text:
Finding molar mass of uranium fluoride using Graham’s law of diffusion
Problem

A sample of uranium fluoride is found to effuses at the rate of $\pu{17.7 mg/h}.$ Under comparable conditions, gaseous iodine effuses at the rate of $\pu{15.0 mg/h}.$ What is the molar mass of the uranium fluoride?

Answer

 $\pu{354 g mol^-1}$

My approach
Perhaps I’m still having issues with significant figures, but I am getting the wrong answer. I know to use Graham’s law of diffusion, which leaves me with:
$$\frac{\pu{17.7 mg h^-1}}{\pu{15.0 mg h^-1}} = \sqrt\frac{\pu{253.8 g mol^-1}}{x}$$
I get $x = \pu{181.70 g mol^-1}.$ I know that the molecular mass of uranium fluoride should be less than iodine, so the answer makes no sense to me.
What am I doing wrong?

Check the molar mass of uranium hexafluoride.

Also check the exact wording of the problem, to see if the flow rates are correctly associated to the two gases.

I agree with both the above - UF6 is 352 g/mol, and you get that if you take 15/17.7 rather than the other way around.  Either you misread or they miswrote.  Smaller is faster; KE = 1/2 mv^2 is the same on average for all gas particles moving through a pore at a given temperature.

Graham's law of diffusion deals with the molar effusion rate rather than mass effusion rate.
This can be confirmed by the description of Graham's Law on Wikipedia.
An excerpt from the aforementioned page,

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$
where:
$r_1$ is the rate of effusion for the first gas (volume or amount per unit time)
$r_2$ is the rate of effusion for the second gas
$M_1$ is the molar mass of gas 1
$M_2$ is the molar mass of gas 2

Now the mathematical part,
$$\begin{align}
r_{\ce{UF6}} &= \pu{17.7 mg/h} = \frac{\pu{17.7 mmol/h}}{M_{\ce{UF6}}} \\
r_{\ce{I2}} &= \pu{15 mg/h} = \frac{\pu{15 mmol/h}}{M_{\ce{I2}}}
\end{align}$$
By Graham's Law,
$$\begin{align}
\frac{r_{\ce{UF6}}}{r_{\ce{I2}}} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{M_{\ce{UF6}}} \cdot \frac{M_{\ce{I2}}}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \cdot \frac{M_{\ce{UF6}}}{M_{\ce{I2}}} \\
\frac{17.7}{15} &= \sqrt{\frac{M_{\ce{UF6}}}{M_{\ce{I2}}}}\\
M_{\ce{UF6}} &= \left(\frac{17.7}{15}\right)^2 M_{\ce{I2}}\\
  &= \pu{353.39 g mol^-1}
\end{align}$$


text:
Would ionic conduction be halted by shielding at the electrodes?
I've done this experiment myself and seen it front of my eyes. Set up a basic circuit with a battery, an LED, and connect the circuit with a salt-water solution and copper wire. The LED lights up.
If a potential difference is applied across two ends of the wire with the battery, positive charges move towards the end with lower potential and negative charges towards the higher potential.
Why does this conduction keep continuing? Shouldn't there eventually be a pile up of Na+ and Cl- on opposite ends to cut off the circuit? This is with a typical 9V battery. Is there a teensy bit of electrolysis going on in the wire that continuously supplies electrons to the circuit?
EDIT: In doing my own reading the next day, I seem to have forgotten the very important fact that water itself is electrolyzing at the surface of the electrodes. The salt water only helps to prevent the resulting electrolysis from quickly shielding the electrodes. The reason I was confused was because I was very skeptical of NaCl being electrolyzed and turns out this is exactly not the case

Without electrolysis, there would be just an extremely short spike of a temporary initial current, until capacitors formed by electrodes charge.
There is no significant accumulation of ions, as even a tiniest such charge separation would cause strong electrostatic forces acting against it.

I don't know if I agree with the last comment there. I'm applying a potential difference across two wires so there is an electric field that pushes charges. In an ideal plasma, these charges move according to this field with the net effect being that they screen the electric field eventually, but there has to be movement initially applied with the potential difference right?

For the first comment, is this just confirming that in my kitchen I am making sodium metal and chlorine gas with a 9V battery?

@cheekylittleduck When you electrolyse salt water, you are making hydrogen gas, and chlorine gas (mainly) in the electrodes. Look up the table of electrode potential, and you will see that it's much easier to reduce H+, or H2O into H2 gas than to reduce Na+ to Na.

@Cheekylittleduck I have said significant. Apply the Coulomb law. Ions separate until creating the ion displacement potential difference cancelling the external one. Unless electrode reactions happen, which cancels this displacement.// No sodium, unless using melted NaCl, or mercury cathode with concentrated NaCl.

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text:
Why do only some metals make protective oxide layers?
Almost all Alkali metals and Alkaline Earth metals react with air to form their respective oxides, but why does only Magnesium forms a protective oxide layer? What properties of magnesium oxide allow it to form a layer?

There is no simple explanation.

Define protective? There are few oxide-forming metals that yield a robust, self-limiting oxide layer. There are some metals that don't oxidize at STP. Then there is everything in between. Much of the reason is the kinetics of moving oxygen to the oxide-metal interface, or metal through the oxide to the surface. Growing silicon dioxide (a self-limiting reaction at STP) requires using steam as the oxygen source to get diffusion to the interface - atomic oxygen doesn't do the trick.

The simple answer is that metals which form sufficiently insoluble oxide layers stifle the very process which caused the oxide layer in the first place.
It's not just a matter of energy of formation of the oxide, but whether the first reaction product formed will be soluble or insoluble, then whether it will be adherent or non-adherent. In the case of the alkali metals, the oxide may be adherent (reaction with oxygen produces a layer that sticks), but is so soluble that it dissolves in water and refreshes the metallic surface. Likewise in the case of the alkaline earth metals - even $\ce{CaO}$, which is almost insoluble, is soluble enough to give a pH of 12.5, and soluble enough to be non-protective.
Now, magnesium oxide is quite insoluble, but probably is on the surface as $\ce{Mg(OH)2}$, a secondary product. You might consider that $\ce{Mg(OH)2}$ is $\ce{MgO}$'s failed attempt to dissolve and go away from the surface. Instead, it precipitates, and not only that, it adheres quite well. So further reaction with water is inhibited. However, if you heat the water, the $\ce{Mg(OH)2}$ becomes more soluble, and corrosion proceeds, although not as fast as sodium reacts. In flowing hot water, where the $\ce{Mg(OH)2}$ could be carried away, corrosion would be faster; in a flask, stirring would exert a similar rate increase.
The effectiveness of $\ce{Mg(OH)2}$ as a stifling passivator may be related to its layer structure:
 (File from Commons, by User:Smokefoot)

Like a clay layer, the hydroxide ions fuzz out of each layer of close-packed magnesium ions, and the metal surface could be like a half-layer of $\ce{Mg(OH)2}$, with $\ce{OH-}$ ions sticking out and bonding to separate layers of $\ce{Mg(OH)2}$, or perhaps the metal itself just hold onto a layer of $\ce{Mg(OH)2}$ - or maybe a little of both. This adherence is necessary to stifle corrosion. The clay layer-like characteristics of $\ce{Mg(OH)2}$ are also evident in milk of magnesia, which is a stable suspension - if the suspension were not stabilized by clay-like interactions, it would settle fairly rapidly and need good shaking before use. (I shake it anyway.)
Corrosion on aluminium is stifled by an adherent, quite insoluble oxide, but not completely: the oxide layer continues to thicken, and corrosion continues, diminishing, but never stopping completely. Nature never sleeps.
Iron rusts with $\ce{H2O}$. Iron oxide could be adherent (it is on stainless steels, when mixed with some chromium), but unfortunately, pure iron oxide tends to form a hydrate which has a greater volume than the oxide, and as it forms, it peels away from the metal surface, leaving bare metal or a very thin layer of oxide, which can still corrode. Chloride ion is especially effective in promoting the continuation of corrosion on iron and steel (even stainless) and aluminium.

pH also plays a role. Low O2 concentration obviously slows corrosion of iron, but corrosion can be further reduced by high pH. Under these conditions iron tends to form magnetite which is much more effective at providing a barrier to further corrosion than the familiar brown rust. On the other hand, high pH dissolves the oxide layer on aluminium, causing it to corrode much quicker.

@LevelRiverSt: I agree! As for Mg(OH)2, low pH totally dissolves it, so there is no protection at all. It's interesting that dilute HNO3 will dissolve iron, but concentrated HNO3 passivates it. And HNO3 won't dissolve aluminum rapidly at all.

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text:
Calculating the pH of a 'buffer' solution?

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$.

The given solution:
Concentration of acid: $\frac{0.030}{0.300} = \pu{0.10 mol dm-3}$
Concentration of base/salt: $\frac{0.020}{0.300} = \pu{0.067 mol dm-3}$
$$\ce{[H+]} = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times 0.10}{0.067} = \pu{1.9 \times 10^{-3} mol dm-3}$$
$$\mathrm{pH} = 2.72$$
Now, here is why I am confused: this calculation is for a buffer solution, where $\ce{HA}$ and $\ce{A-}$ are a conjugate acid/base pair. However, chloroethanoic acid and sodium hydroxide are not conjugate acid/base (are they?) - so why does this calculation work? What is the intuitive way to think about this?

$\ce{NaOH}$ is reacting with the acid to produce it's conjugate base and thus the calculation. Please consider typing the question out instead of using images so that it is searchable. Question in the form of images are quickly closed due to sites policies.

@NisargBhavsar - Thank you, but I am not sure I understand. What is the conjugate base of NaOH!? And thank you for the feedback - I will edit the question now :)

I mean that the hydroxide will react with chloroethaonic acid to form sodium chloroethanoate which is the conjugate base of chloroethanoic acid.

Please do not attach wordy excerpts as an image. Just type it in.

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$.

