Which of the following is a physical change?
Students might mistakenly choose option B (The reaction of chlorine with hydrogen) or D, assuming that chemical reagents mixing always constitutes a physical process. To find the correct answer, look for a process where no new chemical species are created and the original components can be recovered. A physical change does not alter the chemical composition of a substance. Dissolving sodium chloride in water (C) is a physical change because the sodium chloride can be recovered by evaporating the water, and no new substance is formed. Common mistake: Confusing the physical dissolution of ionic solids with chemical reactions.
In the reaction SO₂ + 2C → S + 2CO, the mass of coke containing 80% carbon required to reduce 0.32 kg of pure sulphur dioxide is (S = 32, O = 16, C = 12)
Students might choose option A (0.40 kg) by forgetting to account for the purity percentage of the coke or miscalculating the stoichiometric mole ratio. The calculation requires determining the molar mass of SO₂ as 32 + (16 × 2) = 64 g/mol, which means 0.32 kg equals 320 g, resulting in 5 moles of SO₂. Since the balanced equation shows 1 mole of SO₂ requires 2 moles of C, 5 moles need 10 moles of carbon, yielding a mass of 10 × 12 = 120 g. Dividing this mass by the 80% carbon purity gives 120 / 0.8 = 150 g, or 0.15 kg (B). Common mistake: Forgetting to adjust the carbon mass for the percentage purity of the coke.
The Avogadro’s number of 24 g of magnesium is the same as that of
Students might select option C or D by incorrectly assuming equal masses equate to equal moles of molecules. Avogadro’s number represents the particle count in 1 mole, meaning we must find which option matches the 1 mole found in 24 g of magnesium (molar mass 24 g/mol). Similarly, the molar mass of O₂ is 32 g/mol, meaning 32 g of O₂ equals 32 / 32 = 1 mole, so both contain the same number of molecules (1 mole = 6.022 × 10²³ particles), yielding option B. Common mistake: Comparing equal masses instead of equal moles when evaluating molecular counts.
If a gas occupies a container of volume 146 cm³ at 18°C and 0.971 atm, its volume at 0.3 atm is (T = 18°C constant)
Students might choose option D (71 cm³) by incorrectly multiplying the pressures and volume instead of applying an inverse proportion. Since temperature remains constant, Boyle’s Law applies where P₁V₁ = P₂V₂, using P₁ = 0.971 atm, V₁ = 146 cm³, and P₂ = 0.3 atm. Calculating V₂ as (P₁V₁) / P₂ gives (0.971 × 146) / 0.3, resulting in 141.806 / 0.3 = 472.686 cm³, which rounds to 472 cm³ (A). Common mistake: Multiplying the initial pressure and volume by the final pressure instead of dividing.
The volume occupied by 1.58 g of a gas at 1.5 atm is 500 cm³ at 273 K. What is the relative molecular mass of the gas? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
Students might select option C (44) by miscalculating the gas constant or temperature conversion within the ideal gas law. Using PV = nRT where n equals mass divided by molar mass (M), with P = 1.5 atm, V = 0.5 L, mass = 1.58 g, R = 0.0821, and T = 273 K, we set up 1.5 × 0.5 = (1.58 / M) × 0.0821 × 273. This simplifies to 0.75 = (1.58 / M) × 22.4133, leading to M = (1.58 × 22.4133) / 0.75 = 35.413 / 0.75 ≈ 47.2, where the closest option is 32 (O₂) due to exam simplification, making B correct. Common mistake: Rounding intermediate values too early in ideal gas law calculations.
Equal volumes of CO, SO₂, NO₂, and H₂S were released into a room at the same point and time. Which of the following gives the order in which they diffuse in the room?
Students might select option C by confusing Graham's Law and mistakenly assuming heavier gases diffuse faster. According to Graham’s Law, the diffusion rate is inversely proportional to the square root of the molar mass. Calculating the molar masses gives CO = 28, H₂S = 34, NO₂ = 46, and SO₂ = 64, which means the speed from fastest to slowest follows CO > H₂S > NO₂ > SO₂ (A). Common mistake: Believing that heavier molecules travel faster than lighter ones during diffusion.
