Isolating calcium acetate from water with electrolysis That's a long page, so you can optionally jump to the "Thunderstorm it" part, but a bit of context is always neat :)

Context

Say I'm cooking an egg in vinegar. The egg shell is made of calcium carbonate:

$$\ce{CaCO3 + CH3COOH -> H2O + CO2 + Ca(CH3COO)2}$$

After eating the egg, is left in the cookware: a bit of calcium acetate, lots of water, acetic acid and other unidentified stuff.

It'd be neat to isolate that calcium acetate, as it could be of use sometime. I've seen that we can grow cool looking crystals out of it, maybe I could try that! Anyway chemistry is cool, so let's do it :)

But there will also be loads of water (egg cooking). I could store it all, but that wouldn't be very practical...

Getting rid of the water

I see three ways to get rid of that water.

Boil until water evaporates

Energy-hungry process. I'm a kind of environmental-crazy dude, so that's not an option.

Filtering

While boring, that could be an efficient option, if at all possible. It seems calcium acetate is soluble in water. Is there a way to filter it out without expensive equipment, or even at all?

Electrolysis

Energy-hungry as well, but I have a spare solar panel that produces 6 or 12V and anything between 0 and 4W, so that energy comes for free (albeit quite slowly in Belgium).

Thunderstorm it!

No matter how feasible are the other options, I'm here to experiment, so I'd love to do some electrolysis :) I got time anyway!

Constraints

I have very little knowledge of chemistry (but eager to learn!), and no equipment at all, except for household items. Also, this experiment would take place in living space (probably in my room), so it should be safe to do (e.g. do not produce chlorine gas or other nasty products). And if it's not smelly, it's even better :)

How I would do it

I'll need:

container: a drinking glass should suffice
anode
cathode
some electrolyte in the water
electricity: the solar panel will provide what's needed
time: we all have a pool of that in our life

I still need the anode, cathode and electrolyte. And also, I'd love to understand the process that takes place.

In my younger age, I used copper wire as both anode and cathode, table salt as electrolyte, and it would produce tiny bubbles on one end of the wire, and go green on the other end (do I remember properly?).

In this experiment though, the context is a little different:

I need to "dissolve" a lot of water, so just a copper wire won't be enough
It seems that using NaCl as electrolyte produces chlorine gas instead of oxygen, that's pretty bad

Choosing the electrolyte

Since I cooked the egg in vinegar, I'll probably have some acetic acid left in the water. Would that be a good enough electrolyte? Alternatively, I have baking soda readily available, but that would produce other salts when reacting with the remaining acetic acid...

Choosing the anode and cathode

I'm lost here. First and foremost, I remember my copper wire going green when doing electrolysis. Does that mean I'll "consume" my anode/cathode while doing the electrolysis? That sounds pretty expensive. Also, I'll have to separate calcium acetate from copper oxyde afterwards...

Questions

is the above reflection correct at all?
what's the best choice for electrolyte and anode/cathode materials
what's the exact reaction taking place?
any other way to isolate that calcium acetate?

How to get rid of water to collect calcium acetate?

Evaporate it.

How to do it in an environmentally responsible way?

Electricity from the outlet cost you money and probably used some
combination of coal, natural, gas, and or nuclear to produce it.
Bunsen burner, but it likely uses a fuel that was taken from the ground, made from a crop that that competes against food production for land.
Solar cells to produce electricity. It uses renewable energy but
is relatively expensive and electricity is a high value energy
compared to solar thermal. Using commercial grade solar cells only
converts about 17% of the energy in sunlight into electricity that is then converted into heat to evaporate the water.
Solar thermal is cheap and easy to make with household items like an aluminum or soup can and black spray paint. Further more, it has the potential to be more than twice as efficient as solar cells at providing energy you need to evaporate water.

Electrolysis:

If you insist on electrolysis for experimental reasons, calcium acetate is a salt of acetic acid and calcium from the calcium carbonate in the egg shells. As a salt, like all salts, calcium acetate will dissolve in water forming ions that will conduct electricity. So, there is no need for any electrolyte other than water and calcium acetate.

Furthermore, your end products will be something other than calcium acetate and would depend in part on your anode and cathode, temperature, and applied voltage.


Mechanisms in kinetic reactions In my book is given the mechanism of $\ce{HBr}$ forming and I understand every step of it. We did the calculations of the rate of each reaction, which is easy depending on the sign (-,+) of compounds formed or spent, and knowing those rates we were finally able to find the rate of $\ce{HBr}$ which is well defined by the empirical equation of Bodenstein and Lind. If you know all those reactions, its not hard to find the equations you need, but my questions are :

Is there any way we can know which exactly are those reactions (mechanism)  for other examples?

How would we know if any reaction is a branched one ?

As far as I know, the short answer is no; in general, it is not possible to know the reaction mechanism a priori.

In general, the approach is:

Postulate a reaction mechanism/pathway
Use this to calculate the relevant reaction rate(s)
Compare these rate expressions with the results of experiments

If your proposed reaction mechanism matches the empirical results, it is likely (but you can never be sure!) that the mechanism is correct. If not, a new mechanism should be postulated and tested.

That being said, in many examples it may be possible to make an educated guess of a plausible reaction mechanism - based on your own experience or the experience of others (i.e. from articles etc.) - and then investigate whether the proposed mechanism matches the empirical facts of relevant experiments. Often, the proposed mechanism is more or less correct.

In addition, modern advances in computational chemistry may give useful information regarding likely reaction pathways based on the structure and interaction of the involved molecules. To my knowledge, it is not yet possible to use this in a standalone fashion to figure out the detailed mechanism, but it may help out in terms of making a better educated guess.


Bonding, multiplicity and quantum chemistry for FeP2 I’m trying to do some quantum chemical calculations for the linear molecule $\ce{FeP2}$ in the gas phase as well as its crystal (orthorhombic symmetry like here). I am lead to believe that in both cases iron has an oxidation state of +6, and phosphorous has an oxidation state of −3. My questions are:

Are such oxidation states a reasonable assumption for this molecule?
Would the molecule normally be a singlet (multiplicity of one)?
What quantum chemistry methods would be appropriate for this system?

Admittedly it is vacuum phase, but are you certain about your oxidation state assignments?  "nearly temperature independent susceptibility of -(8.8±0.7)centerdot10-6SI (at room temperature)"

doi:10.1016/j.jpcs.2013.04.015
https://sciborg.uwaterloo.ca/~lfnazar/publications/Electrochem_Solid-State_Lett_6_8_A162-A165_2003.pdf
doi:10.1021/ja301173q
http://www.staff.uni-bayreuth.de/~bt220235/refs/HPR_29_235.pdf
doi:10.1088/0031-8949/4/3/010 (given)


Homemade spectrophotometer to exemplify Beer-Lamberts law I have built a 12-band spectrophotometer, and i'm currently keen on proving Beer-Lamberts law through a small practical example.

The solution that I'm going to test with my spectrophotometer, is orange juice that I thin with water. I naturally know I0 and I1, I.e. the absorption, and the path length that the light has to travel. Meaning that I have 2 unknowns, respectively the concentration and the molar absorptivity.

I am aware that I can figure out the molar absorptivity by doing multiple assays at different concentrations, plot them into a coordinate system and find the slope between the coordinates, which will be equal to epsilon*pathlength.

Unfortunately, I don't know how I find the molarity concentration of the orange juice that I intend investigate. I do however have the weight and the amount. Will that be sufficient to calculate the concentration molarity?

The molar absorptivity is a proportionality constant. You do not know what is the compound responsible by the color. So, this constant will not be in L/mol.cm but will be in terms of the concentration units. For example, if you use the mass to express the concentration (g/L) the constant will be express in this unit L/g.cm. With this system, you can estimate concentrations and so on. To known the molar absorptivity you must know the concentration in mol/L and it is not possible with this experiment.


Two-electron Integrals over Gaussian Plane Waves Is there an efficient method to compute the two-electron integrals over the basis set represented by a product of a Gaussian function $g(r)$ and plane wave:

$\psi(r)=g(r)e^{ikr}$

where $\lambda=1/k$ is approximately one fifth of the width of gaussians.

I'm not sure if anyone has done it, but the closest I can think of is the GPW (mixed Gaussian Plane Wave) method. I think that involves a mapping between Gaussian and PW basis. It's DFT, but just look for the Coulomb integrals.

General idea:

Lippert, Gerald, et al. "A hybrid Gaussian and plane wave density functional scheme." Molecular Physics 92.3 (1997): 477-488.

The implementation can be found here:

VandeVondele, Joost, et al. "Quickstep: Fast and accurate density functional calculations using a mixed Gaussian and plane waves approach." Computer Physics Communications 167.2 (2005): 103-128.


A sample of tap water is found to be 0.025 M in Ca2+. If 105 mg of Na2SO4 is added to 100.0 mL of the tap water, will any CaSO4 precipitate? Given:

Ksp of CaSO4 is 7.10*10^-5
Molar mass of Na2SO4 is 142.04g/mol

My Attempt:

Step 1 - figure out molar solubility of CaSO4

√7.10*10^-5 = 8.43*10^-3M

Step 2 - convert Na2SO4 from mass to moles

105mg = 0.105g
0.105/142.04 = 7.39*10^-4 mol

Step 3 - figure out moles of CaSO4 produced

Since we need 1 mol of Ca2+ & 1 mol of Na2SO4 to form 1 mol of CaSO4
And we have less Na2SO4 than Ca2+

CaSO4 produced = 7.39*10^-4 mol

Step 4 - figure out molarity of CaSO4

7.39*10^-4mol / 100ml = 7.39*10^-3 / L = 7.39*10^-3M

Step 5 - figure out if any CaSO4 precipitates

8.43*10^-3M > 7.39*10^-3M
Therefore NO CaSO4 precipitates

The answer in the book is a Yes, can anyone please explain to me where I have gone wrong?
Thanks!

$K_{sp} =[Ca^{2+}][SO_4^{2-}] = 7.10 \times 10^{-5} M^2$

$0.025M[SO_4^{2-}]  = 7.10 \times 10^{-5} M^2$

$[SO_4^{2-}] = 0.00284M$ (this is the maximum soluble concentration of $SO_4^{2-}$ in the presence of $0.025M Ca^{2+}$)

However, 0.000739 moles of $SO_4^{2-}$ are added to 100 mL of water, corresponding to a concentration of $0.00739M SO_4^{2-}$.

Precipitation occurs because 0.00739M is greater than 0.00284M.


Why does this compound have two benzene rings? If we look at carbons 1 and 2, they have 5 electrons. 3 are directly involved in bonding, and 2 are involved in delocalisation to create the 2 benzene rings. This would mean that they had 5 valence electrons, which isn't possible, and thus there should be only one benzene ring.

