text:
How small is the smallest known carbon ring containing only double bonds?
How small is the smallest known carbon ring containing only double bonds?
The stress from the double bonds makes small rings impossible to make something like cyclohexahexaene so I want to know how large the ring would need to be so the stress is small enough such that the ring can be made.

TL;DR: Isolated derivatives of cyclo-$\ce{C18}$ most likely have the smallest isolated cumulenic 18-membered ring, though upon direct complexation of transition metal in/outside the ring the strain as well as its size can be reduced further.

In the past two decades a research group of Prof. Dr. François Diederich performed numerous studies on cyclo[n]carbons and monocyclic $\mathrm{sp}$-hybridized C-rings. He also published several reviews, e.g. see chapters "Cyclic and Linear Acetylenic Molecular Scaffolding" in [1, p. 43], and more recent "Oligoacetylenes" in [2, p. 443] devoted to versatile chemistry of cumulenes.
As @R.M. referred in the comment, early theoretical work [3] predicted special Hückel-aromatic cyclo-$\ce{C18}$ compound stabilized via two orthogonal $(4n + 2)$ $\pi$-electron systems, but later computational studies predict somewhat controversial electronic structures (from [2]):

Self-consistent field (SCF) calculations with a 3-21G or larger basis set predicted that the cyclic acetylenic $D_{\mathrm{9h}}$ structure 1a with alternating bond lengths represents the ground-state geometry[4].
But optimizations at the Møller-Plesset second-order perturbation theory (MP2) level including valence electron correlations as well as density functional theory calculations favor the cumulenic $D_{\mathrm{18h}}$ structure 1b as the most stable planar monocyclic geometry [5].

More from [1, p. 46]:

Accordingly, cyclo-$\ce{C10}$ is predicted to prefer a cumulenic $D_{\mathrm{5h}}$ structure, cyclo-$\ce{C14}$ is borderline, and the energies of its cumulenic $D_{\mathrm{7h}}$ and polyynic $C_{\mathrm{7h}}$ structures are nearly degenerate. It is clear that this fascinating theoretical controversy over the past years can only be definitively solved with the synthesis and characterization of cyclo-$\ce{C18}$.

cyclo-$\ce{C18}$ appeared to be a very elusive compound to isolate (on macroscoping scale typical products are anthracene and polymers), therefore only its derivatives were synthesized and characterized. Synthetic methods include the following paths: Retro-Diels-Alder; 3-Cyclobutene-l,2-dione route;  transition metal complexation [2]:

Generally, cyclic cumulenes have two types of strains: distortion from linearity imposed by $\mathrm{sp}$-hybridization and distortion from orthogonality of adjacent double bonds (cyclo[2n]carbons). [6]
Attempts to liberate the cyclocarbons strain in cyclic cumulenes is similarly relieved by complexing one of the double bonds directly to a transition metal. For example, like in this zirconocene-hexapentaene complex, where $\ce{C=C=C}$ angle is reduced to approx. $130^\circ$ by complexation with $\ce{Zr}$ [7]:

Unfortunally, up to date there are only mononuclear metal-cumulene complexes isolated in a large scale, and none of them are cyclo[n]carbons, but theoretically polynuclear transition metal complexation can significantly reduce the ring size.
References

Carbon Rich Compounds II, Macrocyclic Oligoacetylenes and Other Linearly Conjugated Systems; de Meijere, A., Ed.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1999; Vol. 201; ISBN 3540653015.
Stang, P. J.; Diederich, F. Modern Acetylene Chemistry; Wiley-VCH: Weinheim, 2008; ISBN 9783527615261.
Hoffmann, R. Tetrahedron 1966, 22 (2), 521–538. DOI 10.1016/0040-4020(66)80020-0.
Diederich, F.; Rubin, Y.; Knobler, C. B.; Whetten, R. L.; Schriver, K. E.; Houk, K. N.; Li, Y. Science 1989, 245 (4922), 1088–1090. DOI 10.1126/science.245.4922.1088.
Hutter, J.; Luethi, H. P.; Diederich, F. J. Am. Chem. Soc. 1994, 116 (2), 750–756. DOI 10.1021/ja00081a041.
Stone, F. G. A.; West, R. Advances in organometallic chemistry; 1998; Vol. 42. ISBN 0120311429.
Suzuki, N.; Hashizume, D.; Yoshida, H.; Tezuka, M.; Ida, K.; Nagashima, S.; Chihara, T. J. Am. Chem. Soc. 2009, 131 (6), 2050–2051. DOI 10.1021/ja8077472.

If you have the time, would you consider updating this since $\ce{C18}$ itself has been made? https://en.wikipedia.org/wiki/Cyclo(18)carbon https://science.sciencemag.org/content/365/6459/1299 Or I could do it one of these days, if you prefer.


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Why is this compound named resorcinol?
Substituted phenols or polyhydroxy phenols have their general name apart form IUPAC names. The general names comes from the source it is found like:-

Catechol (1,2-dihydroxybenzene)- obtained from catechu
Cresol (methylphenol) - obtained from creosote, a component of coal tar
Gallol (1,2,3-Trihydroxybenzene) - synthesized from gallic acid which is obtained from  oak galls, not a compound of gallium which many people get confused.

But what about resorcinol? Why is it named so? Is it resourceful? It is primarily obtained from argan oil. So, it should be named as arganol. But it is not so.
Same thing happens with phloroglucinol. Various sources are there but none of them matches with the general name.

For resorcinol, the naming is covered in a footnote to the Wikipedia article:

"Wir nennen den neuen Körper, da wir gefunden haben, dass er auch aus dem Ammoniakgummiharz erhalten werden kann, Resorcin, um an seine Entstehung aus Harzen und seine Beziehung zum Orcin zu erinnern." (We name the new substance, since we've found that it can be obtained from ammoniated resin gum, resorcin, in order to recall its creation from resin and its relation to orcin [i.e., orcinol].)

Which is itself a quote from  H. Hlasiwetz and L. Barth (1864) "Ueber einen neuen, dem Orcin homologen Körper", Annalen der Chemie, 130 (3) : 354-359.
At some point the ending -ol got added, to reflect the alcohol moieties in the compound.
Then name orcinol itself is derived from its association with orcein, a dye substance derived from orchil lichens.

According to the Oxford English Dictionary, the origin of the word resorcinol is:

Late 19th century: from the earlier term resorcin + -ol.

According to the Merriam-Webster Dictionary, the origin of the word resorcin is:

res- (from Latin resina resin) + orcin, a phenol ($\ce{C7H8O2}$)

It appears that the compound is named the way it is due to the original Latin-derived names for the compound(s) to indicate a phenol derived from resin.

source: Stack Exchange

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Controled burn colored fire and safety
I am a martial artist who uses fire weapons in my shows. I am wondering how I can get the flames to be different colors.
I've done some research about it, and have found that the salts of some metals can be used for the intended purpose (for instance, copper sulphate burns green).
However, I use kerosene as my fuel and am wondering if it is advisable to mix the chemicals with it (kerosene) without burning myself. Also, I would like to know how to obtain the chemicals and what amounts should be added to a given volume of fuel.

I've done some research about it, and have found that the salts of some metals can be used for the intended purpose (for instance, copper sulphate burns green).

You're right! Various melts salts are known to colourize a flame. In the lab and/or for demonstration purposes, this is typically achieved by holding a magnesium oxide stick with a bit of the salt attached into the flame of a Bunsen Burner. There's a wikipedia page on Flame tests that shows some possible effects.

I use kerosene as my fuel[…]

Here's a problem. The salts you could use soluble in water, but insoluble in a solvent as unpolar as kerosene! Consequently, you might want to give a flammable, but more polar solvent a try.
Is it conceivable to change the flammable solvent used in your performances? I reckon that you might achieve the desired effect using ethanol or methanol. With respect to the toxicity of both solvents, I'd definitely try ethanol first!

First, to produce color flames you don't want kerosene. Kerose produces bright flame, while to get a colored flame one need a fluid that produces a colorless flame on its own. I recommend ethanol, preferably dry.
Second is colorant. Lucky for us some strong colorants dissolve pretty well in ethanol. Specifically, green colorant tryethylborate is mixible with alcohol and actually burns on its own. Copper chloride (bluish colorant) Lithium and Calcium chlorides (red colorants) dissolve in ethanol fairly well. Yellow is pretty easy to get, even small addition of sodium salts (such as table salt) gives a bright yellow flame. It is actually a problem, as minor impurities of sodium may contaminate other colors with yellow.
I highly recommend to perform with colored flames only at open air. The substances I mentioned are slightly toxic and would form smoke on burning.
Unfortunately, there is no viable violet flame colorant (violet is usually produced as a mixture of blue and red), and bluish colorant (copper chlorides) cannot produce flame that is bright and saturated at the same time.
To my knowledge, the compounds I mentioned are all purchasable online. While some of them are toxic, none of them is really dangerously so. Breathing in the smoke, though, is unadvised, as well as drinking the solutions.

