Wikipedia:Reference desk/Archives/Science/2016 January 7

= January 7 =

Why does $$\ 80 \cdot 10^{-6} = 8 \cdot 10^{-5}$$?
I'm trying to understand why $$\ 80 \cdot 10^{-6} = 8 \cdot 10^{-5}$$? What is the explanation? 92.249.70.153 (talk) 01:13, 7 January 2016 (UTC)


 * Not sure if this will enlighten you, but (1) Number a x number b = (ten times smaller number a) x (ten times larger number b), or (2) $$\ 80 \cdot 10^{-6} = 8 \cdot 10^{1} \cdot 10^{-6} = 8 \cdot 10^{-5}$$ Clarityfiend (talk) 01:43, 7 January 2016 (UTC)


 * Compare a simplified case: $$\ 80 \cdot 10^{-2} = 8 \cdot 10^{-1}$$. Do they look equal to you?--Denidi (talk) 01:45, 7 January 2016 (UTC)
 * Which is another way of saying 80/100 = 8/10, or 80 x 0.01 = 8 x 0.1. Clarityfiend (talk) 01:54, 7 January 2016 (UTC)


 * (ec)You should review Exponentiation. Here's one way to look at it. Pardon the Fortran-style notation:
 * 80 * 10**-6
 * = (80 * 1/10) * 1/10 * 1/10 * 1/10 * 1/10 * 1/10
 * = (8) * 1/10 * 1/10 * 1/10 * 1/10 * 1/10
 * = 8 * 10**-5
 * ←Baseball Bugs What's up, Doc? carrots→ 01:55, 7 January 2016 (UTC)
 * N.B. The $$ \cdot$$ is the dot product. Sleigh (talk) 01:58, 7 January 2016 (UTC)
 * While the dot sign can be used for the dot product, it can also be used to denoted regular multiplication, as explained in the first bullet point under Multiplication. Usage is determined by context, and in this case, where neither multiplicand nor multiplier are vectors, it is clear that regular multiplication and not the dot product is intended. -- ToE 07:27, 7 January 2016 (UTC)

Also see scientific notation. Bubba73 You talkin' to me? 02:52, 7 January 2016 (UTC)


 * Perhaps the negative exponents are getting you confused. So, let's look at an example that uses only positive exponents:  $$\ 7 \cdot 10^{3} = 70 \cdot 10^{2}$$.  In this example, both expressions mean 7,000.  The first expression means 7 times 1,000 and the second expression means 70 times 100.  Which both mean 7,000 as a final product. Your example with negative exponents is in the same vein.   Joseph A. Spadaro (talk) 08:14, 7 January 2016 (UTC)


 * Also, a suggestion. You might be better served at the Math Help Desk.  Joseph A. Spadaro (talk) 08:17, 7 January 2016 (UTC)


 * The simplest non-mathematically intense way to explain this is that the exponent on the 10 tells you "Move the decimal point that many places to the right" (or to the left if it's negative).  So if you take 80.0 and move the decimal point left 6 times, you get 0.00008 - and if you take 8.0 and move the point to the left 5 times, you get 0.00008...which gives you the exact same result.  It is common to use this notation in one of two ways - either:
 * Choose to make the exponent always be multiple of three (which makes it easier to mentally call it "thousands", "millions", "billions" and so forth - and also to use SI units with "milli-", "micro-", "nano-" and "pico-")...hence 80x10-6.
 * Or choose to always keep exactly one digit to the left of the decimal point at all times...hence 8x10-5.
 * When you mix those two conventions (as presumably happened in your example), the arithmetic is no different, the meaning is exactly the same - but everything becomes much more error-prone for us mere humans. SteveBaker (talk) 17:27, 7 January 2016 (UTC)


 * Or to state what you said more formally: $$80 = 8 \cdot 10^{1} + 0 \cdot 10^{0}$$ in base ten. $$0 \cdot 10^{0} = 0$$ of course; I included it for illustration. Thus, by the commutative property of multiplication and the identities of exponentiation, $$80 \cdot 10^{-6} = 8 \cdot 10^{1} \cdot 10^{-6} = 8 \cdot 10^{1-6} = 8 \cdot 10^{-5}$$. Remember, in base ten, every numeral "place" represents a power of ten. --71.119.131.184 (talk) 21:16, 7 January 2016 (UTC)
 * The fact that $$80 = 8 \cdot 10^{1} + 0 \cdot 10^{0}$$ is a confusing red herring, and is not helpful to the explanation.   D b f i r s   23:53, 7 January 2016 (UTC)
 * Thank you all, I got it thanks to your explanations. I appreciate your help. 92.249.70.153 (talk) 06:55, 8 January 2016 (UTC)

