MrHomeScientist
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The Trouble with Boron
I'm currently working on the next video in my element series, and this time it's boron. I've done this experiment before (see my old video here) and it went pretty smoothly, but my end result was visibly impure. Despite that, I'm pretty proud of that one because to my
knowledge it was the first on YouTube to feature making elemental B.
Anyways now that I've come back to that element for my series, I am revisiting the process. This first section will go over previous work, and then
I'll discuss current issues and what I've found doing research. The main problem I am having is obtaining good purity.
Previous Work
The main reaction is a sort of Goldschmidt (thermite) between boric oxide and magnesium:
B<sub>2</sub>O<sub>3</sub> + 3Mg == 2B + 3MgO
However there are side reactions that produce various magnesium borides:
Mg + 2B == MgB<sub>2</sub>
There are lots of different borides; the above is an example of one possible product.
The post-reaction mass is ground up and subjected to dilute hydrochloric acid to react away the borides and other products to leave elemental boron.
First, the water reacts with any leftover reactants:
Mg (s) + 2H<sub>2</sub>O (l) == Mg(OH)<sub>2</sub> (s) + H<sub>2</sub> (g)
B<sub>2</sub>O<sub>3</sub> (s) + H<sub>2</sub>O (l) == 2B(OH)<sub>2</sub> (aq)
Then, the acid takes care of everything else:
Mg(OH)<sub>2</sub> (s) + 2HCl (aq) == MgCl<sub>2</sub> (aq) + 2H<sub>2</sub>O (l)
Mg (s) + 2HCl (aq) == MgCl<sub>2</sub> (aq) + H<sub>2</sub> (g)
MgO (s) + 2HCl (aq) == MgCl<sub>2</sub> (aq) + H2O (l)
Mg<sub>3</sub>B<sub>2</sub> (s) + 6HCl (aq) == B<sub>2</sub>H<sub>6</sub> (g) +
3MgCl<sub>2</sub> (aq)
The goal being to leave boron as the only remaining solid. It's important to use dilute acid to begin with, because of the borane gas
produced. Borane is pyrophoric, so you want to slow down its production to avoid 'accidents'. I did all of the above in my old video, and everything
went pretty smoothly. Now, however, I'm realizing that my product is even more impure than I thought at the time.
New Work
In filming my new video on the subject, I changed the heating procedure. Instead of lighting it thermite-style, I instead put the reaction mix (4g
each of powdered B<sub>2</sub>O<sub>3</sub> and Mg) in my mini propane furnace to provide the required heat. This led to a bit
of a sudden "explosion" when it reached the right temperature, but still appears to have worked just fine.
I think it's really cool the flame color was green from the boron. You can see it also ejected several pieces of the products, which was scary the
first time when I was right next to it when it happened 
I'm not sure why it would do this. The main reaction doesn't produce any gases, which I'd think are required to do something like that.
I then powderized the reaction products, resulting in 7.35g of material. Based on 4g boric oxide, the theoretical max should be 1.25g B. I then put it
all under dilute HCl and let it stand for several days. After filtering, rinsing, and drying, the mass had decreased to 3.70g. I then put this under
full-strength 9.5M HCl for another few days, filtered and rinsed the products and am currently waiting for it to dry. By eye it looks to still be
quite a bit over the 1.25g maximum. I'm sure the real amount of B produced is quite a bit less, considering the side reactions.
Issues and Research
The whole focus of this new video is on purity. I wanted to make a nicer element sample than my first try. However I've been doing a lot of research
recently and it appears that the magnesium reduction is really not a great path to follow if you want pure B.
This paper has a good overview of this particular reaction scheme:
Elemental Boron & Magnesium Boride synthesis - Review by Neale R. Neelameggham
An excerpt:
| Quote: | | Moisson noted that one of the two magnesium borides formed is as resistant to water and hydrochloric acid as elemental boron causing difficulty in
making pure boron. However, he was able to make 98.30% pure boron using several purification steps. The preparation of crude boron with a 90 to 95%
purity practised even in the twenty-first century is called Moisson Process. |
Unfortunately that means that the acid digestion step will be ineffective at separating out all of the borides. I could find no other
references to the Moisson Process he mentions; if anyone has one I'd be very interested to read it.
