How can I remove the bitartrate anion from an organic bitartrate salt? I have a compound, dimethylaminoethanol (DMAE) bitartrate, which I originally purchased as a supplement and possible smart drug. I bought an absurd amount of it. It didn’t do anything for me as a supplement. I understand that DMAE, not the bitartrate salt, is commonly used in cosmetics, & I’d like to play with it for that purposes, maybe mixing it with glycerin, etc. Is there a simple way to strip off the bitartrate anion and have it precipitate or sublimate or something, leaving the DMAE behind? Or otherwise separating the DMAE out from the salt?

I have a strong preference for any other reactants and reaction products being non-toxic

I see on Wikipedia: "Dimethylethanolamine (DMAE or DMEA) is an organic compound with the formula (CH3)2NCH2CH2OH. It is bifunctional, containing both a tertiary amine and primary alcohol functional groups. It is a colorless viscous liquid. It is used in skin care products."

Now the fact that it is used in skin care products does not make it safe: the SDS  (https://www.fishersci.com/store/msds?partNumber=AC116180010&productDescription=N%2CN-DIMETHYLETHANOLAMINE+1LT&vendorId=VN00032119&countryCode=US&language=en) shows these pictograms:

Uh oh! Oh, well.

Well, let's forget that for a minute. DMAE is a liquid, boiling at 134C. Bitartrate anion is very insoluble if attached to potassium (potassium bitartrate is cream of tartar). If you added a stoichiometric amount of potassium bicarbonate (KHCO3, quite safe) to an aqueous solution of DMAE, you would precipitate almost all of the bitartrate, leaving the DMAE in water. Filter off the ppt and distill off the water and then distill and collect the DMAE. It will probably be a little wet and you will hate the overwhelming fishy smell.

Oh, and use a respirator (from the SDS: "Follow the OSHA respirator regulations found in 29 CFR 1910.134 or European Standard EN 149. Use a NIOSH/MSHA or European Standard EN 149 approved respirator if exposure limits are exceeded or if irritation or other symptoms are experienced.")

Good luck. Actually, I think you will have better luck if you do something else.


What is actually a redox reaction? I need to know what a redox reaction is. How could a redox reaction possibly be identified? Also, I have a confusion whether,

$$\ce{H -> H+  +  e-}$$
$$\ce{e- + Cl -> Cl-}$$
$$\ce{H + Cl -> HCl}$$

is a redox reaction.
Furthermore, could anyone please tell me if ionization and reduction and gaining electron and oxidation are referring to the same phenomena?

A redox reaction is any reaction in which oxidation and reduction are occurring. The word redox is a portmanteau of REDuction-OXidation. So, if we can correctly identify reduction and oxidation, we can identify redox reactants. In practice, this can be a little challenging, so I will focus first on the second part of your questions (definitions) and then go back to the reaction.

  Ionization is any process that produces ions.

The conversion of a hydrogen atom to a hydrogen ion by loss of an electron is an ionization.
$$\ce{H -> H+ + e-}$$
So to is the reaction of hydrochloric acid and water, which produces hydronium ions and chloride ions.
$$\ce{HCl + H2O -> Cl- + H3O+}$$

  Oxidation is a loss of electrons.

The ionization of hydrogen atoms to hydrogen ions is an oxidation.
$$\ce{H -> H+ + e-}$$

So to is the conversion of carbon monoxide to carbon dioxide:

$$\ce{2CO + O2 -> 2CO2}$$

Note that this type of reaction is what gave us the word oxidation - compounds are reacting with oxygen and atoms end up with more oxygen atoms bound to them.

  Reduction is a gain of electrons.

The ionization of chlorine atoms to chloride ions is a reduction.

$$\ce{Cl + e- -> Cl-}$$

So too is the conversions of acetylene to ethane:

$$\ce{C2H2 + 2H2 -> C2H6}$$

Not all oxidations and reductions are ionizations. There is a simple mnemonic in English for remembering oxidation and reduction: OIL RIG (Oxidation Is Loss Reduction Is Gain).

  Oxidation and reduction happen together.

In principle we can write an equation that shows only oxidation:

$$\ce{H -> H+ + e-}$$

but oxidation cannot happen on its own. That electron that was lost needs to be picked up by something else. This reaction that shows only oxidation is called a "half-reaction" and needs to be coupled with a reduction half-reaction to show a valid chemical reaction:

$$\ce{H -> H+ + e-}$$
$$\ce{Cl + e- -> Cl-}$$
$$\ce{H + Cl ->HCl}$$

In the other examples I gave, both processes are occurring. In the oxidation of carbon monoxide, the oxygen molecules are being reduced.

The surest way to identify oxidation and reduction (and thus redox) is to compare the oxidation states of the reactants and the products. Determining oxidation states/numbers is more complex that I want to put here, so start with the Wikipedia article.

For the carbon monoxide example:

                  reactants    products
C                 +2           +4
O (bonded to C)   -2           -2
O2                0            -2

The carbon atoms are being oxidized (oxidation number increased) and the oxygen atoms in $\ce{O2}$ are being reduced (oxidation number decreases). The oxygen atom already attached to carbon did not change its oxidation number.


MO Theory: Why do hydrogen and lithium bond, but hydrogen and helium don't? I've been cracking my head trying to make sense of Molecular Orbital Theory, but all sources I find are simplistic {If anyone has an online course or a book to recommend, I'd really appreciate}. Therefore, my questions are quite simplistic:

1 - Why should atoms in the first row {Hydrogen} interact with atoms in other rows  {Lithium, Beryllium, Boron, etc}? I mean, as a way to answer why two Hydrogen atoms form a Molecular Orbital in the first place, I have the postulate that it is due to Pauli Exclusion Principle: as the 1s Orbitals - and, by extension, its electron - overlap, a new orbital is formed to avoid* having two electrons with the same quantum numbers. But, when Lithium approaches Hydrogen, there is only a significant overlap between 1s and 2s Orbitals. In this sense, there is no threat* to Pauli's principle, given that the approaching electrons do NOT have the same quantum numbers. Which yields the question: how do Lithium and Hydrogen bond?

. * I understand that the universe doesn't care about what we, smart monkeys, think of it. It doesn't see things as "threats". Still, interpreting things this way makes everything easier.

2 - Why won't Helium bond with Hydrogen? The approach of He and H should form bonding and anti-bonding Orbitals, which will be filled with three electrons. This should place two electrons in the Sigma Bond, but only one in the Sigma* Bond. Therefore, the bond order is 0.5! So why HeH isn't observed?
Another way to phrase it: If He2+ can exist as a molecule, why can't HeH?

Thank you very much for your time.
Sorry if I messed something up.

You've asked seemingly simple questions that get into quirky considerations. A molecule of HeH would very unstable:

$\ce{2HeH -> 2He + H2}$

As Oscar points out HeH probably can be made. In all of deep space where molecules could go days before colliding with another particle there must be at least one molecule of HeH. So it can be "made" and it does "exist." But how would you detect it?

Typically when we are talking about "making" something we are talking about making it in a earth bound laboratory. You make enough to at least observe it somehow. For instance a lot of radioisotopes have half-lives less than a second which really doesn't give an experimenter time to do any really "chemistry" with them. So in any practical sense the HeH molecule doesn't exist.


Best Settings for Electrolysis of water I want to electrolyze water as a personal project, but I have found conflicting power recommendations on google. Some sources recommended a higher voltage, claiming that higher amps will just destroy the electrodes, while others insist a higher amperage is required, and extra voltage will be wasted heating up the water. Unable to find one absolute result, I'm turning here- assuming not a crazy amount of amps or volts, what is a good ratio to get the best results? (an explanation of the answer would be appreciated.)

Ideally, it takes 1.23 volts to split water. Less voltage than this will give essentially no current, no electrolysis. If you want more H2 + O2, you raise the potential. (Of course, you use an electrolyte like H2SO4 or NaOH with electrodes that will not corrode. Platinum is OK, also gold. Oh, yeah, also stainless steel for ordinary people.) If the surface becomes covered with bubbles of gas, the effective area of the electrode is reduced, so the current will drop. So you raise the voltage, and bubbles come pouring off. Then the electrolysis uses more energy (volts x amps). 1 amp for 96,500 seconds (27 hours) will give you 1 gram of hydrogen and 8 grams of oxygen. The product is dependent on the current, and the current is dependent on how high you raise the voltage.

It's not as slow as watching grass grow, but even at furious bubbling, it's surprising how slowly the volumes build up.


Conversion of Kc into Kw In deriving the expression for $K_\mathrm w$, we often start with the equilibrium equation of water
$$\ce{H2O <=> H+ + OH-}$$

The $ \rm K_c$ expression is quoted as $$K_\mathrm c = \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} $$ followed on by multiplying the $[\ce{H2O}]$ by $K_\mathrm c$.

The problem I face is I have been taught that the pure liquids aren't included in the expression of $K_\mathrm c$ since they have a concentration of 1. So why are we including in the above example?

"The problem I face is I have been taught that the pure liquids aren't included in the expression of $\mathrm{K}_{c}$ since they have a concentration of 1."

Liquids have a constant concentration (really activity), so in general, it is not helpful to include them in the equilibrium expression. However, that does not mean that the value is 1. Pure water has a concentration of roughly $55.5\ \mathrm{M}$. You can absorb this concentration into the equilibrium constant, since it is a constant. Therefore,

$$\mathrm{K}_{w} = \ce{[H+][OH-]} = K_{\mathrm{c}}\ce{[H2O]}$$

EDIT:

Just for the record, note that the autoionization of water is generally written:

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

since we don't have naked protons in water.

This changes the equilibrium constant slightly:

$$\mathrm{K} = \frac{\ce{[H3O+][OH-]}}{\ce{[H2O]}^{2}}$$

but $\mathrm{K}_{w}$ is still the same.


Why are silicon analogs of alkenes or alkynes so difficult to make? We know that there are silicon analogs of alkanes, but according to Wikipedia, silicon analogs of alkenes or alkynes are virtually unknown. How come?

The typical Si–Si single bond length in a silane is around 2.33 Å.  This is much longer than a typical C–C single bond (~1.53 Å) and helps explain why silicon-silicon single bonds are so much weaker than carbon-carbon single bonds (bond dissociation energy: ~53 kcal/mol for silicon-silicon vs. ~83 kcal/mol for carbon-carbon single bonds).  When we try to form a silicon-silicon double or triple bond, the large separation between the two silicon atoms results in even less effective p-orbital overlap and therefore even weaker π-bonds.  The result is that while disilenes and disilynes are known (see the linked SE Chem posts), they are extremely reactive.  Only when bulky substituents are placed around the double or triple bonds (to provide some "steric protection" against further reaction) can the molecules be isolated and characterized.


