John Cuthber
Resident Experts-
Posts
18386 -
Joined
-
Last visited
-
Days Won
51
Content Type
Profiles
Forums
Events
Everything posted by John Cuthber
-
There's logically an energy requirement too.
-
Wow! Three doctors and they couldn't recognise someone laughing.
-
It's hard to say. Can you post a video of what you mean?
-
What has India got to do with it?
-
Is it safe to directly grow bacteria from hands?
John Cuthber replied to Benjamin Lee's topic in Biology
They typically aren't. However if you culture lots of them up, they would be present in large amounts. -
(Homosexuality) can I apply for artificial insemination programs?
John Cuthber replied to amnesyuck's topic in Genetics
It's a nice, traditional, romantic idea to have a child with the DNA from both parents. However, realistically, it's going to have two people's DNA. And since most people's DNA is practically identical* there won't be much difference whether the second contribution is from one of the couple or from someone else entirely. *(I'm told I share something like 99 % of my DNA with a chimpanzee, so I must share much more than that with any human) -
It's unrealistic for essentially the same reason that it's uneconomic to make gold in a nuclear reactor. It's possible- but the yield is so poor it doesn't cover the cost. This statement "Neutrons are for free in currently existing nuclear plants.." simply isn't true. If you take out too many neutrons you shut down the reactor. You get about 2.5 neutrons per fission, and you need at least one of them to start the next fission reaction. So, at best you can take about 60 % or so of them- call it near 50% to allow for an improbably high efficiency- just to give us a number to play with. It's about 1 "spare" neutron per atom. According to this https://www.physicsforums.com/threads/how-much-uranium-235-does-a-nuclear-power-generator-consume-to-generate-1-5-gw.360052/ a reactor converts something like half a ton of Uranium a year. Each uranium fusion could set free enough neutrons to make about 1 atoms of lithium. And each Li atom weighs about 1/24 as much as the uranium. So at best you could get something like 1/24 times the half a ton or so i.e. about 20 Kg a year of lithium from your reactor (blanketed and moderated entirely with boron). That's a mighty expensive by-product you have, especially once you have to extract it from the boron. There's another simpler argument for why it's clearly impractical. As you say B-10 + n0 -> Li-7 + He-4 + 2.795 MeV So each atom of Li you make produces about 3MeV of energy. If you take that atom of Li and put it in a battery you can use it to store about 3 eV of energy. And the best Li cells have lifespans of something like 10,000 cycles (it's typically a tenth of that). So in it's life time the battery will never store more than 1% (more likely 0.1%) of the energy that was set free by creating its lithium in the first place. Once you run the reactors to make the Li, you don't need the batteries any more- you can just use the energy from the reactors.
-
Well, if her biggest weaknesses have nothing to do with her ability to serve as president, whereas her opponent's biggest weakness is that he really isn't fit to run a bath, never mind a country, you have just explained why everyone should vote for Hillary. Cool, but next time you might like to try doing it without picking on epileptics.
-
Artificial Gravity and Spacetravel
John Cuthber replied to Airbrush's topic in Astronomy and Cosmology
Actually, that is simple and cheap. It's a very good idea. The only problem would be if the rope got hit by something and snapped, but space travel is always going to be horribly risky. -
You can't have got the maths right; the units are wrong. There's only about 10 times more B10 in the crust than there is Li. The neutron capture cross section fo 4He is pretty close to zero. The only way to get enough of it to react would be to put (already scarce) He in lots of huge nuclear reactors. Those rectors would produce enough power to pretty much make the Li cells redundant. The Li cell is great if you are concerned about energy stored per unit volume, or unit weigh. But for load levelling storage you don't need to worry so much about those. Something like this might work better. https://en.wikipedia.org/wiki/Vanadium_redox_battery
-
You have not showed that she lied. The FBI didn't believe that it could show that she lied. The most sensible explanation for the fact that nobody seems able to show that she lied is that she didn't. I presume you are rooting for the guy who tells the truth 4% of the time. http://www.politifact.com/personalities/donald-trump/ Are you saying that Clinton wasn't dishonest enough for your taste or what?
-
If there was a reliable way to grow taller everyone would know about it.
-
Lead is relatively expensive and so the battery business got pretty good at recycling it. I suspect they will do the same with lithium- which will help.
-
Not all humans are attracted to women. Also it's in the Daily Mail...
-
https://en.wikipedia.org/wiki/Proof_by_assertion (spoiler alert- it's not a proof)
-
I work with weather data sometimes. So my working files are pretty nearly random numbers.
-
LOL from the guy who first told him, that's funny.
-
Is it just me or does this sound a little like "I'm going to learn how to fly a plane; I will work on the landing some other time"?
-
Help a non native speaker of English understand this
John Cuthber replied to Alfred001's topic in The Lounge
Undoubtedly, but when a native speaker tells you "Do you realise that's not how we say it" you shouldn't say "Yes it is". -
Help a non native speaker of English understand this
John Cuthber replied to Alfred001's topic in The Lounge
It is, in fact, insulting. Whether you meant it to be or not. So, when (ironically in this thread) a native speaker of the language tells you something is insulting the correct response is not to tell them that it isn't. The correct response is to apologise,find out why it's a problem, and not do it again. -
Help a non native speaker of English understand this
John Cuthber replied to Alfred001's topic in The Lounge
Do you understand that saying "I recomend that you use your brain in such cases." is insullting? -
The stuff in green is too hard to read.
-
Well, this will tell you what micromolar means https://en.wikipedia.org/wiki/Molar_concentration Fundamentally, it's a bit like counting eggs in dozens, rather than one at a time. However, because molecules and atoms are really small chemists use a rather bigger number than 12. You might imagine that we would choose a "simple" number like a million billion billion. That would be fine if we counted molecules- but we don't (usually). We measure out stuff by weight. So we picked a number that's not a dozen, or a million billion billion, we chose 602,214,857,000,000,000,000,000 which seems an odd choice, but it's (ideally) the number of atoms in a gram of hydrogen. It's called Avogadro's number- named after an Italian scientist (strictly it's the number of atoms in 12 grams of carbon- because it's easier to measure that accurately but it's very nearly the same). Why choose that? Well, it's very difficult to weigh single atoms and molecules- but relatively easy to measure how much they weigh relative to one another. It makes sense to use the lightest atom- hydrogen- as a unit of mass. (it's sensible to have a unit that's roughly the same size as the things you are weighing- you don't weigh supertankers in ounces.) We call that unit the amu (atomic mass unit) So we know that the carbon atom, for example, weighs 12 times more than the hydrogen (strictly 12.0107) There are tables of atomic weights like this on the web. http://www.lenntech.com/periodic/mass/atomic-mass.htm The other thing you need to know is that atoms combine together (at least most of the time) in fixed ratios. So, for example, methane always has exactly 4 hydrogen atoms for each carbon atom. So we know that, since the hydrogens weigh 1 AMU and the carbon weighs 12 AMU the methane must weigh 12 +4 =16 AMU. Now let's imagine doing some chemistry- we burn the methane in air to form carbon dioxide and water. CH4 + 2 O2 --> CO2 + 2 H2O Each molecule of methane reacts with two molecules of oxygen to form a molecule of carbon dioxide and two molecules of water (oxygen atoms hang round in pairs- so they are O2 rather than just O. Quite a lot of gases do that) Well, that's fine- we know the numbers of molecules involved- but how could we work out the weights of the materials? The first step is to realise that you can multiply up that equation by any number you like So, for example Each zillion molecule of methane reacts with two zillion molecules of oxygen to form a zillion molecule of carbon dioxide and two zillion molecules of water. It also works (of course) with Avogadro's number which is referred to as L (I have no idea why it's L) So L molecule of methane reacts with two times L molecules of oxygen to form L molecule of carbon dioxide and two times L molecules of water. And by convention, we call that number of molecules a mole so One mole of methane reacts with two moles of oxygen to give one mole of carbon dioxide and two moles of water. Here's the clever bit. We chose L so that it was the number of atoms in a gram of hydrogen. So a mole of hydrogen atoms weighs 1 gram. By adding up the numbers we know that a mole of methane weighs 16 grams We can do the same with the other things in the equation Each oxygen atom weighs (from that table) 16 grams (near enough) So each O2 molecule weighs 32 grams And in the same way you can find that a mole of carbon dioxide weighs 12 + 28(16) = 44 grams and the water weighs 18 grams per mole. So we can see that 16 grams of methane reacts with 64 (that's two lots of 32) grams of oxygen to give 36 (that's twice 18) grams of water and 44 grams of carbon dioxide. So knowing the weights of the molecules in this way lets us calculate the weights of materials involved in a reaction. Instead of having lots of tiny masses to keep track of, we divide by the mass of a hydrogen atom right at the start. It's even more useful when you start doing reactions with chemicals in solution. You make the solutions with odd looking masses of chemicals and in doing so you get solutions where the volumes that react with each other are in simple simple ratios. You do that by making solutions with sensible numbers of moles per litre rather than sensible number of grams. For the stuff you are looking at they chose a two millionth of a mole per litre. I can look on the web and find out about thidiazuron. The formula is C9H8N4OS and I can add the masses of the atoms together to get the mass of the molecule - or I can cheat and find it on line There's a wiki page for the stuff Ihttps://nl.wikipedia.org/wiki/Thidiazuron It's in Dutch I think, but I can still find the molecular weight 220.25 odd grams per mole. So if I wanted to make a solution that has a 2millionth of a mole per litre I could weigh out about 110.125 micrograms of it and dissolve it in a litre of water. Well, I guess I could, but I'd need a good balance to weigh that small a quantity. I could also try weighing out 110.125 miligrams and dissolving it in a thousand litres of water- that's fine if I want a tonne of solution. Or I could take 110.25 mg of it and dissolve it in a litre of water (I suspect it wouldn't dissolve in water at this concentration so I would use a different solvent. One that's a good solvent, but not very toxic. I'd try alcohol) So I's have a solution that was a thousand times more concentrated than I wanted. But I could easily take a millilitre of this with a calibrated dropper, and dilute it to a litre with water. that would give me the right concentration without having to weigh out an absurdly small amount or make up a huge volume of solution.
-
limit to positive infinity of x minus x
John Cuthber replied to babipsylon's topic in Analysis and Calculus
I'm well aware that you think you know what you are talking about. I'm also aware that you made a claim but have not supported it in any way. Perhaps, before you go, you should explain what it is in the first two replies that indicates that they suppose that they are referring to a sequence. You seem to have forgotten that you said "all of written mathematical expressions are INCORRECT in this thread. because x is not a sequence but all members have written supposing so."