What chemicals cause the softer, less-raw taste of purified water sold in bottles?

Water issues tend to be very specific to a certain location, so it is not possible to say exactly for sure what is happening in your situation.

Pure water has no taste.  The taste comes from the impurities.  Impurities, which are almost always present, can partially be removed by the filter shown.  That filter looks like a carbon filter which is effective on chlorine, organic particles, and sediment.  This filter is ineffective on minerals and inorganic compounds.

Minerals give tap water its hard taste. This hard taste is harmless (and may actually serve as a dietary mineral supplement.) Removing the hard taste requires a process like deionization or distillation which is costly. The cost of the water can go up significantly with more mineral removal.

Chlorine, which is used to kill pathogens, can be neutralized with citric acid.  Squeezing a lemon or lime in the water can cut down on a chlorine taste.

Hard water can be chemically softened with something like baking soda; however, this is generally trading one ion for another, so the water will only taste softer.

Try adding a reverse-osmosis filter after the filter shown. The filter shown will treat the chlorine while the reverse-osmosis filter can treat the minerals. (Chlorine may damage a reverse osmosis filter, so it is best to get it out first.)


Why does Co(I) have a 3d8 configuration?

For lighter elements, the shells fill in order. Starting at the transition metals, an outer s orbital may fill before an inner d orbital, so the electron configuration of unioninzed cobalt is written $\ce{[Ar]}4\mathrm s^1\,3\mathrm d^7$, rather than $\ce{[Ar]}3\mathrm d^7\,4\mathrm s^1$.
There is a video diagramming the electron configuration of $\ce{Co}$, $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$, thought it does not explain the reasoning, nor does it cover the less common $\ce{Co^{+}}$ ion, produced by photoionization or as found in some esoteric metal-organic compounds, or in the theoretical $\ce{CoCl}$.
That said, the situation becomes murky for transition elements, and downright turbid for lanthanides and actinides, where f orbitals are added. For example the outermost shells of $\ce{La, Ce and Pr}$ are $\ce5\mathrm d^1\,$, then $\ce4\mathrm f^1\,5\mathrm d^1$, and then $\ce4\mathrm f3$. What happened to the Pr d electron? My understanding is that energy levels are quite close for the larger outer shells, and it is not intuitive which orbital fills first.


Do epoxied rings react with bases

Yes, they can be opened, e.g. with alkoxides.

  what is the mechanism?

In principle, this is a $S_N2$ reaction, with the typical approach of the nucleophile from the back side. The difference to a non-epoxide case is that removal of the leaving group (bond breaking, ring opening) is facilitated by the ring strain.

But the general rules for the reactivity of alkyl-substituted centres in $S_N2$ nevertheless apply: the higher the substitution, the lower the rate. This is attributed to the steric hindrance of the respective centre.

Note that the regioselectivity is different under acidic conditions!


Can butane-2,2,3,3-tetraol be oxidized by potassium permanganate?

You are talking about the dihydrate of biacetyl.  gem-diols are generally not stable, the equilibrium is on the side of the aldehyde/ketone, except when the keto group is next to an electronegative atom.  Glyoxal in aqueous solution is all hydrate and oligomers, as is chloral.

Biacetyl is more hindered than glyoxal, but in aqueous solution one would think they there is still some hydrate in the equilibrium.  What you'd think would happen is that on reaction with permanganate you'd get the cyclic diol and under sufficiently forcing conditions (read: acidic medium) you'd get cleavage to acetic acid.

Biacetyl is awful.  Run away!


Why is chromate stable in basic medium and dichromate stable in acidic medium?

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
The preface of the book says:

The approach: These reactions represent important facets of the elements involved but are typically presented as no more than facts. (Why does boiling white phosphorus in alkali lead to hypophosphite and not phosphate?—Current texts make no attempt to address such questions.) Arrow pushing demystifies them and places them on a larger logical scaffolding. The transformative impact of this approach cannot be overstated. Almost to a person, students who have gone through our introductory course say that they cannot imagine how someone today could remain satisfied with a purely descriptive, nonmechanistic exposition of inorganic main-group chemistry.

Potential concerns: Given the plethora of advantages of a mechanistic approach, it’s worth reflecting why it has never been adopted for introductory inorganic chemistry. A plausible reason is that, in contrast to common organic functional groups, simple p-block compounds such as hydrides, oxides, halides, and so forth, tend to be much more reactive and their vigorous and even violent reactions have been much less thoroughly studied. As good scientists, inorganic chemists may have felt a certain inhibition about emphasizing an approach that has little grounding in experimental fact. This is a legitimate objection, but hardly a dealbreaker, in our opinion, for the following reasons.

Our ideas on main-group element reactivity are not taken out of the blue but are based on parallels with well-studied processes in organic and organoelement chemistry.
Second, it no longer necessarily takes a prohibitive amount of resources to test amechanistic proposal, at least in a preliminary way. Quantum chemical calculations, particularly based on density functional theory (DFT), very often provide an efficient and economical way of evaluating reaction mechanisms.
Third, and perhaps most important, it’s vastly better to be able to formulate a hypothesis on how a reaction might happen than to have no inkling whatsoever about the mechanism

What all of this basically means is that in this approach, you use the classical organic chemistry concepts of electrophiles and nucleophiles to write an arrow-pushing mechanism for the reaction. It will only be a probable mechanism, but often, it's not much far away from the reality, and this methodology can serve to be a pretty great teaching tool for inorganic chemistry, so long as you accept the fact that the mechanism you have made can only be proved by a research paper. But having an idea is also good.
Having got the introduction and disclaimer of this method out of the way, we can now start to explain your phenomenon by arrow-pushing diagrams.
The dichromate anion $\ce{Cr2O7^2-}$ has got a structure consisting of a $\ce{Cr-O-Cr}$ bridge, four $\ce{Cr=O}$ bonds, and two $\ce{Cr-O^-}$ bonds.
Here, the $\ce{Cr=O}$ bonds can be said to be structurally similar to the carbonyl group.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Cr}$ atoms due to the electron-withdrawing effect of two $\ce{Cr=O}$ groups per $\ce{Cr}$. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage as follows:

The $\ce{HCrO4^-}$ formed can  again be deprotonated in the basic medium to form $\ce{CrO4^2-}$
In acidic medium, the $\ce{CrO4^2-}$ anion can be protonated to form $\ce{HCrO4-}$ , which can react with another $\ce{CrO4^2-}$ anion in the following manner to give $\ce{Cr2O7^2-}$ and expelling water as a leaving group:

Any alternate protonation pathways, for example, protonating $\ce{CrO4^2-}$ twice to form $\ce{H2CrO4}$ and then reacting it with $\ce{CrO4^2-}$ might also give the same product with a few extra proton transfers here and there, but the way I have presented thing would be the most probable way, as the reaction is progressing stepwise with each protonation.
Note: Remember that these are just probable pathways we have drawn to reach our product, but there's a good chance they resemble the reality( which can only be determined by kinetic experiments) and as the authors have stated, it is a good learning tool, and not too far off the mark.


There are not two possibilities for low spin and high spin complex for a d9 metal ion.
The only cases where you can have both is for d4, d5, d6 and d7.

Remember that both configurations arise from the possibility of organizing the d electrons in a more stable way, this is, with the lowest possible energy.
You need extra energy for doing two things:

Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, $\Delta_O$
Puting another electron on a semioccupied orbital (pairing energy, due to repulsions)

In the complexes where $\Delta_O$ is higher than the pairing energy, the result will be a high spin configuration (it's easier for the electrons to become promoted to a higher level than pairing them with another electron), and if it is smaller, the result will be the opposite.
But this only applies for d4, d5, d7 and d8.

Now think about d1 d2 and d3:
It's nonsense to put the electrons on the upper level, when they can be all by their own in lower energy orbitals, so there are no high spin/low spin possibilities, but only one possible case, the three electrons one on each $T_2g$ orbital.

Now d8 d9 and d10.
There is once again, only one possibility to put the electrons. For your case, d9, you have nine electrons, so 6 of then will be filling the lower energy orbitals for sure, and the other 3, on the upper orbitals.
You may think, well, there is another possibility, promote one of the lower energy electrons to the hole on the higher energy orbital. But this is again nonsense: you are looking for the most energy favorable situation, so, as long as you only have to add to the system the pairing extra energy, why would you also add the promoting energy instead of filling a semioccupied lower energy orbital?
This would result in a excited state of the complex, but this is not what you're looking for when you talk about high/low spin.
The same applies to d8 and d10, there is only one possible situation, and no high/low spin possibilities.


Is there a universal minimum energy state?

This is an interesting and actually surprisingly hard question, which reaches the edge of our understanding of physics, so really a physicist would be more equipped to answer this.

I'm not sure if there is rigorous mathematical/physical proof that energy must have a minimum. Energy is a continuous, real-valued, state function, which as far as I know makes it mathematically very well-behaved, so it's unlikely that it can display weird behaviour. The extreme value theorem suggests that taking any finite collection of states into account, at least one of them must be a global energy minimum. However, I'm not sure if the Universe has a finite amount of states it can assume (probably not, if it is infinite in size). All that said, even if strictly speaking there is no global minimum, it's likely there is some asymptotically low value. So what is it?

What you're trying to imagine, essentially, is a Universe where there is no kinetics, only thermodynamics. Equivalently, you're thinking about the state of the Universe at $\mathrm{t→+ \infty}$. All finite kinetic barriers are surpassed and all finite distances are traversed. With this much time, a lot of unintuitive events take place. Two very interesting sources to read are "Time without end: Physics and biology in an open universe",  (1979) Rev. Mod. Phys., 51, 447, and "A dying universe: the long-term fate and evolution of astrophysical objects", (1997) Rev. Mod. Phys., 69, 337. There are many processes, varying from almost certain to extremely speculative, which can happen in overlapping time scales, so it is hard to determine exactly which route the Universe (or a toy model of a universe) might take. I'll present several events which may happen as the Universe marches on towards $\mathrm{t→+ \infty}$, and thus which reach closer to the minimum energy state of the Universe, but be aware that there are many uncertainties, especially near the deep end.

Starting with chemistry, allowing systems to evolve for very long periods would have some interesting consequences. Only the most stable compounds could survive, so for example, all nitrogen atoms would eventually recombine to form gaseous $\ce{N2}$. Other very stable molecules would be $\ce{H2O}$, $\ce{CO}$, and $\ce{H2}$. The Universe would turn into a mixture of small molecules. This would take at least something of the order of $10^{65}$ years. For comparison, the Universe is currently $1.38\times 10^{10}$ years old.

But why stop there? Chemical energy is only one of many factors to be taken into account in the total energy, and really it's one of the smallest. For example, nuclear energy scales completely dwarf chemical energy scales, so forget about optimising molecular abundances to minimise energy, because optimising atomic abundances will reach much lower energies. What would really happen is that, given tremendous amounts of time, light/heavy nuclei would overcome the barrier to fusion/fission, even at very low temperature and pressure. As you mention, the ultimate effect would be that all of the atoms in the Universe would fuse/fission into $\ce{^{62}Ni}$, the nucleus with the highest nuclear binding energy per nucleon. This would take on the order of $10^{1500}$ years.

