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katrin [286]
1 year ago
8

In what way do the oceans and the forests perform a similar function in the carbon cycle, under normal conditions?.

Chemistry
1 answer:
ikadub [295]1 year ago
8 0

Both take carbon dioxide out of the atmosphere.

The ocean absorbs much of the carbon dioxide released when fossil fuels are burned. This extra carbon dioxide lowers the ocean's pH through a process called ocean acidification. Both are important elements for all life on earth. Relationship between the water cycle and the carbon cycle. The sun's energy drives both the carbon cycle and the water cycle.

Recall that sunlight water and carbon dioxide combine in green plants via photosynthesis to produce carbohydrates. The ocean also regulates the Earth's climate. They determine the weather by telling temperatures and determining rainfall, droughts, and floods. It is also the world's largest carbon sink, with an estimated 83% of the global carbon cycle circulating in seawater.

Learn more about The carbon cycle here:-brainly.com/question/1194643

#SPJ4

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Answer: This image is the answer to this question.

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At equilibrium the partial pressures of N2O4 and NO2 are 0.35 atm and 4.3 atm. What is the Kp
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Kp = 52.83

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Calculate the average atomic mass of element X using the table below. Then use the periodic table to identify the element. Separ
Ksenya-84 [330]

Answer:

126.8, Iodine

Explanation:

  • mass ×abundance/100
  • (126.9045×80.45/100)+(126.0015×17.23/100)+(128.2230×2.23/100)
  • 102.1+21.7+3=126.8

<em>IODINE</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomic</em><em> </em><em>mass</em><em> </em><em>of</em><em> </em>126.8 or 126.9

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Read 2 more answers
At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO
erastovalidia [21]

Answer : The concentration of NO at equilibrium is 0.9332 M

Solution :  Given,

Concentration of N_2 and O_2 at equilibrium = 0.200 M

Concentration of N_2 and O_2 at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}

K_c=6.25

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

                         N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially             0.200   0.200         0

.800

At equilibrium  (0.200-x) (0.200-x)  (0.800+2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}

By solving the term x, we get

x=2.6\text{ and }-0.0666

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of NO at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of NO at equilibrium is 0.9332 M

5 0
3 years ago
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