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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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Answer:

A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for the amount of a substance. There are, therefore, 6.02 × 10 23 water molecules in a mole of water molecules. Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen.

8 0
2 years ago
What is a matter a wave travels through called?
tresset_1 [31]

Answer:

Medium

Explanation:

It is called a medium

5 0
4 years ago
Please help me answer this question! I don’t understand it at all!
pentagon [3]
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This Should help you
7 0
3 years ago
What is the volume of 400 g of metal with a density of 10 g/mL?
grigory [225]

Answer:

40 mL

Explanation:

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4 0
2 years ago
Tris base has a molecular weight of 121 g/mol. How many grams of tris base would you need to make 250ml of a 200mM SOLUTION
STatiana [176]

Answer:

6.05 g

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

From the question ,

M = 200mM

Since,

1 mM = 10⁻³ M

M = 200 * 10⁻³ M

V = 250 mL

Since,

1 mL = 10⁻³ L

V = 250 * 10⁻³ L

The moles can be calculated , by using the above relation,

M = n / V  

Putting the respective values ,

200 * 10⁻³ M = n / 250 * 10⁻³ L

n = 0.05 mol

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

m = 121 g/mol

n = 0.05 mol ( calculated above )

The mass of tri base can be calculated by using the above equation ,

n = w / m  

Putting the respective values ,

0.05 mol = w / 121 g/mol

w = 0.05 mol * 121 g/mol

w = 6.05 g

3 0
3 years ago
Read 2 more answers
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