text:
Can grignard reagents react with any electrophillic sites?
Yeah I am confused about the above question. Like since textbooks always say how C atom bonded to Mg is very nucleophillic, I was wondering why it does not bond with carbon atom that is single bonded (not the one in carbonyl part) to one of the oxygen atom of ester? Since O atom is more electronegative than C atom, C atom is electrophillic, so shouldn’t grignard reagent also bond with that C atom instead of one double bonded to the other O atom. I always see example where GR reagent attack the carbonyl carbon atom so I was wondering why? Thanks.

To give you an idea of the nucleophilicity of the two carbons, you may have a look at $\ce{^{13}C}$ NMR data (here for example). These data tell you which carbons have less electron density (the ones which are more deshielded).
On methyl formate, you'd find $\delta_\text{COO} = 161.8$ ppm whereas $\delta_\text{OCH3} = 50.0$ ppm. This should give you a little intuition of how much the electron density is reduced on each carbon, due to the effect of the electronegative atoms. Clearly, the ester-bearing carbon is much more affected, and this accounts for the general behaviour that you encounter.
However, as Ben Norris pointed out, this is maybe a simplistic picture. In fact, it's not impossible to break preferentially the $\ce{COO - C}$ bond. This is called hydrolysis by alkyl-oxygen cleavage (see here).
To get back to the question of Grignard reagent, the problem is also explained by HSAB theory. To attack the carbon which is not bearing the ester, you would generally need a "soft" organometallic compound. I can give you an example that corresponds to what you are asking (reference here: Kawashima et al., Tetrahedron 1989, 45, 403.), but note that this is a special case. What drives the reaction is the opening of a strained ring, and the Grignard is modified by addition of copper halide:

I represented this one in particular to show the regioselectivity (breaking of bond represented in red), but the article shows plenty of examples.

The thing is, the carbon with that has a double bond and a single bond with oxygen is the most electrophilic. Therefore the Grignard reagent will always choose that carbon every time instead of the other carbon that has only one single bond with oxygen.

In practical terms, we know Grignard reagents do not react with ethers since ethers are the most common solvents for said reactions. Ethers are not as electrophilic as you think. You cannot just consider the partial charge at an atom, you also need to consider the strength of whatever bond will be broken upon nucleophilic attack.

I totally agree with you. Ethers are generally very unreactive. However his query is on esters not on ethers.


text:
List of functional groups in a molecule
Does anyone know if there's a tool for listing the functional groups present in a molecule?

Norbert Haider from the University of Vienna has been working on the identification of functional groups for quite a while.
You might want to have a look at his checkmol/matchmol tool, which natively reads from MDL, Alchemy, and Sybyl mol files.
Other input formats may be processed after converting them using Open Babel.

RDKit is an option with a python interface. Here's an example using rdkit to find furan groups in 2-Acetylfuran:
from rdkit import Chem
mol = Chem.MolFromInchi("InChI=1S/C6H6O2/c1-5(7)6-3-2-4-8-6/h2-4H,1H3")
furan_smarts = "o1cccc1"
furan = Chem.MolFromSmarts(furan_smarts)
Chem.Mol.GetSubstructMatches(mol, furan, uniquify=True)
>>> ((7, 3, 1, 2, 5),)

The length of the outer tuple is the number of matched substructures, while the integers in the inner tuple are the indices of the molecule's atoms.
Here are some other smarts strings:
aldehydes: [CX3H1](=O)[#6]
esters: [#6][CX3](=O)[OX2H0][#6]
ketones: [#6][CX3](=O)[#6]
amides: C(=O)-N
thiol groups: [SH]

The FP4 fingerprint in Open Babel matches a set of SMARTS patterns for functional groups against a molecule. Usually the results is a binary fingerprint, but the "-xs" option below prints out the names of the matching patterns. Here are the results for acetamide, for example:
> obabel -:CC(=O)N -ofpt -xfFP4 -xs
>
Primary_carbon  Carboxylic_acid_derivative      Amide   Primary_amide   C_ONS_bond      1,3-Tautomerizable

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Why do substances ionise in water to turn into charged species? Wouldn't they prefer to stay as uncharged molecules?
Wouldn't this be more stable if they stayed as neutral molecules?
I've learnt that they will seek to reach the ratio of $[A]^a[B]^b/[C]^c$ that matches the $\ce{Ka}$ or $\ce{Kb}$ value.
But I do not understand why they wouldn't prefer to stay as uncharged molecules as I thought ions are unstable compared to neutral molecules?

Hint: Ions in a vacuum are very different from ions in solution.

Hint2: Take a look at lattice energy of ionic compounds. Take a look at the definition of lattice energy.

Hint3: Not everything ionizes when it dissolves. Salt (sodium chloride) does, but sugar doesn't.

The answers already given here for a slightly different question should (dis)solve your problem: Why don't polar and non-polar compounds dissolve each other?

MaxW, I disagree. The pka of, glucose for example, is 12. The pka of water is 14, or 15.7 depending on your reference. Either way.. Glucose will give up a proton - thus ionization occurs.

Intermolecular forces stabilize the ions of a chemical

Thanks for the answer. But why is this new stability more stable than the NaCl stability from the ionic bond? And so why does the ionisation even occur?

E.g in the question in @Autumn 's comment http://chemistry.stackexchange.com/questions/38260/why-dont-polar-and-non-polar-compounds-dissolve-each-other    it states "when the solute dissolves into the solvent, they are able to be make dispersion forces with each other. The making of these forces releases very little energy. So simply put, very little energy is required to break the forces and very little energy released when making the forces. Hence overall everything balances out and the process occurs."
I don't understand why it gives up a stable scenario to go to another stable scenario

It has to do with the way in which the molecules would need to be arranged, in solution. If NaCl (to keep the example consistent) were to maintain its crystalline structure it would need to maintain a form described as highly ordered. In being dissolved, this "order" is disrupted. Entropy always shifts from order to disorder.

Ah okay. That makes sense then. I didn't think about entropy, only stability.

It is not more stable though. Think of the polarity of water. And the magnetic forces it would exhibit on a salt-crystal. If the crystal were to maintain shape, it somehow would need to "create" energy in a direction that opposes the forces from the water.

Okay. If this is dragging on too long don't worry, but. "Think of the polarity of water. And the magnetic forces it would exhibit on a salt-crystal."  So I have O(delta -) bonded to two H(delta+) in water. And I have Na(delta +) bonded to Cl(delta -) in NaCl. But why would the magnetic force of  O- to Na+  outweigh the magnetic force of Na+ to Cl-?  I would think the Na+ and Cl- force would be stronger as Cl is more electronegative than O? Similarly, why would the forces of H+ and Cl- outweight the forces of Na+ and Cl-? Thanks.

Ah im sorry if i I just confused you. I'll admit that that was a bad example, because it is not a thing that happens in reality - hence why dissolution occurs. But, as a suggestion to you for all of your future as a chemistry student - do not look at a chemical solution as a simple one or two molecule interaction. Because the truth is, in solution, there is so much going on between molecules. Interactions between two specific molecules will occur for a very short time before they move on to interacting with another molecule. This results in a lot of electron-exchange. And again, disorder.

While a picture as in Bob’s answer is sometimes better than a thousand words, it may also help to add the thousand words[1] to explain what the picture is supposed to be showing or why it is relevant.
Water is a very polar solvent. The electronegativity difference between oxygen and hydrogen is rather high and the bent shape of the molecule additionally allows a higher dipole moment. Therefore, you can think of water molecules having a decidedly negative side — the oxygen atom — and a decidedly positive side — the space between the hydrogen atoms. In pure water, the molecules will arrange (at first approximation) in such a way that the positive sites are close to the negative ones and vice-versa.
At second approximation we need to introduce hydrogen bonds, which go one step further. Remember how each single hydrogen atom is strongly positively charged due to the electronegativity difference/the polar bond? Well, because hydrogen does not have any core electrons, this means it can act like a ‘slightly weakened proton’ and interact well with lone pairs on electronegative atoms such as oxygen. Thus, a better description of the intermolecular interactions in water molecules is that hydrogen bonds are formed from each hydrogen to one of another molecule’s oxygen lone pairs. These interactions are pretty strong for intermolecular interactions since they are caused by rather large charge differences (and need to stabilise these).
After having considered the water molecules by themselves, what happens if we submerge ions into them? In solid state, ionic compounds form extensive networks of cations/anions approximately alternating. Each cation will typically only have anions as closest neighbours and vice-versa; therefore each ion’s charge is stabilised by the charges of the opposite ions close by. If you wanted to dissolve this ionic compound into something like hexane, it would not work well. It is just much better for the ions to stick together in their solid-state form and be submerged in the hexane liquid phase — as your question implies.
In water, however, we have the positively charged and negatively charged areas not unlike those of the counterions. Taking a single ion out of the ionic crystal and surrounding it with the appropriate number and orientation of water molecules as shown in Bob’s answer, means that electrostatically almost nothing is gained or lost. True, the water molecules, not being charged particles, can stabilise a little bit less at first approximation but it is still comparable.
At second approximation, we can on the one hand reintroduce hydrogen bonds gonig from water molecules to anions such as chlorine. These may stabile the anions in solution, but note that having a hydrogen bond directed towards oxygen is typically better than one to a ‘typical halide’. The cation side is much more interesting, though: oxygen’s lone pairs can interact with the cationic centre to form a coordination complex. This lowers the energy of oxygen’s electrons at the expense of unoccupied cationic orbitals — a win-win situation for both partners. Thus, we can expect additional stabilisation due to this effect. (To be fully honest, one can and should also consider most ionic crystals as chained local coordination complexes. However, these are typically better stabilised if only one ligand sphere is involved, thereby preferring the solution.)
Thus we have:

A net average enthalpic no-effect concerning charge stabilisation (ionic crystal versus Bob’s picture
A slight net enthalpic gain due to the formation of better coordination complexes
A strong net entropic gain from the fact that a single, highly ordered crystal is broken up into gazillions of minute entities that can move freely in solution (I did not touch this point in this answer, but it is true for all solutions).

Considering all these, the conditions for ionic compounds to dissolve in aquaeous solutions are very favourable. And Bob’s your uncle.

[1]: Well all right. 645 words then. Not quite 1000 but close enough.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why does KMnO4 oxidize iodide to iodine in acidic medium, whereas iodide to iodate in alkaline medium?
$\ce{KMnO4}$ should be a better oxidizing agent in acidic medium, than in alkaline medium, as per my knowledge. However, it oxidizes iodide (oxidation state = -1) to iodine (0) in acidic medium, whereas iodide (-1) to iodate (+5) in alkaline medium. The latter has a higher change in oxidation number, which we would have instead expected for the former. Why is this so?