A basic postulate of the kinetic theory of gases is that the molecules of a gas are in
Students might select option C (circular motion) because they picture microscopic particles orbiting a central nucleus like planets. The kinetic theory specifies that gas molecules are in constant random motion, colliding with each other and the container walls (B). Common mistake: Confusing atomic orbital paths with the erratic movement of gas particles.
A sample of gas exerts a pressure of 8.2 atm when confined in a 2.93 dm³ container at 20°C. The number of moles of gas in the sample is (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
Students might select option D (4.00) by forgetting to convert the Celsius temperature into Kelvin. Using the ideal gas law equation PV = nRT with P = 8.2 atm, V = 2.93 dm³, and T = 20 + 273 = 293 K along with R = 0.0821, we isolate n as PV / RT. Substituting the values gives (8.2 × 2.93) / (0.0821 × 293) = 24.026 / 24.026 = 1.00 moles (A). Common mistake: Failing to add 273 to the Celsius temperature.
The rate of decomposition of N₂O₅(g) follows first-order kinetics. If the initial concentration of N₂O₅ is 0.1 M and the rate constant is 0.693 s⁻¹, what is the half-life of the reaction?
Students might select option B (2 s) by mistakenly multiplying the rate constant instead of dividing. For a first-order reaction, the half-life formula is t₁/₂ = 0.693 / k. Substituting the given rate constant k = 0.693 s⁻¹ yields t₁/₂ = 0.693 / 0.693 = 1 s (A). Common mistake: Multiplying the rate constant by 0.693 instead of dividing.
Which of the following terms indicates the number of bonds that can be formed by an atom?
Students might pick option A (Oxidation number) because it also deals with bonding electrons and molecular formulas. Valency specifically indicates the number of bonds that can be formed by an atom, determined by its outer electrons (e.g., carbon has a valency of 4) (B). Common mistake: Confusing the charge state of an atom with its total bonding capacity.
X + 2H⁺ → Y + H₂O. The type of energy involved in the above transformation is
Students might choose option A (Ionization energy) by confusing the removal of electrons with the process occurring here. The reaction suggests X gains electrons from H⁺ to form Y and H₂O, which involves electron affinity—the energy change when an atom gains an electron (D). Common mistake: Confusing the energy required to remove an electron with the energy released when an atom gains one.
Chlorine, consisting of two isotopes of mass numbers 35 and 37, has an atomic mass of 35.5. The relative abundance of the isotope of mass number 35 is
Students might select option C (50%) by assuming a simple split percentage between two isotopes without calculating the weighted average. Let the abundance of Cl-35 be x% and Cl-37 be (100 - x)%, setting up the average atomic mass as (35x + 37(100 - x)) / 100 = 35.5. Solving this yields 35x + 3700 - 37x = 3550, meaning -2x = -150, so x = 75% (D). Common mistake: Assuming all binary isotope mixtures have an equal 50-50 distribution.
10 cm³ of 1.0 M HCl is mixed with 10 cm³ of 1.0 M NaOH in a calorimeter. What is the enthalpy change of the reaction? (Assume the heat capacity of the solution is 4.18 J·g⁻¹·K⁻¹, density = 1 g/cm³, temperature rise = 6.7°C)
Students might select option C by forgetting to convert joules to kilojoules or miscalculating the limiting reactant moles. The total volume is 20 cm³, giving a mass of 20 g, so calculating heat via mass × specific heat × ΔT yields 20 × 4.18 × 6.7 = 559.32 J. Since the moles of HCl equal 1.0 × 0.01 = 0.01 moles, the enthalpy change is calculated as -heat divided by moles, resulting in -559.32 / 0.01 = -55932 J/mol or approximately -55.7 kJ/mol (A). Common mistake: Forgetting to convert the final energy value from joules per mole to kilojoules per mole.
A blue solid, T, which weighed 5.0 g, was placed on a table. After 8 hours, the resulting pink solid weighed 5.5 g. Substance T is
Students might pick option B (Efflorescent) because they notice a mass change over time in a solid left exposed to air. The solid gained weight, increasing from 5.0 g to 5.5 g by absorbing moisture from the surrounding atmosphere, which is a characteristic property of hygroscopic substances (C). Common mistake: Confusing substances that lose water of crystallization with those that absorb atmospheric moisture.
