JAMB UTME - Chemistry - 2024

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Originalmass2n
  =   Residual mass

Where n = number of activity =  exposuretimehalflife

Given: 

Original mass = 1g, exposure time = 10 minutes , half life = 2 minutes, Residual mass = ?

Substituting all the given parameters appropriately, we have

                     n = 102 $\frac{\mathrm{<br>}}{}$

                     n = 5

Originalmass2n $\frac{\mathrm{<br>}}{}$ =   Residual mass

125
5  =   Residual mass

132 $\frac{\mathrm{<br>}}{}$     =   Residual mass

Residual mass =  132
 or 0.03125g

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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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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.

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

CuO(s) + H₂(g) ⇌ Cu(s) + H₂O(g)

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

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

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

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

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

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The 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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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.

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To determine the percentage of hydrogen in the sixth member of aliphatic alkanes, we first need to understand the general formula for alkanes. Aliphatic alkanes are a class of hydrocarbons with the general formula C_nH_2n+2, where 'n' is the number of carbon atoms.

The sixth member of this series will have n = 6. Therefore, the molecular formula for the sixth member is C₆H₁₄.

To find the percentage of hydrogen, we first calculate the molar mass of C₆H₁₄:

• Molar mass of 1 carbon atom = 12
• Molar mass of 6 carbon atoms = 6 × 12 = 72
• Molar mass of 1 hydrogen atom = 1
• Molar mass of 14 hydrogen atoms = 14 × 1 = 14

Total molar mass of C₆H₁₄ = 72 + 14 = 86

Next, we calculate the percentage of hydrogen:

Percentage of hydrogen = (Molar mass of hydrogen atoms / Total molar mass) × 100

Percentage of hydrogen = (14 / 86) × 100 = 16.28%

Therefore, the percentage of hydrogen in the sixth member of the aliphatic alkanes is 16.28%.

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When a few drops of Millon's reagent is added to an egg-white solution in a test tube and the solution is boiled, the white precipitate turns brick red. This indicates the presence of proteins.

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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.

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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.

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

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

Here's a simplified mechanism for this reaction:

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

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

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

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

C (s) + H₂O (g) → CO (g) + H₂ (g)

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

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A titration is a process used to determine the concentration of an unknown solution by adding a solution of known concentration. The indicator used in a titration is a substance that changes color at the specific pH level of the solution, which usually happens at the equivalence point.

For a titration between a strong acid and a weak base, the solution at the equivalence point is slightly acidic. This is because the salt formed as a result of the neutralization reaction can undergo hydrolysis, producing an excess of hydronium ions (H₃O⁺) which makes the solution acidic.

Among the provided indicators, methyl orange is the most suitable for indicating this type of reaction because it changes color in an acidic pH range of about 3.1 to 4.4. It shifts from red at a pH below 3.1 to yellow at a pH above 4.4.

Therefore, for a titration involving a strong acid and a weak base, methyl orange is the appropriate indicator as it can show the end point effectively when the solution is slightly acidic. The pH at the equivalence point falls within the color change range of methyl orange.

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To determine the 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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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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The compound in question is written as CH₃₃CH₂₂C(CH₃₃)=C(CH₃₃)₂₂, which seems to be intended as (CH₃)₃CH₂CH=C(CH₃)₃. The IUPAC nomenclature of organic compounds follows specific rules to name the compound uniquely such that it is understood universally. Here is a comprehensive breakdown:

1. Select the longest carbon chain that includes the highest-order functional group, which, in this case, is the alkene group (double bond).

2. The longest chain consists of 5 carbons, which gives us the root name "pentene". We choose the carbon chain such that the double bond gets the lowest possible number, starting from the end of the chain closest to the double bond.

3. Number the carbon atoms in the chain from the end closest to the double bond. The numbering direction will determine the position of the double bond and substituents. The double bond starts on carbon 2.

4. Identify and name the substituents attached to the carbon chain. In this case, there are two methyl groups on carbon 3. This means it is dimethyl as there are two of them.

