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beks73 [17]
2 years ago
11

Which wavelength of light is the best choice when attempting to quantitatively relate solution absorbance and concentration?.

Physics
1 answer:
ANEK [815]2 years ago
6 0

The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq)

you should choose a wavelength with maximum absorbance. In this case, you are using the scattered light, not the absorbed light as your signal. So you should avoid wavelengths where there are absorption peaks.

<h3>What is wavelength ?</h3>

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm).

  • The distance between two waves' crests serves as an illustration of wavelength. When you and another person have the same overall mindset and can easily communicate, that is an example of being on the same wavelength.

Learn more about Wavelength here:

brainly.com/question/10750459

#SPJ4

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<h3>How much centripetal force is there in a centrifuge?</h3>

Centripetal force is the force that pushes an item in the direction of its center of curvature. It is fundamental to how a centrifuge operates.

<h3>On a roller coaster, what is centripetal force?</h3>

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6 0
1 year ago
52. How did people spend Life in the<br>hunting age-​
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7 0
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A published hypothesis:
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4 0
2 years ago
A spelunker is surveying a cave. She follows a passage 140 m straight west, then 290 m in a direction 45? east of south, and the
zimovet [89]

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3 0
3 years ago
*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co
sp2606 [1]

Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

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