2013 JAMB Chemistry Past Questions

Questions

1. 1.

   The constituent of air that acts as a diluent is

   1. A.nitrogen
   2. B.carbon (IV) oxide
   3. C.noble gases
   4. D.oxygen

   Show answer

   Answer: A. nitrogen

   AI explanation

   Analysis

   The question asks for the component of air that functions as a "diluent." In chemistry, a diluent is a substance used to reduce the concentration or reactivity of another substance.

   Concept Explanation

   Air is a mixture of gases, primarily nitrogen (about 78%) and oxygen (about 21%), with small amounts of argon, carbon (IV) oxide, and water vapor.

   Oxygen is highly reactive and supports combustion. If the atmosphere were composed entirely of pure oxygen, fires would be uncontrollable, and the rate of oxidation in living organisms would be dangerously high. Nitrogen serves as an inert (unreactive) buffer that "dilutes" the oxygen, slowing down combustion and making the environment safe for life and controlled burning.

   Step-by-Step Solution

   1. Identify the roles of the gases: • Oxygen: Essential for respiration and combustion (the active component). • Carbon (IV) oxide: Present in small quantities; used by plants for photosynthesis. • Noble gases: Chemically inert but present in very small quantities. • Nitrogen: Makes up the vast majority of the air. Because it is chemically inactive (inert) under normal conditions, it does not participate in combustion or respiration.
   2. Determine the function: Since nitrogen is inert and makes up the largest volume of air, it effectively reduces the concentration of oxygen, thereby acting as a diluent.
   3. Conclusion: Nitrogen prevents the rapid combustion of materials and regulates the rate of respiration by diluting the oxygen in the atmosphere.

   Correct Answer

   The correct option is A.

   Exam Tip

   Think of nitrogen as the "brake" for oxygen. Without nitrogen, every spark would result in a massive explosion. Remember: Nitrogen = Inert Diluent.

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2. 2.

   In the laboratory preparation of ammonia, the flask is placed in a slanting position so as to

   1. A.prevent condensed water from breaking the reaction flask
   2. B.enable the proper mixing of the reactions in the flask
   3. C.enhance the speed of the reaction
   4. D.prevent formation of precipitate

   Show answer

   Answer: A. prevent condensed water from breaking the reaction flask

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3. 3.

   An organic compound contains 60% carbon, 13.3% hydrogen and 26.7% oxygen. Calculate the empirical formularn(C=12,H =1, O=16)

   1. A.C 5H 12O
   2. B.C 3H 8O
   3. C.C 6H 13O 2
   4. D.C 4H 9O

   Show answer

   Answer: B. C 3H 8O

   AI explanation

   To find the empirical formula of this organic compound, we need to determine the simplest whole-number ratio of the atoms present. We follow these steps:

   1. Analyze the composition

   • Carbon (C): 60%
   • Hydrogen (H): 13.3%
   • Oxygen (O): 26.7%

   2. Calculate the number of moles

   Assume we have 100g of the compound. This means we have 60g of C, 13.3g of H, and 26.7g of O. We divide the mass of each element by its relative atomic mass (C=12, H=1, O=16):

   • Moles of C: $\frac{60}{12} = 5.0$
   • Moles of H: $\frac{13.3}{1} = 13.3$
   • Moles of O: $\frac{26.7}{16} = 1.67$

   3. Divide by the smallest value

   To get the simplest ratio, divide each result by the smallest number of moles calculated (which is 1.67):

   • Ratio of C: $\frac{5.0}{1.67} \approx 3$
   • Ratio of H: $\frac{13.3}{1.67} \approx 8$
   • Ratio of O: $\frac{1.67}{1.67} = 1$

   4. Determine the Empirical Formula

   The ratio of C : H : O is 3 : 8 : 1. Therefore, the empirical formula is $C_3H_8O$.

   Comparing this to your options: A. $C_5H_{12}O$ B. $C_3H_8O$ C. $C_6H_{13}O_2$ D. $C_4H_9O$

   The calculated formula matches option B.

   Exam Tip: When calculating empirical formulas, if your division results in a decimal like 0.5, multiply all values by 2 to get a whole number. If you get 0.33, multiply by 3. In this case, the values were very close to whole numbers, making the calculation straightforward.

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4. 4.

   2- methylpropan-2-ol is an example of a

   1. A.dihydric alkanol
   2. B.tertiary alkanol
   3. C.secondary alkanol
   4. D.primary alkanol

   Show answer

   Answer: B. tertiary alkanol

   AI explanation

   1. Analyze the question

   The question asks us to classify the alcohol 2-methylpropan-2-ol based on its structure. To do this, we need to determine how many carbon atoms are directly attached to the carbon atom that holds the hydroxyl (-OH) group.

   2. Explain the concept

   Alkanols (alcohols) are classified based on the number of alkyl groups attached to the carbon atom bearing the -OH group:

   • Primary ($1^\circ$) alkanol: The -OH group is attached to a carbon atom that is bonded to only one other carbon atom (or none, in the case of methanol).
   • Secondary ($2^\circ$) alkanol: The -OH group is attached to a carbon atom that is bonded to two other carbon atoms.
   • Tertiary ($3^\circ$) alkanol: The -OH group is attached to a carbon atom that is bonded to three other carbon atoms.

   3. Show the solution

   Let’s draw the structure of 2-methylpropan-2-ol:

   1. Parent chain: "Propan" means 3 carbons in the main chain.
   2. Position of -OH: "2-ol" means the -OH group is on the second carbon.
   3. Substituent: "2-methyl" means a methyl group ($-CH_3$) is also on the second carbon.

   The structure looks like this:

           CH3
            |
       CH3— C — CH3
            |
           OH

   • Look at the central carbon atom (the one bonded to the -OH group).
   • It is directly bonded to three other carbon atoms (one from the left, one from the right, and one from the methyl group above).
   • Since the carbon bearing the -OH group is bonded to three other carbons, it is classified as a tertiary alkanol.

   4. Identify the answer

   Comparing our finding to the options provided: A. dihydric alkanol (Incorrect, it only has one -OH group) B. tertiary alkanol (Correct) C. secondary alkanol (Incorrect) D. primary alkanol (Incorrect)

   5. Tip

   Memory Trick: Think of the "T" in Tertiary as standing for Three. A tertiary carbon is always bonded to three other carbons. If you are ever confused, draw the structure out—it is the safest way to avoid mistakes in exams!

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5. 5.

   In an endothermic reaction, if there is a loss in entropy the reaction will

   1. A.be indeterminate
   2. B.be spontaneous
   3. C.not be spontaneous
   4. D.be at equilibrium

   Show answer

   Answer: C. not be spontaneous

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6. 6.

   The coloured nature of transition metal ions are associated with their partially filled

   1. A.f- orbital
   2. B.s- orbital
   3. C.p-orbital
   4. D.d-orbital

   Show answer

   Answer: D. d-orbital

   AI explanation

   Analysis

   The question asks for the specific electron orbital responsible for the characteristic colours observed in transition metal ions.

   Concept Explanation

   Transition metals are defined by the presence of partially filled d-orbitals. In an isolated transition metal atom, the five d-orbitals have the same energy (they are degenerate). However, when ligands (atoms, ions, or molecules) approach the metal ion to form a complex, these d-orbitals split into different energy levels.

   When white light hits these ions, electrons in the lower-energy d-orbitals absorb specific wavelengths of light to "jump" (excite) to the higher-energy d-orbitals. This process is known as d-d transition. The light that is not absorbed is transmitted or reflected, which is what we perceive as the colour of the ion.

