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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The number is 25. The square root of 25 is 5 and 5x2 is 10 and then 10+5 is 15
.answer:
so the first one is analogous
the second is homologous
and the third is analogous
explanation:
analogous. this means they share a similar function (flight) but do not have the same embryonic origin.
Answer:
Explanation:
A combustion involves the reaction of a fuel with oxygen (O₂). During the reaction of combustion of hydrogen (H₂), H₂ reacts with O₂ to form water (H₂O). The <em>balanced chemical equation</em> is the following:
2 H₂(g) + O₂(g) → 2 H₂O(g)
According to the chemical equation, 2 moles of H₂O are obtained from the reaction of 2 moles of H₂ with 1 mol of O₂. All reactants and products are in the gaseous phase.