Carbon monoxide
Is there any Wikipedia article about mass carbon monoxide poisonings? In particular, there was one steam train that got stuck in a tunnel and the smoke killed everyone -- anyone happen to know when that happened? 2601:646:8E01:9089:F88D:DE34:7772:8E5B (talk) 05:39, 7 January 2016 (UTC)
 * The link is Balvano train disaster. The disaster happened in Italy in March 1944 (just after the fall of the fascist government - the chaos of that era explains why the train was overloaded and running on low-grade coal). Category:Deaths from carbon monoxide poisoning will help you find others, although the only other mass death in that category is the Myojo 56 building fire. Smurrayinchester 09:06, 7 January 2016 (UTC)


 * Thanks! That's what I'm looking for.  (Actually, I've already found it, but thanks for the effort anyway!) 2601:646:8E01:9089:F88D:DE34:7772:8E5B (talk) 13:03, 7 January 2016 (UTC)
 * Just to be helpful, do you want me to add the Myojo 56 building fire to Category:Deaths from carbon monoxide poisoning? 2601:646:8E01:9089:F88D:DE34:7772:8E5B (talk) 13:11, 7 January 2016 (UTC)
 * Good idea! But it's already in there. DMacks (talk) 16:37, 7 January 2016 (UTC)
 * What's also surprising is the number of Carbon Dioxide poisoning deaths...Lake Nyos makes interesting reading. SteveBaker (talk) 17:18, 7 January 2016 (UTC)
 * Actually, there's quite a few more mass deaths from carbon monoxide poisoning, besides Balvano and Myojo 56 -- in fact, many mining disasters involve mass carbon monoxide poisoning. (I've added the Senghenydd Colliery Disaster to the category, for the sake of accuracy.) 2601:646:8E01:9089:F88D:DE34:7772:8E5B (talk) 01:02, 8 January 2016 (UTC)
 * Getting back to Steve's comment about CO2 related deaths, I ran a search, and found this paper by the British health and safety executive on the potential for mass CO2 casualty incidents as carbon is trapped from industrial smoke stacks and other sources to mitigate global warming. Which begs the question "Why not capture the CO2 as carbonate?"  Carbonates are nice, tractable, industrially-useful solids.  The only real way to hurt yourself with some is to drop a 50-pound sack of it on your foot. loupgarous (talk) 07:42, 10 January 2016 (UTC)

Microwave oven entrance plate?
I have a microwave oven. It is 20 years old. It started throwing sparks and shooting small flames. The culprit was a small plastic (?) 'window' the goes over an opening inside the oven. There is a channel that conducts microwaves from the cavitron to the inside of the oven. This window/panel covers the place where this channel enters the interior of the oven. A fingernail sized portion of this cover had become burned, turned black and bubbly.

What is this cover made of? Why is it there? Why did it catch fire? Cpergielx (talk) 19:45, 7 January 2016 (UTC)


 * It's made of plastic or cardboard. My microwave oven has one made of cardboard. It's there to protect the cavitron. Plastic and cardboard are mostly transparent to microwaves, so the microwaves from the cavitron can pass through into the oven chamber, but matter will be blocked. Yours might have gotten contaminated with something, like oil or something metallic, that absorbs microwaves. --71.119.131.184 (talk) 23:57, 7 January 2016 (UTC)


 * I read that some part of the microwave has a lifetime of about 2000 hours so maybe after 20 years it's due replacement anyway. I know the ones at work have gotten way less powerful over the (four) years. --78.148.110.91 (talk) 05:43, 8 January 2016 (UTC)


 * It's the waveguide cover. It may be made of mica.  What you described seems to be a common problem: . --Amble (talk) 09:16, 8 January 2016 (UTC)

Dignitas death experience
According to the WP article, the method used by Dignitas puts people in a coma and takes about 30 minutes for them die. Does this mean they are liable to experience some kind of death dream like that in Jacob's Ladder? I know we can't know exactly what a dying person experiences but what can be inferred from people who came out of comas? --78.148.110.91 (talk) 23:38, 7 January 2016 (UTC)


 * Linking: Dignitas (assisted dying organisation).--Scicurious (talk) 00:02, 8 January 2016 (UTC)


 * Pentobarbital is humane enough for executions, at least in Texas. That does not mean the manufacturers would sell it for this purpose.
 * Regarding the experience of this suicide method: I don't know how this compares to the experience of coma patients. Nor do I know whether they will see their life flashing before their eyes. I'd rather compare this to a barbiturate overdose or a heroin overdose. Some people came back from that and can tell the story.
 * I also wonder why they don't use an elephant dose, and shorten the process to some seconds. Is that for the Jacob's Ladder effect? --Scicurious (talk) 00:22, 8 January 2016 (UTC)


 * For oral administration, it can already be difficult to swallow the necessary dose, according to this interesting article. And remember that a lot of terminal illnesses can interfere with functions like swallowing. Intravenous administration is another option, but of course then you need to insert an IV line, and I suspect there might still be an issue with administering large doses. People who receive large amounts of IV medication often have PICC or "central lines" inserted for that purpose. --71.119.131.184 (talk) 03:27, 9 January 2016 (UTC)


