text:
Required temperature for the reduction of iron oxide using carbon
When reducing iron oxide with carbon, must the substances be set alight or do they only need to be heated. Also, when the iron is formed will it immediately oxidise in the air?

What is the difference between "set alight" and "be heated", really?

The direct reduction with carbon is only possible if the iron oxide ($\ce{Fe2O3}$, $\ce{Fe3O4}$, or $\ce{FeO}$) is present as a liquid.
$$\ce{FeO$_n$ + $n$C -> Fe + $n$CO}$$
Accordingly, the required temperature is very high. Note the melting points of $\ce{Fe2O3}$ (1539 °C), $\ce{Fe3O4}$ (1597 °C), $\ce{FeO}$ (1377 °C), pure $\ce{Fe}$ (1538 °C), and the eutectic of iron containing 4.3 % carbon (about 1150 °C).
In the blast furnace process, solid iron oxides at lower temperatures are mainly reduced with carbon monoxide.
$$\ce{FeO$_n$ + $n$CO -> Fe + $n$CO2}$$
The required carbon monoxide is produced from the exothermic reaction of carbon with oxygen
$$\ce{2C + O2 -> 2CO}$$
and from the endothermic Boudouard rection of carbon with carbon dioxide.
$$\ce{C + CO2 <=> 2CO}$$
A temperature well above 700 °C is required to favour the formation of carbon monoxide. Hence, a part of the carbon has to be burned in order to generate the required heat and to produce carbon monoxide.
Hot fine-dispersed iron would be quickly oxidized in air (for example, steel wool can burn in air) whereas fused iron is only oxidized at the surface of the liquid. Note that the pig iron produced in the blast furnace process has a high carbon content, which significantly lowers the melting point.


text:
How many significant numbers(figures) are in repeating decimal (105-32)×5/9=40.5555...?

Source: Basic Chemistry, Steven S. Zumdahl, Donald J. DeCoste
Referring to the above, rule it seems $(105-32)\times\frac{5}{9}=40.5555\ldots$ has infinitely many significant figures. Am I correct?

You should base the number of significant digits in your answer on the number of significant digits in your starting numbers.

@Shafter How should I base the number of significant digits?

Your example is pure math, where everything is exact. For a real world chemistry problem, you have to give a context, and the error of each starting number. And then do the proper math to calculate the error propagation.

As GumpyCede notes in his answer, if you are calculating the result from the pure numbers $105$, $32$, $5$ and $9$, then yes, the result has an infinite number of significant figures.
If one or more of those values is derived from some sort of measurement(s), though, then the significant figures of those measured value(s) would enter into consideration and the repeating decimal would need to be truncated appropriately.

1457 has four nonzero integers, and captive zeros always count as significant figures. 105 is not defined as 105 in = 266.7 cm, so 105 has three significant numbers, 1, 0, 5.

All numbers (105, 32, $\frac{5}{9}$, and 40.555...) have infinitely many significant figures because there is no uncertain digit.

As for, 105 and 32, when considering captive zero and nonzero integers they have three and two significant figures; the 1 and the 0 and the 5 in the fist, the 3 and the 2 in the second.

source: Stack Exchange

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

sort: votes


text:
Electronegativity (Oxygen and Phosphorus)
What makes an element electronegative. I understand, for example, $\ce{P(OR)3}$ is less basic than $\ce{PR3}$ because the oxygen in $\ce{P(OR)3}$ is more electronegative than P, so it pulls electron density away from the phosphorus, making the lone pair less available, and so, in turn making the phosphite less basic than the corresponding phosphine.
But, what about the oxygen makes it more electronegative than the phosphorus? Is it because of effects of shielding and penetration(?)* that makes oxygen better at pulling electron density. (because it has a higher Zeff)
*I don't think I fully understand what penetration is,so if does have something to do with penetration, could you explain from which orbital will the electrons penetrate to and from, directly referring to oxygen and phosphorus.

I'm afraid you don't understand * for a reason, and I've got no idea what would it supposed to be.

The presence of electron in shells of other electrons is called penetration.

I'm afraid it isn't.

For example, electrons occupying 2s orbitals can be found the 1s shell (and so also the core) because s orbitals have non-zero electron density at the core...but, 2p orbitals cant penetrate to the core, so 2s has a lower energy and experiences less shielding... To be honest, i'm not sure how penetration would be applied here...i was just making a suggestion...

http://chemistry.stackexchange.com/questions/13677/electronegativity-of-potassium-pauling and https://en.wikipedia.org/wiki/Electronegativity looks at least A-R scale is based on Zeff and covalent radius

I just thought, since effects of penetration and shielding increases the effective nuclear charge, (then the oxygen would have a higher Zeff than the phosphorus)  then the lone pair attached to the P would be more attracted to the oxygen...

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


text:
Hot-side and Cold-side rechargeable chemical compound
I am wondering if such a chemical/material exists which, upon triggering, becomes cold on one side and hot on the other in which after all the chemical bonds are broken/built (however it would work), and the chemical begins to move towards thermal equilibrium, it could be 'recharged' using electricity?
What I am investigating is an effect similar to that of the 'Peltier' effect in which one side of a semi-conducting material becomes very cold and the other side very hot! These devices are horribly inefficient though.

It's possible, basically if you have a normal battery-like redox, but with one side being endothermic. I don't know of any case where that happens to a notable degree, though. I doubt any such reaction would be more efficient than Peltiers.

Recharging works via heating, not electricity, but the self-cooling beer keg has been on the market for quite a while. Check your friendly search engine. ;-)


text:
Nomenclature of a fused tricyclic hydrocarbon
What is the preferred IUPAC name of the compound shown below (and its common name, if there is one)?

According to Subsection P-25.1.1 of the  current version of *Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name of the parent structure, which formally results from the ortho- and peri-fusion of benzene and naphthalene, is the retained name ‘phenalene’. In order to indicate the position where no multiple bond is attached, an italic capital ‘H’ preceded by a numerical locant is added to the front of the name. According to Subsection P-25.7.1.3.1, indicated hydrogen may be omitted in general nomenclature and 1H-phenalene may be called just ‘phenalene’. In a
preferred IUPAC name (PIN), however, a locant and the symbol ‘H’ must be cited.

The corresponding (partly) saturated compound can be described by using the prefix ‘hydro’, which indicates the addition of hydrogen.

The four methyl substituents of the compound given in the question are cited as prefixes in substitutive nomenclature, which yields the name 1,1,2,5-tetramethyl-2,3,3a,3a1,4,5,6,6a,7,9a-decahydro-1H-phenalene.

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


text:
Production of hydrogen using vinegar and metal for experiments
If
 I cannot, or do not want to use electrolysis of water to produce
hydrogen, is it a viable alternative to react vinegar (ethanoic acid)
and iron (or steel, or copper) to form iron(II) or copper acetate? (Assume steel wool or copper wires.)
If I tried to fill a balloon with the produced gas (assuming no leaks
apart from balloon), would the hydrogen leak out of the balloon faster
than it would be produced?

No. Iron reacts too slowly to be an option. If you really wish going with acetic acid and be moderately cheap, look for aluminium or magnesium parts or maybe even granules. Don't go with powder unless you know what you are doing, though.

Acetic acid will probably react to slow. If you want to produce hydrogen in the litre scale I would suggest to use diluted hydrochloric acid and zinc flakes or iron. But be aware that it can be difficult to control the reaction. In a laboratory you could use a Kipp's apparatus for this.


text:
Supermarket HDPE plastic bags safety
Supermarket plastic bags are made of high density polyethylene, and I am interested in the decomposition products at high temperature: what type of compounds are released and how dangerous are they? I am interested in a situation where one has placed some kind of heated food inside one of these plastic bags.
The question is, are harmful compounds from the plastic bag passed into the food (food is hot, like baked bread) contained inside the plastic bag?

PE does not decompose at 100°C unless perhaps if kept at such  temperature for an extended time period.

the bag is not "decomposed" i updated question. just hot food inside a plastic bag.

Polyethylene is generally considered suitable for food handling and even for warming food. Unlike some plastics such as polycarbonate, it does not need an additional plasticizer that could be toxic.
That said, the melting point of polyethylene is ~1150C, and I've found that it tends to shrink-wrap around microwaved food. For that reason, polypropylene, m.p. ~1300C, is often used for food containers intended for microwave oven use and for automatic dish-washing.
BTW, both polyethylene and polypropylene absorb some of the natural lycopene coloring of tomatoes or other carotenoid pigments and slowly develop a pale brown discoloration that cannot be removed, but that is of no significance.

Thank you DrMoishe for your answer. The question lays on what does "generally considered" means. Any real test on that subject?

As you wouldn't know what effect to look for, it is hard to make a real test. Ordinary PE does not contain anything that is low molecular weight and could mirate, and it doesn't produce anything in that direction if not heated too far above 100°C for too long.

What you are saying is that HDPE plastic bags shouldn't pass any compounds to the heated food (talking only below 100°C) when touching each other..?

Just the label ink, if used inside-out. There should be no health hazard, but a melted PE bag is a nuisance to remove.


text:
Which of the following alkenes would produce two products when they undergo an addition reaction with hydrogen bromide?
Which of the following alkenes would produce two products when they undergo an addition reaction with hydrogen bromide?

Ethene
But-1-ene
But-2-ene
Hex-3-ene

The correct answer (according to my textbook) is But-1-ene but I do not understand why this is.

Draw products and see importance of symmetry of substrates.

Well as mithoron said.its about symmetry.Let's see how but-1-ene comes out to be the answer
On addition of HBr markownikov rule is followed but lets draw both product.
1.)CH2=CH2 + HBr--->Br-CH2-CH3 + CH3-CH2-Br  (both products are 1-bromoethane)
2)CH2=CH-CH2-CH3 + HBr--->Br-CH2-CH2-CH2-CH3 + CH3-BrCH-CH2-CH3 (notice the                name of both the product. 1st one is 1-bromobutane. 2nd one is 2-bromobutane)
3)CH3-CH=CH-CH3 + HBr---->CH3-BrCH-CH2-CH3 + CH3-CH2-BrCH-CH3 (both products are same i.e both are 2-bromobutane)
4)CH3-CH2-CH=CH-CH2-CH3 + HBr--->CH3-CH2-BrCH-CH2-CH2-CH3 +
                    CH3-CH2-CH2-BrCH-CH2-CH3 (here also both the compounds are same i.e both are 3-bromohexane)
So what distinguishes the 2nd one from these three is that the second molecule  is unsymmetrical.You can find its unsymmetrical by looking at the no:of carbon to both side of double bond.If they are equal then molecule is symmetrical.


text:
How can you find the concentration of copper(II) ions in a solution of 0.5 mole copper(II) chloride and 0.5 mole of copper(II) sulfate?
The question I'm trying to answer is below and the answer is C.