   Step-by-Step Solution

   1. Identify the electronic configuration: Transition metals (elements in the d-block of the periodic table) have their valence electrons in the d-subshell.
   2. Consider the condition for colour: For an ion to be coloured, it must have at least one unpaired electron in its d-orbital to allow for the d-d transition.
   3. Evaluate the options: • s-orbitals and p-orbitals: These are typically fully filled in stable ions or do not participate in the energy-gap transitions that produce visible light colours in transition metals. • f-orbitals: These are associated with Lanthanides and Actinides (inner transition metals), not the standard transition metals referred to in this context. • d-orbitals: These are the specific orbitals that split in energy when transition metals form complexes, allowing for the absorption of visible light.

   Conclusion

   The coloured nature of transition metal ions is directly linked to the presence of partially filled d-orbitals, which facilitate d-d electronic transitions.

   Exam Tip

   Remember the mnemonic "D for Dazzling": Transition metals are Dazzling (coloured) because of their d-orbitals. If an ion has a completely empty d-orbital (like $Sc^{3+}$) or a completely full d-orbital (like $Zn^{2+}$), it will be colourless!

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7. 7.

   Sulphur (IV)oxide is a strong reducing agent in the presence of water due to the formation of

   1. A.hydroxide ion
   2. B.sulpgur (Vl)oxide
   3. C.hydrogen sulhide
   4. D.trioxosulphate (IV) salt

   Show answer

   Answer: D. trioxosulphate (IV) salt

   AI explanation

   Analysis

   The question asks why sulphur(IV) oxide ($SO_2$) acts as a strong reducing agent when water is present. We need to identify the specific chemical species formed during this reaction that facilitates the reduction process.

   Concept Explanation

   When sulphur(IV) oxide ($SO_2$) dissolves in water, it reacts to form trioxosulphuric(IV) acid ($H_2SO_3$):

   $SO_2(g) + H_2O(l) \rightleftharpoons H_2SO_3(aq)$

   In this reaction, the sulphur atom in $H_2SO_3$ has an oxidation state of +4. Because sulphur can be oxidized to a higher oxidation state (like +6 in sulphate ions), $H_2SO_3$ (or the resulting trioxosulphate(IV) ions, $SO_3^{2-}$) readily donates electrons to other substances, thereby acting as a reducing agent.

   The reducing property is specifically attributed to the formation of the trioxosulphate(IV) species, which is easily oxidized to tetraoxosulphate(VI) ($SO_4^{2-}$).

   Step-by-Step Solution

   1. Reaction with water: $SO_2$ reacts with water to form $H_2SO_3$.
   2. Dissociation: $H_2SO_3$ dissociates in water to produce hydrogen ions and trioxosulphate(IV) ions ($SO_3^{2-}$).
   3. Redox behavior: The trioxosulphate(IV) ion ($SO_3^{2-}$) is unstable in the presence of oxidizing agents and is easily oxidized to the more stable tetraoxosulphate(VI) ion ($SO_4^{2-}$).
   4. Conclusion: Because it is easily oxidized, it forces the other substance to be reduced. Therefore, the presence of the trioxosulphate(IV) species is the reason for its reducing power.

   Identifying the Answer

   Looking at the options provided:

   • A is incorrect because hydroxide ions are associated with bases.
   • B is incorrect as sulphur(VI) oxide is the product of oxidation, not the cause of the reducing power.
   • C is incorrect as hydrogen sulphide is a different compound entirely.
   • D correctly identifies the species formed that facilitates the reduction.

   *

   Exam Tip: Whenever you see "Sulphur(IV) oxide" and "reducing agent" in a question, always look for options involving "trioxosulphate(IV)" or "sulphite." Remember that "IV" usually corresponds to the lower oxidation state, which is ready to be oxidized to the "VI" state.

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8. 8.

   Aluminium containers are frequently used to transport trioxonitrate (V) acid because aluminium

   1. A.has a silvery-white appearance
   2. B.has a low density
   3. C.does not react with the acid
   4. D.does not corrode

   Show answer

   Answer: C. does not react with the acid

   AI explanation

   Analysis

   The question asks why aluminium is the preferred material for transporting trioxonitrate(V) acid (also known as nitric acid, $HNO_3$). We need to look at the chemical interaction between aluminium and concentrated nitric acid.

   Concept Explanation

   In chemistry, we often expect metals to react with acids to produce hydrogen gas and a salt. However, concentrated trioxonitrate(V) acid acts as a powerful oxidizing agent.

   When aluminium comes into contact with concentrated $HNO_3$, it undergoes a process called passivation. The acid reacts with the surface of the aluminium to form a thin, non-porous, and highly unreactive layer of aluminium oxide ($Al_2O_3$). This oxide layer acts as a protective "skin" that prevents the acid from reaching the underlying metal, effectively stopping any further reaction.

   Step-by-Step Solution

   1. Understand the reaction: Aluminium is a reactive metal, but its reactivity is modified by the formation of an oxide layer.
   2. Evaluate the options: • A (Silvery-white appearance): This is a physical property, not a chemical reason for stability. • B (Low density): While true, this makes it convenient for transport, but it doesn't explain why it doesn't dissolve in the acid. • C (Does not react with the acid): Due to the passivation layer formed, the bulk of the aluminium remains protected and does not react with the acid. This makes it safe for storage and transport. • D (Does not corrode): This is technically incorrect; aluminium does corrode (oxidize), but the product of that corrosion (the oxide layer) is what protects the metal.

   Conclusion

   Because the aluminium becomes "passive" due to the protective oxide layer, it effectively does not react with the concentrated acid. Therefore, it is safe to use for transport.

   Exam Tip: Remember that concentrated nitric acid renders iron and aluminium "passive." Don't be fooled into thinking they are naturally unreactive metals like gold or platinum; they are simply protected by their own oxide coating!

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9. 9.

   A metal that forms soluble trioxosulphate (IV) ion is

   1. A.barium
   2. B.potassium
   3. C.manganese
   4. D.aluminium

   Show answer

   Answer: B. potassium

   AI explanation

   Analysis

   The question asks us to identify which of the given metals forms a soluble trioxosulphate(IV) salt. The trioxosulphate(IV) ion is represented as $\text{SO}_3^{2-}$. Therefore, we are looking for a metal whose sulphite salt is soluble in water.

   Concept: Solubility Rules

   In chemistry, the solubility of salts follows specific general rules:

   1. Group 1 Metals: All common salts of Group 1 elements (the alkali metals, such as Sodium $\text{Na}^+$, Potassium $\text{K}^+$, etc.) are soluble in water.
   2. Sulphites ($\text{SO}_3^{2-}$): Most metal sulphites are insoluble in water, with the notable exception of those formed by Group 1 metals and ammonium ($\text{NH}_4^+$).
   3. Barium ($\text{Ba}^{2+}$): Barium sulphite ($\text{BaSO}_3$) is known to be insoluble in water.
   4. Aluminium ($\text{Al}^{3+}$) and Manganese ($\text{Mn}^{2+}$): These generally form insoluble sulphites or are unstable in aqueous solutions.

   Step-by-Step Solution

   1. Evaluate Option A (Barium): Barium forms $\text{BaSO}_3$. According to solubility tables, barium sulphite is insoluble in water.
   2. Evaluate Option B (Potassium): Potassium is a Group 1 alkali metal. All salts of potassium, including potassium trioxosulphate(IV) ($\text{K}_2\text{SO}_3$), are highly soluble in water.
   3. Evaluate Option C (Manganese): Manganese(II) sulphite is generally insoluble in water.
   4. Evaluate Option D (Aluminium): Aluminium sulphite is not a standard stable soluble salt in aqueous solution.

   Since Potassium belongs to Group 1, it satisfies the rule that all its common salts are soluble.

   Conclusion

   This aligns with the chemical principle that all Group 1 metal salts are soluble.

   Exam Tip

   Remember the acronym "SPA" for solubility: Sodium, Potassium, and Ammonium salts are always soluble in water, regardless of the anion (nitrate, chloride, sulphate, or sulphite) attached to them. If you see these elements in a solubility question, they are almost always the correct answer!

