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AnnyKZ [126]
2 years ago
15

How is the energy of the universe conserved during the combustion of gasoline in a car engine?

Chemistry
2 answers:
ExtremeBDS [4]2 years ago
8 0

The energy produced during combustion of gasoline in a car engine is converted as kinetic energy.

Explanation

The conservation of energy clearly states that "energy can neither be created nor be destroyed , it can only be transferred from one form to another form".

The combustion of gasoline in a car engine tends to generate heat energy, this heat energy will be transferred as kinetic energy which will lead to motion of cars.

This the energy of the universe will be conserved as the energy created by the combustion is used as kinetic energy for movement of cars.

Marianna [84]2 years ago
6 0
In a car driven by a gasoline combustion engine, heat energy is quickly converted into kinetic energy which results in the motion of the car.

According to the law of the conservation of energy, energy cannot be destroyed or created. It is can only be transformed from one form to another.
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Calculate the average time it took the car to travel 0.25 What is the average time it took the car to travel 0.25
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Explanation:

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You have a racemic mixture of d-2-butanol and l-2-butanol. the d isomer rotates polarized light by +13.5∘. what is the rotation
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L- isomer is considered as the Enantiomer of d- isomer and since the d-isomer optical rotation is + 13.5° so the optical rotation of l-isomer will be the same degree but with opposite sign which equal to -13.5° 

So the degree of rotation of racemic mixture will equal 0° 


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2 years ago
when a calcium atom loses its valence electrona the ion formed has an electron configuration that is the same as an atom of. (1)
andrew-mc [135]
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If the volume of a container of gas is reduced, what will happen to the pressure inside the container?
blondinia [14]
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What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH
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12.78 kJ

Explanation:

The correct balanced reaction would be

2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}

Mass of methanol = 3.24\ \text{g}

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass (32.04\ \text{g/mol})

\dfrac{3.24}{32.04}=0.10112\ \text{moles}

Enthalpy change for the number of moles is given by

\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}

\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}

The change in enthalpy is 12.78 kJ.

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