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ArbitrLikvidat [17]
3 years ago
9

Which waves move by replacing one particle with another? A) light waves B) sound waves C) electrostatic waves D) electromagnetic

waves
Chemistry
2 answers:
ANEK [815]3 years ago
8 0
<span>Answer: B) sound waves</span>

<span> There are 2 classifications of waves according to the media of propagation.</span>

<span> <span>Mechanical Waves – needs a medium, an example is sound waves. Sound waves do not travel through a vacuum. Sound waves move by replacing one particle with another. Particles vibrate and that is why we hear the sounds.</span> <span>Electromagnetic Waves and electrostatic waves are one but the same. They do<span> not require a medium.  – an example is light waves.</span></span> </span>
Bumek [7]3 years ago
7 0

i think it is B) because Sound waves move by replacing one particle with another. Sound waves require a medium to propagate, or move. Light and electromagnetic waves do not require a medium. Electrostatic are not a type of wave.

hope this helps

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tester [92]

Answer:

The state of matter of each compound or molecule is indicated in subscript next to the compound by an abbreviation in parentheses. For example, a compound in the gas state would be indicated by (g), solid (s), liquid (l), and aqueous (aq).

Explanation:

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3 years ago
Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

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As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

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\Delta T_{b} = i\times K_{b} \times m

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i = Van't Hoff factor = 1 (for chloroform)

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Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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