What is the correct chemical formula for monosodium glutamate? Recently, I am interested with home materials and I discover that the chemical symbol for MSG or monosodium glutamate is $\ce{C5H8NO4Na}$.
My problem is shouldn't it be $\ce{NaC5H8NO4}$ as we always write the metal first before the non-metal?

It should, but in organic molecules salts, often the counterion is written latter, to indicate its position on the molecule acting like so, for example, in sodium acetate, $\ce{CH3CO2Na}$, this would be something like a "semicondensated" formula.


How do the differences in the number of protons result in such great differences in elemental properties? I understand(I think) the mass and density aspect, i.e. the more protons you have, the more the element weighs, also the denser the atom is. What about everything else(color, for example)? Elements are not just masses and densities.

Unfortunately, I have to disagree with trb456's answer.  It is both the protons and the electrons that make up chemistry. (Neutrons are considered massive "spectators" though they are what keep protons (all + charged) together)

Bonds are formed due to the attraction of electrons and protons. (cf. Slater, Bader, et al.) (The precise shape and nature of the electron density is additionally determined by constraints such as Pauli Exclusion and electron-electron repulsion.)

So if you think about it, there is a very complex interplay upon mixing electrons up with protons. More positive charge in the nucleus tends to keep electrons drawn tight, but in some ways that is counter-balanced by electron-electron repulsion--and the fact that electrons are attacted to other nuclei as well.

So although we ascribe color ("electronic transitions") and reactivity to electrons, this is only part of the story. The protons are also there, and they provide some governance (via Coulombic interaction) over what electrons are permitted to do.


What is defined by part by weight in the definition of equivalent weight? According to my book, equivalent weight is defined as the ‘number of parts by weight’ of an element that interacts with 1.008 parts by weight of hydrogen. It is dimensionless while gram equivalent weight has gram as units. But in https://en.wikipedia.org/wiki/Equivalent_weight, equivalent weight and gram equivalent weight are same. I’ve two questions: Which one is correct, my book or Wikipedia? And what is ‘number of parts by weight’?

The term "parts by weight" is how you would specify in a recipe the relative proportions of the ingredients. So if the recipe called for one part by weight of flour and two parts by weight of sugar, you could use 1 ounce of flour and two ounces of sugar (to make a small batch) or 1/2 pound of flour and 1 pound of sugar to make a large batch.

So the "recipe" for water calls for exactly 1.008 parts by weight of hydrogen and 8.000 parts by weight of oxygen. If you use too much of either ingredient, there will be an excess of it after the reaction is run.

"Parts by volume" would apply if you were combining liquids and measuring them by volume rather than by weight.


Determining chirality after considering conformational interconversion in a cyclohexane Basically I want to know what does the question mean when it says "taking into consideration conformational interconversions."

I assume it means, when I draw the mirror image of the given compound, the mirror image can be same as the compound itself (the case with option (a) and option (c) ) or it (the mirror image) can be same as confomer of the compound (the case for option (a) and option (b) ), for the compound to be achiral.

Note: option (a) shows both the features because option (a) is same as it's confomer and it is same as it's mirror image as well.

Is my assumption correct?

Your analysis is correct. The stipulation "taking into consideration conformational interconversions" means that conformational interconversion is fast. That's important because compounds that we consider achiral can have chiral conformations. I used the example of butane in a previous question.

In this problem, three compounds are given that have multiple stereocenters, compounds b, c, and d. A common way to decide whether a compound with multiple stereocenters is chiral is to look for planes or points of symmetry within the compound. Compound c has a plane of symmetry, so it is clearly achiral. Neither of compounds b or d has a plane of symmetry in the drawn conformations, which means that those conformations are chiral. However, as you suggest, chair flip of b gives the mirror image of the original drawing, which means the original and mirror image are equivalent, assuming that the two conformations can interconvert.

There are many cases where conformations cannot interconvert, and that leads to very interesting compounds that don't even require a stereocenter to be chiral (again see earlier question).


Electron has volume or not? Sometimes we say that electron has volume, as in orbital we can define it's volume (at least for some fraction, as orbital itself has no limit, it is spread all over the space expect nodes).

On other hand in general chemistry books often spin quantum no. is referred as electron spinning on its axis. But this is absurd, because this magnetic angular momentum is an intrinsic property of the electron. Also that an electron is an elementary particle, it is 1-D (no volume), so it's not possible for an electron to spin on it's own axis.

So what's the matter, do electron has volume or not?

What I think that it's a particle, but we can't describe it's position what we can do is probabilistic approach, we can only describe some region where we can find it more likely or less likely. We defined it's volume, otherwise electron themselves has no volume. Even it is more weird that electron is everywhere in the orbital ;D

Can somebody explain it?

If volume is understood classically, i.e. as the quantity of three-dimensional space enclosed by a closed surface, then quantum systems in general do not have a certain volume. One could, in principle, redefine the notion of volume for a quantum system to be, say, an area chosen so that there is a certain probability (for example, 90%) of finding the system within it, but that would be a completely different quantity, of course. It should also be noted that in quantum theory statements about a system having or possessing certain values of physical quantities has a lot of issues, even when the quantities are well defined. For instance, does a quantum system have (possess) certain values for physical quantities before the corresponding measurements are made?

  On other hand In general Chemistry books often spin quantum no. is
  referred as electron spinning on its axis.

You already know that such picture is an absurd, and you are perfectly right here: electrons are not spinning. So there is no any contradiction there: they don't have a classical volume and they are not spinning.


How to move forward in balancing this equation? I want to get the molecular formula for this reaction:

$$\ce{{$trans$-stilbene} + {$m$-chloroperoxybenzoic acid} -> {$trans$-1,2-diphenyloxirane}}$$

and if you do you get this:

$$\ce{C14H12 + C7H5ClO3 -> C14H12O}$$

but what do I do about the chlorine not being present in the products, is it released as $\ce{Cl2}$?

The reason you cannot balance the reaction is because you did not consider everything that happens. In the Prilezhaev epoxidation, mCPBA is a stoichiometric reactant but only one single atom of mCPBA is transferred to the product. Therefore, these compounds are often written on the reaction arrow rather than as reactants and products:

$$\ce{C14H12 ->[C7H15ClO3] C14H12O}\tag{1}$$

However, that still does not tell you what exactly happened. If you take a close look at the mechanism, you will see that mCPBA remains essentially unchanged except that it loses its peroxy function; the second product is mCBA or meta-chlorobenzoic acid. Thus, a full reaction equation would be:

$$\ce{C14H12 + C7H15ClO3 -> C14H12O + C7H15ClO2}\tag{2}$$

Notice how this balances automatically.


Is the thermodynamic stability against reduction/oxidation of an electrolyte dependent on the electrode material? Consider an electrochemical cell with an inert Pt working electrode (WE) and some electrolyte that irreversibly decomposes upon oxidation. A minimum potential $E_{min}$ (vs. some reference) needs to be applied to the WE to allow oxidation to occur (but most likely a higher value is required to actually measure a current). I would call this potential the thermodynamic stability potential limit (SPL).

Does this minimum potential change when exchanging the WE material with another inert material, e.g. glassy carbon? I think it should not, as all that matters is the electrochemical potential/fermi level in the inert electrode.

Could some type of specific adsorption of electrolyte components change (lower) the thermodynamic SPL? To me this sounds like catalysis, which just influences kinetics, not thermodynamics.

What about adding some other substance? Could its reaction products cause degradation of the electrolyte components in some way?

Technically you are correct. The thermodynamic potential of electrolyte decomposition should not be affected by the electrode composition.

  which just influences kinetics, not thermodynamics.

The word "just" here is doing way more work than you're giving it credit for. There are situation where the kinetics of electrolyte decomposition are so slow that there is effectively no reaction. The stability window of electrolytes commonly used in Lithium-ion batteries, for example, is smaller than the potentials needed to charge the materials. However, once the reaction begins, initial decomposition forms an inter-phasic layer on the anode known as the solide-electrolyte-interphase (SEI). The kinetics of decomposition on this SEI are slow, which allows the battery to be operated at high potentials[1]. In fact, damage to the SEI has been linked to capacity loss in Li-Ion batteries over extended cycling[2].


Is lead radioactive? Is lead (Pb) radioactive? I have heard about it being radioactive but have also heard it referred to as the heaviest stable element. Radioactivity refers to release of alpha, beta, gamma rays.

Important naturally-occurring radioactive lead nuclides are:

$\ce{^{214}Pb}$ $\left(t_{1/2}=26.8\ \mathrm{min}\right)$ from $\ce{^{238}U}$

$\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ from $\ce{^{238}U}$

$\ce{^{211}Pb}$ $\left(t_{1/2}=36.1\ \mathrm{min}\right)$ from $\ce{^{235}U}$

$\ce{^{212}Pb}$ $\left(t_{1/2}=10.64\ \mathrm{h}\right)$ from $\ce{^{232}Th}$

On a geological time scale, their half-lives are very short. Nevertheless, these lead nuclides are members of the $\ce{^{238}U}$, $\ce{^{235}U}$, and $\ce{^{232}Th}$ decay series. Thus, they are constantly reproduced by decay of their respective mother nuclides. Therefore, all environmental samples (including soil, water, air, plants, and animals) naturally contain significant amounts of radioactive lead nuclides. However, in various materials (e.g. residues from mining or oil and gas production, fertilizers, building materials), the concentration of natural lead nuclides can be strongly enhanced by technology.

When lead is freshly refined (chemically purified), most of the mother nuclides are removed and the short-lived lead nuclides quickly decay. Nevertheless, the lead still contains significant amounts of $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$. The presence of $\ce{^{210}Pb}$ is a source of background mainly through the lead X-rays and bremsstrahlung caused by high-energy beta radiation emitted by its decay product $\ce{^{210}Bi}$. This background is a real problem when lead is used as a shielding material for low-background detectors. Therefore, some shields are made from selected lead with certified low $\ce{^{210}Pb}$ content. Preferably, very old lead is used. Since $\ce{^{210}Pb}$ decays with a half-life of 22.3 years, samples of lead that are many decades old are relatively free of this activity. In one extreme example, 2000-year-old lead was salvaged from a sunken Roman ship.


Common Ion Effect - Ionic Equilibrium Question

In which of the aqueous solutions of the following, dissociation of $\ce{NH4OH}$ will be minimum?
A) $\ce{NaOH}$
B) $\ce{H2O}$
C) $\ce{NH4Cl}$
D) $\ce{NaCl}$

My Thoughts
My book says that the answer is option C '$\ce{NH4Cl}$' giving the reason as common ion effect.
But I think that option A '$\ce{NaOH}$' also has a common ion as $\ce{OH-}$.
What should be the right answer and how do we compare which will cause more suppression by common ion effect?
I get why $\ce{H2O}$ will not suppress dissociation of $\ce{NH4OH}$ that much since its equilibrium constant is very low (of the order of $10^{-14}$). But then why not $\ce{NaOH}$ (which is a very strong base and thus will almost completely dissociate into constituent ions)? Which one out of $\ce{NaOH}$ and $\ce{NH4Cl}$ is a more stronger electrolyte?
Is it because $\ce{NH4Cl}$ will form a buffer with $\ce{NH4OH}$?
I am feeling really confused about this. Any help will be highly appreciated!
Final Question
I now understand that both $\ce{NaOH}$ and $\ce{NH4Cl}$ will cause decrease in dissociation of $\ce{NH4OH}$. Hence my final question is this: Which will cause more decrease in dissociation and why?

