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Citrus2011 [14]
3 years ago
11

Describe the organism that scientists believe was responsible for changing the composition of the early atmosphere. What is it c

alled and how did it change the atmosphere? Why was this change to the atmosphere so important?
Chemistry
2 answers:
Yakvenalex [24]3 years ago
6 0

The early atmosphere was most likely principally CO2, with very little or no chemical element. The proportion of chemical element went up attributable to chemical change. The chemical change was conducted from little organisms.

lara31 [8.8K]3 years ago
4 0
The early atmosphere was probably mostly carbon dioxide, with little or no oxygen. <span>The proportion of oxygen went up because of </span>photosynthesis. The photosynthesis was conducted from <span>tiny organisms.
</span><span>cyanobacteria, or blue-green algae. </span><span>
They </span>used sunshine, water and carbon dioxide to produce carbohydrates and, yes, oxygen. This change to the atmosphere was very important because the <span>breathable air we enjoy today was created.</span>
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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
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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:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

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 y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

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:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

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  • brainly.com/question/25475410
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