Electrolysis in aqueous solution – which equations to use to predict product at each electrode? When trying to work out what will be produced at each electrode during electrolysis, I'm confused as to which equations to use for hydrogen and oxygen.

For instance, an electrolytic cell containing aqueous nickel(II) bromide, with a nickel cathode and a platinum anode.

At the cathode:

$$\ce{Ni^2+ + 2e- -> Ni}$$

I have read that this does not occur because $\ce{Ni^2+}$ is a worse oxidising agent than $\ce{H+}$, which makes sense because $\ce{H+}$ is lower down on the left side of the electrochemical series meaning its reduction is more feasible.

Water is therefore reduced at the cathode:

$$\ce{2H+ + 2e- -> H2}\quad E^\circ=0.0\ \mathrm V$$

or:

$$\ce{2H2O + 2e- -> H2 + 2OH-}\quad E^\circ=-0.83\ \mathrm V$$

The second equation, however, has water higher up in the electrochemical series. This would suggest nickel should actually  be reduced in preference to water.

How do I know which of the two "water-reduction" equations to use?

This is even less transparent for the anode, where bromide should be oxidised to bromine, but there are also two equations showing production of oxygen: one below and one above bromide in the series:

$$\begin{alignat}{2}
\ce{O2 + 4H+ + 4e- &-> 2H2O}\quad &E^\circ&=+1.23\ \mathrm V\\[6pt]
\ce{O2 + 2H2O + 4e- &-> 4OH-}\quad &E^\circ&=+0.40\ \mathrm V
\end{alignat}$$

I don't understand which of these equations should be used and how you're supposed to know which to use.

The given potentials $E = 0.00{\text{ V}}$ and $E = -0.83{\text{ V}}$ for the reduction of water apply to different $\text{pH}$.

The thermodynamic relation of the potential $E$ to the composition of the solution is generally known as the Nernst equation:

$$E = E^\circ  + \frac{{0.059{\text{ V}}}}{z}\log \frac{{{{\prod\nolimits_i {\left[ {{\text{ox}}} \right]} }^{{n_i}}}}}{{{{\prod\nolimits_j {\left[ {{\text{red}}} \right]} }^{{n_j}}}}}$$

For the given reaction

$$\ce{2 H+ + 2e- <=> H2}$$

the Nernst equation reads

$$E = 0.00{\text{ V}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{\left[{{{\text{H}}_2}}\right]}}$$

Since

$${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$$

the potential $E$ depends on $\text{pH}$:

$$E = 0.00{\text{ V}} - 0.059{\text{ V}} \times {\text{pH}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{1}{{\left[ {{{\text{H}}_2}} \right]}}$$

Therefore, the potential is $E = 0.00{\text{ V}}$ at $\text{pH} = 0$ and $E = -0.83{\text{ V}}$ at $\text{pH} = 14$ respectively.


Why do sulfur and oxygen form sulfur dioxide? Both sulfur and oxygen have 6 electrons in their outermost shell. Since both only need two electrons to stabilize, why doesn't just one atom of oxygen react with one sulfur atom, forming sulfur monoxide SO instead of sulfur dioxide $\ce{SO2}$?

One atom of oxygen is a fictional species outside of a plasma.  Oxygen is naturally diatomic, and the process by which it oxidizes sulfur is radical-based, leading to both oxygen atoms reacting.  The fact that SO2 forms instead of SO is a result of the [redundant answer] thermodynamics favor SO2 over SO.

Given that simple fact here is a quantum mechanical rationalization: Sulfur can expand its octet, which means in SO2 it has 12 electrons 'around' it.  This allows resonances forms where O- has a significant contribution to resonance, while S still has at least 8 electrons around it.  In SO no such resonance forms exist, meaning the highly electronegative oxygen must share its electrons with with sulfuer to avoid exposing a six-eletron sulfur species.  Hence SO2 should be a better structure than SO overall.


Greater entropy: an atom or macromolecule? A question that appeared on my last chemistry exam was :

  Which of the following has greater entropy A) An atom
  B) A
  macromolecule

The question doesn't specify anything else(i.e. type/size of atom or macromolecule). The macromolecules we learned class were proteins, sugars, nucleic acids(DNA/RNA) or lipids. What is the correct answer and why(hypothetical calculation, if possible) ?

There are many equations that describe the entropy of a system, one is
$$\ce{S = k~ ln ~W}$$
where S is the entropy of the system, k is Boltzmann's constant and W is the number of states a system can exist in.  If we take an atom, it can only move in 3 directions x, y and z - that's it.  If we take a small molecule like ethane, it can also move in these same 3 dimensions but it can also rotate about any one of the atoms, the various bonds can stretch and compress and the various bond angles can expand and collapse.  Clearly, there are a lot more states that ethane can exist in as compared to a single atom, clearly W is larger for ethane than a single atom, clearly S is larger for ethane than a single atom.  As we increase the complexity of the molecule, W and S increase exponentially.  A macromolecule will have a much larger entropy than an atom.

Edit (based on @Gdgames Gamers comment below)

Entropy is a measure of the disorder in a system.  It will always be true that the entropy of substance in the gas phase will always be greater than the entropy of the same substance in the liquid phase, which, in turn, will be always greater than the entropy of the same substance in the solid phase.

Perhaps the person who wrote the exam is assuming that the atom is in the gas phase and the macromolecule is in the solid phase and that based on the phase difference alone the atom has a larger entropy than the macromolecule.  I don't think this is always correct.  First off, the temperature is not specified, nor the state of the macromolecule (is it in solution?).  The macromolecule, depending upon the temperature and state, may or may not translate and rotate through space, but it will vibrate (even at 0K).  I'm just not sure how to compare the entropy contribution from the potential 3n-6 vibrational modes in the macromolecule to the entropy from the 3 translational modes of the atom.  Very poor question by the examiner.


Is there a substance that will eat through and completely dissolve carbon but not damage engine cylinder, piston nor rings? I'm afraid that when I was instructed to be specific in the title of my post, I kind of said it all... Here is the issue:

I have a 2001 Isuzu Rodeo. It has been going through excessive amounts of oil. No blue smoke, no drips whatsoever.

Through an Isuzu forum I find this is a common problem for model years up to and including mine as oil drain holes for the cylinders that were made in the pistons clog with carbon.

In that the vehicle is a 2001, I can well suspect those little holes are well caked and packed.

Though some folks state that they remove the spark plugs, pour a commercial internal auto engine cleaner called B12 into the cylinders and allow an over-nite soak, other auto-techs state that the only way to clear the drain holes is to tear apart the engine and drill the holes in the pistons out or replace them altogether.

Is there a chemical that will eat away and dissolve the carbon whereas after allowing whatever sufficient soak time is suggested I could then change the oil and have a hope for cleared out ports without damage to the metal engine parts?

Thank-you for your thoughts!

If it's pure carbon, then the only way to remove it I can think of is by rigorous oxidation (e.g. by burning in oxygen flow or by reaction with strong oxidants (e.g. conc. $\ce{H2SO4}$, $\ce{HNO3}$)), but this is going to damage the metal parts either, as Ivan said.

However, if the clog is caused by soot and resinous residue from the incompletely burnt fuel, then most likely hot toluene or xylene can dissolve the residue and liberate the carbon particles. Both are widely available solvents which won't damage the metal parts (but avoid contact with rubber tubing) and are also used to remove clogs caused by paraffin wax.

P. S. Both are highly flammable liquids, so do not warm them up using an open fire (e.g. use an improvised water bath)!


How to infer formula from diagram Hello Folks

I am not now nor have I ever been any kind of chemist.

My question is, how is the formula for Cerin, C30H50O2, consistent with this diagram taken from Chemspider?

I can count the 30 Carbon atoms, assuming that each vertex of a hexagon is a single C. I can obviously see the two Oxygen atoms. But when I count the Hydrogens, I can only get to 28. Where are the other 22?

Thank you for your patience.

Each carbon atom (you seem to have identified them) needs four of those stick-like thingies (the lines, sometimes called "bonds"). These include the dashed lines and thick triangles as well as the "double bond" you see at the lower oxygen which looks like this ($=$). That one counts twice. Every time you only see 3 or even 2 for one carbon atom, that means 1 (or 2) hydrogen atoms are implied.


Molecule where an sp3 chalcogen is connected to an sp2 atom I am looking for a small molecule where an $\mathrm{sp^3}$ atom from Group 16 (O, S, Se,...) is connected to an $\mathrm{sp^2}$ or resonant atom of another column?

Background: The universal force field defines the above case in its torsion
rules and I would like to find an example/application of such a structure
to understand it better.

Thanks!

Out of the blue, consider triphenyloxonium cation. It is non-planar thanks to bulkiness of the phenyl groups.

Another possible example is mono-epoxide of allene. Considering oxygen atom as $sp^2$ here is unconvincing since it would involve huge angular strain, much higher than for a structure with $sp^3$ oxygen.


Electrolysis of aqueous lead nitrate When going through one of my chemistry textbooks, I saw that the electrolysis of aqueous lead nitrate led to oxygen being formed at the anode and lead being formed at the cathode.

However, in school, I was taught that only metals less reactive than hydrogen will form at the cathode.  This electrolysis contradicts what I learned, as lead is more reactive than hydrogen, and should remain as ions in solution.

Therefore, my question is: What forms at the cathode during electrolysis of aqueous lead nitrate and why?

Firstly, the reactivity being referred to is not reactivity in the sense of the reactivity series of metals, but rather in the sense of standard electrode potentials (standard reduction potentials).

This comes into play when setting up an electrolytic cell in determining the reaction that will take place at the cathode/anode. In this case, the reaction that requires the least amount of work (or voltage in this case, since voltage is essentially a measure of work) will take place. So, assuming that this is done aqueously, if the reduction/oxidation reaction to take place requires more voltage than the reduction/oxidation of the water, then the water will simply be reduced/oxidized instead. (I think this is where the reactivity of hydrogen comes into play in how you were taught).

So to answer your question more directly, it is actually lead that forms at the cathode.

Edit: To add some detail, the reduction potential of H2O is -0.8277 V, while the lead(II) ion is -0.13. In this case, less energy means closer to zero, the negative being essentially irrelevant. Hope this helped.


Can water be ignited in an ozone atmosphere? According to my calculations, the reaction $\ce{H2O + O3 -> H2O2 + O2}$ should be favourable under standard conditions. However, I am told that the reaction is probably very slow in solution since ozone is poorly soluble in water and the reaction occurs through a radical mechanism.

Water can be ignited in a fluorine atmosphere. Can water also be ignited in an ozone atmosphere? Can this be done at 1 atm?

We are basically talking about a derivation of the $\ce{H2/O2}$ system, which involves a radical mechanism. You maybe want to have a look at this answer of mine going a bit more into detail.