This still isn't the end, though. Given a ridiculous amount of time, much of the nickel could still coalesce and compress into degenerate neutron matter (like neutron stars), releasing even more energy. The kinetic barrier for this is truly colossal, though, so it would take some $10^{10^{76}}$ years.

It's possible to keep going. Sufficiently large chunks of matter could compress further into black holes, releasing even more energy. Depending on the mass of the lightest possible black hole, this process may take "only" $10^{10^{26}}$ years, or up to a scale comparable to neutron matter formation. If the lightest possible black hole has a non-zero mass, then objects lighter than that threshold may be stable indefinitely. The black holes that did form would evaporate almost instantly relative to this time scale, producing an abundance of subatomic particles which would eventually decay into photons and neutrinos.

There's another speculative process which may severely undercut the previous possibilities, however. Currently many types of string theory posit the existence of proton decay. If it does happen, it could happen as quickly as $10^{40}$ years. This would be the end of Chemistry in the Universe; all atoms would eventually decay into photons, neutrinos, electrons and positrons. Black holes would survive this, but they would evaporate by $10^{100}$ years at the latest, quickly resulting in a Universe sparsely filled only with photons and neutrinos.

Last but definitely not least, it may be that even completely empty vacuum is not the minimum energy state of the Universe. It's conceivable that the vacuum we know is less stable than another type of vacuum, separated by an enormous kinetic barrier. If this is the case, then given enough time there could be a vacuum metastability event, a tremendously energetic process in which the Universe as we know it would cease to exist, being replaced by another with entirely different physics.


Finding reaction rate constant for second order equation in Wolfram Mathematica

The general second order equation is
$$\frac{\mathrm d[A]}{\mathrm dt}=-k[\ce{A}][\ce{B}]\tag{1}$$
with $[\ce{A}]_0,$ $[\ce{B}]_0$ the initial amounts. At a time $t$ there are $[\ce{A}]_0-x$, and $[\ce{B}]_0-x$ remaining if amount $x$ reacts. Thus
$$\frac{\mathrm dx}{\mathrm dt}=k([\ce{A}]_0-x)([\ce{B}]_0-x).\tag{2}$$
You should be able to integrate this from $0\to x$ by using partial fractions or directly using computer algebra. The result is
$$\ln\left(\frac{[\ce{A}]_0-x}{[\ce{B}]_0-x}\right)=k([\ce{A}]_0-[\ce{B}]_0)t+\ln\left(\frac{[\ce{A}]_0}{[\ce{B}]_0}\right),\tag{3}$$
so that you can plot (as Maurice has pointed out) the left hand side vs $t$ and obtain the rate constant $k$ from the slope $k([\ce{A}]_0-[\ce{B}]_0)$ in units $\pu{dm^3 mol^-1 s^-1}.$


Beside determining the volumic density (tin: 7.31 g/ccm, aluminum: 2.7 g/ccm, etc. pp) and as asked here on chemistry SE, there is a quick pre-test.  In German it is known as "Leuchtprobe", literally "illumination probe".  Be reminded to protect yourself properly (safety googles, fume hood, gloves).

In a first instance, you have to reduce the tin compound.  For this, a little of your solid under investigation is finely mixed together with some grains of zinc (Zn). In a porcelain dish, to this mixuture is added 5 mL of 20 m% aqueous HCl.  Await this reaction ceased.  Zn will produce hydrogen, and because it is freshly created (statu nascendi), it is of superiour reducing power compared to bottled hydrogen gas, and the solution is cooled to room temperature.

Then dip a cool test tube, partially filled with cold water, into this solution.  While still wetted by the solution, rapidly  place the test tube into the upper part of the non-luminous flame of a Bunsen burner.  Blue fluorescence is an almost specific evidence for tin (caused by $\ce{SnCl2}$, and stannanes), already at low concentrations.

Presence of arsenic may hamper this fluorescence.  Niob-compounds yield some similar fluorescence, too.  Beside using tin-free chemicals and equipment to be used -- and prior to the test of your unknown -- do at least once the test in known absence of tin, and once with a sample with tin to learn how to discern the two instances.


Role of Alcohols(EtOH,MeOH) in nitro reduction used for Metal(Sn,Fe)

LHM...Look at this mechanism that I posted as an answer to an earlier SE Chem question/answer relating to the reduction of nitrobenzene by metals.

In most reactions that occur on metal surfaces, many of the mechanistic details are not fully understood (see @Martin's comments in the earlier post).  Nonetheless, this mechanistic scheme shows several steps where a proton ($\ce{H+}$) is required.  These proton transfers are necessary to complete the reduction process.  This is why these reactions are typically run in solvents, like ethanol, that have available protons that can be donated in order to complete the reaction...ron


Ester as a side product in the synthesis of bromoalkane from alcohol. Is it correct?

Technically it could form, but it would be hydrolyzed. If it did form, it would be in negligible quantities. Lots of other stuff could form, and they do, but in miniscule amounts ... bromine oxidized the alcohol to the aldehyde. The aldehyde undergoes aldol condensation. Another alcohol undergoes elimination. The compounds  react in a hetero diels-alder. So dihydropyrans can also form ... Yes, but not in quantities that are observable.


They don't have values there because they aren't on the pauling scale of electronegativity, as they don't form any compounds with other elements. However, argon and neon can technically form compounds with other elements; it is just extremely unlikely. For example there is a compound called Argon Fluorohydride, HArF, that can exist, but is only stable at around 27K (-246C). I don't think Helium can form any compounds other than with itself.
The other noble gases also can react with fluorine (As well as other elements), and make slightly more stable compounds. Hope that helps :)


Reversible and irreversible cyclic processes

Answer b is incorrect for the reversible Carnot cycle.  The book has given the wrong answer. Answer c is correct if only one cycle of each process is carried out.  You can obviously convert more heat to work if you run process 2 through the same cycle several times, which also counts as a cyclic process.


Are there computer searches for new theoretical minerals in a similar vein to Folding@Home or SETI@Home?

By Ross Snider's request, I have moved my comment into an answer - whilst it does not address computational searches for minerals, it touches on the complexities of actually getting a mineral officially named:

The International Mineralogical Association Commission on New Minerals, Nomenclature and Classification seems to require a physical specimen for consideration for official naming of a new mineral. Under current rules, anthropogenic minerals are also excluded, so if you came up with a hypothetical mineral it seems you would have to also find a naturally occurring specimen to get it named.

To add to this, I am not aware of any software for systematically searching for plausible minerals, however computational packages for the simulation of crystalline structures certainly exist. Determining the electronic structure of a crystal - which is what is really required to make conclusions about geometry, stability and so on - generally involves the use of plane wave QM codes such as VASP, CASTEP or ABINIT, the latter of which is free. The feasibility of mineralogical calculations using these methods is strongly dependent on how crystalline the mineral is and how ideal it is (from a perspective of not being heavily defected or substituted) as they work within the constraints of a periodic unit cell or supercell.


TL;DR Yes, sometimes. However, in general 13C data is much less useful than 1H (and 2D) data and the truth is we can often do without it.

I will limit my answer to the context of organic chemistry. Firstly, I directly answer your question: is 13C NMR ever used instead of 1H NMR? The answer is, yes.

However, I can only think of one case where one would do so. This is when there is severe peak overlap in the 1H NMR, such that one cannot reliably use it to ascertain whether one has (for example) a mixture of compound, or just a single compound. In such a case the 13C NMR (which has much more dispersed peaks and hence nearly no overlap) is more informative than the 1H NMR. This is generally not very common.

I want to go beyond a direct answer, though, and talk about the usefulness of 13C data in general. You are correct about the 1H NMR being more helpful than the 13C. This is generally true, and the main reasons for this are twofold:

13C NMR simply does not give as much information as 1H NMR. The only useful thing that one can obtain from the 13C NMR is the list of chemical shifts – but in turn, these are virtually impossible to interpret for all but the simplest molecules. On the other hand, 1H NMR gives you information about proton environments (shifts), number of protons (integrals), and limited information about molecular connectivity (couplings).
As porphyrin wrote, 13C NMR takes significantly longer than 1H NMR, because (a) the natural abundance of 13C is much lower (b) the gyromagnetic ratio of 13C is smaller.

As such, out of all the NMR data that organic chemists collect, the 13C NMR is by far the least useful and most time-consuming spectrum. On its own, its only real use is to provide a "fingerprint" of the molecule, and perhaps some information about functional groups present. So, in most cases, the 13C NMR is indeed less helpful than the 1H NMR.

The question then becomes, "why do people still take 13C NMRs?" Before going on, though, I should make it clear that people do not take 13C NMRs preferentially over the 1H, but rather in conjunction with it.

It primarily comes back to the fingerprint idea – when everybody tabulates 13C shifts of molecules they make, this gives chemist A the information they need to figure out whether they were successful in making what chemist B previously made. You might argue that the 1H is good enough as a fingerprint, but this is not always true, and anyway the consensus is that two is better than one: all major journals require submission of 13C NMR data for novel synthesised compounds.

The only other real case where the absolute 13C shifts are useful are when one needs to resolve very close peaks in 2D spectra (e.g. HSQC, HMBC), or in compounds with very few protons (thus rendering H–H or H–C correlations in 2D spectra less useful). Even then, it is not true to say that the 13C data is preferred over the 1H data.

All in all, your question is perfectly justified. In most cases, the 13C NMR is less useful as the 1H NMR, and only in some limited cases is it as useful as the 1H NMR.

In fact, there is a pretty recent perspective article by Liermann and Schlörer[1] arguing that the 1D 13C NMR should not be recorded, except for in the special cases described in the previous paragraph. The authors also make the point that the 2D spectra serve as much more useful fingerprints and are also much better at validating a proposed structure for a molecule. Even though in crowded spectra the 13C fingerprint may be more useful than the 1H fingerprint, it's still less useful than the 2D data, where peaks are dispersed in two dimensions.

Reference

Liermann, J. C.; Schlörer, N. E. Data handling in NMR facilities and assignment of NMR spectra in synthetic chemistry labs: Why electronic structure validation should become part of the routine. Magn. Reson. Chem. 2017, 1–7. DOI: 10.1002/mrc.4675.


Transition state and intermediate very close to each other

1 Kcal/mol is indeed quite low. This is around 1 hydrogen bond energy. In general - compute frequencies using freq command in Gaussian or whatever you are using. You should have a single negative frequency. It should be relatively low (abs value compared to lowest positive frequencies). Visualize it using Moleken or a similar tool. If movement is along the reaction coordinate then you are good.

Also, compare all bond lengths and angles to see what changed. Did bonds along the reaction coordinate change the most?

Please use Kcal/mol instead of Hartrees. I realize that Gaussian prints out
Hartrees, but it is easy enough to convert to Kcal/mol.


What is non PV work? How is it related to Gibbs energy?

