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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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In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul
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<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m = molality of barium iodide solution = 0.175 m

m_{solute} = Given mass of solute (barium iodide) = 13.9 g

M_{solute} = Molar mass of solute (barium iodide) = 391.14 g/mol

W_{solvent} = Mass of solvent (water) = ? g

Putting values in above equation, we get:

0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g

Hence, the mass of water that should be added in 203.07 grams

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