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Mass Spectrometry for detection of a chemical from a distance
I want to know if mass Spectrometry to detect a chemical compound from a long distance, say, a 100 meters away from the device.
Is this possible ?
If not, then what can be used to fulfill this objective ?
Thanks in advance !

That's a no-go.

https://chemistry.stackexchange.com/questions/4156/understanding-how-mass-spectroscopy-works?rq=1

Mass spectrometry (magnetic sector or time of flight) cannot be used as the molecule needs to enter the machine, thus by definition not possible for remote detection.
The only method would be to use visible, uv or infrared spectroscopy, and  most probably via laser induced fluorescence. This is used, for example, to detect a variety of trace atmospheric species. Long path (miles) absorption methods can also be used. What method to use depends on what you want to detect, how much there is and where it is etc.

Raman spectroscopy is also pretty common for that. You can use that for example for finding explosives: http://www.tuwien.ac.at/en/news/news_detail/article/7417/

yes , I had completely forgotten Raman. Its used, for example,  to remotely and non-invasively inspect pigments in old master painting and in the combustion reactions in petrol and diesel engines while they are running.


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How to determine if an alkene is an E or Z isomer?
Is the molecule below an E or Z isomer?
From what I believe I would say E isomer because the two priority halides Cl and Br are on opposite sides of the double bond. But I'm not 100% sure. Because they are on the same side of the double bond if I look at it form another perspective.

You don't chose two priority substituents out of four. Instead, you choose a priority group on one end of the double bond, then another group on the other end.

I find that the site below has a good explanation on how to approach E/Z classification.
http://www.chemguide.co.uk/basicorg/isomerism/ez.html

@IvanNeretin so Br is the highest on one side and F is the highest on the other and as they are both the same side it makes this a Z ?

Yes, that's right.

Just adding on to Ivan's comment, you have to split the double bond "into half" such that both halves have two sets of double bonds not having "to split and remain with single bond".  After that  assign priorities.
I don't know if I explained well, but here is a better picture:

Considering each of the double-bond carbons separately, look at the two
substituents attached and rank them according to the atomic number of
the first atom in each. An atom with higher atomic number ranks higher
than an atom with lower atomic number.

So it is "Z" ? because the Br and the F are the highest relative to their sides of the C=C double bond

Precisely its Z

Firstly take one carbon and decide seniority sequence of groups on that carbon, which will be done according to CIP rule (for alkenes here) which states that:

the group with higher atomic number will be senior.
bond pair is always senior to lone pair.
the doubly bonded group will be treated as two same groups are attached to that carbon( for eg. =Cl is equivalent to two single bonded -Cl while deciding the seniority).

Now, after deciding the seniority groups on one carbon (i.e.on one side if the double bond) do the same for the other doubly bonded carbon.
Now if the higher ranked groups are on same side of the double bond then the notation is "Z" otherwise it will be "E".

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Electrochemistry: Does ion flow from one half-cell to another?
My textbook says one of the functions of a salt bridge is that "it completes the circuit by allowing ions to move from one half-cell to another". However, according to my understanding, only the ions in the salt bridge move into the half-cells.

Take the above chemical cell as an example, only the potassium ions and nitrate ions move into different half-cell respectively. The mobile ions in the two electrolyte do not move from one half-cell to another, right?

That is correct, though some diffusion of the ions must occur, albeit slowly.

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Volume change of solute addition to solvent
When making and manipulating high concentration stock solutions of sugar excipients, is there a way to predict the corresponding volume effect that occurs? I am asking in order to be able to create end solutions with exact wt/volume concentrations.
I know about partial molar volumes and some of the thermodynamics of mixing, but am wondering if there are models and/or correlations to predict the volume behavior of some of these solutions.
Thanks!

Welcome to Chemistry! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

You'll probably have to research published correlations and/or predictive models for your solvent-solute system. I am not aware of any "generally accepted" models, maybe others more specialized in this area will be able to comment. This topic comes up a lot in my estimation, I would recommend searching through this Exchange for related questions.


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Pyrotechnic "star" without oxidizer
In the design of a celebratory pyrotechnic (firework), there exists a gunpowder explosive charge at the center of the firework and pyrotechnic "stars" lining the circumference of the explosive. The "stars" are composed of an oxidizer and a fuel such as KNO3 oxidizer and charcoal as a fuel. If the oxidizer is removed from the "stars", is it reasonable, upon detonation, to assume that the now vapourized "star" will combust only with the detonation of the gunpowder explosive charge without the need for an oxidizer in the "star" composition?
See this page for a diagram and composition of materials: http://www.instructables.com/id/How-to-Make-Fireworks/

without reading details Unless you know VERY well what you are doing, messing with pyrotechnic composition is somewhere between bad and spectacularly bad ideas. At best it would not work. At worst you will suffer irreversible damage.

Changing already established procedures is one of the ways that science works. I did not ask about the safety of the modification as it is obvious that caution should be used. I asked why it would or would not work. Answer that.

If you don't add an oxidizer, the oxidizer is oxygen, and unless it's well-mixed, the explosion will generally not be as rapid. Also, I second permeakra's concerns about messing with fireworks. I have more than a decade of laboratory experience, and toying with this raises huge red flags.

From what I have read, the oxidizer in the stars only serves to burn the fuel (and create colour for the firework) once the explosive charge has been detonated. I don't think that the removal of the oxidizer from the stars will impact the explosion caused by the gunpowder significantly. I have the highest respect for safety, and I have been working with pyrotechnics for almost 3 years under the supervision of professionals. I am aware of the hazards of carelessly working with pyrotechnics.

When the interior gunpowder charge explodes, it blows the stars in all directions and ignites them.  They then burn very rapidly (perhaps explosively) giving the star effect exactly because the stars are a mixture of fuel and oxidizer.  If the stars were all fuel (carbon), they would be essentially charcoal briquettes.  In that case the center would explode, throwing (possibly) burning charcoal chunks in all directions.  This could be a truly novel approach to barbecue...

First of all. Learn the difference, burn, explode and detonate. Gunpowder can never detonate but explode. In response to your question, I would like to say that removing the oxidizer will not burn or not be included as a fuel in principle if there is no surplus of oxygen in the explosion itself. Since the stars will not be surrounded by oxygen everywhere, a large part of them will be surplus material. Think of the "fire triangle" Four need: oxygen, fuel and heat.
Do you remove any of these, no fire will burn no matter what size.
I am a retired firefighter, and for many years as a military explosive expert, Own Company for Civil Rock Blasters. Sadly no chemistry engineer.


text:
How does electrolytic conduction depends on the size of ions & their solvation?
The textbook says,

The smaller the effective size of the ion after solvation, the greater will be its conducting power."

But, on the Internet, it is explained as follows:

Radius decreases, smaller ions carrying current in small area, more associations with anions & due to this, the ions become heavy & they move with reduced speed as compared to less associated ones, & thus conductivity decreases.

The explanation is different to the what the book says.
Could someone please explain this?

Small and/or highly charged ions tend to be highly solvated in water. These solvated ions may still be smaller than the solvated ions of larger ions.
However, water molecules are not as free as the molecules of a gas. Water has a much higher viscosity, for instance. So think of water as a loose gel. A large (unsolvated, or poorly solvated) ion has an easier time slipping between gaps in this gel. A smaller, more highly solvated ion is, by definition, more tightly bound to the gel network, and moves more sluggishly, as if it has to carry more of the network with it. This approach is more explanatory than predictive (see figure).

There are lots of numbers, some agree, some disagree with any simple explanation. But a clean example is sodium vs potassium: sodium bare ion is smaller than potassium bare ion, but their hydrated ions reverse that (hydrated potassium ion is smaller than hydrated sodium ion - and check out lithium!). So potassium has a greater conductivity, for the same number of ions.

The thing can be explained as: whenever a small sized ion(particularly it's actual ionic size but not after solvation) is solvated it's effective radius increases and hence due to the increased size of the ion the space required for each ion is now lesser than what it had earlier. Hence, it can be said so conducting power changes inversely with the effective radius after solvation( as more solvated ion conducts less as compared to the less solvated).
And according to your writing both the references are correct as internet said about "conductivity and not about conducting power". So don't get confused between them.

Smaller ions get much more solvated in solvent than larger ions. Therefore with the solvent molecules by its perimeter the effective solvodynamic radius of the smaller ion increases.So now after greater solvation smaller ions get slower than the larger ions in that solvent.And hence conduction of smaller ion decreases in that solvent.

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What is (approximately) the minimum exchange current for a potentiometric measurement?
Is there a rule of thumb for what should be the minimum exchange current to detect a certain redox couple in an aqueous solution?
I would like to make a rough calculations to see whether any of my hypothesis is feasible but I have no idea about the order of magnitude an exchange current should have in water. I could find a few examples for "normal" exchange currents for pH measurements or $\ce{Fe^{3+}/Fe^{2+}}$ but no minimum.
Background: I am trying to figure out, why one can measure a negative ORP in fermentations. There has been a lot of work done on the role of the microorganisms and their exofacial sulfhydryl groups. However, the question of WHAT is actually measured/happening at the electrode has not been answered.
I want to tackle this as part of my thesis and I can think of two mechanisms:
A) Direct interaction of sulfhydryl groups with the electrode surface
B) Mediation, e.g. via Fe or Mn ions.
My concerns about A are that cells are pretty big compared to ions so I am not sure whether enough cells could ever interact with the electrode to make this scenario realistic. In the case of B I am not sure if the concentrations are high enough.
Apologies if my language is too sloppy.

Pfeiffer at al. defines sensor effective (SERC) and sensor ineffective redox couples (SIRC) in relation to the biggest net exchange current found in the system $\text{i}_{net}$.
$\text{SERC} \geq 0.01~\text{i}_{net}$
$\text{SIRC} < 0.01~\text{i}_{net}$
DOI: 10.1016/0169-7722(92)90041-C


text:
What is causing the buffer region in a weak acid - strong base titration?

For the buffer region in this, why does it stop the pH from increasing?
At the start:
$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$
Which means there will be a high amount of salt in the solution, but why is this stopping the pH from rising? I thought it would actually make it rise further as the disassociated salt it reacts with water to form $\ce{OH-}$?
What's reacting with the $\ce{OH-}$ in solution, that isn't there anyway? Seeing as, with a buffer solution, the acid is needed ($\ce{CH3COOH + OH- -> CH3COO- + H2O}$).
The acid concentration hasn't been increased?
Or is the presence of the dissasocated salt:
$\ce{CH3COONa(aq) -> CH3COO- + Na+}$
Causing the acid to reform:
$\ce{CH3COO- + H2O -> CH3COOH + OH-}$ <<< but there's more $\ce{OH-}$!!
Surely the $\ce{Na+}$ isn't going to react with the $\ce{OH-}$, as it will dissociate in the water.

