text:
Can hydrogen and oxygen gasses be turned back into water?
I electrolized water today, and got hydrogen. And I have a bottle of oxygen. Can I combine those to get myself water? I know $\ce{H2O}$ is super pure water, but I just want to try. Is this too idiotic to ask or can it be done?

Yes. If you heat up hydrogen in oxygen, the hydrogen will undergo combustion to give you water: $\ce{2H2 + O2 -> 2H2O}$

You don't need to heat it up. You just need a spark. Don't do it at home unless you have done proper research, this reaction is very exothermic and can potentially be dangerous.

I tried sparking it but the Hydrogen just exploded.i guess the amount H/O was wrong.And i also tried it with a neon transformer and made plasma but it just somehow all disappeared.I trued both in and without vacuum, nothing worked sadly.

You can run the hydrogen gas through a platinum sponge exposed to air - the hydrogen will burn instead of exploding (if done properly). Or use a hydrogen fuel cell for the most 'peaceful' reaction (electricity will be a nice bonus).

As @orthocresol states, do some research first. I did this every year with my students. Do the "burning" in a plastic bottle of no more that 2 mL volume. I've done it with 4 mL and it's scary. You'll get a loud POP and the bottle will show traces of moisture inside. The explosion you experienced was water being created. But the water created was a gas so you probably saw no evidence of it. Also, as a liquid  the water produced would be a very small volume compared to the gases used.

When hydrogen burns, pure water is created; the reaction is exactly the reverse of the electrolysis and shown in the equation below.
$$\ce{2 H2 + O2 -> 2 H2O}$$
If you just have hydrogen, then it will burn with a flame as you might know from a methane gas flame. However, once you have a certain mixture of hydrogen and oxygen gas in your container, the reaction becomes explosive. If you have the exact stoichiometric mixture of $2:1$, the reaction is a very strong (and loud) explosion; a tiny amount created a louder bang in Munich’s chemistry lecture hall than all other explosions the experimentors showed on that day.
The fact that hydrogen burns to give water and only pure water (vapour) was one of the main points used to try and sell hydrogen-powered cars, since water vapour is environmentally friendly and does not cause global heating like $\ce{CO2}$ does.

The electrolysis/explosion difference shows how different conditions for forwards and backwards reaction can be.
It bears repeating: do not attempt the combustion at home unless you know exactly what you are doing!

Good points. Industry everywhere is looking at hydrogen-fueled combustion these days. One problem: hydrogen burns so hot it creates high nitrogen oxide concentrations in air. New combustion technologies have been developed to counteract that.

Water vapor is a green house gas and is responsible for raising the temperature to livable level, changing CO2 is a modifier Also burning H2 in air also forms nitrogen oxides.

@jimchmst water vapour is a greenhouse gas but one that is abundant in the atmosphere as is, that often reaches saturation levels (clouds, fog) and thus creating additional water barely influences global warming. This is especially true since most hydrogen gas is created from water by electrolysis. The same is decidedly not true for carbon dioxide.

Note that the reaction is exothermic so the water you make is going to be a gas. The reaction has a reasonably high kinetic barrier, so just plain heat is not enough. You need some free radicals from a spark or a flame to initialize the reaction.
Do be careful. This reaction is an explosion.


text:
Is there a model for (quality of) two-component-polymerization (dependent on e.g. concentration)?
Sadly my chemistry background is only from high school (I'm a roboticist), but I try my best to explain the problem:
Situation: Lets say I have two water based solutions/suspensions. One contains cells with A-hooks and one contains proteins with B-hooks. Now i mix those two liquids and then the As will attach to the Bs. And so they will form sort of a polymer (not sure if this is the right phrase here)
Question: Now is there a model that tells me how "well" the polymerization works, dependent on the concentration of the two solutions, temperature, affinity of A to B, whatever other parameters there might be. I think a good measure for "good polymerization" would be the distribution of lengths of the chains or connectivity or some other graph metric.
The Background is the idea to build a 3D tissue printer without support structures. A and B refers to Biotin and Streptavidin
Clarifications:

each cell will have several 100 A-hooks, the proteins can be engineered to have an arbitrary number of B-hooks (we would like to optimize that)
to make it work we obviously need to connect each cell with as many different individual proteins as possible and vice versa.

The Carothers equation springs to mind, but from what i read here https://en.wikipedia.org/wiki/Streptavidin, this here is not exactly a polymerisation. Streptavidin simply makes tetramers with biotin, which precipitate?

@Karl Thanks a lot, this Carothers equation looks like a good part of the solution to my Problem.
Concerning the tetramer/polymer issue I will clarify my question in a minute

https://en.wikipedia.org/wiki/Flory-Stockmayer_Theory seams to be another related model, but it's still not general enough

source: Stack Exchange

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sort: votes


text:
IUPAC name of trisubstituted benzene ring
What is the IUPAC name for the compound shown below? I named it as 6-Bromo-3-hydroxy benzonitrile according to the numbering scheme in 1b, since the priority order is $\ce{-CN} > \ce{-OH} > \ce{-Br}$.
However, the answer given in the book is 2-Bromo-5-hydroxy benzonitrile, which implies the numbering scheme in 1a. Please help me to identify whether my answer or the book's answer is correct.

We compare the answer's [1,2,5] and your [1,3,6]. The first numbers (1 and 1) are the same; the second numbers (2 and 3) has the answer smaller than your attempt; therefore the answer is correct.

The priority order only determines which goes into the suffix.

Related (actually duplicate, but the other question is closed and includes proposed answers that are not related to the question): IUPAC name of bromo, cyano, hydroxy substituted benzene ring

Also related: What is the numbering of substituted benzene derivatives according to IUPAC nomenclature?

Numbering of carbons aside, there are two things I want to point out. Firstly, the first letter of the substituent should not be capitalised, so bromo is correct and not Bromo. Secondly, there is no space between the prefixes and the name of the parent compound, so there is no space between hydroxy and benzonitrile.

As -CN has the highest priority, root-word is benzonitrile. Lowest locant is 2 for -Br.
So, the name given is:
2-bromo-5-hydroxybenzonitrile
As the functional group is -CN, with highest priority, no need to look for priority of -OH and -Br; just look for lowest locant.

I'm quite sure that you would typically name this as a substituted phenol.  The name I would get for this is 4-bromo-3-cyanophenol.  Whether or not this is 'strictly' IUPAC I'm not sure.  When I searched this on Google, I found that SigmaAldrich was selling a constitutional isomer of this as 3-bromo-4-hydroxybenzonitrile, which would imply the previous answer is also acceptable.   Frankly, it doesn't strike me as a critically important issue for most applications, since both names are completely unambiguous.

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Why is water an Arrhenius acid?
Is water an Arrhenius acid? We define Arrhenius acidity with respect to water, but what will water itself be? It can donate $\ce{H+}$ as well as $\ce{OH-}$.

https://en.wikipedia.org/wiki/Amphoterism

Is it or is it not?

http://chemistry.stackexchange.com/a/4064/34388

It would be a bronsted acid  acid though,right? @DHMO

As well as a bronsted base.

I gave a test right now, it said water isn't an Arrhenius acid,thus the question

We define Arrhenius acidity with respect to water

That definition is new to me, whatever it is. The definition I know is:

An Arrhenius acid is a substance that dissociates in water to form $\ce{H+}$ ions.
An Arrhenius base is a substance that dissociates in water to form $\ce{OH-}$ ions.

The key concept to answer the question is the self-ionisation of water given in equation $(1)$.
$$\ce{H2O <<=> H+ + OH-}\tag{1}$$
This ionisation happens in water, and the ion product as written in equation $(2)$ is $10^{-14}$.
$$K_\mathrm{w} = [\ce{H+}] [\ce{OH-}] = 10^{-14}\tag{2}$$
Thus, water qualifies as a substance that dissociates in water to form $\ce{H+}$ ions. It also qualifies as a substance that dissociates in water to form $\ce{OH-}$ ions. It is both an Arrhenius acid and an Arrhenius base and thus the only Arrhenius amphoteric compound.

Aren't there other molecules which can donate both $OH^-$ and $H^+$? (I can imagine some organic ones, though I have no examples at hand.) Those would also be Arrhenius amphoteric, right?

@PaŭloEbermann I’m pretty sure there aren’t. Frankly, I don’t know a single organic molecule that dissociates to give $\ce{OH-}$. (Side note: Use mhchem for chemical expressions; it prevents italics and inserts better spacing.)

What about this?

@DHMO What about that?

Well, that claims neither acidic nor basic.

@DHMO It claims that the pH value of water will be neither acidic nor basic. The post itself clearly says that water is both an Arrhenius acid and an Arrhenius base.

"dissociates to give $\ce{OH-}$" pretty sure t-butyl alcohol does.

@DHMO Source, please. (Spoiler: I’m pretty sure you won’t find one.)

What is the first step of the SN1 reaction of t-butyl alcohol?

@DHMO Please start by showing whether tert-butanol takes part in $\mathrm{S_N1}$ reactions. Note: Having $\ce{HI}$ or similar as the nucleophile does not count because the hydrogen-tert-butanol cation has different properties (and also dissociates to give water and not hydroxide).


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Pepto-bismol molar bismuth concentration
I want to measure the Bismuth molar concentration in the OTC drug "pepto bismol".

525 mg bismuth subsalicylate (C7H6BiO4) / 30 mL (17.5 mg/mL)
molecular weight 363.10054 g/mol

If I convert to mM (17.5 mg/mL / 363.10054 g/mol * 1000), I get 48.2 mM. However, this value seems very high. Am I missing something obvious?
I am sorry if this question seems basic to you. I can't find the answer anywhere else though. Thank you so much in advance!

Your calculation is correct. You are not missing anything obvious. Your set up, which converts mL to L and then mg to mmol is correct and you get the correct units.
$$\require{cancel}
\mathrm{\dfrac{17.5\ \cancel{mg}}{\cancel{mL}}\times \dfrac{1000\ \cancel{mL}}{L}\times \dfrac{mmol}{364.10054\ \cancel{mg}} =48.2\ mM}$$
This concentration is not "too high". It is what it is. 48.2 mM is 0.0482 M, which is a pretty low concentration.
For comparison, your 17.5 mg per mL is at most 2% by mass (low-balling the density of the suspension to be around 1). 1 M hydrochloric acid (which perhaps seems like a high concentration) is around 3% by mass. Concentrated HCl is 36% by mass and about 12 M.


text:
How to measure the content of sugar, proteins and fat in meat?
I do a research project for Science Symposium and I need to measure a content of proteins, sugar and fat in a meat. Unfortunately, I have found only the methods that require specials apparatus/device that I do not have in my school. I'm a student of secondary school, so I need a simple method. Could you please recommend me one?

