Study the image. Warm and cold air fronts with points labeled 1 through 4. 1: cold air entering from the left and moving down. 2
2 answers:
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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and and
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
Learn more:
Answer:
<h2>Density = 1.67 g/mL</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 50 g
volume = 30 mL
Substitute the values into the above formula and solve for the density
That's
Wr have the final answer as
<h3>Density = 1.67 g/mL</h3>Hope this helps you