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
Answer:
There are essentially three methods used to produce ethanol:
Manufacture from ethene using steam (the "synthetic" route)
Production from sugars and starches by fermentation, using yeasts.
Production from biomass waste, using bacteria.
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<u>Answer:</u> The solubility of carbon dioxide at 5.50 atm is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:

Or,

where,
are the initial concentration and partial pressure of carbon dioxide
are the final concentration and partial pressure of carbon dioxide
We are given:

Putting values in above equation, we get:

Hence, the solubility of carbon dioxide at 5.50 atm is 
Answer: If the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.
Explanation:
A solution which contains the maximum amount of solute is called a saturated solution. Whereas a solution in which more amount of solute is able to dissolve is called an unsaturated solution.
Now, the number of moles present in 36 g of NaCl (molar mass = 58.4 g/mol) is as follows.

This shows that solubility of sodium chloride is 36 grams per 100 grams of water means a maximum of 0.616 mol of NaCl will dissolve in 100 mL of water.
So, a solution in which number of moles of NaCl are less than 0.616 mol per 100 mL then the solution formed will be an unsaturated solution.
- As 5.8 moles of NaCl dissolved in 1 L (or 1000 mL) of water. So, moles present in 100 mL are calculated as follows.

- Moles present in 100 mL of water for 3.25 moles of NaCl dissolved in 500 ml in water are as follows.

- Moles present in 100 mL of water for 1.85 moles of NaCl dissolved in 300 ml of water are as follows.

Thus, we can conclude that if the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.