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4 years ago

Summarize the process of alcoholic fermentation in yeast

Chemistry

1 answer:

AysviL [449]4 years ago

6 0

Alcoholic fermentation is mainly used by various yeast species to make energy.

If there is no oxygen available, the yeasts have in the alcoholic fermentation another possibility of energy supply. But they can - as compared with cellular respiration - recover substantially less energy from glucose, in the form of adenosine triphosphate (ATP): by complete oxidation, a molecule of glucose provides 36 molecules of ATP, but by alcoholic fermentation only 2 molecules of ATP. These two molecules are obtained in glycolysis, the first step in the chain of reactions for both cellular respiration and fermentation.

The two additional steps of the fermentation, and thus the production of ethanol serve not to make energy, but the regeneration of the NAD + cofactor used by the enzymes of glycolysis. As NAD + is available in limited quantities, it is converted by the NADH reduced state fermentation enzymes to the NAD + oxidized state by reduction of acetaldehyde to ethanol.

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3 years ago

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3 years ago

Consider the following data showing the initial rate of a reaction (A→products) at several different concentrations of A.

OlgaM077 [116]

Answer:

(1) order = 2

(2) R = K [A]²

Explanation:

Given the reaction:

A--------->Product

The rate constant relation for the reaction is given as:

R(i) = K [A]............(*)

Where R(I) is rate constant at different concentration of A.

Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results

0.011 = K [0.15] ⁿ.........(1)

0.044 = K [0.30]ⁿ .......(2)

0.177 = K [0.60]ⁿ .........(3)

Dividing (2) by (1) and (3) by (1)

Gives:

0.044/0.011 = [0.3/0.15]ⁿ

4 = 2ⁿ; 2² = 2ⁿ; n = 2

Similarly

0.177/0.011 = [0.60/0.15]ⁿ

16.09 = 4ⁿ

16.09 = 16 (approximately)

4² = 4ⁿ ; n = 2

Hence the order of the reaction is 2.

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4 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

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3 years ago

How many electrons does the element Zn have?

Rashid [163]

The answer would be 30

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3 years ago