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May 12

Enthalpy of Reaction for Ethylene and Water

I'm working through the problems in my Organic Chemistry text book, and came across one where I'm confused as to what has gone wrong. The question and solution are shown here. The question asks me to calculate the enthalpy of reaction for ethylene and water, and then state whether it is exothermic or endothermic. Using the values stated in the table (and also given in their solution shown in the image I linked) I calculated the enthalpy as +39KJ/mol, and thus characterize the reaction as endothermic. In their solution, they appear to have gotten the sign backwards and have said the result is -39KJ/mol and therefore characterized it as exothermic. I tried checking the known value for the enthalpy of the reaction, and various sources give -44KJ/mol and state it is exothermic. So the solution given in the text conforms closely to the known value but the reasoning seems wrong. One thing to note is that there is no value for the bond dissociation enthalpy of the C-O bond created in ethanol in the table they reference, but both in my answer and in the text book's solution, we simply used the BDE value for the C-O bond in methanol. If this bond has a significantly different BDE to the one we actually need, then it might explain what's gone wrong here. 61.247.39.121 (talk) 00:50, 12 May 2018 (UTC)[reply]

Good point -- the numbers they tot up come out with the opposite sign. To summarize the values:

Source	C=C	O-H	C-H	C-O	total
Theirs	349	497	422	385	+39
Wikipedia	~270*	460	423	n/a	?
Statemaster	261*	460	410	350	-39

Where the sources are your image, Bond-dissociation energy, and www.statemaster.com/encyclopedia/Bond-dissociation-enthalpy (they are blacklisted, but they came up top in a search and have a handy looking table, though I have a suspicion it was once ours if I looked at the history). Note our article, or at least its BDE table as it is now, kind of sucks. Asterisks are inferred by taking C=C minus C-C.

The message I'm getting is that there's a certain amount of sage smoke and tossing of dried chicken bones involved in this computation, though I can't rule out it could give a scientific result in the hands of a competent practitioner. Wnt (talk) 00:56, 13 May 2018 (UTC)[reply]

Thanks for the response. The BDE table they give in the text book is here. I can't see where they cite their source for the data. Presumably they didn't redo weeks of calorimetry work to determine values that can easily be looked up. The text book itself is Organic Chemistry, 8th Edition by Brown, et al. 61.247.39.121 (talk) 05:31, 13 May 2018 (UTC)[reply]

They definitely used values from that table, in which the big disagreement here is C=C. They say CH2=CH2 has double bond energy of 727 kJ/mol, but Statemaster says 611 and Wikipedia has "611-632". That said, this paper works out at 728 kJ/mol (chart 1, page 6, given as 174.1 kcal/mol). (another source doesn't count because it's the book in question). A more serious source is [1] which gives the dissociation of ethylene to "^eqCH2" is 720.96 kJ/mol at 0 K and 730.59 kJ/mol at 298.15 K, with precision of +- 0.24 kJ/mol. The same for ethane is 367.87 at 0 K and 376.66 at 298.15K, with precision of +- 0.19 kJ/mol -- with a caveat that they don't specify what it dissociates to. Which gives a difference of 353.09 to 353.93 ... which vindicates your textbook where the table is concerned, if not the sign of their mathematics. Everyone must be agreeing on something else that is actually wrong! Wnt (talk) 19:49, 13 May 2018 (UTC)[reply]

When all of our errors align so as to give the false impression of a theory that accurately describes our observations of the physical nature of the universe, that's where the real science happens! 202.155.85.18 (talk) 00:26, 14 May 2018 (UTC)[reply]

In another section of the text, this passage appears. The bonds being broken in the enol are a sigma O-H and the pi bond of C=C, and to form the ketone a sigma C-H bond and the pi bond of a C=O must form. From their own table the BDE of the sigma bond being broken is 468kJ/mol and the sigma bond being formed is 439KJ/mol. That's fairly similar I guess, though 29KJ/mol is a big enough difference to drive a reaction. Their table doesn't give any values for C=O bonds, but from elsewhere (after conversion from kcal to KJ), it's apparently 351KJ/mol. Compared to their value of 349KJ/mol for the pi component of the C=C, they're far more similar in energy than the sigma bonds are. I'm starting to wonder if I should find another text book... 202.155.85.18 (talk) 00:50, 14 May 2018 (UTC)[reply]

The Bill

Yesterday I received a letter from my electricity company requesting a meter reading. It included this boxed comment:

Warning: only read your meter if it's safe to do so

Are there any hazards of reading electricity meters consumers haven't been told about (apart from possible shock experienced on opening the resultant bill)? 81.139.216.102 (talk) 09:16, 12 May 2018 (UTC)[reply]

I'm in the UK, I assume you are too. There's a certain injury rate amongst US meter readers, where they're reading US style meters that are mounted on exterior walls, in weatherproofed cases. Some of these are high up, some are obscured by overgrown shrubbery (usually kudzu). The meter reading trade press (yes, that's a thing) sometimes mentions this in their, "We can't believe what those crazy Americans get up to" columns. The UK equivalent are meters in old cellars, with failing steps down to them.

There's also a hazard for reading damaged meters, with failing insulation.

All of these are rare, but as there are so many meters, even the rare becomes common. Andy Dingley (talk) 09:24, 12 May 2018 (UTC)[reply]

Regarding Andy's last point, cf. the totalitarian principle. --76.69.47.55 (talk) 10:26, 12 May 2018 (UTC)[reply]

Condoms failing despite perfect use

Why do condoms sometimes fail to prevent pregnancy despite perfect use? — Preceding unsigned comment added by 172.116.86.16 (talk) 09:33, 12 May 2018 (UTC)[reply]

Every packet comes with an information leaflet. These leaflets explain that the product is not 100% reliable, and they also explain why. 81.139.216.102 (talk) 09:47, 12 May 2018 (UTC)[reply]

Condom#Causes_of_failure. DroneB (talk) 12:12, 12 May 2018 (UTC)[reply]

Bird

Curious about this bird, spotted in Warsaw, Poland. Had yellow or light-colored beak, greenish neck and walked faster than pigeon, not jumped. Not as timid as sparrows. Thanks. Brandmeister^talk 18:24, 12 May 2018 (UTC)[reply]

Common starling? Henry ^Flower 19:50, 12 May 2018 (UTC)[reply]

That is a set of mighty unclear photos. I can safely rule out flamingo, harpy eagle, and wandering albatross, but not much else. :-) Common starling meets your description, at least. Matt Deres (talk) 22:40, 12 May 2018 (UTC)[reply]

Most likely, thnx. Brandmeister^talk 22:49, 12 May 2018 (UTC)[reply]

Agree it's likely to be a starling - they scavenge amongst the sparrows and pigeons in London. Alansplodge (talk) 11:56, 16 May 2018 (UTC)[reply]

May 13

What do these symbols mean?