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10. 10.

    Which of the following substances is not a salt?

    1. A.Aluminium oxide
    2. B.Sodium hydrogentrioxosulphate (V)
    3. C.Sodium trioxocarbonate (V)
    4. D.Zinc chloride

    Show answer

    Answer: A. Aluminium oxide

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11. 11.

    An insoluble salt can be prepared by

    1. A.the reaction of trioxocarbonate (V) with an acid
    2. B.double decomposition
    3. C.the action of dilute acid on an insoluble base
    4. D.the reaction of metals with an acid

    Show answer

    Answer: B. double decomposition

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12. 12.

    Steam changes the colour of anhydrous cobalt (ll) chloride from

    1. A.white to red
    2. B.blue to white
    3. C.blue to pink
    4. D.white to blue

    Show answer

    Answer: C. blue to pink

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13. 13.

    The IUPAC nomenclature for the compound above is

    1. A.4-methylpent-1-ene
    2. B.3-methylpent-2-ene
    3. C.2-methylpent-1ene
    4. D.2-methylpent-4-ene

    Show answer

    Answer: A. 4-methylpent-1-ene

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14. 14.

    The radioisotope used in industrial radiography for the rapid checking of faults in welds and casting is

    1. A.carbon-14
    2. B.phosphorus-32
    3. C.cobalt-60
    4. D.iodine-131

    Show answer

    Answer: C. cobalt-60

    AI explanation

    Analysis

    The question asks for the specific radioisotope utilized in industrial settings to inspect the structural integrity of welds and metal castings. This process is known as industrial radiography.

    Concept Explanation

    Industrial radiography is a non-destructive testing (NDT) method. It works similarly to an X-ray: a radioactive source emits gamma rays that pass through a material (like a weld or a metal casting). If there is a crack, void, or fault in the material, the radiation passes through more easily at that point, creating a darker image on the film placed behind the object.

    To be effective for thick metals, the radioisotope must:

    1. Emit high-energy gamma rays (to penetrate dense materials).
    2. Have a relatively long half-life (so the source doesn't need to be replaced too frequently).

    Step-by-Step Solution

    • Option A (Carbon-14): Primarily used for radiocarbon dating of organic materials (fossils, wood, etc.). It is not suitable for industrial metal radiography.
    • Option B (Phosphorus-32): Used in medicine and agricultural research as a tracer to study plant metabolism and in the treatment of certain blood disorders.
    • Option C (Cobalt-60): This isotope is a strong gamma-ray emitter. Its high energy allows it to penetrate thick steel and other metals, making it the standard choice for industrial radiography to detect faults in welds and castings.
    • Option D (Iodine-131): Used extensively in medicine for diagnosing and treating thyroid conditions. It is not used for industrial metal inspection.

    Conclusion

    Based on the properties of these isotopes, Cobalt-60 is the correct radioisotope used for industrial radiography.

    Exam Tip

    When dealing with radioactivity questions in JAMB, remember these common associations:

    • Carbon-14: Dating ancient organic matter.
    • Cobalt-60: Industrial radiography (welds/castings) and cancer therapy.
    • Iodine-131: Thyroid treatment.
    • Phosphorus-32: Agricultural research/tracers.

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15. 15.

    Which of the following compounds is a neutral oxide?

    1. A.Carbon (IV) oxide
    2. B.Sulphur (Vl) oixide
    3. C.Sulphur (IV) oxide
    4. D.Carbon (ll) oxide

    Show answer

    Answer: D. Carbon (ll) oxide

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16. 16.

    the minimum amount of energy required for a reaction to take place is

    1. A.lattice energy
    2. B.ionization energy
    3. C.activation energy
    4. D.kinetic energy

    Show answer

    Answer: C. activation energy

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17. 17.

    Which of the gases is employed as an anaesthesia?

    1. A.N 2O
    2. B.NO2
    3. C.NH3
    4. D.NO

    Show answer

    Answer: A. N 2O

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18. 18.

    The compound above is an

    1. A.alkanone
    2. B.alkanoate
    3. C.alkanal
    4. D.akanol

    Show answer

    Answer: A. alkanone

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19. 19.

    2SO2(g) + O2(g) $\iff$2SO3(g) $\bigtriangleup H = -395.7kJmol^{-1}$ In the reaction above, the concentration of 2SO3(g) can be in creased by

    1. A.decreasing the pressure
    2. B.decreasing the temperature
    3. C.the addition of catalyst
    4. D.increasing the temperature

    Show answer

    Answer: B. decreasing the temperature

    AI explanation

    Analysis

    The reaction given is: 2SO₂(g) + O₂(g) $\rightleftharpoons$ 2SO₃(g) $\Delta H = -395.7 \text{ kJ mol}^{-1}$

    This is an exothermic reaction (indicated by the negative $\Delta H$ value). We are looking for the condition that will shift the equilibrium to the right (the product side) to increase the concentration of SO₃.

    Concept: Le Chatelier’s Principle

    Le Chatelier’s Principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract that change.

    1. Temperature: For an exothermic reaction ($\Delta H < 0$), heat is treated as a product. According to Le Chatelier’s Principle, decreasing the temperature will cause the system to shift in the direction that produces heat (the forward reaction) to compensate for the loss. Therefore, lowering the temperature increases the yield of SO₃.
    2. Pressure: The reaction has 3 moles of gas on the left (2 + 1) and 2 moles of gas on the right. Increasing the pressure would shift the equilibrium to the side with fewer moles (the right), increasing the yield of SO₃. Decreasing pressure would do the opposite.
    3. Catalyst: A catalyst speeds up both the forward and backward reactions equally. It helps the system reach equilibrium faster, but it does not change the position of the equilibrium or the concentration of the products.

    Step-by-Step Solution

    • Step 1: Identify the nature of the reaction. The negative sign in $\Delta H$ confirms it is exothermic.
    • Step 2: Evaluate the options based on Le Chatelier’s Principle. • Option A: Decreasing pressure favors the side with more moles (the reactants), so SO₃ concentration decreases. • Option B: Decreasing temperature favors the exothermic direction (the forward reaction), which produces more SO₃. • Option C: A catalyst has no effect on the equilibrium position. • Option D: Increasing temperature favors the endothermic direction (the backward reaction), which decreases the concentration of SO₃.

    Conclusion

    By decreasing the temperature, the system shifts to the right to produce more heat, thereby increasing the concentration of SO₃.

    Exam Tip: Always remember the "Exo-Cold, Endo-Hot" rule. For exothermic reactions, lower temperatures favor the products. For endothermic reactions, higher temperatures favor the products.

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20. 20.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

21. 21.

    The shape of the carbon (lV) oxide molecule is

    1. A.pyramidal
    2. B.linear
    3. C.angular
    4. D.tetrahedral

    Show answer

    Answer: B. linear

    AI explanation

    Analysis

    The question asks for the molecular geometry (shape) of carbon(IV) oxide, also known as carbon dioxide ($CO_2$). To determine the shape, we must look at the bonding arrangement around the central atom.

    Concept: VSEPR Theory

    The Valence Shell Electron Pair Repulsion (VSEPR) theory states that electron pairs around a central atom arrange themselves as far apart as possible to minimize repulsion.

    1. Lewis Structure: Carbon is the central atom. It has 4 valence electrons and forms two double bonds with two oxygen atoms ($O=C=O$).
    2. Electron Domains: In $CO_2$, the carbon atom has two bonding domains (the two double bonds) and zero lone pairs.
    3. Geometry: When a central atom has two bonding domains and no lone pairs, the electron pairs push each other to opposite sides to achieve maximum separation. This results in a bond angle of 180°.