Let's consider an aqueous solution, the concentration of which is $\pu{1 M}$ in $\ce{NH3}$ and $\pu{1 M}$ in $\ce{NaOH}$. Thus, following equilibrium would be taken place:
$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NaOH (aq) -> Na+ (aq) + OH- (aq)}\tag2$$
The $\mathrm{p}K_\mathrm{b}$ of equilibrium $(1)$ is 4.75, thus $K_\mathrm{b} = \pu{1.78E{-5}}$. In pure ammonia solution, from equation $(1)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}\tag3$$
If ionized amount at equilibrium is $\alpha$, then
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\alpha \times \alpha}{1-\alpha} = \alpha^2 \ \Rightarrow \ \therefore \ \alpha = \sqrt{\pu{1.78E{-5}}} = \pu{4.22E{-3}}$$
Assumptions: $\alpha \lt\lt 1$, and thus $1-\alpha \approx 1$, and $\alpha  \gt \gt \pu{1.00E{-7}}$ and autoionization of water can be ignored. At the end, since $[\ce{NH4+}] = [\ce{OH-}] = \alpha = \pu{4.22E{-3}}$, both of these assoumtions are correct.
Now consider, if you have $\ce{NaOH}$ in your solution. Since it is a strong base, it completely dissociate according to equation $(2)$. Thus, there is a common ion in this solution: $[\ce{OH-}] = \pu{1 M}$. Hence, from the equation $(3)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\beta \times (1+\beta)}{1-\beta} = \beta \ \Rightarrow \ \therefore \ \beta = \pu{1.78E{-5}} $$
Hence ($\alpha \gt \beta$), the ionization amount of $\ce{NH3}$ in presence of the common ion $\ce{OH-}$ is less than that of the solution when no common ions are present.
In similar way, you can prove the ionization amount of $\ce{NH3}$ is larger in the presence of the common ion $\ce{NH4+}$ Using following equilibria:
$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)}\tag4$$
$$\ce{NH4+ (aq) + H2O <=> H3O+ (aq) + NH3 (aq)}\tag5$$


Why does the "gap" between atomic radii get smaller down the periodic table? The question asks: why is the difference in atomic radii between $\ce{K}$ and $\ce{Na}$ larger than the difference in atomic radii between $\ce{Cs}$ and $\ce{Rb}$? I do have some kind of notion that the gaps between shells decrease as you go down the periodic table, but why would this be the case?

The first thing to remember is that the "shell" model is an over-simplification of quantum chemistry. A better picture is that the electrons represent "clouds" (i.e., electron density).

Your question basically boils down to "why don't 5th and 6th row atoms get much larger?"

Quantum mechanics can give a relatively easy formula for the "average" radius of a given hydrogen orbital

$$\langle r \rangle = a_0 n^2 \left( 1 + \frac{1}{2} \left[ 1- \frac{l(l+1)}{n^2} \right] \right)$$

Here, $a_0$ is the Bohr radius, the expected radius of the hydrogen atom, and $n$ is the principal quantum number and $l$ is the angular quantum number (e.g., l=0 for "s" orbitals and n=5 and 6 for Rb and Cs).

It's pretty obvious that as n gets large for s orbitals (l=0), that formula approximates:

$$\langle r \rangle \approx \frac{3 a_0}{2} n^2$$

(The left side becomes smaller and smaller, so the coefficient gets closer and closer to 3/2.)

So for a hydrogen atom (one electron), the 5s and 6s orbitals in fact get very large.

Your question is about atoms. So we have to consider many electrons. Basically, this is another representation of the lanthanide contraction caused by relativistic effects.

Indeed, in that last link, there's a formula for the atomic radius after relativistic corrections.

For atomic numbers 11 and 19, the relativistic correction is 0.99 (i.e., no real change to the radius). For atomic numbers 37 and 55, it's 0.963 and 0.916, respectively. So ~9% of the "ideal" radius is lost with $\ce{Cs}$.


How to predict if hydrogen atoms could be placed in the interstials of a lattice Palladium hydride is an example of a hydride which contains H atoms in palladium's crystal lattice (which is fcc).

I would like to know if it's possible to predict if another hydride, $\ce{NbH_{x}}$ could exhibit the same behavior.

I know that in an ionic solid, the fraction $$\frac{r_{+}}{r_{-}}$$
Can be used to predict the geometry around a cation.
Can I use this expression for niobium hydride? In the case of niobium, the crystal structure is bcc, so it contains tetrahedral and octahedral interstial sites.

Your intuition is correct; $\ce{NbH_{x}}$ retains the bcc Nb lattice, with the hydrides in the tetrahedral interstital sites, up to x = 2, when the compound switches to a fcc Nb lattice (the Fluorite structure). This is pretty typical behaviour for a transition metal hydride.

I don't think radius ratio rules are going to be useful in predicting the behaviour of these systems; there is a lot of covalency going on (hydride-hydride interactions and Nb-Nb interactions), and the metal lattices will end up being distorted by the correlated interstital occupancies of the hydrides. Hydride has a radius of around 130 pm while Niobium has a radius of around 150 pm so you wouldn't predict the observed behaviour based on radius ratios.

More information on this particular system can be found in Hauck (1976), Ordering of Hydrogen in Niobium Hydride phases.


Is the carbanion formed in benzyne mechanism non-planar? I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

I have given one of these molecules a little spin on the computer. With whatever I tried, I was unable to find a non-planar solution. I really tried my best. Below is an animation (sorry for the poor quality) of an attempt which uses a chair like starting geometry. If there was any non-planar geometry lower in energy, this would have found it.

The level of theory is DF-BP86/def2-TZVPP and the program used is Gaussian 09 rev. D.01.
There is really not much more I can add to the discussion.

  I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

To be absolutely, brutally honest here: I have no clue why your teacher had this idea. If this wasn't a "naked" carbon, we could talk about it, but it is essentially a deprotonated phenyl ring, and as such, you are absolutely right in assuming the lone pair being in phase with the ring. The π system is hardly different from benzene. To fill this post a little bit more, have those fancy π-orbitals.

(I could not find the completely anti-bonding pi-orbital.)

And for good measure, here is the HOMO, which is the lone pair at the carbon:


How to calculate molar mass using the ideal gas law? It is found that 250 ml of a diatomic gas at standard temperature and pressure (STP) has a mass of 1.78 g. What is the diatomic gas?

How can this question be solved? I know that I am supposed to use the ideal gas law

$$pV = nRT,$$

but am not sure how to apply it to this question.

You can first use the ideal gas law to calculate $n$:
\begin{align}
pV &= nRT\\
(\pu{101325 Pa})(\pu{0.250 l}) &= n(\pu{8.314 J K-1 mol-1})(\pu{273 K})\\
\end{align}

Solving for "$n$" gives $n = \pu{0.0111577 mol}$.

Now we use the connection between mass $m$, amount of substance $n$, and molar mass mass $M$:

\begin{align}
  M &=\frac{m}{n}\\
  n &=\frac{m}{M}\\
  \pu{0.0111577 mol} &= \frac{\pu{1.78 g}}{M},
\end{align}

which gives a molar mass of $M = \pu{159.53 g mol-1}$.

Looking at the Periodic Table, we can identify the diatomic gas as being bromine gas ($\ce{Br2}$), which has a molar mass of $\pu{159.8 g mol-1}$.


Glycine in magnesium glycinate To be very specific I'm looking at a supplement that contains 2,000 mg of magnesium glycinate. In that serving it sates there is 2oo mg of elemental magnesium. How much glycine is there?

If there are no other ingredients like citric acid- 1800mg. It is probably magnesium BISglycinate. For each Mg ion, there are two glycine molecules.

After a calculation, if it's MagBisglycinate, for 200mg Mg you should have around 1235.5mg glycine. It probably means there are other ingredients or simply glycine that left unreacted. If it's simple Magnesium glycinate than you should have 617.7mg glycine for 200mg Mg.

Correct me if I'm wrong.


Why they have different flame colour? At school, I learnt about the metal and their flame colour. Here's my observations.

$\ce{K^+}$- Lavender

$\ce{Ba^2+}$- Yellow-Green

$\ce{Ca^2+}$- Yellow-Red

$\ce{Li^+}$- Flame Red

$\ce{Na^+}$- Bright Orange

$\ce{Cu^2+}$- Blue-Green

So, my question is why do the each metal has its own flame colour?

First, remember that electrons cannot exist between principle energy levels (distances). In a flame, electrons from lower principle levels (n=1,2,3... etc) will be knocked up to higher principle levels from the absorbed energy. Electrons do not like to stay in this high energy state and will travel back down to their ground state, and in the process release that stored energy in the form of radiation (often visible light). The frequency of the radiation (color of the light) depends upon how big the fall back to ground state was. Different metals have different attainable energy levels, and so will display different colors when burned.

Some elements only have a few possible emission colors, while other elements are a mixture of dozens of colors ! This principle helps us tell what the composition of stars are, as well as allows us to test for metal contents in food products or other applications.


How to get Jmol to automatically determine and display bond order? Some molecule file formats do not detail bond order and when I open these formats in Jmol implied (by the valency, of course) double bonds are not shown as double bonds but rather as single bonds. Is there a way to correct this behaviour? I tried to answer this question myself with a "calculate bond order jmol" DuckDuckGo search. It returned:

http://wiki.jmol.org/index.php/Jmol_as_editor
http://www.wikihow.com/Calculate-Bond-Order-in-Chemistry
http://wiki.jmol.org/index.php/Support_for_bond_orders

as the top three results. Not a single one mentioned something relevant. 3 seemed most relevant but it seems to talk about how to specify bond orders manually, not getting Jmol to automatically calculate them.

Bond lengths are shown in this screenshot:

The file opened is http://crystallography.net/cod/2236556.cif which is tianeptine hydrochloride.

The current definition of the *.cif file does not care about bond order, rather than indicating connectivity of atoms within a molecule.  This is why representing organometallic complexes (like ferrocenes) drawn in sight of what is important for a crystallographer may differ from the one drawn by an organo(metallic) chemist. Among other data, *.cif include the position of the (hopefully correctly assigned) atoms, but nothing about hybridization ($sp^3$, $sp^2$, ...) -- which is one reason why you offer a drawing of the molecular structure of your product to your crystallographer when it comes to characterization by single crystal diffraction or powder diffraction analysis.  In the style of a the ball-and-stick representation, the above drawing is correct.