Chemists are already accustomed to this type of formula, even though it is not absolutely correct. More exact would be drawing all bonds with double solid/dashed bonds to express that the bond order is between 1 and 2, approximately 1.5, but is almost never done

Rarely can be seen formula with the two circles merged (and maybe dashed also), like

But the electrons are in fact delocalized across all bonds of the polycycle structure in "electron cloud" that looks somewhat like the infinity sign ∞ in this case (naphthalene) above and below the planar molecule. What matters is that the count of π (those from the "second bond" of the double bond) electrons is 4n+2 for n is some natural number (here it is n=2). It is a special feature called aromaticity, which has something to do with advanced topic of quantum chemistry; and it greatly stabilizes the molecule. Any of these formulas is unable to describe reality.


How much water is required to dilute glacial acetic acid to given pH? I have 100% acetic acid (glacial) with a pH of 2.4 and I want to know how much DM water is required to dilute the acetic acid and obtain a solution with a pH of 4.3.
As per observation 0.0001 M is equal to pH 4.3. So how can I implement this calculation using the Nernst equation?

The question is flawed; the $\mathrm{pH}$ of 100 % glacial acetic acid is not $\mathrm{pH=2.4}$. (Note that 100 % glacial acetic acid contains no water; therefore you cannot apply the usual equations for dilute aqueous solutions to calculate the $\mathrm{pH}$ of 100 % glacial acetic acid.)

Also note, since acid dissociation is not a redox reaction, you cannot apply the Nernst equation here.

The quantity $\mathrm{pH}$ is actually defined in terms of the activity of $\ce{H+}$ ions; however, for simplicity’s sake, we may use the old concept of $\mathrm{pH}$ with the definition in terms of the concentration of $\ce{H+}$ ions
$$\mathrm{pH}=-\log_{10}[\ce{H+}]$$
or
$$[\ce{H+}]=10^{-\mathrm{pH}}$$

The equilibrium of acid dissociation is usually written as:
$$\ce{HA <=> A- + H+}$$
where $\ce{HA}$ is a generic acid that dissociates into the conjugate base $\ce{A-}$.
The equilibrium constant $K_\mathrm a$ for this reaction is defined as
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{A-}]}{[\ce{HA}]}$$
Usually, values are given for the logarithmic constant $\mathrm pK_\mathrm a$, which is defined as
$$\mathrm pK_\mathrm a=-\log_{10}K_\mathrm a$$
or
$$K_\mathrm a=10^{-\mathrm pK_\mathrm a}$$

The corresponding equilibrium for acetic acid is
$$\ce{CH3COOH <=> CH3COO- + H+}$$
The value of the logarithmic constant for acetic acid at 25 °C is
$$\mathrm pK_\mathrm a=4.756$$
Thus, the value for the acid dissociation constant is
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]}{[\ce{CH3COOH}]}=10^{-\mathrm pK_\mathrm a}=10^{-4.756}\approx1.75\times10^{-5}$$

The chemical equation shows that, one $\ce{CH3COO-}$ ion and one $\ce{H+}$ ion are produced for every $\ce{CH3COOH}$ molecule that dissociates. This can be represented by subtracting a value $x$ from the original acetic acid concentration $[\ce{CH3COO-}]_0$, and adding this value $x$ to the concentrations of $\ce{CH3COO-}$ and $\ce{H+}$, i.e. when neglecting the $\ce{H+}$ ions from the autoprotolysis of water:

$$\begin{align}
[\ce{CH3COOH}]&=[\ce{CH3COOH}]_0-x\\[6pt]
[\ce{CH3COO-}]&=x\\[6pt]
[\ce{H+}]&=x
\end{align}$$

Thus, the equation for the acid dissociation equilibrium can be rewritten and solved for the original acetic acid concentration $[\ce{CH3COOH}]_0$ as follows:

$$\begin{align}
K_\mathrm a&=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\[6pt]
&=\frac{x\cdot x}{[\ce{CH3COOH}]_0-x}\\[6pt]
&=\frac{x^2}{[\ce{CH3COOH}]_0-x}\\[6pt]
K_\mathrm a\cdot\left([\ce{CH3COOH}]_0-x\right)&=x^2\\[6pt]
[\ce{CH3COOH}]_0-x&=\frac{x^2}{K_\mathrm a}\\[6pt]
[\ce{CH3COOH}]_0&=\frac{x^2}{K_\mathrm a}+x
\end{align}$$

Since $[\ce{H+}]=x$, $[\ce{H+}]=10^{-\mathrm{pH}}$, and $K_\mathrm a=10^{-\mathrm pK_\mathrm a}$, the required acetic acid concentration $[\ce{CH3COOH}]_0$ can be given in terms of the desired $\mathrm{pH}$:

$$\begin{align}
[\ce{CH3COOH}]_0&=\frac{x^2}{K_\mathrm a}+x\\[6pt]
&=\frac{[\ce{H+}]^2}{K_\mathrm a}+[\ce{H+}]\\[6pt]
&=\frac{\left(10^{-\mathrm{pH}}\right)^2}{K_\mathrm a}+10^{-\mathrm{pH}}\\[6pt]
&=\frac{\left(10^{-\mathrm{pH}}\right)^2}{10^{-\mathrm pK_\mathrm a}}+10^{-\mathrm{pH}}\\[6pt]
&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}
\end{align}$$

For example, for the given acetic acid solution with $\mathrm{pH}=2.4$:

$$\begin{align}
[\ce{CH3COOH}]_0&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&=10^{4.756-2\times2.4}+10^{-2.4}\\[6pt]
&=0.907630545\\[6pt]
&\approx0.9
\end{align}$$

and for the desired acetic acid solution with $\mathrm{pH}=4.3$:

$$\begin{align}
[\ce{CH3COOH}]_0&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&=10^{4.756-2\times4.3}+10^{-4.3}\\[6pt]
&=0.000193338\\[6pt]
&\approx0.0002
\end{align}$$

Finally, since
$$c_1\cdot V_1=c_2\cdot V_2$$
you may use the concentrations to calculate how much the given acetic acid solution with $\mathrm{pH}=2.4$ should be diluted in order to obtain the desired acetic acid solution with $\mathrm{pH}=4.3$.


Identifying Donation of an Electron Pair Which one of the following reactions does not involve the donation of an electron pair?

A) $\ce{H+ + CH3NH2 -> CH3NH3+}$

B) $\ce{AlCl3 + Cl- -> AlCl4-}$

C) $\ce{CH3Cl + CN- -> CH3CN + Cl-}$

D) $\ce{Cl2 + 2I- -> 2Cl- + I2}$

The answer is D, but I don't know why.

The equation could be split into two half equations:

$\ce{Cl2 +2e- -> 2Cl-}$

$\ce{2I- -> I2 + 2e-}$

But surely this shows the iodide ion is donating a pair of electrons to the chlorine atom?

I'm sure really what approach I should take for this question.

D donates 2 electrons, not an "electron pair." There is a subtle but important difference here because D is a reaction that takes place via the transfer of single electrons (though in the end, 2 are transferred in total).

So, A, B, and C are examples of Lewis acid-base neutralizations, but D is not. D is classical redox reaction.


Is there any electronic component to water conductivity? Answers to Decrease in temperature of a aqueous salt solution decreases conductivity indicate that the electrical conductivity of salt solutions arises from the mobility of ionic species and therefore the temperature dependence of conductivity is related to viscosity.

Question: Is there any measured or predicted electronic component to water conductivity as well, where the charge carriers are electrons rather than ions?

This could take place via charge exchange (migration of bound electrons) or even perhaps via solvated electrons, or some other mechanism.

In this case I'm interested in both pure water, and salt solutions.

I agree with the commenters that electrical conduction is very unlikely, but it's worth going through some possible mechanisms:

actual solvated electrons: As others have noted, free electrons would be expected to react rapidly with protons, even in a basic solution, so this changes quickly to a scenario of sequential electron transfer between protons, so let's do that next.
Sequential electron transfer between $\ce{H.}$ and $\ce{H+}$: Let's assume the solution is strongly acidic, so protons are abundant, and a proton gets reduced at the cathode to a hydrogen atom radical. Based on the bond dissociation energies, abstraction of $\ce{H.}$ from water to form $\ce{H2}$ and $\ce{HO.}$ is slightly unfavorable, the hydrogen radical would be preferred over the hydroxyl radical. (The H-H bond formed has a BDE of ~ $\pu{105 kcal/mol}$, while the $\ce{H-OH}$ bond broken has a BDE of ~$\pu{120 kcal/mol}$.) The problem is the rate of quenching by reaction of two hydrogen atom radicals to form $\ce{H2}$ (which is water electrolysis produces hydrogen gas). I couldn't find a rate constant for that, but there is a published rate constant for recombination of hydroxyl radicals in water that is around $\pu{10^10 M-1 s-1}$. As you might expect, that's essentially diffusion limited, so the rate constant of hydrogen atom recombination is going to be at least as high. If we optimistically assume that transfer of the electron from $\ce{H.}$ to $\ce{H+}$ has a comparable rate constant, you would still have to have a very low concentration of radical and very short path to travel in order for an electron to make it from a cathode to an anode, but it doesn't seem theoretically impossible.
The third possibility would be sequential transfer of electrons from $\ce{HO-}$ to $\ce{HO.}$. In a strongly basic solution, this also seems like a theoretical possibility given a very short path and a very low concentration of radical, assuming there are no other molecules in solution that can quench the radical.

I'm not suggesting that either of these theoretical possibilities actually ever occurs, just that these are the mechanisms that seem most likely to me.


Radiofrequency transmitter in an NMR experiment: Is there an involvement of (electromagnetic) radio wave? A student in Chemistry StackExchange asked the effect of radio waves on matter, which led to an interesting set of arguments by some users. It was pointed out that in a proton nuclear magnetic resonance (or any other nuclei for that matter), there is no absorption of the photon corresponding to the radio waves. The following reference was cited NMR signal reception: Virtual photons and coherent spontaneous emission which says (in the abstract)

  In portions of the magnetic resonance community, there is a misunderstanding of the process of nuclear magnetic resonance (NMR) signal generation and reception, and even in accepted texts, it is frequently described in terms of absorption and emission of radio waves, or radiation, by a two‐level quantum system. While this explanation can be refuted, for those who do understand that the NMR free induction decay signal is easily explained by Faraday's law of induction, reconciling the presence of an induced electromotive force with an apparent absence of transitions between nuclear energy levels causes conceptual problems.

I had never heard of emission of (electromagnetic radio waves) by the precessing nuclei in an NMR experiment, but a Google search of NMR + "absorption of radio waves" shows the concept of radio wave absorption is very common.

This leads to my main question: What does the RF transmitter do in the NMR experiment and what is practically happening in the RF coil?

(a) My understanding was that the RF transmitter, emits (electromagnetic) radio waves. If the RF coil is being energized with AC current of MHz frequencies, isn't the coil emitting electromagnetic radiation in MHz range? The magnetic field component of the radio waves interacts with the nuclei magnetic dipole.

