How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles of N2

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

Temperature (T) = 28°C

Number of mole (n) = 0.015 mole

Volume (V) =...?

Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

For Pressure:

101.325 KPa = 1 atm

Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

0.98 x V = 0.015 x 0.0821 x 301

Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

Therefore, the volume of N2 collected is 378mL

3 hours because it is a decent distance.

c, liquids evaporate, creating a gas