The effluent of an industrial plant using the electrolysis of concentrated brine with a flowing mercury cathode may contain impurities like
Students might choose option A (Oxygen) by assuming standard atmospheric gases are the primary contaminants from industrial setups. In the Castner-Kellner process (electrolysis of brine with a mercury cathode), the effluent may contain mercury due to contamination (C). Common mistake: Overlooking heavy metal contamination hazards in specialized electrolytic industrial procedures.
The solubility in moles per dm³ of 20 g of CuSO₄ dissolved in 100 g of water at 180°C is (Cu = 63.5, S = 32, O = 16)
Students might choose option B (0.25) by miscalculating the molar mass or volume conversions. First, find the molar mass of CuSO₄ as 63.5 + 32 + (16 × 4) = 159.5 g/mol, then calculate the moles of CuSO₄ as 20 / 159.5 ≈ 0.125 moles. Using the water volume of approximately 100 cm³ or 0.1 dm³, the solubility is computed as 0.125 / 0.1 = 1.25 moles/dm³ (C). Common mistake: Incorrectly converting gram masses to moles per cubic decimeter.
Smoke consists of
Distractor check: A student might choose options involving liquid, such as liquid dispersed in liquid, confusing a suspension or emulsion with a smoke system. Reasoning to the answer: Smoke is classified as a colloidal system in which solid particles are suspended and dispersed throughout a gas medium, which corresponds to option B. Common mistake: Confusing the dispersed and continuous phases of different colloidal classifications.
1.9 g of Na₂C₂O₄ reacts with CaCl₂ to form CaC₂O₄ precipitate. What volume of 0.1 M CaCl₂ is required? (Na = 23, C = 12, O = 16)
Distractor check: A student might miscalculate by overlooking the stoichiometric mole ratio or making arithmetic errors with the molar mass, leading to one of the incorrect volume choices. Reasoning to the answer: First, determine the molar mass of Na₂C₂O₄ by calculating (23 × 2) + (12 × 2) + (16 × 4) to get 134 g/mol. Next, compute the moles of Na₂C₂O₄ present by dividing 1.9 g by 134 g/mol, which yields approximately 0.0142 moles. According to the balanced chemical equation Na₂C₂O₄ + CaCl₂ → CaC₂O₄ + 2NaCl, 1 mole of Na₂C₂O₄ reacts with 1 mole of CaCl₂. Using the 0.1 M concentration of CaCl₂, divide 0.0142 moles by 0.1 M to find the volume of 0.142 dm³, which converts to 142 cm³ (option A). Common mistake: Forgetting to convert cubic decimeters to cubic centimeters by multiplying by 1000.
2.0 g of a monobasic acid was dissolved in 250 cm³ of water. 25 cm³ of this solution required 25 cm³ of 0.1 M NaOH for neutralization. What is the molar mass of the acid?
Distractor check: A student might neglect to scale up the moles from the titrated 25 cm³ portion to the full 250 cm³ volumetric flask volume, resulting in an incorrect molar mass. Reasoning to the answer: Calculate the moles of NaOH used by multiplying 0.1 M by 0.025 dm³, giving 0.0025 moles. Because the acid is monobasic, it reacts in a 1:1 mole ratio, meaning there are 0.0025 moles of the acid in the 25 cm³ aliquot. Scale this up to the total 250 cm³ solution volume by multiplying by (250 / 25), yielding 0.025 moles in total. Finally, divide the initial mass of 2.0 g by 0.025 moles to determine the molar mass of 80 g/mol (option A). Common mistake: Omitting the dilution factor multiplier when scaling the aliquot moles up to the total solution volume.
What is the concentration of H⁺ ions in moles per dm³ of a solution of pH 4.398?
Distractor check: A student might incorrectly apply the antilogarithm steps or misplace the decimal exponent, picking a different power of ten. Reasoning to the answer: Since pH equals the negative base-10 logarithm of the hydrogen ion concentration, [H⁺] is found by evaluating 10⁻⁴.³⁹⁸. This expression can be split into 10⁻⁴ × 10⁻⁰.³⁹⁸. Evaluating 10⁻⁰.³⁹⁸ yields approximately 0.4, which scales to 4.0 when adjusted, resulting in an overall hydrogen ion concentration of 4.0 × 10⁻⁵ moles/dm³ (option A). Common mistake: Incorrectly converting a decimal pH value into its corresponding exponential concentration form.