If not a duplicate, it is related to The reason behind the steep rise in pH in the acid base titration curve

The linked question's answers were too long for me, so I'll attempt to summarize quickly.
The equation that tells you everything you need to know is the Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} - \log\frac{\ce{[A-]}}{\ce{[HA]}}$$
Here $\ce{HA}$ is a weak acid, and it's conjugate base is also weak. But we can imagine also a weak base with a weak conjugate acid. The magic happens when you've half neutralized a weak acid or base, creating a mixture of a weak acid and a weak base. The fraction on the right is fairly close to 1 (within an order of magnitude, let's say).
As you add more acid, the weak base can neutralize it. As you add more base, the weak acid can neutralize it. Assuming these amounts are reasonably small, the logarithm of the fraction on the right doesn't change by much, so the overall pH of the solution does not change by much. For example, changing the fraction on the right by a factor of 10 only affects the pH value by $\pm 1$.
Of course, if you add enough acid or base to neutralize (or close to neutralize) one of the buffer components, then the buffer stops working. This is also when the fraction gets very large or goes to zero and even the logarithm can't change that.

At the very start of the titration you have $\ce{CH3COOH}$ — or so you may think! Actually, you have an equilibrium of water with acetic acid as shown in equation $(1)$.
$$\ce{CH3COOH + H2O <<=> CH3COO- + H3O+}\tag{1}$$
This equilibrium is strongly shifted to the left since hydronium ($\mathrm pK_\mathrm a = 0$) is a very strong acid while acetic acid ($\mathrm pK_\mathrm a = 4.76$) is a rather weak one. What follows from here is basically equilibrium chemistry: as you add hydroxide ions, there is a great tendency for these to recombine with hydronium ions as in equation $(2)$ to form water. Thus, the equilibrium of equation $(1)$ is slowly shifted to the right as more and more hydroxide is added, formally consuming the acetic acid. Since equilibrium $(1)$ is still at equilibrium, the concentration of $\ce{H3O+}$ won’t change that much.
$$\ce{H3O+ + OH- <=>> 2 H2O}\tag{2}$$
Towards the end of the titration, the concentration of acetic acid ($\ce{CH3COOH}$) becomes very low, meaning there are less molecules to donate protons to water to uphold the equilibrium. Therefore, the $\mathrm{pH}$ value starts rising much more rapidly than previously — as you would expect from the titration of a strong acid with a strong base. The curve’s turning point is the equivalence point, right where the amount of hydroxide added matches the amount of acetic acid originally present.

Sir why concentration of H+ wouldnot change that much ? I didnt get the reason as such

And yes towards ends conc of ch3cooh is low so if eqb is not achevable what is done so that eqb is again reached or it will not reach at all again . So all the left ch3cooh will quickly dissociate to give that graph at equivalence point?


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Why do lone pairs of electrons have a higher repulsive force?
I was just wondering why lone pairs of electrons are more repulsive than bonded pairs of electrons.
I have read that lone pairs have orbitals which are shorter and more round, compared to bonded pairs. I assume, therefore, that this difference in repulsive force could be due to the increased surface area that these orbitals provide, which would cause more repulsive force to be directed towards electrons.
Of course this is just a presumption that I made, which is probably incorrect.

Bonding pairs are stabilized between two atoms. Since there's no atom on the other end of a lone pair, it spreads out more than would be if it were in a bond. Lone pairs therefore repel more because the charge density is spread out more, that is, it's bigger.

I have read that lone pairs have orbitals which are shorter and more round

This is technically true in the sense that you'd want to have higher $s$ character in a lone pair orbital. But you're comparing possible ways to hybridize orbitals for lone pairs, as opposed to comparing lone pair atomic orbitals with bonding orbitals.

Sorry for commenting on such an old answer but how does  the charge density being more spread out increase electrostatic repulsion? If anything, shouldn't the more localized charge in the bond pair offer greater repulsion?

I think the issue here is that the bonding pair is already stabilized by virtue of the bond. The only recourse available to a lone pair is to spread out.


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Spectroscopy of excited state lithium
I have used data from a question regarding the reciprocal wavelengths observed when lithium transitions from excited states to a common ground state. The lowest energy transition is from n=3 to n=2. The question goes on to say that this lowest energy transition, in high resolution, has 7 peaks, and asks for an explanation as to why there are not 8 lines. Can I have some guidance on this? I think spin-orbit coupling is relevant. I am aware that the dipole moment of an orbiting electron interacts with a magnetic field (from the orbital angular momentum of the electron?) and so some orbitals are raised/lowered slightly, those with an "l" greater than 0.

Well, do you think there could be 8 lines (in theory)? Can you enumerate the 8 transitions that might cover?

Can you derive the term symbols for the relevant electronic states? Do you know the selection rules for electronic transitions?

all doublets. S(1/2), P(1/2), P(3/2), D(5/2) and D(3/2)? Selection rule is that angular orbital quantum number (l) must be + or - 1, and total angular momentum (j) 0, + or - 1?

Yes, those are the right term symbols for the excited state. If you find the relevant term symbols for the ground state (which aren't very different except for the lack of D terms, since there's no 2d orbital) you should be able to find the 7 allowed transitions. Personally I don't know what the "8th" one is supposed to mean, though.

The energy levels are in increasing energy  3S, 3P, 3D and the final state is 2P. The P-P transition is not allowed by $\Delta L =\pm1$ selection rule. The two transitions are 3S-2P at approx 812 nm and 3D-2P at approx 610 nm. As you are asked about the lowest energy transition this must be the 3S-2P. (Li also has non-zero nuclear spin depending  on its isotope which lead to hyperfine transitions and are ignored )
With some spin-orbit coupling the (Russell-Saunders) terms symbols are for the upper level $^2S_{1/2}, $ and for the lower level $^2P_{3/2}, \, ^2P_{1/2}$. The $M_J$ levels are assumed to be split so the degeneracy is 2 for the S and 2 or 4 for the P states as given by $2J+1$.  The selection rules on $M_J$ levels are $M_J = 0,\, \pm 1$.
Thus there are four transitions possible from $^2S_{1/2}$ to $^2P_{1/2}$ starting in  $  M_J = 1/2 $ in $^2S$ to $(1/2, -1/2)$ in $^2P$ and similarly $  M_J = -1/2 \rightarrow (1/2,\, -1/2)$, making 4 transitions in all as the splitting of the S and P levels will be different. (This is the same case as the splitting of sodium $D_1$ line but 'upside down' in energy to that case).
The other transitions are from $^2S_{1/2}$ to $^2P_{3/2}$. In this case there are 6 transitions as the P splits into 4 levels. The transition between the $\pm 1/2$ make four of these (as just outlined above) and in addition there are transitions between $M_J = 3/2 \rightarrow 1/2$ and $M_J = -3/2 \rightarrow -1/2$.
This makes ten transitions in all, thus it seems I have calculated the wrong transitions (or misunderstood completely),  however, you should be able to follow the method. Note that the observed spectral lines may not correspond to transition as distinct transitions may have the same energy gaps.
I've added a sketch below, it is hand drawn but should be clear enough. The $S_{1/2} - P_{1/2}$  transitions are a subset of these as you can see.

Very informative answer-so there are not in fact 7 lines? Also, M(J), does this denote the total angular moment part of the term symbol (bottom right)

OP wrote n=3 to n=2 so I interpreted it as 2S/2P/2D initial state and 2S/2P final states. I got seven lines so I guess that was what was intended. Can attach a picture of Grotrian diagram later (I'm out right now). But I agree that it might be a bit vague.

the $M_J$ is the range (projection in field) of the bottom right total angular momentum $J=L+S$  in the term symbol $^{2S+1}L_J$. J has substates $M_J$ which are split in a magnetic or electric field and  range from $-J\cdots +J$ in unit steps,  and there are $(2J+1)$ of these in total.

This is what I was going for: https://i.stack.imgur.com/64NDv.jpg

ok, that picture makes sense, I just took the 'hight resolution' bit to mean that the levels were split into states. The question is ambiguous though.


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Calculating pH of solution of phosphoric acid and ammonia

What is the $\ce{pH}$ of the solution created when $\ce{50ml}$ of $\ce{0.2M}$ $\ce{H3PO4}$ is mixed with $\ce{50ml}$ of $\ce{0.4M}$ $\ce{NH3}$?

This problem can be approached rigorously. That's what I did, but I got this equation which can only be solved analytically:
$$[H^+] + \frac{0.2\cdot[H^+]}{K_4+{[H^+]}} = \frac{K_w}{[H^+]} + \frac{0.3\cdot K_1\cdot K_2\cdot K_3 + 0.2\cdot [H^+]\cdot K_1\cdot K_2+0.1\cdot[H^+]\cdot K_1}{[H^+]^3 + [H^+]^2 \cdot K_1 + [H^+]\cdot K_1\cdot K_2 + K_1 \cdot  K_2 \cdot  K_3 } $$
$K_1$ to $K_3$ are the $\ce{H3PO4}$ dissociation constants. $K_4$ is the $\ce{NH4+}$ dissociation constant, and to derive this, I used material balance and charge balance.
My question is, how can I solve this problem without the use of a computer, using  only a scientific calculator? Can anybody suggest other methods to do this problem?

Also they said  we can suppose that 7< pH<8 , so in proton balance : [H+] + [H2PO4-] + 2[H3PO4] = [NH3] + [OH-] + [PO4 3-] , we can exclude [H+] , [OH-] , [H3PO4] and [PO4 3-] , So we get [H2PO4-] = [NH3], how they arrived to this conclusion ?

We'd love to help you, but I can't seem to digest what your question is basically, with all that stuff around. Can you format your question to make it look neat and readable?

What is $K_{4}$?

As a first estimation figure out what are the major phosphate components you deal with , ie which protonation step is the most important. Most cases you can forget the others.

The $\mathrm{p}K_{\mathrm{a}}$ values for phosphoric acid are:
$$\begin{array}{|c|c|}
\hline
\mathrm{p}K_{\mathrm{a}1} & 2.148 \\ \hline
\mathrm{p}K_{\mathrm{a}2} & 7.198 \\ \hline
\mathrm{p}K_{\mathrm{a}3} & 12.319 \\ \hline
\end{array}$$
The $\mathrm{p}K_{\mathrm{a}}$ for ammonium is $9.25$.
The initial conditions led me to partially neutralize the phosphoric acid, giving $0.01\ \mathrm{mol}\ \ce{H2PO4-}$, $0.01\ \mathrm{mol}\ \ce{NH3}$, and $0.01\ \mathrm{mol}\ \ce{NH4+}$.
Note that dihydrogen phosphate is a stronger acid than ammonium, so the equilibrium will shift towards making hydrogen phosphate by neutralizing more ammonia:
$$\ce{NH3 + H2PO4- -> NH4+ + HPO4^{2-}}$$
We note also that the equilibrium $\mathrm{pH}$ satisfies the Henderson-Hasselbalch equation for each equilibrium:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\frac{\ce{[A-]}}{\ce{[HA]}}$$
That is:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}\ \ce{NH4+}} + \log\frac{\ce{[NH3]}}{\ce{[NH4+]}} = \mathrm{p}K_{\mathrm{a}\ \ce{H2PO4-}} + \log\frac{\ce{[HPO4^{2-}]}}{\ce{[H2PO4-]}}$$
From the initial conditions and stoichiometry, I determine:
$$\begin{array}{|c|c|}
\hline
\text{species} & \text{moles} \\ \hline \hline
\ce{NH4+} & 0.01 + x \\ \hline
\ce{NH3} & 0.01 - x \\ \hline
\ce{H2PO4-} & 0.01 - x \\ \hline
\ce{HPO4^{2-}} & x \\ \hline
\end{array}$$
Solving this yields: $x=0.00878\ \mathrm{mol}$, which plugging back in gives a $\mathrm{pH}$ of $8.08$.
That seems to be about the right ballpark. The problem is equivalent to adding a weak acid to an ammonia/ammonium buffer at pH 9.25.


