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anastassius [24]
3 years ago
13

An algal bloom resulting from excess detergent and fertilizer use in the surrounding community is most likely a result of which

of the following?
l. Carbonates
ll. Nitrates
lll. Phosphates

a. l only
b. l and ll only
c. ll and lll only
d. l, ll, and lll
Chemistry
1 answer:
Lilit [14]3 years ago
8 0

Answer:

c. ll and lll only

Explanation:

Algae are plant-like protists capable of photosynthesis in their cells (chloroplast). Like every other living organism that grows and reproduce, algae also requires nutrients for these processes to effectively occur.

Algae makes use of inorganic nutrients including sulphates, phosphates, nitrates etc which are found in fertilizers, detergents etc. In the presence of these compounds, algae tends to grow excessively to form a BLOOM. Hence, according to the question, nitrates and phosphates are the two inorganic nutrients algae makes use of.

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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

Learn more:

  • brainly.com/question/25475410
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