My book (NCERT India) mentions the same oxidation (iodide to iodate) for even neutral/faintly alkaline medium ($\ce{2MnO4- + H2O + I2 -> 2MnO2 + 2OH- + IO3-}$), which should be even weaker than the alkaline $\ce{KMnO4}$. It'd be interesting if an answer elaborated upon this as well.

@GaurangTandon But alkaline $\ce{KMnO4}$ gets reduced to $\ce{MnO4^2-}$, which seems to imply that alkaline $\ce{KMnO4}$ is weaker than neutral/faintly alkaline medium $\ce{KMnO4}$... so if the answer explains it for alkaline medium, I think the same should apply for neutral/faintly alkaline medium as well.

A simple way to explain this is to illustrate with the following Latimer diagrams:

If you calculate the standard electrode potential of the highlighted portions it will be positive, so the reaction is spontaneous under standard conditions.
For acidic conditions, iodine is not converted to iodate because the electrode potential of the reaction would become negative, whereas for basic conditions it is possible.
Note that both the manganate and the iodide ions are both affected by acidic and basic conditions; thus it is important to consider their respective electrode potentials.
(Data was taken from Inorganic Chemistry, Shriver et al., Sixth Edition, Oxford University Press)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How to create 1-100 ppm ethanol gas?
I am trying to use TGS 2602 sensor of air contaminants. I would like to test if my circuit is working as intended. For that purpose I need to generate fumes of one of the gases the can be detected by the sensor.

I thought that I would go for ethanol. I tried holding vodka soaked cotton ball near the sensor - to no avail. Later on I tried spilling vodka around the circuit and placing more vodka soaked cotton balls around. I also covered the whole thing with a cardboard box (with gaps at the bottom between table and the box). Again there was no considerable effect. Vodka is essentially ~40% ethanol and water mix (by volume).
Can you say by that short description that I managed to generate between 1 and 100 ppm ethanol gas? If not, can you propose a robust and safe method of producing such a gas?
Please be aware how the sensor looks. It is enclosed in a metallic case with small holes on the top surface.

Are you certain that the wiring and power supply 5 V DC for sensing and for heating are correct? You might want to look at the datasheet.

@Klaus Warzecha - that is what I am trying to establish. I am using a somewhat more complicated circuit, because I am using digital rather than analog reader. Anyway when I replace the sensor with resistors (~57 ohm for the heater and ~10k or ~100k ohm for the sensor itself) the circuit seems to work - nevertheless I am not 100% sure that it is functioning as intended, therefore I would like to create a testing environment in which I should have prior knowledge that sensor should be sensing something if it was working correctly.

How to do it, and how to do it cheap are two very different problems.

If the 40% alcohol solution came to vapor-liquid equilibrium at room temperature , then the vapor concentration of ethanol should have been on the order of 2%.

Waving a vodka soaked cotton ball near the sensor is sufficient to generate the required fumes. It was the circuit that was problematic - I did a subsequent testing with analog reader (voltmeter to be specific).


text:
How is mercury produced in industry?
Mercury is the only metal that exists in a liquid state at room temperature.

In which form is mercury extracted from its ore?
If it produced as a solid, how and when is it turned into a liquid?

There are many solid compounds that include mercury as one of their constituents. But, pure mercury at room temperature is a liquid as you point out.

tell me ik at room temperature it is liquid tell me in which state we get it from its ore.... that's my question

The classic way is to roast the ore, with the mercury coming off as a vapor that is readily condensed. Note that vaporized mercury is easily taken up by humans and damages the central nervous system.

I'll point out that most Hg now recovered in the US is as a byproduct to prevent pollution rather than being mined for the Hg itself.  https://archive.epa.gov/mercury/archive/web/pdf/byproductmercuryproductioninmodernpreciousmetalsminesinnevada.pdf

Most commonly, mercury $\ce{Hg}$ is separated from its ore $\ce{HgS}$, known as cinnabar. The process used is roasting; this entails heating the raw material in oxygen.
$$\ce{HgS(l) + O2(g) ->[700-800\ ^\circ \mathrm{C}] Hg(g)  + SO2(g)}$$
Mercury is extracted here in the gaseous state. Note that the product is elemental mercury, not an oxide. This is because the oxides decompose at or below $500\ ^\circ \mathrm{C}$: $$\ce{Hg2O -> HgO + Hg,\\ HgO -> Hg + O2}.$$
The mercury vapor is then purified in multiple steps which depend on the required degree of purity. Usually the gaseous mercury is first cleansed from dust, then condensed (bp $356.6\ ^\circ \mathrm{C}$). At this stage, an $80\%$ yield is observed.
The condensed mass is filtered further, washed with bases and acids, and destilled. For ultrapure mercury (less than $10^{-6}\%$ of impurities), electrochemical refinement using mercury electrodes follows.

Hergi Karik, Kalle Truus. ($2003$). Elementide keemia. (pp 331$-$332, 334)

it means cinnabar exist in liquid state so how it is extracted from earth

@Utsavpandey No, cinnabar is a solid ore at standard conditions. The states are written in accordance with the reaction; the sulfide simply melts before $700-800\ ^\circ \mathrm{C}$.

Mercury is extracted from its ore, cinnabar($\ce{HgS}$). According to wikipedia:

Mercury is extracted by heating cinnabar in a current of air and
  condensing the vapor. The equation for this extraction is
$$\ce{HgS + O2 → Hg + SO2}$$
In 2005, China was the top producer of mercury with almost two-thirds
  global share followed by Kyrgyzstan. Several other countries are
  believed to have unrecorded production of mercury from
  copper electrowinning processes and by recovery from effluents.


text:
When does ratiometric measurement account for signal variations?
I've been reading about pH and ion sensors that rely on a ratiometric approach, which supposedly circumvents many problems of intensity-based methods such as dye bleaching, dye leaching, and fluctuations in source intensity. I can think of situations though when this might not be the case.
For example, fura-2 is excited at two independent wavelengths, and the two emissions are detected sequentially. If the excitation source intensity at either wavelength were to fluctuate, the ratio of the emissions would be off from the calibration. Another example would be a FRET sensor in which the acceptor dye leaches at a faster rate than the donor dye; the emission ratio of acceptor to donor would no longer match the calibration.
So when does ratiometric measurement account for signal variations like dye bleaching, dye leaching, and fluctuations in source intensity? Is it only when there is one dye and one excitation source? Am I missing any other requisites? It seems like if there's only a single dye that bleaches/leaches or a single excitation source that fluctuates, the change in dye fluorescence would remain proportional such that the emission ratio remains the same.
Thank you!

So when does ratiometric measurement account for signal variations like dye bleaching, dye leaching, and fluctuations in source intensity? Is it only when there is one dye and one excitation source? Am I missing any other requisites? It seems like if there's only a single dye that bleaches/leaches or a single excitation source that fluctuates, the change in dye fluorescence would remain proportional such that the emission ratio remains the same.

You are correct. Ratiometric readout of optical sensors works best if you use one excitation source and one probe molecule that gives you the two signals to be used.

Another example would be a FRET sensor in which the acceptor dye leaches at a faster rate than the donor dye; the emission ratio of acceptor to donor would no longer match the calibration.

This happend to me before, where donor and acepptor where actually tethered to the same molecule and the acceptor bleached faster then the donor, which also screwed up the measurment.

National Instruments$^{TM}$ gives a good explanation here, in a paper titled "Measuring Bridge-Based Sensors with the Ratiometric Approach, Publish Date: Feb 11, 2016".  Here are a couple excerpts relevant to your question:

1) "The fact that the mV/V ratio stays constant for changes in the excitation voltage increases measurement stability. Small changes in the environment due to temperature differences typically affect excitation voltages. Because the excitation voltage is sensed and fed back to the ADC, any changes play a smaller role in the accuracy and stability of the measurement."
2) "By using the excitation as the source for the ADC reference, the output of the ADC provides measurements that are in units of V/V. This reduces the amount of scaling required to return the appropriate units."

So, the short of it is that noise, particularly that arising from power fluctuations, in either the source or detector in the measurement systems as you described is largely cancelled out when ratioed to a reference system.
This approach is most effective when you are rapidly switching, i.e. via a chopper mirror, between the sample measurement source and detector and that of your reference or blank measurement.  When the signal from the source is then ratioed to that of the reference system, then any noise common to the two is effectively cancelled out, increasing your signal to noise ratio.  This method addresses your examples of "sequential" measurements. In the system describe above, you are effectively taking your reference measurement simultaneously or in parallel rather than sequentially.


text:
What are protecting groups in organic chemistry?
My textbook does not give a very good explanation of what protecting groups are, how they are made, or how they function in Grignard reactions and others. It just gives a couple of examples of protecting groups and that's all.
So my questions are:

How are compounds chosen to be protecting groups? What are the criteria?
I noticed that protecting groups tends to be large molecules, is that always the case?

Wikipedia: https://en.wikipedia.org/wiki/Protecting_group
List of protecting groups and their stability in various conditions: http://www.organic-chemistry.org/protectivegroups/
Please make your own research first.

Functional groups become protecting groups out of practicality.
I would define an ideal protecting group (PG) as a functional group that is

easy to introduce/attach to a given, other functional group (the protection target)
stable/non-reactive, i.e. protecting, under a wide range of reaction conditions
easy to remove by well-defined and mild reaction conditions
with high yields throughout

Clayden/Greeves/Warren/Wothers have referred to PGs as functional groups with an Achilles' Heel.[1] Obviously, the usefulness of a given PG is always dependent on the target molecule and the reactions it will undergo. The "easy" part of the definition also encompasses factors such as price and toxicity.
As a corollary, one can introduce the concept of orthogonal PGs. Since the removal conditions can be very different for different PGs, one can successively attach PGs to, say, the hydroxyl groups of a sugar. By later removing them one by one, a chemist is then able to target them one at a time.[2]
Why do protecting groups tend to be large?
I am not convinced this is entirely the case, because I have seen applications of the methyl group as a PG. However, a larger group can introduce more steric hindrance or a point of attack for the removal that is different from the point of attachment to the protection target.

[1] Clayden, Greeves, Warren, Wothers: Organic Chemistry. Oxford University Press, 2004 (page 635).
[2] Werz, Seeberger, Angew. Chem. Int. Ed., 44, 6135, (2005). DOI:10.1002/anie.200502615.

The term protecting group is not absolutely strictly defined. In general, anything can be a protecting group if it fulfills the following conditions:

it is attached somewhat easily and adequatly selectively to certain specific sites
it is inert towards a sequence of following reactions which would affect the unprotected site; and
it is somewhat easily and adequately selectively removed from those sites to reliberate the original functional groups.