(b) Another user stated "The "transmitter" powers the pulsed radio frequency B1 field. It never decouples from the probe coil to become electromagnetic radiation, i.e. actual photons. " Although this word "decouple" is more of a play of words, can an isolated oscillating B exist without a corresponding oscillating electric field?

I thought to learn from physical chemists and engineers who will have a better picture of the RF transmitters used in the NMR than us analytical chemists. My key question is related the Rf transmitter, not FID.

Thanks.

I finally asked an NMR spectroscopist who was the editor of the Journal of Magnetic Resonance and knows the inside out of NMR spectrometers. He sent a scanned copy of Hoult's paper from 1989. The title explains it all "The magnetic resonance myth of radio waves
." Surprisingly very low citations (15). It is amazing how much nonsense we have been fed with radio waves absorption. Sometimes innocent student questions on these forums can lead to introspection.

Here is the article Radio Waves Myth in NMR. Karl will like it.

The conclusion is that there is no involvement of radio waves, neither in absorption nor in emission (of course). The emission process is easier to understand but the transmission process seems to have confused even the experienced spectroscopists and more MRI people.

This is from another paper of Hoult. "The Origins and Present Status of the Radio Wave Controversy in NMR"

  The origins, history, and present status of the controversy
  surrounding a quantum description of the NMR signal as being due to
  radio waves are traced. With the Principle of Relativity and Coulomb’s
  Law as formal starting points and the minimum of mathematics needed
  for understanding, the derivation of a classical electromagnetic
  theory of signal reception is first given. The agreement between that
  classical theory and a recent NMR experiment is then presented,
  leading to proof that, except for the highest field imaging
  experiments, there is no significant contribution of radio waves to
  the signal.


Is NaHCO₃ a carbonate? I know that $\ce{NaHCO3}$ is sodium bicarbonate, but According to Wikipedia:

  In chemistry, a carbonate is a salt of carbonic acid, characterized by the presence of the carbonate ion, $\ce{CO3^{2-}}$.

$\ce{NaHCO3}$ also has the $\ce{CO3^{2-}}$ group, but is it a carbonate?

That is, are bicarbonates a subset of carbonates?

Carbonates are salts and esters of carbonic acid $\ce{H2CO3}$.

In generic nomenclature you may therefore consider a hydrogen carbonate also a subset of all carbonates. They are sometimes also referred to as primary carbonates.

Another considerable point is, that the structural subunit $\ce{O=C(-O-)2}$is still conserved.

In a stricter sense a hydrogencarbonate is only all forms involving $\ce{HCO3-}$, so it appears to be not a generic carbonate, but a class of its own. This refers to its one $\ce{O-H }$ bond and therefore it being a different structure.

However, another point of view should always consider aqueous solution equilibrium.
\begin{aligned}
\ce{HCO3- + H2O &<=> H3+O + \color{\red}{CO3^{2-}}}\\
\ce{HCO3- + H2O &<=> H2CO3 + {}^{-}OH}
\end{aligned}

You can see that carbonate ions will be present to a certain extend in solution. The chemistry will therefore also to some extend resemble the chemistry of the carbonate.

Chemical properties of bicarbonate compound might differ from the pure carbonates - incorporation of water, solubility, ... .

On another page, in organic chemistry the compound class carbonates refers to structures of the type $\ce{\color{\red}{R}-{CO3}-\color{\navy}{R'}}$, where $\ce{\color{\red}{R},\color{\navy}{R'}$\in$ Alkyl, Aryl}$ - hence esters.

To sum it all up, since a more stringent definition of carbonate is not present to me, I would always consider a bicarbonate a carbonate.


Thermodynamic stability of meta-xylene over ortho- and para-isomers When talking about the example of the alkylation of toluene by chloromethane in the presence of $\ce{AlCl3}$, Hepworth, Waring and Waring (2002) mentioned that:

  At room temperature, a mixture of 1,2-dimethylbenzene and 1,4-dimethylbenzene results, but at 80 °C the product is mainly 1,3-dimethylbenzene. In fact, heating either of the 1,2- or 1,4-isomers in the presence of $\ce {AlCl_3}$/$\ce {HCl}$ results in rearrangement to the more stable 1,3-dimethylbenzene.

This leads me to question: Why would the 1,3-isomer be more thermodynamically stable than the 1,2- and 1,4-isomers? Clearly, there must be an activation energy barrier to overcome as heat is required for the conversion. The methyl group being an ortho/para-director would lead to the formation of 1,2- and 1,4- isomers. Thus, wouldn't these isomers also be the more thermodynamically stable isomers?

Reference

Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

The thermodynamic stability of m-xylene over o-xylene or p-xylene can be deduced by considering the hyperconjugative effect. More specifically, resonance forms with a positive charge on methyl-substituted carbons are more important than resonance forms with a negative charge on them, as the methyl group stabilises the positively charged carbon by hyperconjugation.

In the o- or p- isomers, the hyperconjugation of one methyl group produces negative charge at ortho and para position where another methyl group resides. This is highly destabilising. But, in the m-isomer, hyperconjugation does not destabilise the structure and both methyl groups' hyperconjugation effects act in a cooperative way with each other.


Covalent bond deformation When do covalent bonds of a molecule $M$ extend or compress while no bonds in $M$ are broken? I can conceive of some possibilities but I don't know how common they are:

Temperature changes
Inter-molecular forces around $M$ which may stretch or compress $M$ in various directions
The transition from $M$ into a different conformation isomer (perhaps some stretching and twisting is required to overcome the energy barrier for a large change in conformation?)
Perhaps bonds can store energy like springs?

How likely are the above to happen, and are there other situations which induce bond deformation? Could I also have some references or topics to search on related topics? As far as I know, covalent bonds are relatively rigid and don't deform much.

Edit: one answer. Upon some literature review, it seems that rotation about $\sigma$-bonds and molecular inversion involve temporary bond deformations because the intermediate states (eclipsed conformations for the former, and the planar/linear form for the latter) involve a change in potential energy. After the transformation, the bond lengths revert to the original.

There are two distinct phenomena, which are somewhat related.

Dynamical effect of molecular vibration

Bonds do vibrate like springs, and as you said they can store energy by changing its vibrational state. Like a string, the more energy stored in the vibrational motion, the larger the magnitude of such vibration.  Different from a spring, such vibrations are associated with discrete quantum numbers, and even at absolute zero temperature, such vibration still possess a zero-point energy and can never stop.

Non-dynamical effect that change the statistical average of bond length

Molecular vibration often (but not always) occurs at an extremely short time frame, and the vibration itself is often (again not always) of a very small magnitude, so when we refer to bond length we often mean the average bond length over time.
    Normally a molecule stays near the bond length where the energy is the lowest.  There can be a few reasons for a the average bond length to change while keeping the connectivity of the molecule intact, with some examples:

Each conformational isomers will have different bond lengths.  Usually this is a small number.  Examples of large bond length change between conformational isomers with the same bond connectivity are those with Jahn-Teller effect.
By changing the vibrational state of the molecule, the average bond length can be different.  Usually, because of anharmonicity, the higher the vibrational state, the higher the average bond length.  Temperature change, pressure change, collision or photon excitation (especially infrared light) are the most common causes of a change in vibrational state.
Through inter-molecular interactions or external forces, the potential energy surface of the molecule can be changed so that the energy minimum happens at a different distance.  In this case, the bond length will change, but the energy stored in the bond will not change.  This can be achieved through external electronic field, or intermolecular forces such as electrostatic force from an ion, hydrogen bonds, charge transfer, electron donator/acceptor interaction, etc.


What exactly is lattice energy? I was going through my chemistry textbook (Chemistry, 10th Ed. by Raymond Chang) when I encountered this explanation of lattice energy.

9.3 Lattice Energy of Ionic Compounds
We can predict which elements are likely to form ionic compounds based on ionization energy and electron affinity, but how do we evaluate the stability of on ionic compound? Ionization energy and electron affinity are defined for processes occurring in the gas phase, but at 1 atm and 25 °C all ionic compounds are solids. The solid state is a very different environment because each cation in a solid is surrounded by a specific number of anions, and vice versa. Thus, the overall stability of a solid ionic compound depends on the interactions of all these ions and not merely on the interaction of a single cation with a single anion. A quantitative measure of the stability of any ionic solid is its lattice energy, defined as the energy required to completely separate one mole of a solid ionic compound into gaseous ions (see Section 6.7).

In this text, it seems like the writer is implying that for an ion pair in the gaseous state, ionisation energy (IE) and electron affinity (EA) can be used to measure it's stability. But even an ion pair in the gaseous state is held by an electrostatic force that gives it more stability, so more energy will be needed to separate the ion pair and that energy  would be more than IE and EA. What does the writer mean, and what is the correct definition of lattice energy?

In summary: IE and EA are just partial terms in the formation enthalpy of solid ionic compounds, that includes also LE and atomization enthalpy of elements. Their values alone can give a hint, but do not provide the final word.

Notice that the text says nothing explicit about ion pairs.
IE + EA relate to free gaseous unbound ions. Like:
$$\ce{M(g) -> M+(g) + e-(g)}\tag{IE}$$
resp.
$$\ce{X(g) + e-(g) -> X-(g)}\tag{EA}$$
The lattice energy (LE) is the energy released by forming solid ionic compounds from free ions:
$$\ce{ M+(g) + X-(g) -> MX(s)}\tag{LE}$$
Gaseous ion pairs have already released a part of the lattice energy.
Note that for ion pairs, the pairing energy may be smaller than the sum of IE+EA, especially for big ions, so forming pairs may not be favoured.


Unsaturation of hydrocarbons and derivatives What is the definition of unsaturation?
Do we consider it via baeyers test/ bromine test or
do we need to have atleast one degree of unsaturation?
And if that is so then is benzene an unsaturated hydrocarbon?...cause in benzene it just has the degree of unsaturation. And if we go by that defination...even ketones should be unsaturated

In your question, you are talking about two different things (although they are related). First one is the definition of unsaturation (in organic chemistry). Well, Dictionary.com gives fairly okay definition:

  (An organic compound) having a double or triple bond and capable of taking on elements or groups by direct chemical combination without the liberation of other elements or compounds, as ethylene, $\ce{CH2=CH2}$; undersaturated.

However, this definition would mostly apply to non-aromatic compounds. Since theyaromatic compounds are well stabilized by resonance, they would seldomely undergo addition reaction with reactive reagents like $\ce{Br2}$ to give saturated compounds (even though, theoretically, aromatics are unsaturated compounds). Thus, I'd call the definition of unsaturation of organic compound is the one with one or more $\pi$-bonds (e.g., acetone has one $\pi$-bond between $\ce{C-O}$). The number of $\pi$-bonds is the measure of unsaturation (remember, a triple bond is a combination of two $\pi$-bonds).

