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3 years ago

How do fermentation reactions in oxygen-starved muscle cells and anaerobically grown yeast cells differ?

1 answer:

8 0

In oxygen- starved muscle cells in anaerobic respiration lactic acid is formed while in anaerobic respiration by yeast cells it produces ethanol and carbon dioxide.

Explanation:

Respiration occurring in absence of oxygen in muscles leads to accumulation of lactic acid and very less amount of energy. This is fermentation process as the end product formed is lactic acid.

The process of fermentation takes place in yeast bacteria and oxygen starved cells of muscles.

The yeast can ferment simple sugars like glucose and fructose into ethyl alcohol and carbon dioxide.

In general human respiration is aerobic but during strenuous exercise, oxygen does not meet the demand of working out muscle, so they respire anaerobically and produce lactic acid and 2 ATP i.e very less compared to 38 molecules of ATP in aerobic respiration.

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For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

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