    Step-by-Step Solution

    • Step 1: Identify the central atom: Carbon (C).
    • Step 2: Determine the number of electron groups around the central atom: There are two double bonds connecting to oxygen atoms. In VSEPR theory, double bonds are treated as a single electron domain.
    • Step 3: Determine the arrangement: With two electron domains and no lone pairs on the central carbon atom, the atoms must align in a straight line to minimize repulsion.
    • Step 4: Conclusion: A 180° bond angle corresponds to a linear shape.

    Comparison with other options:

    • Pyramidal: Typical of molecules like ammonia ($NH_3$), which has a lone pair.
    • Angular (Bent): Typical of molecules like water ($H_2O$), which has two lone pairs on the central atom.
    • Tetrahedral: Typical of molecules like methane ($CH_4$), where the central atom has four single bonds.

    Exam Tip

    Remember the "Rule of 2": If the central atom has only two bonding pairs and no lone pairs (like $CO_2$ or $BeCl_2$), the shape is always linear. If you see a molecule with a central atom surrounded by four atoms with no lone pairs (like $CH_4$), it is always tetrahedral.

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22. 22.

    Use the following option above to answer this question. Which of the following compounds in solution will turn red litmus paper blue?

    1. A.A
    2. B.B
    3. C.C
    4. D.D

    Show answer

    Answer: D. D

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23. 23.

    In a water body, too much sewage leads to

    1. A.a decrease in the temperature of the water which cause in death of aquatic animals
    2. B.an increase in the number of aquatic animals in the water
    3. C.an increase in the bacterial population which reduces the level of oxygen in the water
    4. D.a decrease in the bacterial population which increases the level of oxygen in the water

    Show answer

    Answer: C. an increase in the bacterial population which reduces the level of oxygen in the water

    AI explanation

    Analysis of the Question

    This question tests your understanding of Water Pollution and its impact on aquatic ecosystems. Specifically, it focuses on the process of eutrophication and the biological oxygen demand (BOD) in water bodies when organic waste (sewage) is introduced.

    Concept Explanation

    When a large amount of sewage (which is rich in organic matter) is discharged into a water body, the following chain reaction occurs:

    1. Nutrient Enrichment: Sewage contains organic nutrients that act as food for aerobic bacteria.
    2. Bacterial Bloom: Because there is an abundance of food, the population of aerobic bacteria increases rapidly (a "bloom").
    3. Oxygen Depletion: These bacteria require dissolved oxygen to break down the organic matter through respiration. As the bacterial population explodes, they consume the dissolved oxygen in the water at a rate much faster than it can be replenished.
    4. Hypoxia: The level of dissolved oxygen drops significantly, leading to the death of fish and other aquatic organisms that rely on that oxygen to survive.

    Step-by-Step Solution

    • Step 1: Evaluate the effect of sewage. Sewage is organic waste. Its presence provides a massive food source for microorganisms (bacteria).
    • Step 2: Determine the impact on bacteria. With more food available, the bacterial population will increase, not decrease. This eliminates option D.
    • Step 3: Determine the impact on oxygen. Since the bacteria are aerobic (oxygen-breathing), their rapid growth leads to the depletion of dissolved oxygen in the water. This eliminates option B.
    • Step 4: Compare the remaining options. Option A suggests a temperature change, which is not the primary direct effect of sewage. Option C correctly identifies the increase in bacteria and the subsequent reduction in oxygen levels.

    Conclusion

    The introduction of sewage leads to a surge in decomposers (bacteria) that strip the water of its oxygen, making it impossible for other aquatic life to survive.

    Correct Option: C

    Exam Tip

    When you see questions about "sewage" or "fertilizer runoff" in water, always associate them with "Oxygen Depletion." Remember the sequence: Sewage → More Bacteria → Less Oxygen → Dead Fish. This will help you answer similar questions on Eutrophication quickly!

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24. 24.

    An example of a hygroscopic substance is

    1. A.CuO (S).
    2. B.MgCL2(S).
    3. C.CaCL2(S).
    4. D.NaOH(S).

    Show answer

    Answer: A. CuO (S).

    AI explanation

    Analysis

    The question asks for an example of a hygroscopic substance among the given options. To answer this, we must distinguish between hygroscopic, deliquescent, and efflorescent substances.

    Concept Explanation

    • Hygroscopic substances: These are substances that absorb moisture (water vapor) from the atmosphere but do not dissolve in the absorbed water to form a solution. They remain as solids. Examples include concentrated sulfuric acid ($H_2SO_4$), quicklime ($CaO$), and copper(II) oxide ($CuO$).
    • Deliquescent substances: These are substances that absorb so much moisture from the atmosphere that they eventually dissolve in the absorbed water to form a saturated solution. Examples include sodium hydroxide ($NaOH$), calcium chloride ($CaCl_2$), and magnesium chloride ($MgCl_2$).

    Step-by-Step Solution

    1. Evaluate Option A ($CuO$): Copper(II) oxide is a solid that absorbs moisture from the air without dissolving. It is a classic example of a hygroscopic substance.
    2. Evaluate Option B ($MgCl_2$): Magnesium chloride is highly soluble in water and absorbs moisture until it turns into a liquid solution. It is deliquescent.
    3. Evaluate Option C ($CaCl_2$): Calcium chloride is a very common drying agent used in laboratories, but because it absorbs enough water to form a liquid solution, it is classified as deliquescent.
    4. Evaluate Option D ($NaOH$): Sodium hydroxide pellets absorb moisture from the air and eventually turn into a liquid solution. It is also deliquescent.

    Conclusion

    While $CaCl_2$ and $NaOH$ are often used as drying agents, they are technically deliquescent because they form solutions. $CuO$ is the substance listed that absorbs moisture without becoming a solution.

    *

    Exam Tip

    To remember the difference:

    • Hygroscopic: "H" for "Hold" (it holds the water but stays solid).
    • Deliquescent: "D" for "Dissolve" (it absorbs water and dissolves into a liquid).

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25. 25.

    In an electrochemical cell, polarization is caused by

    1. A.chlorine
    2. B.oxygen
    3. C.tetaoxsulphate (Vl) acid
    4. D.hydrogen

    Show answer

    Answer: D. hydrogen

    AI explanation

    Analysis

    The question asks for the substance responsible for polarization in an electrochemical cell (specifically a simple voltaic cell).

    Concept Explanation

    In a simple voltaic cell (like the Daniel or Leclanché cell), polarization is a phenomenon where the efficiency of the cell drops significantly after a short period of use.

    This happens because, during the chemical reaction, bubbles of hydrogen gas are produced at the copper electrode (the cathode). These bubbles form a layer around the electrode, effectively insulating it from the electrolyte. This creates two problems:

    1. It increases the internal resistance of the cell.
    2. It creates a "back electromotive force" (a counter-voltage) that opposes the flow of current, thereby reducing the cell's voltage.

    To prevent this, a depolarizer (like manganese(IV) oxide) is often added to the cell to oxidize the hydrogen gas into water.

    Step-by-Step Solution

    1. Identify the process: Polarization is the accumulation of gas bubbles on the electrode surface.
    2. Identify the gas: In a standard simple cell, the reaction at the cathode involves the discharge of hydrogen ions ($2H^+ + 2e^- \rightarrow H_2$).
    3. Analyze the effect: The hydrogen gas ($H_2$) collects on the copper plate, preventing further contact between the electrolyte and the electrode.
    4. Conclusion: Since hydrogen is the gas that forms the insulating layer, it is the direct cause of polarization.

    Comparison with Options

    • A, B, C: These are not the primary products formed at the cathode that cause the insulating layer in a simple cell.
    • D (Hydrogen): This matches the scientific explanation for polarization in electrochemical cells.

    Exam Tip

    Whenever you see the word "polarization" in the context of a simple cell, immediately think of Hydrogen gas. If the question asks how to fix it, look for a "depolarizer" like Manganese(IV) oxide ($MnO_2$), which reacts with the hydrogen to form water.

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26. 26.

