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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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A 50.0 mL graduated cylinder has a mass of 65.1 g. When it is filled with an unknown liquid to the 49.3 mL mark, the cylinder an
Delvig [45]

<u>Answer:</u> The density of liquid is 1.12g/cm^3

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 65.1 g

Mass of liquid and cylinder combined, M = 120.5 g

Mass of liquid, m_2 = (M-m_1)=(120.5-65.1)g=55.4g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of liquid = 55.4 g

Volume of liquid = 49.3 mL = 49.3 cm^3    (Conversion factor:  1mL=1cm^3 )

Putting values in above equation, we get:

\text{Density of liquid}=\frac{55.4g}{49.3cm^3}\\\\\text{Density of liquid}=1.12g/cm^3

Hence, the density of liquid is 1.12g/cm^3

3 0
3 years ago
If 5,800 j of energy are applied to a 15.2 kg piece of lead, by how much does the temp change if the specific heat of lead is 0.
scoray [572]

Answer:

The answer will be 2.98K

Explanation:

Using the formula:

Q = mc∆T

Q= 5,800 (heat in joules)

m= convert 15.2kg to g which is 15200g (mass in grams)

c= 0.128 J/g °c (Specific heat capacity)

∆T=  what we need to find (temperature change)

5800J = 15200g x 0.128 x ∆T

= 2.98K  

7 0
4 years ago
Which nitrogenous base is NOT found in DNA? *
scoundrel [369]

Answer:

c. uracil

Explanation:

uracil is not found in DNA. the missing base would be thymine

5 0
3 years ago
In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so
gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL

Putting values in above equation, we get:

1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0
3 years ago
What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of<br> 285K?
julia-pushkina [17]

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm

7 0
3 years ago
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