
• Neutrons: These are neutral particles that help to provide a binding force within the nucleus. Neutrons play a vital role in offsetting the repulsive forces between protons by providing additional nuclear forces, which are attractive in nature.

The right balance between the number of neutrons and protons helps in achieving nuclear stability.

• For light elements (like helium or carbon), a neutron/proton ratio near 1:1 is generally stable.
• For heavier elements, more neutrons are needed to stabilize the nucleus, leading to a higher neutron/proton ratio.

An imbalance in this ratio often results in an unstable nucleus, leading to radioactive decay as the nucleus attempts to reach a more stable form. This is why the neutron/proton ratio is a fundamental factor in the stability of the atomic nucleus.

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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.

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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.

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Phosphoric acid is a weak acid that can donate three hydrogen ions in water. Phosphoric acid partially ionizes when dissolved in an aqueous solution.

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The molecular formula C₃H₈O represents organic compounds that contain 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom. Let's elucidate the possible isomers, which are molecules with the same molecular formula but different structural arrangements.

1. Alcohols: One class of compounds that can form isomers for this formula are alcohols, which include a functional group -OH.

a. Propan-1-ol: This is a straight-chain alcohol where the -OH group is on the first carbon. The structure is as follows:

CH₃-CH₂-CH₂-OH

b. Propan-2-ol: This is another alcohol where the -OH group is on the second carbon, giving it a different structure and properties:

CH₃-CH(OH)-CH₃

2. Ethers: This is another class of possible isomers, where the oxygen atom is bonded to two alkyl groups.

c. Methoxyethane: Also known as ethyl methyl ether, it has a structure where the oxygen is in a bridge position between a methyl group and an ethyl group:

CH₃-O-CH₂-CH₃

These are the possible structural isomers for this molecular formula. Therefore, the compound C₃H₈O has three isomers overall:

• Propan-1-ol
• Propan-2-ol
• Methoxyethane

Thus, the answer is three distinct isomers.

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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.

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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.

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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. 

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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.

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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.

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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.

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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.

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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.

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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.

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When discussing the process of substances changing states, some substances can transition directly from a solid to a gas without passing through a liquid state. This process is called sublimation. However, not all substances exhibit this behavior. Let's examine the substances provided:

• Ammonium chloride (NH₄Cl): This substance can sublimate, meaning it can go directly from a solid state to a gaseous state when heated.
• Iodine (I₂): Iodine is known for sublimating as well. When iodine crystals are heated, they turn directly into a violet gas without becoming a liquid.
• Calcium carbonate (CaCO₃): This substance does not sublimate. Instead, when heated, it decomposes, releasing carbon dioxide gas and leaving calcium oxide as a residue. Therefore, it does not directly transition from a solid to a gas.
• Sulfur (S₈): This substance melts into a liquid before it evaporates, so it doesn't sublimate.

In conclusion, calcium carbonate (CaCO₃) is the substance that does not change directly from a solid to a gas when heated, as it undergoes a decomposition process instead.

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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)**.

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The bond between Na and X is most likely to be ionic. Let's break this down simply:

In the equation provided:

Na₂X ⇌ 2Na⁺ + X²⁻

The sodium (Na) atoms become positively charged ions (Na⁺), while X becomes a negatively charged ion (X²⁻). This change in charge occurs because sodium atoms donate electrons to the X atom. The donation of electrons by sodium to X indicates a transfer of electrons, which is a hallmark of an ionic bond.

In an ionic bond, electrons are transferred from one atom to another, resulting in a positively charged ion and a negatively charged ion. These oppositely charged ions attract each other, forming a strong ionic bond.

In summary, since sodium (Na) donates electrons to X forming ions, the bond between Na and X is most likely to be ionic.

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

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

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

Where:

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

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

Ag⁺ + e⁻ → Ag

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

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

Mg²⁺ + 2e⁻ → Mg

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

Given:

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

First, find the number of moles of Ag liberated:

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

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

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

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

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

Finally, calculate the mass of Mg liberated:

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

Rounding this to the closest answer provided:

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

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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.

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When a species undergoes oxidation, it experiences an increase in its oxidation number. Oxidation is a chemical process where a species loses electrons. In terms of oxidation number, electrons have a negative charge, so losing them results in an increase in charge. Thus, the oxidation number of the species becomes more positive or less negative.

To help understand, consider sodium (Na) reacting with chlorine (Cl₂) to form sodium chloride (NaCl):

• Initially, the oxidation number of Na is 0 (it's in a pure element form).
• In NaCl, sodium now has an oxidation number of +1 (because it loses an electron).

This change clearly shows that when sodium is oxidized, its oxidation number increases.

Therefore, the correct explanation is: a species undergoing oxidation will have its oxidation number increase.

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The compound of copper that is commonly used as a fungicide is **Copper(II) sulfate**, which is represented by the chemical formula **CuSO₄**.