Extract with toluene (fat), then extract with water (sugar), dry the rest? The water fraction will also contain sodium & potassium salts.


text:
Why do we use glacial acetic acid in lidocaine synthesis?
The lab sheet says the following :

This reaction is carried out in glacial (anhydrous) acetic acid as the
  solvent. The acetic acid reacts with 2, 6-dimethylaniline to establish
  an equilibrium between it and its acetate salt.

What is the purpose of establishing an equilibrium between 2,6-dimethylaniline and its salt here?
EDIT:
This is the source of the quoted text & the procedure to synthesize lidocaine : https://tigerweb.towson.edu/jdiscord/www/332_lab_info/summer2016chem332labs/webexpt_lidocaine_a.pdf

Please don’t point us to a document but quote and include the contents here.

I think acetic acid is used because it allows you to play funny games with the varying solubility of reactants, products, and by-products. Note that the final product is not soluble in acetic acid/water mixture, but the reactants and salt by-products are. This means that instead of performing an extract in a separatory funnel (extra step, plus you'll need to evaporate solvent and purify), you can just filter the solid product directly from the rest of liquid.
This is the kind of simplifying "magic" you can get when the process chemists get a stab at designing how to run the reaction.


text:
Reaction mechanism in fluoride adsorption to aluminum oxide
Despite a fair amount of research (excluding non-open access journals, to which I have not got access), I cannot seem to find an explanation of the process that takes place when $\ce{F^-}$ adsorbs to activated alumina ($\ce{Al_2O_3}$).
I do know that the process involves both chemisorption (monolayer formation) and physisorption, but since the latter is explained solely by van der Waals forces, I am primarily looking for an explanation of the former.
EDIT: Does adsorption take place only at the aluminium ions? If not, does that mean that all aluminium compounds with oxidation state +3 will be equally good adsorbents?
EDIT: I have found an article (https://link.springer.com/article/10.1023/A:1012929900113) which proposes a mechanism that involves aluminium-fluoride complex formation. If this is the case - why is this even classed as an adsorption reaction at all, and not just a normal chemical reaction? And what role do the $\ce{OH^-}$ ions play? Do they keep the aluminum in the ionic lattice of the alumina?

My impression was that NaOH was used as part of a regeneration process so the alumina can be reused.  Maybe because, as it says in the abstract you site, that absorption best occurs under slightly acidic conditions so under basic conditions the process is reversed?  Just an idea, not an answer ;)

This may be of interest.

Besides the presence of Al and O atoms on an Aluminum Oxide surface, there are also H atoms, which are bounded to the solid by OH bonds, - O --- H+ . Direct evidence for the presence of H atoms, and OH bonds, come from XPS or FTIR (grazing incidence) data. Indirect evidence comes from several sources, but especially from the reactivity of the surface to some compounds, like Silanes, where Si --- O --- R groups react readily with OH. Based on what is said above, I´d say the best explanation for the F– anion interaction with the Aluminum Oxide surface is a chemisorption process, through an indirect process,  bonding of the F– anion to the  H+ of the surface, forming an -O --- H+ --- F– , anion (F–) – to – dipole (+H  --- O– ) complex structure. A real solid – state Hydrogen bonding.


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Why do halogen substituents make molecules more lipophilic?
According to my medicinal chemistry text book, halogens increase a drug’s lipophilicty. This makes no sense to me for two reasons:

Halogens are all quite electronegative and they will form a $\ce{C-X}$ bond with a large dipole moment. Large dipole moments are associated with increased polarity.
As opposed to a methyl group, halogens are capable of donating their lone pair electrons to participate in hydrogen bonding. We see this when looking at water solubility of methyl-halogen compounds:

$$\begin{array}{cc}\hline
\text{Compound} & \text{Solubility} / (\mathrm{g \ L^{-1}})\\ \hline
\ce{CH4} & 0.055\\
\ce{CH3CH3} & 0.0568\\
\ce{CH3F} & 2.295\\
\ce{CH3Cl} & 5.325\\
\ce{CH3Br} & 15.22\\
\ce{CH3I} & 14\\ \hline \end{array}$$

[halogens are far, far larger than hydrogens, and the large, soft electron shell is nicely lipophilic, unlike the tiny, hard, single-electron shell of hydrogen.
Carbon-fluorine bonds are much more polarized, and depending on the other groups in the molecule, actually seem to participate in hydrogen-bonding-type interactions with some enzymes.](http://www.chemicalforums.com/index.php?topic=52247.0)

Halogens, except for fluorine, aren’t all that electronegative. Chlorine is just slightly less electronegative than nitrogen and bromine is even less again. Iodine almost has carbon’s electronegativity. Thus, there won’t be so much of a dipole moment created by substituting a hydrogen with a halogen. This applies even more if a hydroxy group is replaced with a halogen.
To illustrate this, take dichloromethane. Of all the chlorinated methanes it has the highest permanent dipole moment. Yet it is still almost immiscible with water but fully miscible with hexane. Note by the way that your data somewhat reinforces that: the molecule with the highest dipole moment of all your halocarbons, fluoromethane, has the lowest water solubility.
A second point is that halogens come with free lone pairs. Lone pairs are not polarised per se but can easily be polarised in London interactions. Thus, halogens actually increase molecule’s ability to interact with lipophilic, unpolar media by allowing stronger London forces.
Note also that while nitrogen is capable of forming hydrogen bonds with surrounding aquaeous media, chlorine — in spite of its similar electronegativity — cannot since its lone pairs are much more diffuse. Only fluorine is in theory able to accept hydrogen bonds. In principle, this is another consequence of the larger size of higher-period halogens.
In all this, fluorine is somewhat separate and special. But fluorine is special in itself since perfluorinated molecules are typically insoluble in anything but fluorocarbons. Fluorous extraction is actually a technique used by some green chem labs.
To sum it up, just looking at the simple variable polarity cannot explain the solubility of halocarbons adequately. Their size needs to be considered.

1.) Actually Jan, fluoromethanes poor solubility in water has to with its unwillingness to form hydrogen bonds compared with the other halocarbons. In fact, I would bet fluoromethane would also have the lowest solubility in hexane as well (because of dipole moment, cant find data to confirm this though). 2.) Not sure what your point about halogen polarizability is saying. A polar molecule like water should have a much stronger polarizing effect (dipole-dipole bonds) on halogens than nonpolar molecules. London dispersion forces are the weakest of all the intermolecular interactions.

@Nova 2) polariseablility $\ne$ permanent dipole. Permanent dipoles interact with other permanent dipoles preferentially, and something that is ‘just’ large and polariseable does not interact well with permanent dipoles. Like dissolves like.

I severely contest the opinion that chloro- and bromomethane accept hydrogen bonds from water.

Hmm, I thought that due to flourines high electronegativity, it was unwilling to share electron density to form hydrogen bonds. And I thought I- was too disperse to form strong hydrogen bonds. But Br and Cl were in the middle and not too big and not too electronegative.

@Nova I don’t know about your corner of the world, but in my place in school hydrogen bonds are taught as happening between $\ce{N,O,F}$ and a hydrogen bound to one of those, so … ;)

Ok, so how can you explain the order of the halocarbons solubility I have in my question? Why does CH3Br form stronger bonds with water than CH3Cl and CH3F?

@Nova It does not ‘form stronger bonds’, it just ‘dissolves better’; big difference. And to be honest I’m not sure why. I would have expected both to hydrolyse rather rapidly and not really be soluble to a measurable extent.

To quote this page:

That depends on the halogen. The electronegativity difference between carbon and hydrogen isn't that much different from that between carbon and bromine, and is more than the difference between carbon and iodine. It is just oriented in the other direction. In addition, halogens are far, far larger than hydrogens, and the large, soft electron shell is nicely lipophilic, unlike the tiny, hard, single-electron shell of hydrogen.
Carbon-fluorine bonds are much more polarized, and depending on the other groups in the molecule, actually seem to participate in hydrogen-bonding-type interactions with some enzymes.

No copyright infringement intended.

No copyright infringement intended. You've been reading too many YouTube video descriptions....

@hBy2Py That is true.


text:
Why are the secondary alpha hydrogens more acidic than primary alpha hydrogens (keto-enol tautomerization)?
I learned in organic chemistry that secondary/tertiary carbanions are less stable than primary carbanions and therefore primary hydrogens are MORE acidic than secondary/tertiary hydrogens.
But lets take 3-methyl-2-butanone as an example. It seems that the more substituted alpha carbon is more acidic than the nearby primary carbon. In other words, it seems that the keto-enol tautomerization prefers to form a C=C with the secondary carbon over the primary carbon.
Why would a base prefer to deprotonate the secondary carbon over the primary carbon to trigger keto-enol tautomerization over the theoretically more acidic primary carbon?

The acidity really only applies to direct $\ce{C-H}$ acidity. Here, you are generating a more stable double bond.

@Jan When hydrogen is removed by a base, there is a resonance form that contains a double bond between carbonyl carbon and alfa carbon. Wouldn't the stability of this resonance form over the terminal alkene make the central hydrogen more acidic?

@Adroit I think we are saying the same thing but in different ways.

@Jan would you agree that the pKa (being a measure of H3O+ concentration at eqm) of the central hydrogen is lower than that of the terminal one ?

@Adroit Yes, the most acidic position of an aldehyde is the alpha proton.