Thus, the complete name of the compound is 2,3-dimethylpent-2-ene. Here, "2,3-dimethyl" indicates the position and quantity of methyl groups, "pent" indicates the longest chain with 5 carbons, and "-2-ene" indicates a double bond starting at the second carbon.

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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).

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The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is water.

Here's why:

• Formation of Alkali: When sodium (a highly reactive metal) reacts with water, it forms an alkali. The chemical reaction is written as: 2Na + 2H₂O → 2NaOH + H₂↑. This reaction produces sodium hydroxide, which is an alkali.

• Color Change in Copper(II) Tetraoxosulphate (VI): White anhydrous copper(II) tetraoxosulphate (VI) (commonly known as anhydrous copper(II) sulfate) changes to a blue color when it comes in contact with water. This is because water molecules are incorporated into its crystal structure, forming hydrated copper(II) sulfate with the formula CuSO₄·5H₂O, which is blue in color.

Hence, the correct answer is water, as it is the substance that both reacts with sodium to form an alkali and changes the color of anhydrous copper(II) tetraoxosulphate (VI) to blue.

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

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

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

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

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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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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).

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

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

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



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

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



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

Convert ml to liters: 80 ml = 0.080 L



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

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

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

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To understand what y represents in the graph, we need to think about what graphs in chemistry, specifically regarding energy changes in reactions, generally show.

Chemical reaction energy diagrams often depict a reaction's energy change as a curve from the reactants to the products, showing different energy levels throughout the process. The energy required to start a reaction or to transform the reactants into an activated complex (also known as the transition state) is crucial.

The height of this energy barrier is called the activation energy. This is the minimum amount of energy required to start a chemical reaction. The activation energy is represented by the peak in the energy graph between the reactant energy level and the top of the curve.

Therefore, in this context, y represents the activation energy needed for the reaction to proceed. Understanding activation energy is vital as it determines how quickly a reaction will occur. Reactions with a high activation energy tend to happen more slowly because it is less probable that the necessary energy for the reaction to occur spontaneously will be present.

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

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

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

CH₃-CH₂-CH₂-OH

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

CH₃-CH(OH)-CH₃

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

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

CH₃-O-CH₂-CH₃

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

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

Thus, the answer is three distinct isomers.

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

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

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

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

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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**.

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In the process of silver plating a spoon using an electrolytic cell, the correct configuration involves the following:

Cathode: The object to be plated, which in this case is the spoon. In an electrolytic cell, the cathode is where the reduction reaction occurs, and it is the surface on which the metal ions are deposited.

Anode: A rod made of silver. The anode is where oxidation occurs, meaning the silver rod will dissolve into the solution in the form of silver ions. These ions then move towards the cathode to be deposited as a thin layer on the spoon.

Electrolyte: A solution that contains a soluble silver salt (such as silver nitrate, AgNO₃). The silver ions from this salt help in the process of transferring the silver from the anode to the cathode.

Thus, the proper order for silver plating a spoon in an electrolytic cell for industrial production is: "Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt."

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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.

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

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

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

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

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

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

This simplifies to:

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

Rearranging the equation, we find:

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

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

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

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

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

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

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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.

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Water pollutants are substances that, when introduced into the water, cause harm to ecosystems, human health, and the overall quality of the water. Each of the options provided has the potential to be considered a water pollutant, except for one. Let's explain them:

1. Inorganic fertilizers: These are substances mainly composed of synthetic chemicals, including nitrates and phosphates. When these fertilizers enter water bodies, they can lead to nutrient pollution, which causes excessive growth of algae (eutrophication), leading to a decrease in oxygen levels in the water, harming aquatic life.

2. Warm water affluent: This refers to the discharge of heated water into natural water bodies. This heat contamination can change the temperature of the water, affecting the metabolism of aquatic life and leading to thermal pollution.

