JAMB Chemistry 2006 Past Questions & Answers

To determine if a reaction will proceed, we calculate the Gibbs Free Energy change (ΔG) using the formula ΔG = ΔH - TΔS. For the given answer to be correct, we must assume there was a typo and ΔH is positive (+72 kJ/mol or +72000 J/mol). Converting temperature to Kelvin (25°C = 298K), we get ΔG = 72000 - (298 × -106) = 72000 + 31588 = +103,588 J/mol. Since ΔG is positive, the forward reaction is not spontaneous, and the reaction will instead proceed in the reverse direction.

Common mistake: The most common mistake is to assume the reaction is exothermic (negative ΔH), which would lead to a negative ΔG and the incorrect conclusion that the reaction is spontaneous in the forward direction.

To solve this titration problem, first calculate the moles of HCl used: Moles = Concentration × Volume = 0.05 mol/dm³ × $(10/1000)$ dm³ = 0.0005 mol. From the balanced equation, the mole ratio of Na2CO3 to HCl is 1:2, meaning you need half the moles of Na2CO3 to react. Therefore, moles of Na2CO3 = 0.0005 / 2 = 0.00025 mol. NOTE: The provided options are incorrect based on the question's values; option D would be correct if the HCl concentration was 2.0 mol/dm³, which would yield 0.010 mol.

Common mistake: A common mistake is to apply the mole ratio incorrectly, for instance by multiplying the moles of HCl by 2 instead of dividing, which leads to an incorrect answer.

An oxidizing agent is a substance that accepts electrons and gets reduced in a chemical reaction. The calcium ion, Ca²⁺, can accept two electrons to form a neutral calcium atom (Ca²⁺ + 2e⁻ → Ca). Although calcium is a very reactive metal, making its ion a weak oxidizing agent, it still fits the definition because it is capable of accepting electrons and being reduced.

Common mistake: Students often mistakenly choose F⁻ because they confuse an oxidizing agent (which gets reduced by gaining electrons) with a reducing agent (which gets oxidized by losing electrons).

The Leclanche cell, commonly known as a dry cell battery, is a primary electrochemical cell. In this cell, the zinc casing acts as the negative electrode (anode) where oxidation occurs. The positive electrode (cathode) is a central carbon rod where reduction takes place.

Common mistake: Students may confuse this with a Daniel cell, but a Daniel cell uses a copper cathode, not a carbon rod.

Trioxonitrate(V) acid, also known as nitric acid, is a powerful oxidizing agent. When it reacts with most metals, instead of just displacing hydrogen, it forms a very thin, non-reactive layer of metal oxide on the surface of the metal. This process, called passivation, protects the metal from further attack by the acid, effectively stopping the reaction and preventing the continuous production of hydrogen gas.

Common mistake: Students often choose 'Is strongly oxidizing' because it is a correct statement about nitric acid, but 'Renders the metal passive' is the more specific answer that directly describes the effect that stops the reaction.

The heat of combustion of hydrogen is the heat released when one mole of hydrogen gas burns completely in oxygen to form one mole of water (H₂ + ½O₂ → H₂O). The standard heat of formation of water is the heat change when one mole of water is formed from its elements in their standard states, which is the exact same chemical reaction. Therefore, the heat of formation of water is equal to the heat of combustion of hydrogen, which is -285.8 kJ.

Common mistake: Students often mistakenly choose -571.6kJ by balancing the equation to form two moles of water (2H₂ + O₂ → 2H₂O) and doubling the value, forgetting that standard enthalpy of formation is defined per mole of the product.

In the electrolysis of an aqueous solution of CuSO4, the ions present are Cu²⁺, SO₄²⁻ from the salt, and H⁺, OH⁻ from water. At the anode (positive electrode), the anions (OH⁻ and SO₄²⁻) migrate there to be oxidized. According to the electrochemical series, hydroxide ions (OH⁻) are preferentially discharged over sulphate ions (SO₄²⁻). The oxidation of OH⁻ ions produces water and oxygen gas, and the balanced half-equation for this reaction is 4OH⁻ - 4e⁻ → 2H₂O + O₂.

Common mistake: Students often confuse the reactions at the anode and cathode, and might mistakenly choose option A, which is a reduction reaction that would occur at the cathode in an acidic solution.

To find the oxidation number, we use some simple rules. In most compounds, hydrogen (H) has an oxidation number of +1. Since the overall charge of the H₂S molecule is zero, the sum of the oxidation numbers must be zero. If we let the oxidation number of sulphur (S) be 'x', then we have (2 * +1) + x = 0, which gives x = -2.