At constant temperature and pressure, the maximum non expansion work is given by the change in the Gibbs Free Energy.

$$dW_{\text{max}}= dG$$

Proof:

At constant temperature and pressure,

from the definition of enthalpy: $H= U+pV$
first law of thermodynamics: $dU = dq+dw$

$\implies dH = dq+ dw + d(pV)$

Now,

$dG = dH - TdS -SdT \\\implies dG = dq + dw + d(pV)- TdS - SdT$

For isothermal process $dT =0$

$\implies dG = dq+dw+ d(pV)+ -TdS$

When the change is reversible,

$dq = dq_{\text{rev}}= TdS$ and $dw= dw_{\text{rev}}$

$\implies dG = dw_{\text{rev}}+d(pV)$

The work consists of expansion work, which for a reversible change is given by
−pdV, and possibly some other kind of work (for instance, the electrical work of
pushing electrons through a circuit or of raising a column of liquid); let's denote it by $dw_{\text{add},\text{rev}}$

$\implies dG= dw_{\text{add}, \text{rev}}+ Vdp$

For isobaric process, $dp =0$

$\implies dG = dw_{\text{add}, \text{rev}}$

You must be aware the the maximum work is obtained when the process is reversible, if not you may refer to this answer.

$\implies dG = dw_{\text{add}, \text{max}}$

  Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when  $\Delta G$ = Non-$PV$ work.

$\Delta S_{\text{total}}= \dfrac{-\Delta G_{\text{system}}}{T}$

Thus, for a reversible process $\Delta G_{\text{sys}}= 0$

And I have also proved to you above that $\Delta G_{\text{sys}}=$ maximum non expansion (non PV) work.

  I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

PV work is only the work done during the change in volume. There can be different types of works that a system can do during a process, for instance: electrical work, work against gravity, work against friction, magnetic work, shaft work, etc. This is the concept of non expansion or non PV work.

Source:

Physical Chemistry by Atkins (8th ed. )
Wikipedia article on work in thermodynamics


Yes, absolutely. Burning refers to oxidation accompanied by a flame and there are numerous reactions in which oxygen takes no part. Most of these are either with elementary halogens, or compounds thereof.

Here are some examples.

\begin{align}
\ce{2K + F2 &-> 2KF}\\
\ce{2Fe_\text{wool} + 3Cl2 &-> 2FeCl3}\\
\ce{H2O + ClF3 &-> HF + HCl + OF2}\\
\ce{2H2O + ClF3 &-> 3HF + HCl + O2}\\
\ce{2Cl2 + 2OF2 &-> 4ClF + O2}
\end{align}

So in fluorine chemistry, you don't oxygenise it; it flourises you.


How do we get NaClO3 in the reaction between Cl2 and NaOH?

I find that when I'm faced when problems having to do with why the products are formed I make lewis structures.

if you make a lewis diagram of the $$\ce{NaClO3}$$

First off, you will see that a single Cl will be bonded to three oxygen molecules whom all carry full valance shells which were striped from the hydroxide ions in the reactants. (Remember that in solution the Na and OH will be around as free spectators.)

From the structure you will see the reason why there was a formation of the three waters. Since there were three oxygen stripped from the hydroxides and then recombination with more OH to make a reaction like this:

$$\ce{OH- + H+ -> H2O}$$

Also notice that on the products side for every molecule of NaClO3 made there will be a molecule for NaCl made. From there you can speculate that when the Cl-Cl bond breaks three oxygen atoms and a sodium atom will surround one of the Cl while another is tied to a Na molecule simultaneously.


Are there any negatively charged electrophiles, or simply neutral and positive?

Yes, there are many neutral electrophiles.

One large class are the compounds with leaving groups. Some that you have likely encountered are: acids (HCl, HBr, sulfuric acid, etc) which are electrophilic at the hydrogen; unhindered alkyl halides/sulfonates (e.g., methyl iodide for SN2 reaction); dihalides (e.g., Br2 for electrophilic addition to an alkene).

A second large class are compounds with pi bonds to an electronegative atom. Most importantly this class includes the carbonyls (aldehydes, ketones, esters, etc). Michael acceptors fall into this class as well, which have an extended pi system containing an electronegative atom.

Since you are covering electrophilic aromatic substitution, soon you may learn about nucleophilic aromatic substitution, where a highly electron poor aromatic ring (e.g., having multiple nitro groups) bearing a leaving group can be attacked by a strong nucleophile.

In all of these cases, although the compounds are neutral, there is a partial positive charge at the electrophilic site explained by a polarized bond (acids, alkyl halides), resonance (carbonyl), or spontaneously (dihalides).

Neutral electrophiles are less reactive than their positively charged counterparts, which means that the nucleophiles are typically stronger for a reaction to occur.

As for negatively charged electrophiles, I can think of only a couple of cases that could be considered that way. For example, in the reduction of a carboxylic acid with lithium aluminum hydride, the first step is deprotonation of the acid to give a  carboxylate. LiAlH4 is very reactive and adds a hydride to the carboxylate. That could be seen as nucleophilic attack on a negatively charged electrophile, and many people will draw it that way. In reality though, there is something stabilizing the negative charge, either the lithium or aluminum. So it's not really a negatively charged carboxylate that's being attacked. I don't think that there are really any negatively charged electrophiles, but someone may have an example I haven't seen or thought of.


Which has one monochlorination possibility?

Replacing Hydrogen is the only way how chlorination (or halogenation) works.

So in 2,2-dimethylpropane the central carbon (carbon-number 2) is quaternary, meaning all four of its valencies are satisfied by carbon atoms, which implies that there are no hydrogen atoms which can be substituted by chlorine (or any halogen), whereas all the remaining 4 carbon atoms are identical, so substitution on any one of those will result in the same compound (1-chloro-2,2-dimethylpropane).

Whereas in 2-methylpropane(commonly known as isobutane) the carbon-number 2 is only tertiary meaning only 3 of its valencies are satisfied by carbon-atoms and it has one hydrogen atom on it. So on substitution of this hydrogen atom by chlorine the product is 2-chloro-2-methylpropane and on substitution of any one of the terminal (last) carbon atoms would all result in 1-chloro-2-methylpropane. That is two different products.


Is silicon dioxide decomposition possible?

Silicon dioxide is difficult to decompose directly, but it can be broken down in the presence of other reagents at more moderate temperatures.  In particular, the reaction of silica with elemental silicon has been known since 1905:

In 1890, the German chemist Clemens Winkler (the discoverer of germanium) was the first to attempt to synthesize $\ce{SiO}$ by heating silicon dioxide with silicon in a combustion furnace.[1]
$$\ce{SiO2 + Si ⇌ 2 SiO}$$
However, Winkler was not able to produce the monoxide since the temperature of the mixture was only around 1000° C. The experiment was repeated in 1905 by Henry Noel Potter (1869–1942), a Westinghouse engineer. Using an electric furnace, Potter was able to attain a temperature of 1700° C and observe the generation of $\ce{SiO}$.[2] Potter also investigated the properties and applications of the solid form of $\ce{SiO}$.[3,4]

The reaction may be used to remove silica from ores.
Cited references:

C. Winkler Ber. 23, (1890) p. 2652.

J. W. Mellor "A Comprehensive Treatise on Inorganic and Theoretical Chemistry" Vol VI, Longmans, Green and Co. (1947) p. 235.

U.S. Patent 182,082, July 26, 1905.

E. F. Roeber H. C. Parmelee (Eds.) Electrochemical and Metallurgical Industry, Vol. 5 (1907) p. 442.  Link


Low solubility of propanal compared to propanol, propanone, propanoic acid and methyl ethanoate

They're all on the verge of miscibility, methyl ethanoate (methyl acetate) is in fact not miscible according to wikipedia (soluble to 25%).

The acid is the most polar of all the C3 compounds you name. Butyric acid (C4) is also miscible with water, but valeric acid (c5) already not.

n-butanol is already rather badly soluble (8g in 100ml) in water. Same with butanone, etc.

Compared to the ketone, the aldehyde has a longer unpolar part, that is overall less influenced by the polar carbonyl bond. Both are clearly, i think, the least polar of the group, and so the aldehyde just drops beneath the line.

No mystery here. ;-)


Why is the product of sodium and water, sodium hydroxide and not sodium oxide?

In the usual case every student learns, the reaction of sodium with water is the following — I made note to include the states of matter here as they matter (pun intended).

$$\ce{2 Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$

In the typical setup, you have lots of water and little sodium. Therefore, whatever you get will be a solution or a precipitated salt. Alkali metals, however, often form very soluble salts and indeed all alkali hydroxides are well soluble. Thus, we will never get to see any solid by-products; all will happen in the solution phase.

Technically, it is possible for an oxide anion to be formed. However, this is when Brønsted and Lowry’s acid-base theory kicks in. Remember that water is amphoteric and can react as a base to form $\ce{H3O+}$ or as an acid to form $\ce{OH-}$. Well, equation $(2)$ expands that scheme.

$$\ce{H3O+ <=>[$K_\mathrm{a,1}$][H+] H2O <=>[$K_\mathrm{a,2}$][H+] OH- <=>[$K_\mathrm{a,3}$][H+] O^2-}\tag{2}$$

The oxide anion is connected to hydroxide and water by a simple acid-base reaction. From the definition of $\mathrm{p}K_\mathrm{a}$ values and the ionic product of water, we know that

$$\mathrm{p}K_\mathrm{a} (\text{acid}) + \mathrm{p}K_\mathrm{b} (\text{conj. base}) = 14\tag{3}$$

Since $K_\mathrm{w}$ doubles as the acidity constant of water, $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 14$. Likewise it can be shown, that $\mathrm{p}K_\mathrm{a} (\ce{H3O+}) = 0$. This is a difference of $14$ logarithmic units between two subsequent deprotonation steps. The acidity constants of sulfuric acid or phosphoric acid show that while this difference is large, it is not unusual for the conjugate base (if it is acidic) to have an acidity constand orders of magnitude lower than the original acid).

Condensing that into a conclusion, the acidity constant of hydroxide must be even lower still, to the point where we can say practically no oxide anions can be formed in equilibrium. Likewise, considering that there are many more water molecules than hydroxide anions, any oxide accidentally generated will get protonated immediately.

$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)}\tag{4}$$

Therefore, it is not feasible for oxide anions to exist in water. This is known as the nivelling effect: no acid more acidic than $\ce{H3O+}$ can survive in aquaeous solution for extended time and no base more basic than $\ce{OH-}$.

Since the entire reaction produces products in solution, sodium oxide is not a possibility.


Joining @Karl's comment on «what do you want to do with them?» the work to present point groups is work already solved.  For off-line teaching, the freely accessible collection of the 32 point groups by Casas and Estop by (UB Barcelona) may suit your needs:

The model data are provided as interactive 3D .pdf files which a reader like Adobe Acrobat 9.0 may let you toggle on/off the axes / planes, etc.
Beside some background provided here, the authors outline the workflow and tools used (SolidWorks, Adobe Illustrator CS3, and Adobe Acrobat 9 Pro) in an accompanying publication (doi 10.1016/j.cageo.2013.09.004, paywall).  Later, the concept was extended to cover the space groups as to complement the International Tables of Crystallography, too (title page, paywall).


