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Filling up of orbitals
As we know that the orbitals in a sub shell are degenerate, so is it necessary to fill up the electrons starting from x then to y and then z (if in p-subshell) or they can be filled like shown in the photo.

Short answer: it makes no difference. Long answer involves QM.

The very word "degenerate" means they are exactly identical, so there is no difference. If we commonly fill them up left to right, it's just a conventional thing, much like writing left to right.

If you think back to math class one variable is usually x, two use x-y plane and three dimension use x,y and z. Chemists tend to do it that same way.

You are drawing very nice $\mathrm{p}_x$, $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals onto that piece of paper. But if you take an isolated atom, which direction is $x$? Which is $y$? The coordinate system you’re seeing the atom in is just something we planted over its top because humans like things nice and organised. It is entirely arbitrary.
By the way, the direct calculations do not give $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals (strictly speaking not even $\mathrm{p}_z$. They give you $\mathrm{p}_0$, $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the indices referring to the value of $m_l$, the ‘magnetical quantum number’). In spherical coordinates, which are used because it makes the calculations easier, only the $z$ axis is semi-defined by values $\theta = 0$. Because $\psi(\mathrm{p}_0) = f(r, \theta)$ (the wavefunction of one of the p-orbitals is independent of $\phi$), we say that this represents the $\mathrm{p}_z$ orbital. $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the direct solutions of the Hamiltonian) cannot even be mapped to directions as they involve complex numbers; only via linear combinations can we transform them into two real orbitals pointing in an arbitrary $x$ and an arbitrary $z$ direction.
Things change if you introduce an external magnetic field. The atom’s $z$ axis will now align with the magnetic field in some way, giving your system one preferred direction. One, not three. $x$ and $y$ are still essentially undefined. However, in a magnetic field it would make a difference which orbitals the electrons populate.
You might also be interested in checking out the Jahn-Teller effect which occurs in transition metal chemistry to avoid filling some of all degenerate orbitals. In a nutshell it is energetically better for certain complexes to distort themself from a perfect octahedron to create distinct energy levels that can be fully populated, half populated or unpopulated.

The pure answer is, you can unitarily transform them into each other.
In the classical mechanic picture assume a coordinate system you put somewhere onto a map in a way that one car is driving along the y-axis and another one along the x-axis in the same speed and momentum hence same kinetic energy. Nothing stops you from arbitrarily rotating or mirroring this coordinate system so that for example one car is driving along the z-axis and another one along the x-axis.
At least to be honest there is still the possibility that you have negative velocities if you would mirror, but the kinetic energy is still unaffected. And we are interested into the energies.
When starting with px and pz orbitals you could arbitrarily switch your coordinate system and start with py and px orbitals.

Since velocity is essentially a vector, ‘negative’ only has the meaning of velocity pointing in the opposite direction.


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How are odor and electronic structure related to aromaticity?
I know that cyclic planer molecules with $2n + 1$ $\pi$-electrons are aromatic. Also, the color of a compound is due to electronic transitions. But how is it related with aromaticity (smell)? Is there any such relation or is it the chemical property of any compound (matter)?
If it is chemical property of matter then what is its cause?
P.S. My question isn't about nomenclature, its about what is in benzene or any other aromatic compound that make it aromatic i.e. their smell? And what specifically is the roll of electrons in their smell?

I'm not certain I understand your question correctly; are you asking why aromatic compounds are named that way?

@inɒzɘmɒЯ.A.M no, I think he’s asking if there’s a reason in the (electronic) structure of aromatic compounds that makes them smell.

The smell of a compound as we perceive it usually depends on which receptors in the nose will bind the molecule (or not). (There are exceptions for molecules that actually react with these receptors, such as $\ce{HCl}$, but benzene and most aromatics are not one of these exceptions.)
These receptor, being proteins, are usually rather specific for certain shapes, not for bonding patterns. So chlorobenzene might fit into similar receptors as toluene which would lead to at least partly a similar smell, but that is it.
The definition of aromatic is, as Todd pointed out, one based on structure. Some aromatic compounds have similar structures, but there is no way that (say) furane would occupy the same or similar receptors as benzo(a)pyrene would. Therefore, their common feature aromaticity does not lead to a common odour.

But what it has to do with electrons? As you said and from the answer of @Todd what i can understand is that its just a convention that is being followed and nothing related to electrons. Am I right at this???

Yes. There is nothing in the electronic structure that would create a certain smell of any kind.

I can address the part of your query pertaining to odor and nomenclature (more accurately, I can link the UC Davis ChemWiki page that does so and quote it here):

Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (See figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.

You will find (in the linked section on Aromaticity) that presently:

The three general requirements for a compound to be aromatic are:

The compound must be cyclic

Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.

The compound must follow Hückel's Rule (the ring has to contain 4n+2 p-orbital electrons).

Putting it all together: The historical naming convention for molecules that smelled good is responsible for the term aromatic. It turns out that many of the molecules that fall into the aromatic category happen to be rather stable/unreactive. These molecules happen to have certain physical characteristics which, taken together, are indicative of a molecule being classified as aromatic or not (based on present-day definitions not relating to odor/smell).

The theory shape or chemical composition has anything to do with smell can be refuted based on three observations:

Compounds with different shape can smell exactly the same. (creosol and eugenol)
Compounds of different isotopes can smell different. (musk)
Concentrations of compounds can change the odor completely. (phenol)

Since the advent of genetic technologies, a new theory is on the horizon is the underlying mechanism for smell is based on quantum oscillations.  This theory is derived from information gathered from studying the genome which encodes the proteins responsible for the sense of smell.
Todd's answer is correct.  The relationship to smell is a historical artifact from the time chemistry was in its infancy when chemical properties were described using the senses of taste and smell.

Shouldn't be in 1.  "different"? Also some links would be nice.

This information came from one of the TED talks by  Andreas Mershin   https://www.youtube.com/watch?v=sjV7NNwm1GU

also https://www.youtube.com/watch?v=raS_PWgk_iI

source: Stack Exchange

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

sort: votes


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Find the major product when Benzoquinone reacts with HBr

What is the major product when Benzoquinone reacts with $\ce{HBr}$ ?

My effort: I thought it would be an $\text{SN}_{1}$ mechanism, but the carbocation at either double bond wouldn't be stable.

Please don't introduce Mathjax code to your question titles due to some issues $\ldots$ Unless, you're talking about cerium hydrogen bromide.

First, the acid will protonate the carbonyl to give a delocalized carbocation. Where will the nucleophile ($\ce{Br^{-}}$) attack; attack at which position can lead to a stable (aromatic) product by loss of $\ce{H^{+}}$?

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


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Determining the reaction order of acetylsalicylic acid to salicylic acid with a spectrophotometer
How can I determine the reaction order of the reaction acetylsalicylic acid + water to salicylic acid + acetic acid with use of a spectrophotometer?
I thought of measuring the absorbance spectrum of both substances individually, compare the 2 spectra, and then look at which wavelength the absorbance of acetylsalicylic acid exists whereas the absorbance of salicylic acid at that wavelength is negligible.
Lastly, use that wavelength to calculate concentrations of acetylsalicylic acid at certain time intervals and conclude the reaction order out of the change in concentration vs time. (for example making a graph on Excel for Concentration vs time -> linear = 0 order, exponential = 1st order)
Is this possible or am I thinking too hard/easy? Is there any other other way?

I think your reasoning is sound. However, a factor you may need to take into account is that you have a number of possible related species here, each of which has a different UV absorbance spectrum. You have acetylsalicylic acid, acetylsalicylate anion, salicylic acid, salicylate anion, and salicylate dianion (this one will probably have a markedly different UV spectrum from the others). The relative amounts of these species present would be affected by pH, so it would be important to buffer the reaction solution, to maintain constant pH.


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How do biochemists seperate nucleotides from DNA to add to the PCR
How do chemists separate single nucleotides from DNA strands?  How do they add those nucleotides to e.g. a PCR reaction in order to replicate DNA?
I ask this because in my microbiology class, my teacher didn't know the exact answer.
Do you cleave off nucleotides by hydrolysis?  Do you somehow build nucleotides?

I'm not sure the exact process but genomes are deconstructed and reconstructed using enzymes such as taq polymerase, reverse transcriptase and the correct primers ( which target the specific site that you want the enzyme to attach to the dna strand) .  Techniques such as polymerase chain reaction, electrophoresis and lots of bio processing would be used along with other techniques I'm not aware of. I'm not sure how a new sequence is added to a genome there would be too many variables to cover in a practical answer !

When I did PCR and electrophoresis I added the component molecules to the mix like a primordial soup . The nucleotide solution was from sigma aldrich I think. Can't remember if they were synthetic or not. The enzyme uses them to build the new strand. In the PPE reaction the DNA denatures at a certain heat.

I’m not sure what you’re asking. Are you interested in where the nucleotides that one adds to PCR reactions are acquired from?

There's a few technical questions in there.

You just buy the nucleoside triphosphates from your favorite supplier, they're commercially available.
As far as I know the NTPs you can buy are still produced biologically and not synthesized (see this example from Sigma Aldrich, which states a microbial source). So you grow a large amount of bacteria, lyse the cells to get at the contents and then digest the DNA or RNA with an appropriate enzyme that cuts it into single nucleotides. To get the triphosphate from that you have to phosphorylate the monophosphates again with the appropriate kinase.

They have recently discovered that synthetic nucleotides work the same and a much less expensive to produce in high yields.

According to sigma aldrich the nucleotides used for PCR are now produced synthetically.
The low yield and high cost to produce them biological makes it impractical. When produced synthetically they get a high yield at low cost.
They are added to the pcr mixture in a pre prepared solution and usually transferred via micropippette or a machine.
The new gene can be built from these nucleotides.
Using gene therapy techniques the new DNA can be incorporated into a genome.
One, of multiple techniques, uses an adenovirus vector with the new DNA packaged inside to transport and  insert the new DNA to the cell in the form of a nucleic acid polymer which incorporates the genes into the chromosomes.
Experiments were done to compare the efficiency of the synthetic genes to the biological ones with the need for biological nucleotides as a control. Now they know the gene built from synthetic nucleotides shows the same expression as the biological gene of the same code synthetic nucleotides are used. However microbial produced nucleotides can still be used.
Ref:Wikipedia, Gene therapy. Sigma aldrich - Technical support ph - 1800 800 097

Sigma sells multiple different versions, some declared as "from microbial sources" and others without a source, which are likely the synthetic ones.  There doesn't seem to be a huge price difference.

