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Ratling [72]
3 years ago
9

Can someone help me with questions 4-5 21 points

Chemistry
1 answer:
sp2606 [1]3 years ago
8 0

4. describe three ways carbon dioxide was removed from the Earth's atmosphere.

Answer: Forests: Photosynthisis helps clear carbon dioxide naturally, Soils naturally store carbon, but agricultural soils are running a big deficit due to intensive use. Because agricultural land is so expansive, Bio-energy with Carbon Capture and Storage (BECCS) is another way to use photosynthesis to combat climate change. However, it is far more complicated than planting trees or managing soils — and it doesn’t always work for the climate.

5. Explain why there is now 21% Oxygen in the Earth's atomosphere compaired to little or no Oxygen in the Earth's atmosphere 4.5 billion years ago.

Answer: cientists believe that the Earth was formed about 4.5 billion years ago. Its early atmosphere was probably formed from the gases given out by volcanoes. It is believed that there was intense volcanic activity for the first billion years of the Earth's existence.The early atmosphere was probably mostly carbon dioxide, with little or no oxygen. There were smaller proportions of water vapour, ammonia and methane. As the Earth cooled down, most of the water vapour condensed and formed the oceans.

Sorry its soooo long TwT

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A solution is prepared by dissolving 4.66 g of KCl in enough distilled water to give 250 mL of solution. KCl is a strong electro
Leokris [45]

Answer:

Depression in freezing point = 2 X 1.853 X 0.25 = 0.9625

Thus this will be the difference between the freezing point of pure water and the solution.

Explanation:

On adding any non volatile solute to a solvent its boiling point increases and its freezing point decreases [these are two of the four colligative properties].

The depression in freezing point is related to molality of solution as:

ΔTf=iK_{f}Xmolality

where

ΔTf= depression in freezing point

Kf= cryoscopic constant of water = 1.853 K. kg/mol.

i = Van't Hoff factor = 2 ( for KCl)

molality = \frac{molesofsolute}{massofsolvent(Kg)}

moles of solute = mass / molarmass = 4.66 / 74.55 =0.0625

mass of solvent = mass of solution (almost)

considering the density of solution to be 1g/mL

mass of solvent = 250 grams = 0.250 Kg

molality = \frac{0.0625}{0.25}= 0.25

Putting values

depression in freezing point = 2 X 1.853 X 0.25 = 0.9625

Thus this will be the difference between the freezing point of pure water and the solution.

7 0
4 years ago
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