The reaction you are interested about is only a total sum of reactions and is most likely not to occur as other reactions will be favourable. Ozone itself is not a very stable compound and many other molecules may actively catalyse decomposition. There are various publications[1-3] on the mechanism of this reaction:
\begin{align}\ce{
  M + O3 &<=> M + O2 + O\\
  O + O3 &<=> 2O2\tag1
}\end{align}

The range of $\ce{M}$ is in $\ce{O3, O2, CO2, N2, He, etc}$.[1] Although not stated explicitly, there should basically no reason why water is not be able to accomplish the same. Hence the driving force of the total reaction would be entropy, giving the overall equation:
$$\ce{2 O3 ->[\ce{H2O}] 3 O2}\tag2$$

It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}\tag3$$

In presence of Hydrogen donors ($\ce{RH}$ apparently any kind will do) a chain reaction will be initiated.[1] $$\ce{O + RH -> OH + R}\tag4$$

The main chain carriers will be $\ce{HO, HO2}$.
\begin{align}\ce{
  HO + O3 &-> HO2 + O2\\
  HO2 + O3 &-> HO + 2O2\tag5
}\end{align}

The process will be terminated by this (quite unspecific) reaction. This may then also involve $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[1]
$$\ce{2HO2 ->}\text{stable products}\tag6$$

In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}\tag7$$

Steahelin and Hoigné propose several hypothesis in which way ozone will be decomposed, but I will not go into absurd detail.[4] The core assumptions involve $\ce{{}^{-}OH}$ as initiator of a radical chain.
\begin{align}\ce{
  O3 + {}^{-}OH &-> .O2- + HO2. &  HO2. &<=> H+ + .O2-\\
  O3 + {}^{-}OH &-> HO2- + O2   &  H2O2 &<=> HO2- + H+\\
                               &&  O3 + HO2- &-> .OH + .O2- + O2\\
  O3 + {}^{-}OH &-> .O3- + .OH\tag8
}\end{align}

Which is basically followed by another set of radical chain reactions.
\begin{align}\ce{
  O3 + .O2- &-> .O3- + O2\\
  .O3- + H2O &-> .OH + {}^{-}OH + O2\tag9
}\end{align}

Which is all again resulting in $(2)$ again.

In summary the proposed reaction might occur, but will immediately be followed by another set of reactions resulting in the formation of biatomic oxygen.

References

Sidney W. Benson, and Arthur E. Axworthy Jr., J. Chem. Phys., 1957, 26, 1718.
J. V. Michael, J. Chem. Phys., 1971, 54, 4455.
Oliver R. Wulf, and Richard C. Tolman, *Proc. Natl. Acad. Sci. USA, 1927, 13 (5), 272–275.
Johannes. Staehelin, and Juerg. Hoigne, Environ. Sci. Technol., 1982, 16 (10), 676–681.


Least evil way to make soap? I'm not a chemist but something of a tinkerer and adult boy scout.  Currently if you want to make soap, you need fat and lye.  Fat I'm good with, it's easy to get and you can even get it from plants. Lye usually requires wood ash though and that I would rather avoid, if possible. If not possible and burning something is mandatory, then I want the biggest bang for my buck. That means I want to get the most lye I possibly can with the least pollutant release, especially greenhouse gasses and we shouldn't be thinking of just wood. Anything that will get us some low impact soap in the end. We may even want to replace soap as an idea with something else that you can make at home and wash humans, clothing and dishes with.

By far the largest source of lye nowadays is the electrolysis of concentrated solutions of $\ce{NaCl}$ or $\ce{KCl}$ in water, the famous and incredibly industrially important chloralkali process, producing millions of tons of $\ce{NaOH}$ and $\ce{KOH}$ every year. While electrolysis is an energy-intensive process, presumably the electricity could be supplied by a low-carbon energy source such as solar or wind power.

"Replacing soap", in the sense of halting the usage of surfactants, is not a good idea and likely not even possible. Surfactants are too useful a class of substances. In fact, soaps are likely the most environmentally friendly surfactants, given that they're based on a simple hydrolysis and acid-base reaction with common biomolecules, fatty acids and triglycerides. Other surfactants such as alkylbenzenesulphonates (dishwashing liquid), tetraalkylammonium salts (hair conditioner) and perfluoroalcoxylates, among others, can be toxic to aquatic lifeforms and take longer to degrade in the environment. The excessive usage and improper disposal of surfactants is a recognized environmental problem, but tackling soaps won't really get you far in solving the issue.


Finding the equivalent weights of the oxide and the metal when reduced by hydrogen A metal oxide ($\pu{1.80 g}$) required $\pu{833 ml}$ of hydrogen at NTP to be reduced to its metal. What are the equivalent weights of the oxide and the metal?

My attempt:
$$\text{equivalent of metal oxide = equivalent of } \ce{H2}$$
$$\frac{\pu{1.80 g}}{E \ \pu{g/eq}} = \frac{\pu{833 ml}}{\pu{22400 ml/eq}}$$
$$E = \pu{48.4 g/eq}$$
Now, since $$E_\mathrm{oxide} = E_\mathrm{oxygen} + E_\mathrm{metal} $$
$$48.4 = 8 + E_\mathrm{metal} \Rightarrow E_\mathrm{metal} = 40.4$$
where $E$ is equivalent weight.
However, my textbook says that the equivalent weight of the metal oxide is $\pu{24.2 g/eq}$ and that of the metal is $\pu{16.2 g/eq}$ (it does not provide a solution). Is the book's answer correct? If yes, what did I do wrong?

Equivalents are part and parcel of Indian chemical education. I don't think college chemistry is going to be updated soon anytime there. I recently heard a complaint by a company to the chemistry chair in an American University. They said that your students do not understand normality! Please teach these concepts to your undergraduate students. Of course, normality concept sounds Greek to modern students.
Anyway let us assume we are in early 19th century and we are doing a classical experiment of chemical analysis. It is not a bad exercise, then. Balanced equations existed then, so we can write:
$$\ce{M_xO_y + yH2 -> M + yH2O}$$
833 mL of hydrogen gas corresponds to 0.833/22.4 = 0.0372 moles of hydrogen. This concept existed in early 19th century.
So what is the definition of an equivalent weight of anything? It is the weight that reacts with 1 equivalent weight of hydrogen (which is half its molecular weight).
Your first step is wrong. Realize that
No. of Equivalents of $\ce{H2}$ in 833 mL are = 2 x 0.0372.
This implies that equivalents of metal oxide which are present in the reaction are also 2 x0.0372.
Equivalent weight of the metal oxide is then = 1.80 g/ 0.0744 = 24.19 or as per your textbook 24.2
Since the equivalent weight of oxygen is 8, the equivalent of metal is 24.19-8= 16.19.
Now, there is no metal whose atomic weight matches 16.19. So what if the metal is divalent = 32 (this is sulfur, not a metal). Is it possible that this is trivalent metal like Ti so  the atomic weight of that M is 16.19 x3= 48.57 which matches with titanium.
One can do back calculation to double check this using moles. I leave that for an interested reader.


Which site would NaCN attacks (SN2)? Question

Attempt

I thought that the bromine atom farther from the ring would be substituted because of less hindrance, but the book says opposite, why?

The reaction proceeds by SN2, which is highly accelerated at the benzylic position, due to stabilisation of the transition state by conjugation with the aromatic ring.

So you'll have to replace the benzylic bromine.

References: Benzyl chloride: chemtube3d


Respect to other chemicals in reduction/oxidation potentials From what I understand, the standard reduction/oxidation potential table is based on the hydrogen half reaction which makes a reduction and oxidation potential of 0 respectfully.

In a research internship I'm in now, we are looking for more efficient batteries, they are basing potential candidates in respect with magnesium.

My problem is seeing how the 'in respect' part works and how it affects the system. Maybe it's just hard for me to relate to it in my mind so maybe a more common analogy would help?

You cannot define ‘vacuum’ oxidation potentials, or otherwise derive them de novo without having some zero-point. There is, technically, no self-suggesting zero-point around — unlike for example temperature, where $0\,\mathrm{K}$ is equivalent to ‘molecular movement frozen’.

Part of the reason is, that the electron(s) lost or gained during the redox reaction must go or come from somewhere. Therefore, one has chosen the hydrogen cell as a centre of the scale. Why?

Hydrogen is the easiest element with the least electrons.
It’s simple to build.
It actually also comes with a good further distinction: The one between noble and non-noble metals.

You can actually easily recalculate the standard potentials with respect to another half-cell like magnesium: Just add/subtract that cell’s potential with respect to hydrogen to all other potentials. It’s a bit like moving the temperature scale’s zero point from $0\,\mathrm{K}$ to $273\,\mathrm{K} = 0\,\mathrm{°C}$.


Why are strong acids and bases not suitable as primary standards? This is one of the questions with which I have puzzled over, and can arrive at no definite conclusion. Why are strong acids or bases, such as $\ce{H2SO4}$,$\ce{ HNO3}$, $\ce{HCl}$, and $\ce{NaOH}$ not suitable primary standards?

How will their higher or lower pH affect the accuracy of the results of a titration?

strong Bases in general are not suitable as they will react with $\ce{CO2}$ from the air. This is actually more problematic than hygroscopicity, as it means that you also cannot store diluted $\ce{NaOH}$ solution (openly) as standard.
Instead, $\ce{Na2CO3}$ or $\ce{KHCO3}$ are much better alternatives (for strong acids).
The acids are either a gas ($\ce{HCl}$) or are produced from gas. I don't see how you can produce a conentration reliably enough for use as primary standard.
even if you would manage to produce saturated $\ce{HCl}$, it's much more difficult and less reliable than known alternatives (see below).
but that wouldn't work e.g. for $\ce{H2SO4}$ as $\ce{SO3}$ is soluble in $\ce{H2SO4}$ (Oleum).
$\ce{HNO3}$: redox reactions are not only not particularly useful, but actually harmful as they use up acid equivalents:
$\ce{NO3- + 2e- + 2 H+ -> NO2- + H2O}$
also here, a solid alternative (KH-Phthalate) is available. If you don't have that, you may get away with oxalic acid which is still better than the acids suggested here.


Maximum possible hypochlorite concentration from electrolysis? For reference, "electrochlorination" is just the electrolysis of a simple solution of water and table salt (NaCl) followed by reintroducing the chlorine gas produced.

But, the Wikipedia article on "electrochlorination" claims

  The product of this process, sodium hypochlorite, contains 0.7% to 1% chlorine. Anything below the concentration of 1% chlorine is considered a non-hazardous chemical although still a very effective disinfectant. In addition, the sodium hypochlorite produced is in the pH range of 6-7.5. This means that the chemical is relatively neutral in regards to acidity or baseness. Also, at that pH range, the sodium hypochlorite is extremely stable and the electrochlorination extremely effective.

Is the yield really that small? It seems like we should be able to get much more! The electrolysis of NaCl in water
$$\ce{2H+ (aq) + 2Cl- (aq) -> H2(g) + Cl2(g)}$$
upon recombining the chlorine gas
$$\ce{2OH- (aq) + Cl2(g) ->  OCl- (aq) + Cl- (aq) + H2O(\ell)}$$
should lead to a net increase in pH
$$\ce{Cl- (aq) + H2O(\ell) -> OCl- (aq) + H2(g)}$$
as$\ce{\; OCl- (aq)\;}$is the conjugate base of a weak acid.

Consequently, it seems the resulting solution should have a much greater percent active chlorine, due to the hypochlorous stabilization effect of higher pH -- which would only get even higher as the electrolysis (given sufficient external voltage) continues!

A little bit of Wikipedia research reveals that

  Sodium hypochlorite can also be obtained as a crystalline pentahydrate $\ce{NaOCl·5 H2O}$, which is not explosive and is much more stable than the anhydrous compound.5 The formula is sometimes given as $\ce{2NaOCl·10H2O}$. The transparent light greenish yellow orthorombic[10][11] crystals contain 44% NaOCl by weight and melt at 25−27°C. The compound decomposes rapidly at room temperature, so it must be kept under refrigeration. At lower temperatures, however, it is quite stable: reportedly only 1% decomposition after 360 days at 7°C.6[12]

So, it looks like 44% w/v would be an upper bound to any solution of NaOCl in water, regardless of whether it was obtained via electrochlorination or not. This corresponds to approximately 43.1% active chlorine.

