It takes 21.3 days
<h3>Further explanation</h3>
Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
![\tt \dfrac{5000~g}{46~g/mol}=108.7~moles](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B5000~g%7D%7B46~g%2Fmol%7D%3D108.7~moles)
moles of glucose to produce 108.7 moles ethanol :
![\tt \dfrac{1}{2}\times 108.7=54.35](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20108.7%3D54.35)
54.35 moles = 5 hours
moles of 1000 kg of glucose :
![\tt \dfrac{10^6~g}{180~g/mol}=5555.5~moles](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B10%5E6~g%7D%7B180~g%2Fmol%7D%3D5555.5~moles)
So for 5555.5 moles, it takes :
![\tt \dfrac{5555.5}{54.35}\times 5~hours=511.085~h=21.3~days](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B5555.5%7D%7B54.35%7D%5Ctimes%205~hours%3D511.085~h%3D21.3~days)
Below are the choices:
a)0.2168 atm
<span>b)4.613 atm </span>
<span>c)34.60 atm </span>
<span>d467.4 atm
</span>
1 atm = 760mmHg : Therefore:
<span>3,506mmHg = 3,506/760 = 4.613 atm
</span>B is correct answer.
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Answer:
From the graph find the maximum velocity and half it i.e. Vmax/2. Draw a horizontal line from this point till you find the point on the graph that corresponds to it and read off the substrate concentration at that point. This will give the value of Km.