As others have mentioned, kerosene is not the best solvent for the purposes of dissolving salts. Methylated spirits (denatured alcohol) is probably the best, and possibly rubbing alcohol, to give you the desired solubility you want. Caution - alcohols have a much lower flash point, are easier to ignite, are more flammable, harder to extinguish  than kero, and burn with almost undetectable flame. They pose a significant flammability risk, especially if you spill some. You need to weigh these risks up for your performance, and seek some advice from your industry if you have not worked with these solvents before.
You can also consider pre-treating your implements by soaking them in the solution of desired chemical and allowing the solvent to evaporate (possibly for some days) and leave behind impregnated salt, which can then be burnt using other fuels. I have seen this done with pine cones soaked in aqueous solutions and allowed to dry out before being added to a campfire. You would still suffer from the competing flame colour and intensity of your chosen fuel. The colour of a kerosene flame will still probably dominate the subtle colours of your additives.
As for quantities; this will depend on the solubility of the salt in your solvent, and the general rule will be 'as much as dissolves', but will depend on the salt you are using. For instance, the solubility of sodium chloride (yellow flame) in ethanol (and therefore similar to methylated spirits) is less than 1g per litre, while lithium chloride (red flame) dissolves at something like 250g per litre. There are a couple of very good articles about the ins and outs of coloured flames for performance here:

Desire For Fire
Home of POI

Each different flame colour uses a different chemical which in turn has its own toxicity and undesirable residues that need to be considered for your own health (in preparing and performing) and also your audience (who may be breathing the stuff in). Be sure to read and understand the SDS of the chemicals you are using BEFORE you purchase and use them.
Fire colourants and additive kits are available for fire performers, and there are lots of very good sources on the internet for you to purchase and get advice for your performance:
http://www.homeofpoi.com/shop/productDetails/Pack-of-Colored-Flame
http://www.threeworlds.com.au/colour-flame-dye

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Smoke cobweb formations?
One day, my mom decides to light the house on fire, and after we put the fire out, all of the cobwebs in the house are black. Obviously, this is because of the smoke. Later, some renovations guy tells her something about how the cobwebs aren't really cobwebs and when a house fills with smoke, there is some weird chemical formation that makes a black cobweb-like-substance that just happens to show up right where cobwebs would be. Basically, the guy told my mom that those black cobwebs were chemical formations from smoke instead of cobwebs that became black after being in smoke. My mom will not believe me when I say that the renovations guy is wrong and they were obviously cobwebs. I ask her "What happened to the cobwebs we already had?" and "Why would a renovations guy know that it is a smoke formation if there are cobwebs in every house?"
I can't prove to my mom that she's wrong, so I'm asking a bunch of chemistry people to confirm that smoke doesn't stick to walls and form cobwebs.

Awkward. It just so happens that the renovations guy is right. The term you need to google is "smokeweb".
The ionisation process of burning (particularly) synthetic materials like plastics, rubbers and polymers, creates charged smoke particles which can attract towards certain surfaces, and attract to each other creating chains, and ultimately web-like formations. They form under certain conditions called wet smoke, which involves low heat, or smoldering type environments. Smoke webs are typically sticky and pungent and will smudge easily and show up in particular areas where there is limited air flows, typically corners and edges of ceilings, in curtain folds, cabinets etc.
Spider webs will burn up very quickly in a real fire.


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Can I use lye as a cooking leavener instead of baking soda?
I asked this question on the cooking site, but got unsatisfactory answers: https://cooking.stackexchange.com/questions/65520/can-i-use-lye-as-a-leavener-instead-of-baking-soda
I'm interested to try to make waffles using a concentrated food-grade lye water mixture instead of baking soda as a leavening agent. I have three questions related to this subject:

Would this potentially make my waffles fluffier (due to increased CO2 producing reactions or otherwise)?
Would it impact the flavor of the waffles?
Is it safe to ingest waffles that have had a lye water reaction occur in them? (Obviously I should let the lye reaction complete before ingesting).

How long should I let a lye reaction occur before cooking?

Notably, I often use sourdough starter in my waffles, which has an acid component - I've noticed that I get a strong reaction from the baking soda as a result, and it got me thinking that I would get an even stronger result from lye, but I'm worried about taste and ingestion safety.

Typical “food-grade lye” is just an aqueous solution of sodium hydroxide $(\ce{NaOH})$. How do you expect to generate the carbon dioxide $(\ce{CO2})$ that acts as a leavening agent?

I'm not an expert here, but, as a chemist with a strong gastronomic interest (and a penchant for waffles), can suggest some things which may be helpful. Lye is a caustic alkaline solution, traditionally KOH, and more commonly these days NaOH.
First, my understanding of lye as an agent in cooking is limited to its use in pretzels and "laugen" bread and ramen soups. The pretzel making process involves a final dipping of the dough in a lye solution in order to promote the Maillard reaction during cooking; this is not the same as the leavening process. Ramen noodles are a type of "alkaline noodle" and I believe use the lye to break down the gluten to make the flour more pliable and less elastic. Again, not the same as the leavening process.
The use of lye as a leavening agent comes from taking lye (produced by washing fireplace ash) and boiling with lard to produce potash. The purified version, pearl ash, was one of the first chemical leaveners used in the kitchen. Pearl ash acts as a leavening agent by reacting with an acidic ingredient; fruit juice, sour milk or molasses, to produce CO2.
So to answer your questions:

I can't see how it would make your waffles fluffier than conventional modern rising agents, even if you did get your chemistry right. You still need to get your CO2 from somewhere and substituting your bicarb for lye actually removes this element. You would need to rework your entire recipe, I think.
Yes. Without doubt. More than likely it will degrade the taste of what should otherwise be a delicious dish. Alkaline foods are well known to have a bitter after taste (think kale and broccoli). Even scones made with a little too much baking soda can leave your mouth tasting and feeling like you've just had your annual clean and scale at the dentist.
While lye is used as a cooking agent in some instances, even the food grade solutions are extremely caustic, and, undiluted, are actually classed as poisons in some instances. It is corrosive to the body, especially mouth, throat and stomach, and can cause serious damage to the eyes. This may be stating the obvious, but best to assume ignorance and work up from there. Apart form mooncakes, I am not aware of a dough recipe (or anything else for that matter) that incorporates lye within the dough itself. Unless you have experience handling this ingredient in the kitchen, alternatives are recommended. I cannot tell you how you would determine whether your 'lye reaction' is complete within the dough; certainly gauging the completion of 'the rise' would not be sufficient, and the cooking process may need to be adjusted to accommodate this. Mooncake dough, for instance, is allowed to sit overnight, and even for up to 3 days before cooking. Too long to wait for my waffles!

For your interest, there is a very good comparison of various traditional and modern leavening agents here. Interestingly, the author concluded that the best agent overall was baking soda. Just like modern research chemistry in the lab - someone has done the reaction already.

Thanks for your great answer. I make pretzels fairly regularly with lye solution, so I'm comfortable using this hazardous substance to some extent. Regarding safety, I'm still not clear how long lye has to process (or in what way) to become safe. For example, I dip my pretzels in lye and bake them within 15 minutes for 15 minutes. When they come out of the oven they are (empirically) safe to eat. I don't know why - oven heat? Reacting with dough somehow?

The leavening effect of baking soda (sodium bicarbonate: $\ce{NaHCO3}$) is not from its basic/alkali nature, but from the carbon dioxide produced from the decomposition of carbonic acid ($\ce{H2CO3}$) produced from the reaction of the bicarbonate ion ($\ce{HCO3-}$) in the sodium bicarbonate with acids in the food. The more acid, the more vigourously the sodium bicarbonate will react.
Lye (typically sodium hydroxide: $\ce{NaOH}$) is also basic/alkali, but when the hydroxide ion ($\ce{OH-}$) reacts with acids, it forms water ($\ce{H2O}$). No carbon dioxide gas is formed, and thus no leavening of the food is possible. Yes, lye is a stronger base than sodium bicarbonate, but as the leavening doesn't come from the base-ness, you won't get any increase in leavening.


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Would a melted crystal recrystallize back into the same lattice structure when frozen?
Two things that I'm not sure about melted crystals:

For a material that has only one possible crystal structure, would freezing cause it to recrystallize, or could it become amorphous upon freezing?
What would happen upon freezing to a material with more than one possible crystal structure?

The answer to both questions is the same: "It depends". Talking of multiple crystal polymorphs, there must be one of them which is thermodynamically stable under given conditions, but this is not necessarily the one that would crystallize.

If cooling is rapid enough, then crystals don't have time to form, so the substance may be amorphous or glassy (though there may be some localized structure). For example, there are metallic glasses and amorphous water ice. Metallic glasses may have improved wear and corrosion resistance over a crystalline alloy. One interesting intermediate form is a quasicrysal, which can have 5-fold symmetry. Roger Penrose proposed this geometrical tiling before it was found in metallic glasses.
It depends on the conditions in which the material freezes. For example, graphite can be forced into a diamond structure by heat and pressure. A pseudomorph could form if the substrate initializes crystallization in a different form.


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What causes differences in limits of detection?
I'm wondering what causes the limits of detection (LOD) to vary between different compounds, when calculated from the data obtained from one sample.
I've analyzed a 0,1 µg/ml sample, containing fluorene, fluoranthene and benzo[a]pyrene in toluene, using GC-MS. I calculated the LODs to be 0,04 µg/ml, 0,02 µg/ml and 0,03 µg/ml. What could be the reason for the differences between the values?

Have you got any ideas?

None whatsoever. I'm not very familiar with MS and I could easily imagine it's got something to do with how MS works.

Different compounds (from one sample) result in separate signals and very likely different sensitivity from MS measurement. It's related to how easy the ionization is for each compound. In addition, experimental errors always exist. Are you trying to determine the concentrations of those compounds?

authors: Hermia 極緻美學概念店


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Possible causes for flawed results of an EDTA titration for the Ca content of milk
Earlier today I was trying to find the calcium content of different milks as a part of a school assignment. I used this setup as the basis for my experiment:
I was using $\pu{0.1 M}$ EDTA solution as a titrant. The milk was diluted with destilled water and $\pu{8M}\ \ce{NaOH}$ solution to make sure magnesium won't interfere. The only diversion from the procedure above was that eriochrome black T was used as an indicator. After several attempts, starting with $\pu{10ml}$ and reducing the amount of milk with each trial, absurdly large quantities of EDTA were used without any change in color. Lastly, when a sample of $\pu{(0.1± 0.005) ml}$ milk was diluted in $\pu{0.4ml}$ water and $\pu{0.05ml}\ \ce{NaOH}$, the change in color occured after $\pu{68ml}$ of the titrant was added, which is equivalent to there being $\pu{240g}$ of calcium in the sample (fun fact: this would mean that the milk carton weighs 2.4 tons. I haven't weighed it, but I am pretty sure it does not).
Thus my question is: what could the possible mistakes that lead up to these terrible results be? Equipment was rinsed with distilled water before use, and everything else was done according to the procedure, except for the EDTA solution, which was made approximately an hour before the experiment (and it should be left alone for at least 4 hours after creation). Nevertheless, could this impact the experiment in such a severe way?
The teacher was uncertain of what caused it as well.
We chose our own topics, so no other students have done the same experiment.

Eriochrome Black T isn't a suitable indicator for the determination of calcium ions in 8M NaOH. Try with Murexide.

Thanks, I will try that! Could you, however, elaborate on why? It would really help me with my paper, as I need to describe my design in the paper.

