Here are the solutions to the questions from the image. Question 6(a): Determine the oxidation state of F in F₂O₅. Step 1: Assign the known oxidation state for oxygen. Oxygen typically has an oxidation state of $-2$ in oxides. Step 2: Set up the equation for the sum of oxidation states. Let the oxidation state of element F be $x$. In the compound F₂O₅, there are 2 atoms of F and 5 atoms of O. The sum of oxidation states in a neutral compound is zero. $$2(x) + 5(-2) = 0$$ Step 3: Solve for $x$. $$2x - 10 = 0$$ $$2x = 10$$ $$x = +5$$ The oxidation state of element F is $\boxed{+5}$. Question: Compare the electrical conductivity of sodium and aluminum. Sodium (atomic number 11) and Aluminum (atomic number 13) are both metals. • Sodium is in Group 1, having 1 valence electron per atom. • Aluminum is in Group 13, having 3 valence electrons per atom. Electrical conductivity in metals depends on the number of delocalized electrons available to carry charge. Aluminum has more delocalized electrons per atom than sodium. Therefore, aluminum is a better electrical conductor than sodium. Question: What mass of Zinc will be deposited from a solution of Zinc (II) Chloride if a current of 3A is passed through the Zinc (II) Chloride solution during electrolysis for 2.5 hours? (Zn=65, 1 Faraday=96500C) Step 1: Convert time to seconds. Given time $t = 2.5 \text{ hours}$. $$t = 2.5 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min}$$ $$t = 9000 \text{ s}$$ Step 2: Calculate the total charge (Q) passed. Given current $I = 3 \text{ A}$. $$Q = I \times t$$ $$Q = 3 \text{ A} \times 9000 \text{ s}$$ $$Q = 27000 \text{ C}$$ Step 3: Write the half-reaction for zinc deposition. Zinc (II) ions ($Zn^{2+}$) gain electrons to form solid zinc. $$Zn^{2+}(aq) + 2e^- \to Zn(s)$$ This reaction shows that $2 \text{ moles of electrons}$ are required to deposit $1 \text{ mole of Zn}$. Step 4: Calculate the moles of electrons transferred. Given Faraday's constant $F = 96500 \text{ C/mol e}^-$. $$\text{Moles of electrons} = \frac{\text{Charge (Q)}}{\text{Faraday's constant (F)}}$$ $$\text{Moles of electrons} = \frac{27000 \text{ C}}{96500 \text{ C/mol e}^-}$$ $$\text{Moles of electrons} \approx 0.27979 \text{ mol e}^-$$ Step 5: Calculate the moles of zinc deposited. From the half-reaction, $1 \text{ mole of Zn}$ requires $2 \text{ moles of electrons}$. $$\text{Moles of Zn} = \frac{\text{Moles of electrons}}{2}$$ $$\text{Moles of Zn} = \frac{0.27979 \text{ mol}}{2}$$ $$\text{Moles of Zn} \approx 0.139895 \text{ mol}$$ Step 6: Calculate the mass of zinc deposited. Given molar mass of Zn = $65 \text{ g/mol}$. $$\text{Mass of Zn} = \text{Moles of Zn} \times \text{Molar mass of Zn}$$ $$\text{Mass of Zn} = 0.139895 \text{ mol} \times 65 \text{ g/mol}$$ $$\text{Mass of Zn} \approx 9.093 \text{ g}$$ The mass of Zinc deposited is approximately $\boxed{\text{9.093 g}}$. That's 2 down. 3 left today — send the next one.