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3 years ago

Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 5.59×10−3.

Chemistry

1 answer:

Blizzard [7]3 years ago

5 0

Answer:

E° = -0.133 V

Explanation:

In the reaction:

X(s) + Y⁺(aq) → X⁺(aq) + Y(s)

1 electron is transferred from X to Y

Now, using Nernst equation:

E° = RT / nF ln K

Where R is gas constant (8.314 J/molK), T is absolute temperature (Usually 298.15K), n are transferred electrons (1, for the reaction), F is faraday constant (96485C/mol) and K is equilibrium constant (5.59x10⁻³)

Replacing:

E° = 8.314 J/molK*298.15K / 96485C/mol*1 ln 5.59x10⁻³

E° = -0.133 V

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A solution is prepared by mixing 25.0 g H2O and 25.0 g C2H5OH. Determine the mole fractions of each substance.

Veronika [31]

Answer:

Mole fraction H₂O → 0.72

Mole fraction C₂H₅OH → 0.28

Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

25 g . 1 mol/ 18g = 1.39 moles of water (solute)

25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)

Mole fraction solute = Moles of solute / Total moles

Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

Mole fraction H₂O = 1.39 / 1.933 → 0.72

Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28

Remember that sum of mole fractions = 1

8 0

3 years ago

Read 2 more answers

9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

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