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3 years ago

Which of the following is NOT a possible end-product of a fermentation process?- glucose- carbon dioxide- lactic acid- ethanol

Chemistry

1 answer:

MArishka [77]3 years ago

4 0

Answer:

glucose

Explanation:

There are two types of respiration:

1. Aerobic respiration

2. Anaerobic respiration

Aerobic respiration:

It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.

Glucose + oxygen → carbon dioxide + water + 38ATP

Anaerobic Respiration:

It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.

Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide

This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.

Steps involve in anaerobic respiration are:

Glycolysis:

Glycolysis is the first step of both aerobic and anaerobic respiration. It involve the breakdown of one glucose molecule into pyruvate and 2ATP.

Fermentation:

The second step of anaerobic respiration is fermentation. It involve the fermentation of pyruvate into lactic acid or alcohol depending upon the organism in which it is taking place. There is no ATP produced in this step, however carbon dioxide is released.

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Does the iconic compound NaCl have the same particles as the individual elements of sodium and chlorine gas

valentina_108 [34]

Answer:It is not the same

Explanation: the normal sodium atom has an atomic number of 11 while Chlorine has an atomic number of 17. For Nacl; the sodium has lose an electron because it want to attain it octet configuration while chlorine accept the electron release to make it attain its octet configuration. At this point the elements are electrically stable.

Therefore, sodium become 10 and chlorine become 18

4 0

3 years ago

Deposition is a ________________ changing into a ____________________. 1. liquid, solid 2. solid, gas 3. gas, solid 4. liquid, g

ella [17]

Answer:

3. Gas, solid

Explanation:

Deposition is the process of gas changing directly to a solid. It skips the liquid state.

5 0

3 years ago

You wish to prepare a tape-casting slip containing 50 vol% Al2O3 and 50 vol% polyvinyl butyral (PVB) binder. If the density of A

belka [17]

Answer: The mass of PVB required to produce 1000 grams of tape is 213.4 grams

Explanation:

To calculate the mass of aluminium oxide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ .......(1)

For $Al_2O_3$

We are given:

50% (v/v) of $Al_2O_3$

This means that 50 mL of aluminium oxide is present in 100 mL of tape

Calculating the mass of aluminium oxide by using equation 1:

Density of aluminium oxide = $3.98 g/cm^3$

Volume of aluminium oxide = $50mL=50cm^3$ (Conversion factor: $1mL=1cm^3$ )

Putting values in equation 1, we get:

$3.98g/cm^3=\frac{\text{Mass of aluminium oxide}}{50cm^3}\\\\\text{Mass of aluminium oxide}=(3.98g/cm^3\times 50cm^3)=199g$

Mass of aluminium oxide = 199 g

For PVB:

We are given:

50% (v/v) of PVB

This means that 50 mL of PVB is present in 100 mL of tape

Calculating the mass of PVB by using equation 1:

Density of PVB = $1.08 g/cm^3$

Volume of PVB = $50mL=50cm^3$

Putting values in equation 1, we get:

$1.08g/cm^3=\frac{\text{Mass of PVB}}{50cm^3}\\\\\text{Mass of PVB}=(1.08g/cm^3\times 50cm^3)=54g$

Mass of PVB = 54 g

Mass of tape = Mass of aluminium oxide + mass of PVB

Mass of tape = [199 + 54] g = 253 g

To calculate the mass of PVB required to produce 1000 g of tape, we use unitary method:

When 253 grams of tape is made, the mass of PVB required is 54 g

So, when 1000 grams of tape is made, the mass of PVB required will be = $\frac{54}{253}\times 1000=213.4g$

Hence, the mass of PVB required to produce 1000 grams of tape is 213.4 grams

4 0

3 years ago

A 250-ml sample of na2so4 is reacted with an excess of bacl2. if 5.28 g baso4 is precipitated, what is the molarity of the na2so

Oksanka [162]

To get the molarity you need to follow this equation
moles of solute
Molarity (M = -----------------------
Liters of solution

But before you apply that equation you need to find the moles of solute and the liters of solution. Follow this equation

Na2SO4 + BaCl2 = BaSO4 + 2 NaCl

Solution

Moles of BaSO4 = 5.28 g
---------------
233.43 g / mol
= 0.0226 moles
Moles of NaSO4 = 0.0226
0.0226 mole
Molarity = -----------------
0.250 L
= 0.0905 mol / L

So the answer is 0.0905 mol / L

3 0

3 years ago

Atoms of which element have the greatest tendency to gain electrons?

Anastasy [175]

Fluorine.

Because:- Atoms want to become stable, for an atom to become stable, they need 8 valence electrons. Since Fluorine has 7 valence electrons, it only needs one more electron to become stable and have an octet. An octet is when an atom/element has 8 valence electrons. Since Fluorine will need to gain an electron, it will have a negative charge, and become an anion.

8 0

3 years ago

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