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question 9 how can carbon dioxide be used to indicate the presence of a…

Question

question 9

how can carbon dioxide be used to indicate the presence of a carbohydrate in food?

it cannot be.

when carbohydrates age they release carbon dioxide.

carbohydrates turn blue in the presence of carbon dioxide.

yeast can be used to break down carbohydrates in food producing carbon dioxide.

Explanation:

Analyze the question and options

The question asks how carbon dioxide (\(\text{CO}_2\)) can be used to indicate the presence of a carbohydrate in food. We need to evaluate the biological processes associated with carbohydrates and gas production.

Evaluate biological processes

Using our knowledge of Carbohydrates, we know they serve as primary energy sources for living organisms. Yeast, a single-celled fungus, metabolizes simple sugars (carbohydrates) through anaerobic respiration (fermentation).

Connect fermentation to carbon dioxide production

During alcoholic fermentation, yeast breaks down simple sugars like glucose:

$$ \text{C}_6\text{H}_{12}\text{O}_6 ightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2 $$

This metabolic process produces carbon dioxide gas as a measurable byproduct.

Assess the given choices

  • "It cannot be." is incorrect because yeast fermentation provides a viable pathway.
  • "When carbohydrates age they release carbon dioxide." is incorrect because aging alone does not release \(\text{CO}_2\) without microbial action.
  • "Carbohydrates turn blue in the presence of carbon dioxide." is incorrect; iodine turns blue-black in the presence of starch, not \(\text{CO}_2\).
  • "Yeast can be used to break down carbohydrates in food producing carbon dioxide." is correct because yeast fermentation directly converts carbohydrates into ethanol and \(\text{CO}_2\) gas.

Answer:

  • It cannot be.
  • When carbohydrates age they release carbon dioxide.
  • Carbohydrates turn blue in the presence of carbon dioxide.
  • Yeast can be used to break down carbohydrates in food producing carbon dioxide. (Correct answer)