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muminat
3 years ago
13

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.
Which of these reactions is (are) unfavorable?
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
A. A
B. B
C. C
Which of these reactions can be coupled so that overall reaction is favorable?
A. A and B
B. A and C
C. B and C
What is the net change in free energy if one selection from part (b) is coupled so that the overall reaction is favorable?
Chemistry
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

1. Options A and B

2. Options B and C

3.. B. Net ∆G = -16.7 KJ/mol; C. Net ∆G = -14.2 KJ/mol

Explanation:

1. The spontaneity of a chemical reaction depends on its standard free energy change, ∆G. If ∆G is negative, the reaction is favourable, but when it is positive, the reaction is unfavorable.

Therefore, since reaction A and B have ∆G to be positive, they are unfavorable

2. Coupling an unfavorable reaction to a favourable reaction can help the reaction to proceed in the forward direction as long as the net free energy change is negative.

Coupling reaction A and C, as well as reaction B and C will make the reactions to become favourable as net ∆G is negative in both instances.

3. A and C: net ∆G = 13.8 - 30.5 = -16.7 KJ/mol

B and C: net ∆G = 16.3- 30.5 = -14.2 KJ/mol

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Please help me ಥ_ಥ
dexar [7]

Energy is stored in chemical bonds during photosynthesis.

During photosynthesis, the radiant energy from the sun is converted to chemical energy in carbohydrates.

Inorganic materials in the form of carbon dioxide and oxygen combine to form carbohydrates in the presence of radiant energy according to the equation below:

6CO_2 + 6H_2O ---> C_6H_1_2O_6 + 6O_2

The energy is thus, stored in chemical bonds in the carbohydrate and this is what is oxidized during respiration to release the locked energy.

More on photosynthesis can be found here: brainly.com/question/1388366

7 0
2 years ago
Read 2 more answers
At what temp will a gas be at if you allow it to expand from an original 456 mL to 65°C to 3.4 L
alexira [117]
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
                                          V₁T₂ = V₂T₁
Substituting the known values,
                                (0.456 L)(65 + 273.15) = (3.4 L)(T₁)
                                             T₁ = 45.33 K
5 0
3 years ago
Read 2 more answers
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
How many moles of potassium hydroxide are needed to completely react with 2.94 moles of aluminum sulfate
ArbitrLikvidat [17]

Answer:- Third choice is correct, 17.6 moles


Solution:- The given balanced equation is:


Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4


We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.


From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.


It is a simple mole to mole conversion problem. We solve it using dimensional set up as:


2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})


= 17.6 mol KOH


So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.



6 0
3 years ago
Convert 533 cm/s to units of meters per minute. Show the unit analysis by dragging the conversion factors into the unit‑factor s
lukranit [14]

Answer:

319.8 m/min

Explanation:

533 cm/s

We can convert 533 cm/s to m/min by doing the following:

First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:

Recall:

100 cm/s = 1 m/s

Therefore,

533 cm/s = 533 cm/s /100 cm/s × 1 m/s

533 cm/s = 5.33 m/s

Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:

1 m/s = 60 m/min

Therefore,

5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min

5.33 m/s = 319.8 m/min

Therefore, 533 cm/s is equivalent to 319.8 m/min

8 0
3 years ago
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