I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of chemical reactions and buffers. That's probably why OP seems so confused of the solution. The answer elsewhere explains to certain level but has not explore the total picture. Hence, I thought it would be beneficial to OP if I explain it a little further:
$\ce{CH2ClCOOH}$ is a weak acid (about 100 times acidic than acetic acid) with $K_\mathrm{a}$ of $1.3 \times 10^{-3}$ (which is $1.8 \times 10^{-5}$ for acetic acid). On the other hand, $\ce{NaOH}$ is a strong base, which reacts with strong or weak acids irreversibly. When add $\ce{CH2ClCOOH}$ and $\ce{NaOH}$ together, following reaction would take place:
$$\ce{CH2ClCOOH (aq) + NaOH (aq) -> CH2ClCOONa (aq) + H2O (l)} \tag1$$
The net reaction is:
$$\ce{CH2ClCOOH (aq) + OH- (aq) -> CH2ClCOO- (aq) + H2O (l)} \tag2$$
The acid to base mole ratio of the reaction is 1:1. However, added amounts of acid and base are:
Weak acid, $\ce{CH2ClCOOH}$: $\pu{0.50 mol dm-3} \times \pu{0.100 dm3} = \pu{0.050 mol}$
Strong base, $\ce{NaOH}$: $\pu{0.10 mol dm-3} \times \pu{0.200 dm3} = \pu{0.020 mol}$
That makes $\ce{NaOH}$ the limiting reagent (recall your general chemistry knowledge). Thus, $\pu{0.020 mol}$ of $\ce{NaOH}$ reacts with $\pu{0.020 mol}$ of $\ce{\ce{CH2ClCOOH}}$ to give $\pu{0.020 mol}$ of $\ce{\ce{CH2ClCOONa}}$ or $\ce{\ce{CH2ClCOO- + Na+}}$, if you look at the net reaction.
The unreacted amount of $\ce{\ce{CH2ClCOOH}}$ in the mixture ($\pu{0.300 dm3}$ in total volume) is: $\pu{0.050 mol} - \pu{0.020 mol} = \pu{0.030 mol}$.
Therefore, the concentration of acid, $\ce{\ce{CH2ClCOOH}}$: $\frac{\pu{0.030 mol}}{\pu{0.300 dm3}} = \pu{0.10 mol dm-3}$
Simillarly, the concentration of salt, $\ce{\ce{CH2ClCOONa}}$ or $\ce{\ce{CH2ClCOO-}}$: $\frac{\pu{0.020 mol}}{\pu{0.300 dm3}} = \pu{0.067 mol dm-3}$
Since, $\ce{\ce{CH2ClCOOH}}$ is a weak acid, it would ionize in aqueous solution as follows:
$$\ce{CH2ClCOOH (aq) + H2O (aq) -> CH2ClCOO- (aq) + H3O+ (aq)} \tag3$$
Suppose, $\alpha$ amount (in $\pu{mol dm-3}$) of $\ce{\ce{CH2ClCOOH}}$ ionizes at $\ce{\ce{CH2ClCOOH}}$, the equilibrium concentrations of $\ce{\ce{CH2ClCOOH}}$, $\ce{\ce{CH2ClCOO-}}$, and $\ce{\ce{H3O+}}$ are $0.10 - \alpha$, $0.067 + \alpha$, and $\alpha$, respectively.
$$K_\mathrm{a} = \frac{\ce{[H3O+]}\ce{[A-]}}{\ce{[HA]}} $$
$$\therefore \ \ce{[H+]} = \alpha = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times (0.10  - \alpha)}{(0.067  + \alpha)} \approx \pu{1.9 \times 10^{-3} mol dm-3}$$
$$\mathrm{pH} = -\log\alpha = -\log (1.9 \times 10^{-3}) = 2.72$$
On the other hand this can be easily solved using the Henderson-Hasselbalch equation as directed in the other answer elsewhere:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac {\ce{[A-]}}{\ce{[HA]}}\right) = 2.89 + \log \left(\frac {0.067}{0,10}\right) = 2.72 $$

$\ce{CH2ClCOOH}$ is a weak acid; $\ce{NaOH}$ is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid?
They react.
$$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O} $$
A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base.
$$N_1V_1=N_2V_2 $$
At this stage, an equivalence point is reached. However, the limiting reagent in the given question is $\ce{NaOH}$. At the equivalence point, the solution will contain $\pu{0.03 mol}$ of $\ce{CH2ClCOONa}$ and $\pu{0.02 mol}$ of excess reagent $\ce{CH2ClCOOH}$.
Now, $\ce{CH2ClCOONa}$ is a salt of a WEAK acid/STRONG Base while $\ce{CH2ClCOOH}$ is the WEAK acid.
Now, I believe you can refer to the solution given in your book.
Calculate the concentration of the respective constituents in the equilibrium, and then proceed with plugging those values into the Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac {\ce{[A-]}}{\ce{[HA]}}\right)$$
where $\mathrm{p}K_\mathrm{a}$ is the acid dissociation constant of  $\ce{CH2ClCOOH}$.


text:
How does the application of Schrödinger equation to model a particle in a box explain the origin of degeneracy of atomic orbitals?
Is the particle in a box concept analogous to an electron in an orbital? If so, can we apply the equation for the allowed energies of a particle in a box
$$E = \frac{h^2n^2}{8mL^2}$$
to the allowed energies of electrons in orbitals? Are the orbitals degenerate because this equation only includes the quantum number $n$ and not the others and therefore all orbitals in a subshell will roughly be degenerate?
Finally, if this equation can be used to explain the allowed energies of orbitals/electrons, what would $L$ be in this instance?

No, that would be too much of a stretch.

1D or 2D box? 1D has no degeneracy

Electrons in orbitals do not have spatial constrains like particles in a box. There is no sharp cut off of probability density. Particular borders of orbital 3D shapes are just arbitrary convention of probability density threshold and the residual integral probability threshold.

I was referring to 1D. Ok understood. So how would you guys go about answering the question?

I have to ask, what is the source of this question? Curiosity or homework or ...?

This question is on an assignment I was given for university level chemistry. "How does the application of the Schrodinger equation to model system such as particle in a box help us understand the origin of the degeneracy of atomic orbitals". I have been racking my brain all week and can't get it.

My answer to another post tries to explain a link between the problems: https://chemistry.stackexchange.com/questions/149738/schr%c3%b6dinger-equation-and-degeneracy-of-atomic-orbitals/149741#149741 This seems unlikely to be a good question for intro chem though. That's why I asked why you were posting this.

Is the particle in a box concept analogous to an electron in an orbital?

No, but there is an analogy to the $\pi$ systems of dye molecules, and (for the 3D box) to the band structure of nanodots. Some physical chemistry courses have a lab that explore this relationship, e.g. at Saarland university (sorry, in German, but the figures and the math are universally understandable, I hope).

Finally, if this equation can be used to explain the allowed energies of orbitals/electrons, what would L be in this instance?

It would be the length of the conjugated system (for the dyes) or the radius (for the nanoparticles)

Are the orbitals degenerate because this equation only includes the quantum number n and not the others and therefore all orbitals in a subshell will roughly be degenerate?

For a 1D case with degeneracy, you can explore the particle in a ring, Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems?.


text:
Exceeding CPVC in exterior roof coatings
We have a kynar coated metal roof that was primed with an acrylic primer and then coated with a dark blue  acrylic elastomeric paint. It was coated during spring with moderate weather. The roof surface was dry. Within a few days the coating blistered severely.  The blisters were cut out; only the coating was delaminated, the primer was well bonded. Some of the blisters held moisture, most were dry. The blisters were cut out and the roof was reprimed and recoated. The coating blistered again as it had before. It was then revealed that the dark blue coating had been blended without a tint base; the pigment was added directly to a white or gray (or a mixture thereof) paint. We suspect that the CPVC was greatly exceeded when the blue pigment was added to the white and/or gray paint, would this cause the coating to blister? What test needs to be performed to determine the PVC?

Could you please define PVC and CPVC? I guessed they were acronyms for polyvinyl chloride and chlorinated polyvinyl chloride, respectively, but that does not seem to work with “determine the PVC” and “the CPVC was greatly exceeded”.

PVC  in the paint industry is an acronym for Pigment Volume Concentration; CPVC is the Critical Pigment Volume Concentration. (See article at this site -  link below) https://chemistry.stackexchange.com/questions/142980/is-there-an-experimental-way-of-measuring-the-critical-concentration-of-pigment

Ah, now it makes sense! Unfortunately, I know nothing about this, but your link may be useful: maybe post a comment to James Gaidis at the link you provided. He is a helpful fellow.


text:
Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured?
Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule. In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule. Does this mean an actual electron was transferred given that oxidation numbers are fictitious? I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

Yes, electron transfer can happen. Of course, in some cases, the molecule fragments, or there is a transfer of a fragment, which is considered to be "carrying" the electron.

Usually a change in oxidation number is an electron transfer.  But this transfer is not easy to discover. For example,  the air oxidation of carbon to produce $\ce{CO2}$ is due to $$\ce{C +O2->CO2}$$ The electron transfer is not obvious in this case. It is usually forgotten. It may even be ignored, as nobody has ever taken the initiative of explaining the combustion of carbon by the following half-equations $$\ce{C + 2 H2O -> CO2 + 4 H+ +  4 e^-}$$ $$\ce{O2 + 4 H+ + 4 e^- -> 2 H2O}$$ whose sum is $\ce{C + O2->CO2}$

Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule.

Sort of. In calculating oxidation numbers we pretend that covalent bonds are ionic. Perhaps it is valid to say we are pretending the entire molecule is an ionic compound.

In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule.

More accurate, we find the that oxidation states of atoms change going from reactants to products. Charge conservation, coupled with the way we assign oxidation states, guarantees that the sum of oxidation states is constant. So if one atom shows an increase in oxidation state, at least one other has to show a decrease in oxidation state.

Does this mean an actual electron was transferred given that oxidation numbers are fictitious?

It is not clear what "an actual electron" means. Electrons are indistinguishable, you can't paint one red and follow its path during a reaction. If the electron transfer (according to oxidation numbers) is intramolecular, it becomes a bit of a philosophical question whether there was an electron transfer. For example, is there an electron transfer during keto-enol tautomerism?

I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

It is not clear what "physical electron jumps" are.
For some reactions, it is clear that you have an electron transfer because the only thing that distinguishes reactants from products (pairwise) is the number of electrons:
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
If this reaction is run as two half-cells, this is even more clear-cut.
For other reactions, the changes are more subtle. If a iodine radical combines with a methyl radical (as part of a chain termination reaction), it is formally a redox process:
$$\ce{I. + .CH3 -> I-CH3}$$
However, the electronegativities of iodine and carbon are quite similar, so it is not intuitive that an electron transfer happened. As Maurice points out in the comments for a similar case, you would have trouble writing this in terms of half-reactions.
In the reaction of methyl lithium with a carbonyl, the oxidation numbers change. You might not see it as a electron transfer and would rather think of it as methyl transfer. However, because carbon is more electronegative than lithium, the oxidation numbers of the carbonyl carbon and the methyl anion change when a bond is formed between them.

Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured?

If I answer this like a certain politician: it depends upon what the meaning of the phrase "actual electron transfer" is. So it is mostly a question about how we talk about redox reactions rather than a question about the underlying chemistry.

The sum of oxidation numbers is equal to the difference between the total proton count and total electron count.
So yes, it means the real electron transfer, as oxidation means the increase of the sum of oxidation numbers and reduction the opposite.
It also means that while the summary oxidation number is the real thing, oxidation numbers of particular atoms are the arbitrary convention.

