JAMB UTME - Chemistry - 2024

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).

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.

Determine the half-life of a first order reaction with constant 4.5 x 10−3 ${}^{-3}$ sec−1 ${}^{-1}$.

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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.

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.

23892 ${}_{92}^{238}$U  + 10 ${}_0^1$n → 23992 ${}_{92}^{239}$U 

The process above produces 

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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.

Boyle's law can be expressed mathematically as

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

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

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

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

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

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In the fountain experiment, hydrogen chloride gas (HCl) and ammonia (NH₃) are used to demonstrate the creation of a visible 'fountain' due to their high solubility in water. Here's a simple explanation:

When hydrogen chloride gas and ammonia gas come into contact with water, they dissolve very quickly and react vigorously. This is because both gases are very soluble in water. As they dissolve, a vacuum-like pressure is created inside the container where the gases are held, pulling water up into it, creating the 'fountain' effect.

Moreover, when HCl and NH₃ gases react with each other, they form a white, solid product known as ammonium chloride (NH₄Cl), which is a demonstration of how both gases can effectively dissolve and react with not just water, but also with each other.

Thus, the ability of these gases to create a fountain effect is primarily because they are very soluble in water, which allows them to dissolve rapidly and create the pressure differential necessary for the water to be pulled into the container dynamically.

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

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

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

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

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

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

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

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

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

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

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

The volume occupied by 1 mole of an ideal gas at a temperature of 130 ${}^0$C and a pressure of 1.58 atm is 

[ R = 0.082 atm dm3 ${}^3$ K−1 ${}^{-1}$mol−1 ${}^{-1}$ ]

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According to the Ideal gas equation, PV = nRT

Given: P = 1.58 atm, V = ?,  n = 1 mole, R = 0.082,  T= 13 + 273K = 286K

Substituting all the given parameters, 

                   V  = nRTP $\frac{nRT}{P}$

                   V =  1×0.082×2861.58 $\frac{\mathrm{<br>}}{}$

                   V = 14.84 dm3

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

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 number of geometrical isomers of butene are 

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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.

The empirical formula of an organic liquid hydrocarbon is XY. If the relative molar masses of X and Y are 72 and 6 respectively, it's vapour density is likely to be

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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**.

The constituent of petroleum fraction used in surfacing road is

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

Reasons why bitumen is used for road surfacing:

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

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

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

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

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

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

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

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

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

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

Step 1: Determine the molar mass of helium.

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

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

The formula to find the number of moles is:

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

So for helium:

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

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

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

The formula to find the number of molecules is:

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

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

Number of molecules ≈ 1.73 x 10²⁴ molecules

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

In a chemical reaction, surface area of reactants can affect

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

A type of isomerism that ClCH=CHCl can exhibit is 

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

When Calcium ethynide is decomposed by water, the gas produced is 

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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.

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

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

Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver.

[Mg = 24.0, Ag = 108 ]

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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**.

The combustion of candle under limited supply of air forms

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When a candle burns under a limited supply of air, it doesn't get enough oxygen to completely burn the hydrocarbons in the wax. In complete combustion (with enough air), the candle would ideally produce water (H₂O) and carbon dioxide (CO₂). However, under limited air supply, the process is incomplete and results in the formation of soot and carbon monoxide (CO).

Here's why:

• Soot is composed of tiny particles of carbon that are given off when some of the wax does not fully combust. This happens because there isn't enough oxygen to ensure complete combustion.
• Carbon monoxide (CO) is a colorless, odorless, and toxic gas produced alongside soot when the hydrocarbon in the wax doesn't get enough oxygen to form CO₂. Instead of converting to carbon dioxide, carbon remains only partially oxidized, resulting in CO.

In summary, under limited air conditions, the combustion of a candle primarily forms soot and carbon monoxide (CO).

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.

Hydrochloric acid is regarded as a strong acid because it

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Hydrochloric acid (HCl) is regarded as a strong acid because it ionizes completely in water. This means that when HCl is dissolved in water, it breaks down entirely into hydrogen ions (H⁺) and chloride ions (Cl^-). In a solution, there are no molecules of HCl left; only its ions are present.

This complete ionization results in a high concentration of hydrogen ions, which is a key characteristic of strong acids. Because there are more hydrogen ions available, hydrochloric acid can readily participate in chemical reactions, particularly those involving proton transfers, like neutralization reactions with bases.