It is however possible to request Jmol to discriminate between single, double, triple bonds.  One approach to solve such a question is based on tables which relate typical bond lengths to single, double, triple bonds.  To document this, I used the data set indicated by you, as well as Jmol (version 14.13.1, 2017-04-09) under Xubuntu 16.04 LTS with openjdk version "1.8.0_131".  As far as I know, one has to proceed manually, individually for each bond in question.

As a preparation, be sure that Jmol may calculate the bonds, the corresponding menu is accessible via edit -> preferences:

If the option previously was disabled, restart Jmol now.  Subsequently, open your *.cif file and activate the model kit module

which will enable a little drop-down menu on the left hand side, marked by the ribbon in magenta. For now you do not need to access the options there, but hover the mouse over the first, subsequently over the second atom along the bond you are interested.  You will see that first the first atom, then the second atom is encircled.  If you now return hovering from the second atom in question to a position between the two atoms in question (and still on top of the bond in question), and left-click, the bond order may be recalculated, and its display will change if deemed suitable -- as shown in the example of the oxygen atoms of the sulfonamide.

This takes some practice, and frankly, the operator needs to stay vigilant.  In the case of aromatic cycles, the model kit module may correctly recognize the $sp^2$ character of the carbon atoms, but on the expense of hydrogen atoms attached to it.  Hence a result like the following may be expected, where I intentionally halted to "re-assign" the bond order on the last two carbon atoms of this cycle:

Be aware that if you left-click on one of the atoms in question, you may alter its designated element.  In the following, I did so while tapping on the carbonyl oxygen of the acid residue:

In such an instance, you should access the pull down menu of the model kit, and re-assign the atom type. Aberrant hydrogen atoms will be removed by this automatically, too:


Do swimming pools deplete the ozone? I believe that most people know that CFCs deplete the ozone... as does free chlorine. I have read that pool chlorine is not a factor because, it is "rapidly converted to forms that dissolve in water and therefore are removed from the lower atmosphere." and/or "ends up as hydrochloric acid or sodium chloride."

So, the premise is that bleach simply won't react with carbon in the soil to produce ozone depleting chemicals (ODCs)? Or is the mind boggling complexity of chlorine in the environment sufficient to say that it must be gone?

I know that bleach (and/or chlorine gas) reacts with VOCs (like acetone, to make chloroform). Bleach will also release chlorine gas (Cl2) in conjunction with mild acids or even just carbon dioxide. And most importantly, chloramines (which evaporate from the pool) will react with ozone or UVs to produce more bleach.

So I'm wondering, is bleach (especially from swimming pools) really not that bad?

I've heard that question before (which is why I asked). So, to head off any potential urban legends...

There is evidence that the regulations against CFC's are making a difference in the loss of ozone. There's a nice little synopsis of the situation here:  The Ozone Hole


Can you produce activated carbon from toast? On the Skeptics stack exchange we have a question about burned toast being used as a treatment for poison, the idea being that burned toast acts as activated carbon.

We are skeptical due to the fact that gas activation requires heating to high temperature, but don't have the chemistry to rule out toast as a source of activated carbon.

My instinct tells me that:

Toast is not good source of carbon
Activation is not easily done in your average kitchen.

Can someone answer those 2 questions?

Is toast a good source of carbon
Can activated carbon be produced in your average kitchen without training or any special supplies?

Is toast a good source of carbon is quite a broad question without considering the type of toast we're referring to. Even if do stick to a particular toast, what makes a good carbon source sounds a little too vague.

As a side note, recent researches have led to the production of a low cost multi-functional carbon foam from bread. [Source]

As a rough estimate, let's consider we're talking about white bread. 100 grams of white bread consists of 49 grams (i.e. approximately 16%) of carbohydrates. [Source] For the sake of simplicity, let's assume all of the carbohydates to be monosaccharides, specifically glucose. (This definitely isn't the case). The molecular formula of glucose is $\ce{C6H12O6}$. If we calculate the percentage-by-mass composition of carbon in glucose, we get the value as 40%. 40% of 49 grams is 19.6 gram, or in other words, bread consists of (very roughly) 19.6% carbon.

I'm unsure of whether or not that makes it a good source. Reading this article tells me that a rough estimate for the carbon content in wood (it is about some conifers but nevertheless good for rough estimates) is about 45–50%.

The production of activated carbon is undertaken generally in two ways: [Source]

Physical activation: It requires an inert atmosphere of nitrogen or argon along with very high temperatures (600–1200 °C).
Chemical activation: Strong acids, bases or salts and a lower temperature is required (450–900 °C).

Theoretically the chemical method seems appealing but in a practical sense it is not advisable to carbonize materials in a home-kitchen as the fumes evolved contain carbon-monoxide which is extremely toxic.


Does vinegar react with tap water? In an effort to prevent mold in my humidifier, I added a little white vinegar to the tap water in the tank. This creates a white bloom that floats in the water. It is odourless and, despite its appearance, is not slimy. It feels more like paper pulp. After a while it adheres to the sides of the tank as a transparent sort of gel. Is this a known phenomenon? What is the white stuff?

Acidic acid (vinegar) does not react with water, and it is not soluble in water. The gel and white bloom may come from a reaction the acidic acid with minerals such as calcium that may be present in the dehumidifier. the gel could also come from the breakdown of the chitin cell wall of the fungi. If the white bloom appeared within a minute or two of the addition of acidic acid it was most likely a reaction with minerals or other compounds.


Stability of Nitromethane Why is nitromethane not especially reactive? $\ce{NO2}$ is stable by itself, so why does it stay attached to $\ce{CH3}$?

It depends on what kind of reactivity you mean. Nitromethane is indeed flammable and is used as a racing fuel. It can also detonate, especially when mixed with an oxidizer. But it is generally quite stable to atmospheric oxygen at room temperature, perhaps because the nitro group is electron-withdrawing and reduces the likelihood that the hydrogens on the methyl group will react with oxygen or other oxidizers.

In other ways nitromethane is indeed reactive: for example, the nitro group makes the hydrogens on the methyl group acidic, so it reacts with an aqueous base like sodium hydroxide to give the nitromethide anion, which is useful in organic synthesis.

Consider the splitting of nitromethane into methyl radical and nitrogen dioxide. It has a normal sigma bond between carbon and nitrogen, and to break that bond you have to put a lot of energy in. You are right that NO2 is indeed stable on its own, but it is still a reactive, high energy species. Methyl radical is even more reactive, and the energetics of this bond-breaking reaction are overall unfavorable, at least at lower temperatures. Higher temperatures, such as might occur when sparking a detonation, may give the kind of splitting you are talking about, plus a lot of other reactions as well.


What solvents work on aloe vera? I'm looking for a solvent which can fully dissolve dried aloe vera gel. I assume the main problem is the constituent polysaccharides.

Hot water, the "universal solvent", should work nicely.


Is it true that todays crude oil is getting more 'dirty'? I heard that crude oil, which is used to make other products like gasoline, is getting contaminated more and more by atoms like nitrogen and metals, so that the catalysts have more difficulty making it clear for gasoline. If so what causes these contaminations?

We've already extracted most of the easy-to-get, easy-to-use, relatively pure, oil, and are now digging up oil from other sources, like the tar sands of Alberta, which contain a lot more contaminants than most other sources we've drawn from in the past.


What are the functional group names of halogenoalkanes? The functional group name of other organic compounds is easily findable, but that of the halogenoalkanes (or haloalkanes) is not. This makes me wonder if such a name exists for the halogenoalkanes or if they are an exception.

Are there such names? If yes, what are they for the different possible halogens present in a given compound?

Clarification: With functional group name I mean hydroxyl for the alcohols, phenyl for arenes, etc..

They don't have a functional group name similar to that of hydroxyl/phenyl/etc. Instead, you've to name them as if they were an alkyl branch - "haloalkyl" - and that too within parentheses.

Have a look at this simple comparison:

Notice that both 2,3-dimethylbutyl and chloromethyl are both within parentheses, whereas phenyl isn't. Therefore don't mistake "haloalkyl" to be a short name as "phenyl" is. There simply is no such short name, and you've to name it as an alkyl chain instead.


Ion pair definintion in chemistry I read in the journal of organic chemistry:

  A less striking, but nonetheless significant, attenuation of the reactivity of phenyl selenide anion can be achieved by varying the counterion and/or the relative degree of solvation of the ion pair. For example, when valerolactone (3) is allowed to reflux with sodium phenyl selenide in THF/HMPA for $3\,\mathrm{h}$, one obtains an $85\,\%$ yield of the $\unicode[Times]{x3C9}$-phenylselenenyl carboxylic acid 4.

My question is what is the meaning of the bolded expression (solvation of the ion pair) above?

As already said, ion pair is a pair of two ions like NaCl. When the pair separates completely into its constituent ions in the solvent (like Na+ & Cl- in water), it is called dissociation.

After dissociation, if the solvent associates itself and surrounds the ions like this-

(Sodium ion surrounded by water molecules), it is called Solvation.
Solvation stabilises the compound and makes it either more reactive or less reactive.


Calculating yield percentage of magnetite synthesis I’ve made magnetite $\ce{Fe3O4}$ in the lab and now I need to calculate yield percentage. How can I do it? I have these data.

\begin{align}
\ce{FeSO4.7H2O(aq) &-> Fe^2+(aq) + SO4^2- (aq) + 7H2O(l)}\\
\ce{KOH(aq) &-> K+(aq) + OH- (aq)}\\
\ce{KNO3(aq) &-> K+(aq) + NO3- (aq)}\\
\ce{Fe^2+(aq) + 2OH- (aq) &-> Fe(OH)2(s)}\\
\ce{3Fe(OH)2(s) + NO3- (aq) &-> Fe3O4(s) + NO2- (aq) + 3H2O(l)}\\
\end{align}

\begin{align}
\ce{FeSO4.7H2O &->} 4.20~\mathrm{g}\\
\ce{KOH &->} 1.90~\mathrm{g}\\
\ce{KNO3 &->} 0.12~\mathrm{g}\\
\text{Magnetite}~\ce{&->} 0.36~\mathrm{g}\\
\end{align}

I know how to calculate the yield for simple reactions but I don’t know how to do it when there are many reactions involved.

Add up the reactions (double the second one so there is sufficient hydroxide) to get a net reaction with your starting materials as reactants and your desired product as one of the products:

$$\ce{3FeSO4.7H2O(aq) + 6KOH(aq) + KNO3(aq) -> Fe3O4(s) + NO2-(aq) + ...}$$

Then, find the limiting reactant, calculate the theoretical yield from it, and compare it to the amount of product to figure out the percent yield.


