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May 31

Question about coordination compounds

How can I know how many ligands coordinated with a metal? For example, a d-metal cation from a salt dissolved in water: how is possibile to estimate how many water molecules coordinate with metal?--87.9.210.194 (talk) 08:12, 31 May 2020 (UTC)[reply]

See Metal ions in aqueous solution. And there is a relatively recent review from Ingmar Persson on this. However you must also bear in mind that most of the 4d and 5d metals have water easily displaced from the coordination sphere by other ligands and therefore it is difficult to find their cations (try perchloric acid media to avoid this problem, IIRC). Double sharp (talk) 10:03, 31 May 2020 (UTC)[reply]

Perpetuum Mobile

I found one! No of course I didn't. :) But I don't see why it wouldn't work.
Take a (high) column with a heavy gas in it and at the bottom some water with electrodes in it. Hydrogen and oxygen will rise to the top. There, a fuel cell combines the two, producing water. This water now has an increased potential energy because of the height, with which energy might be produced. The electricity from the fuel cell is used for the electrolysis at the bottom.
Of course the fuel cell and the electrolysis wiil not be 100% efficient, but that is independent of the height, so that can't explain it. The length of the electrical wire between the two is dependent on the height, but I doubt that that would explain it.
I chose a heavy gas because else the oxygen might not rise enough. In Earth's atmospere one could use the oxygen at the top, but that might then provide an explanation. I doubt it, but this way I avoided that issue.
So where is the error in my reasoning? DirkvdM (talk) 12:13, 31 May 2020 (UTC)[reply]

You already said it: "Not 100% efficient." ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:31, 31 May 2020 (UTC)[reply]
- That doesn’t really address what the OP wrote. Brianjd (talk) 12:41, 31 May 2020 (UTC)[reply]
I’m not sure what effect the height would have exactly, but I know this:
- The potential energy won’t be recovered by the fuel cell. It will remain with the water (to put it in simple terms) and will be lost when the water falls back to the bottom, where it can undergo electrolysis again.
Therefore, any loss (even if independent of the height) means that the fuel cell won’t produce enough energy to electrolyse the same amount of water. With less water electrolysed, there will be less hydrogen and oxygen to run the fuel cell at the top, meaning it will produce even less energy, and so on.

Brianjd (talk) 12:41, 31 May 2020 (UTC)[reply]

Perhaps I should have said “power” instead of “energy”. Brianjd (talk) 12:43, 31 May 2020 (UTC)[reply]

Why it does not work even in theory is that it takes more energy to convert water to hydrogen and oxygen at a higher pressure than at a lower pressure. A simpler version of this involves water vapour in the earth's atmosphere, which goes up and then condenses to liquid or solid water and then falls as rain or snow. Heat is exchanged in this process. Graeme Bartlett (talk) 12:54, 31 May 2020 (UTC)[reply]

Ah, that could indeed explain it. The gained potential energy depends on how high the hydrogen and (especially) oxygen would rise, which depends on the amount and specific mass of the other gas, which both would influence the pressure on the water. If the efficiency of the electrolysis decreases at the same rate the pressure rises then it would indeed not work.

The article on Electrolysis mentions that the efficiency depends on pressure, but I don't see a further explanation. (Maybe an idea to add that?) So I don't understand why this is the case, but in the real world there is oxygen in the atmosphere. The hydrogen rises higher, so the water could be produced at a higher altitude. Might this be a way to extract energy from the distribution of gases in the atmosphere? DirkvdM (talk) 14:17, 31 May 2020 (UTC)[reply]

@DirkvdM: Regarding your first paragraph, it doesn’t matter, because you don’t get to turn that potential energy into electricity. Please see my comment above. I’ll leave the second paragraph for someone else to respond to. Brianjd (talk) 14:24, 31 May 2020 (UTC)[reply]

Of course the potential energy of the water can be harvested somehow, eg with a turbine. For a perpetuum mobile persé that is not necesary, but utilising excess energy could save the world and make me the new Shell. :) DirkvdM (talk) 15:32, 31 May 2020 (UTC)[reply]

Everyone else is making good comments, but I can’t get over this potential energy thing. How are you going to prevent the gases from resisting the turbine on the way up?

I’m still confused about that second paragraph in your previous comment. Brianjd (talk) 02:14, 1 June 2020 (UTC)[reply]

It turn out that even if you assume zero friction, zero electrical resistance, etc. the best you can do is something that comes so close to moving perpetually that you can't measure the energy loss. Two solid objects orbiting each other alone in deep space would be an example of this. But you can never get a surplus of energy that you can harvest forever. If your calculations say that you can, you missed something. For a great example of something that seems like it might work, see Maxwell's Demon. --Guy Macon (talk) 16:35, 31 May 2020 (UTC)[reply]

Yes, I know a perpetuum mobile can't work out of principle. That was indeed my point. See the first line of my question. DirkvdM (talk) 07:42, 1 June 2020 (UTC)[reply]

There appears to be some confusion about my idea.

First, there is the closed energy system with the electrolysis which produces the hydrogen for the fuel cell, which produces the electricity for the electrolysis, etc, etc. This is the perpetuum mobile, which of course wouldn't work because of friction and efficiency never being 100%.
But then, there is the water that is produced at the top. This increases the potential energy of the water (which might be harvested in some way, eg with a turbine in a tube through whicht the water falls). This was what confused me, but Graeme Bartlett explained that (if I understood it correctly) even under ideal circumstances the gas pressure needed to make the hydrogen rise makes the electrolysis less efficient, giving the energy loss that compensates for the energy gain.
However, then I thought about using the oxygen in the atmosphere instead of the oxygen produced by the electrolysis. With only the lighter hydrogen rising, less gas pressure is needed, making the electrolysis more efficient, circumventing the issue. Of course, this is not a perpetuum mobile, because there apparently is an energy potential in the composition of the atmosphere, which is harvested by this system. Oxygen is produced at the bottom and bonded at the top.

On a side note, the oxygen at the bottom might be used in a green house. Eg for food production. Or just to take CO2 out of the air, helping to solve climate change. DirkvdM (talk) 07:42, 1 June 2020 (UTC)[reply]

Even if the efficiencies of electrolysis and fuel cell are 100%, it won't work. The redox potential of a half-cell depends on the concentrations of the chemicals involved. As the pressure drops, so does the concentration, so that the fuel cell at the top will provide a lower voltage than the electrolysis cell uses. Using atmospheric oxygen won't help, as the concentration of the oxygen at your fuel cell will still be lower than at your electrolysis cell (and even lower than if you had used the pure oxygen from the electrolysis cell, so it will, in fact, be worse). PiusImpavidus (talk) 08:05, 1 June 2020 (UTC)[reply]

Ah, I was already wondering what any effects of the lower air pressure might be. But is this a fundamental problem? Electricity is not a strong point of mine. Could it be solved by just using more fuel cells? Note, though, that this is an open system. If it works at all, then it should keep on working because any used oxygen (or hydrogen) will be replenished (assuming air currents), so the pressure remains constant and nothing fundamentally changes.

Or, from a different angle, can't one simply use a voltage converter? Of course that would introduce more inefficiency.

On a practical note however, with a production of 1 l of water per second at an altitude of 1 km, the produced power would be 1 kg/s * 1000 m * 10 m/s² = 10 kW. Taking inefficiencies into account, that would probably be about 1 kW (or less). So to have the same power production as a medium sized wind turbine (1 MW), 1 m³ of water should be produced per second. That's a lot. The wind turbine might be a more practical thing to build. Ah well, another 'brilliant' idea down the drain. :) DirkvdM (talk) 08:40, 1 June 2020 (UTC)[reply]

Outsourcing of engineering systems integration.

Is the outsourcing of systems integration now common to all engineering industries as it is becoming in civil engineering? As in the top level client just specifies and oversees what they want whilst the integration is outsourced to a tier 1 integration supplier Clover345 (talk) 17:14, 31 May 2020 (UTC)[reply]

All engineering industries? I do a lot of engineering involving putting five cent microelectronics into toys, and the only systems integration we do is when the IT department rolls out an upgrade. Also, we don't have any "top level client" who generates specifications. We have a bunch of children who bug their parents for the latest toy. I have also worked for companies that make hydraulic/pneumatic components. Engineering at those companies is pretty much all systems integration, but is never outsourced because nobody else knows how to, say, put together a pneumatic system that puts out 20 burritos a second. It is important to realize that there are a wide variety of engineering jobs out there. It is always amusing watching someone assume that every computer has megabytes of RAM when I work with systems that have 64 nybbles (not bytes) of RAM. --Guy Macon (talk) 19:02, 31 May 2020 (UTC)[reply]

@Guy Macon: I think it's more common for an 'ordinary user' to expect tens of gigabytes of RAM in a todays computer rather that a few megabytes. :) --CiaPan (talk) 19:21, 31 May 2020 (UTC)[reply]

Well yes, “all” is a broad generalisation but was just wondering if it happens a lot in other industries. Clover345 (talk) 19:45, 31 May 2020 (UTC)[reply]

What's ironic is that that "ordinary user who expects tens of gigabytes of RAM" mentioned above is probaly typing on a keyboard that contains a computer with 4K of RAM, using a mouse that contains 2K of RAM, looking at a monitor that might have 512K of RAM -- but maybe a lot more -- eating toast made with a toaster that has 512 bytes of RAM, drives a car that has several megabytes of RAM, etc. etc. Nowadays even light bulbs have computers inside them. Tiny computer are all around you but (if I do my job correctly) you don't recognize, say, your thermostat or digital watch as being a computer. --Guy Macon (talk) 00:44, 1 June 2020 (UTC)[reply]

Not in automotive. The OEM is the system integrator. In fact GM have or had a formal prototype phase called Integration Prototype, not a bad idea, at least you know what it is for. However sometimes OEMs outsource the complete design and development of models to third parties, in which case I suppose they are the system integrator. For various reasons it's not a great idea unless they are also going to manufacture it. say Magna, Stey_Puch among others do that sort of thing. Greglocock (talk) 23:31, 31 May 2020 (UTC)[reply]

Polyunsaturated fat degradation speed

What's the most unsaturated 22-carbon cis-fatty acid on Template:Fatty acids who's triglycerides take at least 3 seconds to degrade to the naked senses when first touching air? So you have an essentially inert bottle, airtight, fill it with pure triglyceride (without free fatty acid, foreign fatty acid, preservative or any other impurity) that has never touched gas besides neon and its own vapor and break the seal (in the dark if need be) well before a human could tell it's degrading (use a Star Trek transporter if you have to) and (after the surface is touching mostly air instead of neon) taste the surface or sniff it or whichever sense is best, I'm guessing you could tell by flavor before stickiness, sight, stirring resistance, viscosity eyeballing etc. but who knows. How soon can you detect the degradation without instruments if you know what to check? Even a lower bound for double bond count would be nice.

2. Same thing but 20 carbons, 18, 16, 24 and any other biologically important numbers.

3. Why does butter degrade so fast on the kitchen table in both appearance and flavor compared to extra virgin olive oil without the cap when butter has almost no double bonds at all? Sagittarian Milky Way (talk) 20:27, 31 May 2020 (UTC)[reply]

Regarding 3, I believe the reason is olive oil contains high levels of polyphenol antioxidants that, well, inhibit oxidation. They're also what's responsible for much of olive oil's characteristic taste and mouthfeel. Butter is refined and mostly fat. The oil still does degrade over time when exposed to air. --47.146.63.87 (talk) 22:34, 31 May 2020 (UTC)[reply]

It also depends on what we mean by "degradation." Butter, for example, is a mixture of milk fat that is almost entirely saturated, but that's only about 80% of the content. The rest is milk solids, which aren't fat, and water. Water isn't generally something that mixes well with fats, and worse with saturated fats. In butter, it exists as an emulsion, but as you leave it out at room temperature, that emulsion will break overtime and you will be left components that don't want to mix. Get some ghee, which is much more pure milk fat, and you won't see this happen nearly as much. So, is the degradation that we see from the fats degrading, or from other non-fat components degrading or separating from the fats? --OuroborosCobra (talk) 23:12, 31 May 2020 (UTC)[reply]

So olive oil is slower than its double bond amounts imply (which explains why linseed oil is so much faster with not that much more bonds) and butter is not even a real fat but like a fat-water frappé. Sagittarian Milky Way (talk) 20:29, 1 June 2020 (UTC)[reply]

Butter is mostly fat, and the part that is fat, is really fat. There are techniques that process butter to remove most of the water and milk-solid content, such as Ghee and clarified butter. But it is incorrect to say "not even fat". It would be better to say "not only fat". Still mostly fat. --Jayron 32 18:16, 2 June 2020 (UTC)[reply]

I knew about ghee/clarified butter and assumed the unclarified kind wasn't udderly, udderly free of water and milk solids but would've guessed a few percent hydrophilic stuff tops, at least in the modern machine-made stuff. 20 percent surprised me. So I guess if you see water on old butter it isn't all condensation or oxidation products. Sagittarian Milky Way (talk) 00:10, 3 June 2020 (UTC)[reply]