The second thing in your question is degree of unsaturation. That is the most frequent one used in structure elucidation. The degree of unsaturation of given molecule (or Index of $\ce{H2}$ Deficiency) would only give you the sum of double bonds, triple bonds, and/or rings present within the molecule. For example, if the degree of unsaturation of given compound is 1 (e.g., $\ce{C6H12}$), that means the compound is an alkene (with one double bond; e.g., hexene) or a monocyclic alkane (e.g., cyclohexane).

If you look at the saturated linear alkanes, their molecular forrmula is $\ce{C_nH_{2n+2}}$. Take a $\ce{H2}$ away from the molecule, which gives you $\ce{C_nH_{2n}}$ (like $\ce{C6H12}$, one degree of unsaturation), and so on. Thus, benzene ($\ce{C6H6}$) lacks four $\ce{H2}$ molecules (eight hydrogen atoms; $\ce{C6H14-C6H6}$), and its degree of unsaturation is 4 (3 $\pi$-bonds and a ring).


How to predict which complex is formed and which are soluble? How does one come to know (in this table) which metal cations form ammonia complexes and which ones form hydroxy complexes on reaction with $\ce{NH4OH}$
Reference:

J.D Lee - Concise Inorganic Chemistry by Sudarshan Guha

Please be mindful, in my opinion, that your source material may not be of the highest quality.
For example, NH4OH is more accurately represented as aqueous ammonia (that is, NH3(aq), see, for example, this source's comment).
Second, Al(OH)3 does not form a complex with ammonia or, as your reference claims, is 'partially soluble' with aqueous ammonia to any significant extent, and if at all, likely due to the presence of OH-. See, for example, this ebook, Basic Analytical Chemistry
 discussing various metal compounds interactions with aqueous ammonia.
Third, a more accurate depiction of the action of ammonia on Cu(OH)2, for example, is the formation of a complex of the form, $\ce{[Cu(NH3)_x(H2O)y](+)}$, where x + y = 6, with also a possible charge of 2+ for cupric, where the concentration of ammonia influences the value of 'x' (supplying guidance on your question as to which possible complex may form).
Lastly, high strength ammonia is needed to react with some of the listed metal hydroxides and is characteristic of the particular metal. See my suggested source, that may address your question on which salts are (or, are not) soluble.


Which chemical is responsible for making marker ink permanent? As asked in the question and also, a bonus question: which property of a chemical would make it 'permanent' in terms of markers?

The "permanence" is just a reference to it being waterproof.

  In general, the ink comprises a main carrier solvent, a glyceride, a pyrrolidone, a resin and a colorant.

This is mostly the result of solubility. There are other markers sold on the market with labels such as "buff proof" (slang for removal proof), that are really just pens that use a different carrier solvent. With those, typically organic solvents would be the correct solvent for removal.


Electrode potentials vs interfacial potential differences Between an electrode and an electrolyte there exists an interfacial potential difference.

Many authors give the electrode potential to be the electric potential of the electrode relative to the electrolyte. I have a few problems with this, and was wondering whether anyone could help.

We define the electrode potential of SHE to be 0V. It is impossible to measure the interfacial potential difference, however nevertheless one certainly exists between the reference electrode and the solution. Are we saying that we define the interfacial potential difference of the SHE to be zero, and then have electrode potentials of other couples as interfacial potential differences measured relative to this interfacial potential difference?

To my mind, this cannot be right. Don't we just set our zero reference (i.e. one end of our voltmeter) of electric potential at the electrode of the SHE, and measure electric potentials of other electrodes relative to this?

Sergio Trasatti wrote this regarding electrode potentials,

  Equation (4.7) shows that the "electrode potential", as obtained in
  practice, does not measure the electric potential drop between the
  bulk of the metal and the bulk of the solution, but it is a more
  complex quantity which includes both electrical and chemical
  contributions.

Why do so many authors conflate interfacial potential differences and electrode potentials? Or am I missing something here? Thank you!

IUPAC defines the electrode potential of an electrode to be the EMF of a cell in which the electrode in question is on the right hand side of the line representation of the cell and a SHE is on the left hand side.

As such, the electrode potential of the $M^+/M$ couple is

$E^o_{M^+/M} = \phi_{M} - \phi_{R}$

where $R$ refers to the reference electrode, in this case the SHE. This is equivalent to

$E^o_{M^+/M} = (\phi_{M}-\phi_{S}) - (\phi_{R}-\phi_{S})$

Where $\phi_{S}$ is the potential of the solution. Essentially, this means that a relative electrode potential really means

$E^o_{M^+/M} = \Delta \phi_{M^+/M} - \Delta \phi_{H^+/H_2}$

That is, the potential difference of the electrode in question and the solution relative to the potential difference between the SHE electrode and the solution.

It doesn't help the cause that electrochemists refer to potential differences when they use the word potential, however this is just a case of getting to grips with the terminology used in the field!


Identifying the D/L form of any aldose or ketose in cyclic form I understand that the compound above is  a ketose because it has a furanose ring. But, how do we know that it is a "D" form?

  Is there any general rule by which I can identify if any given aldose/ketose is in D/L form?

Your structure is a ketose because it is a hemiketal of a ketone. It is a furanose because it is a 5-membered (furan-oid) ring. Fructose, which your structure is not, is usually drawn as a furanose as in sucrose but fructose itself exists principally in the pyranose (6-membered ring) form.

How do you recognize the D-configuration? Your furanose 1a is in the D-series. Reverse your furanose to the keto form 1b by following the red bonds. Rotate about the $\ce{C4-C5}$ bond by $120^\circ$ as illustrated by the blue arrow in 1b. This rotation produces the Fischer projection 1c having the $\ce{C5}$-hydroxyl on the right. Thus, furanose 1a is of the D-series.

There is another method that can be used with qualification. At any carbon that determines D vs. L in a carbohydrate, as long as the substituents are $\ce{O, CH2OH, CH(OH)C}$ and $\ce{H}$, the D-series will be of the R-configuration. Conversely, the L-series will be of the S-configuration. In furanose 1a, the priorities are $\ce{O>C4>C6>H}$. This determination is easy using this method.

Structures 2a, 2b, and 2c are different representations of $\alpha$-D-glucose. All of the $\ce{C5}$-carbons are, necessarily, of the R-configuration and therefore in the D-series.

Structures 3a and 3b are Haworth renderings of sucrose. Both the glucose (pyranose) and fructose (furanose) moieties are of the R-configuration and in the D-series.


Relative importance of electronegativity and bond length for dipole moment magnitude In this answer to the question "Why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?", the order of the dipole moments was given as $\ce{CH3Cl > CH3F > CH3Br > CH3I}$, with an explanation that bond length is also a contributing factor; specifically, electronegativity dominates for $\ce{Br/I}$ and bond length/order dominates for $\ce{F/Cl}$.

Can more detail be given as to why there is this division between $\ce{F/Cl}$ and $\ce{Br/I}$?
In general, is there a method to determine which effect (electronegativity or bond length/order) is more important for comparing the magnitude of dipole moments?

I need help understanding, in that trend specifically, as to when the length factor dominates, and when the electronegativity factor does?

There is nothing much to understand here. All you need to understand is that this is basically an experimental observation.

We are trying to use qualitative arguments (higher bond length/more electronegativity) to explain the actually calculated values. When we observed a contradiction in our answer for fluoromethane and chloromethane, we corrected our reasoning, by saying that the bond length dominated over the electronegativity difference in this case. And that is true, because this correction is based on the calculated values only.

So, we first try to predict the order by our own qualitative arguments (logically reasoned, of course). If they work, great! If they don't, we look into the factors that we originally deemed to be minor instead. We calculate their values, and see what effect they had. Then, we justify those minor factors by saying that they dominated over the other in the particular case.

Another such interesting case is the order of the basicity of ammonia derivatives in water. It is $\ce{NEt2>NEt3>NEt1}$ (for the ethyl derivatives) and $\ce{NMe2>NMe1>NMe3}$ (for the methyl derivatives). Here, three factors are operating simultaneously! The solvation effect, the steric hindrance of alkyl groups, and the positive inductive effect of alkyl groups - all either reinforcing or hindering each other. In such edge cases, it becomes extremely tough to predict any order solely based on these factors, and hence, computated values are much preferred.

If you prefer to read the theory of those experiments (be warned it's scary), here is one such paper discussing the detailed calculation of the dipole moments of various fluoromethane derivatives.


Why is formic acid considered an acid if it is a reducing agent? My understanding was that since the acceptance of Lavoisier's revised Easter Memoir of 1778, all acids were understood to be oxidising agents.*

However, formic acid is often listed as a reducing agent (i.e. the opposite of an oxidising agent), despite being an acid.

Please can you reconcile this apparent contradiction for me?

* What is meant by "oxidising" or "oxidising agent" has, of course, changed since then, now referring to electron transfer rather than necessarily involving the element oxygen.

A molecule might be an oxidising agent under the electron transfer definition, and simultaneously be a reducing agent under the alternative definition: i.e. that reduction is a loss of oxygen or a gain in hydrogen.

For your specific question, formic acid can be deemed to be an oxidisation agent since it loses a hydrogen during protonation of another molecule. The really silly way to think about this is following deprotonation, formic acid has a higher percentage of oxygen than it did before, i.e. it possesses more oxygen than before.

The case of formic acid being a reducing agent is due to the more modern definition of electron transfer. To see the mechanism of this for clarity, see here. It's clear that the formic acid is donating an electron pair (i.e. itself is being oxidised) and is making the other molecule accept those electrons (i.e. that molecule is reduced). The formic acid is therefore a reducing agent.

So your original issue is due to two different definitions of oxidisation and reduction which still exist to this day, and which are by no means mutually exclusive!

EDIT: Of course, the case of formic acid is more simple because it isn't the acid doing the reducing, it's its conjugate base i.e. formate. So formic acid is an oxidising agent under the proton definition, yet formate is an reducing agent under the electron transfer definition.


Dissociation Energy for CO2 per molecule I'm trying to calculate the energy required to dissociate a single $\ce{CO2}$ molecule into its respective elements, and I'm given that it takes $ 373.6 \times 10^{3} \, \, \mathrm{\frac{joules}{gram}} $ where  $44.01$ grams is the mass of $6.022 \times 10^{23}$ molecules of $\ce{CO2}$.

By dimensional analysis:

$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$

$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot  \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$

or,

$170.43\,\, \mathrm{eV/molecule}$

Is this derivation and the logical assumptions made herein correct?


what is actually supercritical fluid? let’s take an example: if we have a glove-box filled with helium at 5 bars at room temperature which is far beyond its critical point, what actually it would be like ?
would it be like a normal gas ? would it be like a liquid ? or it will be like a condensed vapors ? can this supercritical fluid dissolve salts for example ? would it quench hot metals like water ?
if I put may hands inside this fluid would it feels like putting my hands inside a water ? or it feel like a normal air ?

is supercritical fluid a special types of a liquid ? if no, why then some SCFs used as a solvent ?