Also, note that that the term "active chlorine" measures overall oxidative capability, not necessarily hypochlorite concentration — hypochlorites can easily decompose into chlorates without altering the greater solution's overall oxidative capabilities
$$\ce{3OCl- (aq) -> ClO3- (aq) + 2Cl- (aq)}$$
as demonstrated via their molar-equivalent iodometries
$$\ce{3OCl- (aq) + 6I- (aq) + 6H+ (aq) -> 3I2(s) + 3Cl- (aq) + 3H2O (\ell) \\ ClO3- (aq) + 6I- (aq) + 6H+ (aq) -> 3I2(s) + Cl- (aq) + 3H2O (\ell)}$$
so more accurate measurements of hypochlorite concentration may likely require additional techniques (such as spectroscopy, for example) as well.

Finally, due to the unstable nature of solutions with high [NaOCl(aq)] and their natural tendency to
$$\ce{2OCl- (aq) -> 2Cl- (aq) + O2 (g)}$$
maximizing [NaOCl(aq)] solely via electrolysis/electrochlorination will likely require a dedicated setup that simultaneously:

Limits atmospheric exposure around the anode to prevent loss of chlorine gas while allowing hydrogen gas to escape by the cathode (or, alternatively,
reduce atmospheric oxygen to water)
Keeps the solution at low temperatures (to limit various other decomposition reactions)
Maintains some external "tending" voltage (to preserve the Nernst equilibrium)

For practical applications, this could be a very unwieldy form of storage!


Yama Cold Brew Tower: what are these lab parts / components? When viewing this on a shopping web site, one of the reviewers indicates "This is about 50-60 [dollars] worth of pyrex glass and another 20 [dollars] worth of wood."  The company charges $499 for this setup, so I'd like to find out if there is any truth to this statement by identifying the parts and seeing how much they cost.  Image of the setup below:

I think that you can make your own cheaper Yama cold brew tower however what they are selling to you is the design! What you see in the picture is not chemistry glassware but if you want to make your own with chemistry glassware (this will cost more money because good chemistry glassware is purer and must have many expansive features that normal glassware don't have) I've written the analogous components in the figure. You need an open-top cylindrical dropping funnel with a stopcock valve.

See here and here for reference. You can use a normal funnel and attach to it a stopcock valve if you want. Then there is something similar to a Buchner funnel, I imagine that there is a filter in the bottom, may be included into the funnel, in this case, a 90-150 micron filter should be sufficient. See here for more info. Then there is a glass coil spiral, up to my knowledge this has no practical use is only beautiful and geeky but will cost a lot. Then you have a normal carafe.
(probably a piece of filter paper is placed over the grounded cafe to allow the diffusion of water in the whole volume of the cylinder and not only in the centre).

With some filter paper and a couple of half cut plastic bottles you can do your Yama tower with less of 5 $: you can make a little hole in the screw cap of your first bottle to make drip the first funnel, and use a piece of filter paper in the second.  If you want to use glassware don't use chemistry glassware you don't need it really, try to find normal glassware to adapt to your project and you can make your own cheaper and nice Yama tower.


What is the white substance left behind after boiling down water I was boiling down water in a stainless steel pot as an experiment the other day. When the water was completely vaporised, a white substance was left behind on the bottom of the pot. I tried rubbing it off the bottom, but it would not come off. I'm unsure what this substance is. What is this substance?

These are minerals that are naturally present in "tap" water. Chemically, they are most likely a mixture of calcium carbonate and magnesium carbonate, both of which are only very lightly soluble in water.

As you boil away the water, these dissolved calcium/magnesium carbonates remain behind and their concentrations eventually become greater than their respective solubility limits. At this point they begin to precipitate out of the solution, forming the 'scale' you see on the pot.  Interestingly, while most salts become more soluble at higher temperatures, calcium carbonate is anomalous in that its solubility decreases as temperature increases (see the accepted answer here), which accelerates the precipitation process.

The easiest way to get rid of them is to add a cup of white vinegar and let it sit in the pot. You might need to agitate the scale from time to time. You can boil the vinegar in the pot if you want, but it will stink! Over time, the vinegar will dissolve the precipitate, and you will see carbon dioxide bubbles forming. You can then pour out the vinegar and flush the pot with lots of water.


Relative acidities of alkanes, alkenes, and alkynes How does one explain the trend in bond lengths and acidity of the following hydrocarbons?

$$\begin{array}{ccc}
\hline
\text{Species} & \ce{C-H}\text{ bond length / Å} & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Ethane, }\ce{C2H6} & 1.102 & 50 \\
\text{Ethene, }\ce{C2H4} & 1.085 & 44 \\
\text{Ethyne, }\ce{C2H2} & 1.061 & 25 \\
\hline
\end{array}$$

Shouldn't the shorter bonds between $\ce{C-H}$ be easier to break than longer ones, making ethane the most acidic?

You're right in that bond length, and therefore bond strength does affect acidity (see: $\ce{H2S}$, $\mathrm{p}K_\mathrm{a} = 7$ and $\ce{H2O}$, $\mathrm{p}K_\mathrm{a} = 15.7$). If we defined acidity with the following equation

$$\ce{HX -> H + X}$$

then the bond strength would indeed be the only deciding factor in the acidity of $\ce{HX}$, since the enthalpy change of that process is literally the bond dissociation energy.

However, acidity actually corresponds to the heterolytic dissociation of the $\ce{H-X}$ bond, with both electrons in the bond going to $\ce{X}$:

$$\ce{HX -> H+ + X-}$$

so bond strength isn't the only factor. In the case of ethane, ethene, and ethyne, the most important factor is the type of orbital that the electrons from the $\ce{C-H}$ bond end up in after deprotonation. For ethane, the two electrons end up in a carbon $\mathrm{sp^3}$ orbital; for ethene, an $\mathrm{sp^2}$ orbital; and for ethyne, an $\mathrm{sp}$ orbital.

Now, of these three, the $\mathrm{sp}$ lone pair is the most stable, since the $\mathrm{sp}$ hybrid orbital contains the most s character ($50\%$) and therefore has the lowest energy. So, of the three carbanions, the lone pair in the acetylide ion $\ce{HC2-}$ is most stabilised, making ethyne the most acidic molecule.

The acidity of terminal alkynes is in fact pretty useful from a synthetic viewpoint, since you can deprotonate them with a strong base like $\ce{NaNH2}$, forming a carbon-based nucleophile which can then be used to make new $\ce{C-C}$ bonds.

As for why the electronic stabilisation outweighs the bond strength, it's worth looking at some quantitative data. The $\ce{H-O}$ and $\ce{H-S}$ bonds have mean bond enthalpies of $463$ and $338~\mathrm{kJ~mol^{-1}}$ respectively (source: Physical Chemistry 9th ed., Atkins & de Paula, p 932), which is a difference of $\mathbf{125~kJ~mol^{-1}}$. This can, to some extent, explain the difference in acidities of $\ce{H2O}$ and $\ce{H2S}$.

On the other hand, the dissociation energies of the $\ce{H-C\mathrm{(sp)}}$ and $\ce{H-C\mathrm{(sp^2)}}$ bonds are given in J. Phys. Chem. 1987, 91, 17-19 as $132.6$ and $116~\mathrm{kcal~mol^{-1}}$ respectively. The difference is roughly $17~\mathrm{kcal~mol^{-1}}$, which is equivalent to $\mathbf{69.5~kJ~mol^{-1}}$ - a somewhat smaller value than before. You can see why bond length variations might therefore play a smaller role in the hydrocarbons.

An alternative

Another equivalent way of looking at it is that an $\mathrm{sp}$-hybridised carbon is more electronegative than an $\mathrm{sp^2}$-hybridised carbon, which is in turn more electronegative than an $\mathrm{sp^3}$-hybridised carbon.

As mentioned earlier, an $\mathrm{sp}$ orbital has the greatest s-character, and consequently electrons in a carbon $\mathrm{sp}$ orbital experience a greater effective nuclear charge and are more tightly bound to the nucleus than electrons in $\mathrm{sp^2}$ or $\mathrm{sp^3}$ orbitals.

Of course, the $\ce{C-H}$ bond also involves a contribution from the hydrogen 1s orbital. However, the above means that the electrons in $\ce{C(sp)-H}$ bonds are more strongly attracted to the carbon nucleus than electrons in $\ce{C(sp^2)-H}$ or $\ce{C(sp^3)-H}$ bonds. This means that the $\ce{C(sp)-H}$ bond is more polarised towards carbon, leaving less electron density on hydrogen, and a greater acidity.

The amount of electron density on the hydrogen can in fact be probed using $\ce{^1H}$ NMR spectroscopy. A lower electron density translates into a higher chemical shift. Even though there is a second, unrelated, factor - anisotropic shielding - that serves to decrease the chemical shift of acetylenic hydrogens, they still tend to show up at higher chemical shifts than hydrogens on saturated hydrocarbons (~2 ppm, compared to ~1 ppm). This is a clear indication that the local electron density on a proton attached to $\ce{C(sp)}$ is less than that of a proton attached to $\ce{(sp^3)}$.


Is the definition of anode and cathode failing? We know anode=oxidation=loss of electrons and cathode=reduction=gain of electrons but in the photoelectric effect the electrons are gained at the anode and lost at the cathode of the discharge tube?
References: Anode,
Cathode, Oxidation, Reduction

entropid's answer is correct, but I'll try to express it differently. As Bard et al. state in Electrochemical Dictionary (1), the concept of anode and cathode is related only to the direction of electron flow: The anode is the electrode at which oxidation occurs (in an electrochemical cell) and electrons flow from electrolyte to electrode and the cathode is where reduction occurs and electrons flow from electrode to electrolyte. In a galvanic cell with a zinc anode, the electrons are flowing away from the interface in the electrode. Similarly, in a phototube, electrons emitted by the cathode are collected by the anode and again, electrons are moving away from the interface in the anode.

Where the electrons are actually coming from isn't important, just what direction they flow with respect to the electrode interface. Electrons moving from interface towards electrode = anode, electrons moving from electrode towards interface = cathode.

(1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008, pp 31.


Does the reaction of 1-bromobutane with cyanide ion follow an SN1 or SN2 mechanism? In my textbook it's mentioned that the reaction undergoes $\mathrm{S_N2}$ mechanism. I understand that the carbon attached to the bromine is primary, but both solvents here are polar protic solvents, which favour the $\mathrm{S_N1}$ mechanism. Can anybody explain how to reconcile these?

Agreed, $ \ce{Br-} $ is a good leaving group, so one would expect that an SN1 mechanism is not too difficult to have in a polar protic solvent. But do not forget that the mechanism of any reaction is determined by not only the nature of the leaving group, but also the carbon skeleton. A primary carbocation is extremely difficult to obtain, and will not, in general, be formed unless the corresponding anionic complex is extremely stable. For example, the equilibrium $$ \ce{R-CH2-F + SbF5 <=> R-CH2+ + SbF6-    }$$ lies predominantly towards the products side because the anion $\ce{SbF6-}$ is an extremely stable system (which is why the mixture $\ce{HF + SbF5}$ is considered a super acid.). Had $\ce{AlBr3}$ been used before adding $\ce{KCN}$ to the solution, I would've gone for SN1 mechanism.

So, firstly, the primary carbocation is a difficulty if you consider an SN1 mechanism.

Moreover, the formation of a carbocation would immediately facilitate rearrangements. Since a primary carbocation is unstable, Wagner-Meerwein rearrangements will take place and your end product would have been a substitution at the second carbon instead of the first (terminal) one.

