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madreJ [45]
3 years ago
8

Dr. Ray has developed a new way to remove oil from wetlands in the event of an oil spill. To test his new process, Dr. Ray build

s a scale model of a wetland in his lab and pours a small amount of oil into it. He then observes the rate at which his process is able to remove the oil from the model. Why did Dr. Ray use a model? A. Because dumping oil in a real wetland would be dangerous. B. Because real wetlands are too complicated for his process to work. C. Because his model makes the process more complicated. D. Because real wetlands are too large to be affected by oil.
Chemistry
2 answers:
DiKsa [7]3 years ago
6 0

A scientific model is a pictorial, graphical or theoretical representation of any complex process or idea that is being studied. Any process that cannot be studied directly or that would cause an environmental issue, is studied by building up a model. In the given model, Dr. Ray built a scale model of a wetland to poured some oil to study the rate at which the oil is removed from the model. It would be difficult to study the phenomenon in wetland directly as pouring oil in the wetland would be dangerous to the plant and animal species present in the wetland.

Therefore, the correct option would be A. Because dumping oil in a real wetland would be dangerous.

Vsevolod [243]3 years ago
4 0

Answer:

A. Because dumping oil in a real wetland would be dangerous.

Explanation:

I did the Quiz and got it correct.

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200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486
Dmitriy789 [7]

Explanation:

Mass of the organic compound = 200g

Mass of carbon = 83.884g

Mass of hydrogen = 10.486g

Mass of oxygen = 18.640g

The mass of nitrogen = mass of organic compound - (mass of carbon + mass of hydrogen + mass of oxygen)

Mass of nitrogen = 200 - (83.884 + 10.486 + 18.64) = 200 - 113.01‬

Mass of nitrogen = 86.99g

The empirical formula of a compound is its simplest formula.

It is derived as shown below;

                        C                   H                O                  N

Mass          83.884             10.486        18.64            86.99

molar

mass                12                    1                  16                14

Moles       83.884/12         10.486/1       18.64/16        86.99/14

                   

                    6.99                 10.49              1.17                6.21

Divide

by

lowest      6.99/1.17         10.49/1.17           1.17/1.17            6.21/1.17

                       6                    9                         1                       5

Empirical formula  C₆H₉ON₅

learn more:

Empirical formula brainly.com/question/2790794

#learnwithBrainly

                 

                               

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