I’m pretty sure it’s true
<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it.
Crude appearance.
18 18 0
F --> O + e
9 8 1
Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts.
The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons.
Followed by a right arrow to indicate the direction the reaction is going.
Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons.
Followed by a plus sign to indicate more.
Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>
Other factors that affect how much light<span> is absorbed, reflected and </span>scattered concerns<span> the </span>light<span> itself and how it arrived at the surface of the </span>object<span>, such as the wavelength (color) of the </span>light<span>, and. the angle at which it reaches the surface.</span>
Answer:
Answers are in the explanation.
Explanation:
<em>Given concentrations are:</em>
- <em>SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60M</em>
- <em>SO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M </em>
- <em>And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10M</em>
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In the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are equilbrium concentrations.</em>
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Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are ACTUAL concentrations.</em>
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If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
- Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reaction
- Q = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the left
- Q = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
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