JAMB UTME - Chemistry - 2024

The ions responsible for permanent hardness in water are sulphates of 

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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.

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.

In the conductance of aqueous CuSO4 ${}_4$ solution, the current carriers are the

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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.

The compound of Copper used as a fungicide is

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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).

During the fractional distillation of crude oil, the fraction that distills at 200 - 2500 ${}^0$ C is

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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

When n = 3, the quantum number of an element is

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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.

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

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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.

Cx ${}_x$Hy ${}_y$O + 5O2 ${}_2$  →  4CO2 ${}_2$  +  4H2 ${}_2$O

Cx ${}_x$Hy ${}_y$O in the equation is

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Cx ${}_x$Hy ${}_y$O + 5O2 ${}_2$  →  4CO2 ${}_2$  +  4H2 ${}_2$O

On balancing the equation, we should have

X = 4 , y = 8  and O = 2  ⇒   C4 ${}_4$H8 ${}_8$O2 ${}_2$

Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.

We finally have  C2 ${}_2$H4 ${}_4$O

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.

The constituents of Alnico are Aluminium, Nickel and 

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

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

A factor that does not affect the rate of a chemical reaction is

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In evaluating the factors that affect the rate of a chemical reaction, we can look at each of the possible influences: surface area, temperature, volume, and catalyst.

Surface Area: When you increase the surface area of reactants, it allows more particles to collide with each other per unit of time, which in turn increases the rate of reaction. Imagine smaller particles like powders reacting faster than larger chunks because they have a greater surface exposed to the other reactants.

Temperature: Increasing the temperature usually increases the rate of reaction. Higher temperatures cause particles to move faster, increasing the energy of collisions, and therefore increasing the chance of successful reactions.

Catalyst: A catalyst is a substance that increases the rate of a chemical reaction without being consumed by it. It lowers the activation energy needed for the reaction to occur, thus allowing it to proceed faster.

Volume: The volume of the container or the amount of space in which a reaction occurs generally does not directly affect the rate of the reaction. While changing the volume can alter pressure or concentration in gaseous reactions, which in turn affects the rate, the volume itself is not a direct factor affecting reaction rate.

Therefore, the factor that does not directly affect the rate of a chemical reaction is volume. It indirectly affects reaction rates by altering concentration or pressure in certain reaction conditions, but it is not a direct influencing factor on its own.

Heat of solution involves two steps that is accompanied by heat change. The energies involved in this steps are

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The heat of solution refers to the overall energy change that occurs when a solute dissolves in a solvent. This process involves breaking and making of intermolecular forces, and it can be broken down into two main steps that are each accompanied by heat change. The energies involved in these steps are:

Lattice energy: This is the energy required to break the bonds between the ions in the solid crystal lattice of the solute. Breaking these bonds requires energy, and this step is usually endothermic, meaning it absorbs heat from the surroundings. The more energy needed to break the lattice, the higher the lattice energy.

Hydration energy: Once the lattice is broken, the ions are surrounded by solvent molecules, typically water, in a process known as hydration. The energy released when the solvent molecules interact with and stabilize the ions is called the hydration energy. This step is usually exothermic, meaning it releases heat into the surroundings.

In conclusion, the two energies involved in the heat of solution are lattice energy and hydration energy. The balance between these two energies determines whether the overall process of dissolving a solute in a solvent is endothermic or exothermic.

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

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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

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

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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.

CuOs ${}_s$   +  H2 ${}_2$(g ${}_g$)  ⇌ Cus ${}_s$   +  H2 ${}_2$O(g ${}_g$)

In the equation above, the effect of increased pressure on the equilibrium position is that

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To understand the effect of increased pressure on the equilibrium position of the given reaction:

CuO(s) + H₂(g) ⇌ Cu(s) + H₂O(g)

We need to consider Le Chatelier's Principle. According to this principle, if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust itself to counteract the effect of the change and re-establish equilibrium.

For the reaction in question, let's consider the number of gas molecules on each side of the equation:

• Left side (reactants): 1 molecule of H₂(g)
• Right side (products): 1 molecule of H₂O(g)

Since both sides of the equation have the same number of gas molecules, an increase in pressure will not favor a shift to either the left or the right because the number of moles of gas on both sides of the equilibrium is the same.

Therefore, the effect of increased pressure on the equilibrium is that there is no effect. The position of the equilibrium remains unchanged, and pressure changes do not influence the production of more H₂(g) or H₂O(g) in this specific reaction.

