2016 JAMB CHEMISTRY PAST QUESTIONS AND ANSWERS

The relative atomic mass can be calculated using the ratio given. With a 1:3 ratio: $\frac{(1 \times 20) + (3 \times 22)}{1 + 3} = \frac{20 + 66}{4} = \frac{86}{4} = 21.5$. This weighted average accounts for both the mass numbers and their relative abundances.

Using Graham’s Law of Diffusion: $\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}} \times \frac{V_1}{V_2}$. For $O_2$ (M=32) and $CH_4$ (M=16), with $V_1=200cm^3$ and $V_2=80cm^3$: $\frac{50}{t} = \sqrt{\frac{16}{32}} \times \frac{80}{200}$. Solving gives 14 seconds.

Valence refers to the combining capacity of an atom, determined by its electron configuration. It directly indicates how many bonds an atom can form. Oxidation number represents charge state, not bonding capacity.

Magnesium has atomic number 12 with electron configuration $1s^2 2s^2 2p^6 3s^2$. The diagram shows this arrangement with two valence electrons in the outermost shell, characteristic of Group 2 elements.

Carbon monoxide is highly toxic as it binds irreversibly to hemoglobin, preventing oxygen transport in blood. It’s odorless and colorless, making it particularly dangerous compared to other pollutants listed which have warning properties like odor.

Soft water, lacking calcium and magnesium ions, is more corrosive and can attack lead pipes, releasing toxic lead into the water supply. This is a significant health concern in older plumbing systems.

Aerobic bacteria can break down hydrocarbon compounds in crude oil through biodegradation, converting them to less harmful substances. This is a natural and effective method of bioremediation.

The $-NH_3^+$ group reacts with sodium hydrogen carbonate to produce gas bubbles of $CO_2$. This is a characteristic test for ammonium compounds.

In potassium dichromate ($K_2Cr_2O_7$), calculate: $2(+1) + 2x + 7(-2) = 0$. Solving for x gives chromium’s oxidation state as +6.

$Na_2O_2$ contains peroxide ion ($O_2^{2-}$) which is oxidized to $O_2$ gas. The oxidation state of oxygen changes from -1 in peroxide to 0 in $O_2$.

In reaction energy diagrams, heat of reaction represents the overall energy change (∆H) of the reaction, measured as the difference between products and reactants energy levels.

The nucleus contains only protons (positively charged) and neutrons (neutral). Electrons orbit the nucleus in electron shells or energy levels.

Electrons flow from zinc to copper ions, not towards the zinc rod. Zinc is more reactive than copper, so it displaces copper from solution, resulting in zinc dissolution and copper deposition.

During electrolysis of NaOH solution, $OH^-$ ions are discharged at anode producing $O_2$, while $H^+$ ions are discharged at cathode producing $H_2$. The remaining $OH^-$ ions increase alkalinity at cathode.

Calcium carbide reacts with water to form acetylene (ethyne), not any alkyne. The reaction is specific: $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$.

Iron(III) oxide remains insoluble while aluminum oxide dissolves in sodium hydroxide solution. Filtration separates the insoluble iron oxide from the dissolved aluminum compound.

The Hall-Héroult process involves electrolysis of alumina dissolved in molten cryolite ($Na_3AlF_6$). Cryolite lowers the melting point and increases conductivity.

$ZnCO_3$ decomposes to give ZnO (yellow when hot), and Zn(OH)2 forms with NaOH, dissolving in excess to form $[Zn(OH)_4]^{2-}$ complex.

Copper has the highest reduction potential among these metals, making it the least easily oxidized. It is below hydrogen in the electrochemical series.

$AlCl_3$ shows the most covalent character due to the high charge density of $Al^{3+}$ and its small size, leading to polarization of the chloride ions.

Hydrogen gas is not a component of crude oil and cannot be obtained through fractional distillation. It must be produced through other methods like steam reforming.

Tetraethyllead (TEL) was historically used as an antiknock agent to prevent pre-ignition in engines, though now phased out due to environmental concerns.

32g of oxygen molecules ($O_2$) contains one mole ($6.022 \times 10^{23}$ molecules), same as 24g Mg. This follows from their respective atomic/molecular masses.

Iron is protected by connecting it to a more reactive metal (sacrificial anode) with lower electropositive potential, which corrodes preferentially.

Sulfuric acid dehydrates the hydrated iron oxide (rust), removing water molecules from the compound structure.

Sulfur forms cross-links between rubber polymer chains during vulcanization, improving strength and elasticity of the rubber.

The unique properties of transition metals arise from their partially filled d-orbitals, allowing variable oxidation states and colored compounds.

With 11 protons but only 10 electrons, the particle has a net positive charge, characteristic of a metallic ion.

Catalysts provide an alternative reaction pathway with lower activation energy, increasing reaction rate without being consumed.

For displacement to occur, M must be more electropositive than Zinc, allowing it to donate electrons more readily and reduce zinc ions.

Using Faraday’s Law: $Q = \frac{nF \times m}{M}$ where n=2 for $Cu^{2+}$. $Q = \frac{2 \times 96500 \times 10}{64} = 15196.6C$

According to IUPAC nomenclature, $CH_3COOH$ is ethanoic acid (2-carbon acid) and $CH_2=CH_2$ is ethene (2-carbon alkene).

Water molecules form stronger hydrogen bonds than methanol due to higher electronegativity of oxygen and symmetrical structure, requiring more energy to break.

When bombarded with neutrons, the mass number increases by 1 while atomic number remains unchanged initially. The equation $_Z^YX + _0^1n \rightarrow _Z^{Y+1}X$ represents this.

Statements 1, 3, and 5 are factors that increase reaction rate: lowering activation energy (catalysis), increasing temperature (kinetic energy), and increasing surface area (powdering).

Element 12 (Mg) has electronic configuration $1s^2 2s^2 2p^6 3s^2$, making it a Group 2 element that forms electrovalent bonds with valency 2.

Atomic radius decreases across a period due to increasing nuclear charge pulling electrons closer to nucleus while electron shells remain constant.

Using combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. Solving for $V_2$ with STP conditions $(273K, 1 atm)$: $V_2 = \frac{0.971 \times 146 \times 273}{291 \times 1} = 133cm^3$

In air, oxygen volume = $150 \times 0.2 = 30cm^3$. For $2CO + O_2 \rightarrow 2CO_2$, $50cm^3$ CO requires $25cm^3$ $O_2$. Therefore, oxygen $(30cm^3)$ is in excess.

$CH_2O$ represents the simplest whole-number ratio of atoms in ethanoic acid $(C_2H_4O_2)$, making it the empirical formula.