Classify the following into physical and chemical change: Dissolution of salt in water, Burning of paper, Magnetization of iron, Rusting of iron.

here are the solutions to the chemistry questions:

1. (a) i. Classify the following into physical and chemical change:

• I. Dissolution of salt in water: Physical change
• II. Burning of paper: Chemical change
• III. Magnetization of iron: Physical change
• IV. Rusting of iron: Chemical change

1. (a) ii. Which state is mercury at room temperature?

• Mercury is in the liquid state at room temperature.

1. (b) i. Calculate the relative molecular mass of:

• I. Calcium hydroxide ($\text{Ca(OH)}_2$)
  • Atomic masses: $\text{Ca} = 40$, $\text{O} = 16$, $\text{H} = 1$
  $$\text{RMM of Ca(OH)}_2 = 40 + (16 + 1) \times 2 = 40 + 17 \times 2 = 40 + 34 = \boxed{\mathbf{74}}$$
• II. Lead (II) trioxonitrate (V) ($\text{Pb(NO}_3)_2$)
  • Atomic masses: $\text{Pb} = 207$, $\text{N} = 14$, $\text{O} = 16$
  $$\text{RMM of Pb(NO}_3)_2 = 207 + (14 + 16 \times 3) \times 2 = 207 + (14 + 48) \times 2 = 207 + 62 \times 2 = 207 + 124 = \boxed{\mathbf{331}}$$

1. (b) ii. Define isotope.

• Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons.

1. (b) iii. Consider the following pairs of elements: 6A & 8B, 11C & 16D and 3E & 19F.

• I. Which pair will form ionic compound?
  • Element A (atomic number 6) is Carbon (non-metal).
  • Element B (atomic number 8) is Oxygen (non-metal).
  • Element C (atomic number 11) is Sodium (metal).
  • Element D (atomic number 16) is Sulfur (non-metal).
  • Element E (atomic number 3) is Lithium (metal).
  • Element F (atomic number 19) is Potassium (metal).
  • Ionic compounds form between a metal and a non-metal. The pair that fits this description is 11C & 16D (Sodium and Sulfur).
• II. Give the name of the compound formed in (b)iii I.
  • Sodium (C) forms $\text{Na}^+$ ions, and Sulfur (D) forms $\text{S}^{2-}$ ions. The compound formed is $\text{Na}_2\text{S}$.
  • The name of the compound is Sodium sulfide.

1. (c) i. State Charles' law.

• Charles' law states that for a fixed mass of gas at constant pressure, the volume is directly proportional to its absolute temperature.

1. (c) ii. Calculate the volume of a gas at s.t.p. if it occupies 15 cm$^3$ at 7.50 x 10$^4$ Nm$^{-2}$ and 50.0$^\circ$C.

• Given:
  • Initial volume ($V_1$) = $15 \text{ cm}^3$
  • Initial pressure ($P_1$) = $7.50 \times 10^4 \text{ Nm}^{-2}$
  • Initial temperature ($T_1$) = $50.0^\circ\text{C} = 50.0 + 273 = 323 \text{ K}$
• At S.T.P. (Standard Temperature and Pressure):
  • Standard pressure ($P_2$) = $1.0 \times 10^5 \text{ Nm}^{-2}$ (as given by 1 pascal = $1.0 \times 10^5 \text{ Nm}^{-2}$)
  • Standard temperature ($T_2$) = $0^\circ\text{C} = 0 + 273 = 273 \text{ K}$
• Using the Combined Gas Law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
• Rearrange to solve for $V_2$: $$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$$ $$V_2 = \frac{(7.50 \times 10^4 \text{ Nm}^{-2}) \times (15 \text{ cm}^3) \times (273 \text{ K})}{(1.0 \times 10^5 \text{ Nm}^{-2}) \times (323 \text{ K})}$$ $$V_2 = \frac{30712500 \text{ cm}^3 \cdot \text{K} \cdot \text{Nm}^{-2}}{32300000 \text{ K} \cdot \text{Nm}^{-2}}$$ $$V_2 \approx 9.5085 \text{ cm}^3$$ $$V_2 = \boxed{\mathbf{9.51 \text{ cm}^3}}$$

1. (d) i. State two phenomena that support kinetic theory of gases.

• 1. Diffusion
• 2. Brownian motion

1. (d) ii. Define ionisation energy.

• Ionisation energy is the minimum energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state to form one mole of gaseous ions.

1. (d) iii. Copy and complete Table 2.1 below:

Periodic properties	Down a group	Across the period
Atomic radius	Increases	Decreases
Electron affinity	Decreases	Decreases
Electronegativity	Decreases	Increases