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September 28

Doldrums fact or fiction

When the winds are gone the sea actually has no swells, on a clear day the color of the sky is reflected in the water. At night the same effect, with no clouds or moon, gives one the effect of floating in space.

Is this statement from Doldrums actually true? Somehow this seems dubious to me...I've been under the impression that swells travel long distances across the ocean, thus why swell/waves from a hurricane between Puerto Rico and Bermuda can be a threat in the eastern United States. Ks0stm ^{(T•C•G•E)} 00:00, 28 September 2012 (UTC)[reply]

As originally posted,[1] it's quite obviously someone's personal opinion. I say zap it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:57, 28 September 2012 (UTC)[reply]

I agree. It's also unencyclopedic and poor English ("winds are gone", "actually" and "gives one the effect").--Shantavira|^{feed me} 10:48, 28 September 2012 (UTC)[reply]

The actual original statement was, "When the winds are gone the sea actually has no swells, on a clear day the color of the sky is reflected in the water. At night the same effect, with no clouds or moon, gives you the effect of floating in space. It's quite amazing to experience." Someone afterward tried to make it sound a tad more formal, but it's still just someone reminiscing about it, so I sent it to Davy Jones. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:58, 28 September 2012 (UTC)[reply]

Ketamine in the UK

Why is Ketamine not licensed for prescription in the UK? --TammyMoet (talk) 09:16, 28 September 2012 (UTC)[reply]

Are you sure about that? This suggests it is, or at least was in Scotland in 2009. Here is further evidence. It seems, though, that Ketamine is usually (only?) prescribed by hospital specialists and not local GPs - they want to keep you under observation whilst taking it, at least initially. - Cucumber Mike (talk) 10:43, 28 September 2012 (UTC)[reply]

That explains that then. Thanks. --TammyMoet (talk) 12:16, 28 September 2012 (UTC)[reply]

Van der Waals fugacity as a function of pressure, volume, and a and b. How do I eliminate molar volume?

The only equation for fugacity I can find for VDW requires calculation of the compressibility factor. However, the compressibility factor is often calculated from the molar volume. I am trying to plot fugacity against T and P of a Van der Waals gas, where constants a and b are known. How I do I convert the compressibility factor equation into a function of T,P, a and b? 128.143.175.21 (talk) 11:38, 28 September 2012 (UTC)[reply]

WHY DO PEOPLE ONLY GIVE COMPRESSIBILITY FACTOR AS A FUNCTION OF MOLAR VOLUME AND TEMPERATURE? HOW CAN I FIND IT AS A FUNCTION OF PRESSURE AND TEMPERATURE, IF MOLAR VOLUME IS UNKNOWN? 128.143.175.21 (talk) 11:44, 28 September 2012 (UTC)[reply]

Because of the Kronig-Classius equation - for any ideal gas, the molar volume is equal to:-

Vm = Ro T / P

where Vm is molar volume (m³); Ro is the universal gas constant (Avogradro Number x Bolzman constant) = 8.314 kJ/kmol.K, T is aboslute temperature (K); and P is pressue (Pa). Avogrado's number is the number of molecules in one mole of gas. Boltzman's constant relates kinetic energy to temperature for any gas molecule. Keit60.230.238.139 (talk) 15:53, 28 September 2012 (UTC)[reply]

I'm not dealing with ideal gases. 71.207.151.227 (talk) 14:01, 30 September 2012 (UTC)[reply]

I presume you are the OP again with a different IP address. It would be nice if you adopted a pen-name. I realise you were not talking about an ideal gas. But how real gasses can be handled is to apply ideal gas formulae and then use concepts such as compressibility factors to apply a correction to the ideal gas answer. Hecne the answer I gave is the answer to your question. Keit124.182.175.91 (talk) 02:00, 1 October 2012 (UTC)[reply]