Yes, the electron transfer does take place...oxidation is the loss of electrons and reduction is the gain of electrons. In a redox reaction, the oxidizing agent gains electrons and the reducing agent loses electrons, i.e, actual electron transfer takes place.


text:
How to run electrochemical tests on Microorganism (bacteria in a medium)?
I am trying to run some electrochemical tests(e.g.CV (Cyclic Voltammetry) & EIS (electrical impedance spectroscopy)) using half cell configuration on micro-organisms (e.g.bacteria). The bacteria are cultured in a medium (liquid). The half cell configuration is working, counter, and a reference (AgCl) electrodes.

Usually, it is easy to clip a metal or any thin solid material. However, I don't know how to deal with a solution. Since the beaker will be filled with electrolyte, I am afraid it mixes with bacteria solution causing it to be removed from the electrode. I am avoiding add another material like tape or something as it might alter the measurement.
I am looking for a way to attach the Micro-organism to the working electrode.

Try using a conductivity cell along with an AC source!

Can you pre-grow the organisms on the eletrode as a biofilm?

It is generally good practice to define acronyms when first used. So, please define EIS! (I already know what CV means in your post.)

Yes, i was not aware of what they specifically were and couldn't even find it on google as all top posts were those related to resumes,etc. when i searched for CV....thus, i ended up giving the apparently wrong answer ://

I have proposed an edit, including the full-forms and links to what google searches yield for CV and EIS, i.e, cyclic voltammetry and electrical impedance spectroscopy.

I defined them in the post. I apologize.

it's okay @Anwar Elhadad, I had proposed edit for the same but thank you for your quick action! I wasn't aware of what CV and EIS were before, so gave a pretty general answer that might not suit your experiment, sorry about that

From what I understand, your main issue is connecting a solution in an electrical circuit without changing the composition of the solution due to electrochemical redox reactions.
So, there are two problems here.  Firstly, passing direct current
(DC) changes the composition of the solution. Secondly, a solution cannot
be connected to the bridge like a metallic wire or any other solid conductor.
The first difficulty is resolved by using an alternating current (AC) source
of power. The second problem is solved by using a specially designed
vessel called conductivity cell.

The solution to be connected in the circuit should be put in the conductivity cell, and an alternating current (AC) should be used. The theory behind this is that the two platinised electrodes will keep switching between acting as anode or cathode as the AC source reverses direction 50 times every second, so the ions remain free and move about in the solution and don't get deposited on the electrodes.You can read more about conductivity cells and alternating current if you want. The cell constant,G*,is calculated for the conductivity cell during the manufacture process and is printed on top

The OP specifically mentioned cyclic voltammetry (CV) and the need to attach the micro-organisms to the working electrode. So your conductivity answer is not what was desired.

okay, sorry I wasn't aware of what CV and EIS meant, as it wasn't mentioned in the question....my bad

Now that we have conformation that EIS is electrical impedance spectroscopy (I thought that was what it was, but knowing is better than probably knowing), maybe your answer could evolve to indicate how an EIS cell might be used or how it differs from a typical conductivity cell?

Yes, of course! I am reading up on that and researching right now, and will edit my answer as soon as i find a relevant solution...

Your reference electrode contains chlorine and silver, the ions from these elements can kill bacteria, you should avoid liberating gases too: oxygen is killing anaerobic bacteria. To avoid using metals for your electrodes (they react too much with bacteria: poison, catalyzer etc...) you can use graphite or conductive polymer (without silver). I don't know if bacteria would colonize carbon, but graphite is antiadhesive. You can put your bacteria on a substrate: paper and or gel placed in the liquid with electrodes fixed at 2 edges.


text:
Is this the NMR spectrum of 3-methyl-1-butanol?

It is a $\ce{^1H}$-NMR spectrum and I think the signal near 3,61 ppm is the CH2 group next to the hydroxy group. But which signal is the hydroxy group and which protons are in the multiplet?

Does this answer your question? 1H NMR splitting of 3-methyl-1-butanol

No, this is actually 1-hexanol spectrum.

Welcome to ChemistrySE.
In short: is the spectrum consistent with 3-methyl-1-butanol? No. The triplet splitting and integral of the signal at 0.86ppm is inconsistent with an isopropyl sidechain. This suggests a methyl group attached to a methylene, $\ce{-CH2-CH3}$.
The signal at $\pu{3.6 ppm}$ has a chemical shift symptomatic of an alcohol, as you suggest. It has diagnostic splitting of $\pu{4.2 Hz}$, also consistent with coupling to the alcohol peak. So, this peak is a doublet of triplets..... which makes it look a bit like a quartet. To see this coupling, the sample needs to very dry to eliminate exchange decoupling. So where is the alcohol peak? In the absence of any exchange, and in chloroform (most likely solvent used given the spectrum shown), alcohol peaks usually come in the region of about $1.2$-$\pu{1.5 ppm}$. So possibly under the multiplet. Which would then leave you with 6 protons to account for......or $3 \times \ce{CH2}$ groups. The peak at $\pu{1.5 ppm}$ has coupling that suggests it is attached to the $\ce{-CH2OH}$, giving a $\ce{-CH2CH2OH}$ group. There are 14 protons in all. You've got a $\ce{-CH2-CH3}$, a $\ce{-CH2CH2OH}$ and the rest are probably $\ce{-CH2\!-}$. How might you put them together?
In the real world, identification of complete unknowns are rarely done using $\mathrm{^1H \ NMR}$ alone. Mass Spec, $\mathrm{^{13}C}$, and a range of $\mathrm{2D}$ options are very useful.
And to answer a follow-up question below, the splitting at 1.54 is from the protons on the beta carbon, and is a triplet of triplets. When coupling constants are close (6.5 and 6.8) this will appear as a pseudo-pentet, as shown below:

Do you also know which solvent is used?

Not conclusively. CDCl3 is the most common solvent used for small organics - its residual peak comes at 7.26ppm, and so not visible in this spectrum. Other common possible solvents (dmso, methanol, acetone) are not visible in this spectrum.

Thanks for your answer! One last thing I want to know.... Can you explain the splitting at 1,54Hz?

This is due to the protons on the beta carbon (the middle ones in this group) -CH2-CH2-CH2OH. They are coupled to the terminal CH2 and also the gamma CH2, so will be a triplet of triplets. When the coupling is very close, (6.8 and 6.5), this pattern will look like a pseudo-pentet. I'll put the splitting pattern in the answer above. Feel free to accept the answer to close it off if it addresses what you were after.


text:
Sequential reaction in equilibrium

Derive an equation for the steady-state rate of the sequence of reactions $$\ce{A <=> B <=> C <=> D}$$ with $[\ce{A}]$ maintained at a fixed value and the product $\ce{D}$ removed as soon as it is formed.

I Tried solving this question, but it got really complicated. Therefore, I was not sure whether my method and approach to solving it were correct or not. I tried to use Steady-state approximation for A since it is maintained at a constant concentration as per the question and I considered the concentration of D = 0 all the time. Then, I considered [A] = [Ao - a], [B] = [a - b], [C] = [c] and [D] = 0.

You could try this. As D is removed so its back reaction does not happen, and A is 'maintained'  as fixed then the scheme could be changed to $\overset{k_0}\to A \overset{k_1,; k_{-1}}\leftrightharpoons C\overset{k_2}\to $ where B is 'lost' in the rate constants due to its equilibrium.

Square brackets denote concentration, so $[\ce{A}]$ reads as concentration of compound $\ce{A}.$ Writing $[\ce{A_0}]$ makes no sense as it suggests concentration of another compound $\ce{A_0}.$ Initial concentration should be written as $[\ce{A}]_0.$ Similarly, $[\ce{A_0} - a]$ makes even less sense as you cannot perform math operations on chemical compounds themselves.

Judging by the font and colors, this appears to be a question from one of the Atkins' Physical Chemistry textbooks. Please add a proper reference to it, and also pay attention to the notations the book uses. They are the correct ones, yours are flawed.

$A\xrightleftharpoons[k_{-1}]{k_1}B\xrightleftharpoons[k_{-2}]{k_2}C\xrightleftharpoons[k_{-3}]{k_3}D$
I believe the rate of the overall reaction is $\frac{d[D]}{dt}$ = $k_3[C]$ since D is removed, $k_{-3}$ won't affect the reaction. Now we can apply steady-state approximation to B and C:
Steady-state approximation for B:
$\frac{d[B]}{dt} = 0 = -k_2[B] - k_{-1}[B] + k_1[A]$
$k_2[B] + k_{-1}[B] = k_1[A]$
$[B] = \frac{k_1[A]}{k_{-1}+k_2}$  (1)
Steady-state approximation for C:
$\frac{d[C]}{dt} = 0 = -k_3[C] - k_{-2}[C] + k_2[B]$
$k_3[C] + k_{-2}[C] = k_2[B]$
$[C] = \frac{k_2[B]}{k_3+k_{-2}}$ (2)
We then can substitute (1) into (2) to get
$[C] = \frac{k_2\frac{k_1[A]}{k_{-1}+k_2}}{k_3+k_{-2}}$
$[C] = \frac{k_2k_1[A]}{(k_3+k_{-2})(k_{-1}+k_2)}$ (3)
We can substitute (3) finally into $\frac{d[D]}{dt}$ = $k_3[C]$ to get the expression for $\frac{d[D]}{dt}$:
$\frac{d[D]}{dt} = \frac{k_2k_1k_3[A]}{(k_3+k_{-2})(k_{-1}+k_2)}$
This should be the steady-state rate for the reaction.


text:
Compare reactivity of carbocations and stability of carbanions, radicals, and carbocations

For ranking the carbocations in the image above,
I thought of taking the approach that the 1st molecule, the secondary carbocation was the most reactive because it has no electron donating group, then the carbocation with a terminal NH2 because nitrogen has an electron donating group, then the tertiary carbocation, and lastly the last carbocation has resonance making it very stable and less reactive. Is this the correct train of thought?

Please split your post into two separate questions, questions should focus on one thing at a time, not multiple ones.

Order of stability: 3 > 2 > 1 > 4
I would say that the third structure is the most stable (least reactive) as an octet-complete resonance structure of it exists which is much more stable that the initial carbocation. This is basically what you said, the carbocation is stabilized due to the electron donation tendency of the nitrogen atom in this case.

Then structure 2 then 1 since 2 is tertiary while 1 is secondary. More the alkyl groups, more the positive inductive effect stabilizing the carbocation.
I would say that the fourth structure is the least stable. If you try to draw the resonance structure in which oxygen ends up donating electrons stabilizing the carbocation, you'll notice that the oxygen ends up with a positive charge which is even more unstable than the initial carbocation. Resonance does not help here.

(the structure above should have a positive charge on the oxygen; I wasn't able to draw that using the software available)
Do note that resonance structures don't actually exist and there are better ways to explain more or less the same thing. I've stuck with resonance since you've mentioned that in the question.


text:
How to model 98% ethanol solution concentration graph over time
İ want to plot a graph of ethanol solution concentration over time .So we know that alcohol evaporates quickly in open air but is there a time when the rate of evaporation mathces that of water and concentration stops decreasing or increasing? İ know surface matters but lets assume surface is totally hydrophobic and doesnt interact with solution.Any ideas how to do it?

The concentration drop stops at 95.5 w/w % of ethanol, the composition of azeotrope mixture with the minimum boiling point 78.1 deg C ( b.p. of pure ethanol is 78.4 deg C )

As Poutnik pointed out, if you boil a 98% ethanol solution (2% H$_2$O), then you will lose a few % of ethanol until you get to ~95.5% ethanol; then the rest of the solution will boil off as if it were a pure compound, for the composition will not change. Well, this is strictly true at the high temperature of boiling at 1 atm pressure. At about 0.1 atm, the azeotrope breaks, and so at that boiling temperature (about 35$^o$C, from a graph in https://wps.prenhall.com/wps/media/objects/3311/3391416/blb1105.html),

there is no azeotrope. I suspect that at lower temperatures than 35$^o$C, there may be no azeotrope. Nevertheless, if you start with a 98% ethanol solution, it won't change much until you have exported off most of the ethanol (like ~75% or so).
I would suggest cutting the concentration to about 50% ethanol, put it in a large beaker (~1 liter), stir with a small magnetic stirring bar at low speed to keep the contents somewhat homogeneous, put in a thermometer and a hydrometer with a range of 0.9 to 1.0 and engage a mild cross current of air across the beaker.
You will be able to determine the concentration of ethanol from a diagram like this:

from  https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html using the thermometer and hydrometer. Time is a variable, but is useless for determining any general scientific principle. The conditions of your experiment include air pressure (you can get that), air current (you can describe that - if you don't use any air current, it will take hours or days to do your experiment).
You can read off the volume of solution you have from the graduations on the beaker, and you can then plot the composition of the solution as its volume decreases by evaporation. The measurements are unfortunately not very accurate, but you will be able to make a graph and then spend several pages describing your lab work and potential errors and look up graphs and tables like the ones I have used. A likely error will be that the temperature will not be constant, the liquid level will drop so that the air current will have different strengths.
And as MaxW pointed out, the modeling may not be very "realistic". But I think people will be impressed with your use of scientific equipment and your analysis (which will be of the highest caliber, I'm sure).
While 50% ethanol is a lot less flammable than 98%, it still is a fire hazard, especially in quantities as big as a liter. Not only the liquid, but also the vapor is hazardous, so make sure that there are no flames or sparking devices within about 20 feet of your experiment. And have good overall ventilation: ethanol vapor can put you to sleep or make you dizzy. For extra security, you should plan to have a fire extinguisher handy (and know how to use it).

I don't think that any realistic modeling is possible. There are a lot of details which would made such a system chaotic. For the liquid side of the gas/liquid system consider the differences in evaporating from a thin layer in a very large diameter vessel to evaporation from a extremely long tube of the mixture.
For the gas side how is the diffusion away from the surface to work? Is there an equilibrium being formed between the phases, or is the water/ethanol which evaporates being swept away by some forced ventilation? Also, what is the initial relative humidity of the air? Presumably there would be no ethanol initially.
Also any evaporation would tend to cool the liquid. So what to do about the heat flow into the system?


text:
Second-step activation energy in SN1 reactions
In SN1 reactions, it's known that there are generally two steps, with the first being the R.D.S step involving carbocation formation and the next having a lower activation energy.
Why does the second step require activation energy at all? It's going from being a carbocation to a stable molecule, which will be more stable than the former, no matter how stabilized a cation it is.

Barrierless reactions, i.e. reactions without activation energy, are a thing. I don't really want to comment on whether that step is barrierless. But I do want to point out that your general argument seems to be confusing energies of reactants and products (thermodynamics) with activation energies (kinetics). Just because a product is more stable than a reactant, doesn't mean that there is not going to be some kind of activation energy.

Oh, yes, 'rate'-determining step. Kinetics. I get it, thanks.

But @orthocresol: activation energy is a thermodynamic thing, right? That measures stability, not rate of reaction. Which is what the answer seems to be going with, too.

Well, yes, activation energy is an energy, which has to do with stability. But it is the energy of the transition state that is important. You were talking about the energy of the product, which is not relevant.

Let’s also make sure to separate the concept of energy from the concept of thermodynamics. Thermodynamics is concerned with the equilibrium properties of the system, kinetics with the rates of the reaction. That’s all it is; there isn’t any classification where thermodynamics is concerned with energies, and kinetics with other things. Just because something is an energy doesn’t necessarily mean it’s a thermodynamic property.

Okay, so here, the activation energy involved is related with the stabilities of the intermediate carbocation and the second-step transition state, which are both qualities describing an ongoing reaction, describing how fast it happens- thus they are related to kinetics. The product-reactant energies are the equilibrium properties of the system, determining if the reaction happens, so they're thermodynamics-related. Have I got it right now?

Yes, that's spot on.

The $\mathrm{S_N1}$ reaction has two steps i) the formation of carbocation and 2) the formation of product.

(image taken from libretexts)
In the second step an unstable carbocation reacts with the nucleophile. You are asking why this step has to have an activation energy. You can rationalize that by considering the principle of microscopic reversibility. Think about what would happen if the $\mathrm{S_N1}$ reaction went in reverse—first the nucleophile leaves the product, forming a carbocation, and then the carbocation joins with the leaving group to give the reactant. Obviously when the nucleophile leaves the prdouct, and carbocation forms, there is an activation barrier for that.
So regardless of the themodynamics of the total reaction, whenever you detach a functional group and form a
(carbocation + anion), there has to be an activation barrier i.e. a transition state that is higher in energy than both (carbocation + anion) and the initial state. Which means that if you wish to reattach an anion to the carbocation, there will also be an activation energy.


text:
Why can't NAD+ undergo single-electron reduction?
$\ce{FAD}$ can undergo single-electron reduction to form a stable radical, which can then be reduced again to $\ce{FADH2}$. This is supposedly possible due to resonance stability, where the unpaired electron is delocalised amongst the molecule. In $\ce{NAD+}$ however, the free radical formed is very unstable, so it can only undergo hydride-mediated (or two-electron) reduction.
This sounds simple by itself, but the diagram I found (shown below), in my opinion, contradicts this claim.

It seems to me that $\ce{NAD+}$ is just as capable in forming resonance structures as $\ce{FAD}$; the double bonds are adjacent to the unpaired electron and the lone pair on the carbon (in $\ce{NAD+}$) should be in a p-orbital, which also allows it to be delocalised. It is true that $\ce{FAD}$ is larger and has more aromatic rings, which makes it a better candidate to undergo single-electron reduction; however, $\ce{NAD+}$ should ALSO be able to form radicals, albeit not as often. Contrary to that, it is said that the $\ce{NAD}$ radical is very rare or even non-existent.
The source I mainly used (due to lack of others) is this Chemistry LibreTexts site. If anyone can point out the reasoning behind this oddity, I would appreciate it very much.

The difference between “not as often” and “very rare” is semantic. There is also the issue of evolved function. NAD is not utilized by enzymes that catalyze one electron transfers


text:
Do hydrides, like NaH or CsH, also form H-bond?
I know that when hydrogen acts as a cation, it can form H-bonds with the electronegative ions or groups. But, is this also true for anionic hydrogens? Will they form similar bonds, other than the existing ionic bond, with electropositive ions, like the alkali metals and alkalikne earth metals, or maybe some other electropositive group?

Do the bonds in B2H6 count?

And how are you imagining this? Do you know how hydrogen bonds work, because it sounds like you talk about something else.

Particularly bridging hydride ligand that @IvanNeretin mentions and are a very different thing then hydrogen bonds. There are also agostic interactions, which are like an analogue of hydrogen bonds, but "hydridic" instead of "protic".

@Mithoron.. H-bonds are formed because of the polarisation effect due to unequal sharing of electrons between the ions.. if I'm not wrong, then shouldn't any molecule having a dipole moment show coulombic attraction among its neighbours?

The point of hydrogen bonds is they are something more then dipole interactions.

On the surface, the answer is no, because that's not what we would call a hydrogen bond.
On a deeper level, the answer is... probably still no for NaH or CsH, because they are too ionic for this. Other hydrides, however, actually do have a thing which looks mighty like an "inverted" hydrogen bond (though we don't normally call it that, either).
Really, look at it this way: in $\ce{H2O}$ we have a hydrogen with partial positive charge, which gets attracted to a lone pair of another electronegative atom and forms something like a (3c, 4e) bond.
Now look at another scenario: a hydrogen in $\ce{BH3}$ (or better yet, in $\ce{AlH3}$) bears partial negative charge and gets attracted to a less electronegative atom with an empty orbital, so as to form a (3c, 2e) bond.
Whether to perceive these two scenarios as similar or as opposite is up to you.
So it goes.


text:
Why is the increase in covalent radius from As to Bi not as big as from N to P?
The following is the radius of Group $15$ elements:
$$\begin{array}{c|c}  \hline
 \text{Element} & \text{Covalent Radius }(\pu{pm}) \\ \hline
\ce{N} &75 \\
\ce{P} &110 \\
\ce{As} &121 \\
\ce{Sb} &140 \\
\ce{Bi} &155 \\ \hline
\end{array}$$
Source:
General Properties and Reactions. (2020, August 15). Retrieved April 6, 2021, from https://chem.libretexts.org/@go/page/31730
From the above table it is clear that the increase in radius decreases as we go down the group.

My Attempt to Reason it
The electronic configurations of $\ce{N, P}$ and $\ce{As}$ are:

$\ce{N: [He]} \ \mathrm{2s^2 2p^3}$
$\ce{P: [Ne]} \ \mathrm{3s^2 3p^3}$
$\ce{As: [Ar]} \ \mathrm{4s^2 3d^10 4p^3}$

First let us compare $\ce{N}$ and $\ce{P}$
$\ce{P}$ has $\ce{8 e-}$s and $\ce{8 p+}$s more than $\ce{N}$. The $\ce{8 p+}$s increase the nuclear charge. But, due to the addition of $\ce{e-}$s, the shielding effect also increases i.e., there is more repulsions caused by the core $\ce{e-}$s.
Out of the $\ce{8 e-}$s, $3$ goes into $\mathrm{2p}$, $2$ goes into $\mathrm{3s}$ and $3$ goes into $\mathrm{3p}$. The shielding effect due to $\mathrm{s}$ and $\mathrm{p}$ is appreciable. As a result, the effective nuclear charge is not that high. But due to the addition of extra shell i.e., $3$, the size increases by almost $\pu{35 pm}$.
Now Let us compare $\ce{P}$ and $\ce{As}$
Now in $\ce{As}$, there are $18 \ce{e-}$s and $18 \ce{p+}$s more than that in $\ce{P}$. Out of the $18 \ce{e-}$s, $3$ goes into $\mathrm{3p}$, $2$ goes into $\mathrm{4s}$, $10$ goes into $\mathrm{3d}$ and $3$ goes into $\mathrm{4p}$.
In this case, majority of the $\ce{e-}$s  are in the $\mathrm{3d}$ orbital. The shielding effect due to $\mathrm{3d}$ is not much. In fact, the order of the ability of orbitals to shield is:
$$\mathrm{s} \gt \mathrm{p} \gt \mathrm{d} \gt \mathrm{f}$$
Due to this, the effective nuclear charge is more and therefore the size doesn't increase as much as it increase on going from $\ce{N}$ to $\ce{P}$.
Is my reasoning correct? If not, what is the reason behind this?


text:
Why do green bell peppers become bitter when cooked?
Why do green bell peppers taste sweet raw, but become bitter when cooked (steamed)?
What chemical process is responsible for this? I don't believe the temperatures are high enough for the Maillard reaction

They do not become bitter but do lose some sweetness. The taste of any fruit or vegetable is a complex function of growing conditions, sunshine , soil nutrients, water availability, ripeness, storage, temperature of ingestion. That is what makes growing, choosing, cooking, eating food such a challenge, delight or in your case somewhat of a disappointment.

Thousands of different chemicals, previously spatially separated, are mixed together after the thermal death of cells and kept at high temperatures to react well. The final taste will depend on the original system but will develop by way unpredictable theoretically from basic principles.
The answer for a particular biological source requires detailed knowledge about particulars of green bell peppers biochemistry, which is rare.
Generally, there are basically few, possibly combined, options:

Bitter-tasting compounds - possibly, but not exclusively alkaloids - are already there, but encapsulated in cellular structures and released during cooking.
They are actively created by enzymes released after cell deaths ( like alliin transformed by the enzyme alliinase to allicin that is responsible for the aroma of fresh garlic.
Thermal decomposition of some substances creates such bitter compounds.

Further reading:
bitter-bell-peppers provides rather the opposite opinion:

Why cooked peppers are less bitter
Have you noticed that raw bell peppers taste significantly more bitter when eaten raw (for crudités or in a salad)? That's because most bitter compounds in food break down with heat. The bitterness doesn't go away completely, but it is lessened.

How to Take the Bitterness Out of Bell Peppers:

Select ripe bell peppers. Green bell peppers are harvested before they fully ripen, which causes bitterness. Choose red bell peppers; they are fully mature and slightly sweeter than orange, yellow, or green varieties.
Remove the seeds from the cores. The bell pepper seeds are mostly in the cores, and they are naturally bitter. Slice the top of the bell peppers and place them upright. Use a knife to cut out the cores in a circular motion. Cut open the bell peppers and scrape off the remaining seeds.
Peel the skins. Put your bell peppers in a preheated oven until the skins blister. Let them cool for 10 minutes in a paper bag -- the steam will help loosen the skins. Run cold water over the bell peppers and peel off the skins.
Cook the bell peppers. Add soy sauce, salt, or other spices to give the bell peppers flavor. The cooking process will eliminate some of the bitterness, and the seasoning will help mask the taste.

chemistry of bell peppers:


text:
Why is delta-H negative here?
Very basic question here, but I'm confused why Delta-H seems to be the reverse of what I would expect for bond enthalpy. For example, here's a problem in my textbook where the goal is to find delta-H:
$$\ce{H2(g) + Br2(l) -> 2HBr(g)}$$
Using the table I have, I see that an $\ce{H-H}$ bond has an average enthalpy of 436, $\ce{Br-Br}$ has 193, and $\ce{H-Br}$ has 366. 436+193 = 629, and 2*366 = 732, so the reactants have a total of 629, compared to the products, which have 732.
My intuition (and Google, as far as I can tell) would tell me that, because the H value increased when the reaction occured, the answer would be positive 103. But my textbook says the answer is -103. This makes no sense to me - what am I missing here? Why do all the Delta-H (AKA Change in H) values seem to be negative what you'd expect?

You can also handle the issue this way. In an exothermic reaction the enthalpy is negative which means the bonds formed (negative) outweigh the bonds formed (positive). Just add the two values.

The bond enthalpy is conventionally the enthalpy change of bond breaking, not of bond forming, in the contrary to formation enthalpy of compounds.
This way they are all positive.
The consequence is, the order of products and reactants in computation the enthalpy difference has to be switched.
So if the summary bond enthalpy of products is higher than of reactants, the difference of enthalpy as the reaction enthalpy is negative, not positive, and the reaction is exothermic.
Formation enthalpies:
$$\Delta H_\mathrm{r}=\Sigma  \left( \Delta H_\mathrm{f, products} \right) - \Sigma \left( \Delta H_\mathrm{f, reactants} \right)   $$
Bond enthalpies:
$$\Delta H_\mathrm{r}=  \Sigma   \left(  \Delta H_\mathrm{b, reactants} \right)   - \Sigma  \left(  \Delta H_\mathrm{b, products}  \right) $$


text:
Wurtz reaction for secondary alkyl halide
I have found Wurtz reaction for primary alkyl halide in my textbook. My textbook also says that only primary and secondary alkyl halide shows Wurtz reaction and tertiary alkyl halides doesn't show Wurtz reaction. I tried to write the Wurtz reaction for secondary alkyl halide which is not included in my textbook. I think the product is quaternary alkane. But I faced a problem. It is: Which type of quaternary alkane is produced in this reaction? Vicinal or Geminal? Please help me in this context.

Just join the two molecules by replacing two halogen atoms with a single bond. It will be the second one.

That is the Wurtz reaction

Oh, sorry! I read Chemistry textbook written in Bengali and so I didn't know the spelling of 'Wurtz'. However, I have edited it. Thank you.

@AdityaRoychowdhury Is this an elimination reaction?

You can find the mechanism and more here for Wurtz reaction. It is a coupling reaction, by the way.

@Md.HasibulHossain It is not an elimination reaction. Please read the answer linked in the comment above. Moreover, it is worth mentioning that while Wurtz reaction is generally radical, it may be ionic as well, sometimes. Maybe, it begun with the o sound in bengali so you got confused with the name.


text:
Why does methane have 4-different bond dissociation energies?
I will elaborate on my question here.
In my Lecture Notes it is mentioned that for methane, there are 4 different $\text{C-H}$ bond energies. I plotted a quick graph of the bond energies to analyse the trend.

Here the vertical axis represents the bond dissociation energy while the horizontal axis represents the bond dissociation number. Looking at the graph, there is no clear trend so I decided to turn my attention to the structure.
$\text{CH$_4$}$ has the following structure

It is from here that I am unable to proceed. Here are my questions.

Am I barking up the wrong tree by looking at the structure and how do I account for the changing bond energies
Is there a VSEPR structure for molecules like $\text{CH$_3$}$ $\text{CH$_2$}$ $\text{CH}$? If so how do I figure it out.

All bonds in methane are not broken together. Say one C-H bond is broken by heteroolysis, it is no longer methane. It becomes a methyl cation, where C-H bond enthalpy is different. Likewise, the pKa values of the two hydrogen atoms in sulphuric acid are not same.

But why? It is still a bond between C-H @AdityaRoychowdhury

@SanjayAdhith because molecules are more complicated than localized bonds between parts of them. And even if everithing would be fully localized, which is quite true for molecules as CH4, CH3 ions or radical, and so on, the potential felt by electrons isn't / can't be the same.

VSEPR is consistent with the geometries of the various radicals, but the energetics are more difficult to unravel.
The following article gives this a shot:
https://www.sciencedirect.com/science/article/abs/pii/S2210271X14001285

Also, it looks like the 2nd and 3rd bond dissociation energy values are switched (see for instance the article referenced in the answer).

See Average Bond Energies: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/atomization.html

The dissociation energies are different because bonds don't all break at once
Most chemical processes happen step by step. Even in simple molecules like methane where there are only 4 bonds, bond dissociation almost always occurs stepwise.
The implication of this is that you would not expect each step to involve the same energy (and this applies whatever sort of bond-breaking occurs; some might create a $\ce{H+}$ and a $\ce{CH3-}$ others a $\ce{H.}$ and a $\ce{CH3.}$ ).
The issue is, in general, that the second bond-breaking step is happening to a different molecule to an isolated methane. Whatever the second step is, the starting point is not a complete methane. And, since the structure of a molecule affects the properties of all the bonds in it (though the effect might be small in a large molecule where the bonds are far apart), the second dissociation will, in general, be different from the first (and the bond energy will vary as a consequence).
While, hypothetically, a single concerted breaking of all 4 bonds could happen, this is likely to be vanishing rare or extraordinarily hard to achieve.  If you could somehow do that reaction, then you would measure a different result. But that isn't what chemists are usually talking about when measuring or reporting bond strengths so any such result would be a rare oddity unrelated to most achievable results.

Ah okay that makes sense. Am I right to say that the MO will change every time a bond is broken?

@SanjayAdhith yes. The bonding structure of a molecule changes when atoms are removed.

Bond energies are important, but you have to start somewhere. Actually, I'm surprised that the bond energies for the four dissociations are so similar. But that's a good start.
Now let's get into a bit more interesting compound: ethane. Very interesting!

The table is only part of the information on ethane given in an open access article (Ref 1). Note that the first C-H bond dissociation requires 422 kJ/mol, while the second requires 150 to 465 kJ/mol, depending on which bond you are breaking, because of what product you are making. Each line in the table is worth several paragraphs of explanation.
I admire your desire to tie up your knowledge neatly here - get a lot of string, because it's a very big package!
Ref 1. https://pubs.acs.org/doi/pdf/10.1021/acs.jpca.5b01346   Branko Ruscic, J. Phys. Chem. A 2015, 119, 28, 7810–7837

This is a well put together answer but I am afraid to say that I don't think that it answers OP's question.

@Nisarg: Thanks, you're right. I was answering the question so it satisfied me. In particular, I left out VSEPR calculations because the methane data really just say that the bonding between C and H is pretty uninteresting as far as CH4 goes.  CH4 is too simple. BUT, bond energies help us to understand more complex molecules, and the bigger picture involves looking not just at the original molecule, but also at the final product.


text:
In the aquation of cis-platin, why are the chloride ligands, and not ammonia, replaced with water, despite chloride's larger trans effect?
https://doi.org/10.1021/ic000847w
In this paper, the aquation of cisplatin is examined. When used to treat cancer, the chlorides are substituted with water molecules, which seems to go against the trans rule discussed in my inorganic textbook. Supposedly ligands trans to chloride are more easily replaced than those trans to ammonia.