Are the symbols in this medical formula units of measure? If so, what are they? SpinningSpark 11:21, 13 May 2018 (UTC)[reply]

"f" means fluid. The squiggle means ounce (see the bottom of the table at List of abbreviations used in medical prescriptions or Uncia (unit)). "i" is the roman numeral one, and when there are more than one "i", some styles of writing replace "j" as the last character (ij = 2, iij = 3, iiij = iv = 4). So one fluid ounce of the first two ingredients, and two fluid ounces of the third. PDF file here shows more related symbols. 85.76.74.235 (talk) 12:47, 13 May 2018 (UTC)[reply]

See Apothecaries' system. --catslash (talk) 16:25, 14 May 2018 (UTC)[reply]

May 14

Route of administration of aspirin

When does it make sense to administer aspirins through the rectal route? It is harsh on the stomach, but despite this, I haven't ever heard of anyone taking them this way.--Hofhof (talk) 00:18, 14 May 2018 (UTC)[reply]

When the patient can't swallow, or has other issues with oral administration. See rectal administration. It could also be administered intravenously, but any IV intervention carries risks; you don't want to put in an IV just for some aspirin. (Note: I believe this question is fine for the ref desk. It's not asking how to diagnose or treat something.) --47.146.63.87 (talk) 01:11, 14 May 2018 (UTC)[reply]

Aspirin is harsh on the stomach, but this is caused by the aspirin in the blood; some of this will enter the cells of the stomach well where it knocks out certain enzymes that protect the stomach wall against stomach acid. Count Iblis (talk) 02:09, 14 May 2018 (UTC)[reply]

I've previously read that the route has a strong cultural component as well. Rectal medications are strongly selected against in North America, for example. (Ref). I can confirm from experience (I have a kid who never learned to swallow pills) that adult-strength painkillers are essentially only available to the public (in Canada, at least) in oral pill/tablet form - no syrups, suspensions, or suppositories. Matt Deres (talk) 18:58, 14 May 2018 (UTC)[reply]

The increased risk of gut bleeding still occurs with the rectal route], so that doesn't solve the problem. Klbrain (talk) 21:53, 14 May 2018 (UTC)[reply]

Does monoamine oxidase inhibitors not change the level of acetylcholine?

I was told that monoamine oxidase inhibitors does not change the level of acetylcholine but I'm not sure about that because this site states that "monoamine oxidase, an enzyme that has a key role in the metabolism of the monoamine neurotransmitters acetylcholine, dopamine, epinephrine, norepinephrine, and serotonin." and it seems that it does change its level. Isn't it? 2A02:ED0:6F01:8600:154E:FD5:6431:60CA (talk) 12:53, 14 May 2018 (UTC)[reply]

I think that site is in error. Acetylcholine is not a monoamine and is not metabolized by MAOs. There will be some indirect effects, because levels of those other transmitters affect the release of acetylcholine, but there should not be any substantial direct effect. Looie496 (talk) 14:14, 14 May 2018 (UTC)[reply]

Agree that that site is in error; acetylcholine doesn't have the aromatic ring which is a defining feature of the monoamine neurotransmitter. Also the Guide to Pharmacology doesn't list it as a monoamine oxidase substrate. Klbrain (talk) 21:38, 14 May 2018 (UTC)[reply]

Definition of protein complex

If a single protein associates with RNA (like Cas9) can that be called a complex or is a complex only something comprised of multiple proteins (the article suggests the later but I wanted to be sure) 129.215.47.59 (talk) 17:18, 14 May 2018 (UTC)[reply]

That would be called a nucleoprotein complex. Someguy1221 (talk) 17:31, 14 May 2018 (UTC)[reply]

What sets the northern limit of spring-ripening barley of the Passover-postponement rule phenotype?

I've heard it's not very winter-tolerant. How deep would the snow cover or cold have to get to kill it (or at least prevent it from ripening in the Jerusalem manner?) How much liquid precipitation would it need? (applicable to very continental farming areas) Could you grow it that way in like Bowash or Ireland or is there not enough sun? Sagittarian Milky Way (talk) 20:00, 14 May 2018 (UTC)[reply]

Haven't been able to answer directly but see:-

The Timing of the Barley Harvest in Israel, a comparison of Biblical and modern barley farming.
THE ISCA GRAIN, A ROMAN PLANT INTRODUCTION IN BRITAIN describes finding barley grains which "...may be classified as belonging to a six-row spike {Hordeiim vtdgare L.)" amongst other cereals datable to A.D. 80-130 in Wales, (a climate very similar to Ireland).
Ancient barley DNA gives insight into crop development: "An international group of scientists have analysed the DNA of 6,000 year old barley [found by the Dead Sea] finding that it is remarkably similar to modern day varieties".

Alansplodge (talk) 22:26, 16 May 2018 (UTC)[reply]

Of course barley might be a summer crop in Wales instead of winter like Israel. Sagittarian Milky Way (talk) 23:16, 16 May 2018 (UTC)[reply]

It is a winter crop in Wales now, but in Roman times, who knows? Alansplodge (talk) 17:35, 17 May 2018 (UTC)[reply]

May 15

Space launch, part 2

I have asked this question already (how long does it take at Baikonur to prepare a Soyuz rocket for launch), but it was answered only partially (that it takes 2 days from roll-out to launch). So here's the other part of the question: What is the minimum amount of time it takes from the beginning of payload encapsulation and mating to the rocket until roll-out of the rocket to the pad? I know it can't take longer than 5 days (from the document previously shown), but can they do it any faster? 2601:646:8A00:A0B3:441F:C8FF:AED4:31F7 (talk) 11:46, 15 May 2018 (UTC)[reply]

Finding off-target cleavage sites of Cas9 by dCas9-ChIP: biased technique?

This paper discussing CRISPR-Cas9 features this sentence: "Although a number of studies have employed sequence similarity-based off-target search or dCas9-ChIP to predict off-target sites for Cas9, such approaches cannot assess the nuclease activity of Cas9 in a comprehensive and unbiased manner."

