Here are the solutions to the questions: 8. A student represented an atom of element J using the diagram below. b) Write the valency of element J and the oxidation number of its ion. From the diagram, element J has an electron configuration of 2, 8, 6. To achieve a stable octet, it needs to gain 2 electrons. Valency: 2* Oxidation number of its ion: -2* 9. When the oxide of metal D is heated in the presence of metal S, the oxide is reduced. The oxide of metal S is not reduced when heated together with metal E and with metal D. The oxide of metal E is reduced by metal S and not by metal D. Arrange the three metals horizontally, in their order of decreasing reactivity. Let's deduce the reactivity order: "The oxide of metal D is heated in the presence of metal S, the oxide is reduced." This means metal S is more reactive than metal D. ($\text{S} > \text{D}$) "The oxide of metal S is not reduced when heated together with metal E and with metal D." This means metal S is more reactive than both metal E and metal D. ($\text{S} > \text{E}$ and $\text{S} > \text{D}$) "The oxide of metal E is reduced by metal S and not by metal D." This confirms $\text{S} > \text{E}$. It also means metal D is not reactive enough to reduce the oxide of E, so metal E is more reactive than metal D. ($\text{E} > \text{D}$) Combining these, the order of decreasing reactivity is: S > E > D 10. Calculate the number of nitrogen atoms that are found in 1.4g of nitrogen gas (N = 14, L = 6.023 x 10^23) Step 1: Determine the molar mass of nitrogen gas ($\text{N}_2$). Nitrogen gas is diatomic, so its formula is $\text{N}_2$. Molar mass of $\text{N}_2 = 2 \times \text{atomic mass of N}$ Molar mass of $\text{N}_2 = 2 \times 14 \text{ g/mol} = 28 \text{ g/mol}$ Step 2: Calculate the number of moles of nitrogen gas. $$\text{Moles of } \text{N}_2 = \frac{\text{mass}}{\text{molar mass}}$$ $$\text{Moles of } \text{N}_2 = \frac{1.4 \text{ g}}{28 \text{ g/mol}} = 0.05 \text{ mol}$$ Step 3: Calculate the number of nitrogen gas molecules. $$\text{Number of } \text{N}_2 \text{ molecules} = \text{moles} \times \text{Avogadro's constant (L)}$$ $$\text{Number of } \text{N}_2 \text{ molecules} = 0.05 \text{ mol} \times 6.023 \times 10^{23} \text{ molecules/mol}$$ $$\text{Number of } \text{N}_2 \text{ molecules} = 3.0115 \times 10^{22} \text{ molecules}$$ Step 4: Calculate the number of nitrogen atoms. Each $\text{N}_2$ molecule contains 2 nitrogen atoms. $$\text{Number of N atoms} = 2 \times \text{Number of } \text{N}_2 \text{ molecules}$$ $$\text{Number of N atoms} = 2 \times 3.0115 \times 10^{22} \text{ atoms}$$ $$\text{Number of N atoms} = \boxed{6.023 \times 10^{22} \text{ atoms}}$$ 11. Give the name of the processes that occur when the following substances are left in open watch glasses overnight: a) Concentrated sulphuric (VI) acid Hygroscopy (or absorption of water vapor) b) Sodium carbohydrate decahydrate Assuming "Sodium carbohydrate decahydrate" is a typo and refers to Sodium carbonate decahydrate ($\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$): Efflorescence