Science desk
< May 5	<< Apr \| May \| Jun >>	May 7 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

May 6[edit]

A few chemistry questions[edit]

A few chemistry questions… Please explain how to solve (no calculator allowed).

Assume that gaseous substance A undergoes a first order reaction to form gaseous substance B. At a certain temperature, the partial pressure of gas A drops to 1/8 of its original value in 242 seconds. What’s the half life for this reaction at this temperature?

I just plugged in numbers for the original value, calculated the new value, then looked at the answer choices and figured out the half-life. But how would this be done mathematically (without answer choices)? The answer is 80.7 seconds.

Equal masses of He and Ar are placed in a sealed container. What is the partial pressure of He if the total pressure in the container is 11 atm?

The answer is 10 atm. How?

A 160 mg sample of NaOH (MM=40) is dissolved to prepare an aqueous solution with a volume of 200 mL. What is the molarity of sodium hydroxide in 40 mL of this solution?

The answer is .0200 M. The only explanation I can come up for this is that the molarity in 200 mL is .02 M, but then why wouldn’t it change when looking at it for 40 ml? Does it not change? Molarity changes with volume though?

If .15 mol of K2CO3 and .10 mol of KBr are dissolved in sufficient water to make .20 L of solution, what is the molar concentration of K+ in the solution?

The answer is .40 M. How?

Vessel A containts 32 grams of O2 gas while Vessel B contains 32 grams of CH4 gas. Find the ratio of the pressures of the gas in Vessel A to Vessel B and the ratio of the average kinetic eneriges in vessel A to Vessel B.

For the pressures, I just found the mole ratio was 1:2 and said moles is proportional to atm, so it’s 1:2. That’s correct answer. Is that correct logic?

How do you find the ratio of the average kinetic energies? I thought it was r1/r2 = r_1/r_2 = √(M_2/M_1 ) but that answer would be 1:1.4 yet the answer key says the ratio is 1:1.

Thank you!!! --Jethro B 00:42, 6 May 2013 (UTC)[reply]

Generally, a functional knowledge of simple arithmetic and introductory algebra are considered prerequisites before students begin stoichiometry. Each of these questions are essentially solved by trivial application of simple algebra. (With a little rounding and liberal application of critical thinking, you can solve much more difficult problems than these with no calculator - even pencil/paper are unnecessary for these problems). Other questions are spot-checks to verify that you understand terminology (like the difference between concentration and volume). Can you help us understand your background so we can lead you in the right direction? Nimur (talk) 01:35, 6 May 2013 (UTC)[reply]

If you believe that the stoichiometry and algebra is simple, then please provide some answers for me... I don't have an issue with algebra or stoichiometry. However, you can't just do stoichiometry in each of these problems - they test information across various units and different things need to be applied for each one to understand how to do the stoichiometry. The picture here isn't stoichiometry, but rather the specific units each question deals with, whether it's solubility, gas laws, kinetics, etc...

So if someone could provide an explanation of how to solve them (or some of them), and I note that they are not all the same stoichiometry, I'd appreciate it.

P.S. If you're really interested in my background, you can send me an email - it's not something I will publicly reveal. I will say that I consider myself skilled in the subject area and have proven this in the necessary courses by scoring highest, but there are always specific little things that you either don't know or forget (to put this in context, these are selected from a document of 75 questions, and are the ones I have trouble with. It's just a few, but I'd like to know how to solve them). --Jethro B 02:06, 6 May 2013 (UTC)[reply]

I have no idea how much chemistry you, as an otherwise-anonymous poster, already know. If you're a new student of chemistry, you deserve some directed conceptual explanations. If you're an advanced grade-schooler, you need some detailed help with mathematics you wouldn't know yet. And if you're a physics undergraduate, you need a stern talk about life-choices, and we need to yell "apply the equipartition theorem" at you (for the last problem). Without knowing what type of help you need, it's difficult to help you.

For example, "Equal masses of He and Ar are placed in a sealed container. What is the partial pressure of He if the total pressure in the container is 11 atm?" Look up the atomic mass of Helium and Argon if you don't already know them. Helium's atomic mass is four, and argon's is forty (for the purposes of our discussion, without a calculator, and ignoring some irrelevant decimal places). The ratio 4:40 is simplified to 1:10; and the question gives you a total partial pressure of 11... one plus ten is eleven. The math is alarmingly simple - but only if you already know that partial pressure is proportional to the molar mass ratios. That is a simple fact, but it's one you need to learn somewhere (presumably in a chemistry class). Do you need help with these concepts or do you just need a reminder to apply them?

