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likoan [24]
3 years ago
8

A rechargeable battery initially stores 3.6 kJ of chemical energy. Anna puts the battery in a flashlight, where 2.1 kJ of energy

are transformed into light and heat. Then, Anna puts the battery in a charger, where 6.8 kJ of electrical energy are transformed into chemical energy. How much chemical energy is now stored in the battery?
Chemistry
1 answer:
Tanzania [10]3 years ago
5 0

8.3 kJ of energy is remaining in the battery.

Explanation:

According to conservation of energy, the tendency of energy is to convert it from one form to another. So if the battery has an energy of 3.6 kJ at first and then it has utilized as 2.1 kJ of energy. So after loosing this energy, it will be 3.6-2.1 = 1.5 kJ of energy.

Again it is stated that the battery is recharged with 6.8 kJ of energy.

Then the modified or the latest energy stored in the battery will be the sum of energy present in the battery after using flashlight with the energy added in the battery by charger.

So, the total energy stored in the battery = 6.8+1.5 = 8.3 kJ

Thus, 8.3 kJ of energy is remaining in the battery.

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At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.600 M. N2(g)+O2(g)↽−−⇀2NO(g) If
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Answer:

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Explanation:

Hello,

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9

Now, the new equilibrium condition, taking into account the change x, becomes:

9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}

Thus, we arrange the equation as:

\frac{1}{9} (0.9-2x)^2=(0.2+x)^2\\0.09-0.4x+4x^2=0.04+0.4x+x^2\\3x^2-0.8x+0.05=0\\x_1=0.06

Finally, the new concentration is:

[NO]_{eq}=0.9-0.06=0.84M

Best regards.

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