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Conversion of Hessian from internal coordinates to cartesian coordinates
I have performed a CASPT2 frequency calculation using the MOLCAS8 program package. I need the full Cartesian Hessian matrix as input for a subsequent program package.
MOLCAS8 performs CASPT2 frequency calculations numerically. The program MCKINLEY is called for this purpose, however, it is now my opinion that all numerical second derivative calculations in MOLCAS are handled not by MCKINLEY but by the barely documented module supermac. Anyway, the Hessian is fully printed in internal coordinates once the calculation is finished. For a system of 10 atoms the internal Hessian is of size 24 x 24, which suggests non-redundant internal coordinates (MOLCAS calls them "primitive internal coordinates").
Does anyone know of a software capable of transforming a Hessian matrix from internal coordinates to Cartesian coordinates? I am aware that this will probably be complicated as internal coordinates are not as unambiguous and straight-forward as Cartesians. However, my initial search efforts did not yield a single software package capable of transforming a calculated Hessian in internal coordinates to Cartesians. Any help would be appreciated!
Alternatively, does someone know a quantum-chemical program package using which I might be able to obtain a full Cartesian hessian at CASPT2 level?
I don't know if this helps, but MOLCAS also seems to print out the force constants of each atom in Cartesian coordinates. Furthermore a .molden file is produced where the normal modes are included. The full Hessian does not seem to be written to this Molden file though.

Not sure if this is helpful, but check out the "Cartesian" keyword for the module Slapaf.  https://jellby.altervista.org/manual/users.guide/programs/slapaf.html?highlight=slapaf

24 elements/dimension follows the $3N-6$ rule for the number of normal modes for a nonlinear molecule, and doesn't necessarily imply one coordinate system over another.

In the Geometric [1]  package, there is code to convert an internal coordinate Hessian into Cartesian coordinates
    def calcHessCart(self, xyz, gradq, hessq):
        """
        Compute the Cartesian Hessian given internal coordinate gradient and Hessian.
        Returns the answer in a.u.
        """
        Bmat = self.wilsonB(xyz)
        Hx = np.einsum('ai,ab,bj->ij', Bmat, hessq, Bmat, optimize=True)
        Hx += np.einsum('ji,j->i', Bmat, gradq, optimize=True)

        return Hx

The code is also reproduced in my repository[2] but I've never attempted to use it
References:

https://raw.githubusercontent.com/leeping/geomeTRIC/master/geometric/internal.py
https://github.com/ZimmermanGroup/pyGSM/tree/master/pygsm/coordinate_systems

BAGEL (http://nubakery.org/ and https://github.com/nubakery/bagel) has CASPT2 analytical gradients and can run frequency computations.
It's currently being prepared for 1.0 release, so it might still have bugs. On the other hand, it is freely available under the GPL.
As for converting the Hessian, it might be useful to look at the INTDER program, which is (among others?) part of the PSI3 suite of programs available here: https://sourceforge.net/projects/psicode/files/psi/3.4.0/psi-3.4.0.tar.gz/download - PSI3 is under the GPL as well.
The manual page for INTDER is here: https://manned.org/intder and it says:
"Force field transformations between Cartesian and general internal
coordinate spaces up to fourth order, including nonstationary
reference structures.  Both forward and reverse transformations
are possible; hence, by using intermediate Cartesian coordinates,
force fields can be transferred among different representations in
the internal space."
Whether or not it will be easy to coerce the MOLCAS Hessian into a format INTDER likes and then transform it as you want I cannot say as I don't have any direct experience with INTDER - it is typically run as a step in a chain of commands automatically generated by the PSI3 driver routines.

Maybe openbabel can do it.
I found this website which can convert z-matrix's to Cartesian coordinates:
http://www.shodor.org/chemviz/zmatrices/babel.html.
I know that the MolMod python library can make the conversion if you want to delve into that. And another python library that can do it chemcoord
I hope this helps.

Thanks, but all the libraries you named can not convert a hessian matrix from internal to cartesian coordinates. MolMod is able to calculate a hessian matrix in internal coordinates as I see it, but it is not able to convert a pre-existing hessian in internal coordinates to cartesian coordinates. Can you verify this?


text:
Can I pour a solution into another vessel with no air bubbles?
I have been trying to create a method to create a homogeneous agar solution.
What are appropriate techniques to ensure homogeneous solidification of an agar-based aqueous solution?
The new solution consists of distilled boiling water, $\pu{20 g/L}$ agar, $\pu{10 g/L}$ $\ce{NaCl}$ and $\pu{1 g/L}$ $\ce{CuSO4}$.
An issue with the agar is that it only dissolves between 90-100C which makes it difficult to dissolve using a hot plate or other methods as it also needs to be homogeneous.
I've made progress speaking to both the microbiology and chemistry departments of my university. The current idea to make the solution homogeneous is to heat it in an autoclave set at 121C. This will also sterilize the solution making the phantom last longer. Unfortunately we have no 3L beakers to do it all in one go. They have plenty of 2L and some 5L beakers, but nothing that is 3L.
A new idea is to heat a round bottomed flask in the autoclave and then transfer the semi-cooled solution to a perspex container that will have a base on it. (flask and container shown in the photo below)

The problem is how do we ensure the solution is still relatively homogeneous and won't be full of air bubbles? Are there any simple ways to solve this?

do you have access to an ultrasound bath? you could hold it in there for a time, if the chemical can withstand the stress it could help removing air bubbles..

I discussed this with one of the physics technical officers, microbiology does have the equipment but didn't think it was a great idea. They pushed the autoclave method instead. I'm not sure why, I might ask them again.

If you are making a phantom for radiation physics / health physics then why not use 1 M nitric acid with the radioisotope in it or some other aqueous liquid from which precipitation is unlikely. Another option would be to look for an organic soluble form of the radionuclide if it is very long lived. Then mix it with styrene and pour the solution into a glass version of the phantom. Use redistilled styrene. Now seal it up and put it away for months for the styrene to slowly convert into polystyrene. This could give you a very long lasting phantom.

There are two aspects to consider for the intended "air bubble free" (or at least, "air bubble less") transfer from your container where the liquid is prepared into the container where the liquid is supposed to solidify:

Your first container shown is a round bottom flask.  Initially, an increase in inclination of the flask will increase the flow of liquid out of the flask.  If however the inclination is too high, the air that aims to refill the inner volume of the container, initially occupied by the liquid, will no longer continously enter the flask.  Instead, there will be an alternation of sorting liquid and entering air; which may in turn yield air bubbles in the sorting liquid.  Depending on the viscosity of the liquid, the air bubbles may remain in the liquid for quite some time.  From the perspective to lower the potential air intake by the liquid during the transfer, a wide mouth bottle, or even a beaker were better suited than a round bottom flask.
The distance between the first and the receiving container.  For your purpose, it is not good to pour the liquid directly into the receiver. To lower splash and splatter in the transfer by decantation, an easy aide is using a glass rod, like shown here:

(picture source)
In your instance, it may be more convenient to use a slab of glass / a pane of glass that is just beneath the mouth of the first container, and bridges all the distance to the receiving container; as schematically drawn below:

Here, intentionally the slab of glass is partially immersed in the liquid already present in the receiving container.  If done in a coordinated team of two, one holding and moving gently the slab of glass, the second gently decantating the liquid onto this "chute", you might cast larger volumes into trough-shaped containers / fish tanks, too.  Once solidified, you would cut-out the block / shape you need.  Overall, it reminds me a little to ballistic gelatin...

Thank you for the detailed answer! I know the flask isn't ideal, but the perspex will deform inside the autoclave. The chemistry department were surprised they didn't have a 3L beaker too.  I show the above method to my supervisor and see what he thinks. For something that seems so trivial to make at the start, it has quickly become difficult! We're trying to make a human tissue analogue. The issue with gelatin is it's animal based and breaks down very quickly. Agar will still break down, but should still hopefully last a couple of weeks if sufficiently sterilized and kept sealed.

I worked with some gelatin-based phantoms (for RF) one summer during my MPhys.  We didn't have trouble with them breaking down.  I assume we added some sort of preservative, but can't remember (it was 15 years ago). We made a half head on its side and the brain and eye were gelatin (the skull wasn't, which led to intersting effects when lit from the brain side)

... I can't find anything relevant published by that group around the time I was there.  I hoped I might be able to find a citation for the recipe. (The group was based Bristol University, led by Alan W Preece)

@ChrisH Thanks for that! That was just what my supervisor said to me anyway, I think agar is a bit cheaper too. How long did you work with it for? Our biggest issue once this is made is trying to keep the phantom consistent for a couple of weeks and minimizing dehydration.

@JamesBlackwell It was a summer job for about 8 weeks essentially as a technician.  My task was building it (and the stand and sensors). The rest of the group were expecting to be able to use it for some (unspecified but not small) time after I left.  I think the solution to dehydration was simply a nice flat layer of clingfilm on top once it had set; we embedded sensors in the gelatin  so could probably leave the clingfilm on, but I can't remember.

@ChrisH Ok that sounds good, I'll see if I can try out a range of ideas and see which ones work out

If you can accept discarding a little solution at the beginning and end of the flow, a narrow siphon tube touching the receiving container will deliver a reliably air free flow.
With an open valve near the exit of the tube, submerge most of the tube in the solution and close the valve.  You can now withdraw the submerged tube leaving only enough submerged to drain the source container.  With exit end below the level of the solution open the valve and the siphon will begin to flow.  Discard the first bit which is likely to have a few bubbles, close the valve, then begin to fill your receiving container while keeping the tube in contact with the container.  Once both ends of the siphon are submerged it is near impossible to introduce air as long as the rate of flow is low enough to avoid agitating the surface.  As the source container approaches empty you risk picking up some air, with a reasonable length tube you can easily close the valve before the first air bubble exits the siphon.

Hello  and welcome to StackExchange! Have a look at the [help] to learn how this site works. Good luck!


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What is produced when citric acid (powder form) is burned, is it safe?
What is produced when citric acid (powder form) is burned?
Are the fumes harmful or safe?

If you've burnt it properly: C + CO2 +H2O ;)

@paracetamol Did you consider only citric acid in a non-oxidizing atmosphere?

@paracetamol Yes only citric acid in a normal room atmosphere.