I called and asked them . Your free to do the same.

I'm not saying the microbial sourced ones don't exist.


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Clarification on Recrystallization concept
Any help you could provide me would be greatly appreciated and I thank you for reading (even if you don't comment)
I am in an organic laboratory and as a 3rd chemistry major these concepts usually are pretty basic but recrystallization is kind of leaving me scratching my head. When you dissolve your compound with a small impurity in it, dissolve it in a solvent and heat, I am just confused on this part:

What I understand is that the solubility constant changes with temperature and in the case of organic compounds, solubility almost always increases with increasing temperature. So, if the compound is put in a solvent at room temperature and is heated until it is fully dissolved (solubility constant increased) then it is cooled and the solubility constant decreases and consequently solute starts precipating (crystal formation) My only issue is what is that makes the impurity go away? Thinking through it here, I know I'm so close to the Ah-hah moment but as of right now I'm mental blocked — I suppose.

Wouldn't the impurities remain in the solution as the "pure" material precipitates? And you can separate via filtration.

I don't understand that part. If they existed as solids initially, how are we supposed to know that the impurity doesn't precept. first?

Sorry, that was only one type of instance of how the impurity "goes away". Since you're playing with the solubility properties, then the impurity gets separated anyhow. Does this link help at all? http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/recrystallization

Supposedly the impurity is present in minor amount, so it will not reach its own solubility limit and thus will not start precipitating.

On an unrelated note: Welcome to chemistry.SE! Feel free to take a [tour] of the site. Consult the [help] for any questions on how it works!

You note that you are pretty close to the answer and you are. Let’s assume for a minute (I’m pretty sure the assumption is false, but it’ll get us a good way) that the solubility of the impurity is independent of the amount of desired product dissolved and vice-versa. Let’s also assume we have a mixture of $95~\%$ desired product and $5~\%$ side product which you want to purify by recrystallisation. The desired product has a solubility of $1\,\mathrm{\frac{mol}{l}}$ at $4\,\mathrm{^\circ C}$ and $10\,\mathrm{\frac{mol}{l}}$ at $80\,\mathrm{^\circ C}$; and the side product has the same solubility. And finally, let’s assume that you have $100\,\mathrm{mmol}$ total product (side and desired).
You then add $9.5\,\mathrm{ml}$ of your recrystallisation solvent and heat to $80\,\mathrm{^\circ C}$. That would be enough to dissolve both your $95\,\mathrm{mmol}$ desired product and your $5\,\mathrm{mmol}$ side product. You observe a clear (hopefully boiling because best solubility) solution.*
You then let your solution cool down and put it in the fridge overnight. Due to the decreased temperature, the solubility of both desired product and side product drop. The same volume of solution now dissolves only $9.5\,\mathrm{mmol}$ of both. Since the desired product is supersaturated ($95 > 9.5$), $85.5\,\mathrm{mmol}$ of the desired product will precipitate or crystallise. But the side product is still not supersaturated. We have $5\,\mathrm{mmol}$ of the side product dissolved, but the solvent could dissolve $9.5\,\mathrm{mmol}$. So all of the side product will remain in solution (ideally).
Next step is usually filtration and washing. The filtrate will now contain the saturated product solution which is undersaturated with respect to the side product. The side product didn’t disappear, it merely remained in solution. Wash your pure product, record an NMR spectra and rejoice upon its purity.

Of course, as I stated in the beginning, the assumption that solubilities be independent is wrong. It is also tendencially wrong to assume similar solubilities of desired product and impurity. One would attempt to select a solvent so that the product is not well soluble at low temperatures, but the impurity ideally is. However, the basic principle remains valid: The desired product will form a supersaturated solution and precipitate/crystallise while whichever impurity you have should still be soluble to not co-precipitate.

*: A note on how to do it practically: Usually, you would put your crude product into a flask into an oil bath with a refluxing condenser on top. You would then slowly, dropwise add solvent through the condenser until the solution is boiling and everything is dissolved. One wouldn’t go about saying ‘I should have $x\,\mathrm{mmol}$ so I should need $y\,\mathrm{ml}$ of solvent.’

First of all, I am not sure what you mean by "solubility constant". Perhaps you mean just "solubility", which is the weight of a material that  you can dissolve in a given volume of a specific solvent (at a given temperature).
You are also correct in stating that solubility usually (but not always) increases with temperature. This depends on the choice of solute and solvent.
There are two effects here serving to purify your impure compound. (1) As Ivan has explained above, a small amount of impurity (having a similar solubility) would remain in solution (being below its saturation concentration) while the main ingredient becomes saturated in solution as the solution cools and the solubility decreases, and therefore crystallizes out. (But--this would not work well if the impurity were relatively insoluble in the solvent.) (2) The formation of a regular crystal lattice, which is usually energetically more favorable than an irregular or amorphous solid, tends to favor the formation of crystals composed of a single chemical substance, rather than a mixture--especially if one substance is in large excess. So, if you repeatedly recrystallize a substance (each time, taking the crystals from the previous step and recrystallizing them using fresh solvent) you can sometimes achieve a very high state of purity.

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

I'm not sure why you feel the need to school me, but you should check your soft skills bud. Your explanation added nothing and I'm not going to take solubility constant out of my vocabulary. Why wouldn't something be a constant just because it's temperature dependent? You seem like a real dope

I am sorry, Joshua -- I can see how you would have been offended by the tone of my answer, so I have edited it accordingly. I have occasionally been known to act as a real dope, and my wife can corroborate on this. Please look at my edited answer and decide if it has any value to you on its merits.


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How to remove dichlorometane from water?
I will have water that have been through dichloromethane. Doing so, the dichlorometane (solubility $2~\%$) will contaminate my water by forming an azeotrop which will have a $38~\mathrm{^\circ C}$ boiling point. Would putting that mixture to $40~\mathrm{^\circ C}$ will make that azeotrop leave? Leaving me dichloromethane free water?
If not I could I extract it?

There are two concepts here which should not be mixed:
The solubility of dichloromethane in water is indeed quite low, around 2% as in the question. But this is not the azeotrope, which is formed when there is 1.5% water - 98.5% dichloromethane mixture. That indeed boils at $38^\rm{o} \rm{C}$. See this. It's worth mentioning that you can't possibly bring the azeotrope to $40^\rm{o} \rm{C}$ - because it will boil at a lower temperature by definition, and the vapour will have exactly the same composition as the liquid.
Because what you have isn't the azeotrope, I'd say boiling your mixture (which will most likely happen near the boiling point of pure water) will remove the dichloromethane content.

No. You end up with a lot of distilled water that still contains CH2Cl2.

Boiling with a column on top? Of course you'll always have some minute amount of contaminant remaining but depending on the circumstances I'd think you can get quite far.

No, the methylene chloride will be almost completely in the distillate alright! But if you add a column, the CH2Cl2 has to diffuse through it, that´d also be inefficient. Running the "distillation" with column at  40°C under vacuum would help. Basically sucking the methylene chloride out with a membrane pump. No idea how long that would take.

Given that this was asked almost 5 years ago, by now I would have expected all the CH2Cl2 evaporated anyway :)


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Solubility in carbon tetrachloride

Which of these substances is more soluble in $\ce{CCl4}$?

$\ce{CH3F}$
$\ce{I2}$
$\ce{Na2SO4}$
$\ce{HCl}$

How do I even solve this kind of exercise?
My guess is that $\ce{CCl4}$ is non-polar. Therefore the substance we're gonna need must be non-polar as well. The only non-polar substance is $\ce{I2}$ so that should be the answer.
Is my reasoning fine and, more importantly, is there another way to solve this kind of problem? What if they give me two options that are non-polar?

You are correct in your assumption. For these types of problems it is usually safe to assume like dissolves like and choose the answer with the most similarity in terms of intermolecular bonding. For a general description you can check here.
When you get into questions with two options that have similar intermolecular forces it gets a little hairy. For example, I think a question like, "which is more soluble in carbon tetrachloride: n-Pentane or n-Hexane?" would be fairly challenging to answer without experimental or computational data to compare.

Why, that one is easy: pentane and hexane both would be infinitely miscible with $\ce{CCl4}$. You are right in general, but this example would look better with two solid compounds. Say, naphthalene and $\ce{I2}$. Which is more soluble? I don't know.

That's a great point Ivan, I was grasping for examples, I should have chosen some higher order solid n-alkanes. A relevant paper: for anyone interested http://pubs.acs.org/doi/abs/10.1021/ie00063a008

If this is your introduction to chemistry answer is I2. It gets more complicated in extreme conditions such as ultrahigh pressure or temperature.
I would stick with I2 (nonpolar-nonpolar)
for degrees of polarity check molecular structure, electron orbitals etc.
this and other similar online tools helps too: http://www.gregthatcher.com/Chemistry/ChemicalBondPolarityCalculator.aspx
a lesson on interactions between non-polar molecules:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Overview_of_Intermolecular_Forces/Interactions_Between_Nonpolar_Molecules


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Is that true that if we reduce the Cl in water will reduce the saltiness of water?
I just move to some place near the ocean and I dig a weld and the water taste is a little bit salty (not as salty as sea water but there are salty taste). Then, I bring the water to the lab and there are 2 points that are really high the TDS $3345\,\mathrm{mg/litre}$, and $\ce{Cl-}$ $998\,\mathrm{mg/litre}$.
I already use a softener to reduce the $\ce{CaCO3}$ and magnesium content, and now my water treatment guy that install the first softener suggest me to add another filter that contain resin and can absorb $\ce{Cl-}$ content, he said to me The resin will absorb the $\ce{Cl-}$ content and make the water taste better. He also said that it will need $\ce{HCl}$ or $\ce{(NH4)2 SO4}$ to regenerate the resin.
My question is:
1. Is that true that we can reduce the saltiness in water if we remove the $\ce{Cl-}$ ?
2. What kind of resin this water treatment guy talking about? Is this resin even exist?

In short: 1. True. 2. That's an ion exchange resin; yes, it exists.

In order to remove $\ce{Cl- }$, you need at least an anion exchange resin (or a mixed-bed ion exchanger, which, however, cannot be easily regenerated). Used anion exchange resin (that is loaded with $\ce{Cl- }$) cannot be regenerated with $\ce{HCl}$. $\ce{HCl}$ can be used to regenerate cation exchange resins.

@Loong can you suggest which chemical that need to regenerate anion exchange resin? This conclude that the water guy give me wrong resin.


