▶️ Answer/Explanation
At constant temperature and volume, the pressure of a gas is directly proportional to the number of gas particles present — this is a consequence of kinetic particle theory. More particles mean more frequent collisions with the container walls, producing a higher pressure. Container A contains the greatest number of particles compared to B, C, and D. Since all containers are the same size and at the same temperature, container A produces the highest number of wall collisions per unit time and therefore has the highest pressure.
✅ Answer: A
✅ Answer: A
▶️ Answer/Explanation
On a cooling curve, the sloping sections represent the substance cooling as a single phase, while the horizontal (flat) sections represent phase changes where thermal energy is released at constant temperature. Between W and X the temperature is constant, indicating the substance is undergoing condensation from gas to liquid; once fully liquid, the section between X and Y shows the liquid cooling. Therefore, between W and X the substance is in the liquid state, where particles are close together but move randomly (unlike the fixed positions in a solid). Options A and B are incorrect because U–V shows gas cooling and V–W is the gas-to-liquid phase change (heat released, not absorbed). Option D is incorrect because Y–Z shows the solid cooling after freezing is complete.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
According to Graham’s Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass — lighter gases diffuse faster. The molar masses are: CO₂ = 44 g/mol, HCl = 36.5 g/mol, H₂S = 34 g/mol, NO₂ = 46 g/mol. Hydrogen sulfide (H₂S) has the lowest molar mass of the four gases and therefore its particles move fastest on average at a given temperature, allowing it to diffuse the quickest across the room.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Sulfur has atomic number 16, so S²⁻ has 16 + 2 = 18 electrons, giving the configuration 2,8,8. Argon (atomic number 18) also has 18 electrons and configuration 2,8,8 — it matches. Calcium (atomic number 20) is neutral with 20 electrons (2,8,8,2) — it does not match. Oxide O²⁻ has 8 + 2 = 10 electrons (2,8) — it does not match. Therefore, only argon has the same electronic configuration as S²⁻, giving the answer of 1.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Isotopes are atoms of the same element that differ only in the number of neutrons. Both $^{12}_{6}\text{T}$ and $^{14}_{6}\text{T}$ have 6 protons and therefore 6 electrons, giving them identical electronic configurations and identical chemical properties. Option A is wrong because neutrons play no role in chemical reactions. Option C is wrong because nucleon number (mass number) equals protons + neutrons, which differs between the two isotopes (12 vs 14). Option D is wrong because position in the Periodic Table is determined by proton number, which is the same for both isotopes.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
When we look at the periodic table, we see that lithium is placed in Group I, meaning each atom has exactly one valence electron. To achieve a stable, full outer shell, it naturally loses this single electron, forming a positively charged $\text{Li}^{+}$ cation. On the other hand, nitrogen is found in Group V with five valence electrons, so it requires three additional electrons to complete its octet. By gaining these three electrons, each nitrogen atom becomes a negatively charged $\text{N}^{3-}$ anion. The ionic compound lithium nitride ($\text{Li}_3\text{N}$) forms when three lithium atoms each donate an electron to a single nitrogen atom, perfectly balancing the overall charges.
✅ Answer: A.
✅ Answer: A.