Let's break this down for better understanding:

• Copper(II) sulfate (CuSO₄): This is a blue-colored, crystalline solid that is highly soluble in water, making it easy to apply as a fungicide. It is often used in a mixture called Bordeaux mixture, which is a blend of copper sulfate and slaked lime (calcium hydroxide). This mixture is effective in controlling fungal infections on crops like grapes, melons, and berries.

The other compounds listed do not serve as common fungicides:

• Copper(II) carbonate (CuCO₃): This compound is mainly used as a pigment or for other industrial applications, not as a fungicide.
• Copper(II) nitrate (Cu(NO₃)₂): This compound is primarily used in industrial processes, such as electroplating or as a catalyst, rather than in agriculture.
• Copper(II) chloride (CuCl₂): While it is used in a variety of chemical applications, it is not the standard choice for a fungicide in agricultural practices.

Therefore, the correct and widely used copper compound as a fungicide is Copper(II) sulfate (CuSO₄).

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The relationship between the two compounds is that they are isomers.

To understand why these compounds are isomers, let's break down their structures and definitions:

1. Structures of the Compounds:

• The first compound, CH₃-CH₂-OH, is known as ethanol. This compound is composed of two carbon atoms (C), six hydrogen atoms (H), and one oxygen atom (O).
• The second compound, CH₃-O-CH₃, is known as dimethyl ether. This compound also consists of two carbon atoms (C), six hydrogen atoms (H), and one oxygen atom (O).

2. Definitions:

• Isomers: Compounds that have the same molecular formula (same types and numbers of atoms) but different structural arrangements and properties.
• Allotropes: Different physical forms of the same element (e.g., carbon can exist as diamond, graphite, etc.).
• Isotopes: Atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei.
• Isobars: Atoms of different elements that have the same atomic weight (mass number) but different atomic numbers.

Both compounds have the same molecular formula: C₂H₆O. However, they have different arrangements of their atoms. Ethanol has a hydroxyl group (-OH) attached to an ethyl group (CH₃-CH₂-), while dimethyl ether involves two methyl groups (CH₃-) bonded to an oxygen atom (O). This difference in structure leads to different chemical and physical properties, despite having the same molecular formula. Hence, these two compounds are classified as isomers.

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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.

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In the conductance of aqueous CuSO₄ solution, the current carriers are the hydrated ions.

Here's why:

• CuSO₄ in Water: When copper sulfate (CuSO₄) is dissolved in water, it dissociates into copper ions (Cu²⁺) and sulfate ions (SO₄^2-).
• Role of Water: In an aqueous solution, these ions become surrounded by water molecules. This surrounding by water is known as "hydration."
• Conductivity: The movement of these hydrated copper ions and sulfate ions through the solution constitutes the current in the solution. They are the mobile charge carriers that allow electric current to flow.

The other options can be understood as follows:

• Mobile Electrons: These are the primary current carriers in metallic conductors, not in ionic solutions like CuSO₄.
• Metallic Ions: This term generally refers to ions in a metallic lattice. In an aqueous solution, the relevant ions are hydrated, not metallic.
• Hydrated Electrons: This is not a correct term for describing the charge carriers in an aqueous CuSO₄ solution.

The correct answer is therefore hydrated ions because they enable the conduction of electricity through the aqueous solution.

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To understand the geometrical isomers of butene, we need to explore its structure. Butene has four carbon atoms, and there are various structural forms that butene can take. These structural forms include linear or branched chains, with a double bond present between carbon atoms.

Geometric isomerism is a type of stereoisomerism. It occurs due to restricted rotation around the double bond, leading to different spatial arrangements of groups attached to the carbons forming the double bond. The geometric isomerism primarily occurs in alkenes like butene where the positions of substituents can vary.

Let's consider the different types of butene, focusing on the possibility of geometrical isomerism:

• 1-butene: This is a straight-chain isomer with the double bond starting at the first carbon. It doesn't show geometrical isomerism because the double bond is at the end of the chain, so there are no different spatial arrangements of groups possible.
• 2-butene: Here the double bond is between the second and third carbon. In this case, geometrical isomerism is possible because different groups (or atoms) attached to the carbon atoms can have different spatial arrangements. Specifically:
  • cis-2-butene: Both methyl groups (CH₃) are on the same side of the double bond.
  • trans-2-butene: The methyl groups are on opposite sides of the double bond.
• Isobutene (or 2-methylpropene): This is a branched isomer, and it does not have geometric isomers because the substituents around the double bond do not allow for different spatial configurations on either side of the double bond.

In conclusion, for butene, only 2-butene has geometrical isomers (cis and trans). Therefore, the number of geometric isomers is 2.

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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.

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The main constituent of water-glass is sodium trioxosilicate(IV). Water-glass, also known as liquid glass, is common terminology for a mixture of sodium silicate and water. The primary chemical component in water-glass is sodium silicate, which includes sodium ions (Na⁺) bonded with silicate ions (SiO₄^4-).