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Rotational Barriers - Amide and Hydroxamic Acid
Why the barrier to rotation about the $\ce{C-N}$ bond higher in an amide than it is in a hydroxamic acid? (MO-based explanation preferred)

You can check out molecular orbitals of N-methylacetamide here. Unticking the spin-box will stop the molecule spinning. The important orbitals are $\ce{HOMO} -1$ and $\ce{HOMO} - 2$. (The actual HOMO is not part of the $\unicode[Times]{x3C0}$-system.)
The same site also features a low-level Simple Hückel Molecular Orbit Calculator. I used that to create the four images below of the $\unicode[Times]{x3C0}$-system of N-hydroxyformamide. From left to right the atoms are $\ce{O-N-C=O}$

The lowest $\unicode[Times]{x3C0}$-orbital is always bonding across the entire $\unicode[Times]{x3C0}$-system. In N-methylacetamide, the second $\unicode[Times]{x3C0}$-orbital has a nodal plain almost directly on the carbon, only slightly moved towards the carbonyl oxygen (theory tells me that, I can’t actually see that good in the model). The third $\unicode[Times]{x3C0}$-orbital is empty and thus does not participate in bonding. This means that each bond ($\ce{N-C}$ and $\ce{C=O}$ can be regarded as something between a single and a double bond — in accordance with the Lewis resonance structures. There is considerable overlap to be lost on rotating the system around $\ce{N-C}$.
In the hydroxamic acid, the second and third $\unicode[Times]{x3C0}$-orbitals (the occupied ones) only have very little electron density in the $\ce{N-C}$ bond. In fact, the lowest MO is largest for the $\ce{O-N}$ bond, the second is (weakly) antibonding concerning $\ce{O-N}$ and has the largest contributions on $\ce{C=O}$. The third is again antibonding along $\ce{O-N}$. Altogether this means that most of the electron density is between $\ce{O-N}$ and between $\ce{C=O}$ — the $\ce{N-C}$ bond only carries very little density and is thus much better described by a single bond. Much less resonance is lost when rotating, so the rotational barrier is lower.

I could insert somewhere that you can check ron’s structures for resonance but … w/e ;)

I will sound now really nitpicking, but I was looking more for a qualitative approach, without computations, as this is an exercise from Ian Fleming's Molecular Orbitals. Nevertheless, if no such answer appears I will accept this one.

Qualitative is rather hard imho, because the six-electron $\pi$-system is rather unknown. But maybe I can draw/write one up, too …

In short, this is because the said bond has more "double" character in the former case. And that, in turn, is because the presence of electronegative substituent makes nitrogen less eager to share its lone pair, which otherwise could have contributed to the resonance structure with $\ce{N+=C-O-}$ fragment.


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Best way to remove Aluminium from "silvered" glass?
Requirement is simple(!) - I have 30mm silvered glass balls that are originally Christmas tree decoration that I want to use in a very unconventional manner in a physics experiment. As far as I can tell the Aluminium is external. The balls are exceptionally delicate and the glass probably somewhat stressed i.e. they break easily if scratched.
So, would dilute NaOH be suitable or is there a better solution?
Another requirement is that ideally the treatment does not raise the surface energy of the glass

just for the beauty i would recommend mercury ;)

The classic etch for aluminum in the semiconductor world is 25:5:1 phosphouric-acetic-nitric (for faster etching 25:5:5 can be done). This does not negatively impact field oxides for Si devices, if you take that as any indication of 'surface energy'.

Diluted (5-10%) HCl, HNO3 and H2SO4 should work equally well.

Acid, or just throw it away. Neither are valuable enough to risk injury.

Heating the bulb in a dilute NaOH solution would probably work, yes. Something that might be more 'interesting', and potentially safer (with less hydrogen) though, is to acquire a bit of gallium.
Gallium infiltration is a really fascinating phenomenon. Simply melt a little bit of gallium and dunk your bulb in it, and watch as the aluminum becomes as soft as wet tissue paper. Gallium also melts at a temperature very slightly above that of room temperature, meaning that the the chance of the glass bulb cracking because of heating would be very minimal.


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Chemical bond at Aluminum / Aluminum oxide interface
I am trying to find out what chemical bond holds aluminum oxide to the aluminum on which it forms. The only information I was able to find on the internet was in a Google book which suggested that the bond was ionic in nature.


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Need 28% acetic acid from 98% glacial acetic acid
i need to make a solution of a 28% acetic acid, what i get is 98% glacial acetic acid. If someone could please guide me to make this solution.
I want to know the safety precautions i need to take when using 98% glacial acetic acid.
Also i would like to know safety precautions i need to take when using 28% glacial acetic acid.
Can i store it in PP narrow mouth bottles
How long will this solution stay good and also how to store it.
Thanks for your help.

A lot of the information you request is available with a cursory internet search. If you can narrow the scope of your query, that will be helpful.

As @Todd Minehardt already stated, a lot may be found in the internet.
Beside common sense (incl. appropriate equipment and protective clothing) and requesting help from trained persons (teaching assistant in lab class, chemistry teacher, to mention a few) in case of doubts:  If the chemical is well known and is (commercially) of importance, where available, chemical suppliers provide both the chemical as well as safety relevant data (in material safety data sheets, MSDS).  Often these data are provided in more than one language, too.  As two examples, look-up catalogues by Sigma-Aldrich or Acros Organics.  The wiki page mentioned above even mentions a dedicated search engine.
Sigma, by the way, compiled many useful data here, where the molarity calculator is one possible answer to your third question. In your case, the final volume of the diluted solution is 3.5 times the initial volume of the concentrated volume to be diluted in water.
Fourthly, the diluted acetic acid may be stored (at room temperature) in a bottle of PP.  Charts like this provide an overview. Do not forget to check it prior to use and then regularly for cracks or leaks, as for every container containing chemicals, and to store it safely according to your local law.


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What is the role of acid in keto enol tautomerism
I have heard about an acid acting as a catalyst in keto enol tatutomerism.I have yet to understand the mechanism of the reaction though.pictorial depictions are welcome.

Please see the site's homework policy. You should amend your post to include what work you've done towards an answer.

ill do it next time onwards.I am new to this site.

In acid-catalysed keto-enol tautomerism, a lone pair of electrons is removed from the oxygen of carbonyl group by the acid.

If enol is the starting compound, markovnikov's addition of $\ce{H+}$ occurs.


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Can we use phosphorus triiodide in Hell-Volhard-Zelinsky reaction?
Using the Hell–Volhard—Zelinksy (HVZ) reaction, we can get alpha-chloro or alpha-bromo carboxylic acids:

Is it possible to similarly make alpha-iodo carboxylic acids with the help of HVZ reaction? I could not find any reference of HVZ reaction forming alpha-iodo carboxylic acids.
Though $\ce{PI3}$ is much less stable than $\ce{PBr3}$ and $\ce{PCl3}$, it has a decent shelf life of few months and is actually widely available reagent. But still,  $\ce{PBr3}$ and $\ce{PCl3}$ is seemed to be used in HVZ reaction. Why not $\ce{PI3}$? Is the reaction thermodynamically unfavourable? Is so, what is the reason?
If $\ce{PI3}$ is used instead of $\ce{PBr3}$ or $\ce{PCl3}$, will the reaction proceed to form alpha-iodo carboxylic acids? If so, why?

Update (I was visiting some of my older questions and encountered this question and thought of providing a little more information)
I found this information from a handbook:

Carboxylic acid derivatives can react with iodine without any intermediate enolate ion to produce alpha-iodocarboxylic acids. Alpha-iodocarboxylic acid chlorides can be produced when hexanoic acid reacts with iodine and thionyl chloride at 85°C to give 80% yield of 2-iodohexanoyl chloride. Similarly, butanoic acid reacts with chlorosulfonic acid and iodine to give 94% of 2-iodobutanoyl chloride. These examples are nothing more than the iodine analog of Hell-Volhard-Zelinsky reaction.

Why are these reaction considered as iodine analog of HVZ reaction? Doesn't HVZ reactions strictly uses phosphorus halides?
If these examples are iodine analog of HVZ reaction, why are they producing 2-iodoacyl chloride? They should be producing 2-iodocarboxylic acid.
If it is producing 2-iodoacyl chloride, how to get rid of the chloride atom to get 2-iodocarboxylic acid?

In this site, it is mentioned that only iodine cannot simply iodinate carboxyl acid due to thermodynamic reasons and alternative and noble ways were discovered to overcome this problem. The reactions mentioned were with carboxylic acid dianions, acyl chlorides with iodine in thionyl chloride and iodination with iodine in the presence of cupric salts. There is no mention of phosphorus triiodide where I later found that with phosphorus diiodide, $\ce{P2I4}$, it can produce 1-iodopropanoic acid and not 2-iodopropanoic acid). Is it considered HVZ reaction?

Ketones differ from their reactivity from carboxylic acids and the HVZ is about the later substrate.  This in mind, Denis and Krief published 1981TetLet22 a paper already stating in the abstract «$\ce{PI3}$ and $\ce{P2I4}$ clearly reduce $\alpha$-bromo and $\alpha$-iodo ketones», testing on multiple substrates (in DCM, 25 C, 1-8h, 51-91% yield), stating «The methods proposed can be advantageously compared to the ones already described.»  Their substrates were obtained via epoxide opening (TMS-iodide) and oxdidation ($\ce{CrO3/H+}$).

I doubt if there is enough research done on $PX_3$ or $PX_5$ reagents. Your question is still unanswered properly inspite of being heavily upvoted (my doubt too  +1)

https://pubs.acs.org/doi/abs/10.1021/jo00299a042

Here is a nice starting point (with references) on ways to add iodine in the alpha position: https://chemia.manac-inc.co.jp/en/archives/1461

It is possible to make the acyl iodide from an anhydride and red phosphorus/iodine, which seems to make phosphorus triiodide in situ: mentioned in https://www.thieme-connect.com/products/ejournals/abstract/10.1055/s-1981-29573. Maybe the problem is the direct synthesis of the acyl iodide from the carboxylic acid that does not happen (i.e. blame the intermediate rather than the reagent).

[OP] 1. Why are these reaction considered as iodine analog of HVZ reaction? Doesn't HVZ reactions strictly uses phosphorus halides?

If you look at the mechanism (pasted below from Wikipedia), you see that the phosphorus halides are used in catalytic amounts to make some acyl halide. Once the reaction starts, further acyl halide is made from the reaction intermediate.