However, the term is sometimes used in an even broader sense meaning when ignoring the first bullet point.
Which functional groups you want to protect and which protecting groups you wish to use for the job depends entirely on your desired sequence. In natural product synthesis, it is very common to transform alcohols into silyl ethers — I would want to argue that this is the most common usage of the term. Alcohols have the disadvantages that they are nucleophilic, weakly acidic, can coordinate metal centres and can even act as bases. If you transform the alcohol into a silyl ether, it is no longer acidic, its ligand properties are substantially impeded and it can no longer attack nucleophilicly. Thus, it is possible to perform reactions, e.g. titanium(IV) catalysed aldol additions, which would not be possible with free alcohols.
Sometimes it is also desireable to introduce different protecting groups to similar sites or to selectively cleave only some protecting groups. For example, you may have two primary alcohols and want to turn only one of them into an aldehyde; by clever choice of your reaction sequence and protecting group strategy you will arrive at a point where only one protecting group is easily removed to give only one of the two free alcohols selectively.
While silyl groups — among them the relatively bulky tert-butyldimethylsilyl (TBS), tert-butyldiphenylsilyl (TBDPS) and tetraisopropylsilyl (TIPS) groups are exceedingly common, not all protecting groups need be that large. Methoxymethyl (MOM) is a rather small protecting group that enables the alcohol (then acetalic oxygen) to keep its ligand properties. Even smaller is the methyl group used to protect carboxyl functions or, arguably the smallest, the chlorine atom, used by Woodward in his synthesis of 6-demethyl-6-deoxytetracycline.[1]

Reference:
[1]: J. J. Korst, J. D. Johnston, K. Butler, E. J. Bianco, L. H. Conover, R. B. Woodward, J. Am. Chem. Soc. 1968, 90, 439. DOI: 10.1021/ja01004a041.

A protecting group could, technically, be any functional group. In organic synthesis, if you have a molecule with two or more sites that are reactive to a particular species, but you want to add the new functional group to only one of those sites, then you would add a different functional group to the non-target sites to "block" which ever reaction you want to perform.


text:
Why m-hydroxybenzoic acid has higher acidity then p-hydroxybenzoic acid?
Why  m-hydroxybenzoic acid has higher acidity than p-hydroxybenzoic acid?
I have a reasoning that since hydroxy group is an activating group, then it increases the electron density on its para position. So for  p-hydroxybenzoic acid, the carbonyl carbon has a greater ability to share molecular orbital's electron cloud to its attached hydroxy group. So the hydroxy group doesn't release $\ce{H+}$ easily. But that's not the case for m-hydroxybenzoic acid.
Is it right?

The -$\ce{OH}$ group has two effects, namely +R and -I effect i.e. Resonance and Inductive effect. When -$\ce{OH}$ group is attached to the para position, it has a tendency to delocalize it's electrons towards the aromatic ring (i.e., benzene). Thus the electron density on oxygen atoms of carboxylic group increases and the polarity between $\ce{O}$ and $\ce{H}$ decreases leading to less acidity. Whereas when the -$\ce{OH}$ group is attached to the meta position, +R effect is not applicable and only -I effect is effective. Thus it withdraws the electrons from the ring, and increasing the polarity between $\ce{O}$ and $\ce{H}$ in carboxylic group, making it more acidic.

That's what I said+ something


text:
Can a small amount of scandium be added to gold, platinum or palladium to create a wedding ring?
The number 21 has a special meaning to us, so we would like to make wedding rings with some scandium in it (element 21). I know scandium can be alloyed with aluminium or titanium, but can it also be alloyed with say gold, platinum or palladium without detrimental effects?

Sc would seem to alloy with Au. http://www.sciencedirect.com/science/article/pii/S0925838897000224 Where to get the Sc and how to get the alloy made, then the alloy into ring(s) are different problems.

A careful elementar analysis will probably find the odd scandium atom in a plain wedding ring. Perhaps the printout of the spectrograph will be enough? Will likely be cheaper than finding s.o. who makes you the alloy and forges a ring later.

Coincidentally you now have 21 rep. Congratulations.

Planning to get married?

The go-to place for alloy information is ASM International's Alloy Phase Diagram Database (providing your institution has access).
Scandium-Gold: as noted in the comment by @MaxW, you can get several atomic percent (but less than 10 at.%) Sc in Au. There exist 6 compounds across the diagram which you would want to avoid most likely.
Scandium-Platinum: likely up to about 10 at.% of Sc in Pt. At least several compounds to avoid.
Scandium-Palladium: also order of 10 at.% of Sc. 5 compounds to avoid.
All the compounds should be avoided because (1) they are higher in scandium content, and (2) they are mostly line compounds and unlikely to be malleable.
So, yes, you can get scandium in to a gold-platinum-palladium mix. Find a good jeweler and they can melt it in for you. Scandium is not all that expensive (at least compared to the cost of getting the jeweler to do a custom alloy).


text:
What criteria based on expressing standard rate of reaction?
When given the reaction: $\ce{NO2}$ reacts with $\ce{O2}$ to produce $\ce{N2O5}$, how do I calculate the rate of consumption of $\ce{NO2}$ when the rate of the reaction= 0.02 M/s?
I am confused  about which of these  expressions of reaction rate  can be used to calculate the rate of consumption of $\ce{NO2}$.
a )   r = 0.02 = (1/4) ×  rate of consumption of $\ce{NO2}$
Or :
b)    r = 0.02 = (1/2) ×  rate of consumption of $\ce{NO2}$
Does the rate of the reaction changed when we multiply balanced reaction equation by integer? Why?

I think a good starting point would be to write out the balanced equation for NO2 plus O2 to N2O5.

@airhuff :what happens to the rate of reaction  on doubling the stoichiometry of the equation at constant concentrations?

To your first comment: your equation has 4 N's on the left and 8 on the right; not balanced.  While there are an infinite number of balanced equations, what you want is the lowest factors.  Hint: 4NO2 is a good start but the number of O2 and N2O5 molecules are incorrect. This is necessary to know to figure out how much NO2 is consumed per N2O5 produced.

@airhuff :thank you : We can start with many multiplication of the balanced equation like this :8NO2 +2 O2 -----> 4N2O5 , how to calculate the rate of consumption of NO2?Does it change when  doubling the stoichiometry of the equation?


text:
Job's Plot or Method Of Continuous Variations
I was wondering why in order to find the stoichiometry of a complex using Job's Plot, the two compounds must have the exact same molarity. What kind of effect would it have if molarities weren't the same?
Thanks in advance!

If A and B were reactants, then a peak at a mole fraction of 0.5 A would indicate 1:1 complex, a peak at 0.25 A would indicate a $\text{AB}_3$ complex.

When using for example NiSO4 and Na2EDTA the peak is observed at 0.5A so the complex is 1:1. But why do we have to use the same molarity of the two compounds? Does it have to do with absorption or obedience to Beer's Law?

Huh?!? in my example the mole fraction of A would be plotted from 0% A (100%B) to 100%A (0%B). So the plot doesn't "use the same molarity." The peak occurs at the mole fraction which gives the stoichiometry of the complex.

I agree with you as to where the peak occurs and that it gives us the stoichiometry. In order to be specific I'm going to use the same example of NiSO4 and EDTA 2- .After we apply Job's Plot the peak gives us the stoichiometry which is 1:1. The question is why the solutions of NiSO4 and Na2EDTA have the same concentration before we mix them.

?!? Maybe because it makes the math easy... Assume Ni is 0.1 molar and EDTA is 0.1 molar. Taking $x$ mls of Ni and diluting to 100 mls with EDTA yields a solution which has a mole fraction of $x$% Ni.

I was assuming this, too. It's not exactly a requirement, but it has to do with making things simpler and easier. Thank you again for your help!!


text:
Do Polar Bonds Affect Molecular Shape?

Repulsion between shared electron pairs shapes the molecule in such a
  way that enables the pairs to be as far apart as possible.
  Furthermore, the repulsion between shared and unshared pairs forces
  bonded pairs closer to one another.  These principles of the VSEPR
  theory and orbital hybridization, in addition to Van der Waals
  forces, explain the overall shape of a molecule.

In my textbook,  however, there was little mention to how polar bonds and the partial charges they cause affect the structure of a molecule.  I would expect that poles of the same charge would repel each other and have a profound affect on molecular shape.  However, many compounds make me second guess this assumption, including $\ce{H2O}$.
Do polar bonds at all affect molecular shape? Why or why not?  If so, to what extent?

If I understand correctly, you are saying that by this logic water should be linear like CO2 for example?  If this is the case, think about other forces at play here.  Drawing Lewis dot structures of water, and CO2 for my example of a linear molecule, might illustrate why H2O is not linear and CO2 is.  Let me know if I'm completely off base regarding your question and I'll just delete this ;)

Yes, of course they affect the shape. But only slightly so. More precisely, bond angles can change, however this does not change the general shape predicted by VSEPR.
For example, if two atoms (one with a positive pole, and one with a negative pole)  are connected by another atom, then the two atoms would make an angle that is smaller than the angle predicted by VSEPR.
And also, polar bonds are capable of intramolecular bonding. Such as in protein-folding.  So they definitely contribute to molecular shape.


text:
Optical activity of starch
In this German Wikipedia article about optical activity, I found the beautiful picture of starch granules:

Unfortunately, I have no idea what I am looking at. What are these crosses and why are they blue (i.e., same as the background color)?

You're looking at isogyres (so-called interference figures), which arise from light interference along different optical axes of your sample. They are commonly used to determine the crystal system of a sample when viewed using a petrographic microscope (or similar). See this Wikipedia page for relevant information. The blue color might arise from the use of a mica plate to examine the birefringence, or could be an artifact of the experiment (I'm a bit rushed now and haven't read you link in-depth).

@ToddMinehardt That already seems like a pretty good answer to me, would be great if you found the time to write it up properly :)

The figures you see are isogyres and arise in substances that are optically active, which means that polarized light is transmitted through the material in different ways depending on the orientation of the sample. These figures are called interference figures. I highly recommend that you peruse this excellent write-up on the optical properties of minerals, particularly the sections pertaining to uniaxial and biaxial crystals, birefringence, and how a petrographic microscope works.
From the Wikipedia entry on conoscopic interference pattern:

The figures are produced by optical interference when diverging light rays travel through an optically non-isotropic substance - that is, one in which the substance's refractive index varies in different directions within it.