How to interpret m quantum number values?

OP's answer is technically right while additional details about specific orientations of the real $f$-orbitals mentioned somewhere else are basically wrong. At best such details can be thought of the matter of some convention, but it is better to avoid such conventions.

The thing is that for any value of the azimuthal quantum number $l$ an orbital with a non-zero value of the magnetic quantum number $m$ is complex thanks to $\mathrm{e}^{imf}$ factor. Chemists prefer to work with real orbitals formed by taking linear combinations of complex orbitals. However, such linear combinations are not eigenfunctions of $L_z$, or, in other words, $m$ is no longer a good quantum number (but its absolute value is).

In particular, for both $f_{xyz}$ and $f_{z(x^2−y^2)}$ orbitals $|m|=2$, or $m=\pm2$, but the statement that the former corresponds to $m=-2$ and the later to $m=2$ is technically wrong.


How can value of acid dissociation constant is vary with concentration of acid?

Before we begin, note that $K_a$ is defined in terms of concentration and not amount. Furthermore, $K_a$ is a constant so it must be the same provided the conditions, other than concentration, are the same.

The nature of equilibrium is such that it does not always give the same absolute reaction (i.e. a fixed percentage of reaction, or the reaction does not proceed to the same "position" each time). This is unlike a fixed forward reaction which always ends at 90% completion. In other words, although the 1 molar solution of HA gave 0.9M of $H^+$ and $A^-$, the 2 molar solution would only give out...

$$ K_a = 8.1 = x^2/(2-x) $$

where $x$ is the number of moles of $H^+$ produced. When you solve this, you would get $x=1.66M$


From Roger E. Schirmer "Modern Methods of Pharmaceutical Analysis":

  Solvents are generally ranked by polarity, but polarity is not a uniquely defined physical property of a substance. Hence the relative polarity of a solvent will be somewhat dependent on the method used to measure it....

  Solvent polarity is a complex function of many parameter in addition to adsorption energy. A more recent ranking of solvents by Snyder is based on a combination of parameter such as dipole moment, proton acceptor or donor properties, and dispersion force solvent...

  Snyder's polarity index ranks solvents according to a complex theoretical summation of these properties. As a rule, the higher the polarity index, the more polar the solvent.

Snyder's paper was published in   Journal of Chromatography A, Volume 92, Issue 2, 22 May 1974, Pages 223-230 . Take a look at it to see how the index was calculated.

There are different types of polarity indices. And each of them has different parameters and ways to calculate the polarity.

Unlike polarity, dipole moment is a physical property. Polarity indices often take dipole moments as a parameters when calculating the polarity of solvents.  The reason why there are polarity indices is because dipole moments alone couldn't explain the nature and interactions of solvents.

Part of your confusion stems from the repetition of the word "polar" to describe different phenomena. There is polarity of bonds and there is polarity of solvents. They are different things.


what is the mass of chromium in this sample?

I like to solve such assignments using Python as a scientific calculator.
import scipy.constants

# number of atoms in sample
N_ATOMS_TOTAL = 2.13E+24

# atomic mass of chromium
M_CR = 51.9961 # g/mol

# ratio of chromium atoms to total atoms in K2Cr2O7
CR_RATIO = 2/11

# moles of Cr in sample
n_cr = N_ATOMS_TOTAL * CR_RATIO / scipy.constants.Avogadro

mass_cr = n_cr * M_CR # mass of chromium in g

mass_cr

Let me explain:

You have an unknown mass (in gramm) of potassium dichromate, $\ce{K2Cr2O7}$.

The total number of atoms in this sample is $2.13\cdot10^{24}$.

How can you calculate the number of chromium atoms in the sample?

You need to determine the ratio of $\ce{Cr}$ to all atoms in $\ce{K2Cr2O7}$, and that's $\frac{2}{11}$ atoms.

I know the number of $\ce{Cr}$ atoms now, but how many moles is that?

You know that 1 mol of an element has $6.02214129\cdot10^{23}$ atoms of that element (Avogadro's number).
Divide the number of $\ce{Cr}$ atoms determined before by Avogadro's number to determine the number of moles of $\ce{Cr}$.

I know the number of moles of $\ce{Cr}$ now, but what is the weight?

You know that the standard atomic weight of $\ce{Cr}$ is 51.9961.
Multiply  the standard atomic weight with the number of moles to get the weight.


Why does shampoo lather much more the second time it is applied?

Quoting a reddit post from chemist nallen:

The short answer is that the dirt and oils from your hair compete for the surfactants making them less available to form lather, which is small bubbles.

To better understand the mode of action, you have to know a bit about the formulation of shampoos and the nature of dirt and oils.

Dirts and oils deposit on hair and fibers because they are at least partially hydrophobic (not soluble in water.) The surface of hair, skin, etc... is more hydrophobic than water (well of course!), so these dirts have greater adhesion for the surface than for water (the water actually pushes them out to minimize surface area.) This is why water is ineffective for removal, it won't pull the dirt off the surface by itself. Surfactants associate with the hydrophobic surfaces to make them more dispersible or soluble in water, allowing them to be rinsed off.
So what happens when there is more surfactant than there is hydrophobic dirt surface?

The answer is that the surfactant orients to the surface of the water, because air is hydrophobic, and the water wants to push the hydrophobic portion of the surfactant out of the water molecule matrix. This orientation on the air-water interface is how skin of bubbles is formed. When the water/surfactant solution undergoes sufficient shear mixing, bubbles are formed, and that is lather.
Lather typically forms when there is surfactant in excess of the hydrophobic surface. More dirt means less free surfactant, and therefore less lather.

Now, it's not purely that simple, because the amount of lather that a surfactant develops varies with the type of surfactant, some are better than others.

Lather does not mean that the shampoo is cleaning better, it's purely cosmetic. Typical shampoos are primarily SLES/Betaine, that is sodium laureth sulfate and cocamidopropyl betaine. The SLES is there for cleaning, and the betaine is there for lather, it's specifically added to produce lather because, well, people like it. There are some other benefits, but that's the primary one. If you used SLES by itself is would clean just as well, but people just don't like poor lather, so it would not sell.


When a liquid is supercooled its structure doesn't change significantly, only the particles have less kinetic energy available. As you cool even further, it could occur that particles get caught in local minima and eventually you'll end up with a glass, which can be thought of as a frozen liquid structure. As particles are trapped it will resist shear in an elastic fashion so you have a solid. In this regard it is a barrierless transition. There is no interface whatsoever.

Crystal formation is fundamentally different as the local structure changes first to form a nucleus. This has both energetic and entropic consequences and the aforementioned kinetic barrier is observed. I would also like to add that crystal nuclei are formed all the time but they don't reach the critical size if the temperature is too high.


Are phospholipids capable of forming automatically circles or vesicles?

Phospholipids structure can be described as a charged phosphate head group, with long hydrophobic chains coming off it. When you put these molecules in water, like oil drops, the oils spontaneously conglomerate into sheets on the surface of the water.

How, depending on concentration and size of the phospholipids you can get from sheet like membranes to spherical droplets. In either case the spontaneous formation of sheets or micelles (spheres) results from the entropically driven hydrophobic effect.

The simple way to understand this hydrophobic effect and entropy, is that when two hydrophobic tails between two phospholipids are in close proximity, the hydrophobic tails are able to interact via van der Waals forces. But even more important is that water molecules that surround each hydrophobic region of the molecule, are excluded between them. This means that the entropy of water increases a lot more compared to the loss of entropy from having the phospholipids come together and stay together.

This is a pretty fundamental concept to how proteins fold. Look up hydrophobic effect on google for nice, diagramatic explanation.

Or check intro biochemistry textbook like Nelson and Cox (2012) principles of biochemistry textbook.


How to find the oxidation numbers of individual ions of a transition metal with an average fractional oxidation state

To accurately determine oxidation states in compounds, you need to have a good idea of what the actual underlying bonding model is like. Only in rare cases you can resort to shortcuts.

One such shortcut which is often very helpful for binary compounds is to take one element’s ‘obvious’ oxidation state and use maths to derive the other one. That, for example, works well with ethane (hydrogen must have $\mathrm{+1}$, so carbon has $\mathrm{-III}$), iron(III) chloride ($\ce{Fe^{+III}/Cl^{-I}}$) or manganese dioxide ($\ce{Me^{+IV}/O^{-II}}$).

In some cases, such a shortcut will lead to a fractional oxidation state such as in $\ce{Fe3O4}\ (\ce{Fe^{+\frac83}/O^{-II})},$ $\ce{HN3}\ (\ce{H^{+I}/N^{-\frac13})},$ or $\ce{S4O6^2-}\ (\ce{S^{+\frac52}/O^{-II})}.$ One might be tempted to take the closest common or known oxidation states of the element in question and use maths to derive a potential distribution. That works by accident in $\ce{Fe3O4}$ ($\frac83$ is $2\frac23$ which suggests $\ce{Fe1^{+II}Fe2^{+III}}$) and $\ce{HN3}$ ($-\frac13$ suggests $\ce{N2^{\pm0}N1^{-I}}$) but it already fails with $\ce{S4O6}$ because although $\ce{S3^{+II}S1^{+IV}}$ gives an overall oxidation state of $+\frac52$ so does the correct solution of $\ce{S2^{\pm0}S2^{+V}}$.

Since the simple model already broke down for binary compounds, there is little hope it will be of much use in ternary or even quarternary compounds—especially if there are less leads to go on. So while it might be possible to use the simple shortcut to derive the oxidation states in aluminium potassium sulphate or potassium alum ($\ce{KAl(SO4)2}$) and by extension that of chromium in chromium potassium sulphate or chrom alum ($\ce{KCr(SO4)2}$), it is a lot harder if you have two redox-promiscuous elements (nickel and manganese) to consider.

In short, there is no simple way to derive the oxidation states just by looking at the formula and I myself would have to research what publications say and how they arrived at this conclusion.


How do I cool my water bottle immediately for an emergency use?

Ammonium nitrate, calcium ammonium nitrate, and urea are each used in instant-cold packs. The dissolution of any of these compounds in water is an endothermic process and will thus cool your container of water.

In a practical sense, it doesn't seem likely that you would sacrifice some of your water so the rest of it could be cold in an emergency situation. Of course, the water you use for cooling does not have to be clean and potable.


The only "hydroxyl"-type species whose recombination has been observed to yield  hydrogen peroxide are hydroxyl radicals.

$$\ce{HO^. + ^.OH -> H2O2}$$

In order to generate these hydroxyl radicals from water, a lot of energy has to be provided, typically by radiolysis (short wave uv, $\gamma$, electron beam).

The conceivable steps yielding $\ce{HO^.}$ are either

$$\ce{H2O  -> H^. + ^.OH}$$
or
$$\ce{H2O  -> H2O^{+.} + e-}$$

$$\ce{H2O^{+.} -> H+ + ^.OH}$$


Are allotropes of sulfur two different phases?