From the Handbook of Inorganic Chemistry by Pradyot Patnaik:
| Quote: | | Boron does not react with water at ambient temperatures. The powdered amorphous form, however, reacts slowly at 100°C producing boric acid. The
amorphous metal reacts slowly with dilute mineral acids at ambient temperatures; the crystalline form is inert. The former, however, reacts vigorously
with concentrated nitric acid. The amorphous powder ignites in oxygen at 700°C. |
Powdered amorphous B is what is produced in just about every type of synthesis, or at least the amateur-friendly ones. I considered trying other acids
to get rid of borides, but the above shows that I'll be losing some B to the acid no matter which one I use, and hot sulfuric or nitric are
particularly bad. So acid treatment is not ideal.
Finally I found one other reference: Uses and technological processes for member registrants of boron.doc
This states (emphasis mine):
| Quote: | Amorphous boron is produced in a Moisson process from boron oxide and magnesium. Both raw materials are mixed and ignited to start the strongly
exothermic reaction. An excess of boron oxide is used as reaction moderator. The reaction block is cooled in a water basin and then crushed in a jaw
breaker and a roller crusher. Hydrochloric acid is used to remove remaining magnesium oxide and incompletely reduced boron oxide from the raw boron.
The excess of boron oxide is collected as boric acid. The boron is dried in fluidized bed driers to get boron with boron with 86%. Using a
higher excess of boron oxide another grade with 90% boron content is obtained.
The 86% boron grade can be cleaned further in a thermal treatment step with fluorides to convert insoluble by-products into soluble compounds which
are again removed by washing with hydrochloric acid to obtain boron with 95-07% boron content. |
So it's possible that using an excess of boric oxide may help with purity, although you still run into the problem of some borides not being
digestible by acids.
Finally, considering other possible reaction pathways, Neelameggham also mentions electrolysis:
| Quote: | H Davy in 1809 , and others in the following 40 years, electrolyzed fused borax[sodium borate] to boron [H. Davy, 1809], [R.D.Thomson, 1831], [F.
Wohler, 1856]. The electrolysis surmised the formation of metallic sodium which reduced boron oxide to boron element. Again the product was a powdery
material, containing less than 70% elemental boron along with adhering salts and oxides and other metallic compounds [possibly borides] even after
being washed with hydrochloric acid followed by rinsing with water.
...
Fused salt electrolysis approach using either melts containing borate or fluoborate as boron source with metal or carbon electrodes. Since the
electrolysis took place at temperatures much below the melting point of boron, the deposits were powdery and had to be refined further to get the
purity. Kahlenberg claimed making 100% pure boron powder from a melt of B2O3-K2O-KF [H.H. Kahlenberg, 1925]. Twenty five years later, Cooper showed
that 99.7% pure boron could be obtained from B2O3-KBF4-KF [H.S.Cooper, 1951]. Later studies showed the purity was lower than claimed earlier. Again,
the impurities in the product were oxygen, and electrode material difficult to separate from the boron product. Most of the boron by this technique
was still amorphous and non-crystalline.
|
So electrolysis of fused borax might be something to try and is doable for the amateur. The other exotic salt mixtures mentioned might be difficult to
come by. However, as seems common when electrolysis is mentioned, there is a conspicuous absence of reaction conditions, electrode material, or
applied potential/current. I also don't know what other products such electrolysis would produce. Sodium oxides or hydroxides maybe? This area of
chemistry is relatively new to me.
Summary
The magnesium reduction of boric oxide works to produce elemental boron, but the product is of poor purity and contaminated by magnesium borides. Some
of these are of similar reactivity to the boron itself, and so are very difficult to eliminate completely. I'd also like to try one of the other
routes mentioned by Neelameggham, but that will require no small amount of experimentation.
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woelen
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Interesting experiment. I once did this with H3BO3 instead of B2O3. The latter is hard to obtain and expensive, while H3BO3 is relatively easy to
obtain, especially through eBay.
Did you use commercial B2O3, or did you make it yourself? Making this is hard, heating of H3BO3 yields a glassy mass, which can best be described as
(HBO2)n, a polymeric species, sometimes called metaboric acid. Getting rid of the last amount of water to obtain B2O3 is hard.
I also obtained a lot of impure material. Leaching the resulting material with conc. HCl resulted in a lot of dark grey material (with a somewhat
bluish hue, but also spots with a somewhat brown hue). I suspect it to be boron, but of very low purity. I did not use dilute acid first, but I had no
dangerous production of gases or flames. The material did fizzle, but I think this is simply hydrogen gas, produced by reaction of the acid with small
amounts of left over magnesium.
