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V125BC [204]
2 years ago
10

2. How did small changes in the amount of carbon put into the atmosphere by cellular respiration, and removed

Biology
1 answer:
bearhunter [10]2 years ago
4 0

Impact the amount of carbon in the atmosphere, the biosphere, and the

hydrosphere

Explanation:

The processes of photosynthesis and cellular respiration play a major role in the carbon cycle.

Both these processes are interconnected where the products of a process become reactants of the other process.

Carbon dioxide and water released through cellular respiration become the reactants for photosynthesis which releases oxygen needed for cellular respiration.  

The oxygen released is used by animals and human beings for respiration and releases carbon dioxide which is once again used by plants during photosynthesis. This gaseous exchange happening during these two processes helps to maintain atmospheric carbon dioxide levels stable.

In the biosphere, carbon is stored in the organic form in plants and trees. Plants absorb the external atmospheric carbon dioxide and convert it into organic form and store it as food by the process of photosynthesis.

In the hydrosphere, carbon is available as dissolved in water in the water bodies like ocean, rivers, etc. this carbon is absorbed by marine or aquatic plants to undergo photosynthesis and produce their own food. Again the organic carbon is formed and stored in the form of food which is used by other aquatic organisms.

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What is the volume of 0.25 mol of ammonia gas at 1.00 atm and 273 K? (R= 0.0821 Latm/molK)What is the volume of 0.25 mol of ammo
ICE Princess25 [194]

Answer:

5.6L

Explanation:

Given parameters:

number of moles  = 0.25mol

pressure on gas  = 1atm

temperature  = 273K

Gas constant R = 0.0821Latm/molK

Unknown:

Volume of gas  = ?

Solution:

Using the ideal gas equation, we can solve this problem. The equation is a combination of the three gas laws: Boyle's law, Charles's law and Avogadro's law.

It is mathematically expressed as;

             PV = nRT

where P is the pressure

           V is the volume

          R is the gas constant

          T is the temperature

          n is the number of moles

All the parameters are in the appropriate units and we simply solve for the volume of the gas;

                1 x V = 0.25 x 0.0821 x 273

                      V  = 5.6L

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Ingesting food or water contaminated with viruses, parasites, and protozoa can cause a condition characterized by painful abdomi
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Why elephant herbivore defend your answer
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What is the name for the type of graph shown below? (Image above)
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Cumulative frequency is the right answer 
8 0
3 years ago
This very low concentration of the desired product would be unfavorable for glycolysis. In fact the reaction is coupled to ATP h
abruzzese [7]

Answer:

ΔG'° for the coupled reaction = -16.7 kJ/mol

<em>Note: The question is missing some parts. The complete question is as follows:</em>

<em>The first reaction in glycolysis is the phosphorylation of glucose: </em>

<em> Pi+glucose⟶glucose−6−phosphate+H2O </em>

<em>This is a thermodynamically unfavorable process, with ΔG∘′= +13.8kJ/mol. In a liver cell at 37 ∘C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each.</em>

<em>This very low concentration of the desired product would be unfavorable for glycolysis. In fact the reaction is coupled to ATP hydrolysis to give the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H+</em>

<em>What is the ΔG'° for the coupled reaction?</em>

Explanation:

The coupling of ATP hydrolysis which is a thermodynamically favourable reaction to the phosphorylation of glucose makes it favourable. Since the two reactions constitute a sequential reaction, their standard free energy changes are additive.

For ATP hydrolysis: ATP + H₂O ---> ADP + Pi ; ΔG'°  = -30.5KJ/mol

For phosphorylation of glucose:  Pi + glucose⟶glucose−6−phosphate + H2O  ; ΔG'° = +13.8 kJ/Mol

For the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H⁺ ; ΔG'° = 13.8 + (-30.5) kJ/mol = -16.7 kJ/mol

Therefore, ΔG'° for the coupled reaction = -16.7 kJ/mol

6 0
3 years ago
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