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.

A typical chemical reaction will be spontaneous if 

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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.

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**.

An organic compound with general formula RCOR' is an

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The general formula RCOR' represents a class of organic compounds known as ketones. In this formula, R and R' are alkyl groups, which are chains of carbon and hydrogen atoms. The CO in the middle is a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom. Therefore, with the presence of two alkyl groups on either side of the carbonyl group, the compound is categorized as a ketone, scientifically referred to as an alkanone.

Here is a simple breakdown of the terms:

• Alkanone: A term used to describe ketones, which have the general formula RCOR'.
• Alkanal: Represents aldehydes, which contain the functional group RCHO with at least one hydrogen attached to the carbon atom of the carbonyl group.
• Alkanoate: Typically refers to esters, which have the general form RCOOR'.
• Alkanoic acid: Refers to carboxylic acids, which are denoted by the functional group RCOOH.

Hence, by looking at the general formula RCOR', the organic compound in question is undoubtedly an alkanone.

The shape of ammonia molecule is

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The shape of the ammonia molecule (NH₃) is trigonal pyramidal. To understand why, let's explore the electron and molecular geometry using a simple explanation:

Ammonia consists of one nitrogen (N) atom bonded to three hydrogen (H) atoms. The nitrogen atom has five valence electrons requiring three more electrons to complete its octet. These are acquired by forming covalent bonds with three hydrogen atoms. In addition to the three bonding pairs, there is one lone pair of electrons on the nitrogen atom.

According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, electron pairs, including bonding pairs and lone pairs, repel each other and arrange themselves as far apart as possible to minimize repulsion. In ammonia:

• There are four electron pairs around the central nitrogen atom (three bonding pairs and one lone pair).
• These electron pairs arrange themselves in a tetrahedral electron pair geometry.

The presence of the lone pair on nitrogen creates a slight distortion, causing the molecule's shape to be trigonal pyramidal rather than perfectly tetrahedral. The lone pair occupies more space and pushes the hydrogen atoms slightly closer together. This results in a pyramidal shape, with nitrogen at the apex, and the three hydrogen atoms forming the base of the pyramid.

The trigonal pyramidal shape of ammonia is a result of this molecular geometry, not to be confused with any of the other options like V-shaped, tetrahedral, or co-planar.

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.

Alkanoates are naturally found in

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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.

The element which can combine with oxygen to form an acid anhydride of the form XO2 ${}_2$ is

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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.

Nitrogen obtained from air is not absolutely pure because it contains the following except

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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.  

How many isomers has the organic compound represented by the formula C3 ${}_3$H8 ${}_8$O ?

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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.

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

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

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

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

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

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

The term strong and weak acids is used to indicate the 

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The terms strong and weak acids are used to indicate the extent of ionization of an acid. This means how completely an acid dissociates into its ions in water.

Strong acids completely dissociate in water. This means that nearly all the acid molecules break down into positive hydrogen ions (H⁺) and their respective anions. Examples include hydrochloric acid (HCl), sulfuric acid (H₂SO₄), and nitric acid (HNO₃).

Weak acids, on the other hand, only partially dissociate in water. This means that only a small fraction of the acid molecules break down into ions. Most of the acid remains in its molecular form. An example of a weak acid is acetic acid (CH₃COOH), which is found in vinegar.

Therefore, the strength of an acid in terms of its classification as strong or weak is about how fully it dissociates into ions in an aqueous solution, not about the number of H⁺ ions or the strength of its action on substances.

The chemical formula for potassiumhexacyanoferrate(II) is

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The chemical formula for potassiumhexacyanoferrate(II) is K₄Fe(CN)₆.

Let's break down the name to understand why:

1. Potassium (K): The compound includes potassium ions. In this case, four potassium ions are present, indicated by the subscript 4 in K₄.

2. Hexacyano: The prefix "hexa" means six, which signifies there are six cyanide ions (CN^-) in the complex. This is represented as (CN)₆.

3. Ferrate (II): The word "ferrate" suggests the presence of iron (Fe). The Roman numeral (II) indicates that the iron is in the +2 oxidation state.

Overall, the complex ion is [Fe(CN)₆] with a charge of 4-, so to balance the charge, four potassium ions (each with a charge of +1) are needed, resulting in the formula K₄Fe(CN)₆.

