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Marta_Voda [28]
3 years ago
12

Compare the appearance and pH of the unfiltered and filtered polluted water. What conclusion can be made based on these observat

ions?
Chemistry
2 answers:
Goryan [66]3 years ago
6 0

Answer:

The unfiltered polluted water is not clear and is yellowish in color. The filtered polluted water is clear with a little bit of yellow color. The pH readings are beyond the range in which organisms can grow. Based on these observations, I can conclude that even though water may look clean, it could actually not be clean or it can still be harmful to living things.

harkovskaia [24]3 years ago
6 0

Answer:

The unfiltered polluted water is not clear and is yellowish in color. The filtered polluted water is clear with a little bit of yellow color. The pH readings are beyond the range in which organisms can grow. Based on these observations, I can conclude that even though water may look clean, it could actually not be clean or it can still be harmful to living things.

Explanation:

I just took it on E2020

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Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

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Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

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P_2 = vapor pressure at temperature T_2

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1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

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Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

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