I tried finding the total moles of copper(II) ions first by using mole ratios for copper(II) sulfate and copper(II) chloride, obtaining an answer of 0.42 moles. Then, I used $n=cv$ and obtained the concentration, using the volume in the question in litres, of 0.84, which is incorrect. Now, I'm sure my method incorrect but I can't think of another way to this.

Where are you getting your numbers from? They told you that 0.5 mol of $\ce{CuCl_2}$ and 0.5 mol of $\ce{CuSO_4}$ were placed together. That means you have 0.5 mol + 0.5 mol = 1 mol of copper ions. Divide that by the volume, which is given to you as 0.500 L, giving you 2.0 mol/L. Your numbers don’t seem to match up with the question at all.

@lightweaver Yes but the number of moles in the question are the moles of the two solutions, not the copper ions, I believe.

@lightweaver I gor my number by using the ratio 1mol copper(ii):2mol chlorine and 1mol copper(ii):1 mol $So_4$

The question is: "What is the concentration of  $\ce{Cu^{2+}_{(aq)}}$ ions in the solution in $mol,l^{-1}$.

@aventurin Yes, but does that not still mean we need the moles of the copper ions and not the compounds'?

1 mole copper(II) sulfate contains 1 mole $Cu^{2+}$, and 1 mole copper(II) chloride also contains 1 mole $Cu^{2+}$.

@aventurin I don't mean to be rude but could you clarify or at least refer me to something ehich can show me how this is true because from my learning I've not heard it being true. From what I know,  1 mole, for example, of copper(ii) sulfate cintains 0.5moles of copper, because from the chemical formula of it, CuSO4, we have 1molCu:1mol SO4, which means we have o.5moles of copper if there's 1 mole of copper(ii) sulfate.

I'll point out that one mole of $\ce{CuCl2}$ contains one mole of $\ce{Cu^{2+}}$ ions, but two moles of $\ce{Cl^{-}}$ ions.

1 mole of copper(II) sulphate contains 1 mole of copper ($\ce{Cu^{2+}}$) and one mole of sulphate ($\ce{SO_4^{2-}}$). Or, if you want, 1 mole of copper, 1 mole sulphur, and 4 moles of oxygen.
This is by the definition of the mole as the unit of measurement for the amount of substance.
From this it follows that 0.5 moles of copper(II) sulphate contain 0.5 moles of copper. Similarly 0.5 moles of copper(II) chloride also contain 0.5 moles of copper. Together these are 1 mole of copper.
1 mole of copper ($\ce{Cu^{2+}}$) in $\mathrm{0.5\,l}$ water gives us a concentration of $\mathrm{2\, mol\,l^{-1}}$

Thank you sooooooo much!!!!!! n This made me everyhting completely clear! Couldn't thank you more.


text:
Fe 2+ and MnO4- titration?
As a lab experiment, I am preparing a mixture of iron tablets dissolved in sulfuric acid so they can be titrated against $\ce{MnO4-}$. The tablet contain insoluble filler material.
Why is it important this 'filler' be removed when preparing the iron tablets solution?

Insoluble doesn't mean unreactive. The filler must be something which would be oxidized by permanganate in acidic solution.

Please explain this further, I don't fully understand. Thanks

How? You don't say what the tablets are. There is no information on what the filler is. You didn't provide any details on how the separation was done. // The one thing I do know is that permanganate in acid solution is a very strong oxidizer. For instance cellulose wouldn't dissolve in water, but an acidic permanganate solution would react with it.

Thanks. The tablets were mainly made from Iron ii Salt, and the filler was calcium sulfate (and other substances). ?


text:
If I have an alloy of two materials, will I have a chemical reaction between the two materials
I have always been told that an alloy is a mixture without chemical interference between the materials, but if I look at highly doped silicon (doped with Aluminum via ion implantation and then lattice stabilized via annealing) then my alloy can be written as $Si_x Al_{1-x}$ which is (as I've been told) an alloy. So my question is after annealing when my Al is no longer an interstitial in the lattice but occupies a lattice site, for $x>>0; x<1$, my Al would be $sp_3$ hybridized (which in my opinion) is required for the Si to be a p-type acceptor, if my Al is getting hybridized, isn't this a change in the chemical state of the Al? Or is it only an alloy till my Al dopants behave as an interstitial but once it gets substituted into the lattice it doesn't remain an alloy, in which case graded nanocrystals of the form $Si_x Al_{1-x}$ should not be considered alloys since they are almost always annealed. Kindly help me out here.

Yes, your first sentence is the problem - it is false. Many, nay almost all, alloys are non-ideal, having mixing terms I the free energy that depend on the various concentrations.

Ok, thanks a lot for your answer :). I am not a student of chemistry though, could you please point me to some literature about ideal versus non ideal alloys and their mixing terms...etc etc.

The classic book would be Porter and Easterling, Phase Transformations in Metals and Alloys. I recommend the second edition. This is more materials science than chemistry.

You might be interested in http://chemistry.stackexchange.com/questions/37635/plot-binary-eutectic-from-temperature-of-melting-and-enthalpy/37672#37672 and http://physics.stackexchange.com/questions/196686/what-causes-a-eutectic-point/196692#196692

@Jon Custer: Thanks a lot, I'll post if I have further questions :)


text:
Gases produced by bacterial decomposition of human urine
I know that when urine (more specifically, the urea within) is metabolised by bacteria, ammonia is released. However, because of the high solubility of ammonia, I assume it simply dissolves into the water.
A container of urine will develop gases inside. What are they?

The 112-page report commissioned by NASA in 1971 - Composition and Concentrative Properties of Human Urine - might yield some information.

Of course it somewhat depends on what you eat.  Asparagus is notorious for making urine stink.

Carbon dioxide and ammonia when treated with bacteria along with water vapour

Welcome to Chemistry Stack Exchange. Could you improve your answer by providing more information?

Down voted - There has to be more than only those two compounds in the equilibrium gas phase.

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


text:
Draw a structure for each of the following isomers of unsaturated amine 1
My ochem professor has several practice exams as resources but does not include answers,  I believe I understand the concepts but once I apply them I'm not sure whether I'm right or clueless. Can anyone check my work and if I'm wrong tell me why ?

The format of his exams does not seem to be anywhere else online but if it is can someone provide a link. I really want to understand how to deal with these more complex structures rather than look at Fischer projections and simple problems again

Are you sure that the 2-amino-4-methylhexanoic acid at the bottom left of the page is the enantiomer of the given (2S,4R)-2-amino-4-methylhexanoic acid 1?


text:
HPLC: peak area vs concentration
Many text books say that there is a proportionality between the peak area of the chromatogram and the concentration. However, in my opinion, there should be a proportionality between the concentration and the peak intensity, not the peak area. The peak intensity is increasing proportionally with respect to the absorbance of the substance. Hence, by using the Beer-Lamberant law, we can induce there is a proportionality between peak intenstiy and the concentration.
Is there any mathematical relation between the peak area and the concentration?
Any help will be appreciated.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

You asked the wrong question. The question should be "Why isn't peak intensity as good as peak area to determine concentration using HPLC?"

@MaxW That can be an answer for my question, but I wonder if we can deduce the relation between them mathematically, by using the fact that the area is the integral value of peak intensty with respect to the time.

see http://terpconnect.umd.edu/~toh/spectrum/CaseStudies.html#HeightWidth

Present a correlation for concentration and area used to carry out the validations?

However, in my opinion, there should be a proportionality between the concentration and the peak intensity, not the peak area.

There is a proportionality between both peak area vs. concentration and peak height vs. concentration.

Peak height is proportional to the instantaneous amount of analyte that is transiting the detector.
Peak area is proportional to the sum of all of analyte moleucles that have transited the detector.
From 1 and 2 you might be able to infer the relationship betweeen peak height $h$ and area $A$: $$A = \int{h(t)\;dt}$$
People are usually interested in the total amount of substance injected into the column.  (If they know the injection volume, they can calculate the concentration from this value.)  The total amount of substance would be calculated from the peak area.
The maximal peak height is proportional to the peak area, but only if the peak "shape" is constant.  Here are some scenarios where peak shape will not be constant:

You want to compare injection #2 you made on your column two years ago to injection #2000 that you made recently.  Due to column degradation, the recent injections have much more noticeable peak tailing.  Because the longer tail leads to wider, asymmetric peaks, there will be less total height at the peak maximum relative to "better", symmetric peaks.
You change the flow rate of the HPLC.  All peaks are narrower, and thus, higher.
The scan rate of your detector changes, and the detector reports "counts" or "intensity" rather than counts per time.  This is actually very common for mass spectrometers, but less common for absorbance detectors.  If you scan twice as fast in MS, your peak heights go down by approximately twofold, but you have data points twice as often, so the peak area is relatively unchanged.
Your detector undersamples the peak.  Say your chromatography is nice and the "real" peak shape is nicely gaussian, but is ~ten seconds wide and your detector only gives you a data point every two seconds.  Say that due to very small random drifts in retention time, on some injections one of the data points coincides exactly with the maximum of the "real" peak, but in others, it is a little bit off.  The peak heights in this scenario will vary considerably more than the peak areas. (If the recorded maximum is off-center from the true maximum, there will be two data points that are higher, i.e. "closer" to the maximum than if the data maximum coincides with the true maximum, so integration will partially correct this error.)  Essentially, the peak maximum is a single-point sample from the gaussian distribution of the peak, while the area is a several-point sample from the gaussian distribution, so it has better sampling properties.

Is there any mathematical relation between the peak area and the concentration?

Yes, there is, but it depends on the peak shape.  For perfectly gaussian peaks, $$h_{max} = \frac{A}{\sigma \sqrt{\tau}}$$
where $\tau$ is $2 \pi$ and $\sigma$ is the width of the peak, which is related to the full-width at half-maximum of the peak by $\mathrm{FWHM} = 2 \sigma \sqrt{2 \ln 2}$.
However, for non-gaussian peaks, this relationship does not hold.

Does this mean that I can't use absorbance intensity (maximum peak height) to make my concentration calibration curve if I have a peak that's not bb (like vb, bv, or vv)?


text:
The effects of adding neutrons to an atom
So neutrons are neutral in terms of charge, and adding neutrons to an atom affects its atomic mass. But when neutrons are added to the nucleus, the nuclear radius would be affected. Couldn't that affect how the charge of the nucleus is distributed within the nucleus?
With an atom large enough that, say, 40 neutrons could be added (and still be stable), I would imagine that the nucleus would be slightly rearranged to accommodate this (so that neutrons and protons are distributed throughout the nucleus and there isn't a shell of 40 neutrons surrounding what is the pre-existing nucleus).
I imagine that this redistribution (if it does occur) would dilute the charge coming off of the nucleus (as the protons are farther apart and the nucleus is less dense in terms of charge), which would lead the electron shells to be farther from the nucleus. This would increase atomic radius, affecting a wide-range of chemical characteristics. I could equally see an opposite effect, where the nuclear charge is strengthened (which would decrease atomic radius).
Is this an actual reaction to the adding of nuclear charges? If so, does it occur in large enough quantities to play a significant role in chemistry? Or have I just made a false assumption?

