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:

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Answer:
polyatomic, ionic
Explanation:
unit that contains two or more atoms covalently bonded together but that has an overall charge is called a(n) polyatomic ion. Many ionic compounds contain such units.
Answer:0.300M
Explanation:1) Data:
a) Initial solution
M = 1.50M
V = 50.0 ml = 0.050 l
b) Solvent added = 200 ml = 0.200 l
2) Formula:
Molarity: M = moles of solute / volume of solution is liters
3) Solution:
a) initial solution:
Clearing moles from the molarity formula: moles = M × V
moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol
b) final solution:
i) Volumen of solution = 0.050 l + 0.200l = 0.250l
ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer