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Rufina [12.5K]
3 years ago
13

Which substances are released during cellular respiration?

Chemistry
1 answer:
scZoUnD [109]3 years ago
8 0
Answer is: carbon dioxide and water <span>are released during cellular respiration.
</span>Chemical reaction of cellular respiration:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.
There are 18 oxygen atoms (six in glucose and twelve in six molecules of oxygen) in left side of chemical reaction and also 18 oxygen atoms (twelve in six molecules of carbon dioxide and six in six molecules of water) at the right.
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Arsenate (AsO43-) is structurally and chemically similar to inorganic phosphate (PO43-), and many enzymes that act on phosphate
Lapatulllka [165]

Answer:

A) The effect on the net reaction catalyzed by glyceraldehyde is that

The 1-arsen0, 3-phosphoglycerate will decompose without enzymes hence no ATP will be formed in the reaction ( phosphoglycerate Kinase )

B) There will be  no conversion of ADP to ATP from the conversion of glucose to pyruvate hence No balanced overall equation can be derived

C ) Arsenate is very toxic to most organisms and it is used mostly regarded  as poisons during the formation of glycolysis, it forms  1-arsen0, 3-phosphoglycerate which  hinders the formation of ATP because it is unstable and will hydrolyze quickly, this will also lead to the reduction in oxygen in cells thereby leading to the death of cells

Explanation:

A) The effect on the net reaction catalyzed by glyceraldehyde is that

The 1-arsen0, 3-phosphoglycerate will decompose without enzymes hence no ATP will be formed in the reaction ( phosphoglycerate Kinase )

B) There will be  no conversion of ADP to ATP from the conversion of glucose to pyruvate hence No balanced overall equation can be derived

C ) Arsenate is very toxic to most organisms and it is mostly regarded as  poisons during the formation of glycolysis, it forms  1-arsen0, 3-phosphoglycerate which  hinders the formation of ATP because it is unstable and will hydrolyze quickly, this will also lead to the reduction in oxygen in cells thereby leading to the death of cells

4 0
3 years ago
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
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Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion=\frac{Volume}{Time}

Effusion rate of oxygen gas after time t = E=\frac{4.64 mL}{t}

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = E'=\frac{3.12 mL}{t}

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{E}{E'}=\sqrt{\frac{M'}{M}}

\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

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Answer:

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