You're actually asking two questions. There is both kinetic and thermodynamic acidity. There are times when performing deprotonation at low temperature will yield the kinetically favored conjugate base, i.e., the proton that is most exposed. If you increase the temperature of this intermediate, you can get to the point where proton transfers happen in reaction vessel to create the more stable thermodynamically favored conjugate base. Usually, this is the more substituted one.


text:
Second Peak in X-Ray Diffraction Pattern of bcc Metal
By Bragg’s law,
$$2d\sin\theta=n\lambda$$
and
$$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$
Combining the two gives:
$$\sin\theta=\frac{n\lambda\sqrt{h^2+k^2+l^2}}{2a}$$
By the reflection rule for bcc, there will be destructive interference when $\sqrt{h^2+k^2+l^2}=\sqrt{1}$, on the (100) plane. The first peak will thus occur when $n=1$ and $\sqrt{h^2+k^2+l^2}=\sqrt{2}$, on the (110) plane. However, I’m confused about the second peak. It will be impossible to get $\sqrt3$ while satisfying the reflection rule for bcc, so I’ve moved on to $\sqrt4$, which is equal to 2. However, mathematically there is no difference between having $n=2,\sqrt{h^2+k^2+l^2}=1$ ((100) plane) and $n=1,\sqrt{h^2+k^2+l^2}=2$ ((200) plane). Which one does the second peak represent, then? Does it represent both? Why then is this peak so much smaller than the other peaks in most diffraction patterns that I see? For example, for $\alpha$-iron, the diffraction pattern is

Image Link: http://questions.transtutors.com/Transtutors001/Images/Transtutors001_6804131b-7ca5-47cd-984a-b824a4926c3a.PNG

Have a look at the rules for systematic absences in a powder spectrum, for example, for body centred cubic crystals, reflections interfere destructively only when $h+k+l=$ odd and so these are absent.

@porphyrin I have made a mistake in including the (100) plane. Thank you for pointing that out. I have corrected my question.

The reflection with $n=2,\;(h,k,l)=(1,0,0)$ is the same as $n=1,(h,k,l)=(2,0,0)$. Yes, I said "same", not "similar"; they are one thing, there are no "both". For this reason, you may just as well abandon using $n$ altogether.
As for why the peak is so small compared to others you've seen, let me respectfully suggest that maybe you haven't seen all that many. Indeed, there is no law that all peaks on the diffraction pattern must be about the same size. On the contrary, they are essentially proportional to the squares of Fourier coefficients of electron density, and as such, can be anything from 0 to the maximum. The early textbook exercises usually employ some mercifully simple cubic or tetragonal structures with few well-resolved and well-developed peaks. Not all compounds are like that. Look at the graphite diffraction pattern, for instance: you'll basically see a lonely (002) peak, large as life, and only after the most careful inspection you'll start noticing the rest of them amidst the background noise.
As for the (100) peak, it couldn't be there, since it fell victim to the systematic absences of the bcc lattice. Ditto for (111) and (210). Only those with even $h+k+l$ survived, and that's precisely what you see on the picture.

Thank you. However, I'm not quite understanding your first assertion. Could you further explain why they are exactly the same (perhaps by way of a diagram)?

What diagram? $1\cdot\sqrt{2^2+0^2+0^2}=2\cdot\sqrt{1^2+0^2+0^2}$, there is nothing more to it.

Yes, the math shows that. But I don’t see why physically that’s the case. Shouldn’t the diffractometer detect a constructive maxima arising from the n = 2, (100) case, and another constructive maxima arising from the n = 1, (200) case? They are different planes, so would the second diffraction peak reflect both contributions? You say they are exactly the same thing, but I don’t see how physically that could be.

Just how would it detect these supposed two maxima, if they appear at exactly the same angle? As for the planes, they are but a mathematical abstraction. There are no planes in crystals, much like there are no huge walls along all meridians and parallels on Earth, even though one might get that impression from looking at the maps.

To answer the last part of your question the amplitude of the scatted x-ray is proportional to what is called the structure factor $F_{hkl}$ and the measured intensity to  $|F_{hkl}|^2$.
If the unit cell contains i atoms with a scattering factor of $f_i$ each of the atoms are at the coordinate $ax_i+by_i+cz_i$ within the unit cell of  where a, b, c are the unit cell dimension, and $x, y, z$ are the fractions of a, b, c. For example an atom at the mid-point of a has $x=1/2$. The scattering factors $f_i$ have a complicated form depending on scattering angle but for most use can be considered to be a constant proportional the the number of electrons in each atom.
The structure factor is defined as $$F_{hkl}=\Sigma _i f_i e^{i\phi_i(h,k,l)}$$ where the phase is defined as $\phi_i(h,k,l) = 2\pi(hx_i+jy_i+lz_i)$. (If you are unfamiliar with this notation you can convert to sine and cos using $e^{i\phi} = cos(\phi) + i~sin(\phi)$) where  $i=\sqrt(-1)$
The phase arises because every atom in the unit cell contributes to every reflection and so to every {hkl} but because most atoms are not on an {hkl} they contribute by scattering in proportion to their distance between the parallel planes that make up each {hkl}. As these distances are different this results is a slightly different path length  and this appears as a scattered wave with a different peak position (a phase delay) relative to a scattered wave from an atom on the {hkl} and this alters the total wave amplitude when waves from all atoms are added up.
The calculation is easy with cubic lattices, but very tedious, and starts by drawing out the unit cell (in reciprocal space) and labelling all the planes {hkl} and working out $ x, y, z $ for each atom and doing the sum. The scattering factors for atoms the edge and corners of unit cells have to be  taken as fractions of their values, 1/2 for sides and 1/8 for corners for example. Because $cos(\phi(hkl))$ can vary between $\pm1$ it is easy to see how a variation in intensity occurs.  The rules for cubic lattices are that all {hkl} are present for the primitave cubic, $h+k+l=$ odd are absent in bcc and for face centred lattices for a reflection to be present h, k, l are all odd or all even.


text:
Why is Nitrogen 15 found almost exclusively in marine environments?
Prelude
The following story was told to me earlier this summer
Every year during October, the Pacific rain creates an ideal spawning condition for the wild Pacific salmon in the coastal forests of British Columbia. Of course, when we think of salmon swimming upstream in the Pacific Northwest, some images might come to mind.

Black bears and grizzly bears carry hundreds of kilos of fish out of the water and into the forest to consume. Naturally in the areas proximal to the rivers, fish carcasses are littered throughout, with the bears having ate what they wanted, leaving behind the rest to decompose.
Scientists several years ago took to examining the nitrogen uptake and found that trees in nearby the river received a significant amount of their nutrients from the fish. Through their findings they showed a case that reverses the land to sea flow of nutrients. The study has been used to argue for the dedication of resources to rebuild and maintain wild pacific salmon runs.
Source: David Suzuki Foundation

Chemistry and the Question
The study above uses the abundance of $\ce {^{15}N}$ found naturally in the ocean, and transported via salmon, to support their findings. $\ce {^{15}N}$ has a 0.365% natural abundance, and according to the story, it is almost exclusively found in marine environments. Which leads us to my question: why? Why is this isotope largely found in the ocean and not on the land?

Let's start that the sentence found in the source you provided:

the nitrogen isotope 15, which is found almost exclusively in marine environments

is misleading and simply incorrect.
Now to some basics. When we discuss isotopes in environmental sciences, we never use the absolute values of an isotope content. Instead we use isotope ratios:
$$\text{Ratio (R)} = \frac{\text{heavy isotope abundance}}{\text{light isotope abundance}}$$
However, the ratios themself are not very useful, so the delta ($\delta$) notation is used:
$$\delta[^0\!\!/\!_{00}] = \left(\frac{R_{\mathrm{sample}}}{R_{\mathrm{standard}}}-1\right)\times1000$$
All standards have the delta value equal $0$ by convention. There is an international standard material for each isotope, for example for $\ce{^13C/^12C}$ ratio it is VPDB (Vienna PeeDee Belemnite), and for $\ce{^15N/^14N}$ it is atmospheric nitrogen. The latter standard was chosen because the content of the $\ce{^15N}$ isotope in atmospheric nitrogen is uniform throughout the world and it does not vary significantly over time. Many chemical, biochemical and physical processes cause the so-called isotopic fractionation to occur - it results in the increased (enrichment) or decreased (depletion) isotope ratios.
OK, we know the terminology, so now to the point. In general, the metabolic processes of living organisms cause enrichment of nitrogen. The extent of the enrichment depends on the trophic level (position in the food chain) and dietary sources of nitrogen. The highest enrichment is found in terrestial herbivores, while marine organisms usually show lower $\delta\ce{^14N}$ values. Given those facts the cited sentence makes no sense, and is probably a result of a misinterpretation of some scientific paper.
There is a lot of literature on the subject if you are interested. I can give you one good reference as a starting point:
M.A. Vanderklift, S. Ponsard. Sources of variation in consumer-diet
$\delta\ce{^15N}$ enrichment: a meta-analysis. Oecologia (2003) 136:169–182 DOI 10.1007/s00442-003-1270-z.


text:
How to get the condensed electron configuration for Zinc ion
The question said "Write the condensed electron configuration for $\ce{Zn^{2+}}$I got [Ar] 4s$^2+$ 3d$^8$. Why was this the incorrect answer? Zinc lost 2 electrons and surely those electrons should go from the d orbitals?

By the time the $\ce{3d}$ orbitals are filled with electrons, the $\ce{4s}$ orbitals have become higher in energy, making it easier to remove them. Here are energy level curves for both the neutral and +1 forms of the $\ce{3d}$ atoms, with electron energy on the $y$-axis:

you have done only wrong thing and that is sufficient to make your answer wrong .
Electron will not be removed from 3d subshell but it will be removed from 4s subshell. The reason is whenever we fill electrons we see the value of (n+l){Principal quantum number + azimuthal quantum number}, the subshell having smaller value of (n+l) gets the electroon first, butt when we have to remove an electron,it is removed froom the subshell having greatest value of  n(Principal quantum number).

I don't get it, how do you go from (n+l) to the electron removing from the subshell having greatest value of n?

see,by aufbau principal the electrons are filled in subshells in order of their increasing energy and order of increasing energy can be described with the help of n+l,now as far as use of n is concerned we see that in removing the electrons concept of shells is taken into account(not subshell) and quantum number associated with shells is 'n'.

see (n+l) and aufbau principal on wikipedia .


text:
How many carbons does an alkane need to have an enantiomer?
How many carbon atoms does an alkane need before it is capable of existing in enantiomeric forms?

If we allow our substituents to be isotopes, then 2 is all that is really necessary:
$\hspace{6cm}$
If we exclude such compounds from consideration, then cyclopropanes will give you smallest number of carbons at 5:
$\hspace{5.5cm}$

As a side note, for a non-alkane, only 1 carbon is necessary:
$\hspace{5.7cm}$

If we were not confined to saturated compounds, I'd suggest 1,3-dimethyl allene.

Also, come to think of it, carbon is not necessary at all.

If you are only looking for alkanes, of the form $\ce{C_nH_{2n+2}}$, with no isotopic substitution, then the most logical way is to think back to the definition of a chiral carbon. The IUPAC Gold Book writes:

The traditional name for a carbon atom that is attached to four different entities...

So, start with a carbon in the middle.

The smallest substituent you can put on this carbon is hydrogen (zero carbons). The next-smallest is a methyl (one carbon), and after that you need to use an ethyl group (two carbons).

For the last group, you need to use three carbons, because there is no other two-carbon alkyl substituent apart from the ethyl group. So, you could use a n-propyl group (only one isomer depicted here)

or an isopropyl group (again only one isomer depicted)


text:
How does dissolving LDPE in wax work?
I was looking into machinable wax, and found it's based on dissolving LDPE into melted wax. While I understand the normal salt dissolving in water process, I'm not really sure how plastics can dissolve into wax. Do the plastic chains unravel and get suspended in the wax?
Here's a video of the process.

With a salt the individuals will dissolve into the solvent.  In the case of LDPE the solvent molecules go into the LDPE until the LDPE molecules eventually become solvated.


text:
What is the hybridization of diatomic nitrogen, diatomic anything like oyxgen, bromine; hydrogen and chlorine in HCl?
I'm guessing sp hybridization for nitrogen in diatomic nitrogen based on the lone pair on nitrogen and the triple bond but there is no mixing of s and p orbitals so I want to say that there is no hybridization. Are there any chemicals without hybridization? What are some examples of chemicals without hybridization
Why are lone pairs counted in hybridization if they do not contribute to the mixing of the orbitals, a.k.a. hybridization?

Hybridisation is something you typically deduce from the geometry of the compound you are observing. It is primarily a mathematical concept.
That said, to form the diatomic molecules you asked about, no hybridisation is needed. In fact, you get much better results from assuming unhybridised atoms and then combining their atomic orbitals to molecular ones. Except for hydrogen (which only has 1σ and 2σ), the molecular orbitals of all these diatomic compounds ($\ce{N2, O2, F2, Cl2, Br2, I2}$) are basically identical save the energy levels of the MOs and their populations. Thus, these would all count as unhybridised compounds.
Also, there is no nead to invoke hybridisation to explain the bonding situation in $\ce{HCl}$ either. Instead agan consider a σ bond formed between hydrogen’s 1s orbital and chlorine’s 3p. All of chlorine’s other orbitals will sit there happily filled and unhybridised.


text:
How is the Maxwell-Boltzmann curve obtained?
Is it purely qualitative and determined by experiment?
Or is there some function which defines the familiar graph?

(Image source: Wikipedia)

Related: http://chemistry.stackexchange.com/questions/22040/how-to-get-the-first-step-of-the-derivation-of-the-maxwell-boltzmann-distributio?rq=1

There is a theoretical derivation and it is actually quite long. I wouldn't mind typing it up, but I'm not quite free today :( You can probably find it in any physical chemistry textbook.

There is a lovely experiment showing that the maxwell-Boltzmann curve is correct. The experiment observing how a narrow, horizontal  beam Cs atoms fall under gravity . The original paper is Esterman, Simpson & Stern, Phys. Rev. 19487, v71 p238 but is behind a paywall, the nobel lecture  by stern is at http://www.nobelprize.org/nobel_prizes/physics/laureates/1943/stern-lecture.pdf

@porphyrin - the archive for Physical Review is most decidedly not behind a paywall. Go to https://journals.aps.org/pr/issues/71/4 to find the paper.

@porphyrin - Crud. I take that back - you need an institutional, personal, or US public or highs school library. I remember the archive going up originally intending to be wide open. Sigh - must have been too much mass-downloading abuse. Sadness.

There is, indeed, a function that plots the graph.
There are two common types of graph, one with kinetic energy as the $x$-axis, and one with speed. For the most part, they are qualitatively the same, although there is a slight difference in curvature near the origin. The speed function is much more commonly used, because one can more easily extract useful information from it, such as mean velocities, or mean free paths. It is:
$$f(v) = (4\pi v^2) \cdot \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp{\left(-\frac{mv^2}{2kT}\right)}$$
where $m$ refers to the mass of the particle, $k$ Boltzmann's constant, and $T$ the temperature.
As for the derivation, it is rather long and perhaps you were not really asking for it. If you are interested, I would recommend looking it up in a physical chemistry textbook, for example: Atkins' Physical Chemistry, 10th ed., section 1B(b), or Levine's Physical Chemistry, 6th ed., section 14.4. I am partial towards Levine because he gives a much more detailed explanation than Atkins, who is always very brief.


text:
Cram's rule for nucleophilic addition to carbonyl groups
I recently heard about Cram's rule for predicting the stereoselectivity in nucleophilic additions to carbonyl groups. Can anybody explain what this is and whether it is an accurate rule?

Cram's rule, proposed by Donald Cram in 1952,[1] was the earliest model proposed for rationalising the stereoselectivity of nucleophilic additions to carbonyl groups with α-stereocentres. The model involved assigning each α-substituent a relative "size" (small, medium, and large), then placing the carbonyl oxygen antiperiplanar to the largest of these three groups. The nucleophile then attacks the carbonyl group opposite the larger of the two remaining groups (i.e., the medium group). This is best explained with a diagram:

In the reduction of (S)-3-phenylbutan-2-one with L-Selectride (a bulky source of hydride ion), the anti alcohol is formed as the major diastereomer. Cram's rule rationalises this by placing the phenyl group (RL) anti to the carbonyl oxygen. The hydride nucleophile, drawn here as H− for simplicity, then preferentially attacks over the proton (RS) instead of the methyl group (RM).
Although Cram was able to demonstrate that his rule worked for a number of nucleophilic additions, there are some conceptual issues with it. Most obviously, in the reactive conformation, the phenyl group is eclipsing the methyl group on the ketone; this torsional strain is worsened when the nucleophile adds into the carbonyl, as the resultant product is initially in a staggered conformation.
In 1968, Hugh Felkin pointed out two observations which the Cram model could not explain.[2] One is to do with the reduction of cyclohexanones, which we will not go into here, as nowadays a slightly different model is used to explain reduction of cyclic ketones. The other was about how the stereoselectivity of C=O reduction varied when the ketone was made progressively  more bulky.

According to Cram's model, as the size of R increased, the steric repulsions between R and the phenyl group (RL) should increase, leading to a decrease in stereoselectivity of reduction. However, Felkin showed that the opposite was true: the anti product was formed with higher stereoselectivity when R increased in size from Me to t-Bu.
To cut a long story short, Cram's rule was eventually replaced by the Felkin–Anh model. Here, the large α-substituent RL was instead placed perpendicular to the C=O bond. This leads to a choice of two possible conformations. The nucleophile attacks the carbonyl group along the Bürgi–Dunitz angle, and the attack over RS instead of over RM is favoured:

Like Cram's rule, the Felkin–Anh model correctly predicts the diastereoselectivity observed in this reduction. However, it also allows us to rationalise why the stereoselectivity increases as the size of R increases. A larger R makes the gauche interactions between R and Ph more destabilising, which biases the conformational equilibrium towards the reactive conformer (on the right).
For more detail about the Felkin–Anh rule and other modifications which are necessary (depending on the exact substrate and conditions), see: Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones and the answers therein.

References

Cram, D. J.; Elhafez, F. A. A. Studies in Stereochemistry. X. The Rule of “Steric Control of Asymmetric Induction” in the Syntheses of Acyclic Systems. J. Am. Chem. Soc. 1952, 74 (23), 5828–5835. DOI: 10.1021/ja01143a007.
Chérest, M.; Felkin, H.; Prudent, N. Torsional strain involving partial bonds. The stereochemistry of the lithium aluminium hydride reduction of some simple open-chain ketones. Tetrahedron Lett. 1968, 9 (18), 2199–2204. DOI: 10.1016/S0040-4039(00)89719-1.

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text:
Hydrogen bonding in alcohols vs amines
Consider an alcohol and an amine compound with roughly the same molar mass. If I understand correctly, the boiling point for the alcohol is greater for two reasons:

The $\ce{O-H}$ bond is more polar than $\ce{N-H}$ because oxygen is more electronegative than nitrogen.
The oxygen atom has two lone pairs of electrons to act as hydrogen acceptors, whereas the nitrogen has only one.

The question is, which one of these reasons (if they are valid) contributes more to the increase in boiling point?

Related to your 2nd point, the partially negative oxygen, making only two bonds with neighbors, spacially presents more of itself to neighbors possibly increasing the network of attractions. Also,the partial positives of the amine are distributed over three moities instead of two. This, combined with nitrogen's lesser electronegativity, would make them less positive than those attached to the oxygen.

Nitrogen has one lone pair but three hydrogens. Water has two lone pairs and two hydrogens. Thus water can form a better network of hydrogen bonded molecules.

Both reasons given are valid and contribut to alcohols having higher boiling points than amines, with one caveat. It is not stated whether the amine is primary, secondary or tertiary, and a tertiary amine has no $\ce{N-H}$ bond.
Although the $\ce{O-H}$ bond is more polar than the $\ce{N-H}$ bond and can therefore form somewhat stronger hydrogen bonds, the fact that an alcohol has two lone pairs is the main reason for the higher boiling point.
Even though a primary amine has two hydrogens that can act as hydrogen bond donors, their already weaker $\delta +$ is spread among two hydrogens, making those hydrogen bonds very weak, effectively limiting them to act as a single hydrogen bond donor.
So, in general, an amine can only participate as a single hydrogen bond acceptor and donor (except for tertiary amines), while an alcohol's two lone pairs and single hydrogen allow for a much stronger hydrogen bonding network.


text:
Can the directionality of an SN2 reaction depend on solvent polarity?
Because of the effects of solvation, halide ions have differing nucleophilicity in protic solvents vs aprotic solvents, iodide being the best nucleophile in protic solvents and fluoride being best in aprotic solvents.  For the sake of simplicity, fluoride will be ignored due to its high electronegativity and unusual reactive properties, and chloride will be used instead.
So would that mean that an alkyl chloride will react with HI in a protic solvent to form the alkyl iodide?  And the alkyl chloride will not react with HI in an aprotic solvent?  And by the same token, would an alkyl iodide react with HCl to form alkyl chloride in an aprotic solvent but not in a protic solvent?
I also realize that iodide has a much greater polarizability than chloride and that must be taken into account as well.  Iodide is more a thus a more stable anion than chloride, making iodide much better leaving group than chloride.
But which factor will have a greater overall effect?  Do aprotic/protic solvents affect the leaving group capability (ie although chloride is less polarizable and more unstable relative to iodide, after leaving in a protic solvent the H-bonding "shell" will stabilize the anion and prevent it from reacting with the alkyl iodide formed.  Conversely in aprotic solvent, the chloride ion would not be stabilized after leaving and thus could reattack the alkyl iodide to reform the alkyl chloride)?

https://en.wikipedia.org/wiki/Finkelstein_reaction

@DHMO Exploiting solubility of products would be the more efficient route, but my question is more conceptually oriented, though it does make implications that support my statements.  Acetone is polar aprotic and thus chlorides and bromides would be more nucleophilic than iodide, causing the reaction to favor the alkyl chloride or bromide side over the iodide.  However the reaction is "driven forward" by exploiting the solubility of sodium bromide and chloride salts in acetone.  There wasnt an example of an iodide salt displacing an alkyl chloride or bromide in a protic solvent though.

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text:
Is acetic acid aggressive to rubber?
There seems to be an endless debate about using vinegar as rinse aid in dishwashers, which is explicitly allowed in some models (such as Miele, see p. 25) but usually not mentioned in user's manual. The main contra argument is that acetic acid is somehow destroying the rubber seals, leading to malfunction. On the other hand, common commercial rinse aids also contain various acids (e.g. see here, 5.7% glycolic acid). Also, I haven't heard about rubber being sensitive to acids, at least to weak ones.
Is there a specific property of acetic acid I'm not aware of that makes it aggressive to rubber?
PS. I have asked a related question here.

I think you thing that's missing in your question is what kind of "rubber?" There are all sorts of rubbery plastics that could be used as O-rings or what not, and these substances don't all behave the same way towards acids...

@Zhe If you have an example of a polymer which is sensitive to acetic acid while being tolerant to citric or glycolic acid, that would constitute an answer I'd happily upvote.

No, I don't. I'm not a polymer chemist, but in principle, this is something that could be the case.


text:
How can solids undergo combustion if it only occurs in the gas phase?
According to the smoldering Wikipedia article, flaming combustion occurs in the gaseous phase. I know butane undergoes flaming combustion in a Bunsen burner because it's a gas, but how can wood undergo flaming combustion at all instead of just smouldering? What gas is there to react with oxygen in the flames? Does it smoulder to create carbon monoxide then the carbon monoxide undergoes flaming combustion? Can charcoal also undergo flaming combustion? I've never seen a piece of charcoal without wood inside it undergo flaming combustion, only smoldering.
Also, when oil undergoes flaming combustion, is it its vapour that reacts with oxygen to produce a flame?

I can't remember my exact state of mind at the time of asking the question which I probably wouldn't be able to determine precisely anyway because the brain does everything as a complicated system of discretion. It's okay but it seems a bit silly this question. I'm kind of like "Of course, it's flaming combustion that can only occur in the gas phase." I once learned in scouts that combustion only occurs in the gas phase. Maybe I hadn't quite figured out the answer. It seems so obvious to me now. Flames must be a combustion occuring in the gas phase. I think I once though it was the blackbody.

radiation of the air but now think it's actual the photons released in the chemical reaction and the actual amount of blackbody radiation in the visible light would be very tiny especially for something extremely transparent like air. Combustion can occur with oxygen reacting with a solid but that's smouldering, not flaming combustion.

Combustion is a gaseous phase phenomenon. Oil and gasoline have a high enough vapor pressure at ambient temperatures to produce a gaseous phase of fuel above the liquid. In contrast, hold a lighter up to a piece of wood and try to get it to light. It won't—at least not for quite some time. This is because solid fuels must first undergo endothermic pyrolysis before real combustion can occur. This produces a slew of products which are what truly undergo combustion when solids are burned. Because combustion is exothermic, once a high enough temperature is reached the solid will autopyrolyze, making combustion a self-sustaining reaction.

This is also why a candle uses a wick - candle wax under normal conditions won't sustain a flame.

I read that charcoal is nearly pure carbon. Is wood a sponge like structure that's only a few atoms thick? If not, how can it possibly undergo incomplete flaming combustion to become charcoal instead instead of just forming a passivating charcoal layer which blocks further endothermic pyrolysis?

Is there also a nonporous solid that can undergo flaming combustion for the following reason? Oxygen directly attacks the surface to create an intermadiate product releasing heat in the process and then the intermediate product slowly reacts with oxygen to produce the final product releasing even more heat and it's because the reaction to form the final product isn't very fast that the reaction is preceived as flaming combustion instead of smouldering. Can carbon undergo flaming combustion in that way with carbon monoxide being the intermediate produce if it's hot enough?

Except it's not. Titanium's ignition temperature is below its melting temperature.


text:
Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes?
I am trying to calculate the relationship between the octahedral field splitting parameter ($\Delta_\mathrm{o}$) and the square planar field splitting parameter ($\Delta_\mathrm{sp}$) and thought a good way to start would be to understand why $\Delta_\mathrm{t}=4/9\Delta_\mathrm{o}$ for tetrahedral fields but couldn't find any information on it.
How does one arrive at this particular ratio mathematically?

Related: Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?

I decided to edit and vote for reopening. The linked duplicate asks why one is greater than the other, this one asks about the numerical difference and how it is derived.

The ratio is derived in The angular overlap model. How to use it and why J. Chem. Educ., vol. 51, page 633-640.
First, relative expressions (Shaffer Angular Overlap Factors) are derived for overlap integrals between metal and ligand orbitals as a function of angles.
For ligand sigma orbitals, the expressions are a functions of 2 angles, specifying the direction of the ligands in the coordinate system of the metal.
For pi-x and pi-y ligand orbits, there is a third angle for orientation about the ligand-metal axis.
Then, assuming the ligand-metal distances are the same in both the tetrahedral and octahedral cases, and considering the coordinate geometry of the ligands in the two cases, the relationship:
$$\Delta_{T} = -4/9 \Delta_{O}$$
is obtained.
For another discussion of this ratio in the framework of the angular overlap model, as well as extension to other geometries such as square planar, see A New Look at Structure and Bonding in Transition Metal Complexes, Advances in Inorganic Chemistry and Radiochemistry, Volume 21, pages 113-146.


text:
Is it possible to synthesize magnesium nitrate through an aqueous solution of ammonuim nitrate and magnesium sulfate?
Is it possible to synthesize magnesium nitrate through an aqueous solution of ammonium nitrate and magnesium sulfate? I am attempting to determine whether the sulfate or nitrate radical has a higher electronegativity.

It is not possible as both salts are soluble in water. In order for a reaction between two salts to occur you would need to have some driving force for it, such as a formation of an insoluble solid or gas evolution which is impossible in this case as both the products are also soluble in water. In fact, if you mix ammonium nitrate and ammonium sulfate, you would just end up with a solution of the constituent ions, nothing more.
Also, the sulfate and nitrate are ions, not radicals, and electronegativity is a property of atoms.

Well, technically, if you strip these ions of extra electrons, they would become radicals. (Not that it was relevant to the question at hand, anyway.)

If you will evaporate the resulting solution you will most likely get a mixture of magnesium nitrate, magnesium sulfate, ammonium nitrate and ammonium sulfate. So technically it is a method of synthesis of magnesium nitrate, but totally impractical.


text:
How many significant figures are appropriate when a new digit is created?
In my chem class the other day, we were working through a question about average atomic mass. We were given the following information about two isotopes of boron:
Boron-10: Atomic Mass = 10.013, % in nature: 19.8%
Boron-11: Atomic Mass = 11.009, % in nature: 80.2%
For the weighted average (before sig fig rounding), then, Boron 10 contributes 10.013*.198 = 1.982574 AMU, and Boron 11 contributes 11.009*.802 = 8.829218 AMU.
Now, if we were just finding one of these values, we would report 1.98 or 8.93 AMU (to 3 sig figs). Adding these together, we get 10.81 AMU. However, if we follow "convention" and do all rounding at the end, we get 10.8 AMU (3 sig figs).
My question is, which of these is correct? On the one hand, 10.8 is correct since it has the same number of sig figs as the least accurate given value (19.8% or 80.2%). On the other, why should we go from hundredths to tenths precision just because we got a new digit in the least precise position? One could argue that the "10" is essentially its own digit.
Personally, I agree with the response here (Significant Figures Interpretation) that says sig figs don't make a whole lot of sense. But for the purposes of this question, which is the proper way to express the answer?
Thanks in advance!

The way significant figures is taught is a shortcut for a more exacting method since the exact precision of the original figures isn't typically given. Thus 0.198 is assumed to be +/-0.0005.
Many science papers will use something like 0.1976 +/-0.0014 to indicate the "true" standard deviation of the value. "True" meaning of course the best estimate since the true standard deviation cannot be known. The relative error is thus $\dfrac{0.0014}{0.1976} = 0.709 \%$. This is rather inconvient for calculating the error prorogation since it isn't +/- 1 digit in the least significant digit.
Obviously 10.013*0.198 = 1.982574 AMU is unjustifiable. Having seven significant figures is just wrong. The weighted value for the AMU can't have seven significant figures when the weight per cent is only known to three significant figures. However typically extra significant figures are carried in intermediate calculations and only the final answer is rounded.
Carrying such extra digits in intermediate calculations isn't a problem with a modern calculator. In the "olden days" when a slide rule was used three significant figures in intermediate calculations were the limit. Log tables allowed four significant figures.
So the gist is that since the percentages have three significant figures, then the answer should be rounded to three significant figures. Thus the answer is indeed 10.8 AMU.
Now using a modern computer all the values could be randomly sampled. So the value of 0.198 could for instance be assumed to have a standard deviation of 0.0005 and other variables treated the same way. Then the sampled values could be used for a series of calculations. A mean and a standard deviation of the various final answers could then be calculated. Such grinding is really beyond reasonable human effort but no problem for a computer.