My question is: how could dCas9-ChIP be biased? 129.215.47.59 (talk) 13:43, 15 May 2018 (UTC)[reply]

I have not accessed the paper, so this is a terrible answer, but my first thought is that if the dCas9-ChIP relies on binding, then it won't necessarily match up with where Cas9 actually cuts. It might bind in some way where it can't cut at all, or it might have a site where it has somewhat lower affinity for the uncut DNA sequence but somewhat higher affinity for the transition state. Wnt (talk) 14:23, 15 May 2018 (UTC)[reply]

My reading of that sentence is not that the Cas9 is biased, but that the off-target searches are. Fgf10 (talk) 19:26, 15 May 2018 (UTC)[reply]

Half-cell basics review

This isn't a real test question, but if it had been, maybe I'd keep how reduction potentials and half-cells work clearer in my head. My memory has corroded...

A man is wearing a gold bracelet on one wrist and a platinum bracelet on the other. He is coated with sweat and an unspecified but non-corrosive and non-valuable grime from work which is the same on both sides. A perfect voltmeter is used to measure the potential difference between the bracelets. The reading will be:

a) zero, because neither of these metals reacts with anything unless a voltage is applied.
b) zero, because neither metal has ions in solution.
c) 0.32, because Pt2+ + 2e- has reduction potential of 1.2 and Au3+ + 3e- has reduction potential of 1.52.[2]
d) some other number because I fouled that up, for example using the Au+ + e- reduction potential of 1.83.
e) depends on the composition of the grime and whether it can form compounds with gold or platinum.

The real gold here is if we can explain persuasively why some answers aren't applicable. Wnt (talk) 14:39, 15 May 2018 (UTC)[reply]

None of the above, but e) is closest. It has nothing to do with forming compounds with gold or platinum, but on the potential to be oxidized or reduced by gold or platinum. The half-reactions will depend on 1) what ions are present in the grime and what sorts of half reactions they could undergo and 2) the reduction potentials of those reactions. Regardless, the platinum band will be the anode of the reaction, and gold the cathode if any DOES occur, because platinum has a lower Standard electrode potential, and thus would be more likely to be oxidized than the gold. Because of that, there is no reaction which would cause the gold band to be the anode preferentially over the platinum band. However, WHICH reaction would occur, if any, would depend on what complementary reduction half-reaction is happening at the gold band. --Jayron 32 16:10, 15 May 2018 (UTC)[reply]

Platinum atoms have a proton less than gold atoms (Pt/Au atomic numbers: 78/79) and are less electronegative, which equates to being more electropositive or "noble". Their electronegativities may be compared numerically (Pt/Au: Pauling 2.28/2.54, Allen 1.72/1.92). Therefore to the extent that the man's sweat functions as a shared conductive electrolyte between his bracelets, he becomes the frame of a Galvanic cell where his gold bracelet is the cathode. A high impedance voltmeter will consequently show the platinum bracelet has a positive e.m.f. relative to the gold. Over a long time the relatively "less noble!" gold could even develop sacrificial Galvanic corrosion while the platinum is preserved. Since there are umpteen ways an unspecified grime might react preferentially with one bracelet, the answer e) is indicated. If the "gold" bracelet is in fact only gold plated, its shine may not last long. DroneB (talk) 20:07, 15 May 2018 (UTC)[reply]

From Galvanic series "When two metals are submerged in an electrolyte, while also electrically connected by some external conductor, the less noble (base) will experience galvanic corrosion." The bracelets are not connected by an external conductor, so no corrosion. 114.124.149.159 (talk) 10:31, 16 May 2018 (UTC)[reply]

Galvanic corrosion relies on the flow of a current, and so requires a conduction path outside the cell. However an isolated cell can still show an EMF, without there being current flow (other than that through the voltmeter). Andy Dingley (talk) 11:09, 16 May 2018 (UTC)[reply]

Supernova is loud

I believe the supernova is so loud that if there's air between earth and the exploding star for sound to travel through, we would hear it from many light-years away. Even the loudest recorded sound here on earth, the eruption of Krakatoa in 1883 which is heard for thousands of miles, is merely a pin drop compared with supernova. I judge supernova explosions generate thousands or tens of thousands of decibels. How loud exactly can the supernova be? Planet Star 22:56, 15 May 2018 (UTC)[reply]

Human death by sound is estimated to be 198 to 202. I think the lungs burst or something. So above that. Sagittarian Milky Way (talk) 23:18, 15 May 2018 (UTC)[reply]

Under a thousand though, because logarithms. Sagittarian Milky Way (talk) 23:19, 15 May 2018 (UTC)[reply]

How much air is there in the vacuum of space? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:34, 15 May 2018 (UTC)[reply]

Very little. The Moon's surface has 100,000,000,000,000 times thinner than Earth and that's still in the solar wind. Sagittarian Milky Way (talk) 23:55, 15 May 2018 (UTC)[reply]

So, for all practical purposes, a supernova explosion would not be loud. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:59, 15 May 2018 (UTC)[reply]

Above c. 170 decibels (forgot exact number) if the center of a sine wave was at sea level pressure the bottom would have to be below pure vacuum but you can't go lower than complete absence of atoms. The peak to trough value of a shock wave could still be higher than 2 atmospheres but it's debatable if that's still sound. Sagittarian Milky Way (talk) 00:16, 16 May 2018 (UTC)[reply]

If you were on a close planet it'd briefly do things to your eardrums that could be interpreted as sound but since the things are shock waves that'd kill you on contact and neurons cannot transmit anything faster than about 150 meters/second you will never feel the hit on your body much less hear it. Sagittarian Milky Way (talk) 00:29, 16 May 2018 (UTC)[reply]

If space were filled with air, all kept magically at Earth surface temperature and pressure, it would take almost a million years for a sound wave to travel a distance of one Light-year. (Calculating 9.46x10¹⁵/340/3600/24/365). The sound of Krakatoa erupting was widely reported and happened during the days of early experiments with mechanical recording, see History_of_sound_recording#The_acoustic_era_(1877_to_1925), but it is not known to have been knowingly recorded. The estimated energy of the eruption was a nearly infinitesimal fraction 5.6x10^-27 to 8.4x10^-29 of the net energy released in a Supernova (Calculating in joules: 2x10⁸x4.2×10⁹/1.5 to 100 x10⁴⁴), see Orders of magnitude (energy). DroneB (talk) 00:44, 16 May 2018 (UTC)[reply]