Source: commons.wikimedia.org/wiki/File:HVZhalogenation.png
So while a strict iodine analog of the HVZ reaction would involve a phosphorus iodide, you could make an argument that adding iodine to an enol of a carboxylic acid derivative is somewhat analogous to the HVZ reaction.

[OP] 2. If these examples are iodine analog of HVZ reaction, why are they producing 2-iodoacyl chloride? They should be producing 2-iodocarboxylic acid.

The bromine version of the HVZ reaction also initially produces 2-bromoacyl bromide. So that is not so different.

[OP] 3. If it is producing 2-iodoacyl chloride, how to get rid of the chloride atom to get 2-iodocarboxylic acid?

I would guess you could carefully add water after the intermediate has formed.

[OP] Is it possible to similarly make alpha-iodo carboxylic acids with the help of HVZ reaction?

It seems that alpha-iodination is much slower than alpha-bromination (for the chlorosulfonic acid promoted reaction, there is a 100-fold difference even when the slower reaction is done at 20 degree higher temperature). Also, it is difficult to make the acyl iodide directly (one way is to convert the acyl chloride). So it is possible that the first, catalytic step to form the acyl halide fails in the case of iodine. At least, it is not described in the literature yet.
An aside
There was an alternate reaction mechanism proposed involving a ketene intermediate by Ogata and Watanabe (1979). However, Ogata and Adachi (1982) found that reaction rates of ketenes with iodine were too fast to explain the reaction kinetics found for the alpha-iodination starting with the carboxylic acid, so they updated their mechanistic proposal, to invoke an enol intermediate as well. So it seems that many, if not all, of the alpha-halogenation mechanisms proceed via an enol intermediate, stabilized in various ways.

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text:
Dissociation constants of oxalic acid
For oxalic acid, $\mathrm{K}_{\rm a1}> \mathrm{K}_{\rm a2}$.
My teacher told me that after first deprotonation, $\ce{COO-}$ acts  as an electron donating group and hence reduces the acidic strength of $\ce{HC2O4-}$ (due to the +I effect of $\ce{COO-}$ on the second $\ce{COOH}$).
Why is $\ce{COO-}$ an electron donating group when $\ce{COOH}$ is an electron withdrawing group?

1.25, 4.14 according to https://en.wikipedia.org/wiki/Oxalic_acid suggests it may be even slightly withdrawing.

After loosing $\ce{H+}$ wont it become rich in electrons and act as an $\ce{+I}$ group ?


text:
Basic and nucleophilic character of NaOH (aq)
Sodium Hydroxide (aq) is very effective in plucking off an acidic hydrogen. How effective will it be in substituting chlorine of an alkyl hallide (let us assume that the alkyl hallide is primary) and why ?

Even if, The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factor.
Nucleophilic substitution reaction has $\ce{SN2}$ reaction mechanism.
For $\ce{SN2}$ favors the reaction with primary substituted alkyl halide more than secondary than tertiary, because its effected by steric hinderance. The more crowded it is, the more difficult it is for a nucleophile to attack.
So, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution.
For a given halogenoalkane, to favor elimination rather than substitution, use:
higher temperatures
a concentrated solution of sodium or potassium hydroxide
pure ethanol as the solvent
To favor substitution rather than elimination, use:
lower temperatures
more dilute solutions of sodium or potassium hydroxide
more water in the solvent mixture


text:
Is it possible to have electric field in water without having electrolysis?
Is it possible to have electric field in water (using electrodes with voltage difference) without having electrolysis in the water (or any other reaction)?

Well, if the voltage is lower than needed for any possible electrolysis reaction, there would be no reaction, just the field. So the answer may seem to be yes.
Now, in fact there is a catch. Once you turn on the voltage, even if the reaction is not possible, the ions will rush to the respective electrodes and form the so-called electric double layer. Its thickness, known as Debye length, is usually quite small. As for the rest of the solution, it will be effectively shielded from the electric field.
Whether or not this counts as a positive answer is up to you.

Voltage  may be referred to potential, simply as an ability to perform work (like in French or German, for example).  Only if this potential overcompensates a threshold, the electrolysis (in the meaning of discharge of ions, effective reaction of electrolysis of water) will start with noticeable, and continuous current.  Below this limit, at max the mentioned "catch".

To complete the answer of Ivan Neretin, everywhere on earth there is a electric field around us. And if you have a glasse of water in hand, there is an electric field in it but the potentiel will be still the same in all the water then they will not have electrolysis.
For more information look at feynman lectures on physics.


text:
reference or data on internal hydrogen bond strength
As suggested by the title, I am looking for data on the strength of different internal hydrogen bond configurations. In general, an internal hydrogen bond may form when the donor and acceptor are separated by 3-5 bonds thus forming rings of size 5-7. Below is a picture of the basic examples.

This is an oversimplification that is most relevant to small molecules (not proteins, etc). I'm sure there are situations where other conformational considerations make internal hydrogen bonds of other lengths possible.
At any rate, I am looking for some quantitative data regarding the strength of internal hydrogen bonds of different path distances. This means the same donor and acceptor atom types separated by varying distances like in the picture above. In general I would like to know how much more stable the path 4 configuration is compared to the path 3 and path 5.

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text:
What is the mechanism for the reaction of acetyl chloride and aniline?
I want to devise the synthesis of 1,2-dinitrobenzene without producing a large amount of 1,3 or 1,4 products.
In my synthesis I have a step in which I obtain aniline. I would like to know the mechanism for protecting the amine using $\ce{CH3COCl}$ and triethylamine so I can keep it from reacting in the next step of my synthesis. I'm not really "fluent" with amine chemistry, and I have no good idea at the moment.
If you cannot make picture of mechanism, please explain it with words. I must understand!

For the mechanism you don't need to be fluent with amine chemistry, you just need to know a bit about carbonyl chemistry. That said I think the nitration of acetalinide gives predominantly para product.

After this step I make a sulfonation and after the second nitration. And finally I found my mechanism, I just made a mistake at the begining.

The mechanism is just a regular nucleophilic attack of an amine on a carbonyl, with the triethylamine just there to pick up any loose protons at the end of the reaction.


text:
Is there air in the volume of crystallized sodium hydroxide?
I have a container of crystallized NaOH (98%) which has a volume of 500ml. Now, as per wikipedia, Sodium Hydroxide's density is 2.13g/cm3. This should give the weight of a 500ml container of NaOH at 1065g.
However, when measured, the crystalls weighed in at around 560g (minus the weight of the container). Is it because the NaOH is not in liquid or a homogenous crystal? Is the rest of the volume around the crystals air?

@ChinmayChandak Your comment is misleading - OP has powder, not single ions

@Mithoron understood my mistake. i miss understood the question... :(

Imagine a container with sand; it may seem "full", but surely you may pour in some water, and it will fit in the cavities. Same thing here, except in case of $\ce{NaOH}$ I'd rather not check it directly with water, for the response may be quite intense.
$\ce{NaOH}$ commonly comes in granules; if you'd have one big solid brick inside your container, that would be another story.

Have a look here about the different densities, too.


text:
how to store solvent in laboratory?
I am organising chemicals and solvents in laboratory... being less specific ( as it would a big list of chemicals to provide here in my questions)... I wanted to know... how can we arrange solvents in lab as there would be acids like HCl, h2so4, and at the same time there are other solvents like aniline, ammonia, acetic acid, methanol, benzyl alcohol, esters, mercaptoethanol etc... kindly provide some tips regarding there arrangements.

What do you mean by arrange solvents?

Just put them in a shelf where they can’t fall down and are well vented. Voting to close.

I'm voting to close this question as off-topic because hints on how to store solvents is not part of the scope of chemistry.stackexchange.com.

Also, storage may be (and likely is) location dependent because fire codes vary by country and jurisdiction.  Consult you local safety expert.

All the chemicals should be segregated according to their reactive properties eg. Bases with bases, acids with acids, ionic salts with ionic salts, oxidizing agents by em self etc. The storage should be dry, we'll ventilated and everything labeled correctly and stored in a way that's practical to access without causing a hazard.


text:
How many atoms of helium are in the chamber?
A $22.7 \,\mathrm{L}$ helium-filled chamber at $298\,\mathrm{K}$ has a tiny hole of area $1.63 \times 10^{-5}\,\mathrm{mm^2}$ punched in it through which is leaking helium at a rate of $2.25 \times 10^{15}$ atoms per second. How many atoms of helium are in the chamber?
I'm personally not too sure how to answer this, I'm thinking that it probably uses the root-mean-square speed or collisions with the wall equation ($Z_\mathrm w$). But I'm sort of lost when it comes to substituting the values and so on. (This is also a question on last years Gas Laws test, and I was planning on solving it for practice questions)
What I tried doing was using the Collisions with the wall equation:
https://en.wikipedia.org/wiki/Kinetic_theory#Collisions_with_container
then solved for $N$, but the answer is supposed to be $N = 9.98 \times 10^{21}$, and the closest I've gotten was getting $3.38 \times 10^{22}$, so I'm sort of stuck on the procedure of solving this question.

I absolutely hated this topic. Anyway can you identify which equation you need to use? Once you've done that, it's just a matter of making sure your units are consistent.

You already mentioned the equation for the rate of collisions of gas molecules with a section of wall:
$$Z_\mathrm w=\frac14\frac NV\sqrt{\frac{8RT}{\pi M}}A$$
We know the values of
rate $Z_\mathrm w=2.25\times10^{15}\ \mathrm{s^{-1}}$,
volume $V=22.7\ \mathrm l=0.0227\ \mathrm{m^3}$,
molar gas constant $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$,
temperature $T=298\ \mathrm K$,
molar mass of helium $M=4.003\ \mathrm{g\ mol^{-1}}=0.004003\ \mathrm{kg\ mol^{-1}}$, and
area $A=1.63\times10^{-5}\ \mathrm{mm^2}=1.63\times10^{-11}\ \mathrm{m^2}$.
Thus, the number $N$ of helium atoms is
$$\begin{align}
N&=\frac{4Z_\mathrm wV}{\sqrt{\dfrac{8RT}{\pi M}}A}\\[6pt]
&=\frac{4\times2.25\times10^{15}\ \mathrm{s^{-1}}\times 0.0227\ \mathrm{m^3}}{\sqrt{\dfrac{8\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm K}{\pi \times 0.004003\ \mathrm{kg\ mol^{-1}}}}\times1.63\times10^{-11}\ \mathrm{m^2}}\\[6pt]
&=9.98\times10^{21}
\end{align}$$

Thanks alot, this made sense mainly, the only part from this I didnt understand is, i have the same equation as you provided, but I didnt know that it could be multipled by A (area). Any reason why?