Check out the Wikipedia page on the Spectrochemical series. I think it probably has to do with that.


text:
What force causes an electron to jump from say Na to Cl?
Every explanation I can find says something like: valence electrons in metals have low ionising energies and non-metals ''want'' to complete their valence shell.
But I can't seem to get my head around the ''wants to complete it's valence shell'' force.
Take a neutral sodium atom and a neutral chlorine atom separated by some distance. Now move them closer until the valence electron of sodium makes the jump. What force did it feel if chlorine is neutral ? My only explanation is that the valence electron induces a dipole on the chlorine atom. In other words the presence of the valence electron pushes the electron valence orbital of chlorine away slightly. The positive nucleus of the chlorine atom is not pushed away due to shielding by the other shells in chlorine. Now that the valence shell of chlorine is pushed away slightly the valence electron of sodium begins to feel a slight attraction to the chlorine atom despite it being neutral overall. It is now a dipole. At some point the force of attraction between the valence electron of sodium and the chlorine atom is stronger than to the sodium atom and makes the jump.
Is this a reasonable explanation as to why the electron makes the jump or is it something else entirely ?

Have a look at the energy cartoon for potassium and fluorine: https://chemistry.stackexchange.com/a/128771/79678. As K and F atoms get closer and closer, it is energetically very favorable for ionization to happen. So it happens.

Ionic bonding doesn't really work like that, so believe in any "lies to students" you like.

@Mithoron. What do you mean ? Where are these "lies to the students" ?

Ionisation energy of Na is greater than Cl electron affinity. So it is endothermic process, unless the energy balance is supported by releasing of energy of ion hydration energy , or of the ionic lattice energy.

Also is worth nothing that they are not simply getting Na+ and Cl-. Technically they stick togheter. In other words, at the distance involved in your phrasing, the electrons of both elements are in a field which is due to both nuclei. If you visualize it like this it might look less magic. The "want" must be taken in the right scenario, a new substance is formed. For how the filling of orbitals works and the energy that result is, I am not ashamed to say that there is quite a mystery. But it is how it works. The forces are electromagnetic in nature, but this is just easy to be said.

Think magnetic force (due to electron spin), rather than electric. The hydride ion can also be understood this way.


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Are we technically reducing electrodes/wires for the instant that an electron passes through them?
In lecture I was taught that, in a galvanic cell at standard conditions, platinum is often used as an inert electrode when the species being oxidized or reduced exists only in solution, because platinum is "inert," which means it is very stable at its oxidation state and resists reduction or oxidation. This concept confuses me, because "Reduction" means to lose charge, a.k.a. gain an electron(s). When the electrode has an electron passing through it in order to be transferred to another material, isn't it being reduced, albeit for a very miniscule amount of time?
For example, consider a galvanic cell with platinum electrodes and an iron wire. If we were to examine the cell at an instant and examine the platinum electrodes/wire, we would see a charge of -1 on some of the platinum atoms and iron wire, correct?
What I'd like to imagine is that, while platinum is capable of having a negative charge because of the room for electrons in its d-block, the species being reduced in the half-cell will always have a higher affinity for accepting the electron due to its higher standard reduction potential. This still leaves me confused on why platinum is willing to accept the electron and thus making platinum a medium for electron transfer in the first place, though. This all applies to the iron wire (or any other element that may be used for electron transfer) as well, but I would like to tackle the case of platinum overall.

Think of the Pt electrode as a capacitor plate. The surface gets charged and the charges have to cross the metal-liquid barrier.

Reduction and oxidation are defined as gaining and donating electrons from the valence shell of an atom respectively.
Whereas in the case of electrons flowing in the electrodes and connecting wires the electrons are passing from the interatomic voids and not being released by the wires and the electrode.
(There will be some electrons which will interact with the conductor's atoms but what will happen is that they will get deflected. There will be a slim chance that a few electrons will replace an electron from the conductor's electron cloud but that doesn't change the situation as both the electrons are same in a classical sense. )
If your claim was true than the anode would have got some extra negative charge and the cathode extra positive which would have driven up the voltage higher than that expected from the cell alone.
In a similar way when we connect a wire it doesn't magically change the measured voltage.
Even when we perfom the same experiment with different inert electrodes and different kinds of wires the measured reading doesn't change much. (There are some changes in current flowing through the circuit due to the difference in resistances of different materials)

"Interatomic void" is a property of atoms I've not heard of. If the electrodes and wire are quickly pulled from the solutions during the electrolysis, what happens to the electrons still remaining in the electrodes and wire, since they did not complete their transfer? Are they stuck in this void?

@Ian No electrons will be "left" behind when you pull out the wires. The total electrons in the wire remain constant as the same number of electrons enter and leave the wire. Interatomic void is nothing but the empty space between atoms.

Why is it that none will be left behind? Is it because it's impossible to pull them out at exactly the same time, so the electrons will all be transferred either back into the anode half-cell or the cathode half-cell?

@Ian You need to understand the condition just before pulling out the wires. The electrons are flowing in and out at the same rate at either ends and thus there are no "extra" electrons in the conductor and thus there will be no electrons left behind. For example when a pipe full of water is connected in a loop of pumps working at same rates and we pull the pipe, the pipe will have same water as before it was connected. I am comparing the wire with a filled water pipe as the conductor already has the free electrons which will cause the conduction.

The answer is no. If an electric current is flowing through a piece of metal (platinum or anything else) than the same number of electrons are getting in and getting out of the piece at either ends of the piece. No atoms are reduced or oxidized. The total number of electrons present in the metallic piece does not change. There is no oxidation and no reduction in the metallic piece.


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Why does refrigerant condense in oil?
In a closed system (think of a typical AC system) why does refrigerant tend to condense and settle in oil?
I’ve heard that the vapor pressure of the oil is very low so that causes a pressure difference which draws the refrigerant into the oil and causes it to condense. The refrigerant is more dense than oil so it will tend to settle below the oil. Once enough refrigerant condenses in the oil the vapor pressure will be equal to the vapor pressure in the system so there will no longer be any flow.
I work in HVAC and I’m not satisfied with just knowing the practical side of this effect (refrigerant migration), I’d like to know how exactly this works. Maybe I lack a deep enough understanding of vapor pressure.
I understand that refrigerant and oil have different vapor pressures but I have trouble understanding how this causes flow since the entire system is under the same amount of pressure. Also why does it tend to condense into a liquid? How much refrigerant must condense and settle below the refrigerant before pressures are equalized?
Formulas are appreciated.


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Can allenes be used as motors or generators?
I was thinking about an allene, fucoxanthin:

Potentially, we can draw a resonance structure in which up to eight of the carbon-carbon double bonds have moved one position to the left, a carbocation is present somewhere along the carbon chain, and the allene is replaced by a carbanion.  We could also imagine beginning with the enol form and one extra C=C bond, in which case the positive charge could be on the epoxy alpha carbon.
It is also conceivable to reverse the polarity, with a carbocation at the allene and a carbanion somewhere on the left.
There is a problem, of course, which is that allenes are linear while carbanions are bent. Carbocations should be linear, though I doubt they are as "stiff" as an allene.
To be sure, the bending of these allenes doesn't seem nearly as dramatic as has been observed.  I looked up the crystal structure and worked out the bond angle of the allene for 1337, 1383, 1433, and 1481, getting 2.4, 1.6, 0.8, and 0.0 degrees of bending - I doubt any of those are significant differences.  Still, I wonder if the natural world is always so quiet as a crystal structure.
In general... could this molecule interconvert electrical energy to physical force?  Could electrons passing through the photosynthetic complex physically flap the right side of the molecule as they come and go, or could mechanical force on the right side be used to generate "photosynthetic" energy?

I can't answer your question directly, but this sort of thing seems a bit unlikely. Energy generation in photosynthesis is always by means of H+ ions being pumped against gradient through the cell membrane and then forming ATP as the H+ ions go back via the ATPase enzymes. Most other cases of energy generation or transport in plants or animals happens by some variation of this method.

I'm quite a few bricks shy of a model there, but fucoxanthin generally is part of an antenna complex that promotes electrons to an excited state.  The energy, if not the electron, manages to make its way to chlorophyll.  Fucoxanthin absorbs energy from low-frequency red light, which is much higher energy than ordinary thermal vibration; it would take some clever mechanism to bring enough force to bear.  (I have no idea).  But first ... can allenes do this at all?

That's an interesting question actually. I don't know about allenes, but there are some molecules (stilbene derivatives) that can flip (cis-trans isomerisation) when light shines on it. There is a lot of research going on about how this can be used as a "molecular switch". There are other systems (such as ferrocene coupled with crown ether + group 1 ion) which can similarly change shape when receiving/losing electron. So it is certainly possible to have systems like this.

I'm not sure about this particular molecule (=types of molecules), but there has been quite a bit of research on molecular machines. IIRC excitation-deexcitation mechanisms are also quite popular choices for converting some sort of energy into motion. But no expert here... I've only seen a few talks by Gonzalez about the matter (http://www.columbia.edu/cu/chemistry/groups/gonzalez/).

I do not answer because the question is somewhat unclear. However, in relation to motion, mechanisms similar to what you described exists. Polyenes and conjugated polymers in general are certainly prone to motion and the mechanism relies on counterions intercalation, chain relaxation it the carbocation, etc. Those can be driven by light, electrochemically, and change might be induced by T or surrounding. Google for artifical muscles, cantilever, etc combined with conjugated polymers.


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Equilibrium of zinc ion with ammonia
In my chemistry lab, we are testing the amphoteric properties of this metal. I understand zink can react with an acid or a base to form different compounds in solution based on pH. In one part of the experiment we are testing when ammonia in added gradually to $\ce{Zn(OH)2(s)}$ if it redissolves as $\ce{Zn(OH4)^2-}$ or if it redissolves as $\ce{Zn(NH3)4^2+}$. We are also testing the solution with indicators. So I’m assuming we are using the indicators to determine if there is a pH change because if there wasn’t, the complex ion $\ce{Zn(NH3)4^2+}$ would be using up all the weak base for this formation and none of it (or very little of it) would be able to act in solution and deprotonate water and generate $\ce{OH-}$?

On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

HINT - To figure out which reaction would cause the most pH change edit the question to include the two chemical reactions that can occur when the zinc hydroxide dissolves.