This is an interesting question, but one that may be better posted to Physics.SE, at least in terms of whether such changes actually would occur to the nucleus.

You've just made a false assumption. A charge is a charge; you can't dilute it with neutral matter (or rather, you can, but it would make no difference). From the electron's point of view, a nucleus is but a dot, no matter how many neutrons are there.
Sure enough, isotopes would differ in chemical and physical properties (that's why it is possible to separate them), but only very slightly (that's why it is so difficult).

Some quantum chemistry programs effectively let you simulate this, by letting you specify various models for the nuclear charge (e.g. point, Gaussian, or something else, and with variable parameters). See, for example, the ADF program's docs on the subject.

@Aesin, to be sure I understand, the nuclear charge model affects reactivity negligibly?  It only shows up in certain subtle spectroscopic measurements?

The proton to neutron ratio does effect the stability of the nucleus itself. Unstability in this context meaning that the nucleus would undergo radioactive decay.

@Brian: That's my understanding, yes. Or more specifically, spectroscopic properties involving orbitals with a significant amplitude inside the nucleus.

Possible duplicate of Chemical properties of isotopes

Even if such a phenomenon were to occur, and the resulting nuclei were sufficiently stable to participate in chemistry (they would not be sufficiently stable, in reality), the "inflation" of the charge of the nucleus would almost certainly have a negligible effect on the chemistry.  The nucleus is really really really tiny with respect to the length scales of bonds and the electron cloud.
To give an idea of the scales involved: the largest stable element, uranium, has a nuclear radius of about fifteen femtometers $\left(15~\mathrm{fm}\right)$, per Wikipedia, or $1.5\times 10^{-14}~\mathrm m$.  Bonds and electron clouds are measured in Angstroms $\left(10^{-10}~\mathrm m\right)$.  Thus, the nuclear radius would have to expand by well over $\sim\!\!100$-fold in order for the assumption of a point-charge nucleus to begin to fail.

It is well accepted that the effect of isotopic substitutions on chemistry primarily is due to the change in mass of the nucleus, leading to the kinetic isotope effect.  Another potential isotope effect on reactivity involves the change in nuclear spin, leading to a "magnetic isotope effect" only relevant for certain classes of reactions between species with unpaired electrons (e.g., organic radicals and/or various inorganic species).
Isotopic substitution does affect the behavior of nuclei in certain spectroscopies, however, leading to isotopic shifts.  The isotopic mass change can have noticeable effects on the vibrational spectrum, due to the effect on the local distribution of mass. As described in the link to ADF's documentation provided by Aesin in a comment to the question, these effects can also be seen in the hyperfine structure of atomic spectra.  As well, the NMR isotopic shift of light elements, especially hydrogen, can sometimes be observed.

... and to expand the radius by 100-fold would require $100^3$-fold increase in the number of nucleons since the nucleus is approximately spherical.  (Not entirely true if the nucleus has angular momentum.  But such complication is a but much for this question.)  Now a million nucleons is around 4000-times as many as we find in heavy atoms, like Uranium, so this has never been observed.

@EricTowers Precisely! :-)

But isn't there a chance that the neutrons could have some sort of slight shielding effect?

@DGS Shielding effect of what type?

@Brian Maybe due to polarizability within a neutron, there can be random neutralizing or enhancing effects?

You can't shield a charge (or rather, you can shield it with an opposite charge, but that's another story).

@DGS Agree w/Ivan, the mere presence of a neutron can do nothing to shield the positive charge of protons. Even if a neutron could undergo a local polarization analogous to the induced dipole of a London interaction (another good question for Phys.SE, not here!), on the Angstrom length scale of the surrounding electron cloud a 'polarized neutron' would still appear just as a zero-charge constituent of a ~point nucleus.

The hydrogen atom (a single proton and electron) has the greatest effect from an added neutron, turning it into deuterium.
From http://en.wikipedia.org/wiki/Deuterium:
"Chemically, there are differences in bond energy and length for compounds of heavy hydrogen isotopes compared to normal hydrogen, which are larger than the isotopic differences in any other element."


text:
Dismutation of ethanoic acid
Ethanoic
 (acetic) acid can undergo fermentation by methanogens to create
$\ce{CH4}$ and $\ce{CO2}$. Is there any way that I could perform this
reaction without any microorganisms, specifically with the concentration
 of ethanoic acid found in vinegar, for example with UV light and such?
Also, if this is impossible, what species of bacteria is required?

A UV light would create ozone which would reaction with acetic acid to completely oxidize it to carbon dioxide and water.

Melting acetate with solid alkali should produce $\ce{CH4}$. Not sure if it could be made economically feasible, though.

Bugs are great chemists.  I don't think you'll have any luck beating the methanogens at their game ;)


text:
Mechanism in Acid chloride formation

I want to ask about nucleophilic attack by Cl in Step 2.
To me there seem 4 electrophilic sites and I marked them as (a)~(d).
mechanism I've seen in lecture only deals with site (a).
maybe (c) is less electrophilic and (d) makes this reaction go backwards, so I can get past them. but when Cl- attacks (b), it could break a bond b.w (b) and (d), resulting in completely different product, as well as a bond b.w (a) and (b).
so my question is,
is nucleophilic attack on site(b) by Cl- resulting in breaking bond b.w (b) and (d) possible? if so, to what extent? if not, why?
hope somebody teach me something!!

Just draw the possible products from attack on (b) and breaking either of the (b-d) or (b-a) bonds. Do they look likely to form?

Um... maybe I ignored high electronegativity of O that wouldn't allow breaking these 2 bonds?


text:
Relative stability of carboxylic acid derivatives

This picture depicts the relative stability associated with resonance structure.
My questions are:
1.1. Teacher says anhydride is less stable because of two reasons above.
but it seems there's no reason the resonance of anhydride would be less likely compared to acetate.
1.2. and about reason #2, the actual structure is the hybrid of these resonances so it's hard to agree with this reason.

Why won't Cl give its e- pair to adjacent carbon for resonance structure.

Someone says it's about electronegativity but it contradicts with resonance associated with oxygen. And someone says it's about the high energy orbital of Cl (n=3) but I can imagine a acetyl fluoride and this logic doesn't solve this problem anymore.

Firstly "stabilised" is very ambiguous. "Stable" is always defined with respect to something. The proper phrasing here should probably be stable towards nucleophilic attack.
For the anhydride - yes the actual structure is a hybrid. That just means that both C=O have to "share" the central oxygen's lone pair - in a sense each of them get half resonance each - which means that both C=O's are not very stabilised.
I don't buy the "less likely" explanation. Your teacher is trying to say that there's a "50% chance" that at any one point in time a particular C=O is stabilised, but resonance isn't something that you flip a coin and decide which way it goes.
For the acyl chloride, it is really to do with the leaving group ability of Cl-. See this question for a discussion: Do acyl fluorides exhibit resonance donation or inductive withdrawal?
Essentially the idea is that since $\ce{Cl-}$ is a better leaving group than $\ce{OMe-}$, for the acyl chloride (X = Cl), resonance form 2 contributes more to the actual structure than it does in the ester (X = OR):

As you can see, resonance form 2 makes the C=O more susceptible to nucleophilic attack because of the positive charge associated with the carbonyl group. Therefore, any substituent X that favours resonance form 2 will make the carbonyl group less stable towards nucleophiles.

and in the question you mentioned, why the resonance form III and IV contribute more in case of acetyl flouride than acetyl chloride? is it about # of principal quantum number?


text:
Why do nitro groups preferentially bond through the nitrogen rather than the oxygen?
In many organic and inorganic molecules which contain a nitro group, the bonding takes place through the nitrogen, which gives the nitrogen "four" bonds, but the $\ce{N-O}$ bonds are delocalized across both oxygens which is obviously quite stabilizing.
That being said, I can't help but notice that if the nitro group bonded through the oxygen like $\ce{R-O-N=O}$, then there is not the same situation where nitrogen has over three bonds and the oxygens only have about 1.5 bonds. Additionally, the $\ce{C-O}$ bond is about $360\ \mathrm{\frac{kJ}{mol}}$ compared to about $310\ \mathrm{\frac{kJ}{mol}}$ for the $\ce{C-N}$ bond.
That means that whatever the explanation is, it needs to make up around $50\ \mathrm{\frac{kJ}{mol}}$ in bond energy. I believe the other bonds present should be pretty similar as far as energy goes.
Finally, when the nitro group bonds through the oxygen, presumably there will be less steric hindrance because the group would be roughly linear. I could imagine that assumption being incorrect though.
I imagine that the answer (when bonded to an aromatic molecule) will have something to do with conjugation of the nitrogen lone pair into the pi system. I would think, however, that the oxygen would have a similar effect.
Looking forward to answers.

See, it's not like the nitro group just hops around, thinking "which way should I bond today". The molecules are the way they are, and can't easily change into another form. (In cases when they can, they do.) There are nitro compounds $\ce{R-NO2}$, and there are organic nitrites $\ce{R-O-N=O}$; these are isomers, each with its own unique properties, both relatively stable, synthesized in different ways.


text:
Brown color at end point during titration with phenolphthalein as indicator
We did a titration experiment at school and I did not understand one part of it.
The scenario:

Titrate : White wine (ethanoic acid)
Titrant: $\ce{NaOH}$
Indicator:  Phenolphthalein

What happened: We carried out the titration and the end point was a brown colour!
How does that happen? Why is the color brown and not pink?

I strongly recommend AGAINST closure of this question. It may have come from a school demonstration, but it's a curious observation. Why indeed would phenolphthalein turn brown instead of pink? A Google search yields no obvious explanation, and brown appears nowhere on the phenolphthalein Wikipedia page.

I don't think this can be untangled. (1) There shouldn't be acetic acid in white wine. (2) When titrating did the solution turn pink where the drops of NaOH went into the solution? (3) I assume that the color of the "wine" was yellow. What happens if you titrate with NaOH using no phenolphthalein?

I went looking for a titration of apple cider vinegar (which would be yellow) with NaOH and phenolphthalein on youtube but didn't find one. It could be that the yellow plus pink gives some sort of "dirty orange" that looks brownish right at the end point.

@MaxW "I don't think this can be untangled." Certainly possible, with how complex a solution the white wine is. Still, worth keeping open, I think.

Assuming there are no major color-influencing side reactions, the subtractive color mixing as follows:
#FFA8D9 (Pink, phenolphtaleine) + #DAA520 (Goldenrod, white wine) = #D2691E (Chocolate, result)
actually already gives brown(ish) color. It's hard to pick the right color codes, especially for the white whine, but OP can probably try to perform the same operation with the color he remembered/picked from the photo.

