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Lynna [10]
3 years ago
11

In yeast, ethanol is produced from glucose under anaerobic conditions. A cell‑free yeast extract is placed in a solution that co

ntains 325 mmol glucose, 0.35 mmol ADP , 0.35 mmol P i , 0.70 mmol ATP , 0.20 mmol NAD + , and 0.20 mmol NADH . It is kept under anaerobic conditions. What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under these conditions?

Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answer:

650 mmol.

Explanation:

The equation for the fermentation of one mole of glucose is:

C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺

Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:

C₆H₁₂O₆ + 2 ADP + 2 P i  → 2 EtOH + 2 ATP

With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:

1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH

325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH

Therefore, the maximum amount of ethanol that could be produced is 650 mmol.

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DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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3 years ago
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Answer:

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This task is a joint variation task involving only direct proportionality:

Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.

Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.

In this exercise:

If a ∝ b and a ∝ c, then a ∝ bc

Taking the above three proportionalities,

V ∝ a ∝ b ∝ c

V ∝ a ∝ bc

V ∝ abc

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Answer:

They dissociate into positive and negative ions in the solution.

Explanation:

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