3. Oxygen gas: Oxygen gas is a fundamental component of the Earth's atmosphere and is not considered a water pollutant. In fact, dissolved oxygen is crucial for the survival of aquatic organisms. Rather than causing any harm, adequate levels of dissolved oxygen in water bodies are essential for maintaining healthy aquatic ecosystems.

4. Biodegradable waste: These are organic materials that decompose in the environment. When introduced in large quantities into water bodies, they can consume a significant amount of dissolved oxygen as they decompose, which can lead to depletion of oxygen levels and cause harm to aquatic life, making them pollutants in aquatic ecosystems.

Given the explanations above, oxygen gas is the option that is not a water pollutant. It is vital for the health of aquatic ecosystems, unlike the other options, which can all lead to some form of pollution in water bodies.

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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.

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

Let's break this down for better understanding:

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

The other compounds listed do not serve as common fungicides:

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

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

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To find the molecular formula of a hydrocarbon given its empirical formula and molar mass, you need to compare the empirical formula mass with the given molar mass.

The empirical formula given is CH₃. The molar mass of the empirical formula is calculated as follows:

• Carbon (C): 1 atom × 12 g/mol = 12 g/mol
• Hydrogen (H): 3 atoms × 1 g/mol = 3 g/mol

Total empirical formula mass = 12 + 3 = 15 g/mol

The provided molar mass of the compound is 30 g/mol. To determine how many empirical units are in the molecular formula, divide the molecular mass (given) by the empirical formula mass:

Number of empirical units = 30 g/mol / 15 g/mol = 2

Therefore, the molecular formula is twice the empirical formula:

Empirical formula: CH₃

Molecular formula: (CH₃)₂ = C₂H₆

The correct molecular formula is C₂H₆.

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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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This reaction illustrates dehydration. In chemistry, dehydration refers to the process of removing water (H₂O) from a compound. Let's break down the given reaction to understand this better.

The provided chemical equation is:

C₂H₅OH → C₂H₄ + H₂O

This equation indicates that ethanol (C₂H₅OH) is being transformed into ethylene (C₂H₄) with the production of water (H₂O).

The process involves the breaking of bonds in ethanol and the removal of a water molecule, as follows:

• The ethanol molecule, C₂H₅OH, has the –OH group, which combines with a hydrogen atom from the rest of the molecule to form H₂O (water).
• This results in the formation of C₂H₄ (ethylene), a smaller molecule, and the water molecule is removed.

This reaction is typically carried out under certain conditions, in this case at a high temperature of 1700°C, to facilitate the dehydration process.

Therefore, this is indeed a dehydration reaction as it involves converting ethanol into ethylene by removing water.

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To determine the difference in molecular mass between an alkene and an alkyne, let's first take a look at their general formulas.

Alkene: An alkene is a hydrocarbon with at least one double bond between carbon atoms. For an alkene with six carbon atoms, the general formula is C_nH_2n. Therefore, for 6 carbon atoms, the molecular formula is C₆H₁₂.

Alkyne: An alkyne is a hydrocarbon with at least one triple bond between carbon atoms. For an alkyne with six carbon atoms, the general formula is C_nH_2n-2. Therefore, for 6 carbon atoms, the molecular formula is C₆H₁₀.

Now let's calculate the molecular masses:

Molecular mass of alkene (C₆H₁₂):

1. Carbon (C): 6 atoms x 12 g/mol = 72 g/mol
2. Hydrogen (H): 12 atoms x 1 g/mol = 12 g/mol
3. Total for C₆H₁₂ = 72 g/mol + 12 g/mol = 84 g/mol

Molecular mass of alkyne (C₆H₁₀):

1. Carbon (C): 6 atoms x 12 g/mol = 72 g/mol
2. Hydrogen (H): 10 atoms x 1 g/mol = 10 g/mol
3. Total for C₆H₁₀ = 72 g/mol + 10 g/mol = 82 g/mol

The **difference** in molecular mass between the alkene and alkyne is **84 g/mol - 82 g/mol** = 2 g/mol.

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

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

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

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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.