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alexandr402 [8]
3 years ago
15

If the above two waveforms were sound waves, we would hear the ___________ wave louder. If the above two waveforms were light wa

ves, we would see the ________ wave dimmer.
A) red, red
B) green, red
C) red, green
D) green, green

Chemistry
2 answers:
nalin [4]3 years ago
8 0

Answer:

B) green, red

Explanation:

larisa86 [58]3 years ago
5 0

B) green, red

Explanation:

If the above two wave-forms were sound waves, we would hear the green wave louder.

If the above two wave-forms were light waves, we would see the red wave dimmer.

What determines the loudness of a sound wave?

A sound wave is a disturbance that transmits sound energy from one point to another in a series of rarefaction and compression.

The amplitude of a sound wave determines the loudness of sound.

Amplitude is the vertical displacement a wave undergoes from its rest point. We can see that the green wave-form has a large vertical displacement. This implies a louder sound would be hear.

When the amplitude of sound increases, it is heard more loudly because it is propagated with more intensity. This is why green wave is louder.

Why would red be dimmer?

Still using the same analogy as sound, a light wave is an electromagnetic wave or radiation. Light is a part of the electromagnetic spectrum which photoreceptors in the eye are highly sensitive to.

The more energetic the burst of light waves, the more brighter it is is picked by the receptors in the eye.

Bright light has more energy and photons in it.

The energy carried by light is proportional to the square of its amplitude. The more energy a light carries, the more photons it will have. This makes it have more brightness.

The red wave-form has a lower amplitude and will be dimmer to the eye.

learn more:

light waves brainly.com/question/8032392

Sound waves brainly.com/question/2845448

#learnwithBrainly

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What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

3 0
3 years ago
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the answer is :the concentration the chemicals in the solution because the more ions there are in the solution the higher the conductivity also the more ions there are in the solution the stronger the electrolyte

6 0
2 years ago
Dd is recessive or dominant
slava [35]

Answer:This would be heterozygous, so both dominant and recessive alleles are written.

Explanation:Heterozygous means that the dominant and recessive alleles are written genotypically.

3 0
3 years ago
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The measure of the attraction that an atom has for electrons involved in chemical bonds is known as:
Mariana [72]
Electronegativity is the tendency of atom to attract shared pair of electron towards itself . Its value is out of 4 , and Fluorine is higly electronegative.
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D electronegativity, is the correct answer.
7 0
3 years ago
Please help thank you! And how many significant figures does it have?
zheka24 [161]

Answer:

0.779 cube centimetres

Explanation:

density = mass divided by volume

d=\frac{m}{v} rearrange the equation to solve for volume: v=\frac{m}{d}

plug in the corresponding values:

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