Chemistry entrance exam, 2016 E.C. (2024)

This is a heterogeneous equilibrium: the mixture contains solids and a gas. Pure solids and pure liquids are left out of the equilibrium constant expression because their concentrations do not change during the reaction.

Here CaCO3 and CaO are both solids, so they are dropped. The only species that stays is the gas, CO2. So \(K_c = [\mathrm{CO_2}]\).

Option A is the tempting mistake: it treats the solid CaO like a dissolved or gaseous species and includes it, which is wrong.

Textbook: Chemistry Grade 11, Unit 5 Chemical Equilibrium, section 5.2.4 Equilibrium Expression and Equilibrium Constant (page 228).

A covalent bond forms when two atoms share a pair of valence electrons. Both atoms are attracted to the shared pair, and each atom reaches a stable outer shell. This kind of bond is common between nonmetal atoms, as in H2O, CO2 and CH4.

Option D describes an ionic bond, not a covalent one: transfer of electrons from a metal to a nonmetal makes ions, and the bond is the attraction between those ions.

Textbook: Chemistry Grade 9, Unit 5 Chemical Bonding, section 5.3 Covalent Bonding (page 157).

Use the combined gas law:

\[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]

First change the temperature to kelvin: \(T_1 = 27 + 273 = 300\ \mathrm{K}\).

Solve for \(T_2\):

\[T_2 = T_1 \times \frac{P_2V_2}{P_1V_1} = 300 \times \frac{200 \times 250}{750 \times 100} = 300 \times \frac{50000}{75000} = 200\ \mathrm{K}\]

Change back to Celsius: \(200 - 273 = -73\,^{\circ}\mathrm{C}\). Remember to convert to kelvin before using any gas law; using 27 directly gives a wrong answer.

Textbook: Chemistry Grade 11, Unit 3 Physical States of Matter, section 3.3.2 The Gas Laws (page 153).

Oxides of active metals (like Na2O, K2O and CaO) react with water to give metal hydroxides, which are bases. For example: CaO + H2O gives Ca(OH)2. This is a standard way of preparing bases.

The other options do not produce a base. Option A is a neutralization, which uses up a base to make a salt and water. Option C is a method for preparing volatile acids, not bases. In option D the acid reacts with the carbonate to give a salt, water and CO2 gas.

Textbook: Chemistry Grade 10, Unit 3 Important Inorganic Compounds, section 3.4 Bases (page 142).

All electromagnetic radiation, from radio waves to gamma rays, travels at the same speed: \(c = 3 \times 10^8\ \mathrm{m/s}\) in a vacuum. The speed does not change when the wavelength or the frequency changes, so options A and B are wrong.

Option D is also wrong: since \(c = \lambda \nu\) and c is constant, wavelength and frequency are inversely proportional. When one goes up, the other goes down.

That leaves option C as the intended answer: the speed of the wave is treated as a constant that does not depend on the other properties listed.

Textbook: Chemistry Grade 11, Unit 1 Atomic Structure and Periodic Properties of the Elements, section 1.5.1 Electromagnetic Radiation (page 15).

A saturated monocarboxylic acid has one carboxyl group, COOH, attached to an alkyl group R. Written out, the carboxyl carbon carries a double-bonded O and an OH, so the general formula is R-COOH. Option D, with the OH on the carbonyl carbon, shows this.

The other options are different families: option A (R-CHO) is an aldehyde, option B (R-COOR) is an ester, and option C (R-CO-R) is a ketone.

Textbook: Chemistry Grade 11, Unit 6 Some Important Oxygen-Containing Organic Compounds, section 6.4.1 Structure and Nomenclature of Carboxylic Acids (page 292).