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:
An isotope is defined as an element that has the same number of protons as the common element but different number of neutrons. In this case, a beryllium atom has a molar weight of 10 amu. Thus, there are 4 protons and 6 neutrons. The nuclear symbol of Be-10 is 4 Be10
Answer: 150 kPa
Explanation:
Given that,
Original volume of gas V1 = 30L
Original pressure of gas P1 = 105 kPa
New pressure of gas P2 = ?
New volume of gas V2 = 21L
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
105 kPa x 30L = P2 x 21L
3150 kPa L = P2 x 21L
P2 = 3150 kPa L / 21 L
P2 = 150 kPa
Thus, 150 kPa of pressure is required to compress the gas
Answer:
The anwer is not D the anwer is A
Explanation: