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steposvetlana [31]
3 years ago
7

What is the wavelength range,in nanometer,for infrared light? In what portion of this range does the earth receive IR from the s

un? What are the wavelength limits for the thermal IR range?
Chemistry
1 answer:
andrezito [222]3 years ago
7 0
  • The wavelength range of Infrared radiation is 700 nanometers to 1 millimeter.
  • The sun emits mainly near-infrared which is mainly composed of wavelength below 4 micrometers.
  • The thermal range of infrared ranges between wavelengths 3.5 and 2.0 micrometers

Explanation:

The wavelength range of Infrared radiation is 700 nanometers to 1 millimeter. This also translates to a frequency range of 430 TeraHertz  to 300 Giga Hertz.

Because the sun is a star and is hot in comparison to earth and other planetary bodies, the bigger  range of infrared radiation it emits is in the near-infrared which is mainly composed of wavelength below 4 micrometers.

The earth's surface produces infrared radiation of the mid-infrared range while cooler substances will produce far-infrared range

The thermal range of infrared ranges between wavelengths 3.5 and 2.0 micrometers and is produced by black bodies.

Learn More:

For more on infrared radiation check out;

brainly.com/question/2369243

#LearnWithBrainly

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Which of the following mixture types is characterized by the settling of particles?
torisob [31]
The answer is B. Suspension. Suspension mixtures are composed of two or more materials mixed together wherein the solute particles are usually larger than those found in a solution or colloid. In cases of solid-fluid suspension mixtures, the solid solute particles tend to settle at the bottom of the mixture after some time.
8 0
3 years ago
What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

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Answer:

No

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