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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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The answer is C by vibrating the molecules in the matter
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A chemist prepares a solution of barium chloride by measuring out of barium chloride into a volumetric flask and filling the fla
Romashka [77]

Answer:

30 μmol/L

Explanation:

<em>A chemist prepares a solution of barium chloride by measuring out 8.9 μmol of barium chloride into a 300mL volumetric flask and filling the flask to the mark with water. </em>

<em> Calculate the concentration in μmol L of the chemist's barium chloride solution. Round your answer to 2 significant digits.</em>

<em />

The chemist has 8.9 μmol of solute (barium chloride) and he adds water until the mark of 300 mL in the container, which is the volume of the solution. We will need the conversion factor 1 L = 1000 mL. The concentration of barium chloride in μmol/L is:

\frac{8.9\mu mol}{300mL} .\frac{1000mL}{1L} =30\mu mol/L

3 0
3 years ago
A 0.080 kg robin sits on a perch 6.0 meters above the ground. What is the potential energy of the robin
Nesterboy [21]

Answer:

4.704J

Explanation:

The following data were obtained from the question:

m = 0.080kg

h = 6.0m

g = 9.8m/s^2

P.E =?

P.E = mgh

P.E = 0.08 x 9.8 x 6

P.E = 4.704J

Therefore, the potential energy of the robin is 4.704J

6 0
3 years ago
Consider the gas phase reaction 4HCl + O2 → 2Cl2 + 2H2O. What volume of chlorine gas at STP can be prepared from the reaction of
SOVA2 [1]

Answer:

c

Explanation:

7 0
3 years ago
Fe2O3 + CO → Fe + CO2 If 3 moles of Fe2O3 react with 1.5 moles of CO, how many moles of each product are formed? (3 points)
Dima020 [189]

Answer:

A. 1 mole of Fe.

B. 1.5 moles of CO₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Fe₂O₃ + 3CO —> 2Fe + 3CO₂

From the balanced equation above,

1 mole of Fe₂O₃ reacted with 3 moles of CO to produced 2 moles of Fe and 3 moles of CO₂.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of Fe₂O₃ reacted with 3 moles of CO.

Therefore, 3 moles of Fe₂O₃ will react with = 3 × 3 = 9 moles of CO.

From the calculation made above, we can see clearly that it will require a higher amount (i.e 9 moles) of CO than what was given (i.e 1.5 moles) to react completely with 3 moles of Fe₂O₃.

Therefore, CO is the limiting reactant and Fe₂O₃ is the excess reactant.

A. Determination of the number of mole of Fe produced.

NOTE: The limiting reactant is used to obtain the desired result because it will give the maximum yield of the products since all of it is consumed in the reaction.

From the balanced equation above,

3 moles of CO reacted to produced 2 moles of Fe.

Therefore, 1.5 moles of CO will react to produce = (1.5 × 2)/3 = 1 mole of Fe.

Thus, 1 mole of Fe was obtained from the reaction.

B. Determination of the number of mole of CO₂ produced.

From the balanced equation above,

3 moles of CO reacted to produced 3 moles of CO₂.

Therefore, 1.5 moles of CO will also react to produce 1.5 moles of CO₂.

Thus, 1.5 moles of CO₂ were obtained from the reaction.

6 0
3 years ago
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