Starch is uniaxial and the figures you see are characteristic of uniaxial materials when viewed using a petrographic microscope. See this reference (The Principles of Pharmacognocy. F. A. Flückiger and A. Tscirch. William Wood & Company, New York, 1887), pages 118-119.
As for the color seen in the image you reference: that arises from using a mica (or tint) plate which in turn indicates whether the substance is uniaxial positive or uniaxial negative, which refer to the propagation of light along differing optical axes. However, in your figure, the color of the isogyres and the background are both blue, which indicates to me that the blue color is simply a result of polarized light passing through the substance.


text:
pH notation in text
When referring to units of pH in continuous text (i.e. not in a table), which is the recommended usage?

pH 7 to pH 8
pH 7 to 8

What about when the term "pH" has already been mentioned in the text?

the pH was increased from 7 to 8
the pH was increased from pH 7 to 8
the pH was increased from pH 7 to pH 8
the pH value was increased from 7 to 8

Using pH twice in any of those sentences is redundant.

Additionally, would it be correct to refer to pH as a measure of acidity? Or would this be ambiguous with the acid dissociation constant?

If your readers are chemists then saying that "pH as a measure of acidity" is ludicrous. If you are writing an article for a newspaper about mine water discharge then through the phrase into the article.

In a dilute solution, is not pH an appropriate measure of acidity? Constant references to hydrogen ion activity become tedious.

$$\ce{<-->[He went that way!]}$$ The point is that chemists know pH is a measure of acidity. You don't have to explain it every time you use the term.

Sorry, I misunderstood what you were saying.

I was thinking more as in whether acidity cambiguously refer to either pH or $\mathrm{K_a}$.

E.g. in the ACS style guide, 3rd edition, rather only the pH symbol itself is focused on, with some examples of “half intervals” (or how to name that?) notation:

… Use roman type for pH and pOH; always use lowercase for “p”; always capitalize “H” and “OH”.
 Solutions were titrated to pH >11.
 The UV spectra were measured at pH 6.
 A pOH of <12 was acceptable.

I couldn't find any explicit recommendation for notation of pH ranges. In Measurement of pH. Definition, Standards, and Procedures (IUPAC Recommendations 2002) several “notations” are used in text, e.g.
at p. 2179 (11 of 32):

… between pH 3 and 10 …

p. 2184 (16):

… pH(S) values of … are in the range 3 ≤ pH(S) ≤ 11 …

p. 2199 (31)

… in the pH range 3−10 …

pH 7
The usage is inconsistent with pH being a dimensionless quantity. It would be logical to write "the pH is 7", or "the pH is in the range from 5 to 8". This would correspond to the symbolic notation:
$$\mathrm{pH} = 7$$
$$\mathrm{pH} = [5,8]\quad \text{ or }\quad 5 < \mathrm{pH} < 8$$
where [5,8] indicates a range.
Instead, usage like "at pH 5" are common and sanctioned by the ACS style guide. It's a bit like saying "I am looking for size 11 shoes" instead of "The size of my shoes is 11".
pH 7 to 8
When giving a range of pH values, there is no reason to repeat pH. pH is the quantity, and it is clear that you are giving a range for that quantity. This is different from giving a range of lengths, or a length with a standard deviation. In that case, both lengths, or the length and its standard deviation need to have units, e.g.
$$l_1 = \pu{2.3 cm}\ ..\ \pu{2.4 cm}$$
$$l_2 = \pu{21.7 cm} \pm \pu{0.3 cm}$$
Of course, you can leave out the duplicated unit by using parentheses:
$$l_1 = (\pu{2.3}\ ..\ \pu{2.4})\pu{cm}$$
$$l_2 = (21.7 \pm 0.3)\pu{cm}$$

Shouldn't it be $\mathrm{pH} \in [5,8]$ ?


text:
Why is chromium involved in oxidation of alcohol?
Like what is it about chromium atom that make them able to oxidize atoms? Like can’t any other transition metal like manganese, mercury among several others do the same thing? I always see example of chromium in oxidation of alcohol and i ws curious why? Thanx.

Manganese is also another notable element which has ions that can catalyse oxidation. And strictly speaking, it is the related ions that do the oxidation, not the atoms. As to why this happens, have a look at their half reactions and oxidation states.

Chromium used in the type of chemistry you're talking about is in a +6 oxidation state and can readily accept electrons. Manganese in the form of permanganate (+7 oxidation state) doesn't a great job of oxidizing alcohols though the oxidation is less controlled can leads to by-products that may include some cleavage of adjacent carbon-carbon bonds. Chromium just happens to balance reactivity with selectivity.

Chromium is hardly the only element that is used to oxidise alcohols. Alcohols are really easy to oxidise and apart from chromium, you can also use manganese ($\ce{MnO4-}$ or $\ce{MnO2}$), ruthenium (TPAP), iodine (DMP), sulfur (Swern oxidation and variants), even nitrogen works (TEMPO), and in the Oppenauer oxidation carbon is formally the oxidising agent. The list really does go on, this is just the tip of the iceberg! In all of these reagents the element I named is the one that undergoes reduction.
The only criterion is that, well, it must be an oxidising agent. So, for example, you would probably be hard-pressed to find a calcium-based oxidising agent, since $\ce{Ca^2+}$ is pretty happy being $\ce{Ca^2+}$.
The issue is: saying that "chromium" is involved is a little misleading. Chromium metal is not a very good oxidising agent; neither is Cr(II), and Cr(III) not quite either. You can dump all the $\ce{CrCl3}$ you like into an alcohol, and you probably won't get any ketone. ("Probably" because who knows, you might very well get some ketone if you leave it open to air, since molecular oxygen is a pretty good oxidising agent. There was a slightly infamous incident several years ago where a lab claimed to have discovered a NaH oxidation of an alcohol to a ketone; it turned out that they didn't exclude oxygen thoroughly and that was what was causing the oxidation.)
It's always chromium(VI) that is in the oxidising agents. $\ce{CrO3}$, PDC, PCC, ... all have chromium in the +6 oxidation state. That's strongly oxidising and that's why it sees use as an oxidant. However as said earlier - it's hardly the only element used.


text:
How do we measure the energies of molecular orbitals?
This question came on my mind after a sentence of a teacher who said during a lesson:

If you imagine hydrofluoric acid, to make the MO's diagram because the
  energy gap between the $\text{2s}$ of fluorine and the $\text{1s}$ of
  hydrogen is quite important you can make the hypothesis at first there
  is no interaction between both. But experiments show that the bonding
  MO's of $\ce{HF}$ has an energy a bit different than the energy of the
  $\text{2s}$ of fluorine.

At first it not supprised me that much, I knew it. But know there is one question.

Question's':
How is that possible to measure the energy of a specific MO? Is that also possible if there is no electron in the MO or if it is an anti-bonding MO?

We don't measure them directly (they cannot be as they are not physical ). The quote is most probably about interpreting xps/ups spectrum.

@Greg We were only speaking about complex shapes. It was just in the introduction.

Would you elaborate? What complex shapes are you talking about?

As I said it was in an introduction just to keep some things one mind there are new student at school which never learn about that before that's why teachers need sometime to do things like that. If you read my question, it is not about complex shapes, isn't it? so the answer musn't be about that.


text:
Densest hydrogen storage medium
What compounds or nanostructures contain the most hydrogen atoms per unit of volume in realistic conditions (under 1 GPa). Particularly boron-free compounds because it's to be used in sci-fi nuclear designs.

What I've investigated so far:

Zirconium Tetrahydride (0.229 mol/ml, unstable?)
Thorium tetrahydride?
Ammonium Borohydride (~0.176 mol/ml, unstable?)
Aluminum Borohydride (~0.175 mol/ml)
Borazane (0.152 mol/ml)
Aluminum Hydride (0.148 mol/ml)
Tetramethylammonium Borohydride (0.146 mol/ml)
Solid Ammonia (0.144 mol/ml)
Solid Hydrazine (0.143 mol/ml)
Uranium Hydride (0.136 mol/ml)
Paraffin Wax (0.131 mol/ml)
Liquid Cyclopropane (0.128 mol/ml)
Tetramethylammonium Chloride (0.128 mol/ml)
Beryllium Borohydride (0.125 mol/ml)
Lithium Borohydride (0.122 mol/ml)
Liquid Ammonia (0.120 mol/ml)
Beryllium hydride (0.118 mol/ml)
Sodium Borohydride (0.113 mol/ml)
Magnesium hydride (0.112 mol/ml)
Water (0.111 mol/ml)
Cyclohexane (0.111 mol/ml)
Neopentyl alcohol (0.111 mol/ml)
liquid ethane (0.109 mol/ml)
ammonium fluoride (0.109 mol/ml)
Hexane (0.106 mol/ml)
Neopentane (0.104 mol/ml)
liquid methane (0.105 mol/ml)
ethanol (0.103 mol/ml)
liquid cubane (0.099 mol/ml)
lithium hydride (0.098 mol/ml)
lithium aluminum hydride (0.097 mol/ml)
liquid propane (0.091 mol/ml)
Solid hydrogen (0.087 mol/ml)
Liquid hydrogen (0.070 mol/ml)

Just my two cents.
I guess you already looked these compounds up, but in "Sustainable Energy Systems and Applications", 2011, is written:
"The ammonia-boron compound with chemical formula NH4BH4 appears to be the most “dense” hydrogen storage. "

Zirconium hydride (e.g. $\ce{ZrH}$ or $\ce{ZrH_{1.65}}$, but not $\ce{ZrH4}$) is actually used in some nuclear reactors because of its unusual moderating properties.

As an aside, if you're looking for non-chemisorbed (i.e. the $\ce{H/H2}$ is not bonded to the material), Metal-Organic Frameworks (MOFs) have some of the highest uptakes of the gas -- achieving near-liquid density. NOTT-112 gets ~0.058mol/mL at 77K/60atm if you count the H in the MOF. Obviously not even comparable to bonded H.

Interesting article about storage in palladium http://www.nature.com/nmat/journal/v13/n8/full/nmat4030.html?foxtrotcallback=true

Since it's sf, maybe you can imagine someone figured out a way to keep muons from decaying, in which case muonic H atoms have a radius 200x smaller than regular H atoms, so muonic H2 ought to be about 10^7 times denser than ordinary H2.

Er...but I believe this brings the nuclei so close that there is a reasonable probability of fusion under ordinary conditions, because the one nucleus can tunnel into the other, which could quite mess up the integrity of your storage vessel.

indeed https://en.wikipedia.org/wiki/Muon-catalyzed_fusion


text:
Is cyclopropyl methyl carbanion also exceptionally stable?
I know that cyclopropyl methyl carbocation is exceptionally stable and in fact more stable than benzyl carbocation but I was wondering if it was the same case with cyclopropyl methyl carbanion. Is it also more stable than benzyl carbanion?