Solid allotropes of pretty much anything are different phases. They have different crystal structures, and you can't have a single phase with two different crystal structures inside it.

As for the other criteria, some of them are sufficient but not necessary, and some are difficult to check. The distinct boundary must be there, but that's a knowledge that comes from them being two phases and not vice versa. Physical properties are not all that different, though you can tell one allotrope from another by density. (We used a solution of $\ce{BaI2}$ to this end.)

Chemical properties are even more similar since the molecules are the same. I doubt that a purely chemical test can be devised. But then again, do we really need it?


Why is sulfuric acid used to seal flask with highly reactive chemicals?

Sulphuric acid reacts with dichromate and table salt, forming volatile and harmful chromyl chloride.
\begin{multline}
\ce{K2Cr2O7(s) + 4 NaCl(s) + 6 H2SO4 ->}\\
\ce{2 KHSO4(s) + 4 NaHSO4(s) + 2 CrO2Cl2(l,g) + 3 H2O}
\end{multline}
Therefore it serves as a reagent and at the same time as protection against chromyl chloride escape.

Safety considerations
$\ce{CrO2Cl2}$ hydrolyzes to release $\ce{HCl}$ and hexavalent chromium $\ce{Cr^{\mathrm{VI}}}$.
Acute: Exposure to chromyl chloride vapour irritates the respiratory system and severely irritates the eyes, and the liquid burns the skin and eyes. Ingestion would cause severe internal damage.
Chronic: $\ce{CrO2Cl2}$ can produce chromosome abnormality/aberrations and is a human carcinogen via inhalation. Frequent exposure of the skin to chromyl chloride may result in skin ulcer/ulceration.


Role of alcoholic potassium hydroxide in an E2 reaction

Haloalkanes react with alc. $\ce{KOH}$ mainly via $E_2$ mechanism. But, in some cases, where $\beta-\ce{H}$ isn't present, it proceeds via $E_1$, as in case of $(4)$ one. Let's analyze each option,
Option $(1)$:
Firstly, the $\ce{-Cl}$ departs, and then the base picks up a $\beta-\ce{H}$ from the terminal methyl group (i.e., $\ce{C-5}$), resulting in mono- substituted alkene (Hofmann product as major). Steric factors are significant in selectivity of $\beta-\ce{H}$. Therefore, Zaitsev product will be formed in very minor concentrations, bcoz $\ce{-C(CH3)2Ph}$ is more bulkier substituent than $\ce{^tBu}$.
Option $(3)$:
This one is similar to $(1)$ case, except for the fact that due to less steric hindrance, it will give di- substituted alkene (Zaitsev product).
Option $(2)$: It's clear that this molecule will form a tri- substituted alkene.
Option $(4)$: As hinted above, it will proceed through $E_1$, resulting in a tri- substituted alkene.

Different levels of substitution are shown below, (source: www.chemistrysteps.com)


Coordinates can be accessed using

[OBAtom.GetX(),OBAtom.GetY(),OBAtom.GetZ()]

look at this Github discussion.

If you have a better answer please don't hesitate to answer the question.


Why does chlorine trifluoride ignites concrete and sand?

Concrete, sand, and asbestos are all, in large part, silicon dioxide (SiO₂). It's energetically favorable for available fluorine to take the place of oxygen, producing SiF₄. This occurs even in weak solutions of hydrogen fluoride, and it's why HF solutions are used for etching glass.

ClF₃ has a much greater amount of potential bond energy than HF or even F₂. This means that, when it decomposes, such as upon contact with silicon dioxide, it gives off a lot of energy, which is released as fire.


The intrinsic electron-withdrawing nature of alkyl groups

I wonder if the pKa for water (15.7) is a typo. Usually we think it to be 14.0 (at 25C, anyway). And I wonder how much the gas phase acidity changes over this series.

Acidity in the gas phase means ROH must dissociate to RO- and H+ but doesn't put the proton onto anything else. Since the proton isn't solvated, it must have been knocked loose, say in a mass spectrometer, where the fragmentation potential can be measured. That just seems quite different from our usual idea of "acidity". But OK, it's an observation of a characteristic of a series of related compounds.

The tendency to designate specific places for electrons (LUMO, HOMO) is understandable, but sometimes this description is uncertain and unnecessary. In a gas phase RO- ion, the negative charge is not 100% around the nucleus of the oxygen atom. Surely it will become unsymmetrical and drift where it might find a little positivity, like on the carbons and the hydrogens. Alkyl groups are polarizable and stabilize both anions and cations (https://onlinelibrary.wiley.com/doi/abs/10.1002/ejoc.200700004).

It's simpler to explain the acidity changes in water and in the gas phase by referring to the size-dependent polarizability of the alkyl groups, but using molecular orbitals may have seemed better for publication.


Why does chlorine have a higher electron affinity than fluorine?

Fluorine, though higher than chlorine in the periodic table, has  a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

See this.


HCl is a gas at room temperature and diethyl ether is a low boiler.  Both will have significant vapor pressures and, therefore, exist in the vapor phase above your  solution.  The HCl and diethyl ether in solution will continuously maintain an equilibrium with the vapor phase. As you bubble nitrogen through the system it will continuously sweep the vapor out, removing HCl and diethyl ether.  But since the HCl and diethyl ether in solution continuously maintain an equilibrium in the vapor phase, they will be continuously swept out until all of the HCl and diethyl ether in solution have been removed.


What role does sulfuric acid play in the titration of oxalic acid and potassium permanganate?

The used analytical reaction of purple permanganate to colourless $\ce{Mn^2+}$ occurs under acidic conditions:

$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$

Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:

$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$

Therefore, sulfuric acid is added to make the solution acidic. Actually, excess acid is required since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).

Sulfuric acid is used because it is stable towards oxidation; whereas, for example, hydrochloric acid would be oxidized to chlorine by permanganate.


Thought process behind Rutherford's gold foil experiment

Rutherford conducted his gold foil in order to prove Thomson's plum pudding model but the result was contradicting.
He expected that most of the Alpha particle will show very minor deflections  because the positive charge was evenly spread throughout the atom according to the plum pudding model.


How can we define optical isomerism?What is the example of non enatiomeric optical isomers?

There are, in the sense you are asking the question, three types of optical isomers: +, -, and optically inactive. This is detailed in pages of the textbook you have uploaded.
You should think about optical isomers in how they are determined. Imagine you have a machine which we will call and "optical isometer". To use this machine, you put in any compound in solution, and it will tell you one of three outputs:

"This compound polarises light in the clockwise direction"
"This compound polarises light in the anti-clockwise direction"
"This compound does not polarise light"

Now, our optical isometer doesn't know any chemistry, but it does know how to tell if light is being polarised by our compound in solution. If you fundamentally think of isomers as identical in their molecular formula, but differing in their spatial arrangement of atoms, you need only imagine as optical isomers being different in how the spatial arrangement of atoms polarises light.
Let's imagine we have a solution of a tartaric acid, but we don't know what isomeric form it is in. The options are:

meso-tartaric acid
(+)tartaric acid
(-)tartaric acid
A mixture of (+) and (-)tartaric acid.

Where I think you have been confused by the textbook is that option number 4 doesn't actually describe different spatial arrangements of tartaric acid, rather it describes a mixture of options 2 and 3. So when it says there are four isomeric forms, it isn't strictly correct about that.
Now, how will we determine what isomer of tartaric acid we have in solution? One thing that we can do to find this out would be to use our optical isometer. We put our solution in and it outputs: "This compound polarises light in the clockwise direction". We know that of our four possible options, only (+)tartaric acid polarises light in the clockwise direction. Likewise, we would know it to be (-)tartaric acid if the light were polarised in the anti-clockwise direction.
What if our optical isometer output, "This compound does not polarise light"? Then this means that our compound is optically inactive. We know that both the meso- form and the racemic (mixture of (+) and (-)) don't polarise light in either the clockwise or anti-clockwise direction, and so we would need further analysis to find out for sure!
There are other types of light polarisation, but not in the sense this book is talking about optical isomers. This textbook seems to be a little confusing, but these ideas are actually really well explained on places like wikipedia or khan academy, so next time you get stuck, don't worry too much about the textbook but give other resources a shot!
Hope this helped,
Edan


Plotting a calibration curve for Gas-Liquid Chromatography

If I remember correctly, the amount of your compound isn't proportional to the peak height but to the peak area.

Back in the days, when plotters were just plotters (=no numerical integration of the peaks), people used to cut out the curves and weighted the paper pieces on a balance to determine the peak area.


If done properly, reactions with silanes or siloxanes are effectively permanent, unless removed by serious mechanical abrasion (e.g., scraping the material below the coating) or some sort of etching underneath.

The silanization process involves creating new covalent $\ce{Si-O}$ bonds, which are quite strong.

The silane surface itself should be durable. However, the contact angle may change as material adsorbs on top of it. For example, a colleague has shown that hydrocarbon contaminants change the wettability of many surfaces
I don't think any particular storage is necessary. We store in a vacuum desiccator to minimize the effect of contamination, but as long as you clean the surface, the properties should be consistent.


For an adiabatic isobaric process the work done by the system during a spontaneous expansion is $w= -p_{\text{ext}} \Delta V $.  Since the heat $q=0$ the work is also equal to the change in internal energy, ie $w= \Delta U$. The associated enthalpy change is then

$$\begin{align} \Delta H &= \Delta U + \Delta (pV) \\ &= w + p_fV_f-p_iV_i \\ &=  -p_{\text{ext}}\Delta V+ p_{\text{ext}} V_f-p_iV_i \\ &=  p_{\text{ext}}V_i-p_iV_i\\&=(p_{\text{ext}}-p_i)V_i \end{align}$$

In order for the expansion to proceed spontaneously the initial pressure of the system $p_i$ must exceed that of the surroundings $p_{\text{ext}}$, which means $\Delta H<0$.

Since you are computing $H$, a property of the system, when computing the change in $H=U+pV$ you should use the pressure of the system, not of the surroundings. Before the expansion the pressure of the system (and surroundings) is $p_i$, while at the end of the expansion it is $p_f=p_\text{ext}$, since the system is then again in mechanical equilibrium with the surroundings. Note that previous to the expansion the system and surroundings are also in mechanical equilibrium. The pressure of the surroundings is then suddenly dropped to a lower constant value. The following figure illustrates the process:


Balancing a reaction with unknown products

Water is normally a mix of hydroxide and hydrogen, at very dilute levels.  For example, neutral water (pH of 7) is 10-7 of OH- and H3O+ ions.  Add sodium and it combines with the OH-.  The loss of electrons by sodium are available to the hydrogen ions, which combine to make hydrogen gas.  Sodium in metal form is so reactive, it will sustain this reaction even at the low hydroxide levels in pure water, and even in slightly acidic seawater.


How do you know that Sodium Hydrogencarbonate is amphiprotic?

You have to consider whether or not the species is capable of both losing and gaining a hydrogen.

