Answer:
HgSO₄
Explanation:
% => g => moles => ratio => reduce => empirical ratio
%Hg = 67.6% => 67.6g/201g/mol = 0.34mol
%S = 10.8% => 10.8g/32g/mol = 0.34mol
%O = 21.6% => 21.6g/16g/mol = 1.35mol
Hg:S:O => 0.34:0.34:1.35
Reduce to whole number ratio by dividing by the smaller mole value...
Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4
∴ Empirical Formula is HgSO₄
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:
Answer:
Explanation:
The first step is:
Second step is:
Multiplying second step by 2, and adding both the steps, we get that:
Cancelling common species, we get that: