Answer:
The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.
Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).
However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.
The heat lost is 
The heat lost when the ice is cooled from 400k to 263K can be calculated using the formula of heat transfer.
<h3>Heat Transfer</h3>
This is the heat transferred from a body of higher temperature to a body of lower temperature.
- Q = Heat Transfer
- m = mass = 1277g
- ΔT = change in temperature
We converted the temperature from kelvin scale into Celsius scale and find the change in temperature.
Solving for heat transfer
The heat loss is approximately
Learn more on heat transfer here;
brainly.com/question/16055406
Hello!
We have the following data:
ps: we apply Ka in benzoic acid to the solution.
[acid] = 0.235 M (mol/L)
[salt] = 0.130 M (mol/L)
pKa (acetic acid buffer) =?
pH of a buffer =?
Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 
So:
Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:
![pH = pKa + log\:\dfrac{[salt]}{[acid]}](https://tex.z-dn.net/?f=%20%20pH%20%3D%20pKa%20%2B%20log%5C%3A%5Cdfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%20%20%20)
Note:. The pH <7, then we have an acidic solution.
I Hope this helps, greetings ... DexteR! =)
Hey there!:
That depends on the pH of the water layer. If the water layer is basic or rather just not acidic, it will be in the water layer. If the water layer is acidic, pH 4 or less, it will be in the ether layer. On the question of upper and lower, either has a density of less than one so it will be the upper layer.