If your gas is helium, a is nearly zero and it becomes easy to solve for molar volume. If not, the van der Waals gas equation can be arranged into a cubic equation in Vm. Cubic_equation#General_formula_of_roots gives the roots (1 real, 2 complex), but they're pretty messy. (Vm = Ro T / P is for an ideal gas, which this is not. Or equivalently it assumes the compressibility equals 1.)--Wikimedes (talk) 19:05, 28 September 2012 (UTC)[reply]

Banking your own blood for emergencies

On season 4 episode 11 of Breaking Bad, it's revealed that some of the characters had their own blood drawn, processed, and stored for emergencies. Is this ever done in real life? I can see the advantages of this approach over regular blood bags (no compatibility testing, retaining white blood cells, etc), but do these advantages actually outweigh the hassle? Apparently POTUS doesn't even get this kind of treatment.

I wouldn't be asking if this is any other TV show, but Breaking Bad had extremely accurate portrayal of science throughout so I'm giving the writers the benefit of doubt for now. A8875 (talk) 12:15, 28 September 2012 (UTC)[reply]

See our article on autotransfusion. Gandalf61 (talk) 12:42, 28 September 2012 (UTC)[reply]

Thanks, but I already read that article before asking the question and it doesn't say anything about preoperative autotransfusion. The entire article (except the first sentence) is about intraoperative autotransfusion, which has nothing to do with my question. A8875 (talk) 12:46, 28 September 2012 (UTC)[reply]

Then a quick Google search throws up this page (among many others) which says "Before surgery: You may schedule appointments and donate your own blood. This is known as autologous blood. It will be stored and reinfused as needed during or after your surgery". So, yes, it is done in real life. Gandalf61 (talk) 12:54, 28 September 2012 (UTC)[reply]

That's for planned surgeries. I'm asking about emergency surgeries. Suppose a VIP's life is in possible danger, does it make sense to draw blood from them on a weekly basis indefinitely? I'm sorry if I didn't phrase my question clearly, but I specifically mentioned "for emergencies" in my OP. A8875 (talk) 13:01, 28 September 2012 (UTC)[reply]

I found some use of the term "autologous self stored blood", e.g. here they say it may be useful for people with rare blood types. The blood keeps frozen for a couple of decades - and they'll want money for doing it for you. 88.112.47.131 (talk) 13:51, 28 September 2012 (UTC)[reply]

Thank you very much for the valuable data point. Frozen blood is completely unsuitable for emergencies according to that site: "Because of time required to thaw frozen Blood and deliver it to a where you are, frozen autologous Blood is not practical for emergency use. Thawing takes 60 to 90 minutes. Once your Blood has been thawed, it must be used within 24 hours."[2] So the verdict is still out for non-frozen autologous blood banking. A8875 (talk) 13:58, 28 September 2012 (UTC)[reply]

It makes no sense to me. By definition, in an emergency you need blood that can be got to you in the shortest possible time. As long as it's compatible it doesn't matter who it came from. Finding and then transporting your own blood from wherever it happens to be is bound to take far longer.--Shantavira|^{feed me} 14:07, 28 September 2012 (UTC)[reply]

Which is precisely why I asked the question. Whatever little benefits autologous blood transfusion has is completely overshadowed by the immediate availability of fresh blood from the blood bank. A8875 (talk) 14:24, 28 September 2012 (UTC)[reply]

I agree that the timing might be a problem, but don't agree with "As long as it's compatible it doesn't matter who it came from". People have gotten AIDS and hepatitis via blood transfusions from others, before blood banks were able to test for those conditions in donors. There might very well be other diseases which are currently not screened out of the nation's blood supply. So, if possible, using your own blood would be preferable. StuRat (talk) 02:35, 29 September 2012 (UTC)[reply]

How much biomass is in Canada?

I am looking for the total amount of biomass in Canada in forestry and agriculture.

This is what I have so far:

Forest Biomass: Standing Biomass / Bio-energy stock: The 245 M ha of Timber Productive Forest in

Canada has a biomass carbon stock of about 15,835 Mt C. This resource has an energy

content (566 EJ) that is equal to 69 years of Canada’s current energy demand that is met by

fossil fuels (8.24 EJ/yr).