How is knowing about the symmetry of a crystal useful I am studying the space groups and know that we use them to talk about crystal symmetry, I would like to know what material properties may be produced or predicted from knowing the symmetry or lack thereof in a crystal system of a material?

Properties that are directly dependant on crystal symmetry are

birefringence
piezoelectricity
(obviously) x-ray and electron diffraction pattern
pyroelectricity
nonlinear optical properties
...


Difference between Force Field and topology, and other related questions I am performing some Molecular Dynamics computer simulations and I am learning about the various Force Fields (FF) such as OPLS-AA, OPLS-UA, AMBER, etc.
Many things are still unclear to me. I put these questions together because I think they are strictly related.

1) What is exactly the difference between "Force Field" and "topology", since they define similar parameters?

2) Which parameters can be defined as "belonging" to the FF, and which not?  It seems to me that sometimes topology files recall atoms from the force field and edit some parameters such as charge, mass, bond lengths, elastic constants.

3) Why is water model usually considered separately from the FF, and why is it not included in it?

4) When a molecule does not belong to a known topology and we have to model it, how do we establish the compatibility between our model and the FF used for the other molecules in the same simulation?

Thank you.

1 - "Force Field" (FF) means the set of equations that describe the interactions between atoms and the set of numerical parameters that people using those particular functions have come up with to describe some molecules (typically proteins, Nucleic Acids, lipids, and a few others). Many FFs exist because of different choices in functional forms, in the procedures adopted to obtain the numerical parameters etc etc.
A topology is just the description of a molecule in terms of parameters that can be replaced by the numerical ones contained in the FF of your choice, so that the equations of the FF can be worked out numerically.

2 - see 1.

3 - Water is described separately because there are specific models for water (TIP3P and so on) that work reasonably well, so people stick to these models indipendently of their choices for the rest of the system.
What do you mean by "it is not included"? I can see water parameters in all MD programs I have used.
Usually, FFs are always divided by "optimisation" let's say, meaning that a set of parameters have been optimised for a particular task/family of molecules, and those are given together.

4 - if you want to model a molecule for which you don't have the parameters, then you should fit them according to the procedures that have been used to obtain the other parameters belonging to the FF (good luck). If parameters exist for a similar molecule, you can try to change those to represent your molecule.

In general, it is not a good idea to use mixed parameters, meaning e.g. AMBER for the protein, CHARMM for a DNA in the same simulation.
Choose one FF and if you need parameters for some new molecule, you have all my sympathy (you can find tools to obtain these "pseudo-automatically", though).


Understanding the structure of NCS⁻ Predict the most preferable structure of $\ce{NCS-}.$
$$\Large\underset{(\textbf{1})}{\ce{(:\!\!\overset{\huge.\!\!.}{N}=C=\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{S}})-}}\qquad\underset{(\textbf{2})}{\ce{(\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{C}}=S=\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{N}})-}}$$

My Attempt
A chart for the formal charges:
$$
\begin{array}{l|rr}
\hline
\text{Atom} & (\textbf{1}) & (\textbf{2})\\
\hline
\ce{N} & -1 & -1 \\
\ce{C} & 0 & -2 \\
\ce{S} & 0 & +2 \\
\hline
\end{array}
$$
As a conclusion, in structure 2 the formal charges on $\ce{C}$ and $\ce{S}$ (i.g. bond becomes polar because of $+2$ and $-2$) make it weaker, whereas in structure 1 there are less charge separations. So, (2) must be more a accurate structure.
But according to my book, structure 2 is more stable. Why? Please explain. Is there any other approach to this question?
Edit

I got this statement in favor of structure #2 "is more stable because each atom has non-zero formal charge in the lowest energy state". But i can't understand this statement. Can anyone explain me?"

This is a misprint. Here is a similar problem (OpenStax Chemistry, retrieved from https://opentextbc.ca/chemistry/chapter/7-4-formal-charges-and-resonance/) that makes sense and has a correct answer:

  As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS–, NCS–, or CSN–. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

  Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).

The statement made rationalizing the incorrect structure is also flawed:

  "is more stable because each atom has non-zero formal charge in the lowest energy state"

This is an ion. Some atom has to have a non-zero formal charge because the net charge should equal the sum of formal charges. Why having each atom have a non-zero formal charge should result in a more stable structure is not clear to me. Mentioning the lowest energy state is curious as well. Maybe they meant most relevant resonance structure.

One should also keep in mind that formal charges are mostly an accounting method and do not reflect actual charge distribution (see https://chemistry.stackexchange.com/a/119771).


What repulsion keeps non-hydrogen atoms at a distance: between inner shells or between nuclei? Bonded or not, when two atoms or ions come closer than the bond distance or the van der Waals distance, the energy increases drastically. What is the underlying cause of that repulsion?

The repulsion of the negatively-charged inner-shell electrons
The repulsion of the positively-charged nuclei (or the effective positive charge of nuclei plus inner-shell electrons)
Any classical explanation solely based on electrostatics falls short, and the increase in energy is related to the Pauli exclusion principle.
There is no single underlying cause, it depends on the specific case. Sometimes repulsion of positive charges is the problem, sometimes repulsion of negative charges.

I found a nice figure and the relevant statement in a paper by Frenking and Krapp (Unicorns in the world of chemical bonding models, 2006, https://doi.org/10.1002/jcc.20543):

The crucial term which is responsible for repulsive interactions in chemical bonds except in two‐electron systems such as H2 is the Pauli repulsion. The three terms (a) quasiclassical electrostatic interaction, (b) resonance, and (c) Pauli repulsion give a complete picture of the interatomic interactions which yield a chemical bond.

The figure shows how the distance-dependent repulsion for one scenario including the Pauli repulsion and two scenarios ignoring the Pauli repulsion; the latter do not show the drastic increase in repulsion that is observed.

Source: https://onlinelibrary.wiley.com/cms/attachment/2b2dc015-ca3c-4fd3-b51c-f768a6a924b5/mfig003.jpg
The entire essay is great, connecting the picture of bonding that bench chemists have with the developments of theory and computation in the 90 years since quantum mechanics was first applied to chemistry.


Can I make a buffer with a weak acid without adding it's conjugated base, when the desired pH of the solution is exactly the pKa of the weak acid? An acid will be 50% dissociated when the solution it's in have a pH equal to the acid pKa ('afirmation 1', for later reference).

We also know that a good buffer solution have equal concentrations of an acid and it's conjugated base.

That's why to make a good buffer we look for an acid that have a pKa close to the pH we want to maintain.

So, suppose I have a solution with a pH 6.2, and I want to add a buffer to maintain this pH at 6.2.

Suppose also, that I have a ideal weak acid for this situation: with a pKa of exactly 6.2.

Using the Henderson–Hasselbalch equation, we know that we need to have the same concentration of the acid and it's conjugated base in the buffer. To add the conjugated base, we use a salt of the refereed acid.

So my question is: If we just throw the acid at the solution without the salt, since the solution pH is exactly the pKa of the acid, the acid will dissociate 50% (afirmation 1), and we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.

So, in principle, it's possible to make a buffer without adding the salt?

Is this logic correct? If not, what error I'm making?

Thanks

So your solution is already at a pH equal to the pKa of the weak acid you are about to add. Your hunch is that the pH remains the same after adding acid to a solution.

  [...] we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.

It depends on the concentration of what is already in solution and of what you are adding. If you add a weak acid at very low concentration (e.g. an indicator) to a buffered solution, the indicator's protonation state will be governed by the pH of the solution. On the other hand, if you add a weak acid at very high concentration (e.g. a substance that is supposed to act as buffer) to an unbuffered or hardly buffered solution, the buffer will govern the pH.

In your case, the pH will be off (too acidic), and most of the buffer substance will be in the acid (protonated) state. You would have to adjust the pH by adding a strong base to the amount that deprotonates half of the just added acid to get the pH to match the pKa again.

  So, in principle, it's possible to make a buffer without adding the salt?

Yes, but not in the scenario you suggested. Most buffers are made by starting with a weak base (pH is much higher than pKa of conjugate weak acid) and adjusting the pH with strong acid, or by starting with a weak acid (pH is much lower than pKa of the weak acid) and adjusting the pH with a strong base. In both cases, you effectively get a mixture with the "salt", i.e. you introduce some spectator ions (often chloride or sodium ions).


What is the IUPAC nomenclature for Intermetallic Compounds? IUPAC has a system for naming organic compounds, ionic compounds, and inorganic compounds. I have seen something vaguely suggesting there is one for intermetallic compounds, such as: $$\ce{Fe-Co}$$ What is the system, if it even exists? If it doesn't exist what is the generally accepted way of naming intermetallic compounds?

In your example you have given two metals and a bond-indicating symbol.

Grammar

Hyphens, Plus and Minus Signs, 'em'Dashes and Bond Indicators

'em' Dashes (ALT+0151, 'en' dash would be ALT+0150)

a) Used to indicate metal–metal bonds in polynuclear compounds (e.g. "[Mn2(CO)10 bis(pentacarbonylmanganese)(Mn—Mn)").

b) Used to separate the indvidual constituents in names of (formal) addition compounds (e.g. "3CdSO4·8H2O cadmiumsulfate—water")

Alphabetical Order

a) Intermetallic compounds are sorted alphabetically within the group of cations and anions. Deviations are allowed in order to express structural information.

b) In coordination complexes, ligands are sorted alphabetically and the donating atom is put first (e.g. OH2 instead of H2O).

Further Reading

Therefore, I would write your example as "Co—Fe".

There are a lot more rules that you can read here: https://web.archive.org/web/20181005064417/https://www.iupac.org/fileadmin/user_upload/databases/Red_Book_2005.pdf


Mechanistic explanation of salt lowering temperature of ice slush As we all know, adding salt to ice water lowers its temperature.

I've read plenty of system-level accounts of energy balances, enthalpies, vapor pressures, phase equilibria and freezing points—I understand that chemists' bread and butter is doing energy arithmetic on thermodynamic systems as a whole, but I would like to gain microscopic-level intuition instead of just being handed some thermodynamic abstractions.

I am looking for an intuitive explanation that details, step by step, at an atomic level, what happens at the interface between the ice and the salt water. For each step, I would like to understand:

What energies (kinetic/temperature vs. latent/chemical) are involved, and where?
How exactly (atomically or subatomically) is one type of energy converted into the other?
Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?

The explanation should be clear enough that I should be able to intuitively answer the following, without consulting diagrams:

How does the concentration of salt change the evolution of the system (again, at a microscopic level)
How does adding e.g. ethanol instead of salt change things?
Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)

Water loses kinetic energy when it detaches from the surface of the ice

Water molecule in the solid phase have more hydrogen bonds than in the liquid phase. It takes energy to break these bonds, so only molecules with sufficient kinetic energy will detach (maybe because they just collided with a water molecule that transferred some kinetic energy to it). So the molecule that detaches will have a low kinetic energy as it goes into the liquid phase. Further collisions will quickly ensure that over time, this particular molecule has an average kinetic energy, but overall the temperature will drop from that detaching event.