Common mistake: Students often mistakenly choose S₈, forgetting that the oxidation number of an element in its free or uncombined state is always zero.

The reaction starts with potassium permanganate (KMnO4), which is a deep purple solution. During this redox reaction, the purple permanganate ion (MnO4-), where manganese is in a +7 oxidation state, is reduced to the manganese(II) ion (Mn2+). The resulting product, manganese(II) sulfate (MnSO4), is nearly colorless in solution, and the other products are also colorless. Because the intensely colored reactant is completely used up to form colorless products, the final mixture is a colorless solution.

Common mistake: Students may mistakenly choose 'Purple solution' because they remember that potassium permanganate is purple, but they forget that it is a reactant that gets consumed and converted to a colorless product.

Electrical conductivity in a solution depends on the concentration of mobile ions. Hydrochloric acid (HCl) is a strong acid that completely ionizes in water, producing a large number of ions. Ethanoic acid is a weak acid that only partially ionizes, producing far fewer ions. Therefore, the 2.0 mol dm-3 solution of the strong acid, HCl, will have the highest concentration of ions and conduct the most electricity.

Common mistake: Students might mistakenly choose option A, focusing only on the high concentration (2.0 moldm-3) while ignoring that ethanoic acid is a weak electrolyte that produces far fewer ions than a strong acid like hydrochloric acid.

The relationship between moles, mass, and molar mass is given by the formula: Number of moles = Mass / Molar Mass. To find the molar mass, you rearrange the formula to Molar Mass = Mass / Number of moles. By substituting the given values, you get Molar Mass = 25.0g / 0.4 mole, which equals 62.5 g/mol.

Common mistake: A common mistake is to misapply the formula by multiplying the mass and moles $(25.0 x 0.4 = 10.0)$ or dividing the numbers in the wrong order $(0.4 / 25.0 = 0.016)$.

To separate this mixture, you use the different physical properties of the substances. First, you heat the mixture to make the ammonium chloride sublime (turn from solid to gas), which separates it. Next, you add water to dissolve the sodium chloride (common salt), while the sand remains undissolved. You then use filtration to separate the insoluble sand from the salt solution. Finally, you evaporate the water from the salt solution to recover the solid sodium chloride.

Common mistake: Students often mistakenly choose option D, thinking they should dissolve the soluble salts first, but this mixes the ammonium chloride and sodium chloride, making them much harder to separate from each other.

At standard temperature and pressure (s.t.p.), one mole of any gas always occupies a fixed volume of 22.4 dm³. To find the volume of 2.5 moles of oxygen, you simply multiply the number of moles by this molar volume. Therefore, the calculation is 2.5 moles × 22.4 dm³/mole, which equals 56.0 dm³.

Common mistake: A common mistake is to divide the molar volume by the number of moles $(22.4 / 2.5)$ instead of correctly multiplying them together.

Nuclear fission is the process where the nucleus of a heavy atom, like uranium, is split when it absorbs a neutron. This splitting results in two smaller nuclei that are close in size but not perfectly identical, hence they have 'nearly equal masses'. This process is highly exothermic, releasing a very large amount of energy, and also emits radiation, including more neutrons which can cause a chain reaction.

Common mistake: Students often mistakenly choose option C, assuming the split is perfectly symmetrical into two equal halves, but in reality, the fission products are almost always of slightly different masses.

First, calculate the number of moles of ethyne (C2H2) burned, which is mass divided by molar mass (78g / 26g/mol) giving 3 moles. According to the balanced equation, the mole ratio of C2H2 to CO2 is 2:4, so 3 moles of ethyne will produce $(4/2)$ * 3 = 6 moles of CO2. Finally, convert the moles of CO2 to mass by multiplying by its molar mass (6 moles * 44g/mol), which gives 264g.

Common mistake: Students may mistakenly apply the mole ratio directly to the given mass $(e.g., (4/2) * 78g = 156g)$, forgetting to first convert the mass of ethyne to moles.

The number of protons in an atom's nucleus determines its identity. An element with 11 protons is Sodium (Na), which is a metal. In a neutral atom, the number of protons equals the number of electrons. Since this particle has 11 protons and 10 electrons, it has one more proton than electrons, giving it a net charge of +1, which makes it a metallic ion (Na⁺).

Common mistake: Students may incorrectly choose 'Neutral metal' because they identify the element as a metal from the 11 protons but forget to check the number of electrons to see if the particle is an ion.

To find the element, we first count the total number of electrons from the configuration 1s²2s²2p³. Adding the superscripts gives us 2 + 2 + 3 = 7 electrons. For a neutral atom, the number of electrons is equal to the atomic number. The element with an atomic number of 7 is Nitrogen.