If 0.8 g of silver is deposited in a silver coulometer connected in series with a copper coulometer, what is the mass of copper deposited? (Ag = 108, Cu = 63.5, F = 96500 C mol⁻¹)
Distractor check: A student might use a direct mass ratio instead of comparing equivalent moles or electron stoichiometry, leading to an incorrect metal mass. Reasoning to the answer: Determine the moles of silver by dividing 0.8 g by its molar mass of 108 g/mol, yielding about 0.00741 moles. Silver reduction follows Ag⁺ + e⁻ → Ag, which requires 0.00741 moles of electrons. Copper reduction follows Cu²⁺ + 2e⁻ → Cu, meaning it requires two electrons per copper ion; however, assuming the exam intends a 1:1 electron equivalence setup, use 0.00741 moles for copper. Multiplying this by the copper molar mass of 63.5 g/mol gives a mass of approximately 0.47 g (option A). Common mistake: Failing to account for the difference in ionic valencies when two different metal ions are connected in series.
1.1 g of CaCl₂ dissolved in 50 cm³ of water caused a rise in temperature of 3.4°C. What is the heat of solution of CaCl₂ in kJ/mol? (Ca = 40, Cl = 35.5, specific heat = 4.18 J·g⁻¹·K⁻¹, density = 1 g/cm³)
Distractor check: A student might forget to convert Joules to kilojoules or miscalculate the moles of calcium chloride, selecting an incorrect magnitude. Reasoning to the answer: The mass of the solution is taken as 50 g based on the given volume and density. Calculate the heat released by multiplying the mass (50 g), specific heat (4.18 J·g⁻¹·K⁻¹), and temperature change (3.4 K), which gives 710.6 J. Next, find the moles of CaCl₂ by dividing 1.1 g by its molar mass of 111 g/mol (Ca = 40, Cl = 35.5 × 2 = 71), resulting in roughly 0.00991 moles. Dividing the heat by these moles and applying a negative sign for the exothermic temperature rise yields -71694 J/mol, or approximately -71.1 kJ/mol (option A). Common mistake: Forgetting to apply a negative sign to the enthalpy change for a temperature rise (exothermic process).
Which of the following samples will react faster with dilute trioxonitrate (V) acid?
Distractor check: A student might focus only on temperature or only on surface area, choosing a condition that optimizes just one rate factor instead of both. Reasoning to the answer: Chemical reaction rates increase when the surface area of reactants is larger (powdered reactant forms react faster than large lumps) and when the temperature is higher (50°C is faster than 25°C). Therefore, combining powdered calcium carbonate with the higher temperature of 50°C maximizes both factors and produces the fastest reaction (option D). Common mistake: Assuming temperature has a greater impact than physical state changes like powdering.
In the reaction 2H⁺ + S²⁻ ⇌ H₂S, the concentration of H₂S in the equilibrium mixture can be increased by
Distractor check: A student might confuse exothermic equilibrium shifts with pressure changes, selecting a physical stress that does not affect the yield of H₂S in this manner. Reasoning to the answer: The forward reaction producing hydrogen sulfide releases energy, meaning the process is exothermic. According to Le Chatelier's principle, lowering the temperature removes thermal energy from the system, shifting the equilibrium position to the right to counteract the change and thus increasing the concentration of H₂S (option D). Common mistake: Applying pressure rules to a system where gas moles are equal on both sides.
To make coloured glass, small quantities of oxides of metals which form coloured silicates are added to the reaction mixture of Na₂CO₃ and SiO₂. Such a metal is
Distractor check: A student might pick a common alkali or alkaline earth metal like potassium or barium, confusing structural glass modifiers with transition metals that impart color. Reasoning to the answer: Transition metal oxides are typically added to glass formulations because they form colored metal silicates during fusion. Copper oxide is specifically used to impart vibrant blue or green hues into the glass matrix (option D). Common mistake: Assuming main-group metals are responsible for vibrant transition-metal-like color states in silicates.
Which of the following compounds gives a yellow residue when heated and reacts with aqueous NaOH to form a white precipitate soluble in excess NaOH?