Water gains kinetic energy when it attaches to the surface of the ice

This is just the opposite process, so the opposite happens. As additional hydrogen bonds are formed, the molecule has greater kinetic energy (in the form of vibrational energy). Either the molecule had less than average kinetic energy and now has average energy again, or it had average kinetic energy and will have higher kinetic energy, but quickly dissipate this to other molecules.

When you add salt to the liquid water, less molecules attach to the surface of the ice

Compared to pure water, less collisions of particles in the solution with the surface of the ice will lead to attachment of particles because now, some of these particles are not water molecules but sodium or chloride ions instead (they are not incorporated into the crystal lattice of ice but stay solvated by liquid water). As a consequence, there are less attachment events while the detachment events stay the same. With a net melting of ice, the average kinetic energy of particles decreases (the temperature drops). As a secondary effect, this also affects the rates of attachment and detachment, so once the temperature has sufficiently dropped, the two rates will be the same again.

  What energies (kinetic/temperature vs. latent/chemical) are involved, and where?

In my explanation, I talked about kinetic energy (reflecting the temperature) and energy necessary to break hydrogen bonds (could be generalized as intermolecular interactions, and they underlie the observed bulk latent heat).

  How exactly (atomically or subatomically) is one type of energy converted into the other

There is no chemistry or nuclear physics happening, so it is sufficient to talk about the molecular level. Kinetic energy is converted into potential energy through inelastic collisions. Kinetic energy is redistributed between molecules through elastic collisions.

  Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?

According to microscopic reversibility, all molecular events are reversible. Processes go in a certain direction because it is more probable than in the other direction (2nd law of thermodynamics). The 2nd law of thermodynamics is inherently linked to large numbers of particles, so you need to look at the forest rather than at individual trees.

  How does the concentration of salt change the evolution of the system (again, at a microscopic level).

The higher the concentration of the salt, the lower the concentration of the water in the liquid phase, i.e the fewer events of liquid water attaching to the ice.

  How does adding e.g. ethanol instead of salt change things?

1 mmol/L NaCl has twice the dissolved particles compared to 1 mmol/L ethanol. Otherwise, the effect is the same (colligative property).

  Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)

Ice near the freezing point exhibit surface melting. Once there is a tiny amount of water, more ice will melt. On the other hand, if it is cold enough that even saturated solutions will freeze, a mixture of dry salt and ice will not interact.


Why does radioactive matter decay in series of half-lives? I was studying types of nuclear decay, and I came across the concept of "half-life". And I started wondering why does nuclear matter decay in half-lives? In other words, why does the rate of nuclear decay decrease as a function of time?

Is it because as more atoms decay, the surface area of the radioactive matter decreases, and the number of atoms "exposed" decrease? Does the outer layer of atoms somehow "shield" the inner atoms and stabilize them, preventing them from decay?

Or is there another mechanism at play?

There are fewer decays because there are fewer atoms to decay

The simple reason why the number of decays (strictly, the number of decays per unit time) decreases in simple radioactive decay is because there are fewer atoms left to decay.

Nuclear decay is probabilistic. The probability of any given unstable atom decaying is constant (independent of time or the environment). So the reaction rate (the total number of decays per second) per mole is constant. If this were a chemical reaction we would call it a first-order reaction (since it isn't a chemical phenomenon we can call it a first order process but the mathematics are the same).

In any first order process there is a characteristic time when half the original material will have gone (we call this the half life of the process). So, if we measure time in half-lives at t=1 we will have, say, 1 mole of the material; by t=1 we will have 0.5 moles; at t=2 we will have 0.25 moles of material. If we count the number of emissions per second they will follow the amount of material left. So the radioactivity will decay at the same rate as the material.

We don't need to invoke any interactions between the atoms of the material to explain this.

PS it gets a little more complicated if one radioactive compound decays into another. Then the total radioactivity will depend on two competing decays with different half-lives (but the radioactive emissions may well be different as each has a characteristic energy and expensive instruments can tell them apart giving two separate emission counts). But the mathematics remains the same.


What gives fresh milk a yellow tinge? Every now and then, when I pop over to my grandparents' house, they make sure to give me a glass (or two) full of fresh milk (or "whole milk" if you will) and then proceed to pamper me for the rest of the day.

Now, it suddenly struck me that the fresh milk I get over there actually has this really faint yellow tinge, as opposed to the skimmed milk I get at the store which is almost pure white.

I did try Googling this up, but I can't seem to find anything on what imparts the yellow tinge to fresh milk.

So,

What substance/pigment is responsible for the yellow tinge associated with fresh milk?

The main substances that cause the yellowish color of the milk are carotenoids [1]. The main carotene involved is the beta-carotene coming from the feed that cows eat.

Some studies have been carried on and it has been noticed that the milk with a more yellow tinge was collected during late spring and early summer when carotene levels are at a maximum [2] unfortunately for this reason in many products as mozzarella they add titanium oxide.
The amount of Riboflavin is too little compared to that of beta-carotene and generally gives a slightly green-ish tinge to the milk as reported here and here [3].

P. F. Fox, Fundamentals of Cheese Science,p. 13
Kosikowski, Frank V., and David P. Brown. "Application of titanium dioxide to whiten mozzarella cheese." Journal of Dairy Science 52.7 (1969): 968-970.
Marion Eugene Ensminger,Audrey H. Ensminger,Foods & Nutrition Encyclopedia, 2nd Edition, Volume 1,p. 798


What makes pentane, hexane, heptane, and octane dangerous, while butane and nonane aren't? In particular, how do a methylene more (in the case of the pentane) and a methylene less (in the case of the octane) make the difference in harmfulness to the environment in comparison with, respectively, the butane and the nonane?

I saw that butane and nonane are not classified as dangerous, but I have just done a research and it turns out nonane is harmful to fish, just like octane, but the difference is that the nonane becomes aqueous more difficult because of its low solubility. But I read that its soluble part is still dangerous for aquatic life. Then the question is: Shouldn't it be classified as dangerous anyway?
The propane, as well, appears to be harmful to the fish, while I can find nothing but "not classified as dangerous", as for the butane. Is LDC3's answer the explanation?

Pentane, hexane, heptane, and octane are labelled with the ‘environment’ hazard pictogram (GHS09) since they are considered toxic or very toxic to aquatic life with long-lasting effects (H411 or H410) because of their toxicity to the water flea Daphnia magna in standardized tests.

Nevertheless, nonane is not completely harmless; it still may cause long-lasting harmful effects to aquatic life (H413). However, a pictogram is not required for this hazard category.

According to a comparative study, the toxicity of these substances is mainly determined by their solubility in water and their organism–water partition coefficient.


Phase of reaction between glucose and potassium cyanide to form amygdalin I'm trying to write a detective-fiction novel, and I was wondering if potassium cyanide reacts with sugar only if the glucose for the amygdalin to form is dissolved in liquid or does it react even if the glucose is still sugar, like sugar cubes?
For example if potassium cyanide was mixed in a sugar canister. Will there be amygdalin or will there be amygdalin only if the potassium and sugar from the canister are mixed in a glass of water?
I can't find the answer in google, and I'm not that good with chemistry.

The glycoside amygdalin is a product of enzymatic biosynthesis. Its hydrolysis produces 2 molecules of glucose, benzaldehyde and hydrogen cyanide. Mixing potassium cyanide and sugar ( = sucrose = combined glucose + fructose) does not produce amygdalin. Neither you would get it, mixing glucose, benzaldehyde and hydrogen cyanide.


What is the heat of reaction of weak acid reacting with weak base compared to strong acid reacting with strong base? My book has this line only that "weak acid and weak base reaction will produce lesser heat than constant enthalpy" of strong acid and strong base neutralization which is $\pu{-57.3 kJ/mol}$.

Strong acid and strong base

First, lets see where the enthalpy of $\pu{-57.3 kJ/mol}$ comes from. In aqueous solution, strong acids and bases react completely to form $\ce{H+}$ and $\ce{OH-}$, respectively. The neutralization of a strong acid with a strong base then is always the same reaction:

$$\ce{H+(aq) + OH-(aq) -> H2O(l)}$$

We can calculate the enthalpy of reaction from the enthalpies of formation at a temperature of $\pu{25 ^\circ C}$.

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{H+(aq)}) &= 0 \\
\Delta H_\mathrm{f}(\ce{OH-(aq)}) &= \pu{−230.0 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{H2O(l)}) &= \pu{−285.8 kJ/mol}
\end{align}
$$

Comparing enthalpies of formation for the products with those of the reactants, we get  reaction enthalpy of $\pu{−55.8 kJ/mol}$. The value given in the question is slightly different (maybe for a different temperature, which is not given).

Weak acid and strong base

For this case, the reaction is the following (with $\ce{AH}$ as the weak acid):

$$\ce{AH(aq) + OH-(aq) -> A-(aq) + H2O(l)}$$

The reaction enthalpy for this reaction depends on the nature of the weak acid. Specifically, the answer depends on the difference in the enthalpy of formation of $\ce{A-(aq)}$ vs $\ce{AH(aq)}$. Conceptually, it takes energy to break the $\ce{A-H}$ bond, and the interaction with the solvent water changes when you go from $\ce{AH(aq)}$ to $\ce{A-(aq)}$. Here are two examples:

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) &= \pu{−484.3 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) &= \pu{−486.0 kJ/mol}
\end{align}
$$

Difference is $\pu{-1.7 kJ/mol}$.

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{NH4+(aq)}) &= \pu{−132.5 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{NH3(aq)}) &= \pu{-80.893 kJ/mol}
\end{align}
$$

Difference is $\pu{51.6 kJ/mol}$.

So sometimes the enthalpy is higher, sometimes lower than strong acid/strong base neutralization.

Weak acid and weak base

The weak acid/strong base consideration showed us that once you have a weak acid, the enthalpy depends on the nature of the acid. So can we make a general statement about the enthalpy of a reaction between a weak acid and a weak base? We start by writing down the chemical equation (weak base shown as $\ce{B}$):

$$\ce{AH(aq) + B(aq) <=> A-(aq) + BH+(aq)}$$

The enthalpy of this reaction could be positive or negative (if you read it backwards like this:

$$\ce{BH+(aq) + A-(aq) <=> AH(aq) + B(aq)},$$

you can call $\ce{BH+(aq)}$ the weak acid and $\ce{A-(aq)}$ the weak base, and the sign of the reaction enthalpy switches.