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Why does Sulfur dioxide incompletely freeze when entrained in a helium stream inside tubing dipped in liquid nitrogen?
Background - In a laboratory setting, I use typical elemental analysis (EA) to generate about 50 µg of sulfur in the form of sulfur dioxide (SO2). The EA uses a helium carrier. As an example, a sample of silver sulfide wrapped in tin falls into the reactor set to 1000 °C at the same time as an oxygen pulse, hopefully leading to flash combustion and creation of SO2. The SO2 is carried through a desiccant (Magnesium perchlorate) followed by a quartz chip reactor set to 800 °C, and a GC column to separate out impurities by polarity. The helium flow rate through the EA is typically 80 mL / min. The pressure of helium at the top of the EA is set to approximately 135 kPa. The end detector is a stable isotope ratio mass spectrometer where we measure mass / charge (m/z) 64 and 66 (the hot quartz chip reactor exchanges oxygen atoms so as to force all SO2 to have the same oxygen isotope composition so that our only variability is from sulfur isotopes 34 and 32. The above scenario is routine and can be found in Fry B, Silva SR, Kendall C, Anderson RK. (2002) Oxygen isotope corrections for online δ34S analysis. Rapid Communications in Mass Spectrometry 16, 854-858. doi: 10.1002/rcm.651.
Observation - We were experimenting with analyzing smaller quantities of sulfur by tightening up the SO2 peak using a stainless steal loop dipped in liquid nitrogen for a standard amount of time followed by warming it to room temperature. The tubing is 1/16” OD and allowed to completely cool to liquid nitrogen temperatures before the sample is dropped into the reactor. I observed SO2 at the detector downstream of the loop still residing in liquid nitrogen at helium carrier flow rates above 30 mL / min. I had to reduce the flow rate to about 10 mL / min to completely trap all SO2 in the liquid nitrogen immersed tubing. Based on the phase diagram of SO2, even at this higher helium pressure, it seems like I am still well within the solid phase space.
Question - Are these observations expected and if so please help me understand why?
UPDATE - To add information that addresses the existing answers and comments, I have indeed tried larger diameter tubing, longer tubing, and tubing that loops into the liquid nitrogen back out and back in again which should handle the snow flake example. Also, this approach of freezing out of a helium stream works perfectly at high flow rates with N2O and CO2. My question to this Chemistry Stack then, is what about SO2 prevents easy freezing as I can accomplish with CO2 or N2O?

When you say you dip the stainless steel loop into liquid nitrogen for a specified amount of time, are you confident that the loop is thoroughly cooled before the gas stream is introduced?

Yes. I updated the post. I am absolutely certain the tubing is at liquid nitrogen temperatures before the sample is dropped into the reactor.

In addition to the residence time point in Poutnik's answer, I would add that as the SO2 freezes, you are losing heat transfer area, which isn't much to begin with in 1/16" OD stainless steel tube. I would make the coil longer or use some bigger diameter tubing.

Phase diagrams are about thermodynamics, while you have to consider kinetics as well.
The stream may have too high linear speed at 30+ mL/min to reach equilibrium while still in the loop.
Another  option is dragging  particles of/containing SO2 by the gas stream by mechanical way, not allowing it to settle. Like wind taking snowflakes, not allowing them to land and stay.


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The things you wish you would have done before doing a PhD in Organic Chemistry?
Hello there fellow chemists. I am really interested in doing a PhD in Organic chemistry. I haven't really thought which field I would like to specialize in, but it's definitely on my mind ever since spending a considerable amount of time in the lab during my fourth year research project (note: I am doing a 4 year MSci). I've always had a liking towards Organic chemistry compared to Inorganic and Physical. Apart from my department I haven't been able to find much general information regarding procedures and routes available. I understand that doing a PhD means dedicating 3 or 4 years of your life to research and even though nothing is guaranteed one needs to be prepared and have insight before embarking on a journey.
Please feel free to share your thoughts and comments
Thanks

This question might be more suitable for https://academia.stackexchange.com

@ManRow Probably not suitable enough...


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Saponification - Oil and Alcohol?
I've been reading that a saponification reaction typically involves NaOH or KOH. I was wondering - could anything that contains an OH group cause a saponification reaction? For instance, coconut oil + ethanol? If not, why is a "strong base" needed?

Not everything what has OH is a strong base, the same as not everything what has H is a strong acid.(Otherwise  table sugar would be both ). Strong base is needed as OH- is needed to react with the ester group. Alkohols do not produce OH-.

Coconut oil + EtOH might produce transesterification but you will not get any free acids from it.


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If the current is kept to zero in a potentiometric cell then how do electrons move from one electrode to the other?
If the current is kept to zero then how do electrons move from one electrode to the other electrode? How would  the reduction happen if electrons don't move from one electrode to the other electrode?

As per my knowledge, nothing like potentiometric cell exists. Can you please give details on what cell are you talking about in terms of the electrode reactions?

@NisargBhavsar A potentiometric cell is the arrangement e.g. for potentiometric titrations, using a probe inert electrode ( usually Pt, perhaps glassy carbon too ) and the reference electrode. the probe electrode mirrors potential belonging to the present redox systems.

@Poutnik Ok I got it, the electrode used in the potentiometric titration. Than the current is actually not zero but a very small quantity, just like in the case of Voltmeter. I still can't believe how I didn't see the obvious relation between potentiometric cell and potentiometric titration.

Zero in science is not taken as rigorously as in mathematics. It is seldom exactly zero, but rather any value smaller than some significant or noticeable threshold.
Scientific zero is in data processing context any value with its difference to mathematical zero being statistically not significant.
In the question context, the zero current means a  current small enough not to noticeably disturbing electrode equilibrium.
It is comparable to a voltmeter measuring voltage on particular points of electronic circuits, that has ideally infinite resistance and draws zero current from the circuit. In reality, its resistance is finite and it does draws some current, that is considered negligible, taking as zero for circuit servicing purposes.

Yes i finally understand how the current is kept zero and voltage is measured. It's actually the primary circuit of the potentiometer having a battery of known voltage and an adjustable resistance wire which apply the voltage across the terminals of secondary circuit having unknown voltage so that no current flows. In this way, we measure that voltage  applied by source battery which is actually of unknown battery.

No current needs to flow from an open circuit half cell. Well, no external current needs to flow. An equilibrium is established between the solution and the electrode. This equilibrium involves many transitions, back and forth, but is in equilibrium. There is a net current of zero.
Now, how do you measure a voltage without getting a current? You back it up with an equal and opposite voltage so as to establish a current of zero. Now, I admit that zero is pretty difficult to hit, exactly, but you can come so close that whether the current in or out of the half cell is + or - an electron or two or three, it doesn't change the measured voltage significantly.
A Wheatstone Bridge is the way to get zero current; normally, the galvanometer is very sensitive and resistances to be measured are adjusted for zero current. https://en.wikipedia.org/wiki/Wheatstone_bridge

To measure the voltage of a half cell without drawing current, the half cell is placed in series with the galvanometer and the resistances are adjusted for zero current. Then half cell is disconnected and its voltage is read off the actually unbalanced bridge, which can provide enough current for a non-infinite input resistance voltmeter.

I am glad you have brought this fine piece of the classics. I was aware of the W. bridge, called also potentiometric bridge, but I do not think it is useful for potentiometry or potentiometric cells. The high impedance modern voltmeters based on MOS FET transistors or Operation amplifiers draw lower current that is the residual current of the balanced bridge. They can have easily input impedance $\pu{e12 \Omega}$

The classics are still in use: "The basic principle of the experimental determination of electrolytic conductance is that of the Wheatstone’s bridge." https://qsstudy.com/chemistry/experimental-determination-conductance

Sure, why not ? But it is rather a different case.


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Can you meassure a negative pH value with a pH meter?
I was checking the $\mathrm{pH}$ of $\pu{1 M}$ chromic acid solution, and I got its $\mathrm{pH}$ around -1.7. Yet, $\mathrm{pH}$ of $\pu{0.4 M}$ solution was around -0.6. Could these values be correct? The $\mathrm{pH}$ meter was calibrated before use.
I'm thinking not. I guess $\mathrm{pH}$ meters work in the interval 2-12 $\mathrm{pH}$.

I bet you were outside of the calibration curve and the extrapolation failed. Be also aware activity coefficients go wild for concentrated solutions and they can also affect the electrode functionality.  Saying that, pH scale IS defined for negative values as well.

How about looking in your pH meter's technical documentation?

@Poutnik Is it even possible to meassure the pH in these extreme conditions? I am aware that negative pH exists theoretically. But can a glass pH electrode be even used to do that? I guess the electrode needs to be calibrated to work in the area around 0 pH

Negative pH exists even practically, but measuring it is challenging and not necesserily involves a glass electrode.

I remember the old days when my teacher told me the good practical standard for pH 0.0 is 2 M H2SO4. Is has dissociated mosly just one hydrogen and activity coefficients just climbed back to 1 ( and will climb >1 for more concentrated acid ). But expect the glass electrode would be drifting and it may decrease its life or damage its responses.

You are most likely getting an inaccurate value. As anticipated in Poutnik's comment above is difficult to reliably measure a $\mathrm{pH}$ of very acidic solution but the $\mathrm{pH}$ scale is indeed open, and negative $\mathrm{pH}$ are a real thing.
It is easy and more meaningful to state a very high concentration or activity rather than reporting inaccurate $\mathrm{pH}$. This, together with typical figures found in textbooks, has spread the misconception that $\mathrm{pH}$ is bound in the range 0 - 14, but it is not.
A simple example of a solution with negative $\mathrm{pH}$ is in every lab, namely conc. $\ce{HCl}$, 37% by mass, $\mathrm{pH} = -1.1$.
Edit: For sake of clarity, as for its definition, pH can exceed 14, too. As an example, a saturated NaOH aqueous solution has pH = 15.

Also, activity coeficients for H3O+ in 10-12 M HCl go "through the roof", above 10 or even 20.

@Poutnik indeed in my first lab they were scared of traces of acid or even water in the halogenated waste iron barrel. They claim a kind of superacid due to local extreme activity. Beside, super activity of H+ is at the base of some research in synthesis.

@Alchimista Reminds me of my first lab, they/we were (to a modern standard) extremely callous regarding what goes in the sink directly - but very careful about halogenated solutions and how they were not to go in the sink.

I'm writing a paper on the use of chromic acid on some inorganic material. The concen used were 1-0.01 M and my pH meter said the range is from -0.6 to 1.7, actually. In the literature the pH of the chromic acid solution is often related to the size of the polyanions that can be found. Which is important to us. I don't want to report something incorrect values, maybe I can calculate the pH. My boss asked me to report it. @Alchimista

@marietiara Well that is another aspect. Again a mere calculation would imply to know exactly what is going on on the activity side. But surely it is something to be discussed.


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What is the activation energy of creating molecular hydrogen from atomic hydrogen?
If molecular hydrogen is dissociated, or two hydrogen atoms in space collide, how much energy (in $\pu{eV}$, or perhaps $\pu{kJ/mol}$) does it take for an $\ce{H_2}$ molecule to form?

Zero. On the contrary, energy is released.

I mean, it is a reaction just like any other, and I believe its activation energy is zero.