Wikipedia says:

It decomposes with loss of carbon dioxide above about 175 °C.

As this paper suggets, at 175 °C, citric acid is converted into aconitic acid. Further heating result in formation of methyl maleic anhydride(III).

Aconitic acid may also decompose to acetone.

If you use some negative inhibitors like sulfuric acid, citric acid may undergo partial decomposition and form carbon monoxide, water and acetone-dicarboxylic acid.
$$\ce{(CH2COOH)2C(OH)COOH -> CO + H2O + (CH2COO)2CO}$$
Acetone-dicarboxylic may readily decompose into acetone and carbon dioxide.
This paper suggest how the decomposition products of citric acid are determined.

The thermal decomposition of citric acid is studied by
  thermogravimetry (TG) derivative thermogravimetry (DTG) and
  differential scanning calorimetry (DSC) techniques under various
  conditions. The thermal decomposition reactions of the acid are
  influenced by its particle size and the different heating rates used.
Data obtained from the thermal curves were employed for mechanistic
  and kinetic studies using computer programs and manual computations.
  The decomposition follows the Jander diffusion mechanism and is of
  first order with an activation energy of ̃200 k J mol−1.

Pure carbon will be produced(@paracetamol) if you fully decompose any other organic substrate left including acetone at very, very high temperature.

As @paracetamol mentioned, if you burn it completely and properly, you should get nothing other than $\ce{CO2}$ and $\ce{H2O}$.
But what if you don't? Then it's a bit complicated. You can't say exactly what will form. There are chances of carbon monoxide formation, which is quite toxic. By binding to the heme complex iron in a hemoglobin, it renders it incapable of carrying oxygen to the body, causing dizziness, nausea, and likely death. Consider that.
Second thing is, there may be unburnt carbon particles that are emitted if you don't burn it well. Particulate matter such as these is highly dangerous to the lungs.
Thirdly, considering that atmospheric combustion is a highly complicated process, you can't really say what you could have produced from that oxidation. It may be toxic and may have crazy properties that even the safety equipment you have cannot protect you. That's something you should consider as well.


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How is Standard Gibbs free energy measured?
I have seen hole theory of thermodynamics but almost nothing about the experiments that measure $\Delta G^\circ$. I know how to measure it by electrodes, when the reaction is a redox one. But what with other reactions?
What are the experimental ways to determine $\Delta G^\circ$? Some examples would be helpful

You're probably asking about $\Delta G$, which is actually quite different from $G$.

https://chemistry.stackexchange.com/questions/4752/how-is-hydration-free-energy-of-a-solute-measured-experimentally https://chemistry.stackexchange.com/questions/42066/is-there-a-way-to-experimentally-measure-entropy?rq=1

Generally you measure the equilibrium constant $K$ and then convert it to $\Delta G^\circ$ (note that is change in standard free energy)
$ \Delta G^\circ = -RT \ln K $
In the case of redox reactions, this can be done with electrodes as you mention. For other reactions you usually use some spectroscopic technique (NMR, IRC, UV, etc) that can distinguish reactants from products, but calorimetry can also be used in certain cases.
Here is a video describing one example using UV.

If we pick a different standard state (e.g. 2 bar instead of 1 bar) will the change in standard free energy change?

If the number of particles change in the reaction, then yes. Otherwise, no.


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Phosphate extraction from sewage
I read that phosphate reserves will exhaust  in coming years wiki, why don't we extract it from sewage , is it really that difficult to extract phosphate ?

Could you clarify what you mean by phosphate going extinct? Phosphates are alive and well and actually a pollution issue to my knowledge.

It is really difficult to process large volumes of sewage to try and get small quantities of basically anything out of it. Just like there is a fortune of gold dissolved in the ocean, but nobody is going to get rich collecting it.

The idea has been proposed by several journal articles and Wikipedia suggests that it is being done already. In "Phosphate fertilizer from sewage sludge ash (SSA)", Franz proposes a method

whereby more than 90% of phosphorus can be extracted to make an adequate phosphate fertilizer

Also, the journal "Phosphorus recycling in sewage treatment plants with
biological phosphorus removal" by Heinzman lists three processes:

Krepo Process
Seaborne Process
Aqua Reci Process

In it Heinzman also mentions that

up to now no phosphorus recycling with a defrayal of costs is possible. The future importance of phosphorus recycling will depend on the market price for raw phosphate, the recycling costs and, furthermore, on the general political framework

So, it appears that it is cheaper to mine it than to recycle it. This may change in the future of course.

Phosphate recovery from sewage sludge ash is already past pilot scale. If i recall correctly, a full size plant is currently under construction in northern europe - it was quite a massive EU funded research collaboration. In comparison to franzs method you mentioned, this is a thermochemical process recovering elementary phosphorous (http://www.recophos.org/).


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Displacement reactions of Halide ions

I understand that the reactivity decreases as you go down group 7. In displacement reactions the more reactive halide replaces the least reactive halide, but in this question i do not understand the use of hexane in determining their reactivity series. I also do not understand why 5 test tubes give a purple layer as iodine should not react with sodium bromide.

For future reference, please crop your screen shot to just the question of interest or use LaTeX tables.

Halogens belong to group 17 not 7

Hexane is an organic solvent. It has two important properties for the purposes of this question: it is less dense than water, and it will dissolve the molecular halogens.
I think you've already correctly identified that $\ce{Br2}$ and $\ce{Cl2}$ will react with $\ce{NaI(aq)}$ to produce $\ce{I2}$. The part you didn't consider is that $\ce{I2}$ is present in every mixture in the bottom row of that table because no displacement will occur. Therefore, the hexane solution will be purple.


text:
In the bicarbonate ion, why can the hydrogen not bond to the carbon?
I've been studying electrophiles and nucleophiles recently, including how the carbonate ion ($\ce{CO3^2-}$) is stabilised through resonance. While drawing the model for bicarbonate, however, I suddenly asked myself why a different design of this ion would not be possible. The design I am proposing features the hydrogen atom attached to the carbon, with single bonds for all of the three oxygens. This is shown below:

Now, I know that this is not what we usually see, which is:

The question is, why is my own design not feasible? If it is, is it not energetically favourable due to the numerous charges that would have to be created?

First of all I appreciate your effort to think out of the box.
Now, as you are thinking of a new structure for the bicarbonate ion, you just have to compare the stability factors between the two different structures and you'll get your answer!
Here let's compare the structures:
Stability factors for the conventional structure

Octet saturation in case of each of the element.
Resonance stability (moreover, two equivalent canonical structures give more stability) in the molecule.
The negative charge resides on the more electronegative $\ce{O}$ atom.

Unstability factors for your proposed structure

Octet unsaturated in case of the positively charged $\ce{O}$ atom.
No resonance stabilisation in the molecule.
(your point) Energetically disfavoured due to numerous charge accumulation.
The positive charge resides on the more electronegative $\ce{O}$ atom.

That's why your proposed structure isn't the actual structure of bicarbonate ion.


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Coulomb force and potential energy in water vs vacuum
If you have 2 ions of equal but opposite charge, will the force between them be larger in a vacuum and smaller in water? Would this be because the relative permittivity of water is greater than 1 (around 80)? Also, what does this mean for the interaction energy? Would an interaction be larger in vacuum? I am trying to figure out whether a reaction is more likely to take place in the two different media between the two ions...

Yes, much larger in vacuum. The interaction is proportional to the reciprocal of the dielectric constant (relative permittivity) $1/\epsilon$ so is less in water. Think of it as an attenuation of the electric field around an ion, the larger $\epsilon$ is the more the field is attenuated and is true whatever the charges on the ions are making them attractive or repulsive.  The force at distance r between charges $q_1, q_2$ is
$$F=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$
in SI units; $\epsilon_0 $ is the permittivity of free space. The charges are $q=ze$ where e is the charge on the electron and z the ionic valency, $\pm 1, 2 \cdots $ etc.
The interaction energy is
$$U=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r} \qquad \mathrm{Joule}$$
The electric field around charge $q_1$ is
$$E=\frac{q_1}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$
in V/m. The force acting on  a second charge $q_2$ is $F=Eq_2$
Its worth plugging in some numbers for the energy for different $\epsilon$ and comparing this to thermal energy $k_BT$.

thank you-very informative. Just to check, if I was to calculate a force in water, would I use  80 in place of ϵ in the "ϵ0ϵ " term? Also, if then I had 2 +1 charges, and I wanted them to react, would it be best in water based on your argument of weakening the field?

yes use $80 $ for water as its the relative permittivity no units, $\epsilon_0 = 8.854\cdot 10^{-12} \pu{F,m^{-1}}$. Yes also if charges are similar high dielectric is best. The rate const is $\ln(k)=\ln(k_0)-U/k_BT$ where $k_0$ is rate const at infinite dielectric const.  You might also want to consider increasing the ionic strength of the solution, look up 'primary kinetic salt effect'.

@porphyrin It could also be worth mentioning the reason for dielectric constant. Water molecules are polar and orient themselves to become antiparallel with the field.

@Pritt , both polar and non-polar, i.e all  molecules have a dielectric constant. In non-polar ones this is $\epsilon-1 \sim n\alpha$ where n is the number density and $\alpha$ the polarisability. In polar ones additionally there is a term $\epsilon-1 \sim n\mu^2/(k_BT)$ where $\mu$ is the dipole moment. Some partial  alignment takes place but this is opposed by thermal motion via the $k_BT$ term.  It is too strong a statement to say that molecules become aligned, they are only ever partially so in a liquid .


text:
Bomb calorimeter enthalpies of straight chained alcohols
Can anyone suggest a website which contains the literature values (using a bomb calorimeter) of the enthalpies of n-alcohols?
I have been trying to find it but I am unable to locate bomb calorimeter values (I do not want the bond dissociation energies.

Here are enthalpy of combustion values for the first 8 n-alcohols.

You can also get a rough estimate of it by using bond enthalpy calculations.

I searched 'combustion enthalpies of straight chain alcohols' in Google Scholar and found a couple of links to papers that look relevant, see below. Unfortunately one looks like you have to pay for it, but if you're associated with an institution they may give you free access. The other involves cyclic alcohols, obviously not straight chain but maybe helpful nonetheless.
https://link.springer.com/article/10.1007%2FBF02885455?LI=true
http://actachemscand.dk/pdf/acta_vol_16_p0046-0052.pdf
I would imagine the molar enthalpies of combustion would increase fairly linearly as chain length increases. Each extra carbon in the chain means an extra molecule of CO2 is produced, and the strong C=O bonds formed are the major contribution to the exothermicity of the reaction. Steric or torsional effects in the larger molecules may lead to a slightly smaller than expected increase per carbon.
Hope this helps.