As for the solvent, it is a mixture of $\ce{Et-OH}$ and $\ce{H2O}$ to facilitate the dissolution of both reactants. Because one of them is an ionic compound, and the other an organic compound, getting them both in the same state can only be achieved by the mixture of an organic solvent and water; the organic solvent being such that it is, itself, miscible with water. A polar protic solvent weakens the ionic bonds by screening, and favours ionisation. This doesn't imply that a polar protic solvent will always completely ionise your solute.


Hund's rules for Li, contradiction? Hund's rules basically state that we sould maxiumise $S$ and for a given value of $S$ maximise $L$. So let us take the case of Li. Li has 3 electrons, clearly the first 2 are in the $n=1$ state, with $S=0$ and $L=0$. For our remaining electron it must go into one of the $n=2$ states. No matter which state we put it in $S=1$, so we now use Hund's rule to maximise the value of $L$, which would indicate that we put it into the $p$ state giving $L=1$ rather then $s$ state giving $L=0$. But it does go into the $s$ state, which appears to violate Hund's rule. Why does it do this?

(p.s. my notation may be slightly out)

Hund's rules do not apply to different configurations that arise from the occupancy of orbitals with different energies.

For example, with a $\mathrm{2p^2}$ configuration (e.g. carbon), the allowed term symbols are $^1D$, $^3P$, and $^1S$. From Hund's first rule, you can predict that the most stable one out of these three is the $^3P$ term.

You cannot compare these three terms with a term arising from a different electronic configuration, e.g. a $\mathrm{2p3p}$ configuration, which could give rise to a $^3D$ term which would be "more stable" on the basis of Hund's second rule.

The aufbau principle is much more important here. A simplified diagram would look something like this (don't judge me, it's not to scale and I have no time to calculate the exact splitting):

Term symbols are associated with specific electronic configurations and while we often write term symbols on their own, we should always remember that a term symbol without an electronic configuration makes no sense.

Just to show you why it doesn't make sense: with your lithium example, if you really wanted to, then you might as well promote the 2s electron to a 3d orbital, since that would give you a $^2D$ state. Higher $L$, wonderful! Why not go further and promote it to a 4f orbital, or a 5g orbital? We might as well ionise it...


What would an ammonium sulfide fire look like? I am writing a novel and the climax involves killing my protagonist in an explosion. My science knowledge is poor but I believe my scenario is entirely plausible except possibly a couple of details.

My character is in a small, poorly ventilated room. It has an equipment or instrumentation cart with some 1970s era machinery. Within it is a jar of ammonium sulfide from which the gas is pumped via leaky tubes. The leak has gotten worse and the room is very smelly. He bends over to unplug the equipment, which is not grounded. The smell is worse near the floor. Pulling the plug out of the outlet causes a spark and the gas ignites.

I hope this is perfectly logical so far. Now here are a couple specific questions.

What would that fire look like? I currently have a line of fire in the air back to the source by the jar. It then explodes. I then have gases like hydrogen sulfide produced and there is a larger, secondary explosion which kills the injured character.

Would it be better to say that all of the gas burns in a flash. Not much damage is actually done, but the jar of ammonium sulfide is heated, begins to produce hydrogen sulfide and other pyrophoric gases and then there is a spectacular secondary explosion?

I am not necessarily going to explain the science to the reader, which gives me some wiggle room. But I do want these events to be plausible.

If your container is supposed to contain solid ammonium sulfide $\ce{(NH4)2S}$, it better be cooled to -20 °C or below.

(Typically, ammonium sulfide is only used in aquous solution to precipitate metal cations of cobalt, nickel and others as their sulfides in the Kationentrennungsgang).

Long before the solid ammonium sulfide reaches room temperature, it will start to decompose and ammonia and hydrogen sulfide are liberated.


Why is tautomers not possible in benzoquinone? 1,4-Benzoquinone, commonly known as para-quinone, is a chemical compound with the formula $\ce{C6H4O2}$

When there are alpha Hydrogen's why can't it show tautomerism?

You could sort of devise a tautomer by transferring an $\alpha$ hydrogen from carbon to oxygen.  But if you draw the resulting structure you find a pair of cumulated double bonds.  That tends to be unstable, and since the cumulated double-bonded carbons favor a 180° bond angle putting them into a relatively small ring makes them a lot more unstable.  Your proposed tautomer is way out there in energy.

Tautomerization of a ketone works better when you have an $\alpha$ hydrogen on a saturated carbon atom.


Why aren't lower aromatic organic compounds taught? Why aren't aromatic compounds with 3, 4 or 5 carbon atoms taught at high school level chemistry? We are instead directly taught about aromatic compounds with 6 carbons/benzene rings. Why?

My core question is: is because the lower members are tough to understand? Do they have any strange properties?

Hueckel's rule (i.e. that aromatic compounds have $4n+2$ $\pi$ electrons) places limits on aromaticity in cycles with less than 6 carbons.

As there are only 3 cases to look at, let's enumerate them:

5 carbons: cyclopentadiene ($\ce{C5H6}$) is not aromatic, but the cyclopentadienide anion ($\ce{[C5H5]^{-}}$) is. This species is well known in organometallic chemistry, due to its propensity to act as an $\eta^5$ ligand and form sandwich complexes with transition metals. The most famous of these is no doubt ferrocene, where two cyclopentadienide ligands sandwich an $\ce{Fe^{2+}}$ ion. This ion is didactically problematic and in my experience not covered in high school or early undergraduate curricula because most of its interest lies in its sandwich bonding mode with metal ions, which cannot be easily reconciled with the simplified VB/VSEPR/hybridisation theory used to teach introductory organic chemistry. The same goes for the under-loved tropylium cation ($\ce{[C7H7]^+}$), which has 7 carbons.

4 carbons: cyclobutadiene is antiaromatic, and as such only exists at very low temperatures. If I recall correctly, it possesses two unusually long bonds and two unusually short bonds, quite at odds with the bond equivalency found in benzene. This molecule would have to lose 2 e- to become aromatic. It does so happen that the cyclobutadienyl ligand is known in $\eta^{4}$ metal complexes such as $\ce{(C4H4)Fe(CO)3}$ (1) - I was not aware of these complexes until an expert in the field pointed them out to me.

3 carbons: I was not sure about this, but apparently the cyclopropenyl cation is a thing and is aromatic. Cotton (2) also mentions this being known as a $\eta^3$ ligand. I suspect that it is not taught because it's too weird.

In conclusion:

The smaller aromatic rings are principally known as polyhaptic ligands to metals due to their non-neutrality. This places them outside of the comfort zone of the theories of molecular structure and bonding used in introductory organic chemistry, and as such they are usually left until later.

(1) Cotton, F.A., Wilkinson, G.; Advanced Inorganic Chemistry; Wiley Interscience, 5e, (1988), p. 1170

(2) Ibid, p. 1168


How is the distance between m/z axis values in mass spectrums called? The output data of mass spectrometers is relative intensity (y axis) over m/z (x axis). How is the distance between m/z values on the x-axis called (in case of centroided data and profile/raw data respectively)?

I wanted to back up my comment and make sure it's justified, so I decided to flip through Gross'Mass spectrometry: a textbook [1] and it looks like there are indeed basically two possible variations, including distance between $m/z$ values, as you suggested:

{Distance | difference | mass difference} between {peaks | signals | peak tips}
{Difference in $m/z$ | $m/z$ difference | $Δ(m/z)$}

Both refer to describing isotopic distribution as well as to reading and interpreting the spectra.

Selected parts touching standardization (emphasis mine):

From [1, p. 11]:

  The distance between peaks on that axis has the meaning of a neutral loss from
  the ion at higher $m/z$ to produce the fragment ion at lower $m/z.$ Therefore, the
  amount of this neutral loss is given as “$x~\pu{u}$”, where the symbol $\pu{u}$ stands for unified atomic mass. It is important to notice that the mass of the neutral is only reflected by the difference between the corresponding $m/z$ values, i.e., $Δ(m/z).$ This is because the mass spectrometer detects only charged species, i.e., the charge-retaining group of a fragmenting ion.

Further summary from [1, p. 20]:

  1.7.1 Basic Terminology in Describing Mass Spectra

  […]

  Ranges in spectra or ranges set in operating a mass spectrometer are to be
  referred to in the form of “$m/z~10–100$” or “$m/z~10$ to $m/z~100$”.

  […]

  Neutral losses (rarely called dark matter […]) are exclusively recognized from
  the distance between peaks expressed in terms of the difference $Δ(m/z).$ The
  mass of the corresponding neutral is then given in units of $\mathrm{u}.$

References

Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.


How to identify the anode and cathode in a concentration cell? How do I understand which is the anode and which is the cathode in this problem?

  We have a concentration cell with hydrogen electrodes. It's made up of two half-cells. One half cell contains [BaCl(aq)]=0.5 M, the other one has [BeOH(aq)]=0.5 M. Draw the cell diagram and calculate e.m.f at 25ºC

So I tried this:

$$\ce{Pt}(\ce{H2} \, 1~\mathrm{atm})| \ce{H+,\,Ba^{+2}} \,(0.5 M)||\,\ce{Be^{+2}}\, (0.5 M)|\ce{H2}(g,1~\mathrm{atm})|\ce{Pt}$$
I don't know how I should think to write the correct diagram. Can you please help me?

Let's restate the problem: we have two hydrogen half-cells with other stuff in them. Hydrogen electrodes will have voltages related to their pH, because that is the definition of a hydrogen electrode. Then it goes haywire: does one half-cell contain BaCl2 and the other contain Be(OH)2 at 0.5 M - certainly not BaCl and BeOH. In fact, Be(OH)2 is amphoteric and soluble at only 2 ppm, so even estimating the pH of that cell would be a blind guess; you could guess the BaCl2 solution to be about 7.0, but might be a tiny bit different.

I suspect the question is not one that has actually been created and measured, but is rather a made-up question with no dependence on reality, to ask how to visualize  two hydrogen electrodes in pH 7 (the BaCl2 solution) and pH 13.69 (pH of 0.5 M OH), which could not occur because of the low solubility and amphotericity of Be(OH)2. So the whole cell could be written:

Pt(H2,1 atm)|H+(aq,10e-7 M)||(H+(aq,10e-13.69 M)|Pt(H2,1 atm)

or, just consider both half cells to be at pH 7 and there's no overall cell voltage (my preference).


Can thermite be lit while mixed into butane? If not is there a flammable liquid that would work? I want to create a thermite flamethrower to melt snow since I live in New England and since I am an asthmatic I struggle to shovel and it would be cool, so do you have any idea as to what liquid thermite could be lit in.

Thermite is a solid-solid reaction that I think would be greatly inhibited by any sort of intermediate. In either case, the smoke, the dust and the copious amounts of nitrous oxides produced will not make your asthma any better off than if you were to pick up a shovel.

And then there will likely be other safety-, legal-, environment- and relationship-with-neighbours -related issues that alone would make this a very bad idea.


How do I find the geometry of a tetramer assembly? What's the best way of finding optimal geometry and minimal energy for tetramer with H-bonds? I know where H-bonds should be, but do not know their length.

Should I just draw randomly 4 molecules, freeze H-bonds and then do some calculations (e.g., semi-empirical) to get minimal energy or is it done another way?

All four molecules, as we guess, should assemble in the way so they make kinda 'tunnel'. And is running conformer calculations valid? Or in that case it calculates conformers of individual molecules in tetramer, not the whole structure?

In general, this is a very tricky problem. At its root, it's quite similar to solving the question of predicting a crystal structure -- you want to know the interaction between multiple molecules via hydrogen bonds and other electrostatic non-covalent interactions.

