To much letters and numbers
TLC means Thin Layer Chromatography. It is a method that can best be described as "Affinity-Based" used in the separation of compounds that are in a mixture.
<h3>What is unreacted p-nitrobenzaldehyde?</h3>
Unreacted p-nitrobenzaldehyde is simply an organic aromatic compound that contains a nitro group para-substituted to an aldehyde. in this case, if it is unreacted, that means it is the same as before the chemical reation.
Note that the question is missing key information hence the general answer.
Learn more about TCL at:
brainly.com/question/10296715
Answer:
In aqueous solution the pH scale varies from 0 to 14, which indicates this concentration of hydrogen. Solutions with pH less than 7 are acidic (the value of the exponent of the concentration is higher, because there are more ions in the solution) and alkaline (basic) those with a pH higher than 7. If the solvent is pure water, the pH = 7 indicates neutrality of the solution
Explanation:
PH is a measure of how acidic or basic a liquid is. Specifically, from a dissolution. The acidity of a solution is essentially due to the concentration of hydrogen ions dissolved in it. In reality, the ions are not found alone, but are in the form of hydronium ions consisting of one oxygen molecule and three positively charged hydrogen. PH precisely measures this concentration. And to do it, we can use simple and very visual methods.
Answer:
0.92 kg
Explanation:
The volume occupied by the air is:
![35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L](https://tex.z-dn.net/?f=35.0m%5Ctimes%2035.0m%20%5Ctimes%203.2m%20%5Ctimes%20%5Cfrac%7B10%5E%7B3%7DL%20%7D%7B1m%5E%7B3%7D%20%7D%20%3D3.9%20%5Ctimes%2010%5E%7B6%7D%20L)
The moles of air are:
![3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol](https://tex.z-dn.net/?f=3.9%20%5Ctimes%2010%5E%7B6%7D%20L%20%5Ctimes%20%5Cfrac%7B1.00mol%7D%7B22.4L%7D%20%3D1.7%20%5Ctimes%2010%5E%7B5%7Dmol)
The heat required to heat the air by 10.0 °C (or 10.0 K) is:
![1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J](https://tex.z-dn.net/?f=1.7%20%5Ctimes%2010%5E%7B5%7Dmol%20%5Ctimes%20%5Cfrac%7B30J%7D%7BK.mol%7D%20%5Ctimes%2010.0%20K%20%3D%205.1%20%5Ctimes%2010%5E%7B7%7DJ)
Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:
![5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}](https://tex.z-dn.net/?f=5.1%20%5Ctimes%2010%5E%7B7%7DJ%20%5Ctimes%20%5Cfrac%7B1kgCH_%7B4%7D%7D%7B55.5%20%5Ctimes%2010%5E%7B6%7D%20J%20%7D%20%3D0.92kgCH_%7B4%7D)
Answer:
.000064565
Explanation:
H3O+=10^-pH
- Hope that helps! Please let me know if you need further explanation.