I don't have the time now to do the whole calculation, but I will just direct you towards the proof: compare the pCa at the equivalence point of the titration (approximately 6.3) with the pCa at which the color change of Erio Black T occurs (approximately 3.8).  It is also possible that you don't see the change of color nicely as milk is colloidal.

Milk also has emulsified fat globules, so it is a two-phase system. For such a large over determination of calcium I'd wonder if Eriochrome Black T partitions between the two phases and preferentially prefers the fat globules.

You could test the effect of milk by simply testing an aqueous solution of calcium.

@Kasper How did the experiment go? Did using murexide as the indicator improve your results?

@Marko That sounds like the basis for an answer!


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Does concentration of zero-order reactant affect half-life?

Two substances $\ce{A}$ and $\ce{B}$ react with each other in such a way that $\ce{A}$ is $50\,\%$ consumed in $\pu{33 min}$ and $75\,\%$ consumed in $\pu{66 min}.$ Changing the concentration of $\ce{B}$ has no effect on the results. Which statement is true?

And the answer is:

This reaction is first order in $\ce{A}$ and zero-order in $\ce{B}.$

I understand why $\ce{A}$ is first order but I don't get why $\ce{B}$ is zero order. The equation for the half-life of a zero order reactant is $$t_{1/2} = \frac{[\ce{B}]_0}{2k}.$$
Doesn't this mean that the concentration for a zero-order reactant would affect the half-life?

How can we conclude that $A$  is first order?

Some information is missing. You ask : Which statement is true ? OK. But there are no statements to consider for a choice !...

Changing the concentration of $\ce{B}$ has no effect on the results.

This statement could only mean one thing: $\ce{B}$ does not participate in the reaction. Obviously, if $\ce{B}$ does participate in the reaction, and you change the concentration of $\ce{B}$ to $0,$ the reaction won't happen.

The equation for the half-life of a zero order reactant is $$t_{1/2} = \frac{[\ce{B}]_0}{2k}.$$

Zero order reaction have no half-life. Only 1st order reaction has half-life.

Exactly! The concentration of B during the reaction does not effect the half life. If you see for the zero order reactions, the half life can be written as $\frac{B_{\text{initial}}}{2K}$ in which B & K are constants. So for zero order reactions, half life  does not depend upon the concentration of reactants at any time (after start of the reaction).

But you are changing the initial concentration of B. I thought first order half lives were not affected by the initial concentration.

Let the rate law for the reaction be $$r = k[\ce{A}]^n[\ce{B}]^m$$

Changing the concentration of B has no effect on the results.

What are the results? Is it the rate? or is it the half life?
Finding the value of  $\mathbf{m}$
1. If it is the rate, then
It is obvious that $m$ should be zero.
2. If it is the half life, then
For this to happen, $m$ should be equal to $1$$^*$, because only for a first order or a pseudo first order reaction, half life is independent of initial concentration of reactant.

*This is correct only if A is in excess

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How dangerous are non-soluble trivalent thallium compounds?
I am PhD student in solid state physics. I have been working with a fine powder of a transition metal oxide containing trivalent thallium. The oxide is not soluble in water.
In my university, physicists never learn how to deal with dangerous compounds properly (there's this one guy in my lab who does not believe in wearing gloves...), so I wanted to ask you guys how dangerous you think this compound is.
I'm not dealing with huge quantities of this sample, about 10-100 mg. Perhaps the most dangerous aspect of that particular powder sample is that the grain size is small enough that even the smallest stream of air during sample handling causes a loss of a small amount of powder (= visible grey traces on kimwipes). That probably makes it easy to inhale small quantities without even noticing.
I have occasionally done experiments with that stuff since 2 years now, and I'm not dead yet, nor have I ever noticed any symptoms of thallium poisoning, so I feel like I'm doing fine. I'm just curious how many people I could kill what could happen to me if things go bad (touching the powder with bare skin, inhalation, touching the eye with contaminated gloves because I'm tired, ingestion, etc.)
PS: This is my first time posting on any SE, so if I'm doing it wrong please let me know.

In all seriousness, before I post an answer describing just how toxic thallium compounds are, have you ever read a SDS for any thallium compound, such as Tl2O3?

What is the transition metal oxide?  This can affect the answer

I worked with thallium as a young vacation student in an aerospace company. It was regarded as a significant hazard and beside the usual protective clothing we had urine samples before/after to see if we'd ingested any.


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Fischer Projections confusion
A bit of trouble I am having with grasping the concept of fischer projections for long chained molecules. Take any non-cyclic carbohydrate from a tetrose and upwards and make an actual model of it and what we see is that the carbon backbone follows a zigzag pattern. However, according to the fischer projections, if we take any chiral carbon the vertical bonds go into the plane of the paper and the horizontal out of the plane of the paper. With a real-life model which follows a tetrahedral structure at each carbon atom and a zigzag pattern this just doesn't seem to work. What is going on here??

The short answer is that the Fischer projection is not a zig-zag. If you make a model, the carbon backbone is wrapping onto itself.

This question needs an elaborate answer (preferably one with some diagrams)

The carbon chain of a fischer projection is drawn vertically.
Yes, the vertical and horizontal lines depict 'into the page' and 'out of the page' respectively.
However, when you have a carbon chain with more than 3 carbons, each down carbon on the backbone which would normally be projected away from you (according to the above rule) has to be rotated. Essentially, this creates the 'zig-zag' you would see in other projections (e.g. Natta Projection).


text:
Temperature change when volume halved
For this multiple question:

The correct answer is B. However, I don't understand.
You are halving the volume
so for the original solution
the number of moles of HCl is 0.1 moles
and in the new solution, the number of moles of HCl is 0.05 if we are using the same concentration.
Therefore as there are less number of moles, won't the temperature increase (ie. energy release) get halved?

The correct answer is B in both the case the final concentrations are not changing. Therefore same will be the heat of rise. But heat released will be less but volume of the final solution also got halved. So temp. raise will be same.


text:
paramagnetism vs. dimagnetism
this was a question for my chem class- and for some reason the answer was the last option. Why isn't it $\ce{Mn}$ which has the most unpaired electrons?
Which of the following species would you expect to be most
paramagnetic?
1.  $\ce{Mn}$
2.  $\ce{Mn^2+}$
3.  $\ce{Fe^3+}$
4. Both $\ce{Mn^2+}$ and $\ce{Fe^3+}$ are the same
and more paramagnetic than $\ce{Mn}$
5.  All have the same paramagnetism.

Para-magnetism: Para-magnetism refers to the electronic configuration of an atom having one or more unpaired electron. That means an atom can be called para-magnetic if it has one or more unpaired electron. Now if atom A and B has no unpaired electron they have same para-magnetism (in this case zero). If they both have one unpaired electron still they have same para-magnetism (Here it's one).
Now look at the electronic configuration of the above sated options::
$Mn :: [Ar]3d^54s^2$
$Mn^{2+} :: [Ar] 3d^5$
$Fe :: [Ar]3d^64s^2$
$Fe^{3+} :: [Ar] 3d^5$
So they all have five unpaired electron. That's why they have same paramagnetism.


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5-Chloro-2-Methylphenol vs 2-Chloro-5-Methylbenzene?
Why is the compound called 5-Chloro-2-Methylphenol rather than 2-Chloro-5-methylphenol? The occurence of the highest number 5 is exactly the same, so are they both correct?

The two compound names that you are asking about refer to different compounds. Phenol is numbered as follows:

The structures of 2-chloro-5-methylphenol and 5-chloro-2-methyl phenol are below.

Perhaps you were confused about the 2,6 or 3,5 isomers? In this case, the chloro- substituent should get the lower locant number because it is alphabetically first.


text:
Does mercury amalgamate with Brass or Steel?
Does mercury amalgamate with Brass & Steel ?
Details: I am designing a micro pump to dispense mercury in micro quantities ($\mathrm{300-700}\ \pu{mg}$). I am planning to use 2 check valves with syringe to suck mercury from a mercury reservoir and push it for dispensing.
The problem is that all the high quality check valves are mostly Brass & I am not sure if mercury amalgamates with it.

Brass contains zinc and copper.  Both of those materials will readily amalgamate with mercury and thus it is not recommended to allow brass to contact mercury.
Steel will not form an amalgam and is perfectly fine for use with mercury. In fact is Approved as a container for mercury in transport.

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Drying the solution of two salts
What will be the arrangement of crystals if a solution of two salts, say, Sodium Chloride and Potassium Chromate, is dried? Will one salt crystallize before the other and become the crystallization nuclei of the other? Would both types of crystal form in a chaotic mixture? Or would they interact to form some new lattice?

It depends on their solubility and other factors. You may get a chaotic mixture of all four salts (the original ones plus the products of exchange reaction), or just the original two, or just the other two. Some would interact to form a mixed salt, some would not.

As @Ivan Neretin states, it depends. For example, potassium sulfate and aluminum sulfate form "alum", with a single crystal structure.
Tartaric acid (and its salts) occur in different forms, and even though they have the  same chemical makeup, these stereoisomers can be separated because they form differently shaped crystals. One of Louis Pasteur's first studies was separating the isomers from a mass of crystals.
A good source on growing crystals is Crystals and Crystal Growing by Alan Holden and Phylis Morrison


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Can there be two IUPAC names of a compound?
What is the difference between propan-2-amine and 1-methylethanamine?

in the latter, you have not used the longest carbon-chain for naming it

Generally, since many compounds can have two or more names in accordance with several methods recommended by IUPAC, a compound may be named correctly in more ways than one. Therefore, current IUPAC recommendations include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names.
In particular for primary amines, the rules given in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) are as follows.

P-62.2.1.2 Primary amines, $\ce{R-NH2}$, are systematically named in the following ways:
(1) by adding the suffix ‘amine’ to the name of the parent hydride;
(2) by adding the name of the substituent group $\ce{R-{}}$ to the parent hydride ‘azane’;
(3) by adding the name of the substituent group $\ce{R-{}}$ to the term ‘amine’ used as a preselected parent hydride name for NH3; this method is used only with monoamines.
(…)
Method (1) leads to preferred IUPAC names.