Every other problem had a similarly simple arithmetic answer, as long as you recognized the concept that was being asked. Nimur (talk) 02:52, 6 May 2013 (UTC)[reply]

These questions look like they're on the level of College level chem/AP Chem, which I know. Yes, if you can do what you did with the pressure question - state what concept is applied here and how - that's great. I don't need a detailed explanation, I should be able to understand it. The helium, with the lower mass, would exert more pressure than the Argon? --Jethro B 03:18, 6 May 2013 (UTC)[reply]

Here's my attempt to point you in the correct direction for these questions. An additional overall hint is don't be afraid to start slinging algebraic equations around. You might have an idea of how to solve things "if only I knew T (or V, or ...)". Don't get discouraged - just try representing it by a variable and calculate through algebraically. It's quite possible that the T's or V's will cancel and you'll find you don't actually need to know them to solve.

Equal masses of He and Ar ... as Nimur discusses above - the key point is that partial pressures are distributed like the molar ratio of the gasses (each individual molecule contirbutes equally to the pressure for ideal gasses).
Assume that gaseous substance A undergoes a first order reaction ... You'll need to undestand what a first order reaction is. Drawing from examples of first order reactions, it should be clear what the reaction and stochiometry is. From that, you should be able to calculate final amounts of A & B from the given partial pressures, and from the starting and ending amounts and the rate equation determine the half life.
A 160 mg sample of NaOH ... Questions can have superflous information to catch out those people who are blindly combining numbers. If you're confident in your understanding of what molarity is, the 40 mL shouldn't throw you.
If .15 mol of K2CO3 ... Start by calculating what the molarity of the K2CO3 and KBr would be seperately in the final solution. Then figure out what each would contribute with respect to K+ ions. The final K+ concentration is simply the total contribution for the K+ ions.
Vessel A containts 32 grams... This is a straightforward application of the ideal gas law. You can do PV=nRT for both vessels to find the ratios of pressures. Likewise, you can also write the equation for the average kinetic energy, then take the ratio and cancel like terms. One catch is that they're likely talking about the per molecule or per mole average, rather than a per mass average or something like that, so keep that in mind as you write the expression for the average kinetic energy.

Hopefully that should be sufficient to get you on your way. -- 71.35.116.214 (talk) 04:22, 6 May 2013 (UTC)[reply]

A half life reduces the amount of something by half. The statement that it is a first order reaction merely emphasizes that there is a half-life, i.e. no matter how much is present, half of it is gone in the same amount of time. Half of half of half is 1/8, so the half life is 1/3 the "eighth life".

Helium weighs 4 amu, argon weighs 40. To make equal masses you need 10x as many particles of helium. So there are 10 times as many particles of helium as argon flying around in the gas. Ideally all the particles have the same range of energies, so the helium particles will sock a wall ten times as often as argon particles and therefore be exerting 10x the pressure.

The next one is a dirty trick. An aliquot of a solution has the same molarity as the stock it is taken from. Molarity is moles / volume, so you could make up 0.02 with 160 mg in 200 ml or 32 mg in 40 ml or (easiest for calculation) 800 mg = 0.2 mol in one liter!

0.15 mol of K2something contains 0.30 mol of K. Add 0.10 mol of K from the other and you get 0.40.

Your logic is right. But if you put the two gasses in the same vessel, they won't push a membrane (or the invisible boundary between them) one way or the other, nor will they transfer energy from one to the other because they are more importantly at the same temperature. This implies they're carrying the same kinetic energy. See kinetic theory, which says the energy per particle depends only on the temperature. Wnt (talk) 03:12, 7 May 2013 (UTC)[reply]

For the potassium ion question, it's true that there are 0.40 moles of K⁺ in the solution, but the question asks for molar concentration, not the amount of substance, and gives the volume as 0.20 L. Unless you specify units of moles/200ml (a rather strange choice), you need to divide by the volume. This gives a concentration of 2M, so it seems the answer given in the OP's answer book is incorrect.please correct me if I am missing something obvious here... Equisetum (talk | contributions) 09:39, 7 May 2013 (UTC)[reply]

Reading my answer I feel I need to engage in a little auto-pedantry and point out that you actually divide by the concentration even if you do chose units of moles/200ml, it's just that the concentration is then 1 (if you don't conceptually divide then the units don't come out right, which even five years after doing my last dimensional analysis still makes me nervous). Equisetum (talk | contributions) 09:44, 7 May 2013 (UTC)[reply]

Sorry! I was rushing near the end and didn't notice the incorrect "right answer" was in M not mol. Wnt (talk) 15:57, 7 May 2013 (UTC)[reply]

I want to change a reference on the Elephant Cognition page.[edit]

I posted the following comment on the Elephant Cognition talk page but nothing has been done to change it.