When sodium bicarbonate dissociates in water, it becomes a sodium ion ($\ce{Na+}$) and a bicarbonate ion ($\ce{HCO3-}$). The bicarbonate ion itself is what is amphiprotic, and this is because it can lose the hydrogen it has to become a carbonate ion ($\ce{CO3^{2-}}$), or it can gain a hydrogen to become carbonic acid ($\ce{H2CO3}$).

Many species that contain a hydrogen and an unbonded electron pair are amphiprotic, like $\ce{NH3}$ and $\ce{H2O}$, for example.


As Ed said you are following the correct protocol by diluting the sample. The key reason is for doing so is to avoid a non-linear region in the calibration curve. This is why one would dilute the sample rather than make more concentrated standards. Several interesting effects happen with alkali metals at high concentrations in the flame or plasma. One of them is the concept of self-absorption.

Imagine you are monitoring the resonance line of K emission. Once you introduce lot of potassium atoms in the plasma (by using a concentrated standard) what may happen is that the ground state K atoms may absorb the light emitted by the excited potassium atoms? What is the result? You get less intensity than expected.

Try watching a sodium street lamp with a diffraction grating, or take a CD outside at night. You will see a prominent dark line in the emission spectrum of the lamp like this at 589 nm. This is the same self absorption phenomenon. There are so many Na atoms in the ground state that that they are self absorbing the light emitted by excited sodium atoms in the electrical discharge.


How to calculate the mass of water to add to copper(II) sulfate to form a 4% solution?

A $4~\%$ solution is one that contains $4~\mathrm{g}$ of the solute in $96~\mathrm{g}$ of solvent to give $100~\mathrm{g}$ of solution. Just to get that out of the way.
To calculate the mass of the solute ($m(\ce{CuSO4})$ — not to be confused with $m(\ce{CuSO4 . 5 H2O})$) we can use at least two ways which are mathematically equivalent:

Calculate the amount of $\ce{CuSO4 . 5 H2O}$ — this is equivalent to the amount of $\ce{CuSO4}$. Then calculate the latter’s mass.
$$M(\ce{CuSO4 . 5 H2O}) = 249.72~\mathrm{g\ mol^{-1}}\\[1.2em]
n(\ce{CuSO4 . 5 H2O}) = \frac{m}{M} = \frac{15.0~\mathrm{g}}{249.72~\mathrm{g\ mol^{-1}}} = 6.01 \times 10^{-2}~\mathrm{mol}\\[1.5em]
M(\ce{CuSO4}) = 159.62~\mathrm{g\ mol^{-1}}\\[1.2em]
m(\ce{CuSO4}) = n \times M = 6.01 \times 10^{-2}~\mathrm{mol} \times 159.62~\mathrm{g\ mol^{-1}} = 9.59~\mathrm{g}$$

Use the molar weights of copper(II) sulphate and water to determine the mass percentage of copper(II) sulphate in the salt.
$$\frac{M(\ce{CuSO4})}{M (\ce{CuSO4 . 5 H2O})} = \frac{159.62~\mathrm{g\ mol^{-1}}}{249.72~\mathrm{g\ mol^{-1}}} = 63.9~\%\\[1.5em]
\Longrightarrow m(\ce{CuSO4}) = 15.0~\mathrm{g} \times 63.9~\% = 9.59~\mathrm{g}$$

In any case, we will arrive at a mass of the solute of $9.59~\mathrm{g}$. This is equivalent to the $4~\%$ mentioned at the very top. Therefore:
$$m_\mathrm{tot} = \frac{m_\mathrm{solute}}{0.04} = 240~\mathrm{g}$$
$$m_\mathrm{solvent} = 0.96 \times 240~\mathrm{g} = 230~\mathrm{g}$$
You noted that you already have $5.4~\mathrm{g}$ of water in the copper(II) sulphate hydrate present. Thus, you need to add the difference:
$$m_\mathrm{water} = 230~\mathrm{g} - 5.4~\mathrm{g} = 225~\mathrm{g}$$


What makes Citrate or Citric Acid an acid?

It depends on the pH of the solution.  At higher pHs the carboxyl groups will be dissociated (as drawn).  At lower pHs the equilibrium will shift and most of the COO- will become protonated.

In general, carboxylic acids are relatively weak compared to stronger acids that dissociate almost completely when added to an aqueous solution.  For example, acetic acid, or vinegar, will not burn a hole in cotton clothing, whereas sulfuric acid (the kind in an automobile battery) will easily burn a hole in cotton cloth.

Also, don’t confuse carboxylic acids with their carboxylate salts.  Sodium citrate, where the three protons are replaced with Na+, will act as a base when added to water, and raise the pH.


Conversion of Benzene to 9-Bromo 9,10 Dihydroanthracene

What you are looking for is known as the Wurtz-Fittig reaction - wikipedia article , JACS paper
In this proposed reaction I think you will get a complex mixture (also known as a mess), but you may be able to find some of your desired product in it. Good luck with trying to isolate it cleanly!
If you want a straightforward preparative route then look at the F-C reaction of phthalic anhydride with benzene, the reduction of the resulting 2-benzoyl benzoic acid to 2-benzyl benzoic acid and cyclisation with polyphosphoric acid. This product can be reduced to the dihydroanthracene by Zn/HCl


Could this solution be resolved by titration?

The ammonium nitrate dissociate in water:

$$\ce{NH4NO3 -> NH4+ + NO3-}$$

The $\ce{NH4+}$ and $\ce{NH3}$ form together the ammonia $\mathrm{pH}$ buffer with $\ce{NH4+}$ acidity constant $\mathrm{p}K_\mathrm{a}=9.25$:
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log { \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}}$$

The titration curve as $\mathrm{pH}=f(V)$ is the most flat at $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=9.25$, where the buffer has the biggest capacity.

For a given solution, containing $\ce{NH4+}$ and $\ce{NH3}$, titration with $\ce{HCl}$ or $\ce{NaOH}$ causes  respective acid-base reactions:

$$\begin{align}
\ce{NH3 + H+ &-> NH4+ \\
NH4+ + OH- &-> NH3 + H2O} \\
\end{align}$$

The ammonia buffer causes   $\mathrm{pH}$ being quite stable during titration, until one of the 2 ammonia forms is practically spent.

At that point, the $\mathrm{pH}$ titration curve gets very progressively bent up or down, respectively and the titrator stops the titration.

The content of $\ce{NH4NO3}$ and  $\ce{NH3}$ is then obtained by obvious routine calculation.

$$\begin{align}
m_{\ce{NH3}}&= M_{\ce{NH3}}\cdot c_{\ce{HCl}}\cdot V_{\ce{HCl}}\\
m_{\ce{NH4NO3}}&= M_{\ce{NH4NO3}}\cdot c_{\ce{NaOH}}\cdot V_{\ce{NaOH}}\\
\end{align}$$

Note that at $\mathrm{pH}=13$, $\ce{NH4+}$ ions practically do not exist, as $ \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}=5620$

If we consider $\mathrm{pH}$ of ammonia solution itself, it is
$$\begin{align}
\mathrm{pH}=14 - 0.5 \cdot ( \mathrm{p}K_\mathrm{b}-\log c) \\
13=14 - 0.5 \cdot ( 4.75 -\log c) \\
\end{align}$$

$c_{\ce{NH3}}$ would be 563 mol/L, what is nonsense. Such $\mathrm{pH}$ is not achievable by ammonia alone, but a mineral hydroxide would be present.

If there was initially any ammonium nitrate, it would be converted to ammonia:

$$\ce{NH4NO3 + NaOH -> NaNO3 + NH3 + H2O}$$

In such a case, determination of both compounds by titration is not possible, as there are no ammonium ions and ammonia would be titrated together with hydroxide.


Why 1-ethenyl-3-ethyl-5-ethynylbenzene has that name?

Ordering substituents by complexity has been abandoned long ago (before 1979) in favor of the much simpler alphabetic order. Here, ethenyl < ethyl < ethynyl.

Note, however, that simple substituents are ordered alphabetically before adding a multiplicative prefix, which means that "dimethyl" is ordered as "methyl".

Complex substituents are ordered by their first letter, e.g. isopropyl (I), dimethylphenyl (D)


How to separate sodium iodide dissolved in water?

I summarize some answers given in comments.

Evaporation is really one of the standard methods in any combination of solid (cristalline, amorph, polar, unpolar) and solvent (polar, unpolar, flammable...).
But you must not think of it as the pot heated on a flame.
In modern synthesis you often already work with a nitrogen or noble gas atmosphere in a closed apparature and use varying pressure for controlled destillation.
So it is no problem to just reduce the pressure and evaporate your (organic) solvent.
In your special case you even can think of a pot heated on a flame, there is no risk and disadvantage. It is simply an effective method.

Do you know the concept of polarity? A good rule of thumb is that polar substances mix with polar ones but not with unpolar ones.
So water dissolves salt but not paraffin wax.
Benzene on the other side dissolves parrafin wax but no salt and water and benzene do not mix with themselves.

=================Edit (Addition of Information, because of question in comment)

So let's see how to regain the solvent.
First of all there is not only one boiling temperature. The boiling temperature is always a function of pressure. Approximately it can be described by the Clausius-Clapeyron-Equation.
$$
T_2(p_2)=(\ln(\frac{p_1}{p_2})k+\frac{1}{T_1})^{-1}
$$
Here $k$ is some constant depending on the specific solvent and $T_1$ and $p_1$ is a corresponding pair of known boiling temperature and pressure. For example $100^\circ$C and 1bar for water.
In general the boiling temperature goes down when pressure goes down. (Water boils faster in the Himalaya)

So if you have a typical organic solvent like aceton with a boiling temperature of $56^\circ$C under 1bar it evaporates already at $20^\circ$C when applying a pressure of e.g. 200mbar. (it is not the exact number)
But it also condenses again when cooling down. Using this you can make it liquid again and regain it. Compare this image of a typical distill.
http://www.ssc.education.ed.ac.uk/BSL/pictures/distill.jpg

Summarizing: you use a pump to get into your desired range of boiling temperature and cool the gas in the condenser again to a temperature below the boiling temperature. Depending on your system you have boiling temperatures below $0^\circ$C and cool with liquid nitrogen in the condenser or normal temperature range and cool with water or...


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are placed
immediately in front of the part of the name to which they refer.

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are
placed at the front of their names.

Therefore, the locant ‘3’ in ‘pent-3-enal’ refers to the ending ‘ene’ (not to the suffix ‘al’).
The class name ‘aldehyde’ refers to compounds containing the group $\ce{{}-CH=O}$
attached to another carbon atom. In systematic substitutive nomenclature, the suffix ‘al’ is used for $\ce{{}-(C)HO}$ (see Subsection P-66.6.1). Thus, the aldehyde carbon atom is always a terminal carbon atom.
The characteristic group cited as suffix (here: ‘-al’) receives the lowest locant possible (here: ‘1’). However, terminal locants are omitted in preferred IUPAC names (PINs) for such aldehydes.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Therefore, the PIN for the example given in the question is ‘pent-3-enal’ (not ‘pent-3-en-1-al’).