Since, according to the current recommendations, the substituent group $\ce{(CH3)2CH-{}}$ can be named ‘propan-2-yl’, ‘isopropyl’, or ‘1-methylethyl’ (where ‘propan-2-yl’ is the preferred prefix), the compound given in the question can have the following seven names in accordance with methods recommended by IUPAC. The PIN is ‘propan-2-amine’.

Method (1):
propan-2-amine (PIN)
Method (2):
propan-2-ylazane (propan-2-yl is the preferred prefix)
(1-methylethyl)azane
isopropylazane
Method (3):
propan-2-ylamine (propan-2-yl is the preferred prefix)
1-methylethylamine
isopropylamine

Could you name some amines mostly known as -azane ?

@permeakra My first idea are compounds that include the multiplicative substituent group $\ce{{}-NH-{}}$, such as 2,2'-azanediyldiacetic acid.

nice try, but it is known as iminodiacetic acid.

@permeakra Certainly, that compound is known by many names. The name 2,2'-iminodiacetic acid was in accordance with the 1979 recommendations. However, according to the current recommendations, the preferred name is 2,2'-azanediyldiacetic acid because the preselected prefix for the multiplicative substituent group $\ce{{}-NH-{}}$ is now azanediyl; the name imino is used only for $\ce{{}=NH}$ as a substituent.

Yes it is possible. I suppose IUPAC has also accepted some common names also for compounds as of here. Many other compounds have names, still in use, which are not in accordance with IUPAC guidelines but accepted, e.g., camphor, naphthalene, anthracene, isoprene, etc.


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What is the structure of powdered sulfur?
Quoting this paper:

A variety of allotropic states of elementary sulfur have been identified.
  Solid crystalline sulfur exists either as rings of 6-12 sulfur atoms (cyclohexasulfur, $\ce{S6}$, cyclooctasulfur $\ce{S8}$, etc. or as chains of sulfur atoms (catenasulfur $\ce{S_{\infty}}$).

As said on the same paper, Alpha-sulfur (orthorombic - $\ce{S_{\alpha}}$) is the thermodynamically stable form under ordinary conditions.
What happens with powdered sulfur (also called "dusting sulfur", normally used as fungicide)? Are sulfur atoms still organized in rings or chains?
Can we associate a precise crystalline phase to powdered sulfur?

Powdered sulphur is still the same orthorombic sulphur which is thermodynamically stable under normal conditions.

Powdered sulphur is produced by grinding raw sulphur as it is mined. According to Wikipedia, mining is done as follows:

On Earth, elemental sulfur can be found near hot springs and volcanic regions in many parts of the world, especially along the Pacific Ring of Fire; such volcanic deposits are currently mined in Indonesia, Chile, and Japan. Such deposits are polycrystalline, with the largest documented single crystal measuring 22×16×11 cm.[21] Historically, Sicily was a large source of sulfur in the Industrial Revolution.[22]

Raw sulphur is polycrystalline (meaning many crystals grown together). So basically, powdered sulphur is the same as crystalline sulphur just as very small crystals. Consequently, what you found in that paper for crystalline sulphur also applies to powdered sulphur.


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Does acid anhydride give a positive iodoform test?
I thought that acetic anhydride $\ce{CH3COOCOCH3}$  would give a positive Iodoform test because it has a $\ce{CH3-(CO)}-$ group.But it is given that this does not happen.
Is this true? If so, can anybody explain the reason? Thanks

The iodoform reaction requires enolate formation, which is not favourable for an anhydride.

The hydroxide in the iodoform reaction will hydrolyze acetic anhydride faster than enolization.

@user55119 Exactly, the iodoform test requires aqueous iodine which would clearly hydrolyse the anhydride to form the acid molecules before any enolisation would occur.

According to this source, both acid halides and anhydrides can form enols reasonably readily and can be halogenated. I'd imagine, for anhydrides, this to be related to the potential for intramolecular hydrogen bonding in the enol form (as shown below):

This allows halogation at the alpha position by treatmeant with the halogen. I'd also imagine this process can be catalysed in anhydrous acidic conditions, avoiding hydrolysis.
The same cannot be said for base-facilitated enolate formation. As user55119 implied, the highly reactive $C=O$ is likely to be attacked before the alpha hydrogens (or even if deprotonation occurs first, proceed through an E1cB pathway due to the presence of a good leaving group; implied by Clayden Organic Chemistry), decomposing the anhydride instead of forming the enolate.
Enolate formation in basic conditions is usually required for multiple halogenation, and generally three substitutions are required to form a suitable leaving group for the haloform reaction to proceed. This is why it is a test for a methyl group next to a $C=O$ (as Mathew has said); as there are three alpha hydrogens.
Considering basic conditions are not an option, only mono-halogenation is feasible (perhaps even in acidic conditions) and the haloform test is made virtually impossible.

Note: The anhydride would be hydrolysed first and would form acid which does not give the iodoform test.

Even if we consider that they were not hydrolysed , what would happen ?

$\ce{CH3COO-}$ would be removed and hence $\ce{CHI3}$ will not be formed.
Hence anhydrides will not respond to this test.

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Hydroxide will kick away $\ce{CH3CO2}^-$ group whenever possible, in this case, the very first step

https://web.chemdoodle.com/demos/2d-sketcher this allows you to draw compounds on the web and is a mechanism sketcher.

@AnindyaPrithvi OP has specifically stated that this would not happen but rather this is a hypothetical "if it happens" scenario.

Yes. I am just trying to point out that there is no way the anhydride could give the test. Thanks @Safdar

I didnt say anything about it, also, the 4th step would prefer removing $\ce{CI3}$ and Iodoform will be formed

@AnindyaPrithvi why would that happen? do you have conclusive proof?

@Safdar the point of the fake reaction is to make haloform which it completely fails too, counter question: Does the OP have a conclusive proof to eliminate $\ce{CH3CO2}^-$

In that case is there a reason to not attack the first carbonyl carbon which seems to be more preferable to me?

@AnindyaPrithvi Yes I do have proof. Consider acetic acid and triiodomethane. Obviously acetic acid is the stronger acid. This means that the conjugate base of acetic acid which is acetate ion is weaker than the conjugate base of triiodomethane. Now, if that is true, the acetate ion will have a better leaving tendency. That is why it is eliminated. Also, please don't take the reaction pathway too seriously. There is a very minute chance of such a pathway even being followed when hydrolysis is possible.

my question was directed towards something else, i.e. why did OH not attack the first carbonyl, making a better base due to CI3 indictive effect, your mech mai not be taken seriously but shouldn't atleast be open ended

Iodoform test is used to detect the carbonyl carbon and since acid anhydride is not a carbonyl compound , it does not give this test . The lone pair of oxygen ( the middle one ) and the π bond of carbonyl group are in resonance . When you actually  see the reaction mechanism , you observe that the methyl group attached to the carbonyl should have  acidic hydrogen but this is not so in the case of acid anhydride .

Keep in mind that Iodoform test is not used to detect a carbonyl carbon, but used to detect $\ce{CH3C(=O)-}$ group. Acetic anhydride has a $\ce{CH3C(=O)-}$ group and hence the question. Would you edit your answer accordingly?


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Why are tertiary carbocations so stable?
Today our lecturer provided this data to illustrate the relative stability of carbocations.
\begin{array}{|c|c|c|c|c|} \hline
\text{Substrate} & \ce{t-BuBr} & \ce{i-PrBr} & \ce{EtBr} & \text{MeBr}\\ \hline
\mathrm{Relative~rate~of~S_N1~reaction} & 1.2 \times 10^6 & 12 & 1 & 1\\ \hline
\end{array}
Additionally, in several places I have seen it said that tertiary carbocations are common and stable whereas secondary carbocations are only rarely stable and primary obviously not at all.
What is immediately obvious is that going from a primary to a secondary carbocation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change. The number of bonds available to participate in hyperconjugation increases linearly along the series but the rate clearly doesn't so there must be some other factor at work. The only idea I had was that there might be some similar effect to Y-aromaticity taking place in the trigonal tertiary cation (there are six electrons involved in hyperconjugation after all) but I have no idea whether this is a plausible suggestion.

This might help:-http://chemistry.stackexchange.com/questions/30893/why-are-tertiary-carbocations-the-most-reactive-if-theyre-already-stable

@NilayGhosh The way the passage was worded makes it sound like this is not a case of the alternative reaction ($\mathrm{S_N2}$ being more favourable. Also, I have heard before that secondary cations are not really stable.

@ShankRam I think part of  your comment is missing.

This is just my theory. Please go through it completely and reply your view. Consider a typical $SN^1$ reaction. We have 3 intermediates: primary, sec. and tert. carbocation. Let us say the activation energy of the reactions are E1,E2 and E3. Now, we know that E3<E2<E1, which follows the order of stability of intermediates. From arrhenius equation, we know that rate constant is an exponential function of activation energy. So, even if the tertiary hydrocarbon we have taken has a few hundred kcal less activation energy than primary, the increase in rate constant would be huge.

@bon, what do you think about the above comment?

@ShankRam That sounds plausible.

What is immediately obvious is that going from a primary to a secondary cation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change.

Uh-Oh!
When we see a table reporting that the solvolysis rate for an ethyl and methyl compound are both equal (both 1 in your table), well that's a tip-off that something is grossly wrong.  No one would expect the methyl and ethyl carbocations to have anywhere near similar stabilities. Therefore we wouldn't expect the appropriate ethyl and methyl compounds to solvolyze (if they could) at anywhere near similar rates.
So if we accept that your table has bad data in it, then we really don't know that the change in going from primary to secondary is any different from the change on going from secondary to tertiary.
Carbocation Stability - Background
There are many factors that can affect carbocation stability (inductive effects, resonance effects, neighboring groups), but if we limit our discussion to the stability of saturated hydrocarbon carbocations then we need only consider two of these - inductive and resonance effects.
The carbocation center is roughly $sp^2$ hybridized with an empty p orbital; the alkyl substituents attached to this carbon are roughly $sp^3$ hybridized. There is a flow of electrons through the sigma bonds from the $sp^3$ carbon(s) to the more electronegative $sp^2$ carbon.  We call this an inductive effect and it will stabilize the carbocation center.
Carbocation stabilization can also result from the flow of electrons through p orbitals.  We call this a resonance effect, and in the particular case at hand it is a hyperconjugative resonance effect, as pictured in the diagram below.