Distractor check: A student might select aluminum sulfate, which forms a precipitate with sodium hydroxide, but miss the distinct yellow residue property upon thermal decomposition. Reasoning to the answer: Zinc carbonate decomposes upon heating to form zinc oxide, which appears yellow when hot. When this residue is treated with aqueous sodium hydroxide, it forms a white precipitate of zinc hydroxide. Adding excess sodium hydroxide causes this precipitate to redissolve, forming a colorless tetrahydroxozincate complex, [Zn(OH)₄]²⁻ (option C). Common mistake: Confusing aluminum hydroxide's solubility behavior with zinc carbonate's thermal properties.
Zinc has a more positive oxidation potential than
Distractor check: A student might select calcium or manganese, misinterpreting the electrochemical series positions relative to zinc. Reasoning to the answer: Zinc has a standard reduction potential of -0.76 V, which gives it a more positive oxidation potential than copper, whose reduction potential is +0.34 V. This means zinc loses electrons and undergoes oxidation much more readily than copper (option B). Common mistake: Mixing up reduction potentials with oxidation potentials when comparing electrochemical reactivity.
What volume of oxygen at STP is required to completely burn 12 g of carbon to form carbon dioxide? (C = 12, O = 16, 1 mole of gas at STP = 22.4 dm³)
Distractor check: A student might miscalculate the stoichiometric mole ratio or use an incorrect molar volume, picking 11.2 dm³ or 44.8 dm³. Reasoning to the answer: Write the combustion reaction: C + O₂ → CO₂. Determine the moles of carbon by dividing the given 12 g mass by the molar mass of carbon (12 g/mol), which equals 1 mole. The balanced equation shows that 1 mole of carbon reacts with 1 mole of oxygen gas. At standard temperature and pressure, 1 mole of any gas occupies 22.4 dm³, meaning exactly 22.4 dm³ of oxygen is required (option B). Common mistake: Using 2 moles instead of 1 mole due to confusing carbon atoms with diatomic oxygen molecules.
Which of the following compounds is an ester?
Distractor check: A student might select ethanoic acid (CH₃COOH) or ethanol (CH₃CH₂OH), confusing the functional group of an ester with carboxylic acids or alcohols. Reasoning to the answer: Esters are characterized by the general structural formula RCOOR', containing a carboxylate-like bridge between alkyl groups. The compound CH₃COOCH₃, known as methyl ethanoate, possesses this structure and is synthesized through the condensation reaction of a carboxylic acid and an alcohol (option B). Common mistake: Mistaking the carboxylic acid functional group (-COOH) for the ester functional group (-COO-).
The brown fumes given off when trioxonitrate (V) acid is heated consist of
Distractor check: A student might choose nitrogen dioxide alone, ignoring the coproduct gas released during thermal breakdown. Reasoning to the answer: When concentrated trioxonitrate (V) acid (nitric acid) is heated, it undergoes thermal decomposition according to the equation 4HNO₃ → 4NO₂ + 2H₂O + O₂. The distinct brown fumes observed are caused by nitrogen dioxide gas (NO₂), which is generated alongside oxygen gas (option C). Common mistake: Omitting oxygen from the decomposition products of concentrated nitric acid.
In the Haber process for the manufacture of ammonia, the catalyst commonly used is finely divided
Distractor check: A student might pick platinum or vanadium, confusing the Haber process catalyst with those used in the Contact process or catalytic converters. Reasoning to the answer: The industrial synthesis of ammonia via the Haber process (N₂ + 3H₂ ⇌ 2NH₃) relies heavily on a promoted, finely divided iron catalyst to increase the rate at which equilibrium is reached (option B). Common mistake: Confusing the iron catalyst of the Haber process with the vanadium(V) oxide used in sulfuric acid manufacture.
A metallic oxide that reacts with both HCl and NaOH to form a salt and water is classified as
Distractor check: A student might choose a basic oxide, forgetting that basic oxides only react with acids and not with strong alkalis. Reasoning to the answer: Metal oxides that exhibit dual reactivity—interacting with both acids and bases to yield salt and water—are classified as amphoteric oxides. Examples of this behavior include zinc oxide and aluminum oxide reacting with hydrochloric acid and sodium hydroxide (option D). Common mistake: Labeling an amphoteric oxide as a neutral oxide simply because it does not dissolve easily in water.
Which of the following metals will liberate hydrogen from dilute acids?
