JAMB UTME - Chemistry - 2024

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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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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.

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To determine the IUPAC name of a compound, follow these steps:

1. Identify the longest carbon chain: The compound should be named based on the longest continuous chain of carbon atoms. In this scenario, consider a butane chain (which has 4 carbon atoms) as it appears to feature prominently.
2. Number the carbon atoms: Number the chain in such a way to give the lowest possible numbers to the substituents. The substituents here are a bromo group (Br) and a methyl group (CH₃).
3. Assign locants and organize substituents alphabetically: There seem to be two groups attached to the main chain: a bromo group and a methyl group. When writing the full name, organize them alphabetically (i.e., bromo comes before methyl).
4. Write the complete IUPAC name: After numbering, identify the position of the bromo and methyl groups. If they are both on the second carbon, the name is "2-bromo, 2-methylbutane." But if they are on the third carbon, the name would be "3-bromo, 3-methylbutane."

Hence, by following these steps, if the bromo and methyl groups are both attached to the second carbon (lowest numbering possible), the IUPAC name of the compound is "2-bromo, 2-methyl butane."

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To determine the condition of the solution containing KNO₃ at 30⁰C, let's start by calculating the molarity of the given solution.

The molecular weight of KNO₃ (Potassium Nitrate) is approximately:

• K = 39 g/mol
• N = 14 g/mol
• O = 16 g/mol

Thus, KNO₃ = 39 + 14 + (16 * 3) = 101 g/mol.

Now, to determine the molarity of the given solution:

• 303 g/dm³ of KNO₃
• Moles of KNO₃ = 303 g / 101 g/mol = 3 mol/dm³

Compare with the solubility at 30⁰C:

• The solubility of KNO₃ at 30⁰C is given as 3.10 mol/dm³
• The calculated molarity of the given solution is 3 mol/dm³.

If we compare the values:

• The solution's molarity (3 mol/dm³) is less than the solubility limit (3.10 mol/dm³)

Hence, the solution is unsaturated because it can still dissolve more KNO₃ until it reaches the solubility limit of 3.10 mol/dm³.

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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.

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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.

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Quantum numbers are a set of numbers that describe the position and energy of an electron in an atom. 

When the quantum number is equal to 3, the possible values for the azimuthal quantum number  are 0, 1, and 2:

• l=0, The resulting subshell is an "s" subshell.
• l =1, The resulting subshell is a "p" subshell.
• l =2: The resulting subshell is a "d" subshell. 

The three possible sub-shells when n=3 are 3s, 3p, and 3d.

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The electrolysis of copper(II) tetraoxosulphate(VI) involves the deposition of copper at the cathode. To understand how many moles of copper are deposited when 2 Faraday of electricity is passed through, we need to consider Faraday's first law of electrolysis. Faraday's first law states that the mass (or number of moles) of a substance deposited at an electrode is directly proportional to the quantity of electricity that is passed through the electrolyte.

A Faraday (or Faraday constant) is the charge of one mole of electrons, which is approximately **96500 coulombs** (C). During electrolysis, the chemical reaction occurring at the cathode for copper deposition can be represented by the following equation:

Cu²⁺ + 2e^- → Cu

This equation shows that **2 moles of electrons** (represented by 2e^-) are needed to deposit **1 mole of copper (Cu)**.

If we have **2 Faradays** of electricity, it means we have **2 x 96500 C = 193000 C**. Since **1 Faraday (96500 C)** is required to deposit **0.5 mole** of copper, **2 Faradays** will deposit twice that amount:

0.5 mole of copper deposited per Faraday x 2 Faradays = **1.0 mole** of copper

Thus, when **2 Faradays** of electricity are passed through copper(II) tetraoxosulphate(VI) solution, **1.0 mole** of copper will be deposited.

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

The criteria for spontaneity is as follows:

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

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

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

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

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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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To determine the vapor density of the organic liquid hydrocarbon with the empirical formula XY, we first need to determine the **molecular formula** of the compound, which represents the actual number of atoms of each element in a molecule.

The **relative molar masses** of X and Y are given as 72 and 6, respectively. To find the molar mass of XY, we can add these values together:

Molar mass of XY = Molar mass of X + Molar mass of Y = 72 + 6 = 78 g/mol

Vapor density is defined as half of the molar mass of the compound, since vapor density is often compared to hydrogen, where hydrogen is taken as the standard with a molar mass of 2 g/mol. Therefore, vapor density can be calculated using the formula:

Vapor Density = (Molar Mass of the Compound) / 2

Substituting the molar mass of XY:

Vapor Density of XY = 78 / 2 = 39

Therefore, the vapor density of the hydrocarbon with the empirical formula XY is **39**.

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Biodegradable pollutants are substances that can be broken down by natural processes and microorganisms. However, when they are present in water systems, they can lead to several environmental and health issues. One of the main concerns is their potential to cause ill health. Here's why:

When biodegradable pollutants such as organic waste are introduced into water bodies, they are decomposed by bacteria and other microorganisms. This process consumes dissolved oxygen in the water. As the oxygen levels decrease, aquatic life such as fish and plants may suffer or die due to a lack of oxygen, disrupting the entire aquatic ecosystem.

This situation is known as eutrophication, which can lead to the excessive growth of algae, commonly referred to as algal blooms. These blooms often produce toxins that are harmful to both aquatic life and humans. Furthermore, when this polluted water is used for drinking, agriculture, or recreational purposes, it poses serious health risks to humans. These risks may include gastrointestinal infections, neurological disorders, and skin problems.