Sigma resonance only works on cyclopropyl methyl carbocation (and its cyclopropyl derivatives). So I don't think that cyclopropyl methyl carbanion will be exceptionally stable.

Could you explain why sigma resonance only applies for cyclopropyl methyl carbocation.


text:
Where can I find an extensive list of the preferred chemical nomenclature?
I'm writing text to speech software, and as it stands, I already have a method for reading out mathematical formulas accurately, which I have been using with chemicals as well. Thus, something like:
H2O would be read more or less as "Ayche, sub two, O".
and
CO2 would be read more or less as "Sea, O, sub two"
That is fine here and there, However in a lot of texts this becomes extremely tedious, for example:

In tissue, cellular respiration produces carbon dioxide as a waste product; as one of the primary roles of the cardiovascular system, most of this CO2 is rapidly removed from the tissues by its hydration to bicarbonate ion. The bicarbonate ion present in the blood plasma is transported to the lungs, where it is dehydrated back into CO2 and released during exhalation. These hydration and dehydration conversions of CO2 and H2CO3

https://en.wikipedia.org/wiki/Bicarbonate_buffer_system
In this paragraph alone, you would hear: "Sea, O, sub two" three times. While I am not a chemist, I would assume that most listeners would prefer to hear would prefer to listen to a nomenclature that actually gave a name to the formula in cases like these, thus instead of hearing "Sea, O, sub two" and  "Ayche, sub two, O". -- you would hear the preferred chemical nomenclature,

Carbon Dioxide

and

Water

or

Oxidane

or

Dihydrogen monoxide

Is there a big table or list or resource available of the preferred nomenclature? And how often does this nomenclature depend on context, in the sense where the nomenclature would change depending on whether this was medicine or astronomy (etc...)?

Well, that would be a really big table unless you just made your own with a couple dozen or so compounds that are commonly used in English conversation.  However, it looks like you want it to apply to scientific texts, which certainly complicates things.  There are large molecules, for example proteins or even amino acids which would be horrible to hear anything other than their common names.  For smaller molecules, I'd just say "loose the sub". i.e. "sea oh sub two" sounds very odd but "sea oh two" is perfectly acceptable in most scientific contexts, as well as in everyday conversation.

Out of curiosity, how would your software differ from other text-to-speech software that already exists?

@halcyon Depends what you are speaking of. I use TTS engines, but I do not develop them myself. What I do specifically is I interpret wikipedia pages into SSML, and distribute the audio. Compared to a TTS application where you basically highlight text and play it back, I get rid speech artefacts, tables which are not good for speech, fix abbreviations, reinterpret wiki functions to optimize readable speech, control speed, parse the article into different consumable parts, add audible signals to inform the user of things like "Citation Needed", or for [] and () brackets surrounding text...

@halcyon And a lot more. I'm just starting to sell my audio now. http://frogcast.org/  If you want some free articles to demo, let me know. I'd love some free feedback in exchange, especially from someone like you who could intelligently remark on how chemistry articles are being read back, and what you think could use improving.

@Akiva, I'd love to test your software on some demo articles, and would be happy to provide feedback. I've used a few iOS apps in the past, but the speech artifacts were a bit distracting. Is it just wikipedia pages you've worked on, or can you do also do journal articles?

@halcyon Wikipedia articles are what I focus on, because of its relatively high and consistent standardization, and because its license allows me to sell as derivative works (All my audio is hence licensed as Creative Commons w/ Attribution). However, in query to the idea of other articles, this is something I want to branch out into. The more standardized, the more success I can have. In any case; people will enjoy best articles they are interested in. Name a topic or article and I'll send a bunch over.

@Akiva, here's a Wikipedia article or recent Science paper -
https://en.wikipedia.org/wiki/Voltage-gated_ion_channel
http://science.sciencemag.org/content/355/6328/950.full

@halcyon Here you go: http://frogcast.org/shop/voltage-gated-ion-channel/ -- The price has been set to 0, so you can add to cart, checkout (No charge), and you will be given a download link. It is all creative commons, so feel free to do with it what you wish. Let me know if you want any longer articles or categories of articles.  Thanks again.

@Akiva, great! I'll get back to you within a week or two. How / where would you like feedback? I don't know if continuing this comment thread would be the best choice... ;)

@halcyon great. Reply here, and we can just delete our comments. Let me know if there are any other articles you want. Thanks again.

@Akiva, overall, I thought the reading was great. Better than the 1-2 I've used in the sense that it felt more natural and less jarring. More specific comments: (1) there's melodic noise that plays before reading the ion names - this is distracting.  (2) Cation is mispronounced as "cash-un" instead of "cat-eye-on". (3) Similarly, anion is mispronounced as "an-yun" instead of "an-eye-on". (4) in some places, a dash was read in as minus, such as with "S1-S4" which would be human-read as "S1 through S4."

If you want more test cases / comments, here are additional articles: https://en.wikipedia.org/wiki/Free_energy_perturbation, https://en.wikipedia.org/wiki/Bennett_acceptance_ratio, https://en.wikipedia.org/wiki/Thermodynamic_integration
(chosen to have some general science language such as the cation example).

@halcyon Ah those melodic noises would make sense in more regular articles. They represent ( and ), to help the reader know when something is being read in parentheses. Different noises also exist for [ and ]. I showed it to someone before, and they were able to figure out what it meant without me asking, however others have asked for a demo. Perhaps I can explain it at the beginning...

@halcyon And I will fix the Cation and Anion pronounciations. Stay tuned, I'll have those articles for you in a few minutes.

@halycon Ah, some latex! This will be fun!

@Akiva, I noticed that! They seem to be images in equations though, does your software handle that?

Let us continue this discussion in chat.

@halcyon Yeah, I spent a few months working on a latex to speech interpreter. Listening to it can be jarring because I ramped down the speed for visually impaired people to be given a chance to illustrate it in their mind while it is being read.

Here's a pretty extensive list on Wikipedia.

Dictionary of chemical formulas

I doubt this nomenclature would change much depending on context, given that there is a widely used nomenclature system (IUPAC) (see here).


text:
Does a strong acid's conjugate base act as Brønsted-Lowry base?
Why is the conjugate base of a strong acid not considered a Brønsted-Lowry base? Is it because the base is so weak that it essentially has no ability to accept a proton?
(Quote from my lecture notes: "$\ce{Br-}$ is the conjugate base of a strong acid ($\ce{HBr}$), so it will not act as a Brønsted-Lowry base").

In theory, a Brønsted-Lowry base is a chemical substance which can accept a proton.
But in the case of a strong acid, the acid will lose a proton and form a weak conjugate base. Say there is a strong acid HA.
$$\ce{HA + H2O <=> A- +H3O+}$$
So if we take a certain quantity of HA, most of it will undergo the forward reaction since HA is a strong acid. A- is a weak base and will not want to accept the proton and thus only a small quantity of the conjugate base will do undergo the backward reaction. Hence practically, you could say that the conjugate base of a strong acid is not a Brønsted-Lowry base because of its reluctancy to accept the proton.

In the case of halide ions (setting fluoride aside as its  conjugate acid is weak), they can act as Lewis bases.  They will not, of course, react with solvated protons in aqueous solution but will form complexes with Lewis-acidic transition metal ions.

All compounds with an accessable free lone pair can act as a Brønsted-Lowry base. That explicitly includes conjugate bases of strong acids.
The only question (and difference) remains how basic these bases are, i.e. what their $\mathrm{p}K_\mathrm{b}$ is. Since it is directly related to the $\mathrm{p}K_\mathrm{a}$ by the $\mathrm{p}K_\mathrm{ion}$ of the solvent ($14$ for water), a strong acid will give a weak conjugate base.
However, no matter how weak the base it just takes a strong enough acid to protonate it. Using superacids, even methane $\ce{CH4}$ — a compound that one really would not accuse of having any free lone pairs — could be protonated to give a carbonium ion which quickly decomposed. Its existence was confirmed by the decomposition products the formation of which required the existence of $\ce{CH5+}$.

From my understanding, the Brønsted-Lowry (BL) acid/base theory is in relation to the idea of conjugate acids and bases.
A conjugate base does not always occur when there is interaction between a Lewis acid and a Lewis base, but a conjugate base always occurs when there is a BL acid-base reaction. This is because BL acids-bases are defined by the exchange of protons (H+ ions).
When a BL acid donates a proton to a BL base, the result is a conjugate base (originally the acid) and a conjugate acid (originally the base). And, you must notice that this reaction is in equilibrium! Thus, the reverse reaction is possible. This means that the conjugate base will act like a BL base.
Now, it may be that the initial BL acid that was used was considered a very strong acid, so the conjugate base will be a very weak BL base. Thus, the equilibrium may be favoring the products (K>>>1). As a result, in a closed system, very little conjugate base will undergo the reverse reaction. Nevertheless, at least some should undergo the reverse reaction and it will then act as a BL base!


text:
What happens when you mix Pine Sol with Clorox
I have read all these rumors online that mixing these two solutions will produce the chloride gas. I accidentally mixed these two while doing berthing cleaners in the communal restroom and felt very light headed afterwards.
Pine-sol contains glycolic acid (1.75%) and the other ingredients are C10-12 alcohol ethoxylates, dimethicone/silica/PEG distearates, sodium C14-17 secondary alkyl sulfonates.
My Clorox contains 8.75% sodium hypchlorite.
Are the culprits in producing chloride gas glycolic acid and sodium hypochlorite? I'm not too familiar with chemistry and I need someone who took college organic chemistry to explain this to me. I'm in the military so I can't take physical classes. I would also like to know the mechanism.

There is no reason to believe that there would be a chemical reaction between any of the components of the Pine Sol with sodium hypochlorite that would produce noxious compounds.
The most likely culprit in this case is using Pine Sol without adequate ventilation. As you've indicated, it is loaded with volatile and semi-volatile organics, which could cause the type of central nervous system (CNS) symptoms you described.  This is the reason for the warning on the bottle to only use in well ventilated areas.
Chlorine gas on the other hand attacks the eyes, throat and lungs.  You would not feel CNS effects from chlorine without having noticed these things first.  Also, though mixing anything with bleach is a bad idea, Pine Sol contains only a pretty small amount of a weak acid, which results in the release of a bit more chlorine gas from the sodium hypochlorite.
Please don't heasitate to ask for clarifications in the comments below. I hope I understood the question correctly.


text:
How does PCC not oxidize aldehyde?
I am wondering how does PCC actually work in not oxidizing aldehyde. Like the reaction mechanism I saw online, it seemed as if not having a acidic solvent is what made aldehyde not tranform to carboxylic acid when PCC is present. So the why can't you just use chromium containing compounds without the acidic solvent and then you will be able to isolate aldehyde.