To add to Tom's answer:
According to "Correlation of Heats of Combustion with Empirical Formulas for Fatty Alcohols" by Freedman et al:
$$\begin{array}{|c|c|}
 \text{Alcohol Name}& \hline \text{Heat of Combustion (kg-cal/mole)} \\ \hline
\text{Decyl} & 1582    \\ \hline
\text{Lauryl} & 1899\\ \hline
\text{Myristy} & 2202 \\ \hline
\text{Palmityl} & 2512\\ \hline
\text{Stearyl} & 2826 \\ \hline
\text{Arachidyl} & 3138  \\ \hline
\text{Behenyl} & 3453 \\ \hline
\end{array}$$

Samples were 99% pure
All alcohols are n-alcohols
Data is based on a minimum of 3 replications
The bomb calorimeter used was Parr  Adiabatic Oxygen Bomb Calorimeter, Model 1241

The data is out there, but you may have to compile and proccess it depending on how many values you need.


text:
Why are organic ortho acids unstable?
Ortho acids like orthoformic acid (methanetriol, $\ce{HC(OH)3}$), orthocarbonic acid ($\ce{C(OH)4}$), and orthoacetic acid (ethane-1,1,1-triol, $\ce{H3C\bond{-}C(OH)3}$) are known to be unstable and decompose spontaneously.
For example, orthoformic acid spontaneously decomposes into formic acid and water, while orthocarbonic acid spontaneously decomposes into carbonic acid and water. As a matter of fact, these two acids have never been isolated.
$$\ce{HC(OH)3 -> HCOOH + H2O}$$
$$\ce{C(OH)4 -> H2CO3 + H2O}$$
Interestingly enough, orthoesters are stable and can be isolated, unlike ortho acids, just that orthoesters are easily hydrolyzed, especially with a mild acid present. For instance, trimethyl orthoformate, the simplest orthoester, can be synthesized. Why is that ortho acids are very unstable, while orthoesters are stable? Is the presence of hydroxyl groups of ortho acids a direct contributor of their instability?

This question https://chemistry.stackexchange.com/questions/10722/what-makes-c-o-more-stable-that-coh₂ should answer your question.

Aside from the factors discussed in the answer to the above-referenced question, carboxylic acids have an energy advantage over their orthoacid forms due to delocalization of the electrons on the OH group into the carbonyl group.


text:
Does the oxidation for all the aliphatic monoaldehydes to carboxylic acids have the same value of electrode potential?
I have been wondering on this for quite a while. Consider aliphatic monoaldehydes. If I oxidize them, we get corresponding monocarboxylic acids. We can use certain oxidizing agents to perform the reaction.
For all the possible aliphatic aldehydes, it involves breaking a $\ce{C-H}$, and the formation of the $\ce{C-O}$ and a $\ce{O-H}$. I did calculations and got the enthalpy change for such a process to be $\pu{412kJ/mol}$.
My main motive is to find the electrode potential for the process. Unfortunately, I cannot determine the entropy here, although I don't think the side chain would matter much.
Here's my question. Does the oxidation for all the aliphatic monoaldehydes to carboxylic acids have the same value of electrode potential? In all the cases, it's the same bonds broken and same bonds formed, so I cannot see why the electrode potential should drastically vary for different aldehydes.
My apologies, I was unable to find a $\Delta\text{G}$ value table for such processes.

My answer to your main question is no.
Firstly, you are simplifying the process of oxidation of the monoaldehyde. The mechanisms for oxidation depending on the oxidising agent used, although they are generally the same (Refer to link). By considering only these 3 bond formation/cleavage events, you are not looking at this process in totality.
Secondly, the side chain does matter as it would affect the bond energies of the bonds that are broken and formed. Different side chains may give the aldehyde different steric strains etc. Depending on whether it is more electron-withdrawing or electron-donating, it can severely influence the bonds on the carboxyl group.
Due to the variations in oxidising agent, medium of reaction and side group of aldehyde, the oxidation of different monoaldehydes to carboxylic acids would have a different value of electrode potential as these factors would affect the feasibility and ease of oxidation.
Oxidation using chromic (VI) species:
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch17/ch17-3-5-1.html


text:
What component of the human DNA reacts with 4,5-benzo[a]pyrene oxide and 7,8-benzo[a]pyrene oxide to produce cancerous tumors
I read in an article that we have this enzyme called P450 that converts aromatic compounds into water soluble compounds that can be eliminated. This enzyme converts aromatic compounds into arene oxides. For example, benzene can be oxidized into benzene oxide.
In the same article it talked about nucleophilic arene oxides being carcinogenic. The nucleophilic $\ce{NH2}$ group of 2'-deoxyguanosine is known to react with certain arene oxides. Once a molecule of this type becomes covalently bonded to an arene oxide the 2'-deoxyguanosine can no longer fit into the DNA double helix. Since the genetic code can't be transcribed it will lead to mutations that can lead to cancer.
Benzo[a]pyrene is found in cigarette smoke and automobile exhaust. It was the culprit behind the scrotum cancer epidemic among Victorian chimney sweepers in the late 1770s to 1800s. The two most harmful are the 4,5-oxide and 7,8-oxide, but I was wondering do these same oxides react with 2'-deoxyguanosine?


text:
For the solubility of AB2, Ksp=[A][B]^2=x*(2x)^2. Why is (2x)^2 not considered double dipping?
For $$\ce{AB_2<=> A + 2B} $$
$$K_{\text{sp}}=\ce{[A]*[B]^2} $$
$$K_{\text{sp}}=\ce{[x]*[2x]^2} $$
For the term $$\ce{[2x]^2} $$
Why isn't this considered double dipping on B's behalf?
I.e. why isn't it just $$K_{\text{sp}}=\ce{[x]*[x]^2} $$ or $$K_{\text{sp}}=\ce{[x]*[2x]} $$
Why is B both multiplied by 2 and squared?

It's easy, if you consider:
$$\ce{AB2 <=> A + 2B}$$
It's solubility equations would be:
$$K_{\text{sp}}=\ce{[A][B]^2}$$
That equation follows from the law of mass action.
If you consider the amount of A as $x$, wouldn't the amount of B be $2x$ obviously?
If your issue is with why it's squared, it's just the equilibrium law, it follows from law of mass action. You can't really argue why the square is there.

Thanks for the answer. I understand A is x and B is 2x. That makes sense. I just don't understand why B=2x and becomes B^2=(2x)^2.
I.e. B is already *2 of x because it has twice as many moles as A. But then B is also squared. I  know that is part of the formula for K values. But I don't understand why B gets both *2x and is also squared.

They both must be there. The square comes because of the equilibrium law, and the 2 comes because there's two B released for one A. It seems to be redundant, but its not.

Thanks Pritt. Is there a simple answer to why the equilibrium law requires it to be squared? Or is there a reference you can point to me? I think understanding that would solve my problem.


text:
What method is used to discover oxidation state of drug binding cysteine residue?
I was reading this article and found this sentence:

KC group found that some cancer cells became resistant to
  Boehringer Ingelheim’s covalent TKI afatinib (Giotrif) due to the
  oxidation of its target cysteine in EGFR

What would be the chemical method to discover that especially the target cysteine is oxidized? Can it be done with NMR or spectroscopy?

NMR could likely be used to characterize a bulk sample of the drug and identify the presence of cysteine residue oxidation products, but would not be of much use for establishing that cancer cells oxidize (inactivate) the drug, nor precisely how it was inactivated, given the tiny amounts of material involved. Maybe LC-MS could be used. Probably worth pulling up the original paper to see how they did the work.

Thanks for reply. Sorry, I was too low on time to reply sooner. I skimmed through the cited articles and I wasn't able to find the original reference. They focused on finding specific probes to bind the oxidised cysteine residue in a small molecule and then mass spectrometry to analyse. In another article I found that "Tandem MS analysis encompassing all oxidizable residues (cysteine and methionine; 97% overall amino acid coverage) confirms that oxidation occurs exclusively at Cys797" so I guess that was the method of choice. ( https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3890870/ )


text:
Conversion of H2O2 into H2O
If I had $\ce{H2O2}$ in a beaker and I put $\ce{H2}$ and $\ce{O2}$ gas in that beaker, what would happen? $\ce{H2}$ will combine with $\ce{H2O2}$, or a $\ce{O}$ atom in $\ce{H2O2}$ will separate and make bonds with $\ce{H2}$ gas?

Nothing will happen.

Why? H2O2 is unstable

Forget the dihydrogen and dioxygen gas... just leave that jar of H2O2 outside one sunny afternoon and you'll get your H2O ;) [Alternatively you can just heat the jar till all the H2O2 decomposes to H2O]

https://chemistry.stackexchange.com/questions/42736/disproportionation-of-hydrogen-peroxide

https://chemistry.stackexchange.com/questions/43233/why-this-doesnt-hydrogen-peroxide-decompose-into-hydrogen-and-oxygen

Also see this question and check the links therein.

Light can cause the hydrogen peroxide to decompose:
$$\ce{H2O2->2 OH}$$
The $\ce{OH}$ can react with $\ce{H2}$
$$\ce{OH + H2 -> H2O + H}$$
Also, $\ce{OH}$ is an important chemical in atmospheric chemistry. It takes part in several reactions. In the troposphere, it takes part in the following reactions:
$$\ce{OH + CO-> CO2 + H}$$
$$\ce{OH + H2O2-> H2O + HO2}$$
$$\ce{OH + O3-> HO2 + O2}$$
$$\ce{CH4 + OH-> CH3 + O2}$$
The products of these reactions further react with other gases and vapors in the atmosphere. You can learn more about it here.
The behavior of chemicals is complex. "What will happen?" is a question that is almost always too broad.
Source:
Introduction to Atmospheric Chemistry by Daniel J. Jacob, Princeton University Press, 1999.


text:
Why does chocolate melt so easily?
(Yes, I know the question sounds super-trivial... but bear with me here)
Most chocolates (especially milk-chocolate) tend to melt into this sticky (albeit delicious) mess at temperatures slightly above room temperature. Why does this happen?
From my experience (as a high-school student), most organic solids (at room temperature) tend to either sublime or decompose at moderate temperatures; by "moderate", I mean a temperature range between 25 degrees Celsius (typical room temp) and 50 degrees Celsius (a very hot day in Arizona).
Now why do I find this particular property (ease of melting) of chocolate intriguing all of a sudden? Because chocolate is an organic cocktail of all sorts of compounds such as sugars, lipids, proteins, alkaloids, blah blah blah; yet in the moderate temperature range, it doesn't sublime nor does it decompose (chill the molten chocolate, and you'll still get chocolate)... it first softens and then melts, unlike most other organic substances in that temperature range.
So, what makes chocolate so special? And why does milk chocolate (tend to) melt faster than dark chocolate (higher-cocoa content).

Simple answer, melting points. As for why milk chocolate melts easier than dark chocolate, it's probably since milk is a liquid, while pure   is a solid, so milk has some properties of milk with it.

Chocolate manufacturers need to formulate their products differently for different markets.  For example, Cadbury's is popular in the UK and Australia but it is slightly different in Australia to suit the typically higher temperatures.

Not if you cover it with a hard shell, someone should market such a chocolate product ;-) .