(For more on hyperconjugation, particularly from an MO perspective, see the "Rotational barrier of ethane" section in this Wikipedia article.)
Prediction
The t-butyl carbocation has 3 methyl groups that can stabilize the carbocation center through inductive effects, there are also 9 beta hydrogens that can stabilize the carbocation center through hyperconjugation.
The 2-propyl carbocation has 2 methyl groups that can stabilize the carbocation center through inductive effects, there are also 6 beta hydrogens that can stabilize the carbocation center through hyperconjugation.
The ethyl carbocation has 1 methyl group that can stabilize the carbocation center through inductive effects, there are also 3 beta hydrogens that can stabilize the carbocation center through hyperconjugation.
The methyl carbocation has no inductive or hyperconjugative stabilization.
Consequently we might expect the relative carbocation stability to be
t-butyl > 2-propyl > ethyl > methyl
with the 2-propyl carbocation roughly midway between the t-butyl and ethyl carbocations.
Data
Chemistry is complex enough without adding extra unnecessary variables - like solvent. A relatively straight-forward method to compare carbocation stabilities is to measure the free energy ($\Delta G^o$) of the following equilibrium in the gas phase. The more negative the $\Delta G^o$ observed, the less stable the carbocation.
$$\ce{R^+ + H^- <=> R-H}$$
The following table shows the the results of this investigation and appears to  confirm the above predictions on carbocation stability.
\begin{array}{|c|c|} \hline
\text{carbocation} & \text{Delta} {~G^o} \\ &  \text{(kcal/m)} \\ \hline
\text{methyl} & -314 \\ \hline
\text{ethyl} & -270 \\ \hline
\text{2-propyl} & -252  \\ \hline
\text{t-butyl} & -237 \\ \hline
\end{array}
(data taken from Advanced Organic Chemistry, Part A: Structure and Mechanisms; fifth edition; Authors: Carey, Francis A., Sundberg, Richard J.; table 3.10, page 303)
The differences between the [t-butyl and 2-propyl] and the [2-propyl and ethyl] carbocations is 15 and 18 kcal/m respectively. That the 2-propyl cation is roughly in the middle of the t-butyl and ethyl carbocations is likely fortuitous since the ethyl carbocation is structurally quite different from the other two alkylated carbocations. The ethyl cation is non-classical in nature, involving a 3-center 2-electron bond with a bridging hydrogen as explained in this earlier answer.


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Calculating energy change for calcium carbonate decomposition
The $\Delta H^0$ for the decomposition of $\ce{CaCO3(s)}$ is $178.3~\mathrm{kJ/mol}$ at $25~\mathrm{^\circ C}$ and $1.000~\mathrm{atm}$. What is the $\Delta E$ for the decomposition process?
$$\ce{CaCO3(s) -> CaO(s) + CO2(g)}$$
I'm really not sure how to get from $\Delta H^0$ to $\Delta E$. I'm not sure if the formula $\Delta H=\Delta E+P\Delta V$ is meant to be used. I know if pressure is constant — as it appears to be in this question — $\Delta H=q$. And I also know the equation $\Delta E=q+w$. As there is no work ($w$) being done, does that just mean $\Delta E=q=\Delta H=178.3~\mathrm{kJ}$? I may be extremely off track here, really not sure.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

This is not a homework question - I'm no longer a student. I'm revising my uni chemistry to sit the gamsat test (Australian medical admissions test). I'm just not too sure where to start with this question - what formula do I even use? Thanks!

Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. Don't worry, they're not banned. But, we require a minimal effort. You can just tell us what you did that didn't get answers. We'll take it from there. What do you know about $\Delta H$ or $\Delta E$? Please [edit] your question and add the details.

Ok no worries. Sorry this is the first time I've ever used this site :)

@Nilay Please do not randomly add \ce{...} to every MathJax equation. It is meant only for chemicals and chemical equations. It is not supposed to be used to format units in upright type; use \mathrm{...} around the unit for this.

Would people mind sharing some insight into the question, rather than worrying about the wording or formatting? Thanks...


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What is the meaning when it is said "Temperature of an Oxyacetylene Torch is 3500 degrees Celsius"?
I don't think flame is anything but the product at that temperature but I am not sure. What exactly is flame? And I think the correct statement should be that the exothermic reaction here produces heat at such level that the temperature of the product gas reaches 3500 degrees.
What will be the exact idea behind the statement?

Oxyacetylene is a mixture of acetylene (ethyne) and oxygen (not air). The mentioned temperature of about $T=3500\ \mathrm{^\circ C}$ is the adiabatic flame temperature for the combustion reaction of acetylene and oxygen.
An adiabatic process is one that occurs without transfer of heat. The adiabatic flame temperature is a theoretical temperature, which would be achieved in a combustion system in which there are no heat losses (e.g. no radiation losses from the flame). Note that this cannot be achieved in a real flame; therefore, adiabatic flame temperatures are calculated values.
The heat $Q$ released from the combustion reaction of acetylene and oxygen is the heat of combustion. If all of the energy released by this chemical reaction is used to increase the temperature $T$ of the products ($\ce{CO2}$ and $\ce{H2O}$) with no heat losses, the resulting adiabatic flame temperature can be calculated from the heat capacity $C$ of the products.
Hence, the adiabatic flame temperature represents the maximum possible theoretical temperature for a particular fuel/oxidant combustion.


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Ordinary differential equation for molecule collision
I have no background in chemistry.
But, I have a good background in mathematics and computer programming.
I have planned to simulate the collision of two simple molecules $\ce{Br2}$ and $\ce{H2}$ just for fun and using state space and Ordinary Differential Equation (ODE) to simulate.
I have considered 24 states for simulations of their collision. The state vector is X, and the state space representation is:
X'=f(X,t)

Atom numbering
1: Br
2: Br
3: H
4: H

The 24 states for vector X are:
x1, x2, x3:  position of atom 1 in x, y, z axis
x4, x5, x6:  position of atom 2 in x, y, z axis
x7, x8, x9:  position of atom 3 in x, y, z axis
x10,x11,x12: position of atom 4 in x, y, z axis
x13,x14,x15: velocity of atom 1 in x, y, z axis
x16,x17,x18: velocity of atom 2 in x, y, z axis
x19,x20,x21: velocity of atom 3 in x, y, z axis
x22,x23,x24: velocity of atom 4 in x, y, z axis

To make a simulation, I need to know two things:

Are the mentioned states enough to describe the entire system?  Are there more states to be added?
What are the forces applied to each atom at each moment?

$\frac{d x_1}{d t}=x4$ (derivative of position is velocity)
$\frac{d x_4}{d t}=\frac1{m_{Br}}(???????)$ (derivative of velocity is acceleration)

There are two types of molecular simulation (in a broad sense), namely
1) Monte Carlo simulation
2) Molecular dynamics simulation
And they can be mixed together to give rise to new algorithm. According to your problem statement I am assuming you are trying to use molecular dynamics simulation. There are lots of things you should consider for your simulation system. According to your problem statement your atoms have position and velocity, but the problem is when they will collide each other they will loose energy and will come to a halt after few time steps. Either you can use thermostat to keep the temperature constant (Anderson thermostat and Nose-Hoover thermostat), or you can use constant pressure molecular dynamics simulation. For a beginner I would like Anderson thermostat as this is very simple technique. Then you have to use Newton's second law to determine the force and velocity on each atoms. After that you can use numerical integration technique to find new position based on F=ma formulation. Now you will repeat this task to get the trajectory of individual atoms.
If you want to find the diffusion coefficient of atoms, you can use lattice gas models or Ising models based on random Monte Carlo movement. I wrote a python code for finding diffusion coefficient. Also you can use Markov Chain Monte Carlo (MCMC) method to determice the stochastic evolution of the system.
For your better understanding I would suggest to use Understanding Molecular Simulation by Daan Frenkel & Berend Smit

Thanks a lot. Is Monte Carlo used on atoms? I mean for 4 atoms is Monte Carlo necessary too? My aim is not to simulate a mass of atoms despite I know that is the applicable approach. In addition, are orbitals or their interactions considered? And is there any sample code in this field to give me some intuition?

Please see the attached URL for example code:
http://www.cchem.berkeley.edu/chem195/files.html
http://www.acclab.helsinki.fi/~knordlun/

Usually ab Initio calculations are carried out separately. Because doing molecular simulation with on the fly ab Initio techniques is not worth the time and effort. However Monte Carlo or molecular dynamics techniques are being used to explore the potential energy surface but that's different story.


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What is the use of "pyro" in the naming of organic compound?
Recently I asked a question on the naming of polyhydroxy phenols. One of them was pyrogallol (1,2,3-Trihydroxybenzene) which is  synthesized from gallic acid  obtained from oak galls. (BEWARE- not a compound of gallium).

But why "pyro"? Pyro is usen in naming inorganic acids like pyrosulfuric acid ($\ce{H2S2O7}$) or pyrophosphoric acid ($\ce{H4P2O7}$). It is also observed that the carboxylic acid derivative is gallic acid not pyrogallic acid and the hydroxy derivative is pyrogallol not gallol. Why is it so?

pyro- is not a part of systemtic naming, it is a prefix meaning 'fire' and is used with acids or salt preparable via route involving heating. Obviously, there is often more than one acid preparable such way, but only one usually ends with pyro- prefix. Thus, the prefix is not encourged.

@permeakra please consider submitting your comment as an answer, as this is precisely the reason.... The formation of that product comes from simply heating. The example in the question shows that nicely.  A link to the IUPAC Gold Book may be useful.

This question mainly applies to organic chemistry. For the use of the prefix “pyro-” in inorganic chemistry, see also the questions and answers here, here, and here.

So, as @long and @permeakra stated, the name pyrogallol came from the fact that it is obtained by heating gallic acid. The IUPAC goldbook clearly states that:-

"pyro" is used as a prefix designating compounds formed by heating a compound,
  usually with the elimination of water, carbon dioxide, or other simple
  molecule, e.g. pyroglutamic acid from glutamic acid.