The crux lies in the half-equation for the reduction of PCC. It must have a source of $\ce{H+}$ ions (or  $\ce{OH-}$ ions, for basic conditions) for any reaction to take place in the first place.

The "trick" is in the enviroment, not in the PCC itself.
Alcohols tend to be oxidized to carboxylic acids when they are in aqueous enviroment.
PCC was invented as a workaround to this problem: it works as a reactant in anhydrous enviroment, hence stopping the reaction at the aldheyde stage.
An alternative way to see PCC is to follow its preparation: if you add $\ce{CrO3, HCl}$ and pyridine, PCC is formed.
If you add $\ce{CrO3}$, acid and water, chromic acid is formed.
So, you see that the reagents are similar: the point is in finding at which stage the difference is.
Mechanistically speaking, the difference occurs as follows: in aqueous enviroment, water can form a geminal diol intermediate by attacking the aldheyde. This step adds the second oxygen to the aldheyde that, with further reaction with chromic acid, yelds a carboxylic acid.
When water is not present, this step doesn't occur, and the reaction stops.

Anhydrous PCC oxidations can also form esters as by-products with primary alcohols. Unreacted alcohol forms a hemiacetal with the aldehyde previously produced. As far as PCC is concerned, the hemiacetal is a secondary alcohol. It gets oxidized to an ester. Use dilute suspensions of PCC in dichloromethane; use excess reagent; add alcohol slowly to the oxidant. These steps minimize ester formation.

Good answer. Here's a useful page for anyone who wants more on how this works.


text:
Lattice enthalpy of alkaline earth metals
How lattice enthalpies of alkaline earth metal carbonates is affected on moving down the group?

You could have just opened jd lee or wikipedia

Lattice enthalpies are defined as the change in energy when one mole of crystal lattice is turned into infinite separated gaseous ions. When salts are dissolved in water, we assume the above definition of lattice enthalpy. Lattice enthalpies are affected by two main factors:

Charge (larger charge = stronger enthalpy)
Size of ion (smaller ion = stronger enthalpy)

Now, because you are trying to bond M (alkali earth metals/group 2) to carbonates, then the charge is going to stay constant. This is because carbonates (CO3) will always have a charge of -2 and M will always have a charge of +2. So, we are going to forget about charges for your specific situation.
If you look downwards on a periodic table in G2, the ionic radius of alkali earth metals increases. The smallest, and thus the one that will result in the strongest lattice enthalpy, is Beryllium. The smaller ionic radii allow for tighter packing, which leads to greater lattice enthalpy values.


text:
Does bleach react with disinfectants containing quaternary ammonium salts?
My mom went nuts when we pulled up the laminate to discover mold growing on the subfloor. She sprayed bleach all over it while I was at the store buying Mold Armor mold remover and disinfectant, and found out from calling poison control that this product should not be mixed with bleach. The operator was not sure what I could do to make it so it was safe to apply the mold remover.
Can anyone tell me if when bleach dries, the chemical reaction will be mitigated? I just want to clean my house without killing everyone. The surface is porous, and made of wood/particle board.

"I just want to clean my house without killing everyone"... Don't we all :)

Calling poison control first was the best thing you could do! Thumbs up!
Please note that we're usually reluctant to give medical advice - we don't want to get you or anyone in your house get killed or harmed either.
You might want to call poison control again and/or send them the safety data sheet of the product.
The following is just additional information to give you some direction on why it is probably safe to use the mold remover on the floor previously treated with bleach.
The College of Veterinary Medicine at Cornell University has published Guidelines for the Use of Bleach. According to this document,

Bleach can safely be mixed with quaternary ammonium compounds that contain detergents, such as A-33 and Roccal to create a single solution with both cleaning and disinfecting properties.

So, I'd dare to say that the combined use, as well as subsequent treatment of the floor is safe, but if I were you, I would search again confirmation with official poison control rather than trusting some guy from the internet ;-)


text:
Percent error and significant figures?
For an elementary chemistry class percent error is calculated as:
 [actual-experimental]/actual x 100%  edit:bracket represents absolute value
So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?
EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then,
0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs. -> there are no significant figures so this would mean the calculated 0.0171% error would be zero correct.  If I remember my analytical correctly we will do away with sig. figs. and work with the propagation of error within each measurement but I have not thought about how this would work in a problem like this.
Thank you in advance.

Actually the % error is calculated as:
$\frac{\text{experimental} - \text{actual}}{\text{actual}}\cdot100\%$
That way

if experimental < actual then the error is negative
if experimental > actual then the error is positive

So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?

Not exactly zero...

EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then, 0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs

Correct for subtraction you have used. Addition and subtraction round to the the last common significant decimal place of all the measurements. That would be the 8 in 10.98 in the following example.
So $10.9\bar8 + 0.6754 + 0.4795 = 12.1349 \approx 12.13 $
In actually you should have used the reverse 0.997 - 0.99717 = -0.000171

there are no significant figures so this would mean the calculated 0.0171% error would be zero correct.

Now this gets sticky. You could write that the error was 0% but that would indicate only one part in 100 precision. If you write 0.00% it isn't entirely clear if that is two or three significant figures. But you have claimed that the hundredth place in the % is significant, which it is not.
Remember that the minimum nonzero error possible was +/- 0.001 which yields an error of 0.1%.
So the error would be best written as 0.0% indicating that the error is 0 to 1 part in a thousand.

Reference: Significant Figures at Wikipedia
I'll leave the propagation of errors for a different exercise.

I believe that you are correct in saying that the percent error is zero in this case, which is what I tell my gen chem students who do not know how to statistically evaluate uncertainty.  As the previous answer-er pointed out, "when subtracting,  Addition and subtraction round to the the last common significant decimal place of all the measurements."  The last common sig DP between 0.997 and 0.99717 is the third decimal place, so the result of your subtraction, when rounded to the correct number of sig figs is zero.
Here is the more important part and the rational behind the sig fig rules...
0.997 g/mL implies a precision of +/- 0.001g/ml.  The accepted value of 0.997171 g/mL falls within this range.  So, due to the limitations of the method used to determine the density, the experimental value and accepted value are not distinguishable from each other and that a percent error cannot be determined unless the experiment is redone in a way which increases precision.
FYI, my experience is the most Gen Chem teachers let sig fig errors slide with percent error since this concept can be difficult to understand.


text:
Why is cyclopropane more acidic than propane
Though I found links about ring strain and other things, I actually fail to understand the exact reason for this difference in acidity.

If you think of both compounds as bases and find their conjugate acids, then apply hyperconjugation to find the stronger conjugate acid, you would find the stronger base(which will be the weaker acid).

Indeed, according to the Evans $\mathrm{p}K_\mathrm{a}$ table the cyclopropane $\ce{C-H}$ bond $(\mathrm{p}K_\mathrm{a} \approx 46)$ is more acidic than the $\ce{C-H}$ bond on the central carbon in propane $(\mathrm{p}K_\mathrm{a} \approx 51)$.
We know that, in general, the acidity of $\ce{C-H}$ bonds follows the order $\mathrm{sp \gt sp^2 \gt sp^3}$.  This is because the more $\mathrm{s}$-character in an orbital, the more stable (lower energy) electrons are in the orbital.
The $\ce{C-H}$ bonds around the central carbon in propane are close to being $\mathrm{sp^3}$ hybridized, whereas the $\ce{C-H}$ bonds in cyclopropane are roughly $\mathrm{sp^{2.46}}$ (see this earlier answer to see how the hybridization in cyclopropane was determined).
Now, when we compare the hybridization of the $\ce{C-H}$ bonds in cyclopropane and propane we see that there is more s character in the cyclopropane $\ce{C-H}$ bonds, this explains the increased acidity.


text:
How viable is production of magnesium oxide and sulfuric acid from Epsom salt?
Epsom salt $(\ce{MgSO4})$ is really cheap (less than 0.5 €/kg), whereas $98\,\%$ or better grade magnesia $(\ce{MgO})$ is much more expensive (3 to 4 €/kg).
Is it possible to convert magnesium sulfate $(\ce{MgSO4})$ back to magnesium oxide $(\ce{MgO})$ and sulfuric acid $(\ce{H2SO4})?$ How difficult and economically viable would that be given the necessity of the efficient separation of the products $(\ce{MgO}$ and $\ce{H2SO4})$ from each other?

Yes, at 1124 °C magnesium sulfate decomposes to magnesium oxide without melting producing sulfur trioxide as the side product. [1]
Magnesium sulfate can also be reduced to magnesium oxide using a reducing agent like hydrogen or carbon monoxide. [2]
References

Scheidema, M. N.; Taskinen, P. Decomposition Thermodynamics of Magnesium Sulfate. Ind. Eng. Chem. Res. 2011, 50 (16), 9550–9556. DOI: 10.1021/ie102554f.
Plewa, J.; Steindor, J. Kinetics of Reduction of Magnesium Sulfate by Carbon Oxide. J. Therm. Anal. 1987, 32 (6), 1809–1820. DOI: 10.1007/BF01913973.

Thanks for your answer. In example 2, is it sufficient for me to add as a Carbon source the fuel I use to burn the Magnesium Sulfate? Say for example I use a propane fire to burn the Magnesium Sulfate, is the added Carbon from the propane enough? Or for that matter any other hydro-carbon fuel?

You have to use pure carbon


text:
Are all reactions with ΔG < 0 spontaneous?
Given the formula $\Delta G = \Delta H - T\Delta S$. It is commonly written that all spontaneous reactions necessarily have $\Delta G < 0$. I was curious if the converse is true: Are all reactions with $\Delta G < 0$ spontaneous. If not, why?

Yes they are spontaneous from a thermodynamic point of view. Kinetics is another matter...

Will a reaction with ΔG < 0, involving thermodynamically unstable, but kinetically stable reactants be considered spontaneous, then?

yes spontaneous. Thermodynamics does not involve time at all, just energy changes and heat will be given out in such a reaction. With a catalyst this may be  fast if not can be incredibly slow, eons.  Spontaneous should not be confused with,  as it were, 'it will happen as we watch it'. Kinetically slow, thermodynamically spontaneous is very common, thankfully; its the state we all exists in in an oxygen atmosphere :)

@porphyrin That comment is just good as an answer. Do you please mind writing one? ;) Thanks!

