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I am Lyosha [343]
3 years ago
8

Electrochemistry - Equilibrium

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

Explanation:

The relation between equilibrium constant and Ecell is given below .

E⁰cell = (RT / nF ) lnK  , F is faraday constant T is 273 + 25 = 298 K

E⁰cell  =  1.46 - 1.21 = .25 V

n = 2

Putting the values

.25 = (8.314 x 298  lnK) / (2 x 96485 )

lnK = 19.47

K = 2.85 x 10⁸

2 )

Change in free energy Δ G

Δ G ⁰ = nE⁰ F

n = 4

E⁰ = .4 + .83 = 1.23 V

Δ G ⁰= 4 x 1.23 x 96485

= 474706 J / mol

3 )

E⁰cell = (RT / nF ) lnK

n = 2

1.78 = 8.314 x 298  lnK / 2 x 96485

lnK = 138.638

K = 1.62 x 10⁶⁰

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A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

<h3>What does Beer-Lambert law state?</h3>

The Beer-Lambert law states that for a given material sample, path length and concentration of the sample are directly proportional to the absorbance of the light.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, we can calculate the molarity of the solution using the following expression.

A = ε × b × c

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where,

  • A is the absorbance.
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  • c is the molar concentration.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

Learn more about the Beer-Lambert law here: brainly.com/question/12975133

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How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
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The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

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The volume of the stock solution needed for this dilution will be

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So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
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