▶️ Answer/Explanation
Option C correctly shows the dot-and-cross diagram for ammonia (NH₃). Nitrogen has 5 valence electrons and forms three single covalent bonds with three hydrogen atoms (each contributing 1 electron), leaving one lone pair on nitrogen — giving a total of 8 electrons around nitrogen (one lone pair + three bonding pairs). This satisfies the octet rule for nitrogen and the duet rule for each hydrogen. The other diagrams in options A, B, and D contain incorrect numbers of electrons or bonds around the central atom, making them invalid representations of their respective compounds.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
In a giant metallic lattice, metal atoms release their outermost electrons into a shared “sea,” leaving behind positively charged metal ions called cations. The released electrons are free to move throughout the entire structure and are called delocalised electrons — they carry a negative charge. The metallic bond is the strong electrostatic attraction between the positive cations and the surrounding sea of negative delocalised electrons. There are no anions present in a pure metallic structure, eliminating options A, B, and C.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Calcium forms Ca²⁺ ions and oxygen forms O²⁻ ions; the charges balance in a 1:1 ratio, giving the correct formula CaO. Option B is wrong — cobalt(II) is Co²⁺ and chloride is Cl⁻, so the correct formula is CoCl₂ (not Co₂Cl). Option C is wrong — sulfur dioxide is correctly written as SO₂, not S₂O₂. Option D is wrong — copper(II) is Cu²⁺ and sulfate is SO₄²⁻; the charges already balance in a 1:1 ratio, giving CuSO₄, not Cu(SO₄)₂.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
To determine the mass of magnesium sulfate produced, we first calculate the number of moles of magnesium used in the reaction. Given the molar mass of magnesium is 24 g/mol, a 12 g sample corresponds to exactly 0.5 moles. According to the balanced chemical equation, magnesium reacts with sulfuric acid to form magnesium sulfate in a 1:1 molar ratio, meaning 0.5 moles of magnesium will yield 0.5 moles of $\text{MgSO}_4$. The relative formula mass ($M_r$) of $\text{MgSO}_4$ is given as 120. Multiplying the moles of product by its molar mass ($0.5 \times 120$) gives a final theoretical yield of 60 g. This precisely matches the expected stoichiometric proportions.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Assuming 100 g of Q: C = 40.0 g, H = 6.7 g, O = 100 − 40.0 − 6.7 = 53.3 g. Converting to moles: C = 40.0 ÷ 12 ≈ 3.33 mol; H = 6.7 ÷ 1 = 6.7 mol; O = 53.3 ÷ 16 ≈ 3.33 mol. Dividing each by the smallest (3.33): C = 1, H ≈ 2, O = 1. The simplest whole-number ratio is C:H:O = 1:2:1, giving the empirical formula CH₂O.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Detailed solution:
First, find the number of NH₃ molecules in 2.00 mol: $2.00 \times 6.02 \times 10^{23} = 1.204 \times 10^{24}$ molecules. Each ammonia molecule (NH₃) contains 4 atoms (1 nitrogen + 3 hydrogen). Multiplying: $1.204 \times 10^{24} \times 4 = 4.816 \times 10^{24} \approx 4.82 \times 10^{24}$ atoms.
✅ Answer: (C)
First, find the number of NH₃ molecules in 2.00 mol: $2.00 \times 6.02 \times 10^{23} = 1.204 \times 10^{24}$ molecules. Each ammonia molecule (NH₃) contains 4 atoms (1 nitrogen + 3 hydrogen). Multiplying: $1.204 \times 10^{24} \times 4 = 4.816 \times 10^{24} \approx 4.82 \times 10^{24}$ atoms.
✅ Answer: (C)
▶️ Answer/Explanation
Let’s break this down step by step. The pale yellow-green gas mentioned is chlorine ($\text{Cl}_2$). When we electrolyze concentrated solutions like L (concentrated HCl) and M (concentrated NaCl), the high concentration of chloride ($\text{Cl}^-$) ions means they are preferentially discharged at the anode over hydroxide ($\text{OH}^-$) ions. This reaction produces chlorine gas: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$. On the other hand, in a dilute solution like N (dilute NaCl), the $\text{Cl}^-$ concentration is simply too low. As a result, the $\text{OH}^-$ ions are preferentially discharged instead, producing oxygen gas: $4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-$. Therefore, only solutions L and M will produce the pale yellow-green chlorine gas.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
At the positive electrode (anode), oxidation occurs: hydroxide ions are discharged, giving $4OH^- \rightarrow 2H_2O + O_2 + 4e^-$, producing oxygen gas. At the negative electrode (cathode), reduction occurs: hydrogen ions gain electrons, giving $2H^+ + 2e^- \rightarrow H_2$, producing hydrogen gas. Option C correctly assigns these half-equations to the correct electrodes. Options A and B have the equations reversed or are not valid half-equations.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Detailed solution:
Statement 1 is correct — fuel cells convert the chemical energy stored in hydrogen directly into electrical energy. Statement 2 is incorrect — hydrogen is oxidised at the anode (loses electrons to form H⁺ or reacts with OH⁻), not reduced; it is oxygen that is reduced at the cathode. Statement 3 is correct — the only product of a hydrogen–oxygen fuel cell is water (H₂O), so no harmful atmospheric pollutants are produced. Therefore statements 1 and 3 are correct, giving option C.
✅ Answer: (C)
Statement 1 is correct — fuel cells convert the chemical energy stored in hydrogen directly into electrical energy. Statement 2 is incorrect — hydrogen is oxidised at the anode (loses electrons to form H⁺ or reacts with OH⁻), not reduced; it is oxygen that is reduced at the cathode. Statement 3 is correct — the only product of a hydrogen–oxygen fuel cell is water (H₂O), so no harmful atmospheric pollutants are produced. Therefore statements 1 and 3 are correct, giving option C.
