This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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You're on a roll — Question 9: Step 1: Identify the given values and conditions. Mass of oxygen m = 4.0 g. Molar mass of oxygen M = 32 g/mol. The conditions are S.T.P (Standard Temperature and Pressure). At S.T.P., T = 0^ C = 273.15 K and P = 1 atm = 101325 Pa. Also, at S.T.P., 1 mole of any ideal gas occupies 22.4 L. Step 2: Calculate the number of moles (n) of oxygen. n = (m)/(M) = 4.0 g32 g/mol n = 0.125 mol Step 3: Calculate the volume occupied by the oxygen at S.T.P. Since 1 mole occupies 22.4 L at S.T.P.: V = n × 22.4 L/mol V = 0.125 mol × 22.4 L/mol V = 2.8 L The volume occupied by 4.0 g of oxygen at S.T.P is 2.8 L. --- Question 10: Step 1: Identify the given values and convert them to SI units. Mass of nitrogen m = 18.5 g. Volume V = 5 L = 5 × 10^-3 m^3. Temperature T = 27^ C = 27 + 273 = 300 K. Molar mass of nitrogen M = 28 g/mol = 0.028 kg/mol. Universal Gas Constant R = 8.314 J/(mol·K). Step 2: Calculate the number of moles (n) of nitrogen. n = (m)/(M) = 18.5 g28 g/mol n ≈ 0.6607 mol Step 3: Apply the Ideal Gas Law to find the pressure P. The Ideal Gas Law is PV = nRT. Rearrange to solve for P: P = (nRT)/(V) P = (0.6607 mol) × (8.314 J/(mol·K)) × (300 K)5 × 10^-3 m^3 P = (1647.9)/(5 × 10^-3) Pa P = 329580 Pa P ≈ 3.30 × 10^5 Pa The pressure of the nitrogen gas is 3.30 × 10^5 Pa. --- Question 11: Step 1: Identify the given values and convert temperature to Kelvin. Temperature T = 27^ C = 27 + 273 = 300 K. Boltzmann constant k = 1.38 × 10^-23 J/K. Step 2: Calculate the average translational kinetic energy (KE_avg) of a nitrogen molecule. The formula for the average translational kinetic energy per molecule is: KE_avg = (3)/(2) kT KE_avg = (3)/(2) × (1.38 × 10^-23 J/K) × (300 K) KE_avg = 1.5 × 1.38 × 10^-23 × 300 J KE_avg = 621 × 10^-23 J KE_avg = 6.21 × 10^-21 J The average translational kinetic energy of a nitrogen molecule at 27^ C is 6.21 × 10^-21 J. --- Question 12: Step 1: Identify the given values and convert them to SI units. Temperature T = 40^ C = 40 + 273 = 313 K. Molar mass of helium M = 4.0 Kg/Kmole. This is equivalent to 4.0 g/mol or 0.004 kg/mol. Universal Gas Constant R = 8.314 J/(mol·K). Step 2: Calculate the root mean square speed (v_rms). The formula for the root mean square speed is: v_rms = sqrt((3RT)/(M)) v_rms = sqrt(3 × (8.314 J/(mol)·K)) × (313 K)0.004 kg/mol v_rms = sqrt((7807.74)/(0.004)) m/s v_rms = sqrt(1951935) m/s v_rms ≈ 1397.12 m/s The root mean square speed of a helium atom at 40^ C is 1397.12 m/s. Send me the next one 📸