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0.0274 g of N2 are added to the flask at the same temperature, what is the partial pressure of nitrogen in the flask

User Melena
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Final answer:

Using the Ideal Gas Law and the given mass of nitrogen (0.0274 g), the partial pressure of nitrogen (N2) in the flask is calculated to be 0.00442 atm at 25 °C and a volume of 5.73 L.

Step-by-step explanation:

To determine the partial pressure of nitrogen (N₂) in a flask, we can use the Ideal Gas Law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin. If 0.0274 g of N₂ are added to the flask at a constant temperature, we first need to convert the mass of nitrogen to moles by using the molar mass of nitrogen (28.02 g/mol). The converted mass is 0.0274 g / 28.02 g/mol = 9.78 × 10⁻´ moles of N₂.

The volume of the flask and the temperature have not changed from the initial conditions. Assuming the flask is 5.73 L and the temperature is 25 °C (which is 298 K when converted to Kelvin), we can plug these values into the Ideal Gas Law to find the partial pressure:

P = (nRT) / V

P = ((9.78 × 10⁻´ moles) x (0.0821 L·atm/K·mol) x (298 K)) / (5.73 L)

P = 0.00442 atm

The partial pressure of nitrogen in the flask is 0.00442 atm.

User El Dude
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