I searched a lot but really can’t find any clear answer explains what they ar exactly.

Note: this is not duplicate, I asking about the nature of SCFs while the "duplicated" question was asking about room temperature SCFs which is clearly different.

The name supercritical fluid is fancy but don't be confused by thinking that it this would a very odd looking state of material. Supercritical CO2 is just a state of CO2 at a certain T and P, so it looks very much like gaseous CO2 i.e., it is a transparent clear "fluid" whose density and viscosity is higher than a gas but lesser than that of the liquid. If you are able to see an interface in those transparent cells (e.g. here https://www.youtube.com/watch?v=GEr3NxsPTOA), it means that the liquid phase and the gas phases exist in the system. We are able to see the interface (boundary) when the refractive index is different from the container walls, and the surroundings. In short do not expect a new form of matter in supercritical state. It is just like thinking what does a plasma look like? It very much appears a like an ordinary flame which is extremely hot.

Response to the comment "is supercritical fluid a special types of a liquid ? if no, why then some SCFs used as a solvent ? what if a small pot of liquid heated inside a large pressure chamber to its critical point? would it expanding fill the whole chamber like a gas or stay in its container like a liquid ?"

I am afraid you are mixing several concepts here. The key question is, how does one distinguish a liquid phase and a gas phase? Supercritical fluid is not a liquid because its density and viscosity are not close to liquids, nor it has a definite boiling or melting point (a liquid does, right?). Think of supercritical fluid as a state or condition of a gas where it is above its critical temperature and pressure. That is all. We cannot say supercritical carbon dioxide has a boiling point of x Celsius or a freezing point of y Celsius.

Coming to your second part on SFC extraction: First, let us talk about gases. Can the gases act as a solvent or carrier of molecules? The answer is yes, and this is the very basis of gas chromatography. How do you detect odors? The gas molecules are carrying the odor causing molecules from a far away place to your nose i.e., gases also have some capability of "dissolving" substances. In the same vein, if I use a temperature of Tc, and a pressure Pc for carbon dioxide (or any gas), we can use it for extraction. It just turns out that supercritical carbon dioxide has higher dissolving power than gaseous carbon dioxide and certainly less than liquid carbon dioxide.

Third point: Let us do an imaginary experiment: If you have a sealed container of fixed volume containing nothing but carbon dioxide at very low temperature. Also assume a pressure gauge is also attached. You have filled in enough carbon dioxide in the chamber that the gauge shows a value of 85 atm. Now you start heating the chamber until the temperature of the chamber reaches 35 Celsius. Imagine you can see the insides of that chamber with a camera: You will see nothing because there will be no liquid in that chamber. It will look empty to your eyes. Given the fact you know that an empty glass is not empty, it is filled with gases (nitrogen, oxygen etc). In the same way, our chamber is also filled with a carbon dioxide which our eyes cannot see. We will still call this carbon dioxide as supercritical carbon dioxide because we know from our experimental settings that we are above critical temperature and pressure of carbon dioxide. You can go higher temperature to 45 Celsius, still you are in supercritical state. Thus in a phase diagram, there is a supercritical region.

Don't let the word fluid confuse you. Since we cannot call a gas above its critical temperature and pressure a typical gas, nor we can call it a liquid, it is best to call it a fluid. Hope that clarifies your confusion.


What are the possible magnetic quantum numbers of silver? What are all the possible quantum numbers ml and ms of the electrons in the outermost s orbital of silver? I'm new to this concept. Can anyone walk me through this?

Silver has electronic configuration as [Kr] 4d10 5s1 .
The s orbital ( 5s ) has only one electron. And the s orbital here has n = 5 (as it is in the fifth period) and l = 0 .
The s orbital is symmetric in space (sphere) which has only one orientation and hence has no "different" orientations  each of which is represented by a particular value of ml. So ml can take only one value and that is 0.( If this was a p orbital which could take 3 different orientations, ml would take 3 values -> -1,0,1. In general an orbital with azimuthal quantum number l, ml would take 2l+1 values of ml  ranging from -l to +l )
The quantum number ms represents the spin. It can either be +1/2 or -1/2.
Hence ,  ml =  0 amd ms = +1/2 or -1/2.
Hope this helps!


Is it possible to separate salt from food, or at least convert it into something that doesn't taste salty? A couple months ago, one of my room mates accidentally used a cup of salt to make apple crumble. She asked if there was any way to separate it, probably because I had recently reduced the acidity of a tomato sauce by adding baking soda. I said "probably not", but ever since then I have wondered if there is a way to get the salt out.
At this point, I would settle for a difficult method that requires a lab and ruins the dish, as long as it is still harmless to eat and retains most of it's nutritional value.
Here are some of the ideas I have thought of:
First, scrape off the oat layer, because it is the only hard solid and the salt wasn't mixed with it.
Seperate it by mass:

The apple crumble could be puréed and spun in a centrifuge to separate the salty water, which could be poured off and replaced with sugar water.
It could be atomized and sprayed sideways into a cold vacuum chamber, and then if the salt fell in a consistent radius around the nozzle, we could scrape up the salt, mix the rest back together and eat it.

Alter the the properties of the salt:

Is there a way to encourage the salt to form large crystals, even when it is a relatively minor component of a complicated mixture? If so, those crystals could be filtered out of puréed apple crumble with a coarse filter.

Crazy biochemistry:

Are there proteins or enzymes that could break the salt apart and bind the sodium to something oily that would float up by itself, leaving the chlorine to evaporate? I saw a documentary about bull sharks which mentioned that they have a gland that can absorb salt against the salinity gradient, so that the shark can expel it's salt and move up into fresh water.
Are there bacteria that require lots of salt to survive, and by using it, convert it into something that doesn't taste salty, and wouldn't release tons of sodium in my stomach?

Even if there is a method to extract salt from food, specially from apple crumble, it’ll probably ruin it.

The simplest way I can think of doing this is dissolving the apple crumble in water and blending it. Filter it and then boil the filtrate.

Given that every solid ingredient but salt is organic, when all the water boils off you should be left with salt and burnt carbohydrates, since NaCl boiling point is way above 1000 °C and a stove will reach 400 °C tops. Now, if you ever burnt sugar you know the caramel sticks to the pan, so if you’re lucky just add water again to separate the salt crystals from the burnt carbs and transfer it to a different recipient. Boil the water to get the salt.

You can add the burnt carbs back to the apple crumble but I don’t think it’ll taste like apple crumble again.


Density of moist air Why does moist air have lesser density than dry air? When the amount of vapor inside increases, the mass increases and hence won't the density as well increase?

The composition of dry air is about $78\%$ $\ce{N2}$, $21\%$ $\ce{O2}$ and $1\%$ $\ce{Ar}$.  The molecular weights of these compounds are:
$\ce{N2} = \pu{28 g/mol}$
$\ce{O2} = \pu{32 g/mol}$
$\ce{Ar} = \pu{40 g/mol}$

So, the average molecular weight of dry air is given as:

$\pu{(0.78 * 28 g/mol) + ( 0.21 * 32 g/mol) + ( 0.01 * 40 g/mol) g/mol = 29 g/mol}$

The molecular weight of water is only $\pu{18 g/mol}$.

So, a molecule of air being replaced by a molecule of water results in an air molecule with an average molecular weight of $\pu{29 g/mol}$ being replaced by a water molecule having a molecular weight of only $\pu{18 g/mol}$.

This scenario assumes a constant temperature and pressure when humidifying the air.


Is it possible for a reaction to have an entropy change of zero? If the question seems rather "unscientific", be kind to accept my apology. My textbook is being highly unclear about the topic and I need to know the answer to this question before I go on with my studying.

Yes, entropy change of reversible process is zero.

This will also help


Are groups other than carbonyl reduced by Clemmensen reduction? According to my textbook, Clemmensen reduction reduces only aldehydes and ketones to alkanes. But in this question:

$\ce{Zn-Hg/HCl}$ has reduced both the ketonic and nitro group. Correct answer given is D.

So, what exactly is this reduction's mechanism? Is it an exception? If not, why isn't it stated in the books alongside the carbonyl groups?

I would look at it in the following way: The oxidation state of the carbonyl carbon is +II, while the oxidation state of the nitrogen in the nitro group is +III. So nitrogen is in a higher oxidation state and it is also more electronegative. Thus, isn't it sensible that it gets reduced first before the carbonyl group? In fact, reduction of a nitro group is also possible using much milder reducing agents (e.g. only zinc dust).

In my experience, most questions of chemoselectivity can be answered by "thinking about it". However, if you would still like to have a good book for that, I would recommend Warren "Organic Synthesis: Strategy and Control" which contains a very nice treatment of chemoselectivity problems.


What would the following NGP reaction yield? Question

Attempt
I believe that in first reaction there would be simple Sn2. In the second reaction there Should be NGP. Hence i think the products should be diastereomers. But answer is (A)they are enatiomers. How?. Please help.

Reaction-1 is a simple example of an $\mathrm{S_N2}$ reaction, where the acetate group ($\ce{AcO-}$) substitutes the tosyl group ($\ce{OTs}$) forming the product (1) as follows:

But for Reaction-2, the lone pairs present on the $\ce{NMe2}$ provide anchimeric assistance in the removal of the tosyl group, leading to the formation of a bicyclic intermediate (both rings are not coplanar though). The acetate anion can now open the smaller ring in two ways and hence, lead to two different products labelled (1) and (2) as follows:

Finally,to deduce the enantiomeric relation between (1) and (2), the following transformations can be done:


Extracting iron oxide from ceramics waste In my ceramics class, we use iron oxide in our glazes, but a lot of it goes to waste. We have a bucket full of scrap glaze which contains quite a lot of iron oxide, but is also mixed in with silica, alumina, titanium dioxide, tin oxide, and other substances.

Is there any way to extract the iron oxide or other metal oxides from the scrap bucket?

Assuming the iron oxide is just mixed and has not been fused with the other oxides you could use a magnet to pull it from the other oxides.

If it is fused with the other oxides, you could smelt the iron out with carbon. This will also smelt out the tin, which mixes with iron quite well. If you re-oxidized the iron in a furnace with air the iron will selectively oxidize first leaving tin metal when done. This is likely an impractical and cost-inefficient procedure as iron oxide is incredibly cheap relative to the cost of separation.


Henry's law and O2 concentration in alveoli blood Can Henry's law be applied to find $\ce{[O2]}$ in alveoli blood? This kind of question appeared in an another forum and I've been struggling to find a definitive answer. My initial thought was 'no', because Zumdahl's Chemical Principles book says that

  Henry's law is obeyed most accurately for dilute solutions of gases that do not  dissociate or react with the solvent. $\ce{O2}$ will bind to the haemoglobin in the blood. Given this reaction in the solvent, $\ce{O2_{(g)}}$ in blood does not follow Henry's law.