✅ Answer: (C)
▶️ Answer/Explanation
On a reaction pathway diagram, the enthalpy change (ΔH) is defined as the difference in energy between the products and the reactants — it is the vertical distance between the energy level of P + Q (reactants) and R + S (products). The activation energy is the energy difference from the reactants up to the peak (transition state). The letter C represents the vertical gap between reactant and product energy levels, which is the enthalpy change of the reaction. Other letters label the activation energy or the total energy of the transition state.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
To calculate the enthalpy change for the complete combustion of methane, we first determine the energy required to break the bonds in the reactants. Breaking four $\text{C–H}$ bonds and two $\text{O=O}$ double bonds absorbs $(4 \times 410) + (2 \times 496) = 2632\text{ kJ/mol}$. Next, we calculate the energy released upon forming the products, which involves creating two $\text{C=O}$ double bonds and four $\text{O–H}$ bonds, releasing $(2 \times 805) + (4 \times 460) = 3450\text{ kJ/mol}$. The overall enthalpy change, $\Delta H$, is the energy absorbed minus the energy released: $2632 – 3450 = -818\text{ kJ/mol}$. This negative value indicates the reaction is exothermic.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
A physical change alters the form or state of a substance without changing its chemical composition or producing new substances. Evaporating ethanol (B) is simply a liquid-to-gas phase change — the ethanol molecules remain unchanged (C₂H₅OH). By contrast, cracking (A) breaks C–C bonds to form different hydrocarbons, fermentation (C) converts glucose into ethanol and CO₂, and neutralisation (D) forms salt and water — all of these involve new chemical substances being produced, making them chemical changes.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Increasing temperature raises the rate of reaction through two related effects explained by collision theory. First, particles move faster so they collide more frequently (statement 1 — correct). Second, particles have greater kinetic energy, meaning a larger proportion of collisions have energy equal to or greater than the activation energy, producing more successful collisions (statement 3 — correct). Statement 2 is wrong — temperature does not lower the activation energy; a catalyst does that. Statement 4 is wrong — temperature does not change the number of particles per unit volume (concentration stays constant in a closed system).
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
According to Le Chatelier’s Principle, a decrease in temperature favors the exothermic forward reaction ($\Delta H = -909\,\text{kJ/mol}$), shifting the equilibrium to the right. Additionally, there are 9 moles of gas on the reactant side and 10 moles on the product side. Decreasing the pressure favors the side with a greater number of gas molecules, further shifting the equilibrium to the right. Thus, both changes favor the forward reaction.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Reduction, in terms of oxygen, is defined as the loss of oxygen from a substance. In this reaction, mercury(II) oxide (HgO) loses oxygen to form elemental mercury (Hg), so Hg is reduced. Option C describes oxidation (gain of oxygen), not reduction. Option A is incorrect because by conservation of mass the total mass of products equals the total mass of reactants. Option B is irrelevant to the definition of reduction. Therefore D is the correct reason.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Statement 1 is correct: nitric acid is a strong acid that fully dissociates, producing a high [H⁺] and a low pH of 1.0, whereas propanoic acid is a weak acid that only partially dissociates. Statement 2 is correct: despite propanoic acid having a higher overall concentration (0.4 mol/dm³ vs 0.1 mol/dm³), its partial dissociation means its actual [H⁺] is much lower than that of fully dissociated nitric acid, giving it a higher pH of 2.6. Statement 3 is incorrect: the higher pH of propanoic acid is due to its weak-acid nature (incomplete dissociation), not its higher concentration — concentration alone does not determine pH when comparing strong and weak acids.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