The source [Dubroff, M Dee (August 25, 2010). "Are Elephants Smarter than Humans When It Comes to Mental Arithmetic?". Digital Journal. Retrieved 2010-08-29.] Is just an flake article talking about the journal article [Irie-Sugimoto, Naoko ; Kobayashi, Tessei ; Sato, Takao ; Hasegawa, Toshikazu."Relative quantity judgment by Asian elephants ( Elephas maximus )"Animal Cognition, 2009, Vol.12(1), pp.193-199]. Shouldn't the actual journal article be cited with the link being something like <http://journals.ohiolink.edu/ejc/article.cgi?issn=14359448&issue=v12i0001&article=193_rqjbaem> ? — Preceding unsigned comment added by reku68 (talk) 19:57, 22 April 2013 (UTC) — Preceding unsigned comment added by 129.110.5.89 (talk)

You can edit the article yourself! I'll put the article on my watchlist, and if you mess something up, I'll help fix it. Regards, Looie496 (talk) 14:49, 6 May 2013 (UTC)[reply]

What kind of bug is this?[edit]

Hi there. For the last few weeks, I've been finding little bugs, about one every day or so. I get rid of it, but there's usually one there the next day anyway. It's very rare for me to see more than one at once, but it has happened. They almost always appear in the same area of the room, and whenever I see them they're always on a wall, and rather low to the ground. I don't recall ever seeing them outside of this room, either. I was wondering if anyone could help me identify what it might be? I took a Photo of the bug (apologies for image quality, it's the best I could get out of my old iPhone 3GS camera). I live in England, in case this helps narrow it down. Thanks for any help I might receive. 86.134.231.216 (talk) 12:29, 6 May 2013 (UTC)[reply]

It looks like a shield/stink bug. I don't know which species though. Plasmic Physics (talk) 12:46, 6 May 2013 (UTC)[reply]

To test it, does it stink when you try and catch it, or otherwise disurb it? Plasmic Physics (talk) 12:50, 6 May 2013 (UTC)[reply]

I don't think so. They move so slowly (hell, if I look straight at one it's hard to tell whether it's moving at all - it's only when I look away for a bit and then look back that I can tell it's moved, usually), and they're incredibly tiny (three millimetres top estimate). I can't rule it out for certain though, as I didn't try actively sniffing it or anything like that. If I see another one, I'll update. The last time I had a visitor I asked her, and she guessed at woodworm, though she stressed it was just a guess. I looked at all the insects linked in the woodworm article, though, and they all looked too elongated, whereas these are all rounder. I haven't noticed any wood damage on anything wooden nearby, either. 86.134.231.216 (talk) 13:07, 6 May 2013 (UTC)[reply]

I don't know much about bugs (I'm sure there's a word for the logical study of bug, probably not bugology), but I do know that species can vary considerably in size within the same superfamily. Take Tessaratomidae, they are a famly under stink bug, and they are giants compared to the other families. I would not be surprised if there was a family of dwarf stink bugs, of which this one is a member. I ny case, that's just my opinion. Plasmic Physics (talk) 13:20, 6 May 2013 (UTC)[reply]

The study of insects is entomology. Anyway, the picture has very little detail, but PP's guess of a sheild bug is pretty good. That's good enough for casual purposes, but only narrows it down to ~7000 species... Anyway, as for why this is a good guess: note that the elytra seem to be incomplete, and the pronotum has the general shape we expect in shield bugs. Note that many stink/shield bugs will not display any odor when disturbed. Even the (recently very common) brown marmorated stink bug only rarely produces odors, in my experience-- so that's not a very good criterion. Finally, when trying to ID an insect, remember that things like color, markings and size are not very informative. The pros usually focus on gross insect morphology to get to Category:Orders_of_insects, and then use fine features to get to family or genus. In generally, getting any insect ID down to species is very difficult, even for experts. The exceptions are things like honey bees and monarch butterflies, but even then, there are several close ~~relatives~~ look-alikes that can easily fool amateurs. SemanticMantis (talk) 15:43, 6 May 2013 (UTC)[reply]