However, a quick google search turned up multiple sources that use Henry's law directly to find the dissolved oxygen concentration (which is about $2\%$ of total oxygen):

http://www.umich.edu/~projbnb/cvr/O2transport.pdf
https://www.ncbi.nlm.nih.gov/books/NBK54103/

What I failed to understand is why the dissolved oxygen is linearly proportional to the partial pressure while the oxygen dissociation curve is not. Is there something I'm missing?

Henry's law states that the amount of a gas that dissolves in a liquid
is directly proportional to the partial pressure of that gas.
Oxygen has a larger partial pressure gradient to diffuse into the
bloodstream, so it's lower solubility in blood doesn't hinder it
during gas exchange.  Therefore, based on the properties of Henry's
law, both the partial pressure and solubility of the oxygen and carbon
dioxide determine how they will behave during gas exchange.

I encourage you to read the article below.
Source: Boundless. “Henry's Law.” Boundless Anatomy and Physiology. Boundless, 21 Sep. 2016. Retrieved 16 Oct. 2016


Solvent for caffeine extraction Can anybody please help me find a way to separate caffeine from coffee that uses a solvent other than DCM, ethyl acetate, or diethyl ether?

This an experiment for science class, but I don't have access to any of the above-mentioned solvents.

Can you suggest any alternatives that are easily obtainable?

Following is the list of solvents previously checked for the solubility (From reference 1 below):

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Solvent} & \textrm{Temperature, } \pu{^{\circ}C}  & \textrm{Solubility, } \pu{\%} (w/w) \\ \hline
 \textrm{Water} & 25 & 2.2  \\ \hline
 \textrm{water} & 100  & 66.7  \\ \hline
 \textrm{Ethanol} & 25 & 1.2 \\ \hline
\textrm{ Ethanol} & 60 & 4.5 \\ \hline
\textrm{Ether} & 25 & 0.3 \\ \hline
\textrm{Ethyl Acetete} & 25 & 2.5 \\ \hline
\textrm{Chloroform} & 25 & 18.0 \\ \hline
\textrm{Acetone} & 25 & 2.0 \\ \hline
\textrm{Benzene} & 25 & 1.0 \\ \hline
\end{array}
$$

My suggestion is to use ethanol in $\pu{60 ^{\circ}C}$. However, I'm not sure how ethanol would behave with the presence of tannins. For example, in a separate literature report, $\pu{1.0 mg}$ of tannin was isolated from a $\pu{g}$ of grape skins using 100% ethanol while the amount was increased to about $\pu{5.5 mg}$ when solvent was switched to 100% acetone (Ref.2).

References:

Q. V. Vuong, P. D. Roach, “Caffeine in Green Tea: Its Removal and Isolation,” Separation & Purification Reviews 2014, 43(2), 155-174 (DOI: 10.1080/15422119.2013.771127).
M. O. Downey, R. L. Hanlin, “Comparison of Ethanol and Acetone Mixtures for Extraction of Condensed Tannin from Grape Skin,” South African Journal for Enology and Viticulture 2010, 31(2), 154–159 (DOI: https://doi.org/10.21548/31-2-1412).


Abbreviations in electronic configurations I am teaching my students how to write the electronic configuration of the elements of the periodic table.
For example, following the rules on how to fill the orbitals, the electronic configuration of iron is
$$\ce{Fe}:~\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(4s)^2(3d)^6}.$$
Thus, its abbreviated form is
$$\ce{Fe}:~[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$
However, I has not been easy to find the explanation on why in any periodic table it is written as
$$[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$$
instead of
$$[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$
Same case for many other elements. Why is that?

I would recommend following the convention used by NIST which is to list the subshells in order of principle quantum number $(n),$ and in order of angular momentum quantum number $(\ell)$ for a given $n.$
See also ELECTRONIC STRUCTURE OF THE ELEMENTS (PDF).
The question refers to “rules on how to fill the orbitals”, but I don't think there are any rules, just memorization aids.
See The low-lying level structure of atomic lawrencium $(Z = 103):$ energies and absorption rates for an sample of how difficult it is to calculate the ground state configuration of an atom.  The only rule is quantum electrodynamics.
Also, the configurations are really just designations or leading configurations, with the actually ground state being a mix of configuration.  See for example the above lawrencium reference or A critical compilation of energy levels and spectral lines of neutral boron, where it explained “Configuration and
term labels have little physical meaning for highly mixed levels”. Boron ground state is $95\%$ $\mathrm{(1s)^2(2s)^2(2p)}$ and $4\%$ $\mathrm{(1s)^2(2p)^3}.$  Lawrencium is calculated to be $86\%$ $\mathrm{(7s)^2(7p)}$ and $6\%$ $\mathrm{(6d)(7s)(7p)}.$


How to calculate the equilibrium constant of acetic acid-hydroxide ion equilibrium? I am trying to calculate the equilibrium constant of the following equilibrium:

$$\ce{CH3COOH (aq) + OH- (aq) <=> CH3COO- (aq) + H2O (l)}$$

I am aware that this reaction essentially goes to completion (the position of the equilibrium lies far to the right). However, I still would like to calculate the equilibrium constant.

I know that $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ (I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$). I know that $K_\mathrm{a}(\ce{CH3COOH}) = 1.8\times10^{-5}$ and that $K_\mathrm{w} = 10^{-14}$. However, I am struggling to determine what $K_\mathrm{b}(\ce{OH-})$ is.

I learnt that $K_\mathrm{b}$ is called the "base ionization constant." However, $\ce{OH-}$ is a base, but it is already an ion! So, how can something that is already an ion, ionize? According to this webpage, $\ce{OH-}$ has a $K_\mathrm{b}$ of 1.0. However, what is the equilibrium that led to this number? Is it $\ce{OH- + H2O <=> H2O + OH-}$?

Moreover, if the $K_\mathrm{b}$ of $\ce{OH-}$ is 1.0, is $\ce{OH-}$ considered to be a weak base? Generally, strong bases have a $K_\mathrm{b}$ that is much greater than 1.0. On the other hand, if bases that contain hydroxide ions, such as sodium hydroxide and potassium hydroxide, are strong bases, shouldn't the hydroxide ion itself also be a strong base?

Firstly, to derive this equilibrium constant ($K_\text{eq}$), you don't need to bother about $K_\mathrm{b}$ of $\ce{OH-}$. There is a a far simpler way of doing it.

So from the equation, $\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}$, you can easily write, $$K_{\text{eq}} = \frac{\ce{[CH_3COO^-]}}{\ce{[CH_3COOH][OH-]}} = \frac{\ce{[CH_3COO^-]}[\ce{H^+}]}{\ce{[CH_3COOH]}} \times \frac{1}{\ce{[H^+][OH^-]}}= \frac{K_{\mathrm{a}}(\mathrm{CH_3COOH})}{K_\mathrm{w}}$$ So, here no $K_\mathrm{b}$ of $\ce{OH^-}$ arises and therefore we don't have to bother about it.

Secondly, if we want to find out $K_\mathrm{b}$ of $\ce{OH^-}$, At $\pu{25^o C}$, $\mathrm{p}K_\mathrm{w} = 14$, and from the equation of dissociation of water, $\ce{H2O(l) <=> H+(aq) + OH-(aq)}$, we can consider water as a weak acid, and $K_\mathrm{a} = \ce{[H^+][OH^-]} = K_\mathrm{w}$, so, $\mathrm{p}K_\mathrm{a}= \mathrm{p}K_\mathrm{w}$, and from the relation $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$ (note that in the relation the definitions are that $\mathrm{p}K_\mathrm{a}$ is the acid dissociation constant of a weak acid and $\mathrm{p}K_\mathrm{b}$ is the base dissociation constant of its conjugate base). The conjugate base of water is $\ce{OH-}$, so $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a} = 0$ as, both are same for water. So, $K_\mathrm{w}= 10^{-\mathrm{p}K_\mathrm{a}} = 10^0 =1.$  Now, you can see why $K_\mathrm{b}$ of $\ce{OH-}$ is 1 at $\pu{25^o C}$

Thirdly, yes, $\ce{OH^-}$ can't be said as a strong base. It's a moderately strong base. And, the concept that $\ce{NaOH}$ or $\ce{KOH}$ being a strong base should imply $\ce{OH^-}$ should be strong is a totally wrong concept. There is no logic behind it. Those are ionic compounds and completely dissociates and, those are Arrhenius bases. But to become a base, $\ce{OH^-}$ should accept a proton, and hence its Brønsted base, and there can't be a comparison as such. If you have a weak Brønsted acid, its conjugate base must be strong and vice versa. But here the cases are completely different, and there can't be any direct implication.


Will galvanic corrosion happen on a ring made of gold and bronze? I have recently got a custom-made ring from a jeweler. It’s made from a byzantine-period bronze ring enveloped in a gold bangle. The bronze ring is “glued” onto the gold ring through some kind of resin, as far as he told me.
Would bimetalism with rhe corrosion of the bronze ring happen?
Thank you!

Moist air, rich in an electrolyte (salt particles) or human contact, providing both NaCl and H+ may supply the reagents needed for galvanic corrosion, with dissimilar metals in direct contact. This usually proceeds, albeit slowly, over time.

Note, exposure to fruit juices could be especially problematic, resulting in a matter of days of continuous contact in the presence of moisture and oxygen, in some visible signs of disfiguring pit corrosion.

Nevertheless, I recommend wearing the ring and enjoying it, but perhaps placing it in fresh distilled water (remove O2 exposure) at night after washing it with mildly alkaline soap (to clean off salts and neutralize body acids). If the vessel use is made of Aluminum, a very active anodic metal, the ring should do just fine.


Arrow pushing: why does the O become positive when its lone pair becomes a double bond? In the following image:

The oxygen's top pair of electrons forms a double bond.

But in a covalent bonds, aren't the electrons shared? So won't the oxygen still have these $2$ electrons (that it turned into a double bond) in its outer shell? Leaving its charge as $0$?

So why would its charge change?

I think what is confusing me is what the negative and positive signs mean and how this relates to it's formal charge and its octet.

I know the left structure's oxygen has a formal charge of $0$, and the right hand structure's oxygen has a formal charge of $+1$.

So are plus and minus signs showing formal charge, and not the number of valence electrons?

If so, does the plus sign have nothing to do with the number of valence electrons?

Ah, the confusion that layman’s terms can cause. You are assuming sharing to mean ‘both have the same electrons as before, they just share a little bit of them’ — but that’s not really the nature of a chemical bond let alone the idea behind formal charges. Rather, once a molecule forms the electrons are delocalised across the entire molecule and the Lewis structures we draw are a mere approximation that makes things easier for our brains.