It happened to me as well when I tried to titrate a sulfuric acid based process solution. The brown color at end-point is normally not observed during this very standard test. A pink color appears when some fresh phenolphthalein is being added to the (brown) solution. I believe that in our case, organic contamination was the cause. Since the tested sample in the question is wine, my guess is that the reason is similar. Organic degradation by-products of spoiled or contaminated wine (a bottle used for quite a while in school lab, right?) probably reacted with phenolphthalein. Try using a fresh wine sample.


text:
Can you undo 'blackening' from Chromium after exposure to bleach?
I accidentally used toilet cleaner (instead of bathroom cleaner) containing bleach on a Chromed tap/faucet which has resulted in an black clouding over the surface of the chrome. It doesn't seem to polish out or clean off even with metal polish.
Is there any way to reverse this process or will I have to replace or chrome plate the tap?

From what I've been able to search for, there doesn't seem to be a way to reverse the chemical corrosion that occurred to the chrome metal. Perhaps an electrochemical approach can be taken to electroplate new chromium ions onto the surface.


text:
Why does 1,3-dihydroxybenzene have a higher boiling point than 1,2-dihydroxybenzene?
1,2-dihydroxybenzene (catechol) shows intramolecular H-bonding. So due to strong bonding, it should be difficult to break the bonds in 1,2-dihydroxybenzene and therefore, 1,2-dihydroxybenzene should have a higher boiling point than 1,3-dihydroxybenzene. Why is it not so?

Related: Which among phenol and 1,2-dihydroxybenzene has the higher boiling point?

The answer is in your question itself.Intramolecular hydrogen bonding will only reduce the boiling point as intramoleclar H-bond will not allow one molecule to link with another.Intramolecular H-bond two molecules are linked.For boiling up,the molecule-molecule interaction must be broken and not the bonds within a molecule.

@AbhishekPallipparagopakumar Then what factor decides the difference in boiling point among 1,2dihydroxybenzene and 1,3 dihydroxybenzene

Of course as i mentioned above catechol has intramolecular H-bond whereas resourcinol has ** inter**molecular H-bond.Thus there is no interaction between two molecules in catechol.Whereas there is interaction between two molecules of resourcinol.

Being more volatile implies that the molecules should be set free from mutual interactions. As 1,2-dihydroxybenzene has intramolecular H bonding, it is easy for it to exist independently, but for 1,3-dihydroxybenzene there might be intermolecular H bonding (interaction between two molecules); hence, it is a bit more difficult to break it apart, thus having a higher boiling point than its counterpart.


text:
Lewis structures of S2N2 and [S4]2+
This seems so simple, yet I just can't seem to manage. Apparently $\ce{S2N2}$ and $\ce{S4^2+}$ (which are isoelectronic) obey Huckel's rule (at least to some extent) and I was trying to draw the Lewis structures to show that, but I can't manage to draw them without having a diradical, which doesn't seem very likely to exist. And if there are radicals, it doesn't obey Huckel's rule anymore.
Can someone help me draw their Lewis structures that obey Huckel's rule?

Draw each ring with one double bond and the two atoms having no double bond with two electron pairs instead of one.  Verify that the total electron count agrees with two positive charges on $\ce{S4}$ and zero on $\ce{S2N2}$.  The double bond and the extra electron pairs on the two atoms are the six ring electrons in the Huckel rule.

You can use \ce for formatting chemistry expressions

that works for $\ce{S_4^{2+}}$ but having one double bond doesn't work for $\ce{S_2N_2}$ unless it forms N-N and S-S bonds, rather than interchanging S-N bonds.
The only possible structure I found is having two double bonds on one S atom to the two neighbouring S and N atoms

One thing that you need to note with $S_4N_4$ is that we require the molecule to have zero charge.  In many cases the individual atoms may still have positive and negative charges that cancel out.  See for instance the Lewis structure for sulfur dioxide.  With $S_4N_4$ you will have a positive charge on one sulfur atom, neutralized by a negative charge on nitrogen.


text:
Does crystal field splitting energy Δ increase going down a group of transition metals if the ligands are weak field?

Going down a group, the metal orbitals are more diffuse so there is greter
overlap between the metal and the ligand orbitals.
If there are no π-interactions, i.e. with σ-donor ligands like $\ce{NH3}$, then the $\mathrm{e_g}$ MOs are split further with more metal-ligand overlap, and $\Delta_o$ increases.
If there are π-acceptor interactions, then the $\mathrm{t_{2g}}$ MOs are split further with more metal-ligand overlap, but $\Delta_o$ still increases as it is the gap between the bonding $\mathrm{t_{2g}}$ and antibonding $\mathrm{e_g^*}$.
However with π-donor ligands ($\ce{Cl-}$, $\ce{Br-}$ etc), surely more overlap would cause greater splitting of the $\mathrm{t_{2g}}$ orbitals, which would decrease $\Delta_o$, as it is the energy difference between the antibonding $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$.

So, does $\Delta_o$ still increase going down a group if the ligands are π-donors?

I see where you are coming from, but you forgot that even with π-donor ligands, there is still a σ-bonding present in the complex. In the 4d/5d metals, the $\mathrm{e_g}^*$ orbital is pushed up by a lot due to the σ-type overlap.
The $\mathrm{e_g}$ overlap is always stronger than $\mathrm{t_{2g}}$ overlap because the metal $\mathrm{e_g}$ orbitals point directly at the ligands ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$), whereas $\mathrm{t_{2g}}$ orbitals point in between the ligands ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$).
So, yes, the $\mathrm{t_{2g}^*}$ orbital is pushed up higher by the π-donor, but the $\mathrm{e_g^*}$ orbital is pushed up even higher. Thus, for example, the complex $\ce{[RuCl6]^2-}$ has a $(\mathrm{t_{2g}})^4$ configuration (i.e. low spin) even though $\ce{Cl-}$ is a π-donor ligand.


text:
Troubleshoot anion exchange column (Fe, Cr, Co separation)?
I ran an anion exchange column to separatate Fe, Cr, Co from solution of concentrated HCl.  The column was Dowex 1x8, 200-400 mesh, diameter < 0.5 cm, lenth 7 cm.  A gradient of acid concentrations was used to elute: 12 M (1 mL), 9 M (3 mL), 4 M (3 mL), and water.  The metals were all radioactive isotopes (Cr-51, Co-60, and Fe-59) and Fe (III), Cr(III), Co(II) oxidation states.  I used an NaI(Tl) detector to determine when each of the metal ions eluted.
I intended the Cr to immediately elute as it forms only neutral or cationic species in HCl, then the Co to elute around 9-4M, and the Fe to elute when water was used.
The Cr and Fe did what I expected, but the Co eluted with the Fe (with the water instead of at 9-4M HCl), and the two peaks completely overlapped.
Does anyone know why this behavior might have been observed?  Could I do any better with a cation column instead?

In the original method described by Kraus and Moore (J. Am. Chem. Soc., 1953, 75 (6), 1460–1462), Co(II) was eluted with 4 M HCl (not Co(III)).

@Loong Oops! that was a mistake, the Co was indeed Co (II) when I did it.


text:
Methyl anthranilate sulfuric acid - acid base reaction
When methyl anthranilate is added to sulfuric acid a white precipitate forms. I am assuming this is an insoluble salt produced in the acid base reaction, where the lone pair on the nitrogen will accept a proton.
My question is, does sulfuric acid lose both protons in the reaction, so that there are 2 ions of the protonated methyl anthranilate with one sulfate ion. Or does it form one protonated methyl anthranilate ion with an HSO4- ion?


text:
Why is WF6 stable whereas CrF6 is unknown?
$\ce{F-}$ is a hard base since it is small and relatively polarizable.
Both $\ce{Cr^6+}$ and $\ce{W^6+}$ are hard acids, but shouldn't $\ce{CrF6}$ be favored since it would be a smaller hard acid?

The answer has to do with two things. Note that HSAB theory is dubious at best and doesn't have very useful predictive power, so I am going to avoid talking about it.
(1) The accessibility of the high +6 oxidation state
In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another. On the other hand
In W, the 5d orbitals have two inner radial nodes. The two pockets of electron density close to the nucleus shield the bulk of the electron density, which is far away from the nucleus. So, when you continuously remove electrons from W, the remaining 5d electrons don't drop in energy so rapidly.
This is partly reflected in the ionisation energies for Cr and W (data from NIST Atomic Spectra Database), all units are in eV).

Element
IE1
IE2
IE3
IE4
IE5
IE6

Chromium
6.76651
16.486305
30.959
49.16
69.46
90.6349

Tungsten
7.86403
16.37
26.0
38.2
51.6
64.77

For an alternative perspective, here are Latimer diagrams for Cr, Mo and W (from Shriver et al., Inorganic Chemistry 6ed.); you can see that the stability of the higher oxidation states increases going from Cr to Mo to W (i.e. the reduction potentials are lower)

This trend is hardly limited to the Group 6 elements. $\ce{OsO4}$ features Os(VIII), whereas the highest oxidation state of iron is Fe(VI), which is already extremely difficult to achieve. Likewise Au(III) is pretty normal but Cu hardly goes above Cu(II).
(2) Sterics
Cr(VI) can be made, but the only Cr(VI) compounds that we know of are paired with the oxide ion, viz. $\ce{CrO3}$, $\ce{CrO4^2-}$, $\ce{Cr2O7^2-}$.
So there is likely also an argument here that chromium, being a smaller metal, cannot quite fit six fluorines around it in the same way that tungsten can. It can fit four oxygens, but that's about it.

Technically, iron is known in the +7 oxidation state, but this requires a 4 K matrix.  At +6, however, it's actually used commercially!

Is this "shielding by inner pockets of electrons" by inner radial nodes a general feature of the transition metals, which makes higher oxidation states accessible as we go down the group?

Orthocresol's answer sums up the comparison between $3d$ valence transition metals on the one hand, and $4d$ or $5d$ valence transition metals on the other.  But there is a second aspect to this problem: the $3d$ metals like chromium can still reach high oxidation states with oxide ligands, not so much with fluoride. The key here is not hard/soft basicity but pi electron interactions: oxide ligands are powerful pi donors and can back-donate to the empty, or at least depleted, $3d$ orbitals that such oxidation states as $\ce{Cr(VI), Mn(VII), Fe(VI)}$ imply.  Fluoride ions do not do as well here, and so cannot stabilize high oxidation states as readily in the $3d$ series.


text:
Is it possible to "brute-force" material creation through algorithms?
I know its a bit of a soft question, but lately I've wondered what important properties and calculations would a program that iterates through certain types of molecules and atoms need in order to calculate their individual properties like color, conductivity, reactivity with common substances, etc..
Are there any projects out there that already target something similar to what I have in mind? I know there are a multitude of variables that will come into play, but for the most part say if you were to keep a variety of them constant.

https://books.google.pl/books?id=kG3SBwAAQBAJ&pg=PA324&lpg=PA324&dq=design+alloy&source=bl&ots=xg8CqY-XrK&sig=tbzhIOMAIAwND93L5zx6gBa0b88&hl=pl&sa=X&ved=0ahUKEwj13bPH6vXLAhVLG5oKHXKlAgYQ6AEIXzAM#v=onepage&q=design%20alloy&f=false

This seems related, I would think that there would be more recent books on the subject (this was published in 1977). Surely someone has thought about this before.

