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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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H_3PO_32H_3PO_32H_3PO_3P_2O_3Answer:

B

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This means that, to fully react 3.62 moles of P_2O_3, we would need 3*3.62 or  10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of P_2O_3, 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of P_2O_3.

From the balanced equation, we see that every mole of P_2O_3 reacted gets you 2 moles of 2H_3PO_3. Therefore, 2.1033 moles of P_2O_3 would give you approximately 4.21 moles of H_3PO_3.

5 0
2 years ago
Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c
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6 0
3 years ago
Read 2 more answers
Calculate the vapor pressure at 35ºC of a solution made by dissolving 20.2 g of sucrose (C12H22O11)in 60.5 g of water. The vapor
sukhopar [10]

Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

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Let's replace the data in the formula

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P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

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5 0
3 years ago
g Ammonium nitrate was the oxidizing agent used that damaged the New York World Trade Center in 1993 and the Federal Building in
Cloud [144]

Answer:  51.3\times 10^3 kJ

Explanation:

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass= 50.0 kg =50.0\times 10^3g   (1kg=1000g)

\text{Number of moles of ammonium nitrate}=\frac{50.0\times 10^3g}{80g/mol}=0.625\times 10^3moles

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1 mole of NH_4NO_3 gives = 82.1 kJ of heat

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Thus 51.3\times 10^3 kJ of heat is evolved from the decomposition of 50.0 kg of ammonium nitrate.

7 0
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