Describe the chemical process of fermentation.

Answers

Answer 1
Answer:

Fermentation is a metabolic process. The molecules like glucose are breaks down anaerobically.

What is Fermentation ?

Fermentation is a metabolic process which converts sugars to alcohol, acids or gases. Fermentation is observed in bacteria,yeast cell and also in the muscle of animals.  An Example is lactate fermentation which produces lactic acid.

What is Chemical process of fermentation ?

Fermentation is a chemical process in which molecules like glucose are breaks down anaerobically. Fermentation is chemical change. In chemical changea new substance is formed.

Thus, from above conclusion we can say that Fermentation is a metabolic process. The molecules like glucose are breaks down anaerobically.

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Answer 2
Answer:

Answer:

Fermentation is a chemical process in which complex organic molecules are broken down into simpler compounds like ethanol. This process is catalyzed by certain enzymes. Enzymes are complex chemical catalysts produced by living cells. The sugars used for fermentation are often formed by enzymatic decomposition of starches from corn, potatoes, rice, or grain.

Explanation:


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In a play, the rising action consists of the events that lead up to the a. last line in the scene
b. climax
c. external conflict
d. internal conflict

Answers

The rising action, which follows the expositions, comes just before the climax.
THE ANSWER IS THE RISING ACTION THEN I THINK GO TO INTERNAL CONFLICT THEN THE CLIMAX

How much heat (in kJ) must be added to 1.34 kg of ice at O'C to convert it to water to

Answers

To change 1.34 kg of ice at 0°C to water, you would need to add approximately 448 kJ of heat.

To calculate the amount of heat needed to convert ice at 0°C to water, we use the formula Q = m * Lf. Where Q is the Heat Transfer, m is the mass of the substance (ice in this case), and Lf is the heat of fusion for ice, which is 334 kJ/kg.

Plugging in the values we have: Q = 1.34 kg * 334 kJ/kg = 447.56 kJ.

Therefore, you would need to add approximately 448 kJ of heat to convert 1.34 kg of ice at 0°C to water at the same temperature.

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The heat required to convert 1.34 kg of ice at 0°C to water at 0°C is approximately 450.24 kJ.

When the substance freezes (changes from liquid to solid), the same amount of energy is released back into the surroundings. When a substance changes its phase (solid to liquid or liquid to gas) at a constant temperature, the heat required for this process is known as the heat of fusion.

For ice at its meltingpoint (0°C), the heat of fusion is approximately 336 kJ/kg. Thus for 1.34 kg of ice, we can calculate the heat required by using the formula below:

Heat = mass × heat of fusion

Heat = 1.34 kg × 336 kJ/kg

Heat = 450.24 kJ.

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The complete question is -

How much heat (in kJ) must be added to 1.34 kg of ice at 0°C to convert it to water at 0°C

During a scientific investigation, which step will a scientist perform first?

Answers

Answer:

you need to have a question or a problem first

Generally, which of the three families of elements (metals, nonmetals, or inert gases) has the least tendency to form ionic bonds?

Answers

Answer:

Inert gases

Explanation:

Inert elements have a stable electron configuration meaning their shells/orbitals are full with their requisite number of electrons. Therefore, gaining or losing an electron would take high ionization energy. Therefore they are less likely to be involved in chemical reaction unless a high amount of energy is used. An example of an inert gas is Helium.

Which change is exothermic?(1) freezing of water(2) melting of iron(3) vaporization of ethanol(4) sublimation of iodine

Answers

The answer is (1) freezing of water. The energy of different status is increase from solid to liquid to gas. So the change of liquid to solid is exothermic. The other three are endothermic.

How many moles are in 68.5 liters of oxygen gas at STP?

Answers

We can solve this problem when we use the conditions of a gas at standard temperature and pressure. It has been established that at STP where the temperature is 0 degrees Celsius and the pressure is 101.325 kPa, the volume of 1 mole of gas is 22.4 L. We will use this data for the calculations.

68.5 L ( 1 mol O2 / 22.4 L O2 ) = 3.06 mol O2

Explanation:

It is known that at STP, there are 22.4 L present in one mole of a substance.

Therefore, in 68.5 liters there will be 1 mol divided by 22.4 L times 68.5 L.

Mathematically,      68.5 L * (1 mol O_(2))/(22.4 L)

                              = 3.05 mol

Hence, we can conclude that there are 3.05 moles present in 68.5 liters of oxygen gas at STP.