Are we looking at the same picture? What I see looks like a carpet beetle, family Dermestidae. --Dr Dima (talk) 19:17, 6 May 2013 (UTC)[reply]

Agreed. Looks nothing like a stinkbug. --jpgordon^{::==( o )} 19:33, 6 May 2013 (UTC)[reply]

What something 'looks like', is down to opinion, and by definition, an opinion cannot be false or true. So, it may not look like a stink bug to you, but it does not inherently invalidate my opinion. Plasmic Physics (talk) 01:13, 7 May 2013 (UTC)[reply]

Well, I only said shield bug was a good guess :) Anyway, if you think the elytra are completely covering the hindwings, then I suppose it could be a beetle. I thought the elytra looked incomplete, which would rule out beetles. But the photo is pretty bad. I was mainly trying to point out some of the diagnostic features that we could look for. "Looks like" doesn't hold much water for insect ID. There are several thousand species of beetles and sheild bugs, and many of them will look rather similar when photographed in this manner. SemanticMantis (talk) 21:26, 6 May 2013 (UTC)[reply]

The posterior dorsal section of the insect appears more tappered than rounded, which is why I recognise it as a shield bug. Plasmic Physics (talk) 01:30, 7 May 2013 (UTC)[reply]

That's probably more down to the poor quality of my photograph than anything else. As SemanticMantis said, it's hard to really make any kind of judgement on such a photograph. Not really your fault. Going off the tips from Dr Dima and Jpgordon above, I did a bit of Googling and found quite a number of council websites mentioning carpet beetles, and in particular they tend to mention the Varied carpet beetle, which I think this is. Considering a lot of these pages (and the WP article) mention bird nests, I might have to get on the phone with the council (I live in a council flat) and see what they can find out. Thanks again everyone for your help, and apologies once more for the poor quality of my photograph. 86.134.231.216 (talk) 05:41, 7 May 2013 (UTC)[reply]

Ah, I concur, the dorsal perspective of the varied carpet beetle does indeed look similiar to the photo. Plasmic Physics (talk) 08:01, 7 May 2013 (UTC)[reply]

I also go with Dr Dima and JPG, this image helps to confirm it is likely to be a Varied Carpet Beetle. Richard Avery (talk) 06:47, 7 May 2013 (UTC)[reply]

Meteor trail color[edit]

Why the trail color of this meteor shifts from green on the left to blue? A kind of redshift since it approaches from the left? Brandmeister^talk 16:23, 6 May 2013 (UTC)[reply]

A meteor would need to travel something like a couple of thousand times faster than they do to give visible redshift. I don't see much blue, I see green which comes from the meteor's copper or magnesium, and red which happens when Earth's atmospheric gases are heated. At higher altitudes there is more oxygen and the metal emission dominates, lower altitudes have more nitrogen which gives a deeper red. This green-to-red is not uncommon, see here. The article on meteors mentions color, and there is more at Nasa and good ol' web searches on meteor color. 88.112.41.6 (talk) 17:07, 6 May 2013 (UTC)[reply]

The image was captured with a consumer digital camera, and according to the file's metadata, the image was further post-processed in Adobe Photoshop. Be very very careful drawing scientific conclusions from such images. Digital cameras are incredibly complicated, and if you aren't sure what digital processing has been applied (as well as a very good understanding of all the optical and electronic characteristics of the camera), you should not jump to conclusions about things like color. In other words, a digital camera with color is not the same as a spectrophotometer. Nimur (talk) 17:12, 6 May 2013 (UTC)[reply]

Didn't do google search this time, but thanks anyway, solved. Brandmeister^talk 17:48, 6 May 2013 (UTC)[reply]

Tetrahydrofuran[edit]

If a site is contaminated with tetrahydrofuran (specifically, I am talking about the Seymour Hasardous waste site), would removing all contaminated soil at the site remove all contamination from the site and/or prevent the future spread of contaminants from the site?--149.152.23.33 (talk) 22:34, 6 May 2013 (UTC)[reply]

Removing all of the contaminated soil, by definition, removes all of the contamination as well. 24.23.196.85 (talk) 22:57, 6 May 2013 (UTC)[reply]

Note also that THF is only mildly toxic, highly volatile and rapidly biodegradable -- so digging up all that soil might not even be necessary or worthwhile, it might be just as effective to simply allow the THF to evaporate/leach out/break down over time while taking strict measures to prevent any further contamination. 24.23.196.85 (talk) 23:13, 6 May 2013 (UTC)[reply]