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:
<span>Carrier Gas, Flow Controller, Column, Detector, Recorder
</span>First we have a cylinder containing the
carrier gas. From there, the carrier gas goes to the flow controller, which determines
how much carrier gas we are entering into the column (it doesn’t let more gas
pass through). Then, the carrier gas enters the column, which is the most
important part of the device. The sample enters the column from another place:
the injector. Then, the sample and the carrier gas go together across the
column. The interactions between the sample and the column will determine how
fast each sample component goes through the column, and so: which component
gets out earlier. So, at the end, you will have isolated each substance. Then,
each one passes (alone) through the detector, which measures something about
the sample – this information will let you know which substance it is. Finally,
the recorder provides you with the information the detector has found.
Nowadays, the recorder is a computer. In the “stone age” they just used a rudimentary
printer.
1.) <span>2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g),
Na is oxidizing agent and H is reducing agent.
2.) </span><span>C(s)+O2(g)→CO2(g)
C is an oxidizing agent
3.) </span><span>2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)
Mn is reducing agent and S is oxidizing agent.</span>
Answer:
T = 3.75 K
Explanation:
As we know
PV = nRT
R =8.3144598 J. mol-1. K-1
P = 0.562 atm
V = 83.3 mL
moles in 0.392 g of I2 = 0.392/mass of I2 = 0.392 grams/253.8089 g/mol = 0.0015 moles
Substituting the given values, we get
0.562 atm * 83.3 *10^-3 L = 0.0015 moles * 8.3144598 J. mol-1. K-1 * T
T = 3.75 K