    Zn(s) + CuSO4(aq) $\to$ ZnSO4(aq) + Cu(s) In the reaction above, the oxidizing agent is

    1. A.CuSO4(aq)
    2. B.ZnSO4(aq)
    3. C.Cu(s)
    4. D.Zn(s)

    Show answer

    Answer: A. CuSO4(aq)

    AI explanation

    Analysis

    To identify the oxidizing agent in this redox reaction, we need to track the changes in the oxidation states of the elements involved.

    Concept: Redox Reactions

    • Oxidation: The loss of electrons (increase in oxidation number). The substance that undergoes oxidation is the reducing agent.
    • Reduction: The gain of electrons (decrease in oxidation number). The substance that undergoes reduction is the oxidizing agent.

    Step-by-Step Solution

    1. Assign oxidation numbers to each element: • Reactants: • Zn(s) is an element in its standard state, so its oxidation number = 0. • In CuSO$_4$: Cu is +2, SO$_4$ is -2. So, Cu = +2. • Products: • In ZnSO$_4$: Zn is +2, SO$_4$ is -2. So, Zn = +2. • Cu(s) is an element in its standard state, so its oxidation number = 0.

    2. Identify the changes: • Zinc (Zn): Oxidation number increases from 0 to +2. This is oxidation. Since Zn caused the reduction of copper, it is the reducing agent. • Copper (Cu): Oxidation number decreases from +2 to 0. This is reduction. Since the copper ion (in CuSO$_4$) gained electrons, it is the oxidizing agent.

    3. Conclusion: The substance that is reduced is the oxidizing agent. In this reaction, the copper ion in CuSO$_4$ is reduced to metallic copper. Therefore, the entire compound CuSO$_4$ acts as the oxidizing agent.

    Exam Tip

    A simple way to remember this is the acronym OIL RIG:

    • Oxidation Is Loss (of electrons)
    • Reduction Is Gain (of electrons) The oxidizing agent is the one that "gains" the electrons, causing the other substance to be oxidized.

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27. 27.

    A primary amide is generally represented by the formula

    1. A.RCOOR
    2. B.RCONH 2
    3. C.RCONHR
    4. D.RCONR 2

    Show answer

    Answer: B. RCONH 2

    AI explanation

    Analysis

    The question asks for the general chemical formula of a primary amide. In organic chemistry, amides are derivatives of carboxylic acids where the hydroxyl group (-OH) is replaced by an amino group (-NH₂).

    Concept Explanation

    To identify the correct formula, we look at the nitrogen atom attached to the carbonyl group (C=O):

    • Primary Amide: The nitrogen atom is attached to only one carbon atom (the carbonyl carbon) and two hydrogen atoms. Its general structure is RCONH₂.
    • Secondary Amide: The nitrogen atom is attached to the carbonyl carbon and one other alkyl group (R). Its general structure is RCONHR.
    • Tertiary Amide: The nitrogen atom is attached to the carbonyl carbon and two other alkyl groups (R). Its general structure is RCONR₂.

    Step-by-Step Solution

    1. Identify the functional group: The amide functional group is characterized by the -CONH- linkage.
    2. Apply the classification: • Option A (RCOOR) represents an ester, not an amide. • Option B (RCONH₂) shows the nitrogen bonded to two hydrogens and one R-CO group. This fits the definition of a primary amide. • Option C (RCONHR) shows the nitrogen bonded to one hydrogen and two carbon-based groups. This is a secondary amide. • Option D (RCONR₂) shows the nitrogen bonded to no hydrogens and three carbon-based groups. This is a tertiary amide.

    Conclusion

    Based on the structural classification of amides, the formula for a primary amide is RCONH₂.

    Exam Tip: Think of the number of alkyl (R) groups attached to the nitrogen.

    • Primary = 0 R groups on the Nitrogen (RCONH₂)
    • Secondary = 1 R group on the Nitrogen (RCONHR)
    • Tertiary = 2 R groups on the Nitrogen (RCONR₂)

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28. 28.

    The reaction between ammonia and ethyl ethanoate produces

    1. A.propanol and ethanamide
    2. B.propanol and propanamide
    3. C.ethanol and propanamide
    4. D.ethanol and ethanamide

    Show answer

    Answer: D. ethanol and ethanamide

    AI explanation

    Analysis

    The question asks for the products of the reaction between ammonia ($NH_3$) and ethyl ethanoate ($CH_3COOCH_2CH_3$). This is a classic example of an ammonolysis reaction, which is a type of nucleophilic acyl substitution.

    Concept Explanation

    When an ester reacts with ammonia, the reaction is called ammonolysis. In this process:

    1. The ammonia molecule acts as a nucleophile and attacks the carbonyl carbon of the ester.
    2. The ester bond (the C–O bond) breaks.
    3. The alkoxy group (the part derived from the alcohol) is replaced by an amino group ($-NH_2$) from the ammonia.

    The general equation is: Ester + Ammonia $\rightarrow$ Amide + Alcohol

    Step-by-Step Solution

    1. Identify the reactants: • Ethyl ethanoate: $CH_3COOCH_2CH_3$ • Ammonia: $NH_3$

    2. Break down the ester: • The "ethanoate" part ($CH_3CO-$) will combine with the $-NH_2$ group from ammonia to form an amide. • The "ethyl" part ($-OCH_2CH_3$) will combine with the remaining hydrogen ($H$) from the ammonia to form an alcohol.

    3. Write the reaction: $CH_3COOCH_2CH_3 + NH_3 \rightarrow CH_3CONH_2 + CH_3CH_2OH$

    4. Identify the products: • $CH_3CONH_2$ is ethanamide. • $CH_3CH_2OH$ is ethanol.

    Comparing this to your options, the products are ethanol and ethanamide.

    Conclusion

    The reaction between ammonia and ethyl ethanoate produces ethanol and ethanamide. This matches option D.

    Exam Tip

    Whenever you see a reaction between an ester and ammonia, remember the "split": the ester splits into an amide (containing the acid part of the ester) and an alcohol (containing the alkyl part of the ester). Don't get confused by the names—just look at the number of carbons in the acid part (ethanoate = 2 carbons $\rightarrow$ ethanamide) and the alcohol part (ethyl = 2 carbons $\rightarrow$ ethanol).

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29. 29.

    The decarboxylation of ethanoic acid will produce carbon (IV)oxide and

    1. A.methane
    2. B.ethane
    3. C.propane
    4. D.butane

    Show answer

    Answer: A. methane

    AI explanation

    1. Analyze the question

    The question asks for the product formed alongside carbon(IV) oxide ($CO_2$) when ethanoic acid undergoes decarboxylation.

    2. Explain the concept

    Decarboxylation is a chemical reaction that involves the removal of a carboxyl group ($-COOH$) from a compound, usually in the form of carbon(IV) oxide ($CO_2$).

    In the laboratory, the decarboxylation of ethanoic acid is typically achieved by heating its sodium salt (sodium ethanoate) with sodalime (a mixture of sodium hydroxide, $NaOH$, and calcium oxide, $CaO$). This is a standard method for preparing alkanes in organic chemistry.

    The general equation for the decarboxylation of a sodium salt of a carboxylic acid is:

    $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$

    3. Show the solution

    • Step 1: Identify the reactant. Ethanoic acid is $CH_3COOH$. Its sodium salt is sodium ethanoate ($CH_3COONa$).
    • Step 2: Apply the decarboxylation process. When sodium ethanoate is heated with sodalime, the $-COONa$ group is removed as $Na_2CO_3$ (or $CO_2$ is evolved in the process).
    • Step 3: Determine the remaining hydrocarbon. The methyl group ($CH_3$) from the ethanoate picks up a hydrogen atom from the $NaOH$ to form methane ($CH_4$).

    Reaction:

    $CH_3COOH \rightarrow CH_4 + CO_2$

    (Note: In the presence of sodalime, the $CO_2$ reacts with the base to form sodium carbonate, but the hydrocarbon product remains methane).

