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oee [108]
3 years ago
10

The equilibrium constant of a reaction: A) is not related to the change in free energy of the reaction. B) is the same as the ma

ss action ratio when the reaction is displaced from equilibrium. C) can change if the concentration of reactants and products are changed. D) is related to the change in free energy of the reaction. E) cannot be used to determine whether a reaction will proceed in the direction as written under non-standard conditions.
Chemistry
1 answer:
Taya2010 [7]3 years ago
5 0

Answer:) is related to the change in free energy of the reaction--d

Explanation:

For any reaction that is taking place at any moment the change in Gibbs Free Energy is related to the reaction quotient as

 ΔG=ΔG⁰+RTlnQ  

where R-Universal Gas Constant, T- Temperature in Kelvin, Q is the reaction quotient

Now when the system is in equilibrum,    ΔG⁰ which is the standard Gibb's Free Energy,is then  defined as

ΔG⁰=−RTlnK ,

where K is the equilibrium constant. because  ΔG becomes 0 and reaction quotient Q = K

The equilibrum constant   is related to the change in free energy of the reaction.

because when  ΔG is negative, the value of K is high which leads to a spontaneous. reaction

when   ΔG is positive, the value of K is low, which leads to a spontaneous. reaction in the opposite direction.

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Slav-nsk [51]

Hey there!

325 mL in liters:

325 / 1000 => 0.325 L

1 mole ( Ne ) ------------- 22.4 L ( at STP )

moles ( Ne ) ------------ 0.325 L

moles Ne = 0.325 * 1 / 22.4

moles Ne = 0.325 / 22.4

moles Ne = 0.0145 moles

hope this helps!

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Answer:

A- Speed = distance/time

Explanation:

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Calculate the osmotic pressure (in torr) of 6.00 L of an aqueous 0.958 M solution at 30.°C, if the solute concerned is totally i
cluponka [151]

<u>Answer:</u> The osmotic pressure is 54307.94 Torr.

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iCRT

where,

\pi = osmotic pressure of the solution = ?

i = Van't hoff factor = 3

C = concentration of solute = 0.958 M

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = temperature of the solution = 30^oC=[30+273]K=303K

Putting values in above equation, we get:

\pi=3\times 0.958mol/L\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 303K\\\\\pi=54307.94Torr

Hence, the osmotic pressure is 54307.94 Torr.

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I believe the answer is A
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The higher the temperature of the water, the less oxygen will be dissolved it in.


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