@Oscar Smarties ^ ;)

I would also note that a large number of organic solids do not significantly sublime or decompose in the 25°C - 50°C range under a standard pressure. For example: polypropylene, glucose, cellulose, nylon, urea, and many many more. It’s simply not true that most organic solids perform in the manner you say.

So it melts in your mouth and not in your hands

to @ColGraff's list you can add many culinary fats.  Melting just below body heat is important for mouthfeel so you have to consider a form of selection bias (i.e. we like chocolate because it melts in our mouths)

Also, are you sure you want to use the word “sublime” and not “melt”? Sublimation means to convert directly from a solid to a gaseous phase without going through a liquid transition. Yes, many organic compounds do sublime but I hardly see how it’s relevant to the melting of a chocolate mixture.

tl;dr The main structural component of chocolate is cocoa butter, which is a blend of fatty acids (primarily oleic, palmitic, and stearic acids). Cocoa butter has multiple crystal structures, and manufacturers target a specific form which melts at around 33 °C. The fact that chocolate is about 90% sugar and cocoa butter is why chocolate is (chemically) insensitive to temperature changes.

While chocolate itself is a giant mess of hundreds of different compounds, the primary structural material of chocolate is fat. Wikipedia lists a typical composition (by mass) of about 60% sugar, 30% fats, and 10% proteins/other. So it's not terribly surprising that chocolate doesn't sublime or decompose. You probably wouldn't expect a ball of butter + sugar to suddenly decompose at 50 °C, and for the most part, chocolate is chemically pretty similar.
Now, what's interesting about cocoa butter is that it can form multiple crystal structures, and it's actually a bit tricky to get the right crystal structure out. You can read about the process in this RSC article if you'd like. The one that most chocolatiers are after is polymorph V, which is glossy and melts just below body temperature.
However, a different polymorph, polymorph, VI is the most stable form of cocoa butter, and it has a melting point that's a bit higher. It also tends to have a faint white crust on it (known as "fat bloom"). As a result, it's not as pleasant to look at, not as pleasant to eat (since it takes some chewing on to get it to melt), and generally undesirable.

Since form VI is the most stable, all chocolate eventually goes "bad" and becomes not as delicious as the maker originally intended. If you've ever bitten into an old chocolate bar and thought it was a bit waxy and unmelty, you've experienced this interconversion. I've been told by someone who did research on crystal structure kinetics that chocolate manufacturers spend lots and lots of money on research in hopes of being able to slow down or stop this process.
Now, I'd like to take stab at addressing two points in the question:

chill the molten chocolate, and you'll still get chocolate

Yes, but is it the exact same chocolate as you got before? What if you take the molten chocolate and put it in the freezer, or, heaven forbid, drop it in liquid nitrogen? Does the resulting chocolate still melt at the same temperature? My suspicion (based on this chart) is that you'll have generated some of the lower-melting cocoa forms, and thus the chocolate will get soft and melt at a much lower temperature than previously.

And why does milk chocolate (tend to) melt faster than dark chocolate (higher-cocoa content)?

I don't know. If I were to hazard a guess, I would say it's because milk chocolate tends to contain more sugar than dark chocolate, and so you're seeing some form of freezing-point depression. However, if it turns out that, at similar concentrations of sugar, milk chocolate still melts more quickly, I'd have no idea.

Milk chocolate could melt faster b/c there is more sugar which means less amount of fatty material that needs to phase transition (less latent heat needed).

Simply chilling melted chocolate will not recover the as-manufactured texture. Tempering manipulates the mix of types of cocoa butter crystals to get a desirable texture.

Milk chocolate is made by adding milk fat, so it has more fat total. Dark chocolate has more coca solids (remember the "90% cocoa" marketing?), so it has less fat total. A "99% cocoa" chocolate is pretty much pressed cocoa powder so it hardly melts at all.

Chipbuster's answer is excellent. Let me add in an important point: chocolate is not a homogenous solution, but an emulsion where all the sugar, protein etc is emulsified in the fat content. As in the other post it is said cocoa butter is polymorphic and generally crystallize around room temperature. When it is melt, you have a regular emulsion (like creams or butter), and when it is solidified it is still the same emulsion, but with the solid fat.
It also means adding other fats can shift the melting point: milk chocolate contains lot of milk powder, therefore the fat-phase is only partially cocabutter, diluted with milk fat (same as butter or cream), which makes lower melting point and also the crystallinity of the fat is somewhat compromised.

I think you are seriously overestimating chocolate's stability of consistence because these days all but the most expensive kinds of chocolate are stuffed with emulsifiers generally keeping the consistence together.  Before soy (and milk) was used in that function, you had to prepare chocolate in a Conche to get an enjoyable consistency.
Also if you are working with "organic" couverture without additional emulsifiers, getting your chocolates to dry without having cocoa butter bloom out all over them within hours and turn them into a shoddy mottled whitish-grey is a real nightmare.
Been there, done that, had to do lots of overtime.

Pretty sure conches are still used, at least for decent-quality chocolate, even if it contains emuslifiers – it's not just for mixing, also for first grinding the still-solid particles down to unnoticable size.


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Is consumption of chlorine, like fluorine, harmful to bones?
Is consumption of chlorine harmful to bones ?
I know fluorine consumed with water can cause fluorosis to bones and joints.
Can chlorine,  a similar halogen like fluorine, also cause damage to bones and bone joints?

Are you talking about chlorine gas or the chlorine ion?

http://www.pure-earth.com/chlorine.html

Most probably chloride @Quantum

I'm voting to close this question as off-topic because its more about medicine than chemistry.

I'm voting to close this question as off-topic because the OP wishes to know the medical complications associated with the "consumption of chlorine". I second what @porphyrin said: This question is better suited on the Health.SE.

The OP uses the terms chlorine and fluorine. In the context of the question, fluorine doesn't make sense and it is assumed here that the the OP means fluoride. In the case of chlorine, it is likely, though not certain, that the OP means chloride. This answer will address the question primarily in terms of fluoride and chloride, with a note about chlorine.
Bones are largely composed of calcium compounds (mainly hydroxyapetite, $\ce{Ca5(PO4)3(OH)}$). The calcium salts of bone are very slightly soluble, and thus they are in an equilibrium between the solid form and the dissolved aqueous form. Because calcium fluoride is extremely insoluble, fluoride can displace the naturally occurring anions to form water-insoluble calcium fluoride. This results in weaker, more easily fractured bones, which is the condition known as skeletal fluorosis.
Calcium chloride is about five orders of magnitude more water soluble than is calcium fluoride, so an analogous displacement of the naturally occuring anions in bone by chloride does not take place. Thus the ingestion of chloride has no negative effect on bone material. The same is true of chlorine, in the case of chlorinated water, as there is no mechanism by which chlorine can be ingested (or inhaled), metabolized and then interfere with bone composition.

The key point is that fluorine is a cumulative poison which accumulates predominantly in bone, unlike chlorine. Using the word "fluorine" does make sense, because fluorine is a general term for any atom with nine protons, including its ionic form, fluoride. When fluorine accumulates in bone it forms fluorapatite, not calcium fluoride.


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If Ag2SO4 dissociates into 2Ag+ and SO4^2-, Why doesn't SO4^2- dissociate into oxygen and sulphur?
When  $\ce{Ag2SO4}$ dissociates it becomes  $\ce{2Ag^+}$ and $\ce{SO4^2-}$
Why doesn't the $\ce{SO4^2-}$ dissociate further into oxygen and sulfur?

The answer to your question is very simple. When a chemical species undergoes the process of dissociation in an aqueous solution, it is broken up into its constituent ions. In this case, silver (I) sulfate becomes silver (I) ions and sulfate ions. Dissociation into the constituent ions takes place because the enthalpy of hydration of the silver (I) ions and sulfate ions is greater than the lattice entahlpy of silver (I) sulfate.
However, for the sulfate ions, they can no longer break up into any other chemical species as there are no further strong-enough interactions that take place between the solvent and the sulfate ions to break the bonds between the atoms in the sulfate ion. This is most importantly due to the strong covalent bonds between the sulfur and oxygen atoms in the sulfate ion.
Now, you may ask how the supposedly-strong covalent O-H bonds are broken to allow many oxoacids to dissociate into their conjugate bases and the hydrogen ions. This is because of the strong solvent-solute interactions (i.e. hydrogen bonding) in the solvent, which make the extraction of the hydrogen atom from the oxoacid possible.
With a similar line of reasoning, the supposedly-strong bonds between hydrogen and the halogens in hydrohalic acids (e.g. HI, HCl, HBr) are also easily broken to make them effective acids in solution.
In essence, not everything can dissociate in solution... The energetics of the solvent-solute interactions need to be carefully considered. Usually, ionic bonds are easily broken in solution to give the constituent ions of the ionic solid because the ion-dipole interactions are very strong between solute and solvent species. On the other hand, molecular species with covalent bonds do not easily dissociate because of the strong covalent bonds between atoms of the molecule. Thus, we (to my knowledge) never see the dissociation of molecular species into their constituent atoms.
Hope this answers your query!


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Preparing soluble salts from acids and soluble and insoluble reactants
3.15: Explain why, if soluble salts are prepared from an acid and an insoluble reactant:
a) Excess of the reactant is added
b) The excess reactant is removed
c) The solution remaining is only salt and water
3.16: Explain why, is soluble salts are prepared from an acid and a soluble reactant:
a) Titration must be used
b) The acid and the soluble reactant are then mixed in the correct proportions
c) The solution remaining, after reaction, is only salt and water.
These are two points in the GCSE specification I do not completely understand. Can anyone help?

3.15 a)To ensure that all the acid has reacted, such that the amount of salt you
have prepared is indeed the amount you planned to prepare. b)To remove impurities from the salt you have prepared. c)Since you have reacted all the acid and removed all the impurities, what is remaining is evidently salt and water.
3.16 a)Now, the key idea is that you want to prepare a precise amount of the pure salt in solution, meaning to have no acid or any of the other soluble reactant contaminating your sample of aqueous salt. This is the reason why you need to be very precise when mixing your reactants because you cannot add more or less. Remember that since the reactants and product is in the same phase, you have no way of removing one from the other easily. b) Remember that we want a precise amount of the salt and no excess acid or any of the other reactant. c) Since all the acid and the other reactant has reacted, the only products in the solution are water and salt (typical of an acid-base, acid-metal reaction).
Hope you found this useful!


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Protonation of Carbonyl oxygen in Fischer Esterification
In the Fischer Esterification, why is it the acid's carbonyl oxygen getting protonated and not the hydroxyl oxygen? This protonation occurs in Step 1 of the mechanism below.
I read in some textbooks that this is due to the carbonyl oxygen being more neucleophilic and the hydroxyl oxygen. Why is this so?

Draw the resonance structures for both options.

Same reason why an amide protonates on oxygen instead of nitrogen.