Here's the best general procedure:

Perform a conformer search on an individual molecule. As you guess, a "conformer search" (in this case in Spartan) is not going to sample different configurations of the supramolecular system. Worse, there's no guarantee that the lowest energy conformer is the one that's found in the tetramer. Still, it's better to start with a low energy conformer, since that's the most likely one.
Attempt to find a low-energy dimer configuration by copy/pasting your molecule. Consider how the hydrogen bonds might match. If you have some knowledge of the interactions great. I might consider, for example that the N-C=O combination you've circled in the bottom left corner probably matches up with the two amines above them (e.g., the dimer has one copy flipped "upside down" compared with this depiction).
Optimize the dimer geometry using hydrogen bond constraints and a reasonable computational method. Here, I would not use RM1 or PM3, since both methods do not include hydrogen bonding terms. I would use some flavor of PM6 or PM7 if you want to use semiempirical methods, or I'd try some type of dispersion-corrected DFT method, e.g., B3LYP-D3 or $\omega$-B97X-D.
Copy/paste the dimer to try to build up a tetramer structure.

I know some research groups have tools that somewhat automate this process, but I don't know of any that are currently widely-available and user-friendly. This is very much still an active area of research.

Moreover, as a human, you can often spot "how things fit" better than most automated strategies.

Good luck!


Why do atoms bond? Why are unfilled valence shells energetically unstable? I teach school Physics. And am trying to get a reasonable explanation of the root cause of the energetic instability of unfilled outermost electron shells that leads to atoms bonding. For most of what I teach, I am able to tie things back to the four fundamental interactions/forces. But unfilled valence/outer shells being the reason why atoms bond beats me.  I get Pauli exclusion. I think I get Schrodinger's equation. Yet, to me these don't provide a fundamental explanation of why a shell 'filled' by bonding is more energetically favorable than neutral atom with an unfilled outermost shell. Do I just accept this as fundamental property of our universe?

The idea of a "full" outer shell is a bit of a misnomer, for instance 3rd row has 3d orbitals but doesn't bother to occupy them until 4th row. Does Argon have a full outer shell, even though it has a $3d^0$ configuration?

Generally there's a large energy gap between the p-orbitals in one row and the s-orbitals in the next row, because of the increase in shielding and such factors, which acts as a kind of natural barrier, or a step, to adding any more electrons than what appears to be a 'full' valance shell.

But within any row, the amount of shielding and associated factors (penetration, nuclear charge, etc) remain largely similar, so if the valence shell isn't 'full' there is still sufficient attraction between the electron and a neutral atom that a negative ion may be stable.  Adding more than this, the energy gap is too large to overcome under usual circumstances so it stops at a 'full' shell.

There are always exceptions.  Noble gases lower down can form compounds even though they are "stable" "full" outer shells, and quirks like some transition metals having $s^1d^{10}$ configuration go to show that some of the ideas about electron filling and stable shells are rules of thumb, each individual case really needs to have its energy levels calculated to explain them.


Is there a possible distortion of XeF6 from Oh point group to reduced symmetry? Below the molecular orbital diagram for $\ce{XeF6}$ (from What is the molecular structure of xenon hexafluoride?)

I am trying to figure out if a distortion from $O_h$ can take place and to which (lower symmetry) point group it would be. I know that it will involve the HOMO $a_g$ interacting with the $t_u$ orbital (bonding). Using a descent of symmetry table for $O_h$, I have found that the $a_g$ must remain completely symmetric. The part I am confused about is, looking at the descent of symmetry table, I am looking for an $a_g$ component of $t_u$ in a point group. I am using the following table:

$t_{1u}$ does not $a_{1u}$ in the table. Literature actually suggests $C_{3v}$ as the resulting distortion, but how would I know this when it isn't in the table? and I was wondering where I am going wrong in my approach?

Also, I am curious where the subgroups fit into the table, as $C_{3v}$ is not there but what does this mean?

Now, there are a few things that must be stated very clearly. Firstly, it pays to be very careful when you are talking about point groups. It's careless (and confusing to others) to write $\mathrm{T_{2u}}$ when you mean $\mathrm{T_{1u}}$, or to write $\mathrm{A_{g}}$ when you mean $\mathrm{A_{1g}}$, so please pay attention to this next time, regardless of whether you are writing online or submitting a piece of work. Good communication in science has to be precise and clear!

Secondly, let's use this reduced descent of symmetry table (I have removed some columns and some rows, which aren't relevant now)

$$\begin{array}{ccc} \hline
O_\mathrm h & T_\mathrm d & D_\mathrm{3d} \\
\hline
\mathrm{A_{1g}} & \mathrm{A_1} & \mathrm{A_{1g}} \\
\mathrm{T_{1u}} & \mathrm{T_2} & \mathrm{A_{2u} + E_u} \\ \hline
\end{array}$$

This tells you that if you decrease the symmetry of the system from $O_\mathrm h$ to $T_\mathrm d$, then anything that previously transformed as $\mathrm{A_{1g}}$ will now transform as $\mathrm{A_1}$. Likewise, anything that previously transformed as $\mathrm{T_{1u}}$ will now transform as $\mathrm{T_2}$.

If you are trying to look for potential mixing of orbitals in the lowered symmetry, then you have to look what the irreps in the lowered symmetry are. From what you have written, I sense that you're trying to "look for an $\mathrm{A_{1g}}$ component", which isn't what you should be looking for, because something that transformed as $\mathrm{A_{1g}}$ in $O_\mathrm h$ symmetry may not necessarily transform as "$\mathrm{A_{1g}}$" in the lower symmetry. Point in case: $T_\mathrm d$.

If you are considering the viability of a distortion to $T_\mathrm d$, then you should be looking for an $\mathrm{A_1}$ component in the second row, not $\mathrm{A_{1g}}$. Of course, $\mathrm{T_2}$ does not contain $\mathrm{A_1}$, so the distortion to $T_\mathrm{d}$ isn't favourable.

Now, in $C_\mathrm{3v}$ the correlations are $\mathrm{A_{1g}} \to \mathrm{A_1}$ and $\mathrm{T_{1u}} \to \mathrm{A_1} + \mathrm{E}$. If we add this column to the reduced table above

$$\begin{array}{cccc} \hline
O_\mathrm h & T_\mathrm d & D_\mathrm{3d} & C_\mathrm{3v} \\
\hline
\mathrm{A_{1g}} & \mathrm{A_1} & \mathrm{A_{1g}} & \color{red}{\mathrm{A_1}} \\
\mathrm{T_{1u}} & \mathrm{T_2} & \mathrm{A_{2u} + E_u} & \color{blue}{\mathrm{A_1 + E}} \\ \hline
\end{array}$$

then it is clear that because the $\color{blue}{\text{blue}}$ $\mathrm{A_1 + E}$ shares one irrep in common with the $\color{red}{\text{red}}$ $\mathrm{A_1}$ component, the HOMO–LUMO mixing is allowed in $C_\mathrm{3v}$ symmetry, and this provides a qualitative argument for why distortion to $C_\mathrm{3v}$ symmetry is favoured.

Again, to drive home the point, you shouldn't be looking for whether the $\color{blue}{\text{blue}}$ bit contains $\mathbf{A_{1g}}$. There is no $\mathrm{A_{1g}}$ irrep in $C_\mathrm{3v}$, so of course you won't find it! You should be asking whether the $\color{blue}{\text{blue}}$ bit contains the $\color{red}{\text{red}}$ bit.

How to get those correlations, is quite a fair question. Well, I got it from this diagram in Orbital Interactions in Chemistry, 2nd ed. by Albright et al. The correlation diagram 14.4 should be self-explanatory, hopefully (you can ignore the pictures of the MOs, 14.5 and 14.6).

But if you want to obtain it without flipping through every book in the library, this is how you need to do it. Because high-symmetry point groups like $O_\mathrm h$ contain tens and tens of subgroups, books don't usually list every single possible correlation to lower symmetry, since there are far too many possibilities. The table you were given has a bunch of omitted correlations, listed under "other subgroups" (the implication being, "other subgroups that I couldn't be bothered to put in"). You'll notice that $C_\mathrm{3v}$ is one of those "other subgroups" that weren't explicitly given.

The strategy therefore is to carry out the correlation in two steps. First you go from $O_\mathrm h$ to $D_\mathrm{3d}$, and then you go from $D_\mathrm{3d}$ to $C_\mathrm{3v}$. The tables you are using don't provide that, but a more complete list of tables is provided in the Appendixes of Molecular Vibrations by Wilson et al. The $D_\mathrm{3d}$ to $C_\mathrm{3v}$ correlation is given as (again I removed the irrelevant rows)

$$\begin{array}{cc} \hline
D_\mathrm{3d} & C_\mathrm{3v} \\ \hline
\mathrm{A_{1g}} & \mathrm{A_1} \\
\mathrm{A_{2u}} & \mathrm{A_1} \\
\mathrm{E_u} & \mathrm{E} \\ \hline
\end{array}$$

so now the correlation from $O_\mathrm h$ to $C_\mathrm{3v}$ is easy: $\mathrm{A_{1g}}$ in $O_\mathrm h$ goes to $\mathrm{A_{1g}}$ in $D_\mathrm{3d}$, which in turn goes to $\mathrm{A_1}$ in $C_\mathrm{3v}$. Likewise, $\mathrm{T_{1u}}$ in $O_\mathrm h$ goes to $\mathrm{A_{2u} + E_u}$ in $D_\mathrm{3d}$, which in turn goes to $\mathrm{A_1 + E}$ in $C_\mathrm{3v}$.


Heating 2% solution to get 4% solution I have a 2% solution of Zinc sulfate in water (in heptahydrate form) that I want to turn into 4% solution instead of having to buy 4% solution separately.

Would heating it to 100 degrees Celsius and waiting for extra water to vaporize be the right way to do it?

Thanks.

The best approach would be to crystallize $\ce{ZnSO4.7H2O}$ and then weigh the required amount and prepare a fresh solution.

How:

Slowly evaporate the solution until it is about one-fifth of its original volume. Caution: Do not boil the solution as it may spit.

Allow the concentrated solution to cool until crystals form.

Filter off the crystals and put the filter paper and crystals on a watch glass and dab dry with another piece of filter paper.

Allow it to dry thoroughly (if you're keen on very accurate concentrations)

I assume by $x\%$ solution you mean $x\%$ weight($\pu{g}$)/volume($\pu{mL}$)
You may proceed to prepare the solution by dissolving $4$ grams in 100 $\pu{mL}$ water. In case you need a different volume, you may use unitary method.
Edit:
Crystallization is certinly more tedious than simple evaporation and volume restoration, but is bound to be very accurate. There might be loss of salt in trace amounts during evaporation, adheration to vessel walls et cetera. But if much precision is not your requirement, you may follow the latter, by all means.


Critical separation at which molecule breaks I have the following question.

  The potential energy of two atoms, a distance $r$ apart, is: $$U = -Ar^{-2} + Br^{-10}$$ Given that the atoms form a stable molecule at a separation of $\pu{0.3 nm}$ with an energy of $\pu{-4 eV}$, calculate $A$ and $B$. Also find the force required to break the molecule, and the critical separation at which the molecule breaks.

Molecule is stable at the distance when the first derivative is zero. But how to find the critical separation at which the molecule breaks?

... As far as I know, in the absence of an external influence, the molecule never actually breaks. I mean, most people wouldn't call two atoms a metre apart a molecule as such, but there's no negative curvature stationary point in a potential like this at which you could say "if $r > x$, this molecule is broken and will not reform".


How many particles are necessary to observe a well-defined state of matter? The three ordinary states of matter $-$ solid, liquid, gas $-$ can usually be distinguished by a function of the strength of interparticular forces and distance. A shorter distance and a stronger force results in a solid whereas the opposite will yield a gaseous state. In other words, the states of matter are describing macrostates.