Try to write H2 rather as $\ce{H2}$  ($\ce{H2}$)  than as $H_2$  ($H_2$). The latter is typographically wrong. Element symbols should be upright. Furthermore, the MathJax extension mhchem), implemented in CH SE site, invoked by \ce{}, simplifies chemical formulas and equations a lot.

You can look up the  H$_2$ dissociation energy, it is $D_0 = 432$ kJ/mol but if two H atoms collide they will not form H$_2$ (energy being conserved) as some energy has to be removed by a third body collision to stabilise the newly formed molecule. In total  $  -432$ kJ/mol is released.

As porphyrin pointed out, two hydrogen atoms coming together from infinite distance will tend to bond, if the energy of the bond can be dissipated. This is unlike the situation where two magnets click together and stay together (because that interaction hits an absolute minimum, rather than a roller-coaster upswing as in the diagram. Ref 1). Two hydrogen atoms coming together from infinity would be attracted, gain kinetic energy from the potential energy of their separation, come too close, then repel each other - to infinity again. Over and over!

Perhaps instead of thinking of an activation energy, it is more relevant to consider a deactivation energy, which is all the energy that the bond needs to dissipate. You could ask whether it is possible for the two atoms to come together and emit a photon of the exact energy needed to quell the oscillation. Or even a lower energy photon, just to keep the molecule bonded, even if still in a high energy state. Alas! Selection rules seem to prevent this or at least minimize its likelihood, and recombination of hydrogen atoms is reported to occur thru three-body interactions. Ref 2
Can you do this any other way? Perhaps instead of observing them in free space, you could attach both atoms to a metal surface and let them find each other in small steps. Some metals, like zinc, evolve H$_2$ in acidic solution. Very very pure zinc reacts only very slowly with acids. Ref 3 My interpretation of that situation is that when the first evolution of hydrogen occurs, the hydrogen is in atomic form, and is attached (not strongly bonded) to the surface, whether that is actually metallic zinc or a nanothin layer of ZnO, perhaps as a ZnO  -  H moiety. A more stable state would be for two of these loosely attached hydrogen atoms to bind together to form H$_2$, but movement is inhibited (by a high activation energy). In less pure zinc, the presence of cathodic sites (like a little copper) provides a better site for H$_2$ evolution, lowers the activation energy and the process spreads over the rest of the zinc to allow rapid corrosion. Less pure zinc could be coated with a thin film of mercury to provide another surface for hydrogen atoms to be relatively immobilized on, and this also gives a high overpotential (a high activation energy)for hydrogen evolution. Ref 4
Some metals, noble metals which have no oxide layer, like Pt, Au and Pd, enable rapid migration of H$_2$ across their surface (thus not only a low, near zero activation energy, but also connection with a massive metal substrate that can accept some vibrational energy when two hydrogen atoms get together and dance for a while before settling down. They are good surfaces for evolving H$_2$.
A number of organic reactions were done with atomic hydrogen, using high activation energy to prevent  formation of molecular hydrogen; this was classified as due to nascent hydrogen, a term now in disfavor. Ref 5
Ref 1. http://www1.biologie.uni-hamburg.de/b-online/library/newton/Chy251_253/Lectures/LewisStructures/Dihydrogen.html
Ref 2. I. Amdur,  J. Am. Chem. Soc. 1938, 60, 10, 2347–2355
Ref 3. https://en.wikipedia.org/wiki/Zinc
Ref 4. https://en.wikipedia.org/wiki/Overpotential
Ref 5.https://en.wikipedia.org/wiki/Nascent_hydrogen

I wonder, what if both electrons happen to have the parallel spin, while $\ce{H2}$ needs them to be antiparallel ? Would not they repulse each other and refuse to approach close enough to react ? A photon or the 3rd object can carry away the spin to allow it to switch, but they first have to be willing to overlap.

@Poutnik: we think of the bond in H2 as between two s-orbitals. However there could be a bond between the s-orbital on one hydrogen and a p--orbital on the other. No spin problem, I think. Metastable for sure. Pretty diagrams of the excited states of H2 here: http://www.eso.org/~tstanke/thesis/fig6.html


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When is the n+l rule followed when working with energy of different orbitals?
While going through my book, I came across a problem which gave some options about the order of energy of 3s, 3p and 3d orbitals of hydrogen atom. The correct answer was given as:

3s, 3p and 3d orbitals all have the same energy

I also read somewhere that the order of energy of different orbitals increases as the value of $n+l$ increases where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. So my guess was $3s<3p<3d$ .
What is the reason for this?

The former is true for hydrogen atom, the latter for other atoms.

1s.                                                                                2s.   2p.                                                                              3s.   3p.   3d.                                                                              4s.   4p.   4d.    4f.                                                                                5s.   5p.   5d.    5f.                                                               6s.   6p.   6d.                                                                                7s.   7p.                       8s.

moeller diagram

3s, 3p and 3d orbitals all have the same energy in hydrogen atoms ( or monoelectron atomic ions ) according to the nonrelativistic Schroedinger model.
Fine measurements and relativistic models of hydrogen atoms shows there is slight difference in their energies, related to the fine structure of spectra  and Lamb shift ( Wikipedia Hyperphysics )
For multielectron atoms, s, p, d orbitals have different energies, as there come in the play 2 major effects, which affect electron energies:

Mutual electron repulsion
Screening/shielding of the nucleus charge by other electrons.

oh so as the 2 effects don't take place in the case of mono electron species, we can say that they'll have the same energy. If for a moment we imagine a hypothetical case where these 2 effects don't exist at all, would the energy be same in multi electron species just like it was in the case of Hydrogen?

Yes, we can say they have the same energy as the first approximation.// Yes, they would have had approximately the same energy too, but what is the point of such highly hypothetical thought ?


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What is the explanation of negative temperature coefficient in ignition delay of a fuel?
When studying the ignition delay of a fuel, some fuel have a negative temperature coefficient area (NTC) where an increase of initial temperature of the mixture air/fuel produces an increase of the ignition delay:

What is the explanation of the NTC phenomena? Why does the ignition delay increases when initial temperature increases only for a limited range of temperature? What is happening inside the mixture air/fuel that explains NTC?

Just speculating - there can be creation of relatively stable radicals on tertiary carbon atoms, scavenging oxygen based radicals, but not keeping pace with them for long.


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Why is the disordered CuAu structure more stable at high temperatures?
I know that the disordered CuAu structure is based around FCC whereas the ordered CuAu structure is tetragonal. Why would the FCC structure be more stable at high temperatures?
For context, I was asked to explain this with the aid of a graph. I don't really know where to start

Pretty much all disordered things are more stable at high temperatures. Think of liquids vs solids, to begin with.

That makes sense. But why would a closely packed structure ever be more disordered than one which isn't?

Because every single atomic position in that close-packed structure has some freedom: it may be Cu, or it may be Au. That's a lot of disorder.

Fabulous , I understand this. Thank you!

I'm not a metallurgist or even not a good physical chemist. However, I'd like to explain what's happening in $\ce{AuCu}$ alloy with temperature and any expert can intervene.
According to Ref.1, $\ce{Cu}$ and $\ce{Au}$ are both univalent group IB metals with the atomic size difference ~12%. The $\ce{AuCu}$ alloy forms the tetragonally distorted fcc lattice where alternate (00h) planes contain either $\ce{Cu}$ or $\ce{Au}$ atoms and cause a contraction in c-direction. Resulting tetragonal face-centered structure has c/a ratio of 0.92 $(\frac{c}{a} = \frac{\pu{367 pm}}{\pu{396 pm}} = 0.93)$. In the temperature range ~$\pu{380 ^\circ C}$ to $\pu{410 ^\circ C}$, the superlattice $\ce{CuAu}$-II is formed, which consist of $\ce{CuAu}$-bct unit cells with the antiphase domains along the b-direction (Ref.2). There is a lattice shift of $\frac12(a+c)$ at each five unit-cell length. The superlattice $\ce{CuAu}$-II is described as orthorhombic cell with 10 cells along one of a direction, $oI40$.

Suggested reading: Ref.3 and 4.

References:

Valentina F. Degtyareva, Nataliya S. Afonikova, "Simple Metal and Binary Alloy Phases Based on the fcc Structure: Electronic Origin of Distortions, Superlattices and Vacancies," Crystals 2017, 7(2), 34 (13 pages) (DOI: https://doi.org/10.3390/cryst7020034).
V. Ozoliņš, C. Wolverton, Alex Zunger, "Cu-Au, Ag-Au, Cu-Ag, and Ni-Au intermetallics: First-principles study of temperature-composition phase diagrams and structures," Phys. Rev. B 1998, 57(11), 6427 (DOI: https://doi.org/10.1103/PhysRevB.57.6427).
M. Sanati, L. G. Wang, Alex Zunger, "Adaptive Crystal Structures: $\ce{CuAu}$ and $\ce{NiPt}$," Phys. Rev. Lett. 2003, 90(4), 045502 (DOI: https://doi.org/10.1103/PhysRevLett.90.045502).
S. -H. Wei, A. A. Mbaye, L. G. Ferreira, Alex Zunger, "First-principles calculations of the phase diagrams of noble metals: Cu-Au, Cu-Ag, and Ag-Au," Phys. Rev. B 1987, 36(8), 4163 (DOI: https://doi.org/10.1103/PhysRevB.36.4163).


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Equivalence point pH derivation of a tribasic acid
Considering $\pu{100 mL} \ \pu{0.1 M} \ \ce{H3A}$ and $\pu{0.1 M} \ \ce{NaOH}$ titration curve:

I understood the half equivalence points, but couldn't understand the reason why $\mathrm{pH} = \frac{1}{2}[\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}]$ at equivalence points.
So I tried to derive this and for that I did following:
$$\ce{H3A  -> H2A- + H+}  \quad \mathrm{}K_\mathrm{a1}$$
$$\ce{H2A-  -> HA^2- + H+}  \quad \mathrm{}K_\mathrm{a2}$$
$$\therefore \ \ce{H3A -> HA^2- + 2H+} \quad  K = \mathrm{}K_\mathrm{a1} \cdot \mathrm{}K_\mathrm{a2}$$
So, $\mathrm{}K_\mathrm{a1} \cdot \mathrm{}K_\mathrm{a2} = \frac{[\ce{HA^2-}][\ce{H+}]^2 }{[\ce{H3A}]}$.
This would give $\mathrm{pH} = \frac{1}{2}[\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}]$ but I don't know how $\ce{[H3A]}$ gets cancelled with $\ce{[HA^2-]}$.

Try to search the site, it was discussed several times. For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use  MathJax

Thanks for pointing out @Poutnik

Write down ion balance at the partial  equivalence and consider conditions leading to approximation where concentrations of H3A and HA^2- are equal.

See this and this  Qs/As