Common mistake: Students might mistakenly choose Oxygen if they miscount the electrons to be 8 instead of 7, as Oxygen has an atomic number of 8.

The fundamental definition of a chemical change is a reaction where the original substances are converted into entirely new substances with different chemical properties. This happens because the chemical bonds in the reactants are broken and new bonds are formed to create the products. In contrast, a physical change, like melting ice, only alters the state or form of a substance but does not create a new substance.

Common mistake: Students often mistakenly choose option A because many chemical reactions require heat, but they forget that physical changes like boiling water or melting a solid also require an input of heat.

As you move from left to right across a period in the periodic table, the number of protons in the nucleus increases, which is the 'increase in nuclear charge'. This stronger positive charge pulls the electrons closer, causing a 'decrease in atomic size'. Because the outermost electrons are held more tightly by the nucleus, it becomes more difficult for the atom to lose them, which means its electropositivity (the tendency to lose electrons) decreases.

Common mistake: Students often mistakenly choose option C because they confuse the trend, thinking that a stronger nucleus makes it easier to lose electrons, when in fact the stronger attraction makes it much harder.

Niels Bohr proposed the atomic model where electrons move in fixed, circular orbits, also called energy shells, around a central nucleus. This model is often compared to planets orbiting the sun, which is why it's called the 'planetary model'. While Rutherford discovered the nucleus, it was Bohr who introduced the idea of these specific energy levels for the electrons.

Common mistake: Students often mistakenly choose Rutherford because he discovered the nucleus, but it was Bohr who specifically proposed the idea of electrons in fixed planetary-like orbits.

This question is solved using Charles's Law, which states that V₁/T₁ = V₂/T₂ for a gas at constant pressure. First, convert the temperatures to Kelvin: T₁ = 50°C + 273 = 323 K and T₂ = 80°C + 273 = 353 K. Using the formula, the new volume V₂ = (V₁ × T₂) / T₁ = $(30 × 353)$ / 323 ≈ 32.79 cm³. The increase in volume is 32.79 - 30 = 2.79 cm³. The fractional increase is the increase in volume divided by the original volume, which is 2.79 / 30 ≈ 0.093.

Common mistake: Students often mistakenly calculate the ratio of the final volume to the initial volume (V₂/V₁), which gives 1.09 (Option C), instead of the fractional increase which is (V₂ - V₁)/V₁.

Isomers are compounds that have the same molecular formula but different structural arrangements. For an alkane to have an isomer, it needs at least four carbon atoms to allow for different branching patterns. The molecule C4H10 can exist as a straight chain (n-butane) or as a branched chain (2-methylpropane or isobutane), making them isomers of each other. The other options, C2H6 and C2H4, have too few carbon atoms to form a branched structure.

Common mistake: Students might mistakenly choose C6H6 (benzene), confusing its resonance structures with actual structural isomers.

An aliphatic hydrocarbon is a compound containing only carbon and hydrogen that does not have a benzene ring. An unsaturated hydrocarbon is one that contains at least one carbon-carbon double or triple bond. Propene is an alkene which has a double bond, making it unsaturated, and it is a straight-chain compound, making it aliphatic. Pentane is saturated (only single bonds), Benzene is aromatic (not aliphatic), and Butanol is an alcohol (not a hydrocarbon).

Common mistake: Students often mistakenly choose Benzene because it is unsaturated, but they forget that it is an aromatic hydrocarbon, not an aliphatic one.

First, calculate the molar mass of the hydrocarbon using the formula: Molar Mass = 2 × Vapour Density, which gives 2 × 20 = 40 g/mol. Next, find the number of carbon atoms by dividing the given mass of carbon by its molar mass: 36g / 12g/mol = 3 carbon atoms. The mass of hydrogen in the compound is the total molar mass minus the mass of carbon, so 40g - 36g = 4g, which means there are 4 hydrogen atoms. Therefore, the molecular formula is C3H4, and the only structure that matches this formula is CH3C≡CH (propyne).

Common mistake: Students often correctly find there are 3 carbon atoms but may miscalculate the molar mass or the number of hydrogen atoms, leading them to choose option B (propene, C3H6).

To determine hybridization, we look at the bonds around each carbon atom. The first two carbons (in CH3 and CH2) have four single bonds, so they are sp3 hybridized. The last two carbons are connected by a triple bond (C≡C), and any carbon atom involved in a triple bond is sp hybridized. Therefore, the molecule contains both sp3 and sp hybridized carbon atoms.

Common mistake: Students often mistakenly choose B (Sp3 and sp2), confusing the triple bond which indicates sp hybridization with a double bond that would indicate sp2 hybridization.