5 examples of gases found in the normal home environment include; oxygen (air), nitrogen (most abundant element found in the air), carbon (air), a slight trace of argon and finally, hydrogen. these are- Nitrogen, oxygen, carbondioxide, carbonmonoxide and SO2.
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:
I pick but I'm not sure about it though 1and3
Answer:
Explanation:
Of the numerous sorts of RNA, the three most well-known and most commonly examined are delivery person RNA (mRNA), exchange RNA (tRNA), and ribosomal RNA (rRNA), which are show in all living beings. These and other sorts of RNAs essentially carry out biochemical responses, comparative to proteins.
Density is an intrinsic property, so it is independent of the amount of substance present: one gold coin would have the same density as a solid gold boulder.
So if the density of gold is 19.3 g/cm³, the density of a bar of gold and the pieces into which the bar is cut would all be 19.3 g/cm³.