With clarity in mind, let us define a practical limit to observing different phases.

  Definition. A practical limit of particles is such that a phase transition is concievably observable by experiment. This is the smallest number of individual particles that when probed give us the traditional phase diagrams (omitting different solid phases).

What is the practical limit for most purposes? I am probably looking for a 'number' of the form $10^a\ldots10^{a + \approx2.5}$ where $a$ is a positive integer.

So while an individual particle under standard conditions can be described as gaseous, it does not fit the definition.

An alternative wording (equivalent as far as I can tell) to this question is given by the Clapeyron equation. Apostrophes mark different phases.

$$\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{\Delta H}{T\left(V'' - V'\right)}$$

In which range of the number of particles do the entalphy of transition and difference in volumes of phases become measurable? This does not definitively mean that $\Delta H$ has to be a constant w.r.t. $T$ but rather a function of temperature $\Delta H = \Delta H\left(T\right)$ that has predictive capability.

Computationally:

It depends on the material but in general $10^2 - 10^4$ atoms/molecules is a representative range. I learned this concept as the thermodynamic limit; i.e. the number of particles in a system at which increasing particle number does not change density significantly. For obvious reasons it's also called the macroscopic limit. In Statistical Mechanics we bridge the gap between small "countable" systems and macroscopic "mole-quantity" systems with the thermodynamic limit.

Our lab develops and applies models for small molecules like $\ce{H2, CO2}$, noble gases, hydrocarbons etc. for use in computer simulation. The usual method is to parametrize a model such that phase transitions (i.e. density changes as a function of temperature or pressure) are accurately mapped, and intermolecular potentials are comparable to highly accurate QM calculations (CCS(D/T/Q) methods).

One can reproduce accurate phase diagrams of noble gases using just 40-100 atoms by putting the system in a fluctuating periodic box and changing particle positions until an energy minimum is reached (the computational method is called Monte Carlo) under given $T$ and $p$ -- this is called the isothermal-isobaric ensemble, or $NPT$. You can achieve identical results using the $\mu VT$ (grand-canonical ensemble) wherein chemical potential, volume, and temperature are fixed, but $N$ is allowed to change. For professional model development it is more common to use $10^3 - 10^4$ particles for a good fit.

Below is a snapshot of output from a Monte Carlo program obtaining the density of $\ce{He}$ at STP using 100 atoms (the reported value is 0.1786g/L). You can see that $\ce{He}$ in this model is a near-ideal gas with compressibility factor $Z \approx 1$ :

I think the answer is more appropriately expressed via computational (Statistical-Mechanical) methods because the concept of the thermodynamic limit arose in and is used by SM. In general, the central-limit theorem is where the concept came from, which states that fluctuations to the mean of a quantity (density here, $N/V$) are expressed as $\frac{1}{\sqrt{N}}$, thus for $N=10,000$, the mean density can expectedly vary by $0.01$. For $N=N_A$, the variance would be certainly negligible.


Exo- or endothermic? Clarification required regarding a few cases Are the following processes exothermic or endothermic?

a) When solid $KBr$ is dissolved in water, the solution gets colder.

b) Natural gas ($CH_4$) is burned in a surface.

c) When concentrated $H_2SO_4$ is added to water, the solution gets very hot.

d) Water is boiled in a teakettle.

Attempt at solution: For a), I would say that it is exothermic, since the solution gets colder, which means that it releases contained heat to the surroundings.

b) is exothermic evidently.

c) By the same reasoning, the solution is getting very hot, so it must be endothermic since it absorbs heat.

d) Not sure about this one. I would say endothermic, because the absorbed heat makes it boil.

This is what I answered, and apparently I made atleast two mistakes. So can someone explain to me the flaws in my reasoning?

I think you are confusing terms:

exothermic means that it releases heat, that is energy, because the energy of the products is lower than the one of the reagents.

endothermic means that it absorbs heat (energy) because the products are at a higher level of energy than the reagents.

When a solution becomes hot, it's because it's releasing heat (energy), and likewise if it gets cold it's because the reagents inside are absorbing energy from the surronding environment.

The case of boling water: does it need heat to become steam? Or does it release heat? In other words, do you have to provide heat to the teakettle to make the water boil or it provides heat by itself?

I hope you can get to your answers by yourself now.


What kind of bonds can form between benzene and transition metals in π-complexes? I'm trying to draw the six π-orbitals (3 bonding, which are filled, and 3 anti-bonding) of a benzene ring when it complexes with metal d-orbitals and which metal orbitals have the correct symmetry ($\sigma$, π, $\delta$) to overlap with each of them. I know that metal orbitals will only form a bonding interaction with ligand orbitals if they have matching symmetry but I am not sure which ones will interact.

I think a π bond is formed between $e_{1g}$ of benzene and $d_{xy}$ and $d_{yz}$. Also a $\sigma$ bond is formed between $a_{2u}$ of the benzene and $d_{z^2}$ and $a_{2u}$. No δ bond is formed.

In theory benzene and a transition metal could form σ, π, δ, and even φ bonds as the symmetry allows this.

Here are the orbitals of benzene, calculated at the PBE0/def2-TZVPP level of theory in the D6h  point group with Gamess.

The (valence) orbitals of chromium on the same level of theory in the same point group are as follows:

The linear combination of these orbitals would yield orbitals that have the above mentioned bond symmetries. We will ignore for now that they will most likely not occupied in a complex like bis(benzene)chromium.

benzene a2u + Cr s → σ
benzene a2u + Cr dz² → σ
benzene e1g + Cr dxz → π
benzene e1g + Cr dyz → π
benzene e2u + Cr dxy → δ
benzene e2u + Cr dx²-y² → δ

If we look at f-orbitals (that is of course not very feasible for chromium), we could construct another bond type:

benzene a1g + Cr fx(x2−3y2) → φ

The necessary orbital looks like this:

Of course which of these bonds will form or not depends on the metal in question, the occupation of the orbitals and more. After all, there are also p orbitals which can overlap to form σ and π bonds.

Colour code: Occupied orbitals are shown in blue and orange, while virtual orbitals are shown in yellow and red. Phases have been assigned randomly. The total density of the molecule uses a violet contour.


Sodium Sulfate vs Sodium Sulfite I recently took a test and one of the questions was:

  Which procedures will allow a student to differentiate between solid sodium sulfate and solid sodium sulfite?

  I. Make solutions of each and look for a precipitate when added to $0.10$ M barium nitrate.

  II. Add crystal of each to $0.10$ M HCl and watch for bubbles.

  III. Make solutions of each and test with a pH indicator.

The answer is 2 and 3, however, I incorrectly answered 2 only. I have actually thought a long time about this question.

Why is III also correct, if sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color?

The last statement

  …sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color

is true only for $\ce{Na2SO4}$ as it's formed by both strong base and strong acid and won't noticeably affect pH.
$\ce{Na2SO3}$, on the other hand, undergoes hydrolysis:

$$
\begin{align}
\ce{Na2SO3 + H2O &<=> NaOH + NaHSO3} \\
\ce{2Na+ + SO3^2- + H2O &<=> Na+ + OH- + Na+ + HSO3-} \\
\ce{SO3^2- + H2O &<=> HSO3- + OH-}
\end{align}
$$

resulting in elevated pH (basic solution) up to 9 (according to Wikipedia).


Transformation of virial coefficients from molar volume to pressure I have the following virial equation for a real gas:

$Z = \frac{PV_m}{RT} = 1 + \frac{B}{V_m} + \frac{C}{V_m^2}$

I would like to transform the coefficients $B$ and $C$ so the equation is expressed in terms of pressure:

$Z = \frac{PV_m}{RT} = 1 + B'P+ C'P^2$

My book says that the transformations are:

$B' = \frac{B}{RT}$ and $C' = \frac{(C-B^2)}{(RT)^2}$

How are these derived? I can get the first one by:

$Z(\frac{1}{V_m}) = Z(P)$ and only considering $B$:

$1 + \frac{B}{V_m} = 1 + B'P$,

$\frac{BP}{RT} = B'P$,

$\frac{B}{RT} = B'$.

For the second one I also consider C:

$1 + \frac{B}{V_m} + \frac{C}{V_m^2}= 1 + B'P + C'P^2$,

$\frac{BP}{RT} + \frac{CP^2}{(RT)^2}= \frac{B}{RT}P + C'P^2$,

$\frac{CP^2}{(RT)^2}= C'P^2$,

$\frac{C}{(RT)^2}= C'$.

This result (mine) differs to the one given by the book. They have an additional $-B^2$ term in the numerator. Where is that term coming from?

We want to take this:

$$Z = \frac{PV_m}{RT} = 1 + \frac{B}{V_m} + \frac{C}{V_m^2}+…\tag{1}$$

And transform it into this:

$$Z = \frac{PV_m}{RT} = 1 + B'P+ C'P^2+…\tag{2}$$

We begin by expressing (1) in terms of $\rho_m = 1/V_m$:

$$Z=1+B\rho_m+C\rho_m^2+…\tag{3}$$

We then multiply (2) and (3) by $RT\rho_m$ to get (4) and (5), respectively. Remember that $ZRT\rho=ZRT/V_m =P$, because $Z=PV_m/RT$.

$$RT\rho_m Z = P = RT\rho_m + RT\rho_mB'P+ RT\rho_mC'P^2+…\tag{4}$$
$$RT\rho_m Z= P = RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…\tag{5}$$

Next, we use (5) to substitute for $P$ in (4):

$$P = RT\rho_m + RT\rho_mB'(RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…) +$$
$$ RT\rho_mC'(RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…)^2+…\tag{6}$$

Collecting all the $\rho_m^2$ terms from (5) and (6):

$$RTB=(RTB')(RT)$$
Rearranging leads to the desired $$B'=\frac{B}{RT} \tag{7}$$

Next, collecting all the $\rho_m^3$ terms from (5) and (6):

$$RTC=(RTB')(RTB)+(RTC')(RT)^2$$

Simplifying, rearranging, and making use of (7) leads to the desired
$$C' = \frac{C-B^2}{(RT)^2}\tag{8}$$


Chemical analysis of soapstone Soapstone is a metamorphic mineral with a number of constituent compounds including principally:

silicon dioxide
magnesium oxide
iron oxide
aluminum oxide
calcium oxide
water/carbon dioxide

What would be a typical procedure to analyze a sample of soapstone and determine the proportions of its constituents?

Given its heat resistance to thermal decomposition and the high chemical inertness of soapstone, as noted in this source, for example, to quote:

Soapstone is chemically inert, resistant to heat and durable, yet also malleable...As a metamorphic rock, it is very resistant to heat. In fact, soapstone is often used for casting metals, such as lead, pewter and silver...Because soapstone is primarily talc, which is chemically inert, it is highly resistant to etching and chemical reactions that may mar it’s beauty.

I would suggest placing a small amount of finely powdered soapstone in a vessel with highly alkaline aqueous mix of Sodium hydroxide (NaOH).  Apply conventional heating or, given an iron oxide presence, microwave heating. This is an attempted relatively safe synthesis of, for example, Sodium aluminate.
The proposed reaction is a hot alkaline attack of any of the several metal oxides present, including, for example, aluminum oxide, which could proceed as follows:
$$\ce{Al2O3 (s) + 2 NaOH (s) + 3 H2O (l) -> 2 Na[Al(OH)4] (aq) }$$
Support is provided per this reference, to quote:

All the solutions removed some $\ce{SiO2, Al2O3, and Fe2O3}$ from the samples, but the quantities were small. Sodium hydroxide attacked the kaolin group minerals more strongly...