Notice that this is written as equilibrium. Depending on the $\mathrm{p}K_\mathrm{a}$ values of $\ce{AH}$ and $\ce{BH+}$, this reaction will have its equilibrium more on the left or on the right. If the reaction does not go to completion, the enthalpy (based on amount of starting materials) will be less.

Let's take the neutralization of acetic acid and ammonia as a specific example:

$$\ce{CH3COOH(aq) + NH3(aq) <=> CH3COO-(aq) + NH4+(aq)}$$

The $\mathrm{p}K_\mathrm{a}$ values are separated by more than 4 units, so this reaction goes almost to completion. The reaction enthalpy is:

$$
\begin{align}
\Delta H_\mathrm{r} &= \Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) + \Delta H_\mathrm{f}(\ce{NH4+(aq)}) - \Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) - \Delta H_\mathrm{f}(\ce{NH4+(aq)}) \\
  &= \pu{−486.0 kJ/mol} - \pu{132.5 kJ/mol} + \pu{484.3 kJ/mol} + \pu{80.893 kJ/mol} \\
  &= \pu{-53.3 kJ/mol}
\end{align}
$$

That is pretty close to the strong acid/strong base case.

Back to the question

To prove that

  "weak acid and weak base reaction will produce lesser heat than constant enthalpy" of strong acid and strong base neutralization

you would have to do this calculation for every weak acid and weak base combination out there. I would be careful making a statement like this. I think the idea was that because weak acids and bases dissociate much less, less heat is given off (per amount of starting material). However, when you combine a weak acid and a weak base, sometimes the reaction goes almost to completion. As you can see from the example I gave, there are cases for which the enthalpy is very close to that of the strong acid/strong base combination.

Gibbs energy of reaction

For the Gibbs energy of reaction, the situation is much simpler. Weak and strong are defined by the magnitude of acid dissociation equilibrium constants ($K_\mathrm{a}$, and related $\mathrm{p}K_\mathrm{a}$). The difference in $\mathrm{p}K_\mathrm{a}$ of the acid and the conjugate acid of the base is a direct measure of the standard Gibbs energy of reaction: the larger the difference, the larger the Gibbs energy (with appropriate signs, so the strong acid/strong base combination would have the most negative Gibbs energy of reaction, the strong acid/weak base or weak acid/strong base combination would also have negative values, and for the weak acid/weak base combination if would depend on the specifics).


How to find volume needed from a stock solution I am making a media that is amended with substrates after being autoclaved in order to keep the media anaerobic. In order to do this, we make stocks of the solutions from a known concentration. I am having trouble figuring out what amount of liquid to add to one liter. For example, I make a 100 ml stock of sodium fumarate (160.04 g/mol) using 0.6402 grams in 100 ml of MQ H20. The final desired concentration in the liter media is 40 mM. How do I figure out how much of the liquid stock of sodium fumarate I am supposed to add? Thanks in advance for the help!

This is calculated by equating the moles of the two solutions.

You are using $\pu{0.6402g}$ of sodium fumarate in $\pu{100mL}$ of solution.

$$ \text{No. of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} = \frac{0.6402}{160.04} = \pu{0.0040 moles} $$

$$ \text{The concentration of the solution can be found easily} = \frac{0.0040}{0.1} = \pu{0.04 M}$$
If you require a solution of concentration $40$mM, the stock solution that you have prepared is exactly that.
If you need a more diluted solution, you may use the formula:
$$ M_1 V_1 = M_2V_2$$
where, $M_1$ and $M_2$ are the initial and final concentrations, $V_1$ is the volume of stock you would require for the dilution, and $V_2$ is the volume of final desired concentration.
Let us assume you require $\pu{50 mL}$ of $\pu{10 mM}$ solution, and you have a stock of $\pu{100 mL}$ of $\pu{40 mM}$.
Then
$$
M_1 = \pu{40 mM}\\
V_1 = ? \\
M_2 = \pu{10 mM} \\
V_2 = \pu{50 mL}
$$
Plugging in the values, you will find out that you require $\pu{12.5 mL}$ of stock, and dilute it to $\pu{50mL}$ using the solvent, i.e., add $\pu{37.5 mL}$ of water to $\pu{12.5 mL}$ of stock solution.


Why do they use graphite electrodes in the Hall–Héroult process? In the Hall–Héroult process of extracting aluminium from aluminium oxide, the electrolyte is a liquid mixture of cryolite $(\ce{NaAlF6})$ and aluminium oxide $(\ce{Al2O3}).$ The cathodes and the anodes are made of graphite.
During the electrolysis process, aluminium is deposited at the cathode and oxygen is liberated at the anode. Some of this oxygen reacts with the carbon in the graphite to form carbon-dioxide, thus slowly burning away the anodes. Thus, the anodes have to replaced periodically.
But if this is the case, why do they use graphite anodes at all? Why don't they use anodes of some other, less reactive material? I'm pretty sure there are many out there that would fit the bill.

According to this source, graphite is used as anode material because carbon is easier to oxidize than $\ce{O^2-}$. The anodic oxidation of carbon

$$\ce{C + 2O^2- \rightarrow CO2 + 4e-}$$

proceeds more efficiently than the oxidation of $\ce{O^2-}$ at an inert anode

$$\ce{2O^2- \rightarrow O2 + 4e-}$$

$\ce{C}$ is a much better reducing agent than $\ce{O^2-}$ whose oxidation state is not changed in the reaction with graphite. As a consequence, anodic current density and the deposition rate of aluminium at the cathode will increase.


Gibbs Free Energy Spontaneity Criterion Derivation In Atkins's Physical Chemistry, spontaneity criterion is derived from Clauisius inequality as follows:
Firstly, system is initially at the same temperature T as surroundings. When a change in the system (its state) occurs, there is an corresponding entropy change in the system and a transfer of heat between system and surroundings and Clauisius inequality for this process can be written as: $$dS -\frac {dq}{T} \geqslant 0$$
If process is carried at the same temperature and pressure of a system than $$ dq = dH $$
If we plug dH in Clausius inequality, multiply by T and rearrange what we get is: $$ dG = dH - TdS \le 0$$
Which should prove Gibbs free energy spontaneity criterion for systems at constant p and T. Problem here is that T in the equation for G is of the surroundings not of the system and the whole point of G is to forget about surroundings. Answer you can give to this question is that surroundings and system remain at the same temperature during state change of the system, but if so how can there be any heat transfer between them, for they are already at thermal equilibrium and dq should is zero. When taking Gibbs free energy as spontaneity criterion, system needs to stay at the same pressure and temperature. To keep the system at the same temperature, heat generally needs to be exchanged with the surroundings and if so, heat exchanged with surroundings will change its temperature, so T in Clausius ineqaulity generally is not a constant, it can only be thought as constant if surroundings is uncomparably bigger than the system so that its heat capacity is uncomparably bigger than of system (which in most cases really holds) only in that case T in Clausius inequality can be taken as T of the system. Problem with this is that such an approach gives only approximate solution and equation for G as spontaneity criterion IS NOT AN APPROXIMATION, it holds always regardless of how big system is compared to the surroundings. So, I am not sure this derivation in Atkins's textbook is actually correct for all scenarios not only when surroundings is uncomparably bigger than the system. What are your thoughts?

For a system in contact with a constant temperature reservoir at the same temperature as the system in its initial state, and also in contact with a constant pressure surrounding at the same pressure as the system in its initial state, $$\Delta U=Q-W=Q-P\Delta V-W_{npv}$$where $W_{npv}$ is the amount of non-PV work done by the system.  In addition, from the 2nd law of thermodynamics, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\frac{Q}{T}$ is the amount of entropy transferred to the system from the surroundings across the interface between the system and surroundings (recall that the latter is always at the initial temperature of the material comprising the system).  Note that this does not say anything about how the temperature may be varying with location and time within the internals of the system during the process.  In addition, the parameter $\sigma$ represents the amount of entropy generated within the system due to irreversibilities during the process. Note that $\sigma$ can only be positive (irreversible process) or zero (reversible process)
If we combine these two equations, we obtain:
$$\Delta U=T\Delta S-T\sigma-P\Delta V-W_{npv}$$or$$\Delta G=-W_{npv}-T\sigma$$ According to this, if no non-PV is done by the system on the surroundings during the process, $\Delta G$ must be negative (irreversible) or zero (reversible).


Comparison of two spectra in order to find whether the irradiated sample has significantly different resulting spectra I would like to compare two absorption spectra ( or interferograms) and conclude whether between these two there are statistically significant differences at particular wavelength intervals. At the moment, I have data of two experiments that look like this:
    # A tibble: 6 x 5
      t     x1     y1     x2     y2
  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
1 3999. 0.0124 0.0132 0.0122 0.0113
2 3998. 0.0125 0.0130 0.0122 0.0116
3 3997. 0.0122 0.0131 0.0122 0.0113
4 3996. 0.0121 0.0136 0.0122 0.0114
5 3995. 0.0124 0.0139 0.0122 0.0122
6 3994. 0.0125 0.0141 0.0122 0.0129

The first column represents the wavenumber, the x columns represent the absorbance of sample and the y columns represent the absorbance of irradiated sample ( before and after). I was wondering whether I could compare these data ( x and y) as time series and if so, what could be the method to quantify the differences, if any, between the samples before and after irradiation. Maybe it's already been done and there is somewhere some information as to how to compare the spectra if the wavenumber is interpreted as time ( x axis). I did the t test in R and in both experiments the null hypothesis could not be rejected, although for the second experiment (x2, y2) the p value was much lower than for the first.
If I average the x and y, and then plot both data, I see that there are visible differences at certain wavelength intervals. In R, I used IRISSeismic package and function crossSpectrum, that gave me such an output:
   freq        spec1        spec2 coh        phase             Pxx             Pyy
1   0.0002666667 2.121935e+01 2.152532e+01   1  0.010220252 1.856693e+01+0i 1.883465e+01+0i
2   0.0005333333 7.011069e+00 6.869078e+00   1  0.008984730 6.134686e+00+0i 6.010443e+00+0i
3   0.0008000000 8.385363e+00 8.197039e+00   1  0.011999039 7.337193e+00+0i 7.172409e+00+0i
4   0.0010666667 7.483070e+00 7.272319e+00   1  0.015419031 6.547686e+00+0i 6.363279e+00+0i
5   0.0013333333 4.086251e+00 3.899095e+00   1  0.017551350 3.575469e+00+0i 3.411708e+00+0i
6   0.0016000000 1.537709e+00 1.405562e+00   1  0.008722742 1.345496e+00+0i 1.229867e+00+0i

For all data points the coh index is given as 1. But obviously that doesn't help me to determine, whether the difference is substantial between two data sets.
I tried to average data points between groups (x1, y1 and x2, y2) at each wave number and plotted the graph, it looks like the difference might be significant at certain wavenumber interval. Maybe I could try to express one line ( averaged data points after irradiation of the sample) as a function of the other? But how?
Here is a project with similar experiments: https://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532015001202571 , for statistical analysis they use ANOVA and Tukey's test, but how do I do it for the vectored data?
I cannot find online anything. Here is something of the sort I am looking for: https://www.youtube.com/watch?v=gjKSfILE9nM He compares two spectra in Matlab, but he is using self made toolkit which I can not obtain. How can I do something like this in R? I've read something about modelling the data, but here, I don't have any variables just data of two spectra.