The reaction indeed is violent. I used excess HBO2, relative to the magnesium. I expected that in that case you get no borides and any excess
magnesium reacts away with metaboric acid. In practice, things went less smooth. This probably is due to the fact that this is a solid/solid reaction,
with lots of unreacted material left over.
I did not keep the material. It simply was too impure for my taste and I did not see a possible way for me to purify the boron.
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MrHomeScientist
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I made my (assumed) B<sub>2</sub>O<sub>3</sub> by heating boric acid, stopping when the material stopped bubbling and hardened
into a glass. I had thought this meant complete removal of all water. You say you used boric acid - did you react that directly with magnesium? I
imagine the water driven off of the decomposing boric acid would make it extra violent!
Come to think of it, that may be what happened in my case. If it wasn't completely dehydrated to boric oxide, the last bit of water being driven off
in the furnace could be the gas that caused the mini-explosion.
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woelen
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I did not use boric acid as is. I heated until I got a glassy mass and allowed this to cool down. This I crunched with a clean hammer (which was a
nasty job, the material was quite hard and less brittle than I thought it would be). In this way you get something with empirical constitution close
to B2O3.H2O, (HBO2)n.
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Bert
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With goldschmidt reactions, even if the materials were perfectly dry, there is AIR in between the solids. Under rapid heating to extremely high
temperatures, the air EXPANDS. This alone is quite sufficient to blow some of your reactants out of the container.I
Additionally, why do you think this is a solid/solid reaction? It is certain that you are liquifying the Magnesium at such temperatures, possibly even
boiling it.
Pure solid/solid reactions are quite rare. If truly solid/solid, they usually proceed quite slowly and incompletely or demand extremely fine particle
size for sufficient contact between reactants to go to anywheres near completion.
(Edited due to being a chowderhead and commenting without first looking up all physical data of reactants & products)
[Edited on 5-12-2017 by Bert]
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2. List any points of agreement (especially if they are not matters of general or widespread agreement).
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unionised
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Quote: Originally posted by Bert  |
Also, you may be reaching 3,600 C and boiling some of the Magnesium oxide.
[Edited on 4-12-2017 by Bert] |
Above roughly 1500C you will vapourise B2O3
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MrHomeScientist
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Sheesh, I didn't realize I might be vaporizing some of my reagents. For what it's worth, I lit the furnace burner, placed the crucible in the
furnace about 20-30 seconds later, put on the cover, and about 45-60 seconds later the pictured reaction happened. That'd be a pretty fast rise in
temperature; helped by the strong exotherm of the main reaction too I'm sure.
I found a patent,US 4908196 A, on boric oxide production that does it in stages:
| Quote: | More particularly, according to the present invention boric oxide is prepared by dehydrating boric acid using a method comprising:
(a) gradually heating the boric acid, with the absence or substantial absence of fusion phenomena, at increasing temperature to a value not exceeding
about 150° C., operating at below atmospheric pressure, to eliminate water until the boric acid has been completely or substantially completely
converted into metaboric acid;
(b) gradually heating the metaboric acid obtained in stage (a), with the absence or substantial absence of fusion phenomena, at increasing temperature
to a value not exceeding about 400° C., operating at below atmospheric pressure, to eliminate water until the metaboric acid has been completely or
substantially completely converted into boric oxide;
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They even have a nice, detailed example section:
| Quote: | EXAMPLE
Crystalline boric acid is used with a crystal size of the order of 100 microns and a purity of 99.9%. 4.5 kg of this acid are fed into a stainless
steel reactor fitted with a stirrer, condenser, interspace for circulation of a diathermic fluid and a device for creating a pressure less than
atmospheric.
While slowly agitating the solid mass, the reaction pressure is reduced to 20 mmHg and the temperature is raised to about 115° C., over a time of 30
minutes.
Abundant water vapour development commences at this temperature. The temperature is raised gradually over a time of about one hour to 140° C., the
temperature at which water vapour development tends to substantially subside. Finally, the mass is heated to 150° C. at which heating is interrupted.
3150 g of orthorhombic metaboric acid are recovered from the reactor, with a yield of 98% with respect to the boric acid feed. This metaboric acid is
heated rapidly while stirring to 150° C. in the apparatus used for the first stage, and the temperature is then gradually increased by 2° C. every
10 minutes to about 205° C. During this treatment there is an abundant development of water vapour, which tends to subside beyond 205° C.
The temperature is finally raised from 205° C. to 250° C. over a time of 30 minutes, at which heating is interrupted. 2370 g of boric oxide are
recovered from the reactor with a purity of 99% and a yield of 93% with respect to the boric acid feed of the first stage.