Agricultural biomass: Energy EJ/Year: 1.96

Here is a link to the paper:http://www.biocap.ca/images/pdfs/BIOCAP_Biomass_Inventory.pdf. It was a report prepared for Industry Canada in 2003, so I don't know how much the numbers have changed since then.

Does anyone have any more up to date sources? Thanks. Eiad77 (talk) 19:46, 28 September 2012 (UTC)[reply]

I have no idea for sources whatsoever. But you might look for a workaround, such as an estimate that forests account for XX percentage of the biomass of Canada, and then work backwords from the total. It won't be hugely accurate, but it will be within an order of magnitude. It's the method we used in an undergrad class of mine to estimate the world's available fresh water supply from the volume of Lake Baikal. μηδείς (talk) 20:41, 28 September 2012 (UTC)[reply]

Since agricultural biomass in Canada consists entirely of annual plants, your numbers indicate that it constitutes only a fraction of 1% of the total biomass -- which would be expected, given that roughly half of Canada is covered with evergreen forests. So it really doesn't even come into play. The error in estimating forest biomass must be much larger than the total agricultural biomass. Looie496 (talk) 16:41, 29 September 2012 (UTC)[reply]

That's a good point. I guess it's better to just look at the forest biomass since it will be much more than agricultural biomass. Thanks for the help Eiad77 (talk) 12:37, 1 October 2012 (UTC)[reply]

Chemistry and Molecular Mass

I am having trouble with the following question

The percentage C by mass in the above unidentified substance is 92.3%. It may also contain H and/or O, but no other element. What is the molecular formula of the substance?

My answer was CH24 but it is not right. Any help on how to solve this question would be appreciated. Thank you — Preceding unsigned comment added by 142.58.63.10 (talk) 20:28, 28 September 2012 (UTC)[reply]

Clearly the non-C portion is 7.7% = 1/12.987, which is close to 1/13. Carbon has average mass 12, which means that for every 1 carbon, there's 1 mass unit of something else. So, obviously... this is a diamond doped with about one oxygen for every 16 carbon atoms. Or maybe an -OH side chain for every 17. Or ... well, there are a lot of possibilities; maybe the first part of the question clears it up. 24 hydrogens for every carbon, though, is not among them. But just to be really uncreative - if we had one carbon and one hydrogen you'd have about 7.7% not-carbon. Or two carbons and two hydrogens, or three carbons and three hydrogens ... can you think of a compound (or derivatives thereof) that might have that ratio? (But there are lots of possibilities) Wnt (talk) 21:13, 28 September 2012 (UTC)[reply]

CH24is obviously wrong since carbon has a valence of +/-4, not 24. So there has to be at least one oxygen molecule. You can work out the possibilities. CH3OH is one; CH2(OH)2 or CH3OOH is the next, and so on. I am not going to do the math for you, but that is the way to proceed. μηδείς (talk) 22:33, 28 September 2012 (UTC)[reply]

Unless I seriously slipped up here, I don't see a way for one carbon and 24 hydrogens to have anything but about 33% carbon, not 92.3%. Wnt (talk) 03:09, 29 September 2012 (UTC)[reply]

I didn't bother even to look at the math, the molecule is absurdly impossible given the valences. μηδείς (talk) 19:53, 29 September 2012 (UTC)[reply]

Is there any other information we're missing here? Because the problem you quote says "in the above unidentified substance". It looks like you've omitted some vital information that will help us solve the problem. --Jayron32 06:34, 29 September 2012 (UTC)[reply]

Everyone else here seems to have assumed you know the basics of how to solve this. I'm going to back up a little. First you need to find the atomic weights of the three elements involved: C = 12, O = 16, H = 1. Then you get the following formula for the percentage of carbon, by mass:

   100(12C)
 -------------
 12C + 16O + H

Then you can chart out the various possibilities and see which ones give you an answer near the expected one:

