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rodikova [14]
3 years ago
13

What is the use of liquid electrolyte?​

Chemistry
1 answer:
bogdanovich [222]3 years ago
8 0

Although progress has been made in enhancing the conductivity of solid electrolytes, particularly the polymeric ones, liquid electrolytes are still used in most electrochemical systems. The solvent properties, and dynamics of ion solvent interactions, must be understood in designing new electrolytes.

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Dmitrij [34]
It is going to be D hope this helps
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3 years ago
PLEASE HELP! NO LINKS! CHEMISTRY!
Bingel [31]

Answer:

Solution A is 1,000 times more acidic than Sol. B

Explanation:

for pH values we use scientific notation:

-log10 c (where c is the hydrogen ion concentration) is used to notate pH value (think of it as a unit)

ie:

10^-2 is sol A 10^-5 is sol B

5-2 is 3

10^-3 = 1000

there's a diff of 1,000 between the solutions.

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3 years ago
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Which term best describes the type of bonding in magnesium chloride
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Elements that can be described by the hardness shininess and ductility
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6 0
3 years ago
A sample of an ionic compound NaA, where A- is the anion of a
vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0
3 years ago
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