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Elina [12.6K]
3 years ago
12

The metal which is actually oxidized/reduced in a lithium ion battery is: a. Cobalt b. Lithium c. Manganese d. Zinc

Chemistry
1 answer:
Usimov [2.4K]3 years ago
8 0

Answer:

b. Lithium

Explanation:

<u>Lithium-ion batteries (LIB) is a rechargeable battery</u>. It was first proposed by M stanley  Whittingham in 1970s. These kind of batteries are commonly used in electrical vehicles and portable electronics. It also have military , naval and aerospace applications.

<u> In LIB, the lithium ions move from the negatively charged electrode to the positively charged electrode during the discharge process and move in the reverse direction when the during the charging process.</u> The battery uses an intercalated lithium compound is used to form anode and graphite is used to form cathode.

<u>Thus, Lithium is oxidized/reduced in a lithium ion battery.</u>

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Compare amplitudes, wavelengths, and frequencies of waves
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2 years ago
High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t
Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0
3 years ago
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