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:
20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
Explanation:
Heat is being consumed during vaporization and heat is being released during condensation.
To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.
Molar mass of water = 18.02 g/mol
Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.
So, to vaporize 9.00 g of water,
of heat or 20.3 kJ of heat is being consumed
As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
The volume of the balloon would decrease
Answer:Cant see the picture cause its too big
Explanation:
Answer:
Look at the image pls. The question is there
: