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2 years ago

2. Describe How are the two forms of energy carried by waves and matter related?

Chemistry

1 answer:

Kazeer [188]2 years ago

7 0

Answer:

the energy of matter implies the motion of masses and the energy of the wave has no moving masses

The energy of the particles (matter) is the ability to do some work, therefore the energy can be of motion in the form of kinetic energy or in a configuration of the system called potential energy, the sum of these two energy is constant .

The wave is formed by a disturbance of the medium by matter, therefore for the formation of the wave matter supply energy, for example: in the form of movement, in the form of fluctuation of electric or magnetic field, etc.

The waves after being formed can move away from the matter that formed them, transporting the energy that alternately has kinetic and potential energy, but the total energy is constant.

Therefore the energy in matter is due to the movement of the same and the energy in the wave does not require the movement of matter, so it is a more efficient way of doing work.

In conclusion, the energy of matter implies the movement of masses and the energy of the wave has no moving masses.

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First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

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$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$