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Amanda [17]
1 year ago
12

gelified acetate-based water-in-salt electrolyte stabilizing hexacyanoferrate cathode for aqueous potassium-ion batteries

Physics
1 answer:
Nookie1986 [14]1 year ago
4 0

In recent years, gelified acetate-based water-in-salt electrolyte stabilising hexacyanoferrate cathode for aqueous potassium-ion batteries has gained popularity.

Here, water activity in "Water-in-salt" electrolytes (WiSE) based on potassium acetate (KAc) is investigated using Raman and classical molecular dynamics (MD), demonstrating the significantly reduced water activity of these WiSEs because water can be efficiently coordinated by both the acetate anion and the potassium cation. It is discovered that the overall molecular arrangement resembles the "sponge-like" shape seen in several ionic liquids. According to comprehensive electrochemical, scanning electronic microscopy (SEM), and X-ray photoelectron spectroscopy (XPS) study, such WiSE can also be compatible with Al current collectors with appropriately adjusted composition. However, due to the limited stability of the cathode material in the alkaline environment, the use of inexpensive potassium manganese hexacyanoferrate (KMHCF) in combination with these electrolytes is hampered by poor cycling performance.

Learn more about Electrolyte here:

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onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
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Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA =  1.00 * 10⁻⁴ * 1

Number of moles of HA =  1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of  A⁻ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into  H⁺  and  A⁻, 84% of HA is left

Number of mols of HA left = 0.84 *  1.00 * 10⁻⁴

Number of mols of HA left =  8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ =  1.6 * 10⁻⁵/V

To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

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Answer:

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A girl stands on a moving sidewalk (conveyor belt) that is moving to the right at a speed of 2 m/s relative to the ground. A dog
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To calculate the speed of the dog relative to the ground, we use the formula,

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