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.
Answer:
Explanation:
There are three heat transfers involved.
heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0
q₁ + q₂ + q₃ = 0
m₁ΔH + m₂C₂ΔT + C_calΔT = 0
Data:
m₁ = 2.1 g
m₂ = 280 g
Ti = 25.00 °C
T_f = 26.55 °C
Ccal = 92.3 J·°C⁻¹
Calculations:
Let's calculate the heats separately.
1. q₁
q₁ = 2.1 g × ΔH = 2.1ΔH g
2. q₂
ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C
q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J
3. q₃
q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J
4. ΔH
Answer:
E, only one statement is correct
Explanation:
Absorption is favoured by heating the solution containing the colored impurity.
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A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water!