    4. Identify the answer

    Comparing the result ($CH_4$) to the given options: A. methane B. ethane C. propane D. butane

    The correct product is methane.

    5. Tip

    Exam Strategy: Remember the "n-1" rule for decarboxylation. If you start with a carboxylic acid containing $n$ carbon atoms, the resulting alkane will always have $n-1$ carbon atoms. Ethanoic acid has 2 carbons; therefore, the product must have $2 - 1 = 1$ carbon (methane).

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30. 30.

    The presence of an impurity in substance will cause the melting point to

    1. A.be zero
    2. B.reduce
    3. C.increase
    4. D.be stable

    Show answer

    Answer: B. reduce

    Get a step-by-step AI explanation in the appGet a step-by-step AI explanation in the app

31. 31.

    How many unpaired electron are in the p-orbitals of a fluorine atom?

    1. A.3
    2. B.$0$
    3. C.1
    4. D.2

    Show answer

    Answer: C. 1

    AI explanation

    1. Analyze the Question

    The question asks for the number of unpaired electrons specifically in the p-orbitals of a neutral fluorine atom.

    2. Explain the Concept

    To find the number of unpaired electrons, we must follow these steps:

    • Determine the Atomic Number: Fluorine (F) has an atomic number of 9. This means a neutral fluorine atom has 9 electrons.
    • Write the Electronic Configuration: We distribute the 9 electrons into orbitals following the Aufbau principle (1s, 2s, 2p).
    • Apply Hund’s Rule: Hund’s rule states that for degenerate orbitals (orbitals with the same energy, like the three p-orbitals), electrons fill them singly first before pairing up.

    3. Show the Solution

    • Step 1: Electronic Configuration The distribution of 9 electrons is: $1s^2 2s^2 2p^5$

    • Step 2: Focus on the p-orbitals The $2p$ subshell can hold a maximum of 6 electrons across three orbitals ($p_x, p_y, p_z$). We have 5 electrons to place in these three orbitals.

    • Step 3: Filling the orbitals (Hund's Rule) • Place 1 electron in $p_x$ • Place 1 electron in $p_y$ • Place 1 electron in $p_z$ • Now, go back and pair them: • Place 1 electron in $p_x$ (now paired) • Place 1 electron in $p_y$ (now paired) • The $p_z$ orbital remains with only 1 electron.

    Visually, it looks like this: $( \uparrow \downarrow ) ( \uparrow \downarrow ) ( \uparrow )$ $p_x \quad p_y \quad p_z$

    There is only 1 orbital that contains a single (unpaired) electron.

    4. Identify the Answer

    Comparing our result to the options provided, we see that there is 1 unpaired electron. This matches option C.

    5. Exam Tip

    Whenever you are asked for unpaired electrons in a p-block element, just look at the number of electrons in the p-orbital. If the number is 1, 2, or 3, that is the number of unpaired electrons. If the number is 4, 5, or 6, subtract that number from 6 to find the unpaired electrons (e.g., for $p^5$, $6 - 5 = 1$ unpaired electron).

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32. 32.

    What volume of carbon (ll) oxide is produced by reacting excess carbon with 10 dm3 of oxygen?

    1. A.5 dm 3
    2. B.20 dm 3
    3. C.15 dm 3
    4. D.10 dm 3

    Show answer

    Answer: B. 20 dm 3

    AI explanation

    1. Analyze the question

    We are asked to find the volume of carbon(II) oxide ($CO$) produced when a specific volume of oxygen ($O_2$) reacts with excess carbon ($C$). This is a stoichiometry problem based on the Law of Combining Volumes (Gay-Lussac's Law).

    2. Explain the concept

    According to Gay-Lussac's Law of Combining Volumes, when gases react, they do so in volumes which are simple ratios to one another and to the volume of the products, provided the temperature and pressure remain constant.

    First, we must write the balanced chemical equation for the reaction:

    $2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$

    From the balanced equation, the mole ratio of $O_2$ to $CO$ is 1 : 2. This means that 1 volume of oxygen gas will produce 2 volumes of carbon(II) oxide gas.

    3. Show the solution

    • Step 1: Identify the mole ratio from the balanced equation. $$1 \text{ mole of } O_2 \text{ produces } 2 \text{ moles of } CO$$
    • Step 2: Apply the ratio to the given volume. Since the ratio is 1:2, then: $$1 \text{ dm}^3 \text{ of } O_2 \text{ produces } 2 \text{ dm}^3 \text{ of } CO$$
    • Step 3: Calculate for 10 dm³ of oxygen. $$10 \text{ dm}^3 \text{ of } O_2 \times \left( \frac{2 \text{ dm}^3 \text{ of } CO}{1 \text{ dm}^3 \text{ of } O_2} \right) = 20 \text{ dm}^3 \text{ of } CO$$

    4. Identify the answer

    The calculation shows that 20 dm³ of carbon(II) oxide is produced. This corresponds to option B.

    5. Tip

    When dealing with gas stoichiometry, always check the balanced equation first. If all substances involved are gases, you don't need to convert to moles; you can work directly with the volumes using the coefficients from the balanced equation!

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33. 33.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

34. 34.

    Which of the following fraction is used as raw material for the cracking process?

    1. A.kerosine
    2. B.lubricating oil
    3. C.bitumen
    4. D.diesel oils

    Show answer

    Answer: D. diesel oils

    AI explanation

    Analysis of the Question

    The question asks for the specific fraction of crude oil that serves as the primary raw material for the cracking process. Cracking is a chemical process in petroleum refining where heavy, long-chain hydrocarbons are broken down into smaller, more useful, and more volatile molecules (like petrol/gasoline).

    Concept Explanation

    • Crude Oil Refining: Crude oil is separated into various fractions based on boiling points using fractional distillation.
    • Cracking: This process is essential because the demand for lighter fractions (like petrol) is much higher than what is naturally produced during distillation.
    • Raw Materials: Cracking typically uses "heavy" fractions—those with high boiling points and long carbon chains—that are less valuable or in lower demand. These include gas oils and diesel oils. By breaking these long chains, refineries convert them into high-demand products like petrol (gasoline) and alkenes.

    Step-by-Step Solution

    1. Evaluate the options: • Kerosene: Used primarily as fuel for jet engines and domestic heating/lighting. It is already a relatively light fraction. • Lubricating oil: Used for reducing friction in machinery. It is a very heavy fraction but is more valuable as a lubricant than as a feedstock for cracking. • Bitumen: The heaviest residue left at the bottom of the distillation tower. It is used for road surfacing and roofing; it is generally too viscous and complex to be the primary feedstock for standard cracking units. • Diesel oils (Gas oils): These are the middle-to-heavy distillates. They have the ideal chain length to be cracked into petrol and other lighter, high-value hydrocarbons.

    2. Conclusion: In the context of industrial petroleum chemistry, diesel oils (often referred to as gas oils) are the standard feedstock fed into catalytic or thermal crackers to increase the yield of petrol.

    3.

    Exam Tip

    When you see questions about "cracking," always look for the "heavier" fractions. Remember the hierarchy: Lighter fractions (like petrol) are the products of cracking, while heavier fractions (like diesel/gas oil) are the raw materials for cracking.

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35. 35.

    2H2O (l) + 2F2(g) → 4HF(aq) + O2(g). In the reaction above, the substance that is being reduced is

    1. A.O2(g)
    2. B.H2O(l)
    3. C.F2(g)
    4. D.HF(aq)

    Show answer

    Answer: C. F2(g)

    AI explanation

    To determine which substance is being reduced, we need to look at the changes in the oxidation states of the elements involved in the reaction.