I would then neutralize the soluble salts with $\ce{CO2}$, creating corresponding carbonate (or oxide, in the case of Aluminum ions) precipitates.
Please read the cited reference for more information on procedures (albeit, as alluded to per this 1971 source).


Figure Headings in Laboratory Reports How do I write a figure legend for a thin layer chromatography plate drawing?

Provide all the data that are necessary to reproduce the experiment:

TLC material (silica gel or aluminium oxide)
composition of the mobile phase
special tricks, such as running the TLC in methanol first for 5 mm to squeeze the initial spot to a very narrow band
detection of the separated bands/spots:

$\mathrm{R_f}$ values
do the spots have a colour
do the show fluorescence
which reagents did you use for staining

UPDATE

If you report on a series of measurements, it's usually good practise describe the conditions that didn't change (type of TLC plate, staining solutions, etc.) in the general section of the experimental part of the report/article/thesis.


What is the difference between dynamic and static electronic correlation I am confused. In the static correlation, we use combination of Slater determinants to account for electronic correlation (known as CI or configuration interaction) and the wave function is represented as a sum of various SD where the HF SD is the base function and higher excited SD contributes some parts. Similarly, in dynamic electronic correlation, we use basis set expansion to account for higher excitation orbitals. What is the fundamental difference between these two correlations? Both are essentially expanding the electronic space to allow more relaxation. What am I missing here? Thanks

$\newcommand{\el}{_\mathrm{e}}$In quantum chemistry, when a nomenclature in which one distinguishes between "static" and "dynamic" correlation is used, "correlation" referrers to all the deficiencies of the Hartree-Fock (HF) single-determinantal approach. For instance, the the correlation energy is defined as the difference between the exact (non-relativistic) energy and the HF energy (calculated with a complete basis),
$$
  E_{\mathrm{corr}} = E_{\mathrm{exact}} - E_{\mathrm{HF}} \, .
$$

Now, what are the deficiencies of a Hartree-Fock approach?

First, electrons in this model do not instantaneously interact with each other, as they do in reality, but rather each and every electron interacts with the average, or mean, field created by all other electrons. Classically speaking, each electron moves in a way so that it avoids locations in a close proximity to the instantaneous positions of all other electrons. And the failure of the HF model to correctly reproduce such motion of electrons is the first source of $E_{\mathrm{corr}}$. This type of correlation is called dynamic correlation since it is directly related to electron dynamics.

Secondly, the wave function in the HF model is a single Slater determinant, which might be a rather poor representation of a many-electron system's state: in certain cases an electronic state can be well described only by a linear combination of more than one (nearly-)degenerate Slater determinants. This is the second reason why $E_{\mathrm{HF}}$ may differ from $E_{\mathrm{exact}}$ and the corresponding type of correlation is called static, or nondynamic, since it ist not related to electron dynamics.

Interestingly, both static and dynamic correlation effects can be taken into account by "mixing in" more Slater determinants $\Phi_i$ to the Hartree-Fock one $\Phi_0$,
$$
  \Psi\el(\vec{r}\el) = c_0 \Phi_0 + \sum_i c_i \Phi_i \, ,
$$
Here, if $c_0$ is assumed to be close to $1$ and a large number of excited determinants $\Phi_i$ are added each of which is assumed to give only a small contribution, then the method primarily treats dynamic correlation. And if, on the other hand, it is assumed that there are just a few excited determinants $\Phi_i$ with weights close to that of the reference determinant $\Phi_0$, then the method primarily treats static correlation.

An example of a method that recovers primarily dynamic correlation is Møller–Plesset perturbation theory (MP$n$), while multi-configurational self-consistent field (MCSCF) method primarily takes account of static correlation. Note the word "primarily" here and above. It is almost impossible in principle to keep dynamic and static correlation effects separated since they both arise from the very same physical interaction. Thus, methods that typically cover dynamical correlation effects include at high-order also some of the non-dynamical correlation effects and vice versa.


Reversed size dependence in ion solubility My book says that solubility increases with interionic distance, because the attractive forces between ions gets smaller. However, if one of the ions in a binary ionic system is very large compared to the other, the trend is reversed. For example, CsF has greater solubility than CsCl. Why is this?

This can be explained with some crude thermodynamic models. The lattice enthalpy $\Delta H_\mathrm L$ of an alkali metal halide $\ce{MX}$ is given by

$$\Delta H_\mathrm L = \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

and the hydration enthalpies by

$$\Delta H_\mathrm S(\ce{M}) = -\frac{k_2}{r_\ce{M}}\qquad  \qquad \Delta H_\mathrm S(\ce{X}) = -\frac{k_2}{r_\ce{X}}$$

where $k_1$ and $k_2$ are some positive constants and $r_\ce{M}$ and $r_\ce{X}$ are the ionic radii of $\ce{M+}$ and $\ce{X-}$ respectively.

The solubility is therefore a balance between these two quantities: larger (magnitudes of) solvation enthalpies increase solubility, but larger lattice enthalpies reduce solubility. The solubility $S$ is thus related to the difference of these terms. We ignore the entropic contribution, or to be precise, we assume that $\Delta S$ does not vary too much between different salts $\ce{MX}$.

$$S \sim \frac{k_2}{r_\ce{M}} + \frac{k_2}{r_\ce{X}} - \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

Let us now assume that

$$r_\ce{M} = a+b \qquad \qquad r_\ce{X} = a - b$$

where $a,b$ are more positive constants. Here we have implicitly assumed that $r_\ce{M} \geq r_\ce{X}$, but it works the other way round too, so there is no loss of generality. Thus

$$\begin{align}
S &\sim \frac{k_2}{a+b} + \frac{k_2}{a-b} - \frac{k_1}{2a} \\
&= \frac{2ak_2}{a^2-b^2} - \frac{k_1}{2a}
\end{align}$$

This is clearly a minimum when $b = 0$, i.e. $r_\ce{M} = r_\ce{X}$. So, salts with similar cationic and anionic radii are less soluble, and vice versa.


Speed at which methanol dissolve in water? Suppose that in a container of volume V we pour V1 of methanol and V2 of water (V2>V1), supposing that V2 is big enough that all the methanol will be dissolved, do we have to mix them mechanically or the diffusion will be enough if we want them to mix in seconds?

As already mentioned by @Zhe, the rate of the mixing of methanol in water (or the other way around) depends (among other parameters) on the rate of addition, the geometry of the vessel, if / how the solutions are mixed by a stirrer.  In addition, this process diffusion is dependent on temperature and gradients of concentration.  One approach to quantify diffusion is based on the differential Fick's laws by determination of the diffusion coefficient.

As difficult it may appear, the diffusion of methanol in water was determined experimentally, as here and here, for example.  If you do not have access to a university library, NIH / HHS made a more recent reference available here, and a more general value is given here


Will fluoride evaporate from tap water at the same rate as chlorine Chlorine evaporation is an overnight process from standing tap water in a gallon pan. Does fluoride evaporate in the same way/rate?

Short answer: no, fluoride does not evaporate in the same way/rate as chlorine.

Longer answer: chlorine will be present as covalent, molecular $\ce{Cl2}$. This is, when not in solution, a gas. Chlorine molecules leaving the surface of the water will (on average) not come back (at least unless there is a very high concentration of $\ce{Cl2}$ in the room's air, in which case you have other problems).

Fluoride, on the other hand, is present as $\ce{F-}$ ions (or other, complex ions) along with corresponding positively charged ions (usually $\ce{Na+}$, I believe). Even if the $\ce{NaF}$ were to be removed from solution, $\ce{NaF}$ is a non-volatile salt, and so would not evaporate away.


Why is the boiling point of hydrogen sulfide higher than that of hydrogen chloride? The boiling point of $\ce{H_2S}$ is higher than $\ce{HCl}$. Is this due to the greater number of hydrogen bond that $\ce{H_2S}$  can form compared to $\ce{HCl}$? I found this explanation for the comparison between $\ce{H_2O}$ and $\ce{HF}$ so I tried to use it here also, but I'm not even sure that the intermolecular forces among   $\ce{H_2S}$ and $\ce{HCl}$ molecules can be defined as "hydrogen bond" (which should occur only for  $\ce{F,O,N}$).

So how can the greater boiling point of  $\ce{H_2S}$ compare to $\ce{HCl}$ be explained?

Definition of the hydrogen bond

In 2011, a modern and more general IUPAC definition of what constitutes a hydrogen bond was recommended (Arunan et al., 2011). The definition is stated as such:

  The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a molecular fragment $\ce {X-H}$ in which $\ce {X}$ is more electronegative than $\ce {H}$, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation.

It appears that the definition of then hydrogen bond which includes interactions only with $\ce {F, O or N}$ seems too narrow and restrictive, which I personally feel is indeed the case.

Therefore, it is perfectly fine to say that $\ce {H_2S}$ makes hydrogen bonds with other $\ce {H_2S}$ molecules and that $\ce {HCl}$ makes hydrogen bonds with other $\ce {HCl}$ molecules.

Explaining the boiling points

Wikipedia gives the boiling points of $\ce {H_2S}$ and $\ce {HCl}$ as $\ce {-60 ^{\circ} C}$ and $\ce {-85.05 ^{\circ}C}$ respectively. When rationalising boiling point differences, the first consideration is always the strength of the intermolecular forces between the molecules in the liquid.

The Pauling electronegativities of $\ce {S}$ and $\ce {Cl}$ are $\ce {2.58}$ and $\ce {3.16}$ respectively. Evidently, the electron deficiency in the hydrogen atom bonded to $\ce {Cl}$ is more severe compared to that bonded to $\ce {S}$. This suggests that $\ce {HCl}$ molecules are capable of forming stronger hydrogen bonds. However, it is also important to consider the number of hydrogen bonds formed between molecules, as you have already noted in your question. Each $\ce {H_2S}$ can form 4 hydrogen bonds with other $\ce {H_2S}$ molecules, whereas each $\ce {HCl}$ molecule can only form 2 hydrogen bonds with other $\ce {HCl}$ molecules. Therefore, despite the fact that $\ce {HCl}$ molecules can form stronger hydrogen bonds between themselves, the intermolecular attractions between $\ce {H_2S}$ molecules still prevail as they are twice as numerous. Thus, $\ce {H_2S}$ has a higher boiling point than $\ce {HCl}$.

References

Arunan, E., Desiraju, G. R., Klein, R. A., Sadlej, J., Scheiner, S., Alkorta, I., . . . Nesbitt, D. J. (2011). IUPAC definition of the hydrogen bond. Terminology and nomenclature. Pure Applied Chemistry, 83(8), 1619-1636. doi:10.1351/PAC-REP-10-01-01

Electronegativity. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Electronegativity

Hydrogen chloride. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Hydrogen_chloride

Hydrogen sulfide. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Hydrogen_sulfide


Software that determines whether a molecule can exist and draw it from a formula? I have run calculations that predict atomic configurations.  As a simple example, in a system that contains H and O, I might get a list like:

Configurations for the O atom
O H
0 2
2 6

This would tell me that in my system, there's one O atom bonded to two H atoms and there's also a molecule made up of three O atoms bonded to each other and six H atoms.  My real systems are more complex.

I am not really a chemist and have no idea whether some of the molecules I have calculated actually exist and what they would look like if they did.  Does anyone know if there's software available that might help me with this; i.e., where I can enter a potential formula and the software draws the molecule and tells me whether it exists?

I have so far checked out ChemDraw Professional and Avogadro and it doesn't look like they do what I need.  Thank you for any tips.