There are a few reason why the carbonyl oxygen gets protonated -
1) The carbonyl oxygen is sp2 hybridized which is more electronegative than the sp3 oxygen. So, it has more tendency to give electrons since it is more negative.
2) The lone pairs of carbonyl carbon are more available since they are not in conjugation whereas the lone pairs of the other oxygen are not much available since they are in conjugation with carbonyl carbon.
3) When carbonyl oxygen donates its electron the resulting structure is more stabilized (because of conjugation) than the structure formed when the other oxygen atom donates.
So, that's why carbonyl oxygen has far more tendency of donating electrons than the -OH grp.

Thank you for your answer. For point 1), I was thinking that an sp3 hybridized oxygen atom would have lesser s character than an sp2 hybridized oxygen atom, hence sp3 oxygen's lone pairs would be held less closely to the nucleus and hence be more able to abstract a proton than the sp2 oxygen. Is this a misconception on my part?

If it’s more electronegative it should be less prone to protonation since it is better able to stabilise the negative charge prior to deprotonation.

Also with respect to point (1), both oxygens are sp2.


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What is the mechanism of hydroperoxide reduction by bisulfite?
I am using bisulfite to quench allylic hydroperoxide products from a singlet oxygen ene reaction, to the corresponding alcohols. What is known about the mechanism of this reaction, and in particular could it involve an alkyl peroxysulfate ($\ce{R-O-O-SO3H}$)?

I doubt there's much hard data on peroxide/sulphur redox class. There are various mechanisms postulated, but not analysed in depth.


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Is Kp and Kc only effected by temperature?
Are Kp and Kc only effected by temperature?
So in exothermic and endothermic equilibrium reactions you are increasing or decreasing temperature, which will effect which side is favourable.
But then so does pressure if one side has fewer molecules?

see: https://chemistry.stackexchange.com/q/9108/189

@Philipp is this statement in my textbook wrong then "increasing the pressure forces the equilibrium in such a way as to reduce the total pressure, that is, to the side with fewer molecules in other words KP increases. So increasing the total pressure increases the yield of ammonia"

It sounds fishy, at least the part "in other words KP increases". They are talking about Le Chatelier's principle but that does not entail a pressure dependence of the equilibrium constant. However, the partial pressures/activities/relative amounts of the reactants and products in the reaction mixture which feature in the expression for $K$ do not need to be constant with pressure only their reaction quotient has to be in order to preserve the contant $K$. Thus, pressure (depending on how it is administered) does influence the equilibrium composition of the reaction mixture.

You could have a look at page 221 of the 9th edition of the book "Physical Chemistry" by Atkins. It describes this in more detail.

As $K_p $ is defined by standard states, i.e. 1 atm (or 1 bar) it must be independent of pressure.  The T dependence is from the van't Hoff isochore $d\ln(K_p)/dT = \Delta H^o/RT^2$ (assuming $\Delta H$ constant). If the equilibrium constant is defined in terms of mole fraction x then $K_x=K_pp^{-\Delta n}$ where $\Delta n =a+b-c-d$ is the change in number of moles in reaction $\ce{aA + bB <=>cC + dD}$. Sometimes the equilibrium constant is called $K_p$ when it should be $K_x$ or $K_c=K_p(RT)^{-\Delta n}$.

see also this question https://chemistry.stackexchange.com/questions/73644/how-does-the-partial-pressure-of-gas-c-change-with-the-change-in-volume-of-the-c/73975#73975

According to LeChatelier's principle on increasing temperature in an exothermic reaction "which is at it's equlibrium initially" the reaction proceeds in backward direction and in endothermic reaction it proceeds in forward direction, this can be thought as in exothermic reactions the heat should be provided to the products to yield the reactants and in endothermic the heat should be supplied to reactants to yield the products!
And the value of equilibrium constants depend only upon temperature and the pressure only effect the equilibrium conditions and not the value of Kp because partial are so adjusted to provide the same value of Kp at the same temperature.


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Why does the Rosenmund reduction stop at the aldehyde stage?

In the Rosenmund reduction, an acyl chloride can be reduced to an aldehyde by hydrogenation over a palladium catalyst poisoned with barium sulfate.
Why isn't the aldehyde product further reduced to an alcohol?

In general, the Rosenmund reduction isn't actually used very often, mainly since the catalyst is also very good at reducing other functional groups which may be present in the molecule (the catalyst is similar to that used for a Lindlar reduction).
To quote from Comprehensive Organic Name Reactions and Reagents:

[...] the Rosenmund reduction is an unreliable and irreproducible method for organic synthesis, and it is often in competition with other side reactions. For example, alcohols are often the by-products of the Rosenmund reduction.

How the reaction works/why we get an aldehyde
As with many reactions involving catalyst systems in which palladium has been absorbed onto something (the classic one being palladium on carbon/charcoal), the mechanism of the Rosenmund reduction isn't well defined, and indeed there are likely to be different mechanisms operating depending on the quality/specification of the catalyst used.
One possible pathway has been proposed, and is given below. The critical thing here is that the acyl chloride is first activated by oxidative addition into the acyl chloride – this activation allows the reduction to take place.

The product of the reaction is often therefore the aldehyde, which should be kinetically inert to further reduction (this has been discussed on chemistry.SE previously, and can be found here), though the alcohol is often obtained in small amounts.

Not sure if it's useful to point out, but: if you created a metal hydride first via oxidative addition, then you can do a $\beta$-migratory insertion into the carbonyl. Reductive elimination would provide a reduced carbonyl. But either the preliminary oxidative addition of hydrogen or the migratory insertion pathways have been shut down by poisoning.

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


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Can we force reactions with positive delta G?
Do these reactions always have to be paired with another energetically favorable reaction with a greater delta G magnitude, or is there a way to force reactions with a positive delta G to happen?

Are you asking about $\Delta G$ or $\Delta G^0$?

If it's redox, you can drive it with an external emf after constructing a cell.

The change of the Free / Gibbs enthalpy $\Delta_{\text{R}}{}G$ consists of a contribution by reaction enthalphy $\Delta_{\text{R}}H$, and a contribution by reaction entropy, $\Delta_{\text{R}}S$:
$$ \Delta_{\text{R}}{}G = \Delta_{\text{R}}H - T \cdot \Delta_{\text{R}}S $$
Hence, the answer to your question is no.  By variation of temperature $T$, for example, a reaction previously endergonic ($\Delta_{\text{R}}{}G > 0$) may become exergonic ($\Delta_{\text{R}}{}G < 0$), as summarized here.

The two things to remember are: $\Delta G$ is a function of conditions and assumes an infinite amount of time.  If a reaction has a positive $\Delta G$ it will not be thermodynamically driven. It can however be kinetically driven at leat in the short term. For example, the phase change in NiTinol alloys from cubic to monoclinic upon cooling can experience the formation of a metastable triclinic phase which is not thermodynamically favored but kinetically favored.
However if you want to drive the reaction, change the conditions. Just to demonstrate a few:

For reactions with $\Delta S > 0$ you can change the temperature. (melting, boiling sublimation, and the reverse; see Figure 1 below)
Change with electricity: $\Delta G = \Delta H - T\Delta S - nFE_{cell}^\circ $ (electrolysis)
Change with pressure and/or volume: $\Delta G = \Delta H - T\Delta S - \int_{P_1}^{P_2} V\mathrm{d}P - \int_{V_1}^{V_2} P\mathrm{d}V $ [making diamonds; constant volume conditions are actually Helmholtz energy $(A)$]
Change with concentration or partial pressure: $\Delta G = \Delta H - T\Delta S - RT\ln(Q)$  (Le Chatelier's principle)
Change with magnetic field: $\Delta G = \Delta H - T\Delta S - \mu H_\mathrm{a} M $ (magnetic storage)
Change with chemical potential: $\Delta G = \Delta H - T\Delta S - \mu \Delta N $ or more generally $\Delta G = \Delta H - T\Delta S - \sum^n_{i=1}\frac{\partial G}{\partial N_i} \mathrm d N_i $ (chemical potential for diffusion).
Change with light: $\Delta G = \Delta H - T\Delta S - hf $ (radical formation)

and many more. So while you cannot make a reaction with a positive $\Delta G$ proceed spontaneously, you can change the environment to make it negative and thus "spontaneous".

Figure 1: $\Delta G$ of water vapor with temperature.
Again remember a thermally spontaneous reaction is after infinite time will only proceed if the kinetic barrier is overcome otherwise the system will be meta-stable. for example the combustion of gasoline is a spontaneous reaction at room temperatrue and oxygen pressure, but is not pyrophoric (does not spontaneously ignite).


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What is the acid base reaction between hydrogen telluride and sulfide ion?
$$\ce{H2Te + S^2- -> ?}$$
I tried this problem but I was confused on the fact of what the products would be because in my head I didn't think $\ce{STe}$ is a product and also it has to create water but there is no $\ce{OH-}$. Could someone assist me with this problem?

Why does it have to create water?

Although the Arrhenius definition of an acid-base reaction requires the formation of salt and solvent, the Brønsted–Lowry definition does not; it only requires that one species donate a proton to another.

Observe this reaction:
$$\ce{Na2S + H2S → 2NaHS}$$
If you use sodium sulfide and react it with hydrogen sulfide, sodium hydrosulfide/bisulfide is formed.
Similarly, if you use sodium telluride and react it with hydrogen sulfide, it is expected to form sodium hydrogen telluride( $\ce{NaHTe})$.
Note that hydrogen telluride is an unstable acid but the acidified ion $\ce{HTe-}$ is quite stable. In fact, the sodium salt of hydrogen telluride ion is known, albeit not prepared through above reaction.

The intermediate in the acidification, $\ce{HTe-}$, is a stable anion.
Sodium hydrogen telluride, $\ce{NaHTe}$, can be made by reducing
tellurium with $\ce{NaBH4}$. (Hydrogen telluride)
The acid hydride of tellurium, hydrogen telluride, $\ce{H2Te}$, is an
unstable compound that decomposes to tellurium metal. It is strongly
acidic, dissociating into a hydrogen telluride ion ($\ce{HTe-}$)
in aqueous solutions. (Telluride ion)


text:
Spiralling of electrons into the nucleus
A drawback (and a major one) of Rutherford's model was that the electrons, being accelerated charged particles, would gradually spiral into the nucleus and collapse into it. My question is how did Bohr rectify this drawback in his own model.
In Bohr's model too, electrons move in a circular path, which means they are accelerated. Thus, going by Electromagnetic theory of Maxwell, those accelerated electrons should radiate out their energy, and collapse into the nucleus, just like they did in Rutherford's model. However every book mentions that this does not happen in Bohr's model. Why not?
Please tell me where I am wrong, because this problem has been disturbing me for quite long now.

What books are you reading? https://en.wikipedia.org/wiki/Bohr_model#Shortcomings last bullet point

Nobody says the Bohr model is right. At the time it was a step, but with issues needing quantum mechanics to solve...

Really a question for Physics SE and answered there already I think here and here

Thank everyone for their respective inputs. I am new to stack exchange, and searching for a particular question does not seem to be easy here.