So as per rule, it is pyrogallol and not gallol.


text:
Investigating the displacement of a Se anion by a Sulphide anion
I am working with CdSe Quantum Dots suspended in hexane. I then tested the photoluminescence of my solution. Then I mixed my CdSe dots with a mixture of an organic conducting polymer which was dissolved in chloroform. I am attaching the structure of my polymer below.
My idea was to embed the dots in the polymer matrix. This would obviously quench the resultant luminescence of the dots because the excitons at the surface of the dots would be delocalized by the conducting matrix, but this should not result in a wavelength oriented shift of luminescence, but I am getting a shift. I am attaching the results of my photo-luminescence along with the structure of the polymer chain.
I suspect that the shift of the luminescence observed is due to an attempt by the S radical in the benzene ring of the polymer to displace the Se radical from my CdSe dots, or is there any other way that my dots could get chemically attacked by the polymer? And if so, is there any way to check it? Like a form of spectroscopy for instance?

The polymer to the left is PDMEET and the one to the right is PDMET. The shift in the luminescence image comes here.
.

I'm curious - Is the difference in intensity due to a difference in the concentration of the microdots of CdSe?

Or it may also be due to exciton delocalization  by the surrounding matrix? Is there a way to process it? Also, I am still not sure about the ligand which has been used to cap the dots, probably there is a ligand because the dots are very luminescent even under normal light. But, there is a probability that the quenching is due to lower dot concentration, because I just drop coated the dot polymer mixture from a vial to a slide (hence no uniformity) and then probed it with a 325 nm beam. But, I am more interested in the shift than the quench at present. Could you suggest something?

Color variations in CdSe particles is due to a quantum effect of particle size. It would seem that the surface effects are making the effective size of the particles smaller and shifting the peak to a shorter wavelength. It would be very interesting if shift and quench were related. That is to say that a stronger quench results in a smaller effective particle size.

I'd certainly doubt any sulfur from the polymer replacing selenium in the particle.

Surface effects as in? Could you elaborate a little more? I don't think that quench and shift will be related if the shift is just related to decrements in particle size. Quench would imply either decrease in particle numbers or increase in surface defect state density and hence reduced probability of radiative recombinations, but I will try to get information about the ligands smothering my dots, maybe an attack on the ligands with these two polymers are possible. I will share the info with you, maybe you could make some predictions about their behaviour. Thank you for taking an interest.

I was thinking of some sort of van der Waals forces forming like a rind on a watermelon. If you imagine a capacitor formed by two metal plates, then the charges aren't distributed evenly through the plates but concentrated at the closest surfaces. So the rind would reduce the effective size of the particle. // Sorry but I wouldn't know what to do with more information. I started playing at this site to refresh my knowledge of chemistry. It has been many years since I was a practicing chemist.

Don't apologize. You were of immense help.


text:
Does the initial rate in the reaction of magnesium powder with nitric acid depend on the volume or the concentration of the acid?
For the question regarding the initial reaction rate, I don't quite understand why the answer is D.
If I calculated the number of moles from the volume and conc, I would get A=0.4 moles; B=0.2 moles; C=0.2 moles; D=0.1 moles
As original equation is 0.2 moles of NH3, at 25 degree Celsius, I chose C. But why is the answer D, given that it is 0.1 moles not 0.2 moles.
Thanks in advance and sorry for any wrong tags.

Imagine you have a railroad tank car full of $\ce{HNO3}$; would you expect it to react with the same amount of magnesium any faster? It is concentration, not amount, that is important.

@IvanNeretin, can you explain why I can't use the moles to figure it out?

No, I can't, if my previous comment makes no sense to you.

@IvanNeretin, I don't understand, because concentration is moles per unit volume, and I am also calculating moles in that volume I don't understand how the two are different.

The magnesium is present in form of a solid powder. Therefore, the reaction of magnesium with acid can only occur at the solid–liquid interface between magnesium and the aqueous nitric acid solution. Accordingly, the initial reaction rate depends on the surface area of the solid–liquid interface. For the experiments described in the question, the same mass of magnesium powder is used; therefore, the available surface area of the magnesium powder is approximately the same, too. Hence, as long as the added volume $V$ of the nitric acid solution is sufficient to cover the entire available surface of the magnesium powder, the initial reaction rate does not depend on the added volume of the solution. (Assuming a reasonable geometry of the reaction vessel, a volume of $V=100\ \mathrm{ml}$ should be more than enough to cover a mass of $m=5.0\ \mathrm g$ of magnesium powder.)
Furthermore, the reaction rate depends on the concentration $c$ of the acid at the solid–liquid interface. Therefore, the initial reaction rate depends on the concentration of the added nitric acid solution.
And finally, the reaction rate also depends on temperature $T$. Therefore, the initial reaction rate depends on the temperature of the reactants, i.e. the magnesium powder and the added nitric acid solution.
Hence, in order to obtain the same initial reaction rate, the concentration of the nitric acid solution $(c=1.0\ \mathrm{mol\ l^{-1}})$ and the temperature $(T=25\ \mathrm{^\circ C})$ should be the same during both experiments.


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What is the molecular structure of polymetatelluric acid?
Telluric Acid, $\ce{Te(OH)6}$, dehydrates to form polymetatelluric acid: $\ce{(H2TeO4)10}$. However, I have been unable to locate the exact structure of polymetatelluric acid, and it does not seem entirely obvious.
Are the tellurium centers connected directly to each other or are they separated by oxygen? Are these polymeric bonds single bonds or double bonds or some mixture of the two? My best guess is that the $\ce{Te}$ centers are separated by oxygens with three other oxygens surreounding each $\ce{Te}$ center, two of which are $\ce{-OH}$ groups and the third being a $\ce{=O}$ group: this would make the oxygens between the $\ce{Te}$ centers alternating single and then double bonds — possibly conjugated on some level. Is there any chance polymetatelluic acid has direct $\ce{Te-Te}$ bonds?

Seems stoichiometry isn't precise, so there are differing molecules there. No chance for Te-Te bonds as oxidation state would be different.

It would be ideal to check out x-ray analysis data, but it seems like polymetatelluric acid is a rather amorphous substance and does not form suitable crystals. As of now, I haven't discovered anything relevant neither in ICSD, nor in CCDC.
Another method of structural analysis would be vibrational spectroscopy. There is an article "Ultrarotspektren von Tellursäuren, Telluraten und Antimonaten" [1, p. 165] (in German) containing an extensive research of various $\ce{Te}$-containing substances performed with infrared and Raman-spectroscopy (I reduced the table to the frequencies for polymetatelluric acid exclusively):

\begin{array}{ll}
\hline
\ce{(H2TeO4)_x} &\text{Zuordnung} \\
\hline
450~\mathrm{(m, b)} & \delta~(\ce{TeO}) \\
\hline
600~\mathrm{(st, b, Sch)} & \nu~(\ce{TeO}) \\
720~\mathrm{(sst, b)} & \\
600~\mathrm{(sst, b, Sch)} & \\
\hline
1085~\mathrm{(mst, b)} & \delta~(\ce{TeOH}) \\
\hline
1618~\mathrm{(s, b)} & \delta~(\ce{H2O}) \\
\hline
2360~\mathrm{(m)} & 2\delta~(\ce{TeOH}) \\
\hline
3200~\mathrm{(sst, b)} & \nu~(\ce{TeOH}) \\
3360~\mathrm{(sst, b, Sch)} & \nu~(\ce{H2O}) \\
\hline
\end{array}
Polymetatellursäure $\ce{(H2TeO4)_x}$ ist eine feste amorphe Substanz. In ihrem UR-Spektrum sind alle Banden sehr breit, wie man es bei amorphen Substanzen meist antrifft. Das Tellur ist in dieser Verbindung wahrscheinlich wie im $\ce{Te(OH)6}$ sechsfach koordiniert, wie die annähernd gleiche Lage der $\ce{TeO}$-Valenz- und Deformationsschwingungen in beiden Verbindungen zeigt. Hieraus folgt, daß $\ce{TeOTe}$-Bindungen vorhanden sein müssen. Bei Viererkoordination würden die Valenzschwingungen höher liegen, wie es in den vergleichbaren Molekeln $\ce{IO}$, und $\ce{H5IO6}$ der Fall ist.[...]
Im Gegensatz zum $\ce{Te(OH)6}$ tritt im Spektrum des $\ce{(H2TeO4)_x}$ eine Frequenz bei $\pu{1618 cm-1}$ auf, die nur von freiem Wasser herrühren kann. Ebenso beobachtet man neben der $\ce{OH}$-Valenzschwingung der $\ce{TeOH}$-Gruppen ($\pu{3200 cm-1}$) die Bande $\pu{3360 cm-1}$ des freien Wassers. Aus den Intensitäten dieser Banden ist zu entnehmen, daß $\ce{H2O}$ und $\ce{TeOH}$-Gruppen in größenordnungsmäßig gleicher Menge vorhanden sind. Eine stöchiometrische Konstitutionsformel wird man dieser Substanz wohl nicht zuordnen können.

Briefly, the IR-spectra bands of $\ce{[H2TeO4}]_x$ bands are very broad, suggesting amorphous non-stoichiometric compound (as Mithoron suggested) with polymeric structure where $\ce{Te}$ atom is 6-coordinated. Bands were only assigned for $\ce{Te-O}$, $\ce{Te-OH}$ bonds and $\ce{H2O}$. IR-spectra do not show the presence of $\ce{Te-Te}$ bonds. Most likely there are $\ce{[TeO]6}$ units with edge- or corner-sharing, linked together in a chain-alike fashion.
Reference

Siebert, H. Z. anorg. allg. Chem. 1959, 301 (3–4), 161–170. DOI: 10.1002/zaac.19593010305 (in German).


text:
What factors could lower the voltage of a galvanic cell
Recently I conducted a lab in which the voltage of a galvanic cell between two metals was recorded. Both metals were placed in an solution of their own ions with nitrate and a salt bridge was made by dipping some paper in salt water. The only issue is the voltage recorded in the cell was lower than that calculated using standard reduction and oxidation potentials. Does anyone have any ideas as to why this might be? I known the metals were not super clean although I don't think that would have an affect on the voltage.

Although it seems simple to make a galvanic cell ( i.e. a battery) it is really fairly complex.

The theoretical voltage is given by the Nernst equation.
The actual measured voltage also depends on the current. The equation
$V=iR$ still applies. So various restance sources in the cell cause voltage drops.

