Earth's greenhouse gases trap heat in the atmosphere and warm the planet.
Answer:
Kc = 3.1x10²
Explanation:
At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:
aA + bB ⇄ cC + dD
![Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).
For the reaction given, let's make an equilibrium chart:
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
1.1*10⁻³ 8.2*10⁻⁴ 0 <em> Initial</em>
-x -x +x <em>Reacts</em> (stoichiometry is 1:1:1)
1.1*10⁻³ -x 8.2*10⁻⁴ -x x <em> Equilibrium</em>
x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:
[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M
[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M
[FeSCN⁺²] = 1.8*10⁻⁴ M
Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])
Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)
Kc = 306 = 3.1x10²
<span>Chlorine has 7 valence electrons, but in the diagram, there are 2 chlorine atom making there a total of 14 valence electrons. There are 2 valence electrons in between the two atoms. There are 6 valence electrons around the outside of each individual chlorine atom. The electrons are arranged in groups of two around the outside of each atom.</span>
Answer: 2.75%
Explanation:
![pH=-log [H+]](https://tex.z-dn.net/?f=pH%3D-log%20%5BH%2B%5D)
![3.26 = -log [H+]](https://tex.z-dn.net/?f=3.26%20%3D%20-log%20%5BH%2B%5D)
![[H+] = 5.495\times 10^{-4} M](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%205.495%5Ctimes%2010%5E%7B-4%7D%20M)

initial 0.020 0 0
eqm 0.020 -x x x
![K_a=\frac{[H+][A-]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D)
![K_a=\frac{[x][x]}{[0.020-x]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.020-x%5D%7D)

![K_a=\frac{[5.495\times 10^{-4}]^2}{[0.020-5.495\times 10^{-4}]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5B5.495%5Ctimes%2010%5E%7B-4%7D%5D%5E2%7D%7B%5B0.020-5.495%5Ctimes%2010%5E%7B-4%7D%5D%7D)

percent dissociation = ![\frac{[H^+_eqm]}{[Acid_{initial}]}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B_eqm%5D%7D%7B%5BAcid_%7Binitial%7D%5D%7D%5Ctimes%20100)
percent dissociation=
Thus percent dissociation= 2.75 %