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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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3 years ago
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How does the reaction, A + BC ⇌ AB + C, differ from the reaction, N2 + 3H2 ⇌ 2NH3 + energy, by which ammonia is synthesized?
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Answer:

N₂+3H₂ ⇄2NH₃ is a thermochemical reaction whereas A+BC⇄AB is not.

A+BC⇄AB is a reaction of pure a element with a compound while N₂+3H₂ ⇄2NH₃ is a reaction between two pure elements.

Explanation:

Let A+BC⇄AB be equation i and N₂+3H₂ ⇄2NH₃ be equation ii.

The two reactions differ in that ii is a thermo-chemical reaction whereas i is not. This is because energy is included in reaction ii but not included in reaction i.

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6 0
4 years ago
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8 0
3 years ago
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log 1/[Ox]^2= 0.082/0.0296

log 1/[Ox]^2= 2.77

1/[Ox]^2=Antilog (2.77)

[Ox]^2=1.698×10^-3

[Ox] = 0.0412 M

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