@Mathquestion-eer I can only speculate on differences. Maybe, the result $Z_\mathrm w$ in your equation is defined as ‘rate per area’. You can always check your calculation with the help of the units. When you use coherent units, conversion factors between units are never required ($1\ \mathrm J=1\ \mathrm{kg\ m^2\ s^{-2}}$). If the units cancel out so that the final result has the correct unit, this is a nice indication, but not a proof, that your calculation is correct. Note that the number $N$ is a quantity of dimension one (a quantity of dimension one is often called dimensionless).

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text:
How do I calculate the grams of a product after a reaction?

If you mix $50\,\mathrm{mL}$ of $2.5\,\mathrm{M}$ $\ce{HCl}$ with $100\,\mathrm{mL}$ $2.5\,\mathrm{M}$ $\ce{NaOH}$, you will get sodium chloride. If such reaction occurs only at $50~\%$, how many grams of sodium chloride will you get?

Well, I guess that the reaction is like
$$\ce{HCl + NaOH \to NaCl + H2O}$$
Since the concentration is the same for both reagents ($2.5\,\mathrm{M}$), I imagine that the concentration will be the same for the products?
So we got $2.5\,\mathrm{M}$ of $\ce{NaCl}$. However, to get the grams I first need to know the moles of $\ce{NaCl}$. I know that the concentration is $2.5\,\mathrm{M}$, but what of it? What can I do to get the moles of the resulting $\ce{NaCl}$?

Your idea about the concentration is a good start. Remember that you need moles in the end. How can you calculate moles? Which of the possibilities might be applicable here? And how would that prompt you to continue? ;)

Two notes on MathJax and notation: 1) chloride has a lowercase l; 2) you should include the unit of a value in the $ signs. To get the space either type ~ or \,. The unit itself should be enclosed in \mathrm{...} to force upright type.

The concentration $c$ of a solution is the molar quantity of the solute $n$ devided by the volume of the solution $V$:
$\begin{equation}
c = \frac{n}{V}
\end{equation}$
as you state, for example for HCl 2.5 mol/L (or, equivalent, 2.5 M).  Inversely, if you know volume and concentration of your solution, you may access the molar quantity of your solute.  Actually, there is such an equation for each of the solutions, indepent from each other; you may index the variables accordingly, too.
Yet watch, there is a catch: mixing the two solutions together increases the total volume, and simultaneously decreases the concentration of the specis in question -- without altering the molar quantity of $\ce{Na+}$, nor $\ce{Cl-}$ ions present.  Hence "2.5 M of NaCl" (as measure of concentration) in the solution in common is incorrect.
Because of the initially asked question, one of the solutions contains the limiting reagent.  Figure out which one -- this includes the stoichiometry in the equation of chemical reaction. Knowing the molecular amount $n$ of it, your are able to determine the amount of sodium chloride formed; initially as moles, later (via molecular mass) in grams.


text:
Get the molar mass of an unknown liquid in a dissolution

A dissolution has $10$ g of some unknown liquid and also $90$ g of
water. The freezing point of this dissolution is $-3.3\, ^\text{o}$C.
What is the molar mass of the unknown liquid in g/mol?

Alright. First of all, the formula to calculate the freezing point of a dissolution is
$$T_f - \Delta_f$$
Where
$$\Delta_f = \text{m} \cdot K_f$$
In our case, it would be
$$-3.33 = 0 - (\text{m}\cdot 1.86)$$
The molality m is
$$\frac{\text{moles of unknown liquid}}{\text{kg of water}} = \frac{\text{moles of unknown liquid}}{0.09}$$
We have to solve the equation:
$$-3.33=-\frac{\text{moles of unknown liquid}}{0.09}\cdot 1.86$$
Which yields
$$0.16 = \text{moles of unknown liquid}$$
However, I need the molar mass of this unknown liquid. How do I get it?

Well you know that you have 0.16 moles of the liquid. What else do you know about the liquid?

What am I missing? you got that 0,16 mol of the liquid have a mass of 10 g, so its molar mass is 62,5 g/mol

Your liquid could be  ethanediol $\ce{C2H4(OH)2}$, with a molar mass equal to $62$  g/mol.

The amount of your substance is $\pu{0.16 mol}$. The mass of your substance is $10\ \mathrm g$. So the molar mass is: $M=10\ \mathrm g/0.16\ \mathrm{mol}=62.6\ \mathrm{g/mol}$. With such a molar mass, your substance may be nitric acid $\ce{HNO3}$, whose molar mass is $\pu{63 g/mol}$. But this choice is improbable, because, in aqueous solution, nitric acid is partly dissociated into ions. It could be ethanediol $\ce{C2H4(OH)2}$, with a molar mass of $\pu{62 g/mol}$. So it is hard to guess the nature of your solute, without more information about its formula.

As long as you work with sufficently low concentrations, you do not need the molecular mass of your unkown.  Indeed, this was (is?) your initial question.
Freezeing point depression (salting the footwalk to melt ice in the winter), elevation of the boiling point (adding salt to cooking water that increases it's boiling temperature, too) and osmotic pressure are colligative properties.  What you need is to look up the crysoscopic constant, for water (same page) of $1.853 \mathrm{~K·kg/mol}$.
On the other side, if you already state the equation of $0.16$ moles corresponding to $10\mathrm{~g}$.
Despite advent of mass spectroscopy, such a determination may be still helpful today.

So low that it doesn't really change the melting point? Because if it changes it at all, something that weighs 50 g/mol will have twice the effect of something that weighs 100 g/mol, assuming the same mass.


text:
Gibbs Free Energy and ATP
Gibbs free energy describes the maximum of useful work that can be done by a chemical reaction. My question is(Sorry I am not into chemistry and biochemistry) : during a chemical reaction that creates ATP, we say that ATP stores energy between its chemical bonds . For exemple this is the equation of creating ATP: ADP + Pi + energy ----> ATP and of course this reaction has DG. What is the relation of DG with the stored energy? Is it the same?

No, it can't be the same.  If it were exactly the same, then it would mean that ATP was storing every last iota of available free energy.  Another way of saying that is that ATP synthesis would be 100% energy efficient.  That is, the free energy change $\Delta G$ in the reaction $\ce{ADP + Pi + energy -> ATP}$ would be zero.
But perfectly reversible reactions that have 100% efficiency never happen in the real world.  Even theoretically, they can only happen at infintessimally slow rates.
Actual ATP synthesis isn't 100% efficient.  It's efficiency depends on the details of the particular organism and its biochemistry, etc.  Many organisms show efficiencies of 30 - 45% in the synthesis of ATP.

The useful work (DG) here is what then?

I saw the link. so the other DG is converted into heat right? but the term TDS in gibbs is also a part of the heat that cannot activate or do work on another reaction right?


text:
Getting the molality of a solution given molarity and density

If a solution of $\ce{HNO3}$ has a molarity of 16 and a density of
  $1.42\ \rm{g/mL}$, what is the molality of the solution?

Well, a molarity of 16 means that there are 16 moles of solute for every liter of solution.
Based on density, if we have 1000 milliliters, we have 1420 grams of solution.
To calculate the molality I need:
$$\frac{16}{\text{kilograms of solvent}}$$
Since we have 16 moles, and the molar mass of $\ce{HNO3}$ is 63, we know that we have 1008 grams of solute, meaning there are 412 grams of solvent.
So I calculate the molality to be 0.038.
This answer is wrong because it is not among the options. What did I do wrong?

This appears to be a duplicate of multiple questions.

1 molar = 1 mole/(liter of solution)

@Max Your definition of molal is incorrect. It should be amount of substance of solute per mass of solvent.

1 molal = 1 mole / (kg of water)

If a solution of $\ce{HNO3}$ has a molarity of 16 and a density of
    $1.42\rm{g/mL}$, what is the molality of the solution?

Molarity is an old word for amount concentration and it is defined by the IUPAC Goldbook as

Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by  B. The common unit is mole per cubic decimetre ($\mathrm{mol\,dm^{−3}}$) or mole per litre ($\mathrm{mol\,L^{−1}}$) sometimes denoted by $\mathrm{M}$.

In mathematical terms $$c(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{V(\ce{HNO3(aq)}}.$$

 In this case the concentration is presumably given as $c(\ce{HNO3(aq)})=16~\mathrm{mol\,L^{-1}}$

The molality, $b$, is defined by the IUPAC Goldbook as

Amount of entities of a solute divided by the mass of the solvent.

In mathematical terms this means
$$b(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{m(\ce{H2O})}.$$
The (mass) density, $\rho$, is defined by the IUPAC Goldbook as

Mass of a sample or body divided by its volume.

In mathematical terms this means
$$\rho(\ce{HNO3(aq)})=\frac{m(\ce{HNO3(aq)})}{V(\ce{HNO3(aq)})}.$$

 In this case the density is given as $\rho(\ce{HNO3(aq)})=1.42~\mathrm{g\,mL^{-1}}.$

Now you can go ahead and derive the formula for conversion all by yourself, which I recommend, but you can also check if someone has already done that. I chose to be lazy today, so I checked Wikipedia:
\begin{align}
c &= \frac{\rho b}{1 + bM} &
b &= \frac{c}{\rho - cM}
\end{align}
Now you only have to figure out the molar mass of nitric acid, and you can calculate the result.

 I assume $M(\ce{HNO3})=63~\mathrm{g\,mol^{-1}}$.

Solution:

 You should find $b(\ce{HNO3(aq)})=38.8~\mathrm{mol\,kg^{-1}}$.

So in principle you have done quite good. If you would have used proper notation and included all the units in every step, you would maybe have noticed the mistake all on your own.
(Hover over the hidden fields to reveal their content.)

The concentration $c$, as fraction of molecular amount $n$ per volume $V$ takes into account both the volume of the solvent plus the volume(s) of the solute(s).  In other words, the total volume of the whole solution.  Eventually, to determine the molalilty, you divide the number of moles of the solute per kilogram solvent.


text:
Converting Disolved Oxygen from micromole/L to millimole/kg
I have dissolved oxygen data that is in units of µmol/liter.
I need to convert it to mmol/kilogram.
I have been told by someone who knows more about chemistry than I do (but only a little bit more) that I need to know the mass of my sample as well (which I don't know).
So only knowing my original DO data (in µmol/L) and the density of SW (assume 1.027 g/cm³), how does one convert the data to mmol/kg?

