Answer:
Explanation:
You should allow the solvent to drop to the level of the adsorvent, so it would never run dry.
When you let your sample to run dry it will never finish to flow from the adsorbent depending of it polarity.
Water should not be used because it can dissolve the adsorbent.
You could use another technique to identify the compound, as an infrared or a ultraviolet detector. You can also, if you know the compounds, identify it for the retention time, for example, if you need to detect two compounds, one more polar than the other, and use a polar adsorbent and a non-polar solvent, the first compound to exit the column will be the less polar one, because it will have a bigger interaction with the solvent than the stationary phase (adsorbent) and will go faster, the second will be the more polar one, because it will have a bigger interaction with the stationary phase.
Oxidation reaction is used in redox reaction
<span> under extreme heat or pressure
crystallize from magma
precipitate from a solution reaction of hot mixture with water and a dissolved substance</span>
Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
![Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_%7B3%7D%5D%20%5Ctimes%20%5BH_%7B2%7DS%5D%20%7D%7B%5BNH_%7B4%7DHS%5D%7D%20%3D%5Cfrac%7B9.56%20%5Ctimes%2010%5E%7B2%7D%20%20%5Ctimes%207.62%20%20%5Ctimes%2010%5E%7B2%7D%7D%7B0.196%7D%20%3D3.72%20%5Ctimes%2010%5E%7B6%7D)
Gold is actually very soft. If rings were made of pure gold they would bend and become disfigured. Other elements are added to give the ring its stability.
