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maria [59]
3 years ago
6

Can someone help please.

Chemistry
1 answer:
vovangra [49]3 years ago
5 0
I think that the answer is B
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Compare the evaporation of a contained liquid with that of an uncontained liquid.
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Answer: all other conditions equal, the rate evaporation of a contained liquid will be slower than the rate of evaporation of an uncontained liquid.


Justification:


1) The rate of evaporation increases as the surface area of the liquid (relative to the whole content) increases. This is, the greater the surface is the faster the evaporation.


2) That is so because the higher the surface of the liquid the more the number of particles in the liquid that are in contact with the surrounding air and so the more the particles will escape from the liquid to the air (which is what evaporation is).


3) A liquid contained will take the form of the container, so part of the liquid wil remain below the surface, while an uncontained liquid will spread all over the surface and so pratically all the liquid is in contact witht the air surrounding it.

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3 years ago
Write the parts of the ear in their correct order of transfer of vibrations. to the brain.
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Answer:

The sound waves travel from the outer ear and in through the auditory canal, causing the eardrum, or tympanic membrane, to vibrate. This, in turn, causes the three small bones, known as the ossicles, or the hammer, the anvil and the stirrup, in the middle ear to move.

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How do water, wind, and thermal energy from the sun move about the earth's surface?
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NH4+H2O →NH3+H3O<br><br> Which one is acting weaker base?
Nataliya [291]

Answer:

NH3

Explanation:

In solution, they are in equilibrium. NH4+ acts as a bronsted Lowry acid and donates an H to become NH3, and NH3 acts as a bronsted lowry base and accepts an H. In this pair, NH3 is a weak base, which gets its basic character due to the presence of lone pair of nitrogen and its ability to donate it.

8 0
2 years ago
Suppose that you have a 60.0% solution of NaOH. How many milliliters of water must be added to 30.0 mL of this solution to prepa
defon

Answer:

  • <u>21.4 ml (second choice)</u>

Explanation:

<u>1) Data:</u>

a) C₁ = 60.0% (initial solution)

b) V₁ = 30.0 ml (initial solution)

c) C₂ = 0% (pure water)

d) V₂ = ? (pure water)

e) C₃ = 35.0% (final concentration)

<u>2) Formula:</u>

  • C₁V₁ + C₂V₂ = C₃V₃
  • V₁ + V₂ = V₃ (assuming volume addtivity)

<u>3) Solution:</u>

<u />

a) Substitute values in the first formula:

  • 60.0% × 30 ml + 0 = 35% (30 ml + V₂)

b) Solve the equation (units are supressed just to manipulate the terms)

  • 18 = 10.5 + 0.35V₂

  • 0.35V₂ = 18 - 10.5 = 7.5

  • V₂ = 7.5 / 0.35 = 21.4 ml ← answer    
3 0
4 years ago
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