Yeah, I heard about computer design of alloys, maybe I'll find sth. Generally answer seems to be, yes it is possible.

I found something that seems to be for material design on github (open source software platfor) but it is written in FORTRAN, yikes!.  MSIM3V

Perhaps, I am underestimating the complexity of performing these types of simulations, because it would seem to me that if a close to ideal software of this type was built, there would not be much need for someone to collect actual experimental data.

MOFS are modelled with computational calculations before they are built.


text:
Why can Bi not achieve its 4+ oxidation state?
Bi can reach its 3+ and 5+ oxidation state. Why not 4+? Can you use the inert pair effect here to explain this?

Well bismuth is in group 15 so, +4 oxidation is probably rare. I don't think that has anything to with inert pair effect. Group 15 generally shows +3 and +5 states because they give closed-shell species. +4 would result in an unpaired electron i.e. a radical

Actually bismuth(IV) is known, and in not so exotic conditions at that. In fact both even oxidation states +2 and +4 appear when acidic bismuth(III) solutions are irradiated:

Bi(IV) has been observed; see A. I. Aleksandrov, I. E. Makarov (1987). "Formation of Bi(II) and Bi(IV) in aqueous hydrochloric solutions of Bi(III)". Bulletin of the Academy of Sciences of the USSR, Division of Chemical Science. 36 (2): 217–220. doi:10.1007/BF00959349.

From the reference cited above:

The formation of bismuth(II) and bismuth(IV) under the action of ionizing radiation on hydrochloric acid solutions of bismuth(III) has been shown by low-temperature and pulse radiolysis. The kinetics of bismuth(IV) disappearance at 300°K have a complex nature and would seem to be caused both by the disproportionation of bismuth (IV) and the reaction of bismuth(IV) with bismuth(II), which leads to bismuth(III).

The properties of bismuth (IV), which is formed from the reaction of Cl2 with bismuth (III) agree with previously established relationships on the variation in optical transition energies and the unpaired electron density at the s atomic orbital of the ion for the series of isoelectronic ions mercury (I), thallium (II), lead (III), and bismuth (IV) in the $^2S_{1/2}$ state.


text:
Why does adding PBS to a lipid+propylene glycol solution cause the lipid to precipitate?
I am trying to create an aqueous phospholipid suspension using ~50 mg of 1,2-dibehenoyl-sn-glycero-3-phosphocholine (DBPC), 13 mL of propylene glycol (PG), and 72 mL of phosphate-buffered saline (PBS). I am easily able to get the DBPC to dissolve in the PG when held at 85 degrees (10 above Tm), but as soon as I add any sizable amount of PBS, my solution turns cloudy with chunks of DBPC floating around, implying that the DBPC has precipitated out and aggregated.
Why would this be? I'm basing myself off of this patent, and they do mention that trying to dissolve the lipid directly in water causes aggregation, but why would adding PBS to my already dissolved lipid do this?
Structural photos below:
DBPC:

PG:

That's an example of salting out.


text:
In NMR Spectroscopy why use absorption and dispersion components of the FID spectrum rather than magnitude and phase?
In chapter 4 of Dr James Keeler's book 'Understanding NMR spectroscopy' (http://www-keeler.ch.cam.ac.uk/lectures/Irvine/chapter4.pdf) he explains how to calculate the expected spectrum of a FID signal. In this chapter he analyzes the real and imaginary (absorption mode and dispersion mode) parts of the FID spectrum. This doesn't make sense to me. What is the physical interpretation of absorption mode and dispersion mode? Why not just rewrite the spectrum in polar coordinates and consider the magnitude and phase separately?

One Sentence Answer: The resolution of the real part is better than if you plot the Magnitude.

The FID signal itself is represented/considered as a set of complex points in time. (Keeler mentions this in chapter 4, section 4.1. - same link as you posted in your question.) Upon Fourier transform, the FID will therefore transform correspondingly to a set of complex points in frequency space.
As Keeler said in section 4.2 of the same chapter, the real part of the spectrum is the absorption mode lineshape, and the imaginary part of the spectrum is the dispersion mode lineshape.
From your last question I think you understood this already, but it is worth repeating.
Aside: Why is the FID a set of complex points?
While all the values that we can detect are real (we cannot detect imaginary current in the receiver coil), we can consider values detected in different directions. Taking the static magnetic field of the spectrometer as the $z$ direction, we can measure the $x$ and $y$ components of the NMR signal. Mathematically, the $x$ and $y$ components can be equivalently expressed as the real and imaginary component (and you gain some conveniences with the Fourier transform if you do so)
Detecting the $x$ and $y$ components of the NMR signal is known as quadrature detection (Keeler, section 5.6), and it is used in order to be able to distinguish between frequencies that are offset from the carrier frequency with different sign. As Keeler describes, instead of fitting two coils to detect the $x$ and $y$ components of the FID directly, it is achieved by phase shifting the receiver reference frequency in real spectrometers.
What is the physical interpretation of absorption mode and dispersion mode?
The pure absorption part represents resonant frequencies of the nuclear spins in your sample.
The pure absorption part of the spectrum is what NMR spectroscopists would use as the NMR spectrum. Signals in the absorption mode occur at distinct frequencies for a particular isotope within a particular sample at a particular magnetic field, which is compared to a frequency of a known reference substance (same type of nucleus, same field/spectrometer, etc). The difference between these two frequencies is the chemical shift, in units of parts per million (ppm) Hz. The chemical shift values for most/all of a particular isotope is the parameter that is reported most commonly in NMR experiments.
Just as the imaginary part of the FID is just the real part of the FID with a phase shift, after Fourier transform, the dispersion mode spectral lineshape has in principle the same information as the absorption mode lineshape. Its presence is mostly for signal processing purposes, but mostly done in the background and not displayed to the user. It is also not commonly used as the "result" of the NMR experiment because it can be hard to determine the accurate chemical shift if the signal does not return to baseline on both sides of the signal, which is exacerbated by the long (positive and negative) "tails" either side of the dispersive mode lineshape. The long "tails" also mean that signals are more likely to overlap each other and the resolution is therefore worse. Some more advanced NMR experiments generate a mix of absorptive and dispersive lineshapes "by default" (called a phase twist), and much effort goes into coming up with strategies to recover the absorption mode lineshape. (Example: 2D NMR spectroscopy, see Keeler chapter 7, section 7.6.2)
Why not just rewrite the spectrum in polar coordinates and consider the magnitude and phase separately?
It is possible to do this. But you will get back magnitude/phase lineshapes that is very similar to the real/imaginary (phase-sensitive absorption/dispersion) lineshapes.
The magnitude spectrum can be calculated by $\operatorname{Re}^2 + \operatorname{Im}^2$. We know that the absorptive lineshape should be completely positive and the imaginary part is more or less the same information but half positive, half negative with long tails. The effect of squaring both and adding them together means that you get something that looks mostly like the absorption lineshape with more signal to noise, at a cost of generating positive long tails from the squared imaginary part, which reduce the spectral resolution.
This is actually a (now somewhat old-fashioned?) technique to address the problem of phase twist in 2D spectra.
The phase spectrum can be calculated by $tan^{-1}\frac{\operatorname{Im}}{\operatorname{Re}}$. No one uses this in NMR (as far as I know!), but we can visualising it by considering what it looks like at different points.
We plot the $tan^{-1}(x)$/arctan function. (Axes cross at the origin.)

At absorption lineshape peak, dispersion lineshape is zero: $\frac{\operatorname{Im}}{\operatorname{Re}}$ is zero, arctan is zero.
At large values away from the peak, we assume everything goes to zero:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is zero, arctan is zero.
At a short distance from the peak on the right hand side, absorption lineshape is positive, dispersion lineshape is positive:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is positive, arctan is positive.
At a short distance from the peak on the left hand side, absorption lineshape is positive, dispersion lineshape is negative:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is negative, arctan is negative.

Assuming that the signal in the absorptive/dispersive lineshapes is also centered at zero, you get something like this:

Joining the points all together gives you a curve that looks like the dispersion lineshape (but not completely the same. My guess is that it is it is also a little broader than the phase-sensitive dispersive lineshape).


text:
Dissolving Tylenol Science Experiment - why is it fastest in distilled water?
My son has a science experiment for class to dissolve Tylenol (325mg acetaminophen) in 3 different solutions (125 mL volume each):

Distilled water
Vinegar (1 part white vinegar to 4 parts water)
Salt water (1 tablespoon sea salt)

All solutions were at room temperature.  As far as I could tell, the pill had no fillers and there was no coating on them.  On the bottle, the pills are described as 325mg acetaminophen, and there is no mention of non-medical / filler ingredients.
We used pH test strips, and it seemed the vinegar solutions was a pH of about 6, the water 7 as expected, and the salt water 8 or 9. Edit: it may be the pH was 7 for the salt water, see comment below.
His hypothesis was the Vinegar solution, being acidic would dissolve the fastest.  However, the Tylenol dissolved fastest in the distilled water.  The salt water was slowest, and the vinegar solutions was second.
We also tried pure vinegar, and this dissolved even more slowly than the diluted vinegar used at first.
I should add - it didn't really dissolve in any of the solutions, the pill just came apart into small particles.  Even stirring the solutions afterwards didn't seem to dissolve completely.
I am trying to help my son to determine why distilled water broke up the pill the fastest, but I'm afraid I don't know enough to help.
Any pointers, or websites, where I can help him to research greatly appreciated.

There are always fillers

Gotta look under "inactive ingredients." In the US, they have to provide these. I'm curious as to why the salt water had pH>7. If anything, I would have expected the magnesium ions in particular to lower pH.

@DGS i'm not sure we read the strips correctly.  I looked back at a photograph we took, and they look similar now.   I think the distilled water strip was dry, and the salt water strip was wet when we compared them.

@dave There could be some carbonates in there or something. Not impossible or even unlikely.

