All categories

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.

You might be interested in

Aluminum is an element. Which best describes what makes up a sample of aluminum?

sp2606 [1]

I believe the answer is A.

4 0

3 years ago

Read 2 more answers

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

Which of the following units are commonly used in chemistry? A Kelvin b ampere c fluid ounce d kilogram

sdas [7]

Explanation:

your answer is Kelvin because it is the SI unit of temperature

7 0

3 years ago

Which element would be the most conductive and

Masja [62]

Malleability described the property of physical deformation under some compressive stress; a malleable material could, for example, be hammered into thin sheets. Malleability is generally a property of metallic elements: The atoms of elemental metals in the solid state are held together by a sea of indistinguishable, delocalized electrons. This also partially accounts for the generally high electrical and thermal conductivity of metals.

In any case, only one of the elements listed here is a metal, and that’s copper. Moreover, the other elements (hydrogen, neon, and nitrogen) are gases under standard conditions, and so their malleability wouldn’t even be a sensible consideration.

8 0

3 years ago

Read 2 more answers

3. What is the mass % of a solution that contains 36g KCl in 475g of water?

stira [4]

Answer:

%KCl = 7.05%

%Water = 92.95%

Explanation:

Step 1: Given data

Mass of KCl (solute): 36 g

Mass of water (solvent): 475 g

Step 2: Calculate the mass of the solution

The mass of the solution is equal to the sum of the masses of the solute and the solvent.

m = 36 g + 475 g = 511 g

Step 3: Calculate the mass percentage of the solution

We will use the following expression.

%Component = mComponent/mSolution × 100%

%KCl = 36 g/511 g × 100% = 7.05%

%Water = 475 g/511 g × 100% = 92.95%

7 0

3 years ago