The boric oxide obtained in this manner is in the form of a light, spongy friable mass which is removed without the need for mechanical action to
separate it from the reactor walls.
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So it might be worthwhile to try dehydrating my boric acid as I did previously, then taking the glassy product and subjecting it to higher heat in the
furnace. I won't really have any temperature control, though, so that may be detrimental. Boric acid is easy to come by though. They are trying to
avoid melting it in the patent, but I imagine once it's fully molten that's an indication that all water has been driven off.
Of course, all this still runs into the problem of inseparable borides. I think a different method, or info on that Moisson purification process, is
needed.
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DJF90
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Melting means less surface area which means incomplete dehydration to boric oxide. There is a preparation that is very similar to this in Inorganic
Syntheses.
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Bert
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Yes, quite right.
Reminds me to look up ALL compounds in a reaction's physical data before suggesting what's going on... And stop commenting before my second cup of
coffee too.
Rapopart’s Rules for critical commentary:
1. Attempt to re-express your target’s position so clearly, vividly and fairly that your target says: “Thanks, I wish I’d thought of putting it
that way.”
2. List any points of agreement (especially if they are not matters of general or widespread agreement).
3. Mention anything you have learned from your target.
4. Only then are you permitted to say so much as a word of rebuttal or criticism.
Anatol Rapoport was a Russian-born American mathematical psychologist (1911-2007).
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elementcollector1
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Found this while searching for electrolytic means of making boron:
http://jes.ecsdl.org/content/107/10/817.abstract
97.5% is pretty good purity, but at the cost of molten halide electrolysis - not exactly an amateur setup. Still, if you're considering alternative
routes, this might be worth looking into.
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MrHomeScientist
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Do you have access to the full paper? I'm interested in the details of their electrolysis setup. I'm pretty reluctant to deal with molten fluorides,
though!
=======
Interesting development on the magnesium front: The products of the last digestion in conc. HCl finally dried out, and weighed in at 1.00g exactly.
This is less than the theoretical max of 1.25g, meaning purity is at maximum 80%. It's probably significantly less than that in reality, but the fact
that I reduced it below theoretical mass is encouraging. I might stop there for the purposes of the video, since now I know that B itself will slowly
dissolve in acid too, further hurting my purity.
[Edited on 12-5-2017 by MrHomeScientist]
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elementcollector1
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Sadly, no, I only found the abstract through Google and have no credentials to access the full text. Maybe someone else here can help?
https://www.google.com/patents/US20130045152
The above suggests that boron halides can be reduced by metal borides to yield pure (more pure?) boron, which might allow you to add a 'counteracting'
portion of something like boron triiodide to your thermite mix to react with a suspected amount of magnesium boride. But then the problem becomes
making pure boron triiodide... Of course, if you do accomplish that, any old reducing agent would do the trick - zinc, magnesium, sodium,
etc. Zinc seems interesting, because according to the patent the zinc halide can be distilled off post-reaction.
Wikipedia states that boron triiodide can be prepared by reacting boron with iodine at high temperatures (~210 C), the boiling point of the compound.
There's no citation for this, but another paper claims it can at least happen at higher temperatures, around 600-1000 C.
http://www.sciencedirect.com/science/article/pii/00225088789...
This strikes me as a good amateur way to 'purify' your produced boron from the magnesium approach, by distilling off the triiodide and then reducing
it again. Bit of extra work, but if purity's your goal, it would probably go a long way.
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woelen
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Are you sure? I do not believe
this. I have done experiments with boron of high purity and never observed any reaction with acid, not even when allowed to stand for days.
Your product most likely is less pure than you expected and any oxide, boride or borate is leached out by the acid. Unreacted magnesium metal also
dissolves. Boron remains behind.
If boron really dissolves in hydrochloric acid (or any other non-oxidizing acid), then I learnt something new and I really would like to know a
reference to where this is described.
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MrHomeScientist
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| Quote: Originally posted by woelen | | Your product most likely is less pure than you expected and any oxide, boride or borate is leached out by the acid. Unreacted magnesium metal also
dissolves. Boron remains behind. |
That was the idea behind the process described in the OP, yes, but from what I've read it looks like it's no so easy (see the 'Issues and Research'
section of my first post). Oxide and borate should be easily taken care of, but "one of the two magnesium borides formed is as resistant to water and
hydrochloric acid as elemental boron."