The current must be very low so a high impedance voltmeter must be    used. (Not really a problem these days with modern semiconductor
    voltmeters as opposed to galvanometers.)
The salt bridge offers resistance, so it has a voltage drop.
Impure metals for the anode or cathode can cause a voltage change.
Surface coating can cause a voltage drop.
The acid or base concentration can cause a voltage change.
The ionic strength of the solution can change the voltage.

There are more factors to consider when determining the true voltage of the cell such as the development of overpotentials.
'In electrochemistry,  overpotential is the potential difference (voltage) between a half reactions thermodynamically determined reduction potential and the potential at which the redox event is experimentally observed.
In a galvanic cell overpotential means less energy is recovered than thermodynamics predicts. This extra/missing energy is lost as heat.
Overpotential is specific to each cell design and varies across cells and operational conditions, even for the same reaction.
Regarding the galvanic cell, two possible polarities of overpotential can occur:

the anode may be less negative, supplying less energy than thermodynamically possible.
the cathode is less positive, supplying less energy than thermodynamically possible.

The overpotential increases with growing 'current density', or rate as described by the 'Tafel equation'. Each step of the electrochemical reaction is associated with multiple forms of overpotential. The overall overpotential is the summation of many individual losses.
'Voltage efficiency' describes the fraction of energy lost through overpotential. For a galvanic cell it is the ratio of a cells experimental potential divided by the cells thermodynamic potential converted to a percentile.
Overpotentials can be grouped into three categories, activation, concentration and resistance but  can be divided into many different sub categories also that are not well defined, such as 'polarization overpotential'.'
Ref:overpotential/wikipedia
For more information on 'voltage efficency','Tafel equation','current density' I suggest to 'Google' them or to look at the page I have referenced.
Other factors such as loss of electrolyte and the temperature of the system can also cause loss of voltage.


text:
Can carbon and titanium form an ionic bond?
Can carbon and titanium form an ionic bond because they both have four valence electrons? If titanium transfers four electrons to carbon, we get $\ce{Ti^4+}$ and $\ce{C^4-}$ which both have noble gas configurations.

TiC is interstitial, metallic carbide https://en.wikipedia.org/wiki/Carbide

Yes, carbon can form ionic bonds with Titanium. $\ce{TiC}$ is titanium carbide, an ionic compound with $\ce{C^{4-}}$ (the carbide anion) and $\ce{Ti^4+}$ (titanium cation). It has the same crystal structure as $\ce{NaCl}$ (which probably has to do with the $\ce{1\!\!:\!1}$ ratio of anion to cation).
This is a space filling model of titanium carbide:

Here is sodium chloride for a crystal comparison (click the image for a larger version):

The second model (the $\ce{NaCl}$) shows the ionic bonds as covalent bonds but still, both titanium carbide and sodium chloride have a cubic lattice structure.
However, carbon does not often form ionic bonds with metals since it prefers to be covalent (much like oxygen). In fact carbon has much more covalent character than oxygen does which is why oxides form easily (just $\ce{O2 + element}$ and usually, oxides will form) but carbides don't (a catalyst is needed to form carbides and some oxides like gold oxide).

Is TiC really completely ionic? Or is it "partially" ionic that it still shares electrons some way, like through dative bonds to stabilize the system?

Also, Caters, please cite your sources for where you obtained these images!

TiC is hardly ionic at all. To begin with, it would not hydrolyze.

Not all ionic compounds are soluble in water. Take for example Calcium fluoride. It isn't soluble in a lot of solvents because of the calcium ions but it is most surely ionic. So just because TiC isn't soluble in water does not mean that it is not ionic.

I never said "soluble". I said "hydrolyze", and that's what ionic carbides typically do, because hydrocarbons are much weaker acids than water. This has very little to do with solubility. $\ce{Al4C3}$, for example, hydrolyzes quite readily, but isn't water-soluble in any meaningful sense, nor are the products of its hydrolysis.

But hydrolysis involves water so it would make sense that anything that hydrolyses is soluble in water.

No, hydrolysis does not imply solubility. The reaction could take place at the interface of water and substrate, often resulting in passivation if the hydrolysis products are insoluble or poorly soluble.

See
Kuiying Chen, Sami Kamran, Bonding Characteristics of TiC and TiN, Modeling and Numerical Simulation of Material Science, Volume 3, Number 1, , 2013 PP. 7-11  doi 10.4236/mnsms.2013.31002
From the abstract:

Using ab initio density functional theory calculations, the electron localization function (ELF) of typical transition metal carbide TiC and nitride TiN were computed and analyzed to reveal their nature of the chemical bonds. The ELF approach was initially validated through typical examples of covalent-bonding Diamond (C) and ionic-bonding sodium chloride NaCl. Our results clearly demonstrate the dominantly ionic bonding characteristics of TiC and TiN.

It is better to spend some time discussing what the literature says


text:
Composition and pH of water produced by air conditioners
I would like to know if there are specific studies or research regarding the condensed water that is generated by air conditioners.
Many people feel that this water is acid and corrosive but could not find anything that confirms or denies this belief.
If someone could send me a link or document with a reliable scientific research would help me.

Acid? Why? Condensed water is essentially distilled water.

I suspect some people might think this because all that water dripping out of the AC unit over the course of a few years will begin staining/eroding whatever is in its path. That's just water being water, though.

Distilled water will dissolve carbon dioxide from the atmosphere, which could reduce the pH below 7. Here's a technical document for  water stills

True, but I deliberately tried to avoid this topic. Anyway, this is not what people outside of chemistry mean by "acid".

Additionally, if it's central AC and the blower unit is shared with a high efficiency gas furnace, the condensate from the furnace is quite acidic due to sulfuric acid in the combustion gas. So they may be confusing that with the condensate that comes off during cooling since they usually come out of the same drain pipe.

Pure condensed water is neutral (pH 7). However, water in contact with air is slightly acidic (approximately pH 5.6) due to dissolved carbon dioxide $\ce{(CO2)}$. Remarkably, the same applies to rainwater.
Furthermore, water condensed from air is not pure. The condensing water collects small amounts of various airborne substances (soluble gases and vapours, dust, and even radioactive particles from radon decay). (Note the large volume of air that is passed through the air conditioner while a certain amount of water is condensed.) Usually, these substances do not significantly affect the pH of the water. However, when the air contains large amounts of pollutants, such as sulfur dioxide or nitrogen oxides (e.g. produced in industrial processes), the condensed water can become more acidic (like acid rain, which can have pH values of about 4).
By way of comparison, typical values for black coffee are about pH 5.0–5.1, and for orange juice about pH 3.3–4.2.

What about the unpleasant smell emanating from the shop floor after rinsing the floor with the water  produced by the air conditioner? What is the reason for this smell?


text:
Why do you need to convert copper (II) hydroxide to copper (II) oxide before converting to copper (II) sulfate?
This is the protocol for a practical class I once took. The idea is that copper nitrate is converted to hydroxide, oxide, sulfate and then finally solid copper. I know that $\ce{CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O (l)}$, but can't you avoid converting the hydroxide to oxide altogether and just do $\ce{Cu(OH)2(s) + H2SO4(l) -> CuSO4(aq) + 2H2O(l)}$? Note that $\ce{NaCO3}$ is decanted and washed off before reacting copper (II) oxide with sulfuric acid, but surely you can also do the same thing with copper (II) hydroxide?

I'm not 100% sure, so I'm not posting this as an answer, but I believe it's because the Cu(OH)2 is quite dirty when it's produced.  Cu(OH)2 really holds onto impurities well (here that would be sodium nitrate, carbonate, and hydroxide, largely), and the conversion to CuO cleans it up substantially.  If you started with clean Cu(OH)2, I suspect that you would be able to convert it to the sulfate with sulfuric acid with little difficulty, but it has such a gloopy consistency that getting it cleaned up in hydroxide form is not an insignificant task.

Of course you could have avoided converting the hydroxide to oxide. Also, you could have avoided getting the hydroxide in the first place, and go from nitrate straight to sulfate. Oh wait, and what if we abandon all that and get your solid copper from nitrate in one step? See, it's a practical class, not a real-life technology problem. Its purpose is to get you familiar with certain procedures, and not to get from A to B as fast as one can.

Hmm.. that's absolutely true. But considering the OP has already prepared the hydroxide , then conversion to oxide (and then sulphate) is the best way.

@IvanNeretin well I did ponder if we could get from copper nitrate to copper by just adding zinc like the prac says, but then we would get two solids forming which would be hard to separate. I'm not sure about going from nitrate to sulfate - I think it gives nitric acid instead of just water which may complicate the adding of zinc and removal of zinc by HCl. of course there may be an easier way with different chemicals/equipment but let's assume we don't have access to that. but yeah it's possible they're just adding steps, but I thought there could be a reason

Getting copper from copper nitrate is almost the same as getting copper from copper sulfate; if there is a problem of two solids (which is not that much of a problem, but whatever) in one of them, you'll encounter the same in the other one as well. Yes, they are just adding steps for the sake of teaching, though there are more-or-less real reasons too; these are nicely explained in the existing answers.

I would like to add to what Jason has already mentioned:
1.Copper Hydroxide is comparatively tougher to handle since it contains a lot more impurities ( when we are converting Cu(II)nitrate to hydroxide using NaOH, the impurities could be sodium nitrate, excess sodium hydroxide. )

Copper Hydroxide itself is blue, so the equivalence point (when exact amount of copper hydroxide has reacted with concentrated sulphuric acid) of the reaction would be hard to detect, and so yield would drop (since you would again have to neutralise the excess sulphuric acid before purification or conversion to copper metal).
In contrast to this CuO is a black precipitate and can be easily isolated (comparatively) by a Hirsch funnel (or any such equivalent apparatus). This improves the overall yield of further steps and eases the process of purification.

Hope this helps.

Good points, and actually we don't even neutralise the H2SO4, we just decant it off and wash with water. first year labs for ya

1) Don't expect too much sense in practical chemistry classes, they often use ineffective, but tutoring approach
2) That said, copper hidroxide is very fluffy and adsorbs a lot of water. It is plainly a lot easier to wash compact copper oxide than fluffy copper hydroxide.

I disagree with your first point. It is always a good approach to understand minute details of a practical class, most of the procedures have been standardised with practice. They may not be the MOST effective , but certainly are made with the intent of being effective. If they are not effective,it is also helpful to find out why and improve over it.