 * Maybe? Near-death experience is probably the article you want. Just for general interest, according to the article they use pentobarbital; barbiturate overdose is a fairly standard euthanasia/assisted suicide method. --71.119.131.184 (talk) 00:24, 8 January 2016 (UTC)

Follow-up question: any reason why they don't use nitrogen gas? That should at least be quicker? It sounds pretty mental but I'd probably prefer near instant mechanical destruction - I wouldn't want any neurones in contact with each other - it's the only way to be sure! 78.148.110.91 (talk) 01:55, 8 January 2016 (UTC)


 * The article on the organization says they've used helium gas asphyxiation in the past (inert gas asphyxiation is our article on the method). This verges into speculation (which is discouraged on the Reference Desk), but I think it's probably because it's more involved. You need the gas, breathing equipment, and you need people trained in using it, plus you need to ensure ventilation for the gas so it doesn't build up, which could kill people inadvertently. Administering a drug orally or intravenously is just simpler. --71.119.131.184 (talk) 03:27, 9 January 2016 (UTC)


 * Ack!!! I was given sodium pentothal, not the same, but similar to pentobarbitol, as an anaesthetic to have four impacted wisdom teeth extracted, and was clinically dead for a few minutes.  Assuming the experience is the same, it is horrific.  μηδείς (talk) 02:07, 8 January 2016 (UTC)


 * Do you mean the experience after you woke up, or during the procedure? Scicurious (talk) 02:18, 8 January 2016 (UTC)


 * @ μηδείς "... and was clinically dead for a few minutes" Do you mean with no cardiac or cerebral function? Was this an unforseen occurence? Did the 'horrific' part occur as you went into this state or as you came out? Richard Avery (talk) 07:47, 8 January 2016 (UTC)


 * "Full recovery of the brain after more than 3 minutes of clinical death at normal body temperature is rare." according to Clinical death.Abaget (talk) 09:42, 8 January 2016 (UTC)


 * Apparently, Medeis resuscitated. Scicurious (talk) 15:57, 8 January 2016 (UTC)


 * The doctor couldn't find my pulse for two minutes, I wasn't "declared" dead, so I really shouldn't have said clinically dead. The surgery was completed, and the horrible part was the coming to, although going under wasn't pleasant either.  Like swimming up from the black depths with no air in my lungs. I'll have to ask my dad about the circumstances, since the dentist actually recruited him to help revive me. μηδείς (talk) 17:16, 8 January 2016 (UTC)
 * "the dentist ... recruited him to help revive me" Dude, your daughter ikind of died few minutes ago, if you have a sec to revive her, that'd be nice.

CO2 filter that let's O2 through
Could a mask incorporate a CO2 filter, but let O2 through? After all, O2 is a smaller molecule. --Scicurious (talk) 23:52, 7 January 2016 (UTC)


 * Yes, rebreathers do just this. However they don't filter molecules based on their size. They use principles of chemistry, such as using a chemical that reacts with carbon dioxide but not oxygen. Atoms are tiny. Making filters the size of small molecules like carbon dioxide is something we can't really do very well at present. Now the neat thing is there are quite a few such filters, but they're not man-made. Ion transporters and ion channels in cells often work on the basis of atomic size, as well as other things like electric charge. In the future it's possible we might wind up using bioengineered filters made of human-designed enzymes. --71.119.131.184 (talk) 00:04, 8 January 2016 (UTC)

Zeolites are used to do something similar. http://pergelator.blogspot.com/2010/07/oxygen.html Cpergielx (talk) 05:53, 8 January 2016 (UTC)


 * Because of Graham's Law of effusion, every filter will differentially let through more O2 (MM = 32 g/mol) than CO2 (MM = 44 g/mol). That is, if I had a bag made of a material which which slightly permeable to gas (really tiny holes, slow leak, etc.) and filled it with, say, a mixture of CO2 and O2, as the bag slowly deflated, the O2 would leave faster than the CO2, so slowly the relative concentration of CO2 inside my bag would go up, and O2 would go down.  This is true regardless of the pore size, and is based only on the root-mean-square speed of the molecules at a given temperature:  At the same temperature, the lower mass oxygen molecules are moving faster, so at the same temperature and pressure, more oxygen molecules "hit" a hole in a filter than carbon dioxide molecules.  Even if the holes are many billions of times larger than an individual molecule, in the bulk, oxygen will always leak faster than carbon dioxide.  While it is true this does not meet the requirements of a perfect filter (letting ONLY oxygen through and NEVER carbon dioxide), with any permeable membrane, on the balance oxygen will always pass through the membrane more readily than carbon dioxide.  -- Jayron 32 16:19, 8 January 2016 (UTC)
 * Couldn't you use a zeolite filter to do some of the filtering? Graham's law also assumes non-turbulent diffusion I think. Yanping Nora Soong (talk) 22:21, 8 January 2016 (UTC)