Actually, you are correct - when adding significant figures together, you keep the largest last significant figure between the two numbers. For example, $15.31+2.2=17.5$. On the other hand, when you are multiplying numbers, you keep the same number of significant figures. Using the same numbers, $15.31*2.2=34$. While this decision might appear arbitrary, there is a good reason behind it.
Since we do not know anything past the tenths place about $2.2$, lets say this number is actually $2.2x$ where $x$ is a digit that would cause this number to round down (value between 0 and 4). $15.31+2.2x = 17.5[x+1]$ (where a number in brackets is a digit that is the result of a calculation. It can carry to the next place, but for reasons that I will explain later, we assume that it does not). Since we now know nothing about the hundredths place after adding these two numbers, we cannot include it in our result.
Likewise for multiplication, $15.31 * 2.2x = 33.[6+x][8+5x][2+3x][x].$ Since we know nothing about the value of $x$, we know nothing after the decimal point. Because our next digit is guaranteed to round up, regardless of the value of $x$, we round up to get $15.31*2.2x=34$. This is not a special case. In fact - the number of digits that will not depend on the value of $x$ is always equal to the number of significant figures in that number (this is - again - because we assume that $x$ will not cause a digit to carry), thus giving us our multiplication rule for significant figures.
Now for why it is okay to assume that $x$ will not cause a digit to carry. When you record an experimental value, you are supposed to record one more digit than your device can differentiate (ex. guessing the location of a meniscus between two numbers in a graduated cylinder). As such, the final digit that you use in your calculations is inaccurate anyways. Because of this, it is safe to assume that $x$ did not cause any additional overflow. If we were to assume that x overflowed, then it could cause your doubtful digit to also carry, which would result in numbers that disagree with the measured value.
I hope this helps (I know I went into a lot more detail than you were looking for).


text:
Does a difference in IR signal size from 'normal' provide information on amount of material?
I have an IR spectrum that has signals for my expected product but also signals from what I believe to be some reactant (an alcohol). Typically, an O-H signal is pretty strong, but this O-H signal has an absorbance of only ~35%. Can I make a conclusion about the relative amount of reactant?

No. Frankly, in organic chemistry, IR is usually just used as support for what functional groups you think should be on your molecule. I only obtained IR spectra because you need to include them with supporting information when you want your paper reviewed.


text:
Are IR signals affected by nearby atoms?
E.g. If there is a methyl from an aldehyde vs a methyl at the end of a hydrocarbon chain, will the difference in neighboring atoms have an effect on the location of the IR signal?

I'm a bit confused. Why are you trying to perform structure determination from IR? You get much more information generally if you just use NMR or GCMS.

It's more about whether it is possible to obtain this information than an interest in practical use.

Not so sure about IR, but near-edge x-ray absorption fine structure (NEXAFS) is certainly a well known technique in surface science for determining relative atomic positions precisely through interference of the surrounding atoms on the x-ray emission.

It's not really practical to do this via IR. IR works on bond lengths. So if the difference is negligible, instrument noise is going to dwarf anything useful you'll pull out.

As was already said,

IR works on bond lengths. So if the difference is negligible, instrument noise is going to dwarf anything useful you'll pull out.

There will be a difference, and you can definitely detect it computationally, but by normal experimental means (benchtop FT-IR or ATR, not a gas-phase experiment) background noise might the same order of magnitude as any frequency shift you'd see.
For example, the carbonyl stretching frequency and carbonyl bond lengths calculated using B97-D/6-31G(d):
$$
\newcommand{\AA}{\unicode{x212B}}
\begin{array}{llll}
           & \text{bond length} \, (\AA) & \text{frequency, calc.} \, (\text{cm}^{-1}) & \text{frequency, expt.} \, (\text{cm}^{-1}) \\ \hline
\text{pentanal}   & 1.218 & 1772 & 1727 \\
\text{2-hexanone} & 1.223 & 1753 & 1718
\end{array}
$$
Technically, you can distinguish between these in an experiment, but one would be more interested in the presence/absence of a large aldehyde band. This works due to the sharpness of the carbonyl stretch.
Closer to what I think you're asking about is the influence on vibrational frequencies of groups far from the area of substitution. These particular examples are tough because the normal modes of hydrocarbon-like species tend to couple the motion of many $\ce{C-H}$ stretches together. With that in mind, here are the highest-frequency $\ce{C-H}$ stretching modes, which happen to belong to the terminal $\ce{CH3}$ groups. The bond length is for the hydrogen eclipsed with the carbonyl.
$$
\newcommand{\AA}{\unicode{x212B}}
\begin{array}{lll}
           & \text{bond length} \, (\AA) & \text{frequency, calc.} \, (\text{cm}^{-1}) \\ \hline
\text{pentanal}   & 1.102 & 3044 \\
\text{2-hexanone} & 1.102 & 3038
\end{array}
$$
"Wow, that's amazing, you can resolve it!" Not this time. Compare the full spectra for pentanal/valeraldehyde (top) and 2-hexanone (bottom) and try to perform the peak assignment. Is it it the 2960/2962 pair? 2935/2936?

Experimental values/images taken from the Spectral Database for Organic Compounds.


text:
Why Acetoxy group is ortho/para directing although it shows contradictory behaviour

Why acetoxy group is ortho para directing and not meta directing?

The first line in that image is an incorrect resonance structure. That the ester oxygen would have 4 bonds and a lone pair.
The bottom resonance structure is correct.

Why the first resonating structure is incorrect?? Would you please explain...

Got it...!!!  Thanx a lot..!!!


text:
Is it possible to react chlorine with hydrogen under a UV lamp during the electrolysis of salty water?
I have set up an experiment where I am generating chlorine by electrolysing water saturated in NaCl. I know chlorine gas is actually being generated because I hang wet litmus paper in the gas which quickly bleaches. I also know hydrogen is being generated because if I hold a match to the gas it reacts with a popping sound.
If I have this set up in a closed container, how come I cannot get the hydrogen and chlorine to react under a black light? What am I missing?

I advise you not to attempt a reaction of hydrogen and chlorine gas by light!
Do not attempt it unless you know exactly what you are doing and have seen the experiment be performed by others!
You’re actually rather lucky that one of the following happened:

the black light gave the wrong wavelength
there were too small concentrations of hydrogen gas and chlorine gas to react with each other.

The reaction of hydrogen and chlorine is a radical chain reaction and rather exothermic. There is no change in the amount of substance involved, so entropic effects do not play a role. However, it is an explosive reaction and was often used in the introductory inorganic chemistry course at my university to give off a loud bang (and send the tin can in which the gases were mixed flying).
The reaction is triggered by light energy interacting with chlorine; the first (initiation) step is the generation of chlorine radicals by the following equation:
$$\ce{Cl2 ->[$h\nu$] 2 Cl^.}\tag{1}$$
This chain then propagates according to reactions $(2)$ and $(3)$.
$$\ce{Cl^. + H2 -> HCl + H^.}\tag{2}$$
$$\ce{H^. + Cl2 -> HCl + Cl^.}\tag{3}$$
The light needed for the first step must supply chlorine’s bond dissociation energy; blue (visible) light, not UV light is actually required. But just for your safety please let me restate:
The reaction is explosive! Do not attempt it at home under any circumstance! Do not attempt it unless you know exactly what you are doing!

Im running the experiment for work, I am taking all safety precautions and its very important I find a way to get the two to react. Having said that I appreciate the input, So blue light hey? I still cannot replicate the conditions which lead to explosion, Im wondering are there any possible mechanisms in the electrolysis of brine that could be inhibiting the reaction? for example, moisture? Oxygen Levels? if I turn the voltage up on the electrodes is it possible that I start to electrolyse water instead of salt? releasing H and O2? i need to find a way to get the electrolytic cell to explode.

@MichaelCousins If you’re attempting it in solution, my first guess is that they just won’t react in solution. And yes, if you apply too strong a current, $\ce{O2}$ will be released due to over-something.

yes, but I haven't filled the container completely with water, there is space for the gas to collect. obviously a lot of chlorine will dissolve into the water but surely some gas is released? is there a way to decrease how much chlorine dissolves?

A better question might be: why are you doing this for work? HCl gas is available and relatively cheap. There are safer, more controlled ways to create explosions without UV light.

Yes, the reaction from $$\ce{H2 + Cl2 -> 2HCl}$$ (Hydrogen Chroride) is usually done under the presence of UV light, but did you check if there are enough amount of $\ce{H2}$ and $\ce{Cl2}$ to react?
If the amount of the reacting gas is insufficient, the reaction seldom happens: the probability of collision would significantly decrease.
If you want the reaction to occur, you can collect purer $\ce{H2}$ and $\ce{Cl2}$ (maybe by water displacement) and try the UV lamp on them.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Yes there is enough H2 and Cl2, however; I think there must be O2 in the mix which is inhibiting an explosive reaction (which is what Im going for in this instance.)

Nope; the "explosion" of hydrogen gas is due to the formation of water (H2O) while the proper reaction beween hydrogen and chlorine might not be that "explosive" since it may not cause a flame but just some heat and maybe a gleam of light. Oxygen is not required in the reaction

well i mean detonation, I am running the electrolysis in a closed cell.


text:
In practice, do polar molecules actually align in an electric field?
We have all seen the following diagram (or similar) in our first chemistry class, depicting polar molecules aligned in an electric field. Is this just one of the half-truths of beginner chemistry or is this what really happens? I realize there will still be kinetic motion (spin, etc), but even statistically, is it possible for such a fluid/gas to have a net directional orientation throughout?

I have often seen the diagram depicted with water, but as the top answer on this question explains:

"...the ions will rush to the respective electrodes and form the so-called electric double layer. Its thickness, known as Debye length, is usually quite small. As for the rest of the solution, it will be effectively shielded from the electric field."

Perhaps this would be the same issue with hydrogen fluoride?
Is it possible the Debye length would be longer for a polar covalent molecule like Acetonitrile or Chloromethane?
If there are in fact polar molecules that exhibit this field-orientated behavior on a macro scale, what would be a way that it could be demonstrated?

Your figure is only schematic. Yes they do try to align but thermal motion tends to have the opposite effect so the alignment is only partial; its a competition. For a given dipole, the stronger the field the better the alignment, the higher the temperature the worse the alignment. The time it takes to align is directly proportional to the solution viscosity and molecular volume.

Related question on Physics: Can an electric field align water molecules?

Liquid crystals exhibit enormous field orientation effects, e.g in liquid crystal  displays and devices of all sorts.

I should also add that the relative level of orientation is governed by statistical mechanics. You can probably approximate the relative ratio as $e^{-\beta \Delta E}$ where $E$ is the relative energy difference aligned versus anti aligned. Obviously, the stronger the field, the more aligned they dipoles will be on average.

These are two entirely different and unrelated phenomena, to the point that you'd better unsee that other question and my answer to it, if you can. Ions are very much not like molecules. Ions are charged; molecules are not. Molecules don't form the double layer, nor do they even know what is Debye length. The electric field (though somewhat diminished, by a factor of dielectric constant) penetrates the whole volume of liquid, so all molecules feel it the same way; none is shielded.
Yes, polar molecules get oriented in the electric field, albeit to a very tiny degree. Think of it this way: the molecule still bounces and rotates around, but it spends a little bit longer in the field-aligned orientation than in any other. To orient all molecules in a truly stiff way, like soldiers in formation, the energy of alignment has to overtake the energy of thermal motion, which requires enormous electric fields and is ultimately impossible, because a breakdown would occur way sooner than you get there.

