If this is a true or false question then the answer is true.
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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27
Explicación:
Dado que:
Número de protones en el átomo X = 29
Carga en el átomo X = +2
Si no hay cargo neto;
número de protones = número de electrones
Sin embargo, dado que el átomo X tiene una carga de +2 (dando 2 electrones).
Por lo tanto,
Número de electrones = número de protones - número de carga en el átomo)
Número de electrones = (29 - 2) = 27
V1/T1 = V2/T2
Substitute the value use ratio and proportion. Use calculator.
V1 = (V2 x T1) / T2
1 is initial, 2 is final