But in this alternate magical air-filled universe, the acoustic energy will presumably radiate isotropically, so the energy that reaches Earth is an infinitesimal part of the whole, just like the light energy. -Arch dude (talk) 02:15, 16 May 2018 (UTC)[reply]

This is a science desk, so let's be careful with words like "infinitesimal". In a consistent medium, both sound and light energy are distributed along the surface of an expanding sphere and therefore follow an inverse-square law. In this magical universe filled with (perfectly uniform and dust-free) air, the fraction of the acoustic energy that reaches Earth will be exactly the same fraction as that of the light energy. For example, in comparison to a position 100,000,000 (or 10⁸) kilometers from the supernova (about the Sun–Venus distance), if we are 100 light-years away (or about 10¹⁵ km), then we would receive 10^{2 × (8 − 15)} = 10⁻¹⁴ of both the sound and the light energy. --76.69.47.55 (talk) 06:54, 16 May 2018 (UTC)[reply]

While I agree with your sentiments about technical terms, please note that many words have technical meanings in a technical context but still have general meanings in a general scientific context. "Infinitesimal" is technical in math, but not when used to mean "teensy-weensy". In your sentence, you use the general sense of the terms "consistent", "medium", "energy", "distributed", and "surface", all of which also have technical meanings. -Arch dude (talk) 18:18, 16 May 2018 (UTC)[reply]

... but the usage of those other terms was not inconsistent with the technical meaning, as it was with infinitesimal. If one means extremely small, then one should say so, rather than use terms like infinitesimal or "teensy-weensy". Dbfirs 07:35, 17 May 2018 (UTC)[reply]

If there were air between the earth and any given supernova that has ever been observed from earth, and we ignore all other consequences of such air existing, the supernova's energetic ejection of ionized gas particles would impact upon air particles and transfer kinetic energy to them, with those impacted air particles then in turn impacting nearby air particles and transferring energy to them (according to their mean free path in air). The resulting pressure wave could reach earth. But as it is, with no air between the supernova and earth, the highly energetic ejected particles can just come all the way across space to earth on their own. Once here, having lost essentially no energy due to inelastic collisions, they impact the upper atmosphere, the particles of which can in turn transfer the energy downwards to earth, and once the mean free path shortens enough, this can also form a pressure wave. But that doesn't happen because all observed supernovas have been so far away that the energy for any such pressure wave is insignificant. Having air between earth and the supernova won't abrogate the inverse square law. 202.155.85.18 (talk) 01:55, 17 May 2018 (UTC)[reply]

Are you sure the supernovae of recorded history haven't been deflected before reaching Earth's atmosphere? The stellar winds of the galaxy are slowed to almost nothing by the heliopause but supernovas are much more powerful of course. Sagittarian Milky Way (talk) 02:52, 17 May 2018 (UTC)[reply]

See Local Bubble and Superbubble. Graeme Bartlett (talk) 23:15, 17 May 2018 (UTC)[reply]

May 16

What isomer is this

Is the "major product" shown here a Z or E isomer? To my way of thinking, the phenyl groups will have highest priority on both sides, so the isomer shown is Z, but the text says E. 61.247.39.121 (talk) 21:15, 16 May 2018 (UTC)[reply]

I agree it would be Z. DMacks (talk) 13:19, 17 May 2018 (UTC)[reply]

Thanks. 202.155.85.18 (talk) 00:41, 18 May 2018 (UTC)[reply]

May 17

Leaving group vs Nucleophile

This question relates to nucleophilic substitution and elimination reactions in organic chemistry. If a particular species is a good nucleophile in a given solvent, does that then make it a poor leaving group? One of the characteristics that makes a species a good nucleophile is that it's poorly solvated, thus freeing up its surface and movement for interaction with a haloalkane. Conversely, one of the factors that makes a good leaving group is that it's well solvated, and leaving the haloalkane to form a carbocation is energetically favorable (or at least not very unfavorable). Given both of these facts, it stands to reason to me that if a given nucleophile is poor in a given solvent, then it should be a good leaving group in the same solvent. Further, if leaving group A is displaced by nucleophile B, but A is also a good nucleophile, then it should simply go back and displace B in an endless cycle. Of course an equilibrium of such alternating displacements is established in real world reactions, but the equilibrium position lies far to one side or the other in useful reactions. Despite all this, it seems to be generally accepted for example, that Br is a good leaving group, and Br- is a good nucleophile, and this is often stated without specifying the solvent (which makes it a bit meaningless to me). My text book includes a table of nucleophiles and their strengths in water/ethanol solution. It places Br- as stronger than Cl-. In a question in the chapter it asks a question that essentially boils down to which of the two halides is a better leaving group in ethanol. It says the answer is Br- which seems inexplicable given they already said it is a stronger nucleophile in almost exactly the same medium. 202.155.85.18 (talk) 10:03, 17 May 2018 (UTC)[reply]

You're generally right that good leaving groups make poor nucleophiles, and visa-versa, but there are some exceptions, and the halogens are often exceptions to these rules. For example, see Electrophilic aromatic directing groups, in electrophilic aromatic substitution, halogens are ortho-para directing, but are electron withdrawing groups; all other electron withdrawing groups are meta-directing; and all other ortho-para directing groups are electron donating. The reasoning for this may be closely related to the reason why halogens also buck the trend in nucleophilic substitution trends. --Jayron 32 11:16, 17 May 2018 (UTC)[reply]

Thanks for that response. In that case, if a halogen like Br can be considered both a better leaving group and a better nucleophile in a polar, protic solvent such as water or ethanol, then what do we expect to happen when say sodium bromide is added to a chloroalkane in ethanol? Do we expect the Br to displace chlorine because it's a better nucleophile? If so, does it then immediately form a carbocation because it's also a better leaving group? While I appreciate that the halogens have some unique behaviours, at some point these concepts of good nucleophiles and good leaving groups become mutually inconsistent. 114.124.243.188 (talk) 23:13, 17 May 2018 (UTC)[reply]

And that's why organic chemistry is hard as balls. Real chemistry is messy as heck, because you've got dozens of competing principles, and trying to quantify how each one applies in a particular reaction will make you pull your hair out. This page actually has a good discussion of the solvent effects on nucleophilicity of halides. I think it can help you work out the answers to your questions. --Jayron 32 23:36, 17 May 2018 (UTC)[reply]

You have stumbled upon the property that lets Br⁻ and I⁻ act as effective nucleophilic catalysts; they really are both good nucleophiles and good leaving groups in S_N2 reactions. It often improves yield in S_N2 reactions to pass through an alkyl iodide on the way to the nucleophile you actually want, because an alkyl iodide is more easily substituted than an alkyl chloride (Clayden, Greeves, and Warren's Organic Chemistry discusses this point at the end of Chapter 15, which covers nucleophilic substitution at a saturated carbon atom.) Your reaction will not proceed to completion but will instead reach equilibrium. Other factors (such as the solvent) can drive such halogen exchange reactions to completion by giving things a push in one direction or the other; this is how the Finkelstein reaction works (although that is converting alkyl chlorides or bromides to alkyl iodides). Double sharp (talk) 15:28, 19 May 2018 (UTC)[reply]

P.S. If you can access it, 10.1021/ed074p836 discusses converting alkyl chlorides to alkyl bromides and vice versa. Double sharp (talk) 15:49, 19 May 2018 (UTC)[reply]

What Does Spironolactone have to do with Chronic heart failure that other diuretics doesn't have?