Depends a lot on what the oxygen is dissolved in. If it is in water, then the conversion l → kg is easy.

The O2 is dissolved in seawater.

Oxygen is barely soluble in water, so honestly, treating the saltwater as though it were "pure" saltwater with no dissolved oxygen is roughly equivalent to treating it as heavily oxygenated.  The difference is in the fourth or fifth decimal place.  Note that this would start to fall apart if you are concerned about the solubility of high pressure oxygen in water, but for small quantities of oxygen (such as the micromoles and millimoles you're using) this is likely not a factor.
The conversion is relatively straightforward.  For a solution with concentration $X$:
$$(X\ \mathrm{\mu mol/L}) \times (1\ \mathrm{mmol}/1000\ \mathrm{\mu mol})\times(1\ \mathrm L/1.027\ \mathrm{kg}) = 0.0009737\ X\ \mathrm{mmol/kg}$$

Thanks for that. So to be clear, the conversion factor is 0.0009737? AKA- If I had an O2 measurement of 23umol/L, I can just multiply by 0.0009737 to get 0.00223951 mmol/kg?

@Vinterwoo Yep, that's it, though you'd really only know the first two digits, rounded.  (i.e. 0.0022 mmol/kg)


text:
Why does it not matter what mass of NaOH is used to solve for q?
So we recently did a lab in Chem to prove Hess's law. The reactions we performed were
$$\begin{align}
\ce{NaOH + HCl &-> NaCl + H2O}\tag1\\
\ce{NaOH(s) + HCl &-> NaCl + H2O}\tag2\\
\ce{NaOH(s) &-> Na+ + OH- }\tag3
\end{align}$$
One of our post lab questions asks us to describe what would happen to the $q$ and $\Delta H$ of reaction $\text{(3)}$ if twice the mass of $\ce{NaOH(s)}$ were used.
For the third reaction, 4.1 degrees was the temperature change, 2 grams of $\ce{NaOH(s)}$ were used, and $50\ \mathrm{mL}$ of water were used. This gave me
$$q = (52~\mathrm g \times 4.18 \times 4.1) = -891.176 ~\mathrm J$$
and
$$\Delta H = (-891.176~\mathrm J/0.05~\mathrm{mol}) = -17824 ~\mathrm{J/mol}$$
When I used twice the mass of $\ce{NaOH(s)}$ like the problem asked, my results were
$$q = (54~\mathrm g \times 4.18 \times 4.1) = -925.452 ~\mathrm J$$
and
$$\Delta H = (-925.452/0.1~\mathrm{mol}) = -9225.908 ~\mathrm{J/mol}$$
Considering my results, why does the very next problem say that in reaction $\text{(3)}$, it doesn't really matter what mass of $\ce{NaOH}$ is used? Why would it not matter? My $\Delta H$ and $q$ values were different. I'm convinced that I made an error in my calculations, so I'd greatly appreciate some clarification here.
Also, it asks why is it important that $2\ \mathrm g$ of $\ce{NaOH}$ be used in reaction $\text{(2)}$? Is it because it's reacting with $\ce{HCl}$ instead of water like in reaction $\text{(3)}$?

Were you supposed to calculate the change in enthalpy per mole of reactant?

The temperature rise was exactly the same in both cases?

@JasonPatterson The question doesn't explicitly say that, but I think so.

@chestermiller The question just asks us to assume that twice the mass had been used. It makes no mention of other factors.

Well, you have to think that, if twice as much NaOH dissolved, more heat would be given off and the temperature change would be higher (in this case, about twice as high).

I don't understand your final question. A priori it is not important whether you use $2,\mathrm{g}$ of $\ce{NaOH}$ or any other arbitrary amount. There have to be clues you’re not giving us.

What's the '4.18' mean, and why are you multiplying it by 52 g or 54 g? Units might help, especial ones which include "__ of  X", where X is water/NaOH/Na+/etc. The assumptions you're making here versus the ones the book is making might explain discrepancies.


text:
Conformation identifying cis and trans
How to identify cis and trans in conformers of cyclo hexane.i know that there are two positions axial and equatorial but using this data how to identify cis or trans? Also why are functional groups in equatorial positions more stable?

http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkanes/Properties_of_Alkanes/Cycloalkanes/

Draw the cycle as if it were planar (in fact it is not, but just imagine for a moment), with one hydrogen at each carbon pointing upwards with respect to the plane, and the other downwards. Now, if we introduce two substituents to different carbons on the same side of the plane, that would be the cis isomer, otherwise trans.
As for the stability of equatorial conformers, that's purely for sterical reasons. Draw the cycle, this time in its real geometry, and see what happens if we have two bulky groups in equatorial positions on the same side. Clearly they will be unpleasantly close to each other.


text:
By what mechanism do acids deprotect primary silyl ethers?
Standard protocols report the selective deprotection of primary silyl ethers in the presence of secondary silyl ethers under acidic conditions, as exemplary shown in scheme 1, a reaction by Hartmann and Kalesse.[1]

Scheme 1: Selective deprotection of a primary silyl ether[1]
Numerous other examples can be found in the literature using numerous acidic reagents such as campher sulphonic acid,[2] para-tolylsulphonic acid[3] and even acetic acid[4] or $\ce{HCl}$.[5] Similar conditions can be used in the case of TIPS ethers.[citations needed]
I was wondering what the mechanism of this transformation looks like, and especially why it seems to be selective to primary silyl ethers. Answers that reference papers actually investigating the mechanism are preferred, but others are welcome, too.

List of abbreviations:
b.o.r.s.m: based on recovered starting material
Me: methyl
PPTS: pyridinium para-tolylsulphonate
TBS: tert-butyldimethylsilyl
TIPS: tetra-iso-propylsilyl

References:
[1] O. Hartmann, M. Kalesse, Org. Lett. 2012, 14, 3064. DOI: 10.1021/ol3011387
[2] Y. Mori, K. Yaegashi, H. Furukawa, J. Am. Chem. Soc. 1997, 119, 4557. DOI: 10.1021/ja9701523
[3] S. Kiyooka, K. A. Shahid, F. Goto, M. Okazaki, Y. Shuto, J. Org. Chem. 2003, 68, 7967. DOI: 10.1021/jo034901c
[4] B. H. Hart, S. K. Verma, H. Rapoport, J. Org. Chem. 2003, 68, 187. DOI: 10.1021/jo026499s
[5] A. B. Smith III, B. S. Freeze, I. Brouard, T. Hirose, Org. Lett. 2003, 5, 4405. DOI: 10.1021/ol035697i
[citations needed] I think, you get the picture, I’m not going to go looking for even more …

Fluoride-mediated deprotection of silyl ethers proceeds through a pentavalent silicon pathway:[1]

This mechanism proceeds the same for both primary and secondary silyl ethers, and there's no reason the mechanism for acid-catalyzed deprotection would be any different. Though I couldn't find a mechanism proposed in literature, it seems likely it would proceed through a similar pentavalent intermediate (Nu = nucleophile):

Steric effects are what drive the chemoselectivity of these reactions. Usually the differences in steric environments around the carbinol carbons is enough, but the use of different silyl groups on the two alcohols can further increase yield of the desired product. These steric effects are particularly pronounced in the following transformation, in which a primary TBS ether is selectively deprotected in the presence of a primary TBDPS ether and a secondary TIPS ether:[2]

Reaction conditions can also have a major effect on the chemoselectivity of desilylation reactions. For example, when HF•pyr is used in a THF-pyridine mixture, selective removal of the primary TBS ether in a protected triol was achieved in 84%. After oxidation of the newly deprotected alcohol to a carboxylic acid, the secondary TBS ether was cleaved using HF•pyr without excess pyridine, leading to lactonisation:[3]

Nelson and Crouch have written a review on the selective deprotection of specific silyl ethers in a compound with multiple silyl protecting groups.[4]

References

Kim, S.-K. Chitin and Chitosan Derivatives: Advances in Drug Discovery and Developments; CRC Press, 2013; p 80.
Larivée, A.; Unger, J. B.; Thomas, M.; Wirtz, C.; Dubost, C.; Handa, S.; Fürstner, A. The Leiodolide B Puzzle. Angew. Chem. Int. Ed. 2011, 50 (1), 304–309 DOI: 10.1002/anie.201005850.
Körner, M.; Hiersemann, M. Enantioselective Synthesis of the C8−C20 Segment of Curvicollide C. Org. Lett. 2007, 9 (24), 4979–4982 DOI: 10.1021/ol702092h.
Nelson, T. D.; Crouch, R. D. Selective Deprotection of Silyl Ethers. Synthesis 1996, 1996 (09), 1031–1069 DOI: 10.1055/s-1996-4350.


text:
How is a carboxylic acid acidic?
Carboxyl groups are acids that comprise a carbon double bonded to an oxygen and single bonded to an oxygen-hydrogen. The one oxygen gets the two electrons it wants (so to speak) from the double bond with the carbon, the other gets the two it wants from the bond with the carbon and the hydrogen. The hydrogen is also satisfied by its bond with the oxygen. The only electron stealing opportunity I can see is in the polarity of the hydrogen. Is that what the acidic property of carboxy groups consists in?

The title had a typo and it wasn't clear what the intended meaning was. Please feel free to make an additional edit.

