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Blababa [14]
3 years ago
9

Phosphoglucomutase catalyzes the reaction of glucose 6-phosphate (G6P) to fructose 6-phosphate (F6P). You are starting the react

ion in a test tube with the 0.8M substrate (G6P), and you let the reaction reach equilibrium. The product (F6P) concentration at equilibrium is 0.6M. There are no intermediates in this reaction and no products at the beginning. The Keq for this reaction is:__________.
Chemistry
1 answer:
strojnjashka [21]3 years ago
5 0

Answer: 0.75

Explanation:

Mathematically, the equilibrium constant Keq is the concentration of product divided by the concentration of the reactant.

And since the product is fructose 6-phosphate (F6P) while the reactant is glucose 6-phosphate (G6P):

Keq = [F6P] / [G6P]

Keq = 0.6 / 0.8

Keq = 0.75 (since Keq is almost equal to 1, it means the amount of F6P and G6P in the reaction is almost the same)

Thus, the equilibrium constant Keq for this reaction is 0.75

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Of bond is formed by sharing of electrons - covalent

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A 40.0 L balloon is filled with air at sea level (1 atm @ 25 oC). It is then tied to a rock and thrown into a cold lake and it s
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</span>n= 1.636 moles

The volume at bottom of the lake would be:
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3 years ago
To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w
Elis [28]

<u>Answer:</u> The number of moles of weak acid is 4.24\times 10^{-3} moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}

We are given:

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Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol

The chemical reaction of weak monoprotic acid and KOH follows the equation:

HA+KOH\rightarrow KA+H_2O

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, 4.24\times 10^{-3}mol of KOH will react with = \frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol of weak monoprotic acid.

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