At \(0^{\circ} \mathrm{C}\) a \(1.0\)-\(\mathrm{L}\) flask contains \(5.0 \times 10^{-2}\) mole of \(\mathrm{N}_{2}, 1.5 \times\) \(10^{2} \mathrm{mg} \mathrm{O}_{2},\) and \(5.0 \times 10^{21}\) molecules of \(\mathrm{NH}_{3} .\) What is the partial pressure of each gas, and what is the total pressurelin the flask?

We are given the moles for N2 gas, while for the other two gases, we need to convert the given information into moles. For N2: \(5.0 \times 10^{-2}\) moles For O2: We are given the mass of O2 in milligrams. First, let's convert it into grams and then calculate moles. Given mass of O2= 1.5 x 10^2 mg, and 1 g=1000 mg so, Mass of O2 = \(\frac{1.5 \times 10^{2}}{1000}\) g = 0.15 g Moles of O2 = \(\frac{\text{mass of O2}}{\text{molar mass of O2}}\) Moles of O2 = \(\frac{0.15}{32}\) mol For NH3: We are given the number of molecules of NH3. First, let's convert it into moles. Number of molecules of NH3 = \(5.0 \times 10^{21}\) Moles of NH3 = \(\frac{\text{number of molecules}}{\text{Avogadro's number}}\) Moles of NH3 = \(\frac{5.0 \times 10^{21}}{6.022\times10^{23}}\) mol

Temperature T = 0°C = 273 K Volume V = 1.0 L Ideal gas constant R = 0.0821 L atm / (K mol) For N2: Using the Ideal Gas Law, \[P_{N2} = \frac{n_{N2} \times R \times T}{V}\] \[P_{N2} = \frac{5.0\times10^{-2} \times 0.0821 \times 273}{1.0}\] For O2: Using the Ideal Gas Law, \[P_{O2} = \frac{n_{O2} \times R \times T}{V}\] \[P_{O2} = \frac{0.15/32 \times 0.0821 \times 273}{1.0}\] For NH3: Using the Ideal Gas Law, \[P_{NH3} = \frac{n_{NH3} \times R \times T}{V}\] \[P_{NH3} = \frac{5.0\times10^{21}/6.022\times10^{23} \times 0.0821 \times 273}{1.0}\]

Total pressure inside the flask (P_total) is the sum of the partial pressures of each gas: \[P_{total} = P_{N2} + P_{O2} + P_{NH3}\] Calculate the total pressure using the calculated partial pressures for each gas.