The reference to boron reacting with acid also comes from that first post, from the Handbook of Inorganic Chemistry: "The amorphous metal reacts
slowly with dilute mineral acids at ambient temperatures; the crystalline form is inert."
If you never noticed anything, I suppose it's possible "slowly" is on the order of weeks.
It's taken far longer than I expected to react away the borides; the reaction products spent 3 days under dilute HCl, then another 3 days under
concentrated HCl to get where I am now. I don't want to heat the acid either, as B is specifically stated to react with hot nitric and sulfuric. I did
try heating the hydrochloric, but I foolishly left it uncovered and all the acid escaped!
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MrHomeScientist
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Found another good paper on boron: Boron and Refractory Borides, edited by V.I. Matkovich.
Page 214 starts the section titled Methods of Preparation of Amorphous Boron. I can't copy the text directly, but here's a picture excerpt:
Lots of great information there to digest. The main point, I think, is that an excess of boric oxide is required. It even mentions my experience of
the 'explosion': "Energetic interaction, characterized by a fast temperature increase and blaze appearance, takes place when the finely-divided
reagents are heated to the reaction temperature." Neat!
Also, page 208 mentions sulfuric and nitric acids reacting with B, although HCl is absent.
Finally, here's an excerpt about workup of the products:
"Magnesium boride impurities may be separated from boron by fusing the mixture with a large amount of boron oxide"
Lots to think about...
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MrHomeScientist
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Due to popular demand, I posted my video on boron anyway even though it doesn't produce good quality product: https://www.youtube.com/watch?v=4Y6H5p9wGh4
I hope to use the discussion here along with my research to inform a followup video.
=================
In the meantime, I did an experiment on making boric oxide. Woelen mentioned that without strong heating I may be making metaboric acid,
(HBO<sub>2</sub> n, instead of boric oxide,
B<sub>2</sub>O<sub>3</sub>. Essentially that corresponds to a bit of water remaining in the compound that's difficult to drive
off. I started by heating 10g of boric acid over a butane flame until it melted, and stopping when the mass became glassy and stopped bubbling (as I
did in the video). This yielded 5.42g of material. The target reaction here is:
2H<sub>3</sub>BO<sub>3</sub> == B<sub>2</sub>O<sub>3</sub> + 3H<sub>2</sub>O
So stoichiometry says that full dehydration of 10g should result in 5.65g boric oxide. Strange that I got less than that, but it's pretty darn close.
I then took 6.73g of this material (gathered from several runs), placed it in a metal crucible, and heated to red-orange heat in my propane furnace.
It foamed up a little but calmed down quickly, then bubbled sedately for a few minutes as the remaining water was driven off. After bubbling stopped I
let it cool to yield 6.51g. That's only a 3% difference, which indicates to me that I have gone practically all the way to
B<sub>2</sub>O<sub>3</sub> in the initial heating stage.
After cooling from the furnace, the boric oxide hardened to a glassy puck that I can't get out of the crucible. But the minimal weight change suggests
I don't need to do this step anyway.
The next thing I want to try is mixing this boric oxide with my boron/boride reaction products and heating to molten, to attempt to purify it as
mentioned above. I suppose I'll have to figure out a way to get the glass out of the crucible first though!
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elementcollector1
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Would it be feasible to simply heat the puck back up, and add the other reactants then? Might save you some trouble.
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MrHomeScientist
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Yeah that's possible. I'm somewhat concerned about another "energetic interaction" while adding it to the hot crucible, though! Plus I'll need to mix
things around manually, rather than mixing the powders before loading into the furnace. The good news is everything I'm using is very cheap and
readily available, so I can try it all.
[Edited on 12-11-2017 by MrHomeScientist]
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j_sum1
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I liked the copper trick. The B2O3 came off really cleanly. I wonder how much Cu impurity you actually pick up.
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MrHomeScientist
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Thanks! I was impressed with it too. Some insight from the comments suggested that (1) the molten boric acid acts like a flux and actually dissolves
the copper oxides, rather than reacting them away like I thought, and (2) the dark green color I observed during the second acid digestion could be
from dissolved copper chloride, which is dark green in a high [Cl<sup>-</sup>] environment and pale blue when diluted, which could explain
the disappearance of the color when I was filtering.
Side note: YouTube apparently removed the Annotations feature, which I used to use all the time. I'd add notes from comments like that to help improve
the video over time, but now I can no longer do that. I'd have to use the description, but nobody reads those.
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