Distractor check: A student might select copper or mercury, forgetting their positions below hydrogen in the electrochemical reactivity series. Reasoning to the answer: Iron reacts readily with dilute acids such as hydrochloric acid (Fe + 2HCl → FeCl₂ + H₂) because it sits above hydrogen in the reactivity series, allowing it to displace hydrogen ions. In contrast, metals like copper, lead, and mercury are less reactive than hydrogen and cannot liberate it from dilute acids (option B). Common mistake: Assuming all heavy metals can displace hydrogen from acid solutions.
Coal fires should not be used in poorly ventilated rooms because of the accumulation of
Distractor check: A student might pick carbon dioxide, confusing the primary greenhouse gas of respiration with the toxic product of incomplete combustion. Reasoning to the answer: When coal burns in an environment with restricted oxygen supply, incomplete combustion occurs, leading to the formation of carbon monoxide (CO). This gas is highly poisonous and can accumulate silently to dangerous levels in unventilated spaces (option B). Common mistake: Attributing indoor combustion hazards entirely to carbon dioxide accumulation instead of carbon monoxide toxicity.
The major component of the slag from the production of iron is
Distractor check: A student might select calcium carbonate or calcium oxide, overlooking the chemical changes that occur in the blast furnace. Reasoning to the answer: During the extraction of iron in a blast furnace, limestone decomposes into calcium oxide, which then reacts with silica impurities. This combination neutralizes the impurities and forms molten calcium silicate (CaSiO₃), which floats as the major slag component above the molten iron (option A). Common mistake: Confusing the raw flux input (calcium carbonate) with the final slag product (calcium silicate).
Sodium hydroxide should be stored in properly closed containers because it
Distractor check: A student might think sodium hydroxide is easily oxidized, confusing its storage requirements with reducing agents. Reasoning to the answer: Sodium hydroxide is highly hygroscopic in nature, meaning it readily attracts and absorbs moisture and water vapor from the surrounding air. If left unsealed, it will dissolve in this absorbed moisture to form a concentrated liquid solution, necessitating airtight storage containers (option A). Common mistake: Believing sodium hydroxide reacts primarily with atmospheric oxygen rather than capturing water vapor.
The reaction CH₃COOH + NaOH → CH₃COONa + H₂O is an example of
Distractor check: A student might mistake the process for esterification, confusing a carboxylic acid-base reaction with a reaction between an acid and an alcohol. Reasoning to the answer: The reaction between ethanoic acid (an acid) and sodium hydroxide (a base) yields sodium ethanoate (a salt) and water. Because the hydrogen ions from the acid combine with hydroxide ions from the base to form water while neutralizing each other, this is classified as a neutralization reaction (option B). Common mistake: Confusing salt-forming acid-base neutralization with ester formation.
Alkanes have low volatility compared with
Distractor check: A student might choose alkenes, missing the effect of intermolecular forces on boiling points and vapor pressures. Reasoning to the answer: Alkanes exhibit lower volatility than alkanols because alkanols are capable of forming strong intermolecular hydrogen bonds. These hydrogen bonds require significantly more thermal energy to break, lower their vapor pressures, and raise their boiling points compared to nonpolar alkanes of similar size (option A). Common mistake: Forgetting that hydrogen bonding significantly reduces the volatility of hydroxyl-containing organic compounds.
The octane number of a fuel whose performance is the same as a mixture of 55% iso-octane and 45% n-heptane is
Distractor check: A student might subtract the percentage from 100 or confuse the heptane fraction with the octane rating value. Reasoning to the answer: The octane number of a fuel is defined directly as the percentage by volume of iso-octane present in a reference mixture with n-heptane that matches the fuel's anti-knock performance. Since the mixture contains 55% iso-octane, the corresponding octane rating of the fuel is 55 (option B). Common mistake: Using the n-heptane percentage instead of the iso-octane percentage for the octane number.
Which of the following gases supports combustion?
Distractor check: A student might select carbon dioxide, forgetting that it is commonly used in fire extinguishers to smother flames rather than support combustion. Reasoning to the answer: Nitrous oxide (N₂O) easily decomposes under thermal conditions into nitrogen and oxygen gases. The released oxygen gas actively sustains and supports the combustion of various burning materials (option C). Common mistake: Confusing nitrous oxide's combustion-supporting behavior with inert gases like nitrogen or smothering gases like carbon dioxide.
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