There is a lot in chemistry that cannot simply be explained by Lewis structures, usually requiring a detailed molecular orbital description to rationalise observations.
For example considering a carboxyl group versus a simple hydroxy group. The carboxyl group has a $\mathrm{p}K_\mathrm{a}$ of approximately $5$ while the hydroxy group has a $\mathrm{p}K_\mathrm{a}$ of approximately $15$. In Lewis structures, these two look identical save the neighbouring carbonyl group.
Why can this hydrogen be abstracted easily at all? The hydrogen on a neighbouring carbon (e.g. $\ce{C\mathbf{H}3COOH}$) is not easily abstracted ($\mathrm{p}K_\mathrm{a} \approx 25$ if I recall correctly). The Lewis structure in itself provides no information. You need to additionally consider the concept of polar bonds to explain it: The $\ce{O-H}$ bond is polarised towards oxygen, that means the hydrogen already feels as if it were somewhat like a $\ce{H+}$ and therefore the transition to the true $\ce{H+}$ is easier.
While that explains the acidity per se, it doesn’t explain why one is more acidic than the other. The theory says, that an acid is stronger if its conjugate base is more stable. Since we cannot rationalise the different acidities from looking at the acids, let’s look at the conjugate bases. In one, we just generate an alcoholate much like a hydroxide ($\ce{R-O-}$). Nothing special-looking. The same for a carboxylic acid. No distinction. (However, this clearly shows why $\ce{C-H}$ bonds are usually not acidic: the carbanion is very unstable when compared to an oxyanion.)
Since we cannot find the origin of acidity, we need to expand our theory. Specifically, this is where resonance comes into play as a partial representation of some features of molecular orbital theory in Lewis structures. The carboxylate can resonate between two different structures as shown in the scheme below. Therefore, every oxygen atom only has to take the burden or half a negative charge ($\ce{O^{\frac{1}{2}-}}$). This is much more stable than having a full negative charge — even for an electronegative element such as oxygen. The third structure of the scheme simply displays a way of drawing ‘something in-between’: Both bonds are somewhere between a single and a double one and the charge is located on both atoms simultaneously (but that is not easily drawn with chemical drawing programs).


text:
Can I extract water from diethyl ether to get anhydrous diethyl ether?
I am trying to get anhydrous diethyl ether. So, I am considering using a condensor to heat up diethyl ether and extract water. Would this method work?

No. However you can get your ether very water-poor by drying it over sodium sulphate or magnesium sulphate. Truely dry ether can only be obtained by rigorous drying (e.g. sodium) and subsequent destillation; to be stored under inert gas. Luckily, most suppliers sell dry solvents over molecular sieve with water contents less than 50 ppm. Also, welcome to chemistry.stackexchange.com. Feel free to take a [tour] and visit the [help] for questions about the site.

I generally prefer to pay a little more for small containers of anhydrous ether and then use them up quickly. Once the container is open, your ether is not dry anymore.

I am trying to synthesis phenylmagnesium bromide and I have to use anhydrous ether. Would it matter if I use normal ether for this synthesis?

@EdmondLee The definition of "have to use X" is that it does matter if you don't use X. So I don't understand how you can simultaneously believe that you "have to use anhydrous ether" and that it might be OK to use water-containing ether.

What @DavidRicherby said. Of course, there's a difference between "have to use X or the yield will suffer" and "have to use X or it'll catch fire and explode." If the OP doesn't have convenient access to anhydrous ether, they'll need to figure out what level of "have to" applies here.

No it would not work.  Ether boils at a lower temperature than water.  Plus, it forms an azeotrope with water:  a composition of 98.7 % (by weight) ether boils at 34.2 °C and the vapor phase has identical composition as the liquid, meaning that distillation cannot remove the final 1.3% of the water (at atmospheric pressure).
The standard way to remove water is (i) first to use drying agents such as magnesium sulfate to remove most of the water, and then (ii) if needed, eliminating the final traces of water with small amounts of sodium (or potassium) metal.

Ether is notoriously difficult to remove all traces of water from. Don't even bother unless you have a means to store dry ether without its contacting air, from which it will extract humidity very rapidly. I would recommend using a desiccant such as calcium sulfate (drierite) or a molecular sieve, and let it stand for a few hours. You could then either decant or distil the dry ether, but distilling it with a desiccant filled drying tube on the receiver vent, or under an inert gas blanket, is a must.


text:
Why does technetium not exist in nature?
Why does technetium not occur in nature? It is the only element before bismuth to net exist in nature. What's the reason for that?

It's radioactive.

It's not that big; also, other nuclei, some of which are just a little bit smaller, and some are a little bit bigger, and some are much bigger, are stable. I'd say technetium is a victim of rare coincidence, much like promethium.

This is more suited perhaps for Physics SE, where you would get some nuclear physics folks to discuss nuclear structure and stability. And, you would find this: http://physics.stackexchange.com/questions/40960/why-is-technetium-unstable

Technetium does occur in nature. From the Los Alamos page on technetium:

Technetium was the first element to be produced artificially. Since its discovery, searches for the element in terrestrial material have been made. Finally in 1962, technetium-99 was isolated and identified in African pitchblende (a uranium rich ore) in extremely minute quantities as a spontaneous fission product of uranium-238 by B.T. Kenna and P.K. Kuroda.
Technetium has been found in the spectrum of S-, M-, and N-type stars, and its presence in stellar matter is leading to new theories of the production of heavy elements in the stars.

The journal article describing the 1962 discovery is Technetium in nature. B. T. Kenna and P. K. Kuroda.  Journal of Inorganic and Nuclear Chemistry 26(4) 493-499 (1964).

While minute quantities of Technetium are present in nature they are always the product of recent synthesis from other elements by nuclear decay (or neutron capture).
The reason why Technetium is so rare is that its most stable isotope has a half-life on only about 4m years. Therefore any that was present at the creation of the planet (about 4,500m years ago) will have decayed a long time ago. The short half life implies that the only reason any can be present in nature now is if there is some ongoing process that creates it. And that is what we find.
Almost all technetium in the current world is created via nuclear reactions either as a by-product of nuclear power or deliberately to be used in medicine.

But there is an ongoing process (in Earth), namely spontaneous fission of uranium in pitchblende.  Read Todd Minehardt's answer.

There is a law in nuclear physics, saying that nature prevents two neighbors isobars (nuclides having same mass but differing by only one proton) from being both stable. One of these nuclides must decay through beta-plus or beta-minus into the other one.
Surprisingly the neighbors of Technetium, namely Molybdenum and Ruthenium are made of a great number of stable isotopes, without any holes in the series of masses. So Technétium has no possibility to get at least one stable isotope.

authors: سيف سعد وحيد


text:
Micelles absorb fat, is there an adjective for this?
"Hygroscopic" means the capacity to absorb water. Is there an adjective to describe the same idea except for the absorption of lipids? Like, when you wash your hands with soap and the soap micelles absorb the bacon grease, for example.
I'd like to say that micelles have ____lipid-absorbing_____ properties.

Well... Lipophilic?

Amphiphilic, rather than lipophilic, in the particular case of soaps/surfactants, though amphiphilicity implies lipophilicity. There's no real analogy to hygroscopicity, though, since lipids generally are not found in the atmosphere / gas phase.


text:
Why is there no neutron in protium?
My chemistry teacher once told me, that every atom should have at least as many neutrons as protons, to keep the electrons from crashing into the nucleus.
Protium, however, doesn't even have neutrons to begin with. How does this work? Or was my chemistry teacher wrong about this?

Your chemistry teacher is wrong. On the other hand, protium is the only stable element with no neutrons but this is more to do with the fact that there are no protons repelling each other in the nucleus, rather than stopping the electrons crashing into the nucleus. That problem is solved by recognising that electrons' energies are quantised.

Neutrons have nothing to do with the stability of electrons in atoms. They do contribute to the stability of nuclei but that is a different and complicated subject.

There is no neutron in protium by definition. The question is, why is it so stable and abundant form of hydrogen, I guess.

Your chemistry teacher is making a few simplifications there that make the statement false on a black-and-white true-and-false scale. Protons would repel each other electrostaticly due to their same charges. Neutrons interact with protons by the so-termed strong interaction (because it is stronger than the weak interaction; props to physicists for inventing names) which lead to stability of the nucleus, preventing it from falling apart due to the repulsion.
Hydrogen, specifically the $\ce{^1H}$ isotope, does not need neutrons to stabilise the nucleus because one proton cannot repel itself.
Note that ‘electrons crashing into the nucleus’ is a very strong simplification that you should eliminate from your language. Electrons are waves for all that matters so they are all around the nucleus at the same time. In fact, the highest probability of detecting them is right there, in the nucleus. (On a simpler scale, Bohr once invented ‘orbits’ that you can rationalise much like the orbit of a satellite, which also does not crash onto Earth. But don’t tell a quantum chemist or nuclear physicist that I said that!)
Finally, it is wrong to say ‘at least as many neutrons as protons’. $\ce{^3He}$ is a stable isotope with less neutrons than protons in the nucleus.

@Jan's put up a nice and crisp, answer. My answer is merely a supplement to that ;)
It's because the hydrogen ("protium", if you will) nucleus has no need for neutrons.
What's this "need for neutrons", you ask?
Every element (apart from hydrogen) has multiple protons (it's the number of protons in a given nucleus that allows physicists and chemists to determine that a given atom is a so and so element: the Atomic Number).
Now protons are positively charged, and as you would know, positive (like) charges repel. If you punch in the values for the magnitude of charge on a proton and the distance of separation between two (or more) protons within a nucleus in a certain equation built from Coulomb's Law, you'd find that the force of repulsion between the protons is ridiculously huge! The nuclei, in theory, shouldn't be able to exist...but they do.
Now we know, for a fact, that every 'multi-proton' element (i.e- every element apart from hydrogen) has these neutral particles called neutrons in them. Previously, we didn't factor in neutrons when we brought up the topic of proton-proton repulsion...which makes sense, since they don't have a net charge.
The guys in white lab-coats have already determined the masses of protons and neutrons (outside of the nucleus) and realised that they are (roughly) the same . Plus, we also know (from experiments) how many protons and neutrons constitute each element on the Periodic Table. So it would make sense to assume that if a proton weighs one unit and if a neutrons weighs one unit too, then an atom having $X$ protons and $Y$ neutrons will weigh $(X+Y)$ units. However, (more) experimental evidence suggests that this is not true. In fact, the mass of a given atom (that isn't hydrogen) is less then the sum of the masses of the individual protons and neutrons constituting it (the fancy term for this is "mass defect").
Take for example, the Helium nucleus (we call it the "alpha particle" too):

^ Pulled that off Google Images.
Now this is weird, almost paradoxical. We now have two issues with nuclei (other than hydrogen's):
1) On account of inter-proton repulsion, the nucleus shouldn't even exist. It'd break up and fly in all sorts of directions
2) The mass of a nucleus is less then the total mass of protons and neutrons that constitute it...which is pretty counter-intuitive.
Whoa, that's a real, hot mess we've found ourselves in.
But then...there was hope.
Some chemists and physicists (often, they were the same people) dug up a few notes and found this:
$$\mathrm{E=mc^{2}}$$
Which you might (hopefully) have seen somewhere. It's Einstein's Mass-Energy equivalence relation. Essentially, what it tells us is that any piece of matter can be converted into a corresponding amount of energy.
Click
Wait, protons in a nucleus should fly apart, but they don't. So there must be something holding them together!
Ding!
And a nucleus (with all its protons squeezed together) has less mass than it should have if you counted the mass contributions from protons and neutrons individually. But since mass is conserved, that "missing" mass must've been converted (according to Einstein's relation) to energy!
Ding, Ding!
So, this energy must be what's holding the nucleus together and preventing everything from going poof!
Ding, Ding, Ding!
So in a nutshell;
The "mass defect" isn't an anomaly! It's what keeps the protons from flying apart! The mass-defect arises because the nucleus has neutrons in handy. So if a nucleus has more than one proton in it (again, that only applies to elements other than hydrogen) it'll say "Whoa! Keeping these positively charged guys together's pretty tough! Luckily, I've got these "neutral" guys around, so I'll just borrow some of their mass, turn that to energy, and woo! Now I can pin these guys down easily!"