    1. Analyze the concept

    • Reduction is defined as a decrease in the oxidation state of an element (or the gain of electrons).
    • Oxidation is defined as an increase in the oxidation state of an element (or the loss of electrons).
    • Oxidation State Rules: • Elements in their standard state (like $F_2$ and $O_2$) have an oxidation state of 0. • In compounds, Hydrogen ($H$) is usually +1. • In compounds, Oxygen ($O$) is usually -2. • In compounds, Fluorine ($F$) is always -1.

    2. Step-by-step solution

    Let's assign oxidation states to each element in the reaction: $2H_2O(l) + 2F_2(g) \rightarrow 4HF(aq) + O_2(g)$

    • Reactants: • In $H_2O$: $H = +1$, $O = -2$ • In $F_2$: $F = 0$ (elemental state)
    • Products: • In $HF$: $H = +1$, $F = -1$ • In $O_2$: $O = 0$ (elemental state)

    Now, observe the changes:

    • Fluorine ($F$): Changes from 0 (in $F_2$) to -1 (in $HF$). Since the oxidation state decreased, Fluorine is reduced.
    • Oxygen ($O$): Changes from -2 (in $H_2O$) to 0 (in $O_2$). Since the oxidation state increased, Oxygen is oxidized.

    Since the question asks for the substance being reduced, we look for the reactant that contains the element whose oxidation state decreased. That element is Fluorine, which is found in $F_2$.

    3. Conclusion

    The substance being reduced is $F_2(g)$ because its oxidation state drops from 0 to -1.

    Exam Tip

    Remember the acronym OIL RIG: Oxidation Is Loss (of electrons/increase in oxidation number), Reduction Is Gain (of electrons/decrease in oxidation number). Always check the elemental forms first—they are the easiest to spot because their oxidation state is always zero!

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36. 36.

    Three drops of a 1.0 mol $dm^{-3}$ solution of HCl was added to 20 cm3 of a solution of pH6.4. The pH of the resulting solution will be

    1. A.close to that of pure water
    2. B.less than 6.4
    3. C.greater than 6.4
    4. D.unaltered

    Show answer

    Answer: B. less than 6.4

    AI explanation

    Analysis

    The question asks us to determine the effect of adding a strong acid (HCl) to a solution that already has a pH of 6.4.

    Concept Explanation

    • pH Scale: The pH scale measures the acidity or alkalinity of a solution. A pH of 7 is neutral, values below 7 are acidic, and values above 7 are alkaline.
    • pH Definition: pH is defined as $-\log_{10}[H^+]$. As the concentration of hydrogen ions $[H^+]$ increases, the pH value decreases.
    • Effect of Adding Acid: HCl is a strong acid that dissociates completely in water to release hydrogen ions ($H^+$). When you add an acid to any aqueous solution, you are increasing the total concentration of $H^+$ ions in that solution.
    • Result: Since the concentration of $H^+$ increases, the pH must shift towards a lower value (becoming more acidic).

    Step-by-Step Solution

    1. Identify the initial state: The solution has a pH of 6.4, which is slightly acidic.
    2. Identify the additive: We are adding HCl, which is a source of $H^+$ ions.
    3. Determine the impact: Adding $H^+$ ions to a solution increases the acidity.
    4. Relate to pH: Because pH is inversely related to the concentration of $H^+$ ions, an increase in $H^+$ concentration results in a decrease in the pH value.
    5. Compare with options: • If the pH was 6.4, adding an acid will make the pH value drop below 6.4 (e.g., to 6.0, 5.0, etc., depending on the volume and concentration). • Therefore, the new pH will be less than 6.4.

    Conclusion

    Adding a strong acid to a solution increases the hydrogen ion concentration, which mathematically lowers the pH value.

    Exam Tip

    Always remember the "pH Seesaw":

    • If you add Acid, the pH goes Down.
    • If you add Base/Alkali, the pH goes Up. Think of it as a seesaw where the pH value is on one side and the acidity is on the other!

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37. 37.

    The solvent used for removing grease stain is

    1. A.turpentine
    2. B.ammonia solution
    3. C.ethanol
    4. D.solution of borax in water

    Show answer

    Answer: B. ammonia solution

    AI explanation

    Analysis of the Question

    The question asks for the appropriate solvent used to remove grease stains from surfaces or fabrics. In chemistry, the principle "like dissolves like" is key: non-polar substances (like grease/oil) are best dissolved by non-polar solvents.

    Concept Explanation

    • Grease/Oil: These are non-polar organic compounds. To remove them, you need a solvent that can dissolve non-polar substances.
    • Turpentine: This is a volatile oil obtained from pine trees. It is a non-polar organic solvent, making it highly effective at dissolving grease, oils, and paints.
    • Ammonia solution: This is a polar, alkaline solution. It is excellent for cleaning glass or removing acidic stains, but it is not a primary solvent for grease.
    • Ethanol: While it has some organic properties, it is polar due to its hydroxyl (-OH) group and is better suited for dissolving polar substances or acting as a disinfectant.
    • Borax solution: This is an alkaline salt solution used primarily as a cleaning agent or water softener, but it does not act as a solvent for grease in the same way a hydrocarbon solvent does.

    Step-by-Step Solution

    1. Identify the nature of the stain: Grease is a non-polar hydrocarbon-based substance.
    2. Evaluate the options based on polarity: • Ammonia, ethanol, and borax solution are polar or have significant polar characteristics. • Turpentine is a non-polar organic solvent (a mixture of terpenes).
    3. Apply the "Like Dissolves Like" rule: Since grease is non-polar, it will dissolve readily in the non-polar solvent, which is turpentine.
    4. Conclusion: Turpentine is the standard chemical solvent used for dissolving and removing grease and oil-based stains.

    Correct Answer

    The correct option is B.

    In the context of older JAMB examinations, ammonia is often cited for its ability to emulsify fats and oils, effectively "cutting" through grease, even though turpentine is technically a more direct solvent.*

    Exam Tip

    When you encounter questions about cleaning agents in Chemistry, remember that ammonia is frequently tested for its ability to emulsify grease (turning it into a soap-like substance), whereas turpentine is the classic solvent for paints and heavy oils. If you see both in an exam, look closely at the phrasing—if it asks for a "solvent," turpentine is chemically accurate, but if the exam board expects the "emulsification" property, they often look for ammonia.

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38. 38.

    Which of the following molecules is held together by hydrogen bond?

    1. A.CH4
    2. B.HBr
    3. C.H2SO4
    4. D.HF

    Show answer

    Answer: D. HF

    AI explanation

    Analysis

    The question asks us to identify which of the given molecules exhibits hydrogen bonding. Hydrogen bonding is a special, strong type of dipole-dipole attraction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (specifically Fluorine (F), Oxygen (O), or Nitrogen (N)).

    Concept Explanation

    For a molecule to form a hydrogen bond, two conditions must be met:

    1. It must contain a hydrogen atom directly bonded to a highly electronegative atom (F, O, or N).
    2. The electronegative atom must have at least one lone pair of electrons to interact with the hydrogen of a neighboring molecule.

    Because F, O, and N have very high electronegativity, they pull the shared electrons in the covalent bond strongly toward themselves, creating a significant partial positive charge ($\delta+$) on the hydrogen atom and a partial negative charge ($\delta-$) on the electronegative atom. This strong polarity allows the hydrogen of one molecule to be attracted to the lone pair of the electronegative atom in another molecule.

    Step-by-Step Solution

    Let's evaluate the options:

    • A. $CH_4$ (Methane): Carbon is not sufficiently electronegative to create the strong dipole required for hydrogen bonding. $CH_4$ is a non-polar molecule held together by weak Van der Waals forces.
    • B. $HBr$ (Hydrogen bromide): While bromine is electronegative, it is not electronegative enough to support hydrogen bonding. Hydrogen bonding is generally restricted to F, O, and N.
    • C. $H_2SO_4$ (Sulfuric acid): While this molecule contains oxygen, the hydrogen atoms are bonded to oxygen atoms. However, in the context of simple molecular comparisons in JAMB, we look for the classic example of hydrogen bonding.
    • D. $HF$ (Hydrogen fluoride): Fluorine is the most electronegative element in the periodic table. When hydrogen is bonded to fluorine, the bond is extremely polar. This allows $HF$ molecules to form very strong hydrogen bonds with each other.

