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3 years ago

The less dense areas in a sound wave are called _____.

Chemistry

2 answers:

Rasek [7]3 years ago

4 0

Answer:

rarefaction is the correct answer.

Explanation:

Rarefaction is defined as the reduction in the density item.

Rarefaction takes place mostly when the particles present in the gas become extra scattered out.

Sound need to a medium to travel and the air is the most common medium. As the vibrating object move in the forward direction, the air gets compressed and creates a region of high pressure and the high-pressure area is called compression. This compression begins to move apart from the vibrating object.When the vibrating object moves in the backward direction, it creates less dense areas called rarefaction.

Thus The less dense areas in a sound wave are called rarefaction.

Vitek1552 [10]3 years ago

3 0

The less dense areas in a sound wave is called a rarefaction or rarefactions if there are multiple.

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Explanation:

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4 years ago

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mars1129 [50]

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Explanation :

First we have to calculate the mole of $C_2H_2$ gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $C_2H_2$ gas = 748 mmHg - 23.8 mHg = 724.2 mmHg = 0.953 atm (1 atm = 760 mmHg)

V = Volume of $C_2H_2$ gas = 5.50 L

n = number of moles $C_2H_2$ = ?

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T = Temperature of $C_2H_2$ gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$0.953atm\times 5.50L=n\times (0.0821L.atm/mol.K)\times 298K$

$n=0.214mol$

Now we have to calculate the moles of $CaC_2$

The balanced chemical reaction is:

$CaC_2(s)+2H_2O(l)\rightarrow C_2H_2(g)+Ca(OH)_2(aq)$

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_2H_2$ gas produced from 1 mole of $CaC_2$

So, 0.214 mole of $C_2H_2$ gas produced from 0.214 mole of $CaC_2$

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