I saw the following question "Despite the administration of cardiotonic and thiazide diuretic a patient with chronic heart failure has persistent edemas and the risk of ascites arose. What medication should be administered to enhance the diuretic effect of the administered drugs? A. Manithol B. Furosemide C. Amiloride D. Clopamide E. Spironolactone". The correct answer according to books and sites is Spironolactone. My question is what does it have to do with chronic heart failure that the others don't have? All of them diuretics, some of them are weaker and some of them are more potent diuretics, but I don't understand why they expect from the answerer to answer Spironolactone exactly while the others can be apparently also correct answer. Isn't it? 93.126.116.89 (talk) 13:30, 17 May 2018 (UTC)[reply]

Even without any knowledge of the medications one can grok the reason if one is told the answer is Spironolactone. The others must be cardiotonics or thiazide diuretics, which aren't working. Spironolactone must work on another pathway. Abductive (reasoning) 14:46, 17 May 2018 (UTC)[reply]

) Spironolactone is Spironolactone is K+ sparing diuretic and the rest are cardiotonics as well... I know the pathways of each of this drugs and I don't find any reason for this to be the chosen answer logically. Sometimes there are MCQs that have mistakes that's why I'm asking. 93.126.116.89 (talk) 16:19, 17 May 2018 (UTC)[reply]

Spironolactone is a diuretic, but not a thiazide diuretic! It is a specific pharmacologic antagonist of aldosterone, acting at the collecting tubule, preventing the adaptive process of cell hypertrophy that occurs with long-term loop diuretic use. Circulating aldosterone concentrations are often increased in advanced congestive heart failure, which makes spironolactone yet more effective. Spironolactone opposes the kaliuretic effect of loop or thiazide-type diuretics. And the addition of this potassium-sparing diuretic offsets the potassium-wasting of loop and thiazide-type diuretics. And maintaining normal potassium levels can be important when digitalis (a cardiotonic) is being administered. But the answer is probably to test if you know the empiric evidence: that "spironolactone has been found to increase life expectancy and reduce hospitalization frequency when added to the conventional therapeutic regimen of patients with advanced congestive heart failure and systolic dysfunction."[3]. - Nunh-huh 17:31, 17 May 2018 (UTC)[reply]

The question is really beating the point to death. For CHF patients, spironolactone is preferred. For patients with persistent edemas, spironolactone is preferred. For patients with ascites related to cirrhosis, spironolactone is preferred. So, the question gives the patient all three and asks which is preferred. 209.149.113.5 (talk) 19:51, 17 May 2018 (UTC)[reply]

No, that's not the case. Loop diuretics like furosemide are the mainstay of diuretic therapy in heart failure. Spironolactone is often added to loop diuretics, but it can't be simply stated that it is "preferred" in CHF. It is, as you say, the drug of choice in initial treatment of ascites due to cirrhosis. - Nunh-huh 22:16, 17 May 2018 (UTC)[reply]

I agree that furosemide is more effective diuretic in this context, but note that the question asks which would "enhance the diuretic effect of the administered drugs". Spironolactone, acting downstream (in the collecting duct) of the thiazides in the nephron would prevent the usual distal compensation for the more proximally-acting (distal convoluted tubule) thiazide. I also wonder whether the question might be a little dated, referring to spironolactone rather than newer more selective antagonists like eplerenone, and the thiazides rather than the more recent thiazide-like diuretics like indapamide. I agree with the K⁺/digoxin comments of Nunh-huh. Klbrain (talk) 22:30, 18 May 2018 (UTC)[reply]

Well, yes, but if the question is designed to elicit the fact that adding spironolactone to furosemide results in synergy rather than a merely additive effect, it would benefit from a rewrite. Your point about it being a dated question is well-taken, though my thoughts were that it was perhaps aimed at practitioners in areas where the latest medications are not available. - Nunh-huh 00:26, 19 May 2018 (UTC)[reply]

STO cells; what does STO stand for?

STO feeder cells are a cell line that are used to enable growth of other cells in culture. What does STO stand for? Does it even stand for anything? Everyone just seems to call them STO cells without explaining the letters. — Preceding unsigned comment added by 129.215.47.59 (talk) 15:31, 17 May 2018 (UTC)[reply]

Our STO page suggests it is "SIM Thioguanine/Ouabain-resistant mouse fibroblast cell line", which agrees with what doi:10.1095/biolreprod.103.017467 says about it and its relevance to your context. And clicking on the [characteristics] tab of the webpage you linked agrees that we're all talking about the same thing. DMacks (talk) 15:58, 17 May 2018 (UTC)[reply]

Joseph-Louis Lagrange

Hi,

I've just read the article about Lagrange, who is famous for the Lagrangian points. In the article, it says "before his death at Paris in 1813, in 128 rue du Faubourg Saint-Honoré". As a Frenchman, I'm very familiar with this street (this is our "Pennsylvania Avenue") and I wanted to find out more about his house in the Presidential Palace's street. Problem is, when you google "128 rue du Faubourg Saint-Honoré", there is nothing about Lagrange. Are you sure the information in the article is correct? Ericdec85 (talk) 15:40, 17 May 2018 (UTC)[reply]

He died at the end of the Napoleonic era - Rue du Faubourg Saint-Honoré wasn't really established as a major street yet, the whole area would be redeveloped under Baron Hausmann. So (like so much of Paris) the path of the street is ancient, but the buildings aren't. I would suspect that wherever he lived on this street, there's probably very little left of that specific address today. Andy Dingley (talk) 15:57, 17 May 2018 (UTC)[reply]