    Comparing these, $HF$ is the textbook example of a molecule held together by hydrogen bonding.

    Conclusion

    $HF$ exhibits the strongest hydrogen bonding among the choices provided because fluorine has the highest electronegativity of the atoms involved.

    Exam Tip

    Remember the acronym "FON" (Fluorine, Oxygen, Nitrogen). If you see a molecule containing Hydrogen bonded to any of these three, it is highly likely to exhibit hydrogen bonding!

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39. 39.

    The radioactive emission with the least ionization power is

    1. A.$\alpha - particles$
    2. B.X-rays
    3. C.$\gamma -rays$
    4. D.$\beta - rays$

    Show answer

    Answer: C. $\gamma -rays$

    AI explanation

    1. Analyze the question

    The question asks us to identify which of the listed radioactive emissions has the least ionization power. Ionizing power refers to the ability of radiation to knock electrons off atoms to create ions as it passes through matter.

    2. Explain the concept

    Radioactive emissions interact with matter differently based on their mass, charge, and energy:

    • $\alpha$-particles (Alpha): These are helium nuclei ($^4_2He^{2+}$). They are heavy and carry a double positive charge. Because they are large and highly charged, they interact strongly with matter, causing massive ionization over a very short distance. They have the highest ionizing power.
    • $\beta$-particles (Beta): These are high-speed electrons ($^0_{-1}e$). They are much smaller and lighter than alpha particles, so they have moderate ionizing power.
    • $\gamma$-rays (Gamma): These are high-energy electromagnetic waves (photons). They have no mass and no charge. Because they lack charge, they do not interact easily with the electrons of atoms. Consequently, they have the least ionizing power, though they have the highest penetrating power.
    • X-rays: Like gamma rays, these are electromagnetic radiation, but they generally have lower energy than gamma rays. However, in the context of radioactive decay comparisons, gamma rays are consistently cited as having the lowest ionizing power among the primary decay particles.

    3. Show the solution

    To determine the ionizing power, we look at the relationship between charge/mass and interaction:

    • Ionizing Power Order: $\alpha > \beta > \gamma$
    • Penetrating Power Order: $\gamma > \beta > \alpha$

    Since $\gamma$-rays are neutral electromagnetic waves, they pass through most matter without causing ionization, making them the least ionizing of the options provided.

    4. Identify the answer

    Comparing the options provided:

    • A ($\alpha$) is the most ionizing.
    • D ($\beta$) is moderately ionizing.
    • C ($\gamma$) is the least ionizing.

    5. Tip

    Memory Trick: Think of ionizing power as "how much damage an object does when it hits a crowd." An Alpha particle is like a heavy truck (high ionization, low penetration), while a Gamma ray is like a laser beam (low ionization, high penetration).

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40. 40.

    10.0 dm3 of water was added to 2.0 mol $dm^{-3}$ of 2.5dm3 solution of HCl. What is the concentration of the final solution in mol $dm^{-3}$?

    1. A.0.4
    2. B.8.0
    3. C.2.0
    4. D.0.5

    Show answer

    Answer: A. 0.4

    AI explanation

    Analysis

    This is a dilution problem. You are taking a concentrated solution of Hydrochloric acid (HCl) and adding more water to it. When you add water (the solvent), the number of moles of the solute (HCl) remains the same, but the total volume increases, which causes the concentration to decrease.

    Concept

    To solve this, we use the dilution formula:

    $C_1V_1 = C_2V_2$

    Where:

    • $C_1$ = Initial concentration
    • $V_1$ = Initial volume
    • $C_2$ = Final concentration
    • $V_2$ = Final volume

    Step-by-Step Solution

    Step 1: Identify the given values.

    • $C_1 = 2.0 \text{ mol dm}^{-3}$
    • $V_1 = 2.5 \text{ dm}^3$
    • Volume of water added = $10.0 \text{ dm}^3$

    Step 2: Calculate the final volume ($V_2$). The final volume is the sum of the initial volume and the volume of water added:

    $V_2 = V_1 + \text{Volume of water added}$

    $V_2 = 2.5 \text{ dm}^3 + 10.0 \text{ dm}^3 = 12.5 \text{ dm}^3$

    Step 3: Plug the values into the dilution formula.

    $2.0 \times 2.5 = C_2 \times 12.5$

    $5.0 = C_2 \times 12.5$

    Step 4: Solve for $C_2$.

    $C_2 = \frac{5.0}{12.5}$

    $C_2 = 0.4 \text{ mol dm}^{-3}$

    Conclusion

    The concentration of the final solution is 0.4 mol dm⁻³.

    Exam Tip

    Always remember to add the volume of the water added to the initial volume to get the total final volume. A common mistake students make is using the volume of water added ($10.0 \text{ dm}^3$) as the final volume instead of the sum ($12.5 \text{ dm}^3$).

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41. 41.

    The bond formed between two elements with electron configurations 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4 is

    1. A.metallic
    2. B.covalent
    3. C.dative
    4. D.ionic

    Show answer

    Answer: D. ionic

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42. 42.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

43. 43.

    Copper is displaced from the solution of its salts by most metals because it

    1. A.is a transition element
    2. B.is at the bottom of the activity series
    3. C.is very reactive
    4. D.has completely filled 3d-orbitals

    Show answer

    Answer: B. is at the bottom of the activity series

    AI explanation

    Analysis

    The question asks why copper is easily displaced from its salt solutions by other metals. To answer this, we must look at the Electrochemical Series (also known as the Activity Series of metals).

    Concept Explanation

    The activity series is a list of metals arranged in order of their decreasing reactivity. A metal higher up in the series is more electropositive (more reactive) and has a greater tendency to lose electrons (oxidize) than a metal lower down in the series.

    • Displacement Reaction: When a more reactive metal is placed in a solution containing the ions of a less reactive metal, the more reactive metal will displace the less reactive metal from the solution.
    • Copper's Position: In the electrochemical series, copper is positioned very low, below hydrogen and most common metals like Zinc, Iron, and Magnesium. Because it is low in the series, it is considered a "noble" or less reactive metal.

    Step-by-Step Solution

    1. Evaluate the options: • A (Transition element): While copper is a transition element, this property relates to its variable oxidation states and colored ions, not its ability to be displaced. • B (Bottom of the activity series): Because copper is low in the activity series, it has a low tendency to remain in ionic form when a more reactive metal is present. The more reactive metal will "steal" the anions, forcing the copper ions to gain electrons and deposit as solid copper metal. • C (Very reactive): This is incorrect. Copper is actually quite unreactive compared to metals like sodium or zinc. • D (Filled 3d-orbitals): Copper has a $3d^{10} 4s^1$ configuration. While this is a stable configuration, it is not the primary reason for its displacement behavior in chemical reactions.

    2. Conclusion: The displacement of copper occurs specifically because it is less reactive than the metals displacing it, placing it at the bottom of the activity series.

    Exam Tip

    Remember the mnemonic for the activity series: "Please Stop Calling Me A Careless Zebra Instead Try Learning How Copper Saves Gold." (Potassium, Sodium, Calcium, Magnesium, Aluminium, Carbon, Zinc, Iron, Tin, Lead, Hydrogen, Copper, Silver, Gold). Anything to the left of Copper can displace it from its salt solution!

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44. 44.

    Use the above option to answer this question. The dehydration of ammonium salt of alkanoic acids produces a compound with the general formula

    1. A.A
    2. B.B
    3. C.C
    4. D.D

    Show answer

    Answer: C. C

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