Also, note that the correct function of LCDs depends on the alignment under an electric field.

True, but those are liquid crystals, and that's another story. They are already kinda ordered without any field, much like domains in a ferromagnetic.

Ok. Maybe water is just the oddball because it dissociates into ions; I'm reading that even pure water will self-ionize. Since HF is borderline ionic, would it break into ions and shield the rest of the volume, or would its bond be strong enough to hold it together? It has a high dielectric strength, but so does water supposedly. Is there a way to know or calculate if other polar molecules like Acetonitrile will break down instead of "statistically" orienting in an electric field?

Come to think of it, yes, water would contain ions and thus would form some double layer and have some Debye length. All that goes twice for $\ce{HF}$, to the point that it must have been a trouble to measure its  dielectric constant at all. Still, it is there in the reference books.

Yes, polar molecules can align with a strong field. If the field is strong enough to overpower the local effects and the molecules are mobile.  There is a devise called an ELectret, which is roughly the electrostatic equivalent of a permanent magnet. An Electret can be formed using a polar polymer or wax that is solid at room temperature. The Polymer is installed as the dielectric of a capacitor. The material is melted and a large field potential is applied to the molten dielectric. The field is maintained while the mass solidifies. Once the material is solid, it will retain a permanent electrostatic charge. The piece can be make to act much like a piezo element and is the heart of a Electret Microphone. (A type of capacitance microphone that does not require a battery to charge the capacitor.)
https://sites.google.com/site/polymericmatrixmaterial/how-to-make-an-electret
I believe Carnauba Wax is one material that forms Electrets.  Also, polyvinylidene difluoride (PVDF) has been used. I recall the report of an experimental detonator for a torpedo that used a coil of co-axial cable rapped around the nose-cone. The cable was made with PVDF as the dielectric, and was charged as an Electret. The explosive was triggered by the charge released when the nose-cone was struck and the dielectric was distorted, changing the volume of the charge that could be contained within.
I believe that Electrets are at the heart of "Swiffers" and other types of electrostatic cleaning cloths and air filters. The static fields around the fibers would attract and hold dust.
Homemade Electret microphones became more popular with the Ham radio crowd, during WWII when replacement microphones became difficult to obtain.

Thanks Mike! I had forgotten about electrets. Strange no one sells them commercially for science experiments.  Making one out of wax would definitely be a good demonstration project.


text:
Conversion of formamide to isocyanide
Isocyanides are the only class of organic compounds which possess a divalent carbon. They react like a geminal nucleophilic/electrophilic synthon with oxidation of the C(II) atom to C(IV), which parallels the chemistry of carbenes and carbon monoxide.
Some of the most common methods for synthesizing isocyanides involve the dehydration of formamides. Various dehydrating agents (phosgene, phosphorus oxychloride) are used in the presence of a base (pyridine/TEA). In the follow reaction, however:
$\hspace{2.4cm}$
I'm unable to understand the role of p-TsCl, and can't work out the reaction mechanism.

Carbenes also have divalent carbon.

Just like phosgene or phosphorus oxychloride, p-TsCl is acting as a dehydrating agent. Its affinity for oxygen atoms is fairly apparent when you consider its common use as an alcohol protecting agent. The mechanism is as follows:
$\hspace{2.6cm}$


text:
Alfa-Cypermethrin 97% Techwater solubility
Aloha, I am a farmer in Hawaii and have been having problems with fire ants. I recently purchased a kilogram of Alfa-Cypermethrin 97% Tech, a pesticide that I find is not soluble in water. I would really appreciate it if someone could advise me how to make it water soluble in a 5% solution. I have no knowledge of chemistry.  Thanks for any help you can provide.

Why the vtc as homework?

How much could you dilute the water with say, ethyl alcohol or acetone?  In other words, must your solution contain any water, and if so how much?


text:
Configuration of endocyclic double bond in Vitamin A

I think the double bond inside the ring of vitamin A should have an (E) configuration. Am I wrong? I don't think I'm assigning the priority correctly.

The endocyclic double bond of vitamin A has a (Z) configuration. Here’s why:
First, let’s determine priorities of the atoms connected on the top side of it:

C

C (ring)
C (methyl 1)
C (methyl 2)

C

C
(C) (ghost atom; due to the double bond)
H

So the ‘ring side’ — the atom on the left in the way it is drawn at the moment — has a higher priority than the side chain (C at (1.3) beats H at (2.3)).
What about the bottom side?

C (ring)

C
H
H

C (methyl group)

H
H
H

Here again the ‘ring side’ has a higher priority than the methyl group: (1.1) carbon beats (2.1) hydrogen.
Both left-hand sides have a higher priority than their corresponding right-hand side. Thus, the double bond is (Z) configured.

There are two groups of compounds with Vitamin A, first is retinoids, also know as vitamin A and provitamin A, known as carotenoids which is a vitamin A precursors( which contain at least one molecule of retinol).
Naturally retinoids exist in the form of retinol, retinal and retionic acid and they are derived from animal sources. There are hundered of Carotenioids which are mainly synthesized by plants and only 50 of them can act as vitamin A precursors.
Technically speaking, large number of geometrical isomers are theoretically possible for retinoids and carotenoids but due to steric hinderence many mono-cis and poli-cis isomer are rarely occurred in nature.
All-trans forms are predominant in nature and more stable, but only those cis isomers occur which offer less steric hinderence and thermodynamically stable.

5-cis-retinol geometrical isomer  is rather unstable and do not occur in that configuration in nature.

Two of your molecules lost a double bond. 2) There is no cis-configured extracyclic double bound in your ‘13-cis-retinol’. 3) This does not answer the question.

edited and notified correction has been made. This picture only posted to show that 13-cis isomer is possible and stable,

The pictures still does not show 13-cis in spite of it claiming to do so. The question is still not answered.


text:
Biologically inert water mixture (or other fluid) for heat-exchange purposes
I have a steel chamber submerged in a 14 deg water bath to keep it at constant temperature. The setup needs to operate without intervention for a very long time (weeks or months).
The bath is exposed to open air in the lab and we have been having issues with mold development in the bath despite adding bleach to the water.
Can you suggest a different additive (or a different liquid altogether) to use to make sure the bath stays reasonably sterile?

There are additives for, e.g. chilled water loops, to resist bacterial and biofilm build up (which is why processed chilled water is colored, not clear). And, with bleach, you have to keep adding it, not assume it will last forever.

Hello @jon-custer, thanks a lot for the answer! Could you suggest any brand or product name? I've only been able to find anti algae for aquariums, but I'd prefer something that sterilizes the water as much as possible.

Apparently an acceptable choice of additive is silverised hydrogen peroxide. Only a few 100ppm should be necessary to keep the water sterile and the half life is about 5 weeks.
I'll go with this and update this answer if there are any issues.
PS
This article on overclocking (http://www.overclockers.com/pc-water-coolant-chemistry-part-ii/) has a number of suggestion on water additives to improve the thermal conductivity and  keep the system clean.


text:
Solubility vs Dissociation of HF
Ok, so I understand that solubility is a measure of how well a compound is solvated by the solvent. And dissociation is a measure of how much of a compound splits into its ionized forms.
Does HF have high solubility, but low dissociation (weak acid)? How can that be?
Is it because it forms strong hydrogen bonds in solution and thus forms stable solvation spheres (solubility), but its high electronegativity prevents H+ atoms from leaving its vicinity (dissociation)?

Solubility and dissociation are totally unrelated things, like bulimia and amnesia. Sugar (or better yet, ethanol) dissolves extremely well, but does not dissociate at all. $\ce{AgCl}$ is an example of the opposite kind.

@IvanNeretin - For chemicals which are considered to be ionic compounds solubility and dissociation are somewhat linked.

Yeah, maybe, somewhat. BTW, would you consider $\ce{HF}$ ionic? To me, it isn't (that is, unless dissolved in water, but even then, not quite).

see http://chemistry.stackexchange.com/questions/34818/why-is-hf-the-least-acidic-hydrogen-halide

I think your last sentence answers your own question very well:

It is because it forms strong hydrogen bonds in solution and thus forms
  stable solvation spheres (solubility), but its high electronegativity
  prevents $\ce{H+}$ atoms from leaving its vicinity (dissociation).

Dissociation is one mechanism of dissolution, but not the only one. Also, however, the $\ce{H-F}$ bond strength is not solely determined by electronegativity.

Agreed. Dissociation, and (in most cases) the resulting formation of separated charge pairs, is an important mechanism of solubility for many ionic species in polar solvents.

Electronegativity is one thing, but bond strengths are also important to consider: see for example http://chemistry.stackexchange.com/questions/13609/hydrogen-halide-acidity

@orthocresol, very good point. I kind of lumped bond strength with the difference in electronegativities of H and F in my mind, which isn't right. I made a slight edit to at least address that. Maybe not enough so, but for this specific question I hope that's sufficient.


text:
Why is F2 (fluorine) colourful despite being diamagnetic?
According to Molecular Orbital Theory, $\ce{F2}$ must be diamagnetic, then why is it colourful?
Also, $\ce{O2}$ is paramagnetic but it is colourless in gas phase. Why?

Fluorine is almost invisible to naked eye. See this:http://chemistry.stackexchange.com/questions/33861/why-does-color-of-halogen-deepens-down-the-group

Liquid oxygen is blue. It is just too weakly coloured to be seen in the gas phase.

The colour of a molecule does not depend on the whether the molecule is diamagnetic or paramagnetic. The reason is that the colour is caused by absorption of a photon in the visible part of the spectrum. The photon has enough energy to excite one of the two (spin paired in a diamagnetic molecule)) electrons  in the highest occupied molecular orbital (HOMO) and into a higher energy unoccupied orbital, usually the lowest unoccupied molecular orbital (LUMO). From spin conservation rules the electron should not change spin. A spin change would correspond to a 'forbidden process' which has a very low chance of occurring so leads to exceptionally weak absorption.
The molecule now has one electron in the lower orbital and one in an upper orbital, these form the excited state, and if the spins are paired it is called a singlet state and the spin multiplicity is 1.
(multiplicity is 2*(total spin) + 1:  if the spins are paired the total spin quantum numbers are +1/2+(-1/2)=0. If the spins are unpaired the total spin is 1/2+1/2 and the multiplicity is 3, a triplet state).
In oxygen, which has a triplet ground state, on absorbing a photon an electron is still excited into an upper unoccupied orbital, the spin multiplicity of the excited state produced depends on the relative spins of all the unpaired electrons just as in the case of fluorine.
(As it happens fluorine seems to be a complicated  example as it has a low dissociation energy so absorption of a photon puts the molecule into a repulsive excited state leading to two F radicals as the molecule dissociates. It also absorbs weakly at longer wavelengths than needed to dissociate and this may be due to  a forbidden, spin changing, transition as mentioned above).