Synthesis Golf VIII: 4,6-Diacetylhygrophorone A12

Establishing the correct stereochemistry proved out to be an extremely difficult task, as I have found only one procedure for these systems. This first scheme is a published procedure:

1) Diels-Alder reaction (Paul D. Bartlett, Fred A. Tate
J. Am. Chem. Soc., 1953, 75 (1), pp 91–95);

2) Alpha-allylation and 3) ring closure by anionic addition to ester group to generate two regioisomers which are separated (Siwapinyoyos, Thebtaranonth, J. Org. Chem. 1982,47, 598-599);

4) Epoxidation of the non-conjugated enone and 5) opening of the epoxide by elimination (Chantarasiri, J. Chem. Soc., Chem. Commun., 1990,0, 286-288)

6) Kinetic resolution by enzymatic O-acetylation, followed by 7) hydrolysis of the acetate and vacuum flash pyrolysis at 450 0C and 0.1 mmHg. The acetate is needed in the target molecule, but I guess that it has to be hydrolyzed before the pyrolysis as to avoid side reactions. 8) is epoxidation with dimethyldioxirane and 9) opening of the epoxide with water provides a intermediate resembling the target molecule a lot (Klomkao, Tet. Asymm. 14, 24, 2003, 3885-3889).

From this molecule starts my approach:

10) Acetonide protection of the diol should be selective (syn 1,2 hydroxyl groups react in the presence of anti hydroxyl groups and 1,3 hydroxyl groups). The free hydroxyl group is then acetylated.

11) The acetonide is hydrolyzed. The primary hydroxyl group is oxidized to the aldehyde and the tertiary one is protected as MOM ether.

The following 3 steps are used for protection of ketone in presence of aldehyde:

12) Aldehyde is protected in presence of enone first.

13) Enone is protected by Noyori conditions.

14) Aldehyde is deprotected.

15) Organocopper addition to the aldehyde in presence of the ester to form a mixture of diastereomers (surely it would show diastereoselectivity but this would need to be tried out).

16) If the diastereoselectivity is bad, the mixture of diastereomers is oxidized to the ketone which is then stereoselectively reduced by chelate-control with zinc-borohydride. This could be problematic as the acetonide oxygens could interfere with chelation, but I cannot predict this with confidence. If they don't interfere much the intermediate would look something like in the image attached at the bottom. The hydryde attacks opposite to the acetonide and the alkyl chain. If acetonide oxygens interfere, then a screening of reducing agents is required.

17) Acetylation of the hydroxyl group and MOM and acetal deprotection should provide the target molecule.


Is this home-made pesticide from cigarette butts and battery acid safe?

This seems like a bad idea to me.
The use battery acid (sulfuric acid) seems really terrible. In the past there was an awful problem with coal power plants releasing sulfur oxides into the air which formed acid rain (sulfuric acid). You're proposing making a brew that is orders of magnitude worse than acid rain to put on your garden plants. I think such a strong solution of acid would get rid of the bugs by killing your plants. ;-)
If you goggle nicotine "bug spray" you can find a number of websites that discussing seeping tobacco in water, then filtering the brew to make a spray. In such a solution the nicotine is so dilute that it would be safe to handle. However concentrated solutions of nicotine like used for various e-cigarettes could be dangerous to handle. Lookup nicotine MSDS in google (here is one such MSDS sheet).

(1)So no battery acid is needed.
(2) The various websites state that the sprayed nicotine solution decomposes fairly quickly - a couple of days, not weeks.

Frankly I don't smoke or chew tobacco. Since I'd have to buy the tobacco for such a use, I'd rather just buy a commercial pesticide and follow its use instructions. It is absolutely crazy to me that so many folks so berserk when they see the word "chemical." If you showed a strawberry to most folks they would just identify it as such. But it is nothing but a sack of hundreds of chemicals. So to me all strawberries are "organic strawberries." To someone who is allergic to strawberries one can be deadly. Did you know that there are varieties of lima beans which have a high enough concentration of cyanide that eating them can kill? (dangerous foods) So spraying a nicotine solution may seem "natural" but it is still a chemical.


Inconsistency in Elementary Bi-Molecular Reaction Rate Expressions

If the reaction is
$$\ce{2A -> P},$$
then the rates are defined to be related as
$$
  -\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t}
  = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
$$
and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with the rate constant. With any other usage an explanation should be added so as not to confuse things.
In the general case
$$\ce{aA + bB -> pP + qQ},$$
then
$$
  -\frac{1}{a}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t}
  = -\frac{1}{b}\frac{\mathrm{d}\ce{B}}{\mathrm{d}t}
  = \frac{1}{p}\frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
  = \frac{1}{q}\frac{\mathrm{d}\ce{Q}}{\mathrm{d}t}
  = \frac{\mathrm{d}x}{\mathrm{d}t},
$$
where at any time $t$, $ax$ moles of $\ce{A}$ and $bx$ of $\ce{B}$  are consumed and $px$ of $\ce{P}$ are produced, where $x$ is the extent of reaction.


R,S in Cyclic Compounds - Path of Highest Priority?

If you "collide" without finding some difference, then the molecule is not chiral.  In the example at hand, it appears that you have analyzed it correctly.  First, you look at the 2 carbons adjacent to the chiral center.  They are both $\ce{CH_{2}{'}s}$; no difference, therefore move out to the next carbon on each side of the ring.  Here we find a $\ce{C-CH_{2}-C}$ on one side and a $\ce{C-CMe_{2}-C}$ on the other side.  The latter has higher priority than the former.  Stop at this point, no need to go further, and complete the "1-2-3-4" prioritization.


Diamond filtering - is it real or bogus?

It appears that there is some validity to this technology.  Not surprisingly, the published journal literature available is basically non-existent, except for a handful of patent applications describing something along those lines.

The title of one journal publication "Preparation of a filtering material using nanomaterials for removal of a heavy metals from wastewater" uses a diamond nanopowder to aid in filtering out heavy metals from waste water.  The citation is "Antonenko, L. P., Energotekhnologii i Resursosberezhenie, 6, 2011, 49-54"

Additionally, there is a patent appplication from France (where at least one article I read cited as the technology originating from), title "Apparatus for activation and mineralization of water for domestic use".  In this, they use a variety of powdered gemstones (diamond, ruby, sapphire) as well as precious metals (gold, silver) to filter the water.

Certainly, nano-particle scale materials are effective filtration agents, as most contaminants will not pass through such a small sized filtration device.  The use of some gemstones (certainly diamond) are beneficial for their relative chemical inertness, and certainly silver is well-known for it's antimicrobial activity, and synthesis of gold-nanoparticles or gold-silver coatings are also actively under research


Is temperature double-counted in the Gibbs free energy equation?

One reason we write equations this way is because all of the parameters on the right-hand-side can be measured or computed independently.

Consider an analogous (but more intuitive) equation, the Stokes-Einstein relation for the diffusion coefficient $D$ of a particle in a liquid:

$$D = \frac{k_\mathrm{B} T}{6\pi\eta r}$$

where $\eta$ is the dynamic viscosity, $r$ is the radius of the particle, and $k_\mathrm{B}$ is Botzmann's constant. Here we see an explicit dependence of $D$ on $T$, but there is also an implicit dependence on $T$ through $\eta$ and possibly also $r$. Is this a problem? No, assuming the equation holds and we can determine the parameters accurately at each $T$ of interest. Whether that is a simple thing is another question.

Also, it is fair to wonder to what extent the parameters are fundamentally independent, or, whether by digging into the gory details of atomic structure of matter, we might not find $r$ and $\eta$ to be connected in some other way. But that is another question.


How do I make PDB 1HIG's data usable for producing ball-and-stick models?

This specific PDB file only contains the coordinates of all the $\ce{C_{\alpha}}$ atoms of the protein. It simply does not contain enough information to display an actual 3D structure of the protein.

The file does contain secondary structure assignments, so it should probably be possible to create a cartoon display of the protein, but that would require more than a simple PDB viewing program. But there is no way to create a ball-and-stick model with the data contained in that file.


What this exercise suggests boils down to placing the two electrons on opposite sides of the nucleus. In this way, the repulsion is minimal because the distance between both electrons is maximal. We can safely assume there will be a difference between the energy calculated with this simplified extension to Bohr's model and the experimental value, otherwise Bohr or his peers would have adopted it. The inability to satisfactorily calculate the ionisation energy of Helium with any such model was a major driver for scientists in the early 20th century to adopt a new quantum mechanical description of the atom.
Bohr's theory tells us that the ionisation energy for one-electron atoms varies with the square of the nuclear charge, i.e. $E=Z^2$ in Rydberg units (1Ry = 13.6eV). So the ionisation energies of $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$ are respectively 1, 4, 9, ...
The two-electron model described here can be simplified to a one-electron problem because the particles are assumed to be neatly aligned at all times. To this end, we define an effective nuclear charge $Z_\rm{eff}$ which sums the charge of the protons in the nucleus (+2 for Helium) and that of the other electron (-1), weighed by the square of the distance : since the opposite electron is twice as far as the nucleus, its repulsive force is four times smaller than the attractive force of one proton in the nucleus. So we get $Z_\rm{eff} = 2-\frac{1}{2^2} = \frac{7}{4}$. The ionisation energy then becomes $E=Z_\rm{eff}^2 = \frac{49}{16}$ = 3.06 Ry (or 41.6 eV).
The experimental value is 1.81 (24.6 eV), which is significantly lower.
Apparently the electron-electron repulsion in the Helium atom is (much) stronger than this simple model suggests.


Alas, your problem has no solution, because of the risk of producing $\ce{Cu(OH)2}$ at pH $5$ and $6$. The solubility product of $\ce{Cu(OH)2}$ is $2.2·10^{-20}$. At pH $5$ and $6$, the concentrations of the ion $\ce{OH-}$ are respectively $10^{-9}$ and $10^{-8}$ M. This means that, at pH $5$ and $6$, the maximum molar concentrations of the ion $\ce{Cu^{2+}}$ are $2.2·10^{-2}$ M and $2.2·10^{-4}$ M respectively. So you will never be able to obtain a $0.1$ M solution of $\ce{CuSO4}$ at pH $5$ or $6$.


Electrons jumping from 1 water molecule to another

Electrons can be anywhere in the universe

Kind of: There is a non-zero probability that an electron can be at any distance from some point of origin (for instance, the nucleus of an atom), but there is some distance which corresponds to the place where we're most likely to find it (the place where the probability amplitude is highest) most of the time.

  and where they are is random.

Again, sort of: They are most likely located in certain regions of space for most of the time, but since it's a probability, it's not a certainty. That said, there is a common sense element that can be applied here. Yes, an electron from a hydrogen atom in a flask in your laboratory can be out near Jupiter, but realistically, it's not. And if it's on Jupiter, it's not really that hydrogen atom's electron anymore.

  Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O}^+$ and $\ce{H2O}^−$ to form?

Yes, it is possible (but probable? I don't know, but certainly there are conditions under which such a reaction will occur).


Does motor oil get more viscous when heated & what is the meaning of the SAE motor oil code?

