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3 years ago

Which is a similarity between alcohol fermentation and aerobic respiration?

2 answers:

6 0

Both begin with glycolysis.
Both occur in mitochondria.
Both require oxygen to proceed.
Both end with the electron transport chain.

Alexus [3.1K]3 years ago

3 0

I found this information:

Anaerobic respiration begins the same way as fermentation. The first step is still glycolysis, and it still creates 2 ATP from one carbohydrate molecule. However, instead of ending with glycolysis, as fermentation does, anaerobic respiration creates pyruvate and then continues on the same path as aerobic respiration.

After making a molecule called acetyl coenzyme A, it continues to the citric acid cycle. More electron carriers are made and then everything ends up at the electron transport chain. The electron carriers deposit the electrons at the beginning of the chain and then, through a process called chemiosmosis, produce many ATP. For the electron transport chain to continue working, there must be a final electron acceptor. If that acceptor is oxygen, the process is considered aerobic respiration.

Scientists believe that fermentation and anaerobic respiration are older processes than aerobic respiration.

I hope this help. :)

have a good day!

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3 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

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