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Amanda [17]
3 years ago
15

What condition must be met for a battery to be rechargeable?

Chemistry
1 answer:
Neko [114]3 years ago
3 0

Answer:

D. The electrochemical reaction of the battery must be reversible.

Explanation:

The batteries are based on the production of an electric flux given by a<u> redox reaction</u>. This reaction is <u>spontaneous</u> and is<u> thermodynamically favored</u>.

Thus, when the reactants to the reaction are finished, the flow of current stops and ends. Therefore, when current is administered from another source, the reaction <u>changes its direction</u> and reagents that were previously consumed begin to occur. Therefore the condition for it to be <u>rechargeable</u> is that the reaction can go <u>forward or backward</u>, that is, it is <u>reversible</u>.

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1. chloroplast

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Which is a type of chemical weathering?
Goshia [24]

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The correct answer is D) hydration

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what is the molarity of a solution made by dissolving 18.5 grams of NaCL into enough water to reach 450.0ML​
ValentinkaMS [17]

Answer:

0.702M

Explanation:

Molarity of a solution, which refers to the molar concentration of that solution can be calculated thus;

Molarity = number of moles (n) ÷ volume (V)

Firstly, we convert 18.5 grams of NaCl to moles using the formula; mole = mass/molar mass

molar mass of NaCl = 23 + 35.5 = 58.5g/mol

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4 0
3 years ago
A balloon is launched when the temperature is 15 oC and the pressure is 0.918 atm. Its volume is 11.1 L. It rises in the air unt
ivann1987 [24]

Answer:

The new volume is 7,606.96 Liter.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas in the balloon = 0.918 atm

P_2 = final pressure of gas in the balloon = 0.0012 atm

V_1 = initial volume of gas in the balloon = 11.1 L

V_2 = final volume of gas in the balloon = ?

T_1 = initial temperature of gas in the balloon = 15^oC=273+15=288K

T_2 = final temperature of gas in the balloon = -15^oC=273-15=258K

Now put all the given values in the above equation, we get:

\frac{0.918 atm\times 11.1 L}{288 K}=\frac{0.0012 atm\times V_2}{258 K}

V_2=\frac{0.918 atm\times 11.1 L\times 258 K}{288 K\times 0.0012 atm}

V_2=7,606.97 L

The new volume is 7,606.96 Liter.

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