Volatile organic compounds can be detected by hydrogeologists in the field or labs because of the odor of the vapors emitted from the groundwater and/or soil samples.
<h3>What are volatile substances?</h3>
Volatile substances are substances which can easily vaporize or change to gaseous state.
Volatile substances can either be solids or liquids but are mostly liquids.
Example of volatile substances include ether, petrol, chocolate.
The presence of volatile substances can be detected by the gases they release which may have characteristic odors.
Therefore, volatile organic compounds can be detected by hydrogeologists in the field or labs because of the odor of the vapors emitted from the groundwater and/or soil samples.
Learn more about volatile compounds at: brainly.com/question/25403770
[Co(CN)₆]³⁻ → Yellow
[Co(NH₃)₆]³⁺ → Orange
[CoF₆]³⁻ → Blue
Explanation:
- All the given compounds have octahedral geometry but the ligand in each are different with the same metal ion.
- Ligands strength order: CN⁻ > NH₃ > F⁻
- The ligand CN will act as a strong field ligand so that the splitting is maximum when compared to NH₃ and F⁻
- If the splitting is more, the energy required for transition is more, and the wavelength is inversely proportional to energy.
- So CN complex will absorb at lower wavelength (yellow color)
The volume is 3900cm^3 and the bricks density is 3900 ml
2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x
Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5
The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.
K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>
Coronavirus??? I guessed sorry :/