No, Gibbs energy is only the relevant state function in the special case of constant temperature and pressure.
In the wording of IUPAC:

partial derivative of Gibbs energy with respect to extent of reaction at constant pressure and temperature [is negative] for spontaneous reactions

In the special case of constant temperature and volume, Helmholtz energy is the relevant state function.
Most generally, a process is spontaneous if the entropy of the universe (system + surroundings) increases.


text:
Water with Sodium VS Alcohols with Sodium
Why is the reaction between water and sodium more vigorous than that between ethanol and sodium?
Also, why does the reaction between sodium and an alcohol becomes less vigorous the longer the hydrocarbon chain?

Related: http://chemistry.stackexchange.com/questions/42535

The dominant issue explaining the less vigorous reaction of long-chain (i.e. greater than about 3 carbons) alcohols with metallic sodium is simply a function of the molecular-mass of the alcohol. I will address this aspect last.
At the lower end of the molecular-mass for this series (i.e. water, methanol, ethanol, isopropanol, etc.) the destabilization of the hydroxyl intermediates by the electron-rich hydrocarbon groups (of course absent in water) also slows the progress of the reaction to some degree.
The reaction of sodium metal with water/alcohols is:
1)$$\ce{2Na(s) + 2ROH(l) -> 2NaOR v + H2 ^}$$
where R=H for water, and is the hydrocarbon chain for alcohols.
The inductive destabilization effect mentioned above also largely controls the relative acidities of the alcohols, which in turn are directly related to the reactivity of the alcohol with sodium.  The relative acidities of pure solutions of water and some selected $\ce{C1}$ to $\ce{C3}$ alcohols goes as:
$$\text{water > methanol > ethanol > isopropanol}$$
In the discussion above, we have considered the simple case in which the reactant of interest is also the solvent.  Any comparison of relative concentrations of the different reactants of course requires that we prepare some concentration of the water or alcohol in an appropriate solvent.  It is also desirable to dilute the water or alcohol because of the vigorous, highly exothermic nature of their reactions with sodium.
Because water is the most reactive of the compounds of interest, any sodium added to an aqueous solution of an alcohol would just preferentially react with the water solvent.  And of course you could not measure different concentrations of water in itself!  The solvent of choice in this case is DMSO.  DMSO is not reactive toward metallic sodium, and is miscible with  water, low molecular weight and even very high molecular weight alcohols.  In this system, the relative acidities of neat solutions of the $\ce{C1}$ to $\ce{C3}$ alcohols shown above holds, but with one exception, water:
$$\text{methanol > ethanol > isopropanol > water}$$
Although the acidity of water is low in DMSO due to the poor solvation of the hydroxyl anion (the alkoxy anions are solvated by DMSO in order of increasing hydrocarbon chain length), its reactivity toward sodium remains high.  This is because in the presence of sodium, the hydroxyl anion is the reactive species.  It reacts quickly and does not need to remain solvated for the reaction to proceed.
Of course the effects discussed thus far are also applicable toward higher molecular weight alcohols.  However, once you have a carbon chain of more than 3 or so, these  effects diminish with increasing molecular weight of the alcohol.  Here, the larger the molecular weight of the alcohol, the fewer molecules you have to react for a given weight or volume concentration of the alcohol.  Even if you have a pure alcohol solution, think about the "concentration" of the reactive hydroxyl functional group.  In the case of water, hydroxyl groups are everywhere.  In the case of say a $\ce{C20}$ alcohol, the sodium is mainly surrounded by hydrocarbons with a few reactive hydroxyl groups mixed in and available for reaction.


text:
What is the keto form of a hexahydroxybenzene?

159. The keto isomer of the following compound is:

The answer has been given as (c). I cannot seem to understand why is it not possible for the keto form to only have one OH convert into keto form shown below:

Welcome to Chemistry.se! That is an incredibly awful question and I think your reasoning is correct. Nevertheless I doubt that any 'keto' form would be more stable than the aromatic structure. Since this is multiple choice, you should just rule out the obviously wrong answers and take what's left.

I guess no keto form will ever be stable here, but once the aromaticity is gone, the other enoles would likely also transform into the respective ketones.

If you assume that the formula of the keto form of hexahydroxybenzene should still be C6H6O6, then the answer can only be (c). While your structure qualifies, it was not among the options.

Well the answer is based on logic as the number of double bonds should be conserved

The question is badly phrased because it uses the term ‘the keto form’, implying that there be only one. In the case of hexahydroxybenzene, you are correct in that there is a monoketo form, at least two different diketo forms and potentially even more triketo forms.
All of these polyketo forms are created in the same way as the monoketo form except you do the isomerisation multiple times. So one way of arriving at the answer would be to check whether you can get to any of those molecules by tautomerising the hexahydroxy form often enough.
It may, however, be easier to think in the other direction, i.e. which ketones can I tautomerise to give a fully aromatic ring. Obviously, as your monoketo example shows, a ketone needs an α-hydrogen atom, so a) and b) cannot tautomerise at all and d) can tautomerise only twice, not three times. Upon careful examination of c) you should be able to realise that this can tautomerise three times to give hexahydroxybenzene.
Entirely out of scope of the textbook question is whether this tautomerisation is probable. None of the keto forms provides any advantage over the hexa-ol isomer in no solvent. We cannot argue with better hydrogen bonding and we cannot argue with polarity (both would favour the hexa-ol in all environments I can think of). Thus, the question in itself is somewhat academic. Yet it remains helpful to see how to get from polyketo to poly-ol forms and back.

You said (d) can tautomerise in two ways. How is it possible for (d) to tautomerise ??

in (d) option you can only tautomerise  any 2 of the opposite carbonyls. If you try to tautomerise more than 2 carbonyls, you will violate the octet rule for carbon.

@ABC123 the product after two tautomerisations is tetrahydroxy-para-quinone.

I thought this is a nice test case for the tautomerise option of CREST On the GFN2-xTB level of theory it only found monoketo forms, all of which about 100 kJ/mol higher in energy. I stopped there as I think that diketo forms will be even higher.

@Jan The product after two tautomerisations is not tetrahydroxy-para-quinone, rather 2,3,5,6-tetrahydroxycyclohex-2-ene-1,4-dione, or similar.

@Martin-マーチン My comment above was directed at ABC123 and concerned the tautomerisation of compound (d). Admittedly, I could have written that to make it clearer. (And on paper I just noticed the ortho-quinone is also possible.)

Ah yes, now I see. I did not see the context. Thanks for clearing that up.

@Martin-マーチン No worries ;)

The question in its current form ONE KETO FOMRM implicitly implies a single form , while three keto forms are possible 4 , 3 and c.
Structure 4 can enolize to 3 since enol form 4 can tautomerise to keto form 3.Remember in keto enol tautomerism , equilibrium can be achieved from keto end and also enol end. Energetics would determine the extent of one form over the other.
Similarly structure 3 can enolize to structure (c) or 4.
Tautomers are isomers that differ by the location of a proton and a pi bond.
These are reactions that can occur in acid or in base.

In structure 4 , since both enol and keto forms exist ,tautomerism leads to different structures, Only a single structure is not possible.

I'm afraid, that doesn't completely answer the question. Also option b can have tautomeric structures, as you have shown yourself.

@ Martin - マーチン you are correct ,i will edit my post and answer the question.

The Keto form given in the solution would be the one which is the most stable. The hydrogen of the Hydroxyl group form H bond with the O of the carbonyl group , making this type of structure highly stable. Of course other types of structures are possible and will surely exist , but this is the most likely and most stable form of it


text:
What is entropy change in electrochemistry?
I keep seeing entropy change (or reversible effects) and irreversible effects (Joule heating) when reading about charging and discharging of Li-ion batteries. I understand the Joule heating part (the irreversible), but I have troubles understanding the reversible effects. What's going on in the structure of the electrodes that the process is reversible? As far as I know, reversiblities come from the fact that there is no friction or other losses. If one isolates a single ion, the charging or discharging, in my interpretation, would look roughly like in the figure below:

What happens inside the battery during this charge and discharge that would make one think of reversibility? And why is entropy important in this case? And why would a reversible process release heat?

I think your diagram seems plausible; the physical process happening here is when charging an applied potential difference (voltage) causes lithium ions to leave the crystal lattice and migrate to the cathode. When the battery discharges, the reverse happens, and lithium ions rejoin the anode.
The definition of reversibility is the infinitesimal change in some thermodynamic quantity in the system with respect to its surroundings without increasing the entropy. The electrode is crystalline, so Li+ ions can rejoin the lattice without increasing the configurational entropy, whilst infinitesimally changing the chemical potential.
If the temperature is low enough for crystallisation to be entropically and energetically favourable with respect to dissolving, then the free energy of the system will be reduced by adding an ion to the crystal lattice. The chemical potential induced by charging causes the lithium ions to return to the anode when discharging.

Thank you! Nice explanation of the process. Can you please explain why this reversible process is generating heat (more heat than the joule effect, actually)?

I think the heat generated is the heat of crystallisation. When a lithium ion rejoins the crystal lattice, the creation of new bonds releases energy.

I think it has to do with the equation heat dq=TdS so since gibbs dG=SdT+VdP+mudN we know S or entropy is partialG partialT and also gibbs= -nFE so S=partialE partialT and heat is TdS so TpartialE partialT* nF. In short TdS is the reversible heat

Welcome to chemistry.se! Unfortunately this is just a bunch of formulas with no real explanation.

I meant its called the reversible heat because the reversible heat is defined as TdS so dQrev=TdS=ITPartialEPartialT. Which is what he asked why is this term in the Bernardi Heat Generation equation called the reversible heat becz it is TdS


text:
Why do crown ether look the way they do?
Like, the structure of lines of the rings of crown ether looks very different from regular cycloalkane rings. For example, the oxygen in crown ether is pulled more inside the ring. The line between carbon atoms consecutively bonded look almost horizontal. Why?

What do you mean? 6-crown-2, a.k.a. 1,4-dioxane is drawn exactly like cyclohexane, and the drawings we put on paper only have a very limited correlation to the actual molecular geometry …

Oh okay. Yeah, it does look very similar. I guess what I am asking is as crown ethers become larger, why does ring structure start looking so different. I am trying to understand why they look the way they so, so I can find some rationality while drawing them on paper

Cyclooctadecane, the oxygen-free form (18-crown-0?) of 18-crown-6. Notice any difference except for the missing oxygens? Note again that these are all representations on paper, only loosely connected to actual geometry.

Oh wow. I am new to organic chemistry so I always through that actual structure of ring would be quite closer to drawn structure. But wouldn't drawing them in such different way distort like bond angles and stuff. And I know there are conformation and stuff but but wouldn't drawing them in such vastly different bond angles distort what sturucutre looks like in reality.

Thanks for the previous response though.