Organisation et reglemens de l'institut des sciences lettres et arts, Paris, Janvier 1807 (p. 267): "Liste des membres par ordre alphabetique - LAGRANGE (Joseph-Loius), rue du Faubourg Saint-Honoré, no.128". Alansplodge (talk) 17:49, 17 May 2018 (UTC)[reply]

However, Visiter quartier Saint-Honoré says: "No. 134 : Empire-style hotel built for the mathematician Joseph-Louis Lagrange and acquired in 1846 by the family of the banker Alexandre Aguado" (I couldn't stop Google from translating it). Alansplodge (talk) 18:15, 17 May 2018 (UTC)[reply]

This page also opts for no. 134 - perhaps he moved? Alansplodge (talk) 18:21, 17 May 2018 (UTC)[reply]

Or the number moved. Streets are sometimes renumbered. Hard to say. Regarding the location, the Élysée Palace was built in 1722, so it was certainly there the early 1800s, and in 1813, it was actually owned by Napoleon, according to our article. However, that being noted, that doesn't mean that it was then, or is now, the only building on the street. The analogy to Pennsylvania Avenue is not necessarily all that apt; yes, the White House has a Pennsylvania Avenue address, but that's a fairly long street, and it goes through many different neighborhoods, some of which are sketchy enough I don't think you'd want to be there after dark. Anhyoo, back to the rue du Faubourg Saint-Honoré This is the approximate location of the address today; and it doesn't look like the kind of place where he wouldn't have lived; it's a good mile from the Élysée and there's lots of what looks like late 17th century buildings there. It may even still be there today. --Jayron 32 18:58, 17 May 2018 (UTC)[reply]

This is no. 134 today. It is home to the Chambre de Commerce Italienne en France and a posh hairdresser. Alansplodge (talk) 19:44, 17 May 2018 (UTC)[reply]

Park Avenue in the 80s must've been something. Billionaires and Grand Theft Auto a few blocks apart. Sagittarian Milky Way (talk) 20:59, 17 May 2018 (UTC)[reply]

Can I say that every allergy is autoimmune disease?

If allergy is a wrong way of the body to interpret situations (such as considering nut as pathogens...) can all of allergies be under the umbrella of autoimmune system? And all of the autoimmune diseases are defect of the immune system in which the immune system doesn't interpret substances properly as it should. Isn't it? 93.126.116.89 (talk) 16:13, 17 May 2018 (UTC)[reply]

There's an immune system, but there's not an autoimmune system. Allergies are caused by the immune system (and can be further classified by the type of reaction that occurs—type I, type II, type III, or type IV hypersensitivities.) And yes, autoimmune disease also occurs via the action of the immune system. -Nunh-huh 17:50, 17 May 2018 (UTC)[reply]

No doubt the links above will give more specific information, but for a quick analogy, you can think of autoimmune diseases as being cases where your own tissues are directly hit by "friendly fire" from the immune system — cascades meant to destroy invaders instead destroy your own substance.

In an allergy, on the other hand, the immune system targets "invaders" that don't really need to be destroyed, and you suffer the negative repercussions of being in a "war zone".

For example, the IgE system is triggered by, say, a pollen grain, and it causes mast cells to degranulate, releasing histamine, which provokes local inflammation. If you had a parasitic worm trying to make its way into your flesh, that inflammation would presumably impede the worm somehow. Pollen wasn't going to do that, so you didn't need the inflammation, and it causes you discomfort. But the inflammation isn't the payload; it isn't meant to attack your tissues.

So no, I don't think they're the same thing. There's a reasonably clear distinction. I don't know whether we have it laid out more expertly in our articles. --Trovatore (talk) 18:07, 17 May 2018 (UTC)[reply]

The answer is of course, no: In an allergy, the immune response is directed at an actual pathogen, it's just a benign pathogen, and any damage you suffer is due to the "secondary response" from the immune system. In an autoimmune disease, the auto is the key part: the immune system attacks your own cells as though they were pathogens. Different thing entirely. You can read about these differences at articles like allergy and autoimmune disease. --Jayron 32 18:48, 17 May 2018 (UTC)[reply]

So in other words, you could think of an autoimmune disease as being allergic to yourself. ;-) 2601:646:8A00:A0B3:7508:650A:FEA6:C618 (talk) 01:22, 18 May 2018 (UTC)[reply]

Yanny or Laurel

Is this the only known recording which some English speakers are sure is one word and others are sure is a different word or name? I wouldn’t include recordings which are simply too faint, distorted or drowned out by noise or music to hear clearly, like movie scenes which need captions to be understood. I would also exclude Mondegreens. Edison (talk) 21:10, 17 May 2018 (UTC)[reply]

There are other such auditory illusions, though they are often constructed in far more complex ways. The Tritone paradox is one. If you want to learn more about the effect (the Yanny/Laurel effect or the Shepard tone effect at the center of the Tritone illusion) this is all part of the field of Psychoacoustics. My suspicion is that the Yanny/Laurel thing is someone hacking into the brain's ability to filter sounds and focus on specific speech patterns, known as the Cocktail party effect. This sort of auditory hacking is the the aural equivalent of optical illusions. --Jayron 32 23:50, 17 May 2018 (UTC)[reply]

Maybe you could explain how anyone is hearing "laurel" in the recording in question. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:47, 17 May 2018 (UTC)[reply]

YOU apparently hear "Yanny". Other people hear "Laurel", in fact its pretty close to 50/50. I can only hear "Laurel" in the original recording, and heat nothing of the Yanny. It really is a weird effect, but its real. Just a reminder that what goes on inside your brain (or my brain) is not universal to humanity. All experience is unique. --Jayron 32 23:50, 17 May 2018 (UTC)[reply]

The clip they played on Fallon last night (assuming it's the right one) sounded more like if you started "Larry" with a "Y". I don't see how a trailing "ee" sound could be confused with a trailing "el" sound. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:33, 18 May 2018 (UTC)[reply]

The original sounds to me like Laurel, without question. The New York Times actually has a neat widget to help you hear it both ways, but honestly, it sounds like Laurel and Yarry to me, rather than Laurel and Yanney. https://www.nytimes.com/interactive/2018/05/16/upshot/audio-clip-yanny-laurel-debate.html Someguy1221 (talk) 02:26, 18 May 2018 (UTC)[reply]

This has some discussion and also a video with audio samples modified from the original clip which should enable most people to separately hear both Yanny and Laurel [4]. (The video is here on Twitter [5]) Nil Einne (talk) 02:17, 18 May 2018 (UTC)[reply]

