If 5.15 gof \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a 75.0 - -mL tube filled with 760 torr 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?

Given mass of \({Ag_2O} = 5.15 g\) Molar mass of \({Ag_2O} = 2(107.87) + 16 = 231.74g/mol\) To find moles of \({Ag_2O}\), use the following formula: \(n(Ag_2O) = \frac{mass}{molar mass}\) \(n(Ag_2O) = \frac{5.15 g}{231.74 g/mol} = 0.0222 mol\)

The balanced chemical equation of the decomposition of \({Ag_2O}\) is \({2Ag_2O} \rightarrow {4Ag} + {O_2}\) According to this equation, each mole of \({O_2}\) is produced from 2 moles of \({Ag_2O}\). To find moles of \({O_2}\) produced, use the following formula: \(n(O_2) = \frac{n(Ag_2O)} {2}\) \(n(O_2) = \frac{0.0222 mol}{2} = 0.0111 mol\)

Since we know the initial pressure and temperature, we can use the ideal gas law to find the initial number of moles of \({N_2}\) in the tube. The ideal gas law is given by: \(PV = nRT\) Where: \(P = pressure\) \(V = volume\) \(n = moles\) \(R = ideal gas constant = 0.0821 \frac{L \cdot atm}{mol \cdot K}\) \(T = temperature\) Solving for initial moles of \({N_2}\): \(n(N_2) = \frac{PV} {RT}\) Before we substitute the values, we should convert torr to atm and Celsius to Kelvin: \(32^\circ C = 32 + 273.15 = 305.15K\) \(760\,torr = \frac{760}{760}\,atm = 1\,atm\) Now substitute the values: \(n(N_2) = \frac{1\,atm \cdot 75.0\,mL} {0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 305.15K}\) Since mL and L are not the same unit, we should convert 75.0 mL to L: \(75.0 mL = 0.075 L\) \(n(N_2) = \frac{1\,atm \cdot 0.075LD}{0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 305.15K} = 0.00306\,mol\) Now, let's find the partial pressure of \({O_2}\) using the ideal gas law at the final temperature of \(320^\circ C\): \(320^\circ C = 320 + 273.15 = 593.15K\) \(P(O_2) = \frac{n(O_2)RT}{V}\) \(P(O_2) = \frac{0.0111\,mol \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 593.15K} {0.075 L} = 6.182\,atm\)

The final pressure of \({N_2}\) can be found with the combined gas law: \(P_{1}V_{1}{/}T_{1} = P_{2}V_{2}{/}T_{2}\) Since the volume remains constant: \(P_{1}{/}T_{1} = P_{2}{/}T_{2}\) Solve for final pressure of \({N_2}\) : \(P_{2}(N_2) = P_{1}(N_2) \frac{T_{2}} {T_{1}}\) \(P_{2}(N_2) = 1\,atm \cdot \frac{593.15K}{305.15K} = 1.945\,atm\)

The total pressure inside the tube is the sum of the final pressures of \({O_2}\) and \({N_2}\). \(P_{total} = P(O_2) + P_{2}(N_2)\) \(P_{total} = 6.182\,atm + 1.945\,atm = 8.127\,atm\) So, the total pressure inside the tube after the decomposition of \({Ag_2O}\) is 8.127 atm.