If \(5.15 \mathrm{~g}\) of \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a \(75.0-\mathrm{mL}\) tube filled with 101.3 \(\mathrm{kPa}\) of \(\mathrm{N}_{2}\) gas at \(32{ }^{\circ} \mathrm{C},\) and the tube is heated to $320^{\circ} \mathrm{C}\(, the \)\mathrm{Ag}_{2} \mathrm{O}$ decomposes to form oxygen and silver. What is the total pressure inside the tube assuming the volume of the tube remains constant?

The balanced chemical equation for the decomposition of silver oxide is: \[2 \mathrm{Ag}_2\mathrm{O} \rightarrow 4\mathrm{Ag} + \mathrm{O}_{2}\]

To calculate the number of moles, we need to use the given mass and molar mass for silver oxide and the Ideal Gas Law for nitrogen gas. The molar mass of silver oxide is: \( 2 * 107.87 \) g/mol (for silver) + 16.00 g/mol (for oxygen) = \( 231.74 \) g/mol Now let's calculate the moles of silver oxide: \[\text{moles of Ag}_2\text{O} = \frac{5.15 \;\text{g}}{231.74 \;\text{g/mol}} = 0.0222 \;\text{mol}\] For nitrogen gas, let's convert the temperature from Celsius to Kelvin: \(32{ }^{\circ}\text{C} = 273.15 + 32 = 305.15 \;\text{K}\) Now we can apply the Ideal Gas Law: PV = nRT Solving for n (moles) for nitrogen gas: \[ n = \frac{PV}{RT} = \frac{101.3 \; \text{kPa} * 75.0 \; \text{mL}}{8.314 \;\text{J} \cdot \text{mol}^{-1}\text{K}^{-1} * 305.15 \;\text{K}} \times \frac{1000 \;\text{mL}}{1 \; \text{L}} \times \frac{1 \; \text{Pa}}{10^3 \;\text{kPa}}\] \[ n \approx = 0.00296 \; \text{mol}\]

From the balanced chemical equation, we see that 1 mole of oxygen gas is produced per 2 moles of silver oxide decomposed. So, \[\text{moles of O}_{2}\text{ produced} = 0.5 * \text{moles of Ag}_2\text{O} = 0.5 * 0.0222 \;\text{mol} = 0.0111 \; \text{mol}\]

Now, let's apply the Ideal Gas Law for both gases at the final temperature (320 °C = 593.15 K), considering the partial pressures: \[P_{\text{N}_2} V = n_{\text{N}_2} R T_{\text{final}}\] \[P_{\text{O}_2} V = n_{\text{O}_2} R T_{\text{final}}\] Solving for the final pressures of each gas: \[P_{\text{N}_2} = \frac{0.00296 \;\text{mol} * 8.314 \;\text{J/mol K} * 593.15 \;\text{K}}{75.0 \;\text{mL} * \frac{1000 \;\text{mL}}{1 \;\text{L}}} = \frac{1.463\times10^2 \; \text{Pa}}{10^3} = 146.3 \;\text{kPa}\] \[P_{\text{O}_2} = \frac{0.0111 \;\text{mol} * 8.314 \;\text{J/mol K} * 593.15 \;\text{K}}{75.0 \;\text{mL} * \frac{1000 \;\text{mL}}{1 \;\text{L}}} = \frac{5.733\times10^2 \; \text{Pa}}{10^3} = 573.3 \;\text{kPa}\]

The total pressure is the sum of the partial pressures of the two gases (nitrogen and oxygen): \[P_{\text{total}} = P_{\text{N}_2} + P_{\text{O}_2} = 146.3 \;\text{kPa} + 573.3 \;\text{kPa} = 719.6 \;\text{kPa}\] So, the total pressure inside the tube after heating is 719.6 kPa.