Not more and not less than the distortions already given to linear molecules in the skeletal formula formalism. Angles are typically drawn as $120^\circ$ while angles on non-double bonded carbons are actually closer to $109^\circ$ and bond lengths are standardised rather than having some shorter than others. But I reiterate: these are representations and only meant to loosely represent the actual structure.

Crown ethers don't look like that except in structure drawings
You confuse two completely different things: the drawings we use to describe chemical structures and what molecules actually look like in real chemical environments. We draw crown ethers like this because the drawing accurately describes the connectivity in the structure. In reality crown ethers are flexible and floppy and certainly not flat. we don't add that extra complexity to 2D structure drawing as they get very messy very quickly and obscure the essential features that describe the connectivity in a simple way.
When they are in solution with appropriate ions (for 18-crown-6, potassium ions would be a good example) they form complexes driven by the attraction between the oxygens and the charged ions. These complexes have 3-dimensional structures (with K+ the complex is approximately octahedral as potassium fits neatly into the cavity created by the oxygens when the molecule folds up). It looks a little like this as a 3D structure:

Wikipedia has some good descriptions of this and further pictures.
Crown ethers are not rigid molecules and the actual shapes they take up in 3D space demand a lot on what else is in solution with them. Other, similar, complexing agents are far more rigid. Calixarenes, for example, have a 3D shape like a chalice which is even more misrepresented by a 2D drawing.

Is there a reason why you’re not linking to Wikipedia proper?

@jan No reason other than the fact that I use Wikiwand as it gives neater formatting of wikipedia content (it doesn't alter the content otherwise). And, therefore, including a native wikipedia link would take more effort on my part. And I'm lazy when there is no meaningful advantage in putting extra effort in.

I always thought that crown ethers were drawn like that to emphasize the special coordinating effects of the oxygens on a central ion.


text:
Muriatic acid fumes
hello me and my 6 year old cousin were cleaning sea shells in my garage in 300 ML of water that had 3 drops of muriatic acid in it the cup tipped over and spilled it evaporated pretty quickly are we in any danger from the fumes

Three drops  in 300 mL water? That's not a hell of a lot. Don't worry, but don't forget to wash your hands before you rub your eyes :)

Not sure how old you are, but I'm concerned about how you got the three drops. The container of muriatic acid itself is unsafe. Muriatic acid typically has a lot of iron so it is no good for a chemistry lab. However it is almost as concentrated as concentrated HCl that a chemist would use. Just use vinegar next time and wear safety glasses or goggles to protect your eyes.

Also cleaning sea shells in acid is a silly idea. They are made of calcium carbonate so they will react with the acid.

3 drops of hydrochloric acid (muriatic acid) is probably a volume less than $0.3 ~\pu{ml}$. $0.3 ~\pu{ml}$ acid in $300 ~\pu{ml}$ water is a dilution of $1 : 1000$.
If the acid was concentrated muriatic acid (e.g. $10 ~\pu{mol/l}$) the resulting diluted acid had a concentration of about $0.01 ~\pu{mol/l}$. This concentration is much less than that of gastric acid ($0.05 - 0.1 ~\pu{mol/l})$. The diluted acid itself is not dangerous.
About the fumes: 3 drops of concentrated hydrochloric acid contain about $70 ~\pu{ml}$ of the gas hydrogen chloride ($\ce{HCl}$). If you would completely inhale this volume it would be dangerous. If the same volume is uniformly distributed in a room of $> 35 ~\pu{m^3}$ it would result in an acceptable concentration.
Since you did not mention any pungent odor, you probably did not inhale much of the evaporated hydrogen chloride. So I guess you are not in danger.


text:
Do crown ethers have to have two carbons between the oxygens? If so, why?
I just began learning about ethers and I have noticed that pictures of crown ethers, no matter what size they are, always have two carbon atoms left and right of each oxygen atom before you reach another oxygen atom in the ring. It's almost like it is a poly-ethyl ether. So my question is can you have crown ethers with more than 2 carbon atoms between the oxygen atoms?

Very simple: If they look differently, they are not "crown ethers", by definition. https://en.wikipedia.org/wiki/Crown_ether

The issue isn't whether such compounds exist and might be called "crown ethers": it is whether such crown ethers would be effective as complexing agents for the metal ions such compounds are normally used for.

To be a crown ether, the compound in question must be an ether, i.e. must contain oxygen atoms which are bonded to two carbon atoms neither of which is bonded to another oxygen atom. Thus, if you had just one carbon between two subsequent oxygens you would no longer be dealing with a crown ether; if you wanted, you could call it a crown acetal. But it would have vastly different properties; most notably it would break down in aquaeous solution under acid catalysis.
To the best of my knowledge, there is no formal restriction other than that given in the first paragraph for crown ethers. 20-crown-5, which would have three $\ce{CH2}$ groups between each ether oxygen, would be a valid crown ether in my humble opinion. However, most cations that are coplexed by oxygen prefer five-membered rings for various reasons, resulting in many complexing agents — $\ce{en, dmg, m2g, phen}$ to name just a few examples — having exactly two other atoms between the ligating atoms. Thus, practically all crown ethers one hears of have two carbons between the oxygens.

"must be an ether, i.e. must contain oxygen atoms which are bonded to two carbon atoms neither of which is bonded to another oxygen atom"... By this do you mean that ether by definition is a function group where O is attached to 2  carbon atoms. What about propyl ether then? It is still a ether. May I am reading this wrong?

Nobody would call 20-crown-5 by that name. It'd be cyclopenta (oxy-1,3-propylene). 1,3,5-Trioxan is rather stable in water, as is the linear polyoxymethylene, if you stabilise the end groups.

@Tlama You can built cyclic polyethers with (oxy-1,3-propylene) groups, mixed with ethyleneoxy groups, butylenoxy groups, whatever you like, but they are not usually called "crown ethers".

@Karl Yes to 1,3,5-trioxane, and the full acetal of polyformaldehyde but still less stable than 1,4-dioxane (6-crown-2? ;)) or poly-ethyleneglycol. My point was that even full acetals are decomposed more easily than ethers. Concerning ‘20-crown-5’: Please first show me who actually works with it and what they do with it. I am not aware of that type of cyclic ether being common and hence I consider the discussion which name is common academic.

@Jan POM is a widely used technical polymer. Those acetals are perfectly stable. Trioxan decomposes into formaldehyde if heated to >150°C in the presence of water (says de:wp, without reference). I have no idea what dioxane or PEG do under those conditions.

@Karl PEG is used as an oil bath liquid up to $200~\mathrm{^\circ C}$, if that answers one of your questions ;) Catalytic amounts of acids should greatly accelerate the breakdown speed of polyacetals but not so much of polyethers.

Right, I am merely opposed to your above statement that "it would break down in aquaeous solution". ;-) Btw. processing temperatures of POM are up to 220°C, and it can even be used for food packaging if pH>2.5.


text:
Gases and physical chemistry
PV=nRT. Hence, T=PV/nR.Let's consider a gas containing in a
bulb in a way that  its pressure,volume and number of moles of gas can be changed.So if I want to attain a certain temperature for n moles of a gas in the bulb,I can adjust the PV value in the equation so as to fit to the value of T.Theoretically this can be done with above equation.But is this experimentally possible?

Yes. Related: http://physics.stackexchange.com/questions/17948/why-does-a-gas-get-hot-when-suddenly-compressed-what-is-happening-at-the-molecu

Nonsense. You cannot "change" the pressure actively, except by changing n,V or T. An ideal gas especially does not change it's temperature when compressed. The Joule-Thomson coefficient of an ideal gas is zero.

@Karl Wouldn't that violate conservation of energy though?
http://physics.stackexchange.com/questions/136408/what-happens-to-the-temperature-when-an-ideal-gas-is-compressed

If the system is adiabatic and you apply work to it, that energy has to go somewhere. If I recall, ideal gases do not go through phase changes.

@QuantumAMERICCINO You increase the pressure of an ideal gas by (anti-)proportionally lowering the volume. Exactly antiproportional. No temperature change occurs, because otherwise the result would not be an ideal gas.

@Karl The "nonsense" was slightly harsh.  I understand how you meant it, but nuance doesn't always come across properly in text.

I will try to summarize what Karl is saying about the ideal gases.
Suppose that you measure the temperature of a gas contained in a canister. Suppose that you measure the temperature $T_i$
Using your equation that you derived from $PV=nRT$, we should have $$T_i=\frac{P_iV_i}{nR}$$
Suppose that we then apply work to the canister. Since it is an ideal gas the only way that we can change its energy is through a change in volume ,pressure, or number of moles. So that is what we are going to do. Suppose that we compress the canister so that the pressure of the gas changes. Thus, our new canister will have a new pressure $P_f$ and a new volume $V_f$.
This volume $V_f$, however, is related to $V_i$ through the ideal gas equation $$P_iV_i=P_fV_f$$
Hence we have that $$V_f=\frac{P_iV_i}{P_f}$$
Lets substitute this equation to find the new temperature of our compressed canister:
$$T_{compressed}=\frac{P_f\frac{P_iV_i}{P_f}}{nR}$$
Which gives us the initial equation. Hence the temperature of the ideal doesn't change when it is compressed.
I will post on what happens when you increase the number of moles.


text:
LAMMPS: MD simulation of molecules?
I need to do molecular dynamics simulation of cyclodextrine molecules for a project and I am searching for an open source solution that can be used. I found LAMMPS to be quite well maintained and wanted to use it for the MD simulation. However, after spending now around 10 hours trying to figure out how to specify an input geometry for a LAMMPS calculation, I am stuck. Has anyone experience with doing MD simulations of molecules using LAMMPS?
Given an input structure in standard .xyz format, how do I start a LAMMPS MD calculation on it?

Obviously this discussion was not going anywhere, so I took the liberty to clear it. I understand that you are frustrated with the program, but I am quite certain, that if you don't add more context and what you have tried so far, this question will get closed (It already has 3 CV). I personally think it is also important to mention, that there is no such thing as a standard xyz format.

@Martin-マーチン Thanks : )  Yes, xyz is not really a standardized file format but it is still simple enough that OpenBabel almost never fails on it and that is enough for me. Surely, cml might be a real standard but is also more complicated...

@Martin-マーチン OpenBabel does specify a standard, FWIW. Agree it's not universal or the only way to specify a geometry, though.

This is an old question. I can't see the discussion that happened here, but I think on Matter Modeling Stack Exchange someone would have just given you the input file; especially if you used the software-input-file-request tag: https://mattermodeling.stackexchange.com/questions/tagged/software-input-file-request. There is also a chatroom just for LAMMPS: https://chat.stackexchange.com/rooms/109805/lammps.