In addition, as the pollutants decompose, foul smells may be released, which can affect air quality in the vicinity, although the primary concern with biodegradable pollutants in water is related to how they affect water quality and health.

Therefore, it is crucial to properly manage and treat biodegradable pollutants before they enter water systems to prevent these health hazards. Failure to do so can result in significant environmental and health issues.

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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.

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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 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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Nitrogen obtained from air is not absolutely pure because it contains other gases, including:

• Carbon dioxide: Makes up less than 0.04% of the atmosphere, but is important environmentally
• Argon: A component of air - one of the inert gases (0.93%)
• Water vapour.  

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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.

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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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To determine the half-life of a first-order reaction, you can use the formula:

Half-life (\(t_{1/2}\)) = \(\frac{0.693}{k}\)

where \(k\) is the rate constant of the reaction. For the given problem, the rate constant (\(k\)) is 4.5 x 10^-3 s^-1.

Substituting the value of \(k\) into the formula, we have:

\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}}\)

Perform the division:

\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}} \approx 154\) s

Therefore, the half-life of the reaction is 154 seconds.

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

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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**.

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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).

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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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When water reacts with calcium ethynide, the gas produced is ethyne (also known as acetylene), which is represented by the chemical formula C₂H₂.

The chemical reaction involved is as follows:

CaC₂ + 2 H₂O → C₂H₂ + Ca(OH)₂

Let's break down this process to make it understandable:

• Calcium ethynide (CaC₂), often called calcium carbide, is a compound composed of calcium and the acetylide ion (C₂^2-).
• When water (H₂O) is added to calcium carbide, a chemical reaction occurs.
• The reaction produces **ethyne (C₂H₂)**, a colorless gas with a distinctive odor.
• Another product of this reaction is calcium hydroxide (Ca(OH)₂), which is a solid.

The key point to remember here is that the gas produced is **ethyne (C₂H₂)**, which is useful in various industrial applications, such as welding and as a precursor for other chemicals.

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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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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.

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The alloy known as **permalloy** is composed primarily of **iron** and **nickel**. Permalloy is a well-known magnetic alloy that typically consists of about **80% nickel and 20% iron**. It is renowned for having high magnetic permeability, meaning it can become magnetized easily, which makes it extremely useful in a variety of electrical and magnetic applications, such as transformers, memory storage, and magnetic shielding. The nickel in permalloy enhances the magnetic properties of the iron, giving the alloy its unique characteristics.

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Alkanoates, also known as fatty acid esters, are primarily found in lipids. Lipids are a broad group of naturally occurring molecules that include fats, waxes, sterols, fat-soluble vitamins (such as vitamins A, D, E, and K), and others. One of the main components of lipids is fatty acids and their derivatives, such as alkanoates.

To be more specific, alkanoates can be found in the form of triglycerides, which are the main constituents of body fat in humans and animals, as well as vegetable fat. Triglycerides are composed of glycerol bound to three fatty acids, and these fatty acids are usually present in the form of alkanoates.

Unlike proteins and rubber, which are made up of amino acids and polymers of isoprene respectively, lipids are the primary class of biomolecules where these alkanoate compounds can be found in significant amounts.

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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.

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

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

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

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The petroleum fractions that distill at 200–250°C are naphtha and kerosene,

• Bitumen: has a boiling point value of 5250 ${}^0$C
• Gasoline: Has a boiling range of 40–200°C
• Diesel: Has a boiling range of 250–350°C
• Fuel oil, paraffin, lubricants: Have a boiling range of 300–370°C
• Residue, crude oil: Has a boiling point of more than 370°C

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

In this particular reaction, the balanced chemical equation is:

HCl + NaOH → NaCl + H₂O

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

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

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

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

= 0.20 × 0.020

= 0.004 moles

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

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

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

= 0.004 moles / 0.12 mol/dm³

= 0.03333 dm³

Convert this volume from dm³ to cm³:

0.03333 dm³ × 1000 cm³ / dm³ = 33.33 cm³

Therefore, the volume of HCl required is 33.33 cm³.

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Alkylation of benzene is a part of a reaction class called **Friedel-Crafts alkylation**. In this reaction, an alkyl group is transferred to the aromatic benzene ring, making it a more complex molecule. The catalyst used in this process is **aluminium chloride (AlCl₃)**.

Here's how the reaction typically works:

• Role of Aluminium Chloride: Aluminium chloride acts as a Lewis acid catalyst in this process. It helps in generating a carbocation from the alkyl halide. This carbocation is highly electrophilic, which means it is positively charged and seeks electrons.
• Interaction with Benzene: Benzene has a high electron density due to its conjugated π-electron system, making it nucleophilic, or electron-rich. The carbocation readily attacks the benzene ring to form a new carbon-carbon bond, attaching the alkyl group to the ring.
• Completion of the Reaction: Finally, the aluminium chloride can be regenerated, returning to its initial form, and can be reused in the catalytic cycle. This aspect is crucial as it underscores the functioning of aluminium chloride as a catalyst.

In contrast, the other options wouldn't effectively catalyze alkylation of benzene for the following reasons:

• Palladium: This metal is often used in hydrogenation and coupling reactions but is not suitable for Friedel-Crafts alkylation.
• Nickel: Similar to palladium, nickel is used in hydrogenation but not in this type of reaction.
• Sunlight: It can provide energy for photochemical reactions but does not play the role of a catalyst in the alkylation of benzene.

Therefore, **aluminium chloride** is the catalyst used for the alkylation of benzene in Friedel-Crafts reactions.

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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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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".

Antwoorddetails

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).

Antwoorddetails

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)**.