When determining things such as formal charges, you should stick exactly to the book and perform it strictly as directed. To determine a formal charge, all bonds must be formally cleaved homolytically, i.e. oxygen only gets two electrons back from the double bond (and another from the other single bond). Then count the resulting electrons and compare to the initial state. Since you will arrive at five electrons for oxygen (which is one less than six), it gets a single, formal positive charge. And indeed physically, there is less negative charge density on oxygen than there was before. (But note that not all formal charges correspond well with actual charge density.)

This is a distinct process from determining oxidation states, whereby you must cleave the bonds heterolytically with the electronegative atom taking it all.

In none of these formal processes you can say ‘but it was originally oxygen’s electrons …’ You cannot make electrons wear pink and yellow hats (not even during Winterbash) and thus you cannot distinguish them.


Lagrange Multipliers for the derivation of Maxwell-Boltzmann distribution When deriving the MB-Distribution while taking the number of quantum states $g_j$ into account, the final equation I'd have to solve would be:
$$\ln(g_j)-\ln(n_j)+\alpha+\beta\cdot E_j=0$$
Where $n_j$ is the number of particles at energy level $E_j$. The $\alpha$ and $\beta$ are the applied Lagrange Multipliers restrictions.
I have understood that adding or subtracting the Lagrangian Multipliers should not matter for the end result.
Now, when I continue with this equation, I'd eventually get:
$$n_j=e^\alpha\cdot e^{\beta\cdot E_j} \cdot g_j$$
When I want to derive the probability distribution formula of Maxwell-Boltzmann, I'd have to consider energy to be continuous. Substituting $g_j$ with the energy density formula times $\mathrm dE$, the total number of particles $N$ would then be:
$$\int_0^\infty\frac{V\cdot 2^{2.5}\pi\cdot m^{1.5}}{h^3}\cdot \sqrt E\cdot e^{\alpha}\cdot e^{\beta\cdot E}\cdot \mathrm dE=N$$
However, the result of this integration up to infinity would simply give the outcome $0$. But when you subtract the Lagrange Multipliers in the initial equation, the constants would have negative signs and the integration would give the correct formula for $N$.

But this means that it does matter whether you add or subtract the Lagrange Constants. What am I doing wrong here?

Ignoring multiplicative constants, the required integral is (in general form)

$$ \int \sqrt{x} e^{ax} \mathrm{d} x = \frac{1}{a}\sqrt{x}e^{ax}+\frac{i\sqrt{\pi}}{2a^{3/2}} \mathrm{erf}(i\sqrt{ax})$$

where $$ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}\mathrm{d}t$$

Setting $x=E$, $a=\beta$, and the integration limits as $0$ and $\infty$, gives

$$ \int_0^\infty \sqrt{E} e^{\beta E} \mathrm{d} E = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|_{0}^\infty + \frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|_{0}^\infty = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|^\infty +\frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|^\infty $$

since $\mathrm{erf}(0)=0$.
To make further progress, consider that the function $\mathrm{erf}(x)$ must have a real input $x$ (otherwise it will "blow up" to $\infty$), which implies that $\beta$ must be negative, or, explicitly, that  $\beta = -|\beta| $ (we could also argue that the second term on the r.h.s must be real and come to the same conclusion). With negative $\beta$ we have that

$$\mathrm{erf}(i\sqrt{\beta E})=\mathrm{erf}(i\sqrt{-|\beta|E})=\mathrm{erf}(-\sqrt{|\beta|E})=-\mathrm{erf}(\sqrt{|\beta|E})$$

With this condition we also see that the first term on the r.h.s. is zero and as a result that

$$ \int_0^\infty \sqrt{E} e^{-|\beta| E} \mathrm{d} E =  \frac{\sqrt{\pi}}{2|\beta|^{3/2}} $$


Enthalpy Change Of Formation Enthalpy change of formation means the heat change when 1 mol of the compound is formed from its elements under standard condition.
For example, the enthalpy change for the formation of carbon dioxide gas  is -393.5 KJ/mol

But, I'm confused as the enthalpy change of any element under the standard condition is zero, so shouldn't the enthalpy change of formation of carbon dioxide is also zero as well?

I am missing something but I just couldn't figure out what it is

$$\ce{C(graphite) + O2(g) -> CO2(g)}$$

$$\Delta H^{\varnothing} = -393.5\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$

That is, the reaction is exothermic.

Since the reactants are elements in standard state, the heat of formation for carbon dioxide is the same as the enthalpy change of the above reaction.


What happens when you mix alum with soap? I was handling alum one day and accidentally I touched toilet soap after that. They reacted with each other and formed a rubberlike sticky substance (like dried glue on palm). I am curious what was the chemical reaction and the substance which was formed. I re-tried the same thing with detergent and got similar results.

I think you’ve made a so called “aluminium soap”. It is a compound of the form $\ce{(RCOO)3Al}$ where $\ce{RCOO-}$ is a fatty acid chain. The $\ce{Al^3+}$ ions from the alum you handled would have displaced the $\ce{Na+}$ ions in soap to form the thick gel like aluminium version of soap.

This is similar to how $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions form an insoluble scum (which consists of $\ce{(RCOO)2Mg}$ or $\ce{(RCOO)2Ca}$) when soap is used in hard water.

I assume here that you’re talking about potash alum $\ce{KAl(SO4)2.12H2O}$ as @BenNorris has mentioned in his comment above.

Any other aluminium containing alum will also have the same effect. But chrome alums (containing chromium instead of aluminium) might have a different effect on soap.

These “aluminium soaps” are used as a thickening agent in oils and greases and also in napalm (as Georg has mentioned in his comment above).


What determines how much water of crystallisation a salt has? Is there any reason behind the number of water of crystallisation?

For example, $\ce{LiCl.2H2O}$ has two waters of crystallisation, and $\ce{MgCl2.8H2O}$ has 8 waters of crystallisation. Why does $\ce{LiCl}$ have only 2 waters of crystallisation and not 8?

The water of crystallization in a crystal of any ionic salt can be associated with the size of the cation, magnitude of charge on it, tendency for co-ordination and hydrogen bonds, its lattice structure and other factors involving the anion. But we cant actually theoretically determine why a particular salt has these man water of crystallization. We determine this experimentally in the labs.Then after finding this data for a set of salts we try and link them with the above mentioned factors and try to give a suitable explanation which justifies real data to the greatest possible extent. A theoretical explanation would be very complex and definitely involve way more parameters than what I mentioned above to be accurate for max cases.Even there isn't any 100% full-proof theory or mathematical formula that would allow us to calculate the water of crystallization in any random salt.


How to check which compound will show optical isomerism? So my teacher told that those compounds which have one or more chiral carbon present in them it will definitely show optical isomerism because it has an optically active stereoisomer, but there are many compounds which do not have any  chiral centre like, how will you determine which shows optical isomerism among the following.

(1) https://pasteboard.co/J73FtBo.jpg

(2) https://pasteboard.co/J73FDnl.jpg

(3) https://pasteboard.co/J73FN9N.jpg

In the first image due to the absence of any chiral carbon or steriogenic centre it does not show optical activity.
In the second image there is a plane of symmetry thus it ceases to be optically active(the plane of symmetry is perpendicular to the substituted butane ring).finally Alenes are optically active as the successive π orbitals lie on a different plane.You can definitely get the structure of alene from a reference book and the due to successive π bonds they lie on a different plane (the π orbitals).thus they are optically active.
Thus in compounds where there are no chiral centre check centre of symmetry,axis of symmetry and plane of symmetry as a CHIRAL carbon is essentially an unsymmetrical molecule.Hope you understood:).


Lewis dot structures and exceeding the octet I'm confused about the octet rule.  As far as I understand, you need to complete the valence shell of the elements by sharing electrons.

Now, carbon has 6 electrons which means it has 4 valence electrons, but because it has 6 electrons in total it exceeds the first shell with 4 electrons. The second shell has 4 orbitals with 2 electrons each, so to complete the second shell I need 8 valence electrons, but I had read somewhere that the octet rule holds true until down to the 4th period. How can that be when say, magnesium, has 12 total electrons and thus it exceeds the the second shell with 4 electrons but the third shell needs 18 valence electrons to be complete, not just 8?

Secondly, I read that if the central atom has more then 8 electrons, then that's ok too? What happens if I have a choice of exceeding the 8 electrons by forming another bond with an adjacent atom or sticking those electrons on the opposite side of the adjacent atom?

Is one strictly wrong/preferable?

For the first part of your question, you are correct that Mg has 12 electrons, which are arranged in shells in the pattern 2 - 8 - 2. But magnesium is a typical metal in that it tends to lose electrons to form a cation that then forms solids by ionic bonding. The octet rule predicts that the resulting cation will be stablest if its outermost shell contains eight electrons. And indeed the magnesium ion is $\mathrm{Mg}^{2+}$, two electrons less than the atom, with an electron configuration of 2 - 8.

What would happen if you had asked about a non-metal atom in the third period, say Cl? This has an electron configuration of 2 - 8 - 7, and tends to form a single covalent bond (e.g., to form $\mathrm{Cl}_2$ or $\mathrm{HCl}$), giving a configuration of 2 - 8 - 8 rather than 2 - 8 - 18. The reason for this is that the first eight valence electrons occupy the $s$ (2 electrons) and $p$ (6 electrons) subshells, while any further electrons in the third shell must go into the $d$ subshell. These orbitals are both higher in energy than those from the first two subshells (which is why, as you go through the periodic table, the $4s$ shell is filled before the $3d$) and extend less far from the nucleus, so that they are much less likely to overlap other atoms' orbitals to form a covalent bond. So it is often energetically preferable not to put electrons into this subshell if possible, which is why the octet rule can still apply to elements in this period.

Of course, it is possible to put electrons into this subshell, which is why it is possible for the central atom to have more than an octet in compounds like $\mathrm{SF_6}$. This leads into the second part of your question, which, as Manishearth suggested, looks like a duplicate of Can an atom have more than 8 valence electrons? If not, why is 8 the limit?.


Does urine make concrete hygroscopic? TLDR: Does urine make concrete hygroscopic enough to absorb moisture from the air so much that a spot of humidity is visible? Are uric acid crystals moisture-absorbent? (Urea apparently is according to Wikipedia)
Here is the story:
In my parents' house, there is a mystery that we were unable to solve.
During the renovation, certain spots on the cement floor appeared wet (used to be covered by glued carpet). Even worse, some spots appeared to evolve and sometimes grow.
We assumed the worst. Since this was the ground floor (there is no basement) we thought groundwater might be infiltrating the house. One particularly intriguing spot was near the hinge side of an interior door. The door is in a wall of cored brick next to an interior concrete pillar. The spot on the floor was next to this pillar.
Several investigations have been made:

A dig outside the house to check the depth at which water could be found. It was deep enough.
Checking if there is hydro-isolation on the foundation. There was.
Checking on the outside of the base of the house for humidity. It was dry.
Checking the heating pipes buried in the floor. They were nowhere near and eventually the in-floor pipes were abandoned.

