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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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At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.
andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.


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4 0
3 years ago
PLEASE HELP
Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0
3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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3 years ago
Suppose you lift a 20 kg box by a height of 1.0 m.
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W=mgh W=(20)(9.8)(1) w=196J
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