C   O   H    %C
-   -   -   -----
1   0   0   100.0% 
1   0   1    92.3%
1   1   0    42.9%
1   1   1    41.4%

So, of those, only CH gives the expected result. We could expand the chart to try different ratios of carbon, oxygen, and hydrogen to find solutions with oxygen in it (as some have done above). However, let's just stick with the CH answer (and it's multiples) for now. Is there a molecule with just one carbon and one hydrogen ? No, because carbon needs to have 4 bonds and hydrogen only has 1, so that leaves the carbon atom with 3 unused bonds. OK, let's try C₂H₂. This can work because the 2 carbon atoms can have a triple bond to each other, leaving them each with a single bond left over for a hydrogen:

       
H - C — C - H

So, there's one possible answer, C₂H₂, which is ethyne (acetylene). StuRat (talk) 07:19, 29 September 2012 (UTC)[reply]

Obviously that question has more than one possible answer. Benzene for instance also fits. Not a very good question for a homework in my opinion. Dauto (talk) 18:32, 29 September 2012 (UTC)[reply]

The basics helped me understand the problem better. I did not know how to approach this problem at first. Thank you. — Preceding unsigned comment added by 24.86.170.175 (talk) 19:35, 29 September 2012 (UTC)[reply]

You're quite welcome. I'll mark this Q resolved. StuRat (talk) 20:37, 29 September 2012 (UTC)[reply]

Resolved

C₆H₆ would be another possibility. - ¡Ouch! (^{hurt me} / _{more pain}) 07:05, 1 October 2012 (UTC)[reply]

I believe that answer has benzene already. :-) StuRat (talk) 07:26, 1 October 2012 (UTC) [reply]

Oops. Yes, C₆H₆ is benzene. :( - ¡Ouch! (^{hurt me} / _{more pain}) 08:05, 1 October 2012 (UTC)[reply]

GLP-1 agonists and exercise

Are there any sources which comment on the use of GLP-1 agonists like Victoza or Byetta and their use before or after, as opposed to just with exercise? μηδείς (talk) 20:38, 28 September 2012 (UTC)[reply]

Glancing over the articles, it looks like both these agents are injected once or twice a day, and so should be present at pretty high levels throughout all periods of activity and inactivity. Can you give further information about where you saw mention of the specific timing of exercise to begin with? (obviously exercise is good for diabetics and indeed non-diabetics) Wnt (talk) 21:26, 28 September 2012 (UTC)[reply]

There's no source I can refer to, or I'd look for it. (The question actually arose out of my interest in plant hormones, and having recently read about Gila Monster venom.) I suppose it's a question like the eat before or after you work out? dilemma. If someone works out once a day and takes a GLP-1 agonist once a day, it seems logical to wonder if the highest level of the day or the lowest level of the day affects the quality/benefit of the exercise or if pre- or post-injection exercise affects the efficiency of the hormone or not. Unfortunately, I have not been able to think of good search terms or find any resources that say anything other than people should, of course, exercise. μηδείς (talk) 22:26, 28 September 2012 (UTC)[reply]

Well, this being biology, we really don't know unless the experiment is done, and I'd bet money it hasn't been (I ought to look it up on PubMed, but it doesn't sound likely). A priori you could argue it a few different ways; my guess would be it's more effective to exercise when blood glucose is highest, because it's an available resource and because exercise causes muscles to absorb glucose more or less independently of insulin.[3] I should emphasize that this is purely a guess, and biology has an absolute right to give you an unexpected answer. Wnt (talk) 03:14, 29 September 2012 (UTC) (I eventually did look around in PubMed a bit - the most relevant paper I came up with was [4], but I still can't make up my mind what if anything this implies in regard to your question; only that natural GLP-1 itself has some effect on hepatic glucose production for exercise. Wnt (talk) 03:30, 29 September 2012 (UTC)[reply]

My entirely amateur suspicion is the same as yours, that exercising first, when the glucose levels are presumably highest, would be better. But I am surprised it is so hard to find anything on the topic. It just occurred to me that my own gym has nurses on staff, so if I get an opinion there I will post it here. μηδείς (talk) 17:27, 29 September 2012 (UTC)[reply]