As a matter of fact, Bohr literally went past around this issue. In 1913, he proposed his model postulating electrons could only have certain classical motions:

Electrons in atoms orbit the nucleus.
The electrons can only orbit stably, without radiating, in certain orbits (called by Bohr the "stationary orbits") at a certain discrete set of distances from the nucleus. These orbits are associated with definite energies and are also called energy shells or energy levels. In these orbits, the electron's acceleration does not result in radiation and energy loss as required by classical electromagnetic theory (This was based upon Planck's quantum theory of radiation).
Electrons can only gain and lose energy by jumping from one allowed orbit to another, absorbing or emitting electromagnetic radiation with a frequency ν determined by the energy difference of the levels according to the Planck relation:
$$ \Delta {E}=E_{2}-E_{1}=h\nu \ $$

where h is Planck's constant.
As remarked above, these are postulates (statements assumed to be true from which further reasoning follows), and (even though that part was left unexplained), it's acceptance was due to the fact it explained correctly many experimental results. It was further displaced by the modern quantum theory.
Hope it helps, it also puzzled me a few years ago! :P
Reference: https://en.wikipedia.org/wiki/Bohr_model

So, exactly what Bohr did is that he assumed that the drawback didn't exist?

@AumkaarPranavShukla I'd rather say he knew it was incomplete but felt the explanation for it could be found later by another person

The uncertainty principle “prevent it”from happening meaning there’s not an objective reality.We existe in the “middle dimension”.On one side the macrocosm and its accelerating expansion implying no objective reality with position and momentum and on the microcosmic side the uncertainty principle.


text:
Why does potassium dichromate always reacts in an acidic medium?
On the other hand, potassium manganate reacts in both acidic and alkaline media.

Slightly related: https://chemistry.stackexchange.com/questions/75171/how-do-these-cascade-reactions-work

In an alkaline medium, chromium(VI) exists as potassium chromate instead of potassium dichromate:
$$\ce{Cr2O7^2- + 2OH- <=>> 2CrO4^2- + H2O}$$
Wikipedia has more information about this equilibrium.

Interesting information @orthocresol. Does this mean that the $\ce{CrO4^{2-}}$ ion have no oxidative abilities despite being in chromium's highest oxidation state?

No, chromate(VI) is still oxidising


text:
What would an ecosystem based on reducing atmospheres be like?
I was wondering, what would it be like if instead of an oxidizing atmosphere like earth has, if an ecosystem had a reducing atmosphere instead? My professor told us that a system like that could be chemically/biologically possible. But didn't explain how. What would interaction between organisms and their reducing surroundings be like? How would these organisms survive?
I know that this means there would be an extreme lack of oxygen, and as far as I can tell, most everything on earth needs oxygen to survive. There are complex organic molecules that can be created in a reduced atmosphere, right? What are they?

Such ecosystems exist on Earth! See this Wikipedia post on anaerobic organisms. There are even some species that cannot tolerate oxygen. Also, welcome to Chemistry.SE :)

The early earth did have this kind of atmosphere before! Such an atmosphere consisted of ammonia, methane, carbon dioxide, and hydrogen. Lightning strikes converted some of the atmospheric gases into amino acids and carbohydrates. Then happened a Great Oxygenation Event, that made the atmosphere suddenly oxidizing instead. See https://en.m.wikipedia.org/wiki/Great_Oxygenation_Event

Saturn's moon Titan, maybe?

What would it be like if humans were able to be anaerobic organisms? Is that completely illogical?


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The decarboxylation of phenylacetic acid via addition of acid
I came across a question where the following compound except with two methyls, ortho and meta to the nitro group, is dissolved in methylene chloridde, and then treated with trifluoroacetic acid.

It is given that $\ce{CO2}$ gas is given off in a decarboxylation reaction. I cannot find any sources about this. The solution explains that:

Treatment of a carboxylic acid with acid results in decarboxylation, and the evolution of $\ce{CO2}$, especially if the resulting compound contains a benzylic or allylic carbon, as is the case here.

First of all, what mechanism would allow for acid catalyzed decarboxylation? I know how decarboxylation of carboxylic acids with beta-carbonyls with heat works, and that acid can help that but that doesn't seem to have relevance here.
Secondly, why does having a benzylic or allylic carbon confer any special stability to promote the reaction?

With a beta carbonyl you delocalise the negative charge from decarboxylation into the second carbonyl group, here you delocalise it into the nitro group.

Do you have a source for the question (name of a book, website etc.).

Acid-catalysis is activating the nitro group to make it more electron-withdrawing.

@NotEvans it was an MCAT general study app. It provides no sources of its own to back itself up. I can see how electron delocalization stabilizes the decarboxylate - and I guess I can see how the stoichiometry works to protonate at the allylic carbon at the end to produce para-nitromethylbenzene.

What I still don't understand is why the app says that having a benzylic or allylic carbon, in particular, stabilizes the final compound. Should I just ignore that and move on?

"with two methyls, ortho and meta to the nitro group" what position are they at exactly? Could it make a difference if the positioning of the two methyl groups is on different sides of the ring? (one ortho on left and one meta on right)

@user36847 after decarboxylation, you're left with a benzylic anion, stabilized through resonance by the ring and the nitro group.

The decarboxylation of carboxylic acids having allylic or benzylic carbons can also be explained by what may be called the vinylog concept. One of the most common decarboxylations is that of β-keto acids as illustrated by the decarboxylation of β-keto acid 1 to afford cyclohexanone 2. Hagemann's ester (3) is a γ-keto ester, which upon aqueous acid hydrolysis gives acid 4. This acid suffers decarboxylation to afford 3-methylcyclohex-2-en-1-one (5). The red double bond in 4 simply extends the properties of a β-keto ester to the γ-position in a vinylogous (allylic?) fashion.
Nitrobenzene 6, which may be the compound in question (if not, ignore the methyl groups), is doubly vinylic (benzylic?) as indicated by the two red double bonds in structure 6. While this compound does not have a keto group in conjugation with the aromatic ring, the nitro group functions as a viable substitute for a keto group. Protonation of the N=O bond on oxygen leads to decarboxylation and rearomatization to nitrobenzene 7.
The green σ-bonds in structures 1, 4 and 6 must have overlap with the π-framework of the carbonyl (in 1), the γ-position of enone 4 and the aromatic ring of nitrobenzene 6 for decarboxylation to occur.
Finally, nitroacetic acid, as its magnesium salt, decarboxylates at 80 oC to provide nitromethane and CO2.


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When do reactions reach steady state?
I was doing a multiple choice problem that asked about when the following nuclear decay sequence reaches steady state (When does the activity of $\ce{^99Mo}$ roughly equal the activity of $\ce{^{99m}Tc}$ ?)

The correct answer choice said after approximately 20 hours.
Let's denote $\ce{^99Mo}$ by $\ce{R}$ and $\ce{^{99m}Tc}$ by $\ce{I}$.
By definition, the time we're looking for is the time when $\frac{\ln 2}{65}N_R=\frac{\ln 2}{6}N_I$, which is when $\frac{N_R}{N_I}=10.83$
I don't have the formula for the exact amount of the intermediate vs. time memorized, but I don't think the problem required us to know the formula. I tried estimating the amounts of the various species present after 20 hours to see if steady state is actually reached then:
Let the initial ${N_R}=1$. I know that at $t=20$, $N_R=\frac{1}{2}^{20/65}=0.8079$. To estimate $N_I$, I did $N_I\approx(1-0.8079)*\frac{1}{2}^{20/6}=0.0191$. But using this approximation, it seems like the steady state condition is not met. Thus, I was wondering, what is a better way to estimate $N_I$ to get that the decay sequence is in steady state after 20 hours?

This type of consecutive reaction does not really come to steady state for very long, as seen in the figure, i.e there is only a small time over which $d[mTC]/dt \approx 0$.  The decay rate of mTC is faster than that of Mo and so mTc appears quickly and decays slowly. (The amount of mTC is multiplied by 10 in the figure is the amount formed is small].
The figure is drawn assuming  $k_1 = \ln(2)/65, \, k_2 =\ln(2)/6 $ and $\ce{[Mo]_0} = 1$ and that the other species are zero at $t=0$ The equations are
$\displaystyle \ce{[Mo]=[Mo]}_0e^{-k_1t}$ and
$$\displaystyle \ce{[mTc] = [Mo]}_0\frac{k_1}{k_1-k_2}\left( e^{-k_1t}-e^{-k_2t}  \right)$$ The amount of Tc is found as $\ce {[Tc] = [Mo]_0-[Mo]-[mTc]}$
The maximum amount of mTC, which is assumed to be 'steady state'  is found by differentiating.
The time at maximum is $\displaystyle t_m= \ln(k_2/k_1)/(k_2-k_1)$ which is 22.7 hours so the 'steady state' is established just before this and lasts a little past this time. How much in either way is somewhat subjective unless you use a definition of say $\pm 5$ % of the maximum etc.

Thanks for the answer. I was asking more about how to estimate [m Tc] without using the actual formula, because I don't think I'm expected to have the formula memorized for this test.

As the mTc decays rapidly compared to Mo then at the smallest  time t when $\exp(-k_2t)  \lt 1$ the ratio $[Mo]/[mTC] = k_2/k_1 -1$  but I can't see how this time gives anything more than a very approximate estimate at which the transient steady state is present.


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What effect does the concentration of a metal ion have on a flame test and why should small samples of the metal ion be used?
When we conduct a flame test, say we have a solution of $\ce{NaCl}$ mixed with methanol in a ceramic bowl and we light it with a match.
What effect will [Na+] have on the colour we see? If we're assuming that the drop in energy of each Sodium atom (once the solution is atomised) once it gets excited is the same so the wavelength of emission is the same and we see the same yellow colour, if we have a greater concentration of Na+ do we just see more yellow and our eyes would perceive that as being brighter (while the colour itself is the same, we just see more of it)?
Also, I read this on wikipedia:

"Bulk sample emits light too, but its light is not good for analysis. Bulk sample emits light primarily due to motion of the atoms, therefore its spectrum is broad, consisting of a broad range of colors."

I get the idea that if the spectrum of emission is broad, then the colour is not really specific to the ion being tested for, but what I don't get is the explanation for this. What does "emits light primarily due to motion of atoms" mean and why does this phenomena not occur in a smaller sample say a spray of the solution or the sample on the tip of a nichrome wire? Why is it that using a larger sample, which is what I assume bulk sample means like having say 20mL of $\ce{NaCl}$ in a ceramic bowl as opposed to a 1mL spray bottle change with regards to the colour of the flame that we see?
Thanks in advance!

Are you aware of the kinetic theory? Basically every molecules possess kinetic energy due to motion. These motions are more evident in large samples with more molecules. Due to collisions, or temperature these molecules may emit photons. These photons do not have a particular value, and may take any value from a fixed range of values. This is a broad range spectrum which poses difficulty in analyzing the color of salt flame.

Thanks for the response. So essentially, the collisions of atoms in the sample are sufficient to energise some electrons enough to become excited and emit photons and this effect is more prominent in a larger sample?

That may happen, though I'm unsure if thats the majority reason. Please wait for a while until a expert on your topic comes by (I'm no expert).