Metal oxides are generally basic — Group I metals form ionic oxides that dissolve in water to produce alkaline solutions (e.g. $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$), so the oxide of E (Group I metal) is basic. Non-metal oxides are generally acidic — Group VII halogens form covalent oxides that react with water to produce acidic solutions (e.g. $\text{Cl}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{HClO}$), so the oxide of G (Group VII non-metal) is acidic. Therefore option D correctly describes both oxides.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Insoluble salts are prepared by precipitation: mixing two soluble solutions that contain the required ions, then filtering to collect the insoluble precipitate. In option C, $\text{Pb(NO}_3)_2\text{(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)} + 2\text{HNO}_3\text{(aq)}$; the insoluble $\text{PbSO}_4$ precipitates immediately, is collected by filtration as the residue, washed with distilled water to remove soluble impurities ($\text{HNO}_3$), and dried to give pure $\text{PbSO}_4$. Option B collects the filtrate (wrong — filtrate contains soluble $\text{KNO}_3$, not $\text{PbSO}_4$). Option A would leave mixed salts after evaporation. Option D incorrectly uses titration (for soluble salts) and $\text{Pb(OH)}_2$ is itself insoluble.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Both oxygen and sulfur exist as simple molecular structures, not giant covalent structures. Oxygen exists as diatomic $O_2$ molecules, and sulfur most commonly exists as $S_8$ ring molecules at room temperature — both are simple molecular and held together only by weak intermolecular forces. Giant covalent structures (like diamond or $SiO_2$) have atoms bonded throughout in an extended network. Therefore statement B (“they have giant covalent structures”) is the one that is not correct. All other statements — both are non-metals (A), both have 6 outer electrons as Group VI elements (C), and $SO_2$/$SO_3$ are acidic oxides (D) — are correct.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Going down Group VII (halogens), density increases (Cl₂ is a gas, Br₂ is a liquid, I₂ is a grey-black solid) and reactivity decreases. Tennessine is below iodine, so it is predicted to be a solid (denser than iodine), darker in colour (following the trend from pale yellow-green → red-brown → grey-black, so tennessine would be even darker, i.e. black), and less reactive than iodine. Option B is incorrect because reactivity decreases down Group VII. Options C and D are incorrect because tennessine would be a solid, not a gas, at room temperature.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
Zinc sits above iron in the reactivity series, making it more reactive than iron. Because of this higher reactivity, zinc loses electrons more readily than iron when both are in contact with an electrolyte (such as sea water). This means zinc acts as the anode in an electrochemical cell and is preferentially oxidised (corroded) rather than the iron in the steel hull. As long as zinc remains, the iron is cathodically protected and cannot rust — this is the principle of sacrificial protection.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
In a pure metal, all atoms are the same size and arranged in regular layers that can slide past one another easily, making the metal soft and malleable. When a second element is added to form an alloy, the different-sized atoms disrupt the regular arrangement, preventing the layers from sliding — this is why alloys are harder. Option A is wrong because the hardening is structural, not due to intermolecular forces. Option B is wrong because brass is copper and zinc, not tin (copper-tin makes bronze). Option D is wrong because steel is an important alloy containing iron and carbon.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Although aluminium is actually higher than copper in the reactivity series, it appears unreactive because its surface is already coated with a thin, tough, adherent layer of aluminium oxide ($\text{Al}_2\text{O}_3$) that forms spontaneously in air. This oxide layer acts as a barrier, preventing further reaction with oxygen or other reagents. Copper, which is much lower in the reactivity series, does not form such a protective oxide layer, so it can react with oxygen in air when heated to form the black copper(II) oxide. Option B is therefore incorrect as it states the opposite of the true reactivity order.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
In the blast furnace, carbon (coke) first burns with limited oxygen to produce carbon dioxide (C + O₂ → CO₂). This CO₂ then reacts with more coke in the hotter lower zone to form carbon monoxide: CO₂ + C → 2CO. This is the correct equation for CO formation (not 2C + O₂ → 2CO, which would give CO only if oxygen were severely limited). The iron ore (haematite, Fe₂O₃) is then reduced by carbon monoxide in the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, since the ore is iron(III) oxide, not FeO. Only option D gives both correct equations.
