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rosijanka [135]
2 years ago
7

_______ can travel through solid, liquid, and gaseous materials, whereas _______ can travel only through solid materials.

Chemistry
1 answer:
marishachu [46]2 years ago
7 0

<u>Sound </u>can travel through solid, liquid, and gaseous materials, whereas <u>S-Waves</u> can travel only through solid materials.

<h3>What is Sound ? </h3>

Sound is a form of energy. Sound moves through matter that is solid, liquid, and gas. In a vacuum sound cannot travel. In solids sound can travel more quickly as compared to liquid and gases because in solids molecules are closer together.

<h3>What is S- Waves ? </h3>

S-Waves are shear waves. These are the waves of high frequency and short wavelength. S- Waves can travel only through solid materials.

Thus from the above conclusion we can say that <u>Sound </u>can travel through solid, liquid, and gaseous materials, whereas <u>S-Waves</u> can travel only through solid materials.

Learn more about the Sound here: brainly.com/question/1199084

#SPJ4

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8 0
3 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
3 years ago
Select all statements that are correct:____
topjm [15]

Answer:

A, C and D are correct.

Explanation:

Hello.

In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:

P_{vap}^{solution}=x_{solvent}P_{solvent}

Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.

Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.

Best regards.

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