I believe this website will be of use to you

http://molview.org/

It appears to take names, formulas, smiles etc. If the name/formula/smile ID pops up in the search bar, it will draw it for you. Its database seems quite large, I use it for some pretty big drug molecules.

It draws 2D and 3D images, and it also does single/double/triple bonds.


Controlling Current Density in Electrowinning Cell What are the parameters which govern the current density in an electrowinning cell? What equations relate Current Density to the applied potential difference?

Is it simply just Ohm’s Law J = σE? Where does redox at the electrodes come into play, if this is the case?

The system is complicated by many nonlinear processes.

As the topic is electrode focused, the primary controlled parameter should be the electrode surface current density $\sigma$, especially on the depositing electrode.

the total coltage applied on the electrolytic cell is devided on several potential differences:

$$U_\mathrm{applied} = U_{td}(\sigma) + U_\mathrm{kin}(\sigma) + U_\mathrm{diff}(\sigma) + U_\mathrm{cond}(\sigma)$$

where the particular terms are

electrode potential difference given by reaction thermodynamics
electrode potential difference given by reaction kinetics
electrode potential difference given by near electrode diffusion gradient.
free solution potential difference controlling ion migration.

For practical reasons, I advice using empiric approach :

either constant current power source, if applicable
either manual voltage control, using visual ampermeter feedback.

For further reading, I recommend:

For starters, Electrochemistry on hyperphysics
Wikipedia articles,  following internal and external links, also searching for mentioned topics.
for more advanced matters, textbooks for Physical chemistry.


Which is the more stable enol form? If I understand correctly, when we go from a keto to an enol form, we simply abstract an alpha proton and thereby create a double bond to a carbon-bearing the hydroxyl functional group.

When evaluating the stabilities of the two enol forms, it appears that the enol derivative of 2,4-pentadione is more stable because of intramolecular hydrogen bonding. Both have long-range conjugation, so that cannot be a factor. The only difference really is the hydrogen bonding ... and of course ring strain (a 5-membered ring with two double bonds can't be very stable).

Keto-enol equilibria reflect a delicate thermodynamic balance between the two forms.

The bonding differences between the keto and enol structures shown above are:

keto: C=O double bond, C-C single bond, C-H bond

enol: C=C double bond, C-O single bond, O-H bond

If we look up the bond energies for these bonds we find

keto: 745 + 347 + 413 = 1505 kJ/mol

enol: 614 + 358 + 467 = 1439 kJ/mol

This rough calculation tells us that, generally, the keto form will predominate.  However, the energy difference between the two forms is only 66 kJ/mol, a relatively small number; so it is likely that small differences can cause a dramatic shift in the relative concentrations of the two species.

Some general guidelines for predicting shifts in keto-enol equilibria

Aromaticity and conjugation - The keto-enol equilibrium for phenol (I) lies entirely on the enol side due to thermodynamic stabilization provided by aromaticity. Compound III would be expected to have a higher enol content than compound II due to the extended conjugation present in the enol of III

Substitution - Replacing hydrogen with alkyl groups on a double bond stabilizes the double bond. Therefore, we would expect compound V to have a higher enol content than compound IV

Dipolar repulsion - The carbonyl dipoles in butane-2,3-dione (VI) can reduce their electrostatic repulsion by adopting the geometry pictured with the carbonyl groups oriented away from one another.  Due to the constraints of the 5-membered ring, cyclopentane-1,2-dione (VII) cannot adopt a similar geometry.  In this case, the dipolar repulsion is lessened by increasing the enol content in the equilibrium to a point where the enol predominates.

Hydrogen Bonding - Hydrogen bonding can stabilize the enol form.  If the hydrogen bond is strong enough, and particularly if other factors also stabilize the enol form, the enol form can predominate.  Your molecule, 2,4-pentanedione (IX), is a good example.  There are two possible enol forms, VIII and X. X is by far the predominant enol as it has 1) extended conjugation and 2) substitution on the carbon-carbon double bond.  In addition, X can form a very stable hydrogen bond involving a 6-membered structure between the two oxygen atoms.

Solvent Effects - Especially in cases where hydrogen bonding is involved in stabilizing the enol, the solvent can have a dramatic effect. In benzene, where intramolecular hydrogen bonding predominates, the IX:X ratio is approximately 5:95. In water, where the intramolecular hydrogen bond is replaced by hydrogen bonding to the solvent, the ratio is roughly reversed.

Answer to your question

Cyclopent-2-enone can form two different enols, XI and XIII. Neither enol has any features that would be expected to provide significant enol stabilization.  There is no more conjugation than what is found in the ketone, there is no special hydrogen bonding, and no substitution of the carbon-carbon double bonds.  Indeed, the proton nmr of cyclopentenone is just as expected, no evidence for a significant amount of enol.  On the other hand, 2,4-pentanedione exists primarily as the enol form, at least in non-polar solvents, for the 3 reasons (conjugation, hydrogen bond, double bond substitution) discussed above.

As an aside, you mentioned that a 5-membered ring with two double bonds (a cyclopentadiene) might not be very stable; actually, they are quite stable and quite common, 2 double bonds in a 5-membered ring do not create a lot of strain.


How can elements have the same number of electrons, but different electron configurations? $\ce{Ti}$ and $\ce{Cr^2+}$ are isoelectronic, yet their electron configurations are different, with the former's being $\ce{[Ar] 4s^2 3d^2}$ and the latter's being $\ce{[Ar] 3d^4}$.
How can that be explained?

The reason is very complicated, but in short:
The energy gap between $\ce{3d}$ and $\ce{4s}$ is wider in $\ce{Cr^2+}$ than in $\ce{Ti}$ because $\ce{Cr^2+}$ has more protons.

When you promote an electron from $\ce{3d}$ to $\ce{4s}$, the electron gained the energy thanks to orbitals, but you also lost the energy thanks to repulsion (of electrons in the same orbital). Remember that electrons like to be in states of less energy, not more.

The energy gap between $\ce{3d}$ and $\ce{4s}$ is less in $\ce{Ti}$, so the electron, by promoting, lost more energy thanks to repulsion than it gained thanks to the orbital, so its configuration is $\ce{[Ar] 4s^2 3d^2}$.

In $\ce{Cr^2+}$, the enrgy gap widened, so the electron would gain more energy than its lost by promoting, making its configuration $\ce{[Ar] 3d^4}$.

See also chemguide.co.uk.


Is there a formula or way to determine whether an acid is a strong one or a weak one without conducting experiments in real? I know the common strong and weak acids but if a mineral acid like $\ce{H3BO3}$ springs up, I can't actually tell whether it is strong or weak.
Also can I justify in chemical terms why $\ce{H2SO4}$ is a strong acid but $\ce{H2SO3}$ a weak one?
Why is $\ce{HNO3}$ a strong acid whereas $\ce{HNO2}$ a weak one ?

http://catalog.flatworldknowledge.com/bookhub/2273?e=ball-ch12_s04

The above link shares a list of strong and weak acids but fails to state why one acid is weak while the other strong.

Is there a formula or way to determine whether an acid is a strong one or a weak one without conducting experiments in real?

Yes, there are ways to determine whether an acid is strong or weak without going into the lab yourself. That being said, someone must still go into the lab.

The most straightforward way would be calculating ΔG for an acid-base reaction. Remember, a negative delta G means that a reaction is spontaneous. Since strong acids have a Ka > 1, their ΔG values should be negative.


What is the basic concept of positive and negative charge of an atom? In ionic compounds, e.g. NaCl, chlorine 'steals' one electron from sodium. Because of this, sodium's proton number is increased by one, and the number of electrons in chlorine is increased by one. This makes sodium positively charged and chlorine negatively charged; opposite charges makes them attract to each other.

My question is that the nucleus of the sodium ion is still surrounded by electrons, and so does the nucleus of chlorine. Do those electrons repel, even though we say opposite charges attract? What do the electrons do while they attract?

Sodium's proton number doesn't change else it'd become a different element.

In basic terms, sodium donates an electron to chlorine's outer electron shell; there is no electron 'stealing'. As much as both ions' nuclei are surrounded by electrons, the charges of the ions are based on the ratio of protons to electrons. For example, in an atom, the number of protons and electrons are equal to give no overall charge.

However, in an ion it's different; the sodium ion has 11 protons but only 10 electrons, so has a relative charge of 1+, as there's one more positive charge (proton) than there are negative charges (electrons). On the flip side, the chloride ion has 17 protons and 18 electrons, so has a relative charge of 1-, as there's one more negative charge than there are positive charges.

This then causes the electrostatic attraction between the two ions, which we call ionic bonding.


Is calcium hydroxide or potassium hydroxide more efficient at removing carbon dioxide? In biology, we commonly use $\ce{KOH}$ in respirometers to remove carbon dioxide. A common test in chemistry to detect the presence of carbon dioxide is to bubble it through limewater, turning it milky. This has led me to wonder which is more efficient in removing carbon dioxide—$\ce{KOH}$ or $\ce{Ca(OH)_2}$—when both solutions are sitting in the atmosphere (i.e. they’re just passively absorbing $\ce{CO_2}$ and $\ce{CO_2}$ is not bubbled through either solution). An explanation would be much appreciated.

$\ce{KOH}$ (and $\ce{NaOH}$) absorb $\ce{CO2}$ more quickly and completely than $\ce{Ca(OH)2}$, but they are much more dangerous to work with: caustic to skin, deliquescent (forming a highly concentrated solution that can pour out of the apparatus), etc. For that reason, "lime water" is preferred for classroom use to demonstrate the presence of $\ce{CO2}$.


Does grape juice change to vinegar? If I add cooked grape juice to vinegar, does the juice undergo fermentation and also change to vinegar? Are there other conditions like exposure to air? How long would such a process roughly take?

The process that transforms sugars (e.g. the ones in grape juice) into ethyl alcohol and then this latter into acetic acid is a biological one, that is, performed by micro-organisms. The first step is usually performed by yeasts such as Saccharomyces Cerevisiae, whereas the second step is performed by bacteria of the Acetobacter genus. Thus, your grape juice won't change to vinegar if you mix both liquids. All you will get is a sweeter vinegar, or a sour grape juice.


Oxidation of Vitamin C How is ascorbic acid oxidized into dehydroascorbic acid? What is the reduction reaction in this case? Where are the 2 hydrogen atoms going?

Think of oxidation in a classical sense for a minute or so. Oxidation is the addition of oxygen to a molecule or it can also mean the removal of hydrogen. If you carefully look at the structure of ascorbic acid, you will see that the two -$\ce{OH}$ groups are now converted into carbonyl groups. Those two hydrogens can be lost as a proton, $\ce{H+}$.


How to calculate the latent heat when there is a boiling temperature range? I have a DME (dimethyl ether)- methanol - water mixture with the mol fraction composition: DME 0.4295, methanol 0.1353, water 0.4351.

This mixture has a boiling temperature range: dew point 141.6 °C and bubble point 69.95 °C, and the heat applied for the phase change (condensation) is a latent heat. How can I calculate it?

I know that for pure compounds, during the condensation process, there's no temperature change, so the latent heat is the result of:

Q = mass flow * latent enthalpy (constant value)

For a mixture that has a condensation temperature range, can I use the same equation? If this is not the case, how can I calculate the latent enthalpy for a mixture in function of the temperature?

Conceptually, it is possible to calculate the enthalpy of any mixture of these three compounds in the vapor phase and in the liquid phase.  So heat balances on pieces of equipment can thereby be properly carried out.  But, in the case of such a multi-component mixture, the concept of latent heat ceases to have realistic meaning and applicability in practice.