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mafiozo [28]
3 years ago
5

Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are

thermodynamically unfavorable (nonspontaneous), but proceed when they are coupled with other reactions.
Reaction A: Pi+glucose⟶glucose-6-phosphate+H2OΔG=13.8 kJ/mol

Reaction B: Pi+fructose-6-phosphate⟶fructose-1,6-bisphosphate+H2OΔG=16.3 kJ/mol

Reaction C: ATP+H2O⟶ADP+PiΔG=−30.5 kJ/mol

Select every unfavorable reaction.

reaction B
reaction C
reaction A

Select every set of coupled reactions where the overall reaction is favorable.

reactions B and C
reactions A and B
reactions A and C

What is the net change in free energy if one set of reactions from the previous question is coupled so that the overall reaction is favorable? Note that if you selected more than one set of coupled reactions as favorable in the previous question, you may enter the net change for any one of your selected sets.
Chemistry
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

Unfavorable reactions: Reaction A and Reaction B.

Coupled reactions favorable: Reactions B and C and Reactions A and C.

Net change:

Reactions B and C : -14.2kJ/mol

Reactions A and B : 30.1kJ/mol

Reactions A and C: -16.7kJ/mol

Explanation:

A reaction is thermodynamically favorable (spontaneous) if ΔG < 0. Thus, the unfavorable reactions -ΔG > 0- are:

Reaction A and reaction B.

When reaction C is coupled with reaction B and reaction A the chemical equation is:

ATP + fructose-1,6-phosphate ⟶ ADP + fructose-1,6-bisphosphate

ΔG = 16.3 - 30.5 = -14.2 kJ/mol

ATP + glucose ⟶ ADP + glucose-6-phosphate

ΔG = 13.8 - 30.5 = -16.7 kJ/mol

The coupled reaction of A and B has a change in free energy of:

ΔG = 13.8 + 16.3 = 30.1 kJ/mol

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The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

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