I can hear Laurel quite easily. Run the sound through a graphic equaliser or similar sharp low-pass filter and strip the high frequencies. Normally I hear "Yarry", but take the top end off and it does indeed turn into "Laurel". Andy Dingley (talk) 08:51, 18 May 2018 (UTC)[reply]

Here's a clip in which there is no ambiguity.[6] The "yanny"/"laurel" version must have been technologically inferior. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:48, 18 May 2018 (UTC)[reply]

I can't believe that anyone hears this as "Yanny" ... and given that we know it is a recording of the word "Laurel", the other point of view seems hard to understand. But is the difference a matter of the human ear, or are some people listening on equipment that somehow confuses the sound? Wnt (talk) 02:19, 18 May 2018 (UTC)[reply]

People are able to hear both. Some people after trying hard enough (I even read of someone who tried hard enough and now can only here that), some people semi randomly. So while equipment likely plays a part, clearly the human auditory system (by which I'm including the brain) plays a big part. Nil Einne (talk) 02:27, 18 May 2018 (UTC)[reply]

(edit conflict)The difference is you hear Laurel and someone else hears Yanny. You are not right and they are not wrong, and you are not wrong and they are not right. Both sounds are present in equal measure, and how your own auditory processing system (brain and ears) deals with it is different than someone else. That's just it. It's nothing more than a lesson that your sensory perception is not universal. No one's is.--Jayron 32 02:31, 18 May 2018 (UTC)[reply]

(ec, continuing...) But --- I just RTFA and was directed to this tool. By pulling their slider over two units, I was able to hear "Yanny". Pulling it back one unit (to +1 Yanny) made it sound like "Laurel" again ... but putting it to +1.1 Yanny and then slowly working it back, I could hear that version even at +0.9 Yanny units. But after two minutes, trying again, I had to turn it up to +3 Yanny to hear that version, and tended to lose it around +1.2 or so. The one thing that's clear from my perspective is that there is substantial hysteresis turning up and down ... but it is possible to hear the versions alternately, without adjusting the slider at all, at some point in the intermediate zone, like +1.6 is good. Trying to switch interpretations, sometimes I "fail" and hear the previous one, but usually I make it sound like the version I'm trying to hear at that point. I feel some kind of actual muscular tension/response in my ears doing this mental shift, especially trying to hear the "Y" start ... Wnt (talk) 02:32, 18 May 2018 (UTC)[reply]

And while I was posting that, and hit an edit conflict, with the sound repeating and getting distracted a moment, I somehow got stuck on "Yanny" so hard that I couldn't make it sound like "Laurel" without turning the slider back down to +0.8 or so. But a minute later I couldn't hear "Yanny" without a hard deliberate effort, same setting. This is odd. (Whether my hand is between the speaker and the ear seems to have some effect favoring Yanny, I think, but it isn't that strong) Wnt (talk) 02:37, 18 May 2018 (UTC)[reply]

Sorry to be the bore, but this effect is neither special nor unique to this specific audio recording.

In the general case, you can take any two audible waveforms and combine them using a mixer. For any two waveforms, there exists a wet/dry mix such that any individual will preferentially identify the wet- or dry- as the perceptually dominant sound. This is definitionally true: if we ask participants to classify a waveform as sounding like "A" or "B", there always exists some mix ratio for which there will be an exact 50-th percentile split in any focus group between those who hear "perceptual response A" and those who hear "perceptual response B." In many cases, that split will occur when the mix ratio is also exactly 50% wet/dry; but it could occur at any other value. With a very very trivial bit of standard off-the-shelf technology, and a large-enough focus group, we could construct such an "ambiguous" waveform from any two sounds or phonemes.

Here's a good article on mixing basics from the Music Department at University of Indiana.

And if you're in the mood for meatier treatment of acoustic processing theory, here's Physical Audio Signal Processing, one of my favorite technical books of all time. The chapter on voice synthesis gives a great overview on the history and technology of the vocoder and the art and practice of phoneme synthesis.

"One of the difficulties of formant synthesis methods is that formant parameter estimation is not always easy." This is techno-babble jargon that just means we all think we hear different things when we listen to the exact same waveform.

If you're really interested, another great text is Rabiner and Schafer, Theory and Applications of Digital Speech Processing....

If you start playing games audio filtering, humans may still recognize the sound, but will notice that it's not quite right. One of the easiest ways to disarm the audience, and to make them more receptive to silly audio processing tricks, is to lower the signal to noise ratio by intentionally injecting noise. In fact, a great many of the important historical voice synthesis algorithms begin by injecting pure white noise, and then progressively filtering that noise until it sounds like human speech! The practicality of this is that if you use a modern operating system, you can make the computer talk; and you can change the voice. Ever wonder why there's almost universally an option to "whisper", in almost every configurable text-to-speech synthesizer? That synthesized breathy voice is full-mix white-noise injected into every synthetic phoneme!

Nimur (talk) 03:32, 18 May 2018 (UTC)[reply]

If this is easy to do by mixing audio or filtering white noise or employing psychoacoustics or whatever, then per the original question where is another audio sample which about half the people hear as one word and the rest hear differently, and can’t Imagine how the others can hear what they hear. When this phenomenon first sprang up I seriously wondered if it was a gag and my signifcant other was just pretending to hear Yarry when I heard Laurel. I doubt it is a function of people with more sensitivity to high frequencies hearing Yanni since an 80 year old and someone with significant heating loss heard Yanni or Yarry or Yammy but my child heard Laurel. This makes me wonder if it is genetic. Edison (talk) 04:11, 18 May 2018 (UTC)[reply]

It is rather common. It is more common with digital media. The recording in question is a recording of the word "laurel." That is well documented, including the fact that it was made by an opera singer because the company that hired the speakers preferred opera singers. If you were to listen to the person actually speaking, you would hear laurel. After being digitized and compressed, some people year "yanny." If you go to music recordings, which are being digitized and compressed, you find the same thing. There are some words or phrases that, even without compression, are difficult to understand (e.g. "excuse me while I kiss this guy"). There are also many single words that sound like other words. I don't want to start a long exhaustive list, but everyone knows some lyric where a word just doesn't sound like it should. For me, personally, mp3 versions of Stairway to Heaven really mess up the end. "How everything turns to gold" really sounds like "How everything turns to stone." 209.149.113.5 (talk) 12:09, 18 May 2018 (UTC)[reply]

Here's a multimedia clip of some historical synthesis: Bell Labs' The Voder (1939), including some comparative phoneme modes. Suffice to say, not every listener thought it sounded like it's speaking English. And when it speaks French... well, I'm sure somebody thinks it's intelligible. (Thus began the storied history of annoying robots making barely-intelligible telephone calls, and the many generations of naive scientists who think they invented it, even though people have not wanted this technology for over a century).