@Varun , I once made phosporic acid from red phosphor in a practical class. Isn't it definition of impractical and ineffective ?


text:
How can I choose the correct mechanism for nucleophilic substitution of aryl halides?
I was reading about nucleophilic substitutions of aryl halides, and there are three different mechanisms discussed, namely the SN1, SNAr (addition-elimination), and benzyne (elimination-addition) mechanisms.
However, I am confused as to how to identify which mechanism is operative. How can I tell this from the reaction conditions and reagents?

Before comparing the three main mechanisms, I will first explain why aryl halides typically do not undergo SN2-type reactions. Take bromobenzene as a typical example of an aryl halide. A nucleophile, like hydroxide ion, cannot simply displace the bromide from bromobenzene. Quoting from Organic Chemistry (1st ed., p 589) by Clayden et al.,

In the aromatic compound, the C–Br bond is in the plane of the ring as the carbon atom is trigonal. To attack from the back, the nucleophile would have to appear inside the benzene ring and invert the carbon atom in an absurd way. This reaction is not possible!

In rare cases, SN1 reactions can take place; these characteristically occur when there is an excellent leaving group on the arene. One of the few examples is in the case of an aryldiazonium ion, where the corresponding leaving group is nitrogen gas.

Note that the presence of a diazonium ion does not necessarily imply a SN1 mechanism; there are also other possible mechanistic pathways, mainly involving radical chemistry.

The next possibility is that of an addition-elimination reaction, commonly termed SNAr. This is made possible by electron-withdrawing groups (e.g. nitro or carbonyl groups) when they are ortho and/or para to the halogen, as the negative charge in the intermediate can be delocalised into these electron-withdrawing groups:

Finally, the benzyne mechanism. Consider bromobenzene: it cannot undergo SN2 reactions (as explained above), SN1 reactions (bromide is not a sufficiently good leaving group), or SNAr reactions (no electron-withdrawing group ortho or para to bromine). However, under strongly basic conditions, it is possible to eliminate HBr by abstracting a proton ortho to the halide. Examples of such strong bases include sodium amide (NaNH2), LDA/lithium diisopropylamide (LiNi-Pr2), or potassium tert-butoxide (KOt-Bu).

To summarise the three categories:

If you have a very good leaving group - primarily nitrogen gas - on the aromatic compound, then a SN1 mechanism is possible.
If you have good electron withdrawing groups ortho or para to the halide, then it is likely an addition-elimination or SNAr reaction.
If the above two are not possible, and if you have an extremely strong base such as NaNH2 or KOt-Bu, then the benzyne mechanism may be applicable.


text:
What is the meaning of "equivalent weight" and its formula?
What is meant by "equivalent weight" and how does its formula E=M/V support the definition?

Definition of Equivalent Weight (also known as equivalent mass):
The simplest definition of equivalent mass I can think of is the following:
The mass of a compound which will combine with 8g of oxygen or 1g of hydrogen is said to be the equivalent weight of the compound.
Lets take an example,
$\ce{2Ca + O2 -> 2CaO}$
From basic stoichiometry, you know that 2 moles of calcium react with 1 mole of oxygen to produce 2 moles of calcium oxide. We can say the same statement in terms of masses as 80g of calcium reacts with 32g of oxygen to produce 112g of calcium oxide.
Since 80g of Calcium reacts with 32g of oxygen, 20g of calcium should react with 8g of oxygen.
Now by the definition of equivalent weight mentioned earlier, the equivalent weight of calcium is 20g.
$$E = \frac{M}{V}$$
The formula you were talking about gives the equivalent weight of an element (can be extended to be used with ions).
Equivalent weight = (Atomic mass)/(Valency)
This could be intuitive and obvious for some. Something with valency of 1 would combine with just 1 hydrogen atom, or in other words, a mole of compound with valency 1 would react with 1g of hydrogen. If the element has a valency of 2, then it would love to combine with 2 hydrogen atoms, or in other words, a mole of the element will combine with 2g of hydrogen. By definition, equivalent weight is the mass of the substance that would react with 1g of hydrogen. Therefore you must divide the mass by 2 to get equivalent weight. In fact, this formula based definition of equivalent weight of an element is identical with the definition mentioned earlier in this answer.
Previously, we determined that equivalent weight of calcium was 20g using the other definition. We should obtain the same equivalent weight for calcium by using the formula.
Equivalent weight = (40g)/2 = 20g
Definition of equivalents:
Another terminology you might come across is equivalents.
Equivalents is similar to moles. The number of moles is given by (mass of the compound)/(molecular mass). Similarly the number of equivalents is given by (mass of the compound)/(equivalent weight).
For example, 1 equivalent of calcium contains 20g of calcium. 2 equivalents of calcium contain 40g of calcium.
The utility of equivalent weights is amazing. In any reaction, $N$ equivalents of any compound on the reactants side reacts with N equivalents of another reaction to produce $N$ equivalents of the product(s).
If that sounds complicated, this example should make it clear.
20g of calcium reacts with 8g of oxygen to produce 28g of calcium oxide.
An equivalent statement is, 1 equivalent of calcium reacts with 1 equivalent of oxygen to produce 1 equivalent of calcium oxide.

Thanks and How about finding Equivalent weight in Redox reactions?

There are many definitions for equivalent weight of a substance. You must decide which one to use depending on the situation. In case of redox reactions, equivalent weight is given by dividing molecular mass by the magnitude of change in oxidation state.

Can you please explain How does this formula for Redox reaction verify Equivalent weight?

It is not possible to explain the whole concept in a comment. In fact Redox Reactions is a complete chapter by itself. I found a good link for you which has examples where equivalent weight is calculated on the basis of change in oxidation state. Click Click . Once you learn to calculate the N-factor, you can obtain the equivalent mass by dividing the molecular mass by N-factor.

You can extend the definition to chlorine. BTW nice answer.

Suppose I have AgCl whose equivalent weight I dont know. Can I write the equivalent weight of AgCl as Eq Wt of Ag + Eq Wt of Cl ?

Y.S. is just plain WRONG.
Equivalent weight is the (mass of) number of moles of one reactant in a balanced reaction (or product) divided by the number of moles of another.
In the simple case where we have two reactants which combine 1:1, then the mass of 1 mole of either is its equivalent weight. This means that the same chemical compound can have a wide variety of Eq. Wts. because the reaction it participates in can have all sorts of different stoichiometries. Eq. Wt. should only be used when the reaction is understood - and often is used to compare two different compounds reacting in the "same" reaction eg:
xM + O2 → MaOb
where the mole ratio is x:1 in this specific case if relative to O2 or a:b if relative to the product. In other words, the equivalent weight relates to a specific reaction - and by the way, its possible that the exact same two reactants can have different equivalent weights if they can combine in more than one stoichiometry. I should also note that it is possible that your teacher/professor/lecturer/textbook has decided to use the term in a better (standardized) way, but generally if you don't know (for sure) what the reaction is that is being discussed you should avoid thinking you know what Eq. Wt. means.

What was wrong in my answer?

I'll explain it in simple terms.
If you have
a A + b B->c C
where A, B are the reactants and C is the product and a,b,c are the stoichiometric coefficients of the reaction.
The equivalent weight is in simple terms the mass of one equivalent of a substance.
Now let's see what we mean by one equivalent.
In our reaction let's assume a=2 b=3.
2 moles of A + 3 moles of B react stoichiometrically.
The molar ratio of the reaction is 2:3
Suppose we were to say that A and B react in the ratio of 1 equivalent to 1 equivalent.
The equivalent ratio of the reaction is ALWAYS 1:1.
So we have to define an equivalent for each compound/element depending on the stoichiometry of how it reacts.
So one equivalent of A is in this reaction 1/3rd of a mole and for B, it is 1/2 a mole each of which will be one equivalent.
We notice that these amounts make sure that A and B react stoichiometrically.
Now we need to standardise the definition of one equivalent.
To be more exact one equivalent is the amount of a substance which is needed to react with 1 g of H2.
The number of Equivalents is generally calculated as:
No. of equivalents=n-factor * no. of moles
Where n-factor generally corresponds to the stoichiometric ratio of the other reactant. Or it could be the change in oxidation number of that reactant.Or it could be the (total) valence of the ionic salt(Charge on cation*total no. of cation units present in one molecule/formula unit) or the acidity for a base or basicity for an acid.
The formula for equivalent weight should be simple from this point.
Equivalent weight=Molar weight/n-factor


text:
Strength of substituted Carbon Double Bonds and C-halo bonds

The question asks me to identify the compound which has a stronger double bond. I came up with two kinds of reasoning but they do not agree upon a common answer.
1: $sp^2$ carbons are more electro-negative and they would like to have electron donating groups sitting next to them. So in the compound one , the extra methyl group should increase stability and therefore, one would expect the bond strengths to increase. Thus decreasing the bond length (smaller bond length is more stronger? stronger bonds imply more stability)
2: Since the methyl group partially donates electrons to the $sp^2$ carbon, the $sp^2$ carbon's tendency of pulling the double bond electrons towards itself should decrease. Thus increasing the bond length.
So where am I going wrong?
If you have any good resource/material on strength of bonds, please let me know.
EDIT (another Q):

Which double bond has higher disassociation energy?

Substituted alkene is more stable than unsubstantiated alkene. So I is more stable which means it has higher bond disassociation energy so II must be having lesser bond disassociation energy.
Due to hyper conjugation, the bond order in compound I decreases so it must have lower bond disassociation energy.

Which one is correct among the above two reasoning?

If the electro-negativity difference between the atoms is larger, then the bond will be more ionic and hence should be stronger?

Out of the 2 bromo compounds, the first one has has stronger $\ce{C-Br}$ bond, because the lone pair of $\ce{Br}$ is in conjugation with $\ce{C=C}$.
In first case I think Compound 1 is more stable for the reason that it has more substituted $\ce{C=C}$. In reasoning 2 I think when you think about decrease in bond order, the tendency of pulling electrons of $sp^2$ carbon has not decreased. Me- group itself donates electrons because Carbon is $sp^2$ hybridised. The tendency will not increase or decrease. I think it will remain in the form of slight polarisation.

What do you think?