✅ Answer: D
✅ Answer: D
▶️ Answer/Explanation
Dissolved solutes (like nitrates) lower the freezing point of water below 0 °C (statement 2 ✓) and raise the boiling point slightly above 100 °C — so statement 1 is incorrect. Anhydrous cobalt(II) chloride is blue and turns pink on contact with water, so statement 3 (pink to blue) is backwards and incorrect. Anhydrous copper(II) sulfate is white and turns blue on contact with water (statement 4 ✓). Both test reagents detect the presence of water, not its purity — the river water contains water, so statement 4 is correct. Therefore, only statements 2 and 4 are correct.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Statement 1 is wrong: sedimentation and filtration remove suspended (insoluble) solids, not dissolved solids. Statement 2 is correct: filtration does remove insoluble solids. Statement 3 is correct: activated carbon (charcoal) is used to adsorb chemicals that cause unpleasant tastes and odours. Statement 4 is incorrect: chlorination is used to kill microbes/bacteria, not to decrease the pH. Therefore, statements 2 and 3 are both correct.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
The damp iron wool rusts by reacting with the oxygen in the air sample, causing the water level inside the measuring cylinder to rise as oxygen is consumed. The volume of oxygen used = 150 − 122 = 28 cm³. The percentage of oxygen = (28 ÷ 150) × 100 = 18.67%, which rounds to 19% to the nearest whole number. This is slightly below the normal atmospheric level of 21%, which is consistent with “polluted air” that may contain a higher proportion of other gases. Option C (28%) is a common error from confusing the volume of oxygen removed with the percentage.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
In the catalytic converter the reaction is: 2CO + 2NO → 2CO₂ + N₂. Tracking oxidation numbers: in NO, nitrogen has oxidation number +2; in N₂ it is 0 — a decrease, so nitrogen monoxide is reduced. Carbon in CO has oxidation number +2; in CO₂ it is +4 — an increase, so carbon monoxide is oxidised. This is a redox reaction where NO acts as the oxidising agent and CO acts as the reducing agent.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Just as ethanol ($\mathrm{CH_3CH_2OH}$) is oxidised to ethanoic acid ($\mathrm{CH_3COOH}$), propan-1-ol ($\mathrm{CH_3CH_2CH_2OH}$) is oxidised to propanoic acid ($\mathrm{CH_3CH_2COOH}$). The primary alcohol group (–OH) at the end of the 3-carbon chain is converted into a carboxylic acid group (–COOH) while the carbon skeleton remains unchanged.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Methyl propane (2-methylpropane) has the structure (CH₃)₃CH. It contains two distinct types of hydrogen atoms: one hydrogen on the central (tertiary) carbon, and nine hydrogens on the three equivalent methyl groups. Monosubstitution of the central H gives 2-chloro-2-methylpropane: (CH₃)₃CCl. Monosubstitution of any one of the nine equivalent methyl H atoms gives 1-chloro-2-methylpropane: ClCH₂CH(CH₃)CH₃. Since all nine methyl H atoms are equivalent, they all yield the same product. Therefore exactly 2 structural isomers with formula C₄H₉Cl can be formed.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Statement 1 is an advantage of fermentation (using renewable sugars/glucose from crops), not a disadvantage. Statement 3 is incorrect because fermentation occurs at a low temperature of 25–35 °C, not a high temperature — low temperature operation is actually an advantage. Statement 2 is a true disadvantage: fermentation produces a dilute (approximately 15%) aqueous ethanol mixture that requires energy-intensive distillation to purify. Statement 4 is also a true disadvantage: the batch fermentation process is slow (taking days), compared to the continuous, rapid catalytic hydration of ethene. Therefore statements 2 and 4 are both disadvantages.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Nylon is a polyamide made by condensation polymerisation between a dicarboxylic acid and a diamine. In condensation polymerisation, monomers join together with the elimination of a small molecule — in this case, water (H₂O) is released each time an amide (peptide) bond forms between –COOH and –NH₂ groups. This contrasts with addition polymerisation (e.g. poly(ethene) from ethene), where monomers simply join together with no other product. Therefore the type is condensation and the byproduct is water.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Cr³⁺ forms a green precipitate of Cr(OH)₃ when sodium hydroxide is added, and this precipitate dissolves in excess NaOH to give a dark green solution (Cr³⁺ is amphoteric). Ca²⁺ forms a white precipitate that is insoluble in excess NaOH. Cu²⁺ forms a light blue precipitate insoluble in excess NaOH (though it dissolves in excess aqueous ammonia to give a dark blue solution). Fe²⁺ also forms a green precipitate which is insoluble in excess NaOH and gradually turns brown on standing as Fe²⁺ oxidises to Fe³⁺ at the surface. The key distinguishing feature here is solubility in excess NaOH — only Cr³⁺ gives a green precipitate that dissolves in excess sodium hydroxide.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
The Rf (retardation factor) value is defined as the ratio of the distance moved by the substance to the distance moved by the solvent front, both measured from the starting baseline. Because a substance always moves the same distance or less than the solvent, Rf values are always between 0 and 1. A value of 1 would mean the substance travels with the solvent front, while a value of 0 means it does not move at all. Option A gives the correct formula. Option B is inverted and would give values greater than 1, which is physically impossible. Options C and D are dimensionally incorrect for an identification ratio.
✅ Answer: (A)
✅ Answer: (A)