All of the people who investigated (engineers, constructors) were mystified as the spot kept evolving even though we properly ventilated the area for a significant time.
Even more puzzling, the concrete pillar was dry.
Eventually, believing there might be a spring under the house (and that the pillar was ok due to the hydro-isolation on the foundation) we broke the first layers of the cement floor.
Underneath the first layer, there was a layer of extruded polystyrene thermal insulation. Underneath that, a hydro-isolation layer. And underneath that the concrete slab of the floor. All were dry.
At this point, all other people just gave up and some even suggested we call an exorcist.
However, this meant a return to one of the first theories about these spots.
In the house used to live several cats and unfortunately, there have been plenty of times when some of them peed on the carpet floor. The spot near the pillar was one of the preferred places. This was one of the reasons for completely removing the glued carpet. This theory was initially rejected because the house was ventilated for more than 2 months with the carpet removed and the cats long gone. During this time and more the spots evolved and grew. We believed that since the spot grew noticeably (we traced the edge) and since it didn't smell it could not be because of the cats and it had to be from below (of above, but the roof and the ceiling were ok so we eliminated that possibility).
Now with layers below also dry, the only possible source of humidity left was the air.
So, does (cat) urine make concrete hygroscopic enough to produce a visible wet spot?
I have searched Google and Wikipedia about the moisture absorbant properties of uric acid but I have not found anything relevant. According to Wikipedia, urea is hygroscopic and highly soluble in water. I decided to ask this here since it is about the physical properties of a chemical compound.
The closest I have found is this:
https://pubs.acs.org/doi/10.1021/ja00836a027
What makes these particles hygroscopic?
Salt + Sodium Bicarbonate releasing water in carpet

Very interesting question - and thorough. But I would like to get proof that the spot is due to moisture. #1 Does the spot grow and shrink dependent on the relative humidity at that position? #2 If the spot is misted with a water sprayer, is it sharply defined? #3 If the spot is gently warmed with a space heater, does it disappear? #4 What color is the concrete? Very light (low iron), medium gray (medium iron), or dark, or brown?
If multiple cat pees have been located there, it's possible that attempts at cleaning or deodorizing have also been applied to the carpet, and have soaked thru and into the concrete. At this point, so much could have happened to the cement compounds that you and your parents should consider yourself fortunate that the deviation from perfection is only visual and not oderiferous! I have a spot on an outside patio that is dribbled upon from an aluminum gutter, only occasionally, when it rains (the end cap on the gutter has a tiny leak). The spot is about 4 inches in diameter, and is only visible at certain times (usually high humidity, like when rain has almost completely evaporated). Without a microscopic examination, I can't see anything unusual, but I suspect that some of the $Ca(OH)_2$ in the concrete has been dissolved, or perhaps carbonated more, or both, and the water in the pores at that spot is slower to evaporate.
If the discoloration is obnoxious, you might try painting (or dusting) that area with a non-paint paste of calcium carbonate in water. It would be harmless to the concrete, but provide some lightening. But then the area might be too white...


Graph of compressibility factor vs pressure when real gas is assigned Z=1 Problem:

According to the real gas equation, $Z = 1$ for an ideal gas and $Z$ is variable for a real gas. Suppose, in order to easy our calculations, we fixed $Z=1$ for a real gas and for ideal gas, $Z$ will become variable. $Z$ vs $P$ for an ideal gas will be similar to:

Attempt:
I imagined the answer by thinking the normal graph of $Z$ vs $P$ for a real gas and transposed each point on the locus of the real gas curve on $Z=1$ line. I then imagined where the corresponding point of the ideal gas curve would land.
This gave option (a) as the correct answer, but the book says the answer is (b). I believe then the van der Waals equation must be manipulated to get the answer, but can someone tell me how? Or is the answer given in the book wrong?

For a real gas, $(P+a/V^2)(V-b) = RT$

Now, since we are considering $Z=1$ for real gas, all the real gas equations will be reversed for the ideal gas. Meaning, we had to subtract $b$ for volume correction of real gas, for correction of ideal gas we need to add $b$. Same, with pressure.

The equation then comes out to be: $(P-a/V^2)(V+b) = RT$

Now, for the graph, we can calculate the approximate values of $Z$ (relative to unity) for very low pressure, low pressure and high pressure.

Case 1: Very low pressure: $P \downarrow  \downarrow \implies V  \uparrow  \uparrow $

$(P-a/V^2) \approx P$ and $(V-b)\approx V \implies Z \approx 1$

Case 2: Low pressure $P \downarrow \implies V \uparrow$

$(V+b)\approx V$. On solving, we get $Z= 1+a/(RTV)\implies Z>1$.

Case 3: Very high pressure $P \uparrow  \uparrow \implies V^2\downarrow  \downarrow$

$(P - a/V^2) \approx P$. On solving, we get $Z= 1-Pb/(RT)\implies Z<1$.

The graph would thus be (A), because it is the only one satisfying all these conditions.


Why is dG = dH − TdS? I get that $G=H-TS$ because then: $$\begin{align}\mathrm dG&=\mathrm dH-T\,\mathrm dS-S\mathrm dT\\&=T\,\mathrm dS+V\,\mathrm dp-T\,\mathrm dS-S\,\mathrm dT\end{align}$$
Therefore, by cancelling: $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$ which is the equations for $\mathrm dG$. However, I can’t get this result from using $\mathrm dG=\mathrm dH-T\,\mathrm dS$.
$$\mathrm dH=S\,\mathrm dT+V\,\mathrm dp$$
$$\mathrm dG=\left(S\,\mathrm dT+V\,\mathrm dp\right)-T\,\mathrm dS$$
This does not give me the right equation. Where am I going wrong?

Why is dG=dH-TdS?

It isn't. That would only be true at constant temperature. In general:

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T$$


Proving RTln($a_i$) equal to non PV work done by system,where $a_i$ is activity We know that chemical potential is defined as ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$. Here $a_i$ is the activity of $i^\text{th}$ component of solution. In case of gases instead of $a_i$ it is $f_i$ (fugacity). Activity is the activity coefficient multiplied by concentration or in case of fugacity it is the product of the fugacity coefficient multiplied by partial pressure of the gas.

But now this is not the only definition of Chemical potential. Chemical potential also means the partial molar Gibbs free energy when pressure and temperature of the system are constant. We know that change in Gibbs free energy of a system at constant temperature and pressure also represents Non-PV work done by system.

How can the term $RTln(a_i)$ can be correlated to Non-PV work done by system in order  for the two definitions to be equivalent?

NEW QUESTION AFTER READING YOUR ANSWER(S)

Philipp, Ok after reading all your answers I got now modified question that still remained.First of all we know K and Kc are related but kc can be modified by adding activity coefficients and standard concentrations to give equilibrium constant that should be equal to K.This requires the activity coefficients (That we used with Kc) to always follow relation with fugacity coefficients and RT terms which is not proved in answers there and that is what actually my question is asking.Sorry but I am reluctant here to write maths but I just say refer to the second equation from last in your This Answer

Actually earlier my question has title ways of expressing concentration in activity term

We know that chemical potential is defined as
  ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$.

Be a bit careful with your wording. This equation is not part of the definition of the chemical potential but rather it is derived from it.
The derivation of this equation can be found in this answer of mine.
The definition of the chemical potential is

\begin{equation}
 \mu_{i} = \left( \frac{\partial G}{\partial n_{i}} \right)_{p,T} \ ,
\end{equation}

where $n_{i}$ is the amount of the $i^{\text{th}}$ component in the system.

  We know that Gibbs free energy of a system at constant temperature and
  pressure also represents Non-PV work done by system.

Again this is not quite correct: It is the change in Gibbs free energy of a system during some process, say a chemical reaction or a phase transition, at constant temperature and pressure that can be identified with non-PV work done by the system.

Now considering a general chemical reaction

\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}

the differential change in Gibbs free energy is given by

\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \mu_{i} \text{d} n_{i}
\end{equation}

or introducing the extend of reaction, $\text{d} \xi = \frac{\text{d} n_{i}}{\nu_{i}}$

\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \nu_{i} \mu_{i} \text{d} \xi \ .
\end{equation}

From this it follows that the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$) at constant pressure and temperature is given by

\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ .
\end{equation}

Here you can see that the non-PV work done by the system originates from the differences in the chemical potentials of the reactants.
Substituting the equation for the chemical potential

\begin{equation}
 \mu_{i} = \mu_{i}^{0} + RT \ln a_{i}
\end{equation}

the equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, leads to

\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}

where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced.
This is the relation between the non-PV work, represented by $\Delta G$, and the activity terms. Admittedly, it is not a very graphic explanation but I fear that it simply is a pretty mathematical matter.


Ratio of energy required for electrolysis vs creating water I had an idea a while back for a machine that would be totally green. Powered by a Hydrogen Fuel Cell. A big question came up: Where do I get the Hydrogen? I assumed that I could just electrolysis of water to create Oxygen (which would be released into the air) and Hydrogen (which would be used in the cell). But, then I remembered a key factor of bonds: Creating a bond gives the nuclei a more stable configuration, which releases energy. Breaking a bond requires an input of energy.

Let's say I use electrolysis to split water. This will take energy. Then I use a Hydrogen Fuel Cell to generate electricity. This process gives me the same amount of water that I just split.

Q: Which process uses/gives more energy? Would the above example result in a net gain or loss?

Theoretically, you will get the same amount of energy out of putting hydrogen and oxygen through a fuel cell as it takes to produce them by electrolysis, and you can reach this result by adding up all the enthalpies of formation involved using Hess' Law and realizing that the fuel cell is just the reverse process of electrolysis.
It is therefore impossible to have a net energy gain by this process as it would violate the conservation of energy.
It is, however possible to lose energy during this process and in the real world, energy is lost during both steps.

Electrolysis:

Takes a net input of energy to produce hydrogen and oxygen
Much of the electricity added is wasted to produce heat due to electrical resistance and overpotential losses.
Current large scale electrolysis systems have efficiencies of at best ~70% [1]
Storage of the gases is also energetically wasteful as they need to be compressed. (or perhaps the hydrogen stored as a metal hydride)

Fuel cell:

Produces a net output of energy by consuming hydrogen and oxygen.
Suffers from the same types of electrical losses as the electrolysis process, resulting in sub-100% conversion efficiency

Fuel cells are still useful devices as they can produce electricity without combustion, without the limitation of sunlight, etc., but they require a source of hydrogen.
Most hydrogen is produced from natural gas because producing it by electrolysis is so inefficient and requires energy from somewhere else.
In the future electrolysis combined with fuel cells might become useful for temporarily storing electricity for load balancing solar panels or other green power sources, and there are some projects using hydrogen generation for grid storage purposes where excess power is stored as hydrogen in underground reservoirs, but they use hydrogen combustion generators to get electricity back as fuel cell technology has not advanced sufficiently. Example in Newfoundland