Water can only dissolve a certain amount of solids at a given temperature before it reaches saturation. Distilled water has been boiled and captured in gas phase, almost completely eliminating the presence of dissolved solids.  This lack increases the number of ' water to solid' molecular interactions that will take place and so speeds up the disolving.  The same would be true of any solute.


text:
Will bases typically precipitate silver atoms from silver ions?
In class I learnt that one can dissolve many substances using acid and that this can be predicted using standard electrode potentials. For example,
$$\begin{align}
 \ce{Zn &-> Zn^{2+} + 2e^-}      &  E_0 &=  0.76~\mathrm{V} \\
 \ce{2H^+ + 2e^- &-> H2}         &+ E_0 &=  0~\mathrm{V}    \\\hline
 \ce{Zn + 2H^+ &-> H2 + Zn^{2+}} &  E_0 &=  0.76~\mathrm{V}
\end{align} $$
And so Zinc will dissolve in acidic solutions.
I wondered what bases do and so added up the standard electrode potentials of bases and silver.
$$\begin{align}
 \ce{Ag^+ + e^- &-> Ag}                 &  E_0 &=  0.80~\mathrm{V} \\
 \ce{4OH^- &-> O2 + 2H2O + 4e^-}        &+ E_0 &= -0.40~\mathrm{V} \\\hline
 \ce{4Ag^+ + 4OH^- &-> 4Ag + O2 + 2H2O} &  E_0 &=  0.40~\mathrm{V}
\end{align} $$
This suggests to me that basic solutions will typically precipitate a silver solid from silver ions and produce oxygen gas.
Does this make sense?

What you have written is the precipitation of silver atoms, not ions.

Also such reaction doesn't happen.

@milthoron Explain why.

I think the precipitation of $\ce{Ag2O}$ would be a faster reaction: $$\ce{2Ag^+ + 2OH^- -> Ag2O v + H2O}$$ This reaction causes $\ce{Ag^+}$ to crash out of solution before the reduction can occur. The silver mirror (Tollens) test makes use of ammonia to bring it back into solution, but it doesn't seem to be reduced by NaOH in this case, either. Maybe not enough is used to make a large amount of precipitation? Or maybe the diamminesilver(I) is has a different reduction potential.

@DGS What causes one reaction to occur over the other? Is it mainly just the Arrhenious equation and the activation energy? Is the precipitation of $\ce{Ag2O}$ a faster reaction or does it have a lower activation energy?

Those two things are ways of saying the same thing. I think it's all reaction speed. Acid/base reactions are very fast, typically, so a silver ion would have to simply collide with hydroxide, and then another hydroxide would have to pull off a proton and then you have silver(I) oxide. Precipitation reactions are also quite fast. It doesn't take much energy for two atoms of opposite charge to collide.

I guess you didn't hear about how difficult is to oxidate water. Big overpotential being serious obstacle.

@Mithoron So the oxidation of water, $\ce{2H2O -> O2 + 4H^+ + 4e^-}$, is a necessary part of the reaction mechanism for $\ce{4OH^- -> O2 + 2H2O + 4e^-}$? Makes sense, 4 hydroxide ions colliding in the right way seems a lot less probable then 2 water molecules colliding.

Both equations don't really tell much about mechanisms as they aren't elementary.


text:
Does sulfuric acid formed in a previous step dehydrate a secondary alcohol?
In the following reaction won't the sulphuric acid dehydrate the secondary alcohol after it's formed?

It could, but the addition of water would push equilibrium toward the hydrated product, and the $\ce{H2SO4}$ would likely transfer into the water layer that you made by adding $\ce{H2O}$.
Plus, you would only make stoichiometric sulfuric acid. Depending on the amount of water, the pH of the solution formed could be fairly neutral. Also, usually, the acid (despite being a catalyst) is used in large excess for dehydration reactions.
Therefore, it is possible (given the right conditions) but unlikely.


text:
Adding reduction potentials?
Will it be wrong to add reduction potentials of two intermediate reactions to get reduction potential of net reaction(just like as we do in adding enthalpies and applying Hess's law)?
What is the basic error? What should be the thinking process to avoid these kind of errors in chemistry?

Possible duplicate of How to calculate reduction potentials from given standard potentials for cobalt?

@Mithoron       The answer at there does not explain my question properly

@JM97 please [edit] your question to explain how.

Well, $\Delta G=-nFE$, so as long as you keep count of your electrons (n) properly, there are no reasons to worry.

@IvanNeretin but E is independent of number of electrons

Then $\Delta G$ is.

@IvanNeretin That's an extensive property

That's right. So what?

@IvanNeretin why can't we add E at each stet to get final E just like we add H(Hess law)

Yes we can. Just keep in mind the number of electrons, because it is important. Number of electrons times potential is energy. The energy must be conserved, that's the law. There is no law for conservation of potential, though.


text:
Is (2R,3S)-butane-2,3-diol chiral?

Is the attached compound a chiral or achiral overall? My lecturer said that this is an achiral compound because it has a superposable mirror image but I don't get that. So I tried to figure out the configuration of both stereogenic centers because what I understood is when you have an even number of chiral centers and if they have opposite configuration, say (R,S) or (S,R), they will cancel out each other so that specific rotation is zero and two chiral centers are overall achiral, which gives you a meso compound.

See https://en.wikipedia.org/wiki/Meso_compound

I think a superposable image is  mirror symmetry, ie you can put a mirror somewhere in the molecule and you would be able to describe the molecule with the molecule part being mirrored and the mirror image

Your process in determining chirality is correct. However, you have to check if the compound is meso as the last step. If it is meso, then it is achiral.

Related: Stereo chemistry, optical isomerism, meso isomers

rotate one of the ends by 180 degrees and you will see a plane of symmetry.

The staggered, conformation of (2R,3S)-butane-2,3-diol (1a) that you have shown does not have mirror planes but rather a center of symmetry, which is the reason that this conformation does not rotate plane polarized light. There are two other stable, staggered conformations of the diol, namely 1b and 1c. They are a racemic pair, each of which is chiral. But like any racemate, it is optically inactive. Each of these staggered conformations has also been represented as Newman projections with the S-carbon in the front (1a', 1b' and 1c''). Rotation of 1c'' by 180o about a vertical axis passing through the center of the C2, C3 bond leads to structure 1c', which is the mirror image of structure 1b'. A time tested device for ascertaining whether or not a compound is meso is to pair like groups in an eclipsed conformation  (1a''). While this technique serves its purpose and conformation 1a'' has a plane of symmetry, its population in solution is negligible as is the racemic pair of eclipsed conformations (not shown).

At room temperature this is a free rotation and the staggered forms are transition states. They will still be significantly populated, even though they are not stable.

The molecule you show has two mirror planes. To see this rotate the molecule around the central bond until the OH groups are pointing in the same direction (C-C single bonds usually rotate freely). In this configuration there is a mirror plane through the OH groups along the C-C bond and a  mirror plane perpendicular to the C-C bond.
Build a model of the molecule and rotate it around that central bond if you can't visualise this in your head.

The mirror plane including the $\ce{O-C-C-O}$ atoms would map carbon on hydrogen and vice-versa. The molecule can only have one plane of symmetry after the rotation you propose.

A molecule is chiral when it doesn't exhibit any kind of symmetry.
This particular molecule exhibits centre of inversion symmetry, i.e. it's symmetrical about a particular point.
In this question that point is the midpoint of the bond joining the two substituted carbons.
If you would like to know about the different types of symmetry, Wikipedia has a good page on the topic.


text:
Could we bond hydrogen to the edges of covalent network structures like diamond?
I know that diamond is just carbon atoms structured in a way that differs from graphite and charcoal. The carbon atoms on the edges do not have full valence shells. Would it be possible to attach hydrogen atoms (or other elements) to complete these shells? And if this is possible, would it change the properties of the diamond?

It would be impossible not to attach anything. You just can't have dangling bonds with nothing on them. When you see a diamond, it is already covered with something: maybe $\ce{-H}$, maybe $\ce{-OH}$. And no, this does not affect the properties (most of them).

What is the edge of a diamond like?

What covers the surface is very important to device properties in diamond (and other semiconductors as well). It impacts carrier recombination and surface-induced fields.


text:
nucleophilic attack in Acid chloride formation

First mechanism is the one taught in lecture, and I wondered why the electrons from carbonyl Oxygen wouldn't react with thionyl chloride.
So I tried to draw a second mechanism as much as I could but it won't produce the same product.
So my question is:
Is nucleophilic attack from carbonyl Oxygen less likely. If so, why?
If not, where did I go wrong?

Well, $sp^2$ electrons are less nucleophilic because they are closer to the nucleus. But those oxygens are essentially equivalent because the two tautomers are the same. The first "mechanism" you show isn't really a mechanism, by the way. It's missing several steps.

by the way, why is this question duplicate? it's true I asked about Acid chloride formation twice but these are not the same question as far as i'm concerned.

I'm inclined to agree that these are different questions.


text:
Hands and chirality
I was told by my teacher that human hands are the best examples of chirality as both the hands are mirror images of each other but cannot be superimposed on each other.
My question is if we join hands together as in praying position, won't the hands be superimposed on each other?

See http://chemistry.stackexchange.com/questions/18652/concept-of-non-superimposable-mirror-image-in-chirality

In humans, chirality (also referred to as handedness or laterality) is an attribute of humans defined by their unequal distribution of fine motor skill(movement of muscles)between the left and right hands. An individual who is more dexterous with the right hand is called right-handed, and one who is more skilled with the left is said to be left-handed.  Human hands are perhaps the most universally recognized example of chirality: The left hand is a non-superimposable mirror image of the right hand; *no matter how the two hands are oriented, * it is impossible for all the major features of both hands to coincide across all axes. This difference in symmetry becomes obvious if someone attempts to shake the right hand of a person using their left hand, or if a left-handed glove is placed on a right hand.

Super imposition means when each part of both the surface points to same direction.
Before joining your hands in prayer position, assume that a mirror is kept, the left and right hand can be joined, this is infact why we call them mirror images. But rotate either hand by 180 degrees then you will notice that the thumb of either one would be on other side. Hence they are mirror images that cannot be superimposed.
Infact they are "enantiomers" kind off and mixture of them is a racemic mixture.


text:
Volume changes when mixing DMSO and water
Since I know that a mixture of 500ml $\ce{H_2O}$ and 500ml ethanol gives less than a liter of mixture I wonder how strong this effect is when mixing DMSO and $\ce{H_2O}$. I'd assume it's nonzero - but is it negligible?

See also LeBel, R. G.; I. Goring, D. A. Density, Viscosity, Refractive Index, and Hygroscopicity of Mixtures of Water and Dimethyl Sulfoxide. J. Chem. Eng. Data, 1962, 7 (1), 100–101.

follow up: do we expect this to be significantly different at 20°C, which is a more likely temperatur in real life...?