In summary, the reason HCl is considered strong is due to its ability to consistently and completely ionize in an aqueous solution, not because of its physical state, source, or reactive nature with bases. Therefore, the property that defines it as a strong acid is that it ionizes completely.

An oxide of nitrogen that can rekindle a glowing splint is 

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The ability to rekindle a glowing splint is an indicator of the presence of an oxidizing agent, typically oxygen or a substance that releases oxygen. Among oxides of nitrogen, only a few are capable of doing this.

Nitrogen(I) oxide, commonly known as nitrous oxide (N₂O), is not a strong enough oxidizer to rekindle a glowing splint.

Nitrogen(II) oxide, known as nitric oxide (NO), is not stable in the presence of oxygen and does not have the ability to rekindle a glowing splint because it does not actively release oxygen.

Nitrogen(IV) oxide or nitrogen dioxide (NO₂), can support combustion by releasing oxygen as it decomposes. It is a brown gas and an effective oxidizer.

Dinitrogen tetraoxide (N₂O₄) is in equilibrium with nitrogen dioxide (NO₂). However, at standard conditions, it is not as effective an oxidizer for rekindling a glowing splint as pure NO₂.

In conclusion, the oxide of nitrogen that can rekindle a glowing splint is nitrogen(IV) oxide or nitrogen dioxide (NO₂) due to its ability to release oxygen and support combustion.

The pH of a 0.001 mol dm−3 ${}^{-3}$ of  H2 ${}_2$SO4 ${}_4$ is 

[Log10 ${}_{10}$2 = 0.3]

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The question is asking about the pH of a 0.001 mol dm⁻³ solution of H₂SO₄ (sulfuric acid). To find the pH, we need to understand how sulfuric acid dissociates in water.

Step 1: Dissociation of H₂SO₄
Sulfuric acid, H₂SO₄, is a strong acid and dissociates completely in water in two steps:

1. The first dissociation: H₂SO₄ → H⁺ + HSO₄^-

2. The second dissociation: HSO₄^- → H⁺ + SO₄^2-

For dilute solutions, particularly below 0.1 M, the first dissociation provides the major contribution to the H⁺ concentration. The second dissociation also contributes slightly to the acidity, but for simplicity and due to the dilute nature of this solution, the first step's contribution is primarily considered.

Step 2: Calculate the H⁺ Concentration
Since this is a strong acid and dissociates completely, for every 1 mole of H₂SO₄, we get 2 moles of H⁺. Therefore, for a 0.001 mol dm⁻³ solution of H₂SO₄, the concentration of H⁺ ions will be:

2 x 0.001 = 0.002 mol dm⁻³

Step 3: Calculate the pH
The pH is calculated using the formula: pH = -log[H⁺]

Substitute the H⁺ concentration:

pH = -log(0.002)

We know that log(10^-2) = -2 and log(2) = 0.3 (as provided), so:

pH = -(log(2) + log(10^-3))

pH = -(0.3 - 3)

pH = 3 - 0.3

pH = 2.7

Therefore, the pH of the 0.001 mol dm⁻³ H₂SO₄ solution is 2.7.

The general molecular formula Cn ${}_n$H2n?2 ${}_{2n?2}$ represents that of an

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The molecular formula C_nH_2n-2 represents an alkyne.

To understand this, let's take a look at the characteristics of hydrocarbons, which are compounds made up of hydrogen and carbon:

• Alkanes are saturated hydrocarbons (only single bonds) and have the general formula C_nH_2n+2.
• Alkenes are unsaturated hydrocarbons with at least one double bond and have the general formula C_nH_2n.
• Alkynes are also unsaturated hydrocarbons, but they contain at least one triple bond and follow the general formula C_nH_2n-2.
• Alkanals are not hydrocarbons; they are aldehydes with at least one -CHO group.

The formula C_nH_2n-2 indicates the presence of two fewer hydrogen atoms than in an alkene. This deficiency of hydrogen atoms is characteristic of a triple bond, which is a key feature of alkynes. Therefore, hydrocarbons with this formula must contain at least one triple carbon-carbon bond.

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.

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.