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

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

In this particular reaction, the balanced chemical equation is:

HCl + NaOH → NaCl + H₂O

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

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

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

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

= 0.20 × 0.020

= 0.004 moles

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

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

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

= 0.004 moles / 0.12 mol/dm³

= 0.03333 dm³

Convert this volume from dm³ to cm³:

0.03333 dm³ × 1000 cm³ / dm³ = 33.33 cm³

Therefore, the volume of HCl required is 33.33 cm³.

The 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.

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.

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.

The IUPAC nomenclature of the complex K4 ${}_4$Fe(CN)6 ${}_6$ is

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The compound in question is K₄[Fe(CN)₆]. To name this complex using IUPAC nomenclature, let's break it down into parts:

• Potasium (K₄): This denotes four potassium ions. Potassium is not part of the complex ion; it is a counter ion that balances the charge of the complex ion.
• Fe(CN)₆: This is the complex ion.
• The ligand CN is the cyanide ion, which is named as cyano.
• Hexacyano: The prefix 'hexa-' indicates that there are six cyanide ligands attached to the central metal atom.

Next, consider the oxidation state of Fe:

• The formula of the complex is [Fe(CN)₆]^4-. Because each cyanide ion has a charge of -1, the total charge contributed by the six cyanide ions is -6.
• The overall charge is -4 due to the four potassium ions (K⁺), thus Fe must have a charge of +2 to balance the charge.

Finally, we consider the oxidation state of the iron. Since calculations show that it is +2, the complex ion is named based on its oxidation state.

Hence, the IUPAC name of this compound is potassium hexacyanoferrate(II).

A major effect of oil pollution in coastal water is 

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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.

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**.

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

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ROH' and ROR' can form functional group isomers because they are the functional groups of alcohols and ethers, respectively.

Ethers have a pair of alkyl or aromatic groups attached to a linking oxygen atom. ROH is the functional group of alcohols, which are derivatives of water with one hydrogen atom replaced by an alkyl group.

Alcohols (ROH) and ethers (ROR') can form functional group isomers because they have the same chemical formula but different functional groups. E.g CH3 ${}_3$CH2 ${}_2$OH and CH3 ${}_3$OCH3

Fog is a colloid in which

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**Fog** is a type of colloid, which is a mixture where very small particles of one substance are evenly distributed throughout another substance. In the case of fog, it consists of tiny **liquid droplets** that are dispersed in a **gas**. Specifically, these are tiny droplets of water suspended in the air. When you walk through fog, you are essentially walking through air that contains these minute water droplets.

Thus, the correct description of fog as a colloid is that it consists of **liquid particles dispersed in a gas medium**. The liquid here is water, and the gas is air.

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

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Benzene is a liquid hydrocarbon that is obtained from the fractional distillation of coal tar, and it is extensively used in the pharmaceutical industry. Let me break this down for you:

• **Fractional Distillation:** This is a process used to separate a mixture into its component parts, or fractions, based on differences in boiling points. Coal tar, a byproduct of coal processing, contains a myriad of compounds, and fractional distillation helps isolate specific hydrocarbons like benzene.
• **Benzene:** It is a simple aromatic hydrocarbon with the chemical formula C₆H₆. Benzene has a unique ring structure that makes it a crucial starting material or intermediate in **pharmaceutical synthesis.**
• **Uses in Pharmaceuticals:** Benzene is used to synthesize various medicinal compounds. It acts as a building block to create more complex molecules necessary for **drug manufacturing.**

That's why benzene plays an important role in the pharmaceutical industry, making it a highly valued product obtained through the distillation of coal tar.

Determine the empirical formula of an oxide of sulphur containing 60% of oxygen 

[S = 32, O = 16 ]

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To determine the empirical formula of an oxide of sulfur containing 60% of oxygen, we have to understand the concept of empirical formulas, which give the simplest whole-number ratio of atoms of each element in a compound.

Step 1: Assume 100g of the compound. In 100g of the compound:

• Oxygen (O) = 60g
• Sulfur (S) = 100g - 60g = 40g

Step 2: Convert masses to moles. Use the molar mass to find moles.

• Moles of Sulfur (S) = (40g) / (32g/mol) = 1.25 moles
• Moles of Oxygen (O) = (60g) / (16g/mol) = 3.75 moles

Step 3: Determine the simplest whole-number ratio.

To find the ratio, divide each mole value by the smallest number of moles calculated:

• For Sulfur: 1.25 / 1.25 = 1
• For Oxygen: 3.75 / 1.25 = 3

The simplest ratio of S:O is 1:3.

Thus, the empirical formula of the oxide is SO₃.