Back to the question:

Q-Why doesn't hydrogen have any neutrons?

Ans- Because it doesn't need them.

Q- Why doesn't it "need" them?

Ans- Because in hydrogen, you've only got one proton in the nucleus. So no repulsion, and no need for neutrons. Period.

Note: This answer is a bit of an oversimplification, but it should be sufficient to quench your curiosity ;)


text:
Hand Sanitizer Chemical Changes
I keep a bottle of hand sanitizer on my desk at work, as I touch a lot of things throughout the day. When I used it yesterday, a bit must've been left in the pump track or opening. So I was about to use it and I see that in the opening it looks opaque, kind of like lotion. I press the pump and it releases the opaque portion followed by the normal, gooey sanitizer.
What caused this chemical change? The air conditioning in our building is still on, but at about 6 p.m., the HVAC is turned off. Our lowest temps at night are between the 40s and the 50s.
Please clarify what is going on, particularly if something is wrong with this sanitizer and I need to throw it away.

Thank you, Bon. The ingredients are ethyl alcohol, water, isopropyl alcohol, caprylyl glycol, acrytates/C10-30 alkyl acrylate crosspolymer, glycerin, fragrance, aminomethyl propanol, isopropyl myristate and tocopheryl acetate. I will work on better tags.

I don’t think the sanitiser’s gone bad. At the same time, $99.99999999999\dots~%$ of people who use a hand sanitiser don’t need to. (And probably shouldn’t because it’s bad for the skin, but that’s beside the point.)

Given the list of ingredients in Pat's comment, I would strongly doubt that a chemical reaction has occurred here, and suggest that most likely the volatile solvents ethanol and isopropanol have evaporated, leaving a cloudy residue of the remaining ingredients. Your concern of a possible hazard is appreciated, but I would say no worries.


text:
Definition of Electronic energy
Using Born-Oppenheimer approximation, we separate the nuclear and electronic parts of the wavefunction, and treat them separately. We include the terms that have direct dependence on the electronic positions in the electronic Hamiltonian,
$$\hat{H}_{e} = \hat{T}_e + \hat{V}_{eN} + \hat{V}_{ee}$$
Thus, the total Hamiltonian is given by:
$$ \hat{H} = \hat{H}_{e} + \hat{V}_{NN} $$
So, what is the definition of electronic energy? According to the source mentioned below, electronic energy does not include $\hat{V}_{NN}$, which makes sense. But, all the electronic structure packages quote the Total energy as the electronic energy. The following is quoted from Gaussian website:
Zero-point correction=                  .023261 (Hartree/Particle)
Thermal correction to Energy=           .026094
Thermal correction to Enthalpy=         .027038
Thermal correction to Gibbs Free Energy= .052698
Sum of electronic and zero-point Energies=-527.492585
Sum of electronic and thermal Energies= -527.489751
Sum of electronic and thermal Enthalpies=-527.488807
Sum of electronic and thermal Free Energies=-527.463147

Why is this discrepancy? What is the actual definition of electronic energy?
Source for my argument: http://vergil.chemistry.gatech.edu/notes/quantrev/node31.html

You can find nuclear repulsion energy in gaussian log file under "Nuclear Repulsion Energy"

Electronic energy is indeed an eigenvalue of electronic Hamiltonian, and, as you correctly pointed out, it doesn't include the contribution from repulsions of the fixed nuclei (which has nothing to do with electrons). But sometimes the total energy for a particular nuclear configuration (one that includes the above mentioned interactions contribution) is still (incorrectly) termed "electronic energy" while just an eigenvalue of electronic Hamiltonian is called "pure electronic energy". That is essentially wrong, but it is sort of a custom.
And I would not say that "all the electronic structure packages quote the Total energy as the electronic energy". Did you check all of them? For instance, I quickly checked one of my DALTON calculations, and found this:

@    Final DFT energy:           -210.248305383291
@    Nuclear repulsion:           158.997839473485
@    Electronic energy:          -369.246144856776

So the first one is the (total) energy, then come its two components with the right names.
Same for ORCA:

Total Energy       :         -210.25104538 Eh           -5721.22181 eV

Components:
Nuclear Repulsion  :          158.99783872 Eh            4326.55115 eV
Electronic Energy  :         -369.24888411 Eh          -10047.77296 eV

GAMESS-US (here first two terms = electronic energy)

        ONE ELECTRON ENERGY =    -596.6619030866
        TWO ELECTRON ENERGY =     228.7817683952
   NUCLEAR REPULSION ENERGY =     158.9978510234
                              ------------------
               TOTAL ENERGY =    -208.8822836680

NWChem (middle three terms = electronic energy)

         Total DFT energy =     -210.248032406037
      One electron energy =     -597.377034042581
           Coulomb energy =      258.596296569204
    Exchange-Corr. energy =      -30.465145956066
 Nuclear repulsion energy =      158.997851023405

And I don't want to continue, since I'm pretty sure that most of the programs use the right terms.

Thank you very much for answering to the point. So, it is the 'you know who' package which is notorious! Also, Q-Chem does the exact same thing! It was wrong on my part to conclude that all packages do it this way! Thanks again!


text:
Syn-addition or anti-addition to fumaric acid?
In my textbook, the addition of alkaline $\ce{KMnO4}$ to fumaric acid (trans) is stated as a syn-addition even though $\ce{OH}$ is present on opposite sides in the product.
But,the defination of syn-addition is:

Addition of electrophile and nucleophile on the available sites on the same side of molecule.

Isn’t this contradictory?

Ever heard of rotation around single bonds?

@Mithoron,yes.Molecules can freely rotate around single bond.

I don’t know what molecule you ended up with, but the way I draw it, I end up with (2S,3S)-2,3-dihydroxysuccinic acid as shown below.

The syn-addition step would technically be the addition of the electrophile permanganate to the double bond. From there, the oxygens are in place and are not moved. If you reorder the structure to the identical zig-zag representation, you should arrive at a clear syn-orientation.
Note that there are two rotations in this: the first step includes a rotation of fumaric acid by $90^\circ$ so that I can draw the five-membered ring in a plain. After cleaving off $\ce{MnO4^3-}$ (which further reacts to $\ce{MnO4^2-}$ in basic conditions) I arrive at the syn-product.
With maleic acid, everything would be the same, except for the orientation of the carboxylic groups in the second and third structures. That would lead to the hydroxyl groups pointing in different directions even though we are still dealing with a syn-addition.


text:
How are pi-orbitals delocalized in a porphyrin chain?
According to this paper, the orbitals appear delocalized mostly from left to right, and only along 1-2 units of the chain.
Is that really the case? If so,

Why is the delocalization only along 1-2 units of the chain and not throughout the chain?
Why is there no delocalization in the up-down direction?

What you are seeing are molecular orbitals which, by definition, extend across the entire molecule, restricted only by inherent symmetry. It is important to note and is not clearly visible in your image that you are looking at a $\ce{H2-Fe-Zn-H2}$ system, i.e. from one side to the other the centres of the porphyrin system contain two hydrogens, an iron ion, a zinc ion and another two hydrogens. Replace the zinc with iron or vice-versa and you would get an axis of symmetry with molecular orbitals that are either perfectly symmetric or perfectly antisymmetric to this axis of symmetry.
As it is drawn, the molecule is $C_\sigma$, i.e. there is a single plain of symmetry along the plain of the molecule. All orbitals must be symmetric or antisymmetric to this plain, and since we are looking at a $\unicode[Times]{x3C0}$-system, it will be antisymmetric.
All the rest is the result of a calculation! So we cannot really answer why, all we can say is ‘because that is what maths gives us’. But we can slightly rationalise it. First, there is a difference between the top and the bottom, it is just not as easily visible. This is because the only difference between the top and the bottom is ‘where exactly are the two hydrogens and the double bonds?’ Second, the two ions are clearly different ones with different electronegativities that create entirely different environments. Therefore, it is understandable that one orbital is centred more around one of the two ions and the other around the other. And third, the LUMO$+1$ is clearly delocalised across the entire molecule.

Why do you say that the LUMO+1 is delocalized across the entire molecule if in the drawing it appears discontinuous?

@Sparkler Don’t confuse a nodal plain where the wave changes phase with a discontinuity.

Is that really the case?

That is, in fact, difficult to say on the basis of the study you referenced, since I would not blindly trust the Hartree-Fock results presented there. Try to search for more recent studies done at a higher (DFT or MP2) level of theory.1
But if, for the sake of argument, we pretend for a moment that HF picture presented in the study is qualitatively right, I have to say: you ask the wrong questions! Because, the only right answer to both of your "Why?" questions is: "Because that what's calculations tell us!" We are doing calculations to explain what's going on out there, so you already have the answers. If you want to go further than that, then your probably looking for one of two things:

You either want to go deeper and ask "Why nature works this way?" But this is not a deep scientific question, it is simply not a scientific one at all. Such questions are beyond science, it's a metaphysics already, or philosophy, if you will.
Or you want to go in the opposite direction looking for some simple rationalization of the orbital picture in the usual chemical terms (such as, for instance, resonance, hyperconjugation, steric effects, etc.) I suspects this is also impossible in this case, since all these (over)simplified chemical models usually work well only in the simplest cases. That is, in fact, why chemist do calculations in the first place.

1) About a decade ago (when the paper you quote was published) DFT/MP2 calculations of such relatively big molecules were basically out-of-reach, but with all the progress in RI-DFT/RI-MP2 methods they could be more or less routinely done these days.

Can you please add a few links you'd recommend reading about the theories you mentioned? (RI-DFT/RI-MP2)

@Sparkler RI-MP2, RI-DFT.