C2 ${}_2$H4(g) ${}_{4(g)}$  +  3O2(g) ${}_{2(g)}$ → 2CO2(g) ${}_{2(g)}$  +  2H2 ${}_2$O(g) ${}_{(g)}$

The above equation represents the combustion of ethene.If 10cm3 ${}^3$ of ethene is burnt in 50cm3 ${}^3$ of oxygen, what would be the volume of oxygen that would remain at the end of the reaction?

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Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant. 

                                      C2 ${}_2$H4(g) ${}_{4(g)}$  +  3O2(g) ${}_{2(g)}$ → 2CO2(g) ${}_{2(g)}$  +  2H2 ${}_2$O(g) ${}_{(g)}$

                                         1 mole      :       3 moles

Total volume required:     10 cm3 ${}^3$          50 cm3 ${}^3$

Reacted Volume:              10 cm3 ${}^3$          30 cm3 ${}^3$

Residual volume:               0                    (50 - 30) = 20 cm3

An example of an amphoteric oxide is

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

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

Here is how it works:

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


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

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

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

An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5?   [ C =12, H = 1, N = 14, O = 16]. 

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To find the molecular formula of the compound, follow these steps:

1. Determine the Empirical Formula:

Start by assuming you have 100 grams of the compound. This means you have:

• 53.1 grams of Carbon
• 6.2 grams of Hydrogen
• 12.4 grams of Nitrogen
• 28.3 grams of Oxygen

Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):

• Moles of Carbon = 53.1 g / 12 g/mol = 4.425 moles
• Moles of Hydrogen = 6.2 g / 1 g/mol = 6.2 moles
• Moles of Nitrogen = 12.4 g / 14 g/mol = 0.886 moles
• Moles of Oxygen = 28.3 g / 16 g/mol = 1.769 moles

Next, divide each by the smallest number of moles to get the simplest ratio:

• Carbon: 4.425 / 0.886 ≈ 5
• Hydrogen: 6.2 / 0.886 ≈ 7
• Nitrogen: 0.886 / 0.886 = 1
• Oxygen: 1.769 / 0.886 ≈ 2

This gives us the empirical formula: C₅H₇NO₂.

2. Determine the Molecular Formula:

The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.

Calculate the empirical formula mass:

• 5(C) + 7(H) + 1(N) + 2(O) = 5(12) + 7(1) + 1(14) + 2(16) = 101 g/mol

The molar mass can be calculated from the vapor density:

• Molar Mass = Vapor Density × 2 = 56.5 × 2 = 113 g/mol

Now, find the ratio of the molar mass to the empirical formula mass:

• Ratio = 113 / 101 ≈ 1.12

This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.

Therefore, the molecular formula is C₅H₇NO₂.

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.

What is the vapour density of 560cm3 ${}^3$ of a gas that weighs 0.4g at s.t.p?

[Molar Volume of gas at s.t.p = 22.4 dm3 ${}^3$ ]

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To find the vapour density of a gas, you can use the formula:

Vapour density = (Molar mass of gas) / 2

However, first, we need to determine the molar mass of the gas. One can find the molar mass using the given data:

We know that at standard temperature and pressure (s.t.p.), 1 mole of any gas occupies 22.4 dm³. We need to convert the volume from cm³ to dm³ because the molar volume is given in dm³:

560 cm³ = 0.560 dm³

Now, let's find the number of moles in 0.560 dm³:

The number of moles (n) = Volume of gas (dm³) / Molar volume at s.t.p. (dm³/mol)

n = 0.560 dm³ / 22.4 dm³/mol

n = 0.025 moles

Given that the mass of the gas is 0.4 grams, we can find the molar mass by using the relation:

Molar Mass = Mass / Number of Moles

Molar Mass = 0.4 g / 0.025 moles

Molar Mass = 16 g/mol

Now that we have the molar mass, we can find the vapour density:

Vapour density = Molar mass / 2

Vapour density = 16 g/mol / 2

Vapour density = 8.0

Hence, the vapour density of the gas is 8.0.

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

If the solubility of KNO3 ${}_3$ at 300 ${}^0$C is 3.10 mol/dm3 ${}^3$ a solution containing 303g/dm3 ${}^3$ KNO3 ${}_3$ is likely to be

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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.

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 group VIII elements are the inert gases because they 

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

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

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

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

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