Essentially, when sodium silicate is dissolved in water, it results in a viscous liquid that can be utilized in various applications such as in cements, passive fire protection, textile and lumber processing, and as a sealant. Sodium trioxosilicate(IV) forms a significant part of this mixture as it reacts with other compounds to create a hardened, glass-like structure when it dries. Therefore, when water-glass is mentioned, it is mostly referring to solutions that have sodium trioxosilicate(IV) as their principal compound.

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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.

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The process described appears to depict a nuclear reaction involving a nuclear transmutation. Let's break down the process:

1. The starting element is initially denoted as "23892", which represents Uranium-238. In nuclear notation, "23892" indicates an atomic mass number of 238 and an atomic number of 92.

2. The next step so happens with the element "238"; however, the numbers remain: "92" indicates that the atomic number is unchanged, suggesting no change in the element. This often means a step in between of hypothetical notation.

3. Then there's the occurrence of adding a "U + 10", which again leaves the original atomic number "92".

4. In subsequent steps, it seems that the number "n" transitions to become "23992". The mass number has increased by one unit, turning the initial isotope into "23992", which represents Uranium-239.

The key point here is the transition from Uranium-238 to Uranium-239, which typically happens through the process of a neutron absorption in which a neutron is added, resulting in a change of the mass number. Such a process often leads to the creation of a radioactive isotope.

Therefore, the process described is indicative of producing a radioactive isotope, specifically Uranium-239.

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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.

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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

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

C (s) + H₂O (g) → CO (g) + H₂ (g)

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

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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.

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An air pollutant is any substance in the air, introduced by natural or human activity, that causes harm or discomfort to living organisms, or damages the environment. Let's analyze the substances mentioned:

1. O₂ (Oxygen)
Oxygen is the gas we need to breathe. It's not considered an air pollutant because it is essential for human and animal life, as well as many natural processes.

2. CO (Carbon Monoxide)
Carbon Monoxide is a colorless, odorless gas that is produced by burning fuel (like in cars and factories). This gas can be very dangerous if there is a lot of it, as it can prevent oxygen from entering the bloodstream. Because of its harmful effects, it is considered an air pollutant.

3. H₂ (Hydrogen)
Hydrogen, while a flammable gas, is generally not harmful to the air or to organisms when it is released into the environment. Therefore, it is not considered an air pollutant.

4. O₃ (Ozone)
Ozone is a bit tricky because it is both good and bad. Higher up in the atmosphere, it forms a layer that protects us from the sun’s UV radiation. However, at ground level, it is a harmful air pollutant. Ground-level ozone can cause health problems such as respiratory difficulties, so in this context, it is considered an air pollutant.

In conclusion, the substances that are considered air pollutants in this context are Carbon Monoxide (CO) and ground-level Ozone (O₃).

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When a strong acid reacts with a strong base, the result is the formation of a neutral salt. This reaction is a part of a chemical process known as neutralization.

Let's break it down further:

• A strong acid is an acid that completely dissociates in solution to release hydrogen ions (H⁺). An example of a strong acid is hydrochloric acid (HCl).
• A strong base is a base that completely dissociates in solution to release hydroxide ions (OH⁻). An example of a strong base is sodium hydroxide (NaOH).

During a neutralization reaction, the hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O). Meanwhile, the remaining ions (for example, Na⁺ from NaOH and Cl⁻ from HCl) come together to form a compound known as a salt. This salt does not affect the acidity or basicity of the solution, hence it is considered neutral.

Therefore, the salt formed in such a reaction is a neutral salt, which is what is referred to as a normal salt in the options provided.

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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.

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

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

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

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

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

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

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

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To determine the quantity of electricity required to deposit 180g of Ag (silver) from molten silver trioxonitrate(V), we need to understand the concept of electrolysis. During electrolysis, a metal can be deposited according to Faraday's laws of electrolysis.

The equivalent weight of a substance is calculated by dividing the atomic mass by the valency. For silver (Ag), the atomic mass is given as 108 and the valency of silver in AgNO₃ is 1. This makes the equivalent weight of Ag 108 g/equivalent.

According to Faraday's first law of electrolysis:

Mass of substance deposited = (Equivalent weight × Quantity of electricity (in coulombs) ) / Faraday's constant (96500 C/mol)

Let's calculate the number of equivalents of silver deposited:

Number of equivalents of Ag = Mass of Ag / Equivalent weight = 180 g / 108 g/equivalent = 5/3 equivalents

The quantity of electricity required to deposit 1 equivalent of a substance is 1 Faraday (F) = 96500 C.

Therefore, the total quantity of electricity required:

Quantity of electricity = Number of equivalents × Faraday's constant

Quantity of electricity = (5/3 equivalents) × 1 F = 5/3 F = 1.67 F

Therefore, 1.67 Faraday is required to deposit 180g of Ag from a molten silver trioxonitrate(V).