The numbers in the code have nothing to do with viscosity index. However, your understanding of 'W' for winter is correct. The real code for your car is actually SAE 5W-30 where SAE stands for the Society of Automotive Engineers, which has established a multi-grade numerical code system for grading motor oils according to their viscosity characteristics. I said viscosity characteristics, not viscosity index. According to Wikipedia:

  The original viscosity grades were all mono-grades, e.g., a typical engine oil was a SAE 30. This is because as all oils thin when heated, so to get the right film thickness at operating temperatures, oil manufacturers needed to start with a thick oil. This meant that in cold weather it would be difficult to start the engine as the oil was too thick to crank. However, oil additive technology was introduced that allowed oils to thin more slowly (i.e. to retain a higher viscosity index); this allowed selection of a thinner oil to start with, e.g., "SAE 15W-30", a product that acts like an SAE 15 at cold temperatures (15W for winter) and like an SAE 30 at $\pu{100 ^\circ C}$ ($\pu{212 ^\circ F}$).

viscosity characteristics: Kinematic viscosity is graded by measuring the time it takes for a standard amount of oil to flow through a standard orifice at standard temperatures. The longer it takes, the higher the viscosity and thus the higher the SAE code. Larger numbers in the code are thicker.

The company called Anton Paar has analyzed these SAE coded motor oil for the parameters, Kinematic viscosity and oil density versus elevating temperatures. I have included some of graphs including SAE 5W-40, but failed to find the most common SAE 5W-30:

Finally, for your information, Viscosity Index is a dimensionless number that represents how the viscosity of a hydraulic fluid changes with temperature. The greater the viscosity index, the smaller the change in fluid viscosity for a given change in temperature, and vice versa. Synthetic oils usually have a higher viscosity index than do mineral oils. (Mobile Hydraulic Tips). The following graph illustrate how viscosity index influence the change of viscosity with temperature (Ref.1):

It is also nice to know how change of viscosity index by modifiers affect mono-grade oil to better multi-grade oil:

References:

Tiago Cousseau, “Film Thickness and Friction in Grease Lubricated Contacts: Application to Rolling Bearing Toque Loss,”  PhD Thesis; University of Porto (Universidade do Porto), Porto, Portugal, 2013.


What happens to the solute when a solution freezes?

If you were to freeze the solution in a normal manner, using something similar to a home freezer, a more and more concentrated solution will form as the majority of the solvent freezes. The solute will be found in a small amount of solution that did not freeze, since the solute allows for freezing point depression. So, if you put both of your beakers into the freezer, one beaker will have a completely frozen block of solvent and the other will have a frozen block of solvent with some volume of highly concentrated solution.


Why are some molecules unable to absorb infrared radiation?

In basic terms, for a molecule to absorb radiation there has to be an oscillating dipole being produced. This can occur by nuclear motion (vibrations, rotations) or electronic motion to produce electronically excited states. You ask about IR radiation, which does not have sufficient energy to produce electronic excitation, so this is ignored in the text below.

The restriction on producing a moving dipole exists because radiation consists of oscillating electric and magnetic fields and to allow energy to go from the radiation into the molecule, there has to be an interaction between radiation and the molecule and the origin of this is the oscillating molecule's dipole. (Of course the radiation also has to have the correct energy to bridge the gap between any two energy levels, but this is just another way of saying that the molecular dipole and the radiation oscillate at the same frequency, i.e. are in resonance.)

Thus in any homonuclear diatomic there is no IR absorption because these molecules have no  intrinsic dipole or one caused by vibrations. Also as they have no permanent dipole there is obviously no dipole generated when molecules rotate. Heteronuclear diatomics, HCl for example, all have a permanent dipole and one that is changed  by vibrational motion, thus there is a vibrational and rotational spectrum in all heteronuclear diatomics.

All homo-nuclear diatomics, $\ce{N2}$, $\ce{O2}$ for example, are transparent to IR radiation and this is of prime importance in global warming, since if these molecules obtain energy, say by collision with water or carbon dioxide that has been excited by IR radiation, they cannot radiate the energy away and so must heat the atmosphere by collisions with other molecules.

In a molecule such as $\ce{CO2}$ although it has no overall dipole it does have moving (oscillating) dipoles due to the way it vibrates. One mode is to stretch one bond at the same time as the other contacts (an asymmetric stretch), and so an oscillating dipole is formed and this interacts with radiation and $\ce{CO2}$ has an IR vibrational-rotational spectrum. The bending vibration is also IR active.

A similar effect occurs in methane, the way the molecule vibrates develops oscillating dipoles and these interact with the radiation and so absorb energy. Molecular group theory is usually used to sort out which of the $3N-6$ ways molecules can vibrate can have IR spectra.

There are complicated higher order effect even in homonuclear diatomics which may generate absorption but these effect are absolutely tiny and can be ignored for most every purpose. The relative magnitudes of electric dipole, quadrupole and magnetic dipole intensities are $I_\mathrm{dip}:I_\mathrm{quad}:I_\mathrm{mag} \approx 1:10^{-5}:10^{-5}.$

We can measure the vibrational frequency of homonuclear as well as heteronuclear diatomics by using Raman spectroscopy, which is a scattering effect (no photons are absorbed), and depends on the shape of the 'electron cloud' in the molecule. This is more properly called its polarisability.


Calculating laser wavelength/power to cause emission of light in a gas?

Two-photon excitation using femtosecond lasers that emit in the infrared range is a technique used in living cell imaging via fluorescence microscopy. Advantages of the technique, such as the minimized light scattering (~ 1/$\lambda^4$) using long wavelength for excitation and the fluorescent dyes used are discussed here.

For large volumetric displays, a different technique was/is under development in a joint project of the Japanese National Institute of Advanced Industrial Science and Technology (AIST), the Keio University and an industrial partner.

Here, fluorescence does not play a role and special gases are not needed either.

The images can be displayed in air and are formed - pixel by pixel - by plasma dots generated in the focal point of pulsed infrared lasers with a pulse width of 1 ns and a repitition rate of 100 Hz.


How to relate a reaction barrier to the time the reaction needs to proceed?

Here are some back of the envelope numbers from using the Eyring equation:

$$k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$$

Let's just assume we're at $298\ \mathrm{K}$ for the reaction, and the reaction is relatively simple:

$$\ce{A->B}$$

I constructed the following table by plugging in values. $t_{1/2} = \frac{\ln 2}{k}$

\begin{array}{|c|c|}
\hline \Delta G^{\ddagger}\ (\mathrm{kcal\,mol}^{-1})& k\ (\mathrm{s}^{-1}) & t_{1/2} \\ \hline\hline
15 & 63.4 & 10.9\ \mathrm{ms} \\
20 & 0.0138 & 50.2\ \mathrm{s} \\
25 & 2.98\cdot 10^{-6} & 64.6\ \mathrm{h} \\\hline
\end{array}

The values for 15 and 20 $\mathrm{kcal\,mol}^{-1}$ seem pretty consistent with your rule. The top value is a bit off, but we're working with a very small numbers at this point, and there may be other sources of error that we're not accounting for in the model.


Are viscoelastic materials to be considered solids or fluids?

There are two basic models that describe the two fundamental branches of viscoelastic behaviour:

The Maxwell model is made up of a spring and a dashpot in series: For fast deformations, the dashpot does not really move, and your spring gives pure elastic behaviour. For slow deformations, the spring stays relaxed, and the damper fluid in your dashpot turns all energy into heat.
The Voigt-Kelvin/Kelvin-Voigt model is a spring and dashpot in parallel: Even at the slowest/smallest deformation, the spring never relaxes until you let go and let it return your sample to the original shape. And the faster you try to deform it (e.g. oscillatory), the more heat your damper produces.

The former, e.g. a molten polymer, is a liquid, the latter, e.g. crosslinked rubber, is a solid.


What is the difference between protein domains, motifs, and supersecondary structures?

All supersecondary structures are motifs, but not all motifs are supersecondary structures.

Motifs, in a biological sense, are very similar to fashion motifs. They are patterns that repeat in a lot of different places. Supersecondary structures are motifs that are made of several secondary structures. Some motifs are much smaller (pockets for holding ions are 3-4 amino acids long), or involve no secondary structure interactions (see SLMs).

Regarding the second half of your question, a protein domain (depending on who you ask) is any region or regions of a protein that do something, regardless of how it is folded. Very often, domains can fold correctly on their own. Generally, the required protease/functional assay to determine whether a possible domain is really technically a domain is not usually done. For example, the kinase insert domain found in some tyrosine kinase receptors is considered a domain, despite being a) quite small b) unable to perform its function in isolation (not entirely fair, as its function is to phosphorylate the rest of the receptor).

In sum: protein domains have a couple of competing definitions, so you cannot assume that something referred to as a 'protein domain' is capable of performing its function when separated from the rest of the protein.

Domains can be of nearly any size (due to their functional and not structural classification), but very small and very large ones are rare. They can be made of any number of motifs, or none at all.


Well, very long hydrocarbons can become twisted around each other. This is definitely known in the polymer literature. Consider some branched polyethylene with ~1,000 repeat units. It's not too hard to imagine that multiple chains can become entangled in much the same way as shoelaces or strings when you mix a lot of them in a box.

You are also right that with large molecules, there are more sites for intermolecular interactions. Consider that there is a "combination" effect with lots of sites. That is, to pull two very long linear chains apart, you need to break lots of interactions.

Notice that pulling one atom on the bottom away from its neighbor on the top chain will require the movement of all atoms on the bottom chain. So it's slightly harder to separate two long chains, not only because there are so many interactions, but also because of the combination.

Put simply, the difficulty in separating the two chains is slightly higher than simply the sum of the interactions.


Electrolytic refining of precious metals — why the anode rather than the cathode?

For these metals to deposit, they would have to be present in solution such that they can come into contact with the cathode and be reduced. However, this would first require them to be oxidized, so that they dissolve in solution. Unless you are actually electrolytically refining for gold, the voltage will not be enough to oxidize it.


Kolbe's electrolysis method for getting alkanes

The metal does not play any role other than to form soluble salts here. The metal cations move to the cathode while carboxylate anions move to the anode where oxidation to a $\ce{R-COO·}$ radical takes place. These radicals decompose to $\ce{CO2}$ and $\ce{R·}$ radicals. Two of latter combine to form a C-C bond.


Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently?

What could happen if I would strongly boil it

Mechanistically, nothing different will occur.  The same reaction that occurs at lower temperature will still occur at higher temperature.  The only difference will be the rate of the reaction; the reaction will go faster as you increase the reaction temperature.  In a classroom setting the procedure might involve simmering rather than boiling because

The reaction is usually followed by making uv-vis absorbance measurements every 15, or so, minutes.  Depending upon the reaction temperature and the concentration of base, the reaction might proceed too fast at boiling to allow for convenient measurements (e.g. the reaction might be complete in 10 minutes).
Safety reasons, if something goes wrong (the flask cracks) a warm basic solution won't burn the student as bad as a boiling basic solution would.