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UNO [17]
3 years ago
12

Can you answer the following questions with work, please?A) The conversion of glucose-6-phosphate to fructose-6-phosphate is cat

alyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reactions was allowed to come to equilibrium. If the keq' is 0.50 what is the Delta G (degree prime) in kJ/mol?B) The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Under cellular conditions (37*C), the glucose-6-phosphate is 6.6 uM and the fructose-6-phosphate is 1.3 uM. If the keq' is 0.50 what is the Delta G in kJ/mol? (Use the Delta G (degree prime) from the previous question)
Chemistry
1 answer:
irga5000 [103]3 years ago
3 0

a) dG0= 1.573 kJ/mol

b)  dG = -2.614 kJ/mol

 Explanation:

a)

Write down the values given in the question,

Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reactions was allowed to come to equilibrium the keq is 0.50

To calculate the delta value ,

dG = -RT ln (Keq)

Putting values:

dG0 = -8.314*273*ln(0.5)

dG0= 1.573 kJ/mol

b)

Glucose-6-phosphate is 6.6 uM and the fructose-6-phosphate is 1.3 uM. If the keq' is 0.50.

To calculate the delta value,

dG = dG0 + RT*ln(Q)

Putting values:

dG = 1573 + 8.314*310*ln(1.3/6.6)

Solving we get:

dG = -2.614 kJ/mol

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The concentration of the original calcium ions is 0.005 M

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.

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= 0.0022 moles

Now;

1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by  0.0022 moles of CaCrO4.

Given that the volume of the solution is  0.440 L, the concentration of the solution is;   0.0022 moles/0.440 L

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(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

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And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

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