Explain the trend in atomic radius across a period, state one use of a noble gas, and find the mass of 0.25 moles of CaCO3.

Atomic mass of $\text{Ca} = 40 \text{ g/mol}$
Atomic mass of $\text{C} = 12 \text{ g/mol}$
Atomic mass of $\text{O} = 16 \text{ g/mol}$

$$\text{Molar mass of } \text{CaCO}_3 = 40 + 12 + (3 \times 16)$$ $$\text{Molar mass of } \text{CaCO}_3 = 40 + 12 + 48$$ $$\text{Molar mass of } \text{CaCO}_3 = 100 \text{ g/mol}$$

Mass = Number of moles $\times$ Molar mass

$$\text{Mass} = 0.25 \text{ mol} \times 100 \text{ g/mol}$$ $$\text{Mass} = 25 \text{ g}$$

The mass of 0.25 moles of $\text{CaCO}_3$ is $\boxed{25 \text{ g}}$.

24. a) Explain why carbon forms a large number of compounds (4 marks)
Carbon forms a large number of compounds primarily due to its ability to undergo catenation, which is the formation of strong covalent bonds with other carbon atoms to create long chains, branched structures, and rings. Additionally, carbon is tetravalent, meaning it can form four stable covalent bonds with other carbon atoms and various other elements (like hydrogen, oxygen, nitrogen, and halogens), leading to a vast diversity of organic molecules.

24. b) Give two differences between diamond and graphite (4 marks)
Here are two differences between diamond and graphite:

• Structure: Diamond has a tetrahedral arrangement where each carbon atom is covalently bonded to four other carbon atoms, forming a rigid 3D lattice. Graphite has a layered hexagonal structure where each carbon atom is covalently bonded to three other carbon atoms within a layer, and these layers are held together by weak intermolecular forces.
• Electrical Conductivity: Diamond is a poor conductor of electricity because all its valence electrons are involved in covalent bonding and are localized. Graphite is a good conductor of electricity because each carbon atom has one delocalized valence electron per layer, which can move freely within the layers.

25. A gas occupies $400 \text{ cm}^3$ at $1 \text{ atm}$ pressure. Calculate its volume at $4 \text{ atm}$ (4 marks)

This problem involves a change in volume and pressure at constant temperature, which is described by Boyle's Law: $P_1V_1 = P_2V_2$.

Given:
Initial volume, $V_1 = 400 \text{ cm}^3$
Initial pressure, $P_1 = 1 \text{ atm}$
Final pressure, $P_2 = 4 \text{ atm}$
Final volume, $V_2 = ?$

$$V_2 = \frac{1 \text{ atm} \times 400 \text{ cm}^3}{4 \text{ atm}}$$ $$V_2 = 100 \text{ cm}^3$$

The volume of the gas at $4 \text{ atm}$ is $\boxed{100 \text{ cm}^3}$.

26. a) Explain the difference between simple molecular and giant covalent structures (4 marks)
Simple molecular structures consist of discrete molecules held together by strong covalent bonds within the molecule, but by weak intermolecular forces between molecules. This results in low melting and boiling points, and they are typically gases, liquids, or soft solids at room temperature. Giant covalent structures, on the other hand, consist of a vast network of atoms held together by strong covalent bonds throughout the entire structure. This extensive bonding leads to very high melting and boiling points, and they are typically hard solids.

26. b) State two properties of covalent compounds (2 marks)
Two properties of covalent compounds are:

• They generally have low melting and boiling points because the intermolecular forces between molecules are weak and require little energy to overcome.
• They are typically poor conductors of electricity because they do not have free-moving ions or delocalized electrons to carry an electric current.