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MA_775_DIABLO [31]
3 years ago
7

Which model demonstrates a correct pathway carbon takes in the carbon cycle?

Chemistry
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

C) atmosphere → plants → animals → soil

Explanation:

The third choice provides the correct path through which carbon is cycled in nature.

Carbon passes from the atmosphere to plants then to animals and finally to the soil.

  • Plants uses carbon dioxide from the atmosphere to manufacture their food.
  • The food is made up of giant carbon chains which also provides nourishment for animals.
  • When animals digest plant matter, they obtain energy for their process.
  • The waste is passed into the soil .
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What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of e
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Explanation:

According to Clausius-Claperyon equation,

       ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

The given data is as follows.

         T_{1} = 63.5^{o}C = (63.5 + 273) K

                         = 336.6 K

        T_{2} = 78^{o}C = (78 + 273) K

                         = 351 K

         P_{1} = 1 atm,             P_{2} = ?

Putting the given values into the above equation as follows.    

        ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

       ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]

                      \Delta H = \frac{0.559}{1.466 \times 10^{-4}} J/mol

                                  = 0.38131 \times 10^{4} J/mol

                                  = 3813.1 J/mol

Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.

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