If you want more examples, have a listen to the radio traffic at a busy airport on a website like LiveATC. Especially if there's radio static, the words can be pretty hard to distinguish. By the time the VHF voice radio signal gets received, processed, and digitally compressed for internet transmission, a lot of phonemes sound tricky. "Zero" can sound like "four"; "departure" can sound like "SouthWest". If you spend enough time listening to, or using, voice radio, you'll develop some skills to help reduce collision - but you'll also find times when you need things to get repeated!

If you're looking for more sample::: audio files, here's a whole website dedicated to ultra-low-bit-rate voice compression: Rowetel: open telephony software. Listen to the 700 bit-per-second English speech recording: [7]. The limit of ultra-high compression ratio is the point where each phoneme is "one unit of JND" above un-intelligble!

Nimur (talk) 15:00, 18 May 2018 (UTC)[reply]

Audio files which are simply distorted or degraded and unintelligible but do not present seemingly clear and unconfusable but different perceptions are not what I am looking for. Where is an audio file which some hear as “X” and others hear as an unmistakable “Y” where X and Y are different words or names. I have only seen assertions that it is easy and common, without any examples. In visual science there are many common examples such as the image which is a beautiful young woman or an ugly old woman, or the duck/rabbit, or the Necker cube. The Shepard ascending tone does not sound like a descending tone to half the listeners. Mondegreens do not count. Edison (talk) 02:11, 19 May 2018 (UTC)[reply]

There are some great answers here, but there are some specifics not touched on. Why "Laurel" and "Yanny" but not intermediate forms? I mean, "L" vs. "Y" would seem to be a matter of whether the tongue touches the palate, and so is "R" vs. "N"... but in the opposite direction! So why no "Yannel", "Laurie", "Yaurel", "Lanny"...? How can it be reliable that I, hitting this for the first time, pick out these two and only two combinations, the same as everyone else, when I shift my mind listening to the same ambiguized version of the word repeated over and over? Wnt (talk) 08:33, 19 May 2018 (UTC)[reply]

May 19

What could a human notice if his antipodal hemisphere instantly disappeared?

Without the help of technology or other humans (i.e. Internet, seismometers, Internet feeds of seismometers, grandma tweeting you Internet feeds of seismometers) What's the new (extremely temporary) gravity strength? How close to spherical would the hemisphere centered on you get before you're unconscious? Sagittarian Milky Way (talk) 04:17, 19 May 2018 (UTC)[reply]

This is a bit hypothetical! How would this happen? Blasted by a stream of neutron star fragments? Anyway the gravity would drop a bit so you would notice that first. Next you would be hit with massive seismic waves. Air would disappear at about the speed of sound. You could expect that the remaining hemisphere would absorb so much energy from planetary reconfiguration that it would be turned into a magma ocean. Graeme Bartlett (talk) 06:45, 19 May 2018 (UTC)[reply]

The one thing we can be sure of is you get a 1/30 second grace period due to the speed of light, though a stickler will insist that during that period the Event hadn't happened "yet". Relativity sticklers will also note that the faster you get rid of the mass, by any means, the more you mess with space making gravitational waves; i.e. gravitomagnetic effects become relevant.

Removing one hemisphere implies that material at the center, formerly under no gravitational field, would suddenly be pulled to one side, and hence could release energy by falling. On the other hand, the pressure would reduce from insane to zero in an instant, so it would also push outward. The inner core surface is estimated at 5430 K and the boiling point of iron at atmospheric pressure is 3134 K, so a rather impressive explosion is to be expected. However, note that the core of a half-sphere is still a lot lower than the outer surface. I don't know how you'd begin trying to calculate if the explosion reaches the lip of the crust. To take a wild guess, I'd say figure the heat capacity of liquid iron, i.e. (5430K - 3134K)*heat*core volume, figure out how much iron that can vaporize in moles, figure out how much volume that takes at 1 atm, and see if it fits in the flat side -- but I know that's an equilibrium, but an explosion isn't.

To find the center of a hemisphere, you set up some nested integrals in polar coordinates ... lolno, you go and do a web search and let your skills rust a little longer. [8] It's 3/8 of the way in. That means that if West Berlin is just on the good side of the Event boundary, the gravity that was formerly straight down is now coming from R down and 3/8 R over, i.e. it is off by tan-1 (3/11), or 15.2 degrees. This would be a bad time to have a house on the west side of a lake, though then again, the water might get there just in time to put out the fire from that burning core material, I dunno. ;) But actually that's just an approximation - a planet's gravity seems to come from its center of gravity because it's a sphere, but gravity doesn't come from the center of gravity of the Cavendish apparatus or any other un-planet-like shape. I suppose we could do a set of integrals ... but this one would not be in polar coordinates. But using the approximation, we know that the two halves of the earth pulled equally, summing the cos(+-15.2deg) components of each but cancelling the sin(+-15.2deg) bits, so gravity is at 51.8% or so of what it was. Sounds like a great time to launch a rocket ship, but it better be a fast one. Oh, but if you're at the center of the hemisphere, let's say, somewhere between El Dorado and Cayambe Coca Ecological Reserve in Ecuador, then you are 3/8 closer to the "center of gravity", so by an increasingly dubious approximation you should get 51.8% * 121/64 = 97.9% gravity. Hmmmm. ;) Wnt (talk) 16:55, 19 May 2018 (UTC)[reply]

What to call it?

I'd like to know what to call this condition so I can do some research to understand it better. I'm not asking for medical advice.

I have trouble remembering things I want to do. For some time I've relied on visual clues, like placing objects in appropriate places. I put things in my calendar but sometimes forget to look there (I don't have many calendar events). Sometimes I remember the day but not the time.

On a cruise recently the problem was magnified. The ship each day provided a schedule with a few dozen entertainment opportunities, and I would pick three or four I was interested in. But I had a hard time remembering not only the time and place, but even what the event was. (To tell the truth, I wasn't very committed to most of them.)

I don't experience any other memory problems.

--Halcatalyst (talk) 17:26, 19 May 2018 (UTC)[reply]