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Mamont248 [21]
2 years ago
14

Using the image below, describe in at least 2 paragraphs, what is happening with the arrows. Name and explain the processes and

the changing forms of Carbon. To conclude, predict what the numbers might represent.
Photo Credit:NASA/GLOBE Program

Below is a suggestion for how your paragraphs should be laid out, and which terms to include in each paragraph:
Paragraph one: Places where carbon is being added to the atmosphere (processes to include here could be things like burning, respiration, decay, weathering, and run off).
Paragraph two: Places where carbon is being pulled out of the atmosphere (steps here might be photosynthesis, sinking, diffusion, sedimentation, depositing, and storage).
Finally, come up with a hypothesis (an educated guess) of what the numbers in the arrows might mean?
Chemistry
1 answer:
Eddi Din [679]2 years ago
6 0

Answer: I'm sorry, but we can't see the image from NASA

Explanation:

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1. What is the relation between distance and speed?
Inga [223]

Answer:

option no b is right answer

6 0
2 years ago
An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average
Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = \frac{Displacement }{Change in time} = Δx / Δt = \frac{x2 - x1}{t2 - t1}

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{2100km }{1000km/h}

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{1300km }{800km/h}

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = \frac{x2 - x1}{t2 - t1}

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = \frac{1300 - 2100}{1.625 - 2.1}

Average velocity = 1684.21 km/h

7 0
2 years ago
Pre-lab questions 1. a concentration gradient affects the direction that solutes diffuse. describe how molecules move with respe
VLD [36.1K]

Just like how heat moves from a region of higher temperature to a region of lower temperature, molecules also tend to move from a region of higher concentration to a region of lower concentration. This is called natural diffusion and is naturally happening to reach stability.

7 0
3 years ago
Please help me #6!!!!!!!!
Vesnalui [34]
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
3 0
3 years ago
If the ph of hc3h5o2 is 4.2 and the ka 1.34x10^-5, what is the equilibrium concentration
n200080 [17]
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span> 
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
              = 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴ 
where Ca is the acid concentration 
5 0
3 years ago
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