Why don't you start with the simplest possible approach. I am sure there are very fancy spectral matching statistical algorithms, but simplicity has its own beauty. Comparing an inteferogram (not many features there) might be difficult but what about absorbance rather?
Look at the absorbance difference spectrum: (Spectrum after irradiation)-(Original sample spectrum). How does the plot look like? What is the signal which you are collecting? Sometimes the human eye can save a lot of time rather than doing significance testing on data points and forming null hypotheses.
The features which remain the same will be zero now and the features which change will be visible in the the difference spectrum.
Alternatively, you can take the first derivative of both spectra and smooth it with Savitsky-Golay or study the difference of "first derivatives". Noise will be an issue but I guess you can take care of that.


Reaction of solid silver sulfide on warming with dilute nitric acid Solid silver sulfide is warmed with dilute nitric acid.

I thought that the nitric acid would engage with silver sulfide in a double displacement reaction, as follows: $\ce{Ag_2S {(s)} + 2H+ {(aq)} -> H_2S {(g)} + 2Ag^{+} {(aq)}}$. This double displacement reaction would be driven by the production of $\ce{H_2S}$ gas.

However, my textbook says that the answer is: $\ce{Ag {(s)} + NO_3^{-} {(aq)} + 4H^{+} {(aq)} -> 3Ag^{+} {(aq)} + NO {(g)} + 2H2O {(l)}}$. Does this redox reaction occur because $\ce{HNO_3 {(aq)}}$ is a strong oxidizing agent? Why doesn't the textbook answer contain $\ce{AgS {(s)}}$ in the left hand side of the net ionic equation?

In the future, how do I know when nitric acid will engage in a double displacement reaction to liberate gas, or when it will engage in a redox reaction? Furthermore, if the nitric acid was concentrated, then would $\ce{NO_2}$ be produced?

Nitric acid is a strong oxidising agent and it will oxidise both the reactants and products. Thus, a reaction involving nitric acid is not a simple double-displacement reaction but a redox reaction where nitric acid itself get reduced to a couple of nitrogen oxides depending on concentration of the acid and the other reactants gets oxidised. According to [1]:

  Nitric acid, when of concentration above 5%, dissolves
  precipitated silver sulphide rapidly. Very strong acid yields silver
  sulphate alone, while acid of lower concentration forms some nitrate
  in proportion to its dilution. 15-20%, acid yields the
  maximum, about 95 per cent, nitrate, greater dilution again resulting
  in a smaller percentage. Heating the acid, or prolonging its time of
  action is unfavorable to the formation of nitrate.

Another paper[2] suggests formation of normal silver(I) sulfate through a series of exotic sulfates which is confirmed by the change in color: -

  [...]the acid additions used correspond to 75%, 150% and 200% of the
  amount required stoichiometrically based on the following reaction:

  $$\ce{3Ag2S + 8HNO3 <=>> Ag2SO4 + 8NO + 4H2O}$$

  [...]

  The distinct colors of the various phases involved in this system
  (black silver sulfide ($\ce{Ag2S}$), brown silver sulphide sulphate
  ($\ce{Ag6S3O4}$), white to light yellow silver sulfate ($\ce{Ag2SO4}$)
  and bright yellow sulfur($\ce{S}$)) are a further indicator of this
  behaviour; the leach went from brown ($\ce{Ag6S3O4}$) to white
  ($\ce{Ag2SO4}$) with increased pressure and temperature and some
  residue samples showed evidence of the presence of elemental sulphur.
  This transition from $\ce{Ag2S}$ to $\ce{Ag6S3O4}$ to $\ce{Ag2SO4}$
  shows the progressive oxidation of sulphur in the leach solids as the
  leaching reaction goes to completion.

In conclusion, reaction of silver sulfide and nitric acid will give different products depending on stoichiometry of the reactants, concentration of acid, effect of temperature and pressure etc. As you noted correctly, if the concentration of nitric acid is high enough, it may even produce $\ce{NO2}$ gas.

$$\ce{Ag2S + 10HNO3 → 2AgNO3 + H2SO4 + 8NO2 + 4H2O}$$

References:

Gruener, H. Silver Nitrate Formed by the Action of Nitric Acid on Silver Sulphide. J. Am. Chem. Soc. 1910, 32 (9), 1030–1032. DOI: 10.1021/ja01927a002.
Holloway, P.; Merriam, K.; Etsell, T. Nitric acid leaching of silver sulphide precipitates. Hydrometallurgy 2004, 74 (3-4), 213–220. DOI: 10.1016/j.hydromet.2004.05.003.


Is 1000ml of 'hot-ice' solution dangerous? So I was trying out the sodium acetate trihydrate 'hot ice' experiment, and my 300ml test worked great. So, naturally, more is better right? (cringe)

Now I have a 1000ml in a pyrex beaker cooling down, and I am belatedly wondering if this is going to generate too much heat when I set it off. I'd imagine the beaker itself is safe, I don't imagine it would do anything to the granite countertop, but it is resting on a plastic cutting board, which could conceivably melt.

How much of this can I make before I'm doing something risky?

The heat generated increases directly with the amount of sodium acetate trihydrate used, but the maximum temperature, ignoring surface effects, should not change at all. After all, if the heat in 300 ml is used to raise the temperature of 300 ml of substance, or if 300 kg is used to raise the temperature of 300 kg, it's being spread out proportionately. No matter how much you use, it won't exceed ~58 C

However, do not try this in a container that would not allow room for expansion, e.g. a glass bottle, as in crystallizing, sodium acetate trihydrate might take up more room. See warning at http://chemmovies.unl.edu/chemistry/dochem/DoChem058.html. You could safely do this on a polyethylene cutting board, though, which melts at ~100 C


Acidity order of 4-halophenols How do I theoretically predict the acidity order of the following para-halophenols?
(experimental $\mathrm{p}K\mathrm{a}$ values beside each)

parafluorophenol (9.92)
parachlorophenol (9.41)
parabromophenol (9.17)
paraiodophenol (9.30)

Experimental acidic strength order: parabromophenol > paraiodophenol  > parachlorophenol  > parafluorophenol

At first, I thought of simply using the -I effect to determine the stability of these phenol derivatives. Since -I effect is shown best by fluoro group, therefore charge delocalisation would be best in the fluoro derivative's anion. But then at the same time, there's +M effect of fluoro group too. This answer emphasises on the dominance of inductive effect of chloro group over its mesomeric effect. Is this applicable to fluorine too?

I also thought of intermolecular hydrogen bonding that would be strongest in parafluorophenol decreasing its acidity.

Quora answers here associate the higher acidity of 4-chlorophenol to the presence of vacant d- orbitals in chlorine atom, but "d-orbitals'" usage by a third period element is debatable.

I think that, this acidity order can be justified by considering a combined effect of +R effect, -I effect and intermolecular hydrogen bonding (where it is dominant).

For 4-fluorophenol, the +R effect of $\ce F$ is significant but weaker than -I effect. The electron pair on the $\ce{2p}$ orbitals of $\ce F$ can easily delocalise with the benzene ring because of the stronger $\ce{2p\pi -2p\pi}$ overlap, which makes $\ce {+R} $ effect significant . That's why it increases the electron density on the carbon near the $\ce{-OH}$ group to such an extent which decreases its ability to donate protons, greatly decreasing
its acidity.
Considering intermolecular hydrogen bonding also, the acidity of the compound decreases further due to strongest intermolecular hydrogen bonding between the halogens.

For 4-chlorophenol, the +R effect is negligible due to much weaker $\ce{3p\pi -2p\pi}$ overlap and thus the $\ce{-I}$ effect becomes more prominent. But $\ce {Cl}$ is electronegative enough to form strong intermolecular hydrogen bonding($\ce{HO-C6H4-Cl...H-O-C6H4-Cl...)}$. That's why, it becomes difficult to remove hydrogen atoms from $\ce{-OH}$ group, decreasing its acidity.

For 4-bromophenol and 4-iodophenol, there is no strong hydrogen bonding due to much less electronegative $\ce{Br}$ and $\ce{I}$ atoms, and also +R effect is ineffective due to very weak $\ce{4p\pi -2p\pi}$ and $\ce{5p\pi -2p\pi}$ overlap. Only -I effect is acing for these two compounds in favour of acidity. As -I effect of $\ce{Br}$ is larger than $\ce{I}$, the acidity of p-bromophenol is greater than the p-iodophenol.


Naming of perchloric acid Why is perchloric acid called so? I am aware of the rules, but why is it not called hyperchloric acid? Or is it correct to also refer to perchloric acid as hyperchloric acid?

This nomenclature comes from a long time ago. In the $19$th century, it was decided that the acidic compounds containing H, O and another atom X would be named from the name of this third atom X.
If only one molecule of such an acid was known, the name would be made by adding the ending -ic. Example: Only one acid is known with a Carbon atom, and it is $\ce{H2CO3}$, which is called carbonic acid.
If two acids were known differing in the number of oxygen atoms, the most oxidized would follow the preceding rule. In the least oxidized, the ending would be changed into -ous. Example : In the $19$th century, the two following acids were known containing a nitrogen atom. Well  !$~\ce{HNO3}$ was called nitric acid and $\ce{HNO2}$ nitrous acid.
If more than two acids were known differing in the number of oxygen atoms, the chemists would copy the case of the chlorine atom, where four acids were known : $\ce{HClO, HClO2, HClO3, HClO4}$. In this case, the two acids situated in the middle of the list are named according to the previous method : $\ce{HClO2}$ is chlorous acid, and $\ce{HClO3}$ is chloric acid. For the two extreme cases, it was decided to add prefixes, hypo- for the lowest number, and per- for the highest number. As a consequence, $\ce{HClO}$ became hypochlorous acid, and $\ce{HClO4}$ was called perchloric acid.
Today, the same regulations is applied, but the criterion of the number of oxygen atoms has been replaced by the oxidation number. The suffix -ous and -ic must be used if the corresponding acids have a difference of $+2$ in their oxidation numbers.
It should also be mentioned that the previous nomenclature is valid for acids having only one X atom. If an acid have two X atoms, the prefix di- should be added, and tri- for three X atoms. For example, $\ce{H2S2O5}$ is called disulfurous acid, and $\ce{H2Cr2O7}$ is dichromic acid. There are also other prefixes like ortho- and meta- to distinguish between acids having one X and the same oxidation number, but a difference in the number of H atoms.