Answer :
Final answer:
The partial pressure of the oxygen gas collected is approximately 163.4 mmHg.
Explanation:
The partial pressure of the oxygen gas collected can be calculated using the ideal gas law equation P = (nRT)/V, where P is the pressure, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume.
First, convert the temperature from Celsius to Kelvin by adding 273.15: 23.0°C + 273.15 = 296.15 K.
Next, convert the volume from milliliters to liters by dividing by 1000: 193 ml / 1000 = 0.193 L.
Now we can calculate the partial pressure: P = (nRT) / V. The number of moles of oxygen can be found using the ideal gas law rearranged as n = (PV) / (RT).
By plugging in the values, we get n = (762.0 mmHg * 0.193 L) / (0.0821 L*atm/mol*K * 296.15 K) ≈ 0.01327 moles of oxygen.
Finally, plug the value of n into the partial pressure equation to calculate the partial pressure: P = (0.01327 mol * 0.0821 L*atm/mol*K * 296.15 K) / 0.193 L ≈ 0.215 atm. To convert this to mmHg, multiply by 760 mmHg/atm: 0.215 atm * 760 mmHg/atm ≈ 163.4 mmHg.
Therefore, the partial pressure of the oxygen gas collected is approximately 163.4 mmHg.