If we assume that the density of the mixture is just the average of the individual densities, would get 1 litre of a mixture with a density of 1.04635 kg/l.
However, measured density (1, table 1) is 1.0682.
We thus expect to see a volume that is 0.9795 times the added individual volumes.
In this case: 0.9795 litres.
LeBel, R. G.; I. Goring, D. A. Density, Viscosity, Refractive Index, and Hygroscopicity of Mixtures of Water and Dimethyl Sulfoxide. J. Chem. Eng. Data, 1962, 7 (1), 100–101


text:
Resonance in phenoxide anion
Why is resonance possible in phenoxide anion as when the electron pair of oxygen is involved in conjugation we have 4 pairs of electrons in resonance which should imply that the ring has become anti aromatic and hence no resonance should be possible?

The Huckel's rule does not apply here. It is valid for monocyclic structures without any side groups.

I know that it is valid only for monocylic rings but I am not sure about the side group constrain because I have been given questions as practice where a carbon in a ring was attached to an oxygen by a double bond.

http://chemistry.stackexchange.com/questions/22100/is-the-aromaticity-broken-in-some-resonance-structures-of-para-nitro-aminobenzen

@Mithoron In the phenoxide case,doesn't the compound become anti aromatic owing to 8 electrons and hence resonance shouldn't even be possible?(I am confused about counting electrons while judging it's aromaticity.Is it the number of electrons present in the ring in any resonating (in phenoxide case it will be 6 as in any structure 3 pairs of electrons are present in ring) or the number of electrons in conjugation (which becomes 8 in phenoxide case including 3 pairs of electrons in benzene and 1 pair of electrons from oxygen).

No the electrons outside the ring are not counted


text:
Production of hydrogen through the reaction of aluminium metal and sodium hydroxide
If I have a solution of 25% $\ce{NaOH}$, and I simply add a cut-up aluminium can or some aluminium foil, will it react into hydrogen and sodium aluminate?
In principle this reaction should happen, but it didn't work when I attempted it. Perhaps an oxide layer prevented it, or a protective layer of sodium aluminate was formed. (Perhaps my experiment was faulty.) Should this happen, and what can I do to fix it? (My goal is to generate hydrogen gas.)

Try solid solid Drano (aka Drano Crystals) https://en.wikipedia.org/wiki/Drano#Crystal_Drano Two of the main ingredients are sodium hydroxide (lye), and aluminum.

Oxide layer would not hold against the strong alkali, and sodium aluminate does not form protective layer either. I suppose cans and foil may be covered by some coating, to make them suitable for food grade applications. Try burning it off, or rather use other sorts of aluminium (cut-up wire, maybe). Normally it would react quite vigorously.

You are probably correct, there is a coating on food-grade aluminium (I scraped it off and the experiment succeeded).


text:
Supersaturated gas solution
Can you make a supersaturated gas solution?
As we know solubility decreases when we increase the temperature, So If I do that, will the gas molecules escape the solution or will they stay in the solution and make a supersaturated solution? Or is it just a plain fact that you can't make supersaturated solutions except with solids solutes?
I am kind of leaning towards escaping, because it is hard to keep gas molecules intact with water molecules at high temperatures.

Observe carefully a bottle of cola :)

Every can of soda, coke etc. is supersaturated with carbon dioxide after opening, i.e. relieving the pressure in the can.
The surface tension of a gas bubble adds to the outside pressure, and more pressure always increases solubility of gases, as you know from Le Chatelier.
Same with gas supersaturated with a condensable gas or solvent with too much solute, you have to overcome the surface tension of small particles or droplets before precipitation can take place.


text:
Do Moore's tables really exist?
I recently have a lesson about quantum chemistry, and in one lesson we spoke about hyperfine stucture of elements.
The teacher give us something which looks to be an extract of Moore's tables here for the carbon. He said there are thousands of this tables which give the energy levels of many atoms.
But I search on the net and I only found things about electronic such as transistors et caetera... can you tell me the real name of this tables or how can I find them ?
Thank you in advance

Yes, Charlotte Emma Moore and her tables really exist(ed).
See Atomic Energy Levels, 2nd edition.

Moore's law probably does interfere drastically with such a search...


text:
Ionization Energy and Electron Affinity

This is from my textbook.
But in other reference books I am using and Wikipedia, there is not any mention of absolute zero while defining Ionization Energy and Electron Affinity.
What does it mean? Please Elaborate

At temperatures higher than zero, electrons may gain energy from thermal excitations. https://en.m.wikipedia.org/wiki/Electron_excitation   Because of these thermal excitations the highest occupied state may change. However, I am not sure if temperature is that important. I would think that the temperature that a material has to reach to for electrons to go above the Fermi energy would have to be quite high

I think the definition of the ionization energy is related to the ground state configuration, that is implied at the zero temperature.

This is a practical matter of making sure that there is a common basis for comparing values from all different types atoms and molecules. If it were not the case there would be a range of different values for the same type of atoms or molecules.
Depending on temperature the ionisation energy would be less due to a molecule's rotational and vibrational energy, and vary with temperature because the number of levels populated depends on the temperature. For some types of atoms there may be low lying electronic levels that could be populated at higher temperatures, say at 1000 K in a high temperature gas. Again the measured ionisation energy would depend on temperature.
Also, if this was not done the $\Delta H$ values would in effect have double counted the effect of temperature, once in the ionisation energy and again in the heat capacity term.

If  ionisation enthalpies are measured at absolute 0, then how can  atom be at gaseous  state at the same time ?,   gaseous state is one of the prerequisite for measuring ionsiation energy.

They are defined at absolute zero which means from the lowest energy level, but in practice, as you realise, this is not possible so they are measured however it is possible to do so and then corrected as if at zero K.

The first answer is correct because as its highest occupied state changes,the energy required to remove it valence electron also increases.So one will have to supply the minimum energy for its ionisation which is its characteristic at a particular temperature and the additional energy as its occupancy has changed.

When making definition all cases are to be considered.the necessity to include absolute zero temperature is because of the energy being spent from the energy you give or supplied energy for reaction by reactant to change it's volume and pressure.TsO(delta(r)C(v)(p)dT determines this change.if you compare this to simple equation for enthalpy this term equal to PV. For ideal gas PV=nRT.here T is temperature in absolute scale.So we can see enthalpy of same reaction can variety with value of T . For this absolute zero is taken as the standard to define enthalpy of any reaction like the reaction for electron affinity or ionization.


text:
Is this amide compound electrophilic enough to be attacked by a cyclopentadiene in base?

I am trying to make a fulvene molecule with dimethylamino and phenyl substituents. My worry is that the amide group donates electron density into the carbonyl carbon making it less electrophilic.

Probably this won't work.  I'd try an enamine route instead

Although I have no prove for this, I am almost completely certain, that this will not work.
If you look closely at the cyclopentadienyl anion, what you would like to utilise as the nucleophile, you will notice that it is and aromatic system. (It even completely obeys the Hückel rules.) There are no electrons in an orbital of suitable symmetry, that would allow attacking the π* C-O bond.
In the other molecule, as you have noted, the electrons of the nitrogen will be interacting with the π carbonyl bond. Moreover, this part of the π system is also overlapping with the aromatic system of the phenyl moiety. That will certainly remove quite a bit of charge density from the carbonyl carbon. A nucleophilic attack at that position seems highly unlikely.


text:
Making Ferric Chloride using HCl and Ferrum
I am not chemist: I live in the part of the world that you can't find many things, but I have HCl and bunch of nails. I there way to make ferric chloride from these things?
If this doesn't work what else can I do to make ferric chloride?

You can also see this:- http://chemistry.stackexchange.com/questions/41095/formation-of-ferrous-chloride-or-ferric-chloride

You can find a glass beaker somewhere and simply drop in iron and diluted HCl. That's enough to make $\ce{FeCl2}$.
$\ce{Fe + 2HCl -> FeCl2 + H2}$
That forms hydrogen gas, which is very flammable, so it would be wise to do this outside. You might have to heat or reflux (heat the hell out of it while condensing whatever tries to evaporate out) depending on what concentration you use. Anyway, the resulting solution is gold-colored.
You'd then need to oxidize the Fe from a +2 state to a +3 state. I found this guide that recommends hydrogen peroxide (probably as a safer alternative to potentially stronger oxidizing agents.) It says it can take several days, so that's something to watch for.
Good luck! Again, I'd definitely do this outside or in a lab.
http://www.instructables.com/id/Make-Ferric-Chloride-for-etching-printed-circuit-/
Also, I've heard that not all nails are made of iron, so a density test could help ensure that.

"Heat the hell out of it" is certainly advice I would not want to follow. I think this post lacks a certain level of security instructions, especially since there is a highly flammable gas produced.

Oxidation of Fe from +2 to +3 is not something you need to care much about. Normally it happens by itself, thanks to the oxygen in the air; BTW, that's what may take several days. As for the oxidation with hydrogen peroxide, it occurs instantly.


text:
What is called a salt?
While I was doing experiments in my school days, they gave me different substances with the combination of base and acid radicals, for example ammonium nitrate and zinc chloride and they seldom called it a salt.
But actually a salt is a NaCl as far as I know.
So My Question is: Actually, what is called a salt?
Why do they call the acid and basic radical a salt?

See https://en.wikipedia.org/wiki/Salt_(chemistry)

Actually definition on Wikipedia is rather bad as most combinations of cations of metals and anions of typical acids isn't predominantly ionic.

The IUPAC redbook gives the following definition:

A chemical compound consisting of an assembly of cations and anions.

Red Book, p. 118

@getafix Make it an answer! You might want to link to it though: http://goldbook.iupac.org/S05447.html

I'll just point out that when you go to the grocery store and buy "SALT" you are indeed buying NaCl. It is unfortunate that the name has been used with a different meaning in chemistry, but the chemistry usage is so ingrained historically that it will probably never change.

I think what confuses you  is that 'salt' has two meanings: In the every day sense it means table salt (which has the formula NaCl) but in the chemical sense it means a certain way for particles to hold together and form a tangible substance.
I'd define the latter as any periodic arrangement/lattice (see e.g. here ) of particles held together by ionic bonds.
The positive ion is almost always a metal, the negative one can be anything (so usually either a halogen (those lack only one valence electron to have a full shell, see e.g. Cl) or a (smallish) molecule-ion like SO4^2−)

base and acid radicals

I really don't see much connection between salts and either bases/acids and radicals. Do consider that not all ions are bases/acids and all radicals are highly reactive. It would rather be an edge case to have a salt with a radical in it; most salts have ions with energetically favourable number of electrons (as a rough rule this means all "electron shells" are either full or empty.)
NaCl is a good example.
Because different sized salt crystals still have extremely similar behaviour, the salt formulas express only the ratio of anions and cations. Realize that on the whole salts are neutral.

Why do they call the acid and basic radical a salt?

A single kind of particle is never called a salt, salt means an arrangement of two ions of different polarity. I guess some lab 'recipes' have sloppy language if something is provided as a salt, but only one component is important for the reaction.
Perhaps I can give a better answer I understood what you mean by radical in this context.

