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Margaret [11]
3 years ago
11

In principle, at least one way to increase the concentration of G6P is to drive the equilibrium to the right by increasing the i

ntracellular concentrations of Pi and/or glucose. Assuming a fixed concentration of Pi at 5.0 mM, how high would be the intracellular concentration of glucose have to be to give an equilibrium concentration of glucose−6−phosphate of 0.25 mM (the normal physiological concentration)? Would this route be reasonable, given the maximum solubility of glucose is less than 1 M?
Chemistry
1 answer:
frez [133]3 years ago
4 0

Answer:

.

Explanation:

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What is the pOH of a solution with [OH-] = 1.4 x 10-13?
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A mixture of 14.2 g of H2 and 36.7 g of Ar is placed in a 100.0 L container at
TiliK225 [7]

a) The total pressure of the system is  1.79 atm

b) The mole fraction and partial pressure of hydrogen is  0.89 and 1.59 atm respectively

c) The mole fraction and the partial pressure  of argon is 0.11 and 0.19 atm.

<h3>What is the total pressure?</h3>

We know tat we can be able to obtain the total pressure in the system by the use of the ideal gas equation. We would have from the equation;

PV = nRT

P = pressure

V = volume

n = Number of moles

R = gas constant

T = temperature

Number of moles of hydrogen = 14.2 g/2g = 7.1 moles

Number of moles of Argon = 36.7 g/40 g/mol

= 0.92 moles

Total number of moles =  7.1 moles + 0.92 moles = 8.02 moles

Then;

P = nRT/V

P = 8.02 * 0.082 * 273/100

P = 1.79 atm

Mole fraction of hydrogen = 7.1/8.02 = 0.89

Partial pressure of hydrogen =  0.89 * 1.79 atm

= 1.59 atm

Mole fraction of argon = 0.92 / 8.02

= 0.11

Partial pressure of argon = 0.11  *  1.79 atm

= 0.19 atm

Learn more about partial pressure:brainly.com/question/13199169

#SPJ1

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I believe an atom. may be wrong, science is not my strong suit
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