'The colour of a molecule does not depend on the whether the molecule is diamagnetic or paramagnetic.' That's not what they teach at my school!

I think that you may misunderstand. Both diamagnetic and paramagnetic molecules can absorb a photon's energy and as I try to explain. The extent of this depends only on the energy gap between levels and the magnitude of absorption on selection rules and Franck  Condon Factors.

Magnetic properties is not directly related to the colour: the important thing is the spectrum.
Also, the magnetic properties can change when the molecule is excited.
The electrons get excited by a certain frequency of EM wave (that's when the energy level of the electron comes out.), and when that frequency is of visible ray, it appears as a 'colour'.
PS.
In fact, fluorine gas is not that 'colourful'. It is yellow, but it's very, very pale.

In case of oxygen energy gap between HOMO and LUMO
is large so energy required for this transition lies in UV region which is not visible to us hence it is colourless.
But Fluorine is belongs to visible region hence it is coloured.


text:
Are Meso compounds isomers?
According to my chemistry manual meso forms are "a particular case of of stereoisomers" implying that they are isomers, anyway right after this sentence the manual states that if you rotate the 2 forms of tartaric acids (the green ones in the image) you obtain the same molecule... so they're not isomers.
Is that correct? Are the green forms the same molecule? In this way it would be senseless to speak about isomers...
I'm not asking if these 3 formulas are isomers, because they obviously are. I'm asking whether the green forms (the last 2 forms starting from left) are to be considered isomers (that means they have a different structure) or they are totally (in every aspect) the same molecule with a different drawing.

Everything that's pictured here are isomers. Which of these are different and which are identical is another question.

‘Are meso compounds isomers’ — of what?

Rotate your screen 180 degrees, and you'll see that the left green one is the same as the right green one. They are the very same molecule, one is just drawn upside down. (If you think rotating your screen is somehow an invalid way of thinking about it, you are welcome to rotate your head 180 degrees :D)

Also, the linked duplicate already answers this exactly. Did you read the answer? "[...] the meso-compound and its mirror image are not isomers of each other since they are actually the same compound."


text:
What's the difference between the two equations for work?
I'm studying thermochemistry and I don't get this. What's the difference between  $w = Fd$ and $w = -p \Delta V$?

Aside from a sign convention artifact, they're the same thing.
Pressure is force per unit area:
$$
 P = {F\over A}
$$
If a surface with a given area is moved through a displacement $\Delta x$, then the volume of displacement is the product of $A$ and $\Delta x$:
$$
\Delta V = A \Delta x
$$
So, the $P\Delta V$ work formula can be rewritten:
$$
w = -P\Delta V = -P\left(A \Delta x\right) =
  - \left({F \over A}\right)\left(A \Delta x\right) =
  - F \Delta x
$$
There are a lot of assumptions embedded in the above, and you have to be a lot more careful than I was about defining positive and negative quantities correctly, but on a high level that's all there is to it.

Here's an example to illustrate the difference.
At a pressure of one atmosphere, a musket (gun) is fired due to the expansion of exploding gunpowder.  The musket has a 1 meter barrel, with 10mm inner diameter, and 100 gram musket ball (bullet). The musket ball exits the barrel at a velocity of 100 meters per second.  How much work is done by the expanding gunpowder products by the time the musket ball leaves the barrel?
There are many types of work: gravitation/acceleration, electromagnetic, etc.
PV (pressure-volume) work is just one type of work.

The first equation $w = F\,\mathrm dx$ is the definition of how we define work, which in itself is the energy expanded to move an object of some mass $F$ some distance $\mathrm dx$.
While $w=-P \Delta V$ is an expression that tells us work that has to be done by a gas $w$ to expand into a volume.
The '$-$' is there because we define work done by a system as expanded energy
Remember that work is esentially energy so its ok to compare

Probably mean "expended" and not "expanded".


text:
Why a specific temperature for value of 1 calorie?
Everything in chemistry often comes with a strong reason and facts behind it, but I stuck in definition of $1$ calorie.
My reference book define one calorie as amount of heat required to raise the temperature of one gram of water from $14.5$ degrees Celsius to $15.5$ degrees Celsius.
Why $14.5$ and $15.5$, why temperature such as  $0$ to $1$ or $19$ to $20$ degree Celsius is not chosen?
Please forgive me if this question is too easy and I can't figure out the logic related to it.
I shall be glad to have your suggestions.

Those are close to room temperature. Using 0C runs the risk of getting some enthalpy of the phase transition mixed in.

i cant understand everything after room temperature

What happens to water at zero degrees C? How would that impact a thermodynamic measurement?

Now, the question of why not other temperature ranges - the Gibbs free energy of water is not constant vs. temperature, so the heat capacity varies with temperature. Thus, you must specify a specific temperature range.

That also is the reason it is so difficult to liquefy helium: at low temperatures the heat capacity drops, so that the slightest energy input warms it greatly. See https://en.wikipedia.org/wiki/Superfluid_helium-4#/media/File:Heat_capacity_of_4He_01.jpg

jon but my question is why the specific temperature (you mentioned)is 14.5 to 15.5 degree celSius???

Lone Wolf, please do not readd the clutter that serves on purpose to the actual question. I have removed (and re-removed) it as part of the site’s policy. Instead of writing ‘thank you’ and thus cluttering up the question, say thanks by upvoting and/or accepting answers.

why temperature such as 0 or 1 degree celsius is not chosen??

That has been chosen by some.
According to Physical Laboratory Experiments: Part III.--Heat, 4th edition (1912) :

The following different calories should not be confused:

ZERO DEGREE CALORIE:- The quantity of heat required to raise one gram of water from 0 [degrees] C to 1 [degree] C.

ORDINARY CALORIE: cal.-The quantity of heat required to raise one gram of water from 15 [degrees] C to 16 [degree] C, this being taken as the mean room temperature.

MEAN CALORIE OR ICE CALORIMETER CALORIE: The one-hundred part of heat required to raise one gram of water from 0 [degrees] C to 100 [degrees] C.

...

Heat capacity of liquid water is a continuously varying function of temperature itself, so a particular temperature or temperature interval must be specified.  Various intervals were chosen historically, and when a small interval near 15 degrees C is chosen it is because that was thought of as room temperature 100-200 years ago.

can you provide heat required (in joules) each time,i mean value of zero degree calorie,value of one degree calorie etc.

@VidyanshuMishra http://www.vaxasoftware.com/doc_eduen/qui/caloresph2o.pdf

Some measurements are easier than others. If you try to measure the heat capacity of water at $0\ \mathrm{^\circ C}$, you'll need to be awfully careful that ice melt doesn't alter the accuracy of your measurement. This is not a problem at a higher temperature like $15\ \mathrm{^\circ C}$.

what if i go to 20 or 30 degree celsius??

As @DavePhD pointed out, the heat capacity does vary as a function of temperature. You might get a different value. There's nothing that says you can't report that value or put it in a table, but it might be less useful if that doesn't fit your use case.


text:
Why add specifically hot solvent during recrystallization procedures instead of heating it up after?
Recrystallization methods that I have seen online and in manuals involve adding a minimum amount hot solvent to the solid mixture. Is there a reason I should do this instead of adding an equivalent amount of solvent at room temperature and then heating?
I've tried dissolving table salt and sugar both ways, but I didn't notice a substantial difference other than adding hot solvent directly requires less time to dissolve because it took time to heat the cooler solvent.

You want to dissolve the compound in the minimal amount of hot water possible. If you add lots of cold water to dissolve all of it, then heat it up, then cool it down, it won't precipitate out.

That being said, table salt and sugar are both terrible candidates for recrystallization, albeit for different reasons.

Yes you may add just the right amount of solvent and then heat it together with the solid substance (which would gradually dissolve in the meantime). That should make no difference.

In some cases the material you want to recrystallize is sensitive, and extra heating may result in decomposition, etc

The issue is that, without knowing beforehand the exact solubility of your compound, you do not know what is the "right amount" of solvent. Yeah, if you know the solubility perfectly, fair game, apart from decomposition/side reactions. However, if you're trying to recrystallise something new that you made in the lab (probably a much more common scenario), you don't want to start by guessing how much cold water/ethanol/whatever to add.

Because in certain scenarios (e.g., supersaturated solution) the solute will precipitate out during handling if it is not kept at a high enough temperature.  Imagine, for instance, pouring a solution through a filter only to have the solute crystallize when it comes into contact with the sides of the funnel and the flask.

The goal of recrystallisation is to obtain purified crystals from impure solid, as much as possible. The solute is therefore dissolved in a solvent where the solubility is high at the boiling point, but low (or insoluble) at the room temperature. Additionally the impurities should be soluble in the solvent at room temperature.
The standard recrystallisation workflow would be something like this:

Heat solvent to its boiling point
Pour the boiling solvent slowly onto the impure solid with heating and swirling
When all of the impure solid has dissolved stop adding solvent
Allow the container to return to room temperature, then cool with ice if you wish

You want to extract as much pure crystals as you can, so the impure solid is dissolved in the least amount of hot solvent possible. This is the crucial point—you don't want to have extra solvent, which would mean that when the solution returns to room temperature, there is some of your target compound that remains in the solution (which could have been extracted if you had added less solvent). You also don't want to add less solvent because then the impure solid won't dissolve completely and your obtained crystal would be contaminated with that.
The only way to do this is to keep everything (i.e. pure solvent and the impure solid with hot solvent) at the boiling point of the solvent and add the hot solvent drop by drop until everything has dissolved. (Don't dry heat the impure solid btw, that causes a whole range of other problems!)
If you do this while the solvent is cold, you won't be able to estimate how much of the solvent is required to completely dissolve the solid beforehand (because the solubilities are different). You will either overestimate or underestimate, which would cause problems as mentioned above.
Note that sometimes there are insoluble impurities in the solid, then you would have to filter everything at high temperature. The same logic applies.