The obsolete (19th century) terms “acidic radicals” and “basic radicals” describe anions and cations, respespectively, in salts. They can still be found in some textbooks concerning qualitative inorganic analysis (in particular in India).


text:
Why are anti aromatic compounds unstable?
Just like aromatic compounds, anti aromatic compounds have several resonance structures and some of them are also conjugated ( 1,3-cyclobutene).
Why is it that just having 4n electrons makes them unstable?

The intricacies of aromacity are still being researched. From what I recall, it still not clear as to what exactly aromaticity is. There are  recurring features along aromatic compounds: electron delocalization, stability, and unreactivity. However, these are not completely understood. This is largely due to the fact that electron delocalization and localization is hard to measure directly (some even say that it is not measurable at all).

http://chemistry.stackexchange.com/questions/30927/what-is-the-justification-for-h%c3%bcckels-rule

In conjugated-pi ring systems of N atoms, you typically have a single low energy orbital and then paired (degenerate) energy levels until you have N orbitals total. This has to do with the number of nodes in the conjugated pi system, and the number of ways you can lay them out with N atoms.
Now, when you add in electrons you add two electrons per orbital, so your 4n+2 electrons fill 2n+1 orbitals -- the single low energy orbital and then n complete sets of degenerate pairs.
If you only have 4n electrons, you can fill the single low energy orbital, then 2(n-1) complete sets of degenerate pairs ... but with two "extra" electrons which aren't paired up. This allows those two electrons to "spread" out between the pair of equal-energy orbitals. Those unfilled/partially filled orbitals mean that the anti-aromatic compound is quite reactive.
The other issue is that "stability" is not an absolute property. Stability is always in reference to some other state. The typical anti-aromatic systems you'll encounter tend to have those frontier orbitals as non-bonding or anti-bonding orbitals, meaning the electrons in the orbitals aren't really contributing to the molecular stability. Generally, if you build the energy level diagrams for alternate systems (e.g. a bent ring where you don't get localization) you'll find that - while the lowest energy orbital isn't as low energy as in the ring-conjugated system - the sum of the energies across all orbitals is indeed lower. And it's the total energy of the system which is optimized. (For example, in cyclobutadiene the ring-conjugated form would have two electrons in a low-energy bonding orbital and two in non-bonding orbitals. In contrast, linear butadiene bonding orbitals are not as low energy as for the ring-conjugated cyclo form, but there's two of them. And the sum of the energies from two linear bonding orbitals is lower than what you get from a single cyclo bonding orbital.)
(Final note: this is all high-level generalization. There's more sophisticated levels of quantum mechanical theory where things get more complex.)


text:
Reaction between sodium carbonate and water
In this Wikipedia article, it mentions:

As the concentration of sodium carbonate increases, it undergoes
  hydrolysis to form sodium hydroxide.
  $$\ce{Na2CO3 + H2O → 2NaOH + CO2}$$

Because this occurs in boilers, I can assume that it requires a heat of no more than 100 degrees Celsius. Why is this process not used when commercially producing $\ce{NaOH}$, rather than the chloralkali process?
Possible reasons:

Not enough concentration of sodium hydroxide is formed
The sodium hydroxide is contaminated with significant amounts of sodium carbonate
The reaction is too slow

The three reasons you propose are technically true, but they are not the reason why this is not the process used to produce sodium hydroxide.
But let me first clarify some points (and rant about a bit).
Hydrolysis is a reaction where a compound is split whilst incorporating a molecule of water. The reaction :
$$\mathrm{Na_2CO_3 + H_2O \rightarrow 2 NaOH + CO_2}$$
is not a hydrolysis. It is classified as an acido-basic reaction ($\mathrm{H^{+}}$ exchange) followed by a dehydration reaction. A dehydration is kind of the opposite of an hydrolysis (the true opposite would be called hydration), because is involves the loss of water. The transformation decomposes according to (forgetting about sodium ions):
$$\mathrm{{CO_3}^{2-} + H_2O \rightarrow {HCO_3}^- + HO^-} \\
\mathrm{{HCO_3}^{-} + H_2O \rightarrow {H_2CO_3} + HO^-} \\
\mathrm{H_2CO_3 \rightarrow CO_2 + H_2O }$$
$\mathrm{H_2CO_3}$ is carbonic acid. Carbon dioxide is its anhydride.
The overall transformation actually consumes two molecules of water as acid, and one molecule of water is produced upon dehydration of carbonic acid. But this does not appear in the simplified equation above.
This reaction is actually not thermodynamically favorable ; the reverse transformation is the one that is spontaneous. What drives it here is the fact that the carbon dioxide is gaseous and is driven off the water, into the atmosphere. Unless you work under a $\mathrm{CO_2}$-free atmosphere, it would not be possible to produce sodium hydroxide not contaminated with carbonate.
Now, why is this not used industrially ? The answer lies in the cost of the processes : investments, availability of reactants, price of energy, etc.
Calcinating sodium carbonate at more than 100°C actually gives off sodium oxide ($\mathrm{Na_2O}$), but it also releases $\mathrm{CO_2}$ in the atmosphere, and you have to obtain sodium carbonate first. This is no commonplace mineral. Most of it is actually produced by the Solvay process, which converts two plentiful materials, namely calcium carbonate and sodium chloride into calcium chloride and sodium carbonate :
$$\mathrm{2\,NaCl + CaCO_3 \rightarrow Na_2CO_3 + CaCl_2}$$
The chloralkali process produces sodium hydroxide (and chlorine) directly from sea salt, and the  products are hardly contaminated.

So essentially, among other reasons, making sodium carbonate to perform this reaction, added with the cost of energy to drive the reaction makes it too inefficient compared to the chloralkali process?


text:
Carbocation ring expansion jump

is the given rearrangement possible {major product}? from cyclo butane derivative to cyclohexane derivative.

How did you come up with this?

https://www.youtube.com/watch?v=TzIfHpZiAXY

Q-7 of the video


text:
solubility trends of alkaline earth metals
Why is the solubility trend of group 2 oxalate compounds inconsistent with the actual order in the periodic table?

Related: What is the order of solubility of oxalates in group 2 (alkaline earth metals)?

Because the periodic table has many such apparent oddities. Such oddities are our fault of trying to impose strict simple rules on a messy universe.


text:
Dexamethasone radius of gyration source
I am looking for a source for the radius of gyration of dexamethasone in aqueous solution. Does anyone know where I could get it from? Even better would be an estimation of the Stokes radius, but the radius of gyration will do.


text:
Reaction of iron with steam
While studying about reversible reactions, my professor stated that the following reaction is a reversible reaction:
$$\ce{Fe + H2O <=> Fe3O4 + H2}$$
She also stated that the product $\ce{Fe3O4}$ is formed by a mixture of $\ce{FeO}$ and $\ce{Fe2O3}$. However, there was no mention of the type of reaction or the reason due to which the mixture of $\ce{FeO}$ and $\ce{Fe2O3}$ resulted in the formation of $\ce{Fe3O4}$.
So, my questions are as follows:

What type of a reaction is it? (I think it's a displacement reaction but from what I know, there are ions present in a displacement reaction).
Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$? (According to me, mixtures and compounds are two different things and that we can't combine them.)

We didn't combine anything. Nature did.

Our professor stated that the product is a mixture (excluding hydrogen gas). And mixtures do not form compounds. Then how were the two combined?

Your professor is wrong, or you misunderstood him. $\ce{Fe3O4}$ is a compound and not a mixture.

@Ivan Neretin So $\ce{FeO}$ and $\ce{Fe2O3}$ react to form $\ce{Fe3O4}$? Or the product is directly $\ce{Fe3O4}$?

I suppose the product is directly $\ce{Fe3O4}$.

@Ivan Neretin And what about the kind of reaction? I think its displacement as Iron displaces Hydrogen from its compound. But I have learned that displacement reactions take place when a compound is broken down into its ions and then the element with higher reactivity combines to form a new compound and displace the previous element from its compound. However, I do not think that there is any ion involved in the reaction.

In general this is a redox reaction. redox = reduction and oxidation. The iron is oxidized, and the hydrogen from the water is reduced. In a replacement reaction there wouldn't be any change of oxidation state.

And $\ce{Fe3O4}$ is $\ce{FeO \cdot Fe2O3}$. There is no $\ce{Fe^{4+}}$. So within the compound the iron is in two different oxidation states. It isn't a "mixture" since $\ce{FeO}$ and $\ce{Fe2O3}$ are in a definite 1:1 ratio - sot it is a true "compound." https://en.wikipedia.org/wiki/Iron%28II,III%29_oxide

Single displacement reactions do involve redox -- the element on the reactant side ends up in a combined state in the products and, therefore, must undergo a change of oxidation state.  As to whether or not Fe3O4 is formed directly, that is doubtful since it would require an activated complex of three iron atoms and four water molecules -- very unlikely.  I would guess that there are a number of steps involving various iron oxide and hydroxide intermediates.

Regarding whether or not Fe3O4 is a compound, it surely is as stated by MaxW and Ivan Neretin.  However, the formula FeO⋅Fe2O3  can be misleading in that it might seem to suggest that Fe3O4 is made of FeO and Fe2O3 "molecules" in chemical combination.  Actually, Fe3O4 has a crystal structure (an extended geometrically regular array) of Fe(III), Fe(II) and oxide ions in which the oxide ions exist in a cubic close packed pattern with the Fe(II) ions located at half of the octahedral sites and the Fe(III) ions distributed evenly between octrahedral and tetrahedral sites.

FeO⋅Fe2O3 is the mineral magnetite. It forms nice octahedrons. https://en.wikipedia.org/wiki/Magnetite

What type of a reaction is it?

I would describe it first and foremost as a redox reaction. Iron is oxidised and hydrogen is reduced; the half equations would be:
$$\ce{2Fe -> Fe^2+ + Fe^3+ + 5e-}$$ $$\ce{H2O + 2e- -> H2 + O^2-}$$

Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$?

This is a good observation. $\ce{Fe3O4}$ contains both iron(II) and iron(III) ions and is sometimes written as $\ce{FeO \cdot Fe2O3}$. The dot in the middle signifies that there is some form of bonding between the two compounds (another common example of this is water of crystallisation). $\ce{Fe3O4}$ is a compound and not a mixture because it does not consist of two separate $\ce{FeO}$ and $\ce{Fe2O3}$ phases but rather it is a single crystal structure containing $\ce{Fe^2+}$, $\ce{Fe^3+}$ and $\ce{O^2-}$ ions.

$\ce{Fe3O4}$ has a cubic inverse spinel structure which consists of a cubic close packed array of oxide ions where all of the $\ce{Fe^2+}$ ions occupy half